text:
Wedge Dash Diagram of 1,2-Dichloroethane
I'm in class right now and my teacher put this up on the board as a representation of 1,2-dichloroethane. He also projected a model. I'm thinking ... his board drawing is wrong. Shouldn't the chlorines also be on dashes too? There's no way the chlorines can be in the same plane as the carbon atoms; the model makes it too obvious.

If you look at the model you posted, there's a plane that goes through both carbons and both chlorines. If you make that plane the sheet of paper, then you can connect these atoms with straight lines. Also, keep in mind that there can be rotation about the C-C bond, so there are many possible correct drawings. The only beef I have with the teachers drawing is that the C-C-Cl bond angles are 90 degrees, which is wrong. But, depending on the information that's being conveyed, that might not matter.

I see, thank you! And yes the right angles were a bit jarring.

Your drawing of dichloroethane is another possibility of writing it and there are multiple more ways. Lewis-type structures project a 3D object into a two dimensional plane. It can therefore never be accurate enough and you will find multiple representations of the same molecules in different books.

The teacher's drawing is correct.  Of course there are other possible conformers that could be drawn, but there is nothing wrong with the one that is pictured.   Basic rule:  a dashed wedge indicates a bond projecting behind the plane of the screen, a bold wedge indicates a bond projecting out in front of the plane of the screen, and a simple solid line indicates a bond existing in the plane of the screen.  Also note that this drawing notation can be used for any molecule, not just carbon systems.

Could you elaborate on drawing wedge/dash diagrams correctly or link me to a resource? Thank you!

Does this help?  http://chemwiki.ucdavis.edu/Organic_Chemistry/Chirality/Fischer_Projections

Wait simple solid lines mean existing in the plane of the screen? What about the chlorines? These project into the screen (they are the red balls in the picture above), but these are also coplanar with the carbons - i.e. we can draw a single plane with both the carbons and chlorines on the plane. How about the simple solid lines just mean the ligands are coplanar (if not necessarily in the same plane as the screen)?

The red balls are in the model.  The model is a 3D object.  When we draw a 3D molecule in 2D, a simple solid line means in the plane of the page or screen.  In the teachers drawing, the carbons are also in this plane.


text:
Thermodynamics adding salt to water changes the temperature

$5\:\mathrm{g}$ of an unknown salt are dissolved in $325\:\mathrm{g}$ of water. Both the water and the salt are initially the same temperature. The water's temperature falls by $11.4\:\mathrm{^\circ{}C}$. Explain how it is possible for the salt and water to change temperature even though both substances are initially at the same temperature.

My approach:
We know that $$q = mc \Delta T$$
Since $q, m,$ and $\Delta T$ aren't changed by this action, this process must result in the raising of water's specific heat. Is this correct?

The keyword is that the salt dissolves. Dissolution entails at least two steps:
1) Overcoming solvent-solvent interactions and bonds. An extreme example: your dinner plate doesn't dissolve in your kitchen table. One reason is that there is an extremely high energy barrier to overcoming the hypothetical solvent's (in this case, the kitchen table's) intermolecular forces. You'd have to take an ax to the table to overcome the strong intermolecular forces that hold the molecules of your dinner table together.
On the other hand breaking these solvent-solvent "bonds" or intermolecular forces in water is easy; you can jump into a swimming pool just fine. I wouldn't suggest jumping into a table.
Also, as this step's suggests, this is an energy intensive process or an endothermic process. Energy must be consumed to break these solvent-solvent bonds or interactions.
2) Breaking solute-solute interactions and bonds. This however isn't required. For example, some unionized $\ce{NaCl}$ may be solvated by water. You'll find this more true with some of the less soluble salts. But most of the sodium chloride you toss in water will have dissolved (we'll get to why later), so what is solvated generally isn't $\ce{NaCl_{(s)}}$ or $\ce{NaCl_{(aq)}}$ (this is somewhat misleading terminology) but rather $\ce{Na^+}$ and $\ce{Cl^-}$.
This again is an endothermic process as bonds are being broken.
3) Solute-solvent stabilization (solvation). This steps entails the formation of solute-solvent intermolecular forces. For example, the sodium ion, with its positive charge, may form a hydration shell of water molecules.
This is an exothermic process, as "bonds" or more accurately, stabilizing intermolecular forces are formed.
Couple all three of these processes together and you have the actual dissolution process of $\ce{NaCl}$.
What's interesting about $\ce{NaCl}$ is that even with the third, exothermic step, the dissolution (which comprises all three steps) is still a tiny bit endothermic. So that's what the question is premised on - the addition of sodium chloride to water kicks off a spontaneous process which is endothermic. So one might ask why might this endothermic process still occur. The answer lies with entropy; the dissolution of sodium chloride is not enthalpically favorable (dissolution consumes energy - specifically - from the water) - yet the dissolution is entropically favorable.
Remember that all processes tend toward greater disorder over time, and the dissolution of sodium chloride is a perfect way to increase disorder; we go from unionized sodium chloride to two ions! The two ions are definitely going to have a lot more degrees of freedom than a single, unified molecule.
$$\ce{NaCl_{(s)} + H_2O ->Na^+_{(aq)} + Cl^{-}_{(aq)}}$$

In terms of dissolution processes as in sodium chloride it is better to speak of lattice energy instead of bonding energy.

Good point. Will edit.

You are starting on the right track, but the specific heat of water won't change (at least not by a measurable amount).
You know that the flow of heat is defined by $q=mC{\Delta}T$.
But, you also know that energy is conserved. So if heat is flowing out of the water, where is it going?
This looks like a homework question (if it is, please tag it as one!), so I am reluctant to just give you the answer. I will give you a hint though - the problem states that the salt dissolves - I think that will be important.
As @Dissenter said, dissolution of any salt (ionic compound) can be either exothermic or endothermic, depending on the salt and on the solution. In this case, it must be endothermic.
We know this because ${\Delta}T < 0$, which means $q < 0$ for the water. The heat must have gone somewhere - the only place it could have gone (assuming that it wasn't just lost to the environment) is into the molecules as stored energy (enthalpy). Since the heat transferred from the water is negative (exothermic), that means the enthalpy change for dissolving the salt must be positive (endothermic).

Heat = Energy. When something dissolves it may have positive or negative enthalpy - energy change. See?

@Swedish Architect That's right - in this case the enthalpy of solution is positive (endothermic), and so heat is withdrawn from the water. Ammonium nitrate is a common salt with positive enthalpy of solution. Since this isn't a homework problem I will expand my answer.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
How does the crystal know where to stop stacking the unit cells?
I tried to make different crystals in the cubic system by stacking cubes in a 3D software. However, I need to tell the software where to skip rows or stop stacking the unit cells in a specific direction else it will only make a cube.
How does this work in nature? How does the fluorite "know" where to skip rows in order to make the octahedron crystal? Does it have something to do with Bravais lattices and coordination numbers?
First I thought that the unit cell needs to look like the final shape, but I think I'm wrong there.

The unit cell does not need to look like the final shape, nor via versa.

The ideal crystal is terminated by planes, which are described by
Miller Index. Some of these planes are more favorable than others, and as a result, the crystals do get macroscopic shape which is different from the shape of unit cell. For some discussion on reasons for different stability, I have found this article: iopscience.iop.org/0022-3719/12/22/036/pdf/jcv12i22p4977.pdf

"Real" crystals do not grow by stacking unit cells, but by atoms, ions or molecules adsorbing to the surface of the existing crystallite nucleus.
The different possible surfaces of the crystal (those that are different, depending on orientation towards the unit cell (symmetry!) and current outer layer) each have some probability to start growing (first adsorption into a new plane) and some other to keep growing.
These probabilities (i.e. growth rates) depend on kinetics and thermodynamics, and thus are somewhat hard to predict. They also depend not only on the temperature and pressure, but also on the composition of the solution or melt. Their ratios dictate the macroscopic form of the crystal that grows.
Note that there are a lot more planes in a single crystal than the unit cell has outer faces. Typically the very "odd" planes are not so favourable for growth, but that is the reason why a crystal does not have to resemble the form of its unit cell.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Formal Charge of Terminal Atoms
Would the formal charge of terminal atoms differ from the FC of the non-terminal atoms?
Say you had a sheet of graphite; inside the network covalent solid all the carbons are bound to one another, but the terminal atoms are obviously missing some C-C bonds since they are terminal.

Their formal charges would differ, correct?
What is the consequence of the differing charges?

The terminal carbons may or may not be missing some bonds. In any case, unless there is a charge on the surface/edge, the formal charge will always be zero.
Let's look at a single graphene ring to make it easier to see:
Free radicals

In this first case, there are no atoms attached, and it is pure carbon. However, for the molecule to be electrically neutral, there must be a total of 6*4 valence electrons: 24. If we only have alternating single and double bonds, that gives us 18, so the other 6 have to go somewhere. In this structure I have drawn them as free radicals (although it is extremely unlikely they would be for long). In this case, you can see that the formal charge on each carbon is still zero:
# valence electrons - [$\frac{1}{2}$ # bonding electrons + #non-bonding electrons] = formal charge
4 - ($\frac{1}{2}$*6 + 1) = 0
Triple Bonds

The second image is more likely if we don't allow any other elements to attach. Since the alternating double bonds are are really resonance structures formed by the overlap of p-orbitals, those lone electrons can do the same and form another $\pi$-bond, resulting in a triple bond. This idea is support by this article, which shows that under vacuum, they have observed that graphene sheets can form triple bonds along the edge. Examining each carbon, we again see that the formal charge is zero:
4 - ($\frac{1}{2}$8 + 0) = 0
'Other Stuff'

The third picture is what we commonly assume happens - something (in this case hydrogen) is attached. This would again give a total of four bonds, resulting in a formal charge of zero for each carbon.
4 - ($\frac{1}{2}$8 + 0) = 0
So the answer to your first question:

Their formal charges would differ, correct?

Is most likely "no" unless the surface is charged.
However, your second question:

What is the consequence of the differing charges?

is a good one, considering that regardless of formal charge, the terminal atoms will obviously have a different character.
How exactly it will change depends on the specifics, but in general, we can say that they are "more reactive."

Makes sense. Thank you! I didn't think of the possibility of there existing a lone pair and neither did my teacher it appears.

You're welcome! Please note though, those are not lone pairs - they are free radicals (unpaired electrons). A lone pair would be stable but would result in a net charge. The free radical has no net charge, but would be extremely unstable. I was just showing it so that you could see that the formal charge would be zero in all cases.

Oops. Good point!

While the formal charge of carbon in benzene is zero, the actual charge is much closer to minus one. Unfortunately I cannot access the linked article, but I think that the concept of a triple bond applies to the edges of graphene only. How do they resolve the bonding picture in this case? Overlap must be terrible and $\ce{H4C6}$ is already highly reactive and unstable. How do they keep the structure stable? No Air? Matrix?

Yes, in the article I linked the samples were under vacuum. The benzene ring is meant to represent the edge of graphene sheet - the triple bond would only be on the edge bonds. If I get some time today I'll change the drawing to make it more obvious. Of course, since these are resonance structures, it would really be something between a double and a triple bond along the edges. You make a good point about formal vs. actual charge - something that we often forget when looking at Lewis structures.

I think what you've drawn as the "Triple Bond" structure is really the same as the "Hexaradical" drawn above it.  Even $\ce{o-C_6H_4}$ (o-benzyne) has been shown to exist as a biradical.

@ron - right, the point was just to show that no matter how you look at it, the formal charge is zero. This is really a limitation of the Lewis structures and the concept of formal charges in general.

Yes, the formal charge at such atoms would be different.  Such atoms, typically found at the at the edge of some crystalline or regular solid, are known as "defects" in the solid.  Because they have an incomplete outer shell and lack a stable bonding configuration (bonding vacancies), they usually exist as a radical, anion or cation.  They define an area of extremely high reactivity on the solid's surface and mark the spot where reactions often begin.
Edit: references to charges at defect sites
this link covers inorganic systems, see p.5 for some nice pictures.
this link covers graphene systems, from the "Conclusions" section; "We have performed scanning gate microscopy on graphene and observed such charge inhomogeneity and puddles due to extrinsic doping from possible sources such as metal contacts, graphene edges and surface residues."

For cations and anions the formal charge changes. For radicals it usually does not. However, the actual charge (in terms of electron density) may differ a lot.


text:
Solvent for copper(I) iodide, copper(I) thiocyanate, silver iodide for HTM deposition; safety for methylammonium lead iodide
I'm a student working in a high school laboratory that is looking to pursue research into perovskite cells for next year, but I have some solubility and safety questions.
What would be a good solvent (please be specific) of copper(I) iodide?, copper(I) thiocyanate?  Silver iodide?
I want to deposit these materials out of solution onto the substrate.  Christians et al. in 2014 deposited copper(I) iodide from a 1:39 chlorobenzene/di-n-propyl sulfide solution, but my mentors have rejected chlorobenzene as a substance with high environmental toxicity and difficult disposal procedures.
Ideally, the solvent for any of the three compounds would dissolve the compound well, and then evaporate off with heating, while having no to low toxicity.
Additionally, I have received approval for purchasing methylammonium iodide and lead(II) iodide which I will dissolve in gamma-butyrolactone in 1:1 molar.  There are no safety sheets for the product, methylammonium lead iodide, so my mentors wanted me to see if there were any harmful properties that would come about from making this compound, ie: is there anything more dangerous or something more to watch for from methylammonium lead iodide than methylammonium iodide or lead(II) iodide?
Article: An Inorganic Hole Conductor for Organo-Lead Halide Perovskite Solar Cells. Improved Hole Conductivity with Copper Iodide

Lead Iodide is a little nasty, (Iodomethyl)trimethylammonium iodide is okay.  I'd say your product should be treated at least as hazardous as Lead Iodide. Chlorobenzene is indeed a pest.

Copper (I) iodide is soluble in pyridine, not idea about silver iodide though.

@permeakra If the choices are chlorobenzene and pyridine, I'd choose chlorobenzene every time.

Go for liquid pyridine as solvent it works great for Cu(I) compounds

Tell your mentor that to be concerned about chlorobenzene toxicity on perovskites is riddiculous. The most toxic thing in perovskites is the lead iodide...chlorobenzene´s toxicity is much lower than that. If he is so worried about toxicity, he should not work on perovskites.

Lead is the main toxic factor, if you work in a fume hood and make sure that the solvents used in the process are not going to be inhaled by anyone else and that you wear gloves and wear goggles for protection you are reducing your exposure to the toxic properties of the Lead if absorbed into your skin or accidentally ingested, the reason is this is a very soluble form of Lead. Some might argue lead is used everywhere in the industrial electronics etc but it is the fact that Lead in perovskite solutions and powders is very soluble.
DMF is a solvent that often papers use, quite toxic as is chloroform, need to use them in the fume hood so you don't smell them, their smells are distinct so if you do smell them at least you know that you need to clear up a problem that has been created. When you anneal the perovskite substrate the solvent evaporates and you are left with a film of perovskite crystals. The HTM as you know is also drop casted or spin coated on in most cases and needs to be a solvent that will not react or dissolve the perovskite film.
Most groups use Chloroform and a mixture of others for the different additives for the Spiro HTM (Standard HTM at present) such as TbP. CuI is a HTM that is much cheaper than Spiro and while it is not as good the performance is good enough considering the cost. I just did a google and found https://en.wikipedia.org/wiki/Copper(I)_iodide }
that it is soluble in:
soluble in ammonia and potassium solutions
insoluble in dilute acids
I don't know though if those solvents would affect the HTM film after evaporation or spin coating or react with the perovskite film. Let me know or publish if you find they don't
in any case while doing it
WEAR PROTECTIVE GLOVES AND GOGGLES
The thing about the solvents is something I am not sure of, just make sure that when you do the work you have the relevant safety measures taken, fume cupboard, after using beakers with solvents, wipe/rinse with Acetone and Ethanol then wash the thing that contained your solvent in the relevant area.


text:
Why does pH affect fermentation?
I know that the lower the pH the faster fermentation occurs. Why does this happen?

Read: http://gillesenergies.webs.com/phinfermentation.htm if you're a brewer... :)

That is not a general rule I think. It all depends on the kind of fermentation and microorganism you are dealing with.

pH affects the shape of proteins. In the case of fermentation a collection of enzymes is responsible for the metabolic processes that occur. - an enzyme is a protein which performs a metabolic process. For example sucrase is an enzyme which breaks sucrose down into fructose and glucose. They are kind of like organic catalysts.
Proteins are made of amino acids strung together in a long line, based off DNA. The amino acids bond up into a long polymer. Then cross linking occurs between the R groups on the amino acids. This makes the unique shape of the protein, because the cross-linking turns the protein into a 3-d shape.
Amino acids may be acidic amino acids, or basic amino acids, due to the R group on the amino acids. If the pH is increased, this affects the shape of proteins, by disrupting the bonds in the protein. In the case of fermentation, you say the rate increases when it get's more acidic - when the pH is lower. This is because the organisms - the yeast - producing the enzymes to ferment glucose, have adapted to acidic conditions. This means that through natural selection the enzymes today have been selected because they work best in acidic conditions, which bend the protein into the correct shape to allow fermentation to occur.
Read: http://www.icr.org/article/glycolysis-alcoholic-fermentation/
In this article it says that 10 enzymes turn glucose in pyruvic acid:http://en.wikipedia.org/wiki/File:Pyruvic-acid-2D-skeletal.png
And then two enzymes turn this into ethanol and CO2. The last two enzymes function best in acidic conditions, presumably because they have evolved to work best in the acidic conditions made by the pyruvic cid. However all fermentation stops after the pH drops below about 4.2. The optimum pH is about 4.8 - 5.0. After this we can assume that the pH bends the proteins out of shape too much, and the protein is said to have been denatured.
So to conclude... The rate increases because enzymes are all suited to specific conditions, dictated by natural selection - the conditions they have evolved around - and in the case of fermentation the last 2 enzymes - turning pyruvic acid to ethanol and CO2 - work best in acidic conditions, as they have evolved to those conditions. But increase the pH too much... and the protein denatures, and fermentation stops completely.

In addition to the other answers, another reason for the increase in fermentation rate in a lower PH is that the reaction actually uses up more protons (H+) than are actually produced in glycolysis. Like lactate fermentation the 2 NADH and 2 H+ produced during glycolysis are converted back to NAD+, but in ethanol fermentation there is an earlier step in which a hydrogen atom is added to the pyruvate when it is decarboxylated.
Take a look at this image. The "acetaldehyde" (ethanal) gains 2 hydrogen atoms from the NADH and H+ which is expected, but it has one more hydrogen atom than the pyruvate molecule. Now look at this image of the mechanism of pyruvate decarboxylase. See how in the third part an H+ appears out of nowhere?
In low PH solutions the concentration of H+ is greater. This increases the rate of reaction which, of course, increases the rate of ethanol fermentation.

I don't know that it's correct to say that lower pH increases fermentation rate in all cases. It is true for pH ranges that are close to neutral, but there is a point where increasing acidity will kill the yeast.
In general though, yeasts and molds are more tolerant of acidic conditions than they are of basic conditions, down to a pH of about 5. Bacteria, on the other hand, tend to not be very tolerant of acidic conditions (at least, not the bacteria commonly found in places where we are trying to encourage fermentation.) This article compares the growth and fermentation rates of bacteria to yeast under different conditions, and finds that under mildly acidic conditions (pH of 5.5 to 5.0) the bacterial growth rate was slowed and the yeast growth rate was maximized. Since bacteria competes with yeast for food, the ideal pH is the one that kills the most bacteria without slowing the yeast growth rate down. A pH of 5 is about the same as a cup of black coffee, and is about one hundred times less acidic than orange juice.
I also found this article which describes another interesting effect - the bacteria that was used in the first experiment actually starts producing ethanol as a fermentation product under acidic conditions, instead of just lactose. I suspect that growth rates are a bigger factor, but it is an interesting complication.


text:
Why are acids characterized by giving hydrogen and bases characterized by receiving it when acid and base reactions are double displacement reactions?
One example of this is $\ce{HCl}$ and $\ce{KOH}$. Why would the acid not just give the hydrogen ion in $\ce{HCl}$ to form $\ce{Cl-}$ and $\ce{KOH2+}$?

$\ce{KOH}$ does not exist as a molecule.  In water it is completely dissociated as $\ce{K+}$ and $\ce{OH-}$.  It does not make sense to write $\ce{KOH_2^+}$ for this reason.

This is a good question. Are you sure that KOH does not exist as a molecule. What if the OP were speaking of mixing pure HCl with solid KOH? I would expect that water is made though.

As a solid, KOH is ionic, not molecular.  HCl is molecular.

This is a good question. First, we must be cognizant of what actually exists. If we don't know what actually exists then any chemistry we attempt to do is just a shot in the dark.
If we are talking about solutions, then $\ce{HCl}$ doesn't exist; there isn't much to any unionized hydrochloric acid in the system. Thus it is misleading to write $\ce{HCl_{(aq)}}$, at least for introductory chemistry students, as one might take this to mean there is unionized hydrochloric acid that is solvated by water.
What actually exists in solutions of strong acids (apart from the counterion) is at the bare minimum the hydronium ion, $\ce{H_3^+O}$. The bare hydrogen proton, $\ce{H^+}$ does not exist by itself; it's too reactive. Recent studies have also suggested that the $\ce{H_5^+O_2}$ molecule exists in acidic solution in equal quantity as $\ce{H_3^+O}$. And of course other waters of hydration may exist due to the sheer positive charge density of a bare proton.

Speaking of $\ce{KOH}$, this is a very soluble salt, and unless we have a saturated solution of $\ce{KOH}$, we generally won't have any unionized $\ce{KOH}$. What we do have however in solution is $\ce{K^+}$ and $\ce{HO^-}$.
Therefore, I suggest that before trying to predict the products of any chemical reactions, write out a complete list of what you have in your system - a principle system inventory. Ignore the very minor species such as the higher waters of hydration, such as the $\ce{H_9^+O_4}$ molecule - this likely won't be a big deal as far as introductory-level calculations go, and scientists have yet to even fully characterize the solvation of the hydrogen proton in water themselves. Only upon realizing what's really in the system can we do accurate chemistry.

Now, if we were dealing with pure, liquid $\ce{HCl}$ and solid $\ce{KOH}$, then what we have is simply $\ce{KOH}$ and $\ce{HCl}$. There is nothing to dissolve the potassium hydroxide and no medium for the pure hydrochloric acid to completely ionize in.

Speaking specifically of why $\ce{KOH2+}$ doesn't exist or isn't stable: Note that its components are the $\ce{K^+}$ and $\ce{HO^-}$ ions. Note the opposite charges on these two ions. Remember Coulomb's law about the attraction between electrically charged particles. The opposite charges give $\ce{KOH}$ a substantial lattice energy, allowing it to exist as an ionic solid. If you were to protonate the $\ce{HO^-}$ part of the molecule, however, you'd rob the $\ce{HO^-}$ part of its negative charge. This substantially diminishes the lattice energy of the solid, so the solid wouldn't be stable anymore. Not to mention that this protonation also creates water, and $\ce{KOH}$ is soluble in water.

You have the right idea - many simple acid/base reactions (using the Arrhenius definition) follow the same pattern as a double-displacement reaction in aqueous solution.
The generalized pattern for a double displacement reaction is:
$\ce{AB + CD -> AD + CB}$
Typically, ionic double displacement reactions are driven by the formation of a precipitate:
$\ce{AB_{(aq)} + CD_{(aq)} -> AD_{(aq)} + CB_{(s)}v}$
In this example, the compund $\ce{CB}$ is insoluble in water, and so it precipitates out of solution (indicated by the arrow pointing downward). The reaction goes forward because the $\ce{C}$ ions and $\ce{B}$ ions leave the solution as soon as they meet each other.
In an Arrhenius acid/base reaction, we see a similar pattern:
$\ce{HA_{(aq)} + BOH_{(aq)} -> BA_{(aq)} + H_2O_{(l)}}$
Here, the formation of water drives the reaction - as protons meet hydroxide ions, they form water.
You can see that the pattern is the same, which is good! However, pay careful attention to the details of the pattern:
$\ce{HA}$ corresponds to $\ce{HCl}$, and $\ce{BOH}$ corresponds to $\ce{KOH}$.
In other words, $\ce{A=Cl}$ and $\ce{B=K}$.
Putting those into the pattern above, we get:
$\ce{HCl_{(aq)} + KOH_{(aq)} -> KCl_{(aq)} + H_2O_{(l)}}$
It helps to always think about these as ions when in aqueous solution (as @DavePhD mentioned). For this reason, these reaction patterns are typically taught along with solubility and dissociation of ionic compounds in aqueous solution. If you get in the habit of mentally (or explicitly on paper) dissociating all strong electrolytes, the patterns are much easier to understand.
For example:
This is the ionic equation for the general precipitation reaction pattern:
$\ce{A^+_{(aq)} +B^{-}_{(aq)} + C^+_{(aq)} +D^{-}_{(aq)} -> A^+_{(aq)} +D^{-}_{(aq)} + CB_{(s)}}$
And this is the net ionic equation:
$\ce{B^{-}_{(aq)} + C^+_{(aq)} -> CB_{(s)}}$
For acid/base reactions (only when both are strong):
$\ce{H^+_{(aq)} +A^{-}_{(aq)} + B^+_{(aq)} +OH^{-}_{(aq)} -> B^+_{(aq)} +A^{-}_{(aq)} + H_2O_{(l)}}$
$\ce{H^+_{(aq)} + OH^{-}_{(aq)} -> H_2O_{(l)}}$

Thanks for the response, would it be correct then to conclude that any reaction between an acid and a base, such as KOH and HCL, follows the  Arrhenius definition and that any reaction between a base/acid and water uses the Brønsted–Lowry definition? Also, does the Brønsted–Lowry model replace the Arrhenius model or do they both work for specific problems?

I think it would be correct to say that in any acid/base reaction where the acid is a donating a proton and the base is a hydroxide ion, it is following the Arrhenius model. When water molecules accept a proton to form hydronium ions, they are acting as Bronstead-Lowry bases. The Bronstead-Lowry model doesn't exactly replace the Arrhenius model, it's more like it expands the acid-base concept to include non-hydroxide species as proton acceptors. So we typically start with Arrhenius (since it's easy), move to Bronstead, and then on to the Lewis model from there.


text:
Why is the definition of the mole as it is?
I have asked my teacher, as she was introducing the concept of mole to us, why that number was chosen, instead of more convenient one. She told me that it came from the definition of the mole, that is the number of atoms in 12g of Carbon 12. When I asked why that definition was chosen, she answered me that my question wasn't really a question, and I understood from her stare that asking why? twice in a row was not appreciated.
However, I certainly hope that this definition wasn't chosen randomly and that there is a reason for it being it.
I gave it some thought and I came to believe that it may be for convenience after all. There are 12 nucleons in a carbon atom and if we say that the electrons mass can be neglected, then we get that a mole is the number of nucleon it takes to get one gram of them. However, can we really neglect the mass of electrons, when the atoms get bigger? And if so, why do the element's mass per mole do very rarely end up being integers?
This makes me think there may be another reason, or some things I do not understand. And if there are no reason, then I guess I might just as well be doing religious studies..!

Why is the definition of the moles as it is?

It is a rather arbitary definition that the mole is the number of atoms in 12g of carbon 12.  This has not always been the definition.  For example, prior to 1960, the definition was based upon oxygen rather than carbon-12.
The first standard was based upon 1 gram of hydrogen.  Later, the standard was changed to 16 grams of oxygen being a mole, for convenience because oxygen formed compounds with many other elements.  Eventually it was realized that elements including oxygen have different isotopes of different mass, and the mass of a particular isotope is a more specific and measurable standard than the mass of natural abundance oxygen.  12 grams of carbon-12 matches the old standard of 16 grams of natural abundance oxygen more closely than choosing 16 grams of oxygen 16.

can we really neglect the mass of electrons, when the atoms get bigger?

The mass of a proton or neutron is about 1836 times that of an electron. So depending upon the ratio of neutrons to protons, electrons are at most 1/1836 of the mass of a neutral atom.  If accuracy of more than 1 part in 1836 is desired, it is important to consider the electrons, regardless of whether atoms are big or small.

why do the element's mass per mole do very rarely end up being integers?

No element other than carbon-12 will exactly be an interger.  This is because masses of atoms depend upon number of protons, neutrons, electrons and binding energy.
In other words, considering that carbon-12 has an equal number of protons and neutrons (and electrons), other isotopes with a 1:1 proton/neutron ratio would have essentially integral atomic masses except for binding energy.  For example cadmium 112 is 111.90 instead of exactly 112 because binding energy is slightly more than 0.1 amu, partially offest by a higher number of neutrons than protons, the neutron mass being slightly greater than the combined mass of an electron and proton.
As pointed out by Matt Black, the natural abundance atomic masses further deviate because they are weighted averages of the masses of isotopes of a given element, weighted by their abundance on Earth.

That's a good answer - I would also add that all standards are chosen for two reasons: consistency and convenience, and C-12 meets both requirements better than any other element.

@thomij If you wanted to expand on that a little, it would be a great second answer.

+1 nice answer however I think hydrogen is not related to the concept of mole...

The major reason why elements masses are not integers or even close to them is that most quoted numbers in periodic tables are based on a mixture of isotopes. The errors from binding energy etc. are usually much smaller (parts per thousand so affecting the 2nd or 3rd decimal place).

@matt_black that is right however $H^{1}$ =1.007 825 032 07(10) and hydrogen standard atomic weight is 1.008(1)

@matt_black yes, although for the heaviest elements binding energy can affect the first decimal place.  For example lawrencium-262 is 262.11.

This is a fine answer, but I'd also point out that we want the number of particles in a mole to be the number of atomic mass units in a gram, with the atomic mass unit itself defined so that it's close to the mass of a nucleon. Now, if we define the atomic mass unit  so that it's 1/12th of the mass of a carbon-12 atom, that makes it roughly equal to a nucleon mass, and we get $$\rm 1\ g = 6.022\times 10^{23}\ u$$ and oh, look, it's that magic number that keeps popping up everywhere in genchem.

@GM see the atomic weight tables on pages 546 and 560 of Dalton's 1808 "A New System of Chemical Philosophy" which set hydrogen's atomic weight to 1. https://archive.org/details/newsystemofchemi01daltuoft

@FredSenese but historically, the term "mole" was used before the discovery of the nucleus and nucleons.  For example, this 1908 textbook  says (footnote page 9) "we shall call the molecular weight in grams the mole as has been proposed by Prof. Ostwald".  https://books.google.com/books?id=Ej0JAAAAIAAJ&pg=PA189&dq=mole+chemistry+definition&hl=en&sa=X&ved=0ahUKEwjor6i5_PTfAhVws1kKHa4GCFkQ6AEINzAD#v=onepage&q=mole%20&f=false and Ostwald didn't even believe in atomic theory when he coined the word.

Why is the definition of the moles as it is?
Here what say The International System of Units (SI)

“Atomic  weights” were originally referred to the atomic weight of
oxygen, by general  agreement taken as 16. But whereas physicists
separated the isotopes in a mass  spectrometer and attributed the
value 16 to one of the isotopes of oxygen, chemists  attributed the
same value to the (slightly variable) mixture of isotopes 16, 17 and
18,  which was for them the naturally occurring element oxygen.
Finally an agreement  between the International Union of Pure and
Applied Physics (IUPAP) and the  International Union of Pure and
Applied Chemistry (IUPAC) brought this duality to  an end in 1959/60.
Physicists and chemists have ever since agreed to assign the value
12, exactly, to the so-called atomic weight of the isotope of carbon
with mass number  12 (carbon 12, 12C), correctly called the relative
atomic mass Ar( 12C). The unified  scale thus obtained gives the
relative atomic and molecular masses, also known as the  atomic and
molecular weights, respectively

Atomic Weight' - The Name,
Its History, Definition, And Units
Prepared For Publication By
P. De Bievre' And H. S. Peiser

The unit could be fixed arbitrarily. By choice, it was linked  to 12C
and the macroscopic mass unit, the kilogram of the SI. Nevertheless,
one should note from the  definition below that the magnitude of that
link factor of NA does not affect the amount of substance of  any
entity.

Can we really neglect the mass of electrons when the atoms get bigger?
I would say yes, Neutrons mass is 1.00866491588(49) u for a proton is 1.007276466879(91) u while the electrons at rest have a mass of 0.000548579909 u so even if we take the heaviest atom Oganesson that has 118 electrons we will have as a very approximate weight of the electrons 0.06473242927025999 u.
Why do the element's mass per mole do very rarely end up being integers?
At the root, there is the fact that neither the atomic mass unit u is an integer actually it's 1.660539040(20)×10−27 kg. The kilogram was defined long before the studies of sub-atomic particles so it wasn't designed to be a perfect multiple of the atomic mass unit. However, even if we decide to adjust the kilogram so that the atomic mass unit is 1.00000000000×10−27 kg we will eventually end up with non-integer masses, this is due mainly to two factors:
Nuclear binding energy
The "mass defect" that is related to the fact that: $$E=mc^{2}$$
So the mass is linked to the energy, different elements have different energy due to different Nuclear binding energy and so different mass!
Neutrons heavier than protons
Neutrons weight 1.00866491588(49) u while protons 1.007276466879(91) u so elements with more protons are heavier.
Isotopic abundance? Not really...
Natural abundance (NA) is not at the root of the problem, your observation is valid not only for Standard atomic weight (Ar, standard,) that take into account the isotopic abundance but also for the weight of single isotopes. The periodic table below shows you that also element without isotopes (blue background) have non integer mass. Of course, if you consider the standard atomic weight of an element ( that is the result of a weighted average so you might already expect that is a non integer... ) the isotopic abundance have a greater impact on the result, but if you consider the atomic mass as defined by IUPAC so you are analyzing a single isotope the root of the problem are the reasons described above.

Holden, N., Coplen, T., Böhlke, J., et al. (2018). IUPAC Periodic
Table of the Elements and Isotopes (IPTEI) for the Education Community
(IUPAC Technical Report). Pure and Applied Chemistry, 90(12), pp.
1833-2092. Retrieved 18 Jan. 2019, from doi:10.1515/pac-2015-0703

The biggest reason why the elemental atomic masses quoted in periodic tables are not integers isn't the binding energy: it is the existence of multiple isotopes. For example: chlorine is 35.45, far from an integer but that is because it consists of two major isotopes ~75% chlorine-35 and 25% chlorine-37. The atomic mass of Cl-35 is ~34.969 and this is a result of nuclear binding energy causing deviation from an integer value.

@matt_black thanks for your comment, but the atomic masses are actually of an element and in the periodic table are quoted the standard atomic weight not the atomic mass most of the times. Besides that, the user question is the element mass per mole and not the value reported in the periodic tables.

I used standard atomic mass as a synonym for standard atomic weight. Those are, in effect, the mass per mole of a standardised sample of the element with its natural atomic abundance. This is what even the table you reproduce shows as it is what chemists work with. Your assertion that the non-integer nature isn't (mostly) caused by isotopic abundance is just wrong: binding energies affect mostly the second decimal place. But chlorine, for example, has a quoted mass/weight of 35.45. That's isotopic abundance not nuclear binding energy.

@matt_black I am not saying that it doesn't affect the standard atomic weight that are reported in the periodic table, but I am saying that is not the root of the fact that the weights are not integers and in fact also if we determine the weight of a single isotope we will have a non integer value.

Yes but the binding energy for most elements doesn't affect more than about 5 points in the second decimal place of the atomic weight. No mono isotopic element has a mass that doesn't start with xx.9 or xx.0. But plenty of standard atomic weights do. Isotopic composition is by far the biggest contributor to the non-integer character of atomic weights.

@matt_black thanks for your comment I have adjusted the answer, what I am trying to highlight is that the isotopic abundance is not the root of the fact we have non-integer values. I am not saying that it doesn't have an impact on the standard atomic weight, but up to my knowledge atomic mass is defined for a single atom.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Will heating diatomic oxygen enough break the O=O bonds?
The bond enthalpy associated with a $\ce{O=O}$ double bond is equal to 495 kJ/mol. Does that mean that adding enough kinetic energy in the form of heat will eventually cause the bonds to break and create monoatomic oxygen?

Yes. Adding enough thermal energy to anything will disrupt the bonds.

Yes, adding at least 495 kJ/mol of kinetic energy one way or another (thermally, photochemically by irradiation with photons of that energy, sonication, etc.) will cause $\ce{O2}$ to dissociate into monatomic oxygen.
$$\ce{O2 ->C[energy]\ 2O}$$

Yes. For quantitative information see Gas Phase Reaction Kinetics of Neutral Oxygen Species, particularly section 2.1, especially table 4, which gives the equillibrium constant $K_D$ as a function of temperature.
where
$K_D = \frac{[O]^2}{[O_2]}$
$K_D$ exceeds 1 above ~1000K, in units of particles per cubic centimeter.

It is important to point out that heat is a bulk (statistical mechanical) concept while bond strength is largely quantum mechanical.  If I am not mistaken, heat/temperature can be defined using pictures like molecules striking the edges of a container, or perhaps from energy/entropy relationships that involve Avagadro's number of molecules.
To your question, this means that given a large enough thermal energy, other molecules of $\ce{O2}$ will likely collide with an $\ce{O_2}$ molecule in question -- thereby bringing their nuclear kinetic energy to bear, and perhaps causing a dissociation.
I don't think that an isolated molecule of $\ce{O_2}$ (or anything for that matter) will spontaneously rupture because of heat, because here heat is not a well-defined quantum mechanical concept, as it is mediated by many moving atoms.

In principle yes, but it will need much more than 495 kJ/mol for sure.
The bond enthalpy is based on (experimental) enthalpy of formation. It does not take into account the enthalpy reaction barrier of the $\ce{O2}$ dissociation.

Ah, interesting! I knew it would need more but not much more Approximately how much more would you say?

"much more than 495 kJ/mol for sure"  Hmmm, remember that the molecules have a Boltzmann distribution of energies to begin with.

Did you mean Boltzmann's kT contribution? At room temperature it will be less than 1 kcal/mol, very insignificant to the enthalpy barrier.

I mean there is a large distribution of kinetic energy in the molecules in the sample, such that even at room temperature some will occasionally dissociate.

By enthalpy reaction barrier, do you mean an activation energy? There is no activation energy whatsoever for bond formation; atoms at infinity have zero potential energy, and at any value below that up to a certain equilibrium distance (the bond distance), it's a downhill slope. The curve can be modeled coarsely by a Lennard-Jones potential. Thus, the inverse processes of bond breaking only requires 495 kJ/mol to separate the atoms infinitely.

It sure will. But as soon as the gas cools down it will just re-form into O2 unless you give it something else to react with.
It's possible to get it to react with fluorine at around 700 degrees by heating O2 and F2 through it. You now have monatomic O and F ready to party, the fluorine in particular "losing it's gentle and forgiving nature" when reduced to single life. After cooling (to 90K), you will have a flask full of FOOF. That's assuming you still have an intact lab (hint: please do not ever try this, for your sake or the sake of others in your vicinity --Ed.).

Thank you for your username homage, but it was not appreciated.  Continued misbehavior will result in suspension.

Heating it even further will rip all the electrons away from the nuclei themselves to create a plasma state of free floating positive nuclei and electrons whereby, if enough energy was put into the system, the oxygen nuclei would fuse http://en.wikipedia.org/wiki/Oxygen-burning_process


text:
Balancing Redox Equations
I was taught to balance redox equations with acid/base considerations. Instead of arbitrarily adding $\ce{H^+}$ and $\ce{HO^-}$ ions and memorizing separate rules for acidic and basic solutions, I was taught simply to keep these four things in mind:
1) Charge must be conserved.
2) Mass is similarly conserved.
3) Do what's thermodynamically more favorable. For example, in strongly basic solution, $\ce{O^2-}$ oxygen ions come from not hydronium ion (because that doesn't exist in significant quantity in basic solution) and not from water (because that requires heterolytic clevage of 2 $\ce{H-O}$ bonds but instead $\ce{HO^-}$ because that only requires clevage of only one $\ce{H-O}$ bond.
4) Strongest base reacts with strongest acid.
On the other hand, here's a gem from Chem Wiki by UC Davis:

My question is:
1) How come I've never seen my professor's method mentioned anywhere? I've Googled, looked at scholarly papers; looked in advanced analytical textbooks, etc.
2) How were you taught?
3) Which way do you think is more advantageous?

1) How come I've never seen my professor's method mentioned anywhere? I've Googled, looked at scholarly papers; looked in advanced analytical textbooks, etc.

It is entirely possible that no one has published that approach! It is certainly not common in introductory chemistry textbooks.

2) How were you taught?

I was taught (and currently teach) the "standard" algorithm you found on the Chem Wiki.

3) Which way do you think is more advantageous?

It depends on your goal.
If your goal is to make sure students understand all of the steps and have a good intuition for what is physically happening, then your professor's method makes a lot of sense. This is because you need to already have a clear understanding of what things like "thermodynamically favorable" and "relative acid strength" really mean in order for that approach to work.
On the other hand, if your goal is to make sure students have the functional ability to work through a particular type of problem, regardless of how well they grasp the underlying concepts, then the algorithmic method makes more sense.
And finally, if your goal is to solve a problem quickly without having to think too hard about details, the algorithmic method also makes more sense.

Thank you! I think I prefer my professor's way then since it does foster greater understanding. If I become a professor I'll teach it his way.


text:
What's the Acid Ionization Constant of the Hydrogen Proton?
What is the $\ce{K_{a}}$ value for $\ce{H^+}$. I understand that the hydrogen proton doesn't stick around in solution by itself for very long. Nonetheless, does it have a  $\ce{K_{a}}$ value? Can it have a  $\ce{K_{a}}$? I think it can have one but quantifying it might be hard. Its  $\ce{K_{a}}$ value I think would correspond to this reaction:
$\ce{H^+ + H_2O ->H_3^+O}$
However this doesn't look like a conventional acid/base reaction; we don't have the conjugate base of $\ce{H^+}$ here, unless the definition of a conjugate base isn't what I think it is (acid missing one proton).
So my questions are:
1) Does the above mentioned $\ce{K_{a}}$ value exist.
2) Is my definition of a conjugate base too narrow?

https://chemistry.stackexchange.com/questions/46882/do-bare-protons-exist-even-transiently-in-aqueous-solution

The equillibrium between a proton plus water molecule and a hydronium ion is better considered using the concept of proton affinity.
\begin{aligned}
\ce{H^+(g) &-> H^+(aq)} \\
\Delta E &= −1530~\mathrm{kJ/mol}
\end{aligned}
You could roughly convert to an equillibrium constant using
$$\Delta G = -RT \ln(K)$$

Would $K$ be defined as $$K=\frac{a(\ce{H3+O})}{a(\ce{H+})\cdot{}a(\ce{H2O})}?$$ And then consequently converted into concentrations via $$c(\ce{H+})=\frac{c^\circ}{\gamma}\cdot{}a(\ce{H+})?$$ Ultimately leading to $$K_a\approx\frac{a(\ce{H3+O})}{a(\ce{H+})}?$$ Might be a bit of a stretch?

I don't think $\ce{K_{a}}$ has meaning in that context.  $\ce{K_{a}}$ is defined as the equilibrium constant for the following dissociation reaction $$\ce{HA{<=>}H^+ + A^-}$$
$\ce{H^+}$ can't dissociate; that is, there is no "$\ce{A}$" in the case your considering.

Good point; there is no acid dissociation! Good way of looking at the problem!

@Dissenter ron's point was that no chemical reaction make bare proton. So called dissociation of acid is protonation of another molecules


text:
Graph theory, isomer enumeration and chemical graph theory
Where could we find a good description of graph theory applied to the general problem of isomer enumeration and chemical graph theory?

Isomer enumeration requires group theory, I'd rather say.

Polya enumeration theory is not a part of group theory as far as I know...


text:
What are the possible side reactions when preparing iron(III) naphthenate?
I am trying to prepare some iron(III) naphthenate.
And the raw materials are : ferric(III) chloride, and naphthenate acid,  NaOH, toluene, deionized water
Everything worked well in lab scale: final iron(III) naphthenate dark red in color and soluble in toluene was obtained easily;
However, I encountered problem when scaling it up: it seems there are undesired side reactions, the finish goods looked like black precipitation.
Is that because naphthenate acid was oxidised by Iron(III)? or is there any other side reaction when being scaled up?
The operation:

I first dissolve NaOH and FeCl3 into water separately;
add NaOH aqueous solution into naphthenate acid while stirring at about 60°C;
add FeCl3 aqueous solution into the soldium naphthenate solution slowly;
keep reaction temperature at about 80°C for a while and remove the water;
reflux distillation to obtain finish goods (the maximum temperature might be high)

Are these pure raw materials? Iron(III) Naphthenate is viscous brown liquid.

Yes, all materials are pure. I can obtain desired viscous brown liquid in small vessel (about 500ml); however, it failed when being scaled up to 1$\text{m}^3$; is it because of oxidation?


text:
Number of collisions of particles in a box (or a sphere)
Scenario
Imagine you randomly spread in a cubic space of volume $m^3$ (or in a sphere of volume $V_s$, as you prefer), $n_A$ particles $A$ and $n_B$ particles $B$. Particles $A$ are spheres of volume $V_A$ and are not moving. Particles $B$ have no volume (they are points) and are moving linearly (reflecting at the boundaries of the space (cube or sphere)) at a constant speed $s$.
We can assume that when a particle $B$ meet a particle $A$ (collision), we count a collision but the particle $B$ just go through the particle $A$. Whatever is the model that is easier to calculate.
Question
What function describes the probability that there is $x$ collisions in a time interval $\Delta t$. Or what is the expected the number of collisions?

If A particles are at fixed points in space, the answer will depend critically on their spatial distribution.

If you want a number of collisions in a given time $\Delta t$, then simply from a units perspective, the other answer doesn't make sense, as it produces a number with dimensions of length. You want something that is dimensionless.
Often, physical chemistry textbooks will provide a derivation for collisions per unit volume per unit time. The proofs are a little bit involved and I'd rather refer you to a textbook instead of typing it all out. However, the bottom line is that the collision frequency is given by
$$Z_{AB} = \pi(r_A + r_B)^2\left[\langle v_A \rangle^2 + \langle v_B \rangle^2 \right]^{1/2} \left(\frac{N_A}{V}\frac{N_B}{V}\right)$$

$Z_{AB}$ is the number of collisions between A and B per unit time per unit volume
$r_A$ and $r_B$ are the radii of A and B
$\langle v_A \rangle^2$ and $\langle v_B \rangle^2$ are the mean square velocities of A and B
$N_A$ and $N_B$ are the number of particles of A and B
$V$ is the total volume

For a derivation of this I suggest reading Levine's Physical Chemistry, 6th ed, section 14.7.
For your specific case you've assumed that $r_B = 0$, $v_A = 0$, $v_B = s$. Therefore this simplifies to
$$Z_{AB} = \pi r_A^2 \cdot s \cdot \left(\frac{N_A}{V}\frac{N_B}{V}\right)$$
Obviously, the radius of A, $r_A$ is related to its volume $V_A$ by $V_A = 4\pi r_A^3 /3$. To obtain the number of collisions in a given time you just need to multiply by the total volume and total time.
$$f =  \pi r_A^2 \cdot s\Delta t \cdot \left(\frac{N_A N_B}{V}\right)$$
As you can see, the form of this equation is not all too different from the one in the other answer. The key difference is that the distance travelled $s\Delta t$ must be multiplied by a cross-sectional collision area, equal to $\pi r_A^2$, and not by a volume. The textbooks will explain this well, I will try to elaborate on this later if I have time.

With assumption that $n_A$ and $n_B$, i.e., the number of particles $A$ and $B$, are large and are in random motion, then the probability of a particle $B$ to be found within the volume of particles $A$ is $$\frac{n_A V_A}{V_s}$$
This will hold for every particle $B$. So, when each particle moves $s \Delta t$ distance, the number of collisions it will undergo is
$$\frac{s\Delta t\cdot n_A V_A}{V_s}$$
For $n_B$ particles the number of collisions is
$$\frac{n_B \cdot s \Delta t \cdot n_A V_A}{V_s}$$
With my assumption at top, it will either approach 1 or 0.

I believe this is not entirely correct. The distance $s\Delta t$ must be multiplied by a cross-sectional area, not a volume. Also I don't see why the number of collisions must be 1 or 0.


text:
Balancing redox equations in acidic media: to use H+ or H3O+?
Is it preferable to balance redox equations with  $\ce{H+}$ or $\ce{H3O+}$? I think the latter, because first, the bare hydrogen proton doesn't exist in water solution. So why write down something we know not to exist?
I ask because in this problem, if we balance using both species, $\ce{H^+}$ and $\ce{H3O+}$, we achieve two different equations. Mass is conserved in both equations as well as charge; all the differs is one coefficient.
So are the two equations equivalent in that they are still describing the same reaction? I think so, because sometimes I see textbook authors write these two equations for $K_\mathrm{w}$:
$$\ce{H2O <=> H+ + OH-}$$
$$\ce{2H2O <=> H3O+ + OH-}$$
However, if we were actually going to run an experiment and needed exact quantities, we would probably prefer the second reaction both above and below because the coefficients in these reactions better represent the chemistry of the system right?
For example, which of the following is "better"?
$\ce{ClO_{3}^- + 6I^- + 6H^+ \leftrightharpoons 3I_2 + Cl^- + 3H_2O}$
$\ce{ClO_{3}^- + 6I^- + 6H_3^+O \leftrightharpoons 3I_2 + Cl^- + 9H_2O}$

traditionally it is balanced with $H^+$

This is a question of "style" so there is no correct answer.  In fact, just as $\ce{H^+}$ doesn't really exist in solution, neither does $\ce{H3O+}$ exist in solution.  In reality higher hydrates of $\ce{H^+}$ are what exist in solution. See the first paragraph in this Wikipedia article.  Again, both $\ce{H^+}$ and $\ce{H3O+}$ are just shorthand notations for reality, choose the style that you like best.

I agree; but H3O+ is probably a closer representation of reality than H+ (H+ is Coulombically unlikely - even more so than H3O+).

Yes, $\ce{H3O^+}$ is a closer representation of reality and so if "that" is important to you, then you would use $\ce{H3O^+}$.  However, I could imagine someone else arguing that neither $\ce{H^+}$ or $\ce{H3O^+}$ is correct, so using fewer keystrokes and having a "cleaner, simpler" equation is what's important to me.

I guess writing the latter with a discussion of Coulomb's law and higher hydrates also serves pedagogical purposes.

What is wrong with $H_3O^{+}$? The first paragraph of that Wikipedia article you listed, does say anything about that.

Nothing is "wrong" with $\ce{H3O+}$, but one needs to keep in mind that is a shorthand notation often used to describe reality, but it is not real itself,  As the Wkipedia link says, "The reality is far more complicated, as a proton is bound to several molecules of water, such that other descriptions such as H5O2+, H7O3+ and H9O4+ are increasingly accurate descriptions of the environment of a proton in water."


text:
What is the origin of the transferred oxygendianion in redox reactions?
How come the oxygens transferred in redox reactions are always the $\ce{O^{2-}}$anion?
For example, I have this set of rules, and the rules are implicitly referring to the $\ce{O^{2-}}$  anion, a potent base (otherwise the rules wouldn't make sense; for example, it makes sense that a base in acidic solution is protonated to water, and that a strong base in basic solution is leveled to hydroxide ion).
In other words, how come, when electrons are transferred, such as in this unbalanced reaction:
$\ce{ClO_{3}^- + 6I^- \leftrightharpoons 3I_2 + Cl^-}$
The $\ce{O^{2-}}$ anion is formed (if only to be consumed again)? I remember my prof would say that specifically 3 $\ce{O^{2-}}$ anions "disappear".
Why not the $\ce{O^{-}}$ anion or simply $\ce{O}$? Does this have to do with stability?

I think you are misunderstanding the meaning of superoxide and the rules.
Superoxide is $\ce{O_2^-}$
Superoxide is NOT $\ce{O^{2-}}$ as you have written.
The rules are referring to hydroxide which is $\ce{OH^-}$, not superoxide.
The reason hydroxide is used to balance equations in basic aqueous solution is:
$\ce{H_2O \leftrightharpoons H^+ + OH^-}$
There is hydroxide present in aqueous solution, but not superoxide, peroxide, hydroxyl radial, etc.

What's the origin of this hydroxide ion?

water is in equillibrium with hydroxide and hydronium, according to the equillibrium constant $K_w$.  The rules are for aqueous solutions.  In high pH aqueous solution there is necessarily a high concentration of OH-

But in the case of the chlorate anion, the disappearing oxygen goes where? And the oxygen comes from water?

Well, in acidic conditions this can happen: ClO3− + I− + 2H+ → HOI + HClO2 http://en.wikipedia.org/wiki/Iodine_clock_reaction, but if you are following the rules, oxygen would go to water or hydroxide depending upon pH

water or hydroxide will form from protons or water and the oxygen of the reactant(s)

Okay! I think I understand better now.

One more question - why not the hydroxyl radical? Why isn't it present in aqueous solution?

It's not thermodynamically favorable to form hydroxyl radical from water, it has a positive enthalpy of formation, while water has a very negative enthalpy of formation.  Hydroxyl radial can form in the human body due to iron and copper ions, but it reacts almost immediately when the form.

Okay! And regular old O is unstable too? (Monatomic O).

yes, unless temperature is >3000K

Cool! I guess this is because oxygen wishes to achieve noble gas configuration, and that being isoelectronic with Ne trumps the reactivity brought about by the increase in negative charge density afforded by the extra two electrons?


text:
How to calculate the amount of chlorine emitted yearly by a leak in the air conditioning system?
A leak in the air conditioning system of an office building releases $12~\mathrm{kg}$ of $\ce{CHF2Cl}$ per month. If the leak continues, how many kilograms of $\ce{Cl}$ will be emitted into the atmosphere each year?
My working out is as follows:
\begin{align}
12~\mathrm{kg} &= \mathrm{CHF}_2\mathrm{Cl} \\
\text{Cl mass in compound} &= \frac{35.45}{86.448}\times100\% = 41\% \\
\end{align}
$41\%$ of $12~\mathrm{kg}$ is $4.92~\mathrm{kg}$, so I multiplied it by $12$ to get the annual result.
$$
4.92~\mathrm{kg} \times 12(\text{months}) = 59.04~\mathrm{kg\ year}^{-1}
$$
I don't have access to the answer so I'm not sure whether or not my answer is correct. Also, this process seems a little to simple to be correct. I say this because the question prior to it was very difficult. Can someone please tell me if I am right, and if I'm not, where did I go wrong?

Given the details you provided, your method is correct.

For the sake of correct formalism, here the answer as it should be given in the textbook:
The mass flow rate of the leaking substance, under the assumption of linear scaling, is
$$\dot{m}_{\ce{CHF2Cl}} = 12 ~\mathrm{kg\, month^{-1}} = 144~\mathrm{kg\, a^{-1}}\; .$$
The mass flow rate of chlorine, as contained in the leaking substance, is then calculated as follows:
$$\dot{m}_{\ce{Cl}} = \dot{m}_{\ce{CHF2Cl}} \cdot \frac{M_{\ce{Cl}}}{M_{\ce{CHF2Cl}}} = 144~\mathrm{kg\, a^{-1}} \cdot \frac{35.453~\mathrm{g\,mol^{-1}}}{86.47~\mathrm{g\,mol^{-1}}} = 59.04~\mathrm{kg\, a^{-1}}$$
For the end result to the question "how many kilograms of chlorine will be emitted into the atmosphere each year?", please note that the mass flow rate still has a dependency on time
$$ \dot{m} = \frac{\mathrm{d} m}{\mathrm{d} t}\; , $$
which means that we have to do an integration so we can calculate the mass:
$$ m(\tau) = \int_0^\tau \dot{m}\,\mathrm{d}t = \dot{m}\,\tau$$
For one year, we get
$$ m_{\ce{Cl}}(1~\mathrm{a}) = \dot{m}_{\ce{Cl}}\cdot1~\mathrm{a} = 59.04~\mathrm{kg}\; , $$
which means that your attempt at answering your own question were very good, on spot and totally valid.


text:
Permanent suspension of cocoa powder into water
I have failed in permanently suspending cocoa powder into water.  I have tried heat (212'f) and pressure (40 psi) into a sugar water solution (50/50).  Also, powdered milk and evaporated milk have failed to yield a permanent suspension.  There was another method of Irish moss to coagulate the whole thing together, but that failed as well.  I need the bonding/ emulsifying agent to be edible, delicious and shelf stable.  I am a low level cook and do not have the chemistry knowledge to do this myself.
Thank you for reading.

Do you want a liquid, or something as thick as custard?

liquid and delicious please

You are doing something strange, then. Heat is perfectly adequate. Cocoa powder is mostly starch, and it cooks up like any other custard. The fat remains smoothly suspended in it. No need for emulsifiers.

Might be interesting to ask on cooking as well

Sounds like you need a food-grade surfactant. Maybe you could try lecithin?
The problem is that cocoa powder is mostly made of stuff that is not soluble in water (hydrophobic). So, to create the emulsion, you will need to "package" the powder molecules in something they are soluble in. You could either use a surfactant directly, or build up in layers using fat molecules and proteins, with a surfactant around the whole package.
Lecithin would be a good starting choice since it is FDA approved and used in lots of products already.
I would try this:

Add lecithin in gradually larger amounts until the suspension worked (you will have to try different heat/pressure and emulsification treatments - consider high-pressure homogenization as well.)
Add some whole milk (we are looking for the fat molecules and proteins) and see if you can find an acceptable milk/lecithin ratio that tasted good and that was still stable.

If that doesn't work, your other option is to modify the cocoa powder granule surface so that it is hydrophilic. That means you will need to add polar functional groups. One way to do that is with an acid (citric, for example), but you will add a sour taste which might be hard to get rid of. There is a patent for an approach using enzymes, but I don't know which one they used (it just refers to "starch degrading," "protein degrading," and "cell-wall degrading" enzymes), or how well it works.

Tataric acid (cream of tatar) can be used.  I will give this as my next shot.  Would I need the tartaric acid as the emulsifying agent for the evaporated milk?  Instead of whole milk, evaporated milk was substituted for shelf stability.    I have tried heat and pressure to force the molecules together.  212'f is as hot as the water gets for some reason!  (cook jokes)  40 psi was used and the cocoa powder dropped out of the solution within minutes.  Would I use lecithin along with the heat and pressure?    Sorry if I am misunderstanding what you are trying to convey.

Tartaric acid may or may not work. If it reacts at the surface of the powder and leaves hydrophilic groups exposed, then it will work. The reason I suggested whole milk is that you want to get some fat molecules in there to help with the emulsification. It might not be necessary, but it would likely help. Pressure may or may not help - I don't know much about food chemistry specifically, I am speaking from a general solubility-of-proteins point of view. This link describes some ways of using lecithin you could try first.

Great information from the article!  These are stabilizers and will not act as an emulsifying agent.  I need to suspend it first before stabilizing the solution.

I am thinking it would act as both, but I don't have enough knowledge to tell you for sure.


text:
Why should the sat-temp of refrigerant be below the required temp of the cooled compartment?
I have a question about the refrigerant's pressure in the inlet of an evaporator. Why should the sat-temp of the refrigerant be below the required temp of the cooled compartment ? Shouldn't the refrigerant's temp at the exit of the evaporator be equal to that of the cooled compartment ?
Thanks all.

In order for heat transfer to occur, there must be a temperature gradient. If the temperature of the refrigerant was equal to the temperature of the cooled compartment, no heat would flow from the cooled compartment into the refrigerant to drive the phase change.
This is the reason that convective flow over the evaporator coils is important - it increases the rate of heat transfer by keeping the fluid moving so that the temperature right next to the evaporator coil is always close to the average compartment temperature. A larger ${\Delta}T$ leads to a faster rate of heat flow.


text:
No indicator is suitable when titrating weak acid with weak alkali
This was a statement in my book:
A solution of a weak acid cannot be titrated with a weak base using an indicator to find
the end-point because the pH change is too gradual close to the equivalence point.
Could someone please elaborate the bold text.

Check this link and search for the section "Titration curves for weak acid v weak base". Be sure to click on the "indicator" link in the section's note, too. Does that help?

if there is a weak base and a weak acid, there will be at least two pKs, complicating the titration curve.  The full pH range may be unobtainable by adding a weak base.

An indicator works when the solution's pH is in its range. It's not necessarily the equivalence point.  The point is called the end point where it actually changes color.
If the difference in volume is small between the equivalence and end points, then the indicator can be used.  Since a gradual change implies more volume must be added to obtain the same pH change in other acid base titration, the difference between equivalence point and end point(if you use indicator) will be more and therefore indicator will not be suitable for it.


text:
What are possible hazards of fog machines?
From a chemistry point of view I would like to learn about possible hazards of fog machines.
Such devices, also called “foggers”, create vapor which appears like fog or smoke. It is used in discos, parties, etc.
It works by heating “fog juices” which are made of glycerine and distilled/de-ionized water mix or propylene glycol and distilled water mix.
I don't know the temperature these mixes are heated by the machines.
I know glycerine and propylene glycol are not hazardous, but I would like to know if heating them can pose any risk. I've talked with someone who works in this industry and he couldn't tell if there has been any study to show if the fog produced by these foggers is harmful or not.
So even though these devices have existed for years I would still like to know if they can be harmful and if they should be avoided when possible.
How you can help as chemist is by telling me if vaporizing glycerine and distilled water mix at certain temperature can cause any change to the supposedly harmful substance and make it dangerous or not.

I am not sure that StackExchange is a good forum to seek advice that could potentially put the person who answers in a legal spot. I would suggest writing to the manufacturer of the fog machine.

@AwalGarg Indeed it is chemistry, but could you imagine a circumstance where someone answered "there's no danger with fog machines!" and then someone read this and went and did something stupid, and someone got hurt/killed, then you can bet there would be serious repercussions. I downloaded five PDFs of fog machine operating instructions, and each of them had at least TWO PAGES of warnings and disclaimers.  There is no way an answer given here (even given by a fog machine expert) will cover all these bases. And, what if there are harmful effects after all? Are we running clinical trials?

@Eric I think the assessment of hazards is well on topic here. if we just state where possibly security risks lie, I don't see where we can go wrong. After all, yes, an answer should never say: This is safe.

Given that these machines have been used extensively in the entertainment industry for decades, and one (Rosco Fogger) even has an Academy Award, common sense says that they aren't dangerous.
Breathing in copious quantities of the fog causes irritation, but that applies to a lot of things. Dry ice fog[1] is basically water vapour - fill the room with it and you die from asphyxiation from the CO2

looks better in many cases as it's "fog" and not "smoke"

Extreme heating* of glycerin produces acrolein, which is toxic and highly irritating (I'd say it should be also carcinogenic, but was unable to find any suggestions for it with quick googling). However, the heating should be extreme or an acidic catalyst is needed. Extreme heating of propylene glycol will produce propanal, that is also toxic, but not that much as acrolein.
Free glycerol and propylene glycol are slightly toxic, but their LD50 are high. $(> \pu{10 g/kg}),$ so nothing special.

* Preparative acidic-catalysed dehydration occurs at ~200 °C. Without catalyst glycerol boils without decomposition at 290 °C. Out of blue, I'd say that somewhere around 500 to 900 °C the dehydration will proceed anyway, but can't give a proof just now.


text:
Number of hydrogen atoms in the Sun

Begin by calculating the number of atoms in the Sun. Assume that the Sun is pure hydrogen with a density of $\pu{1.4 g cm^-3}.$ The radius of the Sun is $\pu{7E8 m},$ and the volume of a sphere is $V = \frac 4 3\pi r^3.$

This is what I did, since it worked with similar questions:
$$
N = \frac{\frac{4}{3}\pi(\pu{7E8})^3(1.4)(\pu{6.022E23})}{1.008} = \pu{1.2E51},
$$
but the answer is $\pu{1.2E57}.$ What have I done wrong?

The problem is that you did not include units with your numbers. If you had, you would have easily seen that you are multiplying $\pu{m^3}$ and $\pu{cm^{−3}}$ and coming up with $1$ instead of $10^6.$

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Is CH₃CH₂CH₂Cl saturated or unsaturated?
I believe the molecule $\ce{CH3CH2CH2Cl}$ to be saturated, since there are only single bonds between the carbon atoms, but my workbook says it's unsaturated. Why?

I think your textbook is wrong.

it is definitely a printing mistake in your textbook

What workbook do you have? Is it a published text?

The workbook must be wrong, it is a saturated molecule.

You are correct. Your workbook might be wrong. It is definitely a saturated molecule.
The one quoting wikipedia text must see that the article is talking about saturated hydrocarbon from end of line one , so wikipedia is not wrong it's our interpretation of sentece is wrong.

I refer you to definition 3) via Wikipedia:
         https://en.wikipedia.org/wiki/Saturation_%28chemistry%29
To sum up: a saturated hydrocarbon must have no double bonds, no triple bonds, and no rings (only single bonds), and each carbon is attached to two hydrogen atoms, apart from the end carbons, which are attached to three.
The compound that you're given is not saturated since one of the end carbon atoms has a Cl attached to it.
If it helps, you can think of saturated as meaning that the carbons that make up the chain have to have as many hydrogens as possible attached; the chain is 'saturated' (or completely full) with hydrogen atoms.

Edit:
    Turns out I had a brain misfire: it's absolutely not a hydrocarbon (which would only contain carbon and hydrogen). The very first sentence of definition 3) that I referenced means that your workbook is wrong (and the other replies are correct): it's a saturated compound. My mistake - I erroneously conflated the first two sentences of 3) from Wikipedia while having a simultaneous brain fart.

that's not hydrocarbon


text:
Hess Cycle, determining enthalpy change of formation
I am asked to find the enthalpy change of formation of the following:
$$\ce{N2 + 1/2O2 -> N2O}$$
I am given the following enthalpies of reaction:
\begin{align}
\ce{C + N2O &-> CO + N2} &\Delta H_f =-193 \tag{1} \\
\ce{C + 1/2O2 &-> CO} &\Delta H_f=-111 \tag{2}
\end{align}
How do I calculate enthalpy change of formation of the nitrous oxide?
I thought it would just be adding the two enthalpies together because I substitute the second equation into the first one, but this is wrong.

I took the first equation away from the second equation:
\begin{array}{lll}
  &\ce{C + 1/2O2 &-> CO} &\Delta H_f &= -111 \\
- &\ce{C + N2O &-> CO + N2} &\Delta H_f &= -193 \\
\hline
= &\ce{1/2O2 - N2O &-> - N2} &\Delta H_f &= 82 \\
= &\ce{N2 + 1/2O2 &-> N2O }
\end{array}

If you start by reversing equation 1, you get $$\ce{CO + N2 ->C + N2O~~ {\Delta}H_{f}=+193}$$
Now, add this result with equation 2 and you get $$\ce{CO~+N2~+C~+1/2O2->C~+N2O~+CO~~{\Delta}H_{f}=+193-111=82 }$$
Cancelling like terms that exist on both sides of this equation yields $$\ce{N2~+1/2O2->N2O~~{\Delta}H_{f}=82}$$

Try adding, subtracting, multiplying by real numbers and reversing the equations as necessary to get the desired equation.
For example, since your target equation has $\ce{N_2}$ on the reactant side, I would flip the first given equation so that diatomic nitrogen ends up on the reactant side.
Then note that if you have the same molecule on both sides of the equation in equal quantities, you may remove it from the equation since these species will not affect the enthalpy of the reaction.
(Side note) Catalysts are similarly not consumed by a chemical reaction, yet they definitely affect the rate (kinetics) of the reaction.

What then after flipping the target equation?

As you stated in your question, you can add enthalpies. This correctly implies you can add equations.

You flipped the target equation, but what enthalpies do you add? So far you've only flipped one equation.

I didn't flip the target equation. You add the given enthalpies with signs taken into consideration. Is this a HW question?

its fine i found a way to do it


text:
What is the chemical composition of VOC's off-gassed from polyurethane?
I understand that all polyurethane (PU) materials off gas VOCs. I tried finding the % composition of the gases off-gassed by PU, but I am not able to get satisfactory answer. On the same note, why are these gases off-gassed?
Thanks!

I don't know the composition, but I suspect something bad like benzene.  My understanding is that these are residual solvent that was used during the synthesis of the polymer.

Is there way to find out which gases are emitted? In the first place, how I collect the gas for analysis? What test method can I use? Thanks!

You can pump down (vacuum) on a piece of the material in a Schlenk line, and then collect the leached gasses into a liquid nitrogen trap.  This can be dissolved into a solvent and subjected to GC/MS.

authors: بورحن باسشش سعدی


text:
Acidity of aldehydes
Which is more acidic between methanal ($\ce{HCHO}$) and ethanal ($\ce{CH3CHO}$).
Please explain using General organic chemistry basic concepts.
My Effort:
I saw the stability of the conjugate base.
It would be $\ce{HCO-}$ and $\ce{CH2^{(-)}CHO}$.
Due to resonance I can say that second one is more stabilized and thus a better acid.
But answer says that methanal is more acidic than ethanal.

Hi AJ_ welcome to ChemistryS.E.! It looks like an homework question, can you please show us your efforts, thoughts and your attempts?Please take a look to our homework policy! Thanks a lot! :-)

Formaldehyde/methanal is particularly electrophilic, hence in aqueous solutions it actually exists predominantly as the hydrate (methanediol, $\ce{CH2(OH)2}$). If comparing $pK_a$ values with water as the solvent, this may at least partially account for the seemingly aberrant results.

AJ, is there any additional explanation with the answer? This is a very surprising result.

No there isnt any.

Honestly, I don't see how anyone could have predicted this outcome before seeing the actual $\ce{pK_a}$ data.  If you look up the acidities of methanal and ethanal, you find $\ce{pK_a}$ = 13.27 and 13.57 respectively.  A pretty small difference to begin with. As the picture below illustrates, in the case of methanal we are removing the proton directly bound to the carbonyl carbon to produce the anion pictured below.

Now if someone asked, "would pulling the same proton off from ethanal be easier or harder?", you could answer "harder" because that methyl group (being electron releasing) that is attached to the carbonyl carbon would act to slightly destabilize the anion with the negative charge on carbon.  But it turns out that when ethanal is treated with base a different proton is removed!  The proton that is on the methyl group (said to be alpha to the carbonyl carbon) is the proton that is removed.  Without knowing the actual $\ce{pK_{a}{'s}}$ beforehand, how could you know that the anion involved in the methanal case would be ever so slightly more stable than the resonance structures involved in the ethanal case?
Edit: Thanks to Greg E and Nicolau Saker Neto for pointing out an error in the earlier version of the figure and the corresponding text respectively.

I'm fairly sure that there is no resonance stabilization in the case of methanal. The triple-bonded oxygen violates the octet rule, and I believe the lone-pair would have to reside in a localized $sp^2$ orbital.

Whoops, now I see.  Thanks for pointing that out.  I'll edit my answer.

You edited the figure, but your answer currently still says the methanal conjugate base is resonance-stabilized.

But my doubt still remains. The anion formed from ethanal is clearly resonance stabilized with charge delocalisation thus making it more stable than the ion from methanal.Where did I go wrong? Also thanks for the replies.

Yep, I was surprised too, but them's the facts (the pKa's that is).  All you can say is that for some reason the methanal anion is more stable than the delocalized ethanal anion.  Just because one structure has more resonance structures than a different structure, doesn't necessarily mean that the former is more stable.

I think this reason is induction.

Ron, can you provide a source for those pKa's? What solvent are they measured in? What is the pKa of water on this scale? I'm very surprised by these values.

Here's a link to the Table I used:  http://sites.chem.colostate.edu/diverdi/all_courses/CRC%20reference%20data/dissociation%20constants%20of%20organic%20acids%20and%20bases.pdf   In agrrement with this data, I see that Wikipedia shows pKa=13.3 for formaldehyde and 13.57 for acetaldehyde

Thanks for the insights. If I may ask further, do we often see the deprotonation of the aldehyde hydrogen in organic reaction mechanisms?

@TanYongBoon  They do have $\mathrm{p}K_\text{a}$s so that tells us something, but the carbonyls are extremely reactive and the center of focus for most reactions.

Your explanation of the $\ce {pK_a}$ values is likely incorrect as aldehydes form significant amounts of their geminal diols in aqueous solution. See here: https://chemistry.stackexchange.com/questions/107616/how-can-benzaldehyde-have-a-pka-of-14-9

The answer lies with induction.
The oxygen is better able to stabilize the residual negative charge on the conjugate base of methanal by virtue of residing closer to the negative formal charge.
On the conjugate base of ethanal note that the electronegative oxygen is one carbon removed from the negative formal charge while on the conjugate base of methanal, the oxygen is on the same carbon with the negative formal charge. Thus, the oxygen in methanal will be better able to withdraw electron density from the spot where heterolytic bond clevage occurred. This follows from Coulomb's Law:
${F=k\frac{q_1q_2}{r^2}}$
As we can see from Coulomb's law, force and distance are related in an inverse-square fashion. In other words, increasing the distance of the oxygen nucleus from the site of the negative charge dramatically decreases the electron-withdrawing effects of oxygen.

But the conjugate base of ethanal has a resonance structure that places the negative charge on the oxygen. r=0 for ethanal!

That doesn't mean the negative charge is necessarily there; resonance structures are discrete while the actual molecule definitely does not conform to these discrete forms we may draw on paper. And no r =! zero because there is a distance between the electrons and the nucleus.

If you calculate the electron density of this anion, the electron density will primarily be located on the oxygen. See: https://instruct.uwo.ca/chemistry/283g/Nifty%20Stuff/aldehyde%20enolates_files/image001.gif

Perhaps. I don't know how to compute the electron density so I can't dispute this (or confirm it).


text:
What is difference between aluminium mercury alloy and gallium aluminium alloy?
Aluminum mercury alloy is used as a chemical reagent to reduce compounds, such as the reduction of imines to amines,in this case mercury destroy aluminium oxide layer which protect aluminium to react with water.
Alloy of aluminium and gallium which similar to above alloy make the aluminium more reactive by preventing it from forming an oxide layer.
In both cases mercury and gallium plays a same role of destroying oxide layer on aluminium to react with water but why can't gallium aluminium alloy reduce imines to amines like mercury aluminium alloy ?

Lol, exactly the same question was ask by me on a chemistry forum and never received any replies. I believe you can interchange them in some circumstances, but I don't know the details. This would be especially useful because gallium is a lot less dangerous to the amateur (and perhaps to the professional as well, or at least less toxic waste) chemist. Also gallium is pretty cool as it melts in your hands and solidifies again when restored at room temperature (i.e. removed from your hands).

Oh almost forgot. For the person who answers this question, I'd recommend to be careful (we all like to learn, but there are always shady figures abusing chemistry) as Al/Hg reduction is a very popular method for producing crystal meth and MDMA (for P2P and MDP2P respectively). Seeing that this is the users first question on any SE network site, it could be the case that the questioner seeks a safer alternative to this reduction. Providing this information to someone with the intent to manufacture illegal substances could be illegal depending on where you are from.

Also read http://www.sciencemadness.org/talk/viewthread.php?tid=14173 and probably (if you didn't) https://en.wikipedia.org/wiki/Aluminium_amalgam.

We can rationalize the differences in the reactivity as follows.  Gallium has a higher ionization energy compared to mercury.  During a redox reaction the metal is oxidized and it loses an electron, therefore how "easy" it is for a metal to lose its electon is iimportant. Since gallium is harder to inonze than mercurey, it implies that it is "easier" for mercury to lose its electron compared to Gallium, making mercury a better stronger reducing agent.
I may come back and add some quantitative information later, but i'm going to go eat now. I think looking in a table of standard reduction potentials would be reasonable.

Mercury loses an electron? It gets an electron from $Al(s)$. I'm not quite sure if I can follow your answer. Perhaps you should elaborate on the exact workings of amalgamation.

I cannot think of any reason that it can't. Mercury and gallium should both be acceptable mediators for the electron transfer. In fact, elemental Gallium may even be preferred over salts of Gallium unlike Mercury, where only Mercuric salts can be used (maybe due to mercury having such a low melting point?). Elemental Gallium might actually have an added advantage of having the perfect melting point to mediate electron transfer.
It would actually be quite ironic if all these chemists are taking elemental mercury and making salts of it just to give it the properties of elemental Gallium!
And I'm not even touching on the toxicological superiority of gallium!

This seems very speculative. Do you have any references to support your claims?


text:
What is the structure of RCH=NH?
I know that $\ce{RCH=NH}$ is an imine.
What is it's structure?

Hi! Can you show us your efforts and your attempts, even if this is maybe not a homework q. maybe could be useful to understand where is the problem and the question will be nicer! Thanks a lot! :-)

Have you tried?

Here it is one of the isomers.

The general structure of an imine contains a carbon-nitrogen double bound.

You can add hydrogen atoms or organic groups instead of $R^1$, $R^2$ or $R^3$.
The structure of your formula is:

The structure of an imine is analogous to that of an olefin.  There is a central $\ce{C-N}$ sigma bond along with a pi bond just as in an olefin. In the figure below, the olefin has four substituents attached to the two carbons forming the carbon-carbon double bond.
All 6 of these atoms or groups lie in the same plane, the plane of the screen.  Similarly in an imine, all of the groups attached to the central $\ce{C=N}$ double bond also lie in the plane of the screen, along with the carbon and nitrogen atoms themselves.  Note that while the olefin had four substituents ($\ce{R_1 - R_4}$), the imine has only 3 substituents.  The fourth substituent position in the imine is occupied by a lone pair of electrons on the imine nitrogen.  This gives rise to an important difference between olefins and imines.  Appropriately substituted olefins can exist as cis-trans pairs.  It typically takes energies in excess of 60 kcal/mol to break the pi bond portion of a carbon-carbon double bond and cause cis-trans isomerization.  The barrier for interconversion (called syn-anti isomerization in imines) in an imine is much lower, on the order of 20 kcal/mol, so rotation about the imine double bond occurs much more readily than in the olefin case.


text:
Is NaHCO₃ a carbonate?
I know that $\ce{NaHCO3}$ is sodium bicarbonate, but According to Wikipedia:

In chemistry, a carbonate is a salt of carbonic acid, characterized by the presence of the carbonate ion, $\ce{CO3^{2-}}$.

$\ce{NaHCO3}$ also has the $\ce{CO3^{2-}}$ group, but is it a carbonate?
That is, are bicarbonates a subset of carbonates?

Carbonates are salts and esters of carbonic acid $\ce{H2CO3}$.
In generic nomenclature you may therefore consider a hydrogen carbonate also a subset of all carbonates. They are sometimes also referred to as primary carbonates.
Another considerable point is, that the structural subunit $\ce{O=C(-O-)2}$is still conserved.
In a stricter sense a hydrogencarbonate is only all forms involving $\ce{HCO3-}$, so it appears to be not a generic carbonate, but a class of its own. This refers to its one $\ce{O-H }$ bond and therefore it being a different structure.
However, another point of view should always consider aqueous solution equilibrium.
\begin{aligned}
\ce{HCO3- + H2O &<=> H3+O + \color{\red}{CO3^{2-}}}\\
\ce{HCO3- + H2O &<=> H2CO3 + {}^{-}OH}
\end{aligned}
You can see that carbonate ions will be present to a certain extend in solution. The chemistry will therefore also to some extend resemble the chemistry of the carbonate.
Chemical properties of bicarbonate compound might differ from the pure carbonates - incorporation of water, solubility, ... .
On another page, in organic chemistry the compound class carbonates refers to structures of the type $\ce{\color{\red}{R}-{CO3}-\color{\navy}{R'}}$, where $\ce{\color{\red}{R},\color{\navy}{R'}$\in$ Alkyl, Aryl}$ - hence esters.
To sum it all up, since a more stringent definition of carbonate is not present to me, I would always consider a bicarbonate a carbonate.

I would consider it a carbonate. Chemistry of sodium bicarbonate, or sodium hydrogen carbonate, or simply baking soda, is very similar to that of just sodium carbonate, or $\ce{Na_2CO_3}$.
Adding a strong acid to either $\ce{Na_2CO_3}$ or $\ce{NaHCO_3}$ will effect bubbling. Why? The acid protonates the carbonate or bicarbonate and forms the unstable $\ce{H_2CO_3}$, which is generally referred to as carbonic acid, but is in fact a mixture of carbonic acid and the much less potent acid, hydrated carbon dioxide. $\ce{H_2CO_3}$ decomposes rapidly to water and carbon dioxide. Hence, in the lab, if you wished to test for the present of carbonate in a salt, you could try adding a strong acid and look for bubbling. Copious bubbling is an indication of the presence of a carbonate or a carbonate derivative such as $\ce{HOCO_2^-}$ ($\ce{HCO3^-}$). To verify which carbonate you may simply perform a pH test as the $\ce{CO_3^2-}$ anion is a much strong base than the $\ce{HOCO_2^-}$ ion.


text:
Should binding constants be unitless when deriving fractional occupancy equations from reactions?
It is known that binding (aka association) constants are in fact unitless, as has been discussed here already. However, I'm not a chemist and am confused about when one should or should not use units when working with association constants. One source says:

$K_\text{eq}$ for a reaction with unequal numbers of reactants and products is always given with units, even in published papers.

But why is that? Why not always use unitless values? Is there something inherently wrong with never using units for binding constants?
Consider this example. Let $\ce{A}$ bind to $\ce{X}$ as an n-mer (e.g. $\ce{A}$ can be a transcriptional activator binding to gene promoter). This results in active state, $\ce{X_{A}}$ (e.g. a state leading to gene transcription):
$$\ce{X + nA ->[k_{\text{on}}][k_{\text{off}}] X_{A}}$$
Association constant: $K_\text{A} = \frac{k_\text{on}}{k_\text{off}} $
Assuming equilibrium and law of mass action:
$$K_\text{A} \cdot \text{X} \cdot \mathrm{A^n} = \mathrm{X_A}$$
Now, the fractional occupancy (active states to all states ratio) is:
$$y=\mathrm{\frac{X_A}{X+X_A}} = \frac{K_\text{A}\cdot \mathrm{A^n}}{1 + K_\text{A}\cdot \mathrm{A^n}}$$
As $y$ must be unitless (i.e. it can be interpreted as a probability), for this particular equation, $K_\text{A}$ must have units of $\mathrm{M^{-n}}$ (in general, $\text{concentration}^{-n}$) for the equation to work out, correct (assuming concentration of $\ce{A}$ has units $\mathrm{M}$)? So in this particular case, using unitless $K_\text{A}$ seems wrong, but is it really, or is there something I'm missing?
Now let's extend this equation so that it includes a half-saturation constant $h$, i.e. concentration of $\ce{A}$ required for $y=0.5$ (50% activation). If I'm doing this correctly, we get:
$$y=\frac{K_\text{A}\cdot \mathrm{A^n}}{K_\text{A}\cdot h^n + K_\text{A}\cdot \mathrm{A^n}} = \frac{\mathrm{A^n}}{h^n + \mathrm{A^n}}$$
Note that this is equivalent to the Hill equation for an activator. This generalized equation, unlike the previous one, works just fine regardless of whether or not $K_\text{A}$ is unitless.
Is my understanding of this correct, and does the choice of unitless vs non-unitless binding constant indeed depend on the formulation of a specific equation?

The equilibrium constant does not need to be unitless, because it is depending on its definition. See goldbook:

Equilibrium Constant
  Quantity characterizing the equilibrium of a chemical reaction and defined by an expression of the type $$K_x = \Pi_B x_B^{\nu_B},$$ where $\nu_B$ is the stoichiometric number of a reactant (negative) or product (positive) for the reaction and $x$ stands for a quantity which can be the equilibrium value either of pressure, fugacity, amount concentration, amount fraction, molality, relative activity or reciprocal absolute activity defining the pressure based, fugacity based, concentration based, amount fraction based, molality based, relative activity based or standard equilibrium constant (then denoted $K^\circ$ ), respectively.

The standard equilibrium constant is always unitless, as it is defined differently (goldbook)

Standard Equilibrium Constant $K$, $K^\circ$ (Synonym: thermodynamic equilibrium constant)
  Quantity defined by $$K^\circ = \exp\left\{-\frac{\Delta_rG^\circ}{\mathcal{R}T}\right\}$$  where $\Delta_rG^\circ$  is the standard reaction Gibbs energy, $\mathcal{R}$  the gas constant and $T$  the thermodynamic temperature. Some chemists prefer the name thermodynamic equilibrium constant and the symbol $K$.

In your example both would work if you straighten your definitions, involving concentrations
\begin{aligned}
K_A \cdot c(\ce{X}) \cdot c^n(\ce{A}) &= c(\ce{X_{A}}),\\
y=\frac{c(\ce{X_A})}{c(\ce{X})+c(\ce{X_{A}})} &= \frac{K_A\cdot c^n(\ce{A})}{1 + K_A\cdot c^n(\ce{A})},
\end{aligned}
or activities
\begin{aligned}
K^\circ_A \cdot a(\ce{X}) \cdot a^n(\ce{A}) &= a(\ce{X_{A}}),\\
y=\frac{a(\ce{X_A})}{a(\ce{X})+a(\ce{X_{A}})} &= \frac{K^\circ_A\cdot a^n(\ce{A})}{1 + K^\circ_A\cdot a^n(\ce{A})}.
\end{aligned}
In both cases $y$ remains unitless.
In experimental chemistry it is much easier to observe concentrations instead of activities. Therefore to fist named constant is more often in use. A straightforward use of the thermodynamical equilibrium constant might sometimes prove a little tricky.
For reasonable dilutions one can simply substitute activity with a unitless concentration as given through
$$a(\ce{Y})=\gamma\frac{c(\ce{Y})}{c^\circ},$$
with $\gamma\approx1$ for $c(\ce{Y})\to0$ and $c^\circ=1\:\mathrm{mol/L}$.