Fick's Second Law of Diffusion In the book "Foundations of Materials Science and Engineering" (Smith and Hashemi) the following is written:

For cases of non-steady state diffusion in which diffusivity is independent of time, Fick's second law of diffusion applies, which is

$$\frac{\partial c_{x}}{\partial t}=\frac\partial{\partial x}\left(D\frac{\partial c_{x}}{\partial x}\right)$$

In what cases are materials' diffusivity dependent on time, and in what cases are they not?

Diffusivity is not dependent on time. It is dependent on temperature (and material). Fick's second law holds because the diffusivity, $D$, is constant with respect to depth, $x$, and time, $t$, so it is treated as a constant in the equation.

$$\frac{\partial c_{x}}{\partial t}=\frac\partial{\partial x}\left(D\frac{\partial c_{x}}{\partial x}\right) = D\left(\frac{\partial^2c_x}{\partial x^2}\right)$$

Fick's second law of diffusion applies to non steady state diffusion, which concentration at a depth $x$, $c_x$ is a function of time.


Why constant volume combustion gives higher energy than constant pressure? I have noticed when I have done some combustion equilibrium that enthalpy is higher when doing a constant volume combustion than pressure constant combustion.
Why is it so ? What are the reasons explaining that ?
Is it because at constant pressure combustion the variation of enthalpy is equal to heat transfert $\Delta H = Q $ whereas at constant volume combustion the variation of enthalpy is $\Delta H = Q + V\Delta p$ ?
If so that suppose that heat transfer $Q$ is the same for both constant pressure and constant volume combustions. Is that so ?
Thank you

Combustions of organic stuffs produce a greater number of gaseous moles than in the reactants. When done at constant pressures, a part of the energy produced by the combustion must be used for repelling the outer atmosphere. This is a work and it costs some enegy. This is why a combustion at constant pressure produces less energy than at constant volume.


What are the intermolecular forces between polychloroethylene? Does polychloroethylene have only van der Waals/London forces between its molecules? Surely, if all the chlorine atoms are on one side, due to the fact that chlorine is more electronegative than carbon, shouldn’t it have permanent dipole-dipole forces?

No, but for polymers of any reasonable size, the dispersion (and excluded volume) forces completely overwhelm any dipole-dipole forces, since they grow with the size of the molecule and dipole-dipole forces do not.  Also, bear in mind that the chlorines aren't all on one side of the molecule first because the way it's manufactured tends to put Cls randomly on each side of the chain, and second because the molecule has considerable conformational flexibility.  That's true even in the solid state, which for polymers is usually substantially glassy with interspersed domains of crystal.


Is the P-Cl bond energy same in PCl5 and PCl3? Is the $\ce{P-Cl}$ bond energy same in $\ce{PCl5}$ and $\ce{PCl3}$? Why?

According to wolfram alpha both have $\pu{326 kJ mol-1}$ energy. But my study material says that they are not same. I'm confused. Can someone clarify?

The $\ce{P-Cl}$ bond in $\ce{PCl3}$ is stronger ($\pu{322 kJ/mol}$) than that in $\ce{PCl5}\ (\pu{260 kJ/mol})$. However the $\pu{326 kJ/mol}$ value comes from the average of all $\ce{P-Cl}$ bonds found in all known molecules containing $\ce{P-Cl}$ bonds (including but not limited to $\ce{PCl3}$ and $\ce{PCl5}$).


How to activate charcoal I readily admit to being a pyrophiliac and can generally be found near fire when there is fire. (Fire good, tree bad.)  One of the byproducts of fire, naturally, is charcoal.  I was looking into ways to activate charcoal instead of buying it over the counter - because how cool would it be able to just make your own - but my research so far isn't very promising.

Apparently, I require intense heat and very little oxygen in order to turn the home and garden variety of charcoal into activated charcoal (as one might find in Norit).

Is it possible to activate your own charcoal (preferably in one's backyard), and if so, does anyone have any ideas or suggestions that I can use to turn my meagre amount of (regular) charcoal into activated charcoal?

Note that like most of the colleagues on this site with practical lab experience, I've used activated charcoal once in a while, but I never bothered to make it myself.

Supposed that you live on the countryside with enough space around, you can make your own charcoal in a earth-covered pile or an oven - unless local environmental regulations ban that.

Activation, even when starting from ordinary charcoal is different. The gas activation requires heating (and partial burning) of the charcoal at up to 1000 °C in a stream of air, carbon dioxide or water vapour. I'd rather not do it in my backyard.


Do the names of some elements have anything to do with planets? There are elements like neptunium and plutonium in the periodic table. Did their discovery have anything to do with Neptune and Pluto? Or are they randomly assigned with such names?

Neptunium (93) and plutonium (94) were so named because they followed uranium (92) on the periodic table.

Uranium was named after Uranus. German chemist Martin Heinrich Klaproth discovered uranium in 1789, eight years after William Herschel discovered Uranus. At the time, uranium was the densest element known and Uranus was the farthest planet from the sun known.


What is the symmetry of dxy orbital? Considering the sign of orbital and assuming the $z$-axis as a principal axis, for me, it looks like that it has two perpendicular $C_2$ axes that penetrate the lobes, so I think it is $C_\mathrm{2v}$. But the book (Inorganic chemistry by Miessler and Tarr) says it is $D_\mathrm{2h}$.

Note that $C_\mathrm{2v}$ is a subgroup of $D_\mathrm{2h}$. This means that you're just overlooking some of the symmetry elements. In addition to the $C_2$ axes penetrating the lobes "end to end", you also have a $C_2$ axis perpendicular to the orbital plane through its center.

Your $C_\mathrm{2v}$ guess is also inconsistent because that point group only has one $C_2$ axis, but you said yourself that you already found two.


Is entropy of surroundings a state function? It is a pretty common well known fact that the entropy of a system is a state function ie. it doesn't depend upon the path taken by the process as long as the initial and final states are the same.
I have a doubt whether we can say the same for surroundings or even the universe.
My first thought is that the entropy of the universe would be continuously increasing or at best it would be constant. It can never decrease. Thus the universe can never be at a state where it was before. The talk of state function seems strange here.
Still I am not very sure and I have no idea whether entropy of surroundings is a state function or not.
Any help would be appreciated.
Thanks

In reality, when we talk about the surroundings, what we are referring to isn't really the entire universe, outside what we call "the system."  What we are really referring to is a subregion of the entire universe, wherein it is possible to carry out changes reversibly without encompassing the whole rest of the universe.  So, when we talk about the entropy change of the surroundings, we are not including spontaneous processes occurring light years away in other galaxies.  And, when we say that the entropy change for system plus the surroundings is zero, we are really only including the selected subregion of surroundings chosen to represent our sub-universe.


Band gap of fullerenes From what I have read about fullerenes, the lower fullerenes like $\ce{C60}$ or $\ce{C70}$ have higher bandgaps around 3.5 eV or such, while the higher fullerenes have much smaller bandgaps of the order of 0.5 eV. Initially I thought that this could be explained using electron delocalization on the surface area of the sphere, but later I read that the fullerenes were not necessarily super-aromatic. What could be the reason behind such a dip of the HOMO-LUMO gap?

One of the explanation for answering this question can be provided as, moving from lower to higher Fullerenes the number of atoms in the structure increase, increasing the number of atoms in the system results in reduction of the discreteness of energy states, which results in decrease of the gap between HOMO and LUMO, thus the band gap is reduced.
You can also check for the particle in box model of the Quantum Physics for better understanding.


Shape of Weak-Strong Acid-Base Titration I can understand why in any titration, the pH changes really quickly near the equivalent point.

However, in a titration of either weak base-strong acid or weak acid-strong base, the weak acid/base first has a steep rate of change in pH in the beginning, then slows down at the midpoint, then becomes steep again closer to the equivalence point. Why is this the case? (See picture below). Also, it seems that the weaker the acid/base, the more pronounced the initial quick rate of change.
Also, why is the "midpoint" (not the equivalence point) so special anyways?
You can refer to this picture. The steep drop portion is from 0-10 mL.

Suppose $\ce{NH3(aq)}$ is being titrated with $\ce{HCl(aq)}$. The products are $\ce{NH4+ (aq)}$ and $\ce{Cl- (aq)}$. As the titration proceeds [$\ce{NH3}$] decreases and [$\ce{NH4+}$] increases. The presence of $\ce{NH3}$ and $\ce{NH4+}$ together constitutes a buffer system, in this case a weak base and its conjugate acid. When molarities of both species are relatively large, the buffering capacity is significant.

Halfway to the equivalence point, the molarities of both species are equal, which allows for the greatest possible buffering action. In the very early part of the titration, the molarity of $\ce{NH4+}$ is not great enough to allow much of a buffering effect, so the pH falls rapidly with the addition of $\ce{HCl}$. But more toward the point where [$\ce{NH3}$] becomes equal to the [$\ce{NH4+}$], the buffering action is increasingly able to "resist" changes in pH so the slope of the titration curve becomes more shallow. Past this point, the [$\ce{NH3}$] steadily diminishes to the point that buffering is no longer possible and the pH dives through the equivalence point as we expect.

Past the equivalence point, only $\ce{NH4+}$ is present so no buffering can occur and there are no inflections like the one that occurs at the start of the titration. The importance of the "midpoint" as it is labeled in the illustration is that pH vs. titrant volume analyses are sometimes done specifically to determine the K$_a$ or K$_b$ of a weak acid or weak base, respectively. Continuing with the $\ce{NH3}$ and $\ce{HCl}$ example, the K$_b$ for $\ce{NH3}$ is $[\ce{NH4+}][\ce{OH-}]/[\ce{NH3}]$. Halfway to the equivalence point, [$\ce{NH4+}$]=[$\ce{NH3}$] and we have K$_b = [\ce{OH-}$] or pK$_b$ = pOH.


Regioselectivity in Diels–Alder reaction between 2-nitrobuta-1,3-diene and acrylonitrile Since $\ce{-CN}$ shows strong −I effect, there will be δ+ charge on the carbon adjacent to it. Similarly, $\ce{-NO2}$ shows strong −I and –M effects. So, I proposed that π-electrons are being donated to the electron-deficient parts:

But the answer given in my book has $\ce{-CN}$ group at the third position to $\ce{-NO2}$ group. Can someone please tell me where I have got my concepts wrong?

You pointed out correctly that $-CN$ has a great $-I$ effect but it won't bring a $+\delta$ on adjacent carbon atom, reason being it's instability. Apply electromeric effects on propanenitrile.

If you need help then just comment below :-)


Finding the equivalent point of a weak polyprotic acid when reacted with a strong base I have a graph of a titration of a weak polyprotic acid with a strong base.
I graphed $\mathrm {pH}$ versus amount of strong base added.
Now since the acid is polyprotic, how do I determine which equivalence point is the final one?
Also not sure how to find the initial concentration of the weak acid? I know the volume I put in, as well as the volume of strong base at any given time.

A simple way to determine whether an equivalence point in a titration is the last one is to figure out the volume of titrant added for the first titration point, then realize all subsequent equivalence points will be at exact integer multiples of this volume, under the assumption that the analyte contains only one acid (a mixture of acids won't behave the same). When an integer multiple of the first equivalence point volume fails to be an equivalence point itself, then there will be no further equivalence points.
For example, let us imagine a generic polyprotic acid with formula $\ce{H_{k}A}$, being titrated with a strong monoprotic base such as $\ce{NaOH}$. Starting from the fully protonated acid, let us neutralize each ionizable hydrogen in steps. Let the acid  $\ce{H_{k}A}$ being titrated have an initial concentration $c_{acid}$ in a solution of volume $V_{acid}$, and let the titrant have a concentration $c_{base}$ and a total volume added $V_{base}$.
By definition, to reach the first equivalence point ($V_{base}=V_{eq\ 1}$), we consider the reaction which represents the neutralization of only the first proton:
$$\ce{H_{k}A + OH- -> H_{k-1}A- + H2O}$$
Each one mole of acid  $\ce{H_{k}A}$ must surrender exactly one mole of hydrogen ions $\ce{H+}$, each of which will react with exactly one mole of hydroxide ions $\ce{OH-}$. Thus:
$$n_{\mathrm{released}\ \ce{H+}}=n_\mathrm {acid}=c_\mathrm {acid} \times V_\mathrm {acid}$$
$$n_{\mathrm{added}\ \ce{OH-}}=n_\mathrm {base}=c_\mathrm {base} \times V_\mathrm{base}=c_\mathrm {base} \times V_\mathrm {eq\ 1} $$
$$n_{\mathrm{released}\ \ce{H^+}}= n_{\mathrm{added}\ \ce{OH-}} \rightarrow\  c_\mathrm {acid} \times V_\mathrm {acid} = c_\mathrm {base} \times V_\mathrm {eq\ 1}$$
$$V_\mathrm {eq\ 1} = \frac{c_\mathrm {acid} \times V_\mathrm{acid}}{c_\mathrm {base}}$$
Now, to reach the second equivalence point ($V_{base}=V_{eq\ 2}$), we consider the reaction which represents the neutralization of the first and second protons:
$$\ce{H_{k}A + 2 OH- -> H_{k-2}A^{2-} + 2 H2O}$$
Each one mole of $\ce{H_{k}A}$ must surrender exactly two moles of $\ce{H+}$, each of which will react with exactly one mole of $\ce{OH-}$. The relevant equations are now:
$$n_{\mathrm{released}\ \ce{H+}}=2\times n_\mathrm {acid}=2\times c_\mathrm {acid} \times V_\mathrm {acid}$$
$$n_{\mathrm{added}\ \ce{OH-}}=n_\mathrm {base}=c_\mathrm {base} \times V_\mathrm {base}=c_\mathrm {base} \times V_\mathrm{eq\ 2} $$
$$n_{\mathrm{released}\ \ce{H+}}= n_{\mathrm{added}\ \ce{OH-}} \rightarrow\  2\times c_\mathrm {acid} \times V_\mathrm {acid} = c_\mathrm {base} \times V_\mathrm {eq\ 2}$$
$$V_\mathrm {eq\ 2} = 2\times \frac{c_\mathrm {acid} \times V_\mathrm {acid}}{c_\mathrm{base}} = 2\times V_\mathrm {eq\ 1} $$
Notice how the added volume to reach the second equivalence point is exactly twice that added to reach the first equivalence point. This pattern repeats, i.e. the third equivalence point will happen at $V_\mathrm {base}=3\times V_\mathrm {eq\ 1}$, and the $k$-th equivalence point will happen at $V_\mathrm {base}=k\times V_\mathrm {eq\ 1}$. The equivalence points are equally spaced.
In short, if the titration of a pure polyprotic acid shows the first sudden pH swing after adding $\pu{15 mL}$ of the base titrant, then $V_\mathrm {eq\ 1}$ equals 15 mL, and the only possible next equivalence points are at $\pu{30 mL}$, $\pu{45 mL}$, $\pu{60 mL}$, etc. If during the titration another sudden pH swing is observed at $\pu{30 mL}$, but not at $\pu{45 mL}$, then you can deduce that there is no third equivalence point, and hence the acid is diprotic.


Why does oxalic acid have exactly two water molecules of crystallization? Why does oxalic acid have exactly two water molecules of crystallization?  In other words, in the formula $\ce{C2H2O4\cdot2H2O}$, why is the number before the $\ce{H2O}$ exactly $2$?

I tried to see this from the structure of oxalic acid. Also, I tried to invoke hydrogen bonds. However, I don't see the reason why there are two water molecules of crystallization from this.

Also, with how many precision is it two?

As @Ivan Neretin states, the crystal structure determines how much water can be accommodated. For example, $\ce{CuSO4}$, cupric sulfate or copper(II) sulfate, may be found as a mineral with three, five or seven molecules of bound water:

Bonattite (trihydrate mineral)
Bluestone (pentahydrate)
Boothite (heptahydrate mineral)

Of course, it can also be heated to form the anhydrous salt.

One could say different crystal forms have room for different amounts of water, or the converse: by adding or removing water, the crystal structure is changed.


What will be the major product of of the Wurtz reaction of 3,6-dibromocyclohexa-1,4-diene? According to me, the answer should be (A) because that's a Wurtz reaction, but the answer given in the book is (B).

The answer is (B) because (B) is benzene, which is the prototypical aromatic hydrocarbon. The aromaticity of (B) increases its thermodynamic stability relative to (A), which is called Dewar benzene (mistakenly attributed to James Dewar), which is notoriously difficult to synthesis and isolate. It reverts back to benzene with a half-life of two days.

Generally, if a pathway exists for a reaction to produce an aromatic product, then the reaction likely will do so. In this particular case, the diradical intermediate that might be produced as part of the Wurtz reaction's mechanism, is the triplet excited state of benzene.


Matching the NMR to a compound I was given a molecule that I had to identify the NMR of. The molecule is 2-methylpent-1-ene (shown below):

The correct NMR spectra for this molecule is:

My question is, why is this the right spectra for this molecule. Particularly, how and why did the single peak come up? As far as I know, the only two single peaks are the two that are downfield (aka closest to the $\ce{sp2}$ hybridized carbon).

If you look at the structure closely, you'll find that there are no Hydrogens on the C with the double bond attached. Adjacent to this C is the methyl group. Due to the n+1 rule, the singlet should be due to that Methyl group connected to the "Hydrogenless" Carbon. Hope this simplified answer helps


How to liquefy or extract water from superabsorbent polymer I am finding a good way to liquefy or extract water from superabsorbent polymer. From what I remember it was salt that can liquefy it. Since salt when dissolve will have sodium cation and there for naturalize the polymer chain making it contract.  Is that true?. Thank very much.

This may be true but I suggest another type of explanation.  If a super absorbent polymer absorbs its maximum in distilled water it is called a fully swollen hydrogel.  There are two ways of decreasing the amount of water stored without using heat.  The first one is to increase the salt load of additional water and place the hydrogel within this salted water.  The salt in the water will decrease the total water absorbed by the hydrogel and the hyrogel will shrink, giving up some of its water.  A more complete and full way to make the hydrogel to give up its water is the addition of an acid.  Lowering the pH of the hydrogel to below 3.5 will convert all the sodium carboxylates to the free acid form which does not absorb water.  These hydro gels are pH sensitive and if you plot pH vs absorbed fluid you would find a bell curve that starts around 4.5 or 5.0, tops out around 7.5 and decreases as you increase the pH until around 10.5 to 11.0.


Difference between thermodynamic and kinetic stability What is the difference between thermodynamic and kinetic stability? I'd like a basic explanation, but not too simple. For example, methane does not burn until lit -- why?

To understand the difference between kinetic and thermodynamic stability, you first have to understand potential energy surfaces, and how they are related to the state of a system.

A potential energy surface is a representation of the potential energy of a system as a function of one or more of the other dimensions of a system. Most commonly, the other dimensions are spatial. Potential energy surfaces for chemical systems are usually very complex and hard to draw and visualize. Fortunately, we can make life easier by starting with simple 2-d models, and then extend that understanding to the generalized N-d case.

So, we will start with the easiest type of potential energy to understand: gravitational potential energy. This is easy for us because we live on Earth and are affected by it every day. We have developed an intuitive sense that things tend to move from higher places to lower places, if given the opportunity. For example, if I show you this picture:

You can guess that the rock is eventually going to roll downhill, and eventually come to rest at the bottom of the valley.

However, you also intuitively know that it is not going to move unless something moves it. In other words, it needs some kinetic energy to get going.

I could make it even harder for the rock to get moving by changing the surface a little bit:

Now it is really obvious that the rock isn't going anywhere until it gains enough kinetic energy to overcome the little hill between the valley it is in, and the deeper valley to the right.

We call the first valley a local minimum in the potential energy surface. In mathematical terms, this means that the first derivative of potential energy with respect to position is zero:

$$\frac{\mathrm dE}{\mathrm dx} = 0$$

and the second derivative is positive:

$$\frac{\mathrm d^2E}{\mathrm dx^2} \gt 0$$

In other words, the slope is zero and the shape is concave up (or convex).

The deeper valley to the right is the global minimum (at least as far as we can tell). It has the same mathematical properties, but the magnitude of the energy is lower – the valley is deeper.

If you put all of this together, (and can tolerate a little anthropomorphization) you could say that the rock wants to get to the global minimum, but whether or not it can get there is determined by the amount of kinetic energy it has.

It needs at least enough kinetic energy to overcome all of the local maxima along the path between its current local minimum and the global minimum.

If it doesn't have enough kinetic energy to move out of its current position, we say that it is kinetically stable or kinetically trapped. If it has reached the global minimum, we say it is thermodynamically stable.

To apply this concept to chemical systems, we have to change the potential energy that we use to describe the system. Gravitational potential energy is too weak to play much of a role at the molecular level. For large systems of reacting molecules, we instead look at one of several thermodynamic potential energies. The one we choose depends on which state variables are constant. For macroscopic chemical reactions, there is usually a constant number of particles, constant temperature, and either constant pressure or volume (NPT or NVT), and so we use the Gibbs Free Energy ($G$ for NPT systems) or the Helmholtz Free Energy ($A$ for NVT systems).

Each of these is a thermodynamic potential under the appropriate conditions, which means that it does the same thing that gravitational potential energy does: it allows us to predict where the system will go, if it gets the opportunity to do so.

For kinetic energy, we don't have to change much - the main difference between the kinetic energy of a rock on a hill and the kinetic energy of a large collection of molecules is how we measure it. For single particles, we can measure it using the velocity, but for large groups of molecules, we have to measure it using temperature. In other words, increasing the temperature increases the kinetic energy of all molecules in a system.

If we can describe the thermodynamic potential energy of a system in different states, we can figure out whether a transition between two states is thermodynamically favorable – we can calculate whether the potential energy would increase, decrease, or stay the same.

If we look at all accessible states and decide that the one we are in has the lowest thermodynamic potential energy, then we are in a thermodynamically stable state.

In your example using methane gas, we can look at Gibbs free energy for the reactants and products and decide that the products are more thermodynamically stable than the reactants, and therefore methane gas in the presence of oxygen at 1 atm and 298 K is thermodynamically unstable.

However, you would have to wait a very long time for methane to react without some outside help. The reason is that the transition states along the lowest-energy reaction path have a much higher thermodynamic potential energy than the average kinetic energy of the reactants. The reactants are kinetically trapped - or stable just because they are stuck in a local minimum. The minimum amount of energy that you would need to provide in the form of heat (a lit match) to overcome that barrier is called the activation energy.

We can apply this to lots of other systems as well. One of the most famous and still extensively researched examples is glasses.

Glasses are interesting because they are examples of kinetic stability in physical phases. Usually, phase changes are governed by thermodynamic stability. In glassy solids, the molecules would have a lower potential energy if they were arranged in a crystalline structure, but because they don't have the energy needed to get out of the local minimum, they are "stuck" with a liquid-like disordered structure, even though the phase is a solid.


Can you explain how we get gold(II) sulfide? Can you explain how we get gold(II) sulfide, when gold has oxidation numbers of 1 and 3?

Brauer's Handbook of Preparative Inorganic Chemistry ([1, p. 1062] and references therein) lists the following method for gold(II) sulfide synthesis:

  $$\ce{$\underset{1/10}{}$\qquad$\underset{242.7}{\ce{8 AuCl3}}$ + $\underset{20.2\,l.}{\ce{9 H2S}}$ + $\underset{7.2}{\ce{4 H2O}}$ = $\underset{183.3}{\ce{8 AuS}}$ + $\underset{87.5}{\ce{24 HCl}}$ + $\underset{9.8}{\ce{H2SO4}}$}$$

  A neutral 1-3% solution of $\ce{AuCl3}$ is precipitated in the cold (the temperature must not exceed $\pu{40 ^\circ C}$) with $\ce{H2S}$ or an alkali sulfide. The precipitate is filtered, thoroughly washed with water, and
  treated with alcohol, anhydrous ether, $\ce{CS2}$ and again with ether.
  The product is dried at $\pu{130 ^\circ C}$ over $\ce{P2O5}$.

  Alternate method: A solution of $\ce{AuCl3}$ is added in drops to an
  aqueous solution of sodium dithiosulfatoaurate(I) [...].

  Properties: Formula weight 229.07. Deep black. Insoluble in water and acids; soluble in aqua regia and solutions of potassium cyanide and alkali polysulfides. Resistant to KOH in the cold, decomposes after prolonged boiling, liberating gold. Thermal decomposition begins at $\pu{140 ^\circ C}$.

Reference

Handbook of Preparative Inorganic Chemistry, 2nd ed.; Brauer, G., Ed.; Academic Press: New York; London, 1965; Vol. 2.


Electrolytic Refining of Silver Nitrate I read that the electrolytic refining of Silver is done using dilute silver nitrate as an electrolyte and a small percentage of free $\ce{HNO3}$. The following reactions are possible at the anode:

$$\ce{Ag ->Ag+ + e-}$$
$$\ce{ OH- -> 1/4O2 + 1/2H2O + e-}$$

But the second reaction has $E^\circ = \pu{+0.4V}$ while the first reaction has
$E^\circ = \pu{+0.8V}$. So why does $\ce{Ag}$ of the anode get oxidised? Is it because of a low concentration of $\ce{OH-}$ ions due to poor dissociation? Or is there another reason?

Electrolysis or electrolytic refining is a technique used for extraction as well as purification of metals obtained by other refining methods. In the electrolytic refining process, a block or a strip of impure metal is used as the anode, and a thin sheet of pure metal is used as the cathode. Both the cathode and anode are dipped in an electrolytic cell containing an aqueous solution of the metal salt. Upon the application of direct electric current of a suitable voltage, pure metal is deposited at the cathode (reduction) by the dissolution of impure metal at the anode (oxidation).

Since electrolytic refining id done in aqueous medium, the key to successful refining is the oxidation of the metal should be more facile than the oxidation of water (see the standard oxidation potentials below). Hence the solid metal would dissolve into the solution as cataions, leaving behind many of the impurities, which are inactive to electrolysis. The possible anodic oxidation reactions would be:

$$\ce{Ag <=> Ag+ + e-} \quad E_\mathrm{ox}^\circ = \pu{-0.7996 V} \tag1$$
$$\ce{4H2O <=> O2 + 4H+ + 4e- } \quad E_\mathrm{ox}^\circ = \pu{-1.229 V} \tag2$$

When value of $E_\mathrm{ox}^\circ \ (\ce{Ag})$ is compared with that of $E_\mathrm{ox}^\circ \ (\ce{H2O})$, it is safe to say that $\ce{Ag}$ oxidation at anode is more spontaneous (less negative value) than that of $\ce{H2O}$ (greater negative value) at anode. As a result, impure $\ce{Ag}$ anode will be slowly dissolved into solution as $\ce{Ag+}$ ins as direct current applied.

Similarly, the possible cathodic reduction reactions would be:

$$\ce{Ag+ + e- <=> Ag}  \quad E_\mathrm{red}^\circ = \pu{0.7996 V} \tag3$$
$$\ce{ 2H2O + 2e- <=> H2 + 2OH-} \quad E_\mathrm{red}^\circ = \pu{-0.8277 V} \tag4$$

Since the refining solution is dilute $\ce{HNO3}$ to provide the counter-ion for $\ce{Ag+}$ cation, we should consider possibility of  another reduction reaction at cathode:

$$\ce{NO3- + 3H+ + 2e- <=> HNO2 + H2O} \quad E_\mathrm{red}^\circ = \pu{0.934 V} \tag5$$
$$\text{And, } \ce{NO3- + 4H+ + 3e- <=> NO + 2H2O} \quad E_\mathrm{red}^\circ = \pu{0.957 V} \tag6$$

When the value of $E_\mathrm{red}^\circ \ (\ce{Ag})$ is compared with that of $E_\mathrm{red}^\circ \ (\ce{H2O})$, it is clear that $\ce{Ag}$ reduction at cathode is much more prominent (high positive value) than that of $\ce{H2O}$ (greater negative value) at cathode. However, the competition reduction would happen as well when comparing the value of $E_\mathrm{red}^\circ \ (\ce{Ag})$ with those of $E_\mathrm{red}^\circ \ (\ce{NO3-})$, both of which  are slightly higher positive values. However, if you use appropriately dilute nitric acid to begin with (c.f., Nernst equation), you will be able to avoid nitrate reduction at cathode.


Chemicals in a Rubber Compound - NSR 20, Accel CS, Accel DPG, Anitiox Q, Antioz PMB I have a recipe for a rubber compound and am trying to understand some of the ingredients.

The ones I do not recognize are

Compound A

NSR 20
Accel CS
Accel DPG
Anitiox Q
Antioz PMB

Compound B

SMR 20
Accel NS
Accel DPG
PVI

Compound C

Orox HR
Orac CS
Orac TMTD

Is there a place to look these up on the web?

NSR 20 and SMR 20 are natural rubber (Nigerian Standard Rubber and Standard Malaysian Rubber)

"Accel" and "Orac" mean accelerator.

Accel DPG is diphenyl guanidine.

"Antiox" and "orox" mean antioxidant

"Antioz" means antiozonant

"Orac CS" is likely N-cyclohexyl-2-benzothiazole sulphenamide

"Orac TMTD" is likely Tetramethylthiuram disulfide

"PVI" may refer to pre-vulcanization inhibitor


Why does ATP has a lower entropy than ADP I was confused about why ATP has a lower entropy than ADP when ATP has 1 more phosphate group. According to https://chem.libretexts.org/Textbook_Maps/Biological_Chemistry/Metabolism/ATP%2F%2FADP#Why_is_ATP_hydrolysis_an_exergonic_reaction.3F It states that

  The entropy, which is the level of disorder, of ADP is greater than that of ATP.  Therefore, due to thermodynamics, the reaction spontaneously occurs because it wants to be at a higher entropy level. Also, the Gibbs' free energy of ATP is higher than that of ADP. Naturally, molecules want to be at a lower energy state, so equilibrium is shifted towards ADP.

Can someone please explain to me why does ATP has a lower entropy than ADP?

If you treat the entropy as a "level of disorder" the overall entropy of ADP and phosphate seems to be higher than the entropy of ATP.


Why is osmium(VIII) oxide so toxic? Osmium(VIII) oxide or osmium tetroxide is a volatile chemical compound with the formula of $\ce{OsO4}$. It is highly toxic for supposedly irritating mucous membranes, but I'm not too sure if it is that or because it is a really strong oxidizer.
PubChem rates it as corrosive and highly toxic and according to Wikipedia, the toxic effects of $\ce{OsO4}$ are as follows:

$\ce{OsO4}$ is highly poisonous. In particular, inhalation at concentrations well below those at which a smell can be perceived can lead to pulmonary edema and subsequent death. Noticeable symptoms can take hours to appear after exposure.
$\ce{OsO4}$ will irreversibly stain the human cornea, which can lead to blindness. The permissible exposure limit for osmium(VIII) oxide (8 hour time-weighted average) is $\pu{2 µg/m3}$

How is $\ce{OsO4}$ toxic in nature? What makes it so?

Metallic osmium is harmless but it reacts with oxygen at room temperature, forming volatile osmium tetroxide. Any other osmium compounds are also converted to the tetroxide if oxygen is present. This makes osmium tetroxide the main source of contact with the environment and human body. Exposure to osmium tetroxide cause harm to the skin, eyes, and respiratory system.

So, why makes osmium tetroxide so harmful? (TL;Dr: Because to its oxidization ability)

There is a paper1 which studied the effect of osmium tetroxide in human body and how it can be a lethal bioweapon. The following is the text from the paper:

Osmium tetroxide is highly volatile and penetrates skin readily, and
is very toxic by inhalation, ingestion, and skin contact. Airborne low
concentrations of osmium tetroxide vapor can cause lung congestion and
skin or eye damage.
[...]
Osmium tetroxide is a rapid, indiscriminative oxidizer that does not distinguish between organic tissue and inorganic materials. An
inhalational toxicity study with rabbits proved futile, because of the
rapid reduction of $\ce{OsO4}$ by the skin, hair, mucous membranes,
etc., as well as by the chamber walls. Inhalation, ingestion,
contact with skin and with mucous membranes may all result in severe
consequences. Due to its high vapor pressure (7 mmHg at 20 °C), most
exposures are to vapor. These can cause severe chemical burns to the
eyes, skin and respiratory tract. Very short-term contact with the
vapor may cause lacrimation, accompanied by cough, headaches and
dizziness. $\ce{OsO4}$ may cause irreversible blindness by turning the
cornea black. Symptoms may not be noticeable until several hours after
the exposures. Another severe delayed effect following inhalational
exposure is acute lung injury, which may be followed by
non-cardiogenic pulmonary edema Direct contact with osmium tetroxide
solution will turn the skin black (severe chemical burns due to strong
oxidizing properties). Painful burns or contact dermatitis may
result, depending on the concentration.

The paper compares the LD50 value of osmium tetroxide with other chemical warfare agent and thus proves how lethal it can be as a bioweapon:

The paper also notes that osmium tetroxide can rapidly reduced to relatively inert compounds by e.g. ascorbic acid or polyunsaturated vegetable oils (such as corn oil which contains double/triple bonds. $\ce{OsO4}$ can attack those bonds and can convert to non-toxic vicinal diols).
Reference

Makarovsky, Igor & Markel, Gal & Hoffman, Azik & Schein, Ophir & Finkelstien, Arseny & Brosh-Nissimov, Tal & Tashma, Zeev & Dushnitsky, Tsvika & Eisenkraft, Arik. (2007). Osmium Tetroxide: A New Kind of Weapon. The Israel Medical Association journal : IMAJ. 9. 750-2.
McLaughlin AIG, Milton R, Perry KMA. (1946). Toxic manifestations of osmium tetroxide. Br J Ind Med 3:183-186.


Negative hyperconjugation Compare the acidity of the following compounds: $\ce{CF3OH}$ and $\ce{CCl3OH}$.

My attempt: according to me $\ce{CCl3}$ group will show a stronger Negative hyperconjugation effect than $\ce{CF3}$, and thus would stabilize the anion more effectively.

But the answer is $\ce{CF3OH}$ is more acidic. What is wrong in my approach?

$\ce{CF3OH}$ is more acidic because of the inductive effect of the $\ce{-CF3}$ group: Fluorine is very electronegative and therefore it pulls the electron-density toward itself weakening the $\ce{O-H}$ bond in $\ce{CF3OH}$.
Chlorine is less electronegative than fluorine, thus $\ce{CCl3OH}$ is less acidic than $\ce{CF3OH}$.
Because $\ce{H}$ is even less electronegative, then $\ce{CH3OH}$ is less acidic than all its halogenated derivatives.

A related question can be found here:
Why is CF3COOH exceptionally acidic?

The negative hyperconjugation is a type of resonance effect where there is a donation of electron density from a filled π- or p-orbital to a neighboring σ*-orbital. It does not occur in your molecules.


Accuracy of Quantum Chemistry ML models I am trying to compare the performance of few Quantum Chemistry property prediction ML models. I was looking at the following table from DOI:
10.1039/c7sc02664a

The problem is that it does not contain any units. A preprint (https://arxiv.org/pdf/1909.00259.pdf, Table 4) suggests that the energy units are Hartree. Among energy properties, HOMO, LUMO, and gap have reasonable values for Hartree unit. However, values reported for U0, U, H, and G look absurd. As a comparison, look at MAE for mean value of each property:

These are in [eV] units (1 Ha = 27.212 eV). In other words, for instance MPNN models MAE for U property is 55.78 [eV] vs 8.25 [eV] for the mean baseline. This does not make sense at all. [k cal/mol] sounds more reasonable but then why half the energies in one unit the other half in other unit, and then why there are no units in a paper published by Royal Society of Chemistry. Seems like I'm missing something. Thanks!

Yes, you're exactly right - multiple papers in chemistry ML drop the units.

There are even comparisons (usually by statistics, ML or comp. sci. researchers) where models are compared by "averaging" errors down a column like that. Of course that's meaningless, since you can't average electron volts or Hartree (energies), Debye (dipole moments), and volume (polarizabilities).

Worse, energies are often computed as atomization energies - so for large molecules, they can be enormous...

In my opinion, a more relevant and meaningful statistic would be the Mean Absolute Percent Error (MAPE) which is unitless and easier to understand (e.g., 1% error? 0.01% error?)

That said, the original QM9 paper gives the units Table 3:
Scientific Data (2014) 1, art. 140022


What type of sensor technologies exist to determine the chemical composition of a polymer? Let's say I have several pieces of trash and each is made with a different type of material like:

Plastic bottle : High Density Polyethylene (HDPE)
Cereal box : Paperboard & Epichlorohydrin
Shrink Wrap : Polyolefin
Kitchen utensil : Melamine-resin
Textiles : Urea-formaldehyde
Crown Cork Bottle Cap : Metal and Plastisol lining
Rock
Plasterboard
Coffee cup : Ceramic and Glaze (silica + metal oxides)

My objective is to build a system complete with sensors that automatically determines the material (chemical makeup) of each object. Are there material sensors currently on the market that can identify all of the materials that make up these objects?

You can identify many polymers without any clever sensor technologies at all but by following a number of simple physical tests. Sensor technology might help for some of the very fine distinctions, but if you can make a broad categorisation first with simple physical tests then that is what you should do.

And, of course, most plastic opponents are actually labelled with symbols to aid in recycling so you can often just look to identify the chemical nature of the polymer. For reference the symbols are shown below (V identifies chlorine-containing polymers like PVC):

The steps required to identify polymers, if you don't have a label, involve some simple physical tests. First identify whether the polymer is elastic or hard. Then identify whether it is a thermoplastic or a thermoset (by using a hot object like a soldering iron to see whether it remelts). Then a variety of tests can be applied: does it float in water or not? Does it burn and if so how does it behave and what does the flame look like?

One flowchart for this is available here as a pdf. Another simpler one is here also as a pdf but sufficient to distinguish the major classes of polymer. There are even interactive versions of such flowcharts such as this one (flash).

Included below is another version originally produced by the major, but now defunct, British chemical company ICI:

￼￼


Nitration of naphthalene and anthracene In the nitration of compounds such as naphthalene and anthracene, how can you determine which would be the major product? I tried drawing resonance structures for the intermediate formed because stability of intermediate determines which product will be major, but this became too hard for me as there were too many resonance structures and I couldn't really compare them. Can someone please help?

In the electrophilic substitution of polycyclic aromatics, when drawing resonance structures keeping as many benezene rings intact as possible is important.  Note too that a naphthalene ring isn't as "good" as two separate benzene rings.

Look at the structures below, electrophilic attack at the 1 position in naphthalene (top row of drawing) allows you to draw 2 resonance structures with a benzene ring remaining intact (4 if we count structures with the double bonds simply shifted in the intact benzene ring); attack at the 2 position (middle row in figure) only has 1 resonance structure (2 if we count structures with the double bonds simply shifted in the intact benzene ring) with a full benzene ring intact.  Hence, electrophilic attack at the 1-poisition is preferred in naphthalene.

Applying the same considerations to anthracene, we see that only attack at the central ring position allows 2 full benzene rings to remain intact in the possible resonance structures. Consequently electrophilic attack at anthracene's 9-position is preferred.


Is visible spectroscopy the only non-electronic method of all the spectroscopy method? Is visible spectroscopy the only non-electronic method of all the spectroscopy method?

I know some of the spectroscopy method is possible being used because of electronics.

No, Ultraviolet spectroscopy and Infra-red spectroscopy were practiced without any use of electricity.

Infra-red and ultraviolet light were both known by 1801.

Ultraviolet absorption lines had been observed by 1843.


Do electrons truly flow in circuit? I know this sounds like a stupid question, but do electrons truly flow in a circuit? It doesn't seem logical for electrons to just break away from their atoms to go move through a circuit. If they did, they would probably go through a bunch of reactions and behave oddly (right?). Do the atoms lose/gain electrons to become ions and flow through the circuit? If so, wouldn't they too big to be able to flow easily through a conductor?

The key here is the concept of metallic bonding. or, the sharing of electrons between metal atoms like those in a copper wire (a typical conductor). In this type of atomic bond, electrons no longer belong to one atom or another; they're shared equally amongst all atoms.  As such, one electron truly does go bouncing around between the atoms in the direction of the electric current.

One atom is close enough to another in metallic-bonded substances that an electron is attracted just as much to one nucleus as the next one over, so it's not so "illogical" to think that an electron might "break free" of its original nucleus.

Chemical reactions occur between atoms, not electrons, and atoms in a metal are already metallic-bonded together, so a few stray electrons wouldn't do anything to create new reactions.  However, the electrons do behave oddly!  That's the beauty of metal conductors: their electrons are free to flow wherever an electric or magnetic field dictates, meaning that they can do some pretty cool things!

The nuclei of a conductor don't become ions because every time one electron leaves, another takes its place.  Think of people at a gas station on a busy day: every time one car fills up and leaves, another pulls in almost immediately.  Thus, the metal atoms never have time to become ions, because there's plenty of electrons to keep them company.

As (nearly) always, Wikipedia has a bunch of information on the subject.


Degree of hydrolysis A book which I am reading has this topic on hydrolysis of salts.
Suppose $\ce{NH4Cl}$ is dissolved in water. It is a salt of a strong acid and a weak base. $\ce{NH4+}$ ions being the conjugate acid of a weak base is strong and reacts with $\ce{H2O}$ to give $\ce{H+}$ thus making the solution acidic.
$$\ce{NH4+ + H2O <=> NH4OH + H+}$$
Now for calculating degree of hydrolysis $h$ it takes the initial concentration to be $C$. This means $C(1-h)$ of $\ce{NH4+}$ remains and $Ch$ each of $\ce{NH4OH}$ and  $\ce{H^+}$ is formed. Therefore, $K_h = \frac{Ch^2}{1-h}$. In the next step $1-h$ has been approximated as $1$. While calculating degree of dissociation of weak electrolyte we do a similar thing but that is because degree of dissociation is too small compared to 1 but here, I cannot understand how can degree of hydrolysis be very small compared to 1 when $\ce{NH4+}$ is a strong acid.

I cannot understand how can degree of hydrolysis be very small
  compared to 1 when $\ce{NH4+}$ is a strong acid.

That is wrong.

You also wrote:

  It is a salt of a strong acid and a weak base.

Well, the salt of a strong acid and a weak base, i.e. the conjugated acid of that weak base is a weak acid.

In the case of the ammonium ion, $pK_a=9.25$, very weak indeed.

Also, the hydolysis reaction equilibrium is (because free $\ce{H+}$ don't exist in water and $\ce{NH4OH}$ is old notation, long obsolete):

$$\ce{NH4+ + H2O <=> NH3 + H3O+}$$

$h$ is negligibly small because this equilibrium leans very strongly to the left, so $[\ce{H3O+}]\approx 0$.


How much muriatic acid to dissolve cinder block? Can a cinder block be dissolved in hydrochloric (muriatic acid) ?

Also if it can then how much hydrochloric acid will be needed ?

Keep in mind that a cinder block is likely to contain silicates, these will not dissolve well in hydrochloric acid. The silicates will either not react or they will form silica.

If you want to dissolve silicates then you need to use something which will complex to the silicon such as fluoride (HF). I have seen methods for the digestion of soil samples to get the plutonium out. To obtain the plutonium from a soil sample it is often important to digest down the silicate minerals. To do this you must boil the soil sample to dryness many times with a mixture of hydrofluric acid and nitric acid. This will convert the silicate minerals into silicon tetraflouride which will boil out of the mixture.


Balancing redox equation with oxygen on both sides? The problem is to balance this equation in an acidic medium.

$$\ce{As2O3 (s) + NO3- (aq) -> H3AsO4 (aq) + N2O3 (aq) }$$

However, I am getting stuck balancing the oxygens after dividing them into half reactions.

How do you balance a redox reaction when oxygen is involved in both sides of the equation?

It is always easy to balance redox equations by considering two half reactions. The two relevant half reactions to your redox equation are:
$$ \begin{align}
\ce{As2O3 (s) -> H3AsO4 (aq)} \\
\ce{NO3- (aq) -> N2O3 (aq)}\\
\end{align}
$$

First balance $\ce{As}$ and $\ce{N}$ in both equations. Then, balance $\ce{O}$ in both equations with water and $\ce{H}$ with $\ce{H+}$:

$$ \begin{align}
\ce{As2O3 (s) + 5 H2O (l) -> 2 H3AsO4 (aq) + 4 H+ (aq)}\\
\ce{2 NO3- (aq) + 6 H+ (aq) -> N2O3 (aq) + 3 H2O (l)}\\
\end{align}
$$

Finally, balance the charges by electrons:
$$ \begin{align}
\ce{As2O3 (s) + 5 H2O (l) -> 2 H3AsO4 (aq) + 4 H+ (aq) + 4 e-}\\
\ce{2 NO3- (aq) + 6 H+ (aq) + 4 e- -> N2O3 (aq) + 3 H2O (l)}\\
\end{align}
$$

Now add the last mass and charge balanced two equations in order to cancel electrons (this is easy, because both have 4 $\ce{e-}$s):

$$ \ce{As2O3 (s) + 2 NO3- (aq) + 2 H+ (aq) + 2 H2O (l) -> 2 H3AsO4 (aq) + N2O3 (aq)}$$
Or simply,
$$ \ce{As2O3 (s) + 2 NO3- (aq) + 2 H3O+ (aq)  -> 2 H3AsO4 (aq) + N2O3 (aq)}$$


Strong Acid/Strong Base Titration Bromothymol Blue is an indicator that turns yellow in acid, blue in base.

If I were to titrate NaOH with HCl, what color should I look for at the equivalence point? Should I titrate until the solution turns from blue to yellow or will there be an intermediate colour of green?

I know the salt produced (NaCl) has a pH of 7.

http://antoine.frostburg.edu/chem/senese/101/acidbase/indicators.shtml
This is a link to a webpage that talks all about pH indicators and the different pH ranges if you're looking for more info.

When it comes to your question, the color should go from yellow to green because in your breaker is where you add the indicator fluid and HCl acid and you titrate that with the NaOH base. If you titrate further than your green color you have now made your solution more basic.


to convert ppm gas (N2O) concentrations in microgram nitrogen per gram dry soil basis? What calculations are needed to convert  $\ce{N2O}$ gas concentrations in ppm to micrograms of nitrogen per gram dry soil basis? I measured the concentration of $\ce{N2O}$ in soil samples. I extracted a small gas sample from a container (the soil is kept in this container; the container has an head space of 80 cc). Then I injected the small gas sample in the GC and I had a measurement in ppm. I want to know how many grams of N are present in my ppm measurement, taking into account that the gas comes from 5 grams of soil.

I assume the ppm value you are reporting is ppm by volume (or equivalently for ideal gases, ppm by mole).  If that is true, then:

The total moles of all gas in the container can be calculated from the ideal gas law:

$$ N_{tot} = \frac{PV}{RT}=\frac{(101325 \rm{~Pa})(\frac{80}{1000^2} \rm{~m^3})}{{(8.314 \rm{~\frac{J}{mol\cdot K}})}{(298 \rm{~K}})}$$

Here I have assumed that the temperature was 298 K, or about 25 °C.  Note that the units work out because pascals ($\rm{Pa}$) are equivalent to joules per cubic meter, $\rm{\frac{J}{m^3}}$.  Also, I assume your experiment was done at approximately atmospheric pressure, which is 101325 Pa.

The moles of $\ce{NO2}$ in this 80 cc of gas can be calculated from the ppm measurement:

$$N_{\ce{NO2}}=\frac{\rm{ppm~value}}{1,000,000}\times N_{tot}$$

The mass of the nitrogen, in grams, in the $\ce{NO2}$ gas can be calculated from the formula weight of a nitrogen atom, $\rm{FW}_{\ce{N}}$:

$$m_{\ce{NO2}}={\rm{FW}_{\ce{N}}}\times N_{\ce{NO2}}=14\frac{\rm{g}}{\rm mol} \times N_{\ce{NO2}}$$

The mass of soil used was 5 grams, so the mass ratio of $\ce{NO2}$ to soil, $x_{\ce{NO2}}$  can be calculated by simply dividing the masses:

$$x_{\ce{NO2}} = \frac{m_{\ce{NO2}}}{5 \rm{~grams}}=\frac{m_{\ce{NO2}}\times \frac{1,000,000 \rm{~\mu g}}{\rm{g}}}{5 \rm{~grams}}$$

The last part of the equation above means that if you want the mass ratio in units of micrograms (of $\ce{NO2}$) per gram of soil, you have to multiply by 1,000,000.

Also, the question says you want a "per gram dry soil basis", but you didn't mention if the 5 g of soil you used is "dry".  If it isn't, then you will first have to figure out how many grams of dry soil you added to your reaction, if it wasn't 5.


Structure of Dolomite This the structure of Dolomite.

Can someone explain this structure to me?( like what are the blue, yellow etc atoms are? why are they arranged this way?)

Any help would be appreciated.

I don't really like the picture but let's try to do this with that.
So, you probably know about the NaCl lattice? In there sodium is surrounded by six chlorides, which forms an octahedron. Chloride has six sodium around it to form another octahedron. And you have probably heard that also other compounds crystallize in this lattice, let's say sodium bromide (so increasing the anion in size) or potassium chloride. It does not change the crystal structure.

Now, what about more complicated units. Let's deform that spherical anion to an elongated one. What about calcium carbide? The C-C unit with its triple bond is unlikely cleaved during the crystallization, so we can count the $C_2^2-$ unit as one part. This forms an elongated cell but still, it is a NaCl-type lattice.

If we make it flat instead we end up with something like carbonate for example. As you may no the carbonate unit is flat, it's basically a triangle with the carbon in the center. If we see this unit as one part then we can just form another NaCl-lattice. And as the jump from sodium chloride to potassium chloride was possible, we can assume that if calcium carbonate crystallizes in a deformed NaCl-lattice, then magnesium carbonate will do the same.

And the final step is to combine them now, instead of separating the two phases, 1/2 of the calcium-ions are just exchanged by magnesium to form the dolomite-structure.

So you could say that carbonate forms a cubic packing where 50% of the octahedral sites are occupied by calcium and the 50% by magnesium for example. All carbonate-units are parallel to each other therefore creating one direction in which they do not require that much space causing the deformation of the cell.

The metals are coordinated by (and I had to load this in my solid state software as well to see it) six oxygen on the carbonates, or if you take the center carbon of each carbonate unit also by six carbonates, which forms an octahedron again. On the other hand, each of these center carbons is coordinated by three oxygen in a triangle or if you count the counter ions by six metals in the shape of an octahedron.

In your structure you can start off by looking at the dark atoms, which are the carbons, and connect them to the red ones, so can see it will always form triangles. For all the others it's hard to see in this picture. If you want I can also highlight you the different polyhedra in my software so you can see it better. Because this is just a unit cell so it is cut at some positions creating the feeling that some of the connections are wrong. You'd have to watch several unit cells aligned to see the full structure of it. But if you imagined the yellow and the blue ones to be the same you'd end up with the Calcium carbonate or magnesium carbonate structure.

EDIT: I also added the crystal structure now:

EDIT II: Because there were still some problems I tried to further explain the structure a bit. I don't know how others see this structure but I like this layer explanation and it's also the one you can find in many books. So please just look through the new attachment and tell me if this helps you more.


Is there a reaction between ethanol and aluminum? Is there a potential reaction that would occur between ethanol and aluminum? If so, would it potentially create anything toxic?

The reaction of aluminum and ethanol forms aluminum triethoxide,  $\ce{ Al(OC2H5)3}$, but the reaction is fairly slow at room temperature, particularly if the aluminum is anodized. In the presence of any moisture it decomposes into aluminum hydroxide and ethanol.

The Safety Data Sheet indicates that aluminum triethoxide is both flammable and can cause "severe skin burns and eye damage", but in atmospheric conditions it likely decomposes as quickly as it forms, so the main adverse effect will be the corrosion of the aluminum, covering it with a white crust of $\ce{Al2O3}$ (I've seen this happen to an aluminum lid on a container used for ethanol).

UNH has posted a chart that shows that anodized aluminum is resistant to ethanol attack, but does not state at what temperature, so you may have to experiment to see if it suitable in your application.

BTW, many other alcohols react in a similar manner, such as isopropanol, which forms aluminium isopropoxide with aluminum.


Do melt flow rate and density tell you everything about pure HDPE? I understand that melt flow rate and density can be used to classify HDPE. Does this mean that if I have the melt flow rate and density of an HDPE (with no additives), then I can know its other material properties?

Yes, as a rule the melt flow rate and density do tell you what the other material properties will be. This is why HDPE is often specified by those two properties.


Low-spin / high-spin arguments: why overlooking pairing energy? There are many arguments in inorganic chemistry based on a transition from a high-spin to a low-spin configuration, for example:

$K_{f,3} > K_{f,2}$ for $\ce{Fe(bipy)3}$ complex, because of "a considerable increase in LFSE" as the complex becomes low-spin (from Shriver and Atkins)

However, as the complex becomes low-spin there is also an increase in the pairing energy (or a decrease in the number of Fermi holes), so I don't see why the assumption above that low-spin complexes are always more stable than high-spin is necessarily true (it can just reflect a subtle balance between $\Delta$ and P with not indication of the overall change in energy).

Based on my knowledge, successive bipy ligands would only further enhance the mixing of $\mathrm{t_{2g}}$ SALCs and that is where increasing stabilization is coming from. However, I would predict that this factor would get weaker with each ligand substitution, as the energy match of the occupied $\mathrm{t_{2g}}$ and the π* of an incoming bipy becomes worse.

Perhaps, there is some non-linear dependence of the way the pairing energy is reduced as the orbitals get "smeared" over larger space (i.e. over both the metal and π-acceptors) on the number of $\ce{bipy}$ ligands?

Regarding your question about the non-linear change with ligands, I suspect this isn't that easy of a question to answer, but the data is probably out there somewhere.

Regarding the quote and question in the title, when these kinds of arguments are made, they aren't ignoring the pairing energy, but they also aren't really making any predictions.

Essentially, Ligand Field Theory (LFT) lays out a simple way that one can rationalize the geometry of a particular transition metal complex based on the energy of the d orbitals. I can see that you know this. What this means though is that when they observe whatever change happens for the iron complex, LFT says that this must correspond to some change in the energy of the electrons associated with shifting changes in the energy of occupying orbitals in a specific way.

That is, in LFT the only two terms that describe the energy of a particular electronic arrangement are the ones describing the potential energy of the electrons while ignoring their interactions, and the increase in energy from pairing electrons in orbitals.

So, when any geometrical or similar observation is made, there's really no other explanation than to say the a high-spin complex became low-spin because the orbitals split more ("considerbale change in LFSE") which guarantees this decrease in potential is larger than the cost of pairing electrons.

I agree that it's quite naive to always explain things in this way, but that's why crystal field theory and ligand field theory are not meant as real computational or predictive tools these days. They are useful for connecting geometry, spin, and energy as well as giving a first-order explanation of observations.

It's like when you'll sometimes hear inorganic chemistry students describe the behavior of a catalyst as being due to a "combination of electronic and steric effects." That's true... but those are also the only possibilities, so it doesn't tell you much.


Is the nomenclature "primary", "secondary", "tertiary", and "quaternary" carbon atoms limited to sp3 carbon atoms? In organic chemistry we can classify carbon atoms as primary, secondary, tertiary, or quaternary based on the number of additional carbon atoms bonded to the carbon atom of interest. The definitions in the textbook I teach from (Klein, 1st Ed. Wiley, 2011), as well as in several others I have, and from from several locations around the web, are:

Primary $(1^\circ)$ carbon atom - bonded to one other carbon atom, e.g. $\ce{HO} {\bf \color{red}{\ce{C}}}\ce{H2CH3}$
Secondary $(2^\circ)$ carbon atom - bonded to two other carbon atoms, e.g. $\ce{HO} {\bf \color{red}{\ce{C}}}\ce{H(CH3)2}$
Tertiary $(3^\circ)$ carbon atom - bonded to three other carbon atoms, e.g. $\ce{HO} {\bf \color{red}{\ce{C}}}\ce{(CH3)3}$
Quaternary $(4^\circ)$ carbon atom - bonded to four other carbon atoms, e.g. ${\bf \color{red}{\ce{C}}}\ce{(CH3)4}$

Is this labeling system limited to $sp^3$-hybridized carbon atoms? I do not see anything in the definitions that exclude $sp^2$- or $sp$-hybridized carbon atoms. Nor have I seen or read anything that explicitly excludes these atoms. However, the images in textbooks and around the web all show and discuss only $sp^3$-hybridized carbon atoms in this context. Are the following valid?

The central atom in propene $\ce{CH3 -}{\bf\color{red}{\ce{C}}}\ce{H=CH2}$ is a $2^\circ$ carbon atom because it is bonded to two other carbon atoms.

The carbon atoms in acetylene $\ce{HC#CH}$ are $1^\circ$ carbon atoms because they each are bonded to one other carbon atom.

The ipso carbon in toluene $\ce{C6H5CH3}$ (the carbon atom where the $\ce{CH3}$ is attached) is a $3^\circ$ carbon atom because it is bonded to three other carbon atoms.

To best of my knowledge, the classification primary to quaternary is used for discussing reactivity of hydrocarbons, i.e. alkanes. There the number of substituents matters, as it governs carbocation stability, steric hindrance, etc...

In all other cases, the electronic structure of the molecule rules the reactivity. As you mentioned alkenes, alkynes or aromates, there it makes no sense to discuss such subtle effect as number of neighbouring carbons, when you have $\pi$-system at hand.

The IUPAC Gold Book entry for alkyl groups supports this distinction:

  The groups $\ce{RCH2}$, $\ce{R2CH}\ (\ce{R} ≠ \ce{H})$, and $\ce{R3C}\ (\ce{R} ≠ \ce{H})$ are primary, secondary and tertiary alkyl groups, respectively.


Proton removal during benzyne formation in substituted haloarenes I encountered the following question in an organic chemistry paper -

The reaction follows benzyne mechanism. My answer was B.
However, the answer given in the test solutions was D. The mechanism given was as follows:

My question: The tert-butoxide being a bulky base is more likely to abstract the least hindered hydrogen, which is clearly present at the para-position to $\ce{-OCH3}$. Then the benzyne triple bond would form between the meta and para carbons, following which product (B) would be obtained. However, the given mechanism involves abstraction of the hydrogen from the ortho-position, which has two large groups on either side, and hence shouldn't occur.
Is my reasoning right? Are there any other factors affecting the reaction that decides which hydrogen will be removed?
I am also open to the possibility of the test solution's mechanism being wrong.

I appreciate Waylander's answer, but it did not address the OP's curiocity of why the base abstracts the proton from only ortho-position. Yet, Waylander correctly pointed out that since the aromatic $\pi$-system is at right angle to the triple bond (the second $\pi$-bond of the triple bond), the ability to donate a lone pair to the ring by a substituent (like $\ce{OCH3}$ here), or accept a lone pair from the aromatic nucleus (like $\ce{NO2}$ would do) is not relevant in this case. However, inductive effects of the substituents in aromatic nucleus are still relevant (i.e. electron-withdrawing effects that occur through single $\sigma$-bonds).
However, $\ce{t-BuO-}$ is not as strong a base as $\ce{H2N-}$ in usual cases of benzyne reactions. Thus, it would abstract relatively most acidic available proton from to aromatic nucleus. Compared to para-position, ortho-position is more acidic due to proximity (ortho is one $\ce{C-C}$ bond away compared to para which is three $\ce{C-C}$ bonds away). Following is an example you can find in literature (Ref.1):

References:

José-Antonio García-López and  Michael F. Greaney, "Synthesis of biaryls using aryne intermediates," Chem. Soc. Rev. 2016, 45(24), 6766-6798 (DOI: https://doi.org/10.1039/C6CS00220J).


Formation of 3-chloromethylbenzene VS 3-chloromethylbenzene VS 4-chloromethylbenzene For methylbenzene (with the methyl group at position 1), is it easier for chlorine to substitute the hydrogen at position 2, 3 or 4?

I was thinking somewhere along the lines of different charge densities around the benzene ring, and that the charge density closest to the already existing methyl group has been diminished due to the methyl group itself. However, I am unable to explain it properly enough.

It very much depends on the reagents and reaction type you are using to halogenate the aromatic ring, however, the most common method taught during organic chemistry is electrophilic hologenation, which in this case the methyl will be para directing.

http://en.wikipedia.org/wiki/Electrophilic_halogenation#Scope


Test for glyphosate in water, boolean style Testing for contamination of the ever so popular glyphosate based herbicide "Roundup" in water is a service that many labs provide, one of them is the Microbe Inotec Laboratory . They use an enzyme to check for levels of of contamination in ppb, which, I'm sure, is rather a complicated process.

My question is, is there any way to test for it more easily, without finding the actual amount of glyphosate but only checking whether it's present or not?

The proposed method above most likely uses ELISA (enzyme-linked immunoabsorbent assay). In most modern labs, ELISA is quite a standard procedure, however, each assay is easily worth a couple hundred dollars.

Another method of detection is using HPLC (high-performance liquid chromatography), however those machines can run upwards of a couple hundred thousand dollars and each test is expensive as well.

The EPA has a protocol which they used to detect glyophosphate contamination in water:
https://www.bucksci.com/catalogs/EPA%20Method%20-%20547%20-%20GLYPHOSATE%20IN%20DRINKING%20water%20by%20isocratic%20HPLC%20with%20flourecense%20detector.pdf

What may be of interest to you is an article from the journal of the Royal Chemical Society in which they published an article where they used gold nanoparticles to detect presence of glyphosphate using the naked eye or UV light.

http://pubs.rsc.org/en/content/articlelanding/2013/AY/c2ay26391b#!divAbstract


Are there any guidelines for what constitutes an “organic acid” regarding storage requirements? I am moving a large amount of chemicals to a new lab and I’m curious about any guidelines for chemical separation. I’ve seen suggestions that organic acids and non-organic acid be stored separate from one another, and away from general chemicals. However, where would a long chain fatty acid lie, general chemical, or organic acid? When they get long enough, they become a solid at room temperature. Does storage requirements change for organic acids which are solid?

@Jan is correct.

Organic vs. inorganic is not the main concern for acid storage.  The strength and oxidizing ability of the acids is a much more serious issue.

Your institution/company may already have internal safety guidelines that you should follow.  If not, your country/province/state will have occupational health and safety regulations that you will need to respect.

You can find the American Chemical Society's safe storage guidelines here.


How do we represent a molecule in text form? How do we represent a molecule in text form?

A molecule, as a computer scientist I see it as a connected graph of atoms. A connected graph means that beginning at any atom, there is a path to any other atom in the graph.

The interesting question is, how do we represent these as text?

Specific questions are:

1) Is $\ce{H2O}$ different than $\ce{OH2}$? That is, does order matter, or what does order mean?

2) What does this mean $\ce{NH4NO3}$? Why do we write N two times? Is this the same thing $\ce{NNH4O3}$?

The representation of a molecule in text form actually has a name: the chemical formula (pl. formulae).

The formula we use for everyday purposes is an empirical formula. That article should give you more information, but in a nutshell, it is a simple way of identifying a compound by the ratio of its constituent elements. Since ionic compounds and other compounds such as macromolecules can have huge absolute numbers of atoms which are technically bonded but in practice are subdivided into their simplest components, empirical formulae are useful for unambiguously identifying most compounds. Note, however, that we run into trouble when talking about isomers or other compounds with the same elemental ratio but different structures.

For these compounds, we must disambiguate by using a structural formula, which is basically a graph of the molecule. Translating this into text can be done many different ways, using a formal system for specifying branches and stereochemistry. The three most widely used formats for chemical database applications are SMILES, PDB, and CML. You can get more information on all these from Open Babel (alternately OpenBabel).

There is also a middle ground, using a formula similar to SMILES, variously called a molecular formula or condensed formula. It implies structure by counting every atom and splitting individual elements apart to show functional groups like hydroxyl ($\ce{-OH}$) or amino ($\ce{-NH2}$) bonded to the main structure.

I think this is what your question is really about. Yes, in a condensed formula, the order of elements and functional groups matters, because you are trying to communicate information about a compound's structure with a minimum of overhead. It wouldn't make sense to say that butyraldehyde is $\ce{C2H4CHOCH3}$ because the formyl group is at the end. Therefore we say butyraldehyde is $\ce{CH3(CH2)2CHO}$ to better represent the compound's structure. This helps distinguish isomers (but not stereoisomers). Even using the Hill system, one must still be careful.

I hope this clears things up for you on how chemists use different formulae in different situations, and how a condensed formula works.


What is the role of pH in azo coupling reaction of diazonium with phenol and aniline? Why is it so that azo coupling with phenol needs basic pH (9–10) and with aniline it needs acidic pH (4–5). How does pH participate in the mechanism? I have searched the internet for quite a while nowhere does it explain the actual ‘role’ of the pH. Any light on this matter would be helpful.

Deprotonation of the phenol and protonation of aniline result in species that easily react with a diazonium cation in the intended manner.

Let's have a look at the species involved. On the one hand, there is the diazonium cation:

Deprotonation of phenol yields phenolate, for which a resonance structure with a negative charge in para position to the substituent can be written:

We conclude: Deprotonation of the phenol means activation!

In the case of aniline, the lone pair on the nitrogen atom is the preferred centre for the reaction with the diazonium cation:

This isn't what we want! Protonation of aniline avoids this side reaction.

We conclude: Protonation of aniline means "masking" the wrong centre!


Absorption coefficient for liquid water Wikipedia has this plot of the absorption spectrum of liquid water:

The vertical axis seems weird to me. The Beer-Lambert Law says $T=10^{Ax}$ where $T$ is the fraction of light transmitted, $A$ is the absorption coefficient, and $x$ is the optical distance through the material. If we take the values listed on the vertical axis to be $A$ as I've described it here, the transmittance ranges from 80% transmittance though a 1 nm (!) thick sample at 80 nm wavelength ($10^{-10^8*10^{-9}}$) to 80% transmittance through a 10 m thick sample at 500 nm wavelength ($10^{-10^{-2}*10}$).

Alternatively phrased, the absorption coefficient at 2 um is 10^4. Therefore the transmittance would be $10^{-10^4*0.01}$ through a 1 cm sample. That would be one in $10^{-100}$ photons transmitted through the sample. Given that IR spectroscopy is possible, that sure doesn't seem right to me.

Does the transmission of light through water really vary by 10 orders of magnitude or am I misreading this?

What the plot shows is the reciprocal path length needed for a given absorbance to apply, for example slightly less than 100 m in the visible to $10^{-8}$ m in the uv, as you point out, but this is for $55.345$ molar water which is  very, very concentrated compared to most solutions from which an absorption spectrum is obtained, typically $<< 1 $ molar.

As you also point out the absorbance is $A=\log_{10}(I_0/I)=\epsilon[C]l$ with $\epsilon$ as the extinction coefficient usually in units $\pu{dm^3mol^{-1}cm^{-1}}$ and which depends upon wavelength, and $[C]=55.345 \pu{ mol}$ for pure water and l the path length in cm. What seems to be plotted is
$$\frac{1}{l}=\frac{\epsilon[C]}{A}$$
so is still proportional to the extinction coefficient and so still indicates regions where water absorbs greatly or otherwise. But it is not clear what value for A is used. It may be understood that $A=1$ always applies or that the value is taken at $1/e$ absorption. The values may also be based on $\log_e$ not $10$ so a factor of $2.303$ may be involved. As I'm not familiar with this much detail I can't help any more other than to suggest that  you will have to check by finding a true absorption spectrum or  extinction coefficient and work backwards and converting units as necessary.

In general the amount of absorption a molecule has can vary a huge amount. In water in the visible part of the spectrum absorption is low as there are only weak overtone bands from vibrational transitions and nothing else. At shorter wavelengths electronic absorption begins and these can have extinction coefficients of tens to hundreds of thousands in units of $\pu{dm^3mol^{-1}cm^{-1}}$.

A  concentrated solution of a dye  at, say, $1$ molar with a typical extinction coefficient of $10^4$ would have an optical density of $4$ in a $1$ cm cell or a reciprocal length of $10^6$ $\pu{m^{-1}}$ if $A=1$. Hope this helps.


How to measure concentrations using emission spectrum? Imagine I am using emission spectroscopy to identify the presence of ions in a solution.

Would it be possible to use the intensity of the spectrum to find the concentration of each ion in the solution?

Probably the intensity is directly proportional to the concentration but there might be some scaling factor that I would need to know to find precise values for the concentration of each ion?

You are correct in your assumption. The scaling factor is determined experimentally, this is typically called the 'calibration curve'. Basically you measure the intensity of known concentrations and fit a line through the points. Perhaps this paper can help:
http://science.sciencemag.org/content/202/4364/183


Role of acetate ion in formation of hydrogen sulfide ion from thioacetamide We are using TAA to form metal sulfides. What is the effect of sodium acetate on sulfide ion concentration?

The procedures instruct me to add sodium acetate and to heat the test tube with the metal ion if no precipitate forms after the initial addition of TAA.

From the equilibrium given below, it seems that addition of sodium acetate would actually reduce sulfide ion concentration by driving the equilibrium to the left! That would be antithetical to sulfide salt formation.

However, this cannot be the case; the instructions clearly tell me to add sodium acetate if no precipitate has formed; the instructions are clearly suggesting that sodium acetate helps a precipitate form.

Thinking about it some more, could it be that the addition of sodium acetate basicifies the system by forming hydroxide ion, thereby driving the equilibirum to the right? This seems to make sense, because the addition of sodium acetate is to be accompanied by heating, and wouldn't heating drive the formation of acetic acid and hydroxide ion from acetate ion? And acetic acid is rather volatile, so it'll just leave the solution and not interfere by making the solution too acidic. Am I correct in my thinking? How can I predict the effect of heat on an acid-base reaction? I know heating ammonium chloride causes the ammonium ion to lose its proton. Will heating acetate ion in the presence of water cause it to gain a proton?

You are correct.

Adding sodium acetate to an acidic solution forms a buffer and thus decreases the acidity of the solution without making it basic.

$$\ce{CH3COO- + H3O+ <=> CH3COOH + H2O}$$

This helps to precipitate sulfides $\ce{S^{2-}}$ without forming basic precipitates like hydroxides that might disturb the analysis.

$$\ce{S^{2-} + H3O+ <=> HS- + H2O}$$
$$\ce{HS- + H3O+ <=> H2S + H2O}$$

Heating the test tube just speeds up the hydrolysis of thioacetamide. To keep $\ce{H2S}$ in the system this is sometimes done in a hermetically sealed glass tube.


Can any solute be used to lower the freezing point of water? Let's say I want to lower the freezing point of water by 5 degrees Celsius.
I know that for a fixed volume of water, I will need $x$ moles of solute to do so.

Because freezing point depression is a colligative property, does that mean I can use any solute to lower the freezing point of water as long as the molality of the solution is maintained, or are there other properties of the solute that need to be considered?

The key is in the terminology- the freeze point suppressant must be a solute. Oil won't suppress the freeze point of water because it doesn't dissolve. A small amount of sodium chloride or ethanol will suppress the freeze point of water to -5°C.

The first exception is when the solute stops dissolving. This is called the eutectic point. For example, you can only dissolve 23.3% NaCl in water (by mass). That mixture will freeze at -21.1°C. After that, adding more salt won't reduce the freeze point further.

The second exception is with large amounts. If you mix water and ethanol in a 50/50 mol ratio, which is the solvent and which the solute anymore? You can calculate the freezing point suppression of each with respect to the other and they won't match. Liquid/liquid mixtures also have eutectic points at the minimum freezing point, which is less than the freezing point of either component.

To get the actual freezing point for non-dilute systems, try searching for "solid liquid equilibria", "eutectic point", or "binary phase diagram".


Would a quarternary ammonium electride conduct electricity? If suppose, I crystallised a quarternary ammonium electride could this conduct electricity similar to a metallic conductor? Metals have lattice atoms as charged metal ions are a sea of electrons. This leads me to believe a quartnernary ammonium electrides are quite similar, with lattice points being quartnernary ammonium ions and a sea of electrons.

Is this true? Can I consider quarternary ammonium electrides as metallic conductors?

You can't conclude that by thinking alone; some experiments are necessary.

True, an electron in an electride is kinda "free", in that it isn't connected to any particular atom. But that doesn't mean it can roam the entire structure, free as the wind. In fact, it sits in a potential well formed by the neighboring ions and molecules, only the well is so wide and shallow that it looks more like a puddle. To decide the conductivity question, you want to know how one puddle relates to the others, and the answer may be any of the following:

connected to form a continuous sea, much like in a metal;
connected to form a network of channels;
isolated but still close enough for electrons to tunnel between them;
isolated and spaced wide apart.

All in all, theoretically an electride can be anything from a metallic conductor to an insulator, depending on its composition and structure.


What is the name of C(OOOH)₄? I've been practicing organic nomenclature and I was wondering if anyone could show me the proper way to approach naming this molecule. My guess would be methane-quadra-dioxyanol.

This molecule doesn't have any carbon to hydrogen bonds, so it likely wouldn't even be properly named using organic nomenclature. Tetrachloromethane, for example, is correctly named carbon tetrachloride. If you wanted to throw technicalities to the wind, however, along with the fact that this compound would likely be so incredibly unstable that no one would even bother attempting to synthesize and isolate it, it would probably be named tetratrioxidanylmethane.

You might notice the name bears some similarities to tetrachloromethane. That's because it's four of the same group, the trioxidanyl group (from trioxidane), bonded to a single carbon.


Why is the carbon-fluorine bond stronger than the other haloalkanes? What effect makes  C-F bond  stronger than  other halo-alkanes?

From wikipedia article of carbon-fluorine bond,

  The high electronegativity of fluorine (4.0 for F vs. 2.5 for carbon)
  gives the carbon–fluorine bond a significant polarity/dipole moment.
  The electron density is concentrated around the fluorine, leaving the
  carbon relatively electron poor. This introduces ionic character to
  the bond through partial charges ($\ce{C^{δ+}—F^{δ−}}$). The partial
  charges on the fluorine and carbon are attractive, contributing to the
  unusual bond strength of the carbon–fluorine bond. The bond is
  labeled as "the strongest in organic chemistry," because fluorine
  forms the strongest single bond to carbon. Carbon–fluorine bonds can
  have a bond dissociation energy (BDE) of up to 544 kJ/mol. (This is why) The BDE
  (strength of the bond) is higher than other carbon–halogen and
  carbon–hydrogen bonds. For example, the molecule represented by
  $\ce{CH3X}$ has a BDE of 115 kcal/mol for carbon–fluorine while values
  of 104.9, 83.7, 72.1, and 57.6 kcal/mol represent $\ce{C–X}$ bonds
  to hydrogen, chlorine, bromine, and iodine, respectively.


What are the bond angles in cyclohexane? In the chair conformation of cyclohexane, what are the angles of the triangles defined by:

a carbon atom, an axial hydrogen bonded to it, and the midpoint of a vicinal C-C bond ?
a carbon atom, an equatorial hydrogen bonded to it, and the midpoint of a vicinal C-C bond ?

I have unsuccessfully looked for literature values of these, and my trigonometry is not up to scratch.

The crystal structure shows a strangely distorted molecule with only $C_\mathrm{i}$ symmetry. Values have quite a wide range, $\angle(\ce{CCH})\approx 100-120^\circ$. See below for all values and xyz coordinates to read in with a molecular viewer. Source:
 R. Kahn, R. Fourme, D. André and M. Renaud, Acta Cryst. 1973, B29, 131-138.

A DF-BP86/def2-TZVPP calculation (gas phase, 0 K) gives a highl $D_\mathrm{3d}$ symmetric structure with
\begin{array}{ll}
\angle(\ce{CCH_{ax}}) & =109.1^\circ,\\
\angle(\ce{CCH_{eq}}) & =110.3^\circ,\\
\angle(\ce{CCC}) &=111.5^\circ,\\
\angle(\ce{H_{eq}CH_{ax}}) &=106.4^\circ.\\
\end{array}
 Also Bond length differ very slightly (and impossibly to determine at room temperature and pressure), with
\begin{array}{ll}
\mathbf{d}(\ce{CC}) &=1.537~\mathrm{\mathring{A}},\\
\mathbf{d}(\ce{CH_{ax}}) &=1.105~\mathrm{\mathring{A}},\\
\mathbf{d}(\ce{CH_{eq}}) &=1.103~\mathrm{\mathring{A}}.\\
\end{array}
The tiny differences can be explained with Bent's rule.
In principle the assignment Goeff already gave you is absolutely valid, especially if you take the dynamic nature of the molecules into account that they have at standard temperature and pressure.
Also see xyz coordinates for visualisation below.

Appendix

Crystal structure

\begin{array}{lr}
R(1-2)          &1.528 \\
R(1-3)          &1.519 \\
R(1-4)          &0.884 \\
R(1-5)          &1.146 \\
R(2-6)          &1.055 \\
R(2-7)          &0.929 \\
R(2-10)         &1.521 \\
R(3-8)          &1.100 \\
R(3-9)          &0.944 \\
R(3-11)         &1.521 \\
R(10-12)        &1.519 \\
R(10-13)        &1.100 \\
R(10-14)        &0.944 \\
R(11-12)        &1.528 \\
R(11-15)        &1.055 \\
R(11-16)        &0.929 \\
R(12-17)        &0.884 \\
R(12-18)        &1.146 \\
A(2-1-3)        &110.4 \\
A(2-1-4)        &106.7 \\
A(2-5)        &105.8          \\
A(1-2-6)        &106.6          \\
A(1-2-7)        &118.2          \\
A(1-2-10)       &111.3          \\
A(3-1-4)        &103.0          \\
A(3-1-5)        &114.0          \\
A(1-3-8)        &110.6          \\
A(1-3-9)        &116.9          \\
A(1-3-11)       &112.4          \\
A(4-1-5)        &116.7          \\
A(6-2-7)        &109.5          \\
A(6-2-10)       &110.6          \\
A(7-2-10)       &100.5          \\
A(2-10-12)      &112.4          \\
A(2-10-13)      &109.1          \\
A(2-10-14)      &100.4          \\
A(8-3-9)        &106.9          \\
A(8-3-11)       &109.1          \\
A(9-3-11)       &100.4          \\
A(3-11-12)      &111.3          \\
A(3-11-15)      &110.6          \\
A(3-11-16)      &100.5          \\
A(12-10-13)     &110.6          \\
A(12-10-14)     &116.9          \\
A(10-12-11)     &110.4          \\
A(10-12-17)     &103.0          \\
A(10-12-18)     &114.0          \\
A(13-10-14)     &106.9          \\
A(12-11-15)     &106.6          \\
A(12-11-16)     &118.2          \\
A(11-12-17)     &106.7          \\
A(11-12-18)     &105.8          \\
A(15-11-16)     &109.5          \\
A(17-12-18)     &116.7          \\
\end{array}

C        2.329589000      3.011988000     -0.044238000
C        3.630888000      2.632672000      0.661249000
C        1.367908000      1.836044000     -0.049671000
H        2.524936000      3.113740000     -0.900292000
H        1.933922000      3.936128000      0.506802000
H        3.388552000      2.469096000      1.674854000
H        4.348450000      3.220000000      0.605369000
H        1.030575000      1.607424000      0.972471000
H        0.592407000      1.921052000     -0.581309000
C        4.247092000      1.383956000      0.049671000
C        1.984112000      0.587328000     -0.661249000
C        3.285411000      0.208012000      0.044238000
H        4.584425000      1.612576000     -0.972471000
H        5.022593000      1.298948000      0.581309000
H        2.226448000      0.750904000     -1.674854000
H        1.266550000     -0.000000000     -0.605369000
H        3.090064000      0.106260000      0.900292000
H        3.681078000     -0.716128000     -0.506802000

Computed (DF-BP86/def2-TZVPP)

C       -1.270112000     -0.733300000      0.229506000
C        0.000000000     -1.466599000     -0.229506000
C       -1.270112000      0.733300000     -0.229506000
H       -1.328386000     -0.766944000      1.332131000
H       -2.167276000     -1.251277000     -0.145050000
H        0.000000000     -1.533888000     -1.332131000
H        0.000000000     -2.502555000      0.145050000
H       -1.328386000      0.766944000     -1.332131000
H       -2.167276000      1.251277000      0.145050000
C        1.270112000     -0.733300000      0.229506000
C        0.000000000      1.466599000      0.229506000
C        1.270112000      0.733300000     -0.229506000
H        1.328386000     -0.766944000      1.332131000
H        2.167276000     -1.251277000     -0.145050000
H        0.000000000      1.533888000      1.332131000
H        0.000000000      2.502555000     -0.145050000
H        1.328386000      0.766944000     -1.332131000
H        2.167276000      1.251277000      0.145050000


How is hydration free energy of a solute measured experimentally? I am reading a paper (Shirts, M. R.; and Pitera, J. W. and Swope, W. C. and Pande, V. S. J. Chem. Phys. 2003, 119, 5740-5761).  Near the beginning of the paper (page 5745), the authors state:

  Experimental free energies of hydration for weakly soluble solutes are
  determined from concentration measurements made on two phase systems,
  where one phase consists of a vapor with a partial pressure $P_s$ of
  some solute molecule of type $s$, and the other phase consists of an
  aqueous solution with a number density concentration for solute
  molecules of $\rho_s^{\ell}$.  When such a two phase system is at
  equilibrium with respect to transfer of molecules of type $s$ between
  the phases, the solvation free energy is given by $$\Delta G_{\text{solv}} = kT\ln(P_s / (\rho_s^{\ell} kT)).$$

What does this mean in practice?  For example, suppose that I would like to measure the hydration free energy $\Delta G_{\text{solv}}$ of a small molecule (e.g., the amino acid arginine).  That is to say, I would like to measure the solvation free energy of arginine in water.  (Since arginine has a charged side chain at neutral pH, I would expect $\Delta G_{\text{solv}}$ to be negative, because arginine is likely stabilized by water, which is polar.)

Does the expression for $\Delta G_{\text{solv}}$ above imply that I just need an aqueous solution of arginine?  Then would I need to somehow measure the partial vapor pressure $P_s$ of the arginine vapor which exists above the arginine aqueous solution?  How is such a partial vapor pressure vapor measurement done in practice?  Thanks for your time.

Your interpretation is correct. Theoretically, you will need only an aqueous solution and a fixed volume of gas, and then be able to measure equilibrium concentrations. Although, you may not observe very much vapor pressure without using elevated temperatures (as you noted, the zwitterion is very soluble).

Derivation

When a species $x$ is in equilibrium between two states, $$x_{state1} \rightleftharpoons x_{state2}$$ the equilibrium constant is given by the ratio of concentration of $x$ in each state: $$K = \frac {[x_{state2}]} {[x_{state1}]}.$$ The equilibrium constant relates to Gibbs free energy by the following equation: $$\Delta G = -RT \ln(K)=-RT\ln\left(\frac{x_{state2}}{x_{state1}}\right)=RT\ln\left(\frac{x_{state1}}{x_{state2}}\right).$$ Substituting for an aqueous/vapor phase system yields the cited equation. The literature uses Boltzmann's constant, $k$, in place of the gas constant $R$ because they are using units of molecules instead of moles. (The extra $kT$ term within the logarithm comes from the conversion of pressure to number density concentration using the ideal gas law: $\frac{n}{V}=\frac{P}{RT}$, again using molecular units instead of moles will require $k$ instead of $R$.)

Measurement

Determining the "best" method for gas phase measurement depends on what instrumentation you have available to you and how accurate you need/want to your results. Mass spectrometry would be ideal if it employs a soft ionization technique. If the molecular ion doesn't fraction, you could use the raw counts compared to an internal standard to calculate the concentration.

You don't necessarily need to measure the vapor pressure directly, either. If you have a method for measuring the aqueous concentration, you could calculate the vapor pressure based on decrease in solution concentration: $$\Delta G=RT \ln \left(\frac{[arg_{aq}]_i-[arg_{aq}]_f}{[arg_{aq}]_f}\right)=RT \ln \left(\frac{[arg_{aq}]_i}{[arg_{aq}]_f}-1\right)$$ where $[arg_{aq}]_i$ is the initial aqueous arginine concentration (total arginine) and $[arg_{aq}]_f$ is the final concentration (equilibrium amount).  HPLC (or even UV-Vis if that is all you have available) would be a good method, but you would have to either make a calibration curve or use an internal standard (your buffer may work).

Notes

If you have access to J. Phys. Chem. and/or Biochemistry, this article cites articles (ref. 29, 82) that have experimental data for free energies of hydration for amino acids (all but arginine). They may be helpful to consult for methods.


How do I lower the surface tension of water without using a detergent? For many years I've used a Rainbow vacuum, which uses a reservoir of water as the filter.   While I'm quite pleased with it, I've found that when trying to vacuum cement dust or wood ash not all of the  particles are captured in the water.   I believe a remedy to this would be to lower the surface tension of the water.  Yes, I can always be sure to use hot water, but I'd like to go further.  Obviously I can't use a suds-producing detergent because the vacuum produces lots of turbulence in the reservoir.

What commonly available chemicals could I use that would give me the greatest reduction of surface tension? So far I've tried to research trisodium phosphate, sodium carbonate, and a product called Haemo-Sol which is a non-ionic surfactant... but I can't find surface tension numbers for any of these much less the concentration to use.

The manufacturer suggests adding a tablespoon of vinegar, which I think is
rather lame.

Can anyone make a good guess or point me in the right direction?

Your goal is to reduce the surface tension of the water so that it does not support the formation of large bubbles or inhibit the whetting of the dirt particles.

When the bubbles are small, the dust inside will be more likely to make contact with the liquid surface of the bubble within the lifetime of the bubble. Also, when it does make contact, it will be much more likely to irreversibly adhere to the water if you have reduced the surface tension.

The word "surfactant" is a blend of "surface active agent"$^1$, and this is what you need to interfere with the surface tension of your water.  I believe that you may have written off detergents too hastily. I think the best solution will be to add a small amount of laundry detergent, or possibly non-foaming carpet cleaner. Each of these products are designed to foam to a lesser degree than, say, dishwashing detergent or most hand soaps.

I suspect that you will have to experiment a bit, but that something on the order of a single cc per liter may be sufficient to increase the effectiveness of dust removal without causing excessive suds. Your goal should be to see how much detergent you can add without adversely affecting your vacuum, and decide whether this has resulted in a decreased throughput of dust particles.

As to your title question, "How do I lower the surface tension of water without using a detergent?", what I'm really saying is 1) that most anything that does what you want could probably be deemed a detergent and 2) that a detergent is likely the most practical solution to your problem.

1) Surfactants and Interfacial Phenomena, Fourth Edition, Rosen and Kunjappu, John Wiley and Sons, Inc., 2012


Does ice increase the humidity of its surroundings? Does ice increase the humidity of its surroundings? I'm trying to use the BME280 for an Arduino project regarding locating ice, but I don't know whether the atmosphere is more humid around ice. I'm trying to see whether humidity increases near ice.

It depends on the current humidity.

Water vapor certainly does evaporate from ice, as you can see in a vapor pressure chart. If the humidity is 100%, and the air temperature is the same as that of the ice, then the evaporation and condensation are in equilibrium; i.e. water evaporates as fast as it condenses, so there is no net change in humidity.

However, if the ice is warmer than the air, or the air has less than 100% moisture content at that temperature, then ice will evaporate more than it condenses, raising the humidity.

Conversely, if the ice is colder than the atmosphere, it can lower the humidity.


Hydrolysis of a salt B2X An aqueous solution of soluble salt $\ce{B2X}$ has concentration of 0.1 M.
Given data :

  $\ce{K_b(BOH) = 2 × 10^{–5}}$

  $\ce{K_{a1}(H2X) = 2 × 10^{–7}}$

  $\ce{K_{a2}(H2X) = 5 × 10^{–10}}$

What is the pH of the solution?

My attempt,

$$\ce{B+ + X2- -> BOH + HX-  \,\,...      (1)}$$

The hydrolysis constant of $\ce{HX-}$ comes out to be about $10^{-7}$, so I decided to neglect it (is this correct?)

$$Kh=[HX-][BOH]/[X2-][B+]$$

So  the  value  of  $$\ce{K_h=\frac{Kw}{K_{a2}*K_b}}$$

Upon substituting the values I get the Kh=1..(something i have never seen before)

So this means that for the reaction

$$\ce {B+ + X2- -> BOH + HX-}$$

We will end up with 0.05M of $\ce{B+}$ , 0.05M of $\ce{of X2-}$, 0.05M of $\ce{BOH}$ , and 0.05M $\ce{HX-}$ if we start with 0.1 M of $\ce{B+}$ and 0.1 M of $\ce{X2-}$

Now, I do know that

$$\ce{pH=7+\frac{pK_a + pK_b}{2}}$$

I also know that this is true when the hydrolysis constant is smaller than 1. Therefore I cannot use if for this problem.

(Nevertheless, I disregarded this and just used it. I got the answer to be 9.305)

So I am stuck over here.

I got another idea.

When I now look at the sum, the resulting solution looks like a mixture of two buffer solutions,

Buffer 1: $\ce{BOH/B+}$

Buffer 2: $\ce{HX-/H2X}$

I don't know whether this observation is correct. Even if it is correct, I am not sure whether it be used to solve the problem.

I am posting a solution image regarding same question. 2nd hydrolysis of X2- is neglected as you did. important thing is degree of hydrolysis is not small and should not neglected.


Why are drugs mixed with their sodium salts? I have come across quite a few drugs that are of the form $\ce{HA + NaA}$, where $\ce{A}$ is your target organic compound. A few examples of this are:

Carmicide which is a mixture of Sodium Citrate and Citric Acid
Fucidin which is a mixture of Sodium Fusidate and Fusidic Acid

What advantages are there in conjugating a drug with its sodium salt?
A similar question says that this increases solubility and also makes the drug easier to crystallize and purify. I'm assuming that the same reasoning does not work for organic acids, considering they will dissolve/ionize quite easily. Please clarify and provide further insight into this topic.

A large percentage of pharmaceuticals are sold as salts, and sometimes even as co-crystals, i.e. a drug cation and a drug anion. You are right in the sense that the basic purpose is to make them soluble but this is not the only one. If it is a intravenous drug, salt form may have less pain. A low molecular weight medicine might be in a liquid form yet its salt is a solid. Sometimes you want to make a hydrophobic salt so that its stability is higher in high humidity. Sometimes you wish to improve the taste. There are plenty of factors.

Most likely the reason for a mixture of medicine and its sodium salt is its controlled release and some buffering action. In case of your carmicide which is a mixture of sodium citrate and citric acid, is a buffering system and secondly citric acid alone is very corrosive and very bad tasting (tart). Fucidin is only allowed as a Na-salt.

This review will tell you the principles behind choosing a salt.

Go to Google Scholar and read:
Gupta et al. Salts of Therapeutic Agents: Chemical, Physicochemical, and Biological Considerations, Molecules 2018, 23, 1719; doi:10.3390/molecules23071719

  The choice of a particular salt formulation is based on numerous
  factors such as API chemistry, intended dosage form, pharmacokinetics,
  and pharmacodynamics. The appropriate salt can improve the overall
  therapeutic and pharmaceutical effects of an API. However, the
  incorrect salt form can have the opposite effect, and can be quite
  detrimental for overall drug development.

API= active pharmaceutical ingredient