Changes in defects in a material due to heat treatment Currently i am doing some research on optical properties of  a oxide material and for that purpose i have done some photoluminescence study on it. Because this is an ongoing research i cant divulge much information here. I have used "$Ex$" excitation wavelength for getting emission spectra for my oxide sample. I have prepared two types of oxide samples one is heat treated and the other not heat treated. After taking emission spectra data for both the samples I got peaks similar to this

Samples               Peak wavelength (nm)

Non Heat treated          a, b

Heat treated              b,c,d,e

*a,b,c,d & e are wavelength where emission peaks detected

Both samples (heat treated and non heat treated) has same chemical composition which i confirmed using powder XRD. As you can see from the data the heat treated sample has more emission peaks comparing to non heated samples. I can understand from this data that heat treatment to the sample causes increased defects in the sample. what i thought was that after heat treating a crystalline material; the defect in that material should decrease and not the other way round? Is my rationale wrong. Why heat treated sample has more defects?

It seems, that your oxide can have different crystal polymorphs and probably also other non-crystalline phases.

You synthesize your material as mixture of those, where the composition depends on the preparation process.

Upon heating and slow cooling you probably recrystallize the species to more thermodynamically stable polymorph (new peaks c,d,e) on the expense of less stable one (a).
It can also depend on the speed of cooling, so you could obtain some metastable phases, etc...

Even then, treat your material always as an unkonwn mixture of uknown number of polymorphs and let the spectra tell you in which way the composition shifted.


What is the direction of the net electric dipole moment in hydronium H3O+? Assuming electric dipole moment points towards the negative charge, what would its direction be in hydronium?
Although the bonding electron density is distorted towards the more electronegative oxygen, it already has a formal charge of +1. Assuming the partial charge on oxygen due to distorted shared e.density is less than O.5, wouldn’t the actual electric dipole moment points away from the oxygen(opposite to distortion)? I can’t find any source to check this

In aqueous medium, the proton ($\ce{H+}$) is thought to exist as hydronium, $\ce{H3O+}$. However, in reality, probably bigger complexes of water may also be associated with one proton. In liquid water, everything is interconnected with strong hydrogen bonds anyways, and it is difficult to distinguish the free species. All of this means that it would be difficult to measure the dipole moment of $\ce{H3O+}$ experimentally. I haven't found any reference that did this, and I suspect it is impossible.
So we have to turn to computational chemistry to calculate the dipole moment of $\ce{H3O+}$. For one isolated $\ce{H3O+}$ (i.e. in gas phase):

The z-axis is set along the $C_3$ axis.
$\text{M06-2X/aug-pcseg-2}$ gives a dipole moment of $\text{1.515 D}$ in z-direction. In x and y direction, the dipole values are negligible:
         DX          DY          DZ         /D/  (DEBYE)
    -0.007834    0.000072    1.515546    1.515566

This means the dipole points along +z-axis in the picture. Partial charges(ESP) fitted to the dipole moment are:
 NET CHARGES:
 -------------------------------------
 ATOM                CHARGE    E.S.D.
 -------------------------------------
 O                  -0.3668    0.0000
 H                   0.4561   -0.0000
 H                   0.4553   -0.0000
 H                   0.4554    0.0000
 -------------------------------------

As you can see, the oxygen atoms has a partial negative charge and hydrogen atoms have partial positive charges. So, the formal charges that we assign to the Lewis structure does not have much significance in this case.
(IUPAC convention of dipole moment is that the arrow points from the negative charge towards the direction of the positive charge. This is exactly opposite of what you wrote in the question)
Another method, $\text{MP2/aug-pcseg-2}$ gives $\text{1.507 D}$ along z-axis, which is close to what we got from the DFT calculation.
(Note that I have not enforced symmetry for the calculation, which is why the partial charges are slightly different for each hydrogen. This is a numerical error, because in reality all the hydrogens are equivalent)


Why does (atmospheric) pressure have an effect on solubility? The Wikipedia article titled Solubility table states that

  The table below provides information on the variation of solubility of different substances (mostly inorganic compounds) in water with temperature, at 1 atmosphere pressure.

Why does pressure have an effect of on solubility?

It comes from a combination of Henry's law and Le Châtelier's principle. Henry's law is
$$p=k_hc$$
where $p$ is the partial pressure of the gas, $k_h$ is the Henry's law constant and $c$ is the concentration of the gas (because liquids and solids are not affected as much as gases, if at all). This means that with a greater pressure comes a greater concentration, and with a lower pressure comes a lesser concentration, and vice versa

Le Châtelier's principle states that if the pressure, concentration, temperature or volume of a system changes, something else will change to compensate and swing the whole thing back into equilibrium. If there is a greater pressure, there will therefore be a grater concentration. A greater concentration means there is a greater solubility.

The UC Davis Chem Wiki (I didn't see that DrMoishe Pippik had linked to it until after I posted this!) is rather helpful - much more helpful than Wikipedia, in this case.


Copper brightness change In an electrolysis situation I have identified a slight brightness change on a copper electrode.
What might have happened?
I'm dealing with different experiments and this reaction occurs in some of them.
In this experiment I am doing an electrolysis using a water solution with $\ce{NaCl}$, a Lechanché and two copper bars. The brightness change happens on the anode.
I also get bubbles and a color change in the solution of the cathode, I'm guessing it is due to the reduction of water.

My take on the possible chemistry starting with the electrolysis of salt water, which can produce some chlorine. The latter can further react with water as follows:
$\ce{Cl2 + H2O <=> H+ + Cl- + HOCl}$
The created Hypochlorous acid may react with cuprous (here present as a $\ce{Cu2O}$ on the copper metal electrode). A fenton-type reaction based on $\ce{HOCl}$ could then proceed as follows for $\mathrm{pH > 5}$:

$\ce{Cu(I) (s) + HOCl (aq) -> Cu(II) (aq) + ^.OH (aq) + Cl- (aq) }$

The dissolving of the $\ce{Cu2O}$ results in a de facto cleaning of the copper metal surface, which per a reference to quote:

Copper is a beautiful burnished-gold color when it is clean and well maintained, but like all metals, copper can become discolored when exposed to air and water.

For background on the fenton-type reaction, see "Fenton chemistry in biology and medicine" by Josef Prousek, to quote reaction (15) on page 2330, to quote:

For $\ce{Fe(II)}$ and $\ce{Cu(I)}$, this situation can be generally depicted as follows [20,39]
$\ce{ Fe(II)/Cu(I) + HOX -> Fe(III)/Cu(II) + HO^. + X- }$ (15)
where $\ce{X = Cl, ONO}$, and $\ce{SCN}$.


Physical Explanation for variation of Gibbs Free Energy with Temperature and Pressure Reading Atkins Elements of Physical Chemistry right now, and the book says:
"At constant pressure and entropy, an increase in temperature ($\Delta T\gt0$) results in a decrease in $G_\mathrm m$ (molar Gibbs energy) ($\Delta G_\mathrm m\lt0$)."
"At constant molar volume and temperature, an increase in pressure ($\Delta p >0$) results in an increase in molar Gibbs energy ($\Delta G_\mathrm m\gt0$)."

They have equations to describe this but I'm not really interested in the mathematical derivations (those are understandable) but rather in a logical, physical, and qualitative explanation of why these relations occur.

Given that $\Delta G$ indicates the maximum non-expansion work derivable from a system, what does this mean for the above relations? Does increasing the pressure of the system require work input ($\Delta G$ is positive) while increasing temperature result in the system doing work/an output of work ($\Delta G$ is negative)?

So, given a system (like a gas or something), why does $G$ change the way it does in response to $T$ and $p$ changes, and what does this physically mean for the system?

I'm a high school student (went through AP Chem) who's just trying to delve deeper into chemistry, so the simplest and most "layman's terms" explanation would probably help.

Probably the easiest way to picture it is that, since the Gibbs function is indeed a measure of the potential of the system to transfer energy in certain ways, these transformations affect the energy that can be transferred.

For a certain system at constant pressure and entropy (which therefore cannot absorb thermal energy from its surroundings), an increase in temperature is necessarily linked to an internal transformation - an internal work - that reduces the potential energy available to exert work on its surroundings; that increase in temperature is "using up" part of the potential reversible work as measured by the Gibbs function.

On the other hand, a system at constant volume and temperature which increases its pressure (and which, given these constrains, is therefore receiving some form of energy input from the outside; note that there isn't a pressure-volume tradeoff) will increase the potential reversible work it can exert on its surroundings, as measured by the Gibbs function.

It is important to note the implications of the constant conditions that each of those two cases set, because these constant conditions are telling you which forms of energy transfer are permitted or prohibited in each case.


Osmolarity and osmotic pressure You have discovered a new strain of bacteria called Tankeromyces subtilis that has a special outer membrane that is capable of withstanding high levels of osmotic pressure. These bacteria can survive osmotic pressure as high $\pu{1.3 kJ L-1}$, but not higher. The $(iC)_\mathrm{total}$ value for the inside of these bacteria is $\pu{630 mM}$.
What is the lowest concentration of $\ce{NaCl}$ (in otherwise pure water) in which these bacteria can survive without bursting due to osmotic pressure?
$R = \pu{8.3153 J mol-1 K-1}$ and $T = \pu{296 K}$.

The answer is $\pu{0.10 M}$, but I have no idea how to drive at that answer. However, I do know that it involves this equation:
$$p = iCRT$$

The highest pressure they can handle is $\pu{1.3 kJ L-1}$, or $\pu{13000 J L-1}$. In the formula you use the ideal gas constant $0.08206$ and the temperature given to find the concentration the maximum pressure it can handle.

This comes out at $\pu{0.53 M}$ or $\pu{530 mM}$ and that's the maximum concentration it can handle. To find the minimum, you subtract $530$ from $630$ giving you $\pu{100 mM}$ or $\pu{0.1 M}$, which is the lowest concentration it can handle.


D-Threose: reduction and optical activity I've bumped into this exercise:

  Judging by the mechanism of reaction between D-threose and $\ce{NaBH4}$, do you think the final product will be optically active?

To me looks like a reduction, since $\ce{NaBH4}$ is a lightly reductive agent, but then why the answer is "No, the product is not optically active"?

Is it because of the presence of anomers or there are other reasons?

When you reduce the aldehyde (from the open chain threose, since the cycle cannot be reduced as it has only acetal-like functionality) you introduce a symmetry element (both ends of the molecule become $\ce{CH2OH}$).

The presence of a symmetry element means the reduced molecule is not chiral, but meso.


Ways to change the spontaneity of a reaction I can think of any other way to influence the value of $\Delta G$ than (1) change the temperature (2) providing an external source to overcome the negative cell potential

The relationship between $\Delta G$ and $E_{\mathrm{cell}}$ is $\Delta G =-nFE_{\mathrm{cell}}$

Between $\Delta H$, $\Delta S$ and $\Delta G$ is $\Delta G=\Delta H - T\Delta S$

Sometimes, temperature won't do anything to change the spontaneity. But if we change $\Delta G$ through $\Delta E_{\mathrm{cell}}$ by providing external voltage in excess of the negative voltage, and though $\Delta G$ we are also able to influence enthalpy and entropy(???).

This is the part I am not sure about. Besides the title question, a sub-question is: how in this way can we change the enthalpy and entropy?

Natural variables of $\Delta G$ are $p$, $T$, and $N_i$.

$p$ = pressure

$T$ = Temperature

$N_i$ = number of particles  (or number of moles)

The derivation showing this is listed on the Wikipedia.


Mark all stereocenters in ampicillin Shouldn't the N pointed to by the red arrow also be a stereocenter? I am asking this because I think it's lone pair cannot take part in nitrogen inversion.

After reading the inputs from @Curt F., @PH13, @Martin and @Mithoron, I infer that N won't be involved in nitrogen inversion due to hindrance caused by connected rings and attached sterically demanding groups. Hence, 'N' pointed to by the red arrow will be considered as a stereocenter in ampicillin.


What happens in the reaction of ammonium dichromate and mercury(II) thiocyanate? I found this reaction of $\ce{NH4Cr2O7 + Hg(SCN)2 ->[\Delta]\ ?}$ on youtube, and I'm curious as to how it works. I tried searching online, but didn't find much on it.

Thanks.

This is a combination of two reactions.  The first is the ammonium dichromate vocano and the second is called Pharaoh's snakes.

The volcano

The reaction occurring here is: $$\ce{(NH4)2Cr2O7(s) -> Cr2O3(s) + N2(g) + 4 H2O(g)}$$  The reference above provides several pictures and a description of the chemistry.

Pharaoh's snakes

Again, the reference above describes what is happening, so only a summary of the chemical reactions is given here:
$$\ce{2 Hg(SCN)2->2 HgS + CS2 + C3N4}$$ $$\ce{CS2 + 3O2 -> CO2 + 2SO2}$$ $$\ce{2C3N4->3(CN)2 + N2}$$ $$\ce{HgS + O2->Hg + SO2}$$

Pharaoh's snakes is a rather hazardous demonstration and should not be performed without the proper ventilation, protective gear and waste disposal equipment.


Crucible heating apparatus I have seen some heating apparatus for a crucible use a retort stand with a large metal ring attached to the boss head to support the clay triangle as shown below.

Instinctively, I would have placed the clay triangle on a tripod, however, I see that this has certain advantages, such as being able to adjust the height of the crucible so that it is in the hottest part of the flame.

My question is: What is the name of this piece of laboratory equipment, I have seen it sometimes referred to as an iron ring, but is there a more formal name?

Iron ring is a correct term often used for this piece of equipment, according to the Wikipedia page, also this lab equipment tutorial page and in a very comprehensive pictorial lab equipment guide from Iowa State University.

Additionally, it is also called an iron support ring and a retort ring.


The coldest (the most endothermic) reaction What chemicals can you mix that causes the coldest (the most endothermic) reaction on a $\pu{kJ /kg}$ basis?

There is an endothermic nuclear fusion for the elements heavier than iron. On Earth it's most likely a nuclear fission of light elements, like the one occurring in early "dry" thermonuclear bombs, or a fission processes preceding D-T fusion reaction:

$$\ce{^7_3Li + ^1_0n -> ^3_1H + ^4_2He + ^1_0n - \pu{2.467 MeV}}$$

It's probably not the answer you'd expect, but the absolute values of thermal effects of any chemical reaction would be much lower in comparison.

Special thanks to David Richerby for correcting me!


Transparency and partially deuterated polymers I've been skimming over the literature and have come into some conflicting information about the optical properties of partially deuterated polymers.  From what I've seen, in general deuterated polymers have less attenuation of visible and IR than their non-deuterated equivalents.  But I've seen some papers remarking that partially deuterated polymers also have improved optical properties while others reporting that partial deuteration increases opacity versus non-deuterated equivalents.

Is there some sort of general rule, or does it depend on the polymer in question?  The underlying mechanisms governing transparency in polymers, beyond the basics such as crystallinity, are something I haven't studied in depth.

Deuteration lowers the vibrational frequencies C-H to C-D, but does not affect C-C etc vibrations.  Thus some lowering of vibrational frequencies will make the ir region slightly more transparent by bunching more frequencies together. In the UV deuteration hardy affects the spectrum and so here the polymer should be pretty much the same. Unless there is something in the processing of polymers that is improved by deuteration I can't think of anyhthing else.

EDIT

Having read your comments & looked at the data and thought more about it I now think that its the Franck-Condon factors that are responsible. These measure the strength of a transition and for electronic transitions in visible and uv are all proportional to $exp(-\alpha (\Delta R^2/2))$ where $\alpha = \sqrt(\mu k)/\hbar$ where $k$ is the force constant, (same for CH and CD bonds) and $\mu$  is reduced mass 1.7u for CD and 0.9u for CH so the deuterated bonds will be less intense as the exponential will be smaller for CD than CH. ($\Delta R$ is the displacement between levels in ground state and excited state). In the ir the intensity is proportional to the transition frequency multiplied by the square of $ \int \psi_i x \psi _{i+1} dx$ where x is the displacement from equilibrium bond length and $\psi$ a vibrational wavefunction with i quanta. Evaluating the integral shows that a transition is proportional to $\nu ^3$ where $\nu $ is the transition frequency between levels with i to i+1 quanta. The intensity is thus reduced by substituting D for H $via$ $\nu = \frac {1}{2\pi}\sqrt{\frac{k}{\mu}}$.

As CH/CD vibrations are of higher frequency than CC and the effect should be most pronounced at higher ir frequencies and proportional to the amount of deuterated species. Typically deuterated transition frequencies are 0.715 of the protonated values compared to the expected $1/\sqrt(2)$ =0.707, considered to be due to effects off cubic and quartic terms in the potential. This means that the protonated transitions are 2.7 times more intense than deuterated at approx 2900 cm$^{-1}$, the CH frequency in methane.


Which step below is rate determining What is the rate determining step in the following energy profile? To clarify, the reaction is:

A -> B -> C

The energy of A is greater than B which in turn is also greater than C. The intermediate A-B is higher energy that the B-C intermediate but the energy gap between B and the intermediate B-C is greater than from A to the A-B intermediate. I hope this is clear.

The second step is rate-determining. According to Wikipedia:

  Given a reaction coordinate (energy diagram), the rate determining step can be determined by taking the largest energy difference between any starting material or intermediate on the diagram and any transition state that comes after it. That transition state will then be the rate-determining step of a given reaction. The transition state with highest absolute energy may not necessarily correspond to the rate determining step.


What is an ambident substrate? I had to identify whether a few compounds are ambident substrates or not.
However, as I did not know the term, I took to googling it, but I could not find any explanation for this group anywhere.
I know about ambident nucleophiles and electrophiles, yet I had never heard of ambident substrates.
Please explain what they are.

There is a subsection in March’s Advanced Organic Chemistry [1, pp. 450–451] that provides a straightforward definition and a couple of examples:

10.G.viii. Ambident Substrates
Some substrates (e.g., 1,3-dichlorobutane) can be attacked at two or more positions, and these may be called ambident substrates.
In the example given, there happen to be two leaving groups in the molecule.
Apart from dichlorobutane, and in general, there are two kinds of substrates that are inherently ambident (unless symmetrical).
One of these, the allylic type, has already been discussed (Sec. 10.E).
The other is the epoxy (or the similar aziridine524 or episulfide) substrate.525
Selectivity for one or the other position is usually called regioselectivity.

Substitution of the free epoxide, which generally occurs under basic or neutral
conditions, usually involves an SN2 mechanism.
Since primary substrates undergo SN2 attack more readily than secondary, unsymmetrical epoxides are attacked in neutral or basic solution at the less highly substituted carbon, and stereospeciﬁcally, with inversion at that carbon.
Under acidic conditions, it is the protonated epoxide that undergoes the
reaction. Under these conditions the mechanism can be either SN1 or SN2.
In SN1 mechanisms, which favor tertiary carbons, attack may be expected be at the more highly substituted carbon, and this is indeed the case.
However, even when protonated epoxides react by what is expected to be an SN2 mechanism, attack is usually at the more highly substituted position.526
This result probably indicates signiﬁcant carbocation character at the carbon (ion pairing, for example).
Thus, it is often possible to change the direction of ring opening by changing the conditions from basic to acidic or vice versa.
In the ring opening of 2,3-epoxy alcohols, the presence of $\ce{Ti(O\textit iPr)4}$ increases both the rate and the regioselectivity, favoring attack at C-3 rather than C-2.527
When an epoxide ring is fused to a cyclohexane ring, SN2 ring opening invariably gives diaxial rather than diequatorial ring opening.528
Cyclic sulfates (108), prepared from 1,2-diols, react in the same manner as epoxides, but usually more rapidly:529

524 Chechik, V.O.; Bobylev, V.A. Acta Chem. Scand. B, 1994, 48, 837.
525 Rao, A.S.; Paknikar, S.K.; Kirtane, J.G. Tetrahedron 1983, 39, 2323; Behrens, C.H.; Sharpless, K.B. Aldrichimica Acta 1983, 16, 67; Enikolopiyan, N.S. Pure Appl. Chem. 1976, 48, 317; Dermer, O.C.; Ham, G.E. Ethylenimine and Other Aziridines, Academic Press, NY, 1969, pp. 206–273.
526 Biggs, J.; Chapman, N.B.; Finch, A.F.; Wray, V. J. Chem. Soc. B 1971, 55.
527 Caron M.; Sharpless, K.B. J. Org. Chem. 1985, 50, 1557. See also, Chong, J.M.; Sharpless, K.B. J. Org. Chem. 1985, 50, 1560; Behrens, C.H.; Sharpless, K.B. J. Org. Chem. 1985, 50, 5696.
528 Murphy, D.K.; Alumbaugh, R.L.; Rickborn, B. J. Am. Chem. Soc. 1969, 91, 2649. For a method of overriding this preference, see McKittrick, B.A.; Ganem, B. J. Org. Chem. 1985, 50, 5897.
529 Gao, Y.; Sharpless, K.B. J. Am. Chem. Soc. 1988, 110, 7538; Kim, B.M.; Sharpless, K.B. Tetrahedron Lett. 1989, 30, 655.

Reference

Smith, M. March’s Advanced Organic Chemistry: Reactions, Mechanisms, and Structure, 7th ed.; Wiley: Hoboken, New Jersey, 2013. ISBN 978-0-470-46259-1.


How can I make an alloy wetter? I'm trying to make a Bismuth-Indium based low-melt alloy wetter (specifically, better at wetting to solid metals and glass). I've tried adding gallium, which obviously works, but due to its metal-attacking nature it can't be used here.

I know that antimony and tellurium are often used as surfactants, but I've only read of them acting as surfactants in organic solvents, and can't find anything on whether alloying them with a liquid metal would improve its wetting ability. As I don't have any Sb/Te at present, I'd rather ask if this has any chance of helping before trying them.

If not, is there anything else I could incorporate into the alloy - possibly a non-metal/metalloid - which would improve wetting?

The surface of glass has many Si-O-H and Si-O-Si bonds; deeper inside would be Na and Ca and Al. An active metal might engage those surface atoms. Na might be too reactive (forming a dross of NaOH on the surface of your alloy, or reacting with water at the glass surface). Aluminum or zinc or perhaps magnesium might be active enough to contribute some sort of adhesion to glass.

Tin oxide is bound to glass surfaces for electrical conductivity. Perhaps it could be an intermediate layer, bonding glass and your alloy. And silver is deposited on glass to make mirrors. Your alloy might well stick to a silver deposit. So, while this doesn't actually make your alloy wetter, it could solve your problem of adhering to glass.


Which d-block cation has the maximum hydration enthalpy? Among $\ce{Mn^2+}$, $\ce{V^2+}$, $\ce{Ni^2+}$ and $\ce{Ti^2+}$, which cation is having the highest hydration energy to form the aqua complex $\ce{[M(H2O)6]^2+}$?

I guess $\ce{Mn^2+}$ should be the answer as it has the most stable electronic configuration.

Note that hydration energy is the energy released when water molecules are adsorbed on the surface of the cation. Hence the cation should have a high charge density, which means it should have a small size. So, the smaller, the better.

Here, $\ce{Ni}$ has the smallest atomic radii compared to all other options due to greater effective nuclear charge, so it's ionic radii would also follow the same trend. Hence, $\ce{Ni^2+}$ would have greater hydration enthalpy.


Maximum pressure that can be produced by a combustion reaction Multiple-chamber artillery can launch a projectile with energies much higher than single-chamber guns.  I'm interested in setting a fundamental upper bound on the energy that a chamber (of a known volume) can impart to a projectile.  This would, for example, set a limit on the needed volume for a sub-orbital gun (given a projectile mass).  Since the pressure drop-off (like with single-chamber) isn't a constraint, the kinetic energy would be theoretically limited by pressure times volume.  But you can't possibly achieve infinite pressure with any chemical reaction.

For that limit, some sources claim that pressures of 60,000 psi can be achieved.  Coming from a physics perspective, I understand that a reaction should have a defined energy associated with an individual molecular transition.  We've had related questions on physics stack exchange, but my thinking is that chemical equilibrium may be be a necessary tool.

If we ignore reaction kinematics, I believe that the maximum pressure that a reaction can achieve is limited by two things:

the density of the reactants before reaction and
the energy of the reaction

Given this, we could simply model the reaction with ideal gas assumptions, resulting in simple algebra.  But I think this is wrong.  Higher pressure would push the reaction to incomplete combustion.  From a chemistry perspective, is there any theoretical maximum pressure that we could restrict all combustion reactions to?

According to New Energetic Materials, table 1.2, a list of explosives ranges from 19-48 GPa in pressure achieved.  This is two orders of magnitude greater than the 60,000 psi figure.

As far as maximum theoretical pressure, the question isn't really clear whether the maximum absolute pressure at which combustion can occur or the maximum change in pressure due to combustion is the value of interest.  The answer could be related to the maximum pressure at which chemical bonds exist.


Does precipitated copper carbonate contain impurities I recently performed a double replacement reaction between copper sulfate and sodium carbonate to yield copper carbonate as a precipitate. I then filtered the precipitate, but I noticed that the precipitate still contained a large amount of water.

Would there be a significant amount of sodium sulfate dissolved in this water, making the copper carbonate impure? If so, how would I separate the remaining sodium sulfate from the copper carbonate?

Yes, the water contained in the freshly filtered precipitate will contain dissolved $\ce{Na2SO4}$, as the latter is the soluble by-product of your double displacement reaction. To get rid of it, you need to wash your precipitate thoroughly with distilled water (pH 7). If you really want to make sure that all $\ce{Na2SO4}$ has been removed, you can test samples of the wash water filtrate for $\ce{SO4^2-}$ by adding diluted hydrochloric acid and aqueous $\ce{BaCl2}$. When a white precipitate of $\ce{BaSO4}$ no longer forms, the aqueous sodium sulfate can be considered separated from the $\ce{CuCO3}$ precipitate.


Do we consider the dissociation of water when dealing with ion concentrations of a solution? One mole of hydrated copper(II) sulfate, $\ce{CuSO4 . 5 H2O},$ is dissolved in water. How many moles of ions does the solution contain?
1) 1
2) 2
3) 6
4) 7

The correct answer is 2. Specifically, I'm confused about why the $\ce{OH-}$ and $\ce{H+}$ ions aren't considered.

It dissociates into 7 parts; copper and sulfate ions and five moles of water.

Water is also the solvent, so these don't count.

However the answer "two" is indeed not perfectly right, because the sulfate ion would (partially) react with water, to form

$$\ce{SO4^2- + H2O <=> HSO4- + OH-}$$

A sulfate ion is so to say the opposite of sulfuric acid, which reacts

$$\ce{H2SO4 + 2 H2O -> HSO4- + H3O+ + H2O <=> SO4^2- + 2 H3O+}$$

$\ce{HSO4-}$ is not strongly acidic anymore $(\mathrm{p}K_\mathrm{a} = 2),$ so the sulfate ion logically must react somewhat basic. (OK, $\mathrm{p}K_\mathrm{a} = 2$ is quite acidic, acetic acid has $\mathrm{p}K_\mathrm{a} = 4.8.)$

So the correct answer is slightly larger than two. But from the given possibilities in this multiple-choice test, two is by far the most reasonable.

(Update: The hydrated $\ce{Cu^2+}$ ion might well be acidic enough to overcompensate the basicity of the sulfate. In which case you'd still have more than two equivalents of ions, just it'd be a small amount of $\ce{H3O^+},$ not $\ce{OH^-}.$ And of course there is always the autoprotolysis of water.)


How to derive the Gibbs-Helmholtz equation? Can someone please explain to me how to derive the Gibbs-Helmholtz relationship from $G = H - TS$?

You need to know that

$$\require{begingroup} \begingroup \newcommand{\d}[0]{\mathrm{d}} \d G = V\,\d p - S\,\d T$$

from which you can determine that

$$\newcommand{\pdiff}[3]{\left(\frac{\partial #1}{\partial #2}\right)_{\!#3}} \pdiff{G}{p}{T} = V \qquad \pdiff{G}{T}{p} = -S$$

Therefore (by the quotient rule)

$$\begin{align}\pdiff{}{T}{p}\left(\frac{G}{T}\right) &= \frac{T(\partial G/\partial T)_p - G(\partial T/\partial T)_p}
{T^2} \\[8pt]
&= \frac{T(-S) - G(1)}{T^2} \\[8pt]
&= \frac{-TS-G}{T^2} \\[8pt]
&= -\frac{H}{T^2}
\end{align}$$

as desired (since $G = H - TS$).


E1cb reaction conditions I'm wondering why an E1cb reaction needs a poor leaving group like -OH attached to one of the carbons.

I thought the conjugate base of the leaving group must be stable enough not to de-protonate one of the newly formed alkene's H's. But if this is the case, why can't a good leaving group like Br be used in E1cb which would neither de-protonate the alkene's H's.

I'm wondering why an E1cb reaction needs a poor leaving group like -OH attached to one of the carbons.

In an E1cb reaction, the rate-determining step (RDS) is the $\ce {C-H}$ bond cleavage. According to Carey & Sundberg (2007), p. 552, for the irreversible E1cb reaction (when the hydrogen abstracted is very acidic), the rate of the reaction may increase in the following order of leaving group: $\ce {I^- < Br^- < Cl^- < F^-}$. Why does the reaction rate increase as the leaving group becomes "lousier"?

From kinetics, we known that the rate of a reaction is determined by the slowest step, the RDS. In the E1cb, the RDS involves removal of a proton to form a carbanion. This step would be accelerated if the $\ce {C-H}$ bond is weakened and when the carbanion experiences greater stabilisation of the negative charge. Poor leaving groups, such as $\ce {OH^-}$ and $\ce {F^-}$, are often able to provide inductive stabilisation of the negative charge, e.g. $\ce {-OH}$ and $\ce {-F}$ substituents are electron-withdrawing in nature. This would thus help to speed up the RDS and thus, the overall reaction rate would be faster with these poor leaving groups.

Note: The explanation for this is similar to that for this question: Why are fluorides more reactive in nucleophilic aromatic substitutions than bromides?

  Paraphrased: If this is the case, can E1cb reactions still involve good leaving groups, such as $\ce {Br^-}$?

On the same page, the authors also conclude that:

  E1cb reactions generally require both carbanion stabilization and a relatively poor leaving group

If the leaving group becomes a better one, e.g. if the $\ce {F^-}$ becomes $\ce {I^-}$, then the $\ce {C-X}$ cleavage may be more advanced than the $\ce {C-H}$ bond cleavage, resulting in the reaction becoming E2, E1-like or even E1. Thus, if good leaving groups are present, likely the elimination will not proceed by E1cb but by the other pathways.

Update

I just realised someone had a similar question answered already: Why must the leaving group in E1cb be poor? You can refer to that answer for a different perspective. But ultimately, the essence of the correct explanation is well-contained in my answer above.

Reference

Carey, F. A., & Sundberg, R. J. (2007). Advanced Organic Chemistry Part A. Structure and Mechanisms (5th ed.). Springer.


Silver and milk (or milk products) In the title I am refering to silver spoons, which, when used to consume milk products (like yogurt), get a weird taste. It could be described as umami/meaty, a little bit like onions. I came to the conclusion that it must be the silver, because this taste is never observed when using steel spoons.

My guess was: the silver (or maybe its oxide?) reacts with an ingredient of the milk product.

My question is: Is this documented/known process/reaction or is this some kind of misperception on my part? If the former is true, what exactly is happening? (By)-products?

You're right: the silver is reacting with sulfur compounds in the food to form a tarnish of silver sulfide. This is most commonly observed, in my experience, using silver teaspoons with boiled eggs, which are pretty rich in sulfur.
There are a number of reactions that can take place depending on the sulfur-containing species - the abstract from this paper from the Journal of Chemical Education (unfortunately pay-walled) has a nice example:

Silver and silver-plated objects react with sulfur and sulfur compounds to produce silver sulfide $\ce{Ag2S}$ or tarnish. Contact with materials that contain sulfur compounds, such as hard boiled eggs, mayonnaise, mustard, and rubber bands can cause tarnish. In the air, a silver object can tarnish owing to the reaction of silver with hydrogen sulfide ($\ce{H2S}$). This is a gas found in the air as a result of some industrial processes and the decomposition of dead plants and animals. The reaction of silver with hydrogen sulfide to form tarnish is as follows:
$\ce{2 Ag(s) + H2S(g) -> Ag2S(s) + H2(g)}$

In Sterling silver specifically, you'll often also get copper sulfide formed.
The paper also mentions that you can remove the tarnish abrasively, or through reacting the tarnish with aluminium (via solution, electrochemically):

$\ce{3 Ag2S(s) + 2 Al(s) -> 6 Ag(s) + Al2S3(s)}$

But a more common approach used by many commercial products is to dissolve off the tarnish, usually in sodium thiosulfate or thiourea. This means you end up with slightly less silver on your item, but is otherwise pretty effective and very mild (it's commonly used for museum pieces).


Spontaneity of reversible reactions If a reaction has positive $\Delta G$ then reverse reaction has negative $\Delta G$. How can both occur simultaneously, e.g., in a reversible reaction such as dimerisation of $\ce{NO2}$ to form $\ce{N2O4}$? Do forward and backward reactions occur spontaneously? How can both directions be spontaneous at the same time?
I read in some book that  Haber's process to form ammonia is spontaneous at a temperature below $170\ \mathrm{^\circ C}$ (the author proved this using $\Delta G=$ negative for spontaneous process). This means that above this temperature $\ce{N2}$ and $\ce{H2}$ shouldn't combine to form $\ce{NH3}$. But industrially Haber's process is performed at temperature around $400\ \mathrm{^\circ C}$. Isn't this a contradiction?

You seem to be mixing up thermodynamic and kinetic considerations. All reactions occur spontaneously all the time, and their backward reactions too. If left alone for long enough, a system will eventually reach the point where the forward and backward reactions go at the same rate (and it's not like they both stop altogether!) - that's an equilibrium. Where that equilibrium will be is judged by $\Delta G$; if it is negative, then the equilibrium will be closer to the products.
As for ammonia, that's where the kinetics kicks in. At room temperature $\Delta G$ is in our favor, but the equilibrium would take forever to reach. We raise the temperature so as to make the reaction faster, but that also shifts the equilibrium in the wrong direction. So we raise the pressure, too. This does the trick: now the reaction goes the way we need, and at reasonable rate.


Covalent Compounds In Written Form How do you determine which non-metal is written first in a covalent compound? This has been a confusion for me for quite some time.

Typical convention for organic molecules is to list carbon first, than hydrogen, and then other elements in alphabetical order. The Merck Index, for example, does it this way.


Percentage of trans-stilbene traces in cis-stilbene H NMR This is H NMR spectrum of cis-stilbene. There are some traces of trans stilbene. I have to determine how many trans-stilbene is present, the exact percentage and I have problem with that.

I guess that it has to do with integration of peaks but I'm still not sure is trans-cis isomer ratio 1:10 or am I wrong?

You have to focus on the $\ce{^1H}$ of the double bond.
They are singlets because both molecules are completely symmetrical.
As per the tables, the chemical shift of the cis isomer is $5.25+1.38-0.07=6.56$ ppm.
On the other hand, the chemical shift of the trans isomer is
$5.25+1.38+0.36=6.99$ ppm.

These theoretical values correspond to the peaks of 6.60 and 7.11 ppm.
Since their integrals are of 1 and 1.1571 respectively, the proportion of trans is of:

$\frac{0.1571}{0.1571+1}100=14\%$


Is is possible to do mass spectrometry without coupled liquid chromatography? It's possible to identify a solution with several compounds* without the coupled liquid chromatography (LC) before to isolate the compounds?

On the university where I study, the LC is broken, so all we can do is inject the solution directly on the mass spectrometer. I wonder if I will be able to identify the metabolites? I don't need quantitative analysis, I just want to know what's in there.

(*) To be precise: A bacterial culture, that has gone under organic solid phase extraction, freeze dried and resuspended in methanol.

The practical answer really depends on two things: do you know exactly what you're looking for and can your instrument do tandem MS?

If you know exactly which metabolites you're looking for and they're present in great enough abundance, it may be possible to resolve them from the the background, depending on the resolving power of the instrument and what else is in there (if you have an ICR instrument, it may not be a problem at all, but if it's just a single quadrupole, you may be out of luck).

If you know exactly which metabolites you're looking for (or know that they share some common structural feature that can be knocked off by CID or the like) and your instrument can do tandem MS, you're in a much better position. Tandem MS is often used to avoid needing to do a slow LC step as it's possible to identify two compounds that aren't resolved on a precursor ion scan by looking at their fragment ions: A messy primary ion scan can be dealt with if you know that your analyte has a [M-18] transition or whatever, by CID. Also, a given class of compounds often shares certain CID reactions, which can be picked up with a neutral loss scan.

If you have tandem, but aren't sure exactly what you want to find, it's a bit more work, but it just means you need to look at product ion scans of peaks that are ambiguous and try to work out what they are.


NaOH Normality Slowly Changing Our lab uses a sodium hydroxide solution of a (theoretically) known normality (0.8000 - 1.0000) to titrate product samples for quality control.  However, we suspect that the normality of the $\ce{NaOH}$ left in the burette/tubing connected to our container is changing if we let it sit (our tests come out wrong, and the answers seem to get better if we run a couple of burettes worth of $\ce{NaOH}$ out first).  I expect it's absorbing moisture from the air, but I don't know for sure.

We've tried different container types and we've put a desiccant in line with the tubing to absorb stray moisture (which it seems to be doing, there's been an observable color change), but the situation remains that we need to just throw out up to 200mL of $\ce{NaOH}$ any time we need to run a test.  Does anyone on here have any recommendations for countering this normality drift?

I would be more worried about carbon dioxide (rather than water) being absorbed into the NaOH solution. A 1 molar NaOH solution is not going to be appreciably hygroscopic. But it will absorb CO2: two moles of hydroxide yield one mole of carbonate ion, which, depending on what you are titrating, may act as only deprotonate one mole of acid per mole base - reducing the effective normality of the NaOH solution. You could try protecting any air vents with a drying tube filled with sodium or calcium hydroxide, or use a regulated supply of chemically pure nitrogen as a blanket.


Finding a substance which has a continuous gradient in refractive index I need to find a substance whose refractive index is constantly changing and directly proportional to the square root of the depth. I know I'm looking for a lot, but I'd really appreciate any ideas or advice from anyone. Thanks a lot

It's not entirely true that no such substance exists, though you won't find this property in a homogeneous substance.

It's well known that changing the concentration of solute in a solution can change the index of refraction; a concentration gradient would then provide a changing refractive index, though having this mimic a square root function would be difficult. Typically, the refractive index changes linearly with the concentration of solute, so a nonlinear function may be difficult to achieve.
A similar approach could be applied to solids, forming a solid solution, though again there would be some difficulty in finding two compounds with similar lattice parameters that would allow to a continuous gradient between compositions to product the refractive index gradient.

Your best bet may be with polymers, where changing proportions of the monomers used could change the refractive index; organic dopants can also be used to effect* this gradient in polymers (see, e.g., Polym. Adv. Technol. 2014, 25, 1099). Again, the difficult part would be matching a square-root function, but you could at least have a varying gradient.

*https://xkcd.com/326/


Why do substances ionise in water to turn into charged species? Wouldn't they prefer to stay as uncharged molecules? Wouldn't this be more stable if they stayed as neutral molecules?

I've learnt that they will seek to reach the ratio of $[A]^a[B]^b/[C]^c$ that matches the $\ce{Ka}$ or $\ce{Kb}$ value.

But I do not understand why they wouldn't prefer to stay as uncharged molecules as I thought ions are unstable compared to neutral molecules?

While a picture as in Bob’s answer is sometimes better than a thousand words, it may also help to add the thousand words[1] to explain what the picture is supposed to be showing or why it is relevant.

Water is a very polar solvent. The electronegativity difference between oxygen and hydrogen is rather high and the bent shape of the molecule additionally allows a higher dipole moment. Therefore, you can think of water molecules having a decidedly negative side — the oxygen atom — and a decidedly positive side — the space between the hydrogen atoms. In pure water, the molecules will arrange (at first approximation) in such a way that the positive sites are close to the negative ones and vice-versa.

At second approximation we need to introduce hydrogen bonds, which go one step further. Remember how each single hydrogen atom is strongly positively charged due to the electronegativity difference/the polar bond? Well, because hydrogen does not have any core electrons, this means it can act like a ‘slightly weakened proton’ and interact well with lone pairs on electronegative atoms such as oxygen. Thus, a better description of the intermolecular interactions in water molecules is that hydrogen bonds are formed from each hydrogen to one of another molecule’s oxygen lone pairs. These interactions are pretty strong for intermolecular interactions since they are caused by rather large charge differences (and need to stabilise these).

After having considered the water molecules by themselves, what happens if we submerge ions into them? In solid state, ionic compounds form extensive networks of cations/anions approximately alternating. Each cation will typically only have anions as closest neighbours and vice-versa; therefore each ion’s charge is stabilised by the charges of the opposite ions close by. If you wanted to dissolve this ionic compound into something like hexane, it would not work well. It is just much better for the ions to stick together in their solid-state form and be submerged in the hexane liquid phase — as your question implies.

In water, however, we have the positively charged and negatively charged areas not unlike those of the counterions. Taking a single ion out of the ionic crystal and surrounding it with the appropriate number and orientation of water molecules as shown in Bob’s answer, means that electrostatically almost nothing is gained or lost. True, the water molecules, not being charged particles, can stabilise a little bit less at first approximation but it is still comparable.

At second approximation, we can on the one hand reintroduce hydrogen bonds gonig from water molecules to anions such as chlorine. These may stabile the anions in solution, but note that having a hydrogen bond directed towards oxygen is typically better than one to a ‘typical halide’. The cation side is much more interesting, though: oxygen’s lone pairs can interact with the cationic centre to form a coordination complex. This lowers the energy of oxygen’s electrons at the expense of unoccupied cationic orbitals — a win-win situation for both partners. Thus, we can expect additional stabilisation due to this effect. (To be fully honest, one can and should also consider most ionic crystals as chained local coordination complexes. However, these are typically better stabilised if only one ligand sphere is involved, thereby preferring the solution.)

Thus we have:

A net average enthalpic no-effect concerning charge stabilisation (ionic crystal versus Bob’s picture
A slight net enthalpic gain due to the formation of better coordination complexes
A strong net entropic gain from the fact that a single, highly ordered crystal is broken up into gazillions of minute entities that can move freely in solution (I did not touch this point in this answer, but it is true for all solutions).

Considering all these, the conditions for ionic compounds to dissolve in aquaeous solutions are very favourable. And Bob’s your uncle.

[1]: Well all right. 645 words then. Not quite 1000 but close enough.


How did early chemists measure mass of atoms? Here a book says that Berzelius measured atomic masses using a simple lab and his measurements was so close to modern results. Can someone explain how a person living in 19th century was able to measure atomic masses? Like he took this and those, observed this and that and did something else. Can I repeat his experiments at home too and get his results?

Do I actually understand atomic mass right? For me it is a mass in gramms of one atom of an element.

Also, I can't understand how Gay-Lussac measured the volumes of gases reacted with each other. What does the book mean by saying 2 volumes of Hydrogen? How much are 2 volumes? Are they measured in m^3 or what?

Please try to avoid the use of moles in answers as I don't understand why we need moles actually. So, I think I need to understand questions above first before moving to moles.

Careful measurements were the foundation of chemical theory

Let's take the questions out of order to simplify the answers.

It doesn't matter what yours you use to measure volume as long as you stick to the same units in all your calculations. Cubic widgets is fine as long as you do all your calculations using them.

The key to understanding how chemists came to know about atomic mass is careful measurements of the widths and volumes involved in reactions. Chemists observed that certain reactions always involved the same mass ratios. More interestingly, when gases are reacted they also involve the same volume ratios. When hydrogen and oxygen react to for  water you start with two volumes (2 cubic widgets if you want) of hydrogen and one of oxygen react together to give water. How do we know it is just that ratio? Do the experiment with 3 volumes of hydrogen and one of oxygen and observe that there is still one volume of hydrogen left after the reaction is done.

Those experiments showed that the mass ratios were constant but different to the volume ratios if the reactants were gases. Oh, and we know the weight of the gases because we can measure the weight of a given volume of gas using a vessel of known volume containing a gas at a specific pressure using a carefully constructed vessel. A litre of oxygen weighs about 1.5g which is easy to measure using a good balance and good experimental conditions even in the 19th century.

Chemists hypothesised (and later proved) that the volume of gases were proportional to the number of molecules they contained so the ratio of gas volumes told them something about the number of units of substance (or molecules) involved in reactions. Comparing many reactions led to an understanding of more complicated reactions than hydrogen burning in oxygen. Comparing the known results from gas volume ratios to the known weights of substances involved in liquid and solid reactions extended this idea (with an early glitch caused because chemists didn't realise that many gases consist of diatomic molecules).

The actual number of molecules or atoms in a give volume took a lot longer to determine than the basic principles involved in chemistry (that reactions always involved specific ratios of inputs and outputs). Initial methods involved atomic theory applied to gasses or measuring the charge on a single electron. Later methods involved measuring the mass of a very pure crystal and comparing it with density derived from x-ray measurements to give more accurate values. (see this).

The basic experiments about atomic ratios can be repeated with good experimental technique and apparatus. But you need a very good way to measure weights of gases and volume. Repeatable, but not easy. You won't be easily able to measure the number of atoms in anything with household apparatus, though. That requires fairly sophisticated equipment.


Reagents used to prepare organic ICP standards, from scratch I am trying to figure our how I can make my own calibration standards for organic ICP analysis. I assume that I need organometallic compounds to mix with base-oil/mineral-oil to prep the standards, but I can not figure out what specific compounds are commonly used for the different metals (example: For Na, maybe Sodium isopropylcyclopentadienide?). I have looked online and through some analytical chem texts but no luck. Any source for this information would be greatly appreciated.

Running an ICPOES or ICPMS on organic solvents is not easy, you may have problems such as coke / carbon deposits in the torch.

Also I would like to point out that sodium dicyclopentadienide is very sensitive to oxygen, it decomposes with great ease on contact with air. Ask almost anyone who has made ferrocene in the lab about this question.

I would like to suggest that you consider making an emulsion in water of the organic solution using a detergent such as triton-x100. I have heard of this method for running organic samples in a normal ICPMS but I have never tried using it.

In the ideal world what I would do would be to use a wet combustion (maybe use a acid digestion bomb such as a Parr 45 ml bomb) to convert the organic sample into a aqueous sample.

Another option would be to use neutron activation anaylisis, but these days it is very hard to find a strong neutron source suitable for this type of work. You would have to seal up samples in quartz tubes. I will warn you that even with a nuclear reactor and a good gamma spectrometer this type of work is far from trival.


Carbohydrate configurations I want to study several carbohydrates that includes but not limited to the following molecules:

Glucose
Galactose
Mannose
Fructose
Tagatose
Sorbose

I know that for any carbohydrate, possible list of carbohydrates isomers are

single chain (L and D isomer)
$\alpha$ and $\beta$ pyranose or six member ring structures
$\alpha$ and $\beta$ furanose or five member ring structures.

Now, for ring structures there are several rotamers, mainly gg, gt and tg (g=gauche, t=trans) and also there are several anomers but I am not sure about them. Is there anything I'm missing?

For accurate thorough computational study, I need to identify all the isomers. Is there any single book or document that can guide me to screen the full potential energy surface?

Thanks for your kind help.

There are 32 isomers of C6H12O6, of which you have named 6. The most common one is D-glucose (dextrose). I don't think all of the 16 L-saccharides are found in nature.

α and β refer to the connection between rings, whether the ketal (or aldehyde) oxygen is axial or equatorial.

  Structures made with ChemDraw.


Rank the following radicals in order of decreasing stability Question

  Rank the following radicals in order of decreasing stability

Aromaticity makes a cyclic compund more stable. Here, 1 and 3 are aromatic since they follow Huckels rule(Or am I wrong here?). So, I thought maybe 3 is most stable followed by 1, then followed by 2 and 4(I dont know which is more stable out of 2 and 4). But, that's not the case. 3 is most stable, but the next stable one is 2 and not 1. I don't get how.

Here, 1 and 3 are aromatic since they follow Huckels rule(Or am I wrong here?

Yes you're wrong over here, for applying the Huckel's rule the compound must be a closed circuit of delocalized electrons. In both the cases the circuits are not closed

If you're not aware of Conjugation then give that a read .
Now
The most stable compound is 3 since the extent of conjugation is the greatest i.e 2 pi bonds.

3 is more stable than 4. Even though 4 is aromatic it is an aryl free radical the p orbital containing the radical is out of the plane and can't be stabilized through resonance/Hyperconjugation.
In fact I would say that option 4 is the least stable compound .

The second most stable compound is is 2 since it is stabilized via resonance which is more effective than Hyperconjugation; 1 is stabilized by Hyperconjugation.

3 is more stable than 2 since the extent of conjugation is greater.

The order of stability will be
$$\mathrm{3 > 2 > 1 > 4}$$


Are 1st Electron Affinities positive or negative? According to this ChemistryLibretexts website and this StackExchange answer,  1st electron affinites are negative and 2nd electron affinities are positive.

However, according to this Study.com website, 1st electron affinity (energy released) is positive.

Both contradicting websites address the point that when electrons are added to a neutral atom, energy is released. But the only problem is that one website says that when this exothermic reaction occurs, the 1st electron affinity is negative, while the other states that the 1st electron affinity is positive. Which one is right?

This is sadly, one of those matters of "convention" that chemistry is so plagued by.

For most neutral atoms, the magnitude of $\Delta H$ for the following reaction is negative (i.e. energy is released upon the addition of an electron).

$$\ce{A (g) + e- (g) -> A- (g)} \tag{1}$$

The value of $\Delta U$ for the reaction is, in some places, called the "electron affinity". In other places, $\Delta H$ for the reaction is called the "electron-gain enthalpy".[1]

The convention that is more common (as far as I know) is to define the electron affinity as the negative of $\Delta U$ for the above reaction. This means that electron affinities, as defined in this way, are usually positive. This convention is adopted by most major introductory inorganic chemistry textbooks and is also what I learnt in undergrad. However, if you are supposed to use a different convention for your examinations etc., then do so.

The moral of the story is: in every book you read, always double check what is the sign convention. It can and will differ from book to book.

[1] To make things even more confusing, $\Delta H$ (the change in enthalpy) and $\Delta U$ (the change in internal energy) is used in different contexts. The difference between the two is given by

$$\Delta H(T) = \Delta U(T) + \frac{5}{2}RT$$

which may be understood using Kirchhoff's law since $\Delta H(0) = \Delta U$. (Alternatively there is a derivation in Housecroft & Sharpe's Inorganic Chemistry 4th ed., p 25.) Since $5RT/2$ is only roughly $6~\mathrm{kJ~mol^{-1}}$ at $298~\mathrm{K}$, this contribution is usually negligible compared to the usual magnitudes of electron affinities.


Does Genistein remain in resonance stabilised form in physiological pH or acidic pH? As a follow up, does the Carbonyl group at 4 position group get deprotonated or Hydroxyl group at 5 position get deprotonated? How does the resonance structure look? I have been looking through literature but have not been able to find any reference to this.
Does the deprotonation start with the carbonyl group or does it always have to start with hydroxyl group?
I may not sound clear as I am an amateur when it comes to organic chemistry.

The first acid dissociation constant for genistein ($\mathrm{p}K_\mathrm{a1}$) is $6.51 \pm 0.20$ (Ref.1), mainly because of hydrogen bonding between $\ce{C_{(4)}}$ carbonyl oxygen and $\ce{C_{(5)}-OH}$, similar to that in acetylacetone (see the diagram):

Therefore, in physiological $\mathrm{pH}$ ($\mathrm{pH} \approx 7.3$) it should be partially ionized. If it is in acedic solution, $\ce{C_{(4)}}$ carbonyl oxygen would be partially or fully protonated based on the acid strength of the solution. Furthermore, the energy minimized three-dimensional structure of genistein shows the hydrogen bonding as depicted in following image (See Figure A; Ref.1):

References:

Yoshiyuki Mizushina, Kazuaki Shiomi, Isoko Kuriyama, Yoshihiro Takahashi, Hiromi Yoshida, “Inhibitory effects of a major soy isoflavone, genistein, on human DNA topoisomerase II activity and cancer cell proliferation,” International Journal of Oncology 2013, 43(6), 1117-1124 (https://doi.org/10.3892/ijo.2013.2032).


Easily removable material that sticks to skin I am looking for some kind of material that can be used to easily latch something to someone's finger. This is for a first year engineering project. The idea is to use this material to latch small motors on someone's fingers in the same way that the motor in this picture is attached. The person should be able to move their fingers around without the motor falling off and the motor should be removable with a slightly strong pull. Also the material should create as small a barrier between the finger and motor as possible so that when the motor vibrates, it can easily be felt by the finger.

I'm thinking more along the lines of a sticky polymer sheet that readily sticks to the skin than some sort of adhesive or fluid glue, but I'm open to hearing all ideas.

Are there polymer materials out there (or particular functional groups) that match this description?

I have an engineering solution for you that bypasses the need to find a chemistry solution.

  I'm thinking more along the lines of a sticky polymer sheet that readily sticks to the skin

You want an adhesive bandage. Adhesive bandages are thin, stick to skin, and are easy to remove. In my experience, the less expensive generic brands are both thinner and stickier, so go with a generic brand. If you are worried about the thickness, do not use the central part with the embedded gauze pad. Only use the wings. Use a strong glue to fix your motor to the back of the adhesive wing.


Clarification in a particular application of Le Chatelier principle There is something I don't understand about a particular application of the Le Chatelier principle. The following system is in equilibrium: $$\ce{CH3COOH + H2O <=> H3O+ + CH3COO-}$$ If I add some $\ce{H3O+}$ the reaction will shift left. Now if I keep adding $\ce{H3O+}$, it will keep shifting left until the $\ce{CH3COO-}$ would deplete? And would the $\ce{H3O+}$ will remain at its initial concentration? So I can't understand how a buffer work, because if you have a reaction like above and you are adding acid, the added $\ce{H3O+}$ and the $\ce{CH3COO-}$ will bind together to form $\ce{CH3COOH}$ and water thus keeping the $\ce{H3O+}$ exactly the same. Where am I wrong?

For starters, I wouldn't classify your equilibrium as a buffer. A buffer requires excess weak acid, and it also requires either a strong base or the salt of the weak acid (which you don't have). The equilibrium you stated is the dissociation of CH3COOH into its conjugate base and a conjugate acid.

Say you had the following buffer:

$$CH_{3}COOH + NaOH ⇌ CH_{3}COONa + H_{2}O$$

The equilibrium of this would be:

$$CH_{3}COOH ⇌ CH_{3}COO^{-} + H^{+}$$

And the ionic equation would be:

$$H^{+} + OH^{-} → H_{2}O$$

As per definition, a buffer is a system that minimises pH change on small additions of acid or alkali. How? All down to equilibrium...

If you were to increase $[H^{+}]$:

In terms of the equilibrium, the added $H^{+}$ would react with the base, $CH_{3}COO^{-}$, to form
$CH_{3}COOH$. This is Le Chatelier's principle in the fact that increasing $[H^{+}]$ shifts the equilibrium to the left.

If you were to increase $[OH^{-}]$:

In terms of the ionic equation, the added $OH^{-}$ would react with the $H^{+}$ to form $H_{2}O$. This is Le Chatelier's principle in the fact that increasing $[OH^{-}]$ shifts the equilibrium to the right.

In both of these, it's clear that the addition of $H^{+}$ or $OH^{-}$ would just be an acid-base reaction. The products formed minimise the $pH$ of the system due to their relative strengths compared with the added acids/alkalis.

The $pH$ does change though, but just not as much as it would if it wasn't in the buffer system. A buffer just keeps the equilibrium central.


why does ring flip of L-fructose occur? when drawing the chair conformation for beta-L-fructose, I got the structure above (in the picture). The answer said that the conformation above had a ring flip. Why does a ring flip occur and how can we tell when it occurs?

These ring-flips are called 1C4-4C1 transitions and these happen to all of the pyranose sugars in solution when the conditions allow. However, the abundance of each forms depends on their energy. There is a so called 1,3-diaxial interaction between the axial substituents of a cyclohexane analogue. Two axial substituents on the same side are much closer than two equatorial substituens. If these substituents are relatively big, it will be energetically unfavorable to bring them close together due to steric effects. Therefore the molecule will "choose" the conformation that has less energy, so less "large" substituents on the same side. In your case the OH and Me groups are considered the "large" substituents, while the H atoms (undrawn) are the "small" ones. Therefore the conformation you drawn has an unfavorable Me-OH interaction, which will drive the conformation change from 4C1 to 1C4.


Elimination reactions of vicinal dibromide When a vicinal dibromide undergoes elimination reactions in vigorous basic conditions, say in ammonia with $\ce {NaNH2}$, is the product an alkyne or is it a conjugated diene? I would think that the answer depends on thermodynamic and kinetic factors. Considering bond energies of the $\ce {C-C}$ $\pi$ bonds, perhaps the conjugated diene would be more favourably formed?

After the first elimination, you have a vinylic bromide, so there are two options for the second elimination. The two acidic hydrogens that the base can abstract are one attached to sp2 carbon and one to sp3. Since sp2 hybridized carbon is more electronegative, the base abstracts the hydrogen attached to it, and then bromine leaves being a good leaving group, which gives alkyne as the major product.


What is the best way to convert SMILES strings to file names? I would like to convert SMILES strings into file names. There are many characters to avoid in file names, i.e. \/:*?"<>| or maybe even []()^ #%&!@:+={}'~ (see discussion) for convenience.
I would like to escape such characters that appear in SMILES. However, I could not see a list of allowed characters for SMILES to decide on how to escape them. There might be also some existing convention to do this for databases, so I am all ears for any help.

An alternative is using InChIkeys but since they are not human readable, I want to avoid that option.

Edit: I am mainly interested in small organic molecules with less than 20 atoms.

I personally find SMILES quite unreadable, even for only small compounds. I think they are even less readable when performing substitutions. Given the vast amount of special characters they use, I do not think they are a good way to create a naming scheme.

  An alternative is using InChIkeys but since they are not human readable, I want to avoid that option.

While that is true, InChI strings are human readable, especially for small molecules. In principle only the layers need to be separated by something other than / (underscores would do that).

Other problematic characters are *;?+() which do not fit in [0-9a-zA-Z-.,_], and you would have to find ways around them when using in scripts.

  I am mainly interested in small organic molecules with less than 20 atoms.

If you are really only interested in molecules with less than 20 atoms, general nomenclature should be the easiest to comprehend, and should yield reasonably short file names. For these you probably only have to worry about spaces.


How to apply the Henderson–Hasselbalch equation when adding KOH to an acidic acid buffer? Will the Henderson–Hasselbalch equation be able to calculate the pH when you have an acidic solution and you add a basic solution? Will that be possible or can you only use it for adding an acid to an acidic buffer to calculate the pH, or only a base to a basic buffer? If it can, how is the pH calculated, if you add 0.025 mol of solid KOH to 1 L of an acidic acid buffer solution of pH 3.97.

If the initial solution is just acetic acid and water, and the pH is 3.97, this means that it is a very dilute solution, less than 0.001M acetic acid. See http://depts.washington.edu/chem/facilserv/lecturedemo/pHofAceticAcid-UWDept.ofChemistry.html

and Does the number of H+ ions in solution go up on dilution of a weak acid?

Therefore, of the 0.025 moles of OH-, almost all (more than .024 moles) will remain unreacted.

pH = 14 - pOH

pH = 14 + log(0.025) = 12.4

In a different situation, where you added less strong base than weak acid, the Henderson-Hasselbach equation would be helpful, but it is not helpful in this example.


Can Hartree-Fock mix more orbitals than there are electrons? I'm trying to understand Hartree-Fock well enough to write my own implementation. A point that I seem to be coming to that I'm not sure is correct: the method only produces one-electron orbitals that are linear combinations of the starting orbitals?

That is, if I have $n$ electrons, I start them off in some $n$ orbitals (say, atomic orbitals). Is it correct that the final result will only mix those $n$?

It seems very natural that I might want to include other orbital shapes -- maybe I want the first $m \gg n$ atomic orbitals in my molecule, and let all of those contribute to building lower-energy one-electron orbitals. As best s I understand it, standard HF does not allow for this, because at each step of diagonalizing the Fock matrix it's just using combinations of the existing orbitals -- with no room for the excited states.

Is my understanding correct? Are there other extensions of HF which allow broader mixings?

One possibility that I could imagine would be starting off with $m$ electrons, so that I have $m$ orbitals available. Then I run HF, and at the end I only use the $n$ lowest orbitals. This runs into the problem that I will still be computing Coloumb terms from the $(m-n)$ extra "fake" electrons, which will disturb the shape of my $n$ lowest orbitals. So perhaps the correct approach would be to use $m$ electrons mixing, but when I compute the Fock matrix $F$, I only include $J$ and $K$ terms from the $n$ lowest modes (but letting these affect the energies of all $n$ modes). But that seems messy, because now $F$ isn't symmetric... in the end this seems like a risky path to go down.

Help appreciated, thank you.

First some general considerations

Using more orbitals than electrons is not just possible, but almost essential for Hartree-Fock. Although in practice one can do calculations in a so called minimal basis set, this approach is lacking a lot of accuracy. Usually one adds additional basis functions, which leads to basis sets classified as double zeta (triple, quadruple and so on are also possible).

For open shell systems even a minimal basis set includes more atomic orbitals than electrons are available. For example if you want to calculate a single carbon atom, you would have one basis functions for each of the $1s$, $2s$ and all three $2p$ orbitals. This gives a total of 5 atomic spatial orbitals (=10 spin orbitals), but only 6 electrons. (This is not really a good example, as occupying just one of the three $2p$ orbitals breaks the spherical symmetry of the atom, but I am leaving the scope of your question ...)

Virtual orbitals

Since all molecular orbitals (MOs) are expressed in this basis of atomic orbitals (AOs), it means we will get as many MOs as we put AOs in. If ordered by their energy, we first get a set of occupied MOs until all electrons are distributed. After that comes a set of virtual (or unoccupied) MOs. Those are still soluations of the Roothaan-Hall equations, i.e. they are eigenfunctions of the Fock-Operator, but they do not contribute Hartree-Fock energy.

However, virtual orbitals are important in correlation methods going beyond Hartree-Fock, e.g. Coupled Cluster or Pertubation Theory. Here the virtual orbitals become partially occupied and thus influence the ground state energy.

Equations

The ground state energy in case of closed-shell HF reads

\begin{equation}
  E_0 = 2\sum_a h_{aa} + \sum_a\sum_b 2J_{ab} - K_{ab}
\end{equation}

where the sum only goes over the occupied spatial orbitals, which equals half the number of electrons.
And for the Fock operator we have
\begin{equation}
  f(1) = h(1) + \sum_a 2J_a(1)-K_a(1)
\end{equation}
with the same restrictions to the summation. Since the Fock operator is applied to all MOs, this means the effect (coulomb repulsion) of the occupied orbitals is considered in each virtual orbital, but the virtual orbitals do not affect the occupied ones.

(Equations are taken from the book by Szabo and Ostlund, p. 134)


After following Mithoron's link, I familiarized my self with the writing on the following ChemGuide site (link).

I went into Excel spreadsheet, and had each 'sub orbital' it's own column, and came up with the following.

+---------+---------+----+---------+----+----+---------+-----+
| Shell 1 | Shell 2      | Shell 3           | Shell 4       |
+---------+---------+----+---------+----+----+---------+-----+
| 1s      | 2s      | 2p | 3s      | 3p | 3d | 4s      | SUM |
| 2       | 2       | 6  | 2       | 6  | 0  | 1       | 19  |
| 2       | 2       | 6  | 2       | 6  | 0  | 2       | 20  |
+---------+---------+----+---------+----+----+---------+-----+

This is what the overall table (link) looks like in excel; I can't attach the image directly to my post.

While googling, to get a clear understanding of S,P,D,F I came across ThoughtCo's article (link). This let me know the stability precedence, i.e. electron distribution, more clearer, as follows:

1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f

4s is filled before, 3d. 3s and 3p, can hold a combined maximum of 8 electrons.

The following (link) has a diagram with downward/diagonal arrows, showing how one can better remember the stability precedence.


Which orbital in a subshell gets filled first? Let's say we're filling electrons in subshell 2p. This subshell will have $m_l = -1$, $m_l = 0$, and $m_l = 1$ orbitals. Does the 'filling' of electrons depend on the value of $m_l$? Meaning, does the value of $m_l$ determine if it gets filled first by an electron or not?

The notation [-1,0,1] is given to address the sub shells.The energies of all these orbitals (unless you apply a magnetic field) are identical, therefore the filling is completely arbitrary. That means any of the shells can be filled first. But after 1 shell is filled the next electron goes to any one of the remaining shells. And the next goes to the other one. This is known as the Pauli's Exclusion law. Electrons with like spins are filled first and electrons with the opposite spin are filled next (note: 1 shell can share 2 electrons). This choice of spin is also complete arbitrary.


Does triphenylphosphine show resonance? The lone pair of phosphorus and the double bonds in benzene looks to be in conjugation. Do the molecule show resonance? If so, how?

The triphenylphosphine molecule would show resonance because there is a lone pair on phosphorus and it is in conjugation with the double bonds in benzene. But the lone pair would be in resonance only with one ring at a time because the molecule is non-planar (tetrahedral) having sp3 configuration. Also, we should consider the steric effects on the molecule. The steric hindrance produced by other two benzene rings would cause less resonance effect and the lone pair would be inhibited from going into the other two rings.
So, the resonance in this molecule would be simply like that in aniline.


Adulteration of turmeric with Metanil yellow I read in the local paper that common food products like turmeric can be adulterated. It continued to specify the adulterant which they called metanil yellow. On searching on the Net, I was able to find that metanil yellow is prepared using some basic raw materials like metanilic acid and diphenylamine. The paper continued to lay out procedures on how to find out whether the turmeric you buy from the local market is adulterated. To do this, one must first dissolve half a spoon full of turmeric powder in 20ml of lukewarm water. It then told to add a few drops of hydrochloric acid. Upon doing this, if the aqueous solution turned pink, violet or purple, it indicated the presence of metanil yellow. It continued to state the risks of consuming turmeric adulterated with metanil yellow. One of these risks was that it is a carcinogen. I'm an Indian, so turmeric is included in nearly all of the dishes I eat in a day, so I became very curious. My question is this:

1.) What is the chemical formula for metanil yellow, how is it prepared (in layman's terms), and how does it react with hydrochloric acid to turn pink, purple or violet (a chemical equation would be helpful)?
2.)What are the risks in consuming metanil yellow?

I hope to keep this short.

Metanil yellow is an azo dye synthesized from the coupling of metanilic acid and diphenylamine, as described here. Here's a sketch:

Now, the thing with most azo dyes like metanil yellow is that when they are taken internally, liver enzymes or intestinal flora will usually reduce the azo dye to the components that were used in the diazo coupling. (See this article for more details.) For metanil yellow, the metanilic acid is poorly absorbed from the intestinal tract (and is thus easily excreted), and your remaining worry is the diphenylamine, which is a suspected mutagen and carcinogen (apart from the usual toxicity associated with arylamines). (In short, the metanil yellow itself is not your worry, but the metabolite diphenylamine sure is.)

By way of contrast, the natural coloring agent in turmeric is the phenolic compound curcumin:

As it turns out, both compounds are in fact acid-base indicators, though they have different transition ranges; for metanil yellow, the pH transition range is 1.2-2.3 (red to yellow), while for curcumin, the transition range is 7.8-9.2 (yellow to red-brown). (Data taken from the Handbook of Acid-Base Indicators.) I'm not quite sure where the purple/violet color would come from, but the key point is that natural turmeric would be reddish in an alkaline environment, while metanil yellow will be reddish in an acidic environment; however, vinegar (pH ~ 2.4) is not sufficiently acidic to display the color change for metanil yellow, and thus a stronger acid like hydrochloric acid is needed to display the color change.


Lewis structure of dinitrogen trioxide I have a confusion related to the structure of $\ce{N2O3}.$ I was taught 1 is the correct one, but I believe 2 is better since it has no formal charge on it. Why is 1 correct and 2 wrong?

What really happens in nature has higher priority than our believe what happens or should happen. See Dinitrogen_trioxide with the structure, bond lengths and angles.
Be aware both $\ce{NO}$ and $\ce{NO2}$ are radicals with an unpaired electron.
$\ce{N2O3}$ being( $\ce{ON-NO2}$ ) as 2 paired radicals is the direct analogy
to the dimer $\ce{N2O4}$ being $\ce{O2N-NO2}$, both being formed at low temperatures.


Convert hexane to 1,6-dibromohexane Is it possible by any means or by using some reagents to convert
hexane to 1,6-dibromohexane? The yield should also be good.

In principle, alkanes may be converted to alkyl bromides by photoinitiated bromination.

However, taking statistics (2 $\ce{CH3}$ vs 4 $\ce{CH2}$ groups) and stabilities of the radicals resulting from H abstraction by a bromine radical from your starting material into account, this approach is pretty worthless and will most likely yield 1,6-dibromohexane as a minor product in a complex mixture.

Consequently, your reasonable options are a) a different starting material (and a different synthesis) or b) simply buy 1,6-dibromohexane (around 20 Euro for 100 g).


Why can't you use m-CPBA to synthesise isophorone oxide from isophorone? Apparently you must use sodium hydroxide and hydrogen peroxide because using m-CPBA gives an undesired product

Why doesn't m-CPBA work and what does it do instead? Why does sodium hydroxide and hydrogen peroxide work?

When using MCPBA, the undesired by-product is probably formed due to Baeyer–Villiger oxidation of the ketone to the corresponding ester (here: lactone).


Why is the equilibrium constant increasing in this exothermic reaction? A problem asks, if you have a positive $\Delta S$ (positive change in entropy) for an exothermic (meaning a negative $\Delta H$, or negative change in enthalpy) reaction and temperature increased, what would happen to $K_{\mathrm{eq}}$ and $K_{\mathrm{eq}}/Q$ (where $K_\mathrm{eq}$ is the equilibrium constant and $Q$ is the reaction quotient)?
The answer states that $K_{\mathrm{eq}}/Q$ and $K_{\mathrm{eq}}$ would increase. This is because if you use the thermodynamic equation, $\Delta G = \Delta H - T \Delta S$ (where $\Delta G$ is the change in Gibbs free energy and $T$ is for temperature), as you increase temperature you get a more negative number (since $\Delta S$ is positive and $T\Delta S$ is being subtracted). This would lead to a more negative $\Delta G$. Then, using another thermodynamic equation, $\Delta G = RT\ln(Q/K_{\mathrm{eq}})$ (where $R$ is the ideal gas constant), if $\Delta G$ is decreasing (becoming more negative), then the term $(Q/K_{\mathrm{eq}})$ must also be going down. This means that $K_{\mathrm{eq}}/Q$ must be going up and therefore $K_{\mathrm{eq}}$ increases, as temperature increases for this exothermic reaction.
So my issue is, I thought for an exothermic reaction, where heat can be represented as a product, I thought increasing the temperature would shift the reaction to the left and decrease $K_{\mathrm{eq}}$. Could someone clarify this for me? I tried looking this up with Chemistry Libretexts, but it didn't seem to clarify anything. The table seemed to suggest that we do use heat as a component of the reaction in endothermic or exothermic reactions to predict the $K_\mathrm{eq}$ change.

An exothermic reaction has a reduced equilibrium constant at higher $T$ because while the contribution of the change in the entropy of the system is a fixed quantity (for a small $T$ change), the effect of transferring heat to the surroundings is reduced at higher $T$ (because it causes a smaller change in the entropy of the surroundings). Mathematically this is described by the van't Hoff equation:
$$\left(\frac{\partial \ln K_\textrm{eq}}{\partial T}\right)_p = \frac{\Delta H^\circ}{RT^2}$$
On the other hand the change in $\Delta G$ with temperature is given by $\Delta S$:
$$\left(\frac{\partial\Delta G}{\partial T}\right)_p = -\Delta S$$
Since $\Delta G =\Delta G^\circ +RT\ln Q = -RT\ln (K/Q)$ one can write
$$\left(\frac{\partial\Delta G}{\partial T}\right)_p = \left(\frac{\partial\Delta G^\circ}{\partial T}\right)_p +R\ln Q\\ = -\Delta S^\circ+R\ln Q$$
Therefore if you want to see how $ K_\textrm{eq}$ depends on $T$ you should focus on $\Delta H^\circ$. If you want to see how $\Delta G$ depends on $T$ you should inspect $-\Delta S^\circ+R\ln Q$.
It is somehow unfortunate that the temperature dependence of $\Delta G$ and the equilibrium constant are different, because it leads to endless mix-ups, but it reminds you that they describe different things.


Why should animal fats not be used for cooking? Why we are advised not to use animal fats for cooking oils? I know that vegetable oil contain long unsaturated carbon chains and animals contain saturated carbon chains. What’s the difference if we use this?

For one thing, saturated fats tend to have higher melting points, such that at room or body temperatures, they are often solid (though exceptions obviously exist). Therefore, unsaturated fats and oils will be easier to use since they can be poured.

There is also a longstanding, though possibly unjustified, link between saturated fats and oils, and cardiovascular disease. Avoiding saturated fats is supposedly healthier.
Because livestock are more expensive and slower to grow than plants, vegetable or plant oils are easier and faster and cheaper to obtain.
Finally, animal fats often absorb flavour from the animal meat, giving them an unwelcome taste in many contexts. By contrast, vegetable and plant oils either don't absorb anything or the result is attractive pr characteristic in most cooking contexts. So vegetable or plant oils are preferred for versatility and for flavour.


Computation of pH when an acid and base are mixed in solution I'm doing a basic chemistry course, and we are currently learning how to compute $\text{pH}$ from the acid dissociation constant (using $\left[\text{H}^{+}_{(\text{aq})}\right]=\sqrt{K_{a}\left[\text{HA}_{(\text{aq})}\right]}$) along with computing the $\text{pH}$ of strong bases by assuming full dissociation into $\text{OH}^{-}$ ions, and then using the ionic product of water to calculate the concentration  of protons.

The question I have been given is:

  Calculate the pH of the solution obtained when $14.9\text{ cm}^{3}$ of $0.100\text{ mol dm}^{-3}$ sodium hydroxide solution has been added to $25.0\text{ cm}^{3}$ solution of methanoic acid of concentration $0.100\text{ mol dm}^{-1}$; $K_{a}=1.60\times 10^{-4}\text{ mol dm}^{-3}$

I'm not sure how I should go about solving this problem, I can calculate concentrations of hydrogen and hydroxide ions for each of the solutions but I'm unsure how to combine them (as presumably some of the $\text{OH}^{-}$ ions will react with the $\text{H}^{+}$ ions to form $\text{H}_{2}\text{O}$?)

You are headed in the right direction.  Pretend that the $\ce{H+}$ ions and the $\ce{OH-}$ ions neutralize each other 1 for 1.  That will leave you with only one of those ions left.

The water ionization won't affect your problem as that ionization is repressed by the excess ion present in the solution.

Now you should be able to solve the problem.


Bomb calorimeter enthalpies of straight chained alcohols Can anyone suggest a website which contains the literature values (using a bomb calorimeter) of the enthalpies of n-alcohols?

I have been trying to find it but I am unable to locate bomb calorimeter values (I do not want the bond dissociation energies.

I searched 'combustion enthalpies of straight chain alcohols' in Google Scholar and found a couple of links to papers that look relevant, see below. Unfortunately one looks like you have to pay for it, but if you're associated with an institution they may give you free access. The other involves cyclic alcohols, obviously not straight chain but maybe helpful nonetheless.

https://link.springer.com/article/10.1007%2FBF02885455?LI=true

http://actachemscand.dk/pdf/acta_vol_16_p0046-0052.pdf

I would imagine the molar enthalpies of combustion would increase fairly linearly as chain length increases. Each extra carbon in the chain means an extra molecule of CO2 is produced, and the strong C=O bonds formed are the major contribution to the exothermicity of the reaction. Steric or torsional effects in the larger molecules may lead to a slightly smaller than expected increase per carbon.

Hope this helps.


Is there a way to Determine if an Element can undergo Disproportionation? Can we determine the lowest and highest possible oxidation numbers of an element?
Since if we know this, then it is easy to determine whether the element will undergo disproportionation.

Expanding on my original comment, I disagree with the OP's statement that knowing the possible oxidation states of an element will easily allow you determine if a disproportionate reaction will occur.

To predict such a reaction, thermodynamic data such as the $E_\mathrm{cell}^\circ$ of the two reactions must be known. Latimer diagrams are a really simple and compact way of showing the $E_\mathrm{cell}^\circ$ values for a variety of reduction and oxidation half-equations between different oxidation states of the same element. Below is an example of a Latimer diagram for manganese in acidic medium:

Just from this I can tell the $E_\mathrm{cell}^\circ$ potential for any oxidation or reduction reaction between any pair of manganese species (in fact there is extra information that isn't even required). For example by looking at this diagram I can tell what $E_\mathrm{cell}^\circ$ will be for the reduction of $\ce{Mn^7+}$ to $\ce{Mn^3+}$ (I won't go into how you would do that in this post, I suggest that if you want to know how to, have a look at the website that I have linked).

But wait, there is more! Now the neat thing about Latimer diagrams is that I can easily tell if a particular oxidation state of manganese is stable or not (in other words, will it undergo disproportionate reaction?). All you have to do is follow one very simple rule:

'if the number on the left is lower than the number in the right, the species is unstable to disproportionation'

For example, let's consider $\ce{Mn^3+}$. The $E_\mathrm{cell}^\circ$ for the reduction of $\ce{MnO2}$ to $\ce{Mn^3+}$ is $0.95~\mathrm{V}$ (that is the number on the 'left') while the $E_\mathrm{cell}^\circ$ for the reduction of $\ce{Mn^3+}$ to $\ce{Mn^2+}$ is $1.51~\mathrm{V}$ (this is the number on the 'right'). Since the number on the left is smaller than the number on the right, $\ce{Mn^3+}$ will spontaneously disproportionate to form $\ce{MnO2}$ and $\ce{Mn^2+}$. The reaction is below:

$$\begin{align}
\ce{2Mn^3+ + 2H2O} &\ce{-> MnO2 + Mn^2+ + 4H+} & E_\mathrm{cell}^\circ = +0.56~\mathrm{V}
\end{align}$$

Note how for this reaction $E_\mathrm{cell}^\circ > 0$, hence it is spontaneous.

However these 3 species that are involved don't have to be right next to each other. For example, let's consider $\ce{MnO3-}$. The $E_\mathrm{cell}^\circ$ for the reduction of $\ce{MnO4-}$ to $\ce{MnO3-}$ is $0.419~\mathrm{V}$ while the E cell for the reduction of $\ce{MnO3-}$ to $\ce{Mn^3+}$ is $2.61~\mathrm{V}$. Again, the number on the left is smaller than the number on the right, hence $\ce{MnO3-}$ will disproportionate spontaneously.

This may not be the answer that you were expecting originally, but hopefully this will help you with your original questions.


Effective Bohr Radius in Transition Metal Complexes? I am playing around with a toy model of a transition metal complex where the HOMO are $d$-electron states of predominantly transition metal character. Let's say this is a $d^1$ (or $d^9$) system and the states are split in energy by ligand field effects. The relevant HOMO orbital is some linear combination of $d$-orbitals that correctly respects the symmetry of the transition metal complex.
I want to estimate the transition probability for a quadrupole $d$-to-$d$ transition from HOMO state. In other words, calculate $\langle f\vert r_i r_j \vert i \rangle$, where $\vert i \rangle$ and $\vert f \rangle$ are some $d$-orbitals I care about.
I can roughly estimate this transition amplitude by just using the Hydrogen $d$-orbital wavefunctions with an effective Bohr radius $a_{eff}$. My question is: what effective Bohr radius should I use?
For example, Copper is said to have a 3d-orbital radius of $\sim 0.6$ Angstrom (which is not the same as $a_{eff}$ I believe), but using the Slater estimate of the effective nuclear charge, I would guess $a_{eff} \approx a_0/Z_{eff} = a_0/13.2 \approx0.04$ Angstrom.
What effective Bohr radius should I actually use for very weakly hybridized transition metal orbitals?

If you inspect Table II in Ref. 1, which you will find referenced in the link you provide to the WebElements website, you'll see that AU units are used (also in the linked page), not pm, so the radius of the 3d Cu orbital in Ref. 1 is actually

$$ r_{\textrm{Mann}} = 0.613 \times 0.529 Å = 0.324 Å $$

or 32.4 pm.

As for Slaters approach, it is not enough to simply divide $a_0$ by the effective nuclear charge. You can only do this for an unshielded hydrogenic 1s orbital, otherwise it is generally necessary to account for the principal quantum number (in Slater's method, this may be an effective value not equal to the principal quantum number, see Slater's original article$^2$). The location of the maximum follows from the functional form of a Slater wave function:

$$\psi \propto r^{n^*-1}\exp\left({\frac{Z_{\textrm{eff}}}{n^*a_0}r}\right)$$

In general, if you change the nuclear charge of a hydrogenic atom you can perform the transformation $a_0 \rightarrow a_0/Z_{\textrm{eff}}$ to obtain the new orbitals$^3$.

The effective quantum number is n*=3 for a 3rd shell electron so that

$$r_{\textrm{Slater}} = \frac{n^{*2}}{Z_{\textrm{eff}}}=\frac{9}{13.2}\times0.529 Å = 0.361 Å $$

or 36.1 pm.

The difference in $r_{\textrm{max}}$ obtained by the two approaches is therefore not that dramatically different.

Of course there are two other things that seem to be important here: (1) whether using an accurate value of $r_{\textrm{max}}$ can be expected to provide you with an accurate value of the property you seek and (2) how to obtain an accurate estimate of $r_{\textrm{max}}$for a d-orbital of Cu in the transition metal complex.

References

J.B. Mann, Atomic Structure Calculations II. Hartree-Fock wave functions and radial expectation values: hydrogen to lawrencium, LA-3691, Los Alamos Scientific Laboratory, USA, 1968.
J.C. Slater. Atomic Shielding Constants. Phys. Rev., 36, 57, 1930.
W. Greiner. Quantum Mechanics: An Introduction. Springer 1994.


How to find the largest cyclic substructure with RDKit? I have a bunch of structures and of each I would like to find the largest cyclic substructure.
I'm not a chemist, so I don't know if there is a common terminology for this. I hope, I made it clear what I'm looking for.
Example:

Each structure on the top (1,2,3) as input should return the bottom structure (4).
There is the rdkit.Chem.MCS module which will return the maximum common substructure of a set of molecules.
But that's not what I'm looking for.
Any ideas? And hints where or which module to look at...? Which term to search for?
I'm aware that there is a RDKit mailing list, however, I'm asking here first, because of the possibility to easily attach images. If nobody can help me here, I will try there. Maybe it already goes into "heavier" programming which then would be more a topic for StackOverflow.

In the RDKit Cookbook is an example how to find the indices of ring systems.
RDKit Cookbook Example
Having the indices, it is simple to find the largest ring.
from rdkit import Chem
from rdkit.Chem import Draw
from rdkit.Chem.Draw import IPythonConsole
IPythonConsole.molSize = (400, 300)
IPythonConsole.drawOptions.addAtomIndices = True

m = Chem.MolFromSmiles('C1=CC2C(C=C1)=c1ccccc1=C2C1C=C(c2ccccc2)c2cccc3c4ccccc4nc1c23')
m

# the cook book example
def GetRingSystems(mol, includeSpiro=False):
    ri = mol.GetRingInfo()
    systems = []
    for ring in ri.AtomRings():
        ringAts = set(ring)
        nSystems = []
        for system in systems:
            nInCommon = len(ringAts.intersection(system))
            if nInCommon and (includeSpiro or nInCommon>1):
                ringAts = ringAts.union(system)
            else:
                nSystems.append(system)
        nSystems.append(ringAts)
        systems = nSystems
    return systems

ringSys = GetRingSystems(m)
ringSys

[{16, 17, 18, 19, 20, 21},
 {13, 14, 15, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35},
 {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}]

Highlighting the largest ring for proof.
highlightAtoms = max(ringSys, key=len)
Draw.MolToImage(m, size=(400,300), highlightAtoms=highlightAtoms)


Material that changes solubility when electric potential is applied I am looking for a material witch is easy to deposit in thin films and which changes its solubility if a electric potential is applied.

Do you know of any such materials ?

EDIT: This is for what I need it.
I have a Si Substrate on which there is sputtered SiO2.
Then I have the material (to be found) and then a Si on top. Some parts of the Si are connected electronically to the bottom Si and create a electric potential. The orange material underneath the top "electrodes" should be removed. The aim is to remove the "orange" material underneath the top "electrodes" but keep it where the Si is not connected to the bottom electrode.

How could you do this ?

EDIT2: please Igore the spelling error in the bottom electrode. It should be "connected"

EDIT3: based on Alchimista's answer:

This is a tentative that perhaps does not fit with restrictions posed by the setup of the question, but does offer an example of a class of materials whose solubility (integrity, to be more precise) is easily altered by an electrochemical process.

A number of conjugated polymers can be deposited atop an electrode by
A: electrochemical polymerisation ( this resulting in a film of the polymer in a more or less conductive state ) or

B: as pristine material if available and soluble in some solvent ( spin coating, dipping, dr blading. ..).

The state of the polymer can be electrochemically controlled by the applied potential. ( however the complete neutralisation to its insulating/semiconducting state is generally difficult ).

Overoxidation of these films results in chemical damage associated with removal of the material, directly or by subsequent washing.

3-methylthiophene and poly(3-hexylthiophene) are well known examples for an electrochemical depositable polymer and for a polymer which can be directly cast from its solutions.

In both cases the poly(3-alkylthiophene) (P3AT) film is insoluble in polar solvents and electrolyte solutions. Applying an electrochemical potentially higher than, say, 2 V, overoxidize it damaging the skeleton to such an extent that the material fall out the electrode and / or can be washed out.

Specifically to the question: only case B is viable.

Looking at the figure in the Q, bottom, the orange part will be consisting of neutral P3AT.

Resolution will be not too precise as overoxidation will take place at the edge of the electrode, too.