Contradictory Properties of Mercuric Cyanide It is said that mercuric cyanide [$\ce{Hg(CN)_2}$]  is water soluble and also is a non-electrolyte.

How is this possible?

Mercury cyanide behaves like Mercury chloride $\ce{HgCl_2}$. Both are soluble in water, and both do not dissociate in water. These compounds are not salts. The bonds $\ce{Hg-Cl}$ or $\ce{Hg-CN}$ are more covalent than ionic. When the atom Hg is included in compounds it often behaves as a non-metal.

This is due to a relativistic effect. In all elements with atomic number higher than $70$, the outer electrons move at a speed that is not far from c, the velocity of light in vacuum. Of course everybody knows that electrons do not move like planets in the atom. But the relativistic corrections are significant "as if" the electrons were moving. And one of the effect due to relativity is that the length decreases at velocities going near to c. Pekka Pyykko has shown that this entails a contraction of the relativistic electron orbitals. And this contraction is proportional to the sum of the two quantum number n and l.

Mercury has the configuration {$\ce{Xe}$} $\ce{4f^{14} 5d^7 6s^2}$. The three external orbital have a sum n+l equal to $8$ for $4$f, $8$ for $5$d, and $7$ for $6$s. So the electron $4$f and $5$d (with n+l = $8$) are so much contracted that they nearly disappear in the internal core of electrons. It must be nearly the same for the $6$s electrons. And the mercury atom looks a bit like Xenon when seen from outside. That is why it is "nearly" a noble gas : its boiling point is rather low with respect to its neighboring atoms.

All the chemical properties of Mercury are due to its two $6$s electrons which are rather difficult to extract, as they are partly included in the inner core. These electrons "prefer" making covalent bonds rather than ionic bonds, when included in molecules with $\ce{CN}$ group or $\ce{Cl}$ atoms.

Pekka Pyykko, Relativity and the Periodic System of Elements, Accounts of Chemical Research, Vol. 12, No. 8, (1979), p. 276 - 281.


Protons and the Heisenberg principle The Heisenberg uncertainty principle implies that it is not possible to calculate the position and velocity of a body at the same time accurately. Electrons follow this principle since their orbitals are just probabilities where they may occur. But in the case of protons, we are kind of certain about their position in the atom. So shouldn't they have a very, very, very high velocity to balance the certainty in position?

You are using the Heisenberg uncertainty principle to relate the uncertainty in position $x$ to the uncertainty in velocity $v$.

However, the quantitative version of this principle actually is

$$\Delta x\cdot\Delta p\geqslant\tfrac12\hbar $$

where $\Delta x$ is the uncertainty in position $x$ and $\Delta p$ is the uncertainty in momentum $p$.

Certainly, since
$$p=m\cdot v$$
or
$$v=\frac pm$$
where $m$ is mass and $v$ is velocity, you may also use the uncertainty principle to compare the uncertainty in position $x$ to the uncertainty in velocity $v$.

However, the mass of a proton is $m_\mathrm p=1.672\,621\,923\,69(51)\times10^{-27}\ \mathrm{kg}$[CODATA2018], whereas the mass of an electron is only $m_\mathrm e=9.109\,383\,7015(28)\times10^{-31}\ \mathrm{kg}$[CODATA2018]; i.e. a proton is over a thousand times heavier than an electron.

Therefore, a given uncertainty in position $x$ and the corresponding uncertainty in momentum $p$ result in a smaller uncertainty in velocity $v$ for the proton.


Calculating pH of reaction mixture of silver nitrate and ascorbic acid I am designing an experiment but before I do, I wanted to understand if the following method is right to calculate the expected $\mathrm{pH}$ when I mix $\pu{0.736 M}$ $\ce{AgNO3}$ and $\ce{C6H8O6}$.

I want the $\ce{AgNO3}$ to be the limiting reactant for this reaction. I think chemical equation will be:
$$\ce{2AgNO3 + C6H8O6 <=> 2Ag (s) + C6H6O6 + 2HNO3},$$

which would mean that there would be 2 hydrogen ions.

$$\ce{2Ag+ + C6H8O6 <=> 2Ag (s) + C6H6O6 + 2H+}$$

If I start with $\pu{0.736M}$ $\ce{AgNO3}$ and I plan to have that in $\pu{1L}$ water solution (also intend the $\ce{C6H8O6}$ to be in $\pu{1L}$ water solution), then the moles of hydrogen generated will be:
$$\pu{0.736 mol} \ \ce{H+}$$

This would mean the concentration of $\ce{H+}$ will be
$$\frac{\pu{0.736 mol}}{\pu{2L}} = \pu{0.368M}$$

So the $\mathrm{pH}$ of the solution would be $-\log 0.368  = 0.434$? This number does not seem right since ascorbic is a weak acid, is my logic correct?

For solutions of weak acids and with simplifying assumptions [H+]=[A-], [A-] << [HA ], there is formula:

$$\mathrm{pH}=\frac12(\mathrm{p}K_\mathrm{a} - \log {c})\tag{1}$$

For our case, if the stoichiometric ascorbic acid concentration $0.368/2=\pu{0.184 mol/L}$ is used:

$$\mathrm{pH}=\frac12(4.1 - \log {0.184}) \simeq 2.42\tag{2}$$

If reaction is completed, $\mathrm{pH}$ would be given by $\ce{HNO3}$ only.

Situation would get complicated, if nitric acid starts to react with metallic silver or residual ascorbic acid. These side reactions would decrease mineral and eventually also organic acidity.

$$\ce{3 Ag + 4 HNO3 -> 3 AgNO3 + 2 H2O + NO}$$
$$\ce{3 C6H8O6 + 2 HNO3 -> 3 C6H6O6 + 4 H2O + 2 NO}$$

If more of the ascorbic acid is used, initial $\mathrm{pH}$ is again given by the above equation.

The ending $\mathrm{pH}$ is given by equation:

$$[\ce{H+}]= c_1 + [\ce{A-}]\simeq K_\mathrm{a}\frac{c_2}{[\ce{A-}]}\tag{3}$$

where $c_1$ is the nitric acid concentration, $c_2$ is the residual ascorbic acid concentration. and the right equation side is taken from $(1)$

Concentration of $\ce{H+}$ is equal to sum of $\ce{HNO3}$ concentration $c_1$ + $\ce{H+} from weak acid, that is equal to A- concentration ( HA -> H+ + A-). The latter is then approx equal to simplified extression derived from the acid dissociation constant.

$$c_1 \cdot  [\ce{A-}]  + {[\ce{A-}]}^2 - K_\mathrm{a}\cdot c_2  \simeq 0 \tag{4} $$

$$[\ce{A-}]=\frac{-c_1 + \sqrt{{c_1}^2+4\cdot c_1\cdot c_2 \cdot K_\mathrm{a}}}{2} \tag{5}   $$

From (3):

$$[\ce{H+}]=\frac{c_1 + \sqrt{{c_1}^2+4\cdot c_1\cdot c_2 \cdot K_\mathrm{a}}}{2} \tag{6}  $$

$$[\ce{H+}]=c_1 \cdot \frac{1 + \sqrt{1+ \frac{4\cdot c_2 \cdot K_\mathrm{a}}{c_1}}}{2} \tag{7}  $$

In the case by we use acid concentration twice as much as stoichiometric one, then $c_1 = 2c_2$. As $K_\mathrm{a}=\pu{7.94e-5}$:

$$[\ce{H+}]=c_1 \cdot \frac{1 + \sqrt{1+ 2 \cdot K_\mathrm{a}}}{2} \simeq c_1 \cdot  1.00004 \tag{8}  $$

You can see the dissociation of the weak acid is almost suppressed by $\ce{H+}$ from nitric acid, so we can neglect the weak acid contribution to $\mathrm{pH}$.


What caused this reaction between chlorine and brake fluid? I recently saw a video on Facebook of a couple of people pouring brake fluid into a bottle containing pool chemicals (chlorine pellets, I believe). Now this was a rather strong exothermic reaction, resulting in a fire. At one point during the fire, fire appeared to be vapors burning under pressure (like a propane tank fire). I have included a link here. Unfortunately, it is not the original video that I watched.

My question here is what in the brake fluid is reacting so strongly with the chlorine? And what are the chemical products of this reaction (what's in the smoke)?

Pool Chlorine is calcium hypochlorite and brake fluid is polyethylene glycol. Here is a proposed mechanism for the reaction.


Is elastomeric better with more polymer, or with added calcium carbonate? While comparing elastomeric roof coatings, I notice that one brand is 60% polymer, while the the other is 40% polymer along with 30% calcium carbonate.  Polymer is flexible.  Calcium carbonate is brittle.  With that consideration it would seem more polymer is better, since weather conditions challenge the membrane integrity for water resistance.  However, polymer degrades far more than calcium carbonate.  Perhaps inclusion of calcium carbonate provides some protection against degradation.  So is it better to have more polymer, or some of that replaced with calcium?  (Since polymer is more expensive than calcium carbonate, my instinct suggests there would be no reason to maximize polymer at its higher cost, if it did not improve performance.)

Stuff like calcium carbonate is generally used as a cost cutter, as it is far less expensive than even the cheapest elastomer (polymer). It allows the producer of the formulation to make a profit/make more profit, essentially.

Generally speaking, adding calcium carbonate to an elastomeric formulation will decrease the useful properties of said formulation, including its resistance to weathering. It's resistance to wear and tear (as represented by mechanical properties like tensile strength and abrasion resistance) will also suffer. Calcium carbonate containing formulations, all other things being equal, will also be less elastic and somewhat stiffer.

A polymer that has poor resistance to weathering (like SBR rubber) will generally also not benefit from calcium carbonate as a formulation ingredient. Instead, special anti-oxydant/anti-ozonants will improve this cheap polymer's fight against the elements.

But if a composition does contain significant amounts of calcium carbonate it is not automatically disqualified as suitable: that will largely depend on the end-user's expectations. In other words: cheap can work well too.


Hydrogenation of 1-hexene with Raney nickel For my high school senior chemistry project, I need to create and perform a lab. I've chosen to do the hydrogenation of 1-hexene with the catalyst of Raney nickel.

I'm struggling because I'm not exactly sure how to perform a hydrogenation. Some websites tell me I need nitrogen gas, some don't. I'm most likely performing a batch hydrogenation under atmospheric conditions.

Any advice on the actual procedure of the hydrogenation would be helpful.

To meet your brief of a) performing a hydrogenation and b) using Raney nickel, but performed in the safest manner, I would use aluminium-nickel alloy.

This is cheaper than sponge/Raney nickel and easier to handle i.e
 It's not pyrophoric until activated.

I would advise the use of nitrogen, but not a sweep as it can be detrimental to the procedure I'll outline.

I won't give molar details, but work on about 20% of alloy by mass vs your substrate to start with. I won't quote solvents either as I don't know if you need to isolate your hexane or not, nor how you intend to monitor the reaction.
If you just need to do the reaction, go for a less volatile solvent.

To perform the hydrogenation itself, slowly and in a controlled manner with constant monitoring of temperature,  add 4M NaOH.  This reacts with the aluminium in the alloy to make Raney nickel.  At the same time, hydrogen is generated which the nickel can adsorb and hydrogenate your substrate with.

Bear in mind that the NaOH solution introduces a lot of water,  so if your solvent is immiscible with water, your stirring needs to be all the better.

You could use solvents that mix with a lot of water e.g. 1-butanol can contain up to ~20% water, but from which the bulk aqueous layer can be separated if needs be.

So long as the NaOH solution is added slowly enough, you can control the temperature.  One way to manage is this is to only introduce 25% of your solution to be added at one time. That way if you accidentally add it all, it shouldn't go too crazy and boil/eject the contents uncontrollably.

Once complete, you can decant off the liquid but ensure you leave water behind to cover the catalyst.  Rinse and decant until the pH is 8 or below. Then you have some options:

Dispose of the wet catalyst into bags that go off site from your school for disposal,  using the appropriate school procedures or,
Use dilute (1M) HCl to gradually deactivate the nickel. Again, add very slowly over hours if necessary.  Then follow appropriate waste disposal guidelines for the nickel solution.


How explosive is hydrogen gas? I want to electrolyze water to get oxygen and hydrogen. I want to mix them in a regular balloon and ignite it. How much hydrogen would you need in a 2:1 ratio with air for it to be dangerous? Is this generally a safe experiment?

First, let me say that I've enjoyed many times exploding soap bubbles of about one milliliter filled with hydrolysis gas. That is 1 cubic centimeter.
That will give you a sound that rings in your ears in a decent sized living room. You may wish to use ear protection for the experiment.

50 ml will have an effect in a lecture hall that not only wakes up everyone, but also may make people complain.

Now while the explosive limits of hydrogen in air range from about 18 -- 60 % the flammable limits are from 4 -- 75 %, in oxygen the limit of flammability goes all the way from 4% to 95% read: for practical purposes, hydrogen in oxygen is always at least a flammable mixture.
For comparison, gasoline in air is flammable roughly between 1.5 - 7%.

However, you start with a stochiometric mixture which is ideal for explosion (after all, that's why you do it, right?), and it is in oxygen, not in air. So even if you "only" have a flammable mixture reaching an ignition source (e.g. electric switch), chances are quite high that as the ignition proceeds through the mixture, it can reach a zone where the gas/air mixture is explosive. Also, because hydrogen is so much lighter than air, it tends to accumulate under the ceiling, so even if there is enough air in the room that an ideal mixture would not be explosive, chances are that there is an explosive layer of gas. (Same with gasoline or, practially more relevant: solvent from glue for parquet floor tiles)

The safety relevant points are:

the limits of explosion and flammability are, even in air, extremely wide.
limits of flammability are important even if there happens to be no explosive mixture: hydrogen is a small and light molecule that diffuses fast. If the air you breathe happens to have a hydrogen concentration within the flammability limits, any ignition will burn deep into your throat on its way to the lungs.


Why was Avogadro's number chosen to be the value that it is? I understand that the mole is an extremely useful way of measuring and understanding atomic quantities. I did find some history of the number in its wikipedia article, but I did not find an answer to the question "Why is the number chosen to be what it is?".
I understand that it is not necessary to know the history of the number in order to apply it, but it would seem to me that the number wasn't chosen arbitrarily. Why is it, then, that Avogadro's number was chosen as $6.022\times10^{23}$?

Why was Avogadro's number chosen to be the value that it is?

Your question implies that you already know that it was a choice rather than something derived from first principle. There are some numbers that are derived from first principles. In math, $\pi$ and $e$ are not a choice, but can be derived from their properties. In physics, the fine structure constant is not a choice, but would be measured to get the same value by aliens on a different planet (at least as far as we can tell).

[...] Avogadro's number [...]

To be correct, it is actually Avogadro's constant $N_A$ that we use more often in chemistry (for identities such as $k_B = R / N_A$, where $k_B$ is the Boltzmann constant and R the universal gas constant). Avogadro's constant has the dimension (in the SI unit system) of one divided by amount of substance. So the number in front of the unit will change as you change the unit, just like 30 minutes is the same as 0.5 hours even though 30 is not the same as 0.5.

I also understand that it is not necessary to know the history of the number in order to apply it. But it would seem to me that the number wasn't chosen arbitrarily.

The value of Avogadro's constant is linked to the choices that were made when setting up the metric system. There is one definition, that of the meter, that is linked to our planet (so aliens on another planet would have no reason to choose it). The meter was chosen to be approximately 1/10,000,000th of the quadrant of the Paris meridian (i.e distance from Paris to South Pole to North Pole to Paris divided by 40,000,000). Then, they made an artifact to define the meter exactly. Next, the kilogram was chosen to be approximately the mass of 1/1000th of a cubic meter of water (at some temperature). It is no coincidence that the density of water is very close to 1 kg/L. Again, they made an artifact to define the kilogram exactly.
Once the kilogram was defined, the mole was chosen to contain the same number of particles as (pick your favorite, they are all approximately correct at this point) one gram of protons or neutrons, 16 g of oxygen atoms or 12 g of pure carbon-12 atoms.
How are the definition of kilogram and mole related to Avogadro's constant?
Avogadro's constant tells us how many particles there are per mole of particles. The mole was defined with reference to the gram and fundamental constants such as the mass of a proton or of a single carbon-12 atom, so once these choices are in place, Avogadro's constant is no longer a choice.
Relationship to current definitions of units
The SI units (including meter, kilogram and mole) are defined in a different way today, but their values are still the same up to many significant digits. So you can still trace back the choice of the Avogadro constant to choices made when defining the meter, the kilogram and the mole.


A full verion of the periodic table (For computer game) I am looking for a full set of data on every element on the periodic table. (and if possible those only speculated about)

When I say full I mean everything. Electron configurations, Valence electrons, melting point, Atomic mass, Isotopes and what they decay into and all the rest.

Is there such a database anywhere?

You can find all these information on Wolfram Alpha.  You can use the website directly to obtain these information, and you can call their API to directly fetch the data online from your own program.  Of course, you can also use Mathematica to write your program and query the database with WolframAlpha function.


Nitrogen lone pair electrons on 2-acetyl-1-pyrroline Please consider 2-acetyl-1-pyrroline:
http://en.wikipedia.org/wiki/2-Acetyl-1-pyrroline

What is the molecular geometry of the nitrogen lone pair?  Is the lone pair oriented in-plane with the ring?  How is this distinct from the orientation carbon-acetyl bond, which should presumably point out-of-plane with the ring?

The nitrogen lone pair is in plane with the ring because it is in one of the three $sp^2$ hybrid orbitals of N. The other two hybrid orbitals form $\sigma$ bonds to the adjacent carbons, and all three are in plane, with an angle of 120° between them. The $p_z$ orbital of N which is perpendicular to the ring plain forms the $\pi$ bond of the C=N double bond. The acetyl substituent should also lie in plane with the ring because this would allow energetically favorable conjugation between the two double bonds (C=N and C=O).


How to run crystal simulation on GAMESS? I know GAMESS is not best suited for crystal simulation but I am not ready to start learning another package as VASP. As I heard GAMESS is able to do some calculations in a "terminated" crystal (I mean using an ion in the periphery instead of repeating it multiple times).  Mostly I need to calculate HOMO-LUMO gap / hessian matrix and transitional state of a surface reaction .

I looked in the web but couldn't find a good worked example of using GAMESS for crystals.  Is there anywhere I can find such an example ?

As you said, there are many packages capable of handling calculations of crystals, slabs, etc. with periodic boundary conditions. VASP is only one of dozens. (That's probably a separate question in its own right.)

GAMESS does not include treatment of periodic boundary conditions, so you would need to run a calculation on a "cluster."

Since you mention wanting to treat a surface reaction, you'd use a builder (e.g., Avogadro) to create a large cluster (e.g., Build->Supercell). So rather than one primitive unit cell, you create maybe a 2x2x2 or 3x3x3 slab. If needed, you'll terminate dangling bonds (e.g., adding hydrogens to cap oxygens in $\ce{TiO2}$ for example.. notice the terminal "uncapped" atoms on the back and right of the unit cell).

Finally, after running calculations in GAMESS, you'd need to check for convergence. So you'd want to run calculations with increasing system size and see if the properties converge.

Now, this isn't as strange as it sounds. In some cases ($\ce{MgO}$ and $\ce{TiO2}$ are good examples) the properties require more advanced calculation methods like multi-configuration QM, and this may be easier to do in a quantum chemical package with a cluster, rather than a solid-state crystal code.


How do molecules with muons and electrons bond? Is bonding specifically for and between electrons? Why cant two atoms share muons, which are different particles of same charge, spin and different mass?

Why there aren't muon-electron bonds?

Why is the octet (or eighteen valence electron) rule only for electrons, and not for all particles with similar charge and spin compared to the electron?

You are correct; electrons and muons are fermions with different quantum numbers (specifically, they differ in the electron number and the muon number), so the Pauli exclusion principle does not apply between them (though it of course applies among electrons and muons separately). A somewhat similar case happens with protons and neutrons (also fermions) in the nuclear shell model, which attempts to describe nuclei as containing proton and neutron shells, analogous to electron shells. The proton and neutron shells are filled independently.

Because (negative) muons are the second generation Standard Model equivalent of the electron, whatever electrons do, muons can copy - since there are "electronic" orbitals, there are also "muonic" orbitals. However, there are two main differences.

First, for electron orbitals, it's a good approximation to assume the nucleus is stationary with respect to the electrons (the Born-Oppenheimer approximation), due to the great difference in their masses (the lightest nucleus, a proton, has approximately 1836 times the mass of an electron). Because the muon is approximately 207 times heavier than an electron, now a muon has an appreciable mass relative to a proton (approximately one-ninth), and therefore the BO approximation is considerably worse. This is a scenario in-between a regular atom and positronium, where an electron "orbits" a positron "nucleus", which has the same mass. See Phys. Rep. 1982, 86 (4), 169-216 for a quantitative analysis of BO approximation errors in a simple muonic molecule.

The second and more striking difference is that, again due to the muons being 207 times heavier, muonic orbitals are accordingly around 207 times smaller meaning a muonic orbital has a typical radius of around 0.5-1 pm compared to 100-200 pm for an electronic orbital. The energies are also 207 times larger in magnitude - the 1s electronic orbital in hydrogen has an energy of -13.6 eV, whereas for "muonic hydrogen", the 1s muonic orbital has an energy of -2815 eV. These facts can be determined simply by solving the Schrödinger equation, except inputting a mass 207 times greater for the negatively-charged particle.

As you can see, this leads to a severe mismatch between the realms of electronics and muonics. There is no kind of shared electron-muon bond - at best, there would be a separate electron bond and a muon bond. However, because the energetics involved in the muon bond are so much higher, the system in the ground state is essentially equivalent to just having the muon bond, plus a small correction due to a grossly warped electron bond. The muon bond is formed normally, and the electron bond has to deal with whatever geometry is forced by the muons, however crazy.

As an example, imagine a neutral atom of regular hydrogen and a neutral atom of muonic hydrogen interacting. The muonic hydrogen atom basically pierces into the depth of the electron cloud of the regular hydrogen atom (the electron can hardly repel the muon efficiently, since it is so tightly bound to its nucleus), until both nuclei get quite close. Then the muon latches onto the other proton. The system is stabilised when the two hydrogen nuclei are approximately 0.5 pm apart, and the lone muon forms half of a muonic sigma bond.

From the "point of view" of the muon, the system looks like a singly-ionised muonic dihydrogen molecule ($\ce{\mu-H_2^+}$), with slight corrections due to the electron buzzing around, most of the time far away.

However, from the "point of view" of the electron, it sees a bizarrely elongated nucleus (from the typical ~1 fm sphere to a ~500 fm spindle) with a total charge of +1e (since the muon almost perfectly screens out a full positive charge). This spindly nucleus is still quite small relative to the electron cloud, and so the electron likely behaves similarly to a normal isolated hydrogen atom, with some corrections due to the non-spherical distribution of charge at its "nucleus" (the two separate protons bound by a muon). The electron still provides a slight amount of bonding between the protons, but it is much less than normal due to the odd geometry forced by the muon.

Muonic chemistry in its full glory would be a fascinating (and extremely dangerous!) copy of the electronic chemistry we know, but the two would operate almost completely independently. The muon is actually one of the most stable subatomic particles, with a lifetime of 2.2 µs. That sounds like almost nothing, but it's many orders of magnitude more than what is necessary to observe "chemistry". Unfortunately, it's just too difficult to produce for how ephemeral it is...


Why doesn't diamond conduct electricity but silicon does? Diamond's structure is a large network covalent crystal, where there are strong covalent bonds between neighboring tetrahedral carbon atoms. The explanation I was given was:

  The electrons are held tightly in the covalent bonds of the structure, so they are unable to move around. Thus, diamond cannot conduct electricity.

Why is it then, that in silicon (also a network covalent solid), which has an identical crystal structure to diamond, is known to conduct electricity relatively well and even dubbed a semiconductor? Aren't the electrons held tightly in the covalent bonds between neighboring silicon atoms?

I'm not a solid-state physicist. Yet, this is how I understand this problem, so bare with me.

According to this article:

  Every solid has its own characteristic energy-band structure. This variation in band structure is responsible for the wide range of electrical characteristics observed in various materials. In semiconductors and insulators, electrons are confined to a number of bands of energy, and forbidden from other regions. The term "band gap" refers to the energy difference between the top of the valence band and the bottom of the conduction band. Electrons are able to jump from one band to another. However, in order for an electron to jump from a valence band to a conduction band, it requires a specific minimum amount of energy for the transition. The required energy differs with different materials. Electrons can gain enough energy to jump to the conduction band by absorbing either a phonon (heat) or a photon (light).

This forbidden gap plays a major role in determining the electrical conductivity of material. The materials can be classified into three types based on the forbidden gap. they are:

Insulators: The materials, which does not allow the flow of electric current through them are called insulators. The band gap between the valence band and conduction band is very large in insulators, which is approximately equal to $\pu{15 eV}$. Normally, in insulators the valence band is fully occupied with electrons due to sharing of outer most orbit electrons with the neighboring atoms whereas its conduction band is empty, i.e., no electrons are present in the conduction band.
Semiconductors: The material, which has shown electrical conductivity between that of a conductor and an insulator is called a semiconductor. The band gap of semiconductors is is very small, and it falls between that of valence band and conduction band. The value of this forbidden gap is about $\pu{1-3 eV}$. As a rule of thumb, if the material has $\Delta E_\mathrm{gap} \le \pu{3.0 eV}$, it is considered as a semiconductor (Semiconductors- Band Gaps, Colors, Conductivity and Doping).
Conductors: In a conductor, the valence and conduction bands are overlapped each other. Therefore, there is no forbidden gap in a conductor (the value is generally negative). Thus, the valence band electrons to move into conduction band needs only a little amount of applied external energy.

Experimental values of $\Delta E_\mathrm{gap}$ for diamond and silicon are $\pu{5.48 eV}$ and $\pu{1.17 eV}$, respectively (Ref.1). Keep in mind that band gap get narrowed when going down on a group in periodic table. For example, $\Delta E_\mathrm{gap}$ of germanium is $\pu{0.74 eV}$ (Ref.1).

Therefore, silicon and germanium behave like semiconductors, while diamonds fall into insulator category in general (yet, please refer to  Ed V's comment above for better definition).

References:

Fabien Tran, Peter Blaha, "Accurate Band Gaps of Semiconductors and Insulators with a Semilocal Exchange-Correlation Potential," Phys. Rev. Lett. 2009, 102, 226401, 4 pages (DOI: 10.1103/PhysRevLett.102.226401).


Can we manipulate the rate of water's autoionization? I just got introduced to this concept and these thoughts came into my mind. Is there a way to increase the equilibrium constant of the autoionization of water?

First of all, I'm assuming you mean the equilibrium position. This is not the same as the rate of the reaction.

Sure there is. The easiest way to increase the equilibrium position water's ionization is by heating it.

To find out how much increasing the temperature of water will affect $K_w$, the Arrhenius equation will work for values of $T$ close to $0 °C$. $K_w=Ae^{\frac{-E_a}{RT}}$.

Of course, you'll have to find $A$ from experimental data.

Here is a graph of how changing the temperature affects $pK_w$. For the most part, as temperature increases, you will see an increase in K.


Effect of temperature change in an equilibrium reaction Talking about an exothermic reaction in equilibrium it is said that heat is on the product side, so according to Le Chatelier principle, the reaction should move backward to counteract the increase in the number of products.
But it is also stated that in an exothermic reaction when the temperature increases, the value of equilibrium constant decreases, which means the product is less than reactant, then shouldn't the reaction move in the forward direction?
Please help me with wherever I'm going wrong.

As per Le Chatelier's principle, when the temperature is increased, for an exothermic reaction , the reaction shifts in the backward direction. This is clear from the van't Hoff's equation:

$$ \log\left( \frac{K'_\mathrm{p}}{K_\mathrm{p}}\right) = \frac{\Delta H^0}{2.303R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$$

where $T_1$ corresponds to $K_p$ and $T_2$ corresponds to $K'_p$ . Also $T_2 > T_1$.

This is because $K'_\mathrm{p}$ becomes less than $K_\mathrm{p}$.  Hence the reaction goes in the reverse direction.


Why can alkali metal hydrides usually not act as nucleophiles towards carbonyl groups? Think for example of $\ce{KH}$ or $\ce{NaH}$ in enolization reactions. This remark was made in one of my textbook, but I could not remember why this was actually the case. Why can't $\ce{NaH}$ or $\ce{KH}$, being hard nucleophiles, add to carbonyl groups?

The problem is that neither of these nucleophile are hard. Vanilla hydrides such as potassium hydride or sodium hydride release very soft hydride ions. Hydrogen only has one proton. Giving it two electrons to stabilize as in the case of the hydride anion is a tall order. Note the proton to electron ratio: 1 to 2. Even carbanions which are unstable species don't have such a ratio.

Therefore, the hydride ion has a massive ionic radius and is easily polarized, making it an extremely soft base. Electrophilic carbon atoms in carbonyl groups are usually hard(er).

Plus the 1s orbital of hydrogen is too small to effectively overlap with anything other than another hydrogen. Is your textbook written by Clayden? I distinctly remember Clayden's remark about hydrogen's 1s orbital with regard to its anion's nucleophilicity.

Usual reagents used for hydride delivery/reduction of carbonyl compounds (and related) include LAH, NaBH4, and NaBH3CN, in order of decreasing strength. As far as I know, the lithium ion in LAH plays an integral role in its mechanism of reduction by bonding with the carbonyl oxygen of carboxylic acids.

Also, these aforementioned reducers are in order of decreasing strength because they are listed in order of decreasing partial negative charge on their hydride ions. For example, the nitrile group in sodium cyanoborohydride withdraws electron density from the hydrogens. Hence the reason it is the weakest of the three aforementioned reducers. NaBH3CN will only act on strongly polarized iminium ions rather than plain imines or carbonyl compounds.


Buffers calculating amounts of acid & base Quick run down, I was given pH of buffer 4.5 and then pKa of acid 4.79
I then had to work out volume of acid and volume of base using $\text{pH=pKa}+\log_{10}{\left(\dfrac{[\ce{A-}]}{[\ce{HA}]}\right)}$. Using that I calculated the volumes of acid and base required to make 15 ml of the buffer $\ce{CH3COOH,CH3COONa}$:

$V_{\ce{HA}}=9.7\ \text{ml}$ and $V_{\ce{A-}}=5.3\ \text{ml}$ (both acid and base used to make buffer are of 0.1 M conc, probably come in handy later).

Then I am now asked to calculate the amounts of acid and base in 15ml of my buffer
they give me
Acid: $n_{\ce{HA}} = [\ce{HA}]\cdot V_{\ce{HA}}=0.1\cdot V_{\ce{HA}}$

Base: $n_{\ce{A-}} = [\ce{HA}]\cdot V_{\ce{A-}}=0.1\cdot V_{\ce{A-}}$

Which I don't really get?
I then have to find out the amount of acid was added, some context is that this is n experiment I conducted using a pH probe I added three drops and recorded the pH for a total of 60 drops the final pH reading is 3.52 of an initial reading of 4.38, would I use the simple $\text{pH} = -\log_10[\ce{H+}]$?
They have given me $[\ce{H^+}]\cdot V=0.1\text{ M}\cdot\dfrac{1\text{ ml}}{1000\text{ L}}=...\text{mol}$ which I also don't really know what to do with it.
After that I calculate the new concentrations of acid and conjugate base after all added H+ reacts
initial and final.
you don't have to use my example just explain EVERY SINGLE STEP and guide me through the process :) much appreciated!  I know I am asking a lot in one go but this all correlates with one another, if anyone can help thanks a lot!

I'm going to address only the first part since that was in the original question.

You were given a buffer of pH 4.5 and the pKa of the acid used is 4.79. You worked out the volumes of the acid to be 9.7 mL and the volume of base to be 5.3 mL. You are now asked to get the amount of acid and the amount of base (i.e. the number of moles of each in the buffer).

$n_a$ means the number of moles of the acid.

This is equal to the concentration of the acid multiplied by the volume of the acid, namely:

$n_a = [\ce{HA}].V_a$

Since we know that the concentration of the acid, $[\ce{HA}]$ is 0.1 M and the volume of acid used is 9.7 mL, the number of moles of acid becomes:

$n_a = 0.1\ \text{mol/L}\times 9.7\ \text{mL} \times\dfrac{1\ \text{L}}{1000\ \text{mL}}$

The $\dfrac{1\ \text{L}}{1000\ \text{mL}}$ is only to convert mL to L. Do the same for the number of moles of the conjugate base.

For the second part, you need to realise that the pH you get that way will be the combined pH of the buffer and the acid. You will probably have to use the assumption that 1 drop = 1 mL, unless the question or your tutor already mentioned how to consider the volume of the added acid.


stereochem of epoxide opening Here I have an epoxide opening reaction, but I was wondering that when a nucleophile attacks, How would the stereochemistry work? I have 2 products I predicted and I'm not sure which one would be correct. It's either the first one predicted in the reaction of the second picture I have attached. Or are they actually the same thing?

In these three-membered ring reactions, it's an SN2 substitution reaction that happens. Hence, there is an inversion of the configuration associated.
Edit: Perhaps the mechanism is more complicated than SN2, contrary to Klein's book which states that it's SN2, this online source says that the actual mechanism is between sn1 and SN2 (leaning to SN2 side) to explain the preference of nucleophile to attack tertiary carbon though more sterically hindered.

Reference: See page-431 of Organic Chemistry David-R-Klein


How to determine the number of water ligands attached to Pd in palladium chloride aquacomplex PdClₓ · yH₂O? Normally this can be achieved using UV-vis: there is a strong absorbance around wavelength 220 nm which varies based on the number of ligands. If the palladium concentration is too high, the solution can be diluted.

However, in my case I do not know the concentration of solution as the solution is reactive. I need to measure the palladium ligands after these reactions have occurred. Thus dilution is not an option. Does anyone know if UV-vis can be tweaked or another technique for testing palladium-water speciation in solution?

FYI I get an absorbance value of 4 (the peak is still quite broad at this value) when testing the palladium chloride under the desired conditions.

I hope the following two references would help you find the way. I honestly think Ref.1 shows much more promise, abstract of which states that:

Monomeric palladium(II) chloro aqua complexes of the form $\ce{PdCl_r(H2O)_{4-r}^{2-r}}$ ($r = [0,4]$) were studied both experimentally and theoretically to gain insight on both the stabilities and the nature of palladium-chloride interactions.
The thermodynamic stabilities of these complexes were studied in aqueous solutions from $5$ to $\pu{125 ^{\circ}C}$ with UV-vis spectrophotometry using a quartz flow-through cell. Tentative measurements up to $\pu{200 ^{\circ}C}$  were also carried out in pressurized titanium and gold-lined optical cells but revealed important losses in soluble palladium. The strong ligand-to-metal charge transfer bands of the palladium complexes below $\pu{350 nm}$ were used to constrain the step-wise thermodynamic formation constants at each temperature, using results of singular value decomposition of the spectra over a broad range of palladium:chloride ratios and wavelengths. The temperature-dependent constants were used to obtain changes in enthalpy and in entropy for each reaction. The thermodynamic stabilities of $\ce{PdCl(H2O)3^+}$, $\ce{PdCl2(H2O)2^0}$, and $\ce{PdCl3(H2O)-}$ are larger at higher temperatures, whilst the one of $\ce{PdCl4^2-}$ is smaller. All changes in entropies are positive for the former three species, but negative for the latter, presumably due to a larger solvent reorganization around the doubly charged $\ce{PdCl4^2-}$ species. The room temperature thermodynamic values derived from this study are also in agreement with previously published calorimetric data.
Theoretical calculations on the intramolecular distributions of electrons in the different palladium(II) chloro aqua complexes, using the methods of atoms in molecules and of the electron localisation function, showed $\ce{Pd-Cl}$ and $\ce{Pd-OH2}$ interactions to be of largely closed-shell/ionic nature. These interactions induce an important distortion of the outer core shell electrons of $\ce{Pd}$, as well as stable accumulations of electrons between adjacent $\ce{Pd-Cl}$ and $\ce{Pd-OH2}$ bonds known as ligand opposed core charge concentrations.

References:

J.-F. Boily, T. M. Seward, “Palladium(II) chloride complexation: Spectrophotometric investigation in aqueous solutions from $5$ to $\pu{125 ^{\circ}C}$ and theoretical insight into $\ce{Pd-Cl}$ and $\ce{Pd-OH2}$ interactions,” Geochimica et Cosmochimica Acta 2005, 69(15), 3773-3789 (https://doi.org/10.1016/j.gca.2005.03.015).
L. Espinosa-Alonso, K. P. de Jong, B. M. Weckhuysen, “A UV-Vis micro-spectroscopic study to rationalize the influence of $\ce{Cl-(aq)}$ on the formation of different $\ce{Pd}$ macro-distributions on $\gamma$-$\ce{Al2O3}$ catalyst bodies,” Phys. Chem. Chem. Phys. 2010, 12(15), 97-107 (DOI: 10.1039/b915753k).


Why only the unstable resonating structure proceeds in the reaction? For example, in the given image, I have starred the step where the relatively unstable resonating structure proceeds forward in the reaction.

Why?

Imagine a case where the resonance form with the negative charge localized on the oxygen went forward and attacked the second carbonyl:  That charge needs to go somewhere so the double bond is broken and the electrons go up to the second oxygen (the one that started as the second carbonyl). Those electrons would look to come back and reform the double bond, and in doing so would force one of the groups to leave. And the bond that leaves is the one you just made. This reaction happens with both atoms acting as a nucleophile except only one (the carbanion) goes forward and doesn't immediately leave again.


How is Humic substance destroyed in nature? Humic substance consists of many, to my knowledge poorly understood, species of organic matter, mostly carbohydrates chained in intersting (and also poorly understood) ways. Humic substance is an endproduct of digestion of other organic matter, and is in itself known to be hardly biodegradable. Industrial processes to destroy humic substance or humic acid in wastewater involve UV and strong oxidizers. But how is humic substance destroyed in nature? Given that at least part of the carbon embedded in life ends up as humic substance at one point or other, there must be a natural pathway to destroy it, or several. What is it?

Sunlight is aggressive, as is microbiology.  Ozone, hydroxyl radicals (including Fenton chemistries with peroxides), air pollution in general eventually degrade and destroy near anything organic.  Examine sediments downstream of rivers, including outflows, especially for heavily colored waters.  Sampling vs. linear distance and depth will give you histories of degradative chemistries and biologies.  Note that phenolics are overall good microbiocides re lignin.

Terra preta soils demonstrate the survivability of charcoal.  Humic substances are more functionalized and more reactive.  One can start at the other end, too - weathering of coal.  Coal contains dispersed iron pyrite and other sulfide minerals that speed degradation.  Powdered coal heaps can self-ignite from cumulative oxidation heat.  Carbon dioxide, water, and elemental nitrogen are deep thermodynamic holes.


How to calculate weight of CO₂ in one kg of fuel I'm trying to do homework and I have the answers to the problems but I can't read the explanation because it's illegible and I can't find how it's done in my notes.

Here's the simplest example:

$$\ce{ C + O_2 \to CO_2}$$

It looks like the answer is 44/12 but I don't understand where the 12 comes from at all. How is a question like this approached? (Didn't know what to search in google to get an explanation).

You have 1 mol $\ce{C}$ reacting with 1 mol $\ce{O_2}$ to give 1 mol of $\ce{CO_2}$

What you need now is something to relate weight and mol. This is called molar mass, notated with the letter M, and has the unit g/mol. You'd just look that up on the internet, but it's calculated from the weight of the atoms in each molecule. If a molecule weighs 32.2 units, then it'd have a molar mass of 32.2 g/mol.

The calculations are pretty straight forward. How many moles of fuel per kilo? $1000g / M(\ce{C})$, where M, as noted, indicates molar mass.

You know that 1 mol carbon turned into 1 mol carbondioxide. So the number you just calculated is thus also the moles of carbondioxide. You need to find out how many grams those x mol carbondioxide corresponds to: $n(\ce{CO_2}) * M(\ce{CO_2}) = m(\ce{CO_2})$.

If you're not confident in this, pay attention to the units. n = mol; M = g/mol; m = g. Thus multiplying M by n yields a figure in grams. Is that clear?


Why do some solid chemicals dissolve better in colder water? I just discovered that some solid chemicals, such as $\ce{Na2SO4}$, dissolve better in cold water than hot water from here and would like to know if there is any particular reason as to why. Is it their structure, their charge, etc.?

As with every process, dissolving can be described by the thermodynamical equation:

$$\Delta G = \Delta H - T \Delta S$$

Which the process being spontaneous if $\Delta G$ is negative.

We have two components here, enthalpic and entropic. The enthalpic component $\Delta H$ is a constant of the salt in question. The entropic component $\Delta S$ can usually be considered positive because a highly ordered crystal lattice is broken to give dissolved, much less ordered fragments — hence why for most salts they dissolve easier in hot water.

If the solubility decreases with higher temperatures, then $\Delta S$ must be negative. The salt will then only dissolve if $\Delta H$ is negative enough to give an overall negative result. This seems to be the case for $\ce{Na2SO4}$

It is impossible to predict $\Delta S$ and $\Delta H$ a priori without any computer quantum calculations. One has to resort to experiments to determine whether a salt will dissolve better or worse in the heat.

While I have used the term ‘salt’ most of the time, the same is true for molecules. Take sucrose: I was asked in a physical chemistry lab to guess with my colleague whether sugar would dissolve endothermicly or exothermicly. We decided, that the degree of order be a lot higher in dissolved sugar due to a high amount of water surrounding it rather tightly, that the entropy must decrease, so that the enthalpy has to be negative and sugar has to dissolve exothermically. Hence sugar should dissolve more easily in cold water, too.

According to a comment by Nicolau Saker Neto, our reasoning was false and sucrose displays increasing solubility at higher temperatures. Apparantly the testing faculty member approved of it, though.


How to know which reduction and oxidation reactions occurred? We just had a lab experiment on galvanic cells and we used 0.5 M and 1 M iron(III) nitrate solution at the anode (iron electrode) and 1 M and 0.5 M copper(II) nitrate solution at the cathode (copper electrode) for runs 1 and 2, respectively.

How can I determine which redox reaction occurred? There are many possible reactions $(\ce{Cu^2+ + e- -> Cu+}$ or $\ce{Cu^2+ + 2e- -> Cu}).$ The experimental cell potential we got was 0.48 V and 0.44 V. Temperature that we recorded was 28 °C and 29 °C for runs 1 and 2.

Here, Fe(III) will be reduced to Fe(II), and Copper(0) will be oxidized to Cu(II).
The standard potential for Cu2+/Cu is +0.34 V, and for Fe3+/Fe2+ it is +0.77 V.
Nernst's law gives the potentiels for a 0.5 M solution. It is +0.34 - 0.009 V = 0.33 V pour the Copper electrode, and 0.77 - 0.018 = 0.75 V for the Fe3+/Fe2+ electrode.
With this choice, you get nearly your results.
It you had chosen another system of half-reactions, you will get results which are far away from your measurements.


Why is ethanoic acid more acidic than phthalimide? Why is ethanoic acid more acidic than phthalimide? $\mathrm{p}K_\mathrm{a}$ of phthalimide is 8.3 (source), while that of ethanoic acid is 4.76 (source).

Applying the logic found here would suggest the opposite. This is because both conjugate bases delocalize their negative charge, but while ethanoic acid has two oxygens to delocalize its charge among, phthalimide has two oxygens and a nitrogen (and a benzene ring); it seems that the quality of the delocalization is the same, but the quantity favors phthalimide (as the answers to the question put it). As for the other factors that affect acidity (inductive effects, hybridization, atom size/electronegativity, formal charge, etc.) none seem to come into play.

The deprotonated phthalimide anion is indeed highly delocalized, but the neutral molecule with the N-H proton reattached retains most of that delocalization.  Protonating acetate ion back to acetic acid causes a much greater loss of delocalization.


Why does physisorption (physical adsorption) increase with ease of liquefaction? This is what my textbook says about physisorption

  The amount of gas adsorbed by a solid depends on the nature of gas. In general easily liquefiable gases (i.e. with higher critical temperature) are readily adsorbed. Thus $\pu{1g}$ of activated charcoal adsorbs more sulfur dioxide (critical temperature $\pu{630K}$) than methane (critical temperature $\pu{190K}$).

I couldn't figure out why this should be so. In fact, according to my reasoning it should be the opposite. A gas which is easily liquefiable has higher intermolecular forces (here van der Waals forces) which would bound the gas molecules together and reduce physisorption. Where is my reasoning wrong ?

It can be a probable answer but I am not sure about it.

You are correct to say that easily liquifiable gases have stronger intermolecular forces of attraction (van der Waals forces). As a result of which they are strongly bound to themselves. Consider physisorption, here, the forces between the adsorbate molecules and the adsorbent  are weak van der Waals forces. Once the surface area of the adsorbent is completely occupied, the next layer of adsorbate molecules will be bound weakly. So if the gas molecules have stronger van der Waals force of attraction, they would be adsorbed readily in preference to gas molecules having weaker van der Waals forces of attraction.


Is there a known chemical that makes it possible to control the opacity of a large surface? I have this idea of creating a large domed ceiling/roof for a building that can be made clear to see the actual sky, or made fully opaque/dark grey/black to project an artificial sky onto.

I know that things similar to this effect have been chemically achieved such as UV-activated sun glasses and remote controlled window tint, but I haven't been able to find much information that explains how this kind of thing is done chemically.

What options, chemically speaking, exist for making a large glass surface transform between being clear and opaque?

I'm looking for any known chemical that can do this in the form of a tint applied to a glass surface, or a liquid filling a hollow glass surface, or even a gas filling a hollow glass surface.

To be absolutely specific, I'm describing a chemical that can be either modified or added/removed to/from/within a glass surface to systematically control nearly full opacity or nearly full clarity of a layer applied to or contained within the surface of a curved glass panel.

Note: The opacity transition should be efficiently repeatable, meaning, the effect shouldn't deteriorate significantly with repetition.

There are

thermochromic materials that change color in response to heat
photochromic materials that change color in response to light
and electrochromic materials than change color in response to an applied electric field.

All of these materials are sub-categories of Smart Glass. I suspect that electric field sensitive Smart Glass materials are what you are looking for.  There are several different types of such Smart Glass materials.

Those with suspended particles that are normally randomly aligned, but when an electric field is applied they align an alter the light transmission properties of the material.  These materials have already found commercial application (airplane) and are moving down to lower cost applications (autos)
Electrochromic devices where a molecule undergoes a chemical change between colored and non-colored states when a field is applied.
Liquid crystal materials dissolved in a polymer matrix are normally randomly aligned, but, when a field is applied they align thus altering the light transmission properties of the material.

There are also some non-electrochromic smart glass technologies that are also discussed in the Smart Glass Wikipedia article linked above.  These are generally lower tech and therefore lower cost.  They work based on

Having several aligned windows, but with the alignments in different directions and then manually moving the windows into place to adjust the opacity.
Having two reflective windows with a space in between.  When the space is filled with a material having the same refractive index as the reflective windows, the assembly becomes transparent.
Using multiple polarized panels and moving them into or out of place to adjust opacity.

Which method you choose will depend upon how quick you need the materials to change color, the level of manual intervention you're willing to tolerate and how much your willing to spend.  If you google "Smart Glass" you'll find that this field is already well advanced.


Unusual oxidation numbers for alkaline earth metals? Are there compound of alkaline earth elements showing unusual oxidation numbers? For the sake of the question, every oxidation number different from +2 counts as unusual, most exciting would be oxidation numbers greater than +2.

This answer is an extension to @Ian Bush answer. Not only magnesium, but every group 2 element has a lower oxidation state of +0(beryllium) and +1(for other metals).

Be(0)

A 2016 paper gives an insight to the existence of a zero valent beryllium complex compound i.e it has been verified through computational method but it is yet to be synthesized.

  [...]The team added a single ligand i.e CAAC = cyclic
  (alkyl)(amino)carbene to $\ce{BeCl2}$ in benzene solution to give
  $\ce{(CAAC)BeCl2}$, and obtained the final product,
  $\ce{[Be(CAAC)2]}$, by reducing the beryllium with KC8 in the presence
  of a second equivalent of CAAC ligand.

Be(I)

  High resolution infrared emission spectra of beryllium monohydride and
  monodeuteride have been recorded. The molecules were generated in a
  furnace-discharge source, at 1500 °C and 333 mA discharge current,
  with beryllium metal and a mixture of helium and hydrogen or deuterium
  gases.(source)

Mg(I)

It has been described in @Ian Bush's answer. For more information, refer to these sources(1 and 2). Some examples are $\ce{Mg2RuH4, Mg3RuH3, and Mg4IrH5}$ having the $\ce{Mg-Mg}$ bond and magnesium diboride containing the metastable $\ce{Mg2^{2+}}$ ion(Credit @Oscar Lanzi).

Ca(I)

A Ca(I) Sandwich Complex $\ce{[(thf)3Ca(μ-C6H3-1,3,5-Ph3)Ca(thf)3]}$ is being described in this 2010 paper.

Sr(I)

  The high-resolution infrared spectrum of gas-phase $\ce{SrF}$ was
  obtained in emission with a Fourier transform spectrometer.[...]
  (Source)

Ba(I)

3 isotopes of barium in $\ce{BaF}$ (Source).  In [2018] Wu and Lerner (source) reported a barium(I) in a complex graphite intercalation compound.

I did not had to google each and every element to search for lower oxidation state. The wikipedia article of oxidation element gave a list of all posible O.S of element. Information regarding unusual O.S of any element can be found in the footnote and thus only clicked the relevant elements for the answer.


Sulfates and persulfates as accelerants in black powder? I saw there was a few black powder posts on here so I'd like to pose a question. Would sulfates or persulfates like the oxidixer soduim persulfate be a good replacement for some percentage of the nitrates or usual sulfer content of black powder? Pursulfates are used as an ingredient in flash powder for pyrotechnics and is a strong oxidizer to combustibles, so I assume it would make a good accelerant in some concentration. Would it perform under the pressures and speeds needed for an old fashioned firearm, or would it be too corrosive to be practically used?

Sulphates and persulphates cannot be considered as analogues to chlorates and perchlorates.

Perchlorates are frequently used in professional pyrotechnics, being more reactive than nitrates.
Chlorates are too reactive for safe use.
Sulphates are useless as not acting as oxidant.
Persulphates have limited oxidative oxygen content, compared to nitrates or perchlorates, but as chlorates are too reactive for safe operations.

As a general rule, you must know what you are doing. More reactive mixtures may not only lead to faster wearing of the guns, but also to a gun blowing into your face.

Sulphur cannot be replaced by sulphates/persulphates, as it has the opposite role than oxidants. It would be like if you wanted to replace oxygen for breathing by carbon dioxide, as both contain oxygen.

Gun powder

  Chemical reaction

  Gunpowder does not burn as a single reaction, so the byproducts are not easily predicted. One study showed that it produced (in order of descending quantities) 55.91% solid products:
   potassium carbonate, potassium sulfate, potassium sulfide, sulfur, potassium nitrate, potassium thiocyanate, carbon, ammonium carbonate and 42.98% gaseous products: carbon dioxide, nitrogen, carbon monoxide, hydrogen sulfide, hydrogen, methane, 1.11% water.

  However, simplified equations have been cited.

  A simple, commonly cited, chemical equation for the combustion of black powder is
  $$\ce{2 KNO3 + S + 3 C -> +K2S  + N2 + 3 CO2}$$

  A balanced, but still simplified, equation is $$\ce{10 KNO3 + 3 S + 8 C -> 2 K2CO3 + 3 K2SO4 + 6  CO2 + 5N2}$$


Effect on the partition function upon the mixing of two ideal gases Consider a box that is separated into two compartments by a thin wall. Each
  compartment has a volume V and temperature T. The first compartment contains N
  atoms of ideal monatomic gas A and the second compartment contains N atoms of
  ideal monatomic gas B. Assume that the electronic partition functions of both gases
  are equal to 1. The molecular partition function for each component is given by
  $$q_i = \frac{V}{Λ_i^3}$$

Firstly I am asked to write the total initial canonical partition function which is given in the question as $Q_{initial}=Q_AQ_B$. The algebra is fine, but why would the two separate gases have a total canonical function?

I was then asked to show that after the gases mix (and do not react) that the following is true: $$\frac{Q_{mixed}}{Q_{initial}}=4^N$$.

I got that $Q_{initial}=\frac{q_A^Nq_b^N}{(N!)^2}$. Therefore, the only reasonable expression for the mixed function is:
$$Q_{mixed}=\frac{(2q_A)^N(2q_b)^N}{(N!)^2}$$

However, I cannot see why the individual molecular partition functions will have doubled. Especially since A and B are distinct gases.

The fact that $Q_{\mathrm{initial}} = Q_{A}Q_{B}$ implies that the two subsystems, respectively consisting of gas A and gas B, are independent. Because $\log Q_{\mathrm{initial}} = \log Q_{A} + \log Q_{B}$, it follows that various extensive thermodynamic quantities (e.g., internal energy, free energies, entropy) of the total system are simply given by the sum of the corresponding quantities of the subsystems.

The partition functions are doubled because each gas occupies double the initial volume after mixing, i.e., $V \rightarrow 2V$.


How to balance the equation for the reaction of aluminum bicarbonate and hydrochloric acid? Here's what I have to balance:

  Solid aluminum bicarbonate and hydrochloric  acid combine to form an aluminum chloride solution and carbonic acid.

Or in an equation form:

$$\ce{\text{__ } Al(HCO3)3 + \text{__ } HCl -> \text{__ } AlCl3 + \text{__ } H2CO3}$$

Side Note:

$\ce{Al}$ has a charge of $3+$ and $\ce{(HCO3)}$ has a charge of $1-$. To balance the ionic compound we have to change it to: $\ce{Al(HCO3)3}$.

The underscores before each compound in the chemical equation are just there for number values we can input in order to balance the equation.

My question is: what would be the reactants and products in this chemical equation? Would $\ce{HCO3}$ as a whole compound be a reactant itself?

My question is... what would be the reactants and products in this chemical equation?

You give the equation, you just need to balance it.

  Would HCO3 as a whole compound be a reactant itself?

No, $\ce{HCO3^-}$ is an anion.


Which "exotic salt" can lower water's freezing point by 70 °C? The Medium.com article Mars Phoenix Lander, 10 Years Later shows several remarkable images and discoveries on Mars by the Mars Phoenix Lander circa 2008.
One image (shown below) shows what looks like droplets of liquid water, condensed on the surface of one of the lander's legs.
The article says (emphasis mine):

Shortly after landing, the camera on Phoenix’s robotic arm captured views of blobs of material on one of the landing struts. Over time, these blobs moved, darkened, and coalesced, behaving like droplets of liquid water. The hypothesis here was that these blobs “splashed up” on the struts when the descent thrusters melted the ice exposed upon landing mentioned above.
But if liquid water isn’t stable on the martian surface, how did Phoenix observe liquid water on Mars? The key here lies in salt. If you live anywhere that gets snow, you’re probably familiar with salt as a de-icer for roads, sidewalks, etc. Salt lowers the freezing point of water, allowing it to remain liquid at temperatures lower than that of non-salty water. For example, pure water freezes at 0 °C/32 °F, but ocean saltwater freezes around −2 °C/28.4 °F. While the de-icing salts you get at the hardware store lower the freezing point by a few degrees, more exotic salts can lower the freezing point as much as −70 °C/−89 °F! Phoenix discovered some of these exotic salts in the soil around the lander—in particular, magnesium perchlorate. (note, minor editorial changes have been made)

Question: Which "exotic salt" can lower water's freezing point by 70 °C?
Is it in fact magnesium perchlorate (which was found on Mars) or is it a different salt?

Blobs of possible brine (really salty water) imaged on one of Phoenix’s landing struts shortly after arriving on Mars. Credit: NASA/JPL-Caltech/University of Arizona/Max Planck Institute

I recently got a chance to attend a talk by someone who was working on developing analytical instrumentation on Mars. The interesting story is that the initial results by ion-selective electrode was that Mars soil is full of nitrates. Nobody knew on Earth that the nitrate ion selective electrode is far more responsive to perchlorate than nitrate. After learning this, it was an eye opener for analytical chemists! Now they wish to use chromatography rather than electrochemistry. So this was a good lesson for us on Earth.

The perchlorate ion was discovered in 2008 by the nitrate selective electrode. No specific electrode was attached to detect perchlorate, it was rather an accidental discovery. The Science Report makes a footnote "Detection of Perchlorate and the Soluble Chemistry of Martian Soil at the Phoenix Lander Site" (paper: Science 2009, 325 (5936), 64–67)

A Hofmeister anion ISE was intended to monitor nitrate from a $\ce{LiNO3}$ reference electrolyte that was part of the leaching solution, but was ultimately used for perchlorate detection

  [Footnote] The relative sensitivity of the Hofmeister series ISE to perchlorate
  over nitrate is 1000:1, and substantial quantities of perchlorate will
  overwhelm any other signal. If, as was observed, >1 mM perchlorate
  accounts for the observed signal, it would require >1000 mM nitrate to
  produce the same response. This would correspond to more than the mass
  of the entire sample.

Now that they know it is a perchlorate ion, people did some studies on supercooled brines. See this paper: Toner, J.; Catling, D.; Light, B. The formation of supercooled brines, viscous liquids, and low-temperature perchlorate glasses in aqueous solutions relevant to Mars. Icarus 2014, 233, 36–47 (also available here). They clearly show that if calcium or magnesium perchlorates are slowly cooled, one can get supercooled brines up to -120 Celcius. This is a rather amazing finding. They call it a glassy state.


p-Nitroaniline reaction with sulfuric acid An alternative version of the well-known demonstration carbon snake (the dehydration reaction of sugar by concentrated sulfuric acid forming a column of graphite) consists in mixing  paranitroaniline and sulfuric acid.
This reaction is very vigorous and occurs suddenly, forming a similar column of black solid. However, except from the article "Studies on nitroaniline–sulfuric acid compositions: Aphrogenic pyrostats" (https://onlinelibrary.wiley.com/doi/abs/10.1002/app.1970.070140813), I could not find any relevant information about it. I would like to know if the products of this reaction or its mechanism have already been studied.

I think the abstract of the reference did not give away what's going on the reaction. However, I assume that the full paper should have described what's happening to at least certain extend. Since I don't have access to this paper at the moment, only I can do is speculate. Yet, I'd direct you to reliable source for certain information. Accordingly, when you heat the mixture of p-notroaniline and conc. sulfuric acid:

The only gaseous product you would obtain is $\ce{H2O}$ (with some nitroaniline subliming at higher mole ratios) up to the temperature of $\pu{230 ^\circ C}$. At $\pu{230 ^\circ C}$, it is called pre-expansion stage. Sulfur dioxide has been absent from gaseous products (confirmed by iodine test). It was noted that dehydration and sulfonation are the main chemical processes taken place at this stage.

There are two more stages to follow before you see the carbon snake, after stage of which called oxidative decomposition of the residue. This UCSB article also provides some other interesting excerpts from the original article (Ref.1), and listed below for readers' benefit:

The mole ratio of nitroaniline to sulfuric acid, however, is not critical; voluminous foams have been obtained from mixtures over a mole ratio range of 0.3 to 2.0 if a proper heating schedule is maintained. Sulfur dioxide and water are the principal gaseous products evolved during these expansions.

The foams were good thermal insulators inert to chemicals, e.g., concentrated mineral acids, alkalies, organic and inorganic solvents, and oxidants.

The exact structural nature of the insoluble residue constituting the foam is unknown. All the sulfur in the residue is present as sulfonic acid groups attached to the benzene nuclei. Secondary amino or azinium groups resulting from deammoniative coupling must be present in order to account for ammonium ions and Nitrogen content less than 2 per benzene nucleus. A high nitrogen-to-oxygen ratio indicates that few if any nitro groups remain. Thus, both nitro arid sulfonic acid groups are involved in oxidative processes.

Also, followings are three relevant demonstrations of the reaction: here, here, and here. Also, here is a demonstration of the dehydration reaction of sugar by concentrated sulfuric acid forming a column of graphite for comparison.

It is worth noting that the carbon form created by the p-nitroaniline reaction with sulfuric acid is found to be useful in $\ce{CO2}$ adsorption (Ref.2).

References:

A. C. Poshkus, J. A. Parker, "Studies on nitroaniline–sulfuric acid compositions: Aphrogenic pyrostats," J. Appl. Polym. Sci. 1970, 14(8), 2049–2064 (doi: https://doi.org/10.1002/app.1970.070140813).
Enrico Andreoli, Andrew R. Barron, "$\ce{CO2}$ Adsorption by para-Nitroaniline Sulfuric Acid-Derived Porous Carbon Foam," J. Carbon Res. 2016, 2(4), 25 (9 pages) (doi: https://doi.org/10.3390/c2040025).


High Degrees of Symmetry in Molecules I recently encountered a question on a test:

  Substances whose molecules have a high degree of symmetry will have:

    A. a low specific heat
    B. a high melting point
    C. a high heat of fusion
    D. a low heat of vaporization
    E. strong Van der Waals forces

Looking through a few compounds, I found that $\ce{CO2}$ is symmetric (it's shape is a straight line), $\ce{CH4}$ is symmetric (tetrahedrons seem symmetric).
Through these examples, I was able to eliminate A, B, and C, and have a suspicious feeling about E, so I guessed D and was right.

I was wondering what exactly "high degree of symmetry" meant. Through my math courses, I knew about rotational symmetry (where if you turn any 2-D shape in a plane by n degrees, it will look identical to the shape before the rotation).
Is a high degree of symmetry when the lowest possible value of n for a compound is over some constant, lets say 90 degrees, or is it something else?

I think that the question meant to test your knowledge about the polarity of simple molecules.

The fact that symmetric molecules contain identical bond polarities between the central atom and surrounding atoms, the net dipole moment usually tends to be zero, creating a nonpolar molecule.

In other words, I think symmetry in this case refers to the fact that all of the surrounding atoms are identical.


Calculating concentration of H+ Ions using Ka Value I have this homework question that I am a bit stuck on for an answer. If anyone could help that would be great.

Question:

  For a $\pu{0.2M}$ $\ce{HNO2}$ solution ($\pu{K_a= 4.5 \times 10^{-4}}$) calculate the concentration of the $\ce{H+}$ and $\ce{NO2-}$ ions, $\ce{HNO2}$ molecules and the $\ce{pH}$ of the solution.

If anyone could give any hints or answers that would be great.

Answer:

$K_a = 4.5*10^-4$
Concentration (From ICE Table) of products/reactants:

HNO2 = 0.2 - x
H+ = x
NO2 = x

Therefore:

$$4.5*10^-4 = x^2/(0.2-x)$$
Rearrange:
$$x^2 + x*(4.5*10^-4) - (0.2(4.5*10^-4)) = 0$$
Using quadratic formula:
$x \approx 0.009$

$$pH = -log(10)$$
$$pH \approx 2.05$$.

That should be correct!


Do unsaturated fatty acids with an odd number of carbons exist? Our teacher asked us about fatty acid C21:3.

Does this fatty acid even exist?

C21:3 is a stable fatty acid that could indicate a variety of double-bond isomers.  It is "unusual" in biology because of the odd number of carbons, but that doesn't preclude its stability and accessibility to organic synthesis.  If someone really wanted to make this acid, they could do so easily.
Such a compound, 5Z,14Z,17Z-heneicosatrienoic acid,  has been reported in the natural environment.  It was found in a deep-sea clam, Calyptogena phaseoliformis, that lives in or near hydrothermal vents and lives off of the organic matter produced by chemosynthetic bacteria that live in close proximity to it.  So in addition to being synthetically accessible, it is also made in living organisms.

Thus, for all reasonable interpretations of "exist" in your question, does this fatty acid even exist?, the answer is yes.


Newman projections: how to tell which groups are bulkiest to determine what is the most stable conformation? The scenario

Let's say we have a molecule where in the Newman projection the front carbon has a hydrogen, methyl and an ethyl, and in the back carbon there's a hydrogen, something like a S, Cl or Br (bigger than carbon but only one atom), and something like another ethyl or isopropyl:

My understanding

Methyls/ethyls/isopropyls etc. would cause more steric hindrance than something like S, Cl or Br because they consist of multiple atoms and take up more space, so in the scenario above, the front carbon's ethyl should be 180° away from the back carbon's ethyl/isopropyl, and the S/Cl/Br's position would be dependent on the ethyl from rotation. I think the following should be the most stable conformation:

My question

I've seen some similar examples where (in the same type of scenario) the S/Cl/Br was considered "bulkier" and was put 180° away from a big isopropyl, when something like an ethyl could have been there instead, but was put only 60° away from the big isopropyl (see the first illustration). I'm not sure if these examples were just wrong or if I'm completely misunderstanding how to determine what's bulkiest, so any advice on how to determine that would be super. If a table that ranks most-to-least bulkiest exists, please share.

Well, I think your book is wrong. Read the extract:

  A-Values are numerical values used in the determination of the most stable orientation of atoms in a molecule (conformational analysis), as well as a general representation of steric bulk. The utility of A-values can be generalized for use outside of cyclohexane conformations. A-values can help predict the steric effect of a substituent. In general, the larger a substituent's A-value, the larger the steric effect of that substituent. A methyl group has an A-value of 1.74 while tert-butyl group has an A-value of ~5. Because the A-value of tert-butyl is higher, tert-butyl has a larger steric effect than methyl. This difference in steric effects can be used to help predict reactivity in chemical reactions.

I guess what you're looking for are A-values. The higher the A-value, the more "bulky" the group is considered.

So, the $\ce{-SH}$,$\ce{-Cl}$ and $\ce{-Br}$ groups seem to be less hindering than the $\ce{-iPr}$ group.

References:      A-value, Wikipedia


Entanglement of large molecules In many textbooks and chemistry curricula, we can see the description of large molecules (particularly in the discussion of hydrocarbons and other nonpolar compounds) as having "entanglement." However, I think the term is misleading, since the molecules don't really act like tangled shoelaces or two strings pulling against each other while interlocked. Can we try to clarify what actually occurs that we describe as entanglement?

Some things I think may happen:

The molecules are very large, and thus have many sites for IMFs, leading to high viscosity.
By way of this size, the molecules are much harder pressed for movement and so take a while to move (the distinction is that in the first example the increased size begets more IMFs whereas the second causes "entanglement" just by the bulky molecules' shifting past each other).

Well, very long hydrocarbons can become twisted around each other. This is definitely known in the polymer literature. Consider some branched polyethylene with ~1,000 repeat units. It's not too hard to imagine that multiple chains can become entangled in much the same way as shoelaces or strings when you mix a lot of them in a box.

You are also right that with large molecules, there are more sites for intermolecular interactions. Consider that there is a "combination" effect with lots of sites. That is, to pull two very long linear chains apart, you need to break lots of interactions.

Notice that pulling one atom on the bottom away from its neighbor on the top chain will require the movement of all atoms on the bottom chain. So it's slightly harder to separate two long chains, not only because there are so many interactions, but also because of the combination.

Put simply, the difficulty in separating the two chains is slightly higher than simply the sum of the interactions.


How is it Possible to Measure Electrode Potential? It is known that potential difference on metal/electrolyte or any kind of interphase can't be measured since voltmeter needed to measure it is made of metal which would in contact with metal and solution develop its own potential difference at interphase and as such affect the measurement. To solve this problem, potential is measured with respect to SHE. What I don't understand is that if you want to measure the electrode potential with respect to SHE, voltmeter needs to come in contact with two metal electrodes (one is electrode which potential you want to measure and the other is SHE) which will again create potential difference between the electrodes and voltmeter since two different metals in contact create potential difference between them. Can you explain?

It is the same as measurement of voltage difference at any electronic circuit. The fact the metallic measuring points are attached to electrochemical electrodes is not relevant here.
Yes, there is metal contact potential difference, but it very low and very probably mutually canceling.


Procedures in experimental chemistry An experiment was carried out involving $\alpha$-bromoheptaldehyde dimethyl acetal. The experimental details are given below.

  A solution of enol acetate  (15.6 g, 17.7 mL) in chloroform (20 mL) was cooled
  to 0°C. A solution of bromine (16 g, 5.1 mL) in chloroform (5 mL) was added
  slowly with stirring with the rate of addition being controlled to keep the
  reaction temperature below 10°C. After the completion of addition, anhydrous
  methanol (60 mL) was added to the mixture and it was stirred for 48 hours.
  The mixture was diluted with water (200 mL) and ether (50 mL). The organic
  layer was separated, washed with water (100 mL) and with 5% aqueous
  sodium carbonate (100 mL). The organic layer was dried using sodium sulfate and
  filtered. The solvents were removed by distillation under reduced pressure.
  The residual oil was distilled under reduced pressure in the presence of a
  small amount of sodium carbonate. The fraction boiling at about 118°C at 17
  mmHg was collected to give bromo acetal as a colourless oil (19.5 g).

  (a) Do you think that this reaction should be done using a drying tube, under
  nitrogen or open to the air? Explain your answer.

  (b) Without doing TLC or any other form of spectroscopy or chromatography,
  how would you be able to tell if sufficient Bromine had been added?

  (c) Will the organic layer be the upper or the lower layer? How can you
  quickly check this experimentally?

  (d) Sodium carbonate is added just before the final distillation. Why?

  (e) The procedure calls for removal of the organic solvents by “distillation
  under reduced pressure”. Which one piece of laboratory apparatus would you
  use for this?

  (f) Suggest two serious and different hazards associated with this experiment.

My attempt

(a) It can be done in drying tube, as the reactant have to be prevented from coming into contact with water vapor from the air as the water vapor could potentially react with the reactant.

(b) By observing the colour of the reactant mixture. Since bromine is brown in colour, Once sufficient bromine has been added, there will then be excess bromine if bromine was continued to be added, hence the reactant will turn brown once enough bromine is added.

(c) The organic layer will be at the bottom. Since it is know that the density of the organic layer is more that that of the ether.The density of the organic layer can be measured and compared to that of the ether.

(d) To remove any excess acid that might be present in the mixture.

(e) Rotatory evaporator

(f) Bromine is toxic if inhaled, while methanol is a carcinogen

I am unsure of my answers though. Especially part (c) as I can't decide whether the aqueous or organic layer will be at the top. Could anyone explain this.

I believe that when you say "a solution of enol acetate" you are refering to a solution of the corresponding enol-acetal of hetanal? You should use full names of the compounds used, if not, it might be confusing.

My opinions

  (a) Do you think that this reaction should be done using a drying tube, under nitrogen or open to the air? Explain your answer.
  It can be done in drying tube, as the reactant have to be prevented from coming into contact with water vapor from the air as the water vapor could potentially react with the reactant.

I don't see a reason you would need any inert conditions for this experiment. I think you will be fine if you perform it open to the air.

  (b) Without doing TLC or any other form of spectroscopy or chromatography, how would you be able to tell if sufficient Bromine had been added?
  (b) By observing the colour of the reactant mixture. Since bromine is brown in colour, Once sufficient bromine has been added, there will then be excess bromine if bromine was continued to be added, hence the reactant will turn brown once enough bromine is added.

Indeed. The end point of the reaction is reached when the calculated amount is absorbed and the bromine is no longer decolorized.

  (c) Will the organic layer be the upper or the lower layer? How can you quickly check this experimentally?
  The organic layer will be at the bottom. Since it is how that the density of the organic layer is more that that of the ether.The density of the organic layer can be measured and compared to that of the ether.

This is wrong. The organic layer IS the one which is will be a mixture formed by chloroform, unreacted enol acetate and ether. The other phase will be the one composed mainly by water and methanol.
Methanol+water solution will probably be less dense than chloroform+ether+enolacetate solution, and therefore, organic layer will be the upper one. I cannot tell for sure, since you have mixtures in high proportions of compounds with quite different densities, but and easy way to tell is:
- Separate both phases using an extraction funnel. Put some tap water on the separated phase you suspect is the aqueous phase. It dissolves? That is the aqueous phase. It does not dissolve and forms a two-phased system? It was the organic phase, extract it again.
You can double check adding water to the other phase you extracted, and see that the opposite effect should be observed.

  (d) Sodium carbonate is added just before the final distillation. Why?
  To remove any excess acid that might be present in the mixture.

Yeah. The acetal must be free of acid; otherwise decomposition takes place during distillation.

  (e) The procedure calls for removal of the organic solvents by “distillation under reduced pressure”. Which one piece of laboratory apparatus would you use for this?
  Rotatory evaporator

Correct. Rotovap is your guy.

  (f) Suggest two serious and different hazards associated with this experiment.
  Bromine is toxic if inhaled, Methanol is a carcinogen

Bromine will definitely be your top priority concern. Even more if you have to prepare the solution from pure bromine. Also, it has a very strong unpleasant smell, which gives it the name it has (from Greek: βρῶμος, brómos, meaning "strong-smelling" or "stench")
It is corrosive very toxic, volatile (fumes a lot, MUST be handled on a fumehood).
Chloroform is also toxic; Heptanal is volatile and flammable.

Methanol, even though you should not drink it wouldn't be of an important concern.

To sum up, some of the reactants are of unpleasant odor or have lachrymatory effects it is best to use a good hood.

And also a reference: Organic Syntheses, Coll. Vol. 3, p.127 (1955); Vol. 29, p.14 (1949).


How to make equal density pool dye without diffusion I have a leak I have been struggling to find in my in-ground pool.  I am losing about 1/4" of water per day and I believe the issue is one of the fittings in the bottom of the pool.  There are two main drain fittings and one deep pot that has two pipe plugs that if removed allows groundwater to enter into the pool to prevent the forces of buoyancy from floating the shell out of the ground like a boat when the pool is drained.

I want to make a dye that is has a specific gravity equal to water that has minimal to no diffusion.  It has to be safe because it will be introduced to the pool water.

The only thing I have found that is almost the same gravity of freshwater is beer, or milk.  The problem with those is the diffusion rate.  I was hoping to find something that I could mix with dye that had a specific gravity (I hope that is right) as near to freshwater as possible that wouldn't diffuse when introduced to freshwater.

My intention, hope, is that I would place a blob of dye over these plugged fittings and if there was a leak around the fitting it would eventually suck that dye into it.  It could take 15 minutes or more for the dye to be pulled into the leak, so it has to have a very slow diffusion rate.

Whole milk with food coloring solved my problem.  Whole milk has a specific gravity of 1.01, which allowed it to settle to the bottom around the main drains.  The diffusion rate was extremely slow over regular pool dye, which allowed the dyed milk to linger around the drains for an extended period of time (5+ minutes) to allow for the detection of the slow leak.


Second order reaction with different reactants In a laser photolysis reaction $\ce{C6H5NH2}$ $(C_0 = \pu{60 µM})$ is oxidized to its radical cation. The disappearance of the cation can follow three pathways: (1) Dimerzing of the cation, (2) Reaction between the cation of the initial substance, (3) Reaction between the cation and the solvent. Decide which pathway is most likely to occur.
$$
\begin{array}{l|cccccccc}
\hline
t/\pu{µs}~(\text{efter puls}) & 0 & 5 & 10 & 15 & 20 & 25 & 30 & 35 \\
\hline
[\text{Radikalkatjon}]/\pu{µM} & 20 & 11.12 & 6.86 & 4.47 & 3.00 & 2.06 & 1.43 & 1 \\
\hline
\end{array}
$$

I first checked for the dimerizing of the cation by assuming it to be a second order reaction with the same reactants and plotting $1/[\mathrm{cation}]$ against $t,$ which didn't give a good fit. So I wanted to plot it as a second order reaction with different reactants $(\ce{A + B -> products})$ but I don't understand how that is done.
I should plot $\displaystyle\ln\frac{a-x}{b-x}$ against $t,$ but what exactly is $(a-x)$ and $(b-x)?$
I know that I should set up a reaction like:
$$
\begin{array}{l|ccc}
\hline
& \text{A} & \text{B} & \text{product} \\
\hline
t=0 & a & b & 0 \\
t=t & a-x & b-x & x \\
\hline
\end{array}
$$
And I would assume that my $a = \pu{20 µM}$ and $b = \pu{60 µM}.$ But where do I go from there? All help is appreciated!

The general solution, if the reaction is $aA+bB \to product$, is to allow an amount $x$ to react at time $t$ then there will be concentrations $(A_0-ax)$ and $(B_0-bx)$ of species A and B respectively at time $t$ if $A_0,B_0$ are the concentrations at time zero.  The rate equation is $\displaystyle  -\frac{1}{a}\frac{d(A_0-ax)}{dt}=\frac{dx}{dt}=k(A_0-ax)(B_0-bx)$ which can be integrated ($0\to x, 0\to t)$ to give

$$ \ln\left(\frac{A_0-ax}{B_0-bx} \right)=k(A_0b-B_0a)t +\ln\left(\frac{A_0}{B_0} \right)$$

If the reaction follows this scheme plot the left hand side vs $t$ and this should produce a straight line of slope $k(A_0b-B_0a)$.

Should it happen that concentrations are initially $A_0/B_0 = a/b$ so that species are present in stoichiometric amounts then as $A_0b-B_0a=0$ the  solution above cannot be used. In this case the rate equation becomes $\displaystyle  \frac{dA}{dt}=-bkA^2$ which gives $\displaystyle \frac{1}{A}-\frac{1}{A_0}=bkt$.


Do hydrogen halides (HX) form strong or weak acids? I know that when dissolved in water, hydrogen halides ($\ce{H-X}$), become acids. I know that $\ce{HF}$ forms a weak acid, but can it be generalized that all $\ce{HX}$ acids do so?

No, it can’t be generalised. Amongst the haloacids, only $\ce{HF}$ is weak.

Acid strength can be determined by the stability of the conjugate base (for most acids, the conjugate base is what you get when you remove $\ce{H+}$ from the acid). In this case the conjugate bases are the halides.

As the size of the halide ion increases, the charge is dispersed over a larger volume, and thus the ion is more stable. In this manner, one can say that the order of acid strengths is $\ce{HF<HCl<HBr<HI}$ (I’m not considering the radioactive ones here; they fit in the same trend though).

$\ce{HCl, HBr, HI}$ are all strong acids. $\ce{HCl}$ has a $\mathrm{p}K_\text{a}$ of −7 (the smaller the $\mathrm{p}K_\text{a}$ is, the more acidic it is), $\ce{HBr}$ is approximately −9, and $\ce{HI}$ is also approx −9.

On the other hand, $\ce{HF}$ is 3.2. The $\ce{H-F}$ bond is pretty strong, so it has a smaller tendency to dissociate—making it a weak acid (not to say that it isn’t dangerous).


Site of first protonation of imidazole and aminoimidazole I wish to ask where imidazole and aminoimidazole will be protonated on first protonation.

Imidazole:

Aminoimidazole:

I suppose that the nitrogen (without hydrogen) of the ring is the most probable to be protonated, but I am not sure about it. Do you know where it will be protonated first?

Any link to a study where the issue is solved will be good (I need the information for my thesis).

For imidazole, the lone pair of pyrrole-type nitrogen (the one with hydrogen) is involved in aromaticity of the whole compound, as it contributes two π-electrons in addition to four of those from carbon-carbon or carbon-nitrogen bonds, which makes it 6, which is a classical sign of Huckel aromaticity. In fact, this NH-nitrogen tends more to act as an acidic center than as a basic one, as it loses its hydrogen when treated by strong bases, like $\ce{NaNH2}$.

The pyridine-type nitrogen has its lone pair out of ring conjugation, thus it can act as a acceptor of hydrogen ion, placing its two electrons into a vacant orbital of a $\ce{H+}$.

For reference, you can see any heterocyclic chemistry textbook. I prefer the one authored by Joule, Mills. Or this.

When it comes to 2-aminoimidazole, this publication  (which is unarguably very old) says all three nitrogens are involved in protonation, because the structure is a derivative of guanidine, a strong organic base. Which seems only legit to me if it said that the contribution of pyrrole-type nitrogen is negligible, as aromaticity is ruined in that case. I could also predict that exo-aminogroup gets protonated more likely, as its basicity is somewhat higher: its lone pair is on a sp3-orbital, rather than the lone pair of endo-nitrogen (the one without H), where it is on a sp2-orbital, thus, the contribution of s-orbital is higher and it is closer to the nucleus of nitrogen atom, decreasing its availability to act as a covalent bond acceptor.


Possibility of π–π stacking in non-aromatic systems Is it possible to have π–π stacking for non-conjugated double bond systems?

For example, can the carbonyl groups of two urea molecules participate in π–π stacking? Although urea prefers to stay in hydrogen bonded form.

I would be really happy if you cite some resources (research paper, book, etc.) while answering.

There is a paper titled "Amide-pi interactions between formamide and benzene",  doi: 10.1002/jcc.21212. It has been cited by a paper on a crystal structure on a urea transporter (the interaction is between a peptide bond of a ligand and aromatic side chains of the protein).

Technically, the interaction between the arginine side chain and aromatic groups could also fall under your heading, but it is usually described as a cation pi interaction.


Does Dean Kamen's Slingshot really filter virtually everything? According to Dean Kamen (in the movie http://www.imdb.com/title/tt3242934/) and on the Net (http://www.popsci.com/article/science/pure-genius-how-dean-kamens-invention-could-bring-clean-water-millions) his device that uses pressure-Distillation to purify water, it would work with virtually any water and render medical-grade pure water.

I have a hard time to believe this as:
-chemicals could evaporate before water and so get into the distilled water
-I have no idea how he keeps the distillation chamber clean (prefiltering?) so it does not clog.

So I would like to know what are the limitations of distilling water to purify it?

I know that distilled water cant be drunk as it is without adding salts. I like the project of Mr. Kamen very much, I just want to know the boundaries, as the promises ("Fill in anything that is wet") seem way to optimistic.
Thx

All it does is distill water while keeping heat within the system.  There isn't much detail provided about it but heres what I can say.

  I have a hard time to believe this as: -chemicals could evaporate before water and so get into the distilled water -I have no idea how he keeps the distillation chamber clean (prefiltering?) so it does not clog.

While you are right that chemicals can evaporate with the water this is not usually of concern if the water source is not severly contaminated with volatile chemicals (as opposed to biological and soluble toxins).  If you were to have chemical contamination you could use reverse osmosis or activated charcoal to remove residual contamination.  typically this is not an issue though.

  I know that distilled water cant be drunk as it is without adding salts. I like the project of Mr. Kamen very much, I just want to know the boundaries, as the promises ("Fill in anything that is wet") seem way to optimistic.

Distilled water can be drank.  It may not be the most tasteful water but it does taste better than cholera.  If you want more information his TEDMED Talk is very informative.


Cows milk protein in lactose Does anyone know if the casein/whey cows milk protein can be found within processed lactose?

Some background: our 4-month-old baby was given teething granules. She was then presented with a rash, screaming, etc. When we checked the teething granules (called teetha) we found they contain lactose.

However, from reading it appears that lactose intolerance is extremely rare in the UK and cows milk allergy is much more common.

Yes, process lactose could still contain protein from the milk. It's equally likely that the casein could contain lactose. Whey is much more likely to contain lactose since the whey and lactose are in the water when the cheese is made.

It would be extremely rare for a baby to be lactose intolerant since lactose is the major sugar they consume. As the child ages, the ability to digest lactose decreases. I think I heard that 5% of Caucasians adults can still digest lactose. Only a few people develop lactose intolerance.


Converting units and sig figs (finding molarity)? This may seem very basic, but I'm a little confused.

"What is the molarity of a solution 500 mL of which contains 0.500 moles of HCl?"

Being that 500 has one significant figure, would it be 500 mL = 0.5 L as opposed to 500 mL = 0.500 L? In which case the answer would be 1 mol/L instead of 1.00 mol/L?

In terms of significant figures, you always use the number with least amount of sig figs to determine how many sig figs your answer has. In this case, one number has 3 sig figs, and the other has 1. Therefore your answer should have 1 sig fig and be 1 mol/L, as you said. To make the zeroes in the 500 mL become significant it would need to have a period afterwards like this: 500. mL. Perhaps the writer forgot to do that, however as it is written the answer would be 1 sig fig: 1 mol/L


For a system to undergo a spontaneous change at constant temperature and pressure the Gibbs free energy must decrease, but how? $$\mathrm dG < V\,\mathrm dp - S\,\mathrm dT$$

For a process at constant pressure and temperature ($\mathrm dp = \mathrm dT = 0$), and in the absence of any non-pV work, $\mathrm dG < 0$. Since energy cannot be created or destroyed (and Gibbs free energy is a measure of energy) where does the energy go when this spontaneous change occurs? Likewise for internal energy: $\mathrm dU = T\,\mathrm dS - p\,\mathrm dV$?

Also, why is this even useful? I understand that it's easy to hold temperature and pressure constant in a lab but we only seem to measure enthalpy anyway using $\mathrm dH = C_p\,\mathrm dT$. If it's so easy to keep temperature and pressure constant yet so hard to keep entropy constant then why do we measure enthalpy (which decreases for a spontaneous change when entropy and pressure are held constant) rather than Gibbs free energy. I suspect I am barking up the wrong tree and I am hoping that you'll be able to put me right; the only thought that I had was that the fact that Gibbs free energy muct decrease for a spontaneous change would help to predict whether a spontaneous change would occur when done in a lab - given that $p$ and $T$ will be constant; is this right?

If you want to track where the energy is going, do a first-law analysis (compute heat and work for a given change in internal energy or enthalpy).

If you want to predict whether a process occurs spontaneously or not, or you want to study an equilibrium, the second law is the tool you need. It's convenient to apply the second law by looking at free energy changes, since they let you focus on the system rather than the entire universe.

When we're interested in experimental internal energies or enthalpies we're tracking energy transfers. When we're interested in the Gibbs free energy we're usually answering questions about process spontaneity, or equilibrium conditions, or maybe we want to compute non-PV work. Different tools for different jobs.

Let me edit some of my comments below into my answer, because the comments are scrolling off the page and they make it clearer why you can't really track energy transfers just by looking at the Gibbs free energy:

The Gibbs free energy is an energy in the sense that it is the part of the energy that is "free" to do useful work. It does indeed have units of energy, and $\Delta G$ can be used to compute the maximum  possible nonexpansion work.

It is not an energy in the sense that it is not conserved in a process, the way internal energy and enthalpy are. $G$ is derived partly from entropy. It's quite possible that $G_\text{system} + G_\text{surroundings}$ isn't constant in a process.

If you have an isothermal process, $\Delta G= \Delta H−T\Delta S$. You can see from that equation that it's possible to decrease the Gibbs free energy simply by increasing the entropy, without any energy transfer at all. DavePhD's answer gives a concrete example of that.