What is the order of the following compounds based on polarity? What is the order of the following compounds from more polar to less polar?

  Hydrochloric acid,
  Methanol,
  Hexane,
  Petroleum benzene

I know that $\ce{HCl}$ has the largest polarity. Hexane I guess has the less. I guess the order should be like:
$\ce{HCl}$ > Methanol > Petroleum benzene > Hexane

Is this correct? I'm not sure of the place of petroleum benzene.

The concept of solvent polarity is commonly used to describe solvent effects. For example, the Glossary of Terms Used in Physical Organic Chemistry (IUPAC Recommendations 1994) uses the following general definition (which was originally proposed by Christian Reichardt) of the term polarity:

  When applied to solvents, this rather ill-defined term covers their overall solvation capability (solvation power) for solutes (i.e. in chemical equilibria: reactants and products; in reaction rates: reactants and activated complex; in light absorptions: ions or molecules in the ground and excited state), which in turn depends on the action of all possible, nonspecific and specific, intermolecular interactions between solute ions or molecules and solvent molecules, excluding such interactions leading to definite chemical alterations of the ions or molecules of the solute. Occasionally, the term solvent polarity is restricted to nonspecific solute/solvent interactions only (i.e. to van der Waals forces).

However, the polarity of a solvent is not a physical quantity. Therefore, it is difficult to express quantitatively. Several scales exist that involve various physical properties in order to provide a measure of a solvent’s polarity.

In particular, the relative permittivity $\varepsilon_\mathrm{r}$ is often used as a quantitative measure of solvent polarity:

Hexane: $\varepsilon_\mathrm{r} = 1.8865$
Benzene: $\varepsilon_\mathrm{r} = 2.2825$
Methanol: $\varepsilon_\mathrm{r} = 33.0$
Water: $\varepsilon_\mathrm{r} = 80.100$

(values at 20 °C taken from “Laboratory Solvents and Other Liquid Reagents”, in CRC Handbook of Chemistry and Physics, 97th Edition (2016), William M. Haynes, ed., CRC Press/Taylor and Francis, Boca Raton, FL.)

By way of comparison, the empirical $E_\mathrm{T}(30)$ scale is based on the solvatochromism of Reichardt’s dye (Betaine 30):

Hexane: $E_\mathrm{T}(30) = 31.0\ \mathrm{kcal/mol}$
Benzene: $E_\mathrm{T}(30) = 34.3\ \mathrm{kcal/mol}$
Methanol: $E_\mathrm{T}(30) = 55.4\ \mathrm{kcal/mol}$
Water: $E_\mathrm{T}(30) = 63.1\ \mathrm{kcal/mol}$

(values taken from here)


Is there a cheap, safe liquid that evaporates EXOthermically around room temperature? The heating and AC system in the building where I occasionally work, works like this (basically a thermostat):

The administrators set some temperature that is maintained automatically. Say, in the winter, the outside temperature is 0 °C. The admins will set the desired inside temperature to 20 °C. There is some thermometer behind a little steel panel. It measures it to be cold in the room; therefore the heat comes on. I can trick the heat into coming on even more, by rubbing some alcohol or acetone on the steel panel. The acetone evaporates endothermically, making the steel panel REALLY cold; then the heater thinks that it's REALLY cold in the room and pumps the heat in.

The same system is in place in the summer, but in reverse. However, I can't trick the panel in the same way. If I put acetone on the panel, it'll cool off, and then the A/C unit thinks the room is just fine, because the panel is cool. I'm wondering if there is some liquid that will evaporate exothermically, and heat up the panel, in order to make the system think I need more A/C?

I don't believe that exothermic evaporate exists; here is a list of heat of vaporization for various substances; all positive (endothermic). http://www.engineeringtoolbox.com/fluids-evaporation-latent-heat-d_147.html
But maybe I'll be surprised and it does exist!

I have been holding my laptop's hot backside up to the panel instead, but that's a bit too manual for my taste.

Edit: To be a bit clearer, let me specify that this is just a thought experiment. I'm not actually expecting to find something which doesn't make sense like an exothermically evaporating liquid.

No such liquid, safe or otherwise, can exist. Evaporation is a strictly endothermic process in all cases.

The change in state from liquid to gas is marked by the individual particles gaining enough translational kinetic energy to overcome the mutual attractions present in the liquid phase to "fly free" in the gas phase. It is logically inconsistent for a substance to increase its internal energy and release energy to the surroundings as heat in the same process.

In order to achieve both evaporation and a release of energy, one would have to find a liquid that reacts to (a) release heat and (b) form gaseous products.  The energy required to move from the liquid to the gas phase is substantial; more than likely the only reactions exothermic enough to provide a net release of heat are combustion reactions.

I somehow doubt dumping a flammable liquid into your thermostat, inserting a wick, and lighting it on fire is a satisfactory solution for you.


Converting benzoic acid into meta-fluorobenzoic acid How do I convert benzoic acid to meta-fluorobenzoic acid in not more than 3 steps? I did it but the process was a little too long.
Here is my attempt

Two steps should be enough ;-)

Nitrate benzoic acid. In J. Chem. Educ., 1967, 44, 464
(DOI), the authors added benzoic acid to fuming nitric acid at 0 °C. 3-nitrobenzoic acid is the major isomer (~ 75%). Recrystallize from ethanol.

Your product is an electron-deficient arene and $\ce{-NO2}$ is a decent nucleofuge. Try to replace it by fluorine in a nucleophilic aromatic substitution!

In the case of 4-nitrobenzoic acid, the transformation to 4-fluorobenzoic acid has been described to work with a yield of >95% in a room temperature reaction using DMSO as a solvent and anhydrous tretrabutylammonium fluoride (TBAF) as the reagent. (DOI)


How to determine the specific heat of methanol from calorimeter data? A $\pu{25.95 g}$ sample of methanol at $\pu{35.6 ^\circ C}$ is added to a $\pu{38.65 g}$ sample of ethanol at $\pu{24.7 ^\circ C}$ in a constant pressure calorimeter. If the final temperature of the combined liquids is $\pu{28.5 ^\circ C}$ and heat capacity of the calorimeter is $\pu{19.3 J/^\circ C}$ determine the specific heat of methanol. The heat capacity of ethanol is $\pu{2.46 J/(g * ^\circ C)}$.

\begin{align}
  q_\mathrm{methanol} &= -q_\mathrm{ethanol}\\
  msT &= -msT\\
  \pu{25.95 g} \cdot  (\pu{28.5 ^\circ C} - \pu{35.6 ^\circ C}) \cdot  s
    &= -\pu{38.65 g} \cdot  (\pu{28.5 ^\circ C} - \pu{24.7 ^\circ C})
       \cdot  \pu{2.46 J/(g* ^\circ C)}\\
\therefore s &= \pu{1.96 J/(g * ^\circ C)}
\end{align}
The correct answer, however, is $\pu{2.36 J/(g * ^\circ C)}$.
I think that there is a mistake in my solution, because the heat capacity of the calorimeter is given, and I didn't use it.
Could anyone explain where I have been wrong?

You should have included the heat capacitance of the calorimeter. Think about it, if the calorimeter was at $\pu{24.7 ^\circ C}$ before the methanol was added and then it rose $\pu{3.8 ^\circ C}$ along with the liquids, where did the energy come from to increase its temperature? Since you know that the heat capacitance of the calorimeter is $\pu{19.3 J/^\circ C}$ then you can find out how much energy the calorimeter absorbed to increase in temperature by multiplying its heat capacitance by its temperature increase. I.e. $\pu{19.3J/^\circ C} \cdot \pu{3.8 ^\circ C} = $ "Joules absorbed by the calorimeter".
Then we simply apply the heat lost equals heat gained concept:
$$\begin{multline}
\pu{25.95 g} \cdot (\pu{35.6 ^\circ C} - \pu{28.5 ^\circ C}) \cdot S \\
  = \pu{38.65 g} \cdot (\pu{28.5 ^\circ C} - \pu{24.7 ^\circ C})
    \cdot \pu{2.46 J/(g* ^\circ C)} + \pu{19.3J/^\circ C}
    \cdot \pu{3.8 ^\circ C}
\end{multline}$$
After rearranging, we get $S = \pu{2.36 J/(g * ^\circ C)}$.


Long linear molecule with alternating elements? In the classic freshman physics text Electricity and Magnetism by Purcell, one of his first examples is the computation of the electrical binding energy of a $\ce{NaCl}$ crystal. He models the crystal in terms of perfect ionic bonding and sums up the energy $q_1q_2/r$ pairwise, computing the $r$ values in three dimensions. (Actually he just gives the first few terms explicitly and then says he carried out the computation numerically on a computer.) The result is $U=-0.8738Ne^2/a$, where $N$ is the number of ions, $a$ is the edge length of the cubic lattice, and the units are cgs, so that the Coulomb constant equals 1.

I thought this was a nice example, but it seemed unfortunate that the actual result could only be found through a fairly complicated numerical computation. I thought it would be nice if I could come up with a similar but simpler example for use with my freshman E&M classes, using a linear molecule with charges $\pm e$ arranged like $\cdots +-+-+-\cdots$. A relatively simple calculation gives an energy of
$$-\frac{Ne^2}{4a}\sum_{k=1}^\infty \frac{1}{k(k-1/2)}.$$

I imagine that the sum can be done in closed form, but anyway it's clearly finite and positive, so that the result is a finite binding energy that has the right sign to keep the system bound.

Is there any molecule that actually has this structure, or something close to it? Googling and asking a chemist in the hall resulted in a variety of examples that are not exactly what I had in mind:

There are various 3-ion molecules such as $\ce{BeF2}$.
You can have long carbon molecules, with hydrogens hanging off of them, but they're not exactly linear, and in any case the carbons are all the same element, so there is no ionic bonding.
Acetylene is linear.

Does nature provide anything like the $+-+-$ chain that I'm describing? Do they not occur because they're too unstable with respect to crumpling up? Could they exist on a substrate (the kind of thing people build using scanning electron microscope probes) or as linear impurities in a bulk solid?

For my purposes, I don't think it really matters whether the bonding is perfect ionic bonding, as long as there is some segregation of charge.

Related:

Why does sulfur, but not oxygen, catenate?
Are long oxygen molecules possible?

An example of a linear structure is given by cycopentadienyl lithium, whose structure is explored in this question.  There is some covalent interaction in this polymeric structure, as in most organolithium compounds, but the bonding between lithium and cyclopentadienyl groups is highly polar.


Simple and cheap anti-graffiti layer based non-paint solvent I have this issue with a group of people intentionally damaging a bench. So far, they would paint curses over the bench with permanent marker. However after I removed the curses with ethanol, they damaged the bench physically and threw the trash all around (we used to have a trash can there, with sign "Please do not throw trash around" - they threw this trash can away too).

I would like to keep this question non-personal though, so my question is:

How can I protect a nitro-based paint layer from other paints?

The professional solutions are way too expensive, while the thing I want to do is simple: I'm looking for non-nitro based material and solvent that will form transparent layer over the original paint. In case of damage by nitro-based paints, I could wash the layer under them and then apply it again.

The theory is following: you apply the protective layer over your paint. When somebody applies any paint over it, you've got 3 layers.

It does not matter what the top layer is, as long as it does not dissolve our protection. After this, we apply solvent for the protective layer. Assuming there are microscopic holes in the top paint, the protective layer will dissolve and the top layer will go off with it.

The layer must not be obvious.

I was considering these:

Ethanol-based transparent paint (too expensive, suprisingly)
Wax (parafin) based layer (I don't know how to dissolve wax, it's crystalic so not transparent)

The "professional" solutions here in the US are often fluorinated paints. My understanding is that the graffiti literally slides off and people give up after a few tries.

Given the costs, my suggestion would probably be a thick layer of polyurethane. These are often applied to seal wood floors and can be transparent or stained. If you apply multiple layers (say 2-3 on a floor) the surface is very smooth and slick. It's hard for paint or other stains to stick.


Determining enthalpy of neutralization and heat capacity of calorimeter by overlying with straight lines or "eye-o-metric method" Below is a model example of how to interpret gathered data. Temperature on y-axis, time on x-axis.
We are supposed to read  (temperature change by neutralization) and  (temperature change by resistive heating - to determine heat capacity of calorimeter) by overlaying the selected area with straight lines.

What I don't get is the way the straight lines are put the way they are. Maybe the bottom one: we expect the temperature drop caused by heating the calorimeter will continue by this straight fashion so we must take it into account in subsequent temp change reading.

But what about the upper line? Why read the upper value above the graph and not on its local maximum?

Ah, I understand now. It's of analogous reason to the first case. By adding heat during a time interval we expect that some of it will be absorbed by the calorimeter.


Safe to wash weighing paper? I accidently washed the Fisherbrand weighing paper (cat. 0989812A) with acetonitrile : water (1:8) to remove some sample. The weighing paper appeared to deform a little bit after the wash. I wonder if ACN:H2O mixture dissolved anything from the paper or the paper deformed just because it absorbed some water?

The response to your concern depends on the application. If it were just an academic exercise, forget about it, since you may have already done the experiments. However, the point to keep in mind is that paper based weighing materials should not be rinsed or washed for standard preparation until and until unless it is specifically allowed to do so. Paper can have leachable materials which can contaminate the standard- this applies very well to trace analysis. If you note, Fisherbrand specifically mentions "low nitrogen" content.
The second reason why the paper based weighing paper should not be rinsed is that the bottom of the paper has been in contact with the balance pan (which are often horribly dirty in academic labs due to misuse). Since that paper is thin and permeable to solvents, rinsing the paper is akin to washing the dirty pan into your standard solution rather indirectly. Of course, balances in a professional or research settings are better maintained.
Sometimes it appears that some solids particles are still sticking to the weighing paper after transfer process. In such cases, one uses the "weighing by difference" method. You know the weight of the paper+solid, after transferring, you would weigh the paper again, and record the actual weight transferred to the flask or other container. No washing is required.
Analytical work, where extreme care is required, paper is not used at all for standard preparation. One should use "weighing funnels" which are commercially available and there is no concern in washing or leaching or soluble materials.


Will the concentration of 3% w/v H2O2 be lowered after 3 weeks? I have read online that $\ce{H2O2}$ decomposes if exposed to sunlight and air. If this is the case then will the concentration of 3% w/v of $\ce{H2O2}$ be lowered if it is left out in the open at around 25-degree celsius for a few hours?

Background:
The question arises from a titration experiment that I conducted where I had to verify the concentration of $\ce{H2O2}$ to be 3% w/v. But the titration found the concentration to be lower and other groups in the class found the same.

Hydrogen peroxide is somewhat unstable even when not exposed to air and sunlight. It decomposes to water and oxygen gas over time.
According to this Wikipedia article:

Hydrogen peroxide is thermodynamically unstable and decomposes to form water and oxygen with a $\mathrm{\Delta H}$ of $\mathrm{-98.2}~\pu{kJ/mol}$ and a $\mathrm{\Delta S}$ of $\mathrm{70.5~\pu{kJ/mol}\ce.\pu K}$:
$$\ce{2H2O2 -> 2H2O + O2}$$
The rate of decomposition increases with rising temperature, concentration and pH, with cool, dilute, acidic solutions showing the best stability.

So, not knowing the exact conditions during the 3-week period or other factors, it's not possible to say for certain whether the $\ce{H2O2}$ degraded or whether it just was a little low to begin with. But the potential for degradation over that period of time is certainly there.


Measure ethanol volume fraction in a beverage Given a sample of an alcoholic beverage (be it beer, wine, etc..), what are possible methods to find out the ABV (alcohol by volume) of it with moderate accuracy?

As fas as I know, some methods exist already but are not really suitable in my condition:

Distillation. Impractical because of the apparatus needed
Boiling temperature based. Not accurate if exact condition of sugar is not known
Estimation by measure of specific gravity before and after fermentation. I've found it to be inaccurate with the data I have access to

I was thinking about a process similar to the tritation method used to find water hardness, but I'm not sure what kind of reactions would be involved for this kind of test, nor the nature of the titrant that would be required.

Your question: Given a sample of an alcoholic beverage (be it beer, wine, etc..), what are possible methods to find out the ABV (alcohol by volume) of it with moderate accuracy?

There are a few method used in winery industry as  Karl mentioned a comment in elsewhere. I assume you want to find out better methods to find the alcohol content:

The best technique is the gas chromatography method, which is the most accurate method for determining alcohol content. Gas chromatography has the added benefit of providing additional information about the wine, beer, or any other liquor product (e.g., Ref.1).
Seemingly most popular technique is the determination of ethanol in alcoholic beverages by titrimetric and/or spectrophotometric dichromate methods (e.g., Ref.2-Ref.4).
Another method is enzymatic assay using spectrophotometry: Ethanol, in an alkaline environment, is oxidized through an enzymatic reaction and converted in acetaldehyde in presence of $\mathrm{NAD^+}$, which in turns produces $\mathrm{NADH}$. The increase of the absorbance, measured at $\pu{366 nm}$ (end-point), is proportional to the concentration of ethanol in the sample (e.g., Ref.5).
In addition, you can also use NMR technology to determine the ethanol content in alcoholic beverages (e.g., Ref.6).

References:

M.-L. Wang, Y.-M. Choong, N.-W. Su, and M.-H. Lee, “A rapid method for determination of ethanol in alcoholic beverages using capillary gas chromatography,” Journal of Food and Drug Analysis 2003, 11(2), 133–140.
G. J. Pilone, “Determination of ethanol in wine by titrimetric and spectrophotometric dichromate methods: collaborative study,” J. Assoc. Off. Anal. Chem. 1985, 68(2), 188-190 (https://www.ncbi.nlm.nih.gov/pubmed/3988696).
Arthur Caputi, Masao Ueda, Thomas Brown, "Spectrophotometric Determination of Ethanol in Wine," Am. J. Enol. Vitic. 1968, 19(3), 160-165 (https://www.ajevonline.org/content/19/3/160).
Malinee Sriariyanun, Parita Mutrakulcharoen, Surapun Tepaamorndech, Kraipat Cheenkachorn, Kittipong Rattanaporn, "A Rapid Spectrophotometric Method for Quantitative Determination of Ethanol in Fermentation Products," Oriental Journal of Chemistry  2019, 35(2), 744-750 (http://dx.doi.org/10.13005/ojc/350234).
Amy A. Caudy, "Spectrophotometric Analysis of Ethanol and Glucose Concentrations in Yeast Culture Media," Cold Spring Harbor Protocol  2017, (9), 735-741 (doi: 10.1101/pdb.prot089102).
Meden F. Isaac-Lam, Thomas Brown, "Determination of Alcohol Content in Alcoholic Beverages Using 45 MHz Benchtop NMR Spectrometer," International Journal of Spectroscopy 2016,  Article ID 2526946, 8 pages (https://doi.org/10.1155/2016/2526946).


How right is defining elements with the number of protons? Atomic number is the number of protons a certain atom has. It's the defining attribute of a certain element. But, How right is that?

From the viewpoint of chemistry, the definition is right. Because, for a certain chemical reaction, isotopes of a certain element will react in the same way.

For example, $\ce{H2 + 1/2O2 -> H2O}$
$\ce{ ^2H2 + 1/2O2-> ^2H2O}$

So, it's useful to define elements in terms of their number of protons.

But, other than reactivity in chemical reactions, many properties of an atom differ wildly due to difference of  number of neutrons and electrons eg. half life of radioactive decay, type of radioactive decay, magnetic moment, stability, nuclear quantum states etc. So, even having the same number of protons can't ensure much similarity in some aspects.

I think to a rational perspective, the similarity of reactivity of isotopes in chemical reactions should weigh more than the other nuclear difference between isotopes. So, in that logic, the definition is right.

But, difference in nuclear properties should do matter. So, how right is defining elements with atomic number i.e. number of protons?

Most of chemistry doesn't care a whit about all of the properties you named: they're in the realm of nuclear physics, as noted by user badjohn.  Even radioactive decay is something of a black sheep, included in physical chemistry as a convenient example of a first-order kinetic process. To my knowledge, it's not substantially used in conjunction with any other sub-discipline of chemistry, as it'd be more of an annoyance than anything ("Gah, the cobalt center of my complex keeps changing into nickel!").  It does of course have plenty of applications (medical imaging, nuclear power, etc.), but they're arguably in the realm of physics, with any attendant chemistry being a secondary consideration.

Magnetic moment, neutron count (isotopes), and nuclear quantum states do have some slight influence over certain spectroscopic properties (primarily NMR, as far as I know), but other than the somewhat special cases of the spin isomers of hydrogen and the kinetic isotope effect of deuterium vs. protium I'm not aware of any appreciable effect of these on chemical reactivity.

For the chemist, the identity of a nucleus is sufficiently defined (within an experimentally measurable perturbation, at least) in virtually all cases by its atomic number.


Benzene from cyclopropane-1,2,3-triyltrimethanol in acidic medium What is the mechanism of this reaction? After performing $\ce{E1}$ thrice I was expecting 1,2,3-trimethylidenecyclopropane to be formed instead.

I know it's highly strained but I couldn't adopt any other pathway to reach any other product.

Is the triene product not formed at all? If cyclopropane-1,2,3-trione can exist I think even the triene product can.

This reaction is from my exercise book, couldn't find it on orgsyn.org.

I have not written the steps for carbocation formation, simple dehydration.


Using results of current experiments in silanization, how to achieve a higher contact angle? I've done several silanization tests with different solutions and here is a brief of my results:

DichloroDimethylSilan: Advancing=90, Receding=56

ChloroTrimethylSilan: Adv=91, Rec=70

OctaDecylTrichloroSilan: Adv=108, Rec=90

TetraEthoxySilan: Adv=90, Rec=44

OctaDecylTrichloroSilan+ChloroTrimethylSilan(50%): Adv=89, Rec=74

I didn't dilute any of the above solutions and since I want to achieve Maximum Hydrophobicity I gave enough time (~20 min) to all the above reaction to become fully completed.

I'm wondering whether its possible to get a contact angle higher than 108 that I have now. Please let me know what changes should I made? (any pre-treatment of glass before silanizing, any other solutions, etc.)

It's always good to clean glass surfaces with oxygen plasma or UV ozone treatment before silianizing, but 108° is pretty good with an alkylsilane.
The best you will be able to do with this type of surface treatment is using fluorosilanes.
Silanization agents are basically made of three parts, a functional group, a linker and a coupling group:

The coupling group is what allows the silane to attach to a surface and is found bound to the silicon atom.
Typical couplers are hydrolyzable groups like alkoxy, acyloxy, or halogen.
These don't generally influence the hydrophobicity of the resulting surface directly, though they will affect how well the silane adheres to a given surface, the manner and speed with which it can be applied, and the deposition density.
Alkoxy and acyloxy groups can be applied from aqueous solutions and are generally less reactive, but halogen groups can be deposited from anhydrous solvents or the vapour face quite quickly.
The linker is typically just a short alkyl chain between the silicon atom and the functional group to reduce steric interactions between adjacent silanes.
The functional group is what has the greatest influence on hydrophobicity.
Many different functional groups can be added, some even conferring hydrophilicity instead, if desired.
In general, longer alkyl chains are more hydrophobic, but long fluoroalkane chains are the most hydrophobic and will also not interact easily with organic solvents either so fluorosilane coated surfaces are often known as omniphobic by virtue of being both hydrophobic and oleophobic.

So if you want the maximum contact angle possible from a simple surface treatment, you're looking for something like 1H,1H,2H,2H-Perfluorododecyltrichlorosilane.
It may also be possible to simply spin or spray coat a fluoropolymer like Teflon AF or Fluoropel, depending on your application and surface. All these types of coatings will give contact angles up to about 120° in the best case and to get any higher will require surface patterning to achieve superhydrophobicity.


Total spin number 0 together with $B_2$ state - confusion I'm trying to understand molecular symmetry and I got confused reading the Molpro documentation.

There is an example of a system with $B_2$ symmetry written like this:

$$
wf,10,3,0,
$$

which means, that the system has 10 electrons, $B_2$ symmetry and the total spin number is 0.

If I understard the total spin number well, it is 0 only when all electrons are paired in the orbitals (i.e. the system is in singlet  state).

But, on the other hand, I thought, that when the electrons are paired, then the orbital is "totally symmetric", i.e. $(\cdot)^2 = A / A_g / A_1$.

So, as far as I understand it, $B_2$ symmetry and total spin number 0 is mutually exclusive.

Is it true or do I understand it wrong? If I do, could you, please, explain my mistake and provide some simple example of the abovementioned configuration?

We can see what a $^1B_2$ will look like by consider the MO diagram for water, a $C_{2v}$ molecule with $10$ electrons.

$\hspace{9ex}$

This is the configuration that we might expect for the ground state, which has $^1A_1$ symmetry. We could get a configuration with $^1B_2$ in a couple ways, namely by exciting form $1b_2$ into $4a_1$ or by exciting from either $2a_1$ or $3a_1$ into $2b_2$ without changing the spin of whatever electron we move. These new configurations would still be singlets because we haven't changed any of the spins and would have $B_2$ symmetry since clearly $B_2\otimes A_1=B_2$.

As to your question about spin multiplicity, if we consider a system with a single electron, we account for both the spin up and spin down configuration by determining the spin multiplicity. The spin multiplicity for $S=\frac{1}{2}$ is $2S+1=2$, a doublet, which gives us exactly the number of spins states we would expect.


Is it safe to inhale O2 from electrolysis of a pH 13.7 KOH solution? I'm conducting electrolysis of water with a 0.25M KOH solution in water (about 14 on my pH meter) using graphite electrodes, and I was wondering if the O2 would be safe to inhale, as in an oxygen bar?

No. Untreated and unpurified, the gas coming directly out of your reaction vessel would be unsafe to inhale. It is likely that water vapour containing traces of KOH would be present, and this poses a significant safety risk to the person breathing this vapour. You could test this yourself by measuring the pH of the headspace above your reaction vessel. Hold a piece of indicator paper here for a period, and it will slowly show a change.


Can an ant etch a hole in a tent with formic acid? Given a polymer foil tent, could ant's formic acid help to etch a hole in it? I thought of a polyester or polyamid foil. If one ant's acid is not enough, they come as team.

This is not a theoretical question beacause I have observed it on a camping in South France. I am not just sure whether they just bite through, or indeed use their acid.

https://ibb.co/m5u5B8

I would be VERY surprised if it did. The answer depends on what material the tent is made of. Most tents now days are made of polyester and according to the table below (taken from: http://rfelektronik.se/manuals/Datasheets/solventguide.pdf) polyester has good resistance in formic acid (Note, S stands for satisfactory in table).

However, even if it was nylon, unsatisfactory resistance means that the material will deteriorate in hours to weeks when used with the chemical. I dont think an ant or some ants would be able to produce enough acid to seriously damage the thick fibers of the tent. Maybe if a lot of ants were attacking the same spot one after the other you would eventually dissolve some of the nylon to make a small hole, you never know.

Also, an ant attacking a tent is a very unlikely scenario (ants use formic acid to defend themselves).


How many valence electrons does a copper atom have? The valence electron configuration of $\ce{Cu}$ is $\ce{4s^1 3d^{10}}$. According to this the answer is $11$, but shouldn't it be $1$ since the d shell is already filled up?

Yes, copper only has 1 valence electron. Remember: valence electrons only include the electrons in the highest energy (n) shell.


How is the temperature in which vapor pressure is equal to atmospheric pressure the boiling point of the liquid? I was just learning about vapor pressure and got very confused. Vapor pressure was defined as the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases at a given temperature in a closed system, which was the pressure the vapor exerts on the closed container when it reaches the dynamic equilibrium. However, this is merely a calculation AFTER it reaches equilibrium, and I don't understand how vapor pressure can also be interpreted as the measure of the tendency of a material to change into the vapor state. Also, if it is the lack of enough vapor pressure, or "tendency of the liquid to change to vapor" compared to the atmospheric pressure that's keeping the liquid from vaporizing, how can evaporation occur with lower pressure than the atmospheric pressure(because if the vapor pressure is higher than atmospheric pressure it is already boiling), since evaporation is supposed to be the molecules on the surface of the liquid randomly gaining enough energy to overcome the atmospheric pressure, implying that it has greater pressure than the atmosphere. I feel like I am overthinking the topic, but I didn't just want to give up thinking about it, so please help if possible!

Vapour pressure is not merely calculation, it is easily measureable.
Evaporation is due molecules leaving liquid not by overcoming atmospheric pressure, but by overcoming cohesive intermolecular forces. The saturated vapour pressure is a measure of the rate of this massive leaving of molecules.
Boiling is just a special evaporation case. Above boiling point, vapour is always unsaturated and liquid evaporate at its surface regardless of gas composition. Additionally, random processes create miniature bubbles ( or these may be bubbles of formerly dissolved gasses ), that have the same or higher pressure then external pressure and expands themselves.
The result is more or less massive and turbulent evaporation with creation of vapour bubbles.
Liquids can be frequently overheated above their boiling point, if there are absent "bubble seeds". Once available, e.g. by strirring or repulsion of dissolved gas, creation of vapour bubbles quickly cools liquid down to its boiling point.


Which salts can be used to precipitate chloride ions? We don’t have AgNO3 in the lab. I have no idea what else can be useful.
I am washing a product and I want to make sure the chlorides are washed away.

The only well known method to check for (the absence of) chloride ions is precipitation as AgCl (solubility 2 mg/l at 20°C).

Mercury(I) chloride is similarly badly soluble, but toxic (and unstable, disproportionation) and therefore bad practice.

Lead(II) chloride is already far more soluble (4 g/l) than AgCl, and practically all other simple chlorides are just very well soluble in water.

Silver nitrate is cheap, the internet sells 10 g high purity $\ce{AgNO3}$ for 25€. Just get some.


How to obtain curve energy and non-bonded interaction parameters (e.g. Lennard Jones parameters) with Gaussian? I am parameterizing a force field and would need to obtain non-bonded interaction parameters of atoms (e.g., Lennard-Jones parameters).

My primary idea is to obtain a potential energy curve using the Gaussian 09 software and fit the data on that curve to an equation that has non-bonded interaction parameters of atoms (e.g., Lennard-Jones curve).

What I thought was to put two identical molecules and vary the distance between two atoms of these molecules (e.g., using the scan function on redundant coordinator editor) and, therefore, get the energy curve. What I was trying to do was a simple example to get interaction parameters for the oxygen atoms of two water molecules.

I was taking a test with Gaussian at the theory level B3LYP/6-311G++(d,p). Here is the input file:

%nprocshared=4
# opt=modredundant b3lyp/6-311++g(d,p) geom=connectivity

Title Card Required

0 1
 O                 -3.09468841   -0.10392610    0.00000000
 H                 -2.13468841   -0.10392610    0.00000000
 H                 -3.41514300    0.80100974    0.00000000
 O                  3.39491938    0.10392610    0.00000000
 H                  4.35491938    0.10392610    0.00000000
 H                  3.07446479    1.00886193    0.00000000

 1 2 1.0 3 1.0
 2
 3
 4 5 1.0 6 1.0
 5
 6

 B 1 4 S 10 0.200000

Something I noticed is that the level of theory used (obviously) influences the calculations results. In addition, the initial distance of the water molecules also influences whether the simulation will fail or not (in this example, it works).

Could you kindly tell me if this approach makes sense? Or should I use another software, another type of calculation? Do you have any suggestions on how I could get an energy curve by varying the distance of two non-bonded atoms in the Gaussian? Examples would be appreciated.

With Gaussian and a similar approach I was able to obtain parameters such as bond stretching, angle deformation and dihedral torsion. However, it would be lacking to obtain the non-bonded interaction parameters of atoms.

Things I should consider:

Do a rigid scan instead a relaxed scan (like the code presented).
I need a dispersion-corrected density functional methods.

To start off, you definitely need to use dispersion-corrected density functional methods if you want to obtain reasonable non-bonded interaction parameters. I've seen many cases where B3LYP (with no dispersion) yields a fully-repulsive potential energy scan.

I also agree with your assertion about doing a rigid scan. In principal, a relaxed scan is a good idea, but consider that most force fields separate out the non-bonded interactions. Thus, for parameterizing, you don't really want to change the bond lengths, angles, etc. from a relaxed scan.

There are a few ways to get the parameters. As mentioned above, there's already a similar question with tips on curve fitting.

Multiple papers have also been published on fitting force field parameters from quantum chemical results, e.g:

QubeKit J. Chem. Inf. Model. 2019, 59, 4, 1366-1381 - code at GitHub
QuickFF J. Comput. Chem. 2015, 36, 1015– 1027 - code at GitHub
ForceBalance J. Chem. Theory Compute. 2013, 9, 1, 452-460 and J. Phys. Chem. Lett. 2014, 5, 11, 1885-1891 - code at GitHub
ForceFit J. Comput. Chem., 2010 31: 2307-2316
Parfit J. Chem. Inf. Model. 2017, 57, 3, 391-396 - code at GitHub

Update there's a paper on this very specific problem:

J. Chem. Theory Comput. 2020, 16, 2, 1115-1127

  Here, we utilize a previously described Minimal Basis Iterative Stockholder (MBIS) method to carry out an atoms-in-molecules partitioning of electron densities. Information from these atomic densities is then mapped to Lennard-Jones parameters using a set of mapping parameters much smaller than the typical number of atom types in a force field. This approach is advantageous in two ways: it eliminates atom types by allowing each atom to have unique Lennard-Jones parameters, and it greatly reduces the number of parameters to be optimized.


How to make lead nitrate from pure lead Is it possible to add copper nitrate solution and lead so that a single displacement reaction occurs to yield lead nitrate. I recently melted lead and added cooper nitrate to the molten lead. I then collected the copper nitrate solution I had poured and repeated the process of melting the lead and pouring the solution over it. I figured that eventually the copper will be displaced by the slightly more reactive lead. Will this method yield lead nitrate?

I am not sure of putting copper nitrate to lead. But the common way to produce lead nitrate is to put lead into nitric acid. It will produce lead nitrate and hydrogen gas.

$\ce{Pb + 2HNO_3 -> Pb(NO_3)2 + H_2}$

If you have lead oxide instead of lead it also can priduce lead nitrate by adding to nitric acid

$\ce{PbO + 2HNO_3 -> Pb(NO_3)2 + H2O}$


Froth floatation process - why CuSO4 is activating floating character of ZnS? I read the following :

  "In a mixture of $\ce{PbS}$, $\ce{ZnS}$ and $\ce{FeS2}$, each component is separated from each other in froth floatation process by using the reagents potassium ethyl xanthate, $\ce{KCN}$, $\ce{NaOH}$, copper sulphate and acid. Here potassium ethyl xanthate acts as a collector. $\ce{KCN}$ and $\ce{NaOH}$ depress the floatation property of $\ce{ZnS}$ and $\ce{FeS2}$ particles. Thus, only $\ce{PbS}$ particles go into the froth. Now copper sulphate is added to the mixture which activates floating character of $\ce{ZnS}$ and this time only $\ce{ZnS}$ comes along with froth. The remaining slurry is acidified and $\ce{FeS2}$ floats along with froth."

Can someone help me in exactly understanding the role of copper sulphate clearly along with its specific property associated?

I'm going to consult the literature later, but for me it seems to be as follows.

In the compounds discussed, there are following cations:

$\ce{Fe^2+}$. It is a transition metal with a moderate affinity to oxygen, a slightly larger affinity to sulfur and a high affinity to cyanide.
$\ce{Zn^2+}$. It is a small cation with a high affinity to oxygen, a slightly larger affinity to sulfur and a moderate affinity to cyanide.
$\ce{Pb^2+}$. It is a large cation with a moderate affinity to oxygen and a slightly larger affinity to sulfur.

Xanthate bind to the surface via the sulfur atoms. So, in a basic cyanide solution the binding sites on the surfaces of pyrite and sphalerite are both completely filled with cyanide and hydroxide respectively (please, note that sulfur in xanthate is a lot less nucleophilic than in sulfide form). Copper has a high affinity to sulfur, a lot larger than zinc or iron. So it binds to xanthate and to the sulfur in either pyrite or sphalerite. However, the surface of pyrite is still occupied with cyanide, which binds really strongly there. And in an acidic environment, cyanide is transformed into $\ce{HCN}$ and thus the pyrite surface is open to binding with xanthate.


Theoretical synthesis of dimetinden I tried to find a way to synthesize Dimetinden (N,N-dimethyl-2-[3-(1-phenylethyl)-1H-inden-2-yl]ethan-1-amine) with an Indensynthesis

https://www.thieme-connect.com/products/ejournals/abstract/10.1055/s-0030-1261196

First of all I would like to know if there is anything wrong with the whole synthesis and how about the yield?

Second question would be the mechanism of the inden synthesis, because it looks kind of wrong.

A couple of things, first, inden-1-ol tends to tautomerize into the corresponding 1,3-dicarbonyl.  Second, when adding TsCl to an alcohol, the group formed is -OTs not -OTMS, the latter is a silyl ether.  The reaction with addition of TsCl requires a base such as pyridine, and because of the presence of a 1,3-dicarbonyl, rather than being added to the alcohol, it will be added to the carbon alpha to both the carbonyls.  Third, the reaction of Grignard reagent should be avoided due to loss of yield because of interaction with the carbonyl.  Fourth, the inden-1-ol formation is done in the presence of a base (to form enolates) which is then followed by an acid wash(to form enols which tautomerize to ketones), your mechanism seems okay if enolization occurs instead of tautomerization.  Finally, the correct structure of Dimetindene is below:

https://en.wikipedia.org/wiki/File:Dimetindene.svg

I will briefly explain how I would do it from the indenol you described.  First, would be a Wittig reaction using 1-(pyridin-2-yl)ethylenetriphenylphosphorane, (which can be prepared from 2-acetylpyridine) and the indenol tautomer.  This will yield a beta-gamma unsaturated ketone.  This product can undergo a photochemically induced [1,3] hydride shift to form an alpha-beta unsaturated ketone.  This can be subject to Wolff-Kishner conditions to form Dimetindene.


In Simonini reaction how does the $2:1$ ratio of reagents enable formation of ester as major product? The Wikipedia article on Simonini reaction says:

  The ratio of the reagents play an
  important role in the determination of products, namely, if $1:1$
  ratio of salt and iodine is used alkyl iodide is formed. On the other
  hand, a $2:1$ ratio gives $\ce{RCOOR}$ and $3:2$ ratio gives both the
  products.

For $1:1$ ratio I understood why alkyl iodide is formed (found the mechanism here). But for $2:1$ ratio why is ester formed as major product ?

If you have the extra equivalent of the carboxylate, then this carboxylate will never react with iodine to form the alkyl iodide. Instead, it will do the next best thing, which is $S_{\mathrm{N}}2$ reaction with the alkyl iodide that you did make.

Note that this probably only works if R is primary.


Does a gold detector exist? I see many sites that claim to sell working gold detectors but I am not convinced that they work. Also, does a metal detector detect gold?

You ask:

  Does a gold detector exist?

Absolutely. Here's a metal detector that is fine-tuned for gold detection, available ${\it via}$ Amazon.

That said, if you are looking for something that detects only gold, I'm not sure the best way to proceed in that case. This site appears to be a comprehensive review of some popular precious metal-tuned detectors on the market. The site is maintained by a person with an active interest in prospecting and metal detection. It is stated on the linked page that:

  A quick note to those who know nothing about these machines. These are metal detectors. There is no such thing as a "gold only" detector. These detectors will also find lead, copper, aluminum, and other metals. These units are best used to look for relatively larger pieces of gold at relatively shallow depths. Concentrations of gold dust are not detectable. Some of these units can hit gold that weighs as little as a grain (480 grains per ounce) or less but only at an inch or two. Only the larger nuggets can be found at depths exceeding a foot. Only world class nuggets weighing many ounces can be detected at over two feet. The vast majority of nuggets found are found at inches, not feet.

In any event, you can purchase a device which has been engineered to discriminate base metals from precious ones (such as the model I have linked here).

  Also, does a metal detector detect gold?

Yes - see the reviews for the product offered at Amazon (and many others like it) as well as the informational site I've linked (above). Also see the Wikipedia page on metal detectors for more information, including the issues relating to fine-tuning the devices to discriminate between metals with similar/different properties "detected" by the devices.


Choice of doping elements in semiconductors In a typical semiconducting material, we start with silicon (group IV) and introduce atoms of group V or group III depending on whether we are constructing an n- or p-type semiconductor respectively.

Why is it optimal to choose doping atoms one group away? Why not introduce atoms from groups VI or II, for instance?

Look at the II-VI options from a chemical perspective.

For silicon, Group III boron p-doping would be replaced by Group II beryllium doping, and Group V phosphorus n-doping would be replaced by Group VI sulfur doping.

The beryllium would tend to act like a Lewis base that donates a pair of electrons to some nearby silicon atom, forming a stronger and more localized bond than is possible with the odd electron count created by boron doping. Similarly, the sulfur would tend to act like a Lewis acid and form a localized sulfide bond with some nearby silicon atom, versus the odd electron count created by phosphorus.

So, the main reason for picking elements immediately "next door" (off by one electron) is to avoid introducing paired electrons, which on average will bond more strongly and more locally via a Lewis base-acid process.

That said, there's enough complexity going on there that it's still possible that you could get II-VI or maybe even I-VII doping to work in the right situations.


What is the mechanism for the auto-dissociation of nitrogen tetraoxide How does nitrogen tetraoxide dissociate into nitrosyl ion and nitrate ion?

Is it similar to the auto-dissociation of $\ce{SO2}$ into thionyl ion and sulfite ion by acceptance and donation of oxide ion?

I couldn't find a mechanism for the auto-dissociation of nitrogen tetraoxide.

Dinitrogen tetroxide $\ce{N2O4}$ is a colourless liquid composed of planar dimers of nitrogen dioxide $\ce{NO2}$.$^{[1]\ \text{[a]}\ [2]}$

As Martin noted and is to be expected from the structure of $\ce{N2O4}$, the process

$$\ce{N2O4 <=> NO+ + NO3-}$$

is not very likely to happen. Autoionisation of $\ce{NO2}$ is more common$^{[1]}$.

$$\ce{NO2 -> NO2+ +  e^-\\ NO2 +  e^- -> NO2-}$$

Ionisation energy of $\ce{NO2}$ is $9.8\ \mathrm{eV}$ while its electron affinity equals $1.62\ \mathrm{eV}$.$^{[1]}$ Technically you could look at this as heterolytic bond dissociation $\ce{N2O4}$ but that is probably not what is happening. This autoionisation is, in a way, realised for aqueous solutions$^{[2]}$:

$$\ce{\underset{electron transfer}{NO2 + NO2} + H2O -> \underset{really just HNO3}{NO2OH} +  HNO2},$$
$$\ce{\underset{electron transfer}{NO2 + NO2} + 2KOH -> KNO3 + KNO2 +  H2O}.$$

Table 1. Melting and poiling points of $\ce{NO2}$ and $\ce{N2O4}$. Note the "equality" of melting points.$^{[2]\ [3]}$
$$\begin{array}{|c|c|c|}
\hline
\mathbf{Molecule} & \mathbf{mp\ (\deg  \mathrm{C})} &  \mathbf{bp\ (\deg  \mathrm{C})}\\
\hline
\ce{NO2} & -11.2 & 21 \\
\ce{N2O4} & -11.2 & 22 \\
\hline
\end{array}$$

Table 2. Percentage of  $\ce{NO2}$ and $\ce{N2O4}$ at various temperatures (pressure $1 \ \mathrm{atm}$).$^{[2]}$
$$\begin{array}{|c|c|c|c|}
\hline
\mathbf{Molecule} & \text{below}\  \mathrm{mp}\ (\mathrm{\%}) &  \mathbf{40\ \deg  \mathrm{C}\ (\mathrm{\%})} & \text{over}\mathbf{\  150\deg  \mathrm{C}\ (\mathrm{\%})}\\
\hline
\ce{NO2} & 0 & 31 & 100 \\
\ce{N2O4} & 100 & 69 & 0 \\
\hline
\end{array}$$

$\text{[a]}$ To the best of my knowledge, the enthalpy $-55\ \mathrm{kJ}$ applies at $20 \deg \mathrm{C}$ and $1\  \mathrm{atm}$ for both components in the gas phase. NIST gives $-57.12\ \mathrm{kJ}$ at $25 \deg \mathrm{C}$ and $1\  \mathrm{bar}$. See $\ce{N2O4}$ and $\ce{NO2}$.

$[1]$ N. N. Ahmetov. Anorgaaniline keemia. (1974) (pages 375, 376)

$[2]$ H. Karik, Kalle Truus. Elementide keemia. (2003) (pages 424, 426)

$[3]$ Wikipedia:

Nitrogen dioxide
Dinitrogen tetroxide


Combustion of Rocket-Candy with Fe2O3 We know from Wikipedia that the combustion of rocket candy follow this equation :

$\ce{48 KNO3 + 5 C12H22O11 → 55 H2O + 36 CO2 + 24 K2CO3 + 24 N2}$

We can add small quantity of iron (III) oxide  ($\ce{Fe2O3}$) as a catalyst.

The new equation look like this :
$\ce{KNO3 + C12H22O11 + Fe2O3 -> H2O + CO2 + K2CO3 + N2 + ?}$

What product does the combustion of iron (III) oxide  produce ?

I think you may be missing a key point to catalysis; that the catalyst is not net consumed or altered by the reaction. If you were to put the catalyst in the chemical equation as in your second reaction as a reactant, you would technically need to also put it on the right hand side as a product. However, this is not the way a catalyzed reaction is typically indicated. A common way to indicate that a catalyst is used would be as follows:

$$\ce{48KNO3 + 5C12H22O11 ->[Fe2O3] 55H2O + 36CO2 + 24K2CO3 + 24N2}$$


Classically permitted position values in a quantum harmonic oscillator In  a quantum harmonic oscillator that is in the state with quantum number $n$, what range of the position $x$ is allowed classically? Does it have something to do with Bohr's correspondence principle?

Assume for a minute that instead of having a quantum oscillator with total energy $E$ you had a classical one with the same amount of energy. That energy is the sum of potential and kinetic contributions:
$$E = E_\textrm{p}+E_\textrm{k}.\tag{1}$$
$E_\textrm{k}$ is zero and the total energy equal to the potential energy when the extension (amplitude) is at a maximum, when
$$E = E_{\textrm{p,max}} = \frac12 kx_\mathrm{max}^2\tag{2}$$
which, solving for $x_\mathrm{max},$ leads to
$$x_\mathrm{max} = \sqrt{\frac{2E}{k}}.\tag{3}$$
All that is left to do is to substitute in the right value for $E.$ That is dictated by the expression for the quantized system,
$$E_n = h\nu\left(n + \frac12\right),\tag{4}$$
so that the corresponding classical maximum amplitude is
$$x_\mathrm{max} = \sqrt{\frac{h\nu(2n + 1)}{k}}.\tag{5}$$


What volume of water will dilute an acidic solution back to neutral? Now we are talking about simple dilution, not acid-base neutralization, so we expect asymptotic behavior that never quite gets us back to pH=7; that is, an infinite amount of water would be required. However, most pH probes read to two decimal places, so diluting to reach pH=6.995 is a sufficient end condition. So, given an initial pH of $pH_0$ and an initial solution volume $V_0$, what additional volume $V$ of pure water would be required to dilute the pH to 7.00 (6.995)?

I've wondered this for a while, and today I attempted to solve it. I may have found the solution, but I am skeptical of its implications. If I have 1 L of a solution at pH=2, my derived formula indicates it would require additional volume of about 8.6 million liters of water for the solution to read neutral. Is this reasonable?

My derived formula:

$$V=\Big(\frac{1}{1-10^{-0.005}}\Big)\Big(10^{6.995-pH_0}-1\Big)V_0 \\
\approx \Big(86.36\cdot10^{7-pH_0}-87.36\Big)V_0$$

Derivation:

Moles of $H^+$ from original soln: $n_0 = [H^+]_0V_0$

from added water: $n = [10^{-7}]V$ (via autodissociation)

Total moles: $n'=n_0+n$

Total volume: $V'=V_0+V$

New molarity after addition of $V H_2O$:

$$[H^+]'=\frac{n'}{V'}=\frac{[H^+]_0V_0+[10^{-7}]V}{V_0+V}$$

$$\Rightarrow V=\Big(\frac{[H^+]_0-[H^+]'}{[H^+]'-[10^{-7}]}\Big)V_0$$

(Note that $V \to \infty$ as $[H^+]' \to 10^{-7}M$, an infinite volume is require to exactly reach neutral pH).

Letting $[H^+]'=10^{-6.995}M$ (paragraph 1), $[H^+]_0=10^{-pH_0}$, and multiplying top and bottom by $10^{6.995}$, obtain:

$$V=\Big(\frac{10^{6.995-pH_0}-1}{1-10^{-0.005}}\Big)V_0$$

Question: Is this derivation valid? Are there assumptions I should or should not have made? Would it truly take 8.6 million times the original volume to dilute a pH=2 solution back to neutral pH=7.00?

The very idea of diluting a solution back to (almost) neutral with pure water is unreasonable, precisely because it would require insane amounts of water. That being said, your derivation (though impractical) is not meaningless; it is just slightly wrong.

Like J. Ari said in the comments, there is an equilibrium involving $\ce{H+}$ and $\ce{OH-}$, and it will be shifted by the dissolved acid (that is, the concentrations will shift so the equilibrium constant may stay the same). Therefore you can't assume that water will always contain precisely $10^{-7}$M protons from autodissociation (plus maybe some extra from acid). It doesn't work that way.

How does it work, then?

$$\ce{[H+]\cdot[OH-]=10^{-14}}\\
\ce{[H+]-[OH-]=}c(\rm{HCl})$$
(since the acid is strong, and hence fully ionized even in not-so-diluted solutions)
$$\ce{[H+]=10^{-6.995}}\\
\ce{[OH-]=10^{-7.005}}\\
c\ce{(HCl)=10^{-6.995}-10^{-7.005}=10^{-6.995}(1-10^{-0.01})}\\
V_{total}=V_0\cdot{c_0\over c}\\
V_{added}=V_0\cdot\left({c_0\over c}-1\right)
=V_0\cdot\left({10^{6.995-pH_0}\over1-10^{-0.01}}-1\right)$$
So it is only 4.3 million times the original volume, which is about 2 times less than your estimate.

Not that it made much of a difference, really.


Balancing chemical reactions, issues of Gauss's elimination method I am trying to balance some chemical reactions by using the Gauss' elimination method and I noticed that this method does not work if the creation of a matrix with exactly one degree of freedom is not possible.

In this example

$$\ce{H2O2 + NO3- + H+ → O2 + NO + H2O}$$

I succeeded in creating just 4 equations

$$\ce{\alpha H2O2 + \beta NO3- + \gamma H+ → \delta O_2 + \epsilon NO + \zeta H2O}$$

$$
\left\{
\begin{aligned}
2\alpha+\gamma&=2\zeta &\qquad &(\ce{H})\\
2\alpha+3\beta&=2\delta+\epsilon+\zeta &\qquad &(\ce{O})\\
\beta&=\epsilon &\qquad &(\ce{N})\\
-\beta+\gamma&=0 &\qquad &(\text{electrons})
\end{aligned}
\right.
$$

obtaining

$$(\alpha, \beta, \gamma, \delta, \epsilon, \zeta)=\left(\zeta-\frac{1}{2}\epsilon, \epsilon,\epsilon,\frac{\zeta+\epsilon}{2}, \epsilon, \zeta\right)$$

I know that the solution should be $(3,2,2,3,2,4)$, but, in having two degrees of freedom, I am not able to choose the two free variables in order to obtain what I expect, i.e. $\epsilon=2$ and $\zeta=4$.

Is there any (just one more) equation that I missed?
Is there anything else that I have to add further?
Or do I have to consider that Gauss' elimination fails with problems with a number of free variables $\neq 1$?

Your "should-be" solution,

$$\ce{3H2O2 + 2NO3- + 2H+ → 3O2 + 2NO + 4H2O}$$

is a linear combination of

$$\ce{H2O2 + 2NO3- + 2H+ → 2O2 + 2NO + 2H2O}$$

and

$$\ce{2H2O2  → O2 +  2H2O}$$

Any other linear combination is valid as well.


Is glass an amorphous solid or supercooled liquid? I have been informed that glass is a super-cooled liquid and is also considered to be an amorphous solid.

Can it be both and, if not, what category does it fall into?

You are correct: it's not both.

Why? What you've heard is nonsense. Very old window panes are thicker towards one side because of how they were made at the time. It became common practise to put the thicker edge at the bottom, because it was less likely to exceed the acceptable stress on the thinner area.

Contemporary windows have been found with the thicker edge to one side, or even the top. Gravity on Earth has been consistently towards the core since at least the Middle Ages, so the "flowing glass" hypothesis is false.

Glass is an amorphous solid by definition (per Karl's comment) and is not a super-cooled liquid (per Prabhdeep Singh's comment).


Is this the correct equation for the mean free path? $\lambda=\frac{KT}{\sigma p}$ is the equation given in my textbook but everywhere else I look it is different. Is it wrong? The derivation is as follows:

$$\lambda = \frac{v_{rel}}{z}$$ where $\lambda$ is the mean free path and $z$ is the collision frequency.

Since $z=\frac{\sigma v_{rel}}{KT}$ where $\sigma$ is the area where it is possible for two particles to collide.

Then substituting for $z$ gives the equation $\lambda=\frac{KT}{\sigma p}$. Have I gone wrong somewhere. I often see a factor of root two in the denominator but I don't see where that might come from.

Your derivation is correct. The $\sqrt{2}$ comes from $v_{rel}=\sqrt{2}v_{mean}$, where $v_{rel}$ is the mean relative speed (relative to the other molecules in the gas) and $v_{mean}$ is just the mean speed of the molecules in the gas.

In conclusion: note what the subscript says about the variable!


What is the difference between a ylide and a betaine? I'm not sure what the exact difference between a ylide and a betaine is...

Are both zwitterions?

Something along the lines of: ALL ylides are betaines, but not all betains are ylides(according to wikipedia the cation/anion group can be adjacent or nonadjacent)?

The more definitive source for chemistry terminology is the IUPAC Gold Book.

Betaine

  Originally, the compound betaine, $\ce{(CH3)3N+–CH2C(=O)O−}$ (N,N,N-trimethylammonioacetate), and similar zwitterionic compounds derived from other amino acids. By extension, neutral molecules having charge-separated forms with an onium atom which bears no hydrogen atoms and that is not adjacent to the anionic atom. Betaines cannot be represented without formal charges.

Ylides

  Compounds in which an anionic site $\ce{Y−}$ (originally on carbon, but now including other atoms) is attached directly to a heteroatom $\ce{X+}$ (usually nitrogen, phosphorus or sulfur) carrying a formal positive charge. They are thus 1,2-dipolar species of the type $\ce{R_mX+–Y−R_n}$.

So, ylides are not betaines and betaines are not ylides. Primarily due to the atoms in a ylide that contain the charges must be adjacent, and in betaines they must not be adjacent.

An example of a ylid would be the phosphonium ylid used in Wittig chemistry:

Since it has adjacent positive and negative charges, it is classified as a ylid.

The addition of this ylid to a carbonyl compound would form a betaine (note that the mechanism of the Wittig reaction is now thought to be a concerted [2+2] cycloaddition leading to the oxaphosphetane immediately, without the intermediacy of the betaine):

Note that the ylid has an alternate resonance form (the one with a P=C double bond) which has no formal charges, whereas the betaine cannot be drawn without formal charges.


Why can't saliva clean "stickiness"? Why is it that when you consume something that is sugary-sticky (like candy, chocolate, fruit juice, etc.) and get it on your skin, when you try to clean it off with saliva (particularly, licking/sucking it off) it doesn't work?  What is it about the makeup of saliva that doesn't dissolve the sugar?

So I could have sworn I've cleaned stuff off with saliva before, but I wanted to make sure, so I ran a quick impromptu experiment.

I slathered the underside of my left arm with some maple syrup I had lying around. I waited about 10 minutes for it to dry, then I mentally sectioned off my arm into three parts and cleaned each part off in different ways:

The section closest to my hand was cleaned by putting my other hand under running water and wiping the syrup off (basically, getting my right hand wet and making wiping motions across the arm, then returning the hand to the water, until I felt like I was no longer making any progress).

The middle section was left untouched.

The section closest to my elbow was cleaned by licking my hand, then using it to wipe that section down (essentially the same as the water, except I used my mouth as the source of liquid).

After the cleaning, I let everything dry again for 3 minutes. Then I placed some toilet paper over my entire arm and attempted to remove it. Here are the results:

(Sorry for the horrid image quality)

Unfortunately, I didn't have any skin marking tools on me at the time (I use pretty much only water-soluble inks), so I had to add the lines afterwards.

As we can see, the section of my arm that was cleaned with water and with saliva are pretty much TP free, aside from a small section near my elbow that I think was just me missing some spots. The section cleaned with water was similarly TP-free. The middle section is pretty heavily TP-ed. What you're seeing is what got left behind when I attempted to rip the TP free.

So it seems like saliva actually can clean off stickiness pretty well! This is backed up by the comment that Michiel made, which points out that saliva is 99.5% water.

That being said, running this experiment gave me a possible explanation for why it might seem like it doesn't: it takes a LOT of saliva to do this!

With the water, I was easily able to wash off all of the syrup off of my hand and get a new coating of water on my fingers. Unfortunately, I can't generate that much saliva at once, and so I couldn't wash off the sugar that was already on my fingers that effectively. That made the saliva cleaning take a LOT longer than washing with water, both in terms of how many wipes it took to clean off most of the stuff and how much time it took overall (sadly, I cannot drool at the same rate a bathroom sink outputs water).

There's another possible explanation that springs to mind: when you have syrupy stuff stuck to you, you probably just ate some of the syrupy stuff. That means your mouth probably still has a good bit of sugar in it, which would also make it harder to clean off additional sugar.

Still, based off of this experiment, I feel fairly confident in saying that saliva cleans off sugar stickiness almost as well as regular old water, as long as you have enough of it lying around.

NOTE: I initially tried this with a flavored whiskey, hoping the alcohol would help it dry faster. Terrible mistake! Not only was there not enough sugar in the whiskey to get any sort of real stickiness, most of it basically ran off of my arm right away, so I ended up losing a lot of it. Save your flavored sprits for drinking, use a syrup!


Can carbon be the central atom of a complex? Can it form a quadruple bond with itself? Why does $\ce{[C(NH3)2]}$ not exist as a complex? Here, $\ce{C}$ is carbon and the configuration of $\ce{C}$ in ground state is $\ce{[He] 2s^2  2p^2}$.

Can't the two $\ce{NH3}$ molecules pair the two $\ce{p}$ electrons of carbon and form coordinate bond with carbon?

Why does $\ce{C2}$ (one carbon atom bonded with other through quadruple bonds) not exist? Two $\ce{2p}$ orbitals of $\ce{C}$ should form π bonds and one $\ce{2p}$ and one $\ce{2s}$ should form σ bonds with the other $\ce{C}$ (without hybridisation).

I'd appreciate an answer for a beginner.

Asking multiple unrelated questions at once is considered a bad practice, but let's forget that for a moment.

Your first molecule has more or less satisfactory electronic configuration, but that's not enough for the long-term survival. N would be more than willing to relinquish a proton, and C to accept it, and thus in no time you will have $\ce{H2N-CH2-NH2}$, which then may or may not transform into something else.

Your second molecule has been discussed here before on multiple occasions. Long story short, it does exist, but doesn't have a quadruple bond. (Such bonds are known, but not in $\ce{C2}$; they have a delta component, and that requires d orbitals). The problem with your description is that the sigma-oriented 2p orbital of one carbon will interact not just with the similar 2p, but also with the 2s orbital of the other. If not for this fact, it would work out pretty much as you expected.


Would copper sulfate corrode copper metal over time? I'm performing electrodeposition experiments with copper electrodes (anode and cathode) and a copper sulfate electrolyte solution. For my project, the cells assembled would need to be stored for a period of months. I was wondering if the copper sulfate can corrode the copper metal and how long it would take? The copper sulfate concentrations range from 5 mM to 100 mM.

$\ce{CuSO4}$ will not corrode pure copper appreciably over a few months (though there is some dissolution and redeposition through random processes, such as density gradients and thermal motion, it might just reduce a mirror finish to slightly roughened).

However, if the electrodes or wires are brass, bronze or other alloy, or if there are other metals in contact, such as a tin-lead solder joint, there is a chance of some corrosion. In fact, copper sulfate is used to test passivation of stainless steel, and visible corrosion of iron can take place in minutes.


Why there is no change in water level when salt is added? Let us say we have one glass of water and after that when we add one or two spoon of salt then we notice that salt dissolves in it but when we measure the water level we found that there is no difference between initial reading and final reading. So what's happening over there ? Why water level doesn't changed when salt is added ?

This is an example of electrostriction - see, for example, this paper for a full explanation.

In short, the solvent molecules become more ordered in the vicinity of dissolved (charged) ions and less ordered as we go further away from the dissolved (charged) ions. The increase of solvent density near the ions is offset by the decrease of bulk density in the solvent as distance from the ions increases. This results in an overall net reduction in the observed volume of the solution.

From the Discussion section of the manuscript:

  The interaction of the electrostatic field of an
  ion with water tends to align the dipolar water
  molecules in the direction of the field. In this way
  the field tends to disrupt hydrogen bonded
  structures in liquid water, and to compress the
  water molecules surrounding an ion. These
  electrostatic effect give rise to a shrinkage of the
  water.


Oxidation of an aldehyde - how is this a loss of electrons? I've been reviewing redox reactions and have been using the OIL RIG acronym to classify oxidations vs. reductions. Generally this makes sense, and many resources I've looked at repeat the "Oxidation Is Loss - Reduction Is Gain of electrons" statement.

But, I also know that aldehydes can be oxidized to carboxylic acids:
$$
\ce{RCHO -> RCO2H}
$$

The compound $\ce{RCHO}$ does not lose electrons when oxidized to $\ce{RCO2H}$ -- in fact, it gains a whole new oxygen atom. So, there appears to be a net gain of electrons here. Does OIL RIG not always hold, or am I misunderstanding something?

OILRIG always holds, but you need to be careful at what you are actually looking at. When we say a compound gets oxidised, that is technically not true. In this framework, only atoms get oxidised or reduced.
In the reaction
$$\ce{R-CHO + [O] -> R-COOH}$$
we usually omit the reduction part, because it's always the same: the oxygen.

To identify the whole redox process, use oxidation numbers. Keep in mind that these are simply a bookkeeping tool and the partial charges of the atoms involved might be quite different. Setting $\ce{R {=} CH3}$ or the above reaction it boils down to this:
$$\ce{
\overset{+1}{H}_3\overset{-3}{C}-\overset{\color{\red}{+1}}{C}\overset{+1}{H}\overset{-2}{O}
+ [\overset{\color{\green}{0}}{O}]
->
\overset{+1}{H}_3\overset{-3}{C}-\overset{\color{\red}{+3}}{C}\overset{-2}{O}\overset{\color{\green}{-2}}{O}\overset{+1}{H}
}$$

Now you can see that the carbonyl carbon increases its oxidation state from $\color{\red}{+1}$ to $\color{\red}{+3}$, so formally it lost $\pu{\color{\red}{2} e^-}$. The oxidising agaent, indicated with $\ce{[O]}$, at the same time decreases its oxidation state from $\color{\green}{0}$ to $\color{\green}{-2}$, therefore formally gaining $\pu{\color{\green}{2} e^-}$. The redox reaction is complete.

Let's look at a real world example, the Baeyer-Villiger oxidation:

As you can see, the ketone gets oxidised, more precisely the carbonyl carbon gets oxidised as increases its oxidation state. On the other side, the peroxide oxygens get reduced as they decrease their oxidation state.


Making Cast Iron from Pig Iron In making cast iron from pig iron - the cruder form of iron - coke and limestone are added to pig iron and scrap iron in a furnace. Cast iron is   more refined form of iron and has lesser carbon content as well. What exactly occurs in this process of making cast iron - what reactions happen in the furnace? Why is coke added as well, when the purpose is to decrease the carbon content?

I have quoted the following answers from the UC Davis ChemWiki page on "The Extraction of Iron."

  Why is coke added as well, when the purpose is to decrease the carbon content?

Coke is used primarily as a reducing agent (see below) and comes out as carbon dioxide (and thus does not increase the carbon content of the molten iron).

  What exactly occurs in this process of making cast iron - what reactions happen in the furnace?

The reactions are given below in the narrative of the process.

The Blast Furnace

The common ores of iron are both iron oxides, and these can be reduced to iron by heating them with carbon in the form of coke. Coke is produced by heating coal in the absence of air. Coke is cheap and provides both the reducing agent for the reaction and also the heat source. The most commonly used iron ores are hematite, ($\ce{Fe_2O_3}$), and magnetite, ($\ce{Fe_3O_4}$).

The air blown into the bottom of the furnace is heated using the hot waste gases from the top. Heat energy is valuable, and it is important not to waste any. The coke (essentially impure carbon) burns in the blast of hot air to form carbon dioxide - a strongly exothermic reaction. This reaction is the main source of heat in the furnace.

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

The Reduction of the Ore

At the high temperature at the bottom of the furnace, carbon dioxide reacts with carbon to produce carbon monoxide:

$\ce{C + CO_2 \rightarrow 2CO}$

It is the carbon monoxide which is the main reducing agent in the furnace:

$\ce{Fe_2O_3 +3CO \rightarrow 2Fe + 3CO_2}$

In the hotter parts of the furnace, the carbon itself also acts as a reducing agent. Notice that at these temperatures, the other product of the reaction is carbon monoxide, not carbon dioxide:

$\ce{Fe_2O_3 + 3C \rightarrow 2Fe + 3CO}$

The temperature of the furnace is hot enough to melt the iron which trickles down to the bottom where it can be tapped off.

The Function of the Limestone

Iron ore is not pure iron oxide - it also contains an assortment of rocky material that would not melt at the temperature of the furnace, and would eventually clog it up. The limestone is added to convert this into slag which melts and runs to the bottom. The heat of the furnace decomposes the limestone to give calcium oxide.

$\ce{CaCO_3 \rightarrow CaO + CO_2}$

This is an endothermic reaction, absorbing heat from the furnace. It is therefore important not to add too much limestone because it would otherwise cool the furnace. Calcium oxide is a basic oxide and reacts with acidic oxides such as silicon dioxide present in the rock. Calcium oxide reacts with silicon dioxide to give calcium silicate.

$\ce{CaO + SiO_2 \rightarrow CaSiO_3}$

The calcium silicate melts and runs down through the furnace to form a layer on top of the molten iron. It can be tapped off from time to time as slag. Slag is used in road making and as "slag cement" - a final ground slag which can be used in cement, often mixed with Portland cement.

Cast Iron

The molten iron from the bottom of the furnace can be used as cast iron. Cast iron is very runny when it is molten and doesn't shrink much when it solidifies. It is therefore ideal for making castings - hence its name. However, it is very impure, containing about 4% of carbon. This carbon makes it very hard, but also very brittle. If you hit it hard, it tends to shatter rather than bend or dent. Cast iron is used for things like manhole covers, guttering and drainpipes, cylinder blocks in car engines, Aga-type cookers, and very expensive and very heavy cookware.


Why is a ketone not capable of hydrogen bonding among themselves? The ketone has one oxygen atom in it and there are two lone pairs hanging out, both of which make it seemingly capable of hydrogen bonding. But in reality it is not.

Did some research and found this on chemguide:

  In methoxymethane, the lone pairs on the oxygen are still there, but the hydrogens aren't sufficiently δ+ for hydrogen bonds to form. Except in some rather unusual cases, the hydrogen atom has to be attached directly to the very electronegative element for hydrogen bonding to occur.

Based on the info, I am assuming that the carbon atom in carbonyl group is not electronegative enough to give a partial charge large enough for hydrogen bonding to the hydrogen atoms, say in pentanone.

Short answer: the ketone oxygen can participate in hydrogen bonding

The compound you mention, methoxymethane, is an ether not a ketone.

Ethers don't hydrogen bond very well because the oxygen isn't very polarized.

On the other hand, the carbonyl in a ketone is polarized.  We can draw resonance structures that show this polarization.

(image source)

The negative polarization on the carbonyl oxygen allows it to play the role of hydrogen bond acceptor as illustrated in the following drawing.

(image source)

Edit: If you are asking if the hydrogens on the carbon adjacent to the carbonyl carbon (or any other carbon in the backbone of a simple ketone) can participate as the donor in a hydrogen bond, then the answer to this question is "no". These $\ce{C-H}$ bonds are typically less  acidic and less polar than the $\ce{O-H}$ bond in water.

Therefore, while the oxygen in pentanone can participate in hydrogen bonding with donor hydrogens in other molecules, the hydrogens in pentanone cannot participate in hydrogen bonding.  Pentanone cannot hydrogen bond with itself.


Polypropylene 55 gallon drum Can a polypropylene 55 gallon drum previously filled with chemicals used in the chrome plating process (possibly Nickel Sulfate) ever really be "cleaned out" for re-use? Can polypropylene ever absorb and retain such chemicals, or are these drums practical for this use because they do not?

Polypropylene is hydrocarbon and does not adsorb polar and ionic compounds. So, several treatments with mild hydrochloric acid followed with several treatments with clean water should work. However, I wouldn't store drinking water there without a proven test. For example, leave water there for several weeks and then perform analysis of the water for metal ions. If all concentrations are within safe levels, it should be alright.


Is it possible to produce graphene out of air? In theory you should be able to turn $\ce{CO2}$ to $\ce{CO}$ and then to $\ce{C}$ and connect all the C-atoms together to form graphene.
That should be possible, because plants do something similar, they take $\ce{H2O}$ and $\ce{CO2}$ and link the C-atoms together as $\ce{C6H12O6}$.

Is it possible to produce graphene out of air? Here is a related article from the RCS: "Conjuring graphene oxide from thin air"

A graphene sheet is easily obtained from a block of graphite using adhesive tape. Since graphite itself is essentially a pure grade of coal, it is the result of biological processes in vegetation some millions of years ago. As is pointed out in answers to this Biology.StackExchange question, the majority of a plant's carbon comes from the air via photosynthesis.

So in a sense, we already do that, but we just use the huge stores of graphite from millions of years of high pressure processing beneath our feet rather than start from the beginning.


Identifying Meso compounds (source: iupac.org)

I can't seem to understand why these compounds are meso compounds..... I don't see any plane of symmetry here. I have trouble identifying meso compounds so can you give me heads up on how to identify them

Why don't use the definition just above the image?

  meso-compound

  A term for the achiral member(s) of a set of diastereoisomers which also includes one or more chiral members.

Achiral, i.e. the molecule is superimposable with its mirror image. So, if finding a plane of symmetry, or other symmetry elements in the molecule is difficult for you, a simpler method (and according to the definition) might be drawing a mirror image, i.e. all chiral elements inverted

or unchanged but drawn "right-to-left", i.e. flipped

or flipped vertically

.. (just use any of the mirroring methods) and deciding whether it's the same molecule (stereoisomer), by rotating it, in 3D, in your head. (Some internal rotations around single bonds might be needed as well.)

In the first case, the mirrored structure can be superimposed with the original one by in-plane 180° rotation around an axis perpendicular to the paper plane.

In the second case it can be done by 180° rotation around the in-plane horizontal axis.

In the third case it can be done by 180° rotation around the in-plane vertical axis.


pH effect in the hydrophobic interactions between two polypeptide chains The rise in pH can change the protein conformation by changing the ionic interactions and hydrogen bonds between two polypeptide chains. This is a fact that I understand very well because the change in amino-acid ionization with pH. I would like to know if by changing the pH value the hydrophobic interactions between two chains can also be changed. If yes, what is the mechanism?
Article: a pH Induced increase in hydrophobicity...

Here is one possible mechanism:

In tightly folded, compact, "globular" protein structures, the most hydrophobic bits of the protein tend to be in the core of the structure, away from the surface.

If the pH changes, then by mechanisms you mention in your question, this structure can be disrupted.  If it is side chain interactions among surface-oriented, polar residues that "hold together" the globular structure, then a pH change can disrupt those interactions, allowing the previously isolated hydrophobic "core" to become  exposed.  This core can now bind more surfactant molecules, aggregate with itself, or do any number of other "hydrophobic" things.


Why is sulfuric acid used to seal flask with highly reactive chemicals? I the video https://youtu.be/saANxD0cqy0?t=200 (time 3:20) NileRed states that he uses sulfuric acid to seal off flask with a mixture of pottasium dichromate and table salt. He says that sulfuric acid is used to seal off flasks with highly reactive chemicals and I could not find any reasons why.

I see sulfuric acid as a very reactive chemical, that could potentionally react with substrate causing some minor problems to purity of the final product. I also wonder why he does not use any other less-reactive seal that would not react with substarte what so ever.
Thank you for your contribution

Sulphuric acid reacts with dichromate and table salt, forming volatile and harmful chromyl chloride.
\begin{multline}
\ce{K2Cr2O7(s) + 4 NaCl(s) + 6 H2SO4 ->}\\
\ce{2 KHSO4(s) + 4 NaHSO4(s) + 2 CrO2Cl2(l,g) + 3 H2O}
\end{multline}
Therefore it serves as a reagent and at the same time as protection against chromyl chloride escape.

Safety considerations
$\ce{CrO2Cl2}$ hydrolyzes to release $\ce{HCl}$ and hexavalent chromium $\ce{Cr^{\mathrm{VI}}}$.
Acute: Exposure to chromyl chloride vapour irritates the respiratory system and severely irritates the eyes, and the liquid burns the skin and eyes. Ingestion would cause severe internal damage.
Chronic: $\ce{CrO2Cl2}$ can produce chromosome abnormality/aberrations and is a human carcinogen via inhalation. Frequent exposure of the skin to chromyl chloride may result in skin ulcer/ulceration.


What is relationship between molecular weight distribution of PE and its strength? I'm trying to figure out the relationship between the molecular weight distribution of a polyethylene plastic and its strength and heat sealability.

(The molecular weight distribution is described as being narrow, which apparently is helpful in terms of strength and heat sealability).

I have found an old paper that might answer your question, which is:

"The Effect of Molecular Weight Distribution on Polyethylene Film Properties"
P. J. Perrono, P. B. Lederman, Polymer Egineering and Science, 1972, 12, 5, 340-345 (DOI: 10.1002/pen.760120505).

The properties of polyethylene evaluated in this paper as a function of the molecular weight include film gloss, haze, tear resistance and film impact strength.
This study shows that broadening the molecular weight distribution increases haze and reduces film gloss. Further, it was shown that a linear relationship exists between film gloss and external haze. Both values are measures of surface irregularities in the film which are affected by the drawing characteristics of the polymer. A broader molecular weight distribution results in increased impact strength as measured by the Dart Drop Impact Test. This is, it is believed, a result of the increase in long chain branching of the higher molecular weight fractions of the polymer which cause a higher degree of molecular weight entanglement at the branch sites. In contrast the tear strength is reduced as the molecular weight distribution broadens because of the low molecular weight fraction in the broad spectrum material which tend to decrease resistance to tear.

Hope this helps.


Determining the mechanism followed when alcohols undergo a reaction with sodium bromide and sulphuric acid $$\ce{R-OH + NaBr + H2SO4 -> R-Br + NaHSO4 + H2O}$$
Due to sulphuric acid, the $\ce{-OH}$ group will get protonated and convert into $\ce{-H2O+}$ which is a great leaving group in comparison to the hydroxyl group. Therefore I assumed it would readily form a carbocation and then $\ce{Br-}$ could attack to form the alkyl bromide.
That is, it would follow the SN1 pathway (since the reaction is happening in 2 steps, first the formation of carbocation and then the attack of $\ce{Br-}$) but I also read somewhere that we might need to have a look on the substrate, if it is primary then it would follow SN2.
I am a bit confused after so many conditions and situations being introduced so I was hoping if someone could give me a brief and crisp explanation about this, and is there any particular factor that needs to be focused on before judging this mechanism?

yeah so basically no reaction is purely SN1 or purely Sn2 its always a mixture of both and the overall order which we see is nothing but what that majority of reaction follows so now going by ur question, we do have to check about the substrate to inspect SN2 mechanism but in this case, since the reaction conditions are made in such a way that the overall reaction follows SN1 (as it is acid-catalyzed) hence we would compare the rate of reaction in terms of RDS (FORMATION OF CARBOCATION) and hence the more stable the carbonation is the faster the reaction proceeds.


Why do vinylic carbocations generally not undergo hydride rearrangement from neighbouring sp3 carbon to get more stability? Why do vinylic carbocations generally not undergo hydride rearrangement from neighbouring sp3 carbon to get more stability?

According to me, a rearrangement would lead to an allylic carbocation, which is more stable than vinylic, hence the rearrangement should be favorable.

If you think about protonating 1-propyne on the terminal acetylenic carbon, you get a vinylic cation that could apparently form an allylic cation by a hydride shift. The bond to the hydride that would shift has to be lined up with the empty orbital to which it is migrating.

So far, no problem. But after the shift, the empty $\mathrm{2p}$ orbital of the new carbocation is orthogonal (at right angles) to the $\mathrm{2p}$ orbitals of the adjacent double bond. Therefore, what looks like an allylic carbocation is actually a twisted allylic carbocation.

The $\mathrm{2p}$ orbitals of the double bond can't stabilize the new cation by resonance. Worse yet, the $\mathrm{2p^2}$ carbons of the double bond are relatively electronegative, so they actually destabilize the new carbocation. In other words, the migration is forming another unstable carbocation.

This is why rearrangements in vinylic carbocations don't occur. (There is a Russian Chemical Reviews article [1] that explains all this, but it's hard to get a copy of this, and their PDFs seem to be corrupted.)

References

Shchegolev, A. A.; Kanishchev, M. I. Rearrangements in Vinyl Cations. Russ. Chem. Rev. 1981, 50 (6), 553–564. https://doi.org/10/b7rrk8. (mathnet.ru mirror; PDF (in Russian))


Xe–F bond length in xenon fluorides The order of bond length of $\ce{Xe-F}$ in $\ce{XeF2}$, $\ce{XeF4}$, $\ce{XeF6}$ is?

My thoughts:

As hybridization of $\ce{Xe}$ in $\ce{XeF2}$ is $\ce{sp^3d}$, $\ce{XeF4}$ is $\ce{sp^3d^2}$, $\ce{XeF6}$ is $\ce{sp^3d^3}$.

%$\ce{s}$ decreases from $\ce{XeF2}$, $\ce{XeF4}$ to $\ce{XeF6}$, so bond length should increase, but my book says the opposite, i.e. $\ce{XeF2} > \ce{XeF4} > \ce{XeF6}$.

Yes, your book is correct.
To rationalize it, throw away hydridization and electron exitation and work with concept of three-center four-electron bond. The idea is that in $\ce{XeF2}$ the real structure is a mix of $\ce{F-Xe+...F^-}$ and $\ce{F^-...Xe+-F}$, so the bonding is half-covalent and half-ionic. By packing more fluorine around the xenone atom, we force xenone into state with higher effective atomic charge, so the ionic part of the bonding becomes stronger as $\ce{Xe^{3+}}$ pulls fluorine anions much stronger.

same logic applies to many other cases, in particular sulfur oxides and sulfur fluorides.

For context, I quote experimental bond lengths from paper. Note, the paper itself quotes the values from other sources with slightly different values from different sources.

XeF2 ... 1.977

XeF4 ... 1.94

XeF6 ... 1.94 & 1.850 (two bond types)


Can potassium nitrate and sulfuric acid be used to nitrate organics? If we are doing a classic nitration to produce glyceryl trinitrate, one uses a mix of concentrated sulfuric and nitric acids. Additionally, one can produce concentrated nitric acid by distilling it from a mixture of potassium nitrate and sulfuric acid.
So, is it possible to do a nitration by mixing the glycerol with sulfuric acid and then adding potassium nitrate?
Disclaimer
In case anyone is wondering, I am not going to try this, and I do know how to make many better and safer explosives since I used to be a member of the alt.engr.explosives group way back.

A simple, but dangerous path to nitrating a metal oxide (say MgO) is to treat with NO2 gas (caution: toxic and corrosive), as to quote Wikipedia on Nitrogen dioxide:

NO2 is used to generate anhydrous metal nitrates from the oxides:[10]
$\ce{MO + 3 NO2 -> M(NO3)2 + NO}$

where heating mix of Cu, KNO3 and H2SO4 will generate NO2.
A safer path to nitrating, starts with the introduction of the powerful  radicals, like the hydroxyl radical (•OH). The latter can transform commonly available nitrate salts  (like KNO3) to the nitrate radical (•NO3) especially in the presence of an acid catalyst per the reaction:
$\ce{•OH + NO3- <=> OH- + •NO3}$
A supporting source reference, Nitrate radicals and biogenic volatile organic compounds: oxidation, mechanisms, and organic aerosol. Per the paper's abstract:

Oxidation of biogenic volatile organic compounds (BVOC) by the nitrate radical (NO3) represents one of the important interactions between anthropogenic emissions related to combustion and natural emissions from the biosphere.

Also, see The Reactivity of the Nitrate Radical (•NO3) in Aqueous and Organic Solutions, by
Stephen P. Mezyk, Thomas D. Cullen, Kimberly A. Rickman and Bruce J. Mincher, per the abstract to quote:

ABSTRACT
Rate constants for the nitrate (•NO3) radical reaction with alcohols, alkanes, alkenes, and several aromatic compounds were measured in aqueous and tert‐butanol solution for comparison to aqueous and acetonitrile values from the literature. The measured trends provide insight into the reactions of the •NO3 radical in various media. The reaction with alcohols primarily consists of hydrogen‐atom abstraction from the alpha‐hydroxy position and is faster in solvents of lower polarity where the diffusivity of the radical is greater. Alkenes react faster than alkanes, and their rate constants are also faster in nonpolar solution. The situation is reversed for the nitrate radical reaction with the aromatic compounds, where the rate constants in tert‐butanol are slower. This is attributed to the need to solvate the NO3− anion and corresponding tropylium cation produced by the •NO3 radical electron transfer reaction. A linear correlation was found between measured rate constants in water and acetonitrile, which can be used to estimate aqueous nitrate radical rate constants for compounds having low water solubility.

With respect to the hydroxyl radical, the UV photolysis of a photocatalyst like TiO2 in aqueous KNO3, could be a path to the powerful, but very transient, hydroxyl and nitrate radicals.


New technology that has replaced the old radioisotope What technology has the radioisotope Cobalt 60 replaced in medicine?

E.g.,PET preferred over CT scans

After searching on Google (now there's an idea)...

If you are asking about what technology was used in medicine before Co-60 therapy was developed:

Co-60 therapy was developed in the early 1950's and was seen as an alternative to other forms of X-ray radiotherapy, or radiation therapy. It is a form of external beam radiation therapy. Radiation sources at the time (mostly derived from radium) required very high energy sources, and proved largely ineffective against deep-set cancers. Co-60 therapy was able to penetrate deeper at lower energies, and therefore was more effective against a range of previously untreatable cancer types. The first Co-60 machine was from the The University of Saskatchewan, and they have an excellent site devoted to the history of Co-60 treatment.

If you asking about what technology is now being used in place of Co-60 therapy:

Conventional X-ray sources were improved with the development the linear accelerator which could produce X-ray energies exceeding 8MV. Nowadays, commercial units can produce photons up to 25MV. Here is a useful link to compare linear accelerator and Co-60 radiation sources for medical applications. There are still many developing areas of use, with some experts claiming there is still very much a place for Co-60 therapy. Co-60 therapy can be applied with high precision with low side-effects and is just one of many approaches available, often used in tandem with other medical treatments.


Mass chromatogram : Total Ion Chromatogram (TIC) I am confused about what a Total Ion Chromatogram (TIC)  means. The definition I have says:

  The total ion current (TIC) chromatogram represents the summed
  intensity across the entire range of masses being detected at every
  point in the analysis.

Does this mean it is a (mixture) sample injected into a MS without the preceding chromatographic separation (as is typical in, say, a GC-MS or an LC-MS)?  Are there MS instruments sold today that are not GC-MS's or LC-MS's etc. (i.e. mass spec. sans the preceding seperation?

What's the point in doing this? Wouldn't the resultant spectrum be very noisy since it's essentially the sum over all components of a mixture of unknown proportions.

Is the GC eliminated for speed of analysis?

Does this mean it is a (mixture) sample injected into a MS without the preceding chromatographic separation (as is typical in, say, a GC-MS or an LC-MS)?

No. A chromatogram is the measurement of something after separation by chromatography. So you got chromatography beforehand. In case of direct injection into MS the term is usually used too, if you record over time like you would do using chromatography.

  What's the point in doing this?

Ok so let's imagine I'm running a sample of an unknown mixture and I'm recording m/z 100-1000. Now I don't know at all what to expect, so selecting randomly a specific mass for my chromatogram would, most likely, lead to nothing. What we can do is just integrate over the whole mass range and take the total number of all ions, so we see a peak this means there are more ions than for baseline noise. That should correspond to a substance reaching the detector.

  Wouldn't the resultant spectrum be very noisy since it's essentially the sum over all components of a mixture of unknown proportions.

Depending on the method: yes that might be a problem. And if the noise is very high then even a strong signal of a single m/z wouldn't really contribute to the total number of ions, so we won't see a peak in the TIC.

A way to circumvent this problem is using a base peak chromatogram. Base peak is the highest peak in your spectrum, and you are plotting the intensity of the highest peak over time. Now for noise there might be a large amount of total ions, but the intensity of each m/z is usually small. If you got a specific signal from a substance you will get much higher signals there, so the base peak chromatogram will show a peak.

  What's the alternative?

Collecting data in TIC mode is an alternative to collecting data in SIM mode. SIM stands for "single ion monitoring" or "selected ion monitoring". Let's say that you know your sample contains acetophenone, and the four largest peaks for acetophenone are at 120 (molecular ion), 105 (base), 77, and 51 m/z. Instead of integrating over all ions, you could monitor only those masses (or only one of them). You would know where the acetophenone eluted and could quantify it relative to a standard, but you would know little about the rest of the sample.

By analogy, consider a LC system with a UV/Vis photodiode array detector. The detector can monitor single wavelengths or it can capture a full absorbance spectrum at each time interval. A mass spectrometer collecting data in TIC mode is like a PDA detector collecting a full spectrum. A mass spectrometer collecting data in SIM mode is like a PDA monitoring individual wavelengths.