Safely mixing methanol and chloroform Sorry for what is likely a naive question. I haven't worked with chemicals at all and have only had some brief lab safety instruction.

I need to create a mixture of chloroform and methanol 2:1 for a lipid extraction from bone.

I have read the safety data sheets for both chemicals but can't find any other information about how to safely create this mixture other than doing it in a fume hood. I need about 200 ml total.

If anyone has any information or can direct me to where to look that would be MUCH appreciated.

Thank you!!!

I am sorry to give my opinion here but I feel compelled to do so!

Security officers or the people writing the safety data sheet are motivated by only one thing: they don't want to be held liable! For them, everything is deadly and they don't even try to mitigate that (true) fact according to their audience.

As a result, all the risks are presented at the same level and nobody can really adjust his/her practices according to that.

I vividly remember one of my students at Oxford University. The rule there is(was?) that the student should copy the whole safety sheet on a form submitted to the Safety Officer before he/she can even get the chemical. My student had to do that for a ridiculous amount of cyanide and he was so stressed out that he ended up shaking and putting cyanide everywhere! I mean on the balance, around the balance, on the floor etc... I had to be besides him "It's OK, it's no big deal" while making signs to the others to not come around because there was cyanide everywhere.

So @Heather, I really advise you to ask your co-workers about the true risks in your field. You won't die because you had a whiff of chloroform or methanol, even though you should always work under a hood so you don't have a whiff of nasty materials. Also the safety sheets must be adapted to your own risks.


Can we predict the IMF strength of Acetone vs Glycerol? I was recently assigned a lab where I was to observe several properties of the chemicals Acetone and Glycerol then use my observations to predict their overall Intermolecular Force strength.
The properties I observed were:
Surface Tension: Here, I found that Acetone had a much higher surface tension than Glycerol. (I tested this by seeing how many drops of each substance would stay on a penny: Acetone had 19 and Glycerol had 13)
Viscosity: Here, I found that Glycerol had a much higher viscosity than Acetone. (I did this by dropping a penny through a 500mL graduated cylinder: Took 0.4 seconds for Acetone and 5ish seconds for Glycerol)
Boiling Point: Researching online, I found that Glycerol had a much higher boiling point than Acetone.
Glycerol also took much longer than Acetone to evaporate.
However, this seems to be conflicting evidence. I have learned that a higher surface tension is associated with overall stronger IMFs (suggesting that Acetone has stronger IMFs), but the other pieces of evidence seem to suggest that Glycerol has stronger IMFs.
My question is, can anybody tell me if Acetone's IMFs really are stronger than Glycerol, or is it the other way around? Additionally, if any of my experimental results seem wrong, can somebody please point them out?

So the intermolecular forces would be stronger in glycerol then in acetone. Glycerol can form hydrogen bonds while acetone is not very polar and can’t significantly form h bonds with itself.
The surface tension for glycerol is 63.4 $\frac{mN}{m}$ while acetone is 25.2 $ \frac{mN}{m}$ (both @ 20 C)
I supposed the error in your experiment arises from using pipette to determine the amount of material on the coin. Unless your using very precise pipettes, counting drops isn’t necessarily the best way to quantify the amount of material added. Additionally the density of glycerol is higher then acetone which needs to be accounted for in the test.


Interpreting a formula tattoo I saw

on the Internet and I'm curious as to what compound is represented by it.

The molecule seen in the tattoo is a protein-based mammalian hormone called oxytocin, commonly called "the love hormone" because it is involved in several aspects of sexual reproduction.

Image Source


Are homemade batteries economically practical? Every kid who has been to a science fair knows that you can make a battery with a lemon, a potato, a penny and a nail. I've never done this and I'm not entirely clear on how it works beyond a hazy recollection of undergraduate-level chemistry class; but I gather that you do not need to use a lemon and potato as your power source but can in fact use any number of substances willing to donate or accept electrons.

My question is this: are there any home-made batteries that can produce sufficient electrical energy that its cheaper to make them yourself rather than plug your appliance into the wall or use store-bought batteries

I realize that there are a number of variables here that might make a significant difference in whether its cheaper to use established electricity sources or create your own so for arguments sake lets say that I am trying to:
- charge a cell phone/laptop
- run a microwave long enough to make popcorn
- keep a refrigerator cold

Lets also consider the dual possibilities that I:
- live in a city apartment and have to buy all my fuels and materials
- live on my own far where I can grow as much produce as I want to devote to making batteries

If there are other good fuel sources that could be a logistic variable; let me know in the comments so I can edit the question accordingly.

No: trying to catch up with 100+ years of technological optimizations in your kitchen won't be an economical enterprise.  Store-brought batteries are standardized and relatively inexpensive.  Power from an outlet is even less expensive (by one or two orders of magnitude).

On the other hand,
in many places you can save on electricity by going solar. Even if you can't install solar panels on your roof because you don't own a house, you could buy solar chargers for small appliances if you have a sunny corner in your apartment...


Ionize Hydrogen and Oxygen Gas in Electrolytic Cell? As is well known, water can be electrolyzed into oxygen and hydrogen gasses. However, all redox reactions can be run in reverse. If I were to electrolyze water in 2 closed containers (in order to trap the gasses) and i were to run the cell in reverse using a battery, would the hydrogen and oxygen gasses turn back into ions and, therefore, water? If so how would this work on a physical level? And what path do the ions take to get to each other?

Since the monatomic hydrogen and oxygen have combined, respectively, in H2 and O2 gases, those molecular bonds must be broken in order to get them to recombine as water and produce electricity. That is precisely what a fuel cell does. You can lease a hydrogen fuel cell powered Hyundai Tucson, if you're close to H2 filling stations, and Iceland is looking into setting up a hydrothermal-powered H2 fuel station system.


Does anyone know a database that shows the effect of electrolyte concentration on the production of hydrogen gas during electrolysis of water? I'm currently studying in the International Baccalaureate model. This high school model asks their students to do an individual work called the Internal Assessment. One of the subjects I study is chemistry, and my Chemistry Internal Assessment is based on the production of hydrogen (a source of energy) in the water electrolysis as a way to save electrical energy. I was going to do an experiment to measure the production of hydrogen in the electrolysis of water with various concentrations of four different solutes (NaCl, NaOH, sulphuric acid and sodium bicarbonate, but any electrolyte will do). However, with the Covid-19 outbreak, I cannot go to the laboratory, so in order to get the data I need, I searched on google databases with the information I wanted, but I found barely nothing. These links (https://www.scirp.org/pdf/NR_2013032814044640.pdf) (https://pubs.rsc.org/en/content/articlehtml/2017/se/c7se00334j) are the only useful pdfs I found, and I wonder if anyone could help me giving me some databases with the data I need.

You need to look for the data in a different way. You can use electrical conductivities to get the current flow the solution. From the imagined current flow you calculate how much hydrogen would be generated.


Dependence of the angle of rotation on the wavelength of plane polarized light Wavelength difference is a big deal, I know. It can solely change the whole interaction between the chiral molecule & the light. But I am not sure what's the mechanism by which light of different wavelengths produces different angles of rotation.

So, why and how wavelength matters?

The rotation of plane polarised light by a solution of, say, sucrose depend on the ability of the oscillating magnetic filed of the light to induce an electric dipole moment in the molecule and the ability of the oscillating electric field of the light to induce a magnetic dipole moment.

For these interaction to have any magnitude it is supposed that the electrons in a molecule move in a helical path, or, alternatively that there are two linear dipoles generated by electron motion which are in planes that are at some angle to one another. Although this model is clearly artificial it emphasises the fact that movement of charge must follow somewhat crooked pathways under the influence of the radiation. The theory of this is very complicated but the result from a quantum calculation is relatively straightforwards and is that the molecular rotation M at wavelength $\lambda$ is given by

$$M_\lambda = a\sum_i \frac{\lambda_{0i}^2X}{\lambda^2-\lambda_{0i}^2}$$

where i represents all the electronic states of the molecule and the wavelength $\lambda_{0i} = c/\omega_{0i}$ where $\omega_{0i} = (E_i-E_0)/\hbar$ is the frequency of the $i^{th}$ excited state at energy $E_i$. The parameter $X=\mathrm{Im}(\mu_{m_{0i}}\cdot\mu_{e_{i0}})$ is the value imaginary part of the complex dot product of the induced magnetic and electric dipole moments and a  are a set of constants independent of wavelength.

From this formula it is seen that the polarisation rotation depends on the wavelength. (The dependence of rotation angle on wavelength has been called optical rotatory dispersion.) We can also conclude

that experiments should be performed where the molecule has little absorbance but that the rotation will be larger as an absorption band is approached. (in an absorption band elliptically polarised light can be formed)
The strength of any optical transition is not important as the dot product X depends on induced moments and the angle between them. This also means that weak optical transitions can be as important as strong one in rotation the polarisation. If the induced dipoles are perpendicular then the rotation vanishes as the dot product is zero.
That many excited states i can be involved and their effects can cancel to some extent so the value of the rotation is hard to predict.
The polarisation rotation of two mirror image molecules are equal and opposite in size and if a molecule is identical with its mirror image polarisation rotation must be zero and the term X is zero.
When the wavelength $\lambda < \lambda_{0i} $ i.e. the wavelength crosses a transition then the signal changes sign and this is called the Cotton effect.


What makes oceanic desalination so expensive? What specifically makes oceanic desalination so expensive?

Reason is salt dissolve very easily with water, forming very strong bond. Not only salt there are also other minerals. To break this bond we need technology as well as energy and both of them are very expensive(today).

There are two methods by which we can break this bond:

Thermal distillation:
Thermal distillation involves heat: Boiling water turns it into vapor—leaving the salt behind—that is collected and condensed back into water by cooling it down
Membrane distillation: The most common type of membrane separation is called reverse osmosis. Seawater is forced through a semipermeable membrane that separates salt from water. Because the technology typically requires less energy than thermal distillation, most new plants, now use reverse osmosis.

Today about 10 to 13 billion gallon  water are desalinated worldwide per day. That's only about 0.2% of global water consumption.

Membrane distillation is bit more cheaper than thermal distillation. Even though  large amounts of energy are needed to generate the high pressure that forces the water through the filter. Current methods require about 14 kilowatt-hours of energy to produce 1,000 gallons of desalinated seawater.

So we can conclude that cost of distillation is high because we need notable large amount of electricity to heat water in thermal plant and generate high pressure. Also other technology(like say burner, pumps, etc) are also expensive.

Source: ScientificAmerica and LiveScience

Also note we are trying to decrease the cost of distillation and day by day number of seawater distillation  are increasing.


Can atomicity be infinity? My chemistry teacher told us that graphite or diamond had an atomicity of infinity. However, the definition of molecule in my testbook is

  A molecule is the smallest part of an element or compound which can exist on its own under ordinary condition.

It doesn't make sense to me because if a molecule has an atomicity of infinity, it it is not a discrete compound anymore. It contradicts the definition of molecule.

Introducing the idea of "infinite atomicity" can be useful to highlight the importance of particle size on the properties of metallic or network covalent solids. As the surface to volume ratio of discrete particles becomes sufficiently large, properties of crystalline solids such as melting point, absorption wavelength and reactivity begin to diverge significantly from the limiting values observed for larger crystals. A pronounced change often occurs as the nanoparticle regime is entered, at ~$\pu{ 0.1 \mu m}$, which corresponds to ~$\pu{10^8 atoms/particle}$ or so, well short of infinity, but converging to that limit (presumably heading closer to infinite atomicity will bring you into the regime where gravitational forces become very important, but I am guessing your teacher did not mean this).

But to answer your question, I found the wikipedia useful. On network solids, it states that

  In a network solid there are no individual molecules, and the entire crystal or amorphous solid may be considered a macromolecule.

On atomicity it provides the following comment:

  All metals and some other elements, such as carbon, do not have a simple structure but consist of a very large and indefinite number of atoms bonded together. Their atomicity cannot be determined and is usually considered as 1.


Why are H+ and OH- ions preferentially discharged in the electrolysis of NaCl(aq)? There are 4 types of ions in an $\ce{NaCl(aq)}$ solution: $\ce{H+}$, $\ce{OH-}$, $\ce{Na+}$ and $\ce{Cl-}$. According to the E.C.S. table, $\ce{H+}$ and $\ce{OH-}$ should preferentially discharge at cathode and anode, respectively.

However, according to the Wikipedia page on self-ionization of water, water only dissociates slightly into $\ce{H+}$ and $\ce{OH-}$ in neutral solution. (This is the reason why pure water conducts electricity poorly). Therefore, in the $\ce{NaCl(aq)}$ solution, the concentrations of $\ce{H+}$ and $\ce{OH-}$ ions are extremely small, and by comparison those of $\ce{Na+}$ and $\ce{Cl-}$ are much higher. As a result, it seems to me that $\ce{Na+}$ and $\ce{Cl-}$ should be preferentially discharged because of the concentration effect.

However, my chemistry textbook says that in fact $\ce{H+}$ and $\ce{OH-}$ are preferentially discharged, and hence the main products of the electrolysis are hydrogen and oxygen. So apparently my understanding is wrong. But where?

Water does not need to be dissociated into ions in order to participate in electrochemical reactions. While it is common to refer to electrode reactions as "discharging" this or that species, there is no requirement that the reactant of an electrode reaction must always be an ion.

In neutral solution, it's quite common for the water electrolysis reactions to be written as:

$$
\begin{align}
\text{Cathode:} &\quad \ce{4H2O + 4e- -> 2H2 + 2OH-} \\
\text{Anode:} &\quad \ce{2H2O -> O2 + 4H+ + 4e-}
\end{align}
$$

It's absolutely the case that the electrochemistry written using the ions as reactants in many cases is likely to be more facile:

$$
\begin{align}
\text{Cathode:} &\quad \ce{4H+ + 4e- -> 2H2} \\
\text{Anode:} &\quad \ce{4OH- -> O2 + 2H2O + 4e-}
\end{align}
$$

But it's still far (far) more favorable for $\ce{H2O}$ to be reduced than $\ce{Na+}$, and in many cases more favorable for $\ce{H2O}$ to be oxidized than $\ce{Cl-}$. You're quite right that the concentrations of $\ce{H+}$ and $\ce{OH-}$ ions are extremely low in neutral salt solutions, but that fact is mostly irrelevant to the electrochemistry.


Heterogeneous ion solution I am a college student and for my chemistry of solutions class, I had a lab where we had to titrate orange juice to check the calcium content. However, the experimental results are 20% off the labelled calcium content, and in my report I have to provide a "Tentative explanation of the discrepancy."

Is it possible that the orange juice was not completely homogeneous? What I mean is that maybe some of the calcium and other 2+ ions were not well mixed and there were more/less in our sample. Is that possible or are ions always equally dispersed throughout the solvent when in solution?

The calcium amount may change from one fruit to the next one, or from one day to another one, or from one field to another one. 20 % is not a significant difference between such a measurement and another one. And this may happen for any constituant, not specially for calcium.


Best Way to Store Copper Acetate Solution? I am preparing to copper electroplate an ABS plastic sheet in my home and wanted to know what the best container/environment would be to store the copper acetate solution.
First, I created the solution using copper metal, vinegar and hydrogen peroxide.
Can I simply store this solution in a plastic container?
Should it be closed with an air-tight lid?
Would food-grade LLDPE plastic container be acceptable?
Does the color of container matter? Is black color better than clear color as it reflects more light?
Is it OK to leave it in my home at room temperature or should it be refrigerated?
Thanks!

Electroless copper plating requires a copper salt solution, of course (https://en.wikipedia.org/wiki/Electroless_copper_plating), but there are many other considerations, like the reducing agent and surface preparation.
I think that storage of the homemade copper acetate solution is a minor issue in your project and a plastic container with an air-tight lid should be perfectly fine, without refrigeration, for a few years.
However, for the actual electroplating, your copper solution may not be adequately characterized. There may be excess acetic acid, other stuff from the vinegar (which could be perfectly safe to consume, but not for sensitive chemical reactions). Commercial copper salts are available inexpensively and with high purity, if the project you are working on has significant value.


Rearrangement of double bonds Consider:

I found these rearrangements, but I could not understand what the mechanism is behind this.

When I searched on the Internet I found the Wikipedia page Di-pi-methane rearrangement.

This could be due to hydride shift.

In the first image, the pi-bond shifts on the carbon from which $\ce{H}$ is eliminated. As the pi-bond leaves the first carbon, it becomes positively charged due to lose of electron, and the $\ce{H}$ gets converted to $\ce{H^-}$(hydride ion) as it accepts electron. Therefore, it is called hydride shift. It also stabilises the molecule as the resulting molecule is conjugated. The same can be applied for second and third images.

This is for the second image. The third one you can do it yourself. All the resulting molecules are stabilised due to conjugation.


Conversion of s-trans butadiene to s-cis butadiene No nodes should be introduced in the rotational interconversion between the s-trans and s-cis forms of 1,3-butadiene.

What is the explanation for that, if one of the $\pi$-systems should rotate $180^\circ$? It seems to be that two positive and negative lobes turn upside down.

During the interconversion of s-cis and s-trans 1,3-butadiene, the molecule passes through a geometry where the two ethylenic arms are rotated 90 degrees one from the other.  In this conformation the two ethylenic arms are orthogonal to each other and there is no overlap between them.  In an MO sense they can be treated as two separate ethylenes.  Rotation beyond this conformation allows overlap to reoccur and the phase of the HOMO will be correctly reestablished with only one node occurring between C2 and C3.


Why does the energy level of a particle in a box decrease with length? I want to know the intuition of why L is in the denominator of the equation for the energy of a particle in a box. This helps explain why resonance is stabilizing without using molecular orbitals. Does it have to do with the Uncertainty Principle? As the length increases, uncertainty of position increases, so uncertainty of momentum decreases, consequently meaning that the kinetic energy is lower, lowering the overall energy of the orbital?

In quantum mechanics, particles are waves, and as with any wave the longer the wavelength the less energy is in the wave.  You don't need the box to see this; infrared light has less energy in the photon than ultraviolet (the former makes you warm but the latter can damage your skin) and a slow moving free electron (lower energy) has a shorter associated wavelength than a fast-moving one (higher energy).  In the particle/box problem, the possible wavelengths are set by the dimensions of the box; a bigger box allows longer wavelengths -> lower energies.


How does an ice bath change the rate of an reaction? If I had hydrogen peroxide decompose with the help of manganese dioxide and measured the time taken for the reaction to complete and the volume of oxygen collected. I will get a graph that looks like this:

Now what will happen to the shape of the graph if I placed the reaction vessel in ice?

I thought the following. The graph will rise but at a less rate and less steep than in the graph above. Then it will get to the point where the reaction ceases and doesn't go to completion because there are no more collisions.

Placing the reaction vessel in ice will lower the temperature of the reaction and thus slow down the reaction. Temperature is a measure of the average kinetic energy of the molecules, so lowering the temperature lowers their kinetic energy. In order for reactants to form products, they must collide with a certain amount of energy. Thus, at a lower temperature it will take longer for the same amount of product to be formed.

You also need to keep in mind that at lower temperatures the equilibrium constant will change. Since this is an exothermic reaction, at lower temperatures there will be more product formed (Le Chatelier's principle). There will be more moles of oxygen present in the iced reaction. However, you're measuring the volume of oxygen, and a colder gas will have a lower volume (for the same number of moles of gas). So you've got two opposing trends: the ice will cause more oxygen to be produced (which would normally correlate with a larger volume) but the colder temperature will also cause the gas to shrink. Without more information, such as the temperature of the two reactions, I don't think you can determine whether you will wind up with a larger or smaller volume, so it would be hard to determine the shape of the graph for the colder reaction.


What is the difference between a time of flight mass spectrometer and a spectrometer that uses a magnetic field? Traditional mass spectrometers used electromagnets to deflect isotopes of different masses which allowed only a certain isotope to reach the detector. But how is the TOF MS different besides the fact that it doesn't use magnets? Does it have a different method of accelerating the ions which leads to them moving with a different velocity? If the methods of acceleration are same wouldn't there initially be no need for magnets to deflect the ions? Or is it the detector that works differently?

Both types of mass spectrometers separate different chemical species based on the mass-to-charge ratio.

The time-of-flight mass spectrometer (TOF MS) is based on the ions being released at a given instant, such as when struck by a laser, accelerating the ions electrostatically and measuring how long it take for them to strike the cathode target. This is ideal for a lightweight interplanetary or cometary mission such as Rosetta.

Magnetic and quadrupole MS work by bending the beam magnetically  or electrostatically so as to spread out the target beam along the width of the target. This is like sending a light-beam through a prism to spread out the spectrum. An advantage of this technique is that it can resolve multiple species simultaneously, but at the expense of fine resolution.

An excellent explanation of TOF and other MS is presented by Young, Waite and Miller.


How do I analyse this data? (optimization experiment) I conducted an experiment to find the optimal concentration for sugar in yeast respiration, and this is the data I found:

What do I do from here? Do I just fit a linear model on the increasing portion, another one on the decreasing portion, and see where they meet and say that's the optimal value? If not, do I try to fit another type of model? How do I justify that model?

Some materials (e.g. ionic salts) in certain solvents (e.g. water) actually do decrease volume of the solution in certain concentrations (obviously at some point it inevitably increases).

So you certainly can end up with non-linear functions of displacement from $\Delta_{conc}$.

That said, a negative quadratic may fit your data well, e.g. $ax^2 + bx + c$, with $a < 0$. Maybe even a damped exponential, e.g. $ae^{-bx}$. You can use Microsoft Excel etc. for a quadratic fit or fancy math skills for other functional forms.

Another Ben describes in more detail here but the image he used is worth repeating. Here the concentration of an added substance actually shrinks/increases volume displacement "oddly".


Thermal decomposition of group 2 permanganates Do group 2 permanganates produce manganates via thermal decomposition?

It is observed that true composition of alkaline earth metal manganate is not known. It is proposed to be a mixture of manganite-manganate and has a general formula, $\ce{R3Mn2O8.H2O}$. However the composition/formula varies according to given method of preparation. It is found on some instance that heating the permangantes of alkaline earth metal gives manganates while in other case it gives manganites.

Mangesium permanagate is prone to decomposition and it explode
violently in aqueous solution giving oxygen, ozone and the
manganate?(the reaction product is not defined properly?)
Calcium manganate is formed by decomposing calcium
permanganate.
Strontium permanaganate is also prone to decomposition and
forms either manganite or manganate depending on reaction condition
but either way, strontium manganate is formed by heating
strontium manganite at 1600 °C.
Barium permanganate also decomposes at 320 °C to form barium
manganite but if the permanganate is mixed with barium hydroxide,
barium manganate of composition $\ce{Ba3Mn2O8.H2O}$ is formed.
The reaction of beryllium permanganate decomposition is unknown to
the best of my knowledge mainly due to poor study on the compound.


Why is there a difference in the colours of K₂Cr₂O₇ and K₂CrO₄? In both the compounds the oxidation state of Chromium is +6 so why is there a difference in the colours of their aqueous solutions.
$\ce{K2Cr2O7}$ is red-yellow and $\ce{K2CrO4}$ is bright yellow
I and leaning towards the fact that due to the increased amount of bond pairs in $\ce{K2Cr2O7}$ it becomes harder to remove the d orbital electron and that causes the light emitted to be of a higher wavelength(Lower frequency) and that's why there is colour difference.
NOTE: Even though both dichromate and chromate have the same oxidation state for Cr (i.e. +6) they show different colours. My question is how and why?

I'm really excited for this because I get to reference the almighty color wheel!! Fair warning, this answer is much more qualitative than quantitative, but that's more interesting sometimes anyways.

If you look at the structure of the chromate and dichromate ions next to each other (see here for structures: https://en.wikipedia.org/wiki/Chromate_and_dichromate), the only major difference between the two is that the Cr-O bond joining the two chromate ions (missing an oxygen) is now a single bond. This means that bond will vibrate at a lower frequency, and because frequency and wavelength are inversely related, that bond will absorb a longer wavelength of light.

Now to the color wheel! It is a general chemistry (often unexplained) fact that the color we see is the complementary color of the wavelength of a bond's vibration. Thus, in the case of the chromate ion, we see yellow, and across from yellow is the purple-ish region. That means, if one of the bonds in the chromate ion, and thus two of the bonds in the dichromate ion, were absorbing a longer wavelength like we said earlier, on average we would expect something just longer than purple-ish, like blue, to be absorbed.

The complementary color of blue is red slash orange, and that is in fact the color we see in the dichromate ion!

At the heart of all this is the principle that the colors we see are those wavelengths of light which on average are not absorbed by a large number (on the order of Avogadro's number) of molecules.

An approach like this will only be reliable for very similar molecules like the two we have here.


What is the meaning of the "-osic" suffix on a metal (e.g., cobaltosic oxide)? I just came across the term cobaltosic for the first time in the abstract of Ind Eng Chem Res 52(18): 6076, 2013 (emphasis added):

  A nanocomposite of cobaltosic oxide and nitrogen-doped graphene ($\ce{Co3O4}$/N-G) was prepared by the facile hydrothermal method.

I assume this is a historical name; Wikipedia's article on the cobalt spinel mentions cobaltosic oxide as an "other name", but provides no other historical information.  Google searches on manganosic oxide and ferrosic oxide turn up the sort of results I would expect, though not all consistent:

Manganese

Analogous to cobalt, manganosic oxide appears to be the $\ce{Mn3O4}$ spinel; e.g., see the abstract here (PDF at link).

Iron

The term ferrosic appears to refer (at least) to a mixed-oxidation state hydroxide, $\ce{Fe3(OH)8}$; e.g., here.
However, it is also used to refer to the iron $\ce{Fe3O4}$ spinel (magnetite); e.g., here.

What is the meaning of this "-osic" suffix?

In the course of searching for examples of usage of the "-osic" suffix, I think I found the answer to my own question.  From the last link in the question, a web store selling pigments (emphasis added):

  The source of natural black oxide usually is from an iron ore called magnetite. Magnetite, also known as lodestone, is a ferrous ferric oxide (ferrosic oxide) that is a heavy, black color and opaque.

So, it would appear that the suffix indicates a compound of mixed oxidation state: a combination of -ous and -ic forms of the metal.


Nomenclature of polyatomic ions Is there a way to remember polyatomic ions (i.e $\ce{PO4^{3-}}$, $\ce{SO4^{2-}}$ etc.). For example, our professor will ask on an exam what the equation for iron (III) carbonate.  Obviously the (III) means iron has a 3+ oxidation state. However, if one was not too familiar with the number of oyxgens or charges on specific polyatomic ions, how would you know the symbol for carbonate ($\ce{CO3^{2-}}$)? She gives us a periodic table but I dont think that will help if you dont have carbonate memorized?

Initially the idea was that -ate suffix is reserved for a mononuclear acid with highest oxidation state, while -ite suffix was reserved for a mononuclear acid with second highest oxidation state. The per-...-ate construct was reserved for anion with 'extra' oxygen, typically a peroxocompound and hypo-...-ite construct was reserved for acid extra low on oxygen.

However, since for halogens $\ce{XO4^-}$ anions were obtained rather lately, perhalogenates has this formula and simple halognates are $\ce{XO3^-}$. Mn subgroup (Mn, Tc, Re) inherited this in the sense that highest oxidation state corresponds to per- ... -ate construct

Similarly, ortho- prefix was reserved for mononuclear acids with highest number of hydroxogroups possible, while meta- prefix was reserved for an acid with one water molecule less (typically, polymer/oligomer)

Putting it all together using phosphorous as an example

Perphosphate $\ce{PO5^{3-}}$
Ortophosphate - $\ce{PO4^{3-}}$
Metaphosphate - $\ce{(PO3^-)_n}$
Phosphite $\ce{HPO3^{2-}}$ (since $\ce{H3PO3}$ is actually $\ce{H2[HPO3]}$ for some arcane reasons) Note: depending on the chemistry course you are taking (less advanced courses mainly), some will use $\ce{PO3^{3-}}$ as phosphite
Hypophosphite - $\ce{H2PO2^-}$ (since $\ce{H3PO2}$ is actually $\ce{H[H2PO2]}$

And for chlorine (see above about perhalogenates)

Perchlorate $\ce{ClO4-}$
Chlorate $\ce{ClO3-}$
Chlorite $\ce{ClO2-}$
Hypochlorite $\ce{ClO-}$

Not every degree needs to be present, for example only simple carbonates and percarbonates exist, and sometimes extra stages (usually polynuclear) occur, such as dithionites. This system is widespread in areas heavy on traditions, such as medicine and heavy industry.

There is a gentle pressure from IUPAC to move to much less confusing systematic naming, but the success is limited, especially considering widespread chemicals, which are familiar anyway. It is similar to situation with RJ plugs: everyone recognises phone and Ethernet connectors, but very few care about their full formal name.


Enol content tautomers I have figured out that 4 one will have the highest enol content of all, and the 1 structure would come after 4 in enol content, i.e 4 > 1. Then I got confused between the 2nd and the 3rd structures.

Please give a brief explanation for the answer as well.

Enol content in 3 is greater then enol content of 2

Greater the acidic nature of alpha Hydrogen,greater is the tendency for Tautomeric behavior. H at alpha position in ketone is more acidic then H at alpha position in an ester.Pka values are shown below.The withdrawing power of ester is less compared to carbonyl group in ketone. Since lone pair on oxygen is in conjugation with C=O in ester,its withdrawing power decreases(as shown below).A more acidic α-hydrogen implies a weaker C−H bond. Summarizing,the enol content increases with the acidity of the enolic hydrogen.

$$\ce{carbonyl <=> enol}$$
$$\mathrm{K_{eq}=\frac{[enol]}{[carbonyl]}}$$

The equilibrium between keto and enol form is tipped to enol if H at alpha carbon is more acidic.

Hence enol content in 3 is greater then enol content of 2.

[]

  Pka of cyclohexanone is as per this link: https://www.drugbank.ca/drugs/DB02060

Enol of 4th compound is aromatic and should be most stable.

Therefore extent of enol content must be 4>1>3>2.
As a general guide line the following can be used for predicting enol content.

The enol form of cyclohexanone  has bond angles around double bond other then 120${^o}$ as shown below using avagadro.


Can bone mineral be obtained in pure, crystalline form? Can bone mineral be obtained in pure, crystalline form outside a living organism? Had this been done ever?

Apatite is the chemical name for the mineral found in bone and teeth. It is occasionally found in the form of large crystals, such as this:

BTW, just as bone is made of the mineral apatite, there are critters with skeletons of silica, another common mineral, such as Venus' flower basket!


Soap and rain water, why does it not rinse off? I am in Papua New Guinea at the moment, where rain water harvesting is the most common way of getting water.
When washing hands, the soap feels hard to rinse off, and my skin feels "greasy".

I read around a bit here and here and I think I understand what is said: essentially, sodium stearate in the soap separates into sodium and stearate ions, which then precipitate with other ions (like Calcium, into calcium stearate) and get rinsed away. But in very soft water, there are sodium ions rather than calcium ones, so this reaction is slowed down a lot, resulting in the greasy feeling.
As it turns out, I have access to hard ground water at the same time, and the difference is very obvious: with hard water, soap is rinsed off instantly.

My question now: I assume rain water has in fact almost no ions in it (I'm thinking Sodium ions mostly), and I would therefore assume that the Sodium stearate dissolves very easily, but has no replacement ion to precipitate with, therefore sticks to the skin.

Is this the right explanation or are there other factors at play in the explanation about that greasy feeling?

Yes, this is a good explanation. Rain water is almost pure water, it lacks of bivalent ions such as $\ce{Ca^{2+}}$ and $\ce{Mg^{2+}}$ which helps soap to rinse of. You will get the same effect when a water softener is installed on water distribution.


Acidic cleavage of ether with a 2° alkyl group I know that an ether with a 3° alkyl group and a 1° alkyl group is cleaved by $\mathrm{S_N1}$ mechanism to give a tertiary alkyl halide and a 1° alcohol.

And an ether with a two 1° alkyl groups is cleaved by $\mathrm{S_N2}$ mechanism to give a 1° alkyl halide and a 1° alcohol.

But what happens if a ether with a 2° alkyl group and a 1° alkyl group is cleaved with a halogen acid?

In each of these cases, the ether will generally be cleaved to produce 2 alkyl halides. Acidic clevage can occur via Sn2 or Sn1 depending on the nature of the substrate. A bulky group, such as tert-butyl, will undergo an Sn1 mechanism in which the protonated ether group leaves before the nucleophile attacks. The same process is generally consistant with secondary substrates. I have drawn a mechanism involving both a primary and tertiary substrate to demonstrate this point; however, it is important to note that in the presence of most concentrated halogen acids and heat, ethers will almost always undergo complete acidic clevage in which 2 alkyl halides are generated. I have also included another mechanism involving a secondary and primary substrate. The primary substrate will undergo an Sn2 reaction due to the lack of steric hindrance; however, it is a little less clear with the secondary substrate. Will it undergo Sn1 since Sn2 would be more sterically hindered or will it undergo  Sn2 since the carbocation intermediate in Sn1 is secondary and thus less stable? Since we are using an aqueous halogen acid in this case, it will most likely undergo Sn1 since it is in the presence of a polar protic solvent, which will thus stabilize the carbocation intermediate in the process of forming.


Why do we excite electrons before hybridisation? In case of hybridisation like $\mathrm{sp^3}$ hybridisation of carbon or nitrogen why do we excite electrons before hybridisation? Why couldn't we just mix orbitals without exciting electrons?

Actually number of bonds formed by an atom is generally equal to number of unpaired electrons it has. Like oxygen and nitrogen make two and three bonds because they have these many unpaired electrons in their valence shell.
But, in its ground state, carbon got only two unpaired electrons, so it should make two bonds. But in reality carbon make four bonds. Two satisfy four bonds we need four unpaired electrons. This is possible to consider the excited state.
Hope its clear.


Why does AlCl3 dimerise to Al2Cl6 at low temperatures? I know that it has something to do with HOMOs and LUMOs (I can't what the HOMO/LUMO for AlCl3 is, though).

Perhaps, we can employ the Gibbs free energy equation for spontaneous reactions...

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

Since the dimerisation process involves the formation of bonds, $\Delta H$ should be negative, while $\Delta S$ will be negative too, since two $\ce{AlCl3}$ molecules become one $\ce{Al2Cl6}$. As such, when temperature is low, the magnitude of $T\cdot\Delta S$ will be smaller, and $\Delta G$ will be negative or more negative, thus favouring the dimerisation process.


How much "cleaner" does summer fuel burn? Well, it's that time of year again: the switchover to "summer" fuel and resulting price increases. This EPA regulation system requires gasoline producers to blend additives into the gasoline that reputedly make it burn cleaner.

What exactly does it mean "burn cleaner" and how much "cleaner" is summer fuel than regular gasoline?

Gasoline, diesel fuel, and ethanol-gasoline blends are adjusted seasonally and geographically to ensure proper starting and performance. Winter blends contain higher levels of high vapor pressure compounds like butane to ensure the fuel is sufficiently volatile to ignite in cold temperatures. This in turn makes the gasoline cheaper to produce as it does not have to be as pure as the summer blend$^{[1]}$.

Conversely, summer blends contain lower levels of high vapor pressure liquids to ensure that the fuel doesn't evaporate away in the heat and pollute the atmosphere. This is why summer blends are marginally cleaner than winter blends. A study on winter versus summer emissions near tunnels form the University of Wisconsin collected the following data on non-methane hydrocarbon emissions$^{[2]}$:

As for fuel efficiency, winter-grade gasoline yields $1.7$% less energy on average than the summer. This may seem like a considerable amount, but energy content can vary widely from station to station in either season from $3.4$-$4.8$%$^{[3]}$.

$^{[1]}$ The Oil Drum
$^{[2]}$Univeristy of Wisconsin, Summer and winter nonmethane hydrocarbon emissions from on-road motor vehicles in the Midwestern United States
$^{[3]}$ EPA, Fuel Economy Impact Analysis of RFG


Why might a planar, conjugated molecule be higher in energy than a nonplanar system? I have a small molecule (23 atoms) for which I want to verify parameters in a molecular mechanics force field.

My first step is to look at the potential energy as a function of the dihedral angle, defined by these four atoms:

The results for the dihedral scan, rotating the guanidino tail from 0 to 360 degrees in increments of 5, looks like this:

Alright, something is funky around the 0 and 180 degree structures. Honing in on 180, and performing a QM scan with increments of every 0.1 degrees, I see this:

There is a gap between 177.1-178.2 and between 181.5-182.6, where the QM optimization failed. Checking the plateau and either side of the plateau, what I find is that the plateau structures are all planar, whereas on either side of the plateau, some hydrogens come out of the plane (consistent on both sides of the plateau). For example, this is an overlay of two structures: blue = 178.2° = planar = higher energy. Red = 177.1° = nonplanar Hs = lower in energy. (front view and bottom side view)

Let's assume that the QM calculations are to be trusted (this was run using a few different levels of theory). A mere 1° causes this shift between planarity and nonplanarity. Intuitively, if we think about the ring system being the same in both cases, this makes no sense, as a planar molecule should be better conjugated and thus lower in energy, correct? What other factors might account for this phenomenon?

I reproduced your calculations with PBE/def2-SVP (using Orca). The energy "jump" is even smaller than with MP2, however the causes are probably the same. So here is my answer:

The planar structure is a Transition State, with one negative frequency.
The dihedral you are scanning is probably not the best to describe this twisting.

Now in more detail. Here you can see the Minimum Energy Path (calculated with pDynamo using the NEB method) between the two symmetrically related minima.
 You can see that, as expected, the path is smooth.

The main change between the structures is the pyramidalisation of the terminal NH$_2$. This can be described by the N-C-N-H dihedral, which also changes smoothly as you can see here:

I don't think the reason why the terminal NH$_2$ is not planar is steric problems as @Ivan Neretin suggests. I think its more a question of nitrogen hybridisation, which is between sp$^2$ and sp$^3$. It is the same reason why aniline is not planar. This is discussed in these two questions:

Is nitrogen of aniline sp² or sp³ hybridized?
How to rationalise the resonance structures and hybridisation of the nitrogen in a conjugated amine?

Hope this helps!


What actually happens when 'energy is released'? Thermodynamics classes and equations always discuss a chemical reaction in the form of energy being released or absorbed. This is always hand-waved as just 'energy' from the surroundings.

What's actually happening at a molecular level? When a bond is broken and 'energy is absorbed' to accomplish this, is the kinetic motion of the resulting species lower than the total kinetic motion of the species before it broke? Does something else happen?

When 'energy is released' on a bond being formed, does the new species then have a greater kinetic motion than the combination of the two species alone? Does this energy somehow translate to a change in energy level of an electron and a release of a photon at the subsequent decay?

Thanks in advance.

Thermodynamics works perfectly well by assuming that matter does not need to be made of discrete particles, i.e atoms and molecule are not necessary. It was developed before the structure of atoms and molecules  were known so it cannot inform us about anything at a molecular level. It is concerned primarily about energy and entropy.

If you break a bond in a diatomic molecule then all that can happen is that the two atoms move away from one another with the excess energy used to break the bond  shared between the atoms in the form of kinetic energy. The 'excess energy' is the energy above the dissociation energy. The atoms will be 'hot' and if in a gas (or solvent) of inert molecules will collide with these and their energy will be spread out among all the molecules raising the temperature by just a little bit.

In a polyatomic molecule if the products are lower in energy than reactants the excess energy  will raise the temperature of the surrounding solvent or gas. To break a bond the activation energy has to be overcome, once this happens the product has more potential energy than the reactant, i.e. more vibrational and rotational energy levels are populated than in the reactant, see figure. This is because the potential well of the product is lower than that of the reactant. At the top of the activation barrier is the transition state.

Once in the product potential the extra energy is lost quickly, a few picoseconds, to any solvent surrounding the molecule. Energy is also lost to other vibrational /rotational energy levels in the product molecule. These are vibrations not involved in the bond breaking.  Eventually all this energy is spread out in the surrounding solvent (or gas) raising its temperature. The kinetic energy change in a reaction may not be that large if the energy is taken up by the vibrations and rotations. This is unlike the case when an atom is formed as a product.

The figure shows how the product has more vibrational energy than  the reactant after surmounting the barrier and transition state (blue dot) which is the energy and coordinates at which the molecule changes from reactant to product. The energy difference, leading to heating, is that between the two lowest energy levels in each potential.


Structures for NO+ (Nitrosonium) There are two possible structures for Nitrosonium :

In the first structure, there is a +1 Formal charge on Oxygen, whereas,
In the second structure, there is a +1 Formal charge on Nitrogen.

Q. Which is a better lewis structure?

Q. Are they resonating structures of NO+?

Wikipedia gives the first structure only, so i think that is the only correct structure.

I think structure in which less electronegative atom has a positive formal charge is better, so the first structure is better.

Q. Is the above reasoning correct?

Which is a better Lewis structure?

There is no such thing as a better or a worse Lewis structure. They should be properly referred to as major or minor resonance contributors/forms.

I looked up a textbook to check the rules of determining which resonance structure contributes more. The order is:

Structures that satisfy the octet rule are greater contributors than structures which do not satisfy the octet rule.
As long as the octet rule is not exceeded for second-row elements, the contributing structure with the greater number of covalent bonds contributes more.
When two or more structures satisfy the octet rule with the same number of covalent bonds, the major contributor is the one with the smallest separation of oppositely charged atoms.
Among structural formulas that satisfy the octet rule with the same number of covalent bonds, and in which one or more atoms bear a formal charge, the major contributor is the one in which the negative charge resides on the most electronegative atom.

In the first structure, all atoms possess a complete octet. But in the second structure, nitrogen only has six electrons around it: consequently, by rule 1, the first resonance structure is the major contributor.

  Are they resonating structures of $\ce{NO+}$?

Not exactly.

The actual structure of the nitrosonium cation is somewhere in between the two forms. This means that the N–O bond is somewhere between a triple bond and a double bond. However, since we have established that the first form contributes more to the actual structure, it is going to be closer to a triple bond than a double bond. You could think of it as a 2.9-bond. Molecular orbital theory tells us that the actual bond order in $\ce{NO+}$ is 3, but that's an entirely separate story for another day.

The cation does not resonate between two distinct structures. Let me cite another example. Benzene also has two different resonance structures, both of which contribute equally (since they are identical to each other). Does that mean that the C-C bonds flip between being a single bond and being a double bond? No. All the C-C bonds in benzene are 1.5-bonds, at all times. The common usage of alternating single and double bonds (the Kekule structure) is just us being lazy since many mechanisms are easier to draw that way. Similarly, the $\ce{NO+}$ cation does not flip between the two resonance structures, which represent two extreme ends of a spectrum of possible structures. The actual structure is always in the middle of the spectrum. Which side the actual structure leans towards, is determined by which resonance form contributes more.

At risk of sounding like a broken record - The actual structure is somewhere in between the two resonance forms. It does not oscillate between two different resonance forms. This is a very important concept to understand!

  I think the structure in which the less electronegative atom has a positive formal charge is better, so the first structure is better. Is the above reasoning correct?

Yes and no. There are several methods of measuring electronegativity, and in all of them, nitrogen is less electronegative than oxygen. By your reasoning, this means it would be more favourable to place the positive formal charge on nitrogen, i.e. resonance form 2. So your reasoning is a little bit off there.

On the other hand, you would be correct if there wasn't a difference in the number of bonds between the two resonance forms. As I mentioned at the top of the post, the number of bonds takes precedence over the electronegativity issue.


Interpret Radiodecay Symbolism $\binom{16}{8}$O (n,$\alpha$)$\binom{13}{6}$C  and $\binom{9}{4}$Be ($\alpha$,n)$\binom{12}{6}$C

I had initially thought the products & reactants may be O-16 $\to$ C-13 + He-2 + p$^+$ from alpha decay, but I'm leaning towards this form O-16 + C-13 $\to$ ... now. Any help clarifying & determining products would be much appreciated.


How can endothermic reactions be spontaneous? If endothermic reactions absorb heat from the surroundings, how can they be spontaneous, if heat only flows from high temp to low temp objects?

A spontaneous endothermic reaction can occur when the changes in enthalpy and entropy yield a negative Gibbs free energy. An endothermic reaction can be spontaneous if entropy increases by more than the change in enthalpy.

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

For endothermic r̥eactions, $\Delta H$> 0 and thus if the value of $-T\Delta S$ overshadows the $\Delta H$ value (usually at very high temperatures) then the overall value of $\Delta G$ becomes negative and thus the reaction becomes spontaneous.


Does sorbitol dissolve in ethanol? I know Sorbitol, $\ce{C_6H_14O_6}$ $$\ce {C(OH)H_2 (C(OH)H)_4C(OH)H_2}$$is soluble in water however I did not have much luck finding out whether or not it is soluble in ethanol. My guess is that it isn't however I am not so sure.

Is Sorbitol soluble in ethanol?

Sorbitol's solubility in ethanol seems to be very temperature dependent according to the Merck Index$^{[1]}$

  Freely sol in water (up to 83%); sol in methanol, isopropanol, butanol, cyclohexanol, phenol, acetone, acetic acid, dimethylformamide, pyridine, acetamide solutions; practically insoluble in most other org solvents; quite soluble in hot alcohol, sparingly soluble in cold alcohol

1) O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Whitehouse Station, NJ: Merck and Co., Inc., 2006., p. 1498


How does the temperature of water affect the reactivity of alkali metals acting on it? There are a few experiments comparing the reactivity of alkali metals with cold water. Is there any purpose to use cold water? Does the temperature of water affect the reactivity of alkali metals?

Increasing temperature does indeed increase the reactivity of alkali metals, or most things for that matter. The rate of reaction increases because the average kinetic energy of the molecules increases, and so more collisions can overcome the activation energy for the reaction to take place.

The reason for using cold water rather than hot water for the reactions of alkali metals is that alkali metals are already reactive enough at room temperature. Even fairly small lumps of potassium will burn if you put them in cold water so imagine what it would be like with hot water. This is not something you want to try without very careful precautions.


Why does gas not liquify at a temperature above the critical temperature no matter how much pressure is applied on it? Why? My textbook says that critical temperature is the temperature above which a gas cannot be liquified no matter how much pressure we apply on it. But why? What is so special about this 'critical temperature'? I mean, if we apply a ridiculous amount of pressure, the gas molecules have to come closer to one another and they have to liquefy. What exactly happens to the gas molecules at critical temperature? Do they become immovable and have infinite inertia or what?

The critical point is a point of convergence of all state properties of the respective liquid and gas. It can be considered as the degeneration point, where there is no difference between gas and liquid and this distinguishing does not make sense any more.
It can be also said the supercritical fluid near the critical point is neither gas neither liquid. It is both at the same time. Farther from critical point, like for permanent gases, either gas-like either liquid-like properties are dominant, depending of the state.
By changing  pressure of supercritical fluid, it gradually transforms between  states, where it behaves more like gas or liquid, without evaporation or condensation. It is similar, as if you reach an elevated level via a big step (liquid/gas or evaporation/condensation) versus  along a slope (supercritical fluid). For the latter, there is no upstairs (gas) nor downstairs(liquid), there is just the slope.
Imagine you figurally walk as the substance on the phase diagram ($T$ grows to the right, $p$ grows upwards) around the critical point. If you do it counter-clockwise, you will evaporate as many times as many rounds you do, without ever condensing. If you do so clockwise, you will condensate as many times as many rounds you do, without ever evaporating.


Why oxidation number of oxygen in OF2(oxygen difluoride) is +2? I mean oxygen is electronegative element it should have negative oxidation number
Also how it is able to make bond with fluorine which itself is electronegative?
 Why the oxidation number of oxygen is +2.?

Oxygen is an electronegative element -True.

But oxygen is not the most electronegative element. It has an electro negativity of 3.5 on Pauling scale. That means if oxygen combines with an element which is more electronegative than it will surely possess a positive oxidation state.

Fluorine being the most electronegative element (electro negativity of 4.0 on Pauling scale) will in any case (except in fluorine gas) have an oxidation state of -1. So that's the reason why oxygen has a +2 oxidation state in OF2.

Dioxygen difluoride (O2F2) is another fluoride of oxygen in which oxygen has an oxidation state of +1.

EDIT: Fluorine in its gaseous form (F2) has zero oxidation state.


Vapor pressure lowering I have got this problem in my text book,

  Consider these three
  mercury manometers. One of them
  has 1 mL of water placed on top of the
  mercury, another has 1 mL of a 1 m
  urea solution placed on top of the
  mercury, and the third one has 1 mL of
  a 1 m NaCl solution placed on top of
  the mercury. Identify X, Y, and Z with
  these solutions.

I got the answer in answer sheet is X = ${H_2}O$ , Y = $NaCl$  and Z = ${(NH_2)}_2{CO}$ . Can any one please explain me? Thanks in advance.

The answer hings on the colligative properties of solutes. In essence solutes lower the vapor pressure of the solvent. NaCl splits into two ions (i.e. two solute "particles") and lowers the vapor pressure of water more than urea for which the whole molecule dissolves as one solute "particle."

If there was no liquid the Hg level would be higher than any of the three tubes shown since there would be a pure vacuum above the Hg.

The scale of the relative differences in drawing is obviously greatly exaggerated.


Is the Standard hydrogen electrode potential at T other than 298 K also zero? The standard hydrogen electrode potential by conventional at 298 K is taken to be 0.00 volts.

This is what I have been taught. It talks about SHE at 298 K, so is the hydrogen electrode potential at a temperature other than 298 kelvin also defined to be zero ? If not, how do we calculate at any other temperature?

According to the definition used by IUPAC, the standard electrode potential $E^\circ$ of the standard hydrogen electrode is zero at all temperatures.

  For solutions in protic solvents, the universal reference electrode for which, under standard conditions, the standard electrode potential ($\ce{H+}/\ce{H2}$) is zero at all temperatures.

The absolute electrode potential $E^\circ(\mathrm{abs})$, however, depends on temperature. Its value can be calculated from thermodynamic quantities (e.g. $\Delta G^\circ$). The recommended value for $T=298.15\ \mathrm K$ is $E^\circ(\mathrm{abs})=4.44(2)\ \mathrm V$.


Addition to equally substituted carbons I was wondering whether there is a trick in the following problem:

Since both carbons are equally substituted, shouldn't the products be the same with or without the use
of peroxides?

The use of HBr with peroxide leads to an anti-Markonikov's product(known as the peroxide effect).The carbons are both substituted once, but not equally! One has a methyl group while the other side has a phenyl group. The chemical basis for the Markovnikov's  rule is the formation of a a stable carbocation during the addition process.

So, with just HBr a nomal Markovnikov's addition takes place i.e the Br group gets attached adjacent to the phenyl group while the H on the next carbon(as a benzylic carbocation is more stable).Final product: Ph-CHBr-CH2-CH3.
With the peroxide however , the opposite happens leading to the formation if Ph-CH2-CHBr-CH3.


Number of orbitals in full configuration interaction I am using one quantum chemistry package where I am supposed to assign a number of active orbitals I want to be included in the FCI (full configuration interaction) calculation. For example if I have  ${H_2}$ molecule (singlet) and I use $cc-pV6Z$ basis set of Dunning, what would be the maximum number of active orbitals in the FCI? Can somebody provide me insight into this problem?

In simple terms, Full CI uses all of the possible orbitals (a "full" active space). Thus the number of active orbitals should equal the size of your basis set. If you have 10 basis functions, and then perform a Hartree-Fock calculation, you get 10 orbitals out (some occupied, some virtual).

Complete Active Space (CAS) methods, on the other hand, only use some of the orbitals and then perform a Full CI on that subset of orbitals, leaving the other orbitals (inactive space) alone.

Full CI is CAS in the total orbital / total active space limit.


Hydrolysis of (chloromethyl)cyclopropane: Is but-3-en-1-ol a possible product? This is from one of the practice papers for the JEE , asks for possible hydrolysis products of (chloromethyl)cyclopropane .

Product A happens to be due to stability of cyclopropyl methyl carbocation.
Product B happens to be due to relaxation of ring strain of cyclopropane ring (though A is major even this is possible)
Pathway for C doesn't seem to be anywhere in my books , I tried googling it but found nothing.
So, is but-3-en-1-ol a possible product of hydrolysis of (chloromethyl)cyclopropane?

I'd say, yes it is possible. The pioneering work on interconversion reactions of cyclobutyl, cyclopropylcarbinyl, and allylcarbinyl derivatives has been done by John D. Roberts and coworkers (Ref.1). In these works, they have checked reactions under both $\mathrm{S_N1}$ and $\mathrm{S_N2}$ conditions. They have observed the reactions expected to involve carbonium ion intermediates have shown interconversion, which have been correlated to factors such as carbonium ion stability, ease of interconversion, and stability of possible products. One such interconversion have been shown in the reaction of cyclopropylcarbinyl-,  cyclobutyl-, and allylcarbinylamines with nitrous acid, which gave relevant carbonium ions. Similar approach was recently taken by Arca, et al. 2014 using solid catalysts (Ref.2). They have proposed a bicyclobutonium cation transition state to explain the product formation:

The coorelation to liquid phase reaction is well explanied in their abstract:

Rearrangement of cyclopropylcarbinyl bromide over proton and ammonium-exchanged mordenite ($\ce{H}$-MOR and $\ce{NH4}$-MOR) was studied at different temperatures. The product distribution analysis revealed a similar profile to what is found in solution, with prevalence of cyclobutyl over allylcarbinyl bromide. The activation parameters showed that compared with 80% aqueous ethanol solution, the zeolites present lower enthalpy of activation and higher entropy of activation, but yielding a significantly lower $\Delta G^‡_{298}$. The results may be explained in terms of the ability of zeolites to serve as solid solvents, providing a polar nanoenvironment for ionic reactions to occur.

Therefore, we can assume the reaction in hand have also unedergone with similar mechanism, giving same three products, $\bf{1}$, $\bf{2}$, and $\bf{3}$ (probably in different ratios). The formation of  $\bf{3}$ can also explained by the possible resonance contribution as depicted in above $\color{red}{\text{Red Box}}$. Note that two open resonance structures are essentially the same, but shown because it would make a double contribution due to the symmetry.
It's noteworthy that the open-chain derivatives have been achieved from the reaction of cyclopropylcarbinyl halides with amines (Ref.3).
References:

John D. Roberts, Robert H. Mazur, “Small-Ring Compounds. IV. Interconversion Reactions of Cyclobutyl, Cyclopropylcarbinyl and Allylcarbinyl Derivatives,” J. Am. Chem. Soc. 1951, 73(6), 2509–2520 (https://doi.org/10.1021/ja01150a029).
Hugo A. Arca, Gabriel C. C. Gomesab, Claudio J. A. Mota, “Solid solvents: activation parameters for the rearrangement of cyclopropylcarbinyl bromide on mordenite zeolite,” New Journal of Chemistry 2014, 38(7), 2760-2762 (https://doi.org/10.1039/C4NJ00287C).
Michael B. Smith, Robert T. Hrubiec, Charles A. Zezza, “Reaction of amines with cyclopropylcarbinyl halides: $\mathrm{S_N2'}$ or solvolysis?,” J. Org. Chem. 1985, 50(24), 4815–4821 (https://doi.org/10.1021/jo00224a033).


Contactless non-ferrous metals hardening? Is it possible to work-harden non-ferrous metals without direct contact?
I know induction hardening is used on ferrous metals, but would like to know if a technique exists for non-ferrous metals.

From this link I could find a few non-ferrous systems which show quench-induced or stress-induced martensite . Generally non-ferrous systems dont show this behaviour and they are hardened/strengthened by Precipitation hardening, work hardening, Solid Solution Strengthening.

Also work hardening is basically working/deforming the material plastically to strengthen it. So you can't work harden a material without contact. Induction hardening is used mainly for Case Hardening which is different from hardening(done to bulk of the material). So as long as the non-ferrous alloy is conductive and resistive , Induction hardening is enough for case-hardening.

And if it is hardening , the above mentioned methods are used for non-ferrous. Only the quenching/martensitic transformation is limited to limited systems.


What is the lowest temperature tungsten starts sublimating at? What is the lowest temperature tungsten starts sublimating at? I've tried to search in Google but without any luck.

For example, $\ce{WO3}$ starts sublimating roughly at $\pu{550 °C}$, according to the paper Growth of epitaxial tungsten oxide nanorods.

Well, you see, searching online for "vapor pressure at such-and-such temperature" (or conversely, "temperature to get such-and-such pressure") is often fruitless, and unnecessary too, because pretty decent approximations are known. Here is one:
$$\ln\mathfrak p=-{\Delta H\over R}\left({1\over T}-{1\over T_0}\right)$$
where $\mathfrak p$ is your vapor pressure in atmospheres, $\Delta H$ is the enthalpy of vaporization and $T_0$ is the boiling temperature at normal pressure. For tungsten, the latter two are reported as 774 kJ/mol and 6203 K, respectively. Now you may plug in your desired pressure and solve for temperature.

That's a relatively crude approximation, for the enthalpy will hardly remain constant over such a wide range of temperatures. This page may contain better approximations designed for specific ranges.


Is there a way to generate SMILES from ChemDraw files? I have a set of ChemDraw .cdx files, and I would like to extract smiles strings from it.
Can Open Babel do that or maybe the rdkit in Python? Any other suggestions?

Of course openbabel can do this (at least with release 3.1.0 from here).  To rewrite a file here and then, the GUI on top of the program on top may suffice for you.  However, you would skip a lot of the program's functionality if not trying to interact from the command line: for one, the conversion basically follows the pattern of
obabel -icdx input.cdx -osmi -O output.smi

which actually may be faster than clicking in an interface. For the other, that openbabel offers multiple interfaces to automate file transformations programmatically, i.e. to interact with programming languages like Python and C.  See the current  documentation rather than the a bit outdated on the project's senior page.
Note 1: ChemDraw offers a structure to SMILES string conversion, too.  Be aware that a .cdx file may contain much more information, than a SMILES string may retain.  Seeking for a text-only representation of the molecular structure, axial chirality (P/M) is an example where a SMILES string is less suitable than, for example, an InChI string.
Note 2: Openbabel aims to interconvert chemical formats; on occasion with a visual output like .svg, or .png.  RDKit has a stronger emphasis on computing with the molecules (a.k.a. cheminformatics) as required while interacting e.g., with large sets of data and databases.  There is some overlap of the two programs, and «the trick» is rather to recognize where the two are complementary to each other.


How to reconcile conflict between assumed oxidation state and the most stable multiplicity from electronic structure calculations? Consider either of the small, neutral molecules containing two Zn atoms below. If I were to determine the oxidation state of Zn from this molecule, I would formally assume each Zn to be Zn(I) in order to balance the overall charge (if I give 2- for each O, 3+ for the Al, and 1+ for the H). I also assume that each Zn(I) has 1 unpaired electron as a result of this oxidation state. With this, I would assume that the molecule may have a triplet ground-state (putting aside the possibility of any broken-symmetry behavior for now).

The left and right images are the result of density functional theory geometry optimizations where I set the multiplicity to be 1 and 3, respectively. I find the (closed-shell) singlet structure is lower in energy than the triplet by 28.7 kcal/mol, in contrast with the prior discussion.

How can I reconcile these two conflicting notions? If I were to study reactions involving this molecule, do I assume the structure is a triplet because that's what the 1+ oxidation state on the Zn atoms would likely yield? Alternatively, do I assume that the structure is a singlet because that is what I calculated, even if that does not seem to agree with the oxidation state assumption?

If the appropriate approach is the former, then perhaps the oxidation states of each atom is not what one would formally assign. If the appropriate approach is the latter, then perhaps since Zn does not tend to be in the 1+ oxidation state, there is some unusual change in the electron configuration to stabilize this structure.

Computational details: this is computed using the M06-L functional, I have checked the stability of the wavefunctions, there is minimal spin contamination, both structures are local minima, and the trend holds true for other model chemistries.

Singlet state:   Triplet state:

First reflex: As long as the calculations are otherwise reliable, I don't think you can ignore the possibility of a broken-symmetry/multiconfigurational solution. If such a solution ends up even lower in energy than your closed-shell singlet, I think you have your answer right there.

Alternatively, if the closed-shell singlet ends up exhibiting a lower energy than a BS solution, perhaps this is some sort of cationic Zn analogue of a peroxo species? If this were the case, then the $+1$ oxidation state wouldn't be nearly so surprising.  As you note, the non-$\ce{Zn}$ portion of the system would be $\ce{[AlO3H]^{2-}}$, and thus the species as a whole might be better thought of as $\ce{[Zn2][AlO3H]}$. I have no feel for how realistic this hypothesis is, though Wikipedia notes that organozinc(I) compounds have been prepared, with decamethyldizincocene being the first such compound reported (public domain image; click to enlarge):


Extracting Zinc (oxide) from baby powder I have a bottle of baby powder that is 15% Zinc oxide and the rest is talc. Is there an over-the-counter way of purifying zinc oxide from it? I'm trying to make some sunscreen from it.

This might be not worth it, because you will need special equipment.

Description totally theoretical.

Mixture consist of $Mg_3Si_4O_{10}(OH)_2$ - tale powder, $ZnO$ - zink oxide powder, and starch. 81% to 15% to 4% it's most common variation with 15% zink oxide, there also 10% mixtures, but process will be the same.

You need to get rid of starch. Best choice to convert it to dextrine, because starch in not fully soluble in water. You need to heat it in oven on a foil for about 2 hours, with temperature $200^{\circ}C$. It will have no effect for other components, zync oxide might colorize to orange, but when heat is gone it will be white again. Dextrine will be yellow color and it's fully soluble in water.
Mix new mixture with water, and pour it out. Now you have only $Mg_3Si_4O_{10}(OH)_2$ and $ZnO$ in you mixture. Both of them not soluble in water.
Then you need to obtain or build equipment for sublimation - sublimation apparatus. Because compounds have different melting point Talc - $800^{\circ}C$, zync oxide - $1975^{\circ}C$, you can get them one by one. You will need to freeze and dry your mixture first.
Heat over $800^{\circ}C$ to obtain talc, left over structure will be zinc oxide.

I'm not sure about sublimation from start, with starch inside, but this method in theory should work.


What is the mechanism for the reaction of acetyl chloride and aniline? I want to devise the synthesis of 1,2-dinitrobenzene without producing a large amount of 1,3 or 1,4 products.

In my synthesis I have a step in which I obtain aniline. I would like to know the mechanism for protecting the amine using $\ce{CH3COCl}$ and triethylamine so I can keep it from reacting in the next step of my synthesis. I'm not really "fluent" with amine chemistry, and I have no good idea at the moment.

If you cannot make picture of mechanism, please explain it with words. I must understand!

The mechanism is just a regular nucleophilic attack of an amine on a carbonyl, with the triethylamine just there to pick up any loose protons at the end of the reaction.


Correct Electron Configuration of Ni-28 I have seen two electron configurations for $\ce{_{28}Ni}$ in my chemistry textbook:
$$[\ce{Ar}]~3d^8~4s^2$$
$$[\ce{Ar}]~3d^9~4s^1$$
Which of those is ACTUALLY the correct one (in real life, reality)? Why exactly?

NIST suggests here and here (search for Ni) that $\mathrm{[Ar] (3d)^8 (4s)^2}$, specifically in the state $\mathrm{^3 F_4}$, is the fundamental electronic state for neutral nickel atoms in empty space. However, as Wikipedia describes, if you don't specify the term symbol for the configuration, then the average energy of all terms for the $\mathrm{[Ar] (3d)^9 (4s)^1}$ configuration is slightly lower than the average energy of all terms in the $\mathrm{[Ar] (3d)^8 (4s)^2}$ configuration.

Now, why one configuration has a lower energy than another is a much harder question to answer in close cases such as this, as it involves a precise quantitative analysis of electronic energies for every possible electronic configuration of a nickel atom (a multielectronic atom). The general idea is that even though, for $Z=28$, the configuration $\mathrm{[Ar] (3d)^8 (4s)^2}$ is higher in energy than $\mathrm{[Ar] (3d)^9 (4s)^1}$ if you look at the sum of occupied orbital energies, the latter case has more interelectronic repulsions, which causes its total energy to be slightly higher. A lot of information on this can be found in this previous Chem.SE question and the several nested questions within.

Interestingly, the gas-phase fundamental configuration of the monocation $\ce{Ni^+}$ is $\mathrm{[Ar] (3d)^9}$, which further shows how nickel's electronic configuration is teetering on the boundary between the options.


Why is −78 °C a magic temperature for organic reactions? In many organic reactions that I have seen, running the reaction at $\mathrm{-78\ ^\circ C}$ seems to be quite a popular choice, but I've never seen any explanation for this. What makes this temperature such a popular choice?

Dry ice (solid carbon dioxide) sublimes at −78 °C. Dry ice and acetone are a common cold bath for chemical reactions.  The melting point of acetone is -95 °C so the bath never gets cold enough to freeze the acetone. The bubbling of the carbon dioxide gas as the dry ice sublimes keeps the cold bath well stirred.

Typically, though, the temperature in the flask with an ongoing reaction is at least about 5 °C higher than the one in the cooling bath.  (Except if using a thermocouple, working at a scale of 10...25 mL and less often implies that the thermometer is in cooling bath, and not in the reaction mixture. Thus, this temperature difference isn't recorded.)


Are electron orbital orientations filled in any particular order? This question is about the magnetic quantum numbers and their corresponding orbital orientations, e.g., px, dyz, etc. Are the electrons in any given subshell distributed across the orbitals in any particular order as the number of electrons increases?  Such as m = -1 first, then m = 1, etc.?

No, you cannot say that there is a particular order in which the shells are filled.
Relation between $\mathrm{p_x,p_y,p_z}$ with $\mathbb{m_l}$ values
The magentic quantum number ${m_l}$ represents the projection of the angular momentum $\vec{l}$ of an electron in an orbital, on an axis, usually taken to be the z-axis. The orbitals, and their $\mathrm{m_l}$ values, come from the solution of the Schroedinger's equation for a Hydrogen atom.
Now, the problem is, the solution for the orbitals with ${l=1}$ (i.e. p orbitals) produces one real wavefunction (corresponding to $m_l=0$), and two other complex wavefunctions (corresponding to $m_l=1$ and $m_l=-1$). If the main axis is taken to be z, then the $m_l=0$ orbital has the usual dumb-bell type shape and points in the z-direction, so it is the $\mathrm{p_z}$ orbital. However, the complex orbitals are conjugates of each other, and if you plot the real part, they both look like donuts. (You can find more on this here and here). So the other two orbitals, let's call them $\mathrm{p_1}$ and $\mathrm{p_{-1}}$, do not really correspond to anything we usually see in the chemistry textbooks.
However, wavefunctions themselves don't hold any physical meaning, because we cannot observe them. We can only observe things like electron energy, electron density etc. So, we can take linear combinations of the two complex orbitals to get two new real p orbitals, as long as it conserves the total energy. However, doing this means that the new $\mathrm{p_x}$ and $\mathrm{p_y}$ orbitals are no longer eigenfunctions of the $\hat{L_z}$ operator (i.e. they have no well defined angular momentum along the z-axis, so no well defined $m_l$ value). In most cases angular momentum is not really important, so we can safely use $\mathrm{p_x}$ and $\mathrm{p_y}$.
I suspect that there is a similar case with the d orbitals as well.
Which orbital is filled in first?
Inside one shell (e.g. 2p) the electrons are indistinguishable, so it is not possible to say which orbital is occupied by which electron. However, if a shell is not completely filled, i.e. there are unpaired electrons, then there would be an total orbital angular momentum and total spin angular momentum coming from all of those electrons combined. It is possible to identify states which have different angular momenta, as they have different energies. This forms the basis of the coupling schemes (LS coupling or jj coupling). However, you still cannot say exactly which electrons are residing in orbitals with which $m_l$ values, because multiple arrangements (microstates) can contribute to the same energy level (term).


Spraying aluminium hydroxide nanoparticles onto PLA as fire retardant? Several scientific papers study FR-PLA (fire resistant PLA) in which Aluminium Hydroxide nanoparticles are melt blended into the PLA (example source: Advances in Flame Retardant Poly(Lactic Acid). Would spraying the nanoparticles (dispersed in a liquid -source - maybe non flammable silicon oil) onto a 3D printed PLA help with fire resistance or not make a difference?

I suspect this is too late. The whole point of embedding the nanoparticles inside PLA is that it remains inside until needed (when temperatures rise above ~300 °).

By applying it as a coat, it can be easily scratched or rubbed off because PLA is so soft or flexible. You are also running into the problem of shedding of Al(OH)3 NP everywhere which may be a problem.

A coat of sprayed Al(OH)3 NP is only going to be a few micrometres thick, or you will need to use some glue to make them stick, which complicates thing and changes the appearance of your product. On the other hand, embedded Al(OH)3 NP are present throughout the entire thickness of the product. During fire, your sprayed NP might dehydrate just from the heat of radiation without even coming into contact with fire.

This is a bad solution.


Monobromination of napthalene derivative How does this reaction take place? Why does the bromine attach to that one particular location?

I do know that

since is $\ce{FeBr3}$ is present only 1 $\ce{Br}$ will attach
non-substituted areas of benzene will get $\ce{Br}$

You have electronically asymmetric rings. The right ring is deactivated by the carbonyl group that acts like an acyl group (remember that acyl groups are strong meta-deactivators). The left ring is activated by both $\ce{-CH3}$ and the nitrogen substituent acting as an amine (activating ortho/para directing), so bromination will occur on the left ring (since it is more nucleophilic).

Monobromination is not guaranteed experimentally. Although the right ring is deactivated under the right condition (at higher temperatures) you can end up with more than one $\ce{Br}$ being added. Maybe this question is just showing that a bromination will occur probably in that particular position.


Is the heat capacity infinite for a process at constant temperature? The heat capacity of $\ce{H2}$ at constant $P$ and temperature is ______.

It was explained to me that the answer is infinity with the following reasoning:

  Since heat capacity = $\mathrm{d}H/\mathrm{d}T$ and $\mathrm{d}T$ is zero since constant temperature is part of the question, the answer is infinity.

My attempt: Since $\ce{H2}$ is a diatomic gas, its $C_p$ is $7R/2$ which gives $7~\mathrm{cal \cdot mol^{-1} \cdot K^{-1}}$.

Which method is correct?

If you assume ideal behaviour and a closed system, then since the ideal gas law holds, with pressure and temperature remaining constant volume must also be constant.So that means that nothing is happening to the system, everything is as it is with $dq=0$ and $dw=0$.Since there is no process involved there is no meaning for the term heat capacity.


Do all the angles change in cyclopropane and cyclobutane or it is only the C-C-C angles? Are the angles other than the C-C-C angles in cyclopropane and cyclobutane still 109.5° as the acyclic compound, or what?

The other bond angles can't all be $109.5°$ if the angles in the ring are different from that.  Bond angles around an atom are not independent variables because the amount of space to be spanned is fixed.


confusion about exocyclic double bond I believe that there are only two exocyclic bonds in this structure. But in some text its mentioned as three.
can someone clarify.

By definition (in Woodward-Fieser rule), an exocyclic double bond is a double bond where one of the $sp^2$ carbon atoms of the bond is a part of a ring, while the other $sp^2$ carbon of the bond is not a part of same ring (it could be a part of another ring).

If both the $sp^2$ carbon atoms of the double bond is part of the same ring, then it is called endocyclic double bond. From these definitions, you can understand that exocyclic would stand for a double bond outside the ring and endocyclic would stand for a double bond within the ring. Following examples would illustrate the definition:


How to find the molar volume of 1 gas in a 2 gas system using PV_m=ZRT? Using $PV_m=ZRT$, would I need to use the total pressure of the system or the partial pressure of the gas I am interested in to work out the molar volume?

If you want to get the partial molar volume of a species in a non-ideal gas mixture, you can treat it as an ideal solution.  In this approximation, the partial molar volume of the species is equal to the molar volume of the pure species at the same temperature and pressure as the solution.  Another method is to evaluate the molar weighted averages of the critical pressures, critical temperatures, and ascentric factors for the various species in the mixture, and then treating the mixture as a pure gas.  You then take the partial derivative of the volume with respect to the number of moles of the species under consideration.  I would try it both ways to see how the results compare.


Are there other carbon free gasses that can be used for energy transport, apart from hydrogen? Various gasses are currently used to transport and store energy for heating and electricity generation.

Most of them contain carbon:
Methane, propane, butane, carbon monoxide and natural gas.

It could be of advantage to use gasses that do not contain carbon.
One example for that is hydrogen.

Are there other gases that theoretically could be used, if they could easily be created?
I assume most can not practically used because they are not trivial to generate, but I like to ignore that, and assume they can be created.

Are there other gases that could be used, in principle, as energy carrier?

Yes, ammonia has been cited.

Per a source, 'Ammonia as a Hydrogen Source for Fuel Cells: A Review', to quote:

  Like hydrogen, ammonia is carbon free and can be produced from any energy resource. However there are also some significant advantages in terms of storage and transport.


What is the difference between organosilyl, alkoxysilane and silane compounds? What is difference between organosilyl, alkoxysilane and silanes?

How do these groups make a surface hydrophobic?

Silane is the equivalent of methane for silicon, $\ce{SiH4}$.
Silicon containing organic derivatives of many kinds exist.
Generally speaking, organosilanes are those in which $\ce{Si-C}$ bonds exist, like for example, tetramethylsilane ($\ce{TMS}$, used as a reference on $\ce{{}^{1}H~NMR}$).

Alkoxysilanes are those compounds that contains an organic group bonded to a silicon but not directly, through an oxygen atom instead, like so: $\ce{R-OSi(Me)3}$, common structure on silyl enol ethers.

A surface is hydrophobic if it's repulsions towards water overcome its attractions, and this is the case for many silicon compounds, but not all of them. For example, glass (silicon dioxide, $\ce{SiO2}$) is hydrophilic, due to the possibility of formation of hydrogen bonds between the glass oxygen and water molecules.


What is the triple point pressure for Gallium? I have seen various discussions about the triple point of Gallium determined to a very precise value, so precise that it is used as a reference for NIST scales and measurements.

However, these reference related documents mean 'temperature' only when talking about triple point. Triple-point is defined as a (temperature, pressure) pair, so I fail to understand why despite so many discussions on the triple point of Ga, the pressure value is not easy to find.

I did find a value of $10^{-38} \mathrm{atm}$ (if I'm not wrong) sometime in the past but I can't find the source. However I have two questions related to this value:

Does it even makes sense to talk about such a minuscule value of pressure? Wouldn't this mean one particle every cubic light-year or something? (edit: More like one particle every cubic kilometer, depending on temperature, but still).
Is it safe to say that Gallium melts at absolute zero pressure (perfect vacuum), and never sublimates or desorbs (outgassing of Ga atoms/clusters from solid Ga surface)?

In general, are there other metals that have such a small triple-point pressure that they are guaranteed to never exhibit the process of sublimation/desorption, even in a perfect vacuum? (i.e., they will always melt first before evaporating or boiling).

At the very least your pressure has to be consistent with a whole number of molecules or atoms in your vapor phase.  Suppose you have a pressure of $7×10^{-34}$ atmosphere or (roughly) $7×10^{-29}$ Pa at $300$ Kelvins.  The average volume per atom/molecule is computed by the molecular level version of the Ideal Gas Law, in which the gas constant $R$ per mole is replaced by Boltzmann's constant $k=1.38×20^{-23}$ J/K per unit entity.  Thus

$V=kT/P=(1.38×10^{-23})×300/(7×10^{-29})=6.9×10^7$

cubic meters per vaporized entity.

So unless you are equilibrating your gallium over that much vapor volume, with high probability you do not have in practice even one atom or molecule of vaporized gallium.  In realistic laboratory or even industrual settings the triple point content of gallium vapor is most probably zero.


Gibbs Free Energy of Solutes and Solutions Using the equation $$\ce{CaSO4 (s) <=> Ca^2+ (aq) + SO4^2- (aq)},$$ and $\Delta H/\Delta S$ values, calculate $\Delta G$ at $50~^\circ\mathrm{C}$ when the solution is saturated with $\ce{Ca^2+}$ and $\ce{SO4^2-}$.

My Attempt
$$\Delta G = \Delta G^\circ + RT \ln Q$$
I calculated $\Delta G^\circ$ at $50\ \mathrm{^\circ C}$, But I am confused as to how we know what $Q$ is if there is no reactant then does $Q$ not become $\ce{[Ca^2+][SO4^2- ]/0}$, but this is impossible.

For the reaction
$$\ce{CaSO4 (s) <=> Ca^2+ (aq) + SO4^{2-} (aq)},$$
the appropriate equilibrium constant using activities can be set up as
$$K_\mathrm{eq}=\frac{a(\ce{Ca^2+})\,a(\ce{SO4^2-})}{a(\ce{CaSO4})}.$$

The activity $a$ of a pure substances in condensed phases is approximately one, since their behaviour can be often treated ideally, i.e. the chemical potential $\mu$ at the considered conditions is approximately the same as the chemical potential at standard conditions $\mu^\circ$. This derives directly from the definition of the relative activity: $$a = \exp\left\{\frac{\mu-\mu^\circ}{\mathcal{R}T}\right\}$$

For substances where the solubility is very low, concentrations of the ions are very low, so in a further approximation we can assume that in these solutions the activity coefficient is approximately one, therefore, we can write concentrations instead of activities and obtain for the equilibrium constant the expression, that is also known a solubility product.
$$K_\mathrm{eq} \approx K_\mathrm{sp} = c(\ce{Ca^2+})\,c(\ce{SO4^2-})$$

However, this is completely irrelevant to the question itself. Since the solution is saturated, it also means, that the solid is in equilibrium with the solution, which means that the kinetics of dissolution and precipitation are equal of magnitude. The Gibbs energy has a very distinct value for these equations.

   chemical equilibrium
 Reversible processes [processes which may be made to proceed in the forward or reverse direction by the (infinitesimal) change of one variable], ultimately reach a point where the rates in both directions are identical, so that the system gives the appearance of having a static composition at which the Gibbs energy,  $G$, is a minimum. At equilibrium the sum of the chemical potentials of the reactants equals that of the products, so that:
 $$\Delta G_\mathrm{r} = \Delta G_\mathrm{r}^\circ + \mathcal{R}T\,\ln\,K = 0\\\Delta G_\mathrm{r}^\circ = - \mathcal{R}T\,\ln\,K\\$$
 The equilibrium constant, $K$, is given by the mass-law effect.
 Source: IUPAC goldbook


What are energies in computational methods relative to? As per my understanding, in molecular mechanics (MM) the computed strain energy is based off the strain inherent in the molecule under study relative to an idealized form. For example, cyclobutane to its acyclic form.

However, what about computed values from other methods like PM6 energies, HF energies or MP2 enthalpies. What are these computed values relative to?

Ab initio calculations give the absolute energy of the system of fixed nuclei and moving electrons (all considerations on the type of calculations are implicit such as the Born-Oppenheimer approx.). There isn't any distinction of concepts of absolute energies between softwares when ab initio methods are employed (no ambiguity is allowed as pointed out by Pentavalentcarbon). Maybe the confusion can be illustrated with an example: you can't obtain heats of formation, as an absolute energy, from an ab initio calculation but it serves as a good approximation depending on the method.


Cell potential for redox titration of iron(II) with cerium(IV) Calculate $E_\mathrm{cell}$ for an electrochemical cell consisting of SCE and a platinum wire indicator electrode dipped in a titration vessel containing $\pu{25 ml}$ $(\pu{0.01 M})$ solution of $\ce{Fe^2+}$ ions after addition of $\ce{Ce^4+}$ solution of concentration $\pu{0.01 M}$ of
A) $\pu{5 ml}$
B) $\pu{25 ml}$
C) $\pu{50 ml}$
SCE is attached to the negative terminal of voltmeter. Given data:
$$
\begin{align}
E^\circ(\ce{Fe^2+/Fe^3+}) &= \pu{0.77 V} \\
E^\circ(\ce{Ce^4+/Ce^3+}) &= \pu{1.61 V}\\
E(\text{SCE}) &= \pu{-0.24 V}
\end{align}
$$

I tried doing this, but one thing I don't get over with is that isn't iron ions and cerium ions both oxidising and reducing? Why is there another electrode? Please clarify as much as possible.

Let's start at the beginning, with a solution where $\ce{[Fe^{2+}}] = 0.01$ M. In order to apply Nernst's law, we also need $\ce{[Fe^{3+}}]$, which is not known, but must be very very low, but cannot be zero. We will see that its exact value is without importance. Let's suppose it is  $\ce{[Fe^{3+}}] = 10^{-6}$ M. The redox potential of the solution is : $$\ce{E(Fe^{3+}/Fe^{2+}) = E°(Fe^{3+}/Fe^{2+}) + 0.059 log\frac{[Fe^{3+}]}{[Fe^{2+}]} = 0.77 + 0.059 log(10^{-4}) = 0.53 V}$$ Now this potential cannot be directly measured. The only thing that can be measured is the difference of potentiel between this electrode and a reference, which is here SCE, at - $0.24$ V. So if you dip a platinum electrode and a SCE electrode in this solution, you will measure a voltage between them to be : $$\ce{E_{cell} = 0.53 V -(- 0.24) V = 0.77 V}$$ This not a reliable value, as it supposes that we know the amount of $\ce{Fe^{3+}}$ in the solution. Anyway, let's go on.
First addition of Cerium(IV).
Now you add $5$ mL $\ce{Ce^{4+} 0.001 M}$, that is $50~\mu$moles. This cerium ion will react with $\ce{Fe^{2+}}$ ions according to the equation $$\ce{Ce^{4+} + Fe^{2+} ->  Ce^{3+} + Fe^{3+}}$$ The initial solution contained $250~ \mu $moles $\ce{Fe^{2+}}$. Adding $~50~\mu$mole $\ce{Ce^{4+}}$ consumes $~50~\mu$mole $\ce{Fe^{2+}}$ and produces $~50~\mu$mole $\ce{Ce^{3+}}$ and $~50~\mu$mole $\ce{Fe^{^3+}}$. The final amount of $\ce{Fe^{2+}}$ is $250 - 50 = 200~ \mu$mol. The potential of this solution is, according to Nernst's law : $$\ce{E(Fe^{3+}/Fe^{2+}) = E°(Fe^{3+}/Fe^{2+}) + 0.059 log\frac{[Fe^{3+}]}{[Fe^{2+}]} = 0.77V + 0.059 log\frac{50}{200} = 0.77V - 0.036 V = 0.734 V} $$ The potential of the whole cell (solution + SCE) is : $$\ce{E_{cell} = 0.41 V - (-0.734 V) = 1.144 V}$$ It is worth the trouble mentioning that here we could also have got the same result using the Cerium potential and Nernst's law. In principle. But, to apply Nernst's law to the Cerium couple we would need to know the residual concentration in $\ce{Ce^{4+}}$ which must be very small but not zero. So we have to use the couple $\ce{Fe^{3+}/Fe^{2+}}$.
Second addition of Cerium(IV).
Here you add $25$ mL $\ce{Ce^{4+} 0.001 M}$, that is $250~\mu$moles. The same reaction happens. But here, the amount of Cerium added is equal to the amount of original Iron in the solution. So at the end, there is in solution  $250~\mu$moles of $\ce{Ce^{3+}}$ and the same amount of $\ce{Fe^{3+}}$. The residual amount of $\ce{Ce^{4+}}$ and $\ce{Fe^{2+}}$ are very small, but unknown.
The potential of the solution must be the average of $\ce{E°(Fe^{3+}/Fe^{2+})}$ and $\ce{E°(Ce^{4+}/Ce^{3+})}$, and this is $(0.77$ V + $1.61$ V)/$2$ = $1.19$ V. The potential of the cell after this $2$nd addition is then $$\ce{E_{cell} = 1.19 V - (-0.24V) = 1.43 V}$$
Third edition of Cerium(IV).
Here there is a huge excess of Cerium(IV). There is exactly twice the amount of cerium(IV) necessary to oxidize all Iron(II). The final solution contains the same amount of both Cerium ions. The potential of the solution cannot be calculated with the Iron couple, as the amount of remaining $\ce{Fe^{2+}}$ is unknown and very low. So the potential of the cell must be calculated by Nernst's law applied to the Cerium couple. As the final amount of both ions $\ce{Ce^{3+}}$ and $\ce{Ce^{4+}}$ are equal, the potential of the solution is equal to the standard redox potential, which is $1.61$ V. And the measured difference of potential between the solution and the SCE is : $$\ce{E_{cell} = 1.61 V - (-0.24 V) = 1.85 V}$$
To summarize all these calculations, the measured potential in the cell (solution + SCE) increases steadily from about $0.77$ V, to $1.14$ V, then to $1.43$ V, then to $1.85$ V.


What's the difference between FFKM and PTFE? These two fluorocarbon polymers seem to me very similar, the only difference I see is that FFKM is elastomer while PTFE aka Teflon is more of hard type of plastic.

Isn't FFKM just PTFE with lower average molecular weight, that is the average molecule is shorter chain so its softer? Is the production of these two polymers vastly different or is the process mostly same?

Teflon (PTFE) and Kalrez (FFKM) are different polymers.

PTFE is made from tetrafluoroethylene, and contains only carbon and fluor.
The material is stiff.

FFKM is made from tetrafluoroethylene, vinylidene fluoride and perfluoromethylvinylether, so besides carbon and fluor, it also contains hydrogen and oxygen.
These additional monomers make the polymer an elastomer.