Electride salt preparation In preparation for the Birch reduction, I am trying to prepare an electride salt, $\ce{[Na(NH3)6]+, e-}.$ This salt should be brilliant blue, as far as I know, but I am getting a black sludge when adding sodium to the liquid anhydrous ammonia at −60 °C or so.
Is this because the salt is missing the electron?
How do I remedy this?

As long as your experimental setup did not contain any leakages or impurities through which air or moisture could have gotten into the mixture, the reaction should have worked fine under these conditions. The color of the hexaamminsodium electride solution is actually deep blue and intense, so at higher concentrations it can appear black at first sight (see picture below).


Brown particles in electrolysis of brine This is not a duplicate... pls read the whole question....

I am attempting electrolysis of brine  (tap water + table salt). I am using HB pencil lead (graphite) as my two electrodes. I am using 12 volt- 1.5 amp DC power supply (using an adapter from the wall).
The solution is getting slightly alkaline (~8 pH) and hydrogen is confirmed by pop sound.

However, the whole solution is getting filled with some brown particles (resembling Dirt). If by chance my alligator clips get submerged, the get a brown coating (which I am able to wash it off).

What are this brown particles?

I have a few theories:
1. Pencil lead also contains clay.
2. Something with the water (though it is filtered)

I had attempted this earlier (~3y) and had a similar but a more dramatic result.
Then I used iron/steel electrodes and they became completely brown. The plastic container I used also became completely brown coated. I was unable to wash it off.

If your tap water is groundwater, it will contain a lot of metals that will start rusting if reacting with O2 and Cl (your salt). The reaction causes the brown particles (rust/ Ferric Oxide). The only way to prevent this is to filter the water continuously. This reaction is also common with salt-electrolyzed pools. Here is a forum thread that has exactly the same issue you have.

This was the answer:

  That looks like rust, Ferric Oxide, exactly!
  However, the cell should be Titanium, it contains no iron, but that doesn't mean it can't oxidize. Or "Rust", all metals do, Ti's oxidation would be white and powdery.

  So either there is a non Ti component inside that cell that contains iron, such as steel that is breaking down, or there's something upstream of it, a Ferrous metal that's rusting, thus producing Ferric Oxide and dribbling down the pipe and into the cell. Is there any on the upstream side of the cell?
  If not, then it has to be localized to inside the cell.

  Another though is that you have a very high iron content in your water, mixed with the electrolysis process is causing it to oxidize on the anode, where it combines /w oxygen molecules in the water to form rust, Fe2O3.
  This would keep it localized to the cell, and not be upstream, only at the cell, since that's the catalyst, or source and slightly downstream as the flow of generated Ferric Oxide tapers off.
  This shouldn't effect the Ti plates though, just simply clean them with MA and you're golden!!
  Ti is virtually indestructible, so they should be fine.

  EDIT: Hayward says to submerge it just to the top of the cell, so the wiring harness, the electronics on the cell are not touching the acid, coiling the harness may be helpful prior to immersing. 4:1 dilute, 1 gallon of water to 1 quart of MA. And remember, always, always add the acid to the water, never, ever the other way around.
  The manuf. says to soak for a few, then rinse with a high pressure hose, then repeat if needed, until the plates are clean.
  BTW, Acid attacks Oxides, including Ferric Oxide.

Full thread here

Related Stackposts:
Salt product in water electrolysis


SN1 reaction in a polar aprotic solvent I'm seeking a $\ce{S_N1}$ reaction that happens in a polar aprotic solvent.
Do you know any good simple $\ce{S_N1}$ reactions which satisfy the condition?

I think something like potassium 2-methylpropan-2-olate and 3-bromopentane in $\ce{SO(CH3)2}$ should do the trick.


Is water a better nucleophile than bromide? I am a student of biochemistry. My organic chemistry professor was giving a lecture and he talked about halohydrin formation reactions. At one point, he said that "water attacks instead of bromide because water is a better nucleophile than bromide". He repeated that five times or so.

A classmate argued that, because water is the solvent in the reaction and is in greater proportion, it will attack faster. But that still doesn't mean water is a better nucleophile, right? Bromide itself has a negative net charge, while water has no net charge, only 2 free electrons.

The question is, was my professor wrong or is there something I'm missing? Thank you very much.

That is not true. As you noted bromine has a negative charge (and also lower electronegativity than oxygen) and is therefore the better nucleophile (even under these aqueous conditions). Halohydrins are not even usual the result of bromination in water. Special conditions are required to prevent too much $\ce{Br-}$ from attacking the bromonium cation (e.g. certain pH, adding additional reagents as $\ce{AgOH}$ or using $\ce{HOBr}$, see the section Addition Reactions Initiated by Electrophilic Halogen in W. Reusch's online Organic Chemistry book). You would usually expect the dihalide product since $\ce{Br-}$ is already close to the bromonium cation as it forms.


Is volume expansion possible upon dissolution of solute? I am aware that there is volume reduction upon dissolution of ionic solutes, due to the formation of hydrated spheres of aqueous ions where water molecules aggregate around the solute ions.

But similarly could there be a volume expansion?

Yes. In thermodynamic jargon, the onset of a change in volume under dissolution is known as an excess volume. If the solution takes up less volume than its constituent parts it is said to have a negative excess volume. Otherwise we have a positive excess volume. You might also find the term excess molar volume.
By googling each term we find many results concerning this, which seems to mainly manifest in metal alloys.


Cyanide complexes in Molecular Orbital Theory In molecular orbital theory, please explain why cyanide ion combines with metal from the carbon end and not from the nitrogen end?

In this answer of Martin’s, you can find a molecular orbital diagram of $\ce{CO}$. The corresponding diagram for cyanide, $\ce{CN-}$ is essentially identical, there will only be different orbital energies and very slightly different extends of the lobes.

When forming a coordinate bond to a metal centre, cyanide will primarily attack with its highest occupied molecular orbital, the HOMO, since it is a nucleophile (attacking a positively polarised metal centre). You can also see the HOMO depicted as the bottom image of the following set as taken from Wikipedia:

Note the large lobe extending outwards on carbon’s side. An attack with this lobe is greatly preferred, hence why isocyanides are comparatively rare.


Why and how does carbon hybridize itself according to its need? Sometimes carbon forms sp3 orbital and sometimes sp2 and sp. Also, carbon chains are more stable with sp3 orbitals. So, why and how does carbon hybridize itself to other orbitals?
My question is more about stability. It's not about how it is significant to organic chemistry. For example, Methane which has 'sp3' orbitals is stable because of its sigma bonds. But, ethene, which has a pi bond is not stable and is more reactive than methane. So, why does carbon form this 'pi bond' while on the other hand has a better option of a sigma bond.

Hybridization is part of the language chemists use to speak to one another.  When someone says that a carbon atom is $\ce{sp^2}$ hybridized, it's a shorthand way of telling others

that there is a p-orbital on that carbon
that the angles around the carbon are roughly 120°
that the bonding orbitals (the $\ce{s^1p^2}$ orbitals) have 1 part in 3, or 33%, s-character and 2 parts in 3, or 67%, p-character
that the bonds around that carbon are likely to be a bit shorter than the bonds around an $\ce{sp^3}$ hybridized carbon (because the $\ce{sp^2}$ bonds have more s-character and are therefore closer to the nucleus, e.g. shorter).

So when someone says that a carbon atom is $\ce{sp^2}$ hybridized, they're really saying a lot more.

Using hybridization to describe an atom is similar to using VSEPR or MO theory to describe a molecule.  Hybridization is just another way to try and describe reality; just like the other methods, often it works well, but sometimes it doesn't.

Also keep in mind that there is a continuum of hybridizations available to a molecule all the way from pure s to pure p with everything else in between.

$\ce{s^1p^0...s^1p^1...s^1p^2...s^1p^{2.2}...s^1p^3...s^1p^4...s^0p^1}$

Of course, we normally drop the "1" and the "0" superscripts and write those hybridizations as

$\ce{s...sp...sp^2...sp^{2.2}...sp^3...sp^4...p}$

As to how carbon (or any other element) "decides" which hybrization to use, it adopts the hybridization that produces the overall lowest energy for the molecule. Let's take the case of a carbocation.  Suppose we start with an $\ce{sp^3}$ hybridized carbon in a molecule like $\ce{R_3C-I}$ and subject it to conditions where it ionizes to an $\ce{sp^2}$ hybridized carbocation and $\ce{I-}$.

(image source)

Your question is basically, why did the central carbon rehybridize from $\ce{sp^3}$ to $\ce{sp^2}$? The answer is, the ionized molecule (the carbocation) will reoptimize its geometry (rehybridize) in order to attain the lowest energy possible. In this example note that the carbocation p-orbital is empty - it contains no electrons. So why keep that orbital $\ce{sp^3}$ hybridized, why waste a part of a valuable, low-energy s-orbital in a bond with no electrons in it to stabilize? Instead, why not remove the s-character from the $\ce{sp^3}$ bond making it a pure p-orbital and mix that s-character into the other 3 orbitals (or bonds) to convert them from $\ce{sp^3}$ to $\ce{sp^2}$ thereby lowering the energies of the electrons that are in those 3 bonds. In fact, this rehybridization, or geometry reoptimization, is exactly what does occur in this reaction in order to stabilize the electrons as much as possible and produce a molecule (or ion) with the lowest possible energy.


What is the meaning of the J value in term symbols I understand how to work out the J values of term symbols.

e.g. a d1 configuration gives J values of 5/2 and 3/2. resulting in 2D 3/2

However I dont understand the meaning of it, and why it arises.

For multi-electron atoms where the spin-orbit coupling is weak, it can be presumed that the orbital angular momenta of the individual electrons add to form a resultant orbital angular momentum $\bf{L}$.

Likewise, the individual spin angular momenta are presumed to couple to produce a resultant spin angular momentum $\bf{S}$ . Then $\bf{L}$ and $\bf{S}$ combine to form the total angular momentum $\bf{J} = \bf{L}+ \bf{S}$.


What's that tangy smell on cheap plastic? Most cheap items (from China mostly) have a strong tangy odor to the plastics (or some resin, but I can mostly smell it on plastics). The smell is always consistent over a broad range or products and I get it for at least a decade. So I do not believe this is a one-off thing and I bet most people have experienced it. The only variation is intensity.

On some really bad products that smell impregnates other things and I failed to remove it with soap, baking soda, acid, scrubbing...

I'm curious to know what exactly is that smell.

Apparently it got some attention already, but there's no consensus, and a lot of crywolf(?) e.g. Campaign to Halt the Import of
Chemical-Emitting
Smelly Plastic from China.

Without a sample and a gaschromatograph it is hard to say what exactly it can be but... there are at least three well-known sources of odour in plastics:

Some residual of the monomer that makes up the plastic (that is a polymer).
Some residual of some other substance used during the manufatcturing process (catalyzer, co-polymers, modifiers and so on). For example, PETcan sometime contain small traces of terephthalic acid.
Some odourizing substance that is added to the plastic just to cover others smells.

Usually, tangy smells are a symptom of some kind of  ester


Equilibrium and Completeness of Reaction In class we are currently learning about equilibrium and Le Chateliers principal. The principal makes sense but seems contrary to some other concepts. For example, using stoichiometry to calculate the mass of reactants for a certain amount of products. If we know the reaction is always going to equilibrium where the forward and reverse reactions are happening at the same rate, then how will the reaction ever be "complete" because according to the principal even if the reaction were to be complete and all the reactants were gone, the reverse reaction should be occurring. This is not the case in reactions like burning magnesium.

I'm guessing that not all reactions are reversible and that the principal is correct but in some cases it would take a long time to reach equilibrium.

The basic stoichiometry is done to understand the basic concepts of moles and ratios in reactions. Later on, as the reactions become more complex, alot of them don't go to completion and so for a lot of them, you need equilibrium constants. This being said, not all reactions end partway, some of them do go to completion. The most basic example is disassociation of salt ions in solutions. From basic stoichiometry, you learn that the ratio is one to one (like NaCl -> Na+ + Cl-). So Moles of NaCl added = the moles of Na and Cl ions in water. So basic stoichiometry is sometimes (not always) applicable but is essential for the learning process.


How to buy 1g of enzyme that is sold in units/mg? I am trying to buy 1g of horseradish peroxidase. On sigma-aldrich, it is sold as units/mg. I am confused as to what this unit is. It comes in 50-150, 150-250, and 250-330 u/mg. The way I currently understand it is if I buy 25 KU (kilo-units?) of 50-150 u/mg, than I should have  at minimum 167 mg of the peroxidase. Is this a correct understanding? The price of 25 KU of horseradish peroxidase is $95, so it seems ridiculous to pay that for only 1 2/3g.

Sigma Aldrich defines their units as:

  One pyrogallol unit will form 1.0 mg purpurogallin from pyrogallol in 20 sec at pH 6.0 at 20 °C.

Assuming you're doing a biological imaging experiment, these activity values tell you much more about what you're getting than mass.

From a casual internet search it appears that HRP is pretty expensive. I would consider whether you actually need so much of it for your research.


RCOH and RCHO Nomenclature In terms of nomenclature, is there a difference between RCHO and RCOH when referring to an aldehyde?

$\ce{RCHO}$ is preferred. The problem with $\ce{RCOH}$ is that it gives the misleading impression that the hydrogen is attached to the oxygen. In fact, I don't believe I've ever seen a reference in literature to an aldehyde using the latter representation.


What are chiral environments? I keep seeing the term cropping up, but cannot seem to find a definition for it. My understanding is that it is an environment in which, when the isomers interact with the environment, the two enantiomers can be distinguished. The only example I can think of is with the human body, where one type of enantiomer can form an enzyme-substrate complex and the other cannot. I can't think of any other examples though. Can you create a chiral environment in a lab?

My understanding is that it is an environment in which, when the
  isomers interact with the environment, the two enantiomers can be
  distinguished.

That's exactly right.  Some examples of chiral environments that could be created in a lab would include

Running a reaction in a chiral solvent
Running a reaction in a chiral cavity
Running a reaction with a chiral reagent
Running a photochemical reaction with a polarized light source
Carrying out a reaction on a surface to which chiral molecules have been attached (certain clays have this property)

All of these items are like a glove, and each enantiomer will fit differently.  This different "fit" leads to transition states of different energy as each enantiomer moves along the reaction coordinate towards the product(s).


Placing of lone pairs in axial/equatorial position If the lone pairs of elections are put on axial or equatorial position such as to minimize the intera-electronic repulsion, and for the same reason in $\ce{SF4}$, the election pair is kept on the equatorial position(at 90 degrees to only two domains), why then in $\ce{BrF5}$ the lone pair is kept on the axial position, when it is at 90 degrees with 4 domains?

There is no axial or equatorial position in an octahedral structure. All places are equivalent and you can put the lone pair on position.


What's the difference between aqueous hydrogen chloride and hydrochloric acid? From my understanding, both aqueous hydrogen chloride and hydrochloric acid are written as $\ce{HCl(aq)}$, is there an actual difference between them?

There is no difference.

Actually, it would be more accurate to refer to gaseous (and dry) $\ce{HCl}$ as "hydrogen chloride", and to aqueous $\ce{HCl}$ as "hydrochloric acid", but often the terms are used interchangeably.

The difference is due to the fact that, as you might guess, an acid is a proton donor (it means that it tends to give off $\ce{H+}$ in a solution). On the other side, gaseous and dry $\ce{HCl}$ is simply a diatomic molecule. The dissociation in its ions (and the showdown of its acidic properties) only happen in a polar medium (typically water).

In the paper - Zotikov, V.S. Russ J Appl Chem (2009) 82: 1733. https://doi.org/10.1134/S1070427209090389 - you can find a cool example of the difference between a dry hydrogen halide and a wet hydrogen halide (here, the "terrible" HF is involved):

  In  dry  hydrogen fluoride,  carbon  steel  is  subjected to  minor  corrosion  up  to  $\pu{250^\circ C}$.  In  aqueous  solutions the  carbon  steel  is  stable  only  at  the  concentration  of hydrogen fluoride  above  75  wt  %  and  temperature not exceeding $\pu{50^\circ C}$


How important is it that geometry be optimized at a high level of theory? I've been doing research with a computational chemist for a little while now and in one of our projects we are dealing with a rather large system and because I go to a relatively small university, our resources are limited. Thus, when we began to ramp up the calculations in the project, we opted not to re-optimize our structure at CCSD(T)/def2-ATZVPP despite the fact we are doing the rest of our calculations at that level of theory with that basis set. The reason we chose not to optimize is we have a limited amount of time/resources and we estimated the optimization would take around 45-50 days to complete which would basically hog a computer for that long and stall the other projects we are working on.

I'm wondering how significant of an error, if much at all, that will introduce?

To give a few more details, we are studying vibrational properties of the system using a local mode approach. The structure we have is optimized at MP2/def2-ATZVPP.

Regardless, I'm just looking for a general understanding of how important the optimized structure might be. It seems like it would be less important, but could be quite dependent on what one is studying, than other things based on the fact that no system is truly at that optimized structure, but is only in that geometry on average. Is that correct at all?

Method

Most of the time CCSD(T) would indeed be a huge overkill for geometry optimisation. DFT and MP2 have way better performance/cost ratio. Note that I said DFT and MP2 above, not or MP2; this is well known procedure to compare DFT geometries with MP2 ones to be on the safe side: there should be no big differences.
On the DFT side hybrids might be overkill as well. Try simple pure GGA functionals like BP86 and PBE as well meta-GGA functionals like TPSS first. There might be almost no advantages in using hybrids for geometries.
Include empirical dispersion corrections in DFT calculations even if you do not suspect dispersion forces to be decisive. Just in case.
Employ the density fitting (also known as the resolution of identity) for DFT as well as MP2 to significantly accelerate calculations.

Basis set

With respect to the basis set convergence of geometries is typically reached at the triple-zeta level. This is fairly well-known from excessive benchmark studies for correlation consistent1 and polarisation consistent basis sets2, but Turbomole def2 basis sets should have similar behaviour as well. References are exclusively about DFT, but the same is true for WFT as well.
In practice, though, for DFT calculations (but not necessarily for MP2) a triple-zeta basis might also be an overkill. Try double-zeta def2-SVP first (augmented with diffuse and polarisation functions to your taste). Even Turbomole developers themselves describe DFT/SV(P) and MP2/TZVPP levels of theory as "almost quantitative".

Benchmark

Compare the calculated geometries with experimental data if such are available for your (or at least similar) compounds. At the end of the day, how well does a particular level of theory reproduces experimental geometries is the main criteria of choice.

1) Wang, N. X., & Wilson, A. K. (2004). The behavior of density functionals with respect to basis set. I. The correlation consistent basis sets. The Journal of Chemical Physics, 121(16), 7632–46. http://doi.org/10.1063/1.1792071

2) Wang, N. X., & Wilson, A. K. (2005). Behaviour of density functionals with respect to basis set: II. Polarization consistent basis sets. Molecular Physics, 103(2-3), 345–358. http://doi.org/10.1080/00268970512331317264


Yellowish tinge during titration of oxalic acid with potassium permanganate One of the our chemistry practicals involve finding the molarity of an oxalic acid solution by titrating it with standard $\ce{KMnO4}$ solution.

The oxalic acid solution is heated and little sulfuric acid is added to titration flask before starting the titration.

It was my personal observation that adding the $\ce{KMnO4}$ solution rapidly causes a temporary appearance of a light yellow colour. Slow and gradual addition of $\ce{KMnO4}$ does not lead to this observation.

If sulfuric acid is not added to the titration flask, then also a yellowish tinge appears as we get closer to the end point.

What is the for this yellowish tinge? Is there any intermediate species formed (maybe a complex one) that leads to this observation?

Without knowing the concentrations of the involved substances and the values of some other parameters, we can only guess what has happened during your experiment.

The intended analytical reaction is the reduction of purple permanganate to colourless $\ce{Mn^2+}$:

$$\ce{MnO4- + 8H+ + 5e- <=> Mn^2+ + 4H2O}$$

If you add the permanganate solution too quickly, the reduction may be incomplete because not enough reducing agent (here: oxalic acid) or acid $(\ce{H+})$ is available in the reaction zone:

$$\ce{MnO4- + 4H+ + 3e- <=> MnO2 + 2H2O}$$

The observed colour may be caused by a cloud of finely dispersed particles of dark brown manganese(IV) oxide $(\ce{MnO2})$, which dissolves when the solution is well mixed again so that the intended reaction continues.

The same reaction may occur when you don’t add enough sulfuric acid, since $\ce{H+}$ is consumed during the reaction (8 mol $\ce{H+}$ per 1 mol $\ce{MnO4-}$).


Kinetic isotope experiment used to distinguish radical and concerted mechanisms for the Diels Alder reaction In the Pericyclic Reactions Oxford Primer it is written that:

  the deuterium
  substitution of the hydrogen atoms in the diene and dienophile that
  participate in a possibly Diels-Alder reaction, gives rise to inverse
  secondary isotope effect (as the carbon atoms change from trigonal to
  tetrahedral geometry):

  If both bonds are forming at the same time, the isotope effect when both ends are deuterated is geometrically related to the isotope
  effect at each end.
  If the bonds are being formed one at a time, the isotope effects are arithmetically related.

I don't understand the principle behind this. Why does the isotope effect occur and how does it tell us whether the mechanism is radical or pericyclic? I need an answer much more concise than what is presented in the book.

What Fleming is likely referring to is work by Edward Thornton (J. Am. Chem. Soc. 1972, 94, 1168) in which he attempted to make use of secondary kinetic isotope effects (KIE) to prove whether the Diels Alder reactions being studied were indeed pericyclic, or occurred via a radical pathway.*

The phrasing in your post (which I assume is from the book itself) isn't particularly helpful. By 'geometric', Thornton was actually talking about the geometric mean and the arithmetic mean as ways of considering the isotope effects.

The reaction they actually studied was the retro Diels-Alder of 9,10-dihydro-9,10-ethanoanthrace, with varying levels of deuteration (none, di, tetra) as shown in the figure below.

In order to discuss whats going on, we need to define some terms:

In the reaction, two bonds are being broken, lets call those bonds A and B. Breaking bond A has a KIE of fA and breaking bond B has a KIE of fB
Each molecule (no deuteration, di-, tetra-) has an associated rate constant, lets call these k0, k2, k4 (with the subscript indicating how many deuteriums we have).

We can now consider the calculation that Thornton went through to determine if the reaction is indeed a concerted one:#

In a completely concerted reaction, intuitively fA = fB, since the TS is symmetrical. If the reaction is non-concerted (aka stepwise), fA cannot equal fB unless there is symmetry to the TS.
Considering the di-deuterated molecule, two pathways are possible (breaking A or B first) to give a non-symmetrical intermediate. Thornton defines rate equations for these two as fAk0/2 and fBk0/2. We can do this since k0 is the same no matter which bond is broken in the undeuterated starting material (with the f modifying it to take into account the KIE).
k2 (the rate constant for the retro Diels-Alder of the di-deuterated starting material) can therefore be thought of as the sum of two pathways (break A then B or B then A). Mathematically: k2 = k0(fA + fB)/2. Crucially, this relationship is valid independent of whether the reaction is concerted or stepwise.
k4 (the rate constant for the retro Diels-Alder of the tetra-deuterated starting material) is able to be obtained similarly. Mathematically: k4 = fA fBk0.
At this point, Thornton provides the result of solving the two previous equations simultaneously, which provides a relationship as below:

$$f_\mathrm{A},f_\mathrm{B} = \frac{k_2}{k_0} \pm \left[\left(\frac{k_2}{k_0}\right)^2 - \frac{k_4}{k_0}\right]^{1/2}$$

It is from this relationship that we are able to consider whether the reactions are concerted or stepwise:

if concerted, fA = fB = k2/k0, which means that (k2/k0)2 == k4/k0. This now allows us to measure the rate constants, plug in the numbers, and determine if the reaction is concerted.

*: There are many examples of Diels Alder reactions that are stepwise, via radical mechanisms. The KIE's measured are necessarily small, meaning its often difficult to know if what you're looking at is actually correct. One thing that does hold up generally is that the reactions aren't ionic, due to the lack of any real solvent dependance (if the reactants are soluble in the solvent and you heat it hot enough, they generally proceed at the same rate, suggesting no stabilisation of 'intermediates' in polar solvents).

#: I'm paraphrasing the original paper, which will hopefully name it more digestible, but I still encourage you to read the actual thing, despite it being fairly dense


Ion concentration in acid and base Hello here is a question that is a bit confusing I hope you can help me solve it and understand it:

  Determine the hydronium and hydroxide ion concentrations in a solution
  that is $10^{-4} M \ce{Ca(OH)2} \text{ AND } 10^{-4} M \ce{HCl}$

Note: the "and" is actually capitalised in my book..

So what I understood from this question is that they are both in one solution. Now I thought of two ways to solve it:
A) the concentrations are equal and they are both strong which means they we're neutralise each other completely and end up being a neutral solution and both $\ce{H3O+}$ and $\ce{OH-}$ will be $10^{-7}$.

B) the calcium base has 2 $\ce{OH-}$ while HCl has 1 H+ so the base concentration is double that of the acid which means that the solution is going to be basic but I don't know how can I solve this to find the actual concentrations.

Which way is right and when does an acid and base really react to become neutral?

Your second (B) way of thinking is correct.

In the solution there are twice as many OH- ions as H+ ions.

All the H+ from HCL will react with half the OH- from calcium hydroxide.

Half the OH- will remain ($10^{-4} OH-).

Then use $K_w$ to determine the final concentration of H+.


Are clathrates and interstitial compounds the same thing? In the book$\ce{^{[1]}}$,  it's given that clathrates are compounds in which gases are entrapped within cavities of crystal lattices of certain organic or inorganic compounds. Interstitial compounds are formed when small atoms occupy positions in the interstices of the metal lattice. So I'm not sure what is the difference between the two? Also it's given that only Helium forms interstitial compounds and Argon, Krypton and Xenon forms clathrates. Could anyone help me understand the difference between the two?

Reference

Pradeep, New Course Chemistry, Class XII, Vol. 1

Clathrate compound :

  A clathrate is a chemical substance consisting of a lattice that traps or contains molecules. The word clathrate is derived from the Latin clathratus (clatratus), meaning ‘with bars, latticed’. Traditionally, clathrate compounds are polymeric and completely envelop the guest molecule, but in modern usage clathrates also include host–guest complexes and inclusion compounds. According to IUPAC, clathrates are inclusion compounds "in which the guest molecule is in a cage formed by the host molecule or by a lattice of host molecules."

The famous is the methane in ice clathrate $\ce{4 CH4 \cdot 23 H2O}$
Another case are clathrates of noble gases, found in hunting for noble gas compounds.

Many clathrates are stable just at low temperatures and/or high pressures.

Interstitial compound :

  An interstitial compound, or interstitial alloy, is a compound that is formed when an atom with a small enough radius sits in an interstitial “hole” in a metal lattice. Examples of small atoms are hydrogen, boron, carbon and nitrogen. The compounds are industrially important.

So in some loose sense, the former has more or less stable non metallic lattice, the latter has stable, metallic lattice.


Color of chromate and permanganate I've heard quite a few times that the chromate and permanganate have a $d^3s$ configuration. Also, their colors arise due to a rapid switching of electrons between the oxygen and metal atoms.

I don't really understand the 'rapid switching' part--it's obvious why it can give color, but I fail to see why there is a need for such a switching-what's so special about $\mathrm{Cr}$ and $\mathrm{Mn}$? (I also do not know what the switching exactly is)

An explanation of $d^3s$ would be appreciated, though not necessary.

By "rapid switching" they technically mean Ligand-to-Metal Charge Transfer (LMCT). A more modern framework is Ligand Field Theory. I would have to teach a class to fully explain it in those terms, but I'll try to explain it in terms of this hybridization you brought up.

A chemical bond implies a higher probability of finding electrons between two bound nuclei. Atomic orbitals describe the electron density for individual nuclei. A bond between two nuclei necessitates some physical overlap of the relevant atomic orbitals.

It is in this way that we say that the linear combination of atomic orbitals, say two consisting of a $2s$ and $2p$ orbital, results in a molecular orbital that describes the molecule. We would write the wave function for the complex as:

$$\Psi = C_1 \psi(2s) \pm C_2 \psi(2p)$$

The amount of "mixing" is just a matter of adjusting the coefficients, $C_n$. And so we might say that the bond in this made-up molecule is $sp$ hybridized. $sp^3$ hybridized would mean the following kind of wave function:
$$\Psi = C_1 \psi(2s) \pm C_2 \psi(2p) \pm C_3 \psi(2p) \pm C_4 \psi(2p)$$

Similarly, $sd^3$ could mean a wave function of the following nature:

$$\Psi = C_1 \psi( (n+1) s) \pm C_2 \psi(n d) \pm C_3 \psi(nd) \pm C_4 \psi(nd) $$ where $n=3$ in the case of Mn or Cr.

If one performs the proper molecular orbital calculations on the valence $3d$, $4s$, and $4p$ for manganese or chromium and the $2s$ and $2p$ for oxygen in tetrahedral symmetry then you can draw the molecular orbital energy diagram for your complex once you match your eigenvalues to the correct trace of the appropriate transformation matrix... and you can hear a bunch more jargon on it, or I can explain the main point of our models: why we think MnO$_4$ $^-$ is purple.

The bonds in, specifically permanganate  (I have never actually done the calculations on dichromate, although they should be similar in principle) are between Mn$^{+7}$ and four O$^{-2}$ in the geometry shown here. The electronegativity of oxygen clearly dictates most of the electron density and, as a result, has them sitting on ligands (oxygen) instead of the metal. An electronically excited state can be reached with light absorption in the range of 500-600 nm light due to the relative weakness of the interactions between the ligand and metal. If they were stronger interactions then it would take more energy to promote electrons into a higher energy state and the color would be shifted into the UV. A colorwheel tells you that light absorption in the 500-600nm range should be approximately violet, which is what we see for the color of this particular complex. Anyway, this excited state means some electrons temporarily move from the ligand to the metal, resulting in a LMCT band.


Why must both the critical temperature and pressure be exceeded to achieve the supercritical phase? The characteristics and properties of supercritical fluids and the associated vapor-liquid critical point are well established.  One thing that I've always been curious about, though: what is the physical explanation for why it's necessary for both the critical temperature and critical pressure both to be exceeded simultaneously in order for this phase transformation to occur?

ChemWiki's description ("Critical Point") provides some measure of an explanation:

  [A]bove the critical temperature, it is impossible to condense a gas into a liquid just by increasing the pressure. All you get is a highly compressed gas. The particles have too much energy for the intermolecular attractions to hold them together as a liquid. The critical temperature obviously varies from substance to substance and depends on the strength of the attractions between the particles. The stronger the intermolecular attractions, the higher the critical temperature.

Is there more to the physics than just "[t]he particles have too much energy for the intermolecular attractions to hold them together as a liquid"?

Per an early comment by Mithoron, the fact that increasing pressure on a supercritical fluid can eventually drive it to a solid phase would seem to imply that it's not quite that simple.  See, for example, the phase diagram of carbon dioxide(source):

The critical point is a phase transition of phase transitions.  It's a different beast than your everyday run-of-the-mill phase transition like water boiling to vapor or ice melting to water.

The diagram you included is very misleading in its use of color.  The "yellow" supercritical regime doesn't really exist.  Beyond the critical point, gas is indistinguishable from liquid.  So instead of having a special yellow region, the supercritical region should be some gradient of green (liquid) and pink/orange (gas).  The vapor pressure curve between the triple point and critical point is the curve along which liquid is in equilibrium with gas, i.e. those phases have the same molar free energy $G$.  However, they obviously differ in many other properties, including molar enthalpy $H$, molar entropy $S$, molar volume $V$, etc.  Moving along a trajectory in $PT$ space that crosses the vapor pressure curve results in a discontinuity of those values.  Thus, such a phase transition is a first-order phase transition, as first-order derivatives of $G$ (such as $S=\left (\frac{\partial G}{\partial T}\right )_{P}$ and $V=\left (\frac{\partial G}{\partial P}\right )_{T}$) are discontinuous through the transition.

As you move along the vapor pressure curve closer to the critical point, the size of the discontinuous jump in $S$, $H$, $V$, etc. gets smaller and smaller.  Finally, at the critical point, the discontinuity disappears.  So the existence of a first-order phase transition goes away, and no phase transition at all takes its place.

Say you were to start with liquid $\ce{CO2}$ at a $T$ slightly below and $P$ slightly above the critical point, and move through $PT$ space in a counterclockwise cyclic orbit around the critical point.  As you lowered $P$, there would be a distinct first-order phase transition to gas -- the $\ce{CO2}$ would boil -- but then as you reached the bottom of the circle and started increasing $P$ and $T$, moving past the critical point, there would be no phase transition at all as you went to the yellow region in your chart.  There would similarly be no phase transition as you reached the top of the circle and moved back into "green" territory.  You'd get liquid carbon dioxide again, and when you go again through the circular path through $PT$ space, it would boil -- a distinct first-order phase transition -- again.  But it would never condense, in the sense of a sudden change from gas to liquid.  The change back to liquid $\ce{CO2}$ would be gradual, with no discontinuities.

So to specifically answer your questions:

  what is the physical explanation for why it's necessary for both the critical temperature and critical pressure both to be exceeded simultaneously in order for this phase transformation to occur?

The pseudo-fallacy in the question is using the phrase "phase transformation" to describe critical phenomena.  Passing the critical point is not a first-order phase transition like boiling water under normal conditions.  It's a much higher-order phase transition.  Here's a good chapter with more info.

  The fact that increasing pressure on a supercritical fluid can eventually drive it to a solid phase would seem to imply that it's not quite that simple. See, for example, the phase diagram of carbon dioxide.

Think in terms of free energy landscapes.  Solid, liquid, and gas phases are at the bottom of potential wells or pits in this $G$ surface.  Past the critical point, the liquid "well" and the gas "well" are no longer distinct minima, but are the same.  However that doesn't change the fact that if you move "up" in the free energy landscape by increasing $P$, you will eventually find the still-distinct pit (free energy minimum) for solid.


What reagents or series of reactions would allow for selective hydroxylation of an allylic methyl group outside of a ring? Sample transformation for selective hydroxylation of an allylic methyl group outside of a ring. Selenium dioxide would not work since hydroxylation would be favored inside the ring.

As has been stated by @Beerhunter those benzylic hydrogens are a serious concern . Nonetheless, here is a different offering. To solubilize carboxylic acid 1 both the phenolic group and the acid should be protected. Here, both of the groups are protected as methyl groups. [An ester group and silylated (TBS?) phenol would also suffice.] The double bond of 2 is epoxided to 3. The critical step is the isomerization of the epoxide to an exo-methylene allylic alcohol. There are several reagents that do this transformation. Yamamoto1 has shown that methyl groups are more reactive than methylene groups with the epoxide rearranging reagent diethyl aluminum 2,2,6,6-tetramethyl piperidide although no benzylic methylene examples were provided.
Acetylation of the allylic alcohol affords allylic acetate 4. Grieco2 has reported that bis-acetonitrile PdCl2 isomerizes allylic acetates to their tautomer in what is an equilibrium reaction favoring the more stable isomer. In this instance, the trisubstituted, endocyclic double bond compound 5 is expected. The remaining steps involve removal of the protecting groups of choice to afford the target 6.

1) S. Tanaka, A. Yasuda, H. Yamamoto and H. Nozaki, J. Am. Chem. Soc., 1975, 97, 3252.

2) P. A. Grieco, T. Takigawa, S. L. Bongers and H. Tanaka, J. Am. Chem. Soc., 1980, 102, 7587.


pH scale at various temperatures I am an electrician apprentice in charge of a large number of pH-sensors/transmitters at a power plant and chemical plant.
I do all the pH-instrumentation related maintenance, calibration, adjustments etc.
By studying pH measurements I have learned about the Nernst equation. The transmitter is able to (via an inbuild Pt1000 sensor) automatically keep the temperature variable in the Nernst equation constantly updated.
However, I have also learned that pH bear no meaning unless a temperature of the measured medium is specified. The transmitter is ALSO able to compensate for this variability, however. This is done by heating up a sample of the process medium and noting down the pH difference at various temperatures. This information can be plotted into a matrix in the transmitter.
HOWEVER, since the process medium composition is ever changing at the plant I work at (waste incineration) this matrix can't be made.
Seeing that many of the proces temperatures at the plant are anywhere between 40 and 80 degree celcius, how large of a discrepancy am I looking at? I know that both chemical composition and temperatures dertermine how much, but realistically, what are we talking about? 0,1 difference in pH? 0,5 difference? More? Worst case? Best case?
Thanks in advance.

The pH of acidic solutions does not change much with increasing temperature. The pH of basic solutions does change more significantly with increasing temperature.
Ex. : The pH of $0.05$ M Potassium tetroxalate is $1.67$ at $0$°C to $1.81$ at $95$°C
The pH of $0.05$ M potassium hydrogenophtalate is $4.012$ at $0$°C and $4.23$ at $95$°C.
The pH of pure water is $7.47$ at $0$°C and $6.13$ at $100$°C
The pH of $0.1$ M borax is $9.46$ at $0$°C to $8,83$ at $95$°C.
The pH of $0.02$ M calcium hydroxyde is $13.42$ at $0$°C, and $11.45$ at $75$°C
Ref. : D. R. Lide, Handbook of Chemistry and Physics, CRC Press, 43rd Ed. 1961, and 83rd Ed., 2002.


High Viscosity of Noble Gases Why are the viscosities of noble gases higher than almost all other gases? This seems counter-intuitive since they have much weaker inter-molecular interactions and lower boiling points.

Gas Viscosity Table:
http://www.engineeringtoolbox.com/gases-absolute-dynamic-viscosity-d_1888.html

Viscosity in gasses is generally different than liquids.  Like liquids exhibiting fluidic properties, the viscosity is related to internal friction; however, the friction mechanism is different than inter-molecular interactions experienced by closely packed molecules. The molecules in a gas are far apart resulting in the influence of an entirely different friction mechanism.

The friction mechanism dominating gas viscosity is related to the number of collisions between gas particles. More collisions means more internal friction. In gasses, two dominant factors affect the number of collisions: (1) the mean distance a gas particle has to travel before it collides with another particle and (2) the average velocity of the particle within the gas.

Some noble gasses (He, Ne, Ar) are monatomic particles with a lighter mass than the diatomic elemental gasses. The light mass results in a much higher velocity particles at a given temperature (consider  $E=\frac{3}{2}kT=\frac{1}{2}m \bar{v^2}$) resulting in more collisions. Xenon has a high molecular diameter, so the shear increase in size of the molecule increases the probability of a collision, thus reducing the mean distance it travels before a collision.

More information is available from the following link.


Why should I add lithium bromide to my GPC eluent? I'm running gel permeation chromatography (GPC) of a polar, uncharged polymer (acrylamide-based) in a polar solvent (dimethylformamide). The literature recommends adding a salt to the eluent; for my particular experiment, I'm using 0.05 M LiBr (example). Other salts are reported (Sodium Sulfate, for example) in the literature, both for this eluent/polymer mixture, and for similar polymers and eluents. Why is this done? Most papers simply report the eluent and the additives they used, with no mention of their reasoning. One paper just identifies the eluent, with no mention of any added salts; given the prevalence of the use of salts, I'm not sure if this is an oversight on their part, or if they just didn't use a salt.

If I were working with a polyelectrolyte, I would know that adding a salt would screen the charges along the polymer, reduce the $A_2$ effect, and make my data much easier to interpret. One GPC column manufacturer claims (under "What additives are important...") that dipole-dipole interactions can have a similar effect, but I'm not not sure how much to trust that claim. In addition to confirming or denying that this is the reason, I would also appreciate pointers to any relevant literature.

The answer to my question appears to be in a 1977 paper by Nancy D Hann (DOI: 10.1002/pol.1977.170150604; full citation below). In it, she compares GPC of polyurethane in DMF with and without added LiBr. The chromatograms with no added LiBr show multiple peaks. The chromatagrams with added LiBr show just a single peak. The addition of LiBr also shifts the peak to later elution volume.

Hann explains these results by the polymer having a strong polar interaction with the pure solvent. This causes a higher effective hydrodynamic radius, resulting in the earlier elution time. The polar interactions also case aggregation of the polymer, giving multiple peaks. Adding a salt breaks up the polar interactions, giving just a single, later, peak.

Hann, N. D. (1977). "Effects of lithium bromide on the gel-permeation chromatography of polyester-based polyurethanes in dimethylformamide." Journal of Polymer Science: Polymer Chemistry Edition 15(6): 1331-1339.


The color of resorcinol-formaldehyde resin At school, I did an experiment of synthesizing resorcinol-formaldehyde resin. When I added $\ce{NaOH}$ and gave some heat, the color of resorcinol-formaldehyde resin inside the test tube became reddish-brown. Can anyone tell me why?

The polymerization of resorcinol creates multiple (spaghetti-like) chains containing the aromatic rings of resorcinol.  This is a simultaneous growth in multiple directions. Eventually, the chains trap each other in the solid state. In the analogy used, it is like stirring a pot of spaghetti and then freezing the content.
It may happen that an arene ring is just in proximity of an arene ring of an other chain.  Given a growing polymer chain may be flexible, it may happen that there are intramolecular "close couples", too.  If these arenes are close enough to each other, their $\pi$-electron systems may interact with each other as if they were coupled as one.  This extension of the $\pi$-electron system may move the observable absorption band toward longer wavelengths. Then,

If there are just enough of these "close encounters" per unit of volume with the same arene-arene distance, the perception of color may be intensified (e.g., pale yellow -> bright yellow).

With variation of the arene-arene distance, the intermolecular interaction of these $\pi$-electron systems equally varies. This may yield the absorption of light of different wavelengths / the subtraction of different colors from the white light by these arenes.  Thus it may happen that the color of the polymer is not uniform. If the range of light absorption is wide enough, the polymer may appear just black.


Decrease of volume by half in equilibrium - with specific numbers- how to solve? $$\ce{CO (g) + Cl2(g)<=>COCl2}(g)$$
  We have
  2 mols of $\ce{CO}$, 1 Mol of $\ce{Cl2}$,and 9.765 Mols of $\ce{COCl2}$
  contained in 1 L.
  The system is currently in equilibrium.

  We then decrease the volume by half to 0.5 L

  What will be the new concentration in equilibrium of CO?

I tried calculating at first the constant of concentration;

$Kc=\frac{[\ce{CoCl2}]}{[\ce{CO}]\times[\ce{Cl2}]}$

I found it to be $4.8825$

then I calculated the concentration with the new Volume:

$[\ce{CO}]= \pu{2mol}/\pu{0.5L}= \pu{4 M}$

$[\ce{Cl2}]= \pu{1mol}/\pu{0.5L}= \pu{2 M}$

$[\ce{CoCl2}]= \pu{9.765mol}/\pu{0.5L}= \pu{19.53 M}$

obviously this wasn't in equilibrium so I went to do this:

$\ce{CO + Cl2<=>COCl2}$

$4M-x$, $2M-x$,  $19.53+x$

(as the number of mols in the products is smaller and the system wants to fix the increase in pressure)

When I calculated all of this in the Kc formula it came to me that the Concentration of CO increased, and is now larger than 2 M ,
and in the textbook answers it is stated that it is supposed to be less than 2M, how is it possible?

I would be very glad for help with calculations and technique of solving as well!

I solved the quadratic equation by Wolfram Alpha

The solution indicates that $\ce{[Cl2]}=\pu{1.26926M}<\pu{2M}$,but$[\ce{CO}]=\pu{1.26926M}<\pu{2M}$, if we have
1 mol of $\ce{CO}$, 2 Mols of $\ce{Cl2}$,and 9.765 Mols of $\ce{COCl2}$
contained in 1 L.


Why are elimination reactions in organic chemistry known as beta eliminations? Why are organic elimination reactions known as beta eliminations? Why don't we call them alpha eliminations?

In an elimination reaction we typically have a leaving group (denoted by "X" in the figure below),  The carbon that the leaving group is attached to is referred to as the "alpha" carbon.  If a hydrogen is removed from the "beta" carbon, then we have formed an olefin and the reaction is called a beta elimination.

[reference]

Sometimes an alpha hydrogen can be removed.  In this case both the hydrogen and leaving group were attached to the same carbon and a carbene is formed.


Why 3-chloroprop-1-en-1-ol is unstable? In the treatment of HCl with acrolein 3-chloroprop-1-en-1-ol is the first product which is (my textbook says) unstable and turns into 3-chloropropanal.

Why is this 3-chloroprop-1-en-1-ol unstable? is −I effect of chlorine responsible?

3-Chloroprop-1-en-1-ol is in equilibrium with the aldehyde, 3-chloropropanal (aka $\beta$-chloropropionaldehyde).

According to Organic Syntheses, this compound "is a very unstable substance which polymerizes rapidly, especially in the presence of traces of hydrochloric acid".

But I do not see it being reduced to 3-chloropropanol.

Reactivity/instability

In 3-chloropropanal, or more precisely in its enol form 3-chloropropenol, the chlorine atom is in allylic position (i.e. on a carbon atom adjacent to a $\ce{C=C}$ double bond) which makes it more reactive towards nucleophilic substitution. Another molecule of the enol can then substitute it and the resulting product is

$\ce{Cl-CH2-CH=CH-O-CH2-CH=CH-OH}$

which still has an allylic chlorine atom, so it can react further, hence the polymerization.


Determining weight-percent methanol in water from specific gravity and temperature The company I'm working with uses a fuel mixture of methanol and water. We use a hydrometer to measure specific gravity, in order to test the composition of fuel received from our suppliers. To determine the weight percent of methanol we use a chart provided by The Methanol Institute that correlated specific gravity to weight percent at different temperatures.

Click image for full-size PDF

I'm trying to build a spreadsheet for operators, suppliers, and customers to measure the specific gravity at an arbitrary temperature, enter the temperature and specific gravity measured, and get the weight percent without having to interpret a chart.

In order to do this, I need some sort of formula to calculate the weight percent for any input of specific gravity and temperature  (within a reasonable range anyways).

Here are the approaches I've tried so far:

Fitting a line to the plot above and creating an empirical model. There isn't sufficient resolution in the plot to do this with any useful level of accuracy.
Finding the source information used to create this plot. I called The Methanol Institute and they have no idea where the plot came from, saying it's been used for years and they don't know a source or have any other related data. I tried calling Methanex, a major supplier of methanol which has published similar tables and charts, but I haven't gotten through to anybody so far.
Finding other sources for water/methanol mixes. One of my coworkers found a (printed) table with weight percent methanol in a water/methanol mixture for a wide range of temperatures and covering the weight percent range we're interested in. I scanned and OCR'd the table and created a plot. The table appears to lack sufficient resolution to develop an empirical model as well; with only 3 significant figures the lines appear wavy when plotted:

Calculate using the specific gravity of pure methanol and pure water at a given temperature, then calculate the specific gravity of the mixture using an adjustment for the $\Delta V_{mix}$. This approach seems to work fairly well, but I could only find data for the change in volume due to mixing for 25 °C (page 12 of this PDF from Methanex). This table is "calculated from density and specific volume data" but no additional information is given. A reference table at the end of the document does have the specific volume of methanol at various temperatures, but nothing about change of volume when mixed with water specifically.

I searched at length for methanol/water excess molar volume data. So far I've found data from various sources, but only at 20 and 25 °C, not over a range of temperatures.
Derivation from thermodynamics. I've pulled out my thermo textbook and am reading about equations of state and thermodynamics of mixing. I think given enough work I could probably derive a model from the thermodynamics approach, but I would prefer something based on actual measurements, since theory doesn't always match reality.

The actual question:

Is there an existing resource that would accomplish what I need?
If not, is there a source of good data on water/methanol mixtures at different temperatures that I might be overlooking?
Otherwise, is there a good source of data for excess molar volume over a range of temperatures (at least 18 – 30 °C, preferably wider) that I might be overlooking?
Is there a different approach I should consider?

You are looking for the excess molar volumes of water and methanol.  A quick Google search shows that these quantities have been determined and re-determined dozens of times.  The art is to find which paper has the best values.

The CRC Handbook of Thermophysical and Thermochemical Data and Landolt-Börnstein may have done the sanity check already.


Ring opening of epoxides For the above question, here's what I'd thought:
1.  H+ ion from HCl protonates the oxygen
2. Chloride ion attacks the electrophilic carbon, opening the epoxide ring
3. So the final product formed is a chlorohydrin (which it really isn't)

Also, I didn't know what to do with methanol (guessing it just acts as the solvent, HOWEVER, the product contains a methoxy group)

Could someone please explain the mechanism to get to the actual product, and also highlight the mistake here?

If you think about the quantities involved in this reaction, you'll realize why does the product contain the methoxy group and not the halide. Methanol is present in large excess since it is the solvent of this reaction. Cl- ions will also be present and a very small amount of the chlorohydrin may be formed as well. The reason why methanol is a reagent and not only act as a solvent in this reaction is because its oxygen atom acts as a nucleophile, just like halides or any other molecule containing electronegative atoms (such as O, N, S etc).

I found the following mechanism on Google and it is draw as if the reaction was being performed in water, however you can replace the blue water molecule for methanol or any other alcohol since the mechanism is the same for both.

Whenever you look at a reaction schemes like the one you've posted, you should always address the reactivity of solvents and other reaction components, sometimes they end up reacting with your substrate rather than your desired reagent, this is a very good example of it. So if you intend to synthesize a chlorohydrin, you cannot use a nucleophilic solvent, otherwise, it will react with the epoxide and form the correspondent adduct.

I hope this answers your question.


Is this step justified? This is a question from JEE Advanced  2019 (a highly valued engineering entrance exam in India), but the fact that we are ignoring basic rules of electrophilic substitution is disturbing. Correct me if I am wrong or missing something really important, but according to synergic electrophilic substitution, it should be an ortho attack from methoxy group as we know that an oxygen atom's conjugation would be more contributing than carbon's hyperconjugation and inductive effects combined.
Overall, attack from 2-position (to alkyl group) instead of 3, as done in the question is what I mean to say.

But I also know that it (attack of 3-position) will make the compound to have a bridgehead, will it be stable then? And for overall compound, the stability of the compound, do we have to do a meta attack even in an activated ring?

I'm a little confused what OP's argument is. If you say "it should be a meta attack for methoxide group," then the question is showing an electrophilic substitution at meta-position to methoxy group. But, I understand you are wondering why substitution is at the meta-position to methoxy group rather than on ortho-position to methyl group. If that's what you meant, this is the answer to your curiosity:

The step in question is an example of Friedel-Crafts reaction. More precisely Friedel-Crafts acylation reaction. Here though, the substrate (aromatic nucleus) and the reagent (acyl chloride) are in the same molecule unlike any other typical inter-molecular reactions. In any inter-molecular Friedel-Crafts reaction involving 4-alkylanisole, substitution of acyl group mainly occurs on ortho-position to methoxy group, as you pointed out that "an oxygen atom's conjugation would be more contributing than carbon's hyperconjugation and inductive effects combined." I said here mainly because there would be some minor product from the substitution at ortho-position to alkyl group as well. For example, according to my experience, when Friedel-Crafts acylation reaction is performed in toluene and chlorobenzene, there were mixtures of products including o- and p-chloroacylbenzenes.

Now, we look at the problem in hand. This reaction come down to inter- versus intra-molecular acylation reactions. It's always intra-molecular reaction is much more faster than same type of inter-molecular reactions (here is acylation reaction of both kind) due to proximity. Thus, this phenomenon overcomes the deficiency of activation by methyl group versus methoxy group here, and gives most thermodynamically stable product, 7-methoxytetralone as the major product.

Edit: As OP's request, I include here an example for intra- vs inter-molecular acylation reaction (Ref.1):

Reference:

Yingju Xu, Mark McLaughlin, Cheng-yi Chen, Robert A. Reamer, Peter G. Dormer, Ian W. Davies, "A General Method for the Synthesis of 3,5-Diarylcyclopentenones via Friedel−Crafts Acylation of Vinyl Chlorides," J. Org. Chem. 2009, 74(14), 5100-5103 (https://doi.org/10.1021/jo900696k).


Making ice from carbonated water Sorry if this is not the place for asking this question, but I'm going to do a real life project and don't know where else to ask.

I'm going to prepare and sell snow cones next summer. For making snow cones, I was told by ice cone makers that ice should come from carbonated water.

As in my country is quite difficult to find a proper ice factory that could produce ice with clean water, I'm planning to make my own ice for snow cones.

I was thinking to buy bottles of carbonated water, pour them in many plastic bags and freeze those bags in my refrigirator. In that way I could get carbonated ice.

My question is:

  Do you think that pouring carbonated water in plastic bags for producing ice would be a good idea? Or could I get adverse effects like water loosing sodium ?

There should be no issue with putting carbonated water in plastic bags and freezing them (considering they're clean and sterile, of course). I imagine the benefit of freezing carbonated water instead of regular clean water has to do with the fact that there is much more dissolved gas in carbonated water. As the water freezes, the gas is expelled from the ice (only the water molecules will solidify), creating many pockets inside the solid ice which would weaken its structural integrity. This would likely make it much easier to shave or break into small pieces for a snow cone. I've never frozen carbonated water myself though. You should be a bit careful because as the carbonated water freezes, the carbon dioxide it releases will create pressure in a closed container, so you may end up with torn plastic bags, ruptured bottles, or a damaged freezer in the worst case.


How to interpret m quantum number values? If one electron has the quantum numbers $n = 4,\ l = 3,\ m_l = +2,\ m_s =-1/2$ and another electron has the quantum numbers $n=4,\ l= 3,\ m_l = -2,\ m_s=+1/2$, what are the similarities between the orbitals in which the electron's are found?

My answer is : They have the same number of energy shells. They have the same shape. Their orientations are different and they have different directions of spin.
I checked the answer and it said that the first one lies on the $x$ and $y$ axis and the other one will lie between the $z$ and $y$ axis. But how do they know that? If $m_l= +2$, does that necessarily mean it lies on the $x$ and $y$ axis. And if $m_l = -2$, does that mean it lies between the $x$ and $y$ axis?

OP's answer is technically right while additional details about specific orientations of the real $f$-orbitals mentioned somewhere else are basically wrong. At best such details can be thought of the matter of some convention, but it is better to avoid such conventions.

The thing is that for any value of the azimuthal quantum number $l$ an orbital with a non-zero value of the magnetic quantum number $m$ is complex thanks to $\mathrm{e}^{imf}$ factor. Chemists prefer to work with real orbitals formed by taking linear combinations of complex orbitals. However, such linear combinations are not eigenfunctions of $L_z$, or, in other words, $m$ is no longer a good quantum number (but its absolute value is).

In particular, for both $f_{xyz}$ and $f_{z(x^2−y^2)}$ orbitals $|m|=2$, or $m=\pm2$, but the statement that the former corresponds to $m=-2$ and the later to $m=2$ is technically wrong.


Confusion between moles and molarity units I am a total newbie in the field and i would like some kind of guidance.

So, I have been recently given 2 datasets with measurements regarding several phospholipid fatty acids of plasma HDL. Both datasets consist of the same classes of FAs but with different units of measurement (μΜ and mol%).

Since I have no previous experience with that kind of data, can someone give me a hint about how (and if) it is possible to convert from one unit to the other?

Thanks in advance.

$\ce{M}$ is a shortcut for $\frac{\ce{mol}}{\ce{l}}$. $\ce{mol}\%$ on the other side is defined as $\frac{n}{n_{\ce{tot}}}$. In order to convert between the two you need to know the density $\rho$ and the molar mass $M$ (don't confuse this with the unit) of your solution/mixture. You can then do
$$
c_{\ce{mol}\%} = \frac{c_\ce{M} \cdot V_{\ce{mix}}}{V_{\ce{mix}} \cdot \rho \cdot M^{-1}} = \frac{c_\ce{M}}{\rho \cdot M^{-1}}
$$
$c_{\ce{mol}\%}$ is the concentration in $\ce{mol}\%$ and $c_M$ the one in $\ce{M}$.


Why do chemistry and physics have different sign convention in thermodynamics? In physics class, we write the first law of thermodynamics as $\mathrm dU =\mathrm dQ - \mathrm dW$ and in the physical chemistry class, we write the same law as $\mathrm dU =\mathrm dQ + \mathrm dW$.

The reason being the sign convention is different in both the cases.

In physics we take work done by the system as positive and in chemistry work done on the system.

I realize that this does not cause any change in the actual law of nature but I just want to know why we have different sign conventions. Wouldn't just one convention make life easier?

Is there a historical reason? Or is this is to differentiate between subjects?

This is not a simple physics versus chemistry distinction. I taught Physics for 25 years and saw many examples of either usage in multiple textbooks. In fact, at some point in my tenure, the AP Physics committee swapped conventions on the equation sheet for the AP Exam.

Just my take here: I've always attributed the work-done-by-the-system camp as being more prone to be used by engineering types who want to know "what the system can do for us" in practical applications. On the other hand, work-done-on-the-system seems to foster the view of an experimenter or theoretician operating on a system from without.


what voltage is needed to reduce Silver (I) Oxide in alkaline medium against an inert electrode consider the following electrochemical cell:

silver (I) oxide is anode, platinum electrode is cathode, 6M KOH is
electrolyte

what voltage is needed to be supplied to the cell OR what voltage is generated by this electrolytic cell, with the reaction being:

$\ce{2 Ag2O -> 4 Ag + O2}$

where half reactions are:

$\ce{2 Ag2O + 2H2O + 4e- -> 4Ag + 4OH-}$

$\ce{4OH- -> 2 H2O + 4e- + O2}$

Is the reverse reaction spontaneous in alkaline electrolyte with inert electrode like platinum:

$\ce{4 Ag + 4OH- -> 2 Ag2O + 2 H2O + 4e-}$

$\ce{O2 + 2H2O + 4e- -> 4OH-}$

(the platinum electrode is for oxygen reduction reaction and oxygen evolution reaction)

I am a computer science engineer trying to learn chemistry, excuse me if my equations are not precise

You need to look up the standard electrode potentials of the materials involved.  While wikipedia provides an answer for $\ce{AgO}$ in acidic media, the abstract of this primary publication mentions a value of $(0.599 \pm 0.001)$ V of the $\ce{AgO / Ag2O / OH-}$ electrode, as determined against the H electrode at 25 C in KOH solution.  Note the authors find the potential dependent on the grid material their AgO-electrode was made; beside the expected-to-happen changes in function of temperature of the solution and concentration of $\ce{KOH}$ present.

A generally useful secondary literature reference for physico-chemical data you (still) may carry with you is the CRC Handbook of Chemistry and Physics by CRC Press, including tables like this, too.


Ratio of concentrations I'm studying for my chemistry final and having a hard time with this question on one of the practice exams. Here's the question:

At $20$ degreees Celsius, Henry's Law constant for Ar is half of that for Kr. If a gaseous Ar/Kr mixture containing 20% molar fraction of Kr is placed over water at $20$ degrees Celcius, what is the ratio of concentrations of Ar to Kr in water?

Henry's Law is mathematically expressed by $$p=k_Hc$$ where $p$ is the partial pressure above the liquid, $k_H$ is the Henry constant for the specific gas at the given temperature, and $c$ is the concentration of gas in liquid. If you set up one equation for each gas and divide the two, you get $$\frac{p_{Ar}}{p_{Kr}}=\frac{k_{H, Ar}}{k_{H, Kr}}\frac{c_{Ar}}{c_{Kr}}.$$ You are given both the ratio of partial pressures and the ratio of the Henry's Law constants; the ration of concentrations should be easily calculated from that point.


Rearrangements; Carbocations; E1 Mechanism If the leaving of the leaving group creates a secondary carbocation, and a tertiary carbocation has the possibility of being formed through a 1,2 methyl shift:

1) Will the methyl shift? I'm guessing the answer is an emphatic yes. Tertiary carbocations are more stable than secondary carbocations.

2) Will there still be some secondary carbocation left over? I.e. will there be both major and minor products? I'm guessing yes.

I ask because I was watching a video on E1 elimination and the narrator ignored the possibility of a shift, and I was wondering if this was because he forgot or because it was irrelevant.

1) Will the methyl shift? I'm guessing the answer is an emphatic yes.
  Tertiary carbocations are more stable than secondary carbocations.

Yes, if that is the only way to generate a tertiary carbocation and the C-Methyl bond can more or less line up with the carbocation p orbital, then the methyl would likely shift.  If there are other alkyl groups attached to the carbon initially bearing the methyl group, they could shift too.

  2) Will there still be some secondary carbocation left over? I.e. will
  there be both major and minor products? I'm guessing yes

An equilibrium exists between the various carbocations.  At 300 K, an energy difference of 1.4 kcal/mol results in a power of ten difference in the equilibrium (e.g. from 50:50 to 90:10).  So it really depends on the energy difference between the various carbocations involved, as well as the rate they are captured or proceed to the next step.

Finally, if there is a hydrogen on the carbon adjacent to the carbocation center, then elimination to create a double bond becomes a competing pathway,


Travelling wave ion mobility spectrometry: determination of size? I performed an ESI / travelling wave ion mobility spectrometry / MS experiment and observe $\ce{M + H+}$ at $m/z ~701$ and $\ce{M + K+}$ ions with $m/z ~740$. The drift times are 6.75 ms for $\ce{M + H+}$ and 7.0 ms for $\ce{M + K+}$. My first instinct was that this means that the proton adduct is smaller than the potassium adduct. However, if the mass is taken into account as well, i.e.
$$
   7.0 ~\mathrm{ms} ~/ ~740 = 0.0095 ~\mathrm{ms}
$$
for the potassium adduct, and for the proton adduct
$$
   6.75 ~ \mathrm{ms} ~/~ 701 = 0.0096 ~\mathrm{ms}
$$
which would indicate that the two are very similar in size, the proton adduct being slightly larger. For a linear drift tube it would be very easy for me to calculate the relationship (i.e. Mason-Schamp equation), but for travelling wave IMS I am pretty unsure. Which of the two reasonings is correct?

In terms of practical data analysis standpoint, travelling wave-IMS data can generally be treated similarly as drift tube-IMS for which the technical differences is in resolution (better or worse). However, one point of consideration is that

Low gas pressure in TW IMS means strong fields in E/N terms
and thus substantial heating of ions. This may cause fragmentation
or distortion of macromolecular structures, complicating their
detection and characterization. [1]

Your first instinct to assume that the $\ce{[M + H]+}$ adduct is quite rational, however remember that ion conformation is always in flux and can be influenced by collisions with buffer gas. What you may be experiencing is the proton adduct producing a "looser" structure, resulting in a larger collisional cross section. The $\ce{[M + K]+}$ adduct is heavier in mass but may produce a tighter conformation.
A good example of this is in the fragmentation and mobility of polyethylene glycol (PEG) ions $\ce{[PEG + H]+}$ and $\ce{[PEG + Na]+}$. When carrying out fragmentation studies of PEG, it was found that the sodiated ion produced less extensive fragmentation and required higher collision energy.[2] Ion mobility studies have also found that PEG chelates the $\ce{Na+}$ ion similar to that of crown-ether, producing tightly wound "rigid" structures.[3]
These conditions are highly compound/structure dependent however, it should be something to keep in mind especially in the case of analyzing other molecules such as peptides which respond to adduction in "weird and mysterious" ways.
References

Alexandre A. Shvartsburg and Richard D. Smith, Anal. Chem. 2008, 80 (24), 9689–9699.
Jennifer Gidden, Thomas Wyttenbach, Anthony T. Jackson, James H. Scrivens, and Michael T. Bowers, J. Am. Chem. Soc. 2000, 122 (19), 4692–4699.
Rui Chen, Xinlei Yu, and Liang Li, J. Am. Soc. Mass Spectrom. 2002, 13 (7), 888-897.


Can Aluminum Hydroxide (Al(OH)3) be heated to it's melting temperature before decomposing? On the wikipedia page for aluminum hydroxide the listed melting point is 300 °C (572 °F), but in the same page it states that aluminum hydroxide decomposes at only 180 °C (356 °F). Can aluminum hydroxide be melted without decomposing?

TL.DR: Aluminium oxide (sometimes known as alumina) is made by heating the aluminium hydroxide to a temperature of about 1100 - 1200°C. (Chemguide)

$$\ce{Al(OH)3 ->[\Delta] Al2O3 + 3H2O}$$

As @andselisk said, the temperature Wikipedia mentions is aluminium hydroxide mixture in a fire retardant which might have other compounds. Decomposition of pure aluminium hydroxide has been studied in various papers and its decomposition reaction mechanism, temperature, reaction kinetics has been exhaustively studied.

A paper examines$\ce{^{[1]}}$ the decomposition of aluminium hydroxide at a temperature of 973-1123 K(700 °C-850 °C).

Another paper$\ce{^{[2]}}$ shows aluminium hydroxide being heated from room temperature to 1200 K(926.85 °C) where the exact temperature of decomposition products i.e $\ce{Al2O3}$ is noted:

  [...] we find the first and the smallest endothermic peak at 519
  K, which is due to the partial dehydroxylation of gibbsite
  ($\ce{Al(OH)3}$) and formation of boehmite ($\ce{AlOOH}$). Endothermic
  peak at 585 K corresponds to two processes: (i) transformation of
  gibbsite to phase $\ce{χ-Al2O3}$ and (ii) additional conversion of
  gibbsite to boehmite. The first process is in accordance with the
  results obtained. Another endothermic peak at 815 K is due to
  decomposition of boehmite and formation of alumina $\ce{γ-Al2O3}$. The
  obtained TG curves show clearly three steps of weight loss. The weight
  loss in the first step (about 5 wt.%) is due to the partial
  transformation of gibbsite to boehmite; the second step (about 25
  wt.%) corresponds to decomposition of gibbsite to boehmite, otherwise
  to $\ce{χ-Al2O3}$. The last step of around 3 wt.% relates to the
  formation of $\ce{γ-Al2O3}$. Further observations show that the total weight
  loss is equal to 33 wt.%. All these transformations of gibbsite and
  the chemical composition are confirmed by the XRD analysis.

References

Chemical kinetics and reaction mechanism of thermal decomposition of aluminum hydroxide and magnesium hydroxide at high temperature (973-1123 K) by Ienwhei Chen, Shuh Kwei Hwang, and Shyan Chen, Ind. Eng. Chem. Res. 1989, 28, 6, 738–742 DOI: https://doi.org/10.1021/ie00090a015
http://przyrbwn.icm.edu.pl/APP/PDF/131/a131z3p62.pdf
Bhattacharya, Indra & Das, S. & Mukherjee, P. & Paul, Subir & Mitra, P.. (2004). Thermal Decomposition of Precipitated Fine Aluminium Trihydroxide. Scandinavian Journal of Metallurgy. 33. 211 - 219. 10.1111/j.1600-0692.2004.00686.x.(DOI) (Here aluminium hydroxide is subjected to dehydration upto a temperature of 1440 °C)


Find the maximum number of tripeptides can be synthesized by using three different amino acids I was asked the following question in a myPAT take-home test,

  Find the maximum number of tripeptides can be synthesized by using three different amino acids.

Let those amino-acids (Z) be A, B, C.

So, taking three Z's at a time, we've only one set, ZZZ. So, total 3 tripeptides.

Next, we can take two same and one distinct (eg. AAZ),

For AAZ type, we've four sets, AAB, ABA, AAC, ACA. So, total 12 tripeptides.

Finally, take 3 distinct Z, for that we get total of 6 (i.e., 3!) tripeptides.

I am unsure if I've done it in a valid way, as answer given is 18. Where am I wrong?

There is a misconception that runs through the questions, the answer key and the comments: For tripeptides, the peptide linkage is directional (i.e. the tripeptide has an amino terminal and a carboxy terminal), so ABC is different from CBA.

To get the number of tripeptides combined from three types of amino acids, you have three choices for the first, three independent choices for the second and another three independent choices for the third position, for a total of 3 x 3 x 3 = 27 possibilities.

  [OP] I am unsure if I've done it in a valid way, as answer given is 18.

The answer key is incorrect. I have no way of reading the question that 18 would be the answer, even considering the poor wording of the question.

  [OP] For AAZ type, we've three set, AAB, ABA, AAC, ACA. So, total 12 tripeptides.

There are three choices for Z, two remaining choices for A, and three positions for Z (AAZ, AZA, ZAA). So, total of 3 x 2 x 3 = 18 tripeptides containing two different amino acid types of the initial set of three types.

  [Orthocresol] This question is phrased very poorly. Without reading your working, I would have answered 20×19×18=6840.

This answers how many different tripeptides made from (20) standard amino acids contain no amino acid twice. The question, certainly poorly stated, says "synthesized by using three different amino acids". So your reagents are a set of three amino acids types, and you are making tripeptides from them.

  [Maurice] With two amino-acids, A and B, there is already 8 tripeptides : AAA, AAB, ABA, ABB, BAA, BAB, BBA, BBB. It makes 8 tripeptides with A and C, and also 8 tripeptides with B and C. Plus the tripeptides with ABC, which are : ABC, ACB, BAC, BCA, CAB, CBA. In total : 8 + 8 + 8 + 6 = 30 tripeptides.

That is indeed a math problem, not a biochemistry problem. You can make AAA from A and B, but also from A and C, so you counted AAA twice (and BBB and CCC). So your answer overshot by 3.

  [Rahul Verma] Are the synthesized amino acids valid? If one of the amino acid (say A) is optically active, then can BAB be two tripeptides?

The linkage does not introduce new chiral centers (different from e.g. combining sugars), so each sequence XYZ designates a single molecule (given that you start with enatiomerically pure amino acids).

  [porphyrin] The list is aaa , aab , aac , aba , abb , abc , aca , acb , acc , baa , bab , bac , bba , bbb , bbc , bca , bcb , bcc , caa , cab , cac , cba , cbb , cbc , cca , ccb , ccc which has 27 entries and removing 9 palindromic arrangements (except aaa, bbb, ccc) this leaves 18.

The list is correct, but there is no reason to remove palindromes (9 in the list) or divide non-palindromes by two (18 in the list).

Update

I missed all the comments that came in while I was writing this. They now reflect the misconceptions and how these can be resolved; what is not resolved is the poorly worded question and its incorrect answer key - maybe someone who has the original question can take on the task of sending an erratum.


How can subtilisin still function without its catalytic triad? I read chapter 9 in the book Biochemistry (5th edition), by Berg, Tymoczko, and Stryer (provided in the NCBI site here). It describes the mechanism of action of the chymotrypsin enzyme. The catalysis is performed through the catalytic triad consisting of serine-195, histidine-57, and aspartate-102.
What intrigued me is the following section on the related enzyme subtilisin, which also has a catalytic triad but with different residue numbers:

9.1.5. The Catalytic Triad Has Been Dissected by Site-Directed Mutagenesis
[...] As expected, the conversion of active-site serine 221 into alanine dramatically reduced catalytic power. [...] The mutation of histidine 64 to alanine had very similar effects. [...]
These observations support the notion that the serine-histidine pair act together to generate a nucleophile of sufficient power to attack the carbonyl group of a peptide bond. The conversion of aspartate 32 into alanine had a smaller effect, although the value of $k_\mathrm{cat}$ still fell to less than 0.005% of its wild-type value.

My problem is in what it says next:

The simultaneously conversion of all three catalytic triad residues into alanine was no more deleterious than the conversion of serine or histidine alone. Despite the reduction in their catalytic power, the mutated enzymes still hydrolyze peptides a thousand times as rapidly as does buffer at pH 8.6.

A graph of the relative rates can be taken from this link:

My question is: how can it be that after replacing all the triad residues, the enzyme still possess such a great catalytic activity? What is the mechanism behind that phenomenon?

If the substrate still binds to the protein, other parts of the mechanism are still in place. Specifically, the main chain amide groups that help to stabilize the tetrahedral intermediate are still present. In the absence of serine, water will attack the carbonyl. In the presence of serine and the absence of the histidine, serine will still be the nucleophile, but a less efficient one.

Look up oxyanion hole if this part of the mechanism is unfamiliar to you.


What is the difference between a mixture and a solution? Is there an actual difference, or are they synonymous?

Let's try some examples for mixtures and solutions. Think about the answers yourself before you uncover them ;-)

We cut carrots, celery, and green beans into pieces and throw them into boiling water to prepare a vegetable soup. Mixture or solution?

  It's a mixture, because the pieces of the different vegetables can be retrieved by a physical method, such as picking them with a fork.

Throw a spoonful of sugar into a cup of water and stir until there's no sugar left on the ground of the cup. Mixture or solution?

   This is a solution. It it homogenous; even the tiniest part of the water contains the same amount of sugar and there is no way to get back the sugar (unless you boil the water off).

Take a handful of razorblades and a handful of ants. Throw everything in a bowl. Wear gloves! Shake. Mixture or solution?

   This is a mixture. You can separate the components with a magnet - or by luring the ants out of the bowl with a trace of sugar water.

Melt iron. Add cobalt. Don't mix with bare hands ;-) Let it cool down until it is a solid again. Mixture or solution?

   This is a solution. Solutions do not need to be liquid. They can be solids as well. You can cut the metal block into pieces and mill every piece down to a fine powder and the amount of iron and cobalt in every grain of powder will be the same. Again, you cannot fiddle the components apart by any physical method.


Why is the melting point of lactic acid higher than pyruvic acid I have a quick question about pyruvic acid and lactic acid, which I think I know the answer too but I'm unsure.

  The melting point of lactic acid is said to be higher than that of pyruvic acid despite very similar molecular masses.

Is this because lactic acid has 2 $\ce{OH}$ groups, and pyruvate only has 1, thus there are twice as many hydrogen bonds between lactic acid molcules that must be broken for the substance to melt?

Your analysis is practically correct but not quite — more due to technical reasons. You said:

  Is this because lactic acid has 2 $\ce{OH}$ groups, and pyruvate only has 1 […]?

You probably know the structures of both; they both include a $\ce{COOH}$ group which you apparantly counted as a hydroxy group. You should not do that; the carboxy group ($\ce{COOH}$) has different properties than a lone hydroxy group ($\ce{CR2OH}; \ce{R} \ne \ce{OX}$). The most obvious is the $\mathrm{p}K_\mathrm{a}$ value which differs by about $10$ logarithmic units.

So you should say that lactic acid has higher melting and boiling points becuase it contains a (or: one) hydroxy group while pyruvic acid has a carbonyl group in the same position.


What makes an isotope stable? I have heard many times that when there is the “right balance” of protons and neutrons the atom will be stable, and therefore it will decay to achieve this stability. But what about this supposed balance of protons and neutrons makes it stable? Does it have to do with quarks? Density? I am quite curious as to what causes this.

There is a shell structure within the nucleus also. So the neutron/proton ratio varies from 1:1 for the light elements to 1.5 for the heavy elements.

  Note: I ripped off the image from another website. Not sure where they got it from...


Addition of NaCl and CaCl2 to snow In cold countries, it is a common phenomena to add salt to melt or clear snow. However, only specific salts are being commercially. So my question is that why only sodium chloride and $\ce{CaCl2}$ are used for addition to snow, to melt/clear it?

Salts will lower the melting point of snow or ice.
$\ce{NaCl}$ is chosen because it is cheap.

Edited: Salt with more total ions is more effective. $\ce{CaCl_{2}}$ is the one of the most available salt with 3 total ions. $\ce{MgCl_{2}}$ is another option.
Some crude $\ce{NaCl}$ produced from sea water directly contain $\ce{Ca}$ and $\ce{Mg}$ salt and can be used to melt ice.


How to draw protonated benzene (horseshoe) carbocation structure with ChemDraw? Is there a way to draw the aromatic "horseshoe" (forgot the official name) structure of a protonated benzene cation (arenium) with ChemDraw, and not the cation with double bonds?

Draw cyclohexane. Open Drawing Elements. Select 270o circle. Orient with lasso and adjust size. Insert in cyclohexane. Add H's and +.
ADDENDUM: For a centered structure. Draw cyclohexane, arc and + inside each other. Select them. Go to Object>align. Select L/R centered then T/B centered. Add hydrogens. See also Comments. PS: Clearly, ChemDraw does not recognize the structure as protonated benzene. Can't have everything ;).


What are the g-block's predicted properties? I have been researching for a Physics/Chemistry exam and thought; what will the future periods in the periodic table (periods 8 and above) would entail? Each block contains its own properties that allow for accurate predictions of elements in these blocks. Each block has very unique properties and benefits society in major ways. The g-block is the theoretical block thought to exist in period-8 though no elements have been produced in this period for any sustained time. My question is:

What are the predicted properties of the G-block if any?  Who and what methods have they been predicted by?

I'd also like to know if there are any ways of "stabilizing" these elements so that they take more time to decay.

First of all, from the perspective of chemistry, it is the group (column) that matters more than the period (row), and subshells act at the period level.  Second, the elements beyond the actinides have virtually no significant chemical research owing to their extreme radioactivity.  So there may not be many predicted properties for the g-block beyond the simplistic theoretical assumptions about groups, which are much less relevant for a hyper transitional period like the g-block.

As for you question about potential stability, Glenn Seaborg made some predictions here, which you can read that this Wikipedia page: Island of Stability.  In general, though, these elements are not predicted to be stable.


Why does FeO(OH) have two CAS numbers? Iron(III) oxide-hydroxide, $\ce{FeO(OH)}$ has two CAS numbers, 1310-14-1 and  20344-49-4. What's the difference between them?

Why does it have two CAS numbers? As I know, the CAS number should be unique to each substance.

Wikipedia reference

Why does it have two CAS numbers? As I know, the CAS number should be unique to each substance.

Yes, that’s exactly the principle and also why two CAS reference numbers are associated with $\ce{FeO(OH)}$. SciFinder gives 20344-49-4 to be the ‘boring’ laboratory iron oxyhydroxide, while 1310-14-1 is mentioned as the mineral goethite. Goethite is the α form of $\ce{FeO(OH)}$; β (akaganéite), γ (lepidocrocite) and δ (feroxyhyte) forms exist, too.

Since each of these four minerals has a well-defined structure, much better defined than simple $\ce{FeO(OH)}$, they should (and will) each get a single CAS number for unambiguous referencing. At the same time, the generic CAS number 20344-49-4 also exists for lab-grade non-mineral $\ce{FeO(OH)}$.

So what you are experiencing is not a violation of the one-substance-one-CAS-number rule but rather a confirmation of its strictness.


Are these statements about change in enthalpy contradictory? Just going through my thermodynamics chapter I came across 2 contrasting statements:

Change in enthalpy=enthalpy of (products-reactants) at constant pressure and temperature.

Change in enthalpy=Energy required to break bonds in molecules of reactants -energy
released in the formation of bonds in molecules of products.

So, if I consider these statements individually then these sound correct to me but they are different in the sense that in the first one it is (products-reactants) and in the second one (reactant -products)?
Is the book wrong or am I mistaken in my fundamentals  ?

They are basically same statements. We can use Hess law to elaborate these two statements. For example, we can use the following reaction,
                                 CH4=C+2H2
Here, Delta reaction = Enthalpy of CH4 - Enthalpy of C + 2(Enthalpy of H2).......(1)
We can break the reaction events as,
1) Breaking of 4 C-H bonds
2) Formation of 2 H-H bonds
So, delta H = Energy required to break 4 C-H bonds + Energy required to forming two H-H bonds.......(2)
Now we always take the reactor to be the system and that's why heat input to the reactor is positive and heat going out of the reactor is negative. In the formation of bonds, the energy are released, so the formation energy of H-H is negative with respect to our thermodynamic system.From 2
So,  delta H = Energy required to break 4 C-H bonds - Energy released in the formation of two H-H bonds.......(3)
Now, energy required to break 4 C-H bond= Energy of C + Enrgy of 4 H - energy of CH4
                                        =-Energy of reactants...(4)
Similarly,  energy required to form 2 H-H bond= Energy of 2 H2 - energy of 4 H
                                              =Energy of products (As C is in reference state).........(5)
Adding 4 and 5 yields, equation (1), which proves that both statements are same.


Why 'Enthalpy change' (Delta H) is equal to 'Heat transfer at constant pressure' (Qp)? Why 'Enthalpy change' (∆H) is equal to 'Heat transfer at constant pressure' (Qp)?

∆H = ∆U + ∆pV, here only expansion work done by the sustem is added. If non expansion work is done on the system then it should be,
∆H ≠ Qp.
What's wrong? Explain.

In,Thermodynamics,Work is defined as
$W$= $-\int_{V_1}^{V_2}pdV$. So, if $dV$=0, there is no work done on the system,as $V_1=V_2$.

Coming to the enthalpy,you are right.
$\Delta$H=$\Delta$U + $\Delta$($PV$).But if you apply the product rule then,
$\Delta$H=$\Delta$U + $P$$\Delta$V + $V$$\Delta$P.
At, constanst presurre,$\Delta$P=$0$. so at constant pressure,$\Delta$H=$\Delta$U + $P$$\Delta$V.But,according,to first law of Thermodynamics,$Q$= $\Delta$U - $W$.At,constant pressure, the $p$ in the integral comes out.So,$W$= $-p\int_{V_1}^{V_2}dV$= -$p$($V_2-V_1$) = -$p$$\Delta$V.So, first law in constant presuure becomes,$Q_p$= $\Delta$U - (-$p$$\Delta$V)= $\Delta$U+$p$$\Delta$V= $\Delta$H. Thus, it is justified why Enthalpy change equals Heat change at constant pressure.


How to convert benzene to 1-bromo-3-iodobenzene? Today, my chemistry teacher gave the following organic conversion as homework: benzene to 1-bromo-3-iodobenzene.

I tried something like:

I am not sure about it, can someone review my conversion (as I have got an exam tomorrow)?

It looks pretty good overall. The order of substitution is correct to achieve the required pattern but there are two points I would pick up on.

The final substitution of the diazonium salt is best done with $\ce{CuI}$ rather than $\ce{KI}$ as the copper(I) ions catalyse the reaction. Also heating is not required; the reaction can be run at room temperature (or possible slightly above but not much).
The diazonium salt is drawn incorrectly since it is ionic in nature. There is a triple bond between the nitrogens and a formal positive charge on the middle nitrogen, although in reality the charge is delocalised over both atoms:


Why is p-methoxyphenol more acidic than p-methylphenol? Why is p-methoxyphenol more acidic than p-methylphenol?

According to me, inductive effect really doesn’t matter at para positions and p-methoxyphenol should be less acidic due to the high mesomeric effect of the $\ce{-OCH3}$ group in comparison to the weak hyperconjugative effect of the $\ce{-CH3}$ group.

p-Methoxyphenol – $\mathrm{p}K_\mathrm{a} = 10.1$

p-Methylphenol – $\mathrm{p}K_\mathrm{a} = 10.3$

First off, it is a very small difference in $\mathrm{p}K_\mathrm{a}$ that you are asking about.  In fact, I would say that the two compounds have similar acidity.  It is often difficult to explain such small differences, but what follows is a general approach that can be used to probe the question and to try and answer questions related to equilibria.

In this question we are comparing the effect of two different substituents on an equilibrium.  To perform this comparison we must look at the effect of the substituent on both the starting material (the substituted phenol) and the product (the substituted phenoxide anion).  If the substituent stabilizes the phenol it will tend to shift the equilibrium towards the phenol side making the phenol less acidic; if the substituent stabilizes the phenoxide it will tend to shift the equilibrium to the phenoxide side making the phenol more acidic.

From your other question we see that the p-methoxy group destabilizes the phenoxide anion to a greater extent than the p-methyl group.  Based on this alone, we would say the substituent effect on the phenoxide side of the equilibrium should be such that the p-methoxy substituent pushes the equilibrium more strongly to the phenol side than the p-methyl group will.  This would tend to make p-methoxyphenol less acidic than p-methylphenol.

Looking at the phenol side  of the equilibrium, both the methoxy and the methyl groups are electron releasing through resonance, with the methoxy having a much stronger effect.  The phenolic $\ce{OH}$ group is also strongly electron releasing through resonance. Perhaps having two strongly electron releasing groups (hydroxyl and methoxy) attached to the aromatic ring causes some destabilization due to electron repulsion associated with the high electron density in the ring.  Certainly such an effect would be greater with methoxy than methyl.  If true, then we could argue that the p-methoxy substituent also destabilizes the starting phenol (more then the methyl group) and this would tend to push the equilibrium to the product side making p-methoxyphenol more acidic than p-methylphenol.

If the substituent effects on the phenol and the phenoxide are comparable, then they would cancel and we might expect p-methoxyphenol and p-methylphenol to have comparable acidities and if the substituent effects cancel we would also expect both phenols to have $\mathrm{p}K_\mathrm{a}$'s similar to phenol ($\mathrm{p}K_\mathrm{a}=10$), which they do.

The approach used here is sound.  I think the explanation of the substituent effects on the phenoxide are also sound.  The explanation of the substituent effects on the phenol side is (admittedly) weak.  Nonetheless, this explanation does correctly explain why p-methoxyphenol and p-methylphenol have comparable acidities and why these acidities are close to the acidity of phenol.


What is the appropriate sizing for fiberglass compatible with polyamide? I am looking for a chemical coating for glass A fibers, compatible with polyamide. The fibers are uncoated, 100-filament strands of 22-micrometer "A" glass fibers.

We are producing fiberglass in our lab, and we want to size (coat) them with something which can keep the strands together and at the same time enables the fibers to bond with polyamide. We can use materials like PVA to keep the strands together, but then the fibers do not bond with the polyamide (mixing with polyamide happens after we produce the strands and chop them, with 4mm in length).

I am a beginner in composite world so I would appreciate if you can give me details on how and where to buy such chemical, or how to synthesis it myself if possible?

I think epoxy resin is compatible with polyamide because its chemical structure. Epoxy is compatible with silica (glass) and poyamide functional groups then it could be a good sizing .


In a simple chemical cell Why do the electrons, arising from the oxidation of the anode, flow through an external circuit? Why does it not just flow through the electrolyte and reduce the cathode? Why do the electrons give the wire preference?

The crucial point which explains this is that there are two different forms of conduction happening here.

The first kind of conduction, and the one we are most familiar with is electronic. This is a flow of electrons, and is the conduction which flows in the wire and through the load (a motor, LED etc).

The second kind is the flow which occurs in the electrolyte, and this conduction is ionic. It is a flow of charged particles (ions) within the electrolyte. Even though we often speak of the conductivity of an electrolyte, what is really meant is the electrolyte's ionic conductivity.
This is governed by the mass of the ions, their concentration, temperature and the viscosity of the solution.

Electrolytes do not really conduct electrons at all, which is why there is no 'short circuit' through them. The conductivity arises due to several mechanisms - either the ions themselves move through the solution, or electrons (or protons) hop from one ion to the next, conveying the charge.

This will only happen if there is a means of causing the flow of charge through the electrolyte. When it is open circuit, the electrolyte is overall neutral in charge and no current will flow in it.
If an external circuit is connected, it allows electrons to flow from one electrode to the other, and then the corresponding amount of charge will flow through the electrolyte to balance it out, producing current.