What is the world's finest powder substance? What is the world's finest powder substance that has the smallest granules but remains solid?

Clusters of just a few atoms, such as gold atomic clusters, and quantum dot nanoparticles could be considered very fine powders, though they usually are suspended in another substance to keep the clusters from agglomerating. Particles this tiny have different properties from the bulk material, both because of quantum effects (there are few particles, so there is not a continuous valence band) and because more atoms are at the surface interface than are surrounded by similar atoms in the interior.

See Quantum dot for more information.

I hope this fits your definition of a "fine powder". You might also be interested in pyrophoric iron.

BTW, ruby glass is made using colloidal, i.e. super-finely divided, gold.


Trying to understand the causes and implications of kinks in fatty acid chains I watched this video (The Deal with Fat by SciShow):

https://www.youtube.com/watch?v=mvvx2yQRbzQ

To summarize the main points I want to discuss:

Saturated fats are chains without any double-bonds and tend to result in solid structures because they stack well.
Unsaturated fats have at least one double-bond between carbon atoms, resulting in fewer overall hydrogens. This either results in a kink (cis structure) or a straight lockout (trans) depending on whether the hydrogens around the double bond are on the same side or not.
Kinked fats are harder to stack into lattices so they tend to be liquid, so most unsaturated fats are liquid.

My questions:

I thought there were some trans fats that were healthy / naturally occurring, such as those from certain meat products. What's the difference between the healthy natural trans fats and the trans fats made from partially hydrogenated oils and such?
What exactly causes the kink? Hydrogen atoms repelling each other?
If the hydrogens are repelling each other, then how do the chains even "stack" well in the first place if they are straight? Why wouldn't they all repel each other and remain liquid?
Why do unsaturated fats, which have one double bond and tend to be kinked, straighten out and result in solids when chilled?

No, trans fats rarely occur in nature, and are generally thought to be very unhealthy.
No, like all neutral atoms, hydrogen atoms attract each other, albeit weakly (by London Dispersion Forces). That's why neatly stacked saturated fats are solid. The kink is to do with the structure of a hydrocarbon chain; their structural formula is usually drawn as a straight line, but it's actually more of a zigzag, each carbon at an angle to the last. Loosely speaking, the double bond interrupts that, so you get one more zig than zag, resulting in a kink.
See 2.
They don't need to straighten out to become solid; they just need to attract each other enough to overcome the tendency of warm molecules to keep moving around.


Why does an ideal capacitor give rise to a rectangular cyclic voltammogram (CV)? This question is sort of a sequel to my previous question about cyclic voltammetry (CV).  One of the responses made reference to the fact that an ideal capacitor gives rise to a rectangular cyclic voltammogram.  Can you please help me understand why this is the case?  In other words, why does an ideal capacitor reach a constant current $I$ as soon as a voltage $V$ is applied?

I indeed see nearly ideal CVs in many literature articles (CVs which are rather rectangular with rounded corners).  In other figures, though, I see relative deviation from "rectangles with rounded corners," in that I see abrupt peaks, spikes, or valleys.

For example, below I have plotted two figures from Khomenko, Electrochimica Acta 2005, 50, 2499-2506.  Just very roughly and "hand wavy," what might be the qualitative reason for the "rectangle with rounded corners" behavior of Figure 8 (left) and the "abrupt peaks" behavior of Figure 4 (right)?  Could it be that the sample in Figure 8 (left) is relatively unreactive toward apllied potential, whereas the sample in Figure 4 (right) undergoes redox (Faradaic) reactions -- indicating the presence of so-called pseudocapacitance -- when an external potential is applied?

Please know that I am not looking for an answer specific to the article to which I linked.  I am only asking this question in the context of basic, qualitative aspects of cyclic voltammetry.

The kind of analysis truly required is outside the scope of chemistry and would be best handled in the careful hands of an electrical engineer. I'll give a brief attempt that exploits an easy-to-use circuit simulator.

We have to first think of an equivalent circuit that we can use as a model to determine its behavior. I suggest the following:

where my variable is R1's value for its resistance, which I shall set to 0, 10 and 100 ohm. If we were to watch this happen in time then we would see the following:

Quickly converting these to Current vs Voltage and running two other simulations at different resistances and we get:

These results are due to setting up differential equations and solving them appropriately.

You can play with the circuit I made for yourself here.


I want to make electricity of aluminiun cans i.e. reverse the process of refinning aluminium. Any ideas? Recycling Al cans to produce electricity,i.e. to reverse the process of electrolysis and produce electricity

You could "burn" the aluminum in the presence of oxygen to generate aluminum oxide.  For every two moles of aluminum consumed, a bit less than 1700 kJ of energy would be released.  You could use the energy to drive a turbine and generate electricity.


Why is the disordered CuAu structure more stable at high temperatures? I know that the disordered CuAu structure is based around FCC whereas the ordered CuAu structure is tetragonal. Why would the FCC structure be more stable at high temperatures?
For context, I was asked to explain this with the aid of a graph. I don't really know where to start

I'm not a metallurgist or even not a good physical chemist. However, I'd like to explain what's happening in $\ce{AuCu}$ alloy with temperature and any expert can intervene.
According to Ref.1, $\ce{Cu}$ and $\ce{Au}$ are both univalent group IB metals with the atomic size difference ~12%. The $\ce{AuCu}$ alloy forms the tetragonally distorted fcc lattice where alternate (00h) planes contain either $\ce{Cu}$ or $\ce{Au}$ atoms and cause a contraction in c-direction. Resulting tetragonal face-centered structure has c/a ratio of 0.92 $(\frac{c}{a} = \frac{\pu{367 pm}}{\pu{396 pm}} = 0.93)$. In the temperature range ~$\pu{380 ^\circ C}$ to $\pu{410 ^\circ C}$, the superlattice $\ce{CuAu}$-II is formed, which consist of $\ce{CuAu}$-bct unit cells with the antiphase domains along the b-direction (Ref.2). There is a lattice shift of $\frac12(a+c)$ at each five unit-cell length. The superlattice $\ce{CuAu}$-II is described as orthorhombic cell with 10 cells along one of a direction, $oI40$.

Suggested reading: Ref.3 and 4.

References:

Valentina F. Degtyareva, Nataliya S. Afonikova, "Simple Metal and Binary Alloy Phases Based on the fcc Structure: Electronic Origin of Distortions, Superlattices and Vacancies," Crystals 2017, 7(2), 34 (13 pages) (DOI: https://doi.org/10.3390/cryst7020034).
V. Ozoliņš, C. Wolverton, Alex Zunger, "Cu-Au, Ag-Au, Cu-Ag, and Ni-Au intermetallics: First-principles study of temperature-composition phase diagrams and structures," Phys. Rev. B 1998, 57(11), 6427 (DOI: https://doi.org/10.1103/PhysRevB.57.6427).
M. Sanati, L. G. Wang, Alex Zunger, "Adaptive Crystal Structures: $\ce{CuAu}$ and $\ce{NiPt}$," Phys. Rev. Lett. 2003, 90(4), 045502 (DOI: https://doi.org/10.1103/PhysRevLett.90.045502).
S. -H. Wei, A. A. Mbaye, L. G. Ferreira, Alex Zunger, "First-principles calculations of the phase diagrams of noble metals: Cu-Au, Cu-Ag, and Ag-Au," Phys. Rev. B 1987, 36(8), 4163 (DOI: https://doi.org/10.1103/PhysRevB.36.4163).


IUPAC name of C₆(CH₃)₆²⁺ What is the correct IUPAC name (or least incorrect one, if the current rules can't precisely describe that) of the interesting species $\ce{[C6(CH3)6]^{2+}}$ (which is extensively discussed here), containing exotic hexacoordinate pyramidal carbocation* ?

The structure can be drawn as

or

My guess is methyl(pentamethyl-η5-cyclopentadienyl)carbon(2+).

*) Some structures in the Wikipedia article have the two positive charges placed into the cyclopentadienyl (aromatic) ring, which I think is incorrect.

From the IUPAC Gold Book:

  Carbocations may be named by adding the word 'cation' to the name of the corresponding radical. Such names do not imply structure (e.g. whether three-coordinated or five-coordinated carbon atoms are present).

So since this is the dication of hexamethylbenzene, it would probably  least incorrectly be called hexamethylbenzene dication. At the very least, this seems to be how a lot of people refer to it (or more verbosely, pentagonal-pyramidal hexamethylbenzene dication).


Difference between atomic mass, mass number, atomic weight, and relative atomic mass I think this question is common, but i want a little help.
I'm little confused.
What I concluded.

Relative atomic mass/Atomic weight (both are exactly the same) - It is the mass of atom calculated as taking the sum of all the masses of the subatomic particles and dividing it with mass of (the same technique calculated) 1/12th of C-12 atom.

Atomic mass- It is the mass of the atom calculated as taking the mean according to their abundances in nature, of atomic weight of the different isotopes present in that sample.

Mass Number- It is the mass calculated as the sum of the nucleons , as there is only a slight difference in the mass of the proton and neutron, and the mass of electron is negligible as compared with the mass of the proton and neutron

But this is conflicting with definitions on wikipedia.
Also the relative atomic of carbon should be exactly 12, but its 12.011

Please correct me if I am wrong but in my understanding, the following are the definitions for the aforementioned terms:

Mass number: The total number of nucleons present in the nucleus of an atom(Dimensionless).

Atomic mass: The combined mass of all the nucleons present in the atom of any given element(Unit: amu/u).

Relative isotopic mass: Ratio of atomic mass to one amu(Dimensionless)

Average atomic mass: Weighted average of all the isotopes for the given element(amu/u).

Relative atomic mass/atomic weight: Ratio of atomic weight to one amu(Dimensionless).

Here is the Wikipedia page, explaining that "atomic weight" is simply the older term used for "relative atomic mass":

Relative atomic mass


What is the carbon dioxide content of a soda can or bottle? I sort of know how carbonated beverages are carbonated: a lot of $\ce{CO2}$ gets pushed into the liquid, and the container is sealed. There are at least two things I don't know. First, how much carbon dioxide is actually dissolved in the liquid? Second, what is the resulitng partial pressure of $\ce{CO2}$ in the headspace and the total pressure in the headspace? I'm interested in cans, plastic bottles, and glass bottles. I know from experience that there is some variation among manufacturers even for the same beverage, so I will be happy with general numbers or a good estimate.

General estimates have placed a can of Coca-Cola to have 2.2 grams of $\ce{CO2} $ in a single can. As a can is around 12 fluid ounces, or 355 ml, the amount of $\ce{CO2}$ in a can is:
$$\text{2.2 g} \ \ce{CO2}* \frac{\text{1 mol} \ \ce{CO2}}{\text{44 g} \ \ce{CO2} } = 0.05 \ \text{mol}$$
$$ \text{355 mL} * \frac{\text{1 L}}{\text{1000 mL}} = 0.355 \ \text{L} $$
So here we can see we have about 0.05 mol/0.355 L or about 0.14 mol of carbon dioxide per liter of soda. Of course this value varies by manufacturer, type of drink, container, etc.
Looking at Wikipedia, inside Coca-Cola is:

Carbonated water, Sugar (sucrose or high-fructose corn syrup depending on country of origin), Caffeine, Phosphoric acid, Caramel color (E150d), Natural flavorings
A can of Coke (12 fl ounces/355 ml) has 39 grams of carbohydrates (all from sugar, approximately 10 teaspoons), 50 mg of sodium, 0 grams fat, 0 grams potassium, and 140 calories.

Thus, we can calculate the pressure of $CO_2$ gas using the Ideal Gas equation if we store our coke at, say, 20 Celsius:
$$ P = \frac{nRT}{V} $$
$$ P = \frac{\text{0.05 mol} * \text{0.08206} \frac{\text{L} \cdot \text{atm}}{\text{mol} \cdot \text{K} } * \text{293.15 K}}{\text{0.355 L}}$$
$$ P = 3.39 \ \text{atm} $$
According to this website, http://hypertextbook.com/facts/2000/SeemaMeraj.shtml:

On average, the 12 ounce soda cans sold in the US tend to have a pressure of roughly 120 kPa when canned at 4 °C, and 250 kPa when stored at 20 °C.

$$\text{3.39 atm} * \frac{\text{760 torr}}{\text{1 atm}} * \frac{\text{133 Pa}}{\text{1 torr}} * \frac{\text{1 kPa}}{\text{1000 Pa}} = 342.66 \ \text{kPa}$$
Water vapor exerts it's own partial pressure. Looking at standard tabulated values for water vapor pressure, water exerts a pressure of 17.5 torr at 20 Celsius.
$$\text{17.5 torr}* \frac{\text{133 Pa}}{\text{1 torr}} * \frac{\text{1 kPa}}{\text{1000 Pa}} = 2.3275 \ \text{kPa} $$
Knowing that our total pressure is the sum of all our pressures:
$$P_{total} = \text{342.66 kPa} + \text{2.3275 kPa} = 344.99 \ \text{kPa}$$
Here, we are roughly about 100 kPa off from the data provided by the website. This is just an approximation. A more accurate way would be to calculate the moles of each product inside the soda, and knowing the total pressure or partial pressure of one of the parts, we can calculate the pressures more accurately. However, that information is proprietary. It's their secret recipe!
Headspace:
http://extension.psu.edu/food/preservation/news/2012/why-allow-headspace
We have to determine the volume of the headspace - again, I am not sure of exact data - which is between 1/2 inch to 1 and 1/2 inches depending on the container and what it holds. I will assume that the headpsace occupies 6% of the total volume of the can which is 21.3 mL.
At manufacturing and at storage, the can is at different temperatures. Taking the above data, we'll say it is manufactured at $4^\circ C$ and say, stored at $20^\circ C$
Furthermore, when carbon dioxide is solubilized in water, it forms carbonic acid. I will neglect that as the ionization constant is small.
Assuming our 2.2 grams of carbon dioxide is the maximum amount of carbon dioxide that can be placed inside, some of the carbon dioxide is soluble in water while the rest exerts pressure inside the headspace to force the carbon dioxide inside the liquid. The pressure is necessary inside this closed volume as, once you open the cap, the carbon dioxide tries to achieve equilibrium.
$$ \ce{CO_2 (solution) <=> CO2 (g)} $$
In general, the solubility of gases increases at lower temperatures and decreases at higher temperatures. A notable exception is the noble gases. In regards to pressure, Henry's law states that the solubility of a gas in a liquid is directly proportional to the pressure of that gas above the surface of the solution.
The solubility of carbon dioxide in water shifts according to Le Chatlier's principle.
For the solution to this problem, we would need to know several things about the manufacturing of soda's, or rather, more quantitative data. At manufacturing, the pressure is high and temperature low, so as much carbon dioxide as possible is solubilized in the water.
Once shipped out and stored, the pressure inside the headspace increases as the pressure is decreased from manufacturing, the carbon dioxide then leaves solution and enters the headspace. Increasing temperature also decreases solubility.
If you can provide me with more information, I will be more than happy to help. :)


What meaning do multiple real solutions of a truncated virial equation have, if any? Let's say I truncate the virial equation after the third term for use as my equation of state:

$${P\over\rho RT}=1+B\rho+C\rho^2$$

I have tabulated values for $B(T)$ and $C(T)$.  I know $P$ and $T$ and want to solve for $\rho$.  The equation is cubic in $\rho$ meaning I could potentially have three real solutions.  I suspect that since the virial equation is intended to calculate the densities of vapors or supercritical gasses at relatively low pressures, that the lowest real root is the "real world" value.  Is this correct?  Do the other real values, if they exist, have any meaning?  Can this be generalized to higher order virial expansions?

The virial equation is set up for the gas phase only.  The virial coefficients $B$ and $C$ are found from gas phase measurements.

If you get three solutions, I suspect that it is the largest one that you want.  I'd do a trial solution with something like water vapor to see.

Higher order virial equations are problematic since it is exceptionally difficult to determine higher virial coefficients from experimental data.  Higher order virial equations do have theoretical interest, but that's another story.


Ylides nucleophilicity comparison Why are phosphonate-stabilised ylides used in HWE reaction more nucleophilic than phosphonium ylids used in Wittig reaction? Explanations without rationalisations using d orbitals would be welcome.

It might not be the only factor, but I think that the steric effect of the huge phenyl groups in Wittig ylides might contribute in making them less nucleophile.
If you give a fast look at a "space filling" model of the two molecules, you see that the negatively charged carbon in a Wittig Ylide is much more masked by the phenyl groups

Note that in order for a nucleophile-electrophile interaction to occur, an electrophile should approach to the carbanion's HOMO orbital (approximatively orthogonal to the screen in both the pictures).


Why doesn't a buffer solution change ph(Appreciably?) Consider a buffer solution containing weak acid $\ce{CH3COOH}$ and its salt $\ce{CH3COONa}$:
\begin{align}
\ce{CH3COOH &<=> CH3COO- + H+} \tag{1}\\
\ce{CH3COONa &-> CH3COO- + Na+} \tag{2}
\end{align}

These two reactions take place and now we are majorly left with $\ce{H+}$, $\ce{CH3COOH}$, $\ce{CH3COO-}$, $\ce{Na+}$ in the solution. Since, $\ce{Na+}$ is a spectator ion, we can ignore it.

We know that buffers resist change in pH on adding small quantities of $\ce{H+}$ and $\ce{OH-}$. Let us add $\ce{H^+}$ to this solution.

A buffer solution must contain some things that could remove these $\ce{H+}$ ions, otherwise the pH would change. As we add $\ce{H+}$ to this buffer solution, it will react with $\ce{CH3COO-}$ to form $\ce{CH3COOH}$.

But here's the catch.

$\ce{CH3COO-}$ is getting removed and $\ce{CH3COOH}$ is getting formed. According to Le Chatelier's principle (LCP), the equilibrium should shift towards in the forward reaction (in reaction (1)). Shifting of equilibrium in the forward reaction should increase the concentration of $\ce{H+}$ and hence decrease the pH of the solution.

Now, one might say that $\ce{CH3COOH}$ is a weak acid and it would not decompose much. So, we can actually ignore the amount of $\ce{H+}$ so generated.

Let's "assume" that's true for a while and move ahead.

Let us add $\ce{OH-}$ ions to the solution.

Remember there's $\ce{H+}$, $\ce{CH3COOH}$, $\ce{CH3COO-}$, $\ce{Na+}$ present in the solution. $\ce{OH-}$ can react with either (a) $\ce{H+}$, or (b) $\mathrm{CH_3COOH}$.

Here comes the tricky part.

Let us only consider hydroxide ions reacting with $\ce{H+}$

If $\ce{OH-}$ reacts with $\ce{H+}$, water will be formed.

Here although hydroxide ions are getting removed, $\ce{H+}$ are also being removed from the solution. Now, because of the removal of $\ce{H+}$ shouldn't the solution become less acidic and the pH should increase.

Now, some might say as soon as $\ce{H+}$ is removed, the equilibrium in the first reaction shift towards the right.

But wait, didn't we just ignore that (in the case when we added $\ce{H+}$ in the solution)?

Can anyone explain why is there a contradiction?

P. S.: I have ignored the $\ce{H+}$ and $\ce{OH-}$ from water because it is irrelevant to the discussion.

Having problems with that concept myself, above answers seem mighty complicated, this is how I think it works (warning- only my understanding, could well be wrong)

Point One:
You have a lot of $\ce{CH3COOH}$ (not much decomposition) and a lot of $\ce{CH3COO−}$ (from the salt) and very little $\ce{H+}$

Point Two:
pH works on the amount of $\ce{OH-}$ and $\ce{H+}$ in the system, there are usually only very very small amounts (when looking at weak acids and bases) present compared to the reactant (in water, there is about 500 million molecules to ever one that dissociates) So the point I'm making is adding a few $\ce{OH-}$ or $\ce{H+}$ ions will have a big impact on the pH.

Conclusion drawn:
As you have so much $\ce{CH3COO−}$ to begin with, removing a very very small amount by adding a few $\ce{H+}$ ions and forming a very very small amount of $\ce{CH3COOH}$, is not going to have a noticeable impact on the concentrations, in comparison to the amount already in the solution (both $\ce{CH3COO−}$ and $\ce{CH3COOH}$). You could even say that it is a negligible amount and can be ignored, therefore can say this will not have an effect on the equilibrium.

As the $\ce{H+}$ is removed, the pH is not changed; however as there is so little $\ce{H+}$ ions to begin with, not removing them will have an impact on the pH (see point two).

As already stated, there are so little $\ce{H+}$ ions in the system, removing them (by adding $\ce{OH-}$) will have an impact on equilibrium, you cannot say the amount of $\ce{H+}$ removed is negligible, you cannot ignore Le Chatelier's principle, the equilibrium will shift right to replace the lost $\ce{H+}$.

Note: I'm an A-level student, don't take this as 100% right or even 1% right, this, at best is a calculated guess as to why equilibrium can be ignored when you add $\ce{H+}$ ions and equilibrium can not be ignored when you add $\ce{OH-}$.


How to find pH in the following titration/buffer problem given concentration of base and acid A $100. \text{ mL}$ sample of $0.10 \text{M } \ce{HCl}$ is mixed with $50. \text{ mL of } 0.11 \text{M } \ce{NH3}$. What is the resulting pH? (Kb for $\ce{NH3} = 1.8 × 10–5$)

I was thinking more along the lines of trying to find out the moles of acid and the moles of bases that we have and subtracting the moles and finding the concentrations so I did the following:

I first make the chemical equation:

$\ce{NH3 + H2O <-->OH- + NH4+}$

then I find that there are 0.0055 moles of NH3 and I set up the Kb problem:

$\mathrm{Kb}=\frac{X^2}{0.0055-x}$

then I find $x$ to be $3.15*10^{-4}$

then I subtract that with the moles of $\ce{H+}$ ions from the $\ce{HCl}$ and divide that with $.15$ liters to get $.065$

when I take the negative log of that I get $1.189$ but that is none of the answers.

Why is this and what source of error have I made and is there a concept that I do not understand? Thus I request for assistance.

*I do understand that we can use the Henderson Hasselbach but I was wondering how to solve this without the equation?

This is a titration problem - not necessarily a buffer problem. While you have $0.0055$ moles of $\ce{NH3}$, you have $0.01$ moles of $\ce{H}$ from $\ce{HCl}$ dissociating. $0.01 - 0.0055 = 0.0045$ moles of $\ce{H}$ left. Take H/total L to find M. Take the $-\log$ of $\ce{[H]}$.


Why is (1E,3Z,5E,7E,9Z,11E)-cyclotetradeca-1,3,5,7,9,11-hexaen-13-yne not antiaromatic? The compound shown below was synthesized by Professor I M Dimm who was astonished to find that it was exceedingly stable. Prof. Dimm had expected world renown for making what he thought would be an unstable compound. Dimm reasoned that with six double bonds and a triple bond in the molecule there would be 16 π electrons and thus, no aromatic character ($4n, n=4$, not $4n+2$). Where was his reasoning wrong?

My reasoning so far:
Two of the electrons in the triple bond are not part of the conjugated π system. They are in a π orbital that is perpendicular to the rest of the p orbitals that make the conjugated π system. The conjugated π system thus only contains 14 electrons. $14 = 4n + 2$ for $n= 3$. So this molecule is aromatic.

The textbook answer to this is that the triple bond consists of two orthogonal π systems whose interaction with each other is symmetry-forbidden. So as the attached figure shows, the two p-orbitals marked in red cannot participate in the aromatic system if the molecule’s structure is as drawn; thus the two electrons do not count towards the Hückel rule, thus $4n+2$ is satisfied with $n=3$ and $4n+2 = 14$.

However, that is not the end of the story; merely its beginning. Professor Dimm should take the time to determine the actual crystal structure of the compound he synthesised. You might want to compare this with cyclodecapenta-1,3,5,7,9-ene. While it may look like a perfect Hückel-aromatic molecule, it does not adopt a planar structure due to ‘inverse Baeyer strain’ (the flat all-cis conformation would have $144^\circ$ angles, far away from the theoretical $120^\circ$ $\mathrm{sp^2}$ angles) or steric hindrance (the (1E,3Z,5E,7Z,9Z)-confomation is not flat due to steric strain of the internal hydrogens). A similar case can be made for (1E,3Z,5E,7E,9Z,11E)-cyclotetradeca-1,3,5,7,9,11-hexaen-13-yne as can be seen in the figure below.

The hydrogens coloured in red would overlap far too much to create a stable molecule. It will adopt a non-planar configuration to relieve this steric stress which is a greater source of destabilisation than giving up an aromatic system. We can only speculate on the molecules structure until we actually saw it. Unfortunately, it is also unknown as far as SciFinder is concerned. However, two geometric isomers of dehydro[14]annulene (as the molecule is also called) have been synthesised and isolated:

Interstingly, these compounds that were discovered in the 1960’s[1] show aromaticity by their chemical shifts. The internal hydrogens are shielded and display a chemical shift of approximately $-1~\mathrm{ppm}$ (symmetric isomer) and $0$ and $-1~\mathrm{ppm}$ (unsymmetric isomer). Similarly, the outer hydrogens appear as a group in the aromatic region.[2] Cresp and Sondheimer used low-temperature NMR studies to assign the correct structures.[2] And in 2013, Santander et al. performed calculations to determine the mechanism of interconversion between the two (‘by allowing a dilute ether solution to stand without protection from daylight, or, more rapidly by irradiation of a refluxing benzene solution with a $100~\mathrm{W}$ lamp.’[2]), which turns out to be via an interesting (in my opinion) Hückel-bond shift.[3]

Let’s hope that Professor Dimm actually synthesises the compound and properly reports its structure.

References:

[1]: F. Sondheimer, Proc. Roy. Soc. A, 1967, 297, 173. Taken directly from reference [2].

[2]: T. M. Cresp, F. Sondheimer, Tetrahedron Lett. 1982, 23, 1731. DOI: 10.1016/S0040-4039(00)87202-0.

[3]: M. V. Santander, M. B. Pastor, J. N. Nelson, C. Castro, W. L. Karney, J. Org. Chem. 2013, 78, 2033. DOI: 10.1021/jo302072p.


Why is the segment of the IR spectrum below 1500 wavenumbers called the fingerprint region? Why is the segment of the IR spectrum below 1500 wavenumbers (cm$^{-1}$) considered as the fingerprint region? It is said that "although the entire IR spectrum can be used as a fingerprint for the purposes of comparing molecules, the 600 - 1400 wavenumber range is called the fingerprint region. This is normally a complex area showing many bands, frequently overlapping each other". However, I don't understand why the range from 600 - 1400 wavenumbers can be particularly complex and is unique to the compounds, for instance, why is the fingerprint region of propan-1-ol and propan-2-ol so different from each other given roughly the same appearance for rest of the spectrum?

Spectra taken from http://www.chemguide.co.uk/analysis/ir/fingerprint.html

I guess the naming "fingerprint region" results from the fact that while I can immediately identify amine, alcohol, carbonyl groups, double and triple bonds and so on in the "upper" and "middle" part of the IR spectrum, the peaks observed in the fingerprint region are often arising from vibrations that are very specific to the very compound you are looking at, thus a "fingerprint" of your compound.


Help with molecular dynamics and computational modelling to check validity of hypothesis regarding SN2 reaction mechanisms Polar aprotic solvents have a less hindered electronegative atom pointing out, namely oxygen, fluorine, and nitrogen, all of which have lone pairs besides having a partial negative charge. Now, all the molecules inside a reaction mixture are moving at great speeds. Not all nucleophiles will attack the carbon at 180° to the leaving group since this provides the best overlap between the nucleophile's lone pair and the C–X σ* antibonding orbital. Other nucleophiles running in random directions can maybe bounce off the solvent molecules’ oxygens’ exposed lone pair right into the antibonding orbital, thereby increasing the number of molecules that successfully overlap in some unit time. The nucleophiles are also negatively charged which increases the repulsion while 'bouncing off'. This is illustrated in the picture attached. I am aware of the present explanation which suggests that oxygen or fluorine can solvate the corresponding cations of the nucleophile and it makes sense. This is just another reason that I had thought of(there can be more than one favorable reasons for phenomena?). The number of solvent molecules will obviously outnumber the substrate and nucleophiles, the ones which are roaming around idly can maybe aid this acceleration ( vis-a-vis the repulsion hypothesis) and increase the rate doubly (not doubly but slightly more than the previously anticipated increased rate).

There is a paper that I am currently working on regarding this. I have been made to realize that I'm going to need a molecular dynamics software to check if this phenomenon is statistically more or less likely. However, I'm currently in high school, hence I do not know anything about the same. Can someone please recommend a software and guide me through the basics? Is this something that I can independently work out, or will I need some assistance from a university( or professors)? How much time and work will it take to simulate?

Your hypothesis concerning a "bouncing" contribution to the reaction kinetics would be incredibly difficult to observe or quantify even with the best supercomputers.

The likelihood of a molecule to react in a manner like the SN2 reaction is actually an incredibly rare event at the molecular time scale (femtoseconds 10^-15 s). This may sound contradictory since these reactions are commonplace and occur readily under appropriate conditions (e.g. a few hours heating). However, the two timescales are separated by over 15 orders of magnitude. Many reactions which do approach the transition state can also revert back to the reactant, or in other words abort. So you'd have to run the incredibly long simulations thousands of times in order to see only a small number of reactions. This is the meaning of the pre-factor in the Arrhenius rate  equation

$k = Aexp(\frac{-\Delta G^\ddagger}{k_B T})$

Nevertheless, some of your intuition is correct. A molecule would indeed require momentum along the transition vector in order to overcome the highest energy point (the transition state) and pass over to the product. Otherwise it would return to reactant. Some of this velocity may naturally come from bouncing off a nearby solvent molecule.

Extra Information:

The topic concerning the relationship of transition states and molecular dynamics is actually very complex and a major challenge in the field of chemistry.  The challenge essentially boils down 1) the size of the system and 2) the time length of the simulation.

Fortunately for most practical cases we can simply ignore dynamical effects and study isolated because it is known that solvent interactions are generally weak and reactions closely follow the minimum energy paths.Therefore, the free energy at the transition state ($G^{\ddagger}$) can be approximated in solution using the Harmonic approximation and implicit solvent models like PCM. If you'd like to see how some of these calculations are done, please check out my GitHub on the growing string method (ref 1). These are much simpler to model and actually give a lot of information. Usually the transmission coefficient is assumed to be 1. or 0.5.

If you're interested in some more advanced considerations of the relationship between dynamics and reaction barriers I'd recommend the following references: ref 2,3

References

"Single-ended transition state finding with the growing string method" Journal of Computational Chemistry 2015, 36, 601–611
"QM/MM Protocol for Direct Molecular Dynamics of Chemical Reactions in Solution: The Water-Accelerated Diels–Alder Reaction" J. Chem. Theory Comput. 2015, 11, 12, 5606-5612
https://cims.nyu.edu/~eve2/string.htm


Is chirality a physical or a chemical property? In my opinion, it should be both, because it will affect the way in which a molecule will interact with plane polarized light (physical), but also affect chemical interactions such as receptor binding.

On the other hand, consider $\ce{CHDBrCH3}$, which is chiral, as there 4 distinct groups attached to the central carbon. This would insinuate that chirality is a purely physical property, as despite deuterium having the same chemical properties as hydrogen, we are able to form a chiral molecule.

I would not consider chirality as a property--either physical or chemical--it is part of the definition of a molecule's structure, which in turn gives rise to the properties it exhibits (rotation of polarized light, melting point, reactivity with X, etc.). In your example, the fact that hydrogen and deuterium are almost (but not entirely) chemically equivalent will make the substance behave almost the same as its dihydrogen analog, and the two mirror image isomers nearly indistinguishable by physical or chemical means.


No.of isomeric ethers of formula C3H6O I managed to get 3 ethers:-
1.4-member ring ether(Oxetane)
2.Vinyl ether
3. Epoxy ether

This question was asked in my school exam and the teacher had given its answer as 4..Could anyone please tell what would be the 4th isomer ??

The methyl epoxy ether is chiral, with the chiral center being the middle carbon in the chain, so it has two enantiomers.


Are these reaction equations for the formation of the brown ring complex correct? \begin{align}
\ce{NO3- + 3Fe^2+ + 4H+ &-> NO + 3Fe^3+ + 2H2O}\tag{1}\\
\ce{[Fe(H2O)6]^2+ + NO &-> \underset{\text{(brown)}}{[Fe(H2O)5(NO)]^2+} + H2O}\tag{2}
\end{align}

The above are the reactions given in my book [1] for the qualitative analysis of nitrate ion (formation of the brown ring complex). I think that the reactions above are incorrect. The first one shows formation of $\ce{Fe^3+}$, which should be attacked by excess of $\ce{H2O}$ to give ferrum(III) hexaaquasulphate, not ferrum(II) hexaaquasulphate. Now that should be attacked by $\ce{NO}$ as a ligand and give $\ce{[Fe(H2O)6NO]SO4}$, where the oxidation number of $\ce{Fe}$ is being $3+$.

Here the reactions show that the $\ce{Fe}$ already present gives $\ce{[Fe(H2O)6]^2+}$ and then $\ce{NO}$ acts as a neutral ligand to form the complex. But by the reactions shown in the book, even after formation of the brown ring complex $\ce{Fe^3+}$ is still present in the solution, which according to me shouldn't be because if $\ce{Fe^3+}$ still remains in the solution then water should also have attacked it as a ligand to give $\ce{[Fe(H2O)6NO]SO4}$.

I need some clarity on this. Are the reactions given in the book correct?

Edit: After reading the answers, I do get that the hexaaqua iron(II) will be formed and the $\ce{Fe^3+}$ will be attacked by water to give $\ce{[Fe(H_2O)_6]^3+}$ but why won't $\ce{NO}$ take electrons from the oxidation of iron ($\ce{Fe^2+ -> Fe^3+ + e- }$), make $\ce{NO-}$ and attack the above and substitute one water molecule and give more of the brown ring complex?

NCERT. Chemistry: Textbook for class XII; National Council of Educational Research and Training: New Delhi, 2007. ISBN 978-81-7450-648-1.

According Kinetics, Mechanism, and Spectroscopy of the Reversible Binding of
Nitric Oxide to Aquated Iron(II). An Undergraduate Text Book Reaction
Revisited

The correct structure is $\ce{ [Fe^{III}(H_2O)_5(NO^{-})]^{2+} }$

For many years it was thought that iron was reduced to Fe(I) and NO oxidized to NO+, based upon an observed magnetic moment suggestive of three unpaired electrons, however, the current thinking is that high spin Fe(III) (S=5/2) antiferromagnetically couples with NO- (S=1) for an observed spin of S=3/2.

Other than the above, I agree with Dhanajay Gupta that the equations in the book are correct, that some of the Fe2+ is oxidized to Fe3+ in the first equation forming NO, and NO coordinates to remaining Fe2+ according to the second equation, but the second equation doesn't specify the charge on Fe or NO, just the overall charge of the complex.

In the first equation in the book, it should be understood that the ions are aqueous.  For example H+ doesn't mean there are bare protons, it is understood to mean $\ce{H_3O+}$ and Fe2+ is $\ce{[Fe(H2O)6]^{2+}}$ and Fe3+ is $\ce{[Fe(H2O)6]^{3+}}$.

In the second equation in the book, the inner coordination sphere is explicitly shown.

In the article cited above the mechanism of the second equation is explained. Neutral NO attacks hexaaquairon(II), forming a 7-coordinate transition state, a water leaves, and "This rate-determining displacement of coordinated water is followed by a rapid intramolecular charge-redistribution process to lead to the final $\ce{ [Fe^{III}(H_2O)_5(NO^{-})]^{2+}}$ product".

  but why won't NO take electrons from the oxidation of Iron ($Fe^{2+}
 \rightarrow Fe^{3+} + e^-$), make $NO^-$ and attack the above and
  substitute one water molecule and give more of the brown ring complex?

To get the electron from Fe2+, NO would enter the inner coordination sphere, and when the electron transfer occurs, the brown ring complex is formed at that Fe.  If the NO- then left that Fe, it would destroy one instance of the brown ring complex.  Hypothetically, the Fe2+ could reduce NO via an outer sphere electron transfer, and then NO- coordinate Fe3+, but this is not what the article reports occurring.


Why do Ionic Bonds exist? This is a legitamate question.  If one atom wants to lose an electron and another wants to gain an electron why don't they just transfer the electrons and just part ways?

If a Na+ ion and Cl- ion exist as NaCl,(ionic) it doesn't mean that, that particular Na donated electron to Cl. Na may lose electron somewhere else and Cl may gain electron somewhwere else.
The main point of existence as Na+Cl-, ie NaCl is the electrostatic attraction between them.


Why can oxygen have a coordination number of 3 in a complex between silver and SiO2? I am running an MD simulation of silver with $\ce{SiO2}$. Initially, my silver atom is neutral and oxygen has a coordination of 2 with silicon having a coordination of 4. After the deposition of silver atoms, I find that the coordination of silver atoms ranges up to 12 and oxygen has coordination of 3 where a bond is formed with silver. Since $\ce{Ag-Si}$ bonds are not present, silicon's coordination number does not change.

I want to understand why oxygen is able to have a coordination number of 3 as with 2 bonds with silicon, the valency is already satisfied. I know silver forms a fcc-lattice where the coordination number is 12 but I want to get a more in depth detail of this change of coordination number. The basic physics behind it.

High coordination numbers are really not uncommon in solid-state compounds. For example, consider the structure of $\ce{Li2O}$, an anti-fluorite structure. In fluorite proper, calcium ions form a cubic closest packing structure and the fluoride ions occupy the tetrahedral voids. In $\ce{Li2O}$, it is reversed: the oxide ions form a cubic closest packing and lithium occupies the tetrahedral voids. That gives each lithium a coordination number of 4 and by maths and symmetry oxygen must have a coordination number of 8.

Tricoordinated oxide anions are also well-known in solid-state structures. The prime example would be the rutile structure of $\ce{TiO2}$ with hexacoordinated titanium and tricoordinated oxide ions.

In your specific case, the silver cations are acting as Lewis acids while the oxide anions (which have to lone pairs not directed towards neighbouring silicium atoms) can act as Lewis bases. A coordination number of three points to a somewhat covalent (and not so ionic) structure — which makes sense considering significant fragments of $\ce{SiO2}$ are still there.


How do we know cyclopentadienylmolybdenum tricarbonyl occurs as a dimer? Cyclopentadienylmolybdenum tricarbonyl, $\ce{MoC8H5O3}$, dimerizes to form the cyclopentadienylmolybednum tricarbonyl dimer, $\ce{(MoC8H5O3)2}$.

$\hspace{3cm}$

How was it discovered that it does this?

If you had an sample of a pure substance, but didn't know if it had dimerized or polymerized, what information would you need to find out if it had?

The suggested answers from the textbook are to look at the output of a mass spectrometer, or to observe the macroscopic crystal structure.

However, NMR spectroscopy is also a perfectly acceptable way to go about it.


Pressure-dependence of electrolysis voltage I'm trying to better understand electro-chemistry, so please explain why this second-kind perpetual motion device won't work. I guess it's safe to assume it won't work one way or another.

There is a cycle, which starts with low-pressure water.

The water is pressurized by a compressor. This requires work, but since water is nearly incompressible, the required work is very low: the distance over which the pressurization force is applied is almost zero.
Next, the high-pressure water is split into high-pressure hydrogen and oxygen gas. This requires a certain amount of electrical work.
Next, the high-pressure hydrogen and oxygen gas are expanded through turbines, expanding cylinders or other devices that allow the expansion to apply output work. Heat is added to keep the temperature constant during expansion, to maximize the work output. Note that, unlike the compression stage, expansion deals with gases, which are very compressible; therefore, the amount of work produced is substantial.
To close the cycle, the hydrogen and oxygen gases are recombined in a fuel cell, to end up with the same amount of water, at the same pressure as when we started the cycle. This delivers some electrical work.

Of course there are going to be energy losses; I expect the fuel cell will always yield less energy than is required in the electrolyzer. However, the difference between the compression work and the expansion work can be scaled up arbitrarily by using a higher pressure difference, so at some point this difference in work must yield more energy than is lost in the electrolysis/fuel cell combination. This energy comes from the heat flowing into the expander, but simply turning heat into excess work should not be possible.

So, what is wrong? The only way I see to prevent this to work is if the electrolyzer requires more work per reaction mass (essentially its voltage) than the fuel cell, even if they are both 100% efficient. This would imply that the reaction voltage is pressure-dependent, with a higher voltage at higher pressures. Is this the case? And is this only the case if the output material is more compressible (gas) than the input material (liquid)?

(The same should be true for evaporation/condensation enthalpies, as we could have designed a very similar cycle with evaporation/condensation instead of electrolysis/fuel cell.)

Of course, we're starting by acknowledging a perpetual motion machine is impossible.  The question, then, is how do we understand, through chemical thermodynamics, why your specific set of steps can't constitute a perpetual motion machine.

The answer is that what you're not accounting for is that the electrical energy required for your step 2 is greater than you get back from your step 4, even if we don't have losses.  That's because, everything else being equal, the work needed to create gases against a high pressure is greater than that needed to create gases against a low pressure. I.e., $|w_2| > |w_4|$.

To understand why, let's simplify things by considering the pressure-dependence of the free energy of a pure substance at constant T: dG = VdP.  In step 2, because the pressure is so, high, you have increased the free energies of both the reactants and the products relative to what they'd be at atmospheric pressure.  To get the actual changes, you would need to calculate $\int^{P_f}_{P_i}V(P) dP$ for the reactants and products.  Since the products are gases (oxygen and hydrogen), and the volume of gases is much larger than that of liquids, the free energies of the products would be raised far more than that of the reactant (a liquid).  Thus the separation between the free energies of the reactants and products, (i.e.,  the magnitude of $\Delta G_r$) for step 2 (which is at high pressure), will be substantially greater than it is for step 4 (which is at low pressure).

You could also understand it using Le Chatelier's principle:  The volume of the products is higher than that of the reactants, and thus higher pressure will favor the reactants over the products.


Separating metals out of a magnet I have acquired the crushed magnet out of a motor I was given. If I'm correct that the magnet is $\ce{NdFeB}$ magnet, how could I separate out all the metals into a pure or oxide form? And for hard drive magnets, is there any way to remove the $\ce{Ni}$ coating without having to get nitric acid?  I don't have any proper chemistry equipment at all.

From this paper,  it seems Electrochemical recovery of Neodymium is possible from the Magnet.


Is shielding of electrons really best described as shielding or is it really Coulombic repulsion between electrons? For example, take the 2s electron in lithium - is the reason that it's ionisation energy is lower than would be expected if "shielding" didn't occur at all because the 1s electrons shield the attractive coulombic force from the nucleus or because the 2s electron is repelled by the 1s electrons? Is it possibly a mixture of the two?

To me, shielding seems to be a model to help conceptualize what is actually happening; am I wrong about this? What is actually going on?

The shielding model works as follows:

 (source)

So it is a mixture of attraction and pushing from protons and non-valence electrons, in accord with the Coulomb's law. The prediction of the effective core charge is not an easy task - you will have to use quantum chemistry methods (e.g. Hartree-Fock) to predict those, like its done in this reference.

I think its best viewed as a pragmatic model to boil down the higher, not very graphic, quantum theories, to one number, the effective nuclear charge:
$$ Z_\text{eff} = Z - S$$


What is the CNC bond angle in methyl isocyanate? Methyl isocyanate, $\ce{HC3NCO}$, is a toxic liquid which is used in the manufacture of some pesticides. What is the approximate angle between the bonds formed by the nitrogen atom in a molecule of methyl isocyanate?
(A) $104^\circ$ (B) $109^\circ$ (C) $120^\circ$ (D) $180^\circ$

I answered (A), but the correct answer is (C) according to the marking scheme.
I thought it was a bent-shaped, because  the lone pair of $\ce{e-}$ would repel the bond pairs of electron (similarly to $\ce{H2O}$ or $\ce{SO2}$).
My current assumption is that the $\ce{e-}$ in $\ce{N}$ in $\ce{CH3NCO}$ forms $\mathrm{sp^2}$ hybridization (and therefore should forms a shape like ethene).
I'm not sure if my thinking is correct, so hopefully you can help me understand why (C) is correct and why (A) is impossible.

This is not a good exercise question, because none of the options is actually correct. It basically asks you to blindly apply a model which cannot yield the correct result.
Before continuing any deeper into the matter, let's have a look at the actual structure. I was not able to find a crystal structure, so I ran a quick optimisation on the RI-BP86/def2-SVP level of theory. For this purpose, this is accurate enough.

The bond angle $\angle(\ce{CNC}) = 144^\circ$ is right in between $120^\circ \dots 180^\circ$.
The (pure) VSEPR model must fail for this molecule, as it is  only really working for central atoms with distinct and non-delocalised ligands.
For simple cases it performs very well in describing the approximate shape of a molecule. Once you have gotten this down, you can refine predictions with other tools. Please note, the lone pairs take up more space argument is actually a post-rationalisation and should not be used within the framework of VSEPR theory. More reliable predictions can be made with Bent's rule, which says that more electronegative ligands will concentrate p-orbital contributions towards them, decreasing the bond angle (this is a very, very brief summary).
As far as the VSEPR model goes, one should concentrate on simple coordination patterns:

linear; the bond angles will be close to $180^\circ$
trigonal; the bond angles will be close to $120^\circ$
tetrahedral; the bond angles will be close to $109^\circ$
higher coordinations like trigonal bipyramidal, octahedral, etc.

(TL;DR) From that we see that the central nitrogen has three ligands, so the general coordination around it is trigonal and the bond angle should be close to $120^\circ$. This is the answer, which your book will accept as correct, because it blindly applies the VSEPR model.
Applying a bit of Bent's rule won't help much. Carbon is slightly less electronegative than nitrogen, hence one would expect a slight opening up of the angle (more s-character directed towards carbon), but that's not enough to push it to $144^\circ$.
More importantly, the $\ce{-NCO}$ group is highly delocalised. Another very important concept comes into play: resonance. Ignoring the underlying structure for a moment, these are the most important contributors:
$$\ce{Me-N=C=O <-> Me-N^+\bond{3}C-O^-}\tag1\label{eq:resonance}$$
Within the concept of resonance, the positions of the nuclei will not change, and the structures do not exist on its own. A single structure of a set is therefore always an incomplete description. Please read What is resonance, and are resonance structures real?
The left structure in \eqref{eq:resonance} is basically what we have described above with VSEPR. From the depiction on the right, we would expect a $\angle(\ce{CNC}$ of about $180^\circ$, because triple bonds are coordinated linear. Since these are electronic descriptions of the same structure, we can expect the actual (underlying) structure to be somewhere in between these hypothetical forms.
The observed molecular structure is always the result of a delicate balance of effects. So repulsion of nuclei, electronic repulsion, electronic attraction, dispersion, etc. will all have an effect. Many of these interact quite complicated with each other. Sometimes our used models will yield good approximations, sometimes they don't.
In this particular case, we see that VSEPR alone fails; including resonance will refine the prediction. However, without a more rigorous approach, it is impossible to predict the bond angle correctly, i.e. we do not know the contribution of the right structure without doing some more maths. I used a quantum chemical approach, which is obviously nothing you could do in an exam. This is one of the reasons why I dislike this question.

A quick word on hybridisation:
The concept of hybridisation is independent of VSEPR. Many books will throw them together, but it is actually incorrect. You can make a prediction of the shape of a molecule with VSEPR. After that, you can use the molecular structure to choose the closest matching hybridisation model (if any at all). As such, hybridisation is always a result from a particular structure, never a cause. Therefore, your initial assertion is reverse:

My current assumption is that the $\ce{e-}$ in nitrogen in $\ce{CH3NCO}$ forms sp2 hybridization (and therefore should forms a shape like ethene).

Correct is, because the structure around nitrogen is trigonal, the electronic structure can be approximated with sp2 orbitals.

Why $\angle\ce{CNC} \approx 104^\circ$ cannot be correct
In order for this angle to be achieved, the nitrogen first needs to be surrounded by four ligands (or electron pairs), giving us an angle of about $109^\circ$ to start with. Then more p-character needs to go into the bonds, reducing the angle. Hence the ligands must be more electronegative than the other ligands. A popular example is dichloromethane,  or even water (but that is a bit more complicated). See the Wikipedia article on Bent's rule, our Q&A Utility of Bent's Rule - What can Bent's rule explain that other qualitative considerations cannot?, and multiple other questions with regard to Bent's rule.


Seperating out iron from ferrous sulfate I don't need to necessarily seperate pure iron, but from a ferrous sulfate solution, I am trying to form a precipitate with iron. By getting a precipitation of an iron compound, I can find out the amount of iron in the solution. Any help would be appreciated. I'm completely new and tried finding information online but all of it was beyond my comprehension. I apologise for my ignorance, but help would really be great.

Decomposing ferrous sulfate to get iron(III) oxide should work:
$$\ce{2FeSO4 ->[\Delta] Fe2O3 + SO2 ^ + SO3 ^}$$
Iron content can be measured from iron(III) oxide. Commercially, iron is recovered from waste ferrous sulfate in the form of magnetite ($\ce{Fe2O3}$) by co-precipitation and magnetic separation$\ce{^{[1,2]}}$. Iron content is measured from magnetite. It was observed that after processing the magnetite, the grade of iron in magnetite concentrate increased from 62.05% to 65.58% and the recovery rate of iron decreased from 85.35% to 80.35%.
References

Wang YU, Ying-lin PENG, Ya-jie ZHENG Recovery of iron from waste
ferrous sulphate by co-precipitation and magnetic separation,
Transactions of Nonferrous Metals Society of China, Volume 27, Issue
1, January 2017, Pages 211-219, DOI: https://doi.org/10.1016/S1003-6326(17)60024-4

Yu, W., Peng, Y. & Zheng, Y. Effect of purification pretreatment on the recovery of magnetite from waste ferrous sulfate. Int J Miner Metall Mater 23, 891–897 (2016), DOI: https://doi.org/10.1007/s12613-016-1304-2


Up to which dilution can one assume that 1 mg/l = 1 ppm? For aqueous solutions it is often assumed that 1 milligram per liter equals 1 ppm. Are there any guidelines up to which concentration this shortcut is admissible?
Are there any formulas which calculate the error when using this shortcut?

"Parts per …" are dimensionless entities: to derive $\pu{ppm}$ from mass concentration $γ$ (e.g. $\pu{1 mg L-1})$ one uses density of water $\rho(\ce{H2O}) = \pu{1E6 mg L-1}$ for normalization:

$$\frac{γ}{ρ(\ce{H2O})} = \frac{\pu{1 mg L-1}}{\pu{1E6 mg L-1}} = \pu{1E-6} = \pu{1 ppm}$$

Once the solution is too concentrated, you obviously are not allowed to use $ρ(\ce{H2O})$ and have to use the real concentration of the solution.
However, this occurs at the concentration levels way higher than those to be expressed with $\pu{ppm}$ anyway: that's why "parts per …" are so ubiquitously used in biochemistry. Higher molar masses, lower molar concentrations, primarily aqueous media: no need to account for the error stemming from the difference in the densities of pure water and the solution.

Note that "parts per something" units are deprecated and are not recommended for use, see e.g. Is 1 ppb equal to 1 μg/kg?.


Raman signal of black matte object This is the raman spectrum of the black cap:

I was expecting it to be Polypropylene but the peaks didn't match in wave numbers. Anyone got idea what kind of plastic has that peaks on the side of 1500?

When I tried the raman analyzer on the clear housing.

I got this raman spectrum.

Which matches that of Acrylonitrile butadiene styrene see:https://en.wikipedia.org/wiki/Acrylonitrile_butadiene_styrene

Original message:
I was trying to identify the material of this sediment filter cap whether it is really Polypropylene but i couldnt get a good signal but only noises no matter the exposure and time (my raman spectrometer is 532nm.

Is it really hard to get raman signal out of black matte object, maybe it cant reflect the laser?
I was able to get good raman signal of the blue casing though but not the black cap.

The comments pretty much sum up whats going on, but here is a more in-depth explanation to clear up any confusion for the OP.
There is a lot more behind this but here is the (very) basic principle:
Raman spectroscopy uses monochromatic light to excite molecules. These photons interact inelastically with the molecule (i.e. they gain or lose kinetic energy) and are then scattered. Some detector analyzes the scattered light and determines vibrational energy levels by calculating how much kinetic energy the photon has gained/lost which corresponds to the difference in energy of the vibrational modes.
When a black object is hit with a laser, the photons are absorbed. The absorbed energy is released as heat (This is why black cars are significantly hotter inside than white cars when left in the sun). Maybe some scattering does occur, but who knows if it is detectable. There are some articles on trying to determine the Raman spectra of black objects [Here is one].
Do some research; look at the Journal of Raman Spectroscopy if it peaks your curiosity. I'm not saying it is impossible, but it might require a lot of effort and understanding of the fundamental principles of this technique.


Burning a diamond produces how much energy? How much energy is produced by burning a 1 ct. diamond?

I'm trying to figure out how much energy to build the diamonds in a biological process.  See here: https://worldbuilding.stackexchange.com/questions/15314/are-diamond-berries-possible

Well it looks for me, the question was addressed by a non-Chemist.  Hence my attempt to answer this accordingly.

You have to know the reaction of the combustion, the starting materials, the product(s), and their relative quantities to balance the reaction equation.  Your case is described by $\ce{C} + \ce{O2} \rightarrow \ce{CO2}$, with "$\ce{C}$" for carbon (here: in form of diamond), $\ce{O2}$ for oxygen and $\ce{CO2}$ to yield the product, carbon dioxide. And the reaction equation is already balanced, i.e. the number of atoms per type on the left hand side (prior the arrow) equals the number of atoms per type written on the right, too.

There is a heat called standard enthalpy of formation, describing how much heat is necessary or liberated to form a compound if one (even theoretically) were to prepare this out of elements.  Normally, you look up these tabulated data for all your products and sum this up, and subtract the sum of all these enthalpies of formation of your starting materials.  In your example, by convention elementary gases (like oxygen, $\ce{O2}$) were attributed a zero value.  By means of thermodynamic, diamond is not the most stable form of carbon, and was attributed a value of ${+1.88}{\,\mbox{kJ/mol}}$ (same reference).  Carbon dioxide was attributed a value of ${-393.5}{\,\mbox{kJ/mol}}$ (reference) and the energy balance of the reaction equation equals to ${-391.62}{\,\mbox{kJ/mol}}$, i.e. if you would burn 1 mole of diamond.  (This is a quantity like dozen, used in chemistry.)

Now if we express the quantity of diamond actually burned (mass $m$ of 1 ct) in this above mentioned unity of mole, we need a proportional factor, the molar mass $M$, that happens to be $12{\,\mbox{g/mol}}$ for carbon (reference).
Now we have the pieces.  Hence we calculate the "chemical quantity" ($n$) of carbon deployed as

$n = \frac{m}{M} = \frac{0.2\,\mbox{g}}{12\,\mbox{g/mol}} = 0.017\,\mbox{mol}$
and than multiply this with the above calculated heat of the reaction, i.e
$\mbox{heat} = 0.017\,\mbox{mol} \times (-391.62\,\mbox{kJ/mol}) = -6.527\,\mbox{kJ}.$
So if you burn 1 ct of diamond in oxygen at 25 degree Celsius (about 77 Fahrenheit) and would collect the heat of the product by cooling the $\ce{CO2}$ to the same 25 degrees Celsius, this offers you 6527 Joules (or about 1559 calories).
Notes: The heat calculated here is the reaction enthalpy. The minus sign indicates energy is "liberated" in the course of this reaction. Actually, if all this heat were completely converted into electrical energy, you were able to switch on a light bulb of 60 W for less than 2 minutes (only 108 seconds).  In the beginning of modern chemistry, diamonds actually were burned by concentrated solar power, for example by Lavoisier (here) although to prove their composition.


What is the 'electronic energy', in units of 'negative' hartrees, of a molecule? On a NIST (National Institute for Standards and Technology) page on the molecular details of water and it's various isotopologues, I came across this:
Electronic energy:  -76.4089547246 hartree
There is, really, no other context on the page....
I found the page after typing in 'ionization energies of water'....
I put this here on Chemistry SE rather than Physics SE, I hope that is OK....
I DID look around to try to find an answer elsewhere, before coming here and posting...

The energy is the potential energy of all electrons in the molecule. Sometimes the term also includes the potential energy of the nuclei but in most cases it doesn't.
As reference or zero energy you can imagine the case where the electrons are non interacting or infinitely far apart from each other. In this sense it tells you the energy required to "remove" all electrons, i.e. pulling them away from each other so that they no longer interact with each other.
It can be interpreted as ionization energy where you remove all electrons in one go form the system but that is an theoretical process. The nuclei would push each other apart before you could ionize water to that degree. Normally the ionization energy refers to the removal of one electron. And the second ionization energy normally refers to the removal of a second electron after the system relaxed from the first ionization. And so on for higher ionization. This consecutive ionization process is different from removing all electrons in one go and would yield a different energy than the electronic energy given here.
This question/answer might also be helpuful :
https://chemistry.stackexchange.com/questions/38589/definition-of-electronic-energy#:~:text=Electronic%20energy%20is%20indeed%20an,nothing%20to%20do%20with%20electrons).


Cyclohexane derivative formation of carambolaflavone using a Lewis acid catalyst The scheme below summarizes the results from a study into the optimum conditions needed for a key step in the synthesis of the proposed structure of carambolaflavone, a natural product that has antidiabetic properties. Note that $\ce{Sc^3+}$ acts as a Lewis acid in these reactions.

I'm very confused about how the acid is coordinating to the Benzene derivative here?

@Waylander provided a correct written explanation for your question. I'll expand on his comments by considering the effect of temperature and provide diagrams. Using M+ = Sc+3 to save some space, the Lewis acid  coordinates with the more basic of the two acetate oxygens in 1 to provide the reactive oxonium species 2. [The oxonium species is planar about the double bond but I have left it in the chair conformation as a space-saving device.] This step may be reversible. At room temperature there is not sufficient activation energy to permit substitution of the aromatic ring. Following the red arrows of phenolic species 3, axial attack of the phenol oxygen occurs on the oxonium ion 2 to form, kinetically, compound 4. At 50 oC this reaction is reversible catalyzed by Sc+3 to reform 2 and 3 via the blue arrows. Following the green arrows, now there is sufficient energy to disrupt the aromaticity of 3 and allow for axial attack by carbon on the oxonium ion 2 to form, kinetically, structure 5. An increase of temperature to 70 oC permits coordination of the tetrahydropyran oxygen with the Lewis acid permitting ring opening, bond rotation about the red bond in 6 and ring closure of conformation 7 forming the thermodynamically stable, all equatorial tetrahydropyran ring of 8. [Structures 6 and 7 may also be drawn as benzyl cations with the carbonyl oxygen protonated.]


Identifying the gas given ratio of its specific heats The ratio of the heat capacities $\frac{C_\mathrm p}{C_\mathrm v}$ for one mole of a gas is $1.67$. The gas is:

  a) $\ce{He}$
  b) $\ce{H2}$
  c) $\ce{CO2}$
  d) $\ce{CH4}$

I know how to answer this by analysing the degrees of freedom of the gas molecule. Is there any other way to derive the ratio?

I'd go about such a question looking at the easiest thing first: ideal gases. So, what do you know about the relations between $C_{p}$ and $C_{v}$ (both being the molar heat capacities at constant pressure and constant volume, respectively) for ideal gases? For one thing, it is well known that

\begin{equation}
  C_{p} - C_{v} = R
\end{equation}

with $R$ is the universal gas constant. You can rearrange this equation for the ratio you are after:

\begin{equation}
  \frac{C_{p} - C_{v}}{C_{v}} = \frac{R}{C_{v}} \\
  \frac{C_{p}}{C_{v}} - 1 = \frac{R}{C_{v}} \\
  \frac{C_{p}}{C_{v}} = \frac{R}{C_{v}} + 1 \ .
\end{equation}

Now, you need to find $C_{v}$. Here, the degrees of freedom that you mentioned come into the game, since the equation that helps us out is the equipartition law, stating that each excited quadratic degree of freedom contributes $\frac{1}{2} R T$ to the molar internal energy $U$

\begin{equation}
  U = \frac{1}{2} R T (f_{\text{trans}} + f_{\text{rot}} + 2 f_{\text{vib}})
\end{equation}

with $f_{\text{trans}}$, $f_{\text{rot}}$ and $f_{\text{vib}}$ being the quadratic degrees of freedom for translation, rotation and vibration respectively. It is important to note that only excited degrees of freedom contribute to $U$. Which ones are active depends heavily on temperatur. Vibrations are usually not excited at room temperatur (there are exceptions, e.g. iodine) - they usually need temperatures in the order of $1000 \, \text{K}$ but that varies greatly. Rotations are much easier to excite - I know of no example where they aren't excited at room temperature. Translations take very little energy to excite and can always be considered active. If all degrees of freedom are active you get: For linear molecules $f_{\text{trans}} = 3$, $f_{\text{rot}} = 2$ and $f_{\text{vib}} = 3N - 5$, where $N$ is the number of atoms in the molecule. For non-linear molecules $f_{\text{trans}} = 3$, $f_{\text{rot}} = 3$ and $f_{\text{vib}} = 3N - 6$.
Since $C_{v} = \frac{dU}{dT}$ you get

\begin{equation}
  C_{v} = \frac{1}{2} R (f_{\text{trans}} + f_{\text{rot}} + 2 f_{\text{vib}}) \ .
\end{equation}

Substituting this into the equation for the ratio between $C_{p}$ and $C_{v}$ yields

\begin{equation}
  \frac{C_{p}}{C_{v}} = \frac{2}{(f_{\text{trans}} + f_{\text{rot}} + 2 f_{\text{vib}})} + 1 \ .
\end{equation}

Now, you only have to know how many degrees of freedom each of the molecules has and plug them into the equation.


Why doesn't H₄CO₄ exist as a stable compound? Why doesn't $\ce{H4CO4}$ exist as a relatively stable compound? Its heavier homologue $\ce{H4SiO4}$ seems to exist although it is unstable to some extent but new methods have been able to stabilise orthosilicic acid using choline chloride.

Wikipedia reports that calculations show orthocarbonic acid spontaneously decomposes to the ordinary (meta)carbonic acid $\ce{H2CO3}$ plus water.  However, orthocarbonate esters $\ce{C(OR)4}$ do exist.

The orthoacid may, however, become stable at high pressure [1, with a nontechnical summary given in 2] and possibly exist within Uranus and Neptune.  It is further posited that $\ce{H2CO3}$, ordinarily stable only in aqueous solution, becomes a stable solid in the interiors of icy moons such as Europa, possibly impacting interactions thought to be necessary for the development of subterranean life.

A final note: the swastika-like shape of the orthocarbonic acid molecule has led unfortunately to the name "Hitler's acid".  Actually the swastika long predates the Nazis and retains non-Nazi usages to this day.


Does MOPAC allows calculation of Standard Enthalpy of reaction in this way? I am using MOPAC2016 to calculate $\Delta H^\circ_\mathrm{r}$ (and then $\Delta G^\circ_\mathrm{r}$) for a reaction in aqueous phase, at $\pu{298 K}$. I will explain how, and then the question. Finally I give an example. Suppose this simple reaction:

\begin{align}
\ce{A(aq)<=>B(aq)}
\end{align}
Then, $\Delta H^\circ_\mathrm{r}$ can be obtained as:

$$\Delta H^\circ_\mathrm{r} = \Delta H_\mathrm{f}^\circ(\ce{B}) - \Delta H_\mathrm{f}^\circ(\ce{A})$$ where each of those values is what we get on a thermodynamic calculation (also in an optimization) with the name of H.O.F. (heat of formation, HOF).

Formally the question should be: Is that correct to calculate $\Delta H^\circ_\mathrm{r}$ (and then $\Delta G^\circ_\mathrm{r}$) by using the value of HOF for each reactant?

Example

Here is an example of MOPAC output for a thermodynamic calculation, suppose it is the one for molecule $\ce{B}$:

                    CALCULATED THERMODYNAMIC PROPERTIES
                                          *
   TEMP. (K)   PARTITION FUNCTION   H.O.F.    ENTHALPY  HEAT CAPACITY ENTROPY
                                   KCAL/MOL   CAL/MOLE    CAL/K/MOL  CAL/K/MOL

 298.00  VIB.     0.2155D+06                  8205.6617    53.3353    51.9400
         ROT.     0.3572D+07                   888.2813     2.9808    32.9651
         INT.     0.7697D+12                  9093.9430    56.3161    84.9051
         TRA.     0.4009D+28                  1480.4688     4.9680    42.5424
         TOT.                     -670.232   10574.4119    61.2841   127.4474

   *: NOTE: HEATS OF FORMATION ARE RELATIVE TO THE
            ELEMENTS IN THEIR STANDARD STATE AT 298K
              (=  Standard Enthalpy of Formation).

            Hvib:  Zero-point energy is not included.
                   frequencies of less than zero cm-1 are not included.
            Hrot = (3/2)RT
            Htra = (3/2)RT + pV = (5/2)RT

            Heat capacity is Cp, not Cv

          For more information, see: HTTP://OpenMOPAC.net/Manual/thermochemistry.html

With that data table $\Delta G^\circ_\mathrm{f}$ of the molecule $\ce{B}$ should be:
$$\Delta G^\circ_\mathrm{f}= \mathtt{H.O.F.} - TS^\circ= -670.232\times1000 - 298\times(127.4474- S_{atoms})$$
$S_{atoms}$ will be canceled when calculating $\Delta G^\circ_r$.
On this mopac site, $\Delta H_\mathrm{f}^\circ$ appears to be obtained in a similar way, for the $\ce{NH4+}$ molecule.

Am I correct?

The free energy change of a reaction occurring in solution can be computed using a thermodynamic cycle:
$$\Delta G^\circ_{r,soln} = \Delta G^\circ_{r,gas} + \Delta\Delta G^\circ_{r,solv} $$
$\Delta G^\circ_{r,soln}$ can also be written as
$$\Delta G^\circ_{r,soln} = G^\circ_{soln}(B)- G^\circ_{soln}(A)$$
where
$$G^\circ_{soln}(X) = G^\circ_{gas}(X) + \Delta G^\circ_{solv} $$
and
$$G^\circ_{gas}(X) = E_{gas}(X) + G^\circ_{gas,RRHO}(X)$$
The solvation energy is usually computed using a continuum as the difference in electronic energy with and without the continuum.
$\Delta G^\circ_{solv}(X) = E^\prime_{soln}(X)-E^\prime_{gas}(X)$
The parameters in the continuum model are usually optimized for a particular level of theory, which may not be appropriate for the computation of $\Delta G^\circ_{r,gas}$, so $E_{gas}(X)$ is not necessarily the same as $E^\prime_{gas}(X)$

However, if $E_{gas}(X) = E^\prime_{gas}(X)$ then
$$G^\circ_{soln}(X) = E_{soln}(X) + G^\circ_{gas,RRHO}(X) $$

The parameters in the continuum model are usually determined using the gas phase geometry $(X)$.  But if $E_{gas}(X) = E^\prime_{gas}(X)$ then many people often choose to use the geometry optimized in solution
$G^\circ_{soln}(X_{soln}) = E_{soln}(X_{soln}) + G^\circ_{soln,RRHO}(X_{soln})$.

For semiempirical calculations, the electronic energy corresponds to the enthalpy of formation at 298K, so if one simply makes the substitution
$$E(X) \rightarrow \Delta H^\circ_{f,298}(X)$$
then (at 298K)
$$G^\circ_{soln}(X) = \Delta H^\circ_{f,298}(X) -TS^\circ_{gas,RRHO}(X)$$ $G^\circ_{soln}(X)$ doesn't correspond to a measurable free energy because $S^\circ_{gas,RRHO} \ne S^\circ_{f}$ but it can still be used to compute $\Delta G^\circ_{r,soln}$ since the missing terms cancel.

For temperatures other than 298K
$$G^\circ_{soln}(X) = \Delta H^\circ_{f,298}(X) + (H^\circ_{gas,RRHO}(X,T)-H^\circ_{gas,RRHO}(X,T=298K)) -TS^\circ_{gas,RRHO}(X)$$


Why is the water column empty in ICE table In the ICE table pure solids and liquids have a value of one in the equilibrium expression but when we create an ICE table, why is the $\ce{H_{2}O}$ column empty? Should it not have a value of one?

Presumably water is the solvent for this reaction, which means that it will be in extreme excess and the amount will not change to a measurable extent.


Ab initio method to calculate C-C bond dissociation energy? I'm trying to review how to calculate bond dissociation energy for C-C in ethane, a very simple calculation (or should be).

I get the following Hartree energies for SCF 6-31G(p) for methane, ethane, and H2:

ch4_spe            -40.194639920746,
ethane             -79.228125042573,
h2                 -1.126456057886,

I would think that to get the C-C bond dissociation energy, I would simply do (Perl code)

$energies{ethane} - $energies{ch4_spe} * 2 - $energies{h2}

correct?

The reason that I'm asking is that this give an answer of 2.2876 Hartree using NWChem, which converts to 6006 kJ/mol, which is off by about a factor of 17 from the correct value of about 346 kJ/mol.  I know that SCF isn't that accurate, but it shouldn't be this far off. I optimized the bond lengths using Avogadro.

This is the correct method to calculate the C-C bond energy?

This "experiment" is about determining bond dissociation energy.

The general way to solve this isn't like calculating enthalpies of reaction, as I initially thought, but rather through calculating the energy of two CH3 radicals, and comparing it with ethane.  As another user pointed out, this should be done with unrestricted Hartree Fock and with diffuse basis sets.

Thus, energy(ethane) - 2*energy(methyl radical) is the solution.


Transition Metals' Charges I came across a compound $\ce{[FeO4]^{2-}}$ but $\ce{Fe}$ has +6 charge according to my calculations. How this is possible ? Or it is possible but my periodic table is kind of non-detailed one. (On the table I got $d$-metals possible charges).

If you write out the electron configuration of Fe, you will find it has $3\text{d}^6 4\text{s}^2$ in its valence level, meaning it can theoretically take on any oxidation state from +1 to +8 (and even some negative states, for that matter).

In practice it is extremely uncommon to find any oxidation states other than +2 or +3 for iron - which explains why a simple periodic table containing only the most common naturally occurring states will omit them - but they are possible, just not very stable or long-lived. The polyatomic ion you have observed is called ferrate.


Experiments in chemistry involving use of stereo microscope? I recently got a decent stereo microscope. Besides using the scope to teach my children, I have been thinking of picking up something new to learn myself. I have been interested about chemistry for a long time but never studied it after upper secondary school. I was thinking would there be some experiments that would allow me to learn more microscopy and chemistry at the same time? I have ~10  years of experience in machine vision, signal processing, neuroscience, programming etc., so research-grade problems are welcome too.

For starters, it might be fun to grow different kinds of crystals in shallow dishes and watch them grow in real time under a microscope. Either a low melting solid, melted and resolidified, or growing from a saturated solution. Also, looking at how a metal surface is etched by acid using a microscope would be fascinating.


Are certain logP values eluted when 100% methanol is used as the solvent in a C18 column? I have used a C18 sep-pak to concentrate metabolites in a chemically complex solution. The column was first washed with pure water, 75% methanol:water mix, and finally the compounds eluted with 100% methanol.

What I am wondering is what chemical properties that I can infer from this information? I know I would have washed the very polar compounds away, and in pure methanol, I would expect compounds of a semi-polar nature to be eluted. Can I assume that these compounds would have a specific LogP range? I cannot find an answer online anywhere, so I guess the answer must be no, and if so, why not?

There is one (or more) compound of a biologically active nature in this complex mix. I am wondering if I can use the elution conditions to understand some of the chemical properties of the compounds eluted. Further, I did FT-ICR-MS on this mix, and I would like to look at a list of compound hits from the mass spec databases and understand how feasible a particular hit is based on its polarity.

HPLC is often used to determine lipophilicity in industry, as the original methods to determine logP by partitioning between octanol and water was not particularly amenable to high-throughput assay.

Lipophilicity values measured by HPLC are often reported as cLogP / cLogD (where c is relating to the fact that the value has been determined by some form of chromatography).

Experimentally, a C18 column is used with an acetonitrile/water solvent system, buffered appropriately. Unlike with the octanol/water procedure, different companies run these HPLC assays differently, and as such they aren't as comparable (i.e. if you ran 20 compounds through a GSK assay, and the same 20 compounds through a Pfizer assay, the precise values for cLogD obtained in each case would vary, even if the general trend remained the same).

In terms of you doing this yourself, you'd first need to find a training set of compounds with known and reliable values. If you know paracetamol comes of at x mins and has a  literature logP of y, and ibuprofen comes off at z mins and has a literature logP of w, you can interpolate. When you then run an unknown sample, its retention time using your method can be used to predict the cLogP/D (keep in mind that you'd need a far greater training set than 2 compounds).


Why does an electrochemical cell work? My textbook says that an electrochemical cell works because of difference in reduction potential of two metals which cause one to lose electrons and other one to accept them. If a salt bridge is not used then the solutions of the two metals become charged and the reaction proceeds till the difference in reduction potential is cancelled by the difference in potential due to charges. All this theory seems to work well on paper, but it is never really explained why this reaction occurs.
What I mean is suppose you have a copper and a silver half cell. They are connected. The copper part does not know whether it is connected by a silver half cell or sodium have cell and I see no apparent of this information being communicated between the cells, so how does it know whether to get dissolved in the solution or precipitate out of the solution?
The book says that due to the difference in reduction potential, a potential gradient is set up which causes the charge to flow. I agree with this completely yet I am not able to grasp why this physically happens. What force causes to electrons to move from one electrode to the other? Especially when there is no salt bridge, the electrons move from positive electrode to negative electrode, which is completely opposite to anything I know.
I think that there must be some difference in local phenomena in the solution which spontaneously causes electrons to flow without any information being shared between the two electrodes. eg: At first I thought maybe copper having lesser reduction potential than silver would dissolve and precipitate out faster (basically a fast(?) equilibrium) which is completely independent of the other electrode and this difference speed would somehow cause the electrons to flow. I rejected this idea after a while but I still think difference in some sort of local phenomena in a half cell which is completely independent of the other half cell would drive the reaction forward.
Any thoughts?

Maybe you would understand better with an image, or if you admit that all metals "are wanting" to loose electrons. And in order to do that, they must find another atom which is weak enough or "humble" enough to accept to work "against its will". In other words, some atoms are more "willing" than others. It is like a fight, with a "strong" and a "weak" atom. In your copper-silver cell, copper is more assertive : it imposes its electrons to the "weak" silver atom. Silver cannot emit its own electrons. It is even obliged to accept electrons from copper and have these electrons react with its proper ions. producing more silver atoms. Some atoms are stronger than copper. Zinc for example, opposed to copper, imposed its electrons to copper. This order of "strength" is experimental, and cannot be derived by calculations.
Now to speak more scientifically, you may replace the present notion of strength by the reduction potential. The more the reduction potential is negative, the more the metal has a tendency to loose its electrons.
Anyway, a salt bridge is absolutely necessary for the cell to work. If there is no salt bridge, it is not a cell, and it produces no chemical reaction, and no electric currant.

Complement, required by Manit.
If you want a local phenomena to explain the cell, you may consider the following development. You may admit that copper "wants" to create more $\ce{Cu^{2+}}$ ions in solution. These ions are positively charged. So the copper compartment must attract negative ions to compensate the new positive charges appearing around the copper plate. As a consequence, the new $\ce{Cu^{2+}}$ ions attract negative ions from where they are available, namely in the other compartment. And as $\ce{Ag}$ is not able to create $\ce{Ag+}$  ions with the same force, Silver plate must admit that it looses its negative nitrate ions to send them to the copper compartment. And as a consequence, it must admit that the positive ions $\ce{Ag+}$ should also "disappear" from the compartment. And the only way to make this charge disappear from the compartment is to accept electrons, so that the $\ce{Ag+}$ ion is transformed into metallic silver.
OK ?


What is the cause for thermodynamic reversible and irreversible process? My book defines

  Thermodynamic irreversible process is defined as the process in which the changes take place rapidly, instead of occurring in successive stages with infinitesimal changes in quantities so that the system does not get a chance to attain equilibrium and there is an appreciable difference between the driving and the opposing force.

I have some problems in understanding it.

Which equilibrium are they speaking of?
Why does rapidity of the change cannot help it to attain equilibrium (how does an infinitesimal change in a reversible process help it to reach equilibrium)?
What does they want to mean by driving and opposing force?

Let me start by reiterating what @Phillip said in his comment: this is not a good definition for an irreversible processes - a better definition is:

  An irreversible process is one in which the net entropy generation term is greater than zero.

All real processes are irreversible. The "process speed" comes into play when we want to make approximations for idealized systems. It turns out that one way to convert a process into a theoretically reversible one is to set the condition that it moves very slowly between thermodynamic states, so that at any given step, it is in equilibrium. The reason that this isn't a good definition for a reversible process is that there are cases where a rapid change is reversible, and cases where a very slow change is irreversible. It's just that in general use, the "very slow process" approximation is commonly used to obtain a reversible process.

  Which equilibrium are they speaking of?

Typically they mean equilibrium between the system and surroundings, but it can also mean within the system itself. Equilibrium is another way of saying "maximum entropy" - the matter and energy are as "spread out" as they are going to get. If the entropy is maximized, no more can be generated. In "real life" this means the system is well-mixed and that matter and thermal energy has dissipated throughout the system - there are no temperature or concentration/density gradients. This is easier to visualize using particles instead of energy - imagine an ideal gas in one side of a partitioned chamber, where the partition has a hole in it to allow the molecules to cross into the other section:

This is an example of an internal equilibrium - the particles cannot cross the system boundary, but they can reach an equilibrium within the system. An example of a system in equilibrium with its surroundings would be a container filled with water at room temperature - there are no thermal gradients across the container boundary.

  Why does rapidity of the change prevent it from attaining equilibrium (how does an infinitesimal change in a reversible process help it to reach equilibrium)?

Whenever a system is perturbed, meaning moved away from an equilibrium state, it takes some amount of time to reach a new equilibrium. This amount of time is called the relaxation time. Different systems have different relaxation times, and if you make changes quickly enough, the system won't have time to relax, and therefore won't be in equilibrium. Using a piston in a cylinder as an example, if you compress the piston by a very small amount, then the pressure inside will be in equilibrium with the pressure outside almost immediately.

  What do they mean by driving and opposing force?

Using the piston as an example again, in this case the driving force would be the external pressure on the piston. The opposing force would be the pressure of the gas inside the cylinder. For a very slow process, at any given time the outside pressure and inside pressure would be the same - the volume would change to make sure this is true. For a very rapid change, the gas inside the cylinder would take longer to reach equilibrium, and so the pressure outside would not equal the pressure inside - the driving and opposing forces would not be equal.

It is important to remember that these are approximations that are intended to allow us to convert really difficult (usually unsolvable) differential equations that describe non-equilibrium or dynamic systems into something that is solvable. The trade-off is that we go from a realistic, fully dynamic system to one with constraints - we treat it like a series of tiny jumps between systems that are in equilibrium. In many cases, these approximations don't affect the accuracy of the solution. However, whether or not it will is determined by the time scale of the change relative to the relaxation time of the system.


How to simulate surface tension? I am trying to create a water drop simulation for measuring hydrophobicity of surface. I don't know how to simulate the contact angle which is related by younges equation to the surface tensions for each of the solid-liquid-gas interfaces. I have come across some models for molecular dynamics, which for chemistry novice like me (I'm a EE) are bit advanced. I am looking for simpler method in which I could simulate the interfacial tensions by some sort of derived mathematical model which would be simple enough to act as an objective function to an evolutionary optimizer.

If I understand your question correctly, you want to relate a droplet's geometry to it's environment. The curvature of a liquid droplet is described by the Laplace equation of capillarity, also known as the Young-Laplace equation. This is valid for both a sessile (sitting on a surface) and a pendant (hanging) droplet. I am assuming a static experiment in which the droplet is axisymmetric, with $z$ defined as the axis of symmetry. Such methods of solving for interfacial tension and contact angles are called Axisymmetric Drop Shape Analysis (ADSA) methods. ADSA methods are based on fitting the shape of an experimental droplet to the Laplace equation. This differential equation has been numerically solved in 1883, but thanks to computers and camera's this method is easier and more accurately to apply.

The Laplace equation is given by:

$$\gamma\left(\frac{1}{R_1}+\frac{1}{R_2}\right)=\Delta P\label{eq:laplace}\tag{1}$$

in which $\gamma$ is the surface tension and $\Delta P$ is the pressure difference over the interface. After some work, this may be written as a system of differential equations.
$R_1$ and $R_2$ are the principal radii of curvature.
$R_1$ is called the meridional curvature, which turns in the plane of the screen you are reading from.
Perpendicular to this is the azimuthal curvature $R_2$.

The meridional curvature is defined as:

$$\frac{1}{R_1}=\frac{\partial \phi}{\partial s}$$
The azimuthal curvature is defined as:
$$\frac{1}{R_2}=\frac{\sin{\phi}}{x}$$

`In the absence of external forces, other than gravity, the pressure difference is a linear function of the elevation' (Del Rio and Neumann, 1996):

$$\Delta P= \Delta P_0+\left(\Delta\rho\right)gz\label{eq:rio}\tag{2}$$
$\Delta P_0$ is the pressure at the origin, at which the droplet is not perturbed and $R_1=R_2=R_0$.
Using \eqref{eq:laplace} this leads to the following expression:
$$\Delta P_0 =\gamma\left(\frac{2}{R_0}\right)\label{eq:p0}$$
Combining \eqref{eq:laplace} and \eqref{eq:rio} and filling in the expression for $\Delta P_0$ and the curvatures leads to:

\begin{align}
\gamma\left(\frac{\partial \phi}{\partial s}+\frac{\sin{\phi}}{x}\right)&=\frac{2\gamma}{R_0}+\left(\Delta\rho\right)gz\\
\frac{\partial\phi}{\partial s}&=\frac{2}{R_0}+\frac{\left(\Delta\rho\right)gz}{\gamma}-\frac{\sin{\phi}}{x}
\end{align}
Now the coordinates are transformed to dimensionless coordinates:
\begin{align}
\bar{x}=\frac{x}{R_0}, \bar{z}=\frac{z}{R_0}, \bar{s}=\frac{s}{R_0}
\end{align}

This changes the differential:
\begin{align}
\frac{\partial\phi}{\partial s}=\frac{\partial\phi}{\partial\left(R_0\bar{s}\right)}=\frac{\partial\phi}{R_0\partial\bar s}
\end{align}
Multiplying both sides with $R_0$ gives:
\begin{align}
\frac{\partial\phi}{\bar s}&=2+\frac{\left(\Delta\rho\right)gR_0^2}{\gamma}\bar z-\frac{\sin\phi}{\bar x}\label{eq:phi}
\end{align}

Furthermore the following geometric interpretation leads to the remaining differential equations.
The arc length $s$ follows the curvature of the interface of the drop. $\phi$ is the tangential contact angle, so a change in $x$ over a change in $s$ is equal to the cosine of $\phi$:

\begin{align}
\frac{\partial x}{\partial s} &= \frac{\partial \bar{x}}{\partial \bar{s}} = \cos(\phi)\label{eq:cos}
\end{align}
 Similarly, a change in $z$ over a change in $s$ is equal to the sine of $\phi$:
\begin{align}
\frac{\partial z}{\partial s} &= \frac{\partial \bar{z}}{\partial \bar{s}} =\sin(\phi)\label{eq:sin}
\end{align}
The same holds for the dimensionless coordinates, resulting in the given system of differential equations.

Solving the curvature of the droplet is an initial value problem, with the initial values being $x_0$, $z_0$ and $\phi_0$. You can feed this to your numerical differential equation solver, e.g. Matlab's ODE45 function. With this method you will be able to solve the contact angle and surface tension from a given curvature, or draw a curvature for a given contact angle and surface tension.

Suggested reading on which I have based my answer:

Y. Rotenberg, L. Boruvka, and A. W. Neumann, “Determination of surface tension and contact angle from the shapes of axisymmetric fluid interfaces,” Journal of Colloid and Interface Science, vol. 93, pp. 169–183, May 1983.

Rı́o, O. I. del, and A. W. Neumann. “Axisymmetric Drop Shape Analysis: Computational Methods for the Measurement of Interfacial Properties from the Shape and Dimensions of Pendant and Sessile Drops.” Journal of Colloid and Interface Science 196.2 (1997): 136–147. ScienceDirect. Web.


vertically growing icicle I made some "super" soap bubbles with my kids using the following ingredients:

liquid dish soap
tap water
corn syrup

We left the solution out over night last night here in Maine and the temperature got down to −3 °C. This morning we found that the solution left in the lid had grown a vertical icicle.

There is no trickery here. I stepped outside and found these items as you see them below.

How could this icicle have formed?

It pretty much has to be a supercooling phenomenon.  This page from the University of Toronto Physics department presents an explanation that I think fits the scenario in the question nicely.  The description and photos used suggest that this is likely the case for this question:

  The short explanation is this: as the ice freezes fast under
  supercooled conditions, the surface can get covered except for a small
  hole. Water expands when it freezes. As freezing continues, the
  expanding ice under the surface forces the remaining water up through
  the hole and it freezes around the edge forming a hollow spike.
  Eventually, the whole thing freezes and the spike is left.

  A slightly longer explanation: the form of the ice crystals depends on
  the cooling rate and hence on the degree of supercooling. Large
  supercooling favors sheets which rapidly cover the surface, with some
  sheets hanging down into the water like curtains. These crystallites
  tend to join at 60 degrees and leave triangular holes in the surface.
  Hence, spikes often have a triangular base. The sides of the spike are
  sometimes a continuation of pre-existing subsurface crystallites, and
  can extend from the surface at steep angles.

Notice the triangular shape at the base of the ice spike in the photos in the question, as is also mentioned in the description above.  The spike also appears to possibly have a hollow region through the center.  Additionally, the flexible plastic lid would allow for some pressure buildup before the release through some weak spot in the ice sheet.


Does PEO particle size decrease when dissolved in water? I am looking to make metallic ink for inkjet printing and using PEO 1 wt%  (MW ~600,000) as a binder. The print head I am hoping to use has a nozzle diameter of 80 microns. The metal powder dispersed in the ink is ~100nm. If I dissolve the PEO into deionized water, will the PEO particles be too big to pass through the nozzle?  The PEO purchased is from sigma aldrich (https://www.sigmaaldrich.com/catalog/product/aldrich/182028?lang=en&region=IE)

(Physicist asking a chemistry question!)

Thanks :)

PEO is very soluble in water. At 1 wt% concentration, you should expect it to be individual molecules throughout the solution. A PEO molecule with MW of 600,000 has ~13,600 repeating units. You can imagine them to be roughly ~200 nm blobs calculated based on the radius of gyration of a linear polymer in a good solvent.


Reaction of Boron with Alkali In one of my books it is written that boron does not react with alkali but in another book it is written that it reacts with alkali.
Can any one help in clearing this confusion?

Boron react with $\ce{NaOH}$ but under certain conditions. Boron exist in two form (1) crystalline form (2) amorphous form. Both forms are generally inert to chemical reactions and is unaffected by acids and alkali at normal conditions but at elevated temperature, concentrated oxidizing acids like nitric or sulfuric acids react with boron to yield boric acid. On the other hand, amorphous boron does react with fused alkali at high temperature to obtain trisodium borate:
$$\ce{2B(s) + 6NaOH(s) ->[t > 773K] 2Na3BO3(s) + 3H2(g) ^}$$
High purity, hydrated sodium metaborate(sodium tetrahydroxoborate, $\ce{Na[B(OH)4]}$) can also be produced from boron in a $\ce{NaOH-H2O}$ system. The process applied was called pressure alkaline leaching.
When fused in the presence of air at $\pu{350-400 ^\circ C}$, sodium metaborate is formed. You can find the reaction in this answer by @andselisk.
References

https://onlinelibrary.wiley.com/doi/abs/10.1002/0471743984.vse1217
Yin, B.W & Qin, S.Y & Zhang, Y.F & Zhang, Y (2014). Preparation of hydrated sodium metaborate from boron concentrate via pressure alkaline leaching. Guocheng Gongcheng Xuebao/The Chinese Journal of Process Engineering. 14. 258-265, (link)


Should I use specific heat to calculate how much fuel is required? Question:

  A backpacker wants to carry enough fuel to heat $\pu{3.0 kg}$ of water from $\pu{20 ^\circ C}$ to $\pu{100.0 ^\circ C}$. If the fuel he carries produces $\pu{36 kJ}$ of heat per gram when it burns, how much fuel should he carry? (For the sake of simplicity, assume that the transfer of heat is $100\%$ efficient.)

My hunch is that it involves the specific heat equation $Q_\mathrm p = m  c_\mathrm p  \Delta T$ where $Q$ is heat, $m$ is mass, $c_\mathrm p$ is the specific heat capacity (in this case $\pu{4.184 J/(g ^\circ C)}$) under constant pressure and $\Delta T$, but twice I calculated the wrong answer, and would like a more intuitive understanding of the problem.

$$\Delta H = mc_p\Delta T$$

Where $\Delta H$ is either energy absorbed or released at constant pressure.
Assuming this specific problem refers to $3\ \mathrm{kg}$ of water:
$$
\Delta H = \left(3000\ \mathrm g\right)\left(4.184\ \mathrm{J\cdot g^{-1}\cdot K^{-1}}\right)\left(80\ \mathrm K\right)
$$

That means that the energy required is: $1\,004\,160$ joules.

Now, simply divide those 1 004 160 joules by the 36 000 joules your fuel releases per gram.
$$
m_\text{fuel}= \frac{1\,004\,160\ \mathrm J}{36\,000\ \mathrm{J\cdot g^{-1}}} = 27.89\ \mathrm g
$$