Paper Towel Soaked with Isopropanol/Water Mixture Burns without Damaging the Paper. Why? Video: https://www.youtube.com/watch?v=_qFN1hDZwp8

If you choose not to view the video, in the first demonstration a paper towel is soaked in isopropanol. As expected, the paper towel burns. However, in the second demonstration the paper towel is soaked in isopropanol and water. Surprisingly, the towel burns without being damaged.

My chemistry professor tells me that it has something to do with the energy cost of a phase change $\left[\ce{H2O}\ (l) \longrightarrow \ce{H2O}\ (g)\right]$; however, I am still very confused.

Could someone please elaborate? Thank you.

I think your professor is incorrect, and it has to do with the thermodynamics and kinetics of the combined vaporization/combustion process.

The heat of vaporization of water $\left(40.7\ \mathrm{kJ}\ \mathrm{mol}^{-1}\right)$cite is considerably less than the heat of combustion of isopropanol $\left(1830\ \mathrm{kJ}\ \mathrm{mol}^{-1}\right)$cite.  These numbers indicate that approximately $45$ water molecules must evaporate for each isopropanol molecule combusted in order to hold the temperature constant. This  corresponds to a $\sim2\%$ mol/mol solution, which (a) is not consistent with the $\sim50\%$ v/v solution he actually used, and (b) would probably be too dilute to burn at room temperature the way it does in the video. (Someone more motivated than I could probably confirm this using flash point data.)

I think, instead, the behavior depends on the following:

Vapor-liquid equilibrium stipulates that a liquid phase can never be at a temperature greater than its boiling point$^\dagger$.
It is the vapor of combustible substances that actually burns, not the liquid (see, e.g., this OSHA training document)
Up to a concentration of about $60\%$ alcohol, the vapor phase in equilibrium with a water-isopropanol liquid mixture has a considerably higher concentration of isopropanol as compared to the liquid (source):

For the present purpose, to read the diagram choose a temperature on the $y$-axis and track across the $x$-axis, noting where you cross the red and blue lines.  These correspond to the equilibrium liquid and gas phase compositions, respectively, at that temperature, given a mixture that's not too lean or too rich in isopropanol.  For example, take $82\ ^\circ\mathrm C$: you hit the red line (liquid) at about $20\%$ isopropanol, and the blue line (vapor) at about $53\%$ isopropanol.

Since the boiling point of isopropanol is $82.6\ ^\circ\mathrm C$, #1 means that as long as there's liquid left soaked into the paper (even $100\%$ isopropanol!), the paper can't burn.  You can actually see this in the video: he ignites the paper at around 0:57, but it doesn't actually start to char until about 1:08, ten seconds later.

The reason why the paper does then start to char in spots is that those spots have locally run out of liquid isopropanol to keep them cool.  There remains a cloud of burning gaseous isopropanol all around the paper (clearly!), which is more than enough to burn the now-dry paper.

When there's water mixed with the isopropanol, though, the vapor-liquid equilibrium comes into play.  Per point #3, based on the phase diagram, the isopropanol likes to be in the gas phase a lot more than the water, so once the vapor is ignited and burning (point #2), the heat produced tends to evaporate more of the alcohol than the water.  This feeds the flame for a while, but eventually it sputters out, with plenty of water left on the paper to keep it cooled well below its char point.

$^\dagger$ Mixtures of liquids technically don't have boiling points, as their behavior is more complex, but the concept is useful for making the point.


When are the prefixes bis, tris etc. used in nomenclature of complexs? Are the prefixes bis-, tris- etc. used only for ligands that are bidentate in complex nomenclature?

For naming coordination compounds, the multiplicative prefixes ‘di’, ‘tri’, ‘tetra’, ‘penta’, etc., are used in the case of simple ligands, for example:
$\ce{[PtCl4]^2-}$  tetrachloridoplatinate(2−)
The alternative multiplicative prefixes ‘bis()’, ‘tris()’, ‘tetrakis()’, ‘pentakis()’, etc. are not only used for bidentate ligands. They are generally used with composite ligand names or in order to avoid ambiguity, for example:
$\ce{[Pt(PPh3)4]}$  tetrakis(triphenylphosphane)platinum(0)
The corresponding section in the current version of Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book) reads as follows:

IR-9.2.2.2 Number of ligands in a coordination entity
Two kinds of multiplicative prefix are available for indicating the number of each type of ligand within the name of the coordination entity (…).
(i) Prefixes di, tri, etc. are generally used with the names of simple ligands. Enclosing marks are not required.
(ii) Prefixes bis, tris, tetrakis, etc. are used with complex ligand names and in order to avoid ambiguity. Enclosing marks (…) must be placed around the multiplicand.

The use of multiplicative prefixes is also explained in the IUPAC Technical Report Brief guide to the nomenclature of inorganic chemistry. Pure Appl. Chem. 2015, 87(9–10), 1039–1049 as well as in the corresponding four-sided lift-out document, which is available as supplementary material.
Composite ligand names include ligand names which are substituted or which themselves contain multiplicative prefixes or locants.
For example, the normal prefix is used in ‘diammine’ $(\ce{(NH3)2})$, but the alternative multiplicative prefix is used in ‘bis(methylamine)’ $(\ce{(CH3NH2)2})$ to make a distinction from ‘dimethylamine’ $(\ce{(CH3)2NH})$.
Note that the use of the alternative multiplicative prefixes is not limited to coordination compounds. They are generally used for compositional names in inorganic chemistry when a simple multiplicative prefix may be misinterpreted. For example,
‘tris(iodide)’ is used for $\ce{(I- )3}$ (rather than ‘triiodide’, which is used for $\ce{I3-}$);
‘bis(phosphate)’ is used for $\ce{(PO4^3- )2}$ (rather than ‘diphosphate’, which is used for $\ce{P2O7^4-}$.


Why Does A Brønsted–Lowry Acid Accept Proton from Stronger Acid? This is mechanism of Nitration of Benzene. In the first step, Nitric Acid (pKa = -1.4) accepts a proton from sulfuric acid which is stronger acid (pKa = -9). According to Brønsted-Lowry theory we consider acids dissociate increasing proton concentration and bases dissociate increasing hydroxy concentration. On the other hand bases also accept proton from acids. Therefore should we accept nitric acid as a base in this case ? If not, why does an acid accept a proton from another "stronger" acid ?

Nitric acid is definitely a strong acid, but that does not exclude the possibility of protonating it with a stronger acid. It can be effectively protonated by a stronger acid such as sulfuric acid. Simply because something has "acid" in its name does not mean it has to dissociate into protons and its conjugate base, or protonate something else.

$$\ce{HNO_3 + H2SO4 <=>H2NO3^+ +HSO4-}$$

Note also that even strong acids in their pure forms - i.e. liquid nitric acid or liquid sulfuric acid - can and do auto-ionize to a fair extent. So if nitric acid can react with itself, of course it can react with a stronger acid.


Textbook chemistry vs Real life chemistry and impurities Long ago when I took chemistry courses in high school and lower, chemistry seemed so simple and elegant. All equations and exercises involved pure elements, pure compounds, with a very clean simplified diagrams.

Now that I look at real world, the world is full of impurities. Even water we drink is not pure $\ce{H2O}$, but rather full of "other stuff", and no two samples of water are "absolutely identical".

This makes me wonder. To what extent can you overlook these impurities when doing science, in predicting outcomes of experiments.

For example, you can make naive formulas for how much salt can dissolve in water until it stops dissolving salt. However, in real world,

Pure water does not exist as something easily available, and if it does, from what I read, it doesn't stay pure for very long.
Table salt, while mostly sodium chloride, often contains iodine, and "other stuff" that I am not sure about.

So surely these impurities impact the behavior of your experiments, and depending on the experiment, you can get reactions you never expected or wanted, or behavior can be drastically different.

My question is, how do scientists control experiments when dealing with unpure compounds? How do they know that their reactions to produce "table salt" don't create unwanted chemicals that will blow up in small quantities, or create some new kind of poison that they will only be noticed in a few days? Perhaps create some kind of acid that burns through the glass of their beaker or cause some other hazardous effects. Are they naive and ready to face the unexpected or are there absolute certainties in which they can control it (can you really control everything?)?

Well, scientists usually purchase the purest grade of chemical available if they want to use it for a reaction. (Leading to the absurd case that the less pure reagent grade is often dismissed for the purer analytical grade.) Usually, special care is taken for these chemicals to remove especially those impurities which could hurt most — you really don’t want any palladium in your sample of sodium bromide; that could do some really bad things.

That directly leads to a second point: No chemist would ever use ‘table salt’ or tap water unless in dire need. As you state, the impurities are basically incontrollable.

With that in hand, usually chemists have a good idea of what they just added and how that influenced the reactions. But often some things still go wrong. One of the most famous examples in organic chemistry is the Nozaki-Hiyama-Kishi coupling, which was initially reported as requiring only a $\ce{Cr^{II}}$ salt — usually $\ce{CrCl2}$. However, yields of other groups were much much worse than those originally reported by Nozaki and Hiyama. It wasn’t until eleven years after initial publication that — independently of each other — the original authors and Kishi both discovered that $2~\text{mol-}\%$ $\ce{Ni^{II}}$ are required as a co-catalyst for the success of the reaction. The original chromium salt batches of Nozaki and Hiyama were obviously contaminated by enough nickel(II).

An inorganic example I once heard of in a solid state chemistry course was a crystal structure published with experimentals in a solid state reaction. (I can’t remember the exact compound.) Other authors tried to reproduce the reaction but got much worse yields than the $100~\%$ that are expected in solid state reactions. The solution was that the square unit cell contained a hydride anion in its centre, almost invisible due to its low electron density. This hydride was introduced to the reaction from the silicon grease the authors used for the glass joints.


Realistic conditions for fuel container explosion First off, I'm not actually trying to blow anything up. Also, my apologies if this isn't the right SE to ask. If so, I'd appreciate it if you could point me to the right SE.
I'm writing something where at a point I need a big explosion to happen. I've done some digging and I found out that the old bullet-fired-at-gas-tank trope is essentially just a trope, and that it alone is unlikely to cause an explosion.
The explosion needs to happen near a big mansion on an otherwise uninhabited island. The mansion is powered by an outdoor generator that runs on some kind of fuel, for example propane, though it can be anything else that makes sense and can more easily get me the explosion I need. There are fuel containers
to refill the generator sitting somewhere outside the mansion, and they need to be accidentally blown up during an action sequence.
The ideal explosion would be strong enough to blow up the closest wall of the mansion, so that the fire can spread inside the mansion, and this leads to my first issue: I have no idea if it's realistic to expect such a dramatic effect or not—for all I know, it might take far more fuel than it makes sense to keep in stock.
Secondly, as said just shooting one of the containers won't trigger an explosion. I understand that you need the right mixture of oxygen and fuel, and the containers would have to be exposed to sufficient heat for some time before they actually blow up. I thought that, maybe, if some containers are shot and start leaking, and then the leaked fuel somehow catches fire during the sequence, this might trigger an explosion, because air would start leaking into the containers that got a hole punched in them.
To sum up, my questions are:

Can a fuel explosion realistically blow up a nearby wall and spread a cloud of fire into the mansion so that at least the nearest room is almost completely on fire very quickly, without unrealistically huge amounts of fuel involved?

Can the explosion be caused by a bunch of fuel cans that were hit by bullets, started leaking, and were heated up by the flames that followed when the leaked fuel somehow caught fire? Would this take too long to happen (eg, ten minutes), or could it happen within a minute or two?

The topic of explosives in movie production is a specialty application as one does not generally employ classic explosives!
In movies, the aim is NOT to kill/injure anyone (especially, your famous lead actor). As such, one can select even a flameless explosive, which, for example, could include Ammonium nitrite ($\ce{NH4NO2}$). The latter is primarily a pressure explosive which, per a chain reaction, quickly releases a massive amount of gas. Such explosives find applications, for example, in blasting in coal mines. Apparently, mines can develop a potentially explosive mixture of methane and air, where a flame/spark could result in a massive secondary explosion!
Also, distinct from a battlefield setting, consider low brisance explosives, which do not produce lethal shockwaves.
In essence, your objective should differ significantly from the design/use of military-grade explosives. Instead, aim at much flash and dust, but less lethal. The sound level of the explosion can be addressed in the post-editing of the film.
To accomplish this, you need more research and the opinion of someone who actually specializes in this area.


How to perform Mulliken population analysis by hand with GAMESS output? I did a simple calculation on hydrogen fluoride with GAMESS and I wanted to reproduce the Mulliken population analysis for atomic orbitals.

A molecular orbital is a linear combination of atomic orbitals and if I understand correctly the Mulliken analysis uses coefficients to calculate population (overlap or atomic).

In the molecular orbital part of my GAMESS output, if I want to calculate the population of the 1s orbital of the hydrogen, I'm summing every coefficients squared of the contribution of this AO in all the MO, but I'm not getting the right answer.

Would someone be able to clarify this?

INPUT:

 $CONTRL SCFTYP=RHF RUNTYP=OPTIMIZE COORD=CART ICHARG=0 MULT=1 $END
 $SYSTEM TIMLIM=90  MEMORY=1000000 $END
 $STATPT OPTTOL=1.0E-4  NSTEP=100 $END
 $BASIS  GBASIS=STO NGAUSS=3 $END
 $SCF DIRSCF=.TRUE. $END
 $GUESS  GUESS=HUCKEL  $END
 $DATA
 HF
C1
H     1.0     0.00000     0.00000     0.00000
F     9.0     0.00000     0.00000     1.0000
 $END

OUTPUT:

      ------------------
      MOLECULAR ORBITALS
      ------------------

                  1          2          3          4          5
              -25.9035    -1.4599    -0.5737    -0.4631    -0.4631
                 A          A          A          A          A
1  H  1  S    0.005341   0.150431  -0.533704   0.000000   0.000000
2  F  2  S   -0.994746  -0.250678  -0.078268   0.000000   0.000000
3  F  2  S   -0.022260   0.946703   0.410905   0.000000   0.000000
4  F  2  X    0.000000   0.000000   0.000000  -0.004070   0.999992
5  F  2  Y    0.000000   0.000000   0.000000  -0.999992  -0.004070
6  F  2  Z    0.002673  -0.078256   0.698060   0.000000   0.000000

                  6
                0.5898
                 A
1  H  1  S    1.054350
2  F  2  S    0.080571
3  F  2  S   -0.515861
4  F  2  X    0.000000
5  F  2  Y    0.000000
6  F  2  Z    0.816436

           ----- POPULATIONS IN EACH AO -----
                         MULLIKEN      LOWDIN
          1  H  1  S      0.80774     0.85856
          2  F  2  S      1.99913     1.99848
          3  F  2  S      1.94822     1.88452
          4  F  2  X      2.00000     2.00000
          5  F  2  Y      2.00000     2.00000
          6  F  2  Z      1.24491     1.25844


Is Changing between Allotropic forms a Physical or Chemical Process? When a substance changes its alloptropic form, is this considered a physical change or a chemical change?

Right now, I am thinking that it would depend on what the change was. For example, if graphite was changed to diamond under high pressure, I would say it is a chemical change since new bonds are formed. However the change of iron from a body-centred cubic structure (ferrite) to a face-centred cubic structure (austenite) at high temperatures would be a physical change because only the arrangement of the atoms are changed.

Am I correct?

Well, it's a matter of choice and tradition. We traditionally classify phase transitions (such as melting of ice or evaporation of water) as physical changes. The conversion of ferrite into austenite is a phase transition, so it should formally be classified as a physical change. But yes, it's hair splitting indeed, especially when you consider the phase change from diamond (giant covalent lattice, sigma-bonds only) to fullerene (simple molecular lattice, sigma plus delocalised pi-bonds), which clearly shows some characteristics of a chemical transformation (the change in the bonding nature). Still, even such changes are formally "physical". The traditions do not go easily!


Where does the Fe(III) ion go? Iron(II) sulphate solution is added to a test-tube. Following this, potassium manganate solution is added gradually to the test-tube as well. The observation seen is that the green solution turns pink at first, then potassium manganate turns colourless as manganese(VII) is oxidised to manganese(II).

But Fe(II) is oxidised to Fe(III) ions as well, which are brown in colour, so shouldn’t the solution turn brown instead of colourless?

I have seen these observations experimentally in the lab but still did not understand. I know I’m probably thinking about it the wrong way so any help would be much appreciated.

In solution, the iron(III) will be hexaaqua iron(III), 6 water molecules coordinating the iron ion, with the oxygen atoms being at the vertices of an octahedron.

Hexaaqua iron(III) has a high spin d5 electron configuration.

All the d-d transitions are spin forbidden.

Hexaaqua iron(III) is nearly colorless.


How to determine the theoretical carbonic acid concentration in soda water from a Soda Stream I've seen it cited that commercial carbonation devices like a SodaStream put about 100 psi (700 kPa) of pressure to dissolve gaseous $\text{C}\text{O}_{2}$ in to solution. I've also heard that the reason carbonated water tastes different than the water before it is carbonated is the presence of carbonic acid ($\text{H}_{2}\text{CO}_{3}$). I guess my question is a two-parter:

1) How can one determine the gaseous $\text{CO}_{2}$ concentration knowing only the temperature (assume room temperature) and the pressure (above).

2) How can one determine the carbonic acid concentration knowing only the dissolved $\text{CO}_{2}$ (g) concentration?

Is there a lit. value for the eq. reaction coefficient for this reaction at room temperature?

$$\ce{CO2(g) + H2O(l) <=> H2CO3(aq)}$$

If so, where would I find it? And could it be used to get a ballpark for the carbonic acid concentration?

The equilibrium you describe is quite well documented (especially with entire industries depending on their ability to carbonate drinks), so the Wikipedia page for carbonic acid is often a good place to start. The equilibrium constant, ${K_\mathrm{h}}$, is given by:

$$\ce{\frac{[H_2CO_3]}{[CO_2]}}=\pu{1.7e-3}$$

Now, for your first question:

  How can one determine the gaseous $\ce{CO_2}$ concentration knowing only the temperature (assume room temperature) and the pressure (above).

Well, for starters you should probably check the volume of the container and from there you could work out the amount of $\ce{CO2}$ that is above the solution. But if you are interested in the amount inside solution, you could check this paper published in 1940 which has basically compiled the data for you.

And the second part:

  How can one determine the carbonic acid concentration knowing only the dissolved $\ce{CO2(g)}$ concentration?

Now that you know the equilibrium constant, I'm sure you can work out the amount of carbonic acid.


Dinitrophenyl (DNP) derivatives of amino acids (specifically epsilon-DNP-lysine) I have a pretty basic biochemistry question but am having trouble finding the answer to it:

Normally, DNFB (dinitrofluorobenzene, or 1-fluoro-2,4-dinitrobenzene to be exact) reacts with just the amino terminus of amino acids, creating dinitrophenyl (DNP) derivatives.  However, with lysine, DNFB can react in two places (at the amino terminus and to the epsilon amino group).  The DNP derivative of lysine is referred to as "epsilon-DNP-lysine".

Does "epsilon-DNP-lysine" refer to a product where the DNP is bound only to the epsilon amino group (but not to the amino terminus), or does it refer to a product where DNP is bound simultaneously to the epsilon amino group and to the amino terminus?

Also, can the same thing happen with other amino acids that have an amino group in the side chain?  i.e. can there be epsilon-DNP-arginine, for instance, or is this behavior unique to lysine?

Does "epsilon-DNP-lysine" refer to a product where the DNP is bound only to the epsilon amino group (but not to the amino terminus), or does it refer to a product where DNP is bound simultaneously to the epsilon amino group and to the amino terminus?

Good question.  ε-DNP-lysine usually refers to a lysine in which the dinitrophenyl substituent is bound only to the ε atom, i.e., to the side-chain nitrogen.

The reason is that in lysine-focused derivatization methods, the reaction with DNP can be done prior to protein hydrolysis.  Thus, during DNP derivatization, the vast majority of lysine residues will have no free α-amino terminus because they are part of polypeptide chains.  Only the ε-amino groups are free to react with DNFB.  Derivatization forms DNP-ylated proteins where the peptide bonds are intact.  Later steps in the procedure hydrolyze all the peptide bonds.  During this step, free α-amino groups are created.  But no DNFB is used in these later steps.

I found a paper from 1963 which describes the derivatization procedure in detail.

  ...Care was taken to
  deposit the sample at the bottom of
  the flask. Ten milliliters of bicarbonate
  solution were added, care being taken
  that no meal adhered to the side of the
  flask. The contents were then thoroughly
  mixed by gentle swirling, and the
  suspension was permitted to stand for 10
  minutes. A solution of 0.3 ml. of
  DNFB in 10 ml. of ethanol was then
  added and the contents of the flask
  were thoroughly mixed by gentle
  swirling. The side of the flask was
  rinsed with 3 ml. of absolute ethanol
  and then the contents were shaken in
  subdued light for two hours on a shaker
  with a wrist-like motion. Previous
  work (10) has established that 2 hours
  reaction time is sufficient for substantially
  complete reaction between DNFB
  arid the free epsilon amino groups of
  lysine. alcohol and most of the
  water were removed by evaporation
  under an air stream, and the residue was extracted with four 50-ml. portions
  of anhydrous, peroxide-free diethyl
  ether. The ether in each case was
  removed from the residue by decantation
  and the residue in the flask was dried at ambient temperature by
  aeration.

The ether washings will remove the vast majority of unreacted DNFB.

  ... Two hundred milliliters of
  constant-boiling aqueous HCl nere
  added to the flask, and the resulting
  mixture was boiled overnight at the
  reflux temperature, cooled and then filtered through a sintered glass funnel
  directly into a 250-ml. volumetric flask.
  The filtrate, and washings from the
  filtrate, were made up to 250 mi. with
  distilled water.
  An aliquot (2.0 to 4.0 ml.) of the
  hydrolyzate containing from 0.1 to 0.3
  mg. of epsilon-dinitrophenyllysine...

Now for your other question:

  Also, can the same thing happen with other amino acids that have an amino group in the side chain? i.e. can there be epsilon-DNP-arginine, for instance, or is this behavior unique to lysine?

DNFB is not reactive with arginine.  The arginine side-chain is best viewed as a guanidine, not as an amino group.  Guanidine is considerably less nucleophilic than lysine.


Work done by gas in isothermal process A $20\ \mathrm l$ steel cylinder containing $\ce{N2(g)}$ at $20\ \mathrm{atm}$ and $300\ \mathrm K$ is kept in a big jar containing air at $1\ \mathrm{atm}$ and $300\ \mathrm K$ and sealed. $\ce{N2(g)}$ starts leaking into the jar but reversibly and isothermally. When both
  mechanical and thermal equilibria are stabilized, pressure in the jar was $1.19\ \mathrm{atm}$. Determine work
  done by $\ce{N2(g)}$.

Since there is mechanical equilibrium, the pressures should be equal to $1.19\ \mathrm{atm}$ inside and outside the cylinder with nitrogen. And since temperatures are equal too, should I use this formula for work?
$$W=-nRT\ln\frac{V_{2}}{V_{1}}$$

The compression of the air outside the cylinder by the $\ce{N2}$ escaping from the cylinder takes place reversibly (as far as the air is concerned), but the process of $\ce{N2}$ leaking out of the cylinder through a valve or small orifice is definitely irreversible because of the Joule–Thompson throttling that takes place as the $\ce{N2}$ flows from a high pressure inside the cylinder to a much lower pressure outside the cylinder.  So to calculate the work done by the $\ce{N2}$, one cannot treat the expansion of the nitrogen as reversible.  Instead, one must focus on the reversible work required to compress the air.

Assuming that the $\ce{N2}$ does not diffuse into the air (i.e., the boundary between the air and the $\ce{N2}$ remains well-defined), the final volume occupied by the $\ce{N2}$ is $11 × 20/1.19 = 184.9$ liters.  That means that the volume of $\ce{N2}$ increased by $184.9 - 20 = 164.9$ liters.  This is the volume of $\ce{N2}$ outside the cylinder (in the jar) in the final state of the system.  If $V_0$ represents the volume in the jar outside the cylinder (and also the initial volume of air), the final volume of air is $V_0 - 164.9$.  So, the ratio of the initial volume of air to the final volume of air is equal to the pressure ratio of the air, and given by:

$$\frac{V_0}{V_0 - 164.9} = 1.19$$

Solving for $V_0$ then yields $V_0 = \pu{1033 L}$.

The work done by the air on the $\ce{N2}$ is given by

$$
\begin{align}
W_\mathrm{a} &= n_\mathrm{a}RT\ln{\left(\frac{1033 - 164.9}{1033}\right)} \\
  &= P_0V_0\ln{\left(\frac{1033-164.9}{1033}\right)} \\
  &= 1\cdot 1033\ln{\left(\frac{1033-164.9}{1033}\right)} \\
  &= \pu{-180 L atm}
\end{align}
$$

This is equal in magnitude and opposite in sign to the work done by the $\ce{N2}$ on the air.  So the work done by the $\ce{N2}$ on the air is 180 liter-atm.


How does the relationship between carbonate, pH, and dissolved carbon dioxide work in water? Fish keepers use a table like the one below to determine dissolved $\ce{CO2}$ by observing the color of a pH indicator in water with a known concentration carbonate solution (typically 4 degrees KH ~= 70ppm carbonate).
I'm curious where the equation in the top of the picture comes from though.  Below is my reasoning for wondering where that equation comes from and knowing would (I think) answer all my questions below too.
It appears to me that the equation doesn't work with pure water (water lacking carbonate).
I can think of a few reasons for this:

The equation implies that either pure water cannot hold any $\ce{CO2}$
$\ce{CO2}$ concentration and carbonate concentration are not truly independent
Depending on how the algebra was done, it could be that pH isn't well defined without the presence of carbonate.
It's an approximation that doesn't work everywhere.

My gut says its #4 or #2 but then the table would be kind of worthless unless $\ce{CO2}$'s contribution to carbonate is low relative to the amount added to solution.  If this is the case, I wonder where does the table become inaccurate.

$^{\circ}KH$ or $dKH$ is called (degree of )carbonate hardness, but it is usually about bicarbonates $\ce{HCO3-}$, not carbonates $\ce{CO3^2-}$ . Carbonates are present in water just in traces, unless it is e.g. highly alkaline water from soda lakes of East Africa rift (like where cichlidae from lake Malawi live).
There is the equilibrium for carbon dioxide solution acidity, keeping the expression below constant:
$$K_\mathrm{a1,H2CO3^{*}} = \  \frac{\ce{[H+][HCO3-]}}{[\ce{CO2(aq,total}]}=10^{-\mathrm{pH}}\cdot \frac{\ce{[HCO3-]}}{[\ce{CO2(aq,total}]}=const \cdot 10^{(7-\mathrm{pH})}\cdot \frac{KH}{\mathrm{ppm}  \ce{CO2}},$$
where [] denotes the molar concentration.
Therefore $$\mathrm{ppm}  \ce{CO2} = \frac{\mathrm{const}}{K_\mathrm{a1,H2CO3^{*}}} \cdot 10^{(7-\mathrm{pH})}\cdot KH \approx 3 \cdot 10^{(7-\mathrm{pH})}\cdot KH$$

Pure water in equilibrium with air has theoretical $\mathrm{pH=5.6}$, given by dissolved $\ce{CO2}$ from air. There is established equilibrium
$\ce{CO2(g) <=>[H2O] CO2(aq) <=>[H2O] H2CO3(aq) <=> H+(aq) + HCO3-(aq)}$, but all components are present just in traces.

Addressing feedback:
Carbonate hardness implies there are dissolved bicarbonates $\ce{Ca(HCO3)2}$ resp. $\ce{Mg(HCO3)2}$. These are determined indirectly by titration, that determines bicarbonate content.
This of course fails, if there is sodium bicarbonate or carbonate presence, or if all bicarbonate comes from $\ce{CO2}$. Both cases have carbonate hardness lower or zero, compared to what would say its measurement.
In fact, measuring carbonate hardness ( dKH ) is measuring alkality:
$$\ce{HCO3-(aq) + H3O+(aq) -> 2 H2O + CO2(aq,g)},$$
assuming it is equivalent to content of $\ce{Ca(HCO3)2}$ or $\ce{Mg(HCO3)2}$. But it can be aslo sodium carbonate/bicarbonate, if added to water to increase alkality.

pH of water is determined by this equation:
$$\mathrm{pH} = 6.35 + \log {\frac{[\ce{HCO3-}]}{[\ce{CO2}]}}$$
both in molar concentration.
Calcium carbonate has solubility about 15 mg/L, what makes roughly 1 dKH.
$$\ce{CaCO3(s) <=> Ca^2+(aq) + CO3^2-}$$
$$\ce{CO3^2-(aq) + H2O <=>HCO3-(aq) + OH-(aq)}$$
This solubility increases in presence of dissolved carbon dioxide,which decreases concentration of carbonate:
$$\ce{CO3^2-(aq) + CO2(aq)  + H2O <=> 2 HCO3-(aq)}$$
There is direct relation between pH and relative ratio carbon dioxide : bicarbonate : carbonate.
pH ( in usual aquarium range ) is directly determined carbon dioxide / bicarbonate ratio. If pH change is forced  by other chemicals, the new ratio is established.
BTW, there are just 2 degrees of freedom. You choose 2 of 3 parameters. The third one is determined by the former 2.


Is the bond dissociation enthalpy of Cl-F greater than that of Cl-Cl? In my book, it is mentioned that $\ce{Cl-F}$ has greater bond dissociation enthalpy than $\ce{Cl-Cl}$. Is it true? I know that interhalogen molecules usually have weaker bonds than dihalogen molecules except $\ce{F-F}$ molecule due to high electron-electron repulsion.

From this source, it can be seen that the $\ce {Cl-F}$ bond (BDE $\approx\pu{251 kJ/mol}$) is indeed stronger than the $\ce {Cl-Cl}$ bond (BDE $\approx\pu{243 kJ/mol}$). This can be explained by the partial ionic character of the $\ce {Cl-F}$ bond. The Pauling electronegativity value of fluorine ($\chi = 3.98$) is appreciably higher than that of chlorine ($\chi = 3.16$). Thus, there are significant partial positive and negative charges on the fluorine and chlorine atoms in the $\ce {ClF}$ molecule. Although the difference in orbital energies and size would mean that the covalent interaction between the chlorine $\ce {3p}$ and fluorine $\ce {2p}$ orbital would be less strong compared to that between 2 chlorine $\ce {3p}$ orbitals, the ionic character of the bond overcompensates for this and in fact, results in the $\ce {Cl-F}$ bond being stronger than the $\ce {Cl-Cl}$ bond.


Work equation in thermodynamics As known, work done on the system is: $$\mathrm{d}W = -P_{\text{ex}}\mathrm{d}V $$
However, I have some ambiguity about this equation:

Why is there $P_{\text{ex}}$? Pressure of a gas contained in a cylinder is by definition the force that gas exerts on a piston divided by a area of a piston. But, by the Newton's third law piston exerts the same force on a gas, and so, why is the infinitesimal work not $\mathrm{d}W = -P\mathrm{d}V$?
Some can argue, that we use whatever pressure the thing doing work is pushing against. But I don't accept it. It would be correct if I regarded a gas and a piston as a one system. However, system of interest is only a gas, on which a piston exerts the force with the same magnitude as a gas exerts on a piston, which is $P_{\text{gas}}$.

My calculations:

$$ m\frac{\mathrm dv}{\mathrm d t} = F_{\text{ext}} - F_{\text{gas}}=S(P_{\text{ext}}-P_{\text{gas}}) = S\bigg(P_{\text{ext}} - \frac{nRT}{V_\mathrm{i}-Sx} \bigg)$$
$$ mv\frac{\mathrm d v }{\mathrm dx} = S\bigg(P_{\text{ext}} - \frac{nRT}{V_i - Sx}\bigg)$$
After intergration:
$$ \frac{mv^2}{2} = \int_0^x{SP_{\text{ext}}\mathrm dx} - \int_0^x{\frac{SnRT \mathrm  dx}{V_i - Sx} } = P_{\text{ext}}\Delta V - nRT\ln\bigg(\frac{V_\mathrm{i}}{V_\mathrm{f}}\bigg) $$
Rearrange, we get:
$$ P_{\text{ext}}\Delta V = W = \frac{mv^2}{2} + nRT\ln\bigg(\frac{V_\mathrm{i}}{V_\mathrm{f}}\bigg)$$
There, on the right side, first term is just an energy gain of a piston. The second term is familiar, it's work on the system in reversible process.

So work done on the system is the same as it would be if process was reversible. Why then infinitesimal work is $\mathrm d W = P_{\text{ext}}\mathrm d V$ and not just $\mathrm d W = P_{\text{gas}}\mathrm d V$ ?

In a reversible process, the gas pressure is spatially uniform within the cylinder, and is described globally by the ideal gas law.  However, in an irreversible process, the force per unit area at the piston face is not equal the force per unit area at other locations within the cylinder.  Furthermore, the ideal gas law does not describe the behavior of the gas because viscous stresses contribute to the force per unit area for a rapid irreversible deformation.  So, even though Newton's 3rd law is satisfied at the piston face, unless we specify the force per unit area externally (e.g., manually), we will get the wrong answer if we try to calculate the pressure at the piston face using the ideal gas law.

In applying the equation $W=\int{P_{ext}dV}$ to calculate the work, $P_{ext}$ is supposed to be the force per unit area exerted by the surroundings on your system, at the interface between your system and the surroundings.  So, if the gas is your system, $P_{ext}$ is the force per unit area exerted by the inner face of the piston on your gas (and by your gas on the inner face of the piston).  In your example, if the cylinder is vertical and we do a force balance on the piston, we get:  $$m\frac{dv}{dt}=P_{ext}A-mg$$assuming there is vacuum on the outer face of the piston.  If we multiply this equation by the piston velocity v = dx/dt and integrate, we obtain:$$m\frac{v^2}{2}=\int{P_{ext}dV}-mg\Delta x$$Therefore, the work done by the gas on its surroundings (up until an arbitrary time) is given by:  $$W=\int{P_{ext}dV=mg\Delta x}+m\frac{v^2}{2}$$When the final thermodynamic equilibrium state of the system has been realized, the piston will no longer be moving (including any oscillations of the piston, which will have eventually been damped out by viscous stresses) and the work will then be determined by:$$W=\int{P_{ext}dV=\frac{mg}{A}\Delta V}$$
If, instead of vacuum, there is some constant force external to the piston (say, $P_{atm}$ equal to the atmospheric pressure), the previous results change instead to:$$W=\int{P_{ext}dV=mg\Delta x}+m\frac{v^2}{2}+P_{atm}\Delta V$$when the piston is still moving and $$W=\int{P_{ext}dV=(P_{atm}+\frac{mg}{A})\Delta V}$$at final equilibrium when the piston has been damped to rest.


Qualitative explanation of how COSY works I think I understand 1D Proton NMR with the Fourier Transform, qualitatively.  However, I am struggling to understand 2D COSY NMR.

I am stuck on a few things;

In 1D Proton FT-NMR, the "Pulse" I understand to be the application of a transverse magnetic field, applied at a mixture of frequencies simultaneously, like "noise".  This is so that the chemically different protons in the molecule, who each have a different precession frequency, can all resonate from a single pulse.  Then, we turn off the transverse field and measure the FID at a certain sampling rate.

So my first question is - in 2D COSY NMR, are both "pulses" the same?  i.e., do we subject the sample to a "noise" pulse, then a delay, and then a second "noise" pulse, and then measure the FID?

I think we do, because as far as I have read, the only thing we are varying in the second dimension in the experiment is the length of the delay between pulses.

So to my mind, in 2D COSY, during the delay time, the protons are undergoing FID the same as they do in 1D NMR, even if we are not measuring it at this time.

How, then, does varying the delay time lead to the discovery of "correlation" between protons?  What is it about applying a second pulse to partially relaxed protons, and then measuring the FID, that leads us to discover how the protons are correlated, and to the presence or absence of cross-peaks in our 2D spectrum?

If anyone has a link that explains this, that would be fine.  All my searches on the Internet so far just talk about how to interpret the COSY spectrum, with nothing about how/why it actually works.

In the 1D-proton experiment a hard 90° pulse is applied to the sample before the FID is acquired over some time, t. This set of events is known as the ‘pulse sequence’, and the resulting data is in the time domain, and needs to subjected to a Fourier transform in order to get a 1D spectrum in the frequency domain.

2D experiments are measured much in the same way, except instead of running the pulse sequence once, we run it multiple times, changing some variable and collecting all of the data into one set which will later become the 2D spectrum. [*]

The  standard COSY experiment is one of the simplest 2D experiments to understand. It’s pulse sequence differs from the 1D-proton experiment only by an additional hard 90° pulse. We can explain most (though not all) properties of the COSY experiment using a vector model with no mathematics.

—

Consider a single magnetisation vector, M, which isn't coupled to anything. $\ce{CHCl3}$ would be a good example of this, and leads to a single cross-peak along the diagonal of the  resulting COSY spectrum (a trivial result, but easier to explain).

The experiment consists of 4 steps:

An initial 90° pulse is applied, which transfers M from the z-axis down to the y-axis. This is no different to the pulse used when acquiring a 1D-proton spectrum.
M is allowed to precess for some time, t1. During this time, M ends up somewhere in the xy-plane (having both an x and a y component). The exact position will depend on the length of t1, which is the variable we will differ during the acquisition of our COSY.
A second 90° pulse is applied. This 90° pulse transfers the y-component back onto the z-axis (-z to be precise). The x-component is unaffected (this requires more than a qualitative explanation).
Immediately following the second 90° pulse, we detect. The FID is acquired during t2, which remains constant (just as multiple scans of a 1D proton are acquired for the same length of time). What we are actually acquiring is the residual magnetisation in the x-axis (which wasn't affected by the previous 90° pulse), similar to the standard 1D-proton experiment. If we applied a Fourier transform at this stage, we would indeed just end up with a 1D spectrum. FID in memory to be combined with subsequenc
The above procedure (steps 1-4) is repeated, changing t1 and keeping all other variables constant. For each value of t1, the FID is recorded during t2 - this is the basis of the COSY experiment. The data is combined into a single set, and a Fourier transform applied twice which gives us the 2D-COSY spectrum in the frequency domain. In this simple case, we'd end up with a single cross-peak along the diagonal corresponding to the chemical shift of the proton (in $\ce{CHCl3}$) as expected:

In more complex systems, where multiple protons are present, we observe additional cross peaks in the COSY spectrum off the diagonal corresponding to the coupling between pairs of spins.

The appearance of these additional cross peaks cannot be explained qualitatively using a vector diagram (as per above), but is a consequence of the second 90° pulse that we apply prior to acquisition of the FID.

Where coupling exists, the second 90° pulse doesn't just change the position of the magnetisation vector M, but also causes a population change for the transitions of the spin system through a phenomenon known as coherence transfer.

[*]: There is some potential confusion here, as even in a 1D experiment, we typically run the pulse sequence multiple times in order to improve signal:noise ratio. In this case, all of the variables are kept constant. In a 2D experiment, we run the pulse sequence multiple times, changing some variable in order to build up a 2D data set. We may also run the set of pulse sequences multiple times, to improve signal:noise ratio as per the 1D experiment.


If I can't draw resonance structures for a pi system, does that mean it doesn't exist? Whilst trying to answer this question I came across an interesting situation regarding the relationship between resonance structures and molecular orbitals. There are two places that you can deprotonate this compound. The two different deprotonated forms are shown below along with their resonance structures.

Both compounds have a continuous system of p orbitals which can align in a plane and overlap to form a $\pi$ system. However, in the first example it is not possible to draw resonance structures that delocalise the negative charge over all five atoms that have the available p orbitals. Does this mean that the negative charge is not delocalised over all five atoms? @Martin said in his comment on this answer that valence bond theory and molecular orbital theory ultimately give the same result so this should also apply here. In the molecular orbital description, what is the reason why the negative charge is not delocalised over all five atoms? What do the molecular orbitals for these anions look like?

If there are adjacent, parallel, overlapping p-orbitals in a molecule, the electrons will be delocalized over all such atoms. While the negative charge isn't delocalized over every atom in the conjugated system, it's more than you have shown. What you are failing to consider is:

$\hspace{2.4cm}$

What's important to note about these resonances structures is that the net charge remains constant. These structures are consistent with both the reactivities of the enolate and $\alpha$,$\beta$-unsaturated carbonyl compounds. There negative charge is delocalized between the carbonyl oxygen and its $\alpha$-carbon, and the positive charge is delocalized between the carbonyl carbon (giving rise to 1,2-addition) and its $\beta$-carbon (giving rise to 1,4-addition).


Why is the boiling point of alkyl iodide is more than alkyl fluoride? As far as I know, more the polar character of the organic compound, more will be the energy required to break the bond and thus more boiling temperature. Here in this case, $\ce{R-F}$ is more polar in nature than $\ce{R-I}$. Then, how $\ce{R-I}$ has more boiling point than $\ce{R-F}$?

The main factor affecting the boiling point is the molecular weight, as can be pictured from the noble gases boiling point in the periodic table (the noble gases are those which have the weakest interactions due to their complete outer shell). But note that if you want to consider all the effects altogether, you must use Kelvins instead of Celsius. So going from 44°C to 88°C is not twice as much but a mere 14% increase.

The other factor affecting boiling point is the nature and strength of intermolecular interactions, usually in the order ionic > hydrogen bond > dipole-dipole > Van der Waals.

In the comparison of RF vs RI boiling point, molecular weight is definitely in favor of higher boiling points for RI, at least for small molecules. It is true that the C-F bond is much more ionized than C-I so it will create stronger dipole-dipole interactions but not sufficiently to compensate for the effect of the mass increase.

HTH


Polymers that aren't brittle at the cryogenic temperatures found on Titan? For entertainment purposes, I'm imagining a colony on a non-terraformed Titan, and wondering what is achievable. Hydrocarbons are incredibly plentiful there, while metals and even silicon would be extremely rare and precious, which makes me think that polymers would be used to an even greater extent than on Earth in nearly every form of engineering.
On Titan, the pressure is around $\pu{1.5 atmospheres}$ and the surface temperature is cryogenic at $\pu{76 K}$ on average. This makes me imagine that they would want to use a lot of polymers with extremely low glass transition temperatures (below and around $\pu{76 K}$ at $\pu{1.5 atmospheres}$), so as to have rubbery and non-brittle materials that work outdoors.
Do we know of such materials, and if not, are they physically plausible? I didn't see any examples in tables of glass transition temperatures that I found on the internet, but then again, such polymers would have a lot less application on Earth, if they even exist as solids here.
Of course, these cryogenic temperatures only apply to things that go outside, like suits, vehicles, extraction and construction equipment, buildings, and the necessary pumps and other equipment needed for life support.

According to NASA1 the main factor inhibiting the widespread use of polymer materials in space environments is oxygen erosion, the materials tested so far to be proven effective in space are  Teflon, Mylar, Kapton and Tedlar.
Further experimentation is being conducted on board ISS according to their website2.
I do not believe there are polymers currently capable of handing the cold, pressure, and oxygen erosion of Titan, as such, the probe3 Huygens sent to Titan was constructed of aluminum coated Kapton, which seems to be the industry standard for space environments.
References:

https://www.nasa.gov/centers/glenn/shuttlestation/station/misse.html
http://www.spacestationresearch.com/tag/polymer-erosion-and-contamination-experiment-polymers/
https://en.wikipedia.org/wiki/Huygens_(spacecraft)


Elimination of hydroxyl group while deprotecting acetal Regarding this reaction, other than de-protecting the acetal, will we have an elimination reaction for the alcohol or no?

As I said in the comments, it entirely depends on the conditions. It is easy to select mild conditions to cleave the dioxolane. Elimination would require very strong forcing conditions (concentrated acid and heating). The first site of protonation is the carbonyl, so you would form the hemi-acetal of the product long before elimination or substitution occured at the primary alcohol. If you were foolish enough to heat this with cH2SO4 then you could get elimination of the primary alcohol and generate quite a mess.


When will the Lassaigne's method for detection of nitrogen and Kjeldahl's method for quantitative estimation of nitrogen fail? Lassaigne's method:

The organic compound is fused with dry $Na$ in a fusion tube. The fused mass after extraction with water is boiled and filtered. This filtrate is called sodium extract (S.E.), which is used to detect elements other than $\ce{C}$ and $\ce{H}$ in organic compound.

If the organic compound contains nitrogen then the S.E. will form $\ce{NaCN}$. Further reactions with $\ce{FeSO4 + NaOH}$ and $\ce{FeCl3 + HCl}$ gives blue or green colour, confirming the presence of nitrogen.

This test fails in case of diazo compounds because sodium metal can not extract nitrogen from them. What about the compounds containing nitrogen in the ring or having nitro groups?

Kjeldahl's method:

Nitrogen containing organic compound is heated with conc. $\ce{H2SO4}$. $\ce{N}$ in the compound converted to $\ce{(NH4)2SO4}$. The mixture is then heated with excess of $\ce{NaOH}$. The liberated $\ce{NH3}$ gas is absorbed in excess of standard solution of $\ce{H2SO4}$. From the amount of acid reacted with $\ce{NH3}$, % of $\ce{N}$ can be calculated.

This method is not applicable to diazo, nitro and compounds having nitrogen in the ring. This is because nitrogen of these compounds does not change to ammonium sulphate upon reacting with conc. sulphuric acid.

You are right about the limitations of century old Lassaigne's method. Why are they still teaching it? Modern day CHNSO elemental analysis is done with gas chromatograpy systems. If you are quoting from a book, provide a reference.

This is an excerpt from a 1953 work Sulfur in Organic Compounds, by a Modified. Lassaigne's Method ACTA CHEMICA SCANDINAVICA 7 (1953) 1395

  Objections to Lassaigne's method are: a) low boiling substances evaporate
  before reaction with the sodium, b) diazo-, azo and polynitro compounds evolve nitrogen c) some heterocyclic compounds do not react.


Does a non-terminal alkyne react with sodamide? Hydrocarbon (A), $\ce{C6H10}$, on treatment with $\ce{H2/Ni}$, $\ce{H2}$/Lindlar catalyst and $\ce{Na}$/liquid $\ce{NH3}$ forms three different reduction products (B),(C) and (D) respectively. (A) does not form any salt with ammoniacal $\ce{AgNO3}$ solution but forms a salt (E) on heating with $\ce{NaNH2}$  in an inert solvent. Compund (E) reacts with $\ce{CH3I}$ to give (F). Compound (D) on oxidative ozonolysis gives n-butanoic acid along with other other products. Give structures of (A) to (F) with proper reasoning.

From the formula of (A) and the details given, it is clear that (A) is an alkyne. I could make out that (B) would be an alkane while (C) and (D) would be cis and trans alkene respectively. Now, as (A) does not react with ammoniacal $\ce{AgNO3}$ solution, it is not a terminal alkyne. But, then I got confused as the question told, (A) reacts with $\ce{NaNH2}$ to give a salt. Till now, I knew that only terminal alkynes react with $\ce{NaNH2}$. What is it that am I missing?

By the given question conditions you should easily guess that $(A)$ is $\ce {Hex-2-yne}$. When it is reacted with $\ce{NaNH_2}$ in presence of heat it can isomerise to terminal alkyne by following this mechanism and then $\ce{CH_3I}$ addition will simply follow an $\ce{S_N2}$ mechanism.

Secondly,Ozonolysis will yield an aldehyde which can get oxidised to acid in oxidative medium.


Can carboxylic acids be used in the aldol reaction? A question in a practice paper I did recently showed the aldol condensation of benzaldehyde with ethanoic acid. I think I am right in saying that carboxylic acids do not form enolates because the carboxyl proton is much more acidic and so this reaction could not take place under basic conditions but it is possible to protonate a carboxylic acid and then have it take part in an aldol reaction?

Additionally, is this also possible for acyl chlorides, and if not, why?

Carboxylic acids form enolates

I think I am right in saying that carboxylic acids do not form enolates because the carboxyl proton is much more acidic and so this reaction could not take place under basic conditions...

They do, if you beat them to it:

Dissolve the carboxylic acid in a polar, aprotic solvent, such as 1,2-dimethoxyethane, 1,4-dioxane, tetrahydrofuran, etc.

Add one quivalent of sodium hydride to deprotonate and form the carboxylate.

Add one equivalent of a strong base, such as LDA (lithium diisopropylamide) to deprotonate the carboxylate at the $\alpha$-position.

Add your aldehyde

What about acyl chlorides?
Adding a base, such as Hünig's base (plus catalytic amounts of a Lewis acid) to an enolizable acyl chloride under anhydrous conditions gives rise to a different pathway: a ketene can be formed!
How do ketenes react with aldehydes?
In the presence of a Lewis acid, they form $\beta$-lactones!


Betaine HCl stomach pH It seems betaine HCl is often recommended for those suffering from "low stomach acid" -- which, as I understand, is having too high stomach pH for proper digestion (especially for proteolysis via pepsin). However, I have a few questions then -- though I'm not sure if my reasoning is correct here:

Understandably, one wouldn't want to drink pure or highly concentrated HCl by itself to increase stomach acid! But then, what role does the betaine play?

I'm guessing betaine HCl probably does not dissolve in water to give the same pH as just straight (or concentrated) hydrochloric acid -- since then it would seem just as dangerous as drinking plain HCl!

In that case, if it doesn't decrease the pH as much, making it safe for oral consumption, what value does it bring for "lowering stomach pH" anyway? I mean, one could just drink a little vinegar or citric acid for the same effect?

Or, perhaps, is betaine-HCl just a means of some "delayed release" of HCl to lower stomach pH without hurting the mouth and esophagus during its initial consumption? Betaine may just be a useful amphoteric here, given it's quarternary ammonium and carboxylic acid groups (although not just any amphoteric, such as water, may suffice here!)

Alternatively, could the value of betaine HCl simply be in providing a source of $\ce{Cl-}$ anions, possibly for increased pepsin activity? (I'm not sure if pepsin requires merely low pH or specifically also needs $\ce{Cl-}$ anions as well) Betaine-HCl seems to be often formulated with additional pepsin enzyme(s) as well...

Betaine HCl is usually delivered in capsules which pass through the mouth and esophagus intact.
Regarding the webmd article that says " Even if it were true, betaine hydrochloride wouldn’t help. It only delivers hydrochloric acid but does not itself alter stomach acidity."
This statement seems contradictory. Indeed, I found article the tests stomach ph after betaine hcl usage and confirms that indeed it reduces it:
https://www.ncbi.nlm.nih.gov/pubmed/23980906


How do I decide the parent chain of an organic compound that has a ring as well as a side chain while naming the compound? I think that this is decided by the number of carbon atoms. The part which has greater the number of carbon atoms, is considered the parent chain. But this fact is not leading to the correct result always. Are there some rules that govern this fact besides the number of carbon atoms.

If you have a simple chain with carbon atoms and an aromatic ring, the priority has the ring, not matter how the chain is long. This rule is NOT valid when you have a functional group: if it is in the linear chain, it has the priority, and the numeration starts from the carbon more near at the functional group, for example
$\ce{CH3-CH2-CO-CH2-CH2-CH3}$. The numeration starts from left, so you write hexan-3-one.


Can grignard reagent contain secondary amino group? Is there any reason why this grignard reagent wouldn't work? My best guess is that perhaps the nitrogen could act as a second nucleophilic "head", thereby not giving a good yield on the intended product. Or maybe this isn't stable enough?

I guess it wouldn't even exist. Grignard's reagent is basic enough to easily take off an imine proton, so as soon as you try to make such a thing, it will turn upon itself.

Probably they have a suitable protecting group for this, but here I'm stepping outside my expertise.


Shouldn't the Hofmann product be preferred in this elimination? Shouldn't the exocyclic alkene (Hofmann) be the major product? The base is $\ce{t-BuO-}$ (tertiary butoxide ion), which is pretty bulky. I think the Saytzeff product should be the minor product.

Also, the solution says "E2, some E1". It's a 3° alkyl halide (more sterically hindered than 1° and 2°), so shouldn't it undergo E1 elimination?

Yup, $\ce{t-BuO-}$ is a bulky base.

But is it bulky enough that it can't even abstract a hydrogen from the ring? Nope.

Take a look at this picture of a $\ce{t-BuO-}$ molecules approaching the hydrogen of 1-bromo-1-methyl-cyclopentane.

Does that $\ce{t-BuO-}$ look like it's struggling to pull that hydrogen out? Definitely not. It's doing what it does with ease.


Is aluminium foil pure aluminium? Is aluminium foil pure aluminium? Or does it have other elements in the e.g: tin (Sn) or lead (Pb)? Because when I try to oxidize it through the process of submerging it in water, it doesn't look like it wants to oxidize to aluminium oxide (AlO+). I did this exact same thing with pure aluminium and it oxidized within minutes.

From the Institute of Minerals, Materials and Mining,

  Most aluminium foil is made from pure aluminium, but increasingly alloys are used to improve properties and reduce the thickness required.

If you want to make sure that the Aluminium foil you're using is pure, the only way would probably to ask the manufacturer. Other than that, it could be the case that the foil is either pure or alloyed, depending on how it's made.


Syntax and Typography of Atomic Orbitals For mhchem (i.e. the \ce command you use here), I want to improve the support for atomic orbitals. (Reason: I see it quite often used here.) But only a few things about orbitals are mentioned in IUPAC documents. Many more seem to be "unwritten law".
As I am not a chemist I need your help.

1

The Green Book p. 32 defines the letters s, p, d, f, g, h, i, k ... (omitting j). They have to be typeset upright.

Question a: How far does this "alphabet" reasonably go? (Wikipedia goes up to i).

2

On the same Greek Book page, electron configurations are explained, e.g. $\mathrm{(1s)^2 (2s)^2 (2p)^1}$.

Many other sources write it without parenthesis: $\mathrm{[Kr]\mskip2mu4d^{10}\mskip2mu5s^1\mskip2mu5p^3\mskip2mu5d^4}$

Question a: What values can the left-side number take? (At Wikipedia, I see 1 to 7.) Can it be 2-digit?

Question b: What values can the right-side exponent take? (At Wikipedia, I see 1 to 10.)

3

I think I did not come across the configuration with one-electron orbitals (in Greek). $\mathrm{(1\sigma)^2 (2\sigma)^2 (3\sigma)^2 (1\pi)^1}$. (The Greek letters should be upright, but that's a drawback of MathJax.)

Would this be useful? Or not needed?

4

The examples introduce another notation, "electronic states"  $\mathrm{\cdot\cdot\cdot (2a_1)^2 (1b_2)^2 (3a_1)^2}$

a: What values could the letter take? b: The left-side number? c: The subscripts? d: The right-side exponent?

5

From Wikipedia, I learned that there are notations IUPAC does not mention: $\mathrm{p_x}$, $\mathrm{p_y}$ and $\mathrm{p_z}$. (All subscripts should be upright, because they are labels, no variables.)

But it does not end there. Here and here I saw:
$\mathrm{d_{xy}, d_{xz}, d_{yx}, d_{x^2{-}y^2}, d_{z^2}}$
This was confirmed by this book.

There are alternative styles
here ($\mathrm{{p_x}^1}$).
and
here ($\mathrm{p_{x1}}$).

a: I guess the $\mathrm{d_{z^2}}$ notation (with index z-square) is correct?

Oh, I just found this book that also includes $\mathrm{4f_{x(x^2{-}3y^2)}}$ and $\mathrm{4f_{y(3x^2{-}y^2)}}$.

b: Are there other indexes possible than x, y, z?

c: Can it be just squares? Or are other index-superscripts possible?

d: What about the 3 in the last examples? Could this also be any other numer?

(You see, I just look at the typography of things and don't know their meaning.)

e: Could there be any other forms of indexes?

f: Typographically, what does this dash stand for? Is it a hyphen or a "transition between a higher energy state and a lower energy state"?

6

Finally, there is orbital hybridization, which will lead to notations like $\mathrm{sp, 2sp, sp^2, 2sp^3, s^{0.5}p^3, s{-}s, sp^3{-}sp^3}$.

a: Can the superscript be any other value than 2 and 3? Can the "base" be any other thing than s, p and sp?

b: Is $\mathrm{2sp^3}$, is it $\mathrm{2s}$ and $\mathrm{p^3}$ or is sp one ... thing with a 2 before and a 3 in superscript.

c: Is it a hyphen or a "transition between a higher energy state and a lower energy state"?

Question 1. The Green Book p. 32 defines the letters s, p, d, f, g, h, i, k ... (omitting j). They have to be typeset upright.

  How far does this "alphabet" reasonably go? (Wikipedia goes up to i).

Section 2.6.3 of that version of the Green Book shows the term symbols (Wikipedia) which are capital letters S,P,D,F,G,H,I,K,... and used to indicate the electronic configuration of the valence electrons and for excited states.
The lower case letters are used for the azimuthal quantum number (ℓ) which is one of the four quantum numbers for an electron in an atom. You can typeset $99.9~\%$ of chemistry of even more using only the letters s, p, d and f, corresponding to the left-hand main group elements, the right-hand main group elements, the transition metals and the lanthanides/actinides, respectively. Everything above f is and will be for quite some time not relevant to chemistry outside a very small circle that would know how to deal with stuff.

  Question 2. On the same Greek Book page, electron configurations are explained, e.g. $\mathrm{(1s)^2 (2s)^2 (2p)^1}$.

  Many other sources write it without parenthesis: $\mathrm{[Kr]\mskip2mu4d^{10}\mskip2mu5s^1\mskip2mu5p^3\mskip2mu5d^4}$

  Question a: What values can the left-side number take? (At Wikipedia, I see 1 to 7.) Can it be 2-digit?

  Question b: What values can the right-side exponent take? (At Wikipedia, I see 1 to 10.)

This has two sub-questions.

By a similar reasoning as presented in the answer to 1, I think 8 will be the limit until the not-so-close future. But there is nothing intrinsically preventing it from becoming two-digit at some distant point.
The superscript can take values of up to $4l+2$, where $l$ can be considered ‘computer science counting’ of the letter sequence s, p, d, …; so for s $l = 0$ and thus $4l+2=2$; likewise for p ($l=2$) the value is maximum $4l+2=6$ etc. Note that $4l+2$ is actually $2(2l+1)$: each subshell gives up to $2l+1$ orientation possibilities that each can be populated with two electrons.

  Question 3. I think I did not come across the configuration with one-electron orbitals (in Greek). $\mathrm{(1\sigma)^2 (2\sigma)^2 (3\sigma)^2 (1\pi)^1}$. (The Greek letters should be upright, but that's a drawback of MathJax.)

  Would this be useful? Or not needed?

Yes, this is useful, especially for discussing molecular orbitals. The Greek letters you want to be aware of are σ, π and δ. Maybe a fourth will be added soon corresponding to f orbitals (υ?) but I don’t see that coming anytime soon.

  Question 4. The examples introduce another notation, "electronic states"  $\mathrm{\cdot\cdot\cdot (2a_1)^2 (1b_2)^2 (3a_1)^2}$

  a: What values could the letter take? b: The left-side number? c: The subscripts? d: The right-side exponent?

I’m going to answer that as one.

This notation uses irreducible representations to define orbitals. The bit between the initial number and the superscript can be any of the symbols that appear in Little Bobby Tables (courtesy of Orthocresol). They each represent a type of orbital with a certain symmetry.

Orbitals of the same irreducible representation (i.e. the same centre) are numbered from lowest to highest energy — so in principle the sky is the limit. Practically, however, you will be hard-pressed finding three-digit examples.

The exponent can at maximum be $2n$, where $n$ is the number in the first column of the respective character table (the E column). This number corresponds to a certain symmetry’s degeneracy (i.e. so many orbitals of the same energy) and each orbital can be populated by two electrons.

  Question 5. From Wikipedia, I learned that there are notations IUPAC does not mention: $\mathrm{p_x}$, $\mathrm{p_y}$ and $\mathrm{p_z}$. (All subscripts should be upright, because they are labels, no variables.)

  But it does not end there. Here and here I saw:
  $\mathrm{d_{xy}, d_{xz}, d_{yx}, d_{x^2{-}y^2}, d_{z^2}}$
  This was confirmed by this book. There are alternative styles
  here ($\mathrm{{p_x}^1}$).
  and
  here ($\mathrm{p_{x1}}$).

  a: I guess the $\mathrm{d_{z^2}}$ notation (with index z-square) is correct?

  Oh, I just found this book that also includes $\mathrm{4f_{x(x^2{-}3y^2)}}$ and $\mathrm{4f_{y(3x^2{-}y^2)}}$.

  b: Are there other indexes possible than x, y, z?

  c: Can it be just squares? Or are other index-superscripts possible?

  d: What about the 3 in the last examples? Could this also be any other numer?

  e: Could there be any other forms of indexes?

  f: Typographically, what does this dash stand for? Is it a hyphen or a "transition between a higher energy state and a lower energy state"?

(Somebody please add and correct this.)

It was my assumption that the axis directions should be italicised but I don’t know. To discuss this, please go to the break-out discussion for this question.

p orbitals will always have $x$, $y$ or $z$. d orbitals will have one of the five you noted ($\mathrm{d}_{z^2}$ is correct — as I said, I am not sure on the upright versus italic question but I would have assumed italics). The f orbitals themselves can be labelled by various different polynomials in $x$, $y$, $z$, and $r$. See these two web pages: 1, 2. In general the subscript is just a polynomial.

Which symbols and exponents can appear here is fixed already by the results of the hydrogen-like orbitals — again, for most of chemistry d orbitals are sufficient and you can cover $99.9~\%$ by including f orbitals.

The ‘dash’ is a mathematical minus-sign operator. Yes, it truly is $x^2$ minus $y^2$.

  Question 6. Finally, there is orbital hybridization, which will lead to notations like $\mathrm{sp, 2sp, sp^2, 2sp^3, s^{0.5}p^3, s{-}s, sp^3{-}sp^3}$.

  a: Can the superscript be any other value than 2 and 3? Can the "base" be any other thing than s, p and sp?

  b: Is $\mathrm{2sp^3}$, is it $\mathrm{2s}$ and $\mathrm{p^3}$ or is sp one ... thing with a 2 before and a 3 in superscript.

  c: Is it a hyphen or a "transition between a higher energy state and a lower energy state"?

Sub questions.

Yes. You have $0.5$ as a superscript in your own examples. But, as posts by Martin and Ron somewhere point out, in $\mathrm{sp}^x$ the superscript $x$ can, in principle, be any number.
$\mathrm{2\,sp^3}$ is $\mathrm{\{2\}\{sp^3\}}$ — and there could potentially be another superscript making it $\mathrm{(2sp^3)^2}$.
I don’t know what that hyphen is supposed to be.


Is this sydnone aromatic? Is the following molecule aromatic?

It does seem like it will be aromatic because it is similar to furan. I am confused because of the positive charge on the $\ce{N}$.

As correctly pointed out in the comments, the compound you present belongs to the class of sydnones. According to Hückel's rule, it should be aromatic, as the molecule is planar and contains $4n+2$ π-electrons.

However, calculations performed in 1998[1] that mainly looked at chemical shifts and inferred aromaticity from the magnitude of observed diamagnetic shielding, show that sydnones in fact are not fully delocalized, i.e. they are not aromatic. To quote from the linked publication:

  Mesoionic compounds  are  planar  five-membered  heterocyclic  betaines
  with at least one side chain whose α-atom is also in the ring plane and with dipole moments of the order of 5 D. Electrons are delocalized over two regions separated by what are essentially single bonds. One region, which includes the α-atom of the side chain, is associated with the HOMO and negative π-charge whereas the other is associated with the LUMO and positive π-charge.

The publication is also riddled with images that depict syndones like this:

It ends with another quote from Gilchrist's book "Heterocyclic Chemistry", 2nd edition, 1992, p. 329:

  It does not therefore seem useful or appropriate to classify sydnones and other mesoionic compounds as aromatic, despite the existence of a cyclic array of p-orbitals.

Reference

Simas, A. M.; Miller, J.; de Athayade Filho, P. F. Are mesoionic compounds aromatic? Can. J. Chem. 1998, 76 (6), 869–872. DOI: 10.1139/v98-065.


Why must a voltaic cell take place in a conductive solution? In order for a voltaic cell to function, its electrodes must be placed in a conductive electrolyte solution, most often solutions consisting of their metal ions. While I understand the use of the metal ions (allows for the plating to occur on the cathode), why must the electrolyte be conductive? Thanks!

The electrolyte must be conductive to ions, not electrons. In fact, if there's an easy path for electrons to move between the two electrodes, the cell will immediately self-discharge. The exact reason for needing ionic-conductivity kind of depends on the cell. For example, in lithium-ion batteries, lithium-ions are generated at the anode and consumed at the cathode (or vice-versa). If ions cannot move freely between the two (through the electrolyte), then you will quickly build up an ion-rich region at one electrode and an ion-deficient region at the other at which point the reaction will stop.

It's not even a requirement that the electrolyte be liquid, as long as it conducts ions. Eg. the sodium-sulfur battery uses a form of solid alumina that conducts sodium ions.


Must compounds formed between solids be a solid as well? My teacher gave us a long list of compounds and asked whether they are solid, liquid, or gas at room temperature and normal atmospheric pressure. I only know the more common compounds (eg. carbon dioxide, ammonia) and elements but not the rest.

Can I assume that a compound is solid because its constituent elements are solids at normal temperatures and pressures as well? If so, does the same goes with liquids and gases? If so, what if its constituting elements are not of the same physical state?

If not, then how can I determine the answer? Or must I memorize the physical state of the compounds in order to answer the questions?

No, they don't have to. In short, you have to memorize. The physical state of a compound has nothing to do with the states of its elements.

Examples:
$$\ce{Hg~(liquid) + Br2~(liquid) -> HgBr2~(solid)}$$
$$\ce{C~(solid) + S~(solid) -> CS2~(liquid)}$$

Now, there are some common patterns (an ionic compound is most certainly a solid; a compound with small non-polar molecules is probably a gas), but that's another story.


Basic concept of the first rule of thermodynamics There is a cylinder-piston system with some gas in it, and let's say this system is a closed system.
Say that the system is submerged in a large beaker of water whose temperature is $25\ \mathrm{^\circ C}$. A spark is used to trigger a reaction in the system, and after the reaction is completed, the temperature of the water bath increased to $28\ \mathrm{^\circ C}$, and the piston went inside the cylinder.

By instinct, I think that heat is lost by the system since the surrounding(water)'s temperature increased, and work has been done to the system. However, when I consider the thermal equilibrium, the system's temperature must be $28\ \mathrm{^\circ C}$ as well. In this case, is heat added to the system or not? Furthermore, in the equation $\Delta E = q+w$, what are the signs (plus or minus) of the three terms?

Chemical energy of system has been converted to thermal energy of both  the system and the surrounding water.

Additionally, water did work on system, as the system shrinked down.

$$\Delta E_\mathrm{system}=-C_\mathrm{water} \cdot \Delta T_\mathrm{water}-p \cdot \Delta V_\mathrm{system}$$

For $H=E + p \cdot V$,
$\Delta H=\Delta  E + p \cdot \Delta V + V \cdot \Delta p$

As p is constant,

$\Delta H=\Delta  E + p \cdot \Delta V $

$\Delta H_\mathrm{system}=-C_\mathrm{water} \cdot \Delta T_\mathrm{water}$

where $H$ means enthalpy, energy at the constant pressure.

If we considered other thermodynamic quantities like Gibbs energy or entropy, we would need to examine the underwent reaction and system composition. But if the system energy is considered, its change is related just to exchanged thermal energy and work.

$q$ in the equation $\Delta E = q+w$ means heat transfers to the system, so it is negative. $w$ is work done on system and is positive.  $\Delta E$ is very probably negative, as work value would be much less than transferred heat.


Limitations of Lewis structures Using Lewis structures to describe chemical bonds inside a molecule is still fairly common start in undergraduate education. Lewis structure however can be misleading, or just simply wrong in many cases. Mostly I see this with aromatic systems, Diels-Alder reactions etc.

My question is if there is a rule of thumb when one can safely use Lewis structure and when we should just  forget them? Or should I just check it every single time and compare them with eg. MOs?

OK, here goes. Chemists use a variety of representations of molecules to try to predict their behavior and reactivity, but none of them come close to preparing you for what happens in the laboratory. In that sense, none of them are "safe" to use. Lewis structures, and the curly arrow formalisms, are good for organic chemists to quickly explore possible avenues of reactivity and synthetic pathways. But they are not very good at dealing with chemistry south of the second row of the periodic table, nor at phenomena such as aromaticity, electrocyclic reactions, bulk properties, solvation, steric effects, polarity, polarizability, photochemistry, free radicals, etc. So we add in to the mix some "rules", try to think of molecules as being best described as "multinuclear atoms", and try to improve and promulgate our best tools for computational chemistry. For all that, we still need to approach the laboratory with a degree of humility--things that look good on paper usually do not look so good when you try them out in the real world.


Why does sodium sulfate initially harden instead of dissolving? I was recently drying a product by passing it through a bed of sodium sulfate in a fritted glass Buchner funnel. Once I had finished up working my reaction, in an ill-advised attempt to clean up, I decided to wash the salt out with water before proceeding with acetone. Upon adding water, the salt starting foaming, and became hard as a rock and adhered to the fritted glass. I decided to leave it sitting in the water, and after a while it did finally turn gel-like and was able to be washed out.

My question is, why in the presence of so much water, why did the salt remain as the hydrated salt instead of dissolving, and why did it take so long to dissolve after it formed the large crystal? Also any idea what the bubbling was?

Sodium sulfate forms various hydrates including a decahydrate. The dissolution is exothermic which saturates the solution. When the warm solution reaches the relatively cold frit the supersaturated solutions precipitates. In the glass frit the pores would clog and it would take time to dissolve the sodium sulfate out.


Oxidizing and reducing agent What i know about the above question is that a compound acts as an oxidising as well as reducing agent when the oxidation of central metal is somewhere between it's maximum and minimum oxidation state which is in the case for $NaNO_2,NaNO$.I don't know how to choose between the two.Any more information about the behaviour of other two compounds will be helpful.Thank you

I agree with Ivan Neretin in that the sodium oxidation state is equal in all compounds, and it will not act as oxidizing or reducing agent.

My thoughts:

Na$_2$O will not act as a strong oxidizing or reducing agent, as both ions are in their preferred state.

NaO$_3$, or sodium ozonide, will likely act as an oxidative agent. However, I believe it disproportionates in water: 4 O$_3$ $^-$ + 2 H$_2$O $\rightarrow$ 5 O$_2$ + 4 OH$^-$, so that's kind of a candidate: it goes from oxidation state -1/3 to 0 and -II.

NaNO also seems strange to me - if it's Na$_2$N$_2$O$_2$, I see here that it acts as a reducing agent.

NaNO$_2$ can certainly act as both reducing and oxidizing agent, as the nitrogen atom is in oxidation state +III.

In conclusion, (4) seems to be the safest bet, considering only one answer is expected.


Decomposition of acetonide protecting group in chloroform Today I purified my compound containing an acetonide protected diol. I took an NMR sample and it looked great, but because I have only little amount of the compound, I put the sample back into the flask and evaporated the $\ce{CDCl3}$.

In a while, the solution turned from yellow to brown, so I took another NMR and it looked a little bit more messy and the $\ce{CH3}$ signals from acetonide group had decreased intensity (integral from 6 to around 4.5).

Is it possible that the $\ce{CDCl3}$ is so acidic it causes acetonide removal? The compound survived silica chromatography. Have you ever experienced something similar?

Deuterochloroform (and indeed 'normal' chloroform) are known to be mildly acidic if left standing for any length of time:

  Chloroform: It reacts slowly with oxygen, or oxidising agents, when exposed to air and light, giving, mainly, phosgene, Cl2 and HCl.

  Source: Purification of Laboratory Chemicals, 7 ed.

To compensate for this, its common to add molecular sieves and a small amount of potassium carbonate to a new bottle of deuterochloroform to keep it dry and to minimise any acid that may be present/may be generated as the bottle ages.

If you look up the entry for acetonides in Greens' Protecting Groups in Organic Chemistry, you'll see that acetonices can come off under quite mild conditions, though personally I've never had an acetonide come off in an NMR tube of chloroform (I assume you didn't leave it standing for months in the tube...)

In order to cleave the acetonide under acidic conditions, you generally also need a nucleophile (methanol is common, water would do the trick), which if present should be very obvious on the NMR, and might go some way to explaining how you lost the acetonide.  You should also consider other explanations, for instance you mentioned concentrating the chloroform- are you sure the rotary evaporator hadn't just been used to take off something wet and acidic?

Hopefully you can just put the acetonide back on without any issues, I suggest picking up a new bottle of chloroform-d and adding some sieves and potassium carbonate (you could try NMR-ing your old bottle to see if theres water/methanol etc in it, and then purify appropriately, but the stuff is sufficiently cheap that I wouldn't waste my time).


Which of the following para-substituted benzyl alcohols reacts fastest with HBr? I had a question in a recent test which asked me to pick the compound that would react fastest with $\ce{HBr}$ with the options being:

p-nitrobenzyl alcohol
p-chlorobenzyl alcohol
benzyl alcohol
p-methoxybenzyl alcohol

The answer was p-methoxybenzyl alcohol.

Now what I don't understand is that $\ce{HBr}$ would produce nucleophilic $\ce{Br-}$, which would be looking to attack a region of electron deficiency. In that case option (1) p-nitrobenzyl alcohol would the right answer because the nitro group is an electron withdrawing group, thus making the ring more electron deficient.

Could someone please explain as to why this logic fails?

Inductive effects will play much more of a role in electrophilic aromatic substitution. Here, positive charge on a benzylic position is being delocalized to the aromatic ring, and the question is asking which group will best stabilize this positive charge (via resonance):

Between $\ce{-H}$ (neutral), $\ce{-NO2}$ (which would destabilize the positive charge), $\ce{-Cl}$ (poor at stabilizing positive charge), and $\ce{-OCH3}$ (good at stabilizing positive charge), $\ce{-OCH3}$ is the clear winner.


Electroplating copper yields rough surface? I am attempting to electro-plate vias in a home brew printed circuit board. The medium I am attempting to electroplate is fiberglass bound together with epoxy. This is an intentional insulator.

I drilled a few holes in this medium using a regular electric drill and covered the holes with a graphite based conductive paint by covering the board in the paint and drawing it through the holes using a vacuum. I repeated this in both directions in an attempt to cover the holes evenly. I then cured the paint at about 100C for 20 minutes. In my experiments, this was far sufficient to dry the paint.

I then created a solution of 1000ml distilled water, 250ml of 38% sulfuric acid (battery acid to improve conductivity) 160g of copper sulphate pentahydrate, and 2 ml of polysorbate 20.

I prepared this solution by adding the sulfuric acid to the water slowly (to avoid rapid exothermic reaction between acid and water) then adding the polysorbate and the  the copper sulfate pentahydrate. I then mechanically agitated the copper sulfate pentahydrate crystals until the solution was homogenous.

I ran .8A of constant current through the solution with the board as the cathode and an anode consisting of two electrically common (14ish guage) copper wire on either side of the board. After about 90 minutes this was my result...

I'm wondering why the copper clad had "beads" on it...the entire surface was sanded with 1500 grit sandpaper directly preceding the electroplating.

I'm also wondering why the holes are so "rough". In my testing, every via has come up as 0.0Oohms...or electrically common to the clad. Aesthetically though, this is not pleasing and makes me wonder if my method is truly reliable.

There are few things that you can improve.

First, beads are forming because your copper anodes are not pure copper. Beads are forming around impurities. You can try to remove (filter) impurities by firmly wrapping anodes in paper tissue or better with water filtration polypropylene felt filter so only Cu ions can pass into solution. You can after that collect impurities already present in solution by electroplating few hours on some scrap piece of wire.

Second, holes should be drilled cleanly. You should invest into carbide drill bits and good drill stand.

Third, you should also think about adding Cl ions to solution, NaCl or HCl in small quantities. Plated surface will be shinier...

Best regards,
Pavle


Do aromatic dienes undergo the Diels-Alder reaction? Do aromatic dienes undergo Diels-Alder reactions?

My understanding is that the $\pi$ bonds in aromatic compounds are not as localised as they are in other compounds, such as aliphatic dienes. Therefore, the dienophile shouldn't be as willing to go to these compounds. Am I correct?

Yes, you are correct.  Aromatic compounds are stabilized by their resonance energy. This means that the activation energy for reaction of an aromatic compound will typically be greater than that for a similar reaction of a non-aromatic compound.

That said, aromatics will react as the diene component in a Diels-Alder reaction if

the dienophile is extremely reactive (o-benzyne for example)

if the aromatic system is destabilized (for example by steric interactions or structural constraints) so that the resonance stabilization is decreased

or if the aromatic system is large enough that loss of one aromatic ring does not significantly destabilize the entire system.  Anthracene is such an example where the middle ring will react via a Diels-Alder reaction.

(image source)


Hybridisation of hexaamminecobalt(III) complex Does $\ce{[Cr(NH3)6]^3+}$ have $sp^3d^2$ or $d^2sp^3$ hybridisation?

Oxidation state of $\ce{Cr}$ is $+3$. The electronic configuration is $3d^3$ and the $d$ electrons occupy $t_{2g}$ orbitals. If the electrons from the ligands were to occupy the $e_g$ orbitals the stability of the complex decreases. So shouldn't the hybridisation be $sp^3d^2$?

It has d2sp3 configuration since NH3 is a strong field ligand which causes the electron to pair up and use the inner d orbitals(3d) .


Does NaOH react with K to form NaOK or KOH? I realized that the $\ce{-OH}$ group when it reacts with some metal like $\ce{Na}$ or $\ce{K}$ will replace the $\ce{H}$ element to form $\ce{-OK}$ or $\ce{-ONa}$:

\begin{align}
    \ce{H2O + Na &-> NaOH + H2}\\
    \ce{CH3OH + Na &-> CH3ONa + H2}\\
    \ce{H2O + K &-> KOH + H2}\\
\end{align}

So I am just wondering is that always true? Will $\ce{NaOH + K}$ form $\ce{NaOK}$ or $\ce{KOH}$? Does $\ce{NaOH + Na}$ form $\ce{Na2O}$?

The question: Will $\ce{NaOH + K}$ form $\ce{NaOK}$ or $\ce{KOH}$?

This is a redox reaction regardless of what species $\ce{K}$ would react with. The answer is two-fold: (a) Having potassium added to an aqueous solution of sodium hydroxide; and (b) Having potassium added to molten sodium hydroxide, therefore no water present (or trace amount).

Answer to situation (a) is potassium would essentially react with water (other ions in the solution are $\ce{Na+}$ and $\ce{OH-}$) based on standard potentials of following possible half-reactions.

$$
\begin{align}
\ce{K+ + e- &<=> K} &E^\circ &=\pu{-2.931V}&\\
\ce{Na+ + e- &<=> Na} &E^\circ &= \pu{-2.71V}&\\
\ce{2H2O + 2e- &<=> H2 + 2OH-} &E^\circ &=\pu{-0.8277V}&\\
\end{align}$$

Therefore, the only ionic species in the solution would be $(\ce{Na+ + K+ + OH-})$ in $\ce{H2O}$.

Answer to situation (b) is a little complicated than this. Nowadays, in electrochemical research, use of molten $\ce{NaOH}$ is the choice of the electrolyte. The choice of molten $\ce{NaOH}$ over molten $\ce{Na2CO3}$ is based on a number of advantages of the molten hydroxide electrolyte over others. The first advantage is its higher electrical (ionic) conductivity, which is important in the question at hand. The ionic conductivity of molten $\ce{NaOH}$ exceeds that of molten sodium carbonate by a factor of 1.5 [1]. Thus, it is safe to assume that molten $\ce{NaOH}$ functions as $\ce{Na+}$ and $\ce{OH-}$ ions in molten state. Additionally, $\ce{Ni}$ has already been shown to be a chemically stable material in a molten $\ce{NaOH}$ environment, and hence, nickel containers are used to produce anhydrous $\ce{NaOH}$. Concentrating an aqueous $\ce{NaOH}$ solution in $\ce{Ni}$ containers by heating up to $\pu{500 ^\circ C}$ will produce the desired $\ce{NaOH}$ melt.

According to [1], molecular oxygen is extremely oxidizing in fused hydroxides and reacts with hydroxide ions to produce superoxide and peroxide ions. The concentration of these ions is determined by the following equilibria (Note also that increasing the water vapor pressure would reduce the peroxide and superoxide ion content by favoring the equilibria to LHS, but it's not our concern here):

$$
\begin{align}
\ce{O2 + 4OH- &<=> 2O_2^{2-} &+& 2H2O }\\
\ce{3O2 + 4OH- &<=> 4O_2^{-} &+& 2H2O }\\
\ce{2OH- &<=> 4O^{2-} &+& H2O}
\end{align}$$

For example, see following Abstract:

  A thermodynamical discussion of the $\ce{Pt/O2}$ electrode in molten $\ce{NaOH}$ shows that the redox potential depends on $p_\mathrm{O_2}$, $p_\mathrm{H_2O}$, $a_\mathrm{O_2^{2-}}$ and $a_\mathrm{O^{2-}}$. Potential measurements have shown that the reaction $\ce{O2 + 2e− = O2^{2−}}$ is potential-controlling. From the electrode materials investigated—$\ce{Pt}$, $\ce{Au}$, $\ce{Ag}$, $\ce{Ni}$ and $\ce{Fe}$—the last one only exhibits a positive potential shift by comparison to the other metals; this shift has been interpreted in terms of an $\ce{O2^{2-}}$ impoverishment following corrosion.

  The peroxide which is required for the corrosion reactions can be formed only by chemical reaction of the melt with $\ce{O2}$ , not, however, by anodic oxidation, because peroxide oxidation essentially precedes the oxidation of hydroxyl ions yielding oxygen and water vapour. With anodic and cathodic polarization, peroxide is reacted with limiting current flow, the anodic reaction being less inhibited. Water vapour is cathodically reduced to $\ce{H2}$ and $\ce{OH-}$ —with limiting current flow—before $\ce{NaOH}$ is reduced. $\ce{Pt}$  corrosion is anodically enhanced by water vapour; in this context, current fluctuations are observed. In the pure $\ce{NaOH}$  melt the cathidic reduction may be assumed to involve formation of $\ce{H2}$ and $\ce{Na2O}$ ; not, however, $\ce{Na}$  deposition. [2]

Thus, in a molten hydroxide, one can expect ionic species such as metal cation ($\ce{Na+}$ in this case), $\ce{OH-}$, $\ce{O^{2-}}$, $\ce{O_2^{2-}}$, and $\ce{O_2^{-}}$ anions, and molecular $\ce{O2}$ present in this so-called basic melt in equilibria. Thus, when $\ce{K}$  reacts with any of these (most probably with $\ce{H2O}$ and $\ce{OH-}$ based on their concentrations at equilibrium) to oxidize to $\ce{K+}$, it will stay as mixture of $\ce{KOH}$, $\ce{K2O}$, $\ce{K2O2}$, and $\ce{KO2}$.

Example for ionic nature in molten salt solution molten $\ce{NaCl}$:

References:

Direct electrochemical power generation from carbon in fuel cells with molten sodium hydroxide: Chemical Engineering Communications, 2005, 192 (12), 1655-1670. DOI: 10.1080/009864490896241.
Elektrochemische messungen in NaOH-schmelzen: Electrochimica Acta, 1968, 13 (3), 625-643. DOI: 10.1016/0013-4686(68)87031-8.


How do Shaving Creams, Foams, Soaps ensure cut Hair Fragments that stick & clump between Safety Razor blades are easily washed away? This is my first post on Chemistry SE and I figure it's pretty chemical oriented.

On Sustainability SE we were trying to find "organic & natural" alternatives which are okay for lubrication & moisturizing (don't really care about foaming etc), but so far we have one key problem:

Hair fragments clumping and getting stuck in between 3-4 blades of a safety razor and not easily washing out using water, which is not as much an issue with commercial shave products.

Is it some (one or more) active ingredients/chemicals that make it happen or some other chemical phenomenon by which this works?

Active chemicals are called surfactants. It is well explained in colloid chemistry books and kind of in wikipedia (https://en.wikipedia.org/wiki/Surfactant)

And it is possible to find alternatives for such chemicals in nature, but they won't be as effective as artificial.


Mass of sugar that was added to the water (Raoult's law) Water has a vapour pressure of $15.5~\mathrm{mmHg}$ at $25~\mathrm{^\circ C}$. An unknown amount of the non-volatile sugar glucose ($\ce{C6H12O6}$) is added to $1.31~\mathrm{kg}$ of water and the vapour pressure of the system is found to decrease by $0.13~\mathrm{mmHg}$.  What mass of sugar, in grams, was added to the water?

I'm a little bit confused about this problem.

Here it says that after the sugar was added to water, vapour pressure of the system is found to decrease by $0.13~\mathrm{mmHg}$.

Does it mean that before sugar was added, there was only one element $\ce{H2O}$?

And I used Raoult's law:

$$P_i = x_iP_i^*$$

where $P_i$ is the vapour pressure of the solvent, $x_i$ is the mole fraction of the solvent, and $P_i^*$ is the vapour pressure of the pure solvent.

So can I use this to find the partial pressure before the sugar was added?

I know that $1.31~\mathrm{kg}$ of water is $72.78~\mathrm{mol}$.

And if my assumption is right and before sugar was added there was only water, will the mole fraction before sugar was added just be $72.78$ since there was only $\ce{H2O}$?

Your question

  Here it says that after the sugar was added to water, vapour pressure of the system is found to decrease by $0.13~\mathrm{mmHg}$.

  Does it mean that before sugar was added, there was only one element $\ce{H2O}$?

One component, not element. Water isn't an element. But yes. The question does actually say it: "sugar is added to water".

  I used Raoult's law: [...] So can I use this to find the partial pressure before the sugar was added?

Why do you want to find the partial pressure before it was added? It's already given to you. When there's only one component in the system, the partial pressure is the same as the total pressure, which is the vapour pressure of pure water: $15.5~\mathrm{mmHg}$.

In fact, the partial pressure after it was added is also given to you, it's just $(15.5 - 0.13)~\mathrm{mmHg}$. You already have all the pressure data you need. This question is giving you partial pressure data and asking you to find the mole fraction.

  I know that $1.31~\mathrm{kg}$ of water is $72.78~\mathrm{mol}$.

Yes. Although if you use a more accurate molar mass for water, you will get a more accurate answer.

  And if my assumption is right and before sugar was added there was only water, will the mole fraction before sugar was added just be $72.78$ since there was only $\ce{H2O}$?

No, the mole fraction would be $1$. The mole fraction of a component is the number of moles of that component, divided by the total number of moles.

The answer

The vapour pressure of pure water, which is $P_{\ce{H2O}}^*$ in Raoult's law, is $15.5~\mathrm{mmHg}$.

After you add a certain amount of sugar, the vapour pressure is decreased to $P_{\ce{H2O}} = 15.37~\mathrm{mmHg}$.

So, if you just plug these numbers into Raoult's law, you find:

$$\begin{align}
x_{\ce{H2O}} &= \frac{P_{\ce{H2O}}}{P_{\ce{H2O}}^*} \\
&= \frac{15.37~\mathrm{mmHg}}{15.5~\mathrm{mmHg}} \\
&= 0.9916
\end{align}$$

$x_{\ce{H2O}}$ is the mole fraction of water after sugar is added. So:

$$\begin{align}
x_{\ce{H2O}} &= \frac{n_{\ce{H2O}}}{n_\mathrm{total}} \\
0.9916 &= \frac{n_{\ce{H2O}}}{n_{\ce{H2O}} + n_\mathrm{glucose}}
\end{align}$$

The number of moles of water is:

$$\frac{1310~\mathrm{g}}{18.02~\mathrm{g~mol^{-1}}} = 72.70~\mathrm{mol}$$

This is what I said earlier about using a more accurate value of the molar mass. So:

$$0.9916 = \frac{72.70~\mathrm{mol}}{72.70~\mathrm{mol} + n_\mathrm{glucose}}$$

Rearranging:

$$n_\mathrm{glucose} = 0.6149~\mathrm{mol}$$

The molar mass of glucose is $180.16~\mathrm{g~mol^{-1}}$. So:

$$\begin{align}
m_\mathrm{glucose} &= (0.6149~\mathrm{mol})(180.16~\mathrm{g~mol^{-1}}) \\
&= 110.8~\mathrm{g}
\end{align}$$


Accuracy of FP-LAPW vs Plane Wave Pseudopotential: examples I am trying to learn about FP-LAPW software. I would like to know how well FP-LAPW outperforms Pseudopotential PW calculations. What are some properties/calculations/quantities in which FP-LAPW gives consistently better results than Pseudopotential PW theory? I know that FP-LAPW should give better results on almost everything (after all, it is an all-electron calculation). I am specifically looking for calculations where FP-LAPW gives much better results than Pseudopotential PW.

In all cases FP-LAPW will be better than Pseudopotential + Plane Waves calculations. However, FP-LAPW calculations are better at calculating properties which involve core electrons (e.g. core level shifts, hyperfine structures). Such calculations are often not possible with Pseudopotentials + Plane Waves, or are more complicated.This is because PP+ PW calculations do not treat core electrons explicitly.  Furthermore,  FP-LAPW calculations do not depend on pseupotentials. Although one can test pseudopotentials to check for transferability, there is no guarantee that the pseudopotential works for the system that you are simulating. As a result, "artifacts" due to pseudopotentials can be avoided with FP LAPW.


Is an explosion merely an exothermic reaction? An exothermic reaction is apparently one that releases energy to it's environment. According to Wikipedia, an explosion is a rapid increase in volume & energy.

Does this mean that an explosion is basically an exothermic reaction?

The key parts to the definition of an explosion are "rapid" and "increase in volume".

Exothermic reactions do not necessarily meet these requirements. Exothermic reactions need only release energy to the surroundings by converting chemical potential energy into kinetic energy.

For example, the combustion of methane is very exothermic, with a standard enthalpy change of -890 kJ/mol.

$\ce{CH4(g) + 2O2(g)->CO2(g) +2H2O(g)} \space  \space \space \space \Delta_c H^o=-890 \text{ kJ/mol}$

However, at room temperature this reaction is very slow. Mixtures of oxygen and methane do not spontaneously ignite without a spark source. Also, there is no change in pressure with this reaction, since there are 3 moles of gas on both sides of the equation.

On the other hand, consider nitroglycerin. Yes, it's combustion is significantly exothermic. More importantly, the combustion happens rapidly, needs only the heat of friction or the energy of impact to ignite, and produces significantly more gas than it consumes:

$\ce{4C3H5O9}(l) \ce{+1O2(g)->12CO2(g) +10H2O(g) +6N2(g)} \space  \space \space \space \Delta_c H^o=-5660 \text{ kJ/mol}$

All thermodynamic data were mined from the NIST Chemistry Webbook.


How to write the formula of a mixed metal salt? I suppose the formula of the Sulphate salt of Sodium & Potassium is $\ce{KNaSO4}$ as the Pubchem site says so.
So, is there any rule that in the formulae such mixed metal salts we need to write the metals in the order of electropositivity?

"$\ce{KNaSO4}$" is correct, but not because of electronegativity order, but rather the alphabetical one.
From IUPAC "Red Book", section IR-2.15 ORDERING PRINCIPLES [1, p. 40]:

IR-2.15.2 Alphabetical order
Alphabetical order is used in formulae as follows.
(a) Within the group of cations and within the group of anions, respectively, in formulae of salts and double salts.
Deviations from this rule may be acceptable if it is desired to convey specific structural information, as in Example 5 below.
Examples:

$\ce{BiClO}$ (anions $\ce{Cl-}$ and $\ce{O^2-})$
$\ce{NaOCl}$ (the anion is $\ce{OCl-},$ cf. Section IR-4.4.3.1)
$\ce{KNa4Cl(SO4)2}$
$\ce{CaTiO3}$ (perovskite type)
$\ce{SrFeO3}$ (perovskite type)

Reference

IUPAC “Red Book” Nomenclature of Inorganic Chemistry, 1st ed.; Connelly, N. G., Damhus, T., Hartshorn, R. M., Hutton, A. T., Eds.; IUPAC Recommendations; Royal Society of Chemistry: Cambridge, UK, 2005. (PDF)