Calcium chloride hexahydrate has separated in jar - is it beyond saving? I work in a lab where I'm in charge of ordering new supplies. A graduate student brought up a $500~\mathrm{g}$ amber jar of $\ce{CaCl2.6H2O}$ and pointed out that it had separated into a clear liquid (on top) and a white crystal (on bottom). The cap seemed to be on normally and there is no expiration date on the jar.

Before I decide if I need to order more, I am wondering:

What happened to the $\ce{CaCl2}$?
Can it be saved?

Calcium chloride is incredibly hygroscopic (it sucks in water), this property makes it incredibly useful (for instance when used in a desiccator or in a drying tube to prevent water from getting into a reaction).

The wet $\ce{CaCl2}$ can be turned back into a solid  by heating (ideally under vacuum), however you can end up with a mixture of different compounds (for instance heating very wet $\ce{CaCl2}$ has been known to give $\ce{Ca(OH)2}$).  It can also be recrystallised from ethanol to give the di-hydrate, which in turn may be dehydrated by heating, however fresh $\ce{CaCl2}$ is very inexpensive, and so I would suggest buying a new bottle, and ensuring it is sealed well after use to prevent it taking up moisture from the atmosphere.


Band structure in half-metals In a typical band diagram of a half-metal, the band structures for electrons of up and down spin are different. Why is this so?

You got it reversed: it is not a half-metal properties that make band structure different but it is the band structure that gives rise to half-metal properties.

That said, the proper question would be: why band structure may be different for electrons with different spin orientation?

The answer is that electrons do not exist in vacuum and unpaired electrons with different spin interact with each other through exchange which lead to having unpaired electrons having lower energy when their spins are parallel (please, note: what 'direction' we use for 'up' and 'down' when performing this type of calculation is completely arbitrary). So, when we have a half-filled band, we may expect electrons on it to have parallel spins. In reality it is usually not the case, because exchange interactions are fairly weak.

For more details you probably should repost on Physics.SE or go for specialized literature. You need at least basic understanding of quantum mechanics to understand the details. Some insight may be obtained here.


the molecular formula of an unknown hydrocarbon A student analysed a hydrocarbon X and found it to be containing 43g carbon and 7.2g hydrogen.The relative molecular mass of X was found to be 42.What is the molecular formula of X?

$43\ \mathrm g$ of carbon for $7.2\ \mathrm g$ hydrogen.
Molar mass of carbon is $12\ \mathrm{g/mol}$ and that of hydrogen is $1\ \mathrm{g/mol}$.

Hence there are $\frac{43\ \mathrm g}{12\ \mathrm{g/mol}} = 3.58\ \mathrm{mol} \approx 3.6\ \mathrm{mol}$ of carbon for every $\frac{7.2\ \mathrm g}{1\ \mathrm{g/mol}}=7.2\ \mathrm {mol}$ of hydrogen.

Hence the empirical formula of the compound is $\ce{CH2}$.

Relative molecular mass of $\ce{CH2}$ is $12+2\times1 = 14$. Given relative molecular mass is $42$.

$\displaystyle\frac{42}{14}=3$

Hence the formula of the compound is $\ce{C3H6}$


molecular orbital theory i read in my textbook that the energy diagram of moleculal orbits is different for atoms nitrogen and below them.
my question is which one of the energy diagrams to use in heteronuclear compounds like CO?

(In heteronuclear diatomic molecules, atomic orbitals only mix when the electronegativity values are similar.)
(While MOs for homonuclear diatomic molecules contain equal contributions from each interacting atomic orbital, MOs for heteronuclear diatomics contain different atomic orbital contributions.)
(Orbital interactions to produce bonding or antibonding orbitals in heteronuclear diatomics occur if there is sufficient overlap between atomic orbitals, as determined by their symmetries and similarity in orbital energies.)


Is the negative excess volume of ethanol/water the cause of my math error? I have a binary mixture, say ethanol at $90\%wt$ ethanol, $10\%wt$ water.  According to Perry's Chemical Engineers' Handbook 9th Ed., that mixture has a density of $0.81797g/cc$ (at $20^{\circ}C$).  So my $1L$ of mixture should have a mass of $817.97g$.

Now, I want to create this mixture.  I know the total mass and the mass %'s, so I just multiply them out: we find that we need $736.17g$ ethanol and $81.80g$ water.  Ok, so far so good.  However, say all of my balances are broken, so I only have graduated cylinders to work with.  Easy, just convert those masses back to volumes using their respective pure substance densities: $0.78934g/cc$ for ethanol and $0.99823g/cc$ for water.

However, doing this tells me that I should have $932.6mL$ of ethanol and $81.9mL$ of water, which adds up to a total volume of $1014.6mL$ of binary mixture, so somehow I magicked $14.6mL$ of liquid out of thin air using this conversion algorithm.

Is this due to the negative excess volume of the ethanol/water mixture upon mixing?  That's my current guess, but for the life of me I'm having a hard time figuring out the math here.  It seems like the change in volume should be much smaller: $10\%wt$ water in ethanol is about $10\%mol$ and there's about $4.5mol$ of water in the mixture, so based on this dataset it seems like there should only be a difference of about $3mL$.  Even with all the rounding I just did, it seems like this can't be the only error source.

Is there something wrong with the logic of my algorithm here?  Can anyone point me to the reason I'm seeing this numerical error whenever I run this calculation?

When you mix ethanol and water they take up less space due to the way the molecules fit together, so you will have what is called a negative volume of mixing.  In this case, the 932.6 mL of ethanol and the 81.9 mL of water will give you a 100 mL solution of 90% ethanol.


Effect of pH on positive and negative charges of a zwitterion I read that the isoionic point is defined as the

  pH at which a zwitterion molecule has an equal number of positive and negative charges and no adherent ionic species.

I have two questions regarding this definition.

How does pH affect the number of positive and negative charges of the groups in a zwitterion?
What does "no adherent ionic species" mean?

Consider an amino acid as a zwitterion. It has both a basic amine functionality and an acidic carboxylic acid functionality. At a high pH, the acidic functionality will be largely deprotinated and the ion will have a net negative charge. Oppositely, at low pH, the amine group is largely protinated and the compound will have a net positive charge. Clearly, there is some intermediate pH at which the positive and negative charges are equal (or it's neither protinated nor deprotinated) and the net charge is zero.

To maintain charge neutrality, a charged species must have some oppositely charged ionic species associated to maintain charge neutrality. For example, under acidic conditions when the amine group is protinated, you need some anion present, like chloride for example, to be associated with the compound to maintain charge balance. This counter-ion is also known as an adherent ionic species (adherent meaning adhering to, like when the compound is an undissolved solid).


Acid strengths for HBrO4 vs HIO4, and HBr vs HI So I was looking at factors that control the relative strengths of acids and bases.

We know that $\ce{HI}$ is a stronger acid that $\ce{HBr}$ because $\ce{I}$ is a much larger atom.

But then, I read somewhere that $\ce{HBrO4}$ is stronger than $\ce{HIO4}$, because in this case it's electronegativity that plays the important role.

How do you know whether to use size or EN as a guideline for comparing strengths of acids/bases?

"We know that HI is a stronger acid that HBr because I is a much larger atom"

That is only half the story. It is about the strength of the H-Br versus H-I bond. Since I is larger than Br, the H-I bond is longer, weaker, and therefore more prone to dissociation. See here for more in-depth discussion.

  $\ce{HBrO4}$ and $\ce{HIO4}$

both have the acidic proton bound to oxygen:

The H-O bond is approximately of equal strength in both acids and therefore you turn to electronegativity arguments.


C-N bond lengths in 2-bromo-1,3,5-trinitrobenzene Are all the $\ce{C-N}$ single bonds in 2-bromo-1,3,5-trinitrobenzene the same or different? I am of the view that the $\ce{C_1-N}$ and $\ce{C_3-N}$ bonds are of the same length and they are longer than the $\ce{C_5-N}$ bond, due to steric hindrance. Am I right or am I missing out something?

(source: nist.gov)

By rotating the molecule 180° about the $\mathrm{C_2}$ symmetry axis that passes through the bromine and the nitro group in the 5-position, you can interconvert $\ce{C_1-NO2}$ and $\ce{C_3-NO2}$.  Groups that can be interconverted by a rotational axis must be equivalent.  There is no symmetry element that interconcerts $\ce{C_1-NO2}$ (or $\ce{C_3-NO2}$) and $\ce{C_5-NO2}$, therefore these groups cannot be equivalent. So symmetry tells us that the  $\ce{C_1-NO2}$ and $\ce{C_3-NO2}$ bond lengths are exactly the same, and both are different from the $\ce{C_5-NO2}$ bond length.

Symmetry can only tell us if things are the same or different, it can't tell us the magnitude of any difference.  So I don't know for sure which set of carbon-nitro bonds is longer or shorter, but your argument based on steric strain seems reasonable.  It is also possible that the bulky bromine causes the adjacent nitro groups to twist out of the plane of the aromatic ring, reducing resonance overlap between nitrogen and carbon and also lengthening those adjacent nitro bonds.


Viscosity of branched vs. linear short alkanes I was looking through the CRC Handbook of Chemistry and Physics section on liquid viscosities, trying to explain trends in viscosities. In higher alkanes, the viscosity increases with the degree of branching, which is logical. However, I noticed for lower alkanes, the opposite is true. For example, 1-methylbutane has a viscosity of 0.286 mPa-s at 25 C, whereas its linear counterpart pentane has a viscosity of 0.30 mPa-s at 25 C. The same trend holds for alkanes with 6 carbons. These trends go against my chemical intuition and I cannot think of a sound explanation, which brings me to my question: why is the viscosity lower for branched vs. linear alkanes in the lower alkanes?

Viscosity is related to Van der Waals/London forces between molecules.  This in turn is related to packing and surface area.

A branched 6-chain carbon, tumbling in solution cannot align as much of its surface area with another molecule as the linear hexane.

At some point the larger molecules are becoming able to be 'entangled' and form structures somewhat similar to knots.  When the molecules get this large, viscosity increasing branching becomes more likely.

Finally, when you get to true polymers, the chains are so long that branching again begins exclude other molecules from structure and again linear molecules are more likely to have higher viscosities (although for polymers, these are solution viscosities).


Difference between adiabatic and nonadiabatic tunneling for molecule What is the difference between adiabatic and nonadiabatic tunneling for a molecule in a laser-field?
Does nonadiabatic tunneling necessarily relate to the Conical intersection or curve-crossing?

This is very far from a complete answer, especially w.r.t the part of your question regarding "laser fields". After doing some reading, nonadiabatic tunneling is conceptually the same as adiabatic tunneling except that one allows for coupling to another adiabatic state.

Adiabatic tunneling is the phenomenon where a particle described by a wavefunction is propagating on a potential and begins with less energy than the height of the barrier it approaches. One finds, however, that there is amplitude of the wavefunction on the other side of the barrier despite the particle having insufficient energy to cross the barrier.

Now, in physics and chemistry, nonadiabatic processes typically describe those processes where it is a bad approximation to consider only a single potential energy surface. That is, both surfaces may be adiabatic states, but these states are coupled in some manner. In general, there will always be some small coupling between states that are not too far apart in energy.

Nonadiabatic tunneling just describes the situation that one does not neglect this coupling between states. To a first approximation, nothing will change and one will find the same amplitudes as a function of position and time. If you consider the presence of the higher-energy adiabatic state as a perturbation of the Hamiltonian of the original particle, then the zeroth order picture will just be adiabatic tunneling. The next term will be something like the case where the particle is either reflected or transmitted over the barrier, but it does so by first "visiting" the higher-energy adiabatic state.

Now, with that description in mind, I think it is reasonable to say that nonadiabatic tunneling does not necessarily have to be involved with with conical intersections or curve crossing. Nonadiabatic tunneling is just regular tunneling which allows other states to couple with the original state. Now, of course, conical intersections and curve crossing could very well come into play in nonadiabatic tunneling because this would just be another form of coupling one could introduce.

As an aside, the description I give where the higher-energy state is a perturbation on the original system is probably not a good one in general as one could imagine very strong coupling between the potentials (as would be the case with a conical intersection).


Reaction of acetylene with hypochlorous acid In reaction of acetylene with hypochlorus acid chloro enol is intermediate, which further reacts with hypochlorous acid to form dichloro  ketone. My question is that why chloro enol does not tautomerise to form ketone. And if the chloro enlo further reacts to form dichloro ketone, then this must be the minor product. And mono chloro ketone should be major product . But thisdoes not happen . Why ?

The enol probably does tautomerize but it doesn't last. Some of the enol remains in enol form at equilibrium, and that reacts further with the hypochlorous acid.  This causes some of the keto form to revert to the enol, which then gets consumed again by the hypochlorous acid, and so on.  The selective reaction of the acid only with the enol ultimately pulls all of the keto form back to the enol and then the final, fully reacted product you see.


Regioselectivity in electrophilic substitution of pyrrole In the electrophilic aromatic substitution (EAS) of pyrrole, why is the alpha position preferred over the beta position so much?

The following figure shows the resonance structures we can draw to describe the intermediates produced by electrophilic at the 2- ($\alpha$) or 3- ($\beta$) positions in pyrrole.

image source

You can see that the intermediate produced by attack at the $\alpha$-position can be described by 3 resonance structures.  Whereas attack at the $\beta$-position produces an intermediate that can be described by only 2 resonance structures.

Therefore, the intermediate formed by $\alpha$ attack is more stable and the activation energy leading to it will be reduced.  Consequently, electrophilic attack at the $\alpha$ position in pyrrole is preferred.


When non-polar particles interact through dispersion forces, how fast does a temporary dipole induce a dipole in the neighboring molecule? The current question is inspired by this existing question: At what frequency does a non-polar molecule acquire a dipole able to participate in London Dispersion forces?. In the comments and answers, it says that there is always a dipole (but the time-average of these dipoles is zero), so it does not make sense to ask how long the dipoles last, or how frequently they appear.

I am wondering how long it takes the neighboring molecule to adjust its electron density to the temporary dipole. Also, I wonder if this is the correct picture: that one molecule has the temporary dipole and the other responds with an induced dipole. Or is it more accurate to say that the neighboring atoms find some common groove of swaying their charge distribution.

The upper limit are typically optical light frequencies. That is where the very last plateau is reached in dielectric spectroscopy. The electron shell becomes purely elastic.

I wrote a bit more about DES in my answer here: Difference between relaxation and resonance leading to an absorption spectral feature?


How many electrons are in the d orbitals of NiCl2·6H2O? On Wikipedia, it says that

  $\ce{NiCl2·6H2O}$ consists of separated trans-$\ce{[NiCl2(H2O)4]}$ molecules
  linked more weakly to adjacent water molecules. Only four of the six
  water molecules in the formula are bound to the nickel, and the
  remaining two are water of crystallisation.[4] Cobalt(II) chloride
  hexahydrate has a similar structure.

So nickel chloride hexahydrate has an octahedral symmetry and the ligands are two chloride ions and 4 water molecules? Or is $\ce{NiCl2}$ a solid in water, but separates into $\ce{Ni^2+}$ and $\ce{Cl-}$ ions in water since they are ionic?

Does this also indicate that there are 8 electrons in the $d$ orbitals and that it will form a high-spin complex (I know that the low-spin orbitals look the same though) so that there are two unpaired electrons? I think I'm thrown off by the chloride ions because my textbook only explains complex ions like $\ce{[CoCl4]^4-}$ or $\ce{Fe(CN)6}$ which are easier to understand for me.

The hexahydrate has octahedral symmetry, and it is very probable that this is the case for the solid and dissolved state of the complex.

The compound will not separate into the ions, because the solvation of three ions vs the solvation of one neutral complex is higher in energy, especially considering the fact that opposite charges attract. So in solution state it is a $\ce{[NiCl2(OH2)4]}$ complex with octahedral symmetry.

The $d$-electron count is 8, as it starts with 10 and has a formal oxidation state of II. This is always the case for $\ce{Ni^{II}}$ in any medium but vacuum.

Whether it's a high-spin or low-spin complex depends on the splitting energy of the $e_\text{g}$ and $t_\text{2g}$ molecular orbitals that arise from the $d$ orbitals in combination with the ligand group orbitals, colloquially known as $\Delta_\text{o}$ or $\Delta_\text{oct}$. Have a look at this Wikipedia article on crystal field theory for more details.

In short, there are several factors that influence the magnitude of the splitting:

Oxidation state of central atom
Geometry
Ligand types

The octahedral geometry usually gives good energy splitting.$^\text{[citation needed]}$ In conjunction with a moderate oxidation state compared to Ni(0), and four mid-range splitting ligands (water), it is a safe bet to say that we will have a low-spin octahedral $d^8$ complex: Three pairs of electrons in the $t_\text{2g}$ orbitals and two unpaired electrons in the $e_\text{g}$ orbitals, that is.

Note at this point that the terms high-spin and low-spin are superfluous. The electrons in the $e_\text{g}$ orbitals will always be in a triplet configuration according to Hund's Rule. So it will always be in a "high-spin" configuration from an experimental standpoint.


Finding the value of stoichiometric coefficient x $\ce{2A_{(s)} + xB_{(g)} <=> 3C_{(g)}}$

$\mathrm{K_p}=19.3, \mathrm{K_c}=0.45, \mathrm T=523\pu K$

I work $\Delta n$ to be $1$, So would $x$ be $1$ or $2$? I'm not sure if the coefficient is part of $\Delta n$.

The relationship between $K_p$ and $K_c$ is

$$K_p=K_c(RT)^{\Delta n}$$

where $\Delta n$ is the change in the number of moles of gas molecules over the course of the reaction. The only way you could know this change is from the stoichiometric coefficients. The $\Delta n$ term derives from the laws of mass action in which all partial pressures and/or concentrations are raised to the stoichiometric coefficients.

Consider:

$$\ce{2NO(g) + O2(g) <=> 2NO2(g)}$$
$$K_c = \dfrac{[\ce{NO2}]^2}{[\ce{NO}]^2[\ce{O2}]}$$
$$K_p = \dfrac{P^2_{\ce{NO2}}}{P^2_{\ce{NO}}P_{\ce{O2}}}$$
$$P_x = \dfrac{n_x}{V_x}RT=[\ce{X}]RT$$
$$K_p = \dfrac{([\ce{NO2}]RT)^2}{([\ce{NO}]RT)^2([\ce{O2}]RT)}$$
$$K_p = \dfrac{[\ce{NO2}]^2}{[\ce{NO}]^2[\ce{O2}]}\dfrac{(RT)^2}{(RT)^3}=K_c(RT)^{-1}$$
$$\Delta n = -1 =2-(2+1)$$


Real space term for PPPM vs PME I was under the impression that differences between the various long range electrostatic algorithms, ewald, PME, PPPM, to name a few, lied solely in their treatment of the reciprocal space term. Specifically, I thought all methods looped over a particles neighbor and summed q(i)*q(j)erfc(alphar)/r for the real space calculation. However, this is my impression, which I have yet to validate with others. I became confused when I recently read "Ewald summation techniques in perspective: a survey" whose equation 22 seems to have suggested a different way to calculate the short range potential. Could anyone else chime in here, and either tell me where I am wrong, or I am correct?

Thanks!

There are a number of ways of looking at Ewald's summation, but one is that instead of trying to solve the original problem where the potential is due to a set of point charges you, instead, solve the system for a set of smeared charges and add a correction. Now there are limitations on what are good ways to do the smearing, but there are still quite a lot of possible choices. The most common is to replace the original delta function charge with a Gaussian, the width of which is determined by alpha in your question. This leads to the error function and exponentials you are used to. However it's not the only possible choice. In equation 22 in the paper you refer to a different way of smearing the charges has been chosen, a linear ramp centred on the particle. This is described in equation 21. As the smearing is different the individual terms in the Ewald summation have different forms when compared to Gaussian smearing, but it's no less valid a way of proceeding.

If you want to learn more about this look into Fourier Transforms and the Convolution theorem. A bit of reading on that and the Ewald Summation in all its forms becomes pretty straightforward.


Why does ethanol and water mixture show positive deviation from Raoult's Law? So, my teacher said, "For a solution to show positive deviation from Raoult's Law, it must have a compound which is lacking hydrogen bond and a compound which has hydrogen bond."

This is because the compound not having a hydrogen bond breaks the hydrogen bond of the other compound.

Now, why does ethanol and water mixture show positive deviation? Both of them have hydrogen bonds.

In more generality, Raoult's law is stated as $$P_i = x_iP_\text{vap}$$ and read "the partial vapor pressure of species $i$ is its (pure) vapor pressure multiplied by its mole fraction in the liquid phase." It assumes an ideal-gas gas phase and an ideal-solution liquid phase; an ideal solution is one in which all intermolecular attractions are of equal strength. More explicitly, the $a{-}b$ interactions are of equal strength to the $a{-}a$ and $b{-}b$ interactions, where $a$ and $b$ represent different chemical species in solution.

A positive deviation from Raoult's law occurs when the partial vapor pressure is greater than would be expected from Raoult's law. From a microscopic perspective, this implies that the $a{-}b$ interactions are weaker than the $a{-}a$ and $b{-}b$ interactions.

You should see now that your teacher's statement is a general rule-of-thumb following from the concepts outlined above, but does not necessarily hold true for all situations. Here, water forms hydrogen bonds of greater strength and magnitude than does methanol, and one expects that a water-methanol mixture will therefore possess weaker intermolecular interactions than those in a solution of pure water.


Form the oxides of the element Z=39 Its electronic configuration is $4d^15s^2$ so in normal state it has valence=1 and in energetic state, valence=2 (because our teacher said we count only the electrons in the outer shell firstly) and valence =3. First question: is it true what my teacher said? It was for Z=22 example, where she said it has the valence 2 and  4). Second question when we form the oxide we should use only the valence in the energetic states and not the one in the normal state?

I am not sure that I understand your question, but in terms of establishing the valency in this case the key point is that in d block elements the energy level of the outer s electrons are very similar to that of the d electrons in the shell below, so the valency in this case will be 3.


Finding equilibrium temperature when two things are mixed 10 g of ice at −10 °C is added to 10 g of water at 85 °C. What is the final temperature and amount of ice left in the system? (System is kept inside an ideal insulator)

This question may seem quite easy for many of the readers, but please help me with such questions as I find them tricky. I understand that this is a question which involves the application of the $Q=mc\,\delta T$ formula and that we have to equate the $Q$'s of both ice and water in order to obtain the answer.
Another thing which is troubling me is that the answers to most of such questions that I have encountered is 0 °C, and this question is not an exception.
Please suggest me on how to go about such questions, and provide some conceptual clarity on the topic if possible.

In this type of question, you should bring all the matter into one state and calculate net heat.
In this solution, we will consider heat released as positive heat, absorbed as negative, and the state to which we will bring all the matter be water at 0 °C.

Here we were left with 20 g of water at 0 °C and zero heat if heat was some positive quantity we would use the heat to raise 20 g water to a specific temperature.
By equation $20\times T\times1=\text{heat left}$
$T$ is temperature to which 20 g water is raised
If heat was negative quantity, we would convert part of water at 0 °C to ice at 0 °C by equation
$M\times80=\text{heat left}$
$M$ mass of water which is converted to ice


Mysterious precipitate in reaction of copper sulfate and calcium hypochlorite So I can’t seem to figure this reaction out. I’m pretty new to chemistry and am still learning it in high school so please don’t think I’m stupid. I have reacted 2 solitons (copper sulfate) and ( calcium hypochloride / dry chlorine) and got a light blue  precipitate that resembles copper carbonate. This precipitate is what I cant seem to identify. I don’t think it’s copper chlorate because the precipate is too light blue. Any answers will be helpful thank you in advance.

The creation reaction for copper hypochlorite per Mellor:
$$
CuSO4+Ca(OCl)2 → CaSO4(s)+Cu(OCl)2
$$
However, copper hypochlorite is not stable and on standing breaks down. I believe Mellor is likely accurate on describing the reaction and products:

Reference: "A comprehensive treatise on inorganic and theoretical chemistry", Volume 2, by Joseph William Mellor, page 271. To quote:

"R. Chenevix notes the ready solubility of cupric oxide in chlorine water, and P. Grouvelle found that the soln. obtained by passing chlorine into water with cupric oxide in suspension possessed bleaching properties, and these were retained even after the soln. had been boiled for a quarter of an hour. A. J. Balard found that the distillation of P. Grouvelle's liquor furnished some hypochlorous acid and a green oxychloride, 3CuO.CuCl2.4H20, was formed in the retort. A. J. Balard prepared a soln. of cupric hypochlorite by dissolving cupric hydroxide in hypochlorous acid. It is also made by the action of cupric sulphate on calcium hypochlorite. A. J. Balard found that copper filings are partially dissolved by hypochlorous acid, the soln. after standing some time contains cupric chloride, and deposits a green pulverulent cupric oxychloride."

Link: https://books.google.com/books?pg=PA271&lpg=PA271&dq=zinc+hypochlorite&id=7XoGAQAAIAAJ&ots=R0byHnamKL#v=onepage&q=zinc%20hypochlorite&f=false

So, the cupric oxychloride product maybe mixed with CaSO4.

Note, Copper oxychloride, per Wikipedia, is now referred to as dicopper chloride trihydroxide with the formula Cu2(OH)3Cl (or, more informatively, I would write as Cu(OH)2Cu(OH)Cl).

[I have personally prepared dibasic magnesium hypochlorite on another forum by mixing MgSO4 with NaOCl (aq) and adding NaOH. A white precipitate of dibasic magnesium hypochlorite results (which I have collected and experimented with). It is also NOT Mg(OH)2. Link: https://www.sciencemadness.org/whisper/viewthread.php?tid=26741#pid303723 There is also the reputed compound 'copper carbonate' which actually is also always basic copper carbonate per Wikipedia https://en.wikipedia.org/wiki/Basic_copper_carbonate ].


Which carbocation is more stable, the ethyl- or 1-propyl-carbocation? $\ce{C2H5+ vs \ C3H7+}$
There are two conflicting trends here.

Inductive effect of ethyl will be higher than that of methyl so carbocation attached to ethyl (3 carbons in total) should be more stable.

On the other hand 3 hyperconjugation structures can be drawn for the carbocation attached to methyl (2 carbons in total) and only 2 hyperconjugation structure for the other.

Which effect wins? I searched a lot of books but didn't find anything.

The ethyl carbocation ($\ce{C2H5+}$) has a different structure than what you might have anticipated.  It is a non-classical ion with a bridging hydrogen.
For a description of just what a non-classical carbocation is see this answer.
The hallmark of a non-classical ion is 3 atoms with 2 electrons spread over them.  This is called a 3-center 2-electron bond (hypercoordinate bonding). In the case at hand the 3-atom 2-electron bond involves the 2 carbon atoms and the bridged hydrogen atom.

So there is no "methyl" group present in the ethyl carbocation that can inductively release electrons.  Additionally, the hyperconjugated resonance structure you mentioned has actually been replaced by a hydrogen that is partially bonded to the other carbon atom.

The n-propyl carbocation would likely have a similar non-classical structure - if it existed at all.  This carbocation has not been experimentally observed, it rearranges too rapidly to the much more stable 2-propyl carbocation.

So while we can't answer your question about whether inductive or hyperconjugative effects are more important in this series of compounds, we can assess which cation is more stable.

A variety of experiments suggest that the heat of formation of the non-classical ethyl cation is around 215 kcal/mol (reference, see pp. 68-72). This same reference points out that the heat of formation of the classical 1-propyl cation lies approximately 20 kcal/mol above that of the 2-propyl cation.  The heat of formation of the 2-propyl cation is around 193 kcal/mol.  Adding 20 kcal/mol to that value would place the classical 1-propyl cation around 213 kcal/mol. But, assuming the 1-propyl cation would also be a non-classical ion it would be lower in energy than the classical version, its heat of formation would therefore be less than 213 kcal/mol.

This analysis suggests that the 1-propyl cation would likely have a lower heat of formation than the ethyl cation.  In other words, the 1-propyl cation would likely be more stable than the ethyl cation.


How to calculate air tank capacity from tank displacement, tank fill (gauge) pressure, and specific type of gas? I am trying to calculate the amount of a specific compressed gas that will fit in an air tank of specific displacement, at a specific pressure. This is in reference to compressed air guns.

The inputs are:

Tank Displacement (units: cubic inches or cubic centimeters)
Tank Fill (Gauge) Pressure (units: psi or bar)
Type of Gas:

Dry Air
Carbon Dioxide
Nitrogen
Helium

*For this scenario, Temperature is assumed to be constant @ 20C / 68F / 293.15K.

The output is:

Tank Capacity (units: cubic feet or cubic meters)

In other words... an example of a possible question that could be asked pertaining to this would be: "How many cubic feet of helium will fit in a 122cc tank when filled at 3000psi?"

My question is: What amount of specific gas will fit in an air tank of specific displacement when filled at a specific pressure? (Please provide equation along with an explanation)

I am having trouble coming up with the proper equation to use with the given inputs. I have tried the Ideal Gas Law, Boyle's Law, and the Combined Gas Law. I have no way to verify if the answer I am getting is correct because I'm not sure if I'm using the correct equation.

Note: this is not homework. I am a PHP developer. This is functionality I am implementing for a client's website.

Thanks!

"Quick and Dirty" for gases other than carbon dioxide

You're trying to figure out what volume of gas as STP (standard temperature and pressure) would fill a 122 cc tank at 3000 psi. From ideal gas laws:

$PV = nRT$

but since nRT is constant

$P_1V_1 = P_2V_2$

so

$V_{14.7psi} = \dfrac{(3000\text{ psi})V_{3000psi}}{14.7\text{ psi}} = 204 \times V_{3000psi}$

It takes one tank of gas to fill the "empty" tank, so 203 more tanks of gas can be pumped into the tank full of gas at 14.7 psi to get to $3000 \text{ psi}$.

$ 203 \times 122 \text{ cubic centimeters} = 24.7 \text{ liters}$

1 liter is 0.0353147 cubic feet. So:

$24.7 * 0.0353147 = 0.87$ cubic feet of whatever gas.

I lifted the following figure from this webpage. In basically shows that for a given temperature and amount of gas (number of moles) then the product of pressure and volume, PV, ought to be a constant which isn't observed with "real" gases. The lines show the real behavior and the data points estimates from the Van der Waals equations. The gist here that for most normal gases at 3000 psi (200 atmospheres), you'll only be off by about 20%.

More exact solution using Van der Waals equation for gases other than carbon dioxide

Starting from the Van der Waals equation

    $(P + \dfrac{an^2}{V^2})(V - nb) = nRT$

Where:

P - pressure
V - volume
n - number of moles
T - absolute temperature
R - ideal gas constant = 0.08206    L atm/(K mol)
a - constant
b - constant

T must be an absolute temperature. Typically in science the Kelvin scale is used.

P is conveniently converted to atmospheres where 1 atmosphere = 14 psi.

a,b are looked up in tables of gas constants

In the following table the constant $a$ has units $L^2 \cdot \text{atm} \cdot mole^{-2}$ (assuming 1 bar = 0.987 atm )and $b$ has units $L \cdot mole^{-1}$

   Gas                a             b
   Nitrogen          1.352          0.0387
   Helium            0.0342         0.0238
   Dry Air(a)        1.33           0.0366
   Carbon dioxide    3.593          0.04267

(a) reference

Steps:

(1) To calculate the number of moles of gas at high pressure, you solve the following cubic equation for $n$. (Knowing $P, V, T, a,$ and $b$). At high pressure (hundreds of atmospheres), and well above the critical temperature there will only be one root for $n$ for all the gases listed except carbon dioxide.

    $(\dfrac{ab}{V^2})n^3 - (\dfrac{a}{v})n^2 + (Pb+RT)n -PV = 0 $

(2) To calculate the volume of gas at room temperature and pressure, you solve the following cubic equation for volume, V, using $n$ from step 1. (Knowing $P, n, T, a$, and $b$). At room temperatures and pressure there will only be one root for $V$ for all gases including carbon dioxide.

    $(P)V^3 -(nbP+nRT)V^2 +(an^2)V -abn^3 = 0$

Solution for carbon dioxide

At room temperature and pressure gaseous carbon dioxide gas would follow Van der Waals equation as noted in step 2 above. However with "moderate" pressure carbon dioxide liquefies at room temperature. Thus a carbon dioxide tank is filled to about 85% of its volume capacity with liquid carbon dioxide.

The overall idea is that the liquid carbon dioxide expands to a gas in an airgun. This creates a relatively stable pressure as the airgun is shot. (You do have to allow the carbon dioxide reservoir to warm back to ambient temperature since vaporizing the liquid to a gas cools the reservoir.)

There are two other considerations here. First above the critical point 304.18 K (31.03 °C, 89 °F) and 72.8 atmospheres, carbon dioxide exists as a supercritical fluid. Second a carbon dioxide tank is never filled completely full of liquid. There is always some gas headspace, hence the 85% fill. Carbon dioxide tanks have been documented to explode in a hot car on a summer day for example.

    (Image from Wikipedia)


How full is a gas cylinder? There are 7 different gas cylinders filled with:

nitrogen
oxygen
air
methane
hydrogen sulfide
carbon dioxide
sulfur dioxide

How do I determine how much of the respective gases is left?

Nitrogen and oxygen are in a gaseous state, so I reckon a glance at a pressure indicator will do. Air? The same. How about the last four cylinders which are filled with liquids? They have their vapor pressure, but I am not sure that those pressures can be used to accurately calculate the volume of respective liquids. Should the weight of the cylinder be measured?

Correct.

For the non-liquifying gases (room temperature is far above their critical temperature) knowing the pressure is enough, if you know the volume of your cylinders. (You need a better formula than $pV=nRT$ to calculate the amount of gas at those high pressures. https://en.wikipedia.org/wiki/Van_der_Waals_equation, virial expansion, ...)

Actually the van der Waals equation is not useful, because, although it gives a good  idea why the ideal gas law breaks, it gives very wrong numbers esp. for high pressures. BWR or BWRS are supposed to work well, at the cost of using eight or eleven constants.

For those gases which are partially liquified ($\ce{CO2}$ at $57\ \mathrm{bar}$, $\ce{H2S}$ at $12\ \mathrm{bar}$, $\ce{SO2}$ at $3.5\ \mathrm{bar}$), you need to know the weight (and tare weight) of the cylinders, of course.


Substitute solvent for diethyl ether? I want to run a reaction that needs diethyl ether as anti solvent in the last step (comes from articles).
But I can't use it due to its harmful effects on the body.
Is there any other solvent?

My reaction is about substitution of $\ce{-OH}$ of lignin with BiBB in THF and in other one in $\ce{EtOAc}$ as solvent.

It is strange that you are concerned about diethyl ether (which has been used as an anesthetic), but you are not concerned about THF. From the US Pharmacopoeia:

  Permitted Daily Exposure
  Tetrahydrofuran 7.2 mg/day
  Diethyl Ether 50 mg/day
  Ethyl Acetate 50 mg/day

If you dry the product thoroughly (above 120°C), you should not have any problems with residual solvents.


Rhodizonic Acid and Silver Citrate? I have used some of those lead check swabs from 3M. I used them on some ceramic tile and received strange results. I called the tile manufacturer who was insistent that lead had never been used in their operations. After talking with 3M they asked me to send a picture, which I did. After looking at the picture the 3M lab told me that they did not believe the test to be positive for lead but didn't explain the color I observed.

I realized I had cleaned the floors with a cleaner which contained silver citrate as a preservative. Would the silver ions have reacted with the rhodizonic acid? What color would this have produced? I observed a pinkish color (definitely not red. More pink/purple) in some areas I tested.

I can include a picture if it would help.

$\ce{Ag+}$ reacts with Sodium Rhodizonate to form a Black colored compound - but only in a neutral or weak-acidic environment.

$\ce{Pb^{2+}}$ should give you a Blue-Violet color (or Scarlet if we're dealing with a weak-acidic environment); therefore, I don't think you're dealing with a lead contamination.

source: http://pubs.acs.org/doi/abs/10.1021/i560110a034?journalCode=iecac0


How do you determine how many electrons occupy the anti bonding orbitals? For molecular orbital theory:
How do you determine how many electrons occupy the anti bonding orbitals?
Also, why does hydrogen gas have no anti bonding orbitals- if each atom's electrons form a molecular orbital, then their energy is reduced so where does the extra energy go ?

This is the MO diagram for Hydrogen. As you can see, for hydrogen atom, there is only 1 electron in the 1s orbital. Two hydrogens could come together to form a sigma bond, resulting in lowered energy. This doesn't mean that there is no sigma antiorbital, but rather due to the fact that hydrogen only contributes one electron each leading to a sum of two electrons, there are not enough electrons to fill the antiorbital, hence it is unfilled.

Now, imagine if it was helium instead. Helium has two 1s electrons, therefore, if two He atoms form a bond, 4 electrons has to be placed into the molecular orbitals. Since every orbital can hold a maximum of 2 electrons, two electrons would be in the sigma orbital, and the other two would be in the sigma antiorbital. An orbital and antiorbital would effectively cancel out, therefore, you see Helium exist in the monoatomic state.


If gases assume the volume of the container that they are in, then wouldn't a gas that isn't enclosed in a container have an infinite volume? But if the volume was infinite for gases not enclosed in a container, then how would the ideal gas law even function?

Thanks to anyone who can point out the piece of this puzzle I am not considering!

I give you a case. Suppose you opened an airtight jar containing an ideal gas. Suppose by some mysterious power, you are able to observe the molecules of the gas. What do you think you'll see on opening the jar?
Obviously you'll see the molecules dissipating in space(the correct word would be diffusing). Initially they suppose on opening occupied 100ml, then they occupied 200, then 300... Thus at every point of time the molecules have a finite volume, the value may be very large, but still its finite. Thus we say a gas "will occupy infinite volume" but only after "infinite time".


Why are most glow-in-the-dark things green? I've noticed that many glow-in-the-dark objects you can purchase in the store are green. Occasionally I would come across something that was orange or blue, but mostly it seems that green is the ubiquitous glow-in-the-dark color.

Does this have anything to do with common wavelengths of light emitted from the relaxing of electrons in some of the more common phosphorescent compounds? Are some phosphorescent compounds just more readily available or "safe" for use in toys?

The human eye is most sensitive in green wavelengths. Therefore, if one were to put the same amount of light energy into different wavelengths, the green portion of the spectrum would appear brightest, though it contained no more energy than the rest.

If a phosphor or glow-stick manufacturer is seeking the "greatest bang for the buck", i.e. the most light for the least amount of chemical, green is a good choice.

There is a variety of phosphor colors, such as these listed in Wikipedia:

Red: Yttrium oxide-sulfide activated with europium is a red phosphor used for CRT's.
Yellow: Zinc cadmium sulfide (Zn,Cd)S:Ag, emits yellow light.
Green: Zinc sulfide with copper, ZnS:Cu, emits green light, ~530 nm. It was widely used in CRT oscilloscopes and radar displays because of its long persistence and simple chemistry. It is known as phosphor P31.
Blue: Zinc sulfide with few ppm of silver, ZnS:Ag,emits blue light with peaking at 450 nm. It is known as phosphor P22B and, unlike ZnS:Cu, has a very short persistence, so is suitable for slit-photography CRT's.

For a more complete list of phosphors and their colors, see this useful PDF or Phosphor Technology.

Many stores carry a variety of inexpensive chemiluminescent products, such as glow-in-the-dark necklaces and bracelets, in colors ranging from red through violet.


Discrepancy between the apparent volume of the solution and the volume of the solute arising from the definition of solubility I have solubility $s$ given in $\pu{g/mL}.$ I understand this as mass of solute $m$ divided by volume of solvent $V.$

But how about the volume of the resulting solution? I read online I should pour the solute into a measuring cylinder/cup first, and then add solvent until the volume fits the calculated value $(V = m/s),$ but this seems to be at odds with the definition of solubility.

What is the correct way to understand solubility? I can't know the volume of a solution based on solubility.

I have solubility s given in g/mL. I understand this as mass of solute m divided by volume of solvent V.

Your confusion is arising from mixing the concept of solubility and concentration. Unfortunately, both are expressed the same way. You have to see the context. Most authors (like good scientists) try to write their experiment explicitly so that others can understand and repeat their experiment.

See the examples in this link: Solubility Concepts

Let us say, we find from the literature that the solubility of NaCl in water is 360 g / 1L. It means that if we take 1 L of pure water, we should be able to dissolve 360 g of salt in it. Note by adding 360 g of NaCl, the volume of water may not be exactly what 1 liter anymore. I cannot calculate concentration out of this information.

Now some authors may refer to concentration when they talk about solubility. For very dilute solutions it does not make a difference. So it is always good to see the context.

  I read online I should pour the solute into a measuring cylinder/cup first, and then add solvent until the volume fits the calculated value (V=m/s), but this seems to be at odds with the definition of solubility.

Don't trust the webpages made by amateurs or pages from non-reliable sources. With time you will develop the sense which sources are more reliable. There is a saying that paper never refuses ink. It means one can write whatever on paper, and the paper will never say no, even if it is wrong. The same is true for the web. If we search the web for corona virus, you will find a lot of non-sense. Only reliable sources are to be trusted.


the longevity of amorphous metal I gathered that many of amorphous solids are made by quickly cooling the liquid material, thereby giving it no time to form any long-range order (crystallinity). So it makes a perfect sense that when high heat is applied to amorphous solid, it will regain its crystallinity, and will no longer be glassy.

But is this behavior non-continuous? That is, can sustained low thermal excitation give rise to the crystallinity in amorphous solids given enough time? Or is there a threshold temperature (or a level of thermal/mechanical excitation) below with no excitation can bring back the crystallinity?

The answer to your question is no in general. Because amorphous metal is usually an alloy rather than a pure metal. These alloys are made of atoms of largely different sizes, which results in low free volume and higher viscosity (compared with other metal and alloys) in molten state. The viscosity hinders the atoms movement to form a crystal.

So, the significantly different atomic radii of the components that constitute the amorphous metal (about 12%) favors the high packing density and low free volume. This leads to a complex crystal units with higher potential energy and lower chance of formation.

Accordingly, I think that even if we heat an amorphous metal above a threshold temperature  and let it cool slowly, we won't have a conventional crystalline structure. At best, we get regions of some order.


When an acid is added to water, why does the hydroxide ion concentration decrease? At equilibrium in pure water, we have
$$\ce{[H_3O+][OH-]} = 10^{-14}$$

Since $\ce{H3O+}$ and $\ce{OH-}$ ions are produced in pairs, we may conclude
$$\ce{[H_3O+]}=\ce{[OH-]} = 10^{-7}$$

So far so good. But shouldn't things change when we introduce a new substance into water ? I mean why does the first equation above hold no matter what ?

Also when I introduce $\ce{H2SO4}$ into the water, it doesn't just give a $\ce{H+}$ ion, it also gives $\ce{HSO4-}$ ion. Shouldn't these new negative ions change the behavior of water? Why does my textbook never talk about these new negative ions? Help appreciated. Thanks!

As long as you stay in the dilute regime, that is if water molecules have most of the time to interact just with each other, $K_w$ won't change. Add 20% of anything soluble (salt, acid, alcohol, acetone, ...) and things look different.


Shape Of Cetyl Alcohol Why is cetyl alcohol bent at the 9th position?

This is simply how the chemist chose to draw the molecule in that particular instance, and cetyl alcohol is formally regarded as a straight-chain molecule. Cetyl alcohol has all single bonds and no rings, so it is not structurally locked into any particular conformation or geometry, and every bond is free to rotate.

For very simple saturated molecules like this, where the conformation is not in any way fixed (under typical conditions), the chemist is free to represent the molecule as one of any of its possible conformers.

In reality, the most stable (i.e., lowest energy) conformer should statistically predominate, and this is the one with all groups staggered and the largest groups antiperiplanar (q.v., the IUPAC Gold Book entry on torsion angles for details), the conventional explanation being that this minimizes steric hindrance/strain. The skeletal formula corresponding to this minimum-energy conformer is the simple straight-chain representation (seen here, for example).


Which one of the following is NOT a haloalkane? Which one of the following is NOT a haloalkane?
A) Freon
B) Teflon
C) Iodoform
D) Vinyl chloride

Every compound here seems to me as a haloalkane: Freon, Teflon, iodoform, and vinyl chloride.
My text says the answer is vinyl chloride, but how?

I have went through each option:
A) Freon is $\ce{CCl2F2},$ which is derived from methane
B) Teflon is a polymer of $\ce{-F2C-CF2 -}$ (or) $\ce{(C2F4)_n},$ derivative of 1,2-difloroethene. Here, carbon is saturated, which means no double or triple bond is present.
C) Iodoform is $\ce{CHI3},$ which again is a derivative of methane.
D) Vinyl chloride is unsaturated thus it does not classify as an haloalkane.


pH Testing Without Cabbage/Disposal of Ferric Chloride I have a year old bottle of ferric chloride in my basement for PCB etching that I realized that I was never going to use. What is the best way to make sure that it is neutralized without pH paper or cabbage for testing?

You can add powdered calcium carbonate (limestone) until the liquid component was clear upon settling or until the addition of calcium carbonate did not produce additional bubbling. Calcium carbonate is ideal because it is cheap, neutralizes acids to a non-corrosive pH with no need to measure, and is not itself caustic.


Determining enthalpy and internal energy of formation for potassium superoxide Given $2.14\ \mathrm g$ of $\ce{K(s)}$, determine $\Delta H^\circ_\mathrm{f,m}$ and $\Delta U^\circ_\mathrm{f,m}$ for $\ce{K2O}$.

  We know:

  The calorimeter's constant: $1849\ \mathrm{J\cdot K^{-1}}$
  The mass of water inside it: $1450\ \mathrm g$
  The change in temperature: $2.62\ \mathrm K$
  The end product is $\ce{K2O}$

The process should be: determining mol of $\ce{K(s)}$, which is $0.054\ \mathrm{mol}$. Then we obtain the amount of energy absorbed by the calorimeter/water. Here are the issues. I don't know which one changed its temperature by $2.62\ \mathrm K$ or if the calorimeter's constant already considers the water. Either way, the amount of energy released by each mol of potassium is extravagant; according to the result it is $322\ \mathrm{kJ}$ or $96.8\ \mathrm{kJ}$.

Continuing with this reasoning, the $\Delta U^\circ_\mathrm{f,m}$ should be $755\ \mathrm{kJ}$ or $193\ \mathrm{kJ}$, right? What do we need to obtain $\Delta H^\circ_\mathrm{f,m}$?

Now, when we say that the temperature rose by $\pu{2.62K}$, this is the temperature of the calorimeter and the water together. We are assuming that they have been given time to reach thermal equilibrium (and generally water and copper calorimeters take relatively less time to reach equilibrium, it's pretty safe to neglect the time). Calorimeters can be filled with varied amounts of water, so the calorimeter's constant is for just the calorimeter, not the water in it.

Note that if we were to assume that the specific heat of potassium/oxide were not negligible, we would have to know the specific heats of the two, as well as the temperature at the start of the reaction and the temperature at which they want $\Delta H^o$ to be calculated at. "Standard state" doesn't fix temperature, though when you say $\Delta H^o$, you generally mean $\Delta H^o_{\pu{298K}}$. Using these, you can form a reaction pathand calculate the answer (I can elaborate on this if you want, it's pretty interesting)

With these assumptions, one gets the heat released to the order of 10 kJ, and the energy is to the order of 100 kJ/mol. Not too extravagant, remember that a joule is a rather tiny unit of energy. And formation enthalpies/energies are generally to the order of 100 or 1000 kJ.

Finding the enthalpy change now is pretty easy. Since $\Delta H=\Delta U+ \Delta(PV)=\Delta U+\Delta(nRT)$. Using this and looking at the overall reaction, it shouldn't be too hard :)


Why distinguish between polarity and hydrophobicity? I have yet to see an example of a molecule that's both polar and hydrophobic or that's both non-polar an hydrophilic. If no such molecule exists, why use the terms "hydrophilic" and "hydrophobic" at all if we can just refer to them as being polar or non-polar? If there are such molecules, what are some examples and why do they exhibit non-polarity and hydrophilicty, or polarity and hydrophobicity?

Hydrophilicity/hydrophobicity/lipophilicity/lipophobicity and polarity/non-polarity are terms that overlap to a great extent in simple substances, but which are not strictly equal. The former terms can have a much more involved explanation depending on several factors, even if polarity is usually the largest by far.

Philicity and phobicity are determined by contact angles between substances. It is possible to vary how a same substance gets wetted by exploiting surface patterning, which fiddles with the surface tension between the substances in contact.

Another interesting example I can think of is that some large perfluorocarbons are non-polar, but they are also lipophobic. It is possible to mix water, oil and a perfluorinated solvent, shake, and then watch as three clearly distinct phases separate.


For a benzene shown in a skeletal structure, what does a substituent to the center of the ring mean? I have seen diagrams with both the substituents attached to the benzene ring by the edges and other occasions directly to the delocalized ring. What does signify?

Here the bond connecting the substituent with the center of the ring suggests that the substituent may be located on any of the three positions (ortho-, meta-, para-). From Graphical representation standards for chemical structure diagrams (IUPAC Recommendations 2008), section GR-9.1 Small substituents [1, p. 393] (emphasis mine):

  Parent structures with a substituent that is attached at one of several ring atoms may be depicted by drawing that substituent attached to a bond that extends into the center of a ring. In printed works, it is assumed that the substituent may then be attached to any of the ring atoms as is allowed by normal bonding rules. When working in electronic formats, it may be possible to specify that the substituent
  can be attached to certain of the ring atoms and not to others, depending on the capabilities of the software being used. If more than one substituent may be independently connected to various ring atoms, the bonds attached to those substituents should both extend toward the center of the ring, but should not
  themselves be connected to each other.

References

Brecher, J. Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008). Pure and Applied Chemistry 2009, 80 (2), 277–410. https://doi.org/10.1351/pac200880020277. (Free Access)


Why do most carboxylic acids have high pKa (~5) in spite of having a conjugate base ion that is stabilized by resonance? This is from my textbook:

  Carboxylic acids owe their acidity ($\mathrm pK_\mathrm a$ of about $5$) to the resonance-stabilized carboxylate anions formed by deprotonation.

Why are they such weak acids? If the anions are resonance-stabilized, I would expect them to be much stronger acids.

Stated another way: this screenshot of the textbook page shows a representative protonation/deprotonation equilibrium for a carboxylic acid:

Why is the reaction favoured in the reverse direction if the carboxylate ion is more stable than the reactant?

Everything is relative in acid-base reactions; the equilibrium will favor the side with the weaker acid. A carboxylic acid with $\ce{pKa = 5}$ will not protonate $\ce{H2O}$ significantly because $\ce{pKa(H3O+) = -0.7}$, but it will protonate something more basic like $\ce{NH3}$ because $\ce{pKa(NH4+) = 9.24}$.

Note: the pKa of $\ce{H3O+}$ is debated, see What is the pKa of the hydronium, or oxonium, ion (H3O+)? for more discussion.


Does fluorescein decompose from long term exposure to light? I've had about one liter of fluorescein solution setting on the shelf for at least several years and one day noticed that it suddenly lost its characteristic green color. It's now a light yellowish-green solution that seems to have lost the property of fluorescence altogether.

The container is a reagent bottle with a leak-proof cap so I doubt that the solution reacted with anything in the atmosphere other than maybe 100 mL of air that was in the bottle to begin with.

The bottle was exposed to direct and indirect sunlight on a daily basis.

Does it have to do with the decomposition of the fluorescein, and if so, what are the expected decomposition products?

Yes, Fluorescein decomposes due to exposure to light.

See The Conservative Behavior of Fluorescein for quantitative data.


Strongest chloride complex as predicted by HSAB theory Why won't $\ce{Ni^2+}$ form a complex with $\ce{Cl-}$, while the ions $\ce{Cu^2+}$ and   $\ce{Co^2+}$ form the complexes $\ce{[CuCl4]^2-}$ and $\ce{[CoCl4]^2-}$?

According to the HSAB theory, $\ce{Cl-}$ is considered a hard/borderline base. So I figured the strongest complex would be the one in which the center ion is the hardest acid. The $\ce{Co^2+}$ ion is the largest and the least electronegative. So it should be the softest acid (although they are all considered borderline). It forms a complex with $\ce{Cl-}$. $\ce{Ni^2+}$ should be the hardest acid, and yet it does not form a complex with $\ce{Cl-}$ at all. Why is that?

References:

http://www2.chemistry.msu.edu/courses/CEM812/HSAB_Theory.pdf
R. Pearson, Chemical hardness and density functional theory, J. Chem. Sci., Vol. 117, No. 5, September 2005, pp. 369–377

The link to your handout explains why $\ce{Fe^{3+}}$ (and it is $\ce{Fe^{3+}}$ in $\ce{FeCl4^{-}}$) is a harder acid than $\ce{Co^{3+}}$ or $\ce{Cu^{2+}}$ or $\ce{Ni^{2+}}$.

  Hard acids (in context, HA) are characterized by (s,f blocks, left side of d block in higher OS's)

Iron is the furthest to the left of the four metals you mention. Thus, it has fewer electrons in its $d$ shell to balance the positive charge:

$\ce{Fe^{3+}}$ is $\ce{[Ar]}3d^5$
$\ce{Co^{3+}}$ is $\ce{[Ar]}3d^6$
$\ce{Ni^{2+}}$ is $\ce{[Ar]}3d^8$
$\ce{Cu^{2+}}$ is $\ce{[Ar]}3d^9$

and

  Low electronegativity (χ) of the acidic atom. A value in the range 0.7-1.6 is typical of hard acids;

Iron has the lowest electronegativity of the four metals:

$\ce{Fe}$ 1.83
$\ce{Co}$ 1.88
$\ce{Ni}$ 1.91
$\ce{Cu}$ 1.90

Iron is the hardest Lewis acid of the metal cations you list.

As for $\ce{Ni^{2+}}$, it does form a complex $\ce{NiCl4^{2-}}$ with $\ce{Cl-}$. Here is a Google search for it.


Can hydroxides ionize alcohols? For example, can $\ce{NaOH}$ ionize ethanol into ethoxide?

My textbook (Organic Chemistry, 6th edition, by M. Loudon and J. Parise, Roberts & Company 2016) suggests yes — it uses it in its mechanism for the intramolecular substitution of 4-bromo-1-butanol in the presence of $\ce{NaOH}$.

But I was taught that $\ce{^-OH}$ could only ionize thiols, not alcohols. After all, if simple $\ce{NaOH}$ can create alkoxides, why would chemists bother to use $\ce{NaH}$?

Yes, and no.

Acid and base chemistry is a neat balance between equilibria. As you well know, water dissociates readily by itself. A tendency that can be observed in any compound with $\ce{-OH}$ groups, yet not to a similar extend.
In other words, a hydroxyl group tends to be slightly acidic. Given a base, i.e. sodium hydroxide in this particular case, the following equilibrium will develop:
$$\ce{EtOH + NaOH <<=> EtO- + Na+ + H2O}$$

While the equilibrium will mostly be on the left side, depending on what other substrates are available, a tiny amount of alkoxide might already be enough to drive the reaction forward. In this case it is important to look at what actually drives the reaction.

In the example of 4-bromo-1-butanol that your book uses I am assuming there is a cyclisation happening.
$$\ce{BrH2C(CH2)2CH2OH + NaOH <=>> THF + H2O + NaBr}$$
Unfortunately I cannot quickly find a reference, but sodium bromide is not very soluble in diethylether, so I would assume it's not very soluble in THF either. The driving force here would be having a well enough leaving group and the continuous removal of product, basically making it a one-way reaction.

In some cases you need a higher concentration of alkoxide, and that's where hydrides come into play.
$$\ce{EtOH + NaH <=>> EtO^-\bond{~}Na+ + H2 ^}$$

In this case dihydrogen, as it is a gas, will be removed from the reaction mixture, basically ensuring full conversion to the alkoxide.


Classical reason for good heat conductance in diamond? My teacher in my physics course attributed this effect to phonons, more here, but I am skeptical about this argument, it feels like he is overlooking the whole question -- what about with Silica that has similar structure to diamond with covalent bonds?

My friend attributed this to the small mass of atoms and large Young constant, considering the system to be made of springs -- $v_{sound}=\sqrt{\frac{Y}{\rho}}$. He suggested to think of heat as sound to use the formula.

Silica has less strong bonds to diamond so perhaps this is a reason why diamondis a good conductor of hear and silica is not. Air is good insulator so does it mean that carbon atoms are so closely packed and silicon atoms not? Is this the reason for the difference in conductance?

I think that there is no really good classical reason.  Heat conductivity in a solid depends on bonding in that solid and bonding is inherently a quantum mechanical phenomenon.

For instance, recall that graphite is a good conductor of heat (and electricity) in two dimensions and a fairly good insulator (heatwise and electrically) in the third.  This is due to the layered structure in graphite and that is due to the bonding in graphite.


Formation of zwitterion from amino acids Why it is easier for the carbonyl group to lose a proton to become negatively charged and the amino group to accept a proton to become positively charged?

I know it has to do with the very polar bond due to the oxygen. And the $\ce{H+}$ ions attracted to the lone pair of electrons on $\ce{N}$ in the amino group.

But I don't understand why would an electronegative atom such as oxygen wants to lose a proton.

I was told that they will attract more electrons towards towards them. But that does not help in my understanding of why protons will leave.

The reason amino acids exist largely in their zwitterionic form at biological $\mathrm{pH}$ (usually around 7) is due to the $\mathrm{p}K_\mathrm{a}$ of the constituent groups. It's not that the oxygen 'wants' to lose a proton, but more that at that $\mathrm{pH}$ the equilibrium lies towards the deprotonated state (things are rarely 100% protonated/deprotonated).

In general, carboxylic acids easily lose their proton to form carboxylates (the anionic charge is delocalised over the three atoms making the proton loss 'favourable'), and amines act as bases, to pick up a proton.

At a $\mathrm{pH}$ of 7, the medium is basic enough to cause the carboxylic acid to be deprotonated, and the amine to be protonated. If you look at the diagram below, you can see that increasing or decreasing the pecan shift the equilibrium towards fully protonated (at low $\mathrm{pH}$) or fully deprotonated (at high $\mathrm{pH}$) amino acid.

Ionisation state of amino acids over 0-14 $\mathrm{pH}$ range, taken from Biochemistry, Berg (7 ed.)


Why is Technetium unstable? Technetium is probably the most awkward element of the periodic table for me. It seems to me that Technetium is an exception, in the sense that it doesn't have any stable isotopes, despite having a relatively low atomic number when compared to other unstable elements.

So I thought, maybe there is something special about it. Some exceptional property that causes it to be so unstable with such a low atomic number.

Please tell me. What is it that makes Technetium so unstable?

If you google "why is technetium unstable" you'll see that this question has been around for a while.  Over on physics.stackexchange, it is discussed in Why is technetium unstable?.

I've read some of what others have posted on this topic and the short answer appears to be: because we have such a poor understanding of the atomic nucleus, there is no answer to this question at this time.  Again, the link above would be a good place to start.  One of the answers (3 upvotes) claims to be an answer.  The selected answer (10 upvotes) says there is no answer.


What is the directive influence of the tert-butyl group in electrophilic aromatic substitution? We know that hydrocarbon side-chains on a benzene ring influence electrophilic substitutions by directing the incoming electrophile to attack at ortho and para positions. The following hyperconjugations are shown to explain directive effect of toluene:

However, the situation was different when I tried to analyse for tert-butylbenzene. Tert-butylbenzene has no $\alpha$-hydrogens so the above mechanism cannot be extended to this case. Does the following mechanism work?

Does tert-butylbenzene have any directive influence at all? If we consider only inductive effects, then the incoming electrophile may attack at the meta or para positions (ortho being crowded).

What you've drawn in your bottom figure is an example of carbon-carbon hyperconjugation. The t-butylbenzene and the charge separated structure drawn to its right are resonance structures (so you should use a double-headed resonance arrow to connect them).  Additional resonance structures with the negative charge delocalized into the ortho and para positions of the benzene ring can also be drawn, as well as additional structures using the other methyl groups attached to the central t-butyl carbon.  Based on the carbon-carbon hyperconjugation argument we would expect that

the t-butyl group would be an ortho-para director and
that the t-butyl group would accelerate the rate of electrophilic aromatic substitution compared to benzene itself.

Generally, carbon-carbon hyperconjugation is a small effect (smaller than carbon-hydrogen hyperconjugation) and not that commonly invoked.  The problem that I have with using carbon-carbon hyperconjugation arguments in this case is that our resonance structures involve a high energy methyl cation.

Backing up to toluene for a moment, there are two arguments why the methyl group is an activating ortho-para director. You’ve already outlined the hyperconjugative argument where electron density is shifted from a $\ce{C-H}$ bond to the aromatic ring using pi orbitals. However, in addition to this pi transmission argument there is also electron transfer from the methyl carbon to the aromatic ring inductively through the sigma system. The methyl group is roughly $\ce{sp^3}$ hybridized while the aromatic carbon is roughly $\ce{sp^2}$ hybridized.  An $\ce{sp^2}$ orbital is generally lower in energy than an $\ce{sp^3}$ orbital because it contains more s-character.  Therefore electrons will generally prefer to flow through the sigma bonds from $\ce{sp^3}$ to $\ce{sp^2}$ carbons. So both inductive (sigma) and hyperconjugation (pi) arguments can be used to explain the electron donating ability of a methyl group.

Here is some data on the relative rates for the electrophilic nitration of several aromatic compounds:

benzene=1, toluene=24, t-butylbenzene=15.7 (reference, p. 1060)

We see that t-butylbenzene undergoes electrophilic aromatic nitration slightly faster than benzene and slightly slower than toluene - and the effect is very small.  Further, it is also reported (same reference) that nitration of t-butylbenzene produces ortho, meta and para products in a 12 : 8.5 : 79.5 ratio. Since the bulky t-butyl group will sterically block the ortho position to some extent, the fact that the para isomer predominates and there is more ortho than meta isomer is consistent with the t-butyl group acting as an activating ortho-para director.

Small effects are often difficult to explain, but I suspect that the carbon-carbon hyperconjugation effect (pi transmission) is a small contributor at best here.  Rather, the inductive sigma transmission effect (electron flow from the roughly $\ce{sp^3}$ central t-butyl carbon to the aromatic  $\ce{sp^2}$ carbon) is how the t-butyl group donates electron density and stabilizes the electrophilic aromatic substitution transition state.  In fact, we could even argue that since carbon is more electronegative than hydrogen, the 3 methyl groups surrounding the central t-butyl carbon pull some electron density away from it compared to the 3 hydrogens surrounding the methyl carbon.  This would make the t-butyl carbon slightly less electron releasing than a methyl group in accord with observed relative rates.


Kinetics: Order of Reaction I am confused as to how an order of a reactant (n or m) in the rate law can be negative. This means that increasing the concentration of a reactant would actually decrease its rate of disappearance. How does this make sense? Doesn't a higher concentration always allow for more collisions? Also, when the order is 0, this means that concentration has no effect on the rate of disappearance. How does this make sense chemically/at the atomic level?

In principle, there are no zeroth-order reaction. Experimentally, however, zeroth-order rate law occurs because of the reaction condition. For example, in catalytic reactions such as hydrogenation of an alkene using platinum as a catalyst, the hydrogenation occurs on the surface of platinum. There is a finite surface area of platinum, so at a some point, it doesn't matter at all how much more hydrogen is put in. This is because no reaction can occur if the hydrogen cannot bind to the platinum.

An example of inverse first order reaction can be found here.

Basically, you have a two-step reaction:
$$\ce{A + B <=> C}$$
$$\ce{C <=> D\ (slow)}$$
Increasing concentration of $\ce{C}$ would decrease the rate of reaction.


Why is the amu of chlorine-35 less than 35? My book says that a proton weighs 1.0073u, a neutron weighs 1.0087u, and an electron weighs 0.00055u.

Now, why is the mass of chlorine-35 equal to 34.969? Are there not 17 protons, 18 neutrons, and 17 electrons? I calculated it and it sums to around 35.29. Where did I go wrong?

You need to account for the energy released when nucleons and electrons come together and form a Cl-35 atom. Its called the Binding Energy.

This sort of equation can help to explain :

(Rest Mass Energy of Individual nucleons,electrons*) - (Various Binding Energies) = (Rest Mass Energy of Natural)

*edit


How to electron count bridging ligands? Although I am okay with rationalising bridging halides and $\mu$-2 CO I can't seem to figure out others, including $\mu$-2 hydride, CN- and NO.

When considering the electron count of bridging ligands, this is my rationale so far:

-Neutral bridging ligands contribute to the overall complex the number of electrons they would have when terminally bonded. These electrons are given in any proportion. This means that $\mu$-2 bridging CO gives 1 electron to a metal centre, for example.

-Ionic counting gives even numbers of electrons for ligands; I am unsure if this is still the case for bridging ligands - I suspect not?

Considering NO, I would therefore believe it to be a 2 electron donor to one metal and a 1 electron donor to the second metal.

Considering hydride, more specifically the dimer form of titanocene, I have been taught that each Ti centre is 16 electrons. I have been told, separately, that each hydride contributes half an electron in neutral counting, and 1 electron in ionic. This would lead to a 16 electron complex for "Titanocene" if there is a Ti-Ti bond, however this source: https://link.springer.com/content/pdf/10.1007%2Fs40828-015-0010-4.pdf seems to imply that it is in fact an 18 electron complex, with each individual hydride contributing more than its total electron count? Is this source correct (and my lecturer wrong) in how you would count this complex?

Considering CN-, this is similar to hydride I believe, and should be treated in the same way - is this correct?

In the covalent (L) method, $\mu$-hydrides should be treated as a normal hydride (1-electron ligand) for one metal. The $\ce{M-H}$ bond itself should be treated as a 2-electron ligand for the other method. This explains the two entries for hydride in the link you provided:

Cyano-ligands are isoelectronic to carbonyl ligands, so I'll leave that to you as an exercise to figure out.


Neutralize the smell of alcohol in liquid chalk i'm trying to mix my own liquid chalk, which is a simple mixture of chalk and a high percentage alcohol in a ~2 part chalk 1 part alcohol relation.
Unfortunately the smell is rather intensive and I would like to find a way to neutralize or cover it up.
Any suggestions?

I do not have a chemical backround.

Chemically, chalk is CaCO3. But blackboard chalk is mostly CaSO4.2H2O. If you want to mark something, the aim is to lay a coating of a certain (dry) thickness so as to be visible.

One way to reduce the odor is to reduce the amount of alcohol used by 1) making the paste richer in the chalk by a) just adding more chalk (but that might increase the viscosity too much), so b) grind the paste with a mortar and pestle (this may thin out the paste by causing a better variation in particle size and packing), or c) add some of whichever "chalk" you are not using (i.e., if you are using CaCO3, add some blackboard chalk and grind).

Or, 2) dilute the alcohol with a little water. You probably use alcohol to get a fast dry time, so may not be able to use much water. There are other solvents that have less odor (like heptane), but have other objections. Isopropanol has a higher boiling point (less odor, slower drying, but not much slower) than ethanol (but you are probably not using ethanol).

Or, 3) add a little TiO2. This will dramatically increase the visibility of thinner films, allowing application of much less paste and reducing the amount of alcohol exposure.

Or, 4) add a dispersant to reduce viscosity. Assuming you do not have access to many chemicals, I would suggest a little trisodium phosphate. Or a tiny amount of dishwashing detergent (be careful not to stir too vigorously so as to incorporate air bubbles); this may help wet particles or displace tiny air in pockets between particles. Then you can add more solid.

5) Chemical "masks" are available that claim to cover up obnoxious odors. In my experience they do not cover, but only add extra odor. But adding some delightful fragrance could distract from the alcohol odor to a more pleasing experience. Commercial additives I have used are coconut and cherry - both quite nice. You don't need much. Some food flavorings have enough odor to be useful here. Some people find acetone to be more pleasant than alcohol - a little acetone could change the odor to be more acceptable.

6) Slowing the evaporation by adding wax or high molecular weight alcohols (octadecanol) might reduce the odor while extending the dry time, if that is acceptable. Octadecanol is used to reduce the evaporation rate of water, and may work with alcohol; wax is often used with solvents.

Other than that, I can't think of anything.


Why do batteries specifically vent *hydrogen* in the event of abuse? When a battery is subject to overcharging or overdischarging (including attempts to recharge a primary cell), it may vent hydrogen.  It seems that every type of electrochemical battery I've read about or used specifically emits hydrogen under these circumstances.

Why is hydrogen emitted in all cases, and not some other gas or chemical? What role does hydrogen play in electrochemical batteries that other chemicals can't?

Lead storage batteries use an electrolyte of sulfuric acid and water. On overcharging you are electrolysing the water with emission of $\ce{H2}$ in one electrode and $\ce{O2}$ in the other.

Other batteries use molecules with H and Li. The idea is to use light elements so you can accumulate more charge for the weight.


How does the work that won the 2012 Sustainable Chemistry Award contribute to sustainable chemistry? I'm seeking a lay explanation for how the work of Dr Marc Taillefer that won the 2012 European Sustainable Chemistry Award, contributes to sustainable chemistry.

From the press release, Dr. Taillefer

  is being recognised for his seminal contribution to the field of
  homogeneously catalysed coupling reactions leading to C—C, C—N, C—O, C—P bonds. His team at
  the Institut Charles Gerhardt, ICG (Montpellier, France) is investigating for a decade the
  environmentally sustainable conversion of small molecules into more valuable substances
  catalysed by copper and iron molecular complexes. This renaissance of “Ullmann type arylations” is now often used at the academic or industrial level and avoids the use of more expensive catalysts based on palladium.

The objectives  of the award are (to quote from this press release) to:

  Recognise individuals or small research groups which make an outstanding contribution to sustainable development by applying green and sustainable chemistry.
  Promote innovation in chemistry and chemicals that will deliver clear improvements in the sustainable production and use of chemicals and chemical products.
  Demonstrate that chemistry and chemicals can play a central role in delivering society’s needs, while minimizing and solving environmental problems.

His work is about developping new catalysts based on copper and iron, to replace to traditional catalysts based on palladium. Copper and iron are both very common elements in nature, while palladium is considered a high supply risk (see the 2012 British Geological Survey risk list for details).

The new RSC Visual Elements Periodic Table can be used to check this kind of information.


The Evans auxiliary: rotational freedom of the benzyl group - stereoselectivity still guaranteed? I created the L-Phenylalanine based Evans auxiliary bearing a benzyl group with help of a molecular model set and got into some questions. The structural formula will do:

As I suppose a low rotational barrier, the phenyl group can be aligned off (a) and towards (b) the alpha position of the N-propionyl group. I suppose maximum sterical repulsion for attacking electrophiles for (b).

The phenylalanine and valine based auxiliaries (Bn or i-Pr) are the most commonly involved types of Evans auxiliaries. But why do I get good stereoselectivities for this Bn auxiliary? There are no corresponding carbon-carbon rotational freedoms in the valine species.

The comment left by CHM is very helpful.  You need to consider what this molecule actually looks like in 3D.  That may help your understanding greatly, but to be honest there is considerably more going on here than you would expect.

A couple pieces of information would be very helpful for clarifying this question.  The first is what type of a metal are you using for this synthesis step.  The second is what is the outcome.  Believe it or not, based on those two pieces of information the mechanisms for this reaction can be explained entirely differently.

Since the answer basically breaks down into two possibilities I will attempt to explain both.  One of the most helpful articles I came across in my search was entitled "Transition State Models for Probing Stereoinduction in Evans Chiral Auxiliary-Based Asymmetric Aldol Reactions" J. Am. Chem. Soc., 2010, 132 (35), pp 12319–12330.

In this article they go into depth about how the electronic and steric effects balance out to produce the almost complete stereocontrol throughout the course of this reaction.  The scheme below is what I find the most helpful:

Note how the selectivity is varied based on whether the auxiliary is chelating to the titanium or not.  This is why the metal lewis acid used in these reactions is important.

One final note not shown here is the rational behind the non chelating setup.  The opposing dipoles on "y" and aldehyde oxygen in the Zimmerman-Traxler model help to hold the six member ring in place while the aldol reaction occurs.

The scheme shown is incredibly useful for visualizing this transformation and I would like to compliment the authors for their efforts in this publication.

Hope this helps!


Negative retention for chloride in hydrolysate filtration I tried to fractionate proteins as a function of their size with an ultrafiltration membrane. I observed a higher concentration of chloride in my permeate than in my feed or retentate. The proteins do not pass through the membrane and the filtration has been done at pH = 7. I asked, how is it possible to have a higher concentration in the permeate than in the feed or retentate?

I found this explanation but I'm not sure about it :

  It may be explained by the Donnan effect. Negatively charged proteins are retained in the retentate. To compensate the positive charges from the cations that are on the other side of the membrane, they attract the chloride.

Is it a possible way to explain this issue?

A combination of osmosis and Gibbs-Donnan effect is the correct explanation.

With a semi-permeable membrane such as the one you are using, or a cell membrane, or in dialysis, water can move freely as can chloride and other small ions, however large molecules such as proteins cannot.

A simplified explanation is this: in osmosis, the tendency is for the osmotic pressure and charge to equalise on either side of the membrane. Proteins usually have a negative charge and exert a large osmotic pressure. Thus, to balance this out, small anions will diffuse to the opposite side to where the proteins are. Water will diffuse until the osmolality is equal on either side of the membrane unless there is a hydrostatic pressure forcing it to the other side. In the absence of hydrostatic pressure, you will end up with a higher concentration chloride solution on one side and a higher concentration protein solution on the other side.

Here is a diagram:

If you assume that the small particles you can see are chloride and the big particles are your proteins, you can see that the chloride concentration in the permeate is higher than the retentate and higher even than the feed.


Choosing LC-MS modifier - why only acids? During method selection our chromatographists have options like NH4, TFA, Formic Acid. What's the point of choosing only between acid? Why not use the strongest that guarantees ionization? The sample contains small organic molecules.

And if the analyte is an acid, don't we want to have a base as a modifier? Otherwise how can we ionize it using ESI?

Stanislav, that is a Nobel Prize winning question. In short, nobody knows nitty-gritty details of how the electrospray ionization imparts a charge to the analyte. When Fenn (reinvented) the electrospray phenomenon and used for ionizing large biomolecules he was shocked and gave an analogy that it is like making the elephants fly. He got a Nobel Prize as well and a lot of patent issues in the university. There are three electrospray ionization models, you can read on Wikipedia, https://en.wikipedia.org/wiki/Electrospray_ionization#Ionization_mechanism

The key point in electrospray is that you need an electrolyte (forget about acids or bases). All you need is a conducting liquid between the electrode to generate the electrospray. In that process, the ion is created. There are studies, I am sure, on the effect of pH on the efficiency of ionization. I can tell what to avoid in electrospray: TFA (because it forms ion pairs), hexanes, all other pure non-conducting organic solvents. So you you can get the point that ionization in the solution has little to do with ionization in the electrospray.

Now coming to your chromatography part, which is purely empirical. The reason for avoiding bases in mobile phases is that it they will eat (=dissolve) the silica columns. Silica is not pH stable above 8. Silica based HPLC columns can tolerate low pH like 2.5, and below that the bonded starts to peel off, e.g. if you use a C-18 column and use pH 1, it will become a bare silica in a few days.


Borane as a reducing agent I understand that adding borane to an alkene will produce a hydroborane. Addition of -OOH will form an alcohol.

My lecture notes state that addition of borane to:

A carboxylic acid will form an alcohol
An amide will form an amine
An aldehyde/ketone will form an alcohol
A nitrile will form an amine
An epoxide will form an alcohol
An alkene will form an alkane
An alkyne will form an alkene

I can't find the mechanisms for these anywhere. I'd appreciate seeing a few of them, namely carboxylic acid to alcohol, amide to amine, epoxide to alcohol and alkene to alkane. I should then be able to have a good crack at working out the rest.

Henry Rzepa gives a mechanism in which two borane molecules attach to the carboxyl group of a carboxylic acid.  The boron atoms are attached to the electronegative but electron-rich portions of the carboxyl group (meaning the oxygen atoms) and then transfer hydride ion moieties to the carbon while taking away one of the oxygens.  This exchange of hydrogen for oxygen reduces the carboxyl carbon.  Other carbonyl compounds may be assumed to react similarly.


Does MgCl2 sol coagulate gold sol faster than Fe(OH)3 sol? Here is a question from the former AIEEE 2005

The disperse phase in colloidal iron (III) hydroxide and colloidal gold is positively and negatively charged respectively. Which of the following statements is not correct?
(a) Mixing the sols has no effect
(b) Coagulation in both sols can be brought about by electrophoresis.
(c) Magnesium chloride solution coagulates the gold sol. more readily than the iron (III) hydroxide sol.
(d) Sodium sulphate solution causes coagulation in both sols.

It is a single answer correct, and it is obvious that (a) is false - hence is the answer. But the statement (c) also seems to be false (and therefore should also be an answer).
According to Hardy Schulze law,

Greater the valency of the flocculating ion, greater will be its coagulation power.

Doesn't that mean $\ce{Fe(OH)3}$ sol, having $\ce{Fe^{3+}}$ ions, will coagulate the gold sol more readily than $\ce{MgCl2}$ sol, which contains $\ce{Mg^{2+}}$ ions? Or is there any specific reason why the $\ce{MgCl2}$ sol coagulates the gold sol more readily?

Option (c) is actually correct. Magnesium chloride does coagulate gold sol faster than ferric hydroxide, because ferric hydroxide does not coagulate in the first place.

Wikipedia mentions that ferric hydroxide is insoluble at $\ce{pH}=7$, so there aren't any $\ce{Fe^{3+}}$ ions to coagulate the gold sol anyway.


Atomic orbitals: how is bond strength related to stability, energy, bond length? I have a lot of confusion regarding atomic orbitals and I don't know if what I am having in mind is even correct or not.

The bond dissociation energy of H-H is greater than that of Cl-Cl and as far as I can make out, it is because in H-H, there is an s-s overlap whereas in Cl-Cl, there is a p-p overlap. Because s-s is more compact, therefore more energy is required to break the bond.

Assuming what I wrote above to be correct, this means an s-s bond is stronger than p-p bond. But then, bond strength of p-p is more than that of s-s because of greater extent of overlap.

So there is definitely something/everything wrong in what I am thinking.

In short, how is the bond strength related to other things like stability, energy, bond length (bond length of s-s < s-p < p-p?).

Kindly help because I cannot find any of my queries anywhere else.

There are several factors that affect bond disassociation energies (bond enthalpies) but one quick correlation you can make is bond distance vs bond strength. The interatomic distance of the H2 bond is 74 pm vs 199 pm in Cl2. The H2 bond consists of 1S electrons that are very close to the nucleus, giving a short bond. The valence electrons in Cl2 are 3p (relatively far away). This leads to other properties such as polarizability.

In short your first assumption is incorrect, the trend you are seeing has nothing to do with s or p bonds, but on the bond lengths. Therefore comparing s-s bonds and p-p bonds is meaningless unless you take into account the atomic orbital number and therefore an estimate of bond length.

For your final questions, the stronger the bond, the more stable, the higher energy, and the shorter the bond.

For more info this UF link has good information:http://www.science.uwaterloo.ca/~cchieh/cact/c120/bondel.html


Kinetics of inversion of cane sugar Question 1
The formula for concentration of cane sugar as a function of time is given by
$$k = \left(\frac{2.303}{t}\right) \log\left[\frac{\alpha(0) - \alpha(\infty)}{\alpha(t) - \alpha(\infty)}\right]$$
where $\alpha(t)$ is the optical rotation of the solution at any time t. This was the formula given in my textbook, but I feel it is wrong.
The optical rotation in the formula is the rotation of the solution, which includes rotation due to external factors which is taken care of by the $\alpha(\infty)$ term but it also includes rotation due to glucose and fructose , whereas the formula would only be valid if $\alpha(t)$ denoted only the rotation of cane sugar + external factors.
The reasoning given in the book is that the rotation due to glucose and fructose included in the external factors, but that is just wrong. The fundamental assumption is that rotation due to external factors remains constant and hence gets cancelled out, but if you include glucose and fructose in external factors, then their concentration and hence rotation is changing with time. Also they do not form a racemic mixture, so their rotations do not get cancelled out.
Question 2
Another question given in my textbook was to find the optical rotation if the reaction was 50% complete. Given rotations of cane sugar = 34, glucose = 10 , fructose = -20
The solution was $$\left(\frac{17+5-10}{3/2}\right)$$
Notice the 3/2 which represent the total moles in the denominator. If this is correct then the formula becomes even more incorrect as optical rotation also depends on total number of moles.
What is the correct formula? I am also not able to understand the calculation for total optical rotation as given in second question. Why did they divide by 3/2 and how does the optical rotation depend on the number of moles?

It seems as though your issue with the formula arises due to the fact that there is a feeling of not knowing what is considered to be factors that are internal or if they are external factors. To prove the formula, the method would be as follows.
The formula for the given reaction would be:
$$\ce{C12H22O11 + H2O ->  C6H12O6 (glucose) +C6H12O6 (fructose)}$$
Drawing a modified version of a RICE table with optical rotation, and the case at t=$\infty$, we get:
\begin{array}{|l|c|} \hline
\text{R} &\ce{C12H22O11   &C6H12O6(glucose)  &C6H12O6 (fructose)} &\text{Optical Rotation} \\ \hline
\text{I} & n & 0& 0 & \alpha_{s}n+\mathrm E \\
\text{C} & -x &  x&  x & - \\
\text{E} & n-x &  x & x & \alpha_{s}(n-x)+ \alpha_{g}(x) + \alpha_{f}(x)+ \mathrm E \\
\infty & 0 & n & n & (\alpha_{g}+\alpha_{f})(n) +\mathrm E \\ \hline
\end{array}
Now, for a general first order reaction, the formula would be:
$$k=\dfrac{2.303}{t}\log \frac{n}{n-x}$$
Therefore, we need to find the value of $\dfrac{n}{n-x}$. Using the given data, we get:
$$\frac{\alpha(\mathrm I)-\alpha(\infty)}{\alpha(\mathrm t)-\alpha(\infty)}=\frac{n}{n-x}$$
Therefore the formula given is true, even considering an external factor. You consider glucose and fructose to be internal.
$$k=\dfrac{2.303}{t}\log \frac{\alpha(\mathrm I)-\alpha(\infty)}{\alpha(\mathrm t)-\alpha(\infty)}$$


What makes scorpions glow under UV light? What substance (or class of substances) is responsible for the neon blue-colored fluorescence we observe when we shine UV light on scorpions?

Do note, I want to know what substance makes them glow and not why they glow... the latter being more suited for the Bio.SE.

As mentioned in this comment, which links to this paper, and this comment, which links to this Wired article, there are at least two molecules:

$\beta$-carboline:

4-methyl-7-hydroxycoumarin:

$\tiny\text{Yes I poached those comments, they could have been answers.}$


Does autoionization occur in steam? It is well known that H2O spontaneously autoionizes to H+ and OH- when it is in liquid phase. I was wondering whether it does the same in gas phase, either to the same extent as in liquid phase or at all.

Yes, water is ionized to some extent in all phases. One study even extends to 1,000 K!

A concise discussion of the change in ionization with temperature at Chemguide also explains the variation of pH of neutral water with temperature. [The concept of pH 7 being neutral is based on Kw (in mol2 dm-6) of ~10-14 at ~300 K.]

In general, the higher the temperature, the more a compound is ionized.


Diffusion Phenomena : Why does diffusion occur? Why do things diffuse that is why do they go from a place of higher concentration to a place of lower concentration?

The short answer
They don't.
The long answer
Molecules are very fast. For example, a single molecule in the air around you can be as fast as 500 km/h. The reason that you do not feel them is because they are too small.
Also, you need to understand that there are very, very, very many molecules in a "small" volume.
So, until this point, you have many molecules which travel very fast.
Suppose a gas is divided to high and low concentration areas. The area with high concentration would have more molecules, and let's say 50% of them is travelling towards the area of low concentration (they all move randomly, and in a straight line unless collision happens).
Let's say, 50% of the gas molecules in the area of low concentration is also travelling to the area of high concentration.
Then, you would have a net movement of molecules from high to low concentration.
The keyword here is net. Some molecules do go from low to high concentration. However, they are outnumbered.
After they reach equilibrium, there is still gas molecules travelling between the two regions, just that they are equal in number.
Extra
Of course, because my explanation being not very formal and strict, there would be some errors in my explanation, but I hope that you understand it.


Why does the Gibbs Free Energy change determine spontaneity of a reaction? Here is a doubt that arose in my mind while studying the chapter Thermodynamics of the 11th standard. (It might be silly, but I would love it if you try to understand what I mean to ask before down voting. Also, I want the doubt cleared.)

We have been told that if the value of $\Delta G$ or Gibbs' free energy change is negative or zero the process is spontaneous or in equilibrium, respectively. If otherwise, it is non-spontaneous.
Why? What does $\Delta G$ exactly signify that if it is positive the process won't happen by itself. I mean, I know what $\Delta G$ is equal to, but why does spontaneity of the reaction depend on its sign?

By the second law of thermodynamics, your main goal is to ensure that $$\Delta S_{\mathrm{universe}} \geq 0$$

$$\Delta S_{\mathrm{universe}} = \Delta S_{\mathrm{sys}} + \Delta S_{\mathrm{external}}$$

$\Delta S_{\mathrm{sys}}$ is measurable on your system. $\Delta S_{\mathrm{external}}$ is a bit more challenging, since you're not going to count the microstates of the external system, so you compute indirectly by heat transfer to the surrounding. Thus,

$$\Delta S_{\mathrm{external}} = -\frac{\Delta H}{T}$$

Combine:

$$T\Delta S_{\mathrm{universe}} = T\Delta S_{\mathrm{sys}} - \Delta H$$

or

$$\Delta G = -T\Delta S_{\mathrm{universe}} = \Delta H -T\Delta S$$

And with the sign flip, you're looking for $\Delta G \leq 0$.

Importantly, I should point out that this is not a closed system in the sense that we assume constant-pressure and not constant-volume.

If you're looking at a process that is constant-volume, you need to replace $\Delta H$ (which has pressure-volume work compensation) with $\Delta U$ (which does not) and replace $\Delta G$ (Gibbs free energy, constant pressure) with $\Delta F$ (Helmholtz free energy, constant volume).