Alkane deprotonation by organolithiums Butyllithium isomers are all stored in alkane solvents. Theoretically, equilibrium should (given enough time) lead to formation of mostly n-hexyllithium in the case of  n-BuLi stored in (mostly) n-hexane, I believe. I understand that the kinetics of deprotonating an alkyl proton are extraordinarily slow, apparently even in the case of the much stronger t-BuLi in pentane (which isn't merely thermodynamically neutral; it should be thermodynamically favorable for the lithium to exchange with the proton of a less substituted carbon).

I was wondering, though, does anyone know of any instance where a significant portion (or any observable portion, really) of something like t-BuLi has been noted to have reacted with alkane solvents?

All searches I've attempted have given me completely unrelated results, and even attempting to sift through them, I have come away with no mention of the idea, even.

Also, would the addition of some kind of lithium-binding reagent such as TMEDA or 15-crown-5 increase or decrease the rate here (assuming no Brønsted reaction between BuLi and the ligand)?

I ask because BuLi will typically deprotonate a heteroatom before doing nucleophilic addition, but will do nucleophilic addition (to a carbonyl) before deprotonating most "softer" carbon acids, suggesting alkyllithiums are pretty hard bases. I'm assuming making a more "naked" carbanion would make a harder base (more charge-driven than "HOMO-LUMO overlap"-driven), and I'm wondering if that could make it slower to react with an alkyl proton (though, an unstabilized carbanion is a fairly hard nucleofuge, so I could see the reaction going faster, too).

I would bet activation energy for this sort of reaction is quite high, and since primary and tertiary carbons aren't hugely different in stability the sort of H for Li exchange reaction would be very slow. For nBuLi, the reaction would practically not happen, and for sBuLi and tBuLi intramolecular rearrangement (proton shift) is more likely than reaction with solvent, resulting in nBuLi and iBuLi respectively.

Agreed that elimination and precipitation of LiH will be faster than alkyl exchanges in these solution, though I've never checked an old bottle of tBuLi for any precipitate so this is entirely speculative.

Investigating this would be quite challenging, as loss of base power in practice usually happens from slow exposure to O2 and H2O,which would resolve Li2O and LiOH respectively, also basic precipitates.


Electrochemical characterization of benzenesulfonamide I'm trying to elucidate current peaks for benzenesulfonamide, using an acetic acid-sodium acetate 0.1M buffer at pH 5.5 as electrolyte. When doing cyclic voltammetry, I get a voltammogram similar to the platinum (Electroanalytical Chemistry and lnterfacial Electrochemistry, 43 (1973) 9-36), because I use platinum as a working electrode. So, I can not see any current peak related to benzenesulfonamide, and I have not found any article related to this. I've been thinking about using a glassy carbon electrode instead of platinum.

I'd change the solvent to acetonitrile (don't forget the quarternary ammonium salt) and give a glassy carbon electrode a try.

It should be possible to observe the (irreversible) reduction of the sulfonamide.

And once you have optimized the conditions, please perform a final run with ferrocene as an internal reference.

Edit:
Let me explain the last suggestion. The literature is full of electrochemical data of organic compounds measured in non-aqueous solvents, mostly acetonitrile.

It was once common (bad) practise to report the potentials vs SCE (saturated calomel electrode), although the latter, an aqueous(!) reference electrode, was atually never used in the measurement. Often rather exotic conditions plus a shaky conversion of data turned good experimental work into a pile of data of limited value.

Using an internal standard is the obvious solution to obtain solvent-independent electrochemical data for publication.

Note that the use of ferrocene as such a standard isn't my idea! It was originally proposed by GRITZER and KUTA in Pure Appl. Chem. and is a IUPAC recommendation nowadays.


List of functional groups in a molecule Does anyone know if there's a tool for listing the functional groups present in a molecule?

Norbert Haider from the University of Vienna has been working on the identification of functional groups for quite a while.
You might want to have a look at his checkmol/matchmol tool, which natively reads from MDL, Alchemy, and Sybyl mol files.
Other input formats may be processed after converting them using Open Babel.


Can someone explain why the electrical conductivity of magnesium chloride decreases with increasing concentration? I was under the impression that the electrical conductivity should increase initially, then begin to plateau, then fall past the saturation point for strong electrolytes. However, from this, I see that the conductivity of magnesium chloride does not follow this ( it is decreasing with increasing concentration) , BUT the data could be wrong though. Would someone mind explaining what I should expect when I conduct the experiment myself?

You simply started your data range at too high a concentration to see the initial increase.  Semantic Scholar gives data for the conductivities of several aqueous electrolyte solutions at 25°C, including magnesium chloride.  The strongly increasing portion of the magnesium chloride curve goes up to about 10%, which corresponds to about one molar (where your data happens to start).


How to derive the steady-state solution for simple two-step reaction with differential equations? Consider this simple two step reaction, a variant of a Michaelis-Menten type of problem, where $\ce{A}$ and $\ce{B}$ reversibly bind to make $\ce{AB}$, and $\ce{AB}$ and $\ce{C}$ reversibly bind to make $\ce{ABC}$:

\begin{align}
\ce{A + B  &<=> AB}\\
\ce{AB + C &<=> ABC}
\end{align}

Assume all the rate constants are equal to $k$ for simplicity, and that $[\ce{A}]_0$, $[\ce{B}]_0$, $[\ce{C}]_0$ are initial values for $\ce{A}$, $\ce{B}$, $\ce{C}$, and $[\ce{AB}]_0 = [\ce{ABC}]_0 = 0$.

What is the proper way to derive the steady-state concentration of $\ce{ABC}$ as a function of initial values? here is an attempt to get $\mathrm{d}[\ce{AB}]/\mathrm{d}t$:
$$\frac{\mathrm{d}[\ce{AB}]}{\mathrm{d}t} = ([\ce{A}][\ce{B}]k + [\ce{ABC}]k) - ([\ce{AB}]k + [\ce{AB}][\ce{C}]k)$$

The first parenthesis of the sum are the two ways to make $\ce{AB}$ and second parenthesis are the two ways to lose $\ce{AB}$.
We can set $\mathrm{d}[\ce{AB}]/\mathrm{d}t = 0$ and try to solve for $[\ce{ABC}]$, but I am not sure how this helps and where initial values come in.

Based on the discussion in comments, the OP is interested in the equilibrium concentration of the species ABC in this reaction scheme:

$$\ce{A + B + C <=>[k_1][k_2] AB + C <=>[k_3][k_4] ABC}$$

Equilibrium is reached when the rates of the forward and backward reactions of each step are equal. Thus, we have have two equations describing the equilibrium:

$$k_1\ce{[A][B]}=k_2\ce{[AB]}$$
$$k_3\ce{[AB][C]}=k_4\ce{[ABC]}$$

We also know from mass conservation that
$$\ce{[A]_o}=\ce{[A] + [AB] + [ABC]}$$
$$\ce{[B]_o}=\ce{[B] + [AB] + [ABC]}$$
$$\ce{[C]_o}=\ce{[C] + [ABC]}$$

These five equations can be manipulated to find an expression for $\ce{[ABC]}$ in terms of $\ce{[A]_o}$, $\ce{[B]_o}$, and $\ce{[C]_o}$.


What is the difference between hand wash and body wash soap? I do not like body wash always with strong favor. I like hand wash soap to wash my body. Are they difference at all?

Thanks for the comments, so that I add the following information
I am really sorry that I posted the question using my phone at my office parking lot right before I went to work this morning, so not much detail was added

To my limited knowledge, Body wash and hard wash soaps are detergent, they both are OK to use on skin. If that is the case, why body wash soap is body wash soap? What I mean is that from the chemistry point of view, what makes body wash soap difference from hand wash soap?

(My sister-in-law works at Unilever, and she makes these soaps)

These are all detergents.  The primary differences are harshness (hand soaps can be harsher, as hands are often more soiled), lathering, concentration, perfuming/scents, and marketing.

(Emphasis on the marketing.)


Why are intensive properties not additive? The extensive properties are additive. An intensive property is a ratio of two extensive properties.  So I think we can add intensive properties.

Its because intensive properties like temperature, refractive index, density are same for all the portions of the container of gas. You can't say that if the temperature of one portion is x and and other portion is also x, then the total temperature is $2x$.

But extensive properties, they are different for different portions of the container like mass, volume, length. So they can be added.

It doesn't matter if the intensive property is a ratio of two extensive properties. If two extensive properties are in ratio then if first extensive property changes then the second one should change.


How closely should the energy components of ab initio SCF results at a stationary nuclear configuration preserve the virial theorem? When running a SCF geometry optimization for water molecule, I found that the energy components of the optimized geometrical structure appear to deviate from Virial theorem sightly:

Kinetic energy = 75.424942151953331
Total energy = -75.585962823645531

How should this discrepancy be interpreted? Is it an indication of inaccuracy in the geometrical optimization algorithm of the software being used?

I hope I am not mistaken about this, but if I'm wrong then I'll learn so no big deal.

I believe this is a result of the fact that the electrons do not experience a true $r^{-1}$ potential due to electron-electron repulsion. Namely, we should expect that the electrons will repel each other which will cause the apparent attraction to the nuclei to be weaker than $r^{-1}$.

In other words, for a general potential,
$$
V(r)=r^{n}
$$
we have,
$$
\frac{dV(r)}{dr}=nr^{n-1}
$$

Because the Virial theorem can be stated as,
$$
\langle T\rangle=\frac12\langle\vec{r}\cdot\nabla V\rangle
$$
But for a potential which is less than coulombic, we have,
$$
\langle T\rangle=-\frac{n}{2}\langle V\rangle=-nE_{tot}
$$

That is to say, the electrons in water are feeling an effective potential of the form $V(r)=r^{-.99787}$ or so.

Or, in the language of the book you linked, the trial wavefunction of the form being tried for the hydrogen atom does not evaluate to a parameter $\zeta=1$.

Again, I am not completely confident that I nailed this answer, but for systems that have no obvious analytic form of the potential energy, there is no reason I see that we have should have $\langle T\rangle=-E_{tot}$ be exactly true. That is not to say the virial theorem is not satisfied, it certainly is, only that it is not as clean as we are used to.

I was a bit loose with some constants in my equations, please forgive me :)


Calculate molarity given density and mass percentage Problem

Calculate molarity of a solution of $\ce{H2SO4}$ with density $\pu{1.198 g/cm3}$ and containing $27~\%$ mass of $\ce{H2SO4}.$

Answer

 $\pu{3.374 M}$

My approach
I converted the density to g/l. Then i took 27% of the density and converted it into no of moles. Thus, came moles per litre, which is actually molarity. Yet, the answer key that my teacher had provided had a different answer compared to mine.
Is my method wrong or is there any calculation mistake?

Solution
By definition, molarity $c$ is the amount of substance $n$ per volume of solution $V$:
$$c = \frac{n}{V}\label{eqn:1}\tag{1}$$
The volume $V$ can be found using total mass of solution $m_\mathrm{tot}$ and its density $\rho$:
$$V = \frac{m_\mathrm{tot}}{\rho}\label{eqn:2}\tag{2}$$
The amount of substance $n$ can be found using mass of solute $m$ and molar mass of solute $M$.
Knowing mass fraction $\omega,$ one can switch to the mass of solution $m_\mathrm{tot}$ already used in \eqref{eqn:2}:
$$n = \frac{m}{M} = \frac{\omega m_\mathrm{tot}}{M}\label{eqn:3}\tag{3}$$
By plugging both \eqref{eqn:2} and \eqref{eqn:3} into \eqref{eqn:1}, one receives the final expression:
$$
\require{cancel}
\begin{align}
c &= \frac{\omega \cancel{m_\mathrm{tot}}\rho}{M\cancel{m_\mathrm{tot}}} \\
  &= \frac{\omega\rho}{M} \\
  &= \frac{0.27\times\pu{1.198 g cm-3}}{\pu{98.08 g mol-1}} \\
  &= \pu{3.30E-3 mol cm-3}~\text{or}~\pu{3.30 mol L-1}\tag{4}
\end{align}
$$
Notes

There is little to no point of converting anything prior to calculations as if the dimensional analysis is done right, you'll get the answer anyways.
Density is an intensive property: you cannot use mass fraction to determine the density of solute alone.
Regarding "no of moles": first, don't use non-standardized abbreviations like "no", and second, it's amount of substance, not "number of moles".
"Number of moles" is an appropriate term in a dialogue with your neighbor about the population of talpids in your garden, but not in science.
Don't forget to define each physical quantity properly and always try to solve the problem algebraically first, and plug in the values at the end when you have the ready-to-use equation for the quantity you were asked to find.
Keep the track of significant figures (slang: "sigfigs"), e.g. you are given the density with four sigfigs, so it would make sense to also use molar mass with four sigfigs as well.
Yet due to two sigfigs of mass fraction, the accuracy of the final answer is reduced.


How can the dipole moment of carbon monoxide be rationalised by molecular orbital theory? Despite the fact that oxygen is much more electronegative than carbon, the bond in $\ce{CO}$ presents a weak dipole moment.
This observation can easily be explained using the concept of "dative bond", that is, one bond is formed with two electrons from oxygen, producing a polarization $\ce{O\bond{->}C}$ which equilibrates the expected polarization $\ce{C->O}$.
I would like to know if the molecular orbital model could be used to explain this phenomenon.
Here's the diagram for $\ce{CO}$:

Unfortunately, nothing in the bonding situation in carbon monoxide is easily explained, especially not the dipole moment. According to the electronegativities of the elements, you would expect the partial positive charge to be at the carbon and a partial negative charge at oxygen. However, this is not the case, which can only be explained by molecular orbital theory. A complete analysis of this can be found in Gernot Frenking, Christoph Loschen, Andreas Krapp, Stefan Fau,  and Steven H. Strauss, J. Comp. Chem., 2007, 28 (1), 117-126. (I believe it is available free of charge.)
Responsible for the dipole moment is the highest occupied molecular orbital, a $\pmb{\sigma}$ orbital, which has its largest coefficient at the carbon atom. In first order approximation, this orbital can be considered the lone pair of carbon. All other valence orbitals are more strongly polarised towards the oxygen. The orbital that can in first order approximation be considered as the oxygen lone pair has almost only s character and therefore contributes only little to the dipole moment.
\begin{align}
\ce{{}^{\ominus}\!:C#O:^{\oplus}} &&
\text{Dipole:}~|\mathbf{q}|=0.11~\mathrm{D} &&
\text{Direction:}~\longleftarrow
\end{align}

I have reproduced the MO scheme of carbon monoxide for you below. Please note, that the blue/orange coloured orbitals are virtual (unoccupied) orbitals, which should be taken with a grain of salt.

There are two possible decomposition schemes to explain the bonding, both of them involve donor-acceptor interactions. The term "dative bonding" should be avoided here, it is better to use it only for bonds, that consist purely of donor-acceptor interactions, as for example in $\ce{H3N\bond{->}BH3}$.

Below, the two decomposition schemes are reproduced from figure 6 (b & c) in the linked paper. Please note, that this decomposition does not include hybridised orbitals.

The left decomposition is a better description, since it retains the $C_{\infty{}v}$ symmetry of the molecule. We can see a donor-acceptor $\sigma$ bond and two electrons sharing $\pi$ bonds.
In the right configuration we assume an electron sharing $\sigma$ bond, an electron sharing $\pi$ bond and a donor-acceptor $\pi$ bond.

It is very important to understand, that the concept of a dative bond, that you are trying to employ here is only right by coincidence. The reason that the dipole moment is oriented towards the carbon is only to find in the weakly bonding HOMO, the lone pair of carbon.


What happen to the Salt in Water Electrolysis I haven't found yet what happen to the Salt Ions in Water during Electrolysis and is there an Electron Flow or Ion Flow.
I have expierienced strong magnetic Forces between the Electrodes but couldnt measure an outside Magnetic Field.

When a battery is connected to an electrolytic cell one electrode becomes positively charged the other negatively charged. The ions in the liquid are attracted to the electrode with the opposite charge and move through solution. Ions with a negative charge move to the positive anode (anions), Ions with a positive charge migrate to the negative cathode (cations).

Ionic solids produce electrolytes. When an ionic solid is dissolved in water the aqueous ions are free to move and they carry the current. The anions of the salt and water give up electrons which flow through the circuit. The anions are oxidized at the anode while the cations of the salt and water recieve the electrons and are reduced at the cathode.

Cations will cluster around the cathode and anions will cluster at the anode. It is usually easier to displace hydrogen atoms from water molecules in the aqua ion than to displace the cation from the system  so electrons from the cathode react with water molecules , and hydrogen is displaced.

The result is decomposition of water molecules:

2H-OH(l) + 2e- → H2(g) + 2OH-(aq)

The solution becomes alkaline around the cathode due to the hydroxide ions being formed. If a metal is above hydrogen in the electrochemical series, hydrogen will always be produced at the cathode during electrolysis of solutions of its salts. If the ionic salt consists of a metal below hydrogen in the electrochemical series, electrolysis of its salt solution will always produce metal at the cathode.

Both ion flow and electron flow are occurring.

The intensity of an electromagnetic field reduces significantly the further away from the source.


How can I attempt to inhibit the hydrolysis/degeneration of aspartame? I've surfed the internet, looking for some possible inhibitors for the hydrolysis process of aspartame, but couldn't find any. I'm wondering if such a thing is possible and if yes, how should I proceed?

I'm writing an essay about how the rate of degradation of aspartame could be slowed down and wanted to attempt using some compounds to inhibit the reaction.

Aspartame is hydrolysed in both acid and alkaline solution with the methyl ester being the first site hydrolysed, so control of the pH of the aqueous solution is important. Addition of buffer solutions would enable control of the pH, keeping it close to neutrality.

https://www.chemguide.co.uk/physical/acidbaseeqia/buffers.html


Calculating the number of ligands of complex ions in d- block How do you know the number of ligands surrounding a metal ion?

Example: the obtained complex of $\ce{Ni^2+}$ ion in $\ce{NH3}$ solution is written as $\ce{[Ni(NH3)6]^2+}$ but $\ce{Cu^2+}$ in $\ce{NH3}$ solution is $\ce{[Cu(NH3)4]^2+}$.

So how do you find the number for different metal ions like $\ce{Ag^2+}$?

1st-row $d$-element + 2nd-row $p$-element - usually octahedron except for $\ce{Ni(II),Cu(II),Co(I)}$ (often a square or a distorted octahedron with two weakly bound ligands) and $\ce{Zn}$ (usually a tetrahedron). 1st-row $d$-element + 3+-row $p$-element - usually a tetrahedron.

2nd/3rd row $d$-element are somewhat more complicated as they allow higher coordination number up to 9 ($\ce{ReH9^{2-}}$), but the situation is so much influenced by electron configuration, ligands and metal-metal interactions here that simple rules loose a lot of meaning.

In short, the coordination number is usually a compromise between 18-electron rule and size of ligands. For example, $\ce{V(CO)6}$ should dimerize, giving c.n. of 7 to vanadium, but can't do so because of size constrains. Somewhat similar happens with bulky alcoholats of titanium, that tend to include $\ce{Ti(IV)}$ in octahedral coordination, but lower c.n. is observable in case of tritox (tris-tert-butyl) alcoholat.

Though it is generally not recommended to guess the c.n., it is always better to google for some reference compounds.


Which bond is stronger: N=O or C=N? Which bond is stronger?  N=O or C=N?
I tried comparing it using electronegativity difference. But, I don't know if that is the correct method. Can anyone explain it?

The $\ce{C=N}$ bond has a dissociation energy of $\pu{615kJ/mol}$, while $\ce{N=O}$ has an energy of $\pu{607kJ/mol}$. The difference isn't much, so its difficult trying to comparing the energies without being given the exact values.

References:

Yoder, Claude. “Wired Chemist.” Common Bond Energies (D, www.wiredchemist.com/chemistry/data/bond_energies_lengths.html.


Mixing isopropyl alcohol and propylene glycol What would happen if I mixed isopropyl alcohol and propylene glycol? I need to winterize the water base of my basketball hoop, and already added some isopropyl alcohol, but I discovered that propylene glycol would be better.

Also, I know isopropyl alcohol is toxic to ingest, but will it contaminate my soil if I drain it?

The two chemicals can be mixed without any reaction between them. However, you should add enough water to avoid fire hazards after the winter. Take a look at this table to find out, what mixture should suffice.

You should never pour solvents onto your soil or down the drain. Contact local authorities to find out, how you can safely dispose of that mixture. You can, of course, reuse the mixture and store it in some plastic cans if you need to empty the basketball hoop.


Why can’t lanthanum through lutetium and actinium through lawrencium all be in group 3? In 2015, IUPAC established a task force to “deliver a recommendation in favor of the composition of group 3 of the periodic table.”  Not much about their decision-making process has been made known to the public.
It is noteworthy that the task force must select one of two compositions for group 3:

the elements Sc, Y, Lu and Lr, or
the elements Sc, Y, La and Ac.

This seems to call into question the conceptual validity of arrangements that depict La–Lu and Ac–Lr as falling under the umbrella of group 3 by leaving a gap, such as the one currently published on PubChem (as opposed to arrangements that insert the f-block between groups 3 and 4, such as the one currently on RSC.org).
If all the lanthanides, including lutetium, are so similar, and if all the actinides, including lawrencium, are so similar, why was a third option—“the elements Sc, Y, La through Lu, and Ac through Lr”—disqualified as a recommendation for the composition of group 3 by IUPAC?

Since I am neither clairvoyant nor able to travel back and forth in time, I can only guess what IUPAC was thinking.  But there is no guessing that the inner transition elements do not completely fit in with Group 3 chemically.
One major chemical difference is in the oxidation states the elements may achieve, and thus the stoichiometry of compounds with nonmetaks that have strong ionic character.  Whereas the Group 3 elements overwhelmingly form compounds in the +3 oxidation state, other oxidation states are more likely for at least some inner transition elements.  While in some other groups there is more variance linked oxidation states, this is little seen in Group 3 whereas, below, variance in oxidation states is more common with inner transition metals.  Especially note the higher oxidation states in the examples that follow, which would not be expected in a group generally defined by elements normally having only three valence electrons.
Among lanthanides, cerium achieves a +4 oxidation state commonly enough to be called "ceric" and be available as a reagent for redox reactions, as in ceric ammonium nitrate and cerium(IV) sulfate.  Cerium can also form +4 as well as +3 ion cores in the metal by either including or excluding its $4f$ electron in the metallic band structure, leading to a transition between two FCC phases [1].  At the opposite end, europium accesses the +2 oxidation state much more readily than a true Group 3 element, leaving half-filled valence shell, and thus overlaps the heavier alkaline earth metals in its chemical properties.  Wikipedia reports that

In anaerobic, and particularly geothermal conditions, the divalent form is sufficiently stable that it tends to be incorporated into minerals of calcium and the other alkaline earths.

With actinides, oxidation states departing from the +3 predominance of Group 3 are more common.  Well-known examples include thorium forming a highly refractory oxide in the +4 oxidation state and uranium achieving oxidation states as high as +6 when combined with fluorine.
Reference
1.
Johansson, B., Luo, W., Li, S. et al. Cerium; Crystal Structure and Position in The Periodic Table. Sci Rep 4, 6398 (2014). https://doi.org/10.1038/srep06398


Arrange the following in increasing order of acidity: water, ammonia, ethyne and ethane I know that ethyne is more acidic than ethane because of more s character. For water and ammonia, I used the fact that oxygen is more electronegative than nitrogen and will be able to stabilize the charge better. So water will be more acidic than ammonia.Now how do I compare ammonia and ethyne?

The order of acidic nature would be: water > ethyne > ammonia > ethane.

Since oxygen has highest electronegativity amongst the atoms of the compounds, so its hydrogen will be most acidic and hence will be best acid among them.

Ethyne contains $\mathrm{sp}$-carbon and as a result the electronegativity of the carbon in ethyne is more than $\mathrm{sp^3}$ due to increased $\mathrm{s}$-character (50%) in the bonds between the carbon atoms.

The electronegativity of the carbon increases to a great extent when  and turns out to be 3.2, which is greater than the electronegativity of nitrogen which is 3.04. Hence the hydrogen of ethyne is more acidic than that of $\ce{NH3}.$

Ethane does not possess any acidic hydrogen hence it is least acidic.


If no oxygen is present, will cut avocados stop browning? Oxygen makes cut avocados brown, or at least it is necessary to facilitate the rapid browning reaction we see within 24 hours of cutting the avocados.

If oxygen was completely removed from the air surrounding the cut avocado, will the browning stop?

If a small fraction of oxygen remained, will the browning slow? In other words, is the rate of browning proportional to the percentage of oxygen in the surrounding air?

I never studied chemistry so my understanding comes from here:

http://www.compoundchem.com/2014/08/03/why-do-avocados-turn-brown-the-chemistry-of-avocados/

Though I'm sure that it's possible for them to brown due to substances other than oxygen, in general, yes, if you remove oxygen from the cut surfaces of an avocado, it will not turn brown, and yes, if you reduce the quantity of oxygen, browning will slow.  The same works for apples, bananas, and pretty much any other fruit that browns upon exposure to air.

It's not particularly practical for intact avocados, but one method for preventing guacamole from browning is to put plastic wrap right against the surface of the dip, preventing air contact.  Additionally, if you look at packaged guacamole, in addition to the browning resistance caused by acidity, the amount of oxygen in any given container is limited to a few small bubbles.  The oxygen in those bubbles is quickly consumed and the remaining gas does not oxidize the guacamole further.

You could conceivably test this with a vacuum chamber and keep normal slices of avocado green for a prolonged time.  You also could use a food vacuum sealer that bags and evacuates food for long term storage, though ripe avocados are delicate enough that they will probably be crushed into paste by the process.


Can you measure an IR spectrum of a sample inside a glass vial? If I have a sample of some compounds dissolved in water inside a glass vial (let's say quartz). Would it be possible to get a usable measurement through this glass vial without too much absorption by the glass itself?

As EdV said, when you have aqueous samples, Raman spectroscopy is the way to go. The problem is not the container (which should not be glass etc),the problem is water. Its absorption will hide the absorption of everything.
If you ever open an FTIR instruments, you will see that there is nothing which is made of glass or quartz because they absorb too strongly in the useful IR range.


Nomenclature of Dihydroxyacetone Phosphate Why the dihydroxyacetone phosphate is called that way when it has only one OH ?

Because it is the phosphoester of dihydroxyacetone. One of the -OH groups was used in the esterification reaction.

See Here


Balancing reduction equations for galvanic cell Problem: Compute the potential of the following galvanic cell: $$(\ce{Mg (s)} \mid \ce{Mg^2+ (aq)}\ (0.30\ \mathrm{M}) \mid \mid \ce{Sn(OH)4^2- (aq)}\ (0.10\ \mathrm{M}); \mathrm{pH} = 9 \mid \ce{Sn (s)})$$

Attempt at solution: I know we have to use the formula $$E_\text{cell} = E - \frac{0.0592}{n} \log Q$$ The anode is situated on the left; this means electrons are going away there. So we have for the oxidation half-reaction: $$\ce{Mg -> Mg^2+ + 2e-}$$ The potential for this is $E_1 = 2.36\ \mathrm{V}$. Then I’m not sure how to write my reduction half-reaction. I just wrote: $$\ce{Sn(OH)4^2- + 2e- -> Sn(s)}$$ and $E_2 = -0.91\ \mathrm{V}$. But this doesn’t seem correct, because if I add both reactions to get the sum, my charges aren’t balanced anymore. Also, I wonder what happens with the $\ce{(OH)4^2-}$ ? It’s not on the right side of the arrow anymore.

To construct reduction equations there are a set of steps you can follow.

For a balanced equation, the number of atoms of each element, and the total charge, must be the same on each side of the equation.

Firstly identify the main reactants and products which you have correctly done in this case as: $$\ce{[Sn(OH)4]^{2-}(aq) -> Sn(s)}$$

Under acidic conditions the only things we can add to balance the reaction are $\ce{H=}$, $\ce{H2O}$ and $\ce{e-}$. Under alkaline conditions we can add $\ce{OH-}$ instead of $\ce{H+}$. Usually, when dealing with alkaline reactions the best way of dealing with them is to treat them as acidic and then neutralize the acid later. However, in this instance, since the only reactant involves hydroxide ions, we can immediately balance the hydrogens and oxygens just by adding four hydroxide ions to the products: $$\ce{[Sn(OH)4]^{2-}(aq) -> Sn(s) + 4OH- (aq)}$$

Then balance the charges to get the overall equation: $$\ce{[Sn(OH)4]^{2-}(aq) + 2e- -> Sn(s) + 4OH- (aq)}$$

The alternative method which works for any alkaline reactions, which are sometimes less obvious than the example above, is to balance the reaction under acidic conditions and then neutralize the acid.

Since we only have one option for balancing the oxygens we add $\ce{H2O}$ first: $$\ce{[Sn(OH)4]^{2-}(aq) -> Sn(s) + 4H2O(l)}$$

Now balance the hydrogens: $$\ce{[Sn(OH)4]^{2-}(aq) + 4H+(aq) -> Sn(s) + 4H2O(l)}$$

Now the charges: $$\ce{[Sn(OH)4]^{2-}(aq) + 4H+(aq) + 2e- -> Sn(s) + 4H2O(l)}$$

If this were in acidic solution then this would be the final balanced equation. However, since we are in alkaline solution, we add $\ce{OH-}$ ions to cancel out the $\ce{H+}$, forming $\ce{H2O}$. $$\ce{[Sn(OH)4]^{2-}(aq) + 4H2O(l) + 2e- -> Sn(s) + 4H2O(l) + 4OH- (aq)}$$

Finally we can cancel the excess $\ce{H2O}$ to give: $$\ce{[Sn(OH)4]^{2-}(aq) + 2e- -> Sn(s) + 4OH- (aq)}$$

This is the same result as we got earlier by the quicker method.


Expressing solubility in binary solvents with non-additive volumes I have experimentally obtained data that shows the molarity of a saturated solution   in mixed solvents, with different proportions of solvents.

The issue here is that the two solvents (water/ethanol) have non-additive volumes, and expressing the solubility as $\mathrm{mol}$ $\mathrm{L^{-1}}$ thus seems a bit odd, especially when comparing solubilities between different solvent ratios. Would mole fraction be a better alternative? If so, why? And is it possible to convert from the "old" amount-per-volume answer to the new, better one?

Also, I include a graphical representation of my results, with solvent composition on the x-axis and solubility on the y-axis. Would it be appropriate to use mole fraction (or whatever other unit is better suited) on both axes: the mole fraction of solvents on the x-axis and that of the solute in the mixed solvent on the y-axis?

For solvent mixtures, especially in mixture or solubility design of experiments,  I work in mass for the solvents.  It's unaffected by temperature,  more accurate (very advantageous if small values are being used) and should get around your problem. You can quote your initial results as %w/w


What is wrong in this reaction? What is wrong in this reaction?

I cannot find anything wrong, please help. Any hints would be appreciated.

The problem is with the free $\ce{NH}$. Grignard reagents are strong bases ($\mathrm{p}K_\mathrm{a}\sim 50$), strong enough to deprotonate $\ce{R2NH}$. Any Grignard that formed would be immediately quenched by the free $\ce{NH}$. You need to protect, this is typically done with tBOC, more discussion here.


Is elemental lead as toxic as lead compounds? On the internet, nobody mentions about lead's oxidation state when talking about lead poisoning. I assume it's Pb(II) and Pb(IV), as chelation therapy is suggested as a treatment. Also, its accumulation in bones and teeth suggest its ionic state. But what about elemental lead? How is it transformed in human body? Is it as toxic as Pb(II) and Pb(IV) compounds?

Pb(IV), in all likelihood, will quickly get reduced to Pb(II) once it gets into a human body. So as far as poisonings go, these two can be considered equal.
Elemental lead will partially dissolve in stomach, but mostly go out undigested. So it is way less harmful than any compound of Pb(II). There were people in the old times who survived bullet wounds and used to carry bullets in their bodies for decades, with little or no apparent ill effects.
Then there is a wholly different story of the covalent compounds of Pb. They are more toxic than ionic Pb(II), and some are capable of penetrating through the skin. In particular, the notorious $\ce{Pb(C2H5)4}$ has caused our civilization a great deal of harm.
So it goes.


How to determine force constant of C-C bond in ethanol? How would one determine the theoretical force constant of the C-C bond in ethanol?
We are asked to calculate the frequency of the C-C bond (theoretical), but to do this we need a force constant, which is not provided to us.
I cannot find a source of this information. The only way I see to calculate the force constant is from the frequency of the bond, which is what we are asked to calculate.
This professor has a reputation of omitting important details

Use the fundamental stretch function below to determine what is the frequency of the C-C bond and what is the wavenumber?
$$\bar{\nu}\mathrm{(cm^{-1})}=1303\sqrt{f\left(\frac{1}{m_1}+\frac{1}{m_2}\right)}$$
where $f$ is the force constant in N/cm, and $m_1$ and $m_2$ are the masses in atomic mass units (u). Compare your calculated values with the experimental value. Provide the reference.

You can estimate the force constant (as I have done here) and calculate the frequency of the stretch or, as you request, go the other way. Regardless, you have 2 unknowns and thus the problem you present isn't solvable. Below, I work it from the assumed value of the force constant: to go the other way, you'll need to locate the absorption spectral line for the C-C stretch in ethanol and go from there. Either way, you'll need to make some assumptions regarding reduced mass.

From page 496 in Organic Chemistry by Brown, Iverson, Anslyn, and Foote, the formula

$$\tilde{\nu} = 4.12\;\sqrt{{k\over\mu}}$$

is given, where $\tilde{\nu}$ is the frequency in wavenumbers, $k$ is the force constant in units of dyne/cm, and $\mu$ is the reduced mass (in amu, or atomic mass units). They state that for a single bond, the force constant can be taken as $5\cdot 10^{5}$ dynes/cm (an approximation, and we do have to start somewhere). The 4.12 differs from the constant in your given equation due to unit conversion between dynes/cm and N/m.

Now, on to your problem. The reduced mass term is (generally)

$$\mu = {m_{1}m_{2}\over m_{1} + m_{2}}$$

which is for two masses. I posit that in ethanol, one carbon has a "mass" of 15 amu (that's the carbon with 3 hydrogens attached to it) and the other has a "mass" of 31 amu (the carbon plus 3 hydrogens and the oxygen). To go beyond this involves a significant amount of advanced work that is beyond the scope of your problem (but if you're interested, Molecular Vibrations by Wilson, Decius, and Cross is the place to start).

We calculate

$$\mu = {15\times 31\over 15+31} = 10.1$$

and substitute into the first equation

$$\tilde{\nu} = 4.12\;\sqrt{{5\cdot 10^{5}\over 10.1}} = 916\;\mathrm{cm}^{-1}$$

From some notes for a class at Caltech, it appears we can expect a carbon-carbon single bond to appear anywhere in the range of 800 to 1200 cm$^{-1}$.

The calculated number for the C-C stretch here (towards the lower range we'd expect) is due to our "bloated" reduced masses, and that makes intuitive sense (if we do the same calculation for ethane, with "lighter" reduced masses for each carbon, we get $\tilde{\nu} = 1063$ cm$^{-1}$).


Rate of SN2 reaction depending on following different conditions in case of primary halides Most important points to remember when thinking of $\mathrm{S_N2}$ reactions are that they are nucleophilic reactions, a pentavalent transition state is formed, depends on steric hindrance, strength of the bonds, nucleophilic and substrate used i.e. what kind of leaving group is attached to it, etc. I know and understand these.

I also know that in the following conditions the rate of $\mathrm{S_N2}$ increases, but I don't seem to understand the reason behind it.

1) The presence of unsaturation on beta-$\ce{C}$ in primary halide increases $\mathrm{S_N2}$ reaction rate.

I really don't seem to understand this one, I know for the alpha $\ce{C}$ it would be difficult because the halide and the carbon will acquire partial double bond character which would make it very difficult to break.

2) The presence of carbonyl group at alpha carbon is most suitable for $\mathrm{S_N2}$ reaction.

My reason: it might be because of tautomerism, i.e., due  to its conversion to enol form it will increase the reactivity.

3) Presence of hetero $\ce{O,N,S}$ atom at alpha carbon in primary halides increase $\mathrm{S_N2}$ reaction.

This too, I fail the understand.

I know this question was from awhile ago, but I'll answer anyway.

a) The key to understanding this is the concept of the transition state. Below is an image taken from Khan Academy

For an $\mathrm{S_N2}$ reaction to take place, there must be enough activation energy to reach the transition state. In this type of reaction, the transition state comprises an alpha carbon in a pseudo-$\mathrm{sp^2}$ state. The nucleophile and the leaving group are directly above and below the alpha carbon in this state, with atoms bound to the carbon in a horizontal plane.

The reason this reaction proceeds more quickly with an unsaturated beta-carbon is the pi-electrons from the unsaturated carbon stabilize the transition state. This lowers the activation energy of the reaction, making it more likely to happen with each molecular collision.

c) I'm assuming you're comparing $\ce{N, O}$ and $\ce{S to C}$. $\ce{N, O,}$ and $\ce{S}$ are more electronegative than carbon, which will impart a more positive-like characteristic on the alpha carbon, making it more susceptible to attack from the nucleophile.


Is the pressure in an HPLC system the same throughout? I'm not sure if I'm articulating myself well enough, but here's my problem: I have a current HPLC system, where we typically have backpressure readouts of around 15 MPa. We are looking to get a new HPLC, and the rep said its DAD detector has a maximum pressure of 6 MPa. My questions are:

Is the pressure at the pump outlet the same as the pressure at the column inlet and outlet?
What pressure is actually being shown on the system readout when looking at the pump? (I'm assuming just the pressure at the pump outlet)
Is the pressure at the detector usually significantly less than at the pump outlet?
Assuming everything transfers from one system to our new one, and we end up with the same types of backpressures, will I have issues with this new detector?

I figured this out. The pressure displayed is between pump and column inlet. There's a pressure gradient from the column inlet to the outlet, so the pressure reaching the detector is significantly less than the readout.


What will be produced when sucrose and potassium bitartrate react with each other? What would happen if sugar and cream of tartar react with each other? Would it create an ester?

$$\ce{KC4H5O6 + C12H22O11 -> KC16H21O14 + 3H2O}$$

Is this correct? If not, why?

Rather than "cream of tarter" I'm going to stick to the pure chemical potassium bitartrate and tartaric acid, and I'm going to speak from the point of view of a chemist about the difference between potassium bitartrate and tartaric acid (rather than a cook or winemaker).

First a simple answer to your questions and then the details. Simplistically speaking, when you add potassium bitartrate to an aqueous solution of sucrose you will get some hydrolysis (producing glucose and fructose) and some ester formation. The key word here, however, is some -- and the amount depends strongly on the reaction conditions. At room temperature, for example, you're likely to get essentially no reaction within, say a day. More details below.

Tartaric acid is a dicarboxylic acid (a weak acid). Each of its carboxyl groups (-COOH) can donate a proton (H+ ion), and when this happens it's "acting as an acid". This is important to your question because it is the H+ ion that acts as a catalyst to sucrose hydrolysis and not the tartaric acid itself. For this situation, chemists call it a specific acid catalyzed reaction. By contrast, if a specific molecule such as tartaric acid were involved in the catalytic step chemists would call that "general acid catalysis".

Acidity of molecules are gauged by assigning a pKa value. If there are 2 acidic groups in the molecule (as with Tartaric acid) then it has 2 pKa values. Adding KOH to a solution of Tartaric acid will give you potassium bitartrate. At 25C, the pKa for removing Tartaric acid's first H+ is about 3.0 and the pKa for removing potassium bitartrate's H+ is 4.3. This makes both of them what chemist's call "weak acids" because they don't release anywhere near as much H+ into a solution as a strong acid (like HCl).

The other reaction, ester formation, is also specific acid catalyzed and so would also occur as I mentioned, but to a small extent. Esters are formed in acidic solutions whenever carboxyl (-COOH) and alcohol (-OH) groups are present, as is the case here. Note that there are many -OH groups here, on both molecules.

To understand the extent of each of these reactions it's important to consider the role of water. Water is a reactant in the hydrolysis reaction and this tends to drive the reaction forward. By contrast, water is a product of the ester formation reaction. This is especially important for the ester reaction since it is highly reversible. In other words, whenever molecules collide to form an ester, it's much more likely that that ester will collide with the far more abundant water molecules and revert back to the starting materials, resulting in no net reaction.

Usually one thinks of these reactions happening somewhere near room temperature. Under these conditions neither of the above reactions happens to a great extent; they are both generally very slow. In order to push them forward you have to do things like increasing the temperature, adding more concentrated reaction materials or catalysts, or removing reaction products. Of course, in this case increasing the concentration of the reaction materials also increase the amount of catalyst (H+), but since they are such weak acids, adding HCl (or another strong acid) would be much more effective.


Acidity of aluminium trihalides Would the acidity of the aluminium trihalides be the same as that for the boron trihalides?

The interesting thing about the acidity of the boron trihalides is that they actually increase in acidity from fluorine to iodine. This is due to the presence of stronger pi bonding in boron trifluoride compared to the other boron trihalides, which means that stronger partial pi bonds would have to be broken when boron accepts an electron pair from a base.

Would we observe this trend in acidity in the aluminium trihalides as well?
Based on the same reasoning, we should be able to right?

I have chanced upon the answer in an inorganic chemistry textbook (Essential Trends in Inorganic Chemistry by D. M. P. Mingos) I was reading. It turns out that this trend only holds for the boron trihalides and does not apply to the aluminium trihalides. However, the book did not provide the explanation for the exception of boron.

My thinking is that perhaps, backbonding is not present in the aluminium trihalides, possibly due to incompatibility of the p atomic orbitals for pi bonding between aluminium and the halogen atoms. Does anyone else have any insights to share?


Is it possible to make a drug that liquefies heart plaque to treat heart disease? Is it possible to make a drug that liquefies heart plaque to treat heart disease without damaging other parts of the body?  If so, would the liquefied plaque be eliminated as regular fluid is through the kidneys?  I believe that if this is possible, it would definitely be worthwhile to make as it could save many lives!

I heard there are drugs that can dissolve a blood clot.

I also heard of a technique being developed in Israel that dissolves heart plaque with an ultrasonic probe.

It looks like there are some efforts already in this area, but it appears they halt rather than dissolve the plaque:
http://heartdisease.about.com/cs/coronarydisease/a/plaquebuster.htm

Plaques are continually being removed from arteries by natural mechanisms within the body.  Statin use, when combined with aggressive dietary changes, can slow down the rate of plaque deposition to the point that the rate of plaque removal is actually higher than the rate of plaque deposition, and, as a consequence, plaque thinning can be observed.

Reducing plaques faster (plaque liquefication) has some potential, life threatening drawbacks.

We've all seen how ice cubes, when frozen together into a clump and then placed in a water bath, dissolve.  Some ice cubes break off from the main structure and float away to dissolve separately.  The same thing can happen with a large plaque deposit.  If a piece of the plaque breaks off during the liquefication process and continues to travel through the blood stream it can lead to downstream blockages, particularly in the brain (stroke).  Rapid dissolution of plaques would increase the risk for such adverse events.
While plaques form more or less everywhere as relatively uniform deposits, the real concern is areas where plaques have grown to large size and begin to block arteries.  It has been found that these larger deposits are initiated by events related to damage of the arterial wall at that specific location. The damage leads to infection (that's why highly sensitive C-reactive protein [hs-crp] measurements are routinely made in conjunction with cholesterol measurements).  The body mobilizes the immune system to fight the infection; microphages and white blood cells are directed to these infection sites.  Once this happens deposition of other materials (lipids, calcium, etc.) start at the infection site and abnormal plaque growth results.  Because of the 1) initial damage, 2) the infection and 3) infection fighting, the arterial wall becomes abnormally thinned at the site.  To some degree, the plaque actually serves to stabilize the thin wall preventing wall rupture and leaks.  Rapid removal of the plaque would increase the chances for the wall to rupture,  producing adverse events and increasing mortality.

So while it is likely that drugs could be designed to remove the plaque extremely fast (e.g. liquefication), the risks might far outweigh the benefits.

  would the liquefied plaque be eliminated as regular fluid is through
  the kidneys?

A major component of plaque is lipoprotein.  Lipoprotein is transported by blood as insoluble particles of varying density and size.  Lipoprotein itself is an assembled complex of proteins and fats.  The fats are various esters of glycerol including esterified cholesterol.  The lipoprotein transports cholesterol and fats, via the bloodstream to cells where these compounds are needed for biosynthesis. After the lipoprotein has released the transported molecules it returns to the liver where it is "reloaded" and sent out again. Lipoproteins are not eliminated (unless damaged) from the body, but are rather recycled.

If a large amount of lipoprotein were suddenly released into the body, I'm sure some would be eliminated through the kidneys.  Certainly the lipoprotein receptors would sense this event and stop synthesis of new lipoprotein until systemic balance was restored. it seems likely that adverse events like kidney damage, gallstones, bile duct clogging, etc. would occur.

Since statins were first released onto the market, drug companies have been considering where to go next.  I have to believe that plaque liquefication was among the options considered.  While the reduction of plaque build up (statins) have their share of problems, I would guess that these problems are minor compared to the possible problems that can result from liquefication (discussed above).  Indeed, with the recent announcements (see here for example) of monoclonal antibodies achieving plaque reductions superior to statins, it seems that drug companies are continuing to focus on avoiding the problem in the first place, rather than fixing things (liquefication) down the road.


Is there an equivalent unit cell that can describe a bcc unit cell where all points are the same element? Similar to how an NaCl-structure can be described as just a primitive cubic structure if all the lattice points were the same element or ion. I'm asking if there any such alternative for the CsCl-structure.

You certainly can construct what is called a primitive cell containing one atom for a bcc lattice.  There could be more than one way, but here is my method.

Start with the usual cubic representation.  Pick any face and connect the midpoints of its edges to make a smaller square.  Do the same with the opposite face.  This gives a pair of squares which are then the bases of an elongated square prism, whose lateral edges connect the two opposing faces you "contracted".  This cell, which contains just the central atom of the original bcc structure, is a primitive cell.  In this cell the lateral edges, which we did not "contract", are about 1.41 (actually, the square root of 2) times as long as the basal edges.

How does this replicate to make the lattice? If you just scoot the primitive cell over to "catch" a neighboring atom on either side of the lateral faces, you find that you must move the adjacent cell "up" or "down", in the direction perpendicular to the bases, to get the second atom centered properly.  The "up"/"down" move is half the height of the prism.  But once you see this, you can repeat this process and get all the atoms involved in the entire cubic representation.

Last question:  I said the primitive cell is an elongated square prism.  It no longer has cubic symmetry.  Where did the cubic symmetry go?  The answer is you can't make a primitive cell in the bcc lattice with the full cubic symmetry.  Instead, note that when you modified the cube you had the choice of three pairs of opposite faces to "contract".  Thus there is really a combination of three primitive cells with different orientations.  The cubic symmetry emerges from the combination rather than any one primitive cell.


Finding partial derivatives for an ideal gas and van der Waals gas How do I determine $\left({\partial P\over\partial V_\mathrm m}\right)_T$ for an ideal gas and also for a van der Waals gas? Note that $T$ is the only variable which is held constant, and that's my trouble. I've only derived these equations in cases where two variables are held constant. I don't know where to start.

Use the equation of state, express $P$ as a function of $T$ and $V$, and find the derivative with respect to $V$, assuming $T$ to be constant.


Why does KMnO4 oxidize iodide to iodine in acidic medium, whereas iodide to iodate in alkaline medium? $\ce{KMnO4}$ should be a better oxidizing agent in acidic medium, than in alkaline medium, as per my knowledge. However, it oxidizes iodide (oxidation state = -1) to iodine (0) in acidic medium, whereas iodide (-1) to iodate (+5) in alkaline medium. The latter has a higher change in oxidation number, which we would have instead expected for the former. Why is this so?

A simple way to explain this is to illustrate with the following Latimer diagrams:

If you calculate the standard electrode potential of the highlighted portions it will be positive, so the reaction is spontaneous under standard conditions.

For acidic conditions, iodine is not converted to iodate because the electrode potential of the reaction would become negative, whereas for basic conditions it is possible.

Note that both the manganate and the iodide ions are both affected by acidic and basic conditions; thus it is important to consider their respective electrode potentials.

(Data was taken from Inorganic Chemistry, Shriver et al., Sixth Edition, Oxford University Press)


What is Resonance, exactly? I know the that it(Resonance) is the delocalisation of pi electrons but how and why are those electrons delocalised? What is the driving force that causes the effect?

Resonance is a conceptual tool to help make sense of the real observed structures of molecules
Resonance is not a real physical phenomenon. That is the first important thing to realise. It is merely a tool to bridge the simple view where all bonds consist of integral numbers of electrons and the real structures of molecules.
The simple view of bonds often fails to represent the "real" observed structure of molecules. The archetypal example is benzene. A simple view of the bonding of benzene has the molecule consisting of alternating single and double bonds (which might be called cyclohexatriene). This theoretical molecule has the bonds neatly grouped into simple bonds that use up all the available electrons into either single bonds (one pairs of electrons) and double bonds (two pairs of electrons). But this would imply that the bond lengths alternate. But we know the structure of benzene and the bond lengths are all the same. We can explain this with molecular orbital theory when we recognise that some of the bonds are delocalised, but that gets fairly sophisticated fairly quickly and chemists prefer easier ways to keep track of the electron counts when writing structures on paper and explaining reaction mechanisms.
One halfway house that saves the ability to write structures where simple electron counts are possible is the idea of "resonance". Instead of needing to invoke molecular orbital theory all the time we can write the structure of benzene as a combination of two structures where the double and single bonds alternate. We pretend that the real structure is a "resonance hybrid" of the two hypothetical structures. This is like the mathematical average of the two competing structures.
More complex molecules can also be represented as combinations of (more complex) resonance structures. We often say that the structure consists of resonance hybrids. But resonance is not a real physical phenomenon. The different structures are not real independent structures that rapidly interchange: they are merely a convenient model so chemists can talk about electron counts and draw structures without having to invoke much more complex bonding theories.
So there is no such thing as "resonance" as a physical phenomenon. What drives the structures of molecules is the way orbitals in atoms combine together in molecules and that is more complicated than simple views of bonding can represent. The underlying driving force is quantum mechanics and energy minimisation of the possible ways atomic orbitals interact in molecules. And that is sometimes far more complicated than simple views of bonding can easily represent.


How to show a band structure for water? This question is generated by my attempt to answer this one.  I thought I could demonstrate that electronic conduction in water is possible because of the impact of an electric field on the band structure, but I am being hit with comments questioning the existence of band structure in liquids.  I thoughtot was self-evident, but I was asked for references showing water has a band structure and what I am coming up with does not appear to be working (the comments are getting more upvotes than my answer).  So I need help.  Can anybody come up with more clear-cut references demonstrating that water, or liquids generally, has such a band structure?

do Couto et al have published an article [1] characterizing the density of states (band structure) of water using DFT theory combined with MC sampling. Extrapolating from clusters to bulk water, the HOMO-LUMO gap is 8.55 eV (Table 3).

From the abstract:

  By extrapolating the results for larger clusters the threshold energy
  for photoelectron emission is $9.6±0.15 eV$ free clusters and
  $10.58±0.10 eV$ embedded clusters. Our results  for  the  electron
  affinity $V_0=−0.17±0.05 eV$ and  adiabatic  band  gap
  $E_{G,Ad}=6.83±0.05 eV$ of liquid water are in excellent agreement
  with recent information from theoretical and experimental works.

From the introduction:

  It is generally accepted that water canbe  described  as  a  very
  large  band-gap  amorphous semiconductor.11,17 However, it is not
  obvious that the band-gap of liquid water can be uniquely associated
  with an optical vertical excitation process, where the gap is
  defined simply as the highest occupied molecular orbital-lowest
  unoccupied molecular orbital HOMO-LUMO energy difference. As recently pointed out
  by Coeet  al.,3 the reactive nature of electronically excited water
  molecules,18 the reorganization of water molecules around charged
  species in liquid phase, and the observed photophysics of anionic
  defects known as the anion problem 4 strongly indicate that an
  adiabatic route can be defined for accessing the conduction-band edge
  in liquid phase. Moreover, given that the time scale ofthe solvent
  relaxation is much larger than the vertical process, the adiabatic
  band gap of liquid water cannot be determined vertically.

Reference
[1] Cabral do Couto, Estácio, and Costa Cabral J. Chem. Phys.123, 0545102005


Maximum amount of P4 $\ce{2Ca3(PO4)2 +6SiO2 +10C->6CaSiO3 +P4 +10CO}$

What is the maximum amount of P4 that can be produced from 1.0 kg of phosphorite if the phosphorite sample is 75% $\ce{2Ca3(PO4)2}$ by mass

The wording is confusing. I do not know what to do. Would you start by finding the limiting reactant?

First you find the number of moles of $\ce{Ca3(PO4)2}$ reacting. You have 75% of 1 kg, i.e. 750 grams, of calcium phosphate. The rest of reagents are assumed to be in excess as we have to find the maximum amount of $\ce{P4}$ obtained, hence reaction should not have other limiting reagents.


What does addition of iodine and base do this case? I figured that the first reaction just adds a bromine to the meta position but I don't know what step ii) does. The only thing I can think of is the base gets rid of the alpha hydrogen next to the carbonyl. Then perhaps the negative carbon now attacks iodine? But that seems a bit off to me.

$\ce{I2}/\ce{OH-}$ will perform the iodoform reaction on the the product of previous reaction. The base deprotonates an alpha hydrogen, and the resulting enolate nucleophile attacks iodine. After three substitutions, the $\ce{CI3}$ group is a passable leaving group, and a nucleophilic acyl substitution occurs. The reaction is done in basic conditions, so the carboxylic acid is deprotonated. The leaving group is a strong base, so it is protonated. $\ce{CHI3}$ will be formed along with the conjugate base of m-bromo-benzoic acid. Here is a generic version:

$$\ce{RC(O)CH3 + 3I2 + 4OH- -> RCO2- + 3I- + 3H2O + CHI3}$$


How can chlorine have an oxidation number of 3 in chlorite? I am studying redox reactions right now and I'm a bit confused about something:
$\ce{ClO2^-}$ is a compound ion and in a compound ion the sum of oxidation state of participating ions has to be equal the charge on compound since $\ce{ClO2^-}$ has a whole charge of $-1$ the oxidation state sum of $\ce{Cl}$ and $\ce{O2}$ must be $-1$, right? Oxygen has an oxidation state of $-2$ so $\ce{O2}$ will have
\begin{align}
2\cdot -2 &= -4\\
-4 + x &= -1\\
x &= -1 + 4\\
x &= 3\\
\end{align}

Chlorine has to have an oxidation state of $+3$.
But if you look at the periodic table chlorine is in 7th period which means it has 7 electrons in valence so it gains one electron while reacting.
Getting oxidized by definition means losing electrons so an oxidation state of $+3$ means chlorine is losing 3 electrons. I just can't get my head around why would chlorine do that? Or am I calculating the oxidation state wrong?

It is indeed in the +3 oxidation state. Keep in mind, though, that that does not mean chlorine lost three electrons. That polyatomic ion is not made up of smaller ions.


When considering Van der Waals forces, why do dipoles form? Imagine two atoms, and only consider the Van der Waals force. The electron cloud will jitter due to its quantum mechanical nature- some of these jitters forming dipoles, some not. However, on average they form dipoles. Why?
This question is equivalent to asking (although it is useful here to ask in another way): why is the potential energy of a dipole lower than a non-dipole?

What determines which electrons are going to move to the middle of the atoms to form a dipole, and which ones move to the outside of an atom (I assume that a dipole is formed like this $e^-..ion...........e^-..ion$ for the centre of positive charge shifted rightwards of the centre of negative charge): given that the jitters are instantaneous, and photons travel only at the speed of light, how are dipoles formed quickly enough before the next jitter?
Am I falling into a rut because I am not considering the wave nature of electrons?

on average they form dipoles.

Not quite. On average they form the nice (electrically neutral) electron orbitals we know and love. However, when they happen to be in a dipole state, there is an attraction between the two atoms, and that is evidently more than enough to overcome the randomly occurring opposing dipole states. That's just mathematics.

  why is the potential energy of a dipole lower than a non-dipole

It isn't. But quantum mechanics allows electrons to visit higher energy states, with a probability dependent on the energy difference.

  What determines which electrons are going to move to the middle of the atoms to form a dipole, and which ones move to the outside of an atom

Nothing. The  "movement" of electrons is just a random statistical fluctuation in the location of the electrons' centers of charge. Quantum-mechanical events like this are inherently nondeterministic.  And strictly speaking, you can't tell electrons apart anyway. There is no "which electron," at one point in time you have one electron here and another there, the next they may have switched places. Or not, you can never know.

  (I assume that a dipole is formed like this $e−..ion...........e−..ion$ for the centre of positive charge shifted rightwards of the centre of negative charge)

Don't think of those as being electrons, just the statistical average of the negative charge distribution.

  given that the jitters are instantaneous, and photons travel only at the speed of light, how are dipoles formed quickly enough before the next jitter

First, They're not quite instantaneous (that would have them basically everywhere possible at once), and not quite jitters.

In summary, don't think of electrons as particles with a distinct position moving around, just think of them as waves, with a peak that occasionally forms, and that peak causes the electric arrangement of two nearby atoms to, on average, be mildly attractive.


Does fluorine ever form a double or triple bond? Does fluorine ever form a double or triple bond?
I wonder if seeming lack of such higher order bonds is related to the electronegativity of fluorine.

The compound $\ce{NF}$, which is isoelectronic to $\ce{O2}$ is known, has been isolated in matrices and characterised and subjected to calculations. Like for dioxygen, three different states of this molecule are known: one triplet and two singlet states. Without performing any sophisticated analysis of its orbitals, we can expect a bond order of 2, and thus a double bond.

Harbison performed calculations on this compound. He came to the conclusion that the most stable triplet state is best described using only a single $\ce{N-F}$ bond. The two singlet states which require full electron pairing, however, display a much shorter $\ce{N-F}$ bond distance and are thus better described by an $\ce{N=F}$ double bond. Adding formal charges would lead to:
$$\ce{\overset{-}{N}=\overset{+}{F}}$$

Reference:

G. S. Harbison, J. Am. Chem. Soc., 2002, 124, 366–367. DOI: 10.1021/ja0159261.


Reaction of secondary alcohol with ammoniacal silver nitrate (Tollens' reagent) What are the products of the reaction of 2-butanol with silver nitrate?

For the moment, I am a bit stumped as I only have $\ce{AgNO3}$ as an oxidizing agent to form a carboxylic acid from an aldehyde. Additionally, the lack of $\ce{H2O}$ indicates that this is not the reaction I should be looking for. The only reaction that seems feasible to me is an SN2 mechanism where the nitrate anion acts a nucleophile and the hydroxide ion as the leaving group.

Here is my proposed mechanism much in the style of Tollen's reaction with an aldehyde.

This question refers to the use of Tollens' Reagent (ammoniacal silver nitrate). The oxidising species here is the diamminesilver(I) complex $\ce{[Ag(NH3)2]+}$.

Tollens' reagent is usually used as a test for aldehydes, with which it gives a characteristic silver mirror:

It is often said that Tollens' reagent does not react with alcohols. However, from personal experience, gentle heating of a primary alcohol with Tollens' will cause a small amount of oxidation and result in a fine black precipitate. Again, in my experience, secondary alcohols (such as in the question) do not react with Tollens' reagent and so I would predict that no reaction will take place here.

EDIT: Your proposed mechanism certainly looks plausible but whether it will take place in reality I can't say for certain (my experience of Tollens' reagent says no).


Is dipole moment a vector? Isn't it a misconception to assume that dipole moments are similar to vectors in their behavior? My reasons are as follows:

Let us take an example: Methane ($\ce{CH4}$)

Clearly, the hydrogen atoms pushing their electrons into the more electronegative carbon only enforce the effects of each other. But, we assume for no particular reason that dipole moments must behave like vectors. We then go on to add the vectors and finally decide that the net dipole moment is $0$. Why?

we assume for no particular reason that dipole moments must be behaving like vectors

Ah, but there is a reason. Consider the interaction of a molecule with the scalar potential

$$
E_{\text{int}} = \int \rho(\mathbf{r})\phi(\mathbf{r}) \, \mathrm{d}^3 \mathbf{r}
$$

where the integral over all space is turned into an expansion:

$$
\int \rho(\mathbf{r})\phi(\mathbf{r}) \, \mathrm{d}^3 \mathbf{r} = Q^{[0]}\phi^{[0]} - \sum_{n=1}^{\infty} \frac{1}{n!} \mathbf{Q}_{j_{1}j_{2}...j_{n-1}}^{[n]}\cdot\mathbf{E}_{j_{1}...j_{n-1}}^{[n-1]}.
$$

Now to identify some terms. $\phi^{[0]} = 1$, so $Q^{[0]}\phi^{[0]} = E_0$, the interaction-free or field-free energy. $\mathbf{E}^{[n]}$ is the internal electric field or electrostatic potential generated by the atoms to the $n$th total order, and the part that would be the derivative term in a Taylor expansion is the set of electric multipole moments:

$$
\mathbf{Q}_{j_{1}...j_{n-1}}^{[n]} = \int r_{j_1} r_{j_2} ... r_{j_n} \rho(\mathbf{r}) \, \mathrm{d}^3 \mathbf{r}.
$$

The first-order term is the (electric) dipole moment:

$$
\mathbf{Q}_{i}^{[1]} = \int r_{i} \rho(\mathbf{r}) \, \mathrm{d}^3 \mathbf{r} = \mu_i.
$$

The second-order term is the (electric) quadrupole moment:

$$
\mathbf{Q}_{ij}^{[2]} = \int r_{i} r_{j} \rho(\mathbf{r}) \, \mathrm{d}^3 \mathbf{r}.
$$

Convert the continuous electric dipole integral back into a finite sum by discretizing all space into atomic positions and representing the interaction of the electrostatic potential at zero strength with the atomic density as the atomic charge:

$$
\mu_i = \sum_{a}^{N_{\text{atoms}}} \sum_{i\in\{x,y,z\}} r_{ia}q_{a}.
$$

The set of atomic coordinates is nicely represented by each atom as a vector pointing from the Cartesian origin (0,0,0) to the atom $a$'s position $\mathbf{r} = \vec{r} = (r_x,r_y,r_z) = (a_x,a_y,a_z)$ or however you want to notate it, leading to

$$
\vec{\mu} = \sum_{a}^{N_{\text{atoms}}} \vec{r}_{a}q_{a},
$$

which can be seen as a vector of length 3 from the reduction $r_{ia} \rightarrow \vec{r}_{a}$. The quadrupole moment can be seen as a matrix describing the non-uniformity of the internal electric field, or its gradient.

  Dipole moment is a concept introduced by man ryt? So why are they vector quantities? Or better: How did man know that they are vector quantities? How was that deduced?

The dipole moment is a concept "introduced" by us in only two ways that come to mind. The first is that the electric field interaction is truncated to first order in the summation, which is mathematically inexact, even if higher-order moments happen to be zero. Nature includes this interaction to infinite order. Related to the first point, take a molecular sample and apply a single, uniform, finite electric field to it. The above answer deals with a system's permanent or internal moments, but one can also consider the dependence of the energy on an external electric field

$$
E(\mathbf{F}) = \sum_{n=0}^{\infty} \frac{1}{n!}E^{(n)}(\mathbf{a})\cdot(\mathbf{F}-\mathbf{a})^{n},
$$

where the electric field is $\mathbf{F} = \vec{F} = (F_x,F_y,F_z)$ and $\mathbf{a}$ is the expansion point. In practice, we always expand around $\mathbf{a} = \mathbf{0}$, so it is a Maclaurin series:

$$
E(\mathbf{F}) = \sum_{n=0}^{\infty} \frac{1}{n!}\mathbf{E}^{(n)}(\mathbf{0})\cdot\mathbf{F}^{n}.
$$

Assume that $E(\mathbf{F})$ is well-approximated by the first 3 terms of this expansion ($0 \leq n \leq 2$) and truncate it:

$$
E(\mathbf{F}) \approx E^{(0)}(\mathbf{0}) + \mathbf{E}^{(1)}(\mathbf{0})\cdot\mathbf{F} + \frac{1}{2}\mathbf{E}^{(2)}(\mathbf{0})\cdot\mathbf{F}^{2}
$$

I've been rather sloppy, but this is another way of writing the first derivation, except we are using external applied fields $\{\mathbf{F}\}$ rather than the internal fields $\{\mathbf{E}\}$. For a single applied electric field, only up to the first term $\mathbf{E}^{(1)}(\mathbf{0})$ survives, which is the induced electric dipole moment. I bring this up because assuming the applied field is strong enough that the induced dipole is stronger than the permanent dipole, the total electron distribution will most likely move uniformly or linearly with the field direction, so it makes physical sense to consider just the dipole (up to a point). (Unrelated, $-\mathbf{E}^{(2)}$ is the molecular polarizability $\alpha$.)

I would also say the above shows that we don't define the dipole moment as a vector, it appears naturally that way from mathematics. I also wouldn't call these "thought inventions".

So yes, the dipole moment is represented as a vector, the quadrupole moment is represented as a matrix, the octupole moment is represented as a 3rd-order tensor, etc.


Dissolving cast iron in nitric acid Decades ago my job was to find the silicon content in cast iron gravimetrically. The first step was dissolving the sample in hot Nitric acid. Did this turn all of the constituents in the sample, except the silicon, into water soluble salts? Was the silicon then bound as silicon dioxide?

No, nitric acid is a sufficiently hard acid to dissolve silicon metal without precipitating in low concentrations. In fact, nitric acid is used to dissolve silicon in ICP Standards. The oxide reacts slowly with water and water-based solutions - nitric acid included, but the acid does not increase the kinetics appreciably, otherwise glass would be unsuitable as a vessel for nitric acid. The element in an alloy, however, can be dissolved more easily like that in cast iron.


What is the lethal dose ( LD50 ) of coffee? Of course, there are many types of coffee, so I would like a range of lethal dose amounts for the average human, from the mildest of lattes to the most hardcore of espressos.

Pure caffeine's lethal dose is easily calculable, but what about coffee?

According to one reference:

LD$_{50}$ for coffee = 118 servings at 240 ml each; and

LD$_{50}$ for espresso = 175 shots (volume not specified),

for a 75 kg person.

A much more comprehensive quantitative description is given here for the LD$_{50}$ of coffee for an 80 kg person (113 cups at 250 mL each, 135 mg caffeine per serving). They account for half-life of elimination of caffeine, and arrive a the conclusion that one can't actually reach the LD$_{50}$ (15,360 mg) for caffeine due to metabolism, elimination, and the limit on how much water can be processed by the kidneys.

Finally, as noted in the last reference and hinted at above, the amount of water a person would imbibe on the way to reaching the 113 cup total (about 30 L) would likely kill the person (water intoxication).


What might serve as an initial starting photocatalyst for this large water-splitting solar simulator? Question: What might serve as an initial starting photocatalyst for this large water-splitting solar simulator? Surely there must have been some planned experiments!

The Gizmodo article Insane Light System Blasts the Energy of 10,000 Suns and Phys.org's Let there be light: German scientists test 'artificial sun' both describe the large, powerful concentrated light source called Synlight which has been built by the German Aerospace Center (DLR).

This large (~350 kW!) solar simulator - built from 149 high intensity xenon arc lamps, all collimated/focused to a single 20x20cm area - will be used to test and develop materials and catalysts that will allow photochemical splitting of water to yield hydrogen gas, to be scaled up later and used with a solar concentrator instead of electrically powered lamps. Hydrogen could be used as fuel directly, or to synthesize hydrocarbon fuels when combined with carbon dioxide or monoxide.

Water is transparent over most of the sun's useful spectrum, and this is for the obvious reason that the wavelengths reaching the ground are those not already absorbed in the atmosphere. Thus a photocatalyst will be necessary for a high yield of hydrogen gas from concentrated solar light.

From the Gizmodo article:

  When the array is focused onto a single spot, it heats metal to 1,475 degrees Fahrenheit (800 degrees Celsius), which is then sprayed with water vapor. The metal reacts with the oxygen in water, and hydrogen remains. With further heating, the oxygen is once again separated from the metal.

There are certainly easier ways to heat a metal plate than building a giant solar simulator, so I assume the point is to provide an intense source of photons with a somewhat similar spectrum to sunlight as a test platform for various catalysts. But the quote suggests the surface will react with the oxygen released, and need to then be cycled to higher temperature to release it. This does not sound like a very practical photocatalytic process.

So far I have not found anything clearer about what kinds of catalysts are likely to be investigated, but there must be some potential candidates. Based on known photocatalytic water splitting mechanisms, or if someone can find a discussion by DLR, what would be a starting point, or proof-of-concept photocatalyst that might work here?

below: Images from here, photo credit: DLR (top), DPA (bottom). The shape of each reflector is likely to be close to an ellipsoid with the xenon arc at one focus and the sample near the other. This way a ~100 cm diameter reflector can illuminate a much smaller 20x20cm sample.

[

Vitamin B-12 is a photocatalyst, see, for example, https://pubs.rsc.org/en/content/articlelanding/2018/ra/c7ra13037f#!divAbstract and also https://www.ncbi.nlm.nih.gov/pubmed/29271445 , but it has not been applied to water splitting!

What has been used is TiO2, see this source to split H2O into H2 and O2.

Mechanics: The action of light on TiO2 creates electrons (e-) and electron holes (h+). Some reactions:

$\ce{H+ + e- -> .H}$

$\ce{.H + .H -> H2}$

$\ce{OH- + h+ -> .HO}$

$\ce{.HO -> H+ + .O-}$

$\ce{.O- + h+ -> .O}$

$\ce{.O + .O -> O2}$

One of the less desirable reaction is:

$\ce{.H + .HO -> H2O}$


TM hydride chemical shift Why is the chemical shift for hydrides in TM complexes so strongly shielded, to the point where it's negative?

Initially, I thought it was because hydrides are $\ce{H-}$, but isn't the charge meaningless once it coordinates to a metal centre?

You essentially answer this question yourself. The chemical shifts of transition metal hydrides are often largely negative because they are highly shielded. That's what chemical shifts are a measure of - the amount of shielding from the external magnetic field. A very small hydride nucleus buried in a great big fluffy cloud of d-shell electrons of a transition metal is incredibly highly shielded from the base magnetic field.

The negative number of the chemical shift is arbitrary; what is relevant is that chemical shifts to the right are more shielded, and those to the left are less shielded.

Read the information about chemical shifts on this great online resource.


What is "Cresol Soap"? I recently came across a suggestion in an old (first published 1893) book that a solution of ~1% cresol soap in water could be used as a way to store a particular kind of root without allowing it to dry out or rot.

I've tried searching, but have been unable to determine what cresol soap might be. The closest I can find are several shady poorly-translated websites selling Lysol as cresol soap, or possibly a mixture of the two.

Does anyone have a clue what this book could be referring to?

According to an 1894 tract with the title Merck's Market Report and Pharmaceutical Journal: An Independent Monthly Magazine Devoted to the Professional and Commercial Interests of the Druggist, Volume 3, "cresol soap" was also called "crelium" or "crelium soap".
The google n-gram viewer suggests that "cresol soap" is a historical term, whose usage began in the 1890s, peaked dramatically during World War I and its immediate aftermath, and declined substantially afterwards.
A 1915 British medical report argues for using cresol soap as a vermin control agent in troop populations.  It gives a recipe as:

10 gallons of water
1.5 pounds of soft soap
Jeyes' fluid 1.5 ounces

This raises the question of what "Jeyes' fluid" is. Amazingly, Jeyes' Fluid is apparently still a commercial product, and contains a mixture of 4-chloro-m-cresol, isopropanol, terpineol, and "tar acids".
"Cresol soaps" are apparently still an item of commerce somewhere in the world, as a 2011 applications report form an analytical equipment vendor illustrates the analysis of a "cresol soap" sample, although a specific source is sadly not mentioned.  The primary detected constituents were m-cresol and p-cresol.

I suspect there are many other recipes that involve mixing soap, water, and crude cresol preparations of various kinds.  In the era of cresol soap's invention and popularity, coal was far more ubiquitous than now, and the cresol sources were probably crude fractions of distilled coal tar that were enriched in various isomers of cresol.  In those days awareness of potential toxicity from cresols was also not as developed as now.  The "tar acids" in Jeyes' fluid probably include a variety of cresol isomers.


Is the flame temperature of a burning fuel affected by pre-heating the fuel? This page on Adiabatic temperatures helpfully lists the theoretical maximum temperatures that flames will reach under certain conditions for a number of fuels. They assume initial atmospheric conditions of 1 bar and 20 °C.

However, I am remain confused about what would happen if the fuels (and air) were pre-heated to temperatures close to these maximums before they were mixed and ignited.

Would the flame temperature then be higher than these listed maxima?

Would the reaction be slowed down by the higher temperature? (I understand that this might happen for exothermic reaction in equilibrium with an endothermic one, but I don't think that describes burning fuel.)

Is there some other mechanism that prevents the flame temperature from exceeding these minima?

This question was inspired by discussion on a Skeptics.SE question similar to this closed Chemistry.SE question about jet fuel and steel beams. I think it is sufficiently focussed and apolitical to be on-topic here. My background level is an ancient undergraduate science degree; I have forgotten more chemistry than I remember.

Yes and no: preheating a mix a little bit may increase the final temperature somewhat, but adding even more initial heat has a decreasing effect on the final temperature.

There are two competing factors that limit the final temperature in a closed system:

The total amount of energy as the sum of the initial amount of thermal energy and the chemical energy, divided by the thermal mass [mass times heat capacity] of the system, and
The tendency for a reaction to be forced backwards by applying additional energy.

For example, $\ce{2H2 + O2 -> 2H2O}$ + 286 kJ/mole, but water dissociates more as the temperature rises. At 2,200 °C water is about 3% dissociated. At 3,000 °C  it is about 50% split, so half the chemical enegy is unavailable.

BTW, thermodynamically, a heat engine produces power based on the difference in temperature, so preheating a fuel-air mixture actually decreases efficiency, though it might be needed to vaporize the fuel (e.g. in Otto-cycle engine in cold weather).


How to calculate the pH of a buffer after adding HCl? In a certain $\pu{1 L}$ buffer with $\pu{0.25 M}$ $\ce{NH3}$ and $\pu{0.4 M}$ $\ce{NH4Cl}$, $\pu{0.1 mol}$ of $\ce{HCl}$ has been added to the solution. What is the $\mathrm{pH}$?
$K_\mathrm{a}(\ce{NH4+}) = \pu{5.56e-10}$, and $K_\mathrm{b}(\ce{NH3}) = \pu{1.8e-5}$.

The answer says that at last there will be $\pu{0.5 mol}$ of $\ce{NH4+}$ formed from the equation:
$$\ce{NH3 + H+ -> NH4+}$$
So $\pu{0.1 mol}$ of additional $\ce{NH4+}$ is formed and the total amount of $\ce{NH4+}$ is $\pu{0.5 mol}$.
However, I do not agree with this statement and I believe that there is still a limited power of $\ce{NH3}$ accepting $\ce{H+}$ (from $\ce{HCl}$).
My theory is that $\ce{NH3}$ accepts $x$ amount of protons and will form $x$ amount of $\ce{NH4+}$, with the original $K_\mathrm{b}$. Meaning
$$K_\mathrm{eq} = \frac{[\ce{NH4+}]}{[\ce{NH3}][\ce{H+}]}$$
which I thought was $1/K_\mathrm{b}$. Is my theory correct?

I'm not sure I follow your argumentation, but I guess you don't really need $K_\mathrm{b}$ at all since addition of strong acid ($\ce{HCl}$) influences the dissociation of the one buffer component - the weak acid ($\ce{NH4+}$):

$$\ce{NH4+ <<=>[K_\mathrm{a}] NH3 + H+} \qquad K_\mathrm{a} = \frac{[\ce{NH3}][\ce{H+}]}{[\ce{NH4+}]}$$

Henderson–Hasselbalch equation applied to this buffer system before the addition of acid allows to find initial $\mathrm{pH}$ (not required by the problem, I do this solely for demonstration):

\begin{align}
\mathrm{pH} &= \mathrm{p}K_\mathrm{a} + \log{\frac{[\ce{NH3}]}{[\ce{NH4+}]}} \\
&= -\log{(\pu{5.56e-10})} + \log{\frac{\pu{0.25 M}}{\pu{0.40 M}}} \\
&= 9.05
\end{align}

Once the strong acid ($\ce{HCl}$, assuming complete dissociation) is added, the equilibrium shifts accordingly:

\begin{align}
\mathrm{pH_1} &= \mathrm{p}K_\mathrm{a} + \log{\frac{[\ce{NH3}] - [\ce{HCl}]}{[\ce{NH4+}] + [\ce{HCl}]}} \\
&= -\log{(\pu{5.56e-10})} + \log{\frac{\pu{0.25 M} - \pu{0.10 M}}{\pu{0.40 M} + \pu{0.10 M}}} \\
&= 8.73
\end{align}

You would've needed $\mathrm{p}K_\mathrm{b}$ though when a strong base (e.g. $\ce{NaOH}$) were added.


Reconciling the units of pH calculation As part of a simulation I am developing I have an acid in aqueous solution represented as a concentration field of mass density (units g/L) at various points in space (denoted $[S]$). I want to calculate the pH of the solution. I am attempting to use the method for weak acids outlined here, but am struggling with the dimensionality of the variables,

$$[H]^2 + K_a [H] - K_a C_a = 0.$$

Looking at the quadratic equation, and knowing that $K_a$ has no units, it implies that $[H]$ and $C_a$ are also dimensionless. Does this mean that $[H]$ is in fact the activity of the the solution, and not the $H^+$ concentration? And is the relationship between $C_a$ and $[S]$ $$C_a = \gamma \frac{[S]}{M(S)c^\ominus},$$

where $\gamma$ is the activity coefficient, $M(S)$ is the molar mass of the acid, and $c^\ominus = 1$M is the standard amount concentration?

Followup question:

The solution to the quadratic equation is

$$[H] = \frac{-K_a \pm \sqrt{K_a^2 + 4K_a C_a}}{2},$$

thus $[S] = 0 \implies [H] = \frac{-K_a \pm K_a}{2} = -K_a$ or $0$.

As $pH = -\log_{10}\left([H]\right)$ and $K_a > 0$  there is no real solution, but logic would dictate that $pH = 7$ in this case. Where am I going wrong?

$K_a$ is not unitless (in this case). Some equilibrium constants may have their units totally cancel, but the acidity constant has units of concentration:

$$K_a = \dfrac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}\implies\dfrac{\frac{\text{mol}}{\text{L}}\cdot\frac{\text{mol}}{\text{L}}}{\frac{\text{mol}}{\text{L}}}=\frac{\text{mol}}{\text{L}}$$

I have used moles per liter (molarity) as it is a common concentration unit.

In this case, $K_a$ and $[\ce{H+}]$ both have concentration units, as does $C_a$, so your initial equation $[\ce{H+}]^2 + K_a [\ce{H+}]+ K_a C_a =0$ is valid (at least for units.

If you recast $K_a$ and $C_a$ in terms of activities, then you need to replace $[\ce{H+}]$ with $\alpha_{\ce{H+}}$ (which is unitless).


Endo rule and stereoselectivity in the Diels-Alder reaction Do you agree with the stereocenter with the $\ce{-CF3}$ and $\ce{-CN}$ substituents in the product?

source

I thought that since $\ce{-CF3}$ is a better electron withdrawing group than $\ce{-CN}$, it would be placed opposite the diene, giving the product in which $\ce{-CF3}$ points up and $\ce{-CN}$ points down.

However, I believe my proposal above is wrong, because probably the substituent with the highest electron-withdrawing property, the $\ce{-CF3}$ should be placed endo because of secondary orbital interactions. Therefore, I expect the product above should be the correct one.

source

When I try to do the endo-product of the following reaction I came to the other syn product with absolute stereochemistry (S,R) instead of the (R,S)-product shown here. So I'm wondering, why we get endo selectivity in the first reaction and exo selectivity here instead.

Do you agree with the stereocenter with the CF3/CN in the product?

No, I would have expected the other isomer, the one with the $\ce{CN}$ and $\ce{NO2}$ groups pointing down in your drawing.

Rationale:

When exo and endo isomers can form in the Diels-Alder reaction, there is often a preference for formation of the endo isomer.  Although this endo preference is often referred to as the "endo rule", it is not a "rule", just a general, and rather weak, preference for endo product formation.  In fact, in most cases both isomers are formed, just more of the endo isomer (see figure below).

The explanation most often advanced for the endo preference is "secondary orbital interactions." The following diagram illustrates the two possible transition states (TS) in the reaction of cyclopentadiene and an $\alpha,\beta$-unsaturated carbonyl compound.  One TS leads to the exo product, the other TS to endo.  Note how the endo TS has the unsaturated carbonyl bond tucked under the double bonds in the cyclopentadiene ring; the exo TS does not.  It is speculated (1) that the overlap of the carbonyl unsaturation with the diene unsaturation somehow stabilizes the endo TS making endo product preferred.  The accompanying table illustrates how weak the endo preference can be.

image source

In your top example, this endo preference for a TS with overlapping unsaturation would lead me to suspect that the product with the unsaturated nitro and cyano groups (the $\ce{CF3}$ group is not unsaturated) oriented away from the 3-carbon bridge and tucked towards the double bond would be preferred.

Also, keep in mind that the endo preference applies to cases were the Diels-Alder reaction is run under kinetic control.  As the following example shows, the exo product is generally thermodynamically preferred (less steric crowding). So if the reaction is run under conditions (high temperature) where the initially formed kinetic (endo) product reverts to starting material, eventually the thermodynamically favored product (exo) will be formed eliminating the kinetic preference for the endo product.

(1) see here for an alternate explanation for the endo preference


How to reverse copper electroplating? We did a lab where we built a simple electrolytic cell. In a beaker was a solution of $\ce{CuSO4}.$ A copper strip (anode) and a copper wire (cathode) was place in it. A wire connected the copper strip to the negative DC power source and a wire connected the copper wire to the positive DC power source. The power source was turned on and the copper wire was electroplated.

Considering that both electrodes are the same material, would it be possible to create a galvanic cell by reversing the reaction?

I think it would not be possible. Since galvanic cells use spontaneous reactions to create electrical energy, you would have these two copper electrodes in a solution of $\ce{CuSO4}$ separated by a salt bridge and connected by a wire. Nothing would happen because both coppers are the same and neither one of them can attract electrons more than the other.

The galvanic cells from the same material are possible. They are called concentration cells and they use two electrolytes of the different concentration of the same electrolyte.

But their voltage is low. In the case of $\ce{CuSO4}$

$$U = 0.0295 \log\frac{c_1}{c_2}\ \mathrm V$$

Note that the electrode naming is based on the direction of the electron flow, and their mutual polarity are switched for electrolysis versus cells (primary cells or by convention discharging mode of secondary cells ).

Anode = electrons flow upward (anabasis), an anode is the place of oxidition. (Xenophon, Anabasis, "The way upwards")

Cathode = electrons flow downwards, a cathode is the place of reduction (katabasis - the way downwards)

During electrolysis, the anode is the positive electrode and the cathode is the negative electrode.

During the cell discharging, the anode is the negative electrode and the cathode is the positive electrode.


Relative stability of keto- enol form of acetyl chloride Why is the keto form of acetyl chloride more stable even though the enol form is stabilized by the hydrogen bonding?

The stability of keto-enol compounds depends on the solution they are present in.

Keto compounds are stabilized due to inter-molecular hydrogen bonding when present in a polar solvent (for example water).

Enol compounds are stable when present in non-polar solvents, in which intra-molecular hydrogen bonding stabilizes them.


Why does a matchstick burning in a container cause suction? As you see in the picture, a burning match in a container causes suction and this is a traditional therapeutic method to relieve bone pain.

What is the reason? I think it should be related to gas changes ($\ce{O2}$ being used for combustion and etc.), but I am not sure about it.

When the container is open to the atmosphere and has the match burning in it, the temperature of the air inside is higher than the temperature of the surrounding air.  In order for pressure to be equal in the container and outside it, the density of the air in the container is less.  When the match is removed and the container placed on the person, two things happen.  One, no air can enter or leave the container, and two, the air in the container begins to cool.  When the air in the container cools its pressure drops.  The atmosphere pushes on the air in the container indirectly through the person, creating what is called suction.

This is similar to drinking through a straw.  You are not sucking the liquid through the straw as much as you are removing the opposing force that prevents the atmosphere from forcing the liquid through the straw.  The opposing force being the air pressure in the straw.


Titration with Ca(OH)2 In the laboratory, $20.0$ milliliters of an aqueous solution of calcium hydroxide, $\ce{Ca(OH)2}$, was used in a titration. A drop of phenolphthalein was added to it to indicate the end point. The solution turned colorless after $20.0$ milliliters of a standard solution of $\pu{0.050 M}$ $\ce{HCl}$ solution was added. What was the molarity of the $\ce{Ca(OH)2}$?
  (A) $\pu{0.010 M}$
  (B) $\pu{0.025 M}$
  (C) $\pu{0.50 M}$
  (D) $\pu{0.75 M}$
  (E) $\pu{1.0 M}$

I did not know how to answer such kind of question using formulas, that is why I applied logic. At first, I wrote such a reaction:

$$\ce{\underset{\pu{20 mL}}{Ca(OH)2} + \underset{\pu{0.05 M}}{\underset{\pu{20 mL}}{HCl}} = end point}$$

Then, I understood that if the volume of both reactants is equal, the molarity of $\ce{HCl}$ must be higher than that of $\ce{Ca(OH)2}$, as $\ce{HCl}$ changes the nature of $\ce{Ca(OH)2}$.

As I knew that the molarity of $\ce{Ca(OH)2}$ is just a bit less than that of $\ce{HCl}$, I chose B, and this variant turned out to be the right one.
In the book, I have such explanation:

  In the titration, the reaction is:
  $$\ce{2HCl + Ca(OH)2 = CaCl2 + 2H2O}$$
  The acid to base ratio is $2 : 1$, or (moles acid used) = 2(moles base used), so $M_\mathrm{a}V_\mathrm{a} = 2M_\mathrm{b}V_\mathrm{b}$, where $M$ is the molarity and $V$ is the volume expressed in liters. Then
  $$M_\mathrm{b} = \frac{M_\mathrm{a}V_\mathrm{a}}{2V_\mathrm{b}}$$
  $$M_\mathrm{b} = \frac{\pu{0.05 M} \times \pu{0.02 L}}{2}$$
  $$\pu{0.02 L} = \pu{0.025 M}$$

I do not understand the explanation. Although it comes up with the same variant as I did, if I had such formula I would use it this way $2M_\mathrm{a}V_\mathrm{a} = M_\mathrm{b}V_\mathrm{b}$ as we have 2 moles of acid.

Then, the answer would be $0.1$. The same formula is used on this forum. Can you help me understand what is wrong here please?

(1) Meaning of the textbook solution

$$\ce{2HCl + Ca(OH)2 = CaCl2 + 2H2O}$$

This reaction clearly tells us that

$$\frac{moles_{acid}}{moles_{base}}=2$$

moles=molarity* volume(in litres)

Therefore, $$\frac{moles_{acid}}{moles_{base}}=\frac{M_{a}V_{a}}{M_{b}V_{b}}$$

(2) What is wrong with your interpretation?

You wrote $2M_\mathrm{a}V_\mathrm{a} = M_\mathrm{b}V_\mathrm{b}$.

$2M_\mathrm{a}V_\mathrm{a} = M_\mathrm{b}V_\mathrm{b}$ actually implies that more number of moles of base need to be taken. For neutralisation, for every one mole of base, we require two moles of acid since each calcium hydroxide will release 2 hydroxide ions (for which two H+ are required).


Structure of a metal in gas phase I'm going to use aluminium in my example, but this is applicable to all metals.

Aluminium has delocalised electrons. It has aluminium cation -electron attraction. It's easier to visualize the atomic structure of aluminium in solid form.

But if it's in the gas phase, will delocalised electrons return to the atom itself, or is there going to be an electron and cation mixture (electrons floating around)?

So can aluminium (or any other metal) conduct electricity as a vapour, and what would its conductance be compared to its solid form?

Metal vapors do conduct electricity, as in the mercury vapor lamp (also used in fluorescent lamps), but, since the atoms are isolated or in small (molecular) clusters, conduction is ionic, not metallic. Since electrons are far less massive (and therefore more mobile) than the heavy ions, they are responsible for the majority of the conducted current.

Electrons are forcibly ripped away from the mercury atom (well, raised to higher levels) -- the first ionization potential is 10.39V, and its second is 18.65V, so in a mercury vapor lamp or mercury vapor rectifier, the potential drop of at least 20 volts implies more than one of the outer electrons are lost from many of the atoms.

Note that ionized gases are much less conductive than solid or liquid metals. Consider a 70 watt sodium vapor lamp, with an arc length of perhaps 10 cm and a potential drop of ~18 volts. The finest commercially available wire, 40 gauge, would have a potential drop of just ~1.3 volts for metallic copper wire, or ~4 volts for sodium, which has 1/3 the conductivity of metallic copper (if they made fine sodium wire), at the same length and current (~3 A).
See the useful chart explaining sodium-vapor ionization .