text:
Does nitrogen deaeration need to be performed in a fume hood?
I have a couple electrochemical cells I will need to sparge (deaerate) with nitrogen prior to and possibly during my experiment (the solution is .1 M NaOH). Does this need to be done in a fume hood?
From what I've read, it seems that $\ce{N2}$ gas isn't a huge danger, but if you're in a confined space it can cause a severe lack of oxygen. If that is the case, would a normal-sized classroom count as a confined space? Also, from what I understand about deaeration, the nitrogen bubbles escape the solution with impurities, but also with oxygen, so maybe in my case the lack of oxygen wouldn't be a problem.
This may be a simple question, but it has been difficult for me to find standard procedures for sparging a liquid with an inert gas, and I want to make sure I'm taken all the necessary precautions and steps.

You're just using a stream of nitrogen. You can't have such a high flow that it will blow the liquid out of your container. As such, the risk of asphyxiation is quite low. On the other hand, if you work with a relatively dangerous liquid (your base is pretty dilute), you are at risk of blowing it out of the container and onto you, so you should do this in a fume hood with appropriate protective clothing and goggles.

Adding some nitrogen to the atmosphere (already 70% nitrogen) is not a big deal.

Not to count that if you deairate, you release air

As implied in the comments, you have nothing to fear by sparging your solution with nitrogen in the classroom. Even a good size tank only contains a few cubic meters of compressed gas, while your classroom likely contains a few hundred cubic meters of air. Thus, even if you could instantly empty the tank the effect would be negligible. And you will be using low enough flow rates that there will be no fear of a "localized" depletion of oxygen.
The safety issue that I need to mention, though you are almost certainly already aware, is regarding the sodium hydroxide solution. Even at $\pu{0.1 M}$ it's not something that you want in anyone's eyes, so protection against splatter from the sparging solution would be prudent.


text:
What properties of carbon dioxide make it a greenhouse gas?
Why is it that heat from the Earth reflects back off carbon dioxide and other greenhouse gases but not gases like nitrogen or oxygen?

Just a note: it doesn't reflect heat, it reflects light. It's just a rather low frequency light that isn't visible to our eyes, and is produced by low-temperature (room-temperature-ish) thermal radiation. And strictly speaking, reflection isn't a good term - it actually absorbs the radiation and emits it again after a while in a random direction. The heating effect comes from the fact that there's a somewhat-less-than-50% chance of the radiation coming back down, which makes the average surface air temperature higher.

According to the Intergovernmental Panel on Climate Change (IPCC):

"Greenhouse gases are those that absorb and emit infrared radiation in the wavelength range emitted by Earth."

In order for a molecule to absorb and emit in the infrared (IR) region, its chemical bonds must rotate and vibrate in a manner that affects something called the molecule's dipole moment. It turns out that due to the symmetry of diatomic molecules like $\ce{O2}$ and $\ce{N2}$, this process cannot happen, and thus these types of molecules cannot absorb in the infrared, which is the wavelength range in which heat is radiated.
Because $\ce{CO2}$ and other greenhouse gases do not have this kind of symmetry, they can vibrate at specific frequencies within the IR in a manner that affects the molecule's dipole moment, and thus absorb this radiation resulting in the transfer of heat. Much of the heat radiated from the Earth's surface is of the proper wavelength (energy) to be absorbed by these gases, so the criteria given by the IPCC for a greenhouse gas are thus met.
As I'm not certain of your level of understanding of chemistry, I tried to direct my answer somewhere between purely technical and purely non-technical. If I've left anything unclear to you, please don't hesitate to leave a question in the comments.

Could you elaborate a little on the dipole moment and explain it in layman's terms?

@Charlie. Perhaps airhuff will add to the clear answer. However, in short : an electrical dipole moment arise when you have a charge separation. Think of a + and a - separated by some distance, there is as a dipole along the line between them (orientation is conventional). In molecules this arise by partial charge separation, dictated by both the chemical nature of the atoms/fragments involved and the geometry. In symmetrical biatomic molecules there cannot be a dipole, as prerequisite to interact with IR em waves. CO2 is symmetric and linear, however one of the vibration is not. O< C> <O ;)

https://chemistry.stackexchange.com/questions/19119/is-carbon-dioxide-ir-inactive

Hmm, so why is the symmetric methane a greenhouse gas?

@PaŭloEbermann, its fundamental vibrations result in changes in its dipole moments. You can see its fundamental vibrations here: http://www2.ess.ucla.edu/~schauble/MoleculeHTML/CH4_html/CH4_page.html

When heat is leaving earth it leaves as infrared radiation. Greenhouse gases are gases that are able to absorb this infrared radiation. If we look at the infrared emission spectrum from Earth[1]:

We can see that between $\pu{400 cm-1}$ and $\pu{700 cm-1}$, a lot of the infrared radiation is absorbed by $\ce{CO2}$. Gases like $\ce{N2}$ and $\ce{O2}$ don't have any absorption in the infrared region, and are therefore not considered greenhouse gases.
[1] http://www.xylenepower.com/Global%20Warming%20Explanation.htm

I see. Do we know why greenhouses gases are able to absorb the radiation though? Do they have specific properties?

@Charlie that is related to the electronic structure of the molecules. The electronic structure dictates which ("specific") wavelenghts of light a molecule can absorb.

also, how does the absorption of infrared create the greenhouse effect? Surely it would be a reflection of infrared that causes it?

It is actually not reflection. The gas absorbs infrared coming from earth (because it is the main source) and emits it in all directions, so a part gets back to earth instead of escaping into space. So in net effect, it works as an insulation.

The 'heat is not reflected off the carbon dioxide' but instead the gas absorbs some of the ir radiation from the sun and that also radiated from the earth. The fact that carbon dioxide (or methane) can absorb and emit infra-red radiation is only part of the process for the greenhouse effect. Equally important is the fact that oxygen and nitrogen molecules do not absorb and emit ir radiation. (They have no permanent or transient oscillating dipole and so have no 'handle' to interact with the oscillating electric field of the radiation).
Some fraction of the ir absorbed by the $\ce{CO2}$ will be radiated into space and some towards the earth where it will be absorbed by trees, soil, oceans etc, and some of this energy can then be re-radiated back out into space as ir radiation. The $\ce{CO2}$ will also absorb some of this radiation.  However, there is a chance that before a $\ce{CO2}$ can radiate it will collide with an oxygen or nitrogen molecules and so these molecules can gain some energy from this process. However, they  cannot radiate as they have no permanent or transient dipole, and become hotter (more vibrational & rotational quanta) and after collision with other molecules (mainly oxygen & nitrogen) the extra energy is spread out among many molecules and the effect is to heat up the atmosphere.

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text:
How to name this quintuply-bonded chromium dimer?
Depicted below is a chromium compound with a metal–metal quintuple bond. It boasts two identical bridging ligands, derivatives of 1,3-diphenylbenzene. The aim of this question is to figure out

the proper name.

Proper means without losing information. When answering, application of the 2005 Red Book and 2013 Blue Book is desired.

Effort

Identify central atom, derive ending: chromium
Naming unsubstituted ligand, derive ending: 1,3-diphenylbenzene $\longrightarrow$ 1,3-(diphenyl)phenyl
Allow for chromium's probably higher priority $\longrightarrow$ 2,6-(diphenyl)phenyl
Add in the isopropyl groups $\longrightarrow$ 2,6-{[2,6-di(1-methylethyl)]diphenyl}phenyl
Look for stereoisomers of ligand: probably none
Count central atoms: two
Identify and denote ligand connectivity to central atoms: 1κ$C^1$,2κ$C^7$
Exploit symmetry, count ligands, prefix di already used: two $\longrightarrow$ bis
Combine: bis(2,6-{[2,6-di(1-methylethyl)]diphenyl}phenyl-1κ$C^1$,2κ$C^7$)chromium
Account for the bridging nature of ligands: add μ
Notice metal–metal bonding: add $(Cr$—$Cr)$
Probably (***) count number of bonds between metals: five $\longrightarrow$ $(5\ Cr$—$Cr)$
Combine:

bis(μ-2,6-{[2,6-di(1-methylethyl)]diphenyl}phenyl-1κ$C^1$,2κ$C^7$)chromium$(5\ Cr$—$Cr)$

Change parentheses to allow for two chromium atoms: add []

Final proposed name (***):

bis[(μ-2,6-{[2,6-di(1-methylethyl)]diphenyl}phenyl-1κ$C^1$,2κ$C^7$)chromium]$(5\ Cr$—$Cr)$

Is this correct? Where have I gone wrong?

*** Point 12 conflicts with the Red Book. See IR-10.2.3.5 which says

For the purpose of nomenclature, no distinction is made between different metal–metal
  bond orders.

This is at odds with imposed criterium of preserving information, hence was ignored. ***

Assembling the chemical formula will most likely be covered in a follow-up question.

It was a great effort. However, in the proposed name

bis[(μ-2,6-{[2,6-di(1-methylethyl)]diphenyl}phenyl-1κC1,2κC7)chromium](5 Cr—Cr)

I can see several issues:

In you point 12: Forcing the metal-metal bond multiplicty in the name is interesting, however I wouldn't go against the rules. See the example with the multiple (triple) Cr-Cr bond, in the rules:

IR-10.2.5.1 The eta (η) convention
  (…)
  Example 26.

(μ-2η4-buta-1,3-diene-1,4-diyl-1κ2C1,C4)carbonyl-1κC-bis[(η5-cyclopentadienyl)chromium](Cr—Cr)

not "…(3 Cr—Cr)".
The purpose of the rule seems to be to just specify (only) the connectivity, i.e. the count of such and such metal-metal connections.

IR-9.2.5.6 Trinuclear and larger structures
  (…)
   Example 6.

  octacarbonyl-1κ4C,24κC-bis(triphenylphosphane-3κP)-triangulo-diironplatinum(Fe—Fe)(2 Fe—Pt)

One could speculate that they, to denote the bond order as well, intend to use a multiplied Em dash character variants, which are non-existent presently.
Another motivation might have been that the bond nature / bond-order is often disputed in organometallics.
Your name can denote a different isomer (though maybe impossible), with the two ligands oriented the same way, not say antiparallel, or head-tail, but head-head. I.e. one of the ligand flipped, and connected to chromiums the other way.
From the rules:

IR-10.2.5.1 The eta (η) convention
  (…)
  Example 24.

(μ-1η5-cyclopenta-2,4-diene-1,1-diyl-2κC)(μ-2η5-cyclopenta-2,4-diene-1,1-diyl-1κC)bis[(η5-cyclopentadienyl)hydridotungsten]

not bis(μ-1η5-cyclopenta-2,4-diene-1,1-diyl-2κC)bis[(η5-cyclopentadienyl)hydridotungsten].
I.e. repeating the ligands name, with the 1/2 metal locants swapped. Which, however, is obviously inconvenient, maybe the rules will be improved in the future to somehow support relative locants, or so.
2,6-{[2,6-di(1-methylethyl)]diphenyl}phenyl part (you construction in points 3, 4) is weird.
The standalone ligand residue would be named according to the organic chemistry nomenclature:

2,6-bis[2,6-di(1-methylethyl)phenyl]phenyl, more correctly
2,6-bis[2,6-di(propan-2-yl)phenyl]phenyl

Inorganic nomenclature does not have so strict "preferred name" concept. Table IR-10.1 lists isopropyl as an acceptable alternative name for the propan-2-yl. So I would go with 2,6-diisopropylphenyl instead of 2,6-di(1-methylethyl)phenyl or , to not complicate the name further.
Problematic locant for the kappa notation in the ligand 2,6-{[2,6-di(1-methylethyl)]diphenyl}phenyl-1κC1,2κC7. Maybe a prime symbol / apostrophe could be used like in biphenyls, terphenyls..
I think it should be rewritten to more verbose, but unambiguous 2-(2,6-diisopropylphenyl-1κC1)-6-(2,6-diisopropylphenyl)phenyl-2κC1 (both locant superscripts after κC's are now 1, but its not a problem).

And one note:

Even though the rules state

IR-9.2.4.3 Comparison of the eta and kappa conventions
The eta convention (Section IR-10.2.5.1) is applied in cases where contiguous donor atoms
  within a given ligand are involved in bonding to a central atom. Thus, it is used only when
  there is more than one ligating atom, and the term η1 is not used.

in the literature(web) for the structure with a more complicated ligand

featuring the similar 2,6-di(iPr)Ph···Cr bond, the following shortcut is used

Ar′Cr(μ2:η1,η3-N3(1-Ad))CrAr′

which contains the η1 (and, it interestingly relates to the NNN part of the ligand on the right, not to the phenyl). But I'm not gonna use it in the proposed name.

So my candidate name is
[μ-2-(2,6-diisopropylphenyl-1κC1)-6-(2,6-diisopropylphenyl)phenyl-2κC1][μ-2-(2,6-diisopropylphenyl-2κC1)-6-(2,6-diisopropylphenyl)phenyl-1κC1]dichromium(Cr—Cr)
Finally, lets mention the paper in which this interesting compound was described: Nguyen, T.; Sutton, A. D., Brynda, M.; Fettinger, J. C.; Long, G. J.; Power, P. P. Synthesis of a Stable Compound with Fivefold Bonding Between Two Chromium(I) Centers. Science 2005, 310 (5749), 844–847.

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text:
Why is fluorine a gas, but lithium isn't?
$\ce{F2}$ is a gas, but pure lithium is a solid metal. Why is this? Lithium is much lighter, just 3 protons and 3.9 neutrons (average). Fluorine is much heaver, 9 protons and 10 neutrons.

Nevermind F2 - how about Tungsten Hexafluoride?  It's 13 times denser than air and has a molecular weight of almost 300g/mol.  Being a gas has very little to do with how heavy the atoms or molecules are - it's entirely about how sticky they are.

Or when it comes to naturally appearing elements, just think of radon.

@J... If I'm doing my math right, Tungsten Hexafloride is more dense than aluminum!  Thanks for showing me amusing compounds!

@CortAmmon Definitely not more dense than aluminum.  WF6 is 13g/L (or 13mg/cm3).  Aluminum is 2.7g/cm3 (or 2700mg/cm3) - about 200 times denser than WF6.  Still, it's a pretty heavy gas.  You could fill a fishtank with it and float aluminum foil boats on top.  If you look around youtube you'll find interesting experiments with SF6, which is also a very dense gas.

@J... Ahh thank you.  I thought it was too good to be true (I had copied the wrong density from Wikipedia's page)

@J... It would make a worthy question on this site. For now, I will stick to F2 vs Lithium and worry about more complex chemicals later. If you have an answer, please post it because I still don't understand.

@J... I'm certainly no chemist, but wouldn't it therefore indirectly indeed have to do with mass, since lighter molecules have more velocity for the same temperature, and so tend to be "less sticky"?  And that'd explain the propensity for the lightest atoms/molecules to trend towards gaseous?

@JeopardyTempest No, not really.  It's more about the electron configuration.  Electrostatic forces dominate by far at this level.  You can't think about this like popcorn or billiard balls.

So the reason Sulfur is solid (and not diatomic) and Oxygan is gas diatomic is difference in electron configuration?

@JeopardyTempest Yes, largely.

A lithium atom has one valence electron, easily lost (shared), so it is connected to other atoms by a metallic bond. This is a  bit like the shell game where a pea (electron) is hidden under a walnut shell... the uncertainty of where it is at any instant implies, in a quantum sense, that all the atoms share it, and are held together. This bond is so strong that the next element, beryllium, has a boiling point of $\pu{3243 K}$.
Fluorine, on the other hand, has an almost complete shell and each atom forms a tight covalent bond with just one other fluorine atom. These molecules of $\ce{F2}$ keep their electrons to themselves and do not associate with each other.
The difference in bonding is like the difference between bricks mortared together in pairs, or mortared together in a three-dimensional structure, and is responsible for the difference in mp and bp.

Excellent physical analogy with the bricks & mortar. I will totally be swiping it.

This bond is so strong that... So are you saying, in general, metallic bonds are much stronger than covalent bonds? These molecules of F2 keep their electrons to themselves and do not associate with each other. And are you saying that one F can only bond to one other F, while Li can bond to many other Li (which would make a large heavy crystal)?

Under ordinary circumstances, yes. There are, as usual, some extra-ordinary circumstances that can produce different outcomes.

@DrZ214  Have a read - Why fluorine is a gas. Further reading - Van der Waals Force.

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text:
Why doesn't the enthalpy change of a reaction affect the rate of reaction?
For example, an exothermic reaction will release energy to provide energy to overcome the activation energy. So, why doesn't the enthalpy change affect the rate of reaction?

If the reaction heats up the environment the rate will change.

Your usage of the word "affect" is a misnomer. You can't take a reaction, change its enthalpy and see what would happen to the rate.

@Ivan Neretin I believe they mean that the heat produced by a reaction (the enthalpy change) would affect the surrounding temperature, which would affect the reaction rate.

The definition of the enthalpy change (heat of reaction) is based on the temperature being the same in the initial and final states thermodynamic equilibrium states.

The rate of a chemical reaction is directly dependant to the activation energy of that particular reaction. These two values are inversely proportional. The higher the activation energy, the lower the rate. As it has been mentioned though, the rate can be increased, in such cases, by increasing the temperature. All of the above is described as kinetics, which shows how fast the reaction will reach its end phase.
Enthalpy is linked with thermodynamics, which shows what kind of entities will be present at the end of the reaction. At what degree the reactants will turn to products.
I don't know your chemical background, so I tried to make this as simple as possible.

" the rate can be increased, in such cases, by increasing the temperature. " Not necessarily since the energy of activation is also temperature dependent.

Isn't activation energy the same at all times? For a reaction to happen you need a certain amount of energy to overcome the transition phase. I thought that this amount is fixed for a given reaction and by changing the temperature you change the amount of molecules that have this appropriate energy and thus increasing the rate of it.

The gibbs free energy of activation, the one that counts if we are talking about transition state theory, is temperature dependend. $\Delta G^{\ddagger}= \Delta H^{\ddagger}-T \Delta S^{\ddagger}$. For reactions with negative $\Delta S^{\ddagger}$ the free energy of activation rises with temperature, however as you said also the average energy of the molecule rises. In most cases this leads to faster reactions.

I agree with what you've mentioned  but I think that activation energy is fixed. It's like a energy barrier for the reactants to overcome in order to become products. I've just made a small google search and they've mentioned the same thing.

Ok first we need to be clear if we are talking about Arrhenius (which I don't like since it's an empirical relationship) or transition state theory. However, in both cases the activation energy (Arrhenius) or free energy of activation(transition state theory) is temperature dependend. I've explained the transition state theory case, here's an answer explaining the temperature dependency of the activation energy and why we usually get away with taking it as a constant: https://chemistry.stackexchange.com/a/75680/37313

I see that even at those answers there is a small confusion. I refer to activation energy not the free energy of activation. There is for example cases such as this: https://chemistry.stackexchange.com/questions/29611/activation-energy-and-temperature

Let us continue this discussion in chat.

The enthalpy change of a reaction is just the difference in the bond ( potential energy) stored within the products compared with that within reactants. However, the activation energy is simply the amount of energy you need to break the bonds and start the reaction. To clarify the specific case you suggested: the only reason that an exothermic reaction would proceed faster than a hypothetical endothermic reaction of the same activation energy would be that the exothermic reaction is releasing energy that raises the temperature. However, if you could keep the temperature constant that would not happen and they would proceed at the same rate.

Your characterization of the activation energy is not quite correct. For an S$_N$2 reaction bonds form and break simultaneously, but the transition state is reached before any bond is "broken". Similarly for a Diels-Alder reaction.

an exothermic reaction will release energy to provide energy to overcome the activation energy

That can happen.
The temperature may change and therefore the thermal energy available to molecules to overcome the barrier may change.
In an isothermal reaction any excess thermal energy in the products (energy which results in the products being hotter than the reagents) is eventually transferred to the surroundings (the rest of the universe) as constant temperature is enforced. The thermal energy due to the enthalpy change is therefore lost and unavailable to any remaining reacting molecules.
In an adiabatic reaction (thermally insulated) heat cannot dissipate into the surroundings. If the reaction is exothermic heat becomes available to the reagents and may help them overcome an activation barrier.
Note that enthalpies or activation energies can depend on temperature. However when discussing activation it is typically assumed that these are constants. Also, they are regarded as thermodynamic properties so that they do not depend on how a transition point or products are formed (the details of the path to reach them).
In the following energy diagram, that means that whatever happens between the fixed energy levels on the way there does not affect where the levels are.

Reaction kinetics are usually studied at constant temperature in such a manner that the heat evolved or absorbed doesn't change the temperature. If the temperature changes the rate changes so the reaction enthalpy change makes a difference. The other difference is more subtle: as a reaction progresses the product concentrations increase and the reverse reaction rate increases The difference between the activation energies is the energy change of the reaction. The smaller the overall energy change the faster the reactions can reach equilibrium or in the case of a high energy intermediate a steady state. This can have profound effects on the overall kinetics and is exemplified in the complications found in enzyme kinetics.

Why doesn't the enthalpy change of a reaction affect the rate of reaction?
States of mater can affect the enthalpy change and rate of reaction. Enthalpy change is closely related to thermodynamics and states of matter could have an outside force preparing reaction kinetics changes, witch can some times be observed by measurement or approximation (such as gasses or fuels) the application of the fuel is a Reaction kinetic in the chain if physics and engineering. Physics and Engineering equations can determine variations in the application on fuel witch may be seen as an outlier in the application of only the chemistry equations. enthalpy change will not occur without reaching a threshold or a flash point. other types of changes may occur for various reasons similar to static electricity and or/ plutonic radiation theory.

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text:
Application of the definition of enthalpy in a modified way
My book gave me this equation at constant temperature and constant volume:
$$\ce{2NH3(g) -> N2(g) + 3H2(g); \Delta U = 92.3 kJ}$$
And it says here enthalpy and internal energy are the same as volume is constant. But I have a thought that its pressure is not constant and hence enthalpy change will be different, because enthalpy by definition is
\begin{align}
\mathrm dH &= \mathrm dU + \mathrm d(pV) \\
\mathrm dH &= \mathrm dU + \mathrm d(nRT)
\end{align}
For this equation we will get:
$$\Delta H = 92.3\ \mathrm{kJ} + 2RT$$
Am I correct with my thinking? If not, why?

You're correct, the book is wrong. $\Delta(pV) = p_2V_2 - p_1V_1$ and if $V_2 = V_1$, the $\Delta(pV)$ term reduces to $V(p_2 - p_1)$, which is not equal to zero unless $p_2 = p_1$ also.


text:
How the Nernst equation is applied for electrodeposition?
For using the Nernst equation
$$E = E_0 - \frac{RT}{nF}\ln\frac{[\ce{Red}]}{[\ce{Ox}]}$$
the concentrations of $\ce{Red}$ and $\ce{Ox}$ species are required. For the reaction
$$\ce{Cu+ + e- <=> Cu}$$
in which the $\ce{Red}$ part leaves the solution and has no concentration in the electrolyte. How is the Nernst equation used to calculate the potential at different stages when the reaction proceeds?

For a metal electrode immersed in a solution of its own salt, the concentration of the reduced form as a solid phase remains constant ($[\ce{M^0}] = \mathrm{const}$), therefore:
$$E = E_0 + \frac{RT}{nF}\ln[\ce{Ox}]$$


text:
Cauchy-Schwarz inequality for integral derivatives
The Cauchy-Schwarz inequality for regular two electron integrals is given as:
$$\left(\left.ab\right|cd\right)\leq\sqrt{\left(\left.ab\right|ab\right)}\sqrt{\left(\left.cd\right|cd\right)}$$
Now this can be differentiated with respect to $x$:
$$\frac{\partial}{\partial x}\left(\left.ab\right|cd\right)\leq\frac{\partial}{\partial x}\sqrt{\left(\left.ab\right|ab\right)}\sqrt{\left(\left.cd\right|cd\right)}$$
By using the chainrule for $\frac{d\sqrt{f(x)}}{dx}=\frac{df(x)}{dx}\frac{1}{2\sqrt{f(x)}}$, now giving:
$$\left(\left.\frac{\partial a}{\partial x}b\right|cd\right)+\left(\left.a\frac{\partial b}{\partial x}\right|cd\right)+\left(ab\left|\frac{\partial c}{\partial x}d\right.\right)+\left(ab\left|c\frac{\partial d}{\partial x}\right.\right)\leq\frac{\sqrt{\left(\left.cd\right|cd\right)}}{2\sqrt{\left(\left.ab\right|ab\right)}}\left[\left(\left.\frac{\partial a}{\partial x}b\right|ab\right)+\left(\left.a\frac{\partial b}{\partial x}\right|ab\right)+\left(ab\left|\frac{\partial a}{\partial x}b\right.\right)+\left(ab\left|a\frac{\partial b}{\partial x}\right.\right)\right]+\frac{\sqrt{\left(\left.ab\right|ab\right)}}{2\sqrt{\left(\left.cd\right|cd\right)}}\left[\left(\left.\frac{\partial c}{\partial x}d\right|cd\right)+\left(\left.c\frac{\partial d}{\partial x}\right|cd\right)+\left(cd\left|\frac{\partial c}{\partial x}d\right.\right)+\left(cd\left|c\frac{\partial d}{\partial x}\right.\right)\right]$$
By using the symmetry of the two electron integrals $(ab|cd)=(cd|ab)$, it can simplified to:
$$\left(\left.\frac{\partial a}{\partial x}b\right|cd\right)+\left(\left.a\frac{\partial b}{\partial x}\right|cd\right)+\left(ab\left|\frac{\partial c}{\partial x}d\right.\right)+\left(ab\left|c\frac{\partial d}{\partial x}\right.\right)\leq\frac{\sqrt{\left(\left.cd\right|cd\right)}}{\sqrt{\left(\left.ab\right|ab\right)}}\left[\left(\left.\frac{\partial a}{\partial x}b\right|ab\right)+\left(\left.a\frac{\partial b}{\partial x}\right|ab\right)\right]+\frac{\sqrt{\left(\left.ab\right|ab\right)}}{\sqrt{\left(\left.cd\right|cd\right)}}\left[\left(\left.\frac{\partial c}{\partial x}d\right|cd\right)+\left(\left.c\frac{\partial d}{\partial x}\right|cd\right)\right] \tag{1}$$
Now by comparing terms it can be seen that:
$$\left(\left.\frac{\partial a}{\partial x}b\right|cd\right)\leq\frac{\sqrt{\left(\left.cd\right|cd\right)}}{\sqrt{\left(\left.ab\right|ab\right)}}\left(\left.\frac{\partial a}{\partial x}b\right|ab\right)$$
$$\left(\left.a\frac{\partial b}{\partial x}\right|cd\right)\leq\frac{\sqrt{\left(\left.cd\right|cd\right)}}{\sqrt{\left(\left.ab\right|ab\right)}}\left(\left.a\frac{\partial b}{\partial x}\right|ab\right)$$
$$\left(ab\left|\frac{\partial c}{\partial x}d\right.\right)\leq\frac{\sqrt{\left(\left.ab\right|ab\right)}}{\sqrt{\left(\left.cd\right|cd\right)}}\left(\left.\frac{\partial c}{\partial x}d\right|cd\right)$$
$$\left(ab\left|c\frac{\partial d}{\partial x}\right.\right)\leq\frac{\sqrt{\left(\left.ab\right|ab\right)}}{\sqrt{\left(\left.cd\right|cd\right)}}\left(cd\left|c\frac{\partial d}{\partial x}\right.\right)$$
For geometrical detrivatices $x$ would be a cartesian direction of a nuclie. Now my question is, for geometrical derivatives when would be the right point to use the Cauchy-Schwarz inequlity for screening the integrals?
I think I could either $(1)$ and screen the total derivative $\frac{\partial}{\partial x}(ab|cd)$, or use the last four equations and screen the four different (up to three for geometrical derivatives) contributions individually.
Also what would be a reasonable threshold for derivatives?
For regular integrals $10^{-10}$ or lower is often used.

Unfortunately, I think the formula derivation is faulty. As far as I can tell, while you can apply a derivative to both sides of an equality and it will still hold, but an inequality can change when a derivative is applied. A small example would be a horizontal line and a vertically raised parabola; the parabola is always larger than the horizontal line, but it's derivative goes from being smaller, to equal, to large. Related:https://math.stackexchange.com/questions/110044/what-are-the-regularity-conditions-for-differentiating-an-inequality

This is indeed a way that pre-screening of two-electron integrals is done, and if you thought of this yourself, then I'm very impressed! Though I doubt that is too much of a compliment.
Unfortunately, Tyberius is correct that you cannot be sure this inequality will hold after taking the derivative. Fortunately, the Cauchy-Schwartz inequality holds for any arbitrary vectors which have an inner-product. This means that you can simply switch the order you are doing things and it will work.
For what comes below, I will basically replicate some work from a paper by Reinhart Alrichs et al., and will use their notation as well [1].
In what follows, a superscipt $i$ represents a derivative in any of the cartesian $x$, $y$, or $z$ directions.
The Cauchy-Schwarz inequality states that,
$$\tag{1}\label{eq:1}
|(ab|cd)|\le Q_{ab}Q_{cd}
$$
where,
$$\tag{2}\label{eq:2}
Q_{ab}=(ab|ab)^{\frac{1}{2}}
$$
and the same for $Q_{cd}$.
If we take the derivative in the $i$ direction of $(\ref{eq:1})$, we have,
$$\tag{3}\label{eq:3}
(ab|cd)^i=(a^ib|cd)+(ab^i|cd)+(ab|c^id)+(ab|cd^i)
$$
which is contained in the work you have done, but with some extra stuff from the original inequality.
Each of these terms, however, are still a part of the original space and hence still have a well-defined inner-product. This means we can apply the Cauchy-Schwarz inequality directly to these derivatives as a means of screening the integrals.
That is,
$$\tag{4}\label{eq:4}
|(a^ib|cd)|\le Q_{a^ib}Q_{cd}
$$
and so one for the other derivatives.
The advantage of this is that one can give an upper bound for the size of these integrals based on calculations which have to be done for the other integrals anyways, and some simple to compute auxiliary quantities. See [1] for details on these auxiliary quantities (eqs. 10-19). I do not fully understand this part of the procedure, so I'm not restating it here.
Next, you asked where one should do the screening. The authors of [1] advocate what they call a "global screening" methodology. This means that all of the quantities required for calculating the upper-bound are pre-computed (before any integral loops but after any checks for symmetry), and compared to some pre-specified threshold, say $10^{-10}$. Although we do have to compute some new things, these are much cheaper than the integrals themselves and thus save quite a bit of time.
Ref. [1] also goes into detail about some test cases as well as doing the same screening for second-derivatives and hence the application of this pre-screening to Hessians.

A quite different approach which is very interesting to read is given in ref. [2], which came after and improves on [1]. This method, quite naturally I think, takes advantage of a generalization of Cauchy-Schwarz known as Holder's Inequality. For our purposes, the relevant form of Holder's inequality is (see the section titled Lebesgue Measure or ref. [2]),
$$\tag{5}
\int|f(x)g(x)|dx\le\left(\int|f(x)|^mdx\right)^{\frac{1}{m}} \left(\int|g(x)|^n dx\right)^{\frac{1}{n}}
$$
for $\frac{1}{m}+\frac{1}{n}=1$.
Notice, however, that while this strengthens the inequality by giving a class of possibilities for different $m$ and $n$, it weakens the inequality in that we may only use one variable $x$. Thus, to apply this to 2-electron integrals, the authors of ref. [2] first integrate out one of the variables, which gives us the same inequality for our physical situation as,
$$\tag{6}
\int|(ab|cd)|\le\left(\int|V_{ab}|^md\textbf{r}\right)^{\frac{1}{m}} \left(\int|(cd|cd)|^n d\textbf{r}\right)^{\frac{1}{n}}
$$
where $V_{ab}$ is the potential due to electron 1. Thus, in this case the screening must always take place after the first coordinate is integrated-out. This method, however, is very flexible in that it apparently gives tighter bounds (a smaller maximum), and also has many possible solutions. Ref. [2] discusses four of these possible solutions, and compares to the Schwarz pre-screening method, which is a special case of what they derive. Again, some special quantities must be pre-computed, but this is not such a big deal for the time it saves.
Ref. [2] does not discuss derivatives, though I assume the same method ought to possible as in the first part, only things might be complicated by the fact we need we need to integrate out a coordinate. Maybe not in the first derivative case, but in the second derivative case, this would present a problem as far as I can see.

[1] Horn, H., Weiß, H., Haser, M., Ehrig, M., & Ahlrichs, R. (1991). Prescreening of two‐electron integral derivatives in SCF gradient and Hessian calculations. Journal of computational chemistry, 12(9), 1058-1064.
[2] Gill, P. M., Johnson, B. G., & Pople, J. A. (1994). A simple yet powerful upper bound for Coulomb integrals. Chemical physics letters, 217(1-2), 65-68.

Very nice answer! Glad to see this can actually work.

Do you by any chance know what is the name of the paper of reference 14 in "Gill, P. M., Johnson, B. G., & Pople, J. A. (1994). A simple yet powerful upper bound for Coulomb integrals. Chemical physics letters, 217(1-2), 65-68." ?

@ErikKjellgren I just looked and I see it's a reference to a paper to be submitted. I'm not sure what paper this would be. You can look at his google scholar page from around 1994, but nothing seemed to fit what they are talking about in the paper.


text:
How do water based solutions (cream, gels) etc permeable through hydrophobic skin?
Most skin care products or the sort are oily, lipid based.
Our skin is hydrophobic. I reckon the reason is to be easily permeable through the skin.
So how do these skin care or any products which are water based said to have higher absorption and provide high nutrient uptake, replenish skin?
How I see it, that it's like a slight burst in the cell since a hypertonic solution seep into the hypotonic cell(!?). And still confused how such product could replenish skin.
I am keen on the chemistry behind this concept of how a water based skin care could actually replenish the skin. In the discussion, it is better not to include, point at any specific products you loathe or adore them.

Thanks @Gordon.  I got rid of your comments because we try to avoid medical advice on the site.

Don't worry. I certainly understand. The only reason I even commented was to let your members know that deeper penetration by topical cosmetic ingredients is not necessarily a good thing.  Consumer manufactures try to avoid it, frankly.


text:
Which bond is stronger: N=O or C=N?
Which bond is stronger?  N=O or C=N?
I tried comparing it using electronegativity difference. But, I don't know if that is the correct method. Can anyone explain it?

Electronegativity difference has hardly anything to do with bond strength.

Electronegativity differences does affect the energy of the valence orbitals involved in making the bonds (i.e. the higher the electronegativity of the element, the lower the energy of the valence orbitals). Thus, it is indeed an important consideration. However, I think the effect is only minor. More important considerations would be bond order and also type of hybridisation of the orbitals used.

The $\ce{C=N}$ bond has a dissociation energy of $\pu{615kJ/mol}$, while $\ce{N=O}$ has an energy of $\pu{607kJ/mol}$. The difference isn't much, so its difficult trying to comparing the energies without being given the exact values.
References:
Yoder, Claude. “Wired Chemist.” Common Bond Energies (D, www.wiredchemist.com/chemistry/data/bond_energies_lengths.html.

Is there any theoretical way to guess bond energies?

I would think the $\alpha$-effect would make the $\ce{N=O}$ bond weaker. Not sure if this is the primary factor though.


text:
Iron-platinum refining
If I have an iron-platinum alloy, can I put this into a solution of $\ce{HCl}$ to dissolve the iron, leaving a $\ce{Pt}$ precipitate?
I know that Fe dissolves in $\ce{HCl}$ to form $\ce{FeCl2}$ and $\ce{H2}$, and I know that $\ce{Pt}$ does not dissolve in $\ce{HCl}$.
But will the $\ce{Fe-Pt}$ bonding affect the Fe dissolution?

Related: https://chemistry.stackexchange.com/questions/62787/how-to-separate-alloys/62812#62812

Given the broad solubility ranges and ordered low temperature phases, the general answer is no - those are alloys, not phase separated.

It depends on the concentration of Pt in your alloy.
If the concentration of Pt is to high the HCl will not or not sufficently attack the alloy.
In this case you could try to melt your alloy with some additional Fe, to lower the Pt concentration and then hit it with HCl again.
You could also try to boil it in HCl to increase the reactivity of the acid (do it in a fumehood or outdoors).


text:
Why do higher-mass isotopes have higher melting and boiling points than lower-mass isotopes?
Different isotopes of the same element have the same number of electrons, so the nature of bonding between different isotopes of the same element should be the same, yet their melting points and boiling points vary. This would suggest that the interparticle forces and distances are different for different isotopes of the same element. Why is this exactly?
My best guess is that is has something to do with the kinetic energy (KE) formula
$$E_k = mv^2/2$$
If two samples of different isotopes of the same element were kept under the same pressure and temperature (so the KE of the particles in the different samples would the same), the particles of the heavier isotope would have a lower velocity than the particles of the lighter isotope, as a greater mass means a lower velocity (where KE is equal). Does it have something to do with the fact particles of a heavier isotope move more slowly when KE is equal?

Sounds about right, though your reasoning would look better with some quantum hand-waving around it.

Related: https://chemistry.stackexchange.com/questions/81235/freezing-point-of-oxygen-18-water/81254#81254

There's probably a primary kinetic isotope effect as well.

As you already mention in your question, melting and boiling of a substance is related to the inter-particle forces. You also correctly assumed that the same number of electrons results in identical (electronic) bonding properties, which basically is a different formulation of the Born-Oppenheimer Approximation (as in the BOA the nuclear mass is assumed to be infinite, different isotopes should have the same electronic structure).
So what is the difference then?
Well, it's in the vibration between the different particles. Consider a number of (neutral) particles in a certain phase (it can be solid, liquid or gas) at a certain distance with respect to each other. Because of dispersion forces between the particles, they interact and to a first approximation, every particle experiences a harmonic force $F(x)=-k(x-x_e)$. The magnitude of this force depends on the distance between the particles and on the electronic structure of the particles and in principle not on the mass. However, as you might know, a harmonic force in quantum mechanics results in quantized energy levels given by
$$
E_v=h\nu_e(v+\frac{1}{2}),
$$
where $E_v$ is the vibrational energy of the $v$th level and $h$ is Planck's constant. What is important to realize is that the energy of the lowest state is not equal to zero but has a value of $h\nu_e/2$, which is referred to as the zero point energy. Of course the potential is not really harmonic but rather anharmonic so that at a certain distance between the particles the force becomes negligible and the particles dissociate (or melt or evaporate). The lower the zero point energy, the larger the binding energy. The constant $\nu_e$ is given by $\nu_e=\frac{1}{2\pi}\sqrt\frac{k}{\mu}$, where $k$ is the harmonic force constant and $\mu$ is the reduced mass. The constant $k$ is the same for the different isotopes as it depends on the electronic potential between the particles but the reduced mass is of course different for the different isotopes. As you can see from the definition of $\nu_e$, the heavier species have a smaller zero point energy and thus a larger binding energy and will require higher temperatures to melt and boil.

This isn't always true, at least not for molecules (maybe it is for monoatomic species?)  13C-methanol has a higher vapor pressure than 12C-methanol, for example.

My answer assumes that isotopic substitution does not change the vibrational frequencies in the molecules too much. Methanol is an interesting molecule due to its internal rotation which is very sensitive to a mass substitution. I found this paper which explains the strange effect for the 13C methanol (the same happens when substituting the H from the hydroxyl group).

Thanks for that link, it is very interesting and relevant to a somewhat related question over here

@Paul Melting and boiling is not related to chemical bonds. Lower zero point energy means more energy is needed to break a chemical bond.

@ado_sar, please read my answer carefully: I'm not referring to covalent bonds in my answer. Most intermolecular potentials also support vibrational states. Because of the higher masses of the molecules the vibrational frequencies are lower.


text:
Nodes of orientation of electron orbitals

As you can see the orientation of 2 p orbital there is a nodal plane in between the electron density pictures.... So doesn't the electron move from left to right side in2 p X orbital.
I think that electrons moved around the nucleus in circular paths

Electrons don't move in circular paths. That's just an oversimplified model used to explain emission spectra.

You are trying to connect the particle and wave picture into one thing. Don't  do that! Those are two different perspectives on the same thing. The electron does not "move around in its orbital". The orbital is the electron (in its wave function description).


text:
Physical gelation result in a single massive polymer chain?
Physically crosslinking PVA polymers using the freeze-thaw method (Peppas 1975) results in regions of interconnected PVA polymers through crystalline regions formed when frozen and which persist after thawing. These physical crosslinks are sufficient to produce a homogenous hydrogel of predictable bulk characteristics. The formation of Jello gelatin uses a similar phenomenon.
My question is:
Are these crystalline regions only available to form at the free ends of the polymer chains, or can these lamellae form along any portion of the polymer chain?
If crystallization were only able to occur at free ends of the chain, this would result in a massively long polymer chain (assuming tip to tail occurred for many polymer chains in the solution) and as such the bulk polymer might be defined as a tangled network.

The physical alignments can occur anywhere. Keep in mind that these regions are both inter- and intra- molecular. (CH4 can't align with itself, but CH3(CH2)1000CH3 can and does.)


text:
discover of the neutron, what happen to 2+ charge
James Chadwik found the neutron by shoot α rays to a Beralium foil. α rays are equal to the He²⁺ and he saw that particles without a charge is out from that. He understood the neutrons are those particles without the charge.
But my question is this.
This is the equation that the chadwik found.
Be + He²⁺= C +neutron
In this equation the left side has a 2+charge, but in right there is no any charge. So, What happen to the 2+ charge when this reaction occurs?

For nuclear reactions, usually occurring at very high energies (compared with typical electron binding energies) one just ignores the electrons - they are going to do what they will do, and have zero impact on what the nucleons are doing. So, that would usually be written $^{9}Be (\alpha, n) ^{12}C$. The alpha was just a way to get an energetic-enough alpha to enter the nucleus and react.

Electrons have no importance whatsoever in the reaction.

When $\ce{He^{2+}}$ ions hit a beryllium target, then the target will become charged. However beryllium is an excellent conductor. So if you electrically ground the target then the charge will disparate. // You have to realize that energetic $\ce{He^{2+}}$ ions will leave a trail of ions as the $\ce{He^{2+}}$ travels through the beryllium. This isn't like a chemical reaction. Also remember that you got the $\ce{He^{2+}}$ ion by stripping off two electrons. Those electrons are floating around somewhere in the universe...

Corrected: "James Chadwik identified the neutron by exposing a Beryllium foil to  α-rays (He+2) and showing that the previously discovered emitted collision product were massive neutrally charged particles." The experimental set-up he used (see:http://rspa.royalsocietypublishing.org/content/royprsa/136/830/692.full.pdf) did not determine where any electrons came from, so while electrical charge can be assumed to balance, he couldn't claim any direct knowledge of the source or fate of any electrons in his experiment. For us, today, balancing a nuclear reaction's charge distribution on both sides of the arrow is simply "bookkeeping", it implies facts that were not as well supported in 1932 as they are today. He avoided the temptation to over-interpret his experimental data.


text:
Can lead-acid batteries release CO? Or can a CO sensor detect gasses other than CO?
I recently found myself troubleshooting a CO alarm on a houseboat. Inside the cabin, a Fluke CO-220 was reading 40 ppm. In the generator engine compartment, the CO level was zero (while the generator was running). We eventually found the source of the CO to be a large lead-acid "house" battery which was warm and leaking:

The OL on the display of the CO-220 indicates a value of >1000 ppm CO. I watched it count up through 600, 800, 900 before indicating OL.
Mechanics replaced the batteries, and ensured the battery box was properly sealed and vented. After this, the CO levels were at zero.

Can lead-acid batteries leak carbon monoxide?
This article discusses the chemical reactions of a lead-acid battery, and as far as I can tell there is no carbon involved whatsoever. So I believe the answer here to be "no".

Can a Fluke CO-220 detect gasses other than carbon monoxide?
The CO-220 manual says nothing about any gasses other than CO. In the specifications:

Sensor type | Stabilized electrochemical Gas-specific (CO)

I contacted Fluke and they responded with:

I do not know what gas batteries would put off and the CO-220 has a senor [sic] for CO only. Hope this helps.

I assume the answer here is also "no".

So then, what was going on?

You're probably picking up hydrogen gas, which is produced when lead-acid batteries are overcharged at high charging voltages (a danger in its own right). This article details a situation similar to yours: charging a lead acid battery in a golf cart (in a confined space) sets off a $\ce{CO}$ alarm, and typical sensors are activated by $\ce{CO}$ at levels of 150 ppm for 30 minutes and/or 300 ppm $\ce{H2}$ for 30 minutes. Because $\ce{CO}$ is not produced by battery charging, the offending gas is thus identified as hydrogen. After venting, the levels recede.
Also see this post from Hawk Environmental, which details how hydrogen gas can create false positives for carbon monoxide. This post goes into a little more detail about the hydrogen/carbon monoxide sensor issue, and the author states that:

Lead-acid batteries produce Hydrogen when charging.
Carbon Monoxide detectors use something called a "Metal Oxide Semiconductor (MOS)" sensor, which detects a variety of gases including Hydrogen.
A MOS sensor calibrated for CO will give a false positive in the presence of Hydrogen gas at ~10% of the actual value. So, that ~85ppm reported of CO was in reality ~850ppm of Hydrogen gas.

Finally, in an article in the September 1996 issue of Fire Engineering, entitled "Other gases may set off $\ce{CO}$ detectors" (here), it is stated that

CO detectors are electrochemical sensors, which detect a variety of specific gases. In most cases, the gas of interest reacts with the chemical solution in the sensing electrode to produce ions (charged particles) that move through the solution to an electron-accepting electrode, completing an electric circuit. The number of ions produced and the magnitude of the current developed are proportional to the concentration of the gas being measured.

and that

Interfering gases of similar molecular size and chemical reactivity may produce false positive response. Major interferents that can affect $\ce{CO}$ detectors are acetylene, dimethyl sulfide, ethyl alcohol, ethylene, hydrogen cyanide, hydrogen sulfide, isopropyl alcohol, mercaptan, methyl alcohol, propane, nitrogen dioxide, and sulfur dioxide.

I suggest you confirm/deny the above by contacting Fluke and asking them specifically if the sensor does not detect hydrogen gas at all and, if so, at what level. I strongly suspect that the well-documented behavior I've found online and summarized here is the cause of what you have observed in this case.

Thank you for this excellent answer! Given the fact that the battery was quite warm, and the visible yellowish, cauliflower-looking growth on the negative terminal, I'm also wondering if it could have been hydrogen sulfide being detected; it is one of the interferents listed. The article I linked to says "hydrogen sulfide (H2S) is a possible by-product of over-charging and battery decomposition." I did smell a somewhat "rotten" smell in the vicinity, but coincidentally, trash bags were stored in an adjacent hatch of the boat, so I didn't suspect sulfur.

@JonathonReinhart - Absolutely, hydrogen sulfide is a likely culprit as well. Feel free to "accept" my answer if it has satisfied your query.

@JonathonReinhart  might be interesting to do electrolysis on a bit of water.  Collect a wee bit of H2 and release it near your Fluke.  Or if you have access to gas cylinders and can spare a bit, could test it that way.

Hemoglobin, the active compound in red blood cells That carry oxygen to the cells through the circulatory system and remove CO2 from them has a higher affinity for carbon monoxide than for oxygen, or it binds with the CO so tightly that the hemoglobin can no longer absorb or release O2 or CO2.  This is why CO is poisonous.  I was wondering if this aspect of hemoglobin can be used for a more effective CO sensor?


text:
Black Sludge Appearing When Making Iron Oxide Through Electrolysis
I am trying to make iron-oxide through electroylsis to make thermite. I'm using nails for the iron on the positive electrode in salt water. While there definitely is a good amount of rust collecting, especially at the surface, after dissolve multiple nails, there are also these black sludge-like solids forming:

I can pick them out of the solution if i'm careful, but they are very brittle
and break up and dissolve almost immediately if I  use too much force. Additionally, if I stir the up or shake the container, more of this stuff comes up from the bottom, most of it dissolved in a black-green watery, solution and you can't even see the rust anymore. If you give it time to settle, though you can see the rust again.
My question is what is this stuff, and will it make the rust impure? Do I need to get rid of it to make the thermite? Will boiling the solution to get rid of the water like I had planned be enough?
Thanks in advance.

If you don't know that rust isn't of much use in thermite, you shouldn't even plan such thing. Real thermites are very dangerous in use.

@Mithoron what do you mean? Thermite is a mixture of fine iron-oxide and aluminum particles. Or am I wrong?

@Mithoron the example picture on wikipedia uses iron oxide: https://en.wikipedia.org/wiki/Thermite#/media/File:Thermite_mix.jpg

sigh Oxides in rust are hydrated, and water can make the mixture useless or to kill the user.

@Mithoron but heating them removes the water no? Like any other hydrate? I was planning on baking them in the oven.

Yes, it does. I still think that no private person should even try to make it in the first place, though.

Since there are safety issues here, please consult with an expert (live, not via the internet) if you have even the slightest doubt about the proper methodology.  As @Mithoron mentioned, this can be extremely dangerous.

The black stuff is your iron oxides. FeO and Fe3O4 are both black, Fe2O3 is red. I would guess you are making some complex iron oxide/ hydroxide/ hydrate. I have seen XRD analyses of many aqueous iron corrosion products , they are typically a complex mix.

Cool, thanks! So you don't think that the chlorine is messing anything up? These should be fine to use for thermite as long as they are just metal oxides, right?


text:
What would be the unit for the calculation of the rate of reaction?
In a reaction between magnesium and hydrochloric acid, I measured the change in mass of the magnesium strip (Initial weight: 0.03g). I calculated the the time for the reaction to reach completion. A sample set of data we collected for during our investigation was:
With 8 molar of HCl the reaction took an average of 15.49 seconds to completely evolve into hydrogen gas.
To calculate the rate of reaction, I used the formula:

I suppose, change in mass would be 0.03 g as we went to the completion of the reaction, and change in time would be noted as the average time 15.49. My question is, what is the unit supposed to be in this formula?

I think you are making this too complicated. According to the units you've given and the formula you've used, your units would just be mass/time or g/s. Or, the official SI unit for mass is actually kg, so if you want that you would convert to kg/s.

The formula has been stated and you simply need to look out for the units of the quantities
 from the formula. Mass(g)/Time(s). Which is g/s.


text:
What are some chemical engineering bottlenecks for a perfectly synthetic meat?
Recently, the Impossible Burger has become more and more popular. It seems to be a very big effort and even making it's way to the top restaurants. Of course, this has been something that scientists have been working on for decades and is a mix of sciences such as gastronomy, chemistry and others.
In such experiments as the Impossible Burger, I keep hearing neuroscientists and flavour experts trying to replicate the smells, the texture and other characteristics by what I understand it to be brute testing (although I am sure they do some biological and chemical modelling to supplement the engineering process). What it got me thinking about was what are some bottlenecks to making a PERFECT replica of meat. As a mechanical engineer graduate myself, my knowledge is limited on the issue. What holds back chemical engineers and scientist to identify the contents of meat and exactly produce a synthetic copy? We have techniques like IR spectroscopy and many other advanced methodologies for identifying chemicals. Is it a manufacturing complexity or is it just another case of expensive computational costs and insufficient observational techniques to fully figure out the structure of meat?


text:
Organic nomenclature of substituted diamines
Nomenclature of substituted amines is done with the N-locants, with the N being used to indicate that a group is connected to the methyl group. I was wondering, therefore, on how to properly name a substituted diamine.
I found this source from the University of Calgary, here, which recommended to use N, N', N'', etc locants for compounds with multiple substituted amines. However, their examples are all with symmetrical diamines, and it is apparent that using the N, N' locants only works because of the molecule's symmetry. Thus, my question is: how can we name substituted diamines properly, such as the one below?

Would it be N,N,N'-trimethyl,3,5-diaminopent-1-ene or N,N',N'-trimethyl,3,5-diaminopent-1-ene, or something else?

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), superscript arabic numbers are used to differentiate the nitrogen atoms.

P-16.9.2 Superscript arabic numbers, which are the locants of the parent structure, are used to differentiate the nitrogen atoms of di- and polyamines, di- and polyimines, di- and polyamides, except for geminal amines, imines, and amides.

In particular, note that the superscript arabic numbers correspond to the locants of the parent structure. Since the parent structure of the compound that is given in the question is pent-4-ene-1,3-diamine, the locants are 1 and 3. Therefore, the complete name for the substituted compound is N1,N3,N3-trimethylpent-4-ene-1,3-diamine.

If a molecule WAS geminal, how would it be named and/or numbered?


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Accuracy of quantum chemistry calculations on iodine neglecting relativistic effects
Triiodothyronine is a molecule with three iodine atoms. If we attempted to calculate its properties with ab initio methods like Hartree-Fock or more complicated ab initio methods which included electron correlations, would neglect of relativistic effects lead to significant error in estimates of its chemical properties?

If you really want to stick with HF, then the error due to missing electron correlation will be larger then the relativistic effects.
For iodine you may want to use an Effective Core Potential (ECP) to reduce computational cost. One nice thing about these ECPs is, that they may already include some relativistic effects.

Thanks for that point. I'm interested in whether relativistic effects would still be important if I took into account electron correlations.

It's easy to check - as a first approximation you can use the Stuttgart/Cologne ECP's that come with and without relativistic corrections with the standard non relativistic Hamiltonian:
www.tc.uni-koeln.de/PP/index.en.html


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Are halogens a further classification of non-metals, or are they another group themselves?
I've searched many pages in the internet, and many just classify halogens
(Grp 7A) as non-metals. However, my teacher classified them as another group in the periodic table, and when asked which groups are classified under non-metals, she written the correct answer to be group 3A-6A. I'm very confused as I thought non-metals include halogens?
Thanks in advance!!:)

You thought right.

@IvanNeretin So halogens are non-metals and are not another group?

That's right. There is no another group. Non-metals include every element which is not a metal.

Halogens are merely just a subset of non-metals. Consider this periodic table:

Clearly halogens do come under the fluoroscent colored elements, which signifies non-metals. There are several other groups, which have their own unique name. For instance, the oxygen group members are called chalcogens, and the nitrogen group members are called pnictogens.
However, there is something special about halogens though. They are infact the most non-metallic elements known, considering their great tendency to gain electrons. This also makes the halogen group the only group other than inert gases to have only non metals in them (ignoring Ununseptium of course).
In short, halogens are a subset of non-metals and not an entire class by themselves. But they are, the most important subset of non metals.


text:
Meisenheimer complex in Preparation of 2-Naphthol from 2-Naphthalenesulfonic acid
I'm trying to figure out the addition-elimination mechanism for the preparation of 2-naphthol from 2-naphthalenesulfonic acid. I know that the $\ce{SO3^2-}$ group is functioning as the leaving group, but I don't know how to draw the Meisenheimer complex.
I also can't find anything on the internet about the Meisenheimer complex of naphthalene, I tried to draw it, but this became too hard for me as there were too many resonance structures. Can someone please help?


text:
Can 1,2-dioxetanedione (active ingredient in glow sticks) undergo stimulated emission (and thus form a laser)?
In common glowsticks, diphenyl oxalate reacts with hydrogen peroxide, forming 1,2-dioxetanedione. The dioxetanedione is unstable, but direct decomposition to carbon dioxide is a forbidden transition, so this can only happen in a reaction with another molecule. If the other molecule is a suitable fluophore dye, the dye molecule will become exited and then emit a photon of light. (The color depending on the fluophore used.)
Some types of lasers make use of forbidden transitions: atoms or molecules are brought into an excited state such that the transition to the ground state is forbidden and thus slowed down. This causes the exited state molecules to collect, forming a population inversion. The excited molecules can then undergo stimulated emission in a resonance cavity, producing the laser effect.
Now my question is if 1,2-dioxetanedione can also be made to undergo stimulated emission if the glow stick chemical reaction takes place in a suitable optical resonance cavity. (It looks to me like it should, but my chemistry knowledge is not enough for this.)
Alternatively, if the 1,2-dioxetanedione cannot undergo stimulated emission, could a suitably chosen dye molecule?

I meant that your peroxide isn't fluorophore so isn't fit for dye lasers.

https://chemistry.stackexchange.com/questions/45177/what-chemicals-are-mixed-together-to-cause-chemoluminescence/45179#45179

The activated complex 1,2-dioxetanedione can't undergo simulated emissions as it isn't the emitting molecule even in glow-sticks: it is the source of the activation energy for other dyes.
But most emitting dyes can be used in suitably configured lasers by providing energy as photons (=light) of a higher energy and a suitable wavelength in a suitable device. Whether they would make good lasers depends on a whole bunch of other factors (like how easy it is for them to absorb light, whether the subsequent transitions are forbidden or not (can't easily get population inversion if they are not) and various quantum efficiencies of the relevant conversions in the dye molecule.)
Dye lasers have been known since 1966. They have some advantages over more conventional lasers as they are often wavelength tunable (as dyes tend to have broad emission lines) and can often achieve very short pulses. Rhodamine 6G is a common laser dye but the dyes in a glow-stick may not be as good in lasers, though, in principle, there is no reason why not.

A nice idea but it is very unlikely indeed that you will be able to make a dye lase either as a amplifier or in a cavity. The energy is transferred from the chemical reaction to the dye very slowly  compared to how quickly a typical visible dye molecule fluoresces; usually the fluorescence lifetime is a few nanoseconds. Thus a population inversion will not be built up.
To get dye lasers to work with flash lamps is very hard, more often a Q switched doubled YAG laser is used or for continuous lasing an argon ion laser or cw YAG, in either case a few watts of light power (not electrical power from the wall plug which is greater) is needed. You just could not get this from the chemical reaction in the light stick even if put into a cavity.

Would a different choice of "dye" with a longer fluorescence lifetime be possible? And the chemical reaction could be made faster by increasing chemical concentrations and increasing pH, which would also increase the number of dye molecules in an excited state. Also I'm not asking for a very high powered laser.

In principle yes, but in practice no; its virtually impossible by orders of magnitude to get a population inversion chemically in a dye: Its hard even with high powered flash lamps or other lasers where the energy is put directly into the dye. (In a few special situations chemical lasers do exist and work in the IR).

Does that also take 4-level lasers into account? From the wikipedia description I got the idea that 4 level lasers made it very easy to achieve a population inversion because the lower band was guaranteed to always be (nearly) empty.

yes dyes form 4 level lasers; its all relative and 'easy' usually means relative to a 3 level laser. In  a dye the lifetime is nanoseconds so that you have to get the energy into the excited state in a comparable time. I don't recall that a dye laser has been made using a chemical reaction only to produce excited states.


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Inductive effect on stability of free radical
How does inductive effect affect the stability of free radical?
I can't seem to understand whether the carbon with the odd electron is deficient or not.
That is, whether a +I group, like alkyl, will stabilize it or not? I know that an adjacent alkyl group like $\ce{-CH3}$ will anyway stabilize it by hyperconjugation.

Generally, we consider radicals to be electron deficient species.

+I groups stabilize free radical and -I groups destabilize free radical.

Related: https://chemistry.stackexchange.com/q/164828

TL;DR - Free radicals, which are carbon intermediates with an extra electron, are stabilized by electron-donating groups and destabilized by electron-withdrawing groups.

Explanation:

Stability Factors: Free radicals, being electron-deficient, benefit from neighboring atoms providing electron density. The more alkyl groups present on the carbon, the greater the stability, following the order: $\mathrm{\dot{C}H_3 < 1^\circ < 2^\circ < 3^\circ}$.

Electron Configuration: Analyzing the electrons in an alkyl free radical reveals three bonding pairs and one unpaired electron. This results in a trigonal pyramidal geometry due to hybridization to an $\mathrm{sp^3}$ state.

Delocalization (Resonance): Similar to carbocations, delocalization or resonance plays a role in stabilizing free radicals.

Sources: NCERT, class notes, and Master Organic Chemistry.


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Ethyl Acetate: how to store
I'm a 3D printing hobbyist and a few months ago I bought some Ethyl Acetate after learning online that it might have a similar effect on PLA as acetone has on ABS. From the tests I made, I was either sold a bad batch or the reaction speed is next to 0 thus making the ethyl acetate useless to smooth out PLA. I have thus placed it in a drawer to test it more at a later time. The liquid was kept inside its original plastic bottle that was then placed inside a glass jar to ensure that none of the vapors got out. Today I opened the drawer again to have a look at it and I noticed a very strong odor consistent with that of the ethyl acetate. Also the label on the plastic bottle separated as if the glue has been dissolved away. As ethyl acetate is highly inflammable I'd like to ask what would be a correct way to store it without leaking the vapors.

In a lab, we would store this in a dedicated flammables cabinet that vents vapors.

Why not use an airtight container.  A plastic bottle and a jar are not airtight.  How about an erlenmeyer with a rubber stopper?  (Assuming it doesn't react with rubber)

Just keep it as it is. Unless you won't stock more and more, there are not big concerns if, as you wrote, it is matter of a bottle. Treat it like it would be acetone, to mention a solvent we are all more familiar with.

@Mad Scientist: never do that with volatile solvents. Unless you are sure of stopper / solvent compatibility. Rubber stopper can swell crashing your flask. A bottle is usually lots robust, though a rubber stopper is still not well suited. Besides this, if you really have to seal an erlenmayer in course of your experiment,and you have no Cork just rubber, a diy fit is to protect your stopper with aluminium foil.

In laboratories, ethyl acetate is usually stored in special ventilated cupboards that are furthermore typically designed to withstand fire. This is because you will typically store not just one but many flammable solvents/reagents in that type of cupboard.
At home, you typically do not have that possibility. However, the basic advice remains the same: store in the original container (if possible) and in a well-ventilated area away from spark sources. The original bottle should, in theory, be airtight enough — remember that the ethyl acetate must have somehow found its way to you in that bottle without having lost too much content.
I would strongly advise against storing in a closed drawer. As you noticed, a few months were enough for enough solvent to evaporate but stay in the vicinity for you to smell it. The amount of solvent you lose per unit of time is typically very small and if it can diffuse away it does not do much harm. But the closed drawer prevented it from efficiently diffusing away and allowed it to concentrate.


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Hellmann-Feynman Forces with Hartree-Fock
The energy in the Hartree-Fock approximation is given as:
$$E_{HF}=\left<\psi_{HF} \left| \hat{H} \right| \psi_{HF} \right>=\sum_{i,j}P_{i,j}H_{i,j}^{core}+\frac{1}{2}\sum_{i,j,k,l}P_{i,j}P_{l,k}(ij||kl)+V_{NN} \tag{1}$$
The geometrical derivative of the Hartree-Fock energy can be shown to be [1]:
$$\frac{\partial E_{HF}}{\partial X_A}=\sum_{i,j}P_{i,j}\frac{\partial H_{i,j}^{core}}{\partial X_A}+\frac{1}{2}\sum_{i,j,k,l}P_{i,j}P_{l,k}\frac{\partial (ij||kl)}{\partial X_A}-\sum_{i,j}Q_{i,j}\frac{\partial S_{i,j}}{\partial X_A}+\frac{\partial V_{NN}}{\partial X_A} \tag{2}$$
The Hellmann-Feynman Theorem states:
$$\frac{dE}{d X_A}=\left<\psi \left| \frac{d\hat{H}}{d X_A} \right| \psi \right>$$
The Hellmann-Feynman Theorem implies that the energy derivative only depends on the parts of the Hamiltonian that have a dependency on the derivative. This leads to the geometrical derivative being equal to:
$$\frac{dE}{d X_A}=\left<\psi \left| \frac{d\hat{V}_{NN}}{d X_A} + \frac{d\hat{V}_{Ne}}{d X_A} \right| \psi \right> = \left<\psi \left| \frac{d\hat{V}_{NN}}{d X_A} \right| \psi \right> + \left<\psi \left| \frac{d\hat{V}_{Ne}}{d X_A} \right| \psi \right>=\left<\psi \left| Z_A\sum_B\frac{Z_B(X_B-X_A)}{R^3_{AB}} \right| \psi \right> + \left<\psi \left| -Z_A\sum_i\frac{X_i-X_A}{r^3_{iA}} \right| \psi \right>$$
Now the Hellmann-Feynman theorem can be applied to find the derivative geometrical of the Hartree-Fock energy:
$$\frac{\partial E_{HF,\ce{Hellmann-Feynman}}}{\partial X_A}=\left<\psi_{HF} \left| \frac{\partial\hat{H}}{\partial X_A} \right| \psi_{HF} \right>=\sum_{i,j}P_{i,j}\frac{\partial V_{Ne,ij}}{\partial X_A}+\frac{\partial V_{NN}}{\partial X_A} \tag{3}$$
I tried to implement equation $(2)$ and equation $(3)$, and used them to calculate, the force between H and Li for different distances, with cc-pVDZ and cc-pVTZ. and got the following:

Hartree-Fock refers to equation $(2)$ and Hellmann-Feynman refers to equation $(3)$. I have read that the Hellmann-Feynman theorem is only valid for exact solutions, but should become a better and better approximation, when going towards the Hartree-Fock limit. As can be seen in the picture, equation $(3)$ performs worse with the larger basisset cc-pVTZ. This now leads me to my question, were am I wrong in my understanding of the application of the Hellmann-Feynman theorem to the Hartree-Fock approximation?
[1] Attila Szabo, Neil S. Ostlund; Modern Quantum Chemistry: Introduction to Advanced Electronic Structure Theory; equation C. 12

Related reading: https://books.google.com/books?id=1Ijmiw_EchIC&pg=PA53&lpg=PA53&dq=pulay+force&source=bl&ots=G3lTd0ilt2&sig=5lY5zGXztnJDV-_crVJUCnmq5lI&hl=en&sa=X&ved=0ahUKEwjomoO9gd3VAhXHxlQKHVxFB444ChDoAQgjMAg#v=onepage&q=pulay%20force&f=false

You may wish to see the paper: Bakken, V., Helgaker, T., Klopper, W., & Ruud, K. (1999). The calculation of molecular geometrical properties in the Hellmann—Feynman approximation. Molecular Physics, 96(4), 653-671.
This makes it clear that a complete basis is needed for the Hellmann-Feynman theorem to hold. Also, one thing which I don't quite understand, but seems to be necessary is that the AOs are not supposed to remain centered on the atoms as the geometry is distorted. Thus, you are always calculating in reference to the original geometry, and convergence is necessarily slower.

@jheindel "AOs are not supposed to remain centered on the atoms as the geometry is distorted" Without doing out the math to prove this, it would ensure that any derivative of an AO w.r.t. a nuclear coordinate is zero.

@pentavalentcarbon I agree that the concept is weird and I don't really understand it. I just brought it up because it's unusual so it would have been something easy to do wrong assuming the paper is right. I don't think it's necessarily zero though. All that having the AOs uncentered does is cause you to evaluate the derivative further away from the peak of the gaussian. I don't see why this should be zero.

@ErikKjellgren While in general you would expect improvement as the basis set is enlarged, this isn't necessarily a guarantee so I don't think it is too surprising to see slightly higher errors.

It is not at all surprising that the forces you get from application of the Hellmann-Feynman theorem do not match the true force. The reason is the appearance of Pulay forces: the atomic orbital basis set depends on the nuclear coordinates, and changing the position of the nuclei also changes the basis set. This yields the contribution you're missing in the plots.
cc-pVDZ and cc-pVTZ are still pretty small basis sets. I would expect that if you add diffuse functions, the result improves quite a bit, since the basis set covers more degrees of freedom. Furthermore, to get something meaningful you'd have to compare e.g. cc-pV5Z vs cc-pV6Z, which are closer to the complete basis set limit.
However, already in your plots you can see that the agreement is better from cc-pVDZ to cc-pVTZ when you look at the far-range behavior. A better way to plot these would be to plot the difference, i.e. the Pulay force.
I would also imagine that decontracting the basis sets helps, because if you imagine just moving the electron density but not the basis functions, then you will see that the description of the core orbitals is impoverished quite a bit as the contraction coefficients are wrong.
Actually, the result in cc-pV6Z is probably still not very close to the Hellman-Feynman limit. When you move the nucleus, 1s core orbitals start to gather p, f, etc character. Thus, when you move the nuclei but not the basis functions, your original basis set should have e.g. very tight p functions that are able to describe this effect.
If you want to toy around with this, you should look at calculations in larger and larger primitive basis sets. Even-tempered exponents should work fine. Start with a small set of s functions, and study how the result changes when you add steeper or more diffuse functions. Next, look at the p shell: which exponent changes the result the most? And then again start adding steeper and more diffuse functions. Repeat ad nauseam. (This is called completeness-optimization, and I've published an automatic procedure for this.)


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Aluminium foil for cooking
I have heard so many times that aluminium pots should not be used to cook sour (acid) foods and I would like to know the chemical  reason for that claim.
I also would like to know if the aluminium foils are not suitable only for cooking sour foods, or if this extends to storing sour (acid) food such as lemon juice?

The reason is that Al will react with acids. This applies to packaging as well as cooking.

Why not Al?

Frankly, Al oxidised in thin air jus like that...think how badly it would react with acidic food.

This and cooking.stackexchange have had few more curious question like yours.

While acidic foods do react with aluminium slightly, the main reason not to use aluminium utensils for those foods isn't the danger from dissolved aluminium salts (which is very low) but the damage done to the aluminium utensils (anodised aluminium surfaces are worn away, though they can be regenerated by using other foods).

related https://chemistry.stackexchange.com/questions/80941/would-swallowing-aluminum-foil-be-considered-dangerous

Have a look at this article that explains the toxicity of Aluminum ions.

Aluminum is also a poor choice for salt containing foods ( if wet).


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Why does copper sulfate disassociate into Cu+2 and SO4-2 when dissolved in water?
I'm trying to understand why exactly copper sulfate disassociates into $\ce{Cu^{2+}}$ and $\ce{SO4^{2-}}$ ions? How is it calculated? I'm trying to find information but it seems as if it is treated as a given. I'm trying to understand what happens exactly? What happens to the electrons in the process? Where do they come from? How to arrive to the charge of the ions?
Thank you.

https://chemistry.stackexchange.com/questions/41694/why-arent-the-copperii-ions-in-cuso4-precipitated-by-hydroxide-ions-from-wate

Those are the ways of salts: metals are cations, and acid residues are anions. "Calculated" is not the right word; the information has been around for quite a while now, and everybody just got used to it. So will you.

But what happens in the process? how do we arrive to it? I want to know. I don't like "just treated as a given". I want to know.

Put two wires in the solution and connect them to a battery. The one which is connected to (-) will get covered with metal. So there must have been positive ions of Cu.

Yes. How do these ions appear? the +2 and -2? This question is driving me nuts.

They were there from the very beginning. By current understanding, solid CuSO4 consists of ions.

I think I got it. I makes sense. I thought some type of reaction took place when dissolved in water, but no. It is there from the very beginning, you are right.

You are getting used to it, just like I said.

Ivan Neretin gave quite a few points, I'd like to elaborate a bit more.
Basically $\ce{CuSO4}$ already exists as $\ce{Cu^{2+}}$ and $\ce{SO4^{2-}}$. It's anhydride form exists as an orthorhombic arrangement of ions. The opposite charge attraction balances the like-charge repulsions and the crystal stays bound.
When you dissolve it into water, a few things happen. First of all, with water being quite a polar molecule, tends to surround ions. This is called hydration of ions. Basically this just separates the ions from the tightly packed lattice and spreads them evenly along the water sample. Hydration is the main reason why the $\ce{Cu^2+}$ ion is quite stable in aqueous medium.


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Correlation between resonance and coordinate covalent bond
So I was listening to an online chemistry lecture the other day, and the theme was "bond".
In the 'coordinate covalent bond' part, the teacher said "Examples of the coordinate covalent bond can be $\ce{SO_2}$, $\ce{HNO_3}$, etc." and I could figure out that both molecules have resonance structures: between $\ce{N}$ and $\ce{O}$ atoms in $\ce{HNO_3}$, between $\ce{O}$ and $\ce{S}$ atoms in $\ce{SO_2}$. Also I am aware of the resonance structures in the $\ce{C_6H_6}$(benzene) molecule, so I looked up the Wikipedia page and could find this sentence:
"..., while the VB description involves a superposition of resonance structures."
So I became curious about this: Is there any close correlation between resonance and coordinate covalent bond? In other words, is it safe to view a molecule as having coordinate covalent bond when it has resonance structure in it?

Correlation? Safe? Your question doesn't make much sense...

I think what you mean by resonance is formal charges. Not resonance. I once thought of this too.

Coordinate covalent bonds and Resonance are very very vastly different things. You may have coordinate compounds without resonance, and you may have resonance involving compounds without any coordinate covalent bonds. There's no connection.
Coordinate Covalent Bonds (a.k.a Dative Bonds):
These bonds are just like any other covalent bond. They involve sharing of electrons in between the two atoms that make up the bond. The only difference between a regular covalent bond and a coordinate one is that both the electrons belong to one atom in case of coordinate covalent bonds. Head over to Wikipedia's page on Coordinate Covalent Bonds to learn more, if you're interested.
Resonance:
This is, as I mentioned, vastly different from dative bonds. This is simply a delocalization of $\pi$-electrons along a larger stretch of the molecule as opposed to being confined to being in-between two atoms. Resonance occurs when two $\pi$-systems are in conjugation with each other.
For example, consider butadiene:

There are two $\pi$-systems that are in conjugation with each other. As a result, the two systems kind of overlap and become one bigger system.
You might wonder, why do these molecules undergo resonance anyway? Well it's all based on a very simple fact: Like charges repel. Electrons in the $\pi$ molecular orbitals are no exception either. As opposed to being confined to being between two atoms, they prefer to spread over the molecule when given the chance. If you put a 100 misanthropes into a large hall, would they all form two clumps or evenly spread throughout the hall?

Anyway, now you can see, there isn't any relationship between dative bonds and resonance. There are compounds which have no dative bonds, but possess resonance, eg. benzene, butadiene, pyridine, et cetera. There are compounds that have dative bonds, and yet no resonance, eg. ammonia-boron adduct, tetraaquacopper(II), et cetera.

Thanks for the detailed explanation! As many pinpointed my question as unclear, I also felt the same and you fixed it.

Based on the examples you have given, that does appear to be the case but if you look for more examples you'll find that this is not particularly true. Two well known compounds can be used to demonstrate:
(1) Carbon Monoxide: Coordinate Covalent Bond from O -> C, but no resonance structure
(2) Benzene: Resonance Structure but no Coordinate Covalent Bonds


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Coming up with a Mathematical Model to Predict how long a redox reaction can last
The description of my problem is as follows:
I will like to find out how can I obtain a mathematical equation which could give me a surface plot of the current output as a function of both concentration and time from an electrochemical cell which uses the following redox equation:
O2 + 2Pb -> 2PbO
I have the following information:

Mass of Pb - 10g
A fixed concentration of oxygen gas - 20.9% over volume of air in room

Assumptions made:

Unlimited supply of oxygen
The gas is supplied at constant room temperature and pressure (25 degree Celcius and 1 atm)

Tentatively, my plan was to use Farday's Law of Electrolysis (https://en.wikipedia.org/wiki/Faraday%27s_laws_of_electrolysis) to calculate the current which flows as a result of the transfer of electrons from the redox reaction. My hypothesis is:

The magnitude of the current generated by the following reaction will be proportional to the rate of oxygen consumption.
The magnitude of the current generated will decrease over time as the rate of reaction decrease (from the decrease in concentration of Pb).

My problem is that the plot I obtained shows me that the current vs time relationship is a reciprocal one under fixed concentration of oxygen gas.

This does not agree with the experimental results I obtained which shows that current output is steady and starts decreasing after a long period of time (a few months at least). Can someone help point out my the flaw in my argument? What should I do instead to obtain the plot I described?

It's a surface reaction - Pb isn't in solution, otherwise it could similar kinetics to  what you've got.

In addition to the redox reactions the transport of material towards the electrodes also needs to be considered. This involves using Fick's equations for diffusion to calculate the concentration of species at distances away from the electrodes as reaction starts and reaches constant conditions. The steady state equations are of the form
$$ D_{ox} \frac{\partial C_{ox}}{\partial x}= k_cC_{ox}-k_aC_{rd} \\
   D_{rd} \frac{\partial C_{rd}}{\partial x}= -k_cC_{ox}+k_aC_{rd}$$
These lead to the Cottrell equation for the limiting flux (current density)  $j=-nFC^o_{ox}\sqrt{D_{ox}/\pi t}$. To get what you want you will probably need to consult some specialist text books on electrochemistry.

The Cottrell equation still won't explain why the reaction produces a steady current for a period of time before decreasing. Also, experimental results show that the decrease in current approximates to a sigmoid function reflected at the y-axis, which is counter-intuitive to Cottrell's equation as it also gives an inverse relationship with time. May I know which part of the specialist textbook do you recommend me to go through? The book which I am using now is Electrochemical Methods_ Fundamentals and Applications by Bard and Faulkner

You sound as though you understand this problem very well :)  I have used 'Principles of Electrochemistry', Koryta, Dvorak & Kavan, publ Wiley.


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Looking for a low boiling point solvent for MAPI precursors (Lead Iodide (II) and Methylammonium Iodide)
Things I have thought of till now:

Acetonitrile with HI additive
Acetonitrile with DMSO additive

The problem with the first would be that HI is in an around 55% aqueous solution and water doesn't play well with the perovskite. Meanwhile, the second might end up needing enough DMSO to drastically increase the boiling point. I am not sure about the second and intend to try it out and will post updates when I am done.


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Refractive index of barium sulfate
What is the refractive index of barium respectively where do I find it? I only find values for alpha, beta and x-ray radiation like on https://pubchem.ncbi.nlm.nih.gov/compound/barium_sulfate#section=Decomposition
I need it for all wavelengths or at least at 455 nm. Thank you in advance!

The index of refraction of $\ce{BaSO4}$ at 589 nm can be found in "Index of Refraction of Inorganic Crystals", in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.
Separate values are given for the indexes of the orthorhombic crystals:
$\begin{align}
n_x &= 1.6362\\
n_y &= 1.6374\\
n_z &= 1.6480
\end{align}$

At first thanks a lot! Unfortunately this is at another wavelength and there might be a (huge) difference due to the wavelength. Are anywhere some databases about materials and their properties which are common to look at?

Have a look for the Sellmeier equations, this will give you an indication of how the refractive index varies with wavelength. You may then have to 'guestimate' a value but it will probably not be so much in error .

@porphyrin Maybe I was too fast: How do I estimate it with the help of the Sellmeier equation? I only have the wavelength and B and C are missing(?)

well, you know that as your wavelength is shorter than wavelength used in std tables (Na D line) so refractive index will be larger. (a) choose two materials with similar refractive index one above and one below whose params are known and interpolate. (b) if (a) is not possible, as curves are all similar shape take one for, say, quartz or some glass etc and set value at Na D line to your value for BaSO4 and then work out refractive index at your wavelength.

continued ...  Try this with different solids and average, probably wont be so far in error. If you could make a crystal prism out of BaSO4 then you measure directly. (If you have access to fast lasers or time correlated single photon counting you could measure delay time through crystal directly).

Hi, thanks a lot! Sorry, I was busy with other stuff but will try this soon! Thanks again!


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Platinum-coated razor blades for making a fuel cell
My daughter wants to construct a fuel cell for a school project. Budget is tight and platinum as preferred catalyst expensive. I know that Gillette razor blades are coated with platinum. Would there be a way to use old razor blades for making a fuel cell?

Umm...Looks to me like "Platinum" is used as a marketing/sales tool. See Gillette's description of their Platinum-Plus Double-Edged Blade, which is made from stainless steel.

I've been in the Gillette factory near London. I can positively confirm they do coat their razor blades with real platinum, tiny amounts however.

I've also been in their other factory in Berlin. Same story. The amount is very very small. From memory they use less than a kg to coat millions of blades. Been a number of years ago however. At that time the first gen Gillette sensor came out.

@Kemmer Happen to know why they'd put platinum in their razors? Given, it's a minute quantity, but I don't see much point in doing so :/

Not sure why they did it. Only thing I can think of is to increase corrosion resistance.

I think polymers like Teflon are put over the Pt , so it may not be very exposed. Pt and Teflon both provide corrosion protection ; bare steel blades are dull after one use due to corrosion , not wear.


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What does stepped Ni(111) surface mean?
I think (111) surface refers to a flat surface of atoms that intersect with x,y,z axis at 1.

https://en.m.wikipedia.org/wiki/Miller_index

Stepped surfaces generally refer to surfaces (of a particular index) which have 'steps' like a flight of stairs. A cursory search on google images can yield some useful examples, such as this one:

As you can see from the example, a stepped surface will be a surface which includes steps which go from one layer to the next.


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Reduction of 1,4-benzoquinone in the synthesis of 1,5-diazocane

Synthesize 1,5-diazocane using only benzene as a source of
  carbon atoms, and ammonia as the only source of nitrogen atoms.

My attempt: Noticing that the target compound has split the benzene ring, I attempted to split the benzene ring as well. I thought of two options:

H2/Pt at high temperature and pressue
Creation of phenol and subsequent oxidation

As the first process would create cyclohexane and be difficult to manipulate, I went with the second option. To synthesize phenol from benzene, I used a nitration, followed by reduction to create aniline, then used nitrous acid to create the benzenediazonium salt, and decomposed the salt in the presence of  water to create phenol. After the oxidation of phenol I got 1,4-benzoquinone.
Now, realizing that I would need to create 1,3-propandial in order to use $\ce{NH3/NaBH3CN}$ to create 1,5-diazocane, I attempted to make 1,3-propandial from 1,4-benzoquinone. Ozonolysis of 1,4-benzoquinone produces 2-oxo,1,3-propandial, which has an extra carbonyl group. Because of the many carbonyl groups involved in this compound, I was thinking of removing the carbonyl groups directly from benzoquinone, and doing an ozonelysis from there. However, I was not sure if any standard carbonyl reactions, such as the Wolff-Krishner or the Clemmensen reaction could be applied to remove the carbonyls, seeing as they are heavily resonance stabilized.
Therefore, I was wondering if these reactions do proceed on benzoquinone, and if not, then what other plausible ways of synthesizing 1,5-diazocane exist?

I think the idea with benzoquinone is good, you're overcomplicating it, though.

This is my approach, start with reaction of benzene and ozone in the presence of zinc and water to form glyoxal. React two equivalents of glyoxal with two equivalents of ammonia to form two equivalents of the monoimine. Next convert some glyoxal into formyl chloride, react two equivalents of formyl chloride with two equivalents of imine, make the product undergo Pinacol Coupling (Magnesium, high temperature), and then dehydrate and reduce.

Think double Beckmann rearrangement.

@ron: In doing some research on my answer, I found an attempted double Beckmann on the bis-oxime of 1,4-cyclohexanedione (arduously prepared from benzene via 1,4-dimethoxybenzene?). Beckmann (29%) and Schmidt (HN3) on the diketone (40%). Rothe, M.; Timler, R. Chemische Berichte, (1962), 95, 783.

@user55119, Yes, that's the reference.  The double Beckmann from 1,3-cyclohexanedione has also been reported to yield only the desired precursor bis-amide, 1,5‐diazocine‐2,4‐dione. The preparation of  1,3-cyclohexanedione and 1,4-cyclohexanedione seem reasonably straightforward.

Birch reduction of benzene provides 1,4-cyclohexadiene (2), which yields malonic acid upon ozonolysis and subsequent oxidation with peroxide and formic acid to form malonic acid 3 [2 --> 3 was found via Chem. Abstr.]. The half acid chloride of malonic ester 5 is well-known. In addition, 1,3-dibromopropane (6) is also accessible from malonic acid (3), or for that matter, from diethyl malonate (4) or the acid chloride 5. Moreover, 1,3-diaminopropane (8) is available from malonic acid (3) in addition to treating dibromide 6 with an excess of NH4OH.
At this juncture the temptation would be to alkylate diamine 8 with dibromide 6 to synthesize 1,5-diazacane (13). But this type of reaction is fraught with problems such as polymerization and overalkylation. Boc protected hydrazine 7 has been shown1 to provide pyrazolidine (9). Although hydrazine can be prepared from ammonia via chloramine and ammonia via the Olin Raushig process, it is unlikely to be conducive to the academic laboratory. However, diamine 8 has been transformed into pyrazolidine (9) with bleach via a more facile intramolecular process.2 To assure the use of ammonia, the dibromide 6 can be treated with excess ammonia to provide diamine 8.
Pyrazolidine (9) can be acylated with malonyl acid halide 5 followed by reduction to alcohol 11. Ring closure to tetrahydro-1H,5H-pyrazolo[1,2-a]pyrazole (12) involves conversion of alcohol 11 to the bromide, neutralization of the intermediate hydrobromide salt with base and heating to effect facile intramolecular alkylation. A second option is to alkylate hydrazine 9 with dibromide 6, a process that is not as susceptible to polymerization as the previously discussed process. Finally, reduction of the N-N bond in hydrazine 12 is accomplished by hydrogenation3 to afford the target molecule 1,5-diazacane (13).
ADDENDUM: If ester 10 were to cyclize, LiAlH4 reduction would form hydrazine 12.

1) Watanabe, H.; Kuwata, S.; Koyama, S. Bull. Chem. Soc. Japan, 1963, 143.
2) El Hajj, A.; Bougrine, A. J.; Le, D. M.; Pasquet, V.; Delalu, H. Reaction Kinetics, Mechanisms and Catalysis, 2016, 117, 429.
3) Stetter, H.; Spangenberger, H. Chemische Berichte, 1958, 91, 1982.

Tempting, but no.  Only benzene and ammonia are supposed to provide the carbon and nitrogen atoms you use.

Good catch, OL. Forgot to add that dibromide 6 gives diamine 8 with ammonia. See new last sentence in second paragraph. Benzene provides all of the carbons. Overall, a sensible approach. Clearly, one wouldn't use benzene to make the product. An academic exercise.

@Oscar Lanza; See above reply and 3 --> 8, which uses ammonium hydroxide.

Other than "ammonium hydroxide" being a misnomer ($\ce{NH3(aq)}$ would be more accurate), now it's all good.

Here is a relatively straightforward synthesis of 1,5 diazocane that uses the Beckmann rearrangement as a key step. The Beckmann rearrangement is a useful synthetic tool for converting ketones to amides.  If the carbonyl is in a ring, then conversion to the amide will enlarge the ring by one nitrogen atom.

The novel twist in the following synthesis is to perform a double Beckmann rearrangement in the penultimate step.

heating benzene with concentrated sulfuric acid containing excess sulfur trioxide (oleum) will lead to the disulfonated product, benzene-1,3-disulfonic acid
heating the disulfonic acid with potassium carbonate produces resorcinol (see here)
treating resorcinol with hydrogen gas at low pressure over a rhodium catalyst yields 1,3-cyclohexanedione (see here)
hydroxylamine can be prepared by oxidizing ammonia with hydrogen peroxide to produce ammonium nitrite which can then be converted by a number of methods to hydroxylamine
the hydroxylamine is used in standard fashion to afford the bis-oxime (various oxime stereoisomers are likely produced, but this does not matter, all oxime stereoisomers will undergo the Beckmann rearrangement in the next step)
the bis-oxime is converted to the bis-amide through a double Beckmann rearrangement (see here for this specific conversion; note that the double Beckmann could afford isomers but only the desired isomer is produced; see the preceding link for the authors explanation as to why this is so)
finally, standard reduction of the bis-amide with $\ce{LiAlH4}$ and aqueous workup gives rise to the target molecule

The double Beckmann could also be run starting with 1,4-cyclohexanedione (readily prepared from benzene, see here).  1,4-cyclohexanedione has the advantage of symmetry - the first Beckmann rearrangement can produce only one isomer.  Although the second Beckmann rearrangement could produce 2 isomers, only the desired 1,5-diazocane-2,4-dione was isolated (see here).