Bond length comparison in substituted phosphorus pentahalide What would be the comparison between $\ce{P-F}$ and $\ce{P-Cl}$ bond length in  phosphorus tetrafluoride chloride $\ce{PF4Cl}$ and phosphorus trifluoride dichloride $\ce{PF3Cl2}$ in the equitorial hybrid orbital arrangements?

While determining the bond length between the two compounds I am using Bent's rule, since in the case of $\ce{PF4Cl}$ there are two $\ce{F}$ atoms in equatorial region and thus they decrease their %s character in turn resulting in a larger bond length and since there are two $\ce{F}$ atoms here the %s would be relatively less as compared to the $\ce{P-F}$ bond in $\ce{PF3Cl2}$, so this decrease in %s character in flourine atom should result in increase of %s character in $\ce{P-Cl}$ bond and reducing the bond length and $\ce{P-F}$ bond length would be greater in $\ce{PF4Cl}$.

  8. Bond lengths (pm) in substituted phosphorous pentahalides are given below:

  $$
\begin{array}{lcc}
  & \ce{PF4Cl} & \ce{PF3Cl2} \\
\ce{P-F_{eq}} & x & y \\
\ce{P-Cl_{eq}} & a & b
\end{array}
$$

  $$\text{(I)}~x > y \qquad \text{(II)}~y > x \qquad \text{(III)}~a > b \qquad \text{(IV)}~b > a$$

  Choose correct code:

  (a) Only I
  (b) II and IV
  (c) I and III
  (d) Only IV

So if I'm assuming things right  my answer about the $\ce{P-F}$ bond length is right, but here in the option for $\ce{P-Cl}$ bond length is that it's greater in $\ce{PF4Cl}$. Why? Can someone please explain?

According to (1) and (2), the values in your table would be as follows (all values in Å):
$$\begin{array}{c c c}
& \ce{PF4Cl} & \ce{PF3Cl2} \\
\ce{P-F_{eq}} & 1.541 & 1.546 \\
\ce{P-Cl_{eq}} & 1.999 & 2.004 \\
\end{array}$$
we have $x < y$ and $a < b$, which means that (b) is the correct answer

According to Bent's rule:

atomic s character concentrates in orbitals directed towards electropositive substituents

Bonds with larger s-character are shorter in nature. Keeping these two points in mind and analyzing the structures, in $\ce{PF4Cl}$, almost all the s-character is directed towards $\ce{Cl}$, which means that $\ce{P-Cl}$ bond length in $\ce{PF4Cl}$ is less than that in $\ce{PF3Cl2}$. Similarily, considering that the $\mathrm{sp^3d}$ hybridization can be considered as $\mathrm{sp^2+pd}$, in the equatorial plane, the s-character shared between the two fluorine atoms in $\ce{PF4Cl}$ would be more than the s-character given to the single fluorine atom in $\ce{PF3Cl2}$, hence the shorter bond length.
Another perspective is that the above set of arguments are largely hand-waving, as Bent's rule is at best empirical and the bond length differences are $\pu{\approx 5 pm}$, which is very small. An easier question would be to compare bond lengths between $\ce{PCl5}$, $\ce{PF3Cl2}$ and $\ce{PF5}$, but for this question, I don't have a 'solid' explanation per se, other than the lengths from the references and the elaboration of Bent's rule above.
References:

Leiding, Jeff, et al. “Bonding in PF2Cl, PF3Cl, and PF4Cl: Insight into Isomerism and Apicophilicity from Ab Initio Calculations and the Recoupled Pair Bonding Model.” Theoretical Chemistry Accounts, vol. 133, no. 2, Feb. 2014, p. 1428. doi:10.1007/s00214-013-1428-7.
French, Richard J., et al. “Dichlorotrifluorophosphorane (PCl2F3): Molecular Structure by Gas-Phase Electron Diffraction and Quadratic Force Field.” Inorganic Chemistry, vol. 24, no. 18, Aug. 1985, pp. 2774–77. doi:10.1021/ic00212a014


Heisenberg's Uncertainty Principle's inequality In the uncertainty principle , whats the significance of the inequality sign ?
Why is it not equal to over there ?
Why do we say greater than equal to instead of just equal to ?

As andynitrox said in the comments :

  The HUP just gives a lower bound in the uncertainty of Δx and Δpx, it is of course possible to measure both quantities with a higher uncertainty than given by the HUP, but not less.

The inequality exists to show that, the uncertainty is atleast this much and in real-life cases, its much larger than this . (source)(provided by Mithoron)


Reasons behind the different colors of a solution containing Iodine Q) Iodine dissolves in a variety of solvents, forming solutions that are
purple, brown or any intermediate shade. This property is due to:

Hydrolysis of iodine
Formation of crystalline adducts
Charge transfer between $\ce{I-I-}$ and the solvent
Charge transfer between $\ce{I-I+}$ and the solvent

I am completely aware of which concept does option 3 and 4 allude to. I know about the formation of $\ce{I3-}$ when iodine is dissolved in solution of iodides. This fact, according to me, is what option 2 refers to.

But the answer given is 3

Can someone provide an explanation as to what concept is involved here and how can we predict the answer? I read about charge transfer complexes, but the Wikipedia page was too complicated for me to understand, and also the application of that to the question under consideration was not given.

Iodine in an inert solvent is purple.  Electron donation alters that.  Linear starch (amylose)-iodine is dark brown.  Branched starch (amylopectin)-iodine is red.  The Wikipedia page shows iodine in chlorinated solvent (very weak electron donation) then addition of strong electron donor triphenylphosphine.

If an electron donor is reactive, then the interacting iodine molecule looks like iodonium (elecron accepter).  Initial interation is (4), post-mixing is (3).

Put a tiny iodine crystal into a melting point capillary, evacuate under vacuum, flame closed.  You will see a very pale purple vapor at equilibrium.  Crack the seal and the vapor is suddenly deeper purple.  Why?  Take a sealed cap and an evacuated sealed cap, and insert the vapor volumes into the beam of a green laser pointer. interesting. Take care that the optical geometry does not reflect back into your eyes.


How to find the volume of the products in methane combustion? The combustion of methane gas, $\ce{CH4}$, gives as products $\ce{CO2}$ and $\ce{H2O}$, both in the gas phase. If $1$ liter of methane is burned in the presence of $10$ liters of $\ce{O2}$. Suppose all volumes measured under the same temperature and pressure conditions and optimal behavior for all gases.
  $$\ce{CH4 (g) + 2O2 (g) -> CO2 (g) + 2H2O (g)}$$
  What is the volume of the products formed?

I know that for a gas, the volume is proportional to the amount of substance, that is, maintaining the temperature and pressure conditions, doubling the volume doubles the volume of moles.

The answer to the exercise is 3 liters.

One mole $\ce{CH4}$ reacts with 2 moles $\ce{O2}$, producing 1 mole $\ce{CO2}$ and 2 moles $\ce{H2O}$.
So 1 liter $\ce{CH4}$ produces 1 liter $\ce{CO2}$ and 2 liter $\ce{H2O}$, if $\ce{H2O}$ is taken in the gaseous phase.


What is the simplest approach to balance a complex reaction equation? Consider the redox reaction:

  $$\begin{multline}
\ce{ a K4Fe(CN)6 + b Ce(NO3)4 + c KOH ->\\ d Ce(OH)3 + e Fe(OH)3 + f H2O + g K2CO3 + h KNO3}
\end{multline}$$

The book's answer for the coefficients $a, b, c, d, e, f, g,$ and $h$ is $1, 61, 258, 61, 1, 36, 6$, and $250$. The final answer for the problem is the sum of the coefficients which is $674$.

I have tried balancing it by using the general methods such as oxidation number method, ion-electron method, etc. However, they seem to be giving me incorrect equations. The answer given by the book is perfectly correct. I am sure of it, as I multiplied and checked each atom on both the sides, and they checked out completely. Also, the coefficients don't have a common factor other than 1.

What is the simplest way to solve this problem?

$\ce{ a K4Fe(CN)6 + b Ce(NO3)4 + c KOH ->\\ d Ce(OH)3 + e Fe(OH)3 + f H2O + g K2CO3 + h KNO3}$

Because there is one $\ce{Fe}$ in $a$ and one in $e$, we know that $a=e$.

To better see things, I rewrite the formulae.

For $\ce{Fe}$: $a=e$
For $\ce{Ce}$: $b=d$
For $\ce{C}$: $6a=g$
For $\ce{N}$: $6a+4b=h$
For $\ce{K}$: $4a+c=2g+h$
For $\ce{H}$: $c=3d+3e+2f$
For $\ce{O}$: $12b+c=3d+3e+f+3g+3h$

Now I replace $e, d, g, h$ with their counterpart, leaving just $a, b, c, f$. I get

For $\ce{K}$: $4a+c=2(6a)+(6a+4b) \Rightarrow c=14a+4b $
For $\ce{H}$: $c=3b+3a+2f$
For $\ce{O}$: $12b+c=3b+3a+f+3(6a)+3(6a+4b) \Rightarrow c=39a+3b+f$

Replacing $c$.

For $\ce{H}$: $14a+4b=3b+3a+2f \Rightarrow 11a+b=2f$
For $\ce{O}$: $14a+4b=39a+3b+f \Rightarrow -25a+b=f$

Combining at $f$.

$11a+b=-50a+2b \Rightarrow 61a = b$

Now I assume $a=1$ and get

$a=1$
$b=61a=61$
$c=14a+4b=258$
$d=b=61$
$e=a=1$
$f=-25a+b=36$
$g=6a=6$
$h=6a+4b=250$

And, finally

$a+b+c+d+e+f+g+h=674$

The last step could also be achieved by taking $a+b+c+d+e+f+g+h$ and replacing $c, d, e, f, g, h$, then replacing $b$, which should eventually lead to $674a$.


What is an example of a very low temperature endothermic reaction? Shakashiri describes the reaction of solid Ammonium Thiocyanate and solid Barium Hydroxide Octahydrate resulting in an endothermic reaction leading to subzero temperatures. I've performed this experiment and measured temperature as low as about $-10\ ^\circ\text{C}$.

Are there known chemical reactions that can lead to even lower temperatures?

Yes there are. In Vogels famous textbook on practical organic chemistry there is a section about the cooling effect of certain salt (mixture) solutions in water.

  If ice is temporarily not available, advantage may be taken of the
  cooling effect attending the solution of certain salts or salt
  mixtures in water. Thus a mixture produced by dissolving 1 part of
  NH4C1 and 1 part of NaNO3 in 1-2 parts of water causes a reduction in
  temperature from 10 to — 15 °C to — 20 °C; 3 parts of NH4C1 in 10
  parts of water from 13 to -15°C; 11 parts of Na2S2O3.5H2O in 10 parts
  of water from 11 to - 8 °C; and 3 parts of NH4NO3 in 5 parts of water
  from 13 to — 13°C

(Taken from Vogels 5th edition, p. 70)


How to make artificial sea water My 7th grader is embarking on a long-term science project and I'm on tap to "help." He is showing a strong interest into finding ways to get drinking water to the masses. His idea is that we need to find inexpensive and efficient desalination processes. So, we are exploring current desalination processes to see what makes it "expensive" and "inefficient."

However, we live in the Rocky Mountains and don't have access to proper sea water. We started to explore simply adding salt to tap water and researching how much. Then, we discovered "Artificial Sea Water" or "ASW" is a common creation in science. Because sea water contains more than just salt. It contains chemicals like fluoride, strontium, boron, bromide, inorganic carbon, potassium, calcium, sulfate, magnesium, chlorine and of course, sodium. I am not a chemist, so this is starting to look daunting.

So, here are my questions to counsel my son:

To experiment with desalination, is adding salt to tap water
sufficient or do we need to investigate finding official ASW?
How do we test the salinity of water? Do I really need to invest
hundreds of dollars on a salinity meter?

If you look at the (average) composition of seawater:

you will find that NaCl is the main solute of this complex solution. Its concentration is 10 times higher than that of the next constituents (magnesium and sulfate). So, it seems to me like a simple NaCl solution is a good approximation for your experiments.

Regarding testing salinity of water, I don't have a definite answer on that, I do have two suggestions to offer:

Measure salinity by electric conductance of the solution, e.g. by making a simple conductometer and testing it with solutions of known concentration of salt.
Measure salinity by the quantity of salt (solid residue) once all water is evaporated. You can speed the evaporation by heating up.


How are organic compounds with radioactive atoms synthesized? We often see reactions such as these in textbooks to highlight how reaction mechanisms work:

How are the radioactive substrates required for these mechanisms synthesized in the first place? There seems to be no easy way of just 'substituting' a radioactive atom in place of a non-radioactive one (especially for Benzene-14C, it might be easier to make a radioactive sodium salt using $\ce{^{24}Na}$). Also, Once the substrate is synthesized, how is it separated and purified? I'm assuming that the physical properties of these radioactive substrates would be very similar to their non-radioactive counterparts, which would make it very difficult to do so.

The current methods employed for synthesis fall into the following three categories1:

Chemical synthesis
Biochemical methods
Isotope exchange reactions

Chemical Synthesis:
Most chemical syntheses involving $\ce{^14 C}$ are done with Grignard reagent and $\ce{^14 CO2}$ or $\ce{^14 CN-}$. As an example, Benzene-14 is synthesized as such2:

Basic preparatory methods such as the above have been worked out long ago. An example of a more complex synthesis would be preparation of 18F-Fluorodeoxyglucose:

Biochemical methods
A few examples of these, as mentioned in (1), are:

High yields of [U-14C]sucrose, up to 70% based on $\ce{^14 CO2}$, are obtainable in detached leaves after a few hours photosynthesis using tobacco or Canna indica leaves for example.
DNA labelled either with $\ce{^14 C}$ or with $\ce{^3 H}$ is prepared by feeding E.coli with the correspondingly labelled thymine precursor. Similarly, by feeding [$\ce{^35 S}$]sulphate to E.coli or to yeast cells, isolation and hydrolysis of the proteins yields L.[$\ce{^35S}$]methionine and L-[$\ce{^35 S}$]cystine in radiochemical yields of 30% and 10% respectively.

Another example is the synthesis of C-14 labelled Phenylpyruvic acid3

Isotope exchange reactions
These are fairly common for atoms that can easily be exchanged, such as replacing $\ce{^23 Na}$ for $\ce{^24 Na}$, which was mentioned in the question. Another common mechanism for synthesis of Hydrogen isotope compounds is Hydrogen-Deuterium exchange. A similar mechanism also allows hydrogen-tritium exchange4.
Purification of radiolabelled compounds
This is mainly done through Chromatography1, as the amount of radiolabelled substrate required is generally quite small. High-performance liquid chromatography is used to separate the compounds. The chemical identity of the compounds is  confirmed by co-chromatographic behaviour with the unlabelled compound.
For further information, refer to the references (1) and (3), which go into a lot more detail with regards to the synthesis of these compounds.
References:

Anthony Evans, E. “Synthesis of Radiolabelled Compounds.” Journal of Radioanalytical Chemistry, vol. 64, no. 1–2, Mar. 1981, pp. 9–32. doi:10.1007/BF02518337.
Schmid, K., et al. “A Simple Method of Preparing Labeled Benzene.” Advances in Tracer Methodology, edited by Seymour Rothchild, Springer US, 1966, pp. 37–44. doi:10.1007/978-1-4684-8625-4_5.
Atzrodt, Jens, and John Allen. “Synthesis of Radiolabeled Compounds for Clinical Studies.” Drug Discovery and Evaluation: Methods in Clinical Pharmacology, edited by Hans Gerhard Vogel et al., Springer Berlin Heidelberg, 2011, pp. 105–18. doi:10.1007/978-3-540-89891-7_12.
Englander, S. W., and J. J. Englander. “[3] Hydrogen—Tritium Exchange.” Methods in Enzymology, vol. 49, Academic Press, 1978, pp. 24–39. doi:10.1016/S0076-6879(78)49005-6.


How is the Nitrate Ion (NO3) formed? I understand how the nitrite $\ce{NO2}$ ion can be formed and have a negative charge, but the nitrate $\ce{NO3}$ ion is confusing me. I made some simple drawings to try to explain what I don't understand:

Does the "electron from outside" that the oxygen receives, as indicated in the drawing, comes from the nitrogen atom? If so, then the N atom would be left with only 4 valence electrons instead of 5, thus being able to form another covalent bond with the extra oxygen, forming $\ce{NO3}$. Is this the correct line of thinking?

An organiker would look at it this way.

An inorganiker might look at it as literal N(+5) with two double-bonded oxygens.  It does redox as N(+5). Coordination structures are varied,


How is the process of converting calcium carbonate (CaCO3) into calcium oxide (CaO)? I need to provide accurate information in relation to the production process of calcium oxide from calcium carbonate.

$$\ce{CaCO3 -> CaO + CO2}$$

From wikipedia article of calcium oxide

  Calcium oxide is usually made by the thermal decomposition of
  materials, such as limestone or seashells, that contain calcium
  carbonate ($\ce{CaCO3}$; mineral calcite) in a lime kiln. This is
  accomplished by heating the material to above 825 °C (1,517 °F), a
  process called calcination or lime-burning, to liberate a molecule
  of carbon dioxide ($\ce{CO2}$), leaving quicklime.

  $$\ce{CaCO3(s) → CaO(s) + CO2(g)}$$

  The quicklime is not stable and, when cooled, will spontaneously react
  with $\ce{CO2}$ from the air until, after enough time, it will be
  completely converted back to calcium carbonate unless slaked with
  water to set as lime plaster or lime mortar.


Could you charge a recharable battery with a voltaic cell? In class we were learning about voltaic cells and my chemistry teacher joked that after modern society falls, we'll be the only ones who can recharge batteries with simple voltaic cells. And this got me thinking, is this possible? Could I create a salt bridge cell, attach wires to a store bought rechargeable battery, and charge it? I'm really curious and I can't ask my chem teacher as he's in the hospital recovering from an emergency appendix removal

Any source of electricity of the required voltage and current could be used to charge a rechargeable ("secondary") cell. A home-made "primary" cell would work (though you might need a few in series to provide the needed voltage).

For that matter, you can generate DC to charge the cell many other ways:

Using a thermopile, e.g. from copper and iron wire
Magnetohydrodynamically, e.g. with flowing salt water and a magnet
Using a photovoltaic panel...


(Energetics) Potential Energy and Kinetic Energy This question came up when I was reviewing the energetics unit. In my notes I wrote "systems with a lower potential energy are more energetically stable, because they have a higher kinetic energy and higher kinetic energy means higher level of disorder." Is this correct? If so, does it mean potential energy stored in chemical bonds are converted to kinetic energy during a chemical reaction?
Is energetically "stable" the same thing as energetically "favourable"?

In classical physics, kinetic energy and gravitational potential energy are almost interchangeable.  You do work on a ball and throw it vertically into the air.  Neglecting friction with the air all the kinetic energy becomes potential energy.  However, it then begins to fall, converting the potential energy back into kinetic energy.  Again, with no friction, the kinetic energy is exactly the amount that you began with.

However, we have to be careful to distinguish between ordered kinetic energy and random kinetic energy.  The ball, thrown into the air has both ordered and random kinetic energy.  The energy that becomes potential energy was the ordered form where-as the random thermal motions of all the individual atoms, which add up to no NET motion of the ball, remained the same throughout the entire flight of the ball.

But did they?  Again and again we have to say in these situations: Ignoring friction.  Friction does in fact exist when there is air present, and friction is very relevant to this discussion.  Friction actually converts ordered kinetic energy of motion into random kinetic energy of various particles (air molecules and ball molecules).

The creation of random motion from ANY other form of energy increases entropy and reduces the amount of energy that is available to do useful work.

I'm not sure that totally answers the question.  I think that the statement needs to clarify that random motion and its associated kinetic energy is, as your note claims, a preferred state over potential energy.


Smith Hieftje correction method I completely don't understand how this method works. My book says that this method works by the principle of self-absorption which occurs by the emitted radiation of hollow cathode lamp when it works on high intensity of electricity, due to which we have more atoms who lead to an extension of emission lines. This is how self-absorption occurs.  Does anyone have any proper explanation for this?

What does this self-absorption actually mean?

How does that lead to a correction?

The Smith-Hieftje background correction is basically another means of "chopping" the signal. The amplifier locks onto the chopper frequency to cancel out the background noise. "Locking on" essentially means that the amplifier will measure the difference between the two signals.

So one way to chop the "light" from the emission tube would be to insert a "fan blade" to mechanically chop the light on and off.

The Smith-Hieftje background correction is a bit more complicated. Basically what happens is that this emission tube has two parts. A heater to vaporize the source of the atoms in the tube and an electron beam to stimulate an excited state. In this type of emission tube the heater is pulsed high and low. When the heater is low it produces enough metals atoms so that the atomic line emission is created. But when the heater is on high heat then it vaporizes so many atoms that the vapor cloud of metal atoms self-quenches the emitted atomic line emission. The high density cloud of atoms also means that the atoms bump into one another. So instead of the atoms in the cloud all having about the same kinetic energy they have a range of kinetic energies which results in line broadening.


How much does the quality of glassware matter, really? Discussing about the pros and cons of glassware from different manufacturers for chemistry experiments easily becomes a heated debate. Well, that's how it seems if you read the chemistry related blogs, discussions, questions, etc. over the internet. How much does it really matter what kind or type of glass one is using? How big the differences are across different manufacturers?

It would be nice if the answers also elaborated on the differences between laboratory and home chemistry.

In addition to differences in the absorption spectrum, different types of glass are more or less able to deal with temperature or mechanical shocks. But that is less about manufacturer: you'll find different types of glassware from the same manufacturer for different purposes (e.g. Fiolax and Duran are both by Schott).

As @LDC already commented, some glassware is thin walled and more mechanically fragile. OTOH, thick walled glass is more likely to break with a temperature shock. So e.g. for test tubes you have to decide whether you want to have them

suitable for local heating => thin wall borosilicate glass
suitable for centrifuge => thick wall borosilicate glass
cheap => soda glass
UV transmittant => quartz
suitable for huge temp shocks => quartz

If this question is about whether your borosilicate glass ware is called Duran or  Pyrex (or Jenaer or Ilmabor or Rasotherm) I don't see that much of a difference.


Modeling perturbations of a quantum mechanical system I'm wondering how to properly perturb a quantum mechanical system.

I'm looking for a way to express the return to equilibrium of such a perturbed state, where the perturbation is in the Hamiltonian. The unperturbed Hamiltonian is $H_{\text{u}}$ and the perturbed Hamiltonian is $H_{\text{p}}$.

I can compute the eigensystem for both $H$ and thus find the relevant energy spectra and eigenfunctions, and can add the time-dependence factor $e^{-iEt}$, but I do not know how to generate the final system from its component, isolated systems. I would guess an inner product to be of use here, but...

What theory/mathematical methods do I need to learn to perform such simulations?

Something I want to model is a molecule in a polarized state, that is, with an asymmetric electron distribution.

I don't understand exactly what is your question:

If you start in an eigenstate of $H_p$, say $\varphi_j$ and then your Hamiltonian changes like $H(t) = H_u + f(t) (H_p - H_u)$ where $f(0) = 1$ and $f(T) = 0$ -$T$ being the final time of the perturbation- then in order to follow the evolution of $\psi(t)$ all you need to do is to solve the time dependent Schroedinger equation
$$i\hbar \partial \psi(t) / \partial t = H(t) \psi(t)$$
where $\psi(0) = \varphi_j$. There are many numerical procedures to do so. For instance if you are interested, as it seems, in following the dynamics in the basis of the eigenstates of $H_u$, then you expand $\psi(t) = \sum_k a_k(t)
\phi_k$, where $\phi_k$ are eigenstates of $H_u$, and you solve a set of ordinary differential equations.

If $f(t)$ is very fast ($T << (\Delta E)^{-1}$, where $\Delta E$ is a characteristic energy of the system, say the energy difference between adjacent eigenstates) then you can use the sudden approximation; if $f(t)$ is very slow you can use the adiabatic approximation, if $H_p - H_u$ is small you can use first order perturbation theory, etc.

In any case it is important to realise that in Quantum Mechanics there is no relaxation unless you couple your initial state to a continuum and there is no return to equilibrium unless you do the same (couple to continuum) and work with the proper formalism that allows you to define the (thermal) equilibrium, which are the density matrices, unless by equilibrium you are referring to the initial state. Again, in ordinary Quantum Mechanics $\psi(t)$ will oscillate and will not relax to the initial state $\varphi_j$ (or $\phi_0$); at most, it will return to the initial state as part of an oscillation, called revival.


Mass spectrum analysis When finding ions which correspond to certain fragments on a mass spectrum; there are multiple possible ions that can correspond to the same fragment;
How should I know which one is the correct one?

Generally speaking: knowing the history of your sample (e.g. earlier steps of synthesis allowing / excluding elements that could numerically fit into the list of suggested fragments) and additional spectroscopic characterisation (UV-Vis, IR, Raman, NMR, etc.) may be helpful.

The problem you describe may be attenuated by using a mass spectrometer of higher resolving power.  Instead of submitting the sample to the detector behind a smaller LC-MS coupled analysis, providing a purified sample to a larger, tuneable sector mass spectrometer, for example.

An important alternative is coupling of several mass spectrometers along one line, where after initial ionisation primary fragments are separated from each other, and their corresponding subsequently generated daughter fragments are eventually characterised, like in a tandem MS/MS:

(source)

An overview about the later provides this wikipedia entry.


How do I arrive at correct furanose structures when starting from a Fischer projection? While numbering the furanose ring from the left are we supposed to place the atoms on right of the Fischer projection to the top? Isn't it just the opposite to what we do while numbering from the right?

In the image the atoms/groups lying on the right of the 2nd Fischer structure is present to the bottom of every furanose carbon of the 3rd image (chronologically). So aren't they supposed to be present at the bottom even if we start from the left (referring to structure A)? But I find Wikipedia says they are supposed to be at the top.

A Fischer structure is one way of depicting structures such as sugars or amino acids to quickly draw them and have a definite way of seeing stereochemistry on every carbon. The rule behind Fischer structures is that the highest oxidised atom should be at the top or as close as possible to the top. The following convention is used to turn tetrahedral carbons into a square environment:

This allows you to assign absolute configurations to every carbon atom.

When instead drawing the same structure as a ring like in your lower images, you just need to take care that every carbon has the same configuration. (Note that your drawing of β-D-fructose is wrong; it should be drawn with a very wavy line like in my image below, because of the constraints of Fischer projections.)

Once you have gotten there, you are free to rotate the ring around as much as you wish while retaining the absolute configuration at every stereocentre. Note especially, that a rotation of $180^\circ$ of my structure of β-D-fructose will lead to you one on the left. If you don’t believe me, print it out, turn the paper around and then nudge the forwards-pointing part of the ring downwards below the tetrahydrofuryl-oxygen.


Are the bonds within bismuth pentafluoride (BiF5) covalent or ionic? When googling $\ce{BiF5}$ most results call it bismuth pentafluoride such as Wikipedia.

I was under the impression that, since it is an ionic-bonded compound, Roman numerals should be used to identify the charge of bismuth, followed by fluorine with an ide ending such as the following:

  Bismuth(V) Fluoride

So, I am wondering whether the bonds are either not ionic or that I've misidentified it?

For all nomenclature purposes, one should refer to the official IUPAC recommendations, which in this case are laid down in Nomenclature of Inorganic Chemistry — IUPAC Recommendations 2005, the ‘Red Book’. The relevant section for this question is section IR-5: Compositional Nomenclature, and Overview of Names of Ions and Radicals
The introduction of this section (IR-5.1) states:

The simplest type of compositional name is a stoichiometric name, which is just a reflection of the empirical formula (Section IR-4.2.1) or the molecular formula (Section IR-4.2.2) of the compound. In stoichiometric names, proportions of constituent elements may be indicated in several ways, using multiplicative prefixes, oxidation numbers or charge numbers.

The following details on Stoichiometric Names of Elements and Binary Compounds (section IR-5.2) detail how to derive a stoichiometric name:

The stoichiometric name of the compound is then formed by combining the name of the electropositive constituent, cited first, with that of the electronegative constituent, both suitably qualified by any necessary multiplicative prefixes (‘mono’, ‘di’, ‘tri’, ‘tetra’, ‘penta’, etc., given in Table IV). The multiplicative prefixes precede the names they multiply, and are joined directly to them without spaces or hyphens. The final vowels of multiplicative prefixes should not be elided (although ‘monoxide’, rather than ‘monooxide’, is an allowed exception because of general usage). The two parts of the name are separated by a space in English.

Thus, the stoichiometric name of $\ce{BiF5}$ is bismuth pentafluoride.[1]
In the later section Generalized Stoichiometric Names, guidance is given on how to name compounds based on one constituent’s charge or oxidation number (IR-5.4.2.2).

It is possible to provide information on the proportions of the constituents in names by using one of two other devices: the charge number, which designates ionic charge, and the oxidation number, which designates oxidation state. […]
[…]
The oxidation number (see Sections IR-4.6.1 and IR-9.1.2.8) of an element is indicated by a Roman numeral placed in parentheses immediately following the name (modified by the ending ‘ate’ if necessary) of the element to which it refers. The oxidation number may be positive, negative or zero (represented by the numeral 0). An oxidation number is always non-negative unless the minus sign is explicitly used (the positive sign is never used). Non-integral oxidation numbers are not used for nomenclature purposes.

Thus, according to this nomenclature, $\ce{BiF5}$ (determination of bismuth’s oxidation state is unambiguous) is bismuth(V) fluoride.

These two different styles of naming binary compounds are equivalent. In no way do they include any statement concerning the bonding situation. In fact, these names explicitly do not contain any information concerning structure or connectivity: dinitrogen tetraoxide is the correct name for both $\ce{O2N-O-N-O}$ and $\ce{O2N-NO2}$ — and the generalized name nitrogen(IV) oxide is also correct for $\ce{NO2}$.
Notably, usage of one or the other nomenclature cannot be used to distinguish between covalent or ionic bonding. As a case in point, $\ce{CCl4}$ — a rather clearly covalently bonded molecule — can be referred to as:

carbon tetrachloride
carbon(IV) chloride

And vice-versa, $\ce{BaCl2}$ — a very ionic compound — can be referred to as:

barium dichloride
barium(II) chloride
barium chloride[1]

Therefore, neither bismuth pentafluoride nor bismuth(V) fluoride make any reference to the type of bonds present in the compound; it is up to the chemist to use their previously acquired knowledge to realise whether bonds are predominantly covalent or predominantly ionic or something else.

Note:

Multiplicative prefixes need not be used in binary names if there is no ambiguity about the stoichiometry of the compound […]. The prefix ‘mono’ is, strictly speaking, superfluous and is only needed for emphasizing stoichiometry when discussing compositionally related substances […].

(Section IR-5.2)
Therefore, bismuth pentafluoride is permitted alongside monobismuth pentafluoride; and barium chloride is permitted, since only barium dichloride is reasonable.


Why does a hydride like NaH act as a base, but a hydride like NaBH4 act as a nucleophile? A TA talked about this today in the context of ketones.

If you stick an acetone and $\ce{NaBH4}$ together in a protic solvent, you get 2-propanol.

If you stick acetone and $\ce{NaH}$ together, you get an enolate. What's the difference? They're both hydrides.

First of all, borohydride as a group is not same as hydride. So the mechanisms are very different.

Because a simple $\ce{H-}$ anion is more basic than an enolate $\ce{C=C-O^{-}}$(product of $\alpha$ hydrogen deprotonation of a ketone), it just deprotonate before reduction will happen. Deprotonated ketones (enolates) will no longer be a substrate for reduction.

Borohydride $\ce{BH4-}$ is not as basic as enolate, so deprotonation hardly happen. Thus a reduction will be the predominant reaction.


Why can't the strength of superacids be measured in water? I learned about acid strength, that the strength of an acid increases with it's degree of ionization when solvated. So, in water, a strong acid is one where $\ce{[H_3O^+]}$ is large, which is equal to a low pH: $\mathrm{pH=-log[H_3O^+]}$.

Considering extreme cases, such as superacids, I have found out that other methods are used to measure their acidity (methods I don't really understand). My question is why is it impossible to simply get super high concentrations of $\ce{[H_3O^+]}$ in aqueous solutions of superacids, and use this to determine the acid strength. Also, is pH used as a measure of acidity outside of aqueous solutions?

I have come over the leveling effect, but I don't think I fully understand it. The way I understand it (for the case with water as solvent) is that basically any acid in water will protolyze $\ce{H2O}$ to $\ce{H3O+}$, making this the effective acid. I don't understand why this would affect the measured pH, as it is $\ce{[H_3O^+]}$ you are measuring.

Any acid-base reaction is always an equilibrium:

$$\ce{HA^1 + (A^2)- <=> (A^1)- + HA2}\tag{1}$$

and for each pair of acids $\ce{HA^1}$ and $\ce{HA^2}$ you could calculate a $K_\mathrm{a}$ value to determine one acid’s strength with respect to the other. This $K_\mathrm{a}$ value is typically calculated according to equation $(2)$ if $\ce{(A^2)-}$ (which does not have to feature a negative charge; I just wanted to avoid different descriptions for the two acids) is the solvent.

$$K_\mathrm{a} = \frac{[\ce{HA^2}][\ce{(A^1)-}]}{[\ce{HA^1}]}\tag{2}$$

If $\ce{(A^2)-}$ is $\ce{H2O}$ (i.e. $\ce{HA^2}$ is $\ce{H3O+}$) and the acid $\ce{HA^1}$ is a strong acid or a superacid, then this equation is of little use; the equilibrium $(1)$ will be strongly shifted towards the product side, the concentration $[\ce{HA^1}]$ will be very small and thus $K_\mathrm{a}$ very large. With ‘very small’, I mean that $[\ce{HA^1}]\approx10^{-10}$ — acid $\ce{HA^1}$ is said to be fully dissociated.

Exact measurements of $K_\mathrm{a}$ become very hard if not impossible because you would need to determine extremely minute concentrations. The concentrations $[\ce{HA^2}]$ and $[\ce{(A^1)-}]$ are of little help since:

$$[\ce{HA^2}] = [\ce{(A^1)-}] \approx c_0(\ce{HA^1})\tag{3}$$

This is the levelling effect of water: no matter how strong the acid, there is no acidic species stronger than $\ce{H3O+}$ that can survive in aqueous solution for extended time. Most importantly, consider two very strong acids with different $K_\mathrm{a}$ — e.g. one with $\mathrm{p}K_\mathrm{a} = -15$, the other with $-10$. They have a very different acidity, yet the concentrations of all relevant species in aqueous solutions will be nigh identical.

Therefore, a new, more expandable way to measure and compare the strengths of very strong acids was required.


What is the joke that is shown in this picture? There is supposedly a joke shown in this picture relating to periodic trends.
Does anyone know what it is? And could they also explain where the elements on the eye exam came from?

As commented by @long, the elements shown are not completely randomised. The elements in all the rows, except in the 3rd row ($\ce{Na, K, Ca}$), are arranged in same order as they are given in periodic table.

After studying them on periodic table for 15 minutes, it seemed to have no meaning. There are no periodic trends. $\ce{H, Li, Na, K, Rb}$ belongs to IA group, $\ce{Be, Ca, Sr}$ belongs to IIA group. And they are arranged in a random way, having no meaning.

The picture is clearly a joke showing that when scientist have eye check-up, they have to read elements instead of alphabets.


Is there an uncertainty associated with the value 0 K for absolute zero? When I say absolute zero, I’m not talking about the hypothetical temperature 0 K; I’m talking about the temperature at which a thermodynamic system has the lowest energy.

Everywhere I look, sites maintain that this should be –273.15 ºC, but I can’t find any uncertainty quoted with this prediction. Could someone shed some light on this?

To add a comment that I left on theorist’s answer:
I know that the Celsius scale is defined to be an easy conversion from the kelvin scale. What I mean to ask is the following: The Boltzmann constant $k_{\rm B}$ has a fixed value; the reference temperature $273.15\;\rm K$ is also a fixed value used in the relation $t/\mathrm{°C}=T/\mathrm{K}-273.15$. Therefore, how can we know that a thermodynamic system has the lowest energy at exactly $0\;\rm K$? If the triple point of water $T_{\rm TPW}=273.16\;\rm K$ has a standard relative uncertainty of $3.7\times10^{-7}$, then why shouldn’t $T_{\rm A0}$, especially since it has never been achieved? Something has to give and not be by definition, so to speak.

As another answer explains, absolute zero is defined in the Kelvin temperature scale as precisely $\pu{0 K}$. But this is not the entire story, as all measurements have an associated uncertainty (all thermometers are imprecise), see e.g. the wikipedia article describing the International Temperature Scale of 1990 and the Provisional Low Temperature Scale of 2000. For instance the second wikipedia article links to a International Committee for Weights and Measures' document describing the uncertainties associated with the calibration of the temperature scale based on measurements of the melting pressure of $\ce{^3He}$.
Recently the SI value of the Kelvin was changed. This is explained in the wikipedia:

The definition of the kelvin underwent a fundamental change. Rather than using the triple point of water to fix the temperature scale, the new definition uses the energy equivalent as given by Boltzmann's equation.
Previous definition: The kelvin, unit of thermodynamic temperature, is 1/273.16 of the thermodynamic temperature of the triple point of water.
2019 definition: The kelvin, symbol K, is the SI unit of thermodynamic temperature. It is defined by taking the fixed numerical value of the Boltzmann constant k to be $1.380649×10^{−23}$ when expressed in the unit $\pu{J⋅K−1}$, which is equal to $\pu{kg⋅m2⋅s−2⋅K−1}$, where the kilogram, metre and second are defined in terms of h, c and $\Delta ν_{Cs}$.
The kelvin may be expressed directly in terms of the defining constants as:
$$\pu{1 K} =  \frac{1.380 649 \times 10^{23}}{(6.626 070 15 \times 10^{-34})(9 192 631 770)} \frac{h \Delta \nu _{Cs}}{k}$$

So what does this mean for the definition of $\pu{0 K}$ ? That doesn't change, that is an absolute, thermodynamic definition. However, evidently the triple point of water is now no longer exactly $\pu{273.16 K}$. You can think of this as shifting the uncertainty from the size of a step to the number of steps that take you from absolute zero to the triple point. But both before and after the redefinition, $\pu{0 K}$ has meant $\pu{0\pm  0 K}$, that hasn't changed.

$^\ast$you can't actually reach absolute zero, so you can only extrapolate to 0 K and provide an uncertainty for this extrapolation, i.e. you could estimate how far your measured temperature is from absolute zero, and provide uncertainties associated with your estimate.


Why does hyperconjugation help for ring cleavage? The question was to rationalize it the ring cleavage happens concerted or non-concerted.

The papers "The Mechanism  of  the Thermal Decomposition  of
1-Pyrazolines and Its Relationship  to
Cyclopropane Isomerizations" (Crawford 1965/1966; Original papers: here and here) have shown the loss of $\ce{N_2}$ is supposed to be concerted.

I think the C-C/C-D(H) bonds of the central position of the ring are involved in a hyperconjugative fashion, so its hybridization does matter.  The out-of-phase linear combination of the $\sigma^\star$ orbitals looks like a p-orbital, therefore I expect a through-space interaction.

I've drawn the the out-of-phase combination as following:

Now my problem is, that I don't see why the out-of-phase combination is involved in hyperconjugation, because I think the overlap is best in the educt, less good in the transition state and has no orbital overlap to the product because the out-of-phase $p$ orbital is kind of perpendicular to the $sp$ Walsh orbitals which lies in the C-C-C plane.

How can hyperconjugation help to understand the observed kinetic isotope effect results?

Summary:

How does the FMO picture (HOMO-LUMO) with all involved bonds for hyperconjugation looks like?
Can the ratio 66%:34% be explained?
What does reactant 3 make compared to 2?
How can the ratio 1.07/1.20 be explained?

Do you have any ideas?

This is a bit late, but my instinct tells me you form the 1,3 diradical. The fusion to a cyclopropane seems pretty reasonable. The alkene is formed from a 1,2-hydrogen atom shift (I admit, this is a strange abstraction), but the formation of the alkene (and the fact that this is flash vacuum pyrolysis) make me think this is not completely unreasonable. Since the mechanism involves breaking the H/D bond, you would expect some selectivity 1:1 versus 2:1. The fact that there is essentially no primary KIE observed is consistent with diradical formation as the r.d.s.

The other reaction, I suspect, forms the cyclopentene with the deuterium substitution on at the 3- position. The secondary KIE would be consistent with diradical formation.


Can we really see the bonds? I was thinking is there really bond present at microscopic level or atoms/molecules are just nearby and are connected with force which is not visible(like gravitational force between earth and sun) and we make bonds just for understanding.

All credit to Zhang et al. "Real-Space Identification of Intermolecular Bonding with Atomic Force Microscopy" Science Vol. 342  no. 6158  pp. 611-614.

Yes, direct images of bonds, not only covalent bonds but also intermolecular hydrogen bonds have been recorded.

It is the electron density that is being observed, covalent and hydrogen bonds involving high electron density between the atoms.

Scanning Tunneling Microscopy can also be utilized to directly observe the electron density of bonds.


UV/Vis spectroscopy transition reference There are mainly fundamental transitions happening during excitation and UV/Vis spectroscopy (as n- > σ* ,π->π* ,n- > π ,..).  I am looking for a good reference that shows these transitions for molecules (for example shows n- > σ* for acetone , π- > π* for benzene,... ).

Actually I am talking about tables. I can do guesswork for most of the molecules based on their structures but definitely I need references for publication.

Update: Mostly interested in these molecules: Acetone, Acetaldehyde , Benzene , Naphthalene, Anthracene , Methylene Blue, Phenanthrene , 1-Butanol

Here are a number of resources that may be of use for your research.

The UCLA website presents tables and data in their webpage Theory of Ultraviolet-Visible (UV-Vis) Spectroscopy, including reference tables of common functional groups, aromatics and compounds often used in the laboratory.  (Many compounds you listed are in the tables).

A reference that may be of use is Spectroscopy of Organic Compounds (Kumar, 2006) which include several tables that include information of the compounds you list.

The book Spectroscopy of Organic Compounds (Kalsi, 2007), from page 12 provide several tables of information related to your listed organic compounds, including details of the electronic transitions. The chapter "Ultraviolet (UV) and Visible Spectroscopy" (starting page 9) has a subsection "The Designation of Various Transitions" which details the electronic transitions using examples in context (from page 11, summarised in a diagram on page 12).  Examples of solvents are discussed from page 23 onwards.  (All available online).

Transitions are explored in a bit of detail in chapter 4.3 of the UC Davis ChemWiki The Dynamic Chemistry E-textbook, Section 4.3: Ultraviolet and visible spectroscopy


Why is zinc more reactive than copper? In terms of electronegativity, from what I understand electronegativity increases going across the period, so surely this should mean that zinc less readily loses its outer shell electrons than copper? Zinc has a greater nuclear charge but the outer shell electrons are in the same shell, so should the outer shell electrons in zinc not experience a stronger attraction to the nucleus? I have a feeling that this has something to do with which sub shells the electrons are removed from in copper and zinc...

You have to think about the whole process.  When a metal loses electrons to make a metal ion the following happens:

The metallic bonds holding the metal atoms together are broken.
The metal atom loses the electrons.
The resulting metal ion is hydrated.

In your analysis you are only focusing on step 2.  The enthalpy and entropy of the entire process factor into the reduction potential.


Potassium vs sodium bicarbonate in fire fighting Why is it that potassium bicarbonate reduces flashback as opposed to sodium bicarbonate when fighting fires?

From what I understand, high heating of bicarbonate salts decomposes them into carbon dioxide, which helps suffocate the fire, and the corresponding metal hydroxide. Could it be potassium hydroxide that is interacting differently with the fire versus sodium hydroxide that causes this issue?

To answer this question, firstly we must understand both how fires are started, and how they are perpetuated. It is common knowledge that in order for fire to exist there must be heat, fuel, and an oxidizing agent (usually oxygen). Once started, a fire is then self perpetuated by a radical chain reaction. Removing any of these three ingredients or interrupting the radical chain reaction can be effective ways of extinguishing a fire. Both sodium and potassium bicarbonate are effective at performing several of these actions, though there are slight differences between them.

Differences between $\ce{NaHCO3}$ and $\ce{KHCO3}$

  It is widely believed that fireextinguishing powders can function as both energy-absorbing materials and solid surfaces on which free radicals can be destroyed. Heat may be absorbed by the heat capacity of the solid [$C_p(\ce{KHCO3})=90.05\:\mathrm{J\:mol^{-1}\:K^{-1}}$ vs. $C_p(\ce{NaHCO3})=87.61\:\mathrm{J\:mol^{-1}\:K^{-1}}$], the heat of fusion, the heat capacity of the liquid, heat of dissociation from breaking of chemical bonds [$\Delta H_\mathrm{decomp}(\ce{KHCO3})=1063.28\:\mathrm{kJ\:mol^{-1}}$ vs. $\Delta H_\mathrm{decomp}(\ce{NaHCO3})=995.72\:\mathrm{kJ\:mol^{-1}}$], and heat of vaporization. All of these contribute to the total endothermicity of the fireextinguishing powder.

  From a chemical aspect, it has been found that potassium salts are more effective than sodium salts, and iodide anions are more effective than chloride anions. Presumably, there is a catalytic path for destruction of free radicals, such as $\ce{H}$, $\ce{O}$, and $\ce{OH}$, utilizing the potassium in the salts. It must be remembered that any powder that has a chemical fireextinguishing capability will also have a heat-absorbing (endothermic) capability.$^{[1]}$

The Chain Radical Mechanism

  Combustion of hydrocarbons is thought to be initiated by hydrogen atom abstraction (not proton abstraction) from the fuel to oxygen, to give a hydroperoxide radical ($\ce{HOO}$). This reacts further to give hydroperoxides, which break up to give hydroxyl radicals. There are a great variety of these processes that produce fuel radicals and oxidizing radicals. Oxidizing species include singlet oxygen, hydroxyl, monatomic oxygen, and hydroperoxyl. Such intermediates are short-lived and cannot be isolated.$^{[2]}$

The Chain Radical Breaking Mechanism

  The mechanism by which chemical inhibition occurs when certain powders are added to fuel-air flames is not completely understood. Solid particles may scavenge chain propagating species by surface adsorption or reaction; or alternatively, the solids may vaporize in the combustion zone to produce gaseous products that react with active species via homogeneous gas reactions.

  Most investigators seem to favor the latter mechanism, although direct experimental verification is very lacking. If homogeneous gas reactions are, in fact, responsible for the inhibition, then for cases in which potassium salts are the inhibiting agents, the chemical reactions

  $$\ce{KOH(g) + H->K(g) + H2O}\:\:\:\:\:\Delta H = -33.2\:\mathrm{kcal\:mol^{-1}}$$
  $\hspace{78 mm}$and/or

  $$\ce{KOH(g) + OH -> KO(g) + H2O}\:\:\:\:\:\Delta H = -1.7\:\mathrm{kcal\:mol^{-1}}^{[3]}$$

[1] Fire Extinguishing Powders

[2] Oxygen Fueled Combustion Reaction Mechanism

[3] Flame Inhibition by Potassium Compounds


Major product of radical fluorination vs bromination I understand what structure is formed by using $\ce{Br2}$ to brominate this alkane, and I know that $\ce{F2}$ will be less selective in its abstraction of hydrogen. However, how can I know where $\ce{F2}$ will primarily react with this alkane? I just know that $\ce{F2}$ will react more readily and be less selective.

There is nearly no thermodynamic aspect to this reaction. The activation energy for any alkane hydrogen abstraction by a fluorine radical is essentially 0$^{[1]}$. This means that there is an equally likely chance of fluorinating at any given position. In this case, we need only know the proportions of homotopic hydrogen. Homotopic hydrogen are labelled with the same colors in the molecule:

The proportion in decreasing order is 9:6:1:1:1. That means that we should expect about the following yield of monofluorinated products:

As an aside, you would never realistically expect to obtain a monofluorinated product. The fluorination of alkanes is violent, extremely fast, and only gets faster upon fluorination.

$^{[1]}$ Chemistry StackExchange, Why are free radicals so reactive?


What acid can dissolve thorium dioxide? In one of my projects (which is about alpha-voltaic cells, don't worry, I know that thorium isn't a strong alpha source, it's only for measurements) I would need to get water soluble thorium compound from a small sample of thorium dioxide  for purification with selective reactions and further measurements. Would any mineral acid work? I found some sources and publications on the internet by google searching but the biggest amount of information I could get is that

  "...Several hot acids can solve it..."

Which is not nearly as accurate as I would need it because I don't want to spend the budget of the project on expensive acids which doesn't even work instead of better equipments. (Their price is from their necessary purity)

Thorium dioxide can be dissolved in the mixture of hydrofluoric acid and hydrochloric acid, as described by Kluger and Lieser in 1981.1

Proceed with caution since hot condensed acid are known to be highly corrosive and hydrofluoric acid is also highly toxic and penetrating.

Kluge, E.; Lieser, K. H. Dissolution of ThO2 in HCl/HF-mixtures. Radiochim. Acta 1981, 29 (2–3), 167–168. DOI: 10.1524/ract.1981.29.23.167.


The Solubility of Tyrosine I find very little data on the solubility of tyrosine, other than that it has a very low solubility at room temperature for an AA. (1)
Does anyone know different ways to dissolve fair amounts of it in water (preferred) or an organic solvent?
I suspect I could increase its water solubility by decreasing the $\mathrm{pH}$ and increasing temperature, but I'm not sure what is the best way to do it, which would be the most effective, and by how much it would increase the solubility.

(1) https://pubchem.ncbi.nlm.nih.gov/compound/Tyrosine#section=Solubility

According to Product  Information of Sigma-Aldrich for L-Tyrosine, the solubility is $\mathrm{pH}$-dependent:

Preparation Instructions: This product is soluble in $\pu{1 M} \ \ce{ HCl}$ ($\pu{100 mg/ml}$), with heating. The solubility in water ($\pu{25 ^\circ C}$) is $\pu{0.45 mg/ml}$ in the pH range $3.2-7.5$ (Ref.1). At $\mathrm{pH}$ $1.8$, the solubility is $\pu{2.0 mg/ml}$; at $\mathrm{pH}$ $9.5$, the solubility is $\pu{1.4 mg/ml}$; and at $\mathrm{pH}$ $10$, the solubility is $\pu{3.8 mg/ml}$.

The above claims can be justified by the study of the solubility of tyrosine in the acidic and basic conditions (Ref.2), the summery of which states that:

Measurements have been made of the solubility at $\pu{25 ^\circ C}$ of tyrosine in hydrochloric acid and in sodium hydroxide solutions varying from $0.001$ to $\pu{0.05 M}$, and also in distilled water. The $\mathrm{pH}$ of the saturated solutions was measured with the hydrogen electrode. The following values for the ionization constants of tyrosine have been obtained from the measurements: $k_\mathrm{b} = 1.57 \times 10^{–12}$, $k_\mathrm{a1} = 7.8 \times 10^{–10}$, $k_\mathrm{a2} = 8.5 \times 10^{–11}$. The changes in solubility with $\mathrm{pH}$ can be satisfactorily explained by the use of these ionization constants.

According to plot of solubility versus $\mathrm{pH}$ of their results, it was concluded that the highest solubility of tyrosine can be achieved in the $\mathrm{pH}$ range of $1-2$ or in the $\mathrm{pH}$ range of $9-10$. For example, the concentration of tyrosine in $\pu{0.05 M}$ $\ce{HCl}$ is found to be $\pu{0.0165 M}$ with the $\mathrm{pH}$ of solution being $1.450$. And also, the concentration of tyrosine in $\pu{0.0498 M}$ $\ce{NaOH}$ is found to be $\pu{0.0358 M}$ with the $\mathrm{pH}$ of solution being $9.953$. The least solubility of tyrosine is shown to be with in the $\mathrm{pH}$ range of $3-8.5$ (Ref.2).
Note: The extrapolation of the acidic side of above graph has shown the solubility increases up to about $\pu{0.04 M}$ at $\mathrm{pH} = 1$ (Ref.2).
References:

T. A. Brown, In Molecular Biology: LabFax; 1st Edition; Bios Scientific Publishers: Oxford, England, 1991, p. 29 (ISBN: 1‐872748‐00‐7).
David I. Hitchcock, "The solubility of tyrosine in the acid and in alkali," Journal of General Physiology 1924, 6(6), 747–757 (https://doi.org/10.1085/jgp.6.6.747)(PDF).


Why don't elements having the same valency as Carbon bond so readily? I haven't been in touch with Chemistry as a subject of study for several years now. Still- if I remember correctly, Si, Ge, Sn, Pb all have same number of electrons as Carbon in their outer orbits.

So why is Carbon so ubiquitous? I believe the number of electrons an element can give or accept decides how readily it bonds with other elements. For instance, inert gases have no room for extra electrons and hence, are inert.

So why don't other elements in the column of Carbon bond as easily? In other words, why aren't we made up of Pb primarily instead of C?

I remember asking this question to my high school chemistry teacher, but as far as I can remember, he wasn't able to give a satisfactory answer. Forgive me for asking this , if this question is too naive. But I genuinely wondered a lot about it back in my school days and still don't know the answer.

It's not that they don't bond readily, it's that they can't form as stable of networks as carbon. Carbon has many different allotropes, while silicon has only a few. As you can see, carbon forms uniform hexagonal sheets with massive $\pi$ systems as in graphene and many of its other allotropes, while silicene is incapable of remaining flat and is bumpy. This is a result of the bond distances and p-orbital overlap.

$$
\begin{array}{|c|c|}\hline
\text{Bond} &\ \Delta H ^\circ_f \text{ (kcal mol}^{-1}) &\ \text{Bond Length (pm)} \\
\hline
\ce{C-C} &\ 82.70 &\ 154 \\
\hline
\ce{Si-Si} &\ 53.06 &\ 233 \\
\hline
\ce{Ge-Ge} &\ 44.93 &\ 241 ^{[1]} \\
\hline
\end{array}
$$

As you can see, as the atoms get larger, the bond length increases. $\pi$ bonding is completely reliant on the overlap of adjacent p-orbitals, so the farther away these orbitals are, the weaker the bonds.

Bond dissociation enthalpy also increases with atom size. This leads to an overall decrease in the stability of larger silicon or germanium compounds. Tin and lead are metals, which have much different chemistry altogether on account of their electronegativity and occupied d-orbitals, in addition to their size.

${[1]}$ Bond Enthalpy Table


How to estimate reaction products of competitive binding? I have two reactions,

$$\ce{A + C \rightleftharpoons AC}$$

$$\ce{B + C \rightleftharpoons BC}$$

I'd like to estimate the relative steady state proportions

$$\frac{[\ce{BC}]_{SS}}{[\ce{AC}]_{SS}+[\ce{BC}]_{SS}}$$

and

$$\frac{[\ce{AC}]_{SS}}{[\ce{AC}]_{SS}+[\ce{BC}]_{SS}}$$

without numerically solving any differential equations, given $[\ce{A}]_{init} = [\ce{B}]_{init}$ and thermodynamic properties: $\Delta G, \Delta H, $ and $\Delta S$.

I'm hoping that there is a way to generalize this for more than two reactions in a linear way. It is for a DNA microarray optimization problem in which computational speed is important. Any thoughts?

Why would you have to solve differential equations for this problem?  What algebra have you tried so far?

If C is in excess, then there is no competition!  You should be able to model the reactions as just $\ce{A <=> AC}$ and $\ce{B <=> BC}$.  The equilibrium equation would be (for A) $K_a=\frac{ac}{A_{init}-ac}$, and you can apparently already calculate $K_a$ from the thermodynamic parameters, so that is an equation where the only unknown is $ac$, the amount of AC formed.  The equation is linear!

$ac=A_{init}\frac{K_a}{1+K_a}$

A similar result holds for B of course.

Once you can calculate $ac$ and $bc$, you should be able to calculate the parameters you want to know very easily.

If C is not in excess, the equilibrium equations will not be linear and the balance on C will couple the two equilibria (competition happens!).


What is a non-classical carbocation? What is a non-classical carbocation?
How is it different from a classical carbocation?

I am confused as I have come across this term many times on Chem.SE but there seems to be nothing for my level of understanding on the Internet!

Here is a picture of a "classical" carbocation, there is an electron deficient carbon bearing a positive charge.

There are many examples of "non-classical" carbocations, but the 2-norbornyl carbocation is among the best known.

Labeling experiments have shown that the positive charge resides on more than one carbon in the 2-norbornyl ion.  Early on, the data was explained by equilibrating classical ions, but soon another possibility emerged - one involving a single non-classical ion.

The problem comes down to: are the equilibrating classical ions ground state structures with the non-classical ion serving as the transition state, or is the non-classical ion the ground state? This debate went on for a very long period of time, but now most agree that the non-classical structure is the ground state in the 2-norbornyl system.  In fact, a recent, and difficult to obtain, crystal structure for the 2-norbornyl cation has been published proving that the ion exists with the non-classical geometry (thanks to Klaus for finding this reference, see his comment below).

A key difference between classical and non-classical structures is the bonding.  As illustrated above, a classical ion has a carbon with a sextet of electrons and 3 other bonds.  The non-classical ion, on the other hand, involves 3 carbons with 2 electrons spread over them.  This is called a 3-center 2-electron bond (hypercoordinate bonding) and is a clear marker for a non-classical ion.  Notice if you count all of the bonds to any of these 3 carbon atoms (solid and dashed lines) you count 5!  Sounds strange, but such "hypercoordinate bonding" is a permitted consequence of the 3-center 2-electron bond.

Here is a link to a simpler non-classical carbocation discussed here yesterday.


Identifying pentane isomers Question
Which is not considered an isomer for pentane?
A) $\ce{CH3CH(CH3)CH2CH3}$
B) $\ce{CH3CH2CH(CH3)CH3}$
C) $\ce{CH3CH2CH2CH2CH3}$
D) $\ce{CH3CH=CHCH2CH3}$

I knew A and B were obviously isomers. I knew D was not an isomer, because the definition of isomers in my book states that isomers "are two DIFFERENT compounds with the SAME molecular formula" and that structural isomers are compounds with SAME molecular formula but DIFFERENT structure.
I was very confused with the choice C since it was IDENTICAL and did not match the definition. What confused me more is that there were practice questions in the book that gave like 4 examples of compounds and asked if  they are isomers or identical compounds.
So, are identical compounds considered isomers or not?

Identical compounds are not considered to be isomers. However, the following explanation should clear up any confusion from your question.

A and B are both isopentane and C is n-pentane (normal pentane). These make up two of the three isomers of pentane, the third being neopentane ($\ce{CH3CH(CH3)2CH3}$). The question is referring to pentane as $\ce{C5H12}$, a group of isomers in which n-pentane ($\ce{CH3CH2CH2CH2CH3}$) is included and not identical to.


How to find the mass of the solute given a concentration as a mass fraction? Formalin is a $37\%~\mathrm{W/W}$ solution of methanal in water. What mass of water must be added to $250~\mathrm{g}$ of methanal to make formalin.

Here's what I've done so far. It's wrong though. According to my book answer is $430~\mathrm{g}$ $\ce{H2O}$.

$$\frac{Cw}{w} = \frac{\mathrm{m\ solute}}{\mathrm{m\ solution}} \cdot 100\%$$

$$37 \% = \mathrm{\frac{m\ solute}{250} \cdot 100\ \%}$$

$$\mathrm{\frac{37}{100} = \frac{m\ solute}{250}}$$

$$\mathrm{0.37 \cdot 250 = m\ solute}$$

$$\mathrm{m\ solute = 92.5\ g}$$

Where have I went wrong? Am I completely off or at least close?

You have been given a mass fraction, $w_{\ce{CH2O}}=37\%~\mathrm{w/w}$, and the mass of the solute, $m_{\ce{CH2O}}=250~\mathrm{g}$. You calculated the mass of the solution wrong.
The mass fraction is defined as
$$w_i = \frac{m_i}{m}\tag1$$
with $m$ being the total mass and therefore
$$m = \sum_i m_i.$$
The latter is also the mass of your solution, the sum of the solute and the solvent.
$$m=m_{\ce{H2O}}+m_{\ce{CH2O}} = m_{\ce{H2O}} + 250~\mathrm{g}$$
Substitute what you know into $(1)$ and transform:
\begin{align}
w_{\ce{CH2O}} &= \frac{m_{\ce{CH2O}}}{m_{\ce{H2O}}+m_{\ce{CH2O}}}\\
0.37 &= \frac{250~\mathrm{g}}{m_{\ce{H2O}} + 250~\mathrm{g}}\\
m_{\ce{H2O}} &= \frac{250~\mathrm{g}}{0.37}-250~\mathrm{g} = 426~\mathrm{g}
\end{align}


What is the appropriate method to visualize electrostatic potentials? Using a molecular viewer like PyMOL, I understand that its often incorrect to simply look at positive and negative residues and use those residues to calculate the charge density and instead, you should be using something like APBS to calculate the electrostatic potential.

What is the correct way of going about this?

One way to do it is to perform a quantum chemistry calculation of the electrostatic field generated by the molecule. Another is to determine (or assume) partial atomic charges for each of the atoms of the system, and use a Poisson solver (as I believe APBS is) to calculate the electrostatic field generated by this point charge distribution.


Best way to remove Aluminium from "silvered" glass? Requirement is simple(!) - I have 30mm silvered glass balls that are originally Christmas tree decoration that I want to use in a very unconventional manner in a physics experiment. As far as I can tell the Aluminium is external. The balls are exceptionally delicate and the glass probably somewhat stressed i.e. they break easily if scratched.

So, would dilute NaOH be suitable or is there a better solution?
Another requirement is that ideally the treatment does not raise the surface energy of the glass

Heating the bulb in a dilute NaOH solution would probably work, yes. Something that might be more 'interesting', and potentially safer (with less hydrogen) though, is to acquire a bit of gallium.

Gallium infiltration is a really fascinating phenomenon. Simply melt a little bit of gallium and dunk your bulb in it, and watch as the aluminum becomes as soft as wet tissue paper. Gallium also melts at a temperature very slightly above that of room temperature, meaning that the the chance of the glass bulb cracking because of heating would be very minimal.


Influence of solubility on the complexation constants at different pH values Given some small chemical compounds whose solubility varies with pH, and in the case of compounds that can form complexes, I have seen that the complexation constant varies with pH.

I think there might be two main reasons for this:

Chemical structure of the host system is affected by pH (for instance, protonation) and therefore it affects complexation constant;
It might be related to the small chemical compound, since its solubility might vary according to pH.

So I guess that in this case there are many chemical/physical equilibria involved, in this case mainly phase solubility and protonation for both species. I wonder what is the best approach for describing this, perhaps one can use different equilibrium constants for the mentioned equilibria, and from there derive a global constant that depends on pH. How do these two factors (protonation and phase/solvation) influence the value of the complexation constant?

As a particular case, I am mostly interested in complexation of flavonoids with cyclodextrins.

It is possible to calculate out all of the complexation and speciation reactions using the concepts of Aquatic Chemistry. I’m a PhD student in environmental chemistry and I TA a course in Aquatic Chemistry. Usually, when pH matters it’s because the protonation state controls the chemistry and can be described using equilibrium constants. I bet that your solubility issue is related to the deprotonated form being an anion and thus more soluble in water than the protonated, neutral form (very common for organic compounds). The protonation state can be calculated from the $\mathrm{p}K_\text{a}$:

$$
\ce{L- + H+ <=> HL}\\
K_\text{a}=10^{-\mathrm{p}K_\text{a}}=\dfrac{[\ce{L-}][\ce{H+}]}{[\ce{HL}]}
$$

Where $K_\text{a}$ is the equilibrium constant, and L is your compound. I would guess that your flavanoids and cyclodextrins might have multiple $\mathrm{p}K_\text{a}$s, you can write an expression like this for each one.

Complexation reactions can be written in a similar way. Let’s say that $\ce{Y}$ is the molecule your $\ce{L}$ molecule complexes with, and it is neutral when protonated ($\ce{HY}$).

$$
\ce{L- + HY <=> HLY-}\\
K=\dfrac{[\ce{L-}][\ce{HY}]}{[\ce{HLY-}]}
$$

Where $K$ is the equilibrium constant for this complexation reaction. Keep in mind that which species (protonation state) is in the expression matters, the value for the equilibrium constant is only valid for those species. Above, I’ve written that the deprotonated form of L ($\ce{L-}$) complexes with the protonated form of Y ($\ce{HY}$). This is where pH comes in. pH controls how much of L is deprotonated and how much of Y is deprotonated and able to participate in complexation, and $K$ controls how much of the available $\ce{L-}$ and $\ce{HY}$ form the complex. Complexes themselves can also have $\mathrm{p}K_\text{a}$s the one above ($\ce{HLY-}$) could probably be protonated to a neutral form $\ce{H2LY}$.

You might be able to guess that the mathematics gets a little complex at this point. It is possible to do these calculations by hand, but there are computer programs that can handle them. The most easily accessible one is Visual MinTEQ. The other hard part is having the data for the equilibrium constants, there may be data for your system, but you’ll have to look.

Just to summarize. Complexation constants don’t change with pH (they are, in fact, constant) it’s just that the speciation matters, and speciation is controlled by pH. The ‘global constant’ you refer to is pH, not something else.


Behaviour of element 111 Is element 111 considered to act as an eka-aurum?
Being under the same column of group 11, which usually provides enough evidence for correlation of properties for an eka-element, would this element have the similar properties such as conductivity or inertness with gold?

With the understanding of the correlation between group 11 being minimal to me, is anyone able to predict perhaps the colour, ductility, or other physical properties of this element with reason?

As far as I understand, there has been very limited experimental investigation of the properties of roentgenium (Z=111), if at all, simply due to the fact that so few atoms have been produced and they decay so quickly. It is possible to do some chemistry with even single atoms in seconds, but as you might expect it's an extremely arduous endeavour and is riddled with massive error bars. For now, just the synthesis of the element seems to dominate experimental roentgenium research.

There has, however, been ample theoretical investigation of roentgenium's chemical properties, both by extrapolation of behaviour for group 11 elements and by ab initio computational methods, at varying levels of approximation. One difficulty is that roentgenium displays strong relativistic effects, which make accurate calculations harder. Of course, atomic properties are easier to predict than molecular properties, which are themselves much easier to predict than condensed phase properties. So while estimates for ionization energy for atoms may be relatively accurate, physical properties of the pure materials such as ductility or melting point are to some extent little more than informed guesses.

Here is a list of some predicted basic properties of the period 7 $d$-block elements, including roentgenium (source, p. 1691):

That source expands upon some of the general properties of roentgenium. Its electronic configuration is predicted to be $\mathrm{[Rn](5f)^{14} (6d)^9 (7s)^2}$, which is dissimilar to the configuration for gold $\mathrm{[Xe](4f)^{14} (5d)^{10} (6s)^1}$. This is because relativistic effects cause a strong stabilization of $s$ orbitals (now more accurately referred to as $s_{1/2}$ due to spin-orbit coupling) by pulling them closer to the nucleus, while simultaneously destabilising the $d$ orbitals (as well as splitting the $d$ subshell into two groups of degenerate orbitals, two $d_{3/2}$ orbitals and three $d_{5/2}$ orbitals) by pushing them away. This means complete population of the $s$ subshell becomes preferable as relativistic effects get stronger. For example, see the effect here (ibid, p. 1667) in the triad Nb/Ta/Db in group 5.

Roentgenium is expected to be a noble metal, like gold, and in fact from its predicted standard reduction potential, is more noble than gold ($\mathrm{Au^{3+}(aq) + 3\ e^{-} \longrightarrow Au^0(s), \ \ \Delta E^0=+1.52\ V}$, for comparison). That said, once the roentgenium atom is ionized to Rg(III), it can reach higher oxidation states more easily due to less stable filled $6d_{5/2}$ orbitals, and so Rg(V) compounds are expected to be more stable. Interestingly, though in the 6th period transition metals gold (group 11) shows the local "relativistic maximum" with respect to stabilization of the $6s$ subshell, in the 7th period the maximum stabilization of the $7s$ subshell shift to copernicium (group 12) rather than roentgenium. This means roentgenium is expected to be less noble than copernicium.

With respect to physical properties, not much is predicted about roentgenium. It will likely be a very dense metal, with good electrical conductivity. The question about its colour is an interesting one. Wikipedia mentions it is expected to be silvery, even though roentgenium displays stronger relativistic effects, which is likely the cause of the colour of gold. This may be related to the switch in electronic ground state between gold and roentgenium. Unfortunately the source for the claim on Wikipedia is no longer accessible.


Nucleophillic addition vs SN2 The following question came in Paper 2 of JEE Advanced 2014-

Here we have $\ce{CH3-}$ as the nucleophile. There are 2 possible attacks:

SN2 type attack on the chlorine group.
Nucleophilic addition on the carbonyl.

The answer given is option D, which will happen by the 2nd pathway(nucleophilic addition).

Why is nucleophilic addition given a preference over SN2? I feel that nucleophilic addition should rather decrease the reaction rate because the carbon goes from sp2 to sp3, decreasing the bond angle and hence increasing the repulsion

To begin this short discussion one has to differ between Sn2 and addition to the carbonyl group, as the former is driven by orbital interactions while the latter is driven by coulomb interactions.
This can be explained with a closer look at the carbonyl function.
due to the highly polarized π-bond between carbon and oxygen, carbon is rendered quite positive. It results that any nucleophile bearing a negative charge will be attracted to attack it. In addition the reaction of a carbanion and a ketone or aldehyde is irreversible!
there is fundamentally no way in which the carbanion could ever be considered as a leaving group.

On the other hand Sn2 reactions appear on saturated carbon (in the best case methyl or primary) with at least one good leaving group (e.g. a halide or tosylate). Often the sigma bond connecting carbon and the leaving group is not even polarized, at least not significantly. Just consider the electronegativity of iodine, the BEST leaving group for SN2.
That is why orbital interactions are more dominant in thess kind of reactions. It can be easily noticed that a potential nucleophile has to donate its electrons into a sigma antibonding orbital in order to  react with the carbon electrophile. These kind of orbital can be considered as very high in energy as it is the counterpart of the sigma bonding orbital which is quite stable.
Therefore the best nucleophiles for nucleophilic attack on saturated carbon is usually a second or third row nucleophile like sulphur or phosphorus (Wittig reaction).

Therefore for a carbanion the irreversible addition to the carbonyl group happens much faster, resulting in a alkoxid. The new formed alkoxid is in close proximity to the chlorine substituted carbon and therefore likely to form the five membered ring .

Any references that support everything written above can be found in Clayden, Organic chemistry second edition.
Specifically the chapter nucleophilic substitution at saturated carbon will help.

Best regards,
John


Why is a drop of chloroform stretched/pulled apart on a water surface? I distributed drops of chloroform (with a lipid in them) on water (in a Langmuir-Blodgett trough). The chloroform drops form a layer on the water (so that the lipids are distributed on the water surface) and I need to explain why. (Afterwards, the chloroform evaporates, leaving back a lipid monolayer.)
I was told that the water pulls each chloroform drop apart and that this has to do with the fact that the surface tension of water to air is greater than the sum of the surface tension of chloroform to air and surface tension of chloroform to water.
Can someone explain how this works in more detail?

As you pointed out, the surface tension of water-air interface is greater than the surface tension of chloroform-air interface. At $25~^\circ\mathrm{C}$, the surface tension of water-air interface is $71.99\pm0.05~\mathrm{mNm^{-1}}$ and that of chloroform-air interface is $26.67~\mathrm{mNm^{-1}}$ (source).
Now, let's understand why a drop of chloroform placed over water gets puller apart forming a layer. Surface tension is the force per unit length. More the surface tension, more is the force per unit length and vice versa. Surface tension is similar to tension in a stretched rubber sheet. As you place a drop of chloroform on water, the system looks as shown in the following diagram:

As we seen earlier, surface tension of water-air interface is greater than that of chloroform-air interface. So in the boundary separating chloroform and water there exists an imbalance of force. The force vectors are the red arrows in the diagram above and their strength is denoted by their length and weight (thickness).
Due to this, the drop of chloroform expands to form a layer on top of water. This can also be looked in the nature's tendency to lower energy. The surface energy of chloroform-air interface is lesser than that of water-air interface. So, a lower energy configuration is preferred.


How much gas is inside a pressurised gas cylinder? Given a 19kg Liquid Petroleum Gas (LPG) cylinder, with this specification:

LPG inside the cylinder weighs 19kg.
The internal volume of the cylinder is 44.5 litres.
The LPG is a mixture of propane gas (60%) and butane (40%).

I would like to calculate the volume of LPG contained in the cylinder. This is the volume the pressurised gas will uncompress to, if it escapes the cylinder.

I have looked at the Ideal Gas Law:

Pressure x Volume = Number of Moles x Gas constant x Temperature

But I cannot use it as I do not know about the pressure inside the cylinder and volume is the unknown I am interested in.

How do I proceed with this problem?

The cylinder contains $\pu{19kg}$ of LPG, $\pu{60\%}$ of which is propane (presumably by mass) so we have $\pu{11.4kg}$ propane and the rest is butane.

The molar mass of propane $\ce{C3H8}$ is $\pu{3*12+8*1 = 44 g/mol}$, so we can work out how many moles of propane there are. Then do the same for butane.

Now we use the ideal gas law:

  But I cannot use it as I do not know about the pressure inside the cylinder and volume is the unknown I am interested in.

You are not interested in the pressure inside the cylinder. You are interested in what the volume is when the LPG uncompresses. Therefore the pressure you need to use is atmospheric pressure.

If you apply the formula for the ideal gas law $PV=nRT$ you will find that $1$ mole of gas occupies about $\pu{24 L}$ at $\pu{25^oC}$ ($\pu{298K}$) and atmospheric pressure. That is a good shortcut to remember for cases like this.

Note that the ideal gas law only works when the conditions are in the range where the gas behaves like an ideal gas. You know the volume of the cylinder, but you cannot use the ideal gas law to find the pressure inside the cylinder, because the gas behaves non-ideally at high pressure (It liquifies!)


When is the carbaldehyde suffix correct? Is benzene carbaldehyde a correct name for Benzaldehyde? I'm (always) getting confused between common and IUPAC names. If the above name is correct then is it a common one or IUPAC? And why don't we use it more often?

Both names, ‘benzenecarbaldehyde’ and ‘benzaldehyde’, are unambiguous and describe the same compound. However, the preferred IUPAC name (PIN) is ‘benzaldehyde’.
Concerning aldehydes, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-66.6.1 Systematic names of aldehydes
Aldehydes are systematically named in three ways:
(1) substitutively, using the suffixes ‘al’ for $\ce{-(C)HO}$ and ‘carbaldehyde’ for $\ce{-CHO}$;
(2) by changing the ‘ic acid’ or ‘oic acid’ endings of retained names of carboxylic acids into ‘aldehyde’; the nomenclatural properties of acids are transferred to aldehydes; thus, preferred names of aldehydes correspond to preferred names of acids, and carboxylic acids that are not substitutable generate nonsubstitutable aldehydes;
(3) by using the prefixes ‘oxo’, denoting $\ce{=O}$, or ‘formyl-’, denoting the substituent group $\ce{-CHO}$.

In particular,

P-66.6.1.1.3 The suffix ‘carbaldehyde’ is used when the $\ce{-CHO}$ group is attached to a carbon atom of a ring or ring system, or to a heteroatom.

For example, ‘cyclohexanecarbaldehyde’ is a PIN.

However, according to P-66.6.1.2.1, the PIN for the compound given in the question is ‘benzaldehyde’. (This name is analogous to the preferred name of the corresponding acid ‘benzoic acid’, see P-66.6.1 (2) above.)

Nevertheless, according to P-66.6.1.2.2, the PINs for phthalaldehyde and terephthalaldehyde are the systematic names ‘benzene-1,2-dicarbaldehyde’ and ‘benzene-1,4-dicarbaldehyde’, respectively. (These names are analogous to the preferred names of the corresponding acids ‘benzene-1,2-dicarboxylic acid’ [not ‘phthalic acid’] and ‘benzene-1,4-dicarboxylic acid’ [not ‘terephthalic acid’], respectively.)


How do electrostatic analysers act as kinetic energy analysers? Consider a mass spectrometer. The ions produced are accelerated across a potential V.

Now, the kinetic energy of an ion with charge q is given by qV. When it enters the analyser with electric field strength E (orthogonal to the trajectory), the radius of the trajectory is given by

r = 2K/qE, where K is the kinetic energy.

Substituting with qV, we get

r = 2V/E

The mass and charge terms disappear! So how exactly is this a kinetic energy analyser? Because as per this equation, all ions, irrespective of charge, mass and kinetic energy must have the same trajectory. Where is the flaw in my reasoning?

The point is that all ions produced in the ion source will not have a kinetic energy outside of the acceleration region equal to $q V$. When produced the ions will most probably be produced at (slightly) different positions in the acceleration field, with a small $\Delta V$ potential deviation from the average acceleration potential, and with a thermal kinetic energy distribution ($E_{kin,thermal}$).

In these conditions, the real kinetic energy of the ions as they exit the ion source will be slightly different:

$$E_{kin,real} = q V + q \Delta V + E_{kin,thermal}$$

and the purpose of the electrostatic analyser is to separate the ions depending on their actual kinetic energy and thus compensate for the distribution of kinetic energies of the ions exiting the ion source.


How do indicators work when they mixed with each other? Methyl orange, bromothymol blue and phenolphthalein indicators were mixed together to form a solution.

Over what pH range would the solution be yellow?

I cannot understand the question. If two or more different indicators are mixed, do they still work as indicators? Because each their colours will blend with each other and result to formation of a new colour, how can we know what the color will be?

In general, the chemistry of one indicator does not usually affect another as far as I know, but the colors will mix as you would expect. It gets tricky here because the question shows the colored ranges but leaves a blank in the transition pH ranges, but I would guess the answer they're looking for is from 3-7.5


Does someone know the name of this equipment? I think it is used to measure force or something like that. I broke it and have to buy a new one.

There are different equipment used to measure force like pivot balance, strain gauge, spring balance, etc.

By seeing the image, we can easily figure out it is spring balance.


Why does HF generate a carbocation with an alkene I was reading of different ways to perform Friedel Craft alkylations.
One of the methods I came across involved an alkene in the presence of H+ or a mineral acid like $\ce{HF}$. Here $\ce{H+}$ or $\ce{HF}$ caused an electromeric shift of the pi electrons and the $\ce{H+}$ bonded there resulting in a carbocation.
However I felt like the bond strength of $\ce{HF}$ is so high that it couldn't be possible to cleave it for the donation of $\ce{H+}$ (If that's what happens. Correct me if I'm wrong).
Why does it still generate a carbocation?

However I felt like the bond strength of $\ce{HF}$ is so high that it couldn't be possible to cleave it for the donation of $\ce{H+}$

HF has the highest single bond energy and hence you were right in expecting that it may not dissociate.However in concentrated solutions, due to the formation of stable hydrogen bonded complex $\ce{[HF_2]-}$, the reaction is pushed in the forward direction making it  a resonably strong acid (in fact strongest among corresponding concentrated solutions of all the hydrogen halides).So, you could generate a carbocation with that.However, as generally dilute solutions are used which are economically feasible, more stronger mineral acids in dilute solutions  like $\ce{HCl, HBr}$ and $\ce{HI}$ are used.
EDIT:
As suggested by Oscar Lanzi in the comments, acids like $\ce{HCl, HBr}$ and $\ce{HI}$ are safer to handle when compared to $\ce{HF}$ as the reactions of $\ce{HF}$ are associated with highly negative enthalpy changes [highly exothermic]


What happens when you 'burn' ash? This popped into my my mind while watching a match burn (don't ask me what I was doing with a burning match...).

Now as I know it, 'ash' is what you call the residual, grey-black powdery material left over following the combustion of wood.

I'm under the impression that ash is largely just carbon, with little or no organic constituents present along with it (since I'm pretty sure that pure powdered carbon is black, not grey-black).

Now I heated what remained of the matchstick with, well...another matchstick and found out, unsurprisingly, that the ash underwent practically no visible change. So I scooped up some ash into a crevice in a concrete block and then blow-torched it for about 2 minutes. Same result. However, while I was heating it this time, it glowed orange, it subsided as soon as I turned off the torch.

Thinking about it, if I did heat it strong enough it should decompose completely to black colored elemental carbon. If that is the case, to what temperature should you heat it?

I don't think I've considered everything there is to consider in this situation, which is why I've adopted a tentative tone while typing out this question.

Could there be other, side-reactions/effects as well? What would they be?

Additionally if anyone happens to know; what is it that contributes to the grey color that ash normally assumes?

The principal component of wood ash appears to be calcium carbonate. (Wiki entry, "wood ash".) Other components include compounds of potassium and phosphorus. If you heat the calcium carbonate strongly enough, it will decompose into CO2 and calcium oxide. Apparently, strong heating is likely to result in less carbon instead of a larger percentage. The remaining calcium oxide is pretty stable as to temperature although addition of water will readily convert it to calcium hydroxide.

Calcium carbonate is white. White mixed with the black of any carbon present would result in a grey color.


Why do compounds with similar properties have different Stitch IDs? I'm not from the Pharmacy or Chemistry background. I have been wondering why certain drug compounds have similar properties but each of them have different Stitch IDs in PubChem?

For example:

Stitch ID 9782 Betamethasone
Stitch ID 5743 Dexamethasone

Bethamethasone and dexamethasone are different chemicals, hence they behave different, therefore they have different identification numbers. The numbers you cite are not STITCH IDs, and as far as my research goes these do not exist. Instead they are CID - PubChem Compound Identification numbers.

As you can see the molecules are diastereomers. The stereo centre of the steroid differs at carbon-16. When you look them up with STICH 4.0 you will see differences with their predicted functional partners. While the first one is nearly identical, the second partner for betamethasone is proopiomelanocortin, while for dexamethasone it is ABCG2.

For more information on STITCH (‘search tool for interactions of chemicals’) I recommend: Michael Kuhn, Christian von Mering, Monica Campillos, Lars Juhl Jensen, and Peer Bork, Nucl. Acids Res. 2008, 36 (Database issue): D684–D688. (Open Access); PMCID: PMC2238848.

Image sources:

National Center for Biotechnology Information. PubChem Compound Database; CID=9782, https://pubchem.ncbi.nlm.nih.gov/compound/9782 (accessed Apr. 20, 2016).
National Center for Biotechnology Information. PubChem Compound Database; CID=5743, https://pubchem.ncbi.nlm.nih.gov/compound/5743 (accessed Apr. 20, 2016).


Relativistic effects and lanthanide contraction So I understand than the lanthanide contraction is due to poor shielding of the 4f electrons which decreases the radius.
However, if Im not mistaken the relativistic effects lead to a contraction of the core orbitals (s,p) while the f and d orbitals expand. So what I do not understand how do they match, one of them decreases the radius and the other one should expand it as the f orbitals expand right?
They usually say relativistic effects are one of the reasons for the lanthanide contraction
Thanks

The lanthanide contraction is caused by two effects

The 4f orbitals are very diffuse and therefore result in poor screening of the electrons further out, those in the n=5 and n=6 orbitals
relativistic effects

This Wikipedia article presents a nice discussion of the lanthanide contraction.  The article estimates that the first effect, the screening effect, is the major factor, with relativistic effects playing only a minor role (~10%) in the observed contraction.

  what I do not understand how do they match, one of them decreases the
  radius and the other one should expand it as the f orbitals expand
  right?

That's correct, the outer s- and p-orbitals contract due to the lanthanide effect, while d- and f-orbitals expand (see this SE Chem post for a more detailed description).  In the lanthanide series, the 6s electrons are further from the nucleus than the 4f electrons, therefore they (the 6s electrons) determine the radius.  A contraction of the 6s orbital radius results in a contraction of the observed lanthanide electron radius, even though the 4f electrons have moved slightly further away from the nucleus.


Why does Fe(II) have electrons in the 3d subshell? The electron configuration for $\ce{Fe^2+}$ in my coursebook is given as:
$$ 1\mathrm s^2\, 2\mathrm s^2\, 2\mathrm p^6\, 3\mathrm s^2\, 3\mathrm p^6\,3\mathrm d^6$$
But shouldn't the $4\mathrm s$ subshell take precedent over the $3\mathrm d$ subshell according to Aufbaus principle?

You might want to compare iron(II) not with neutral iron, but with neutral chromium.
Both iron(II) and neutral chromium have 24 electrons, but there the similarity ends:
$\ce{Fe^{2+}}:  [\text{Ar}]3d^6$
$\ce{Cr^{0}}:  [\text{Ar}]3d^54s^1$
Think of the Bohr model.  According to this model the energy level of electrons depends only on the shell number $n$, so we would expect 24 electrons to follow the $[\text{Ar}]3d^6$ configuration.  In real life that happens exactly only for single-electron atoms where there are only electron-nucleus interactions.  When there are electron-electron interactions they could fill the shells not in order, like the chromium atom described above.
But in a multielectron atom, if you add more nuclear charge you make the electron-nucleus interaction stronger, and the Bohr-model configuration becomes more favorable.  One might suppose that dozens of added protons might be needed to get chromium's $[\text{Ar}]3d^54s^1$ to the Bohr-predicted $[\text{Ar}]3d^6$.  Instead, the energies of the two configurations are so closely spaced that two extra protons, converting neutral chromium to ferrous iron, is enough.


Why does a NOx container get cold when NOx is released? and What is the relationship of the vapor pressure curve with the liquid-vapor dome? The temperature at which $\ce{NO_x}$ is in equilibrium with its liquid and vapor phase at 1 atm is about –84 °C. Does that mean $\ce{NO_x}$ exists as a liquid at that temperature? What about it's vapor phase then, because it's vapor and liquid phase should be at equilibrium, right?

Also, it's that the reason why compressed $\ce{NO_x}$ when released from its closed and pressurized container (e.g. 50 bar) at 20 °C becomes a gas while its temperature drops? Therefore, the compressed $\ce{NO_x}$ was in it's liquid state at 50 bar, right?

If that's the case, then what is the relationship between the vapor pressure curve and the liquid-vapor dome? How would you use the saturated liquid-vapor tables to predict the vapor pressure for your substance? And which specific volume would you chose to define your substance: its liquid specific volume or its vapor specific volume?

The vapor pressure curve describes the relationship between temperature and pressure for which the pure substance can exist as liquid and vapor, together in equilibrium.  So it is not necessarily all liquid or all vapor.  Both can be present in different proportions.  As far as specific volume is concerned, your tables should list both the vapor specific volume and the liquid  specific volume.  The specific volume of the combination of vapor and liquid is proportional to the mass fraction of each.  (So you need to know the mass fractions of vapor and liquid).


Esterification with tertiary alcohol I saw a different mechanism for esterification of tertiary alcohol in a book.

They say that first the $\ce{OH}$ is protonated and it leaves to form the tertiary carbocation. Then the oxygen's lone pair from the carboxylic acid attacks to form the ester.

Is this mechanism actually correct?

Yes, the mechanism is correct and is different from the normal mechanism because here the more stable tertiary carbocation formation.


Lowest CAS Registry Number on record I've been looking for a while and I still can't get a straight answer. I just need to know the lowest CAS Registry Number on record and the corresponding chemical.

I work at CAS and recently answered a similar question on Twitter (@CASChemistry). In fact, there are 8 pre-50-00-0 CAS Registry Numbers in our database:

35-66-5
35-67-6
36-51-1
36-88-4
37-71-8
37-82-1
37-87-6
38-26-6

So, the lowest CAS Registry Number on record is 35-66-5 for "Benzacridine, 7,9-dimethyl- (9CI)".


Is magnesium sulfate basic, neutral or acidic? A question in my college test was whether $\ce{MgSO4}$ is neutral, acidic or basic. I was told to solve this kind of problem by hydrolyzing any ions that do not "come from" a strong acid or a strong base, and if nothing is hydrolyzed, then the salt is neutral. So by this logic, as $\ce{Mg^2+}$ comes from $\ce{Mg(OH)2}$ and $\ce{SO4^2-}$ comes from $\ce{H2SO4}$, I answered "neutral".

My professor marked it as wrong, and said that the salt $\ce{MgSO4}$ is basic. How can you explain this?

Looking at the following references, it seems that most agree that a water solution of $\ce{MgSO4}$ is slightly acidic.

"The pH of an aqueous magnesium sulfate solution is related to the molarity of the MgSO4. Typically, the pH is between 5.5 and 6.5 due to magnesium's affinity for hydroxide ion (OH-). As the sulfate goes into solution, hydroxide anions associate with the magnesium, increasing the relative ratio of H+ to OH-. This shift results in more acidic solutions."

"The pH of a pure magnesium sulfate solution is approximately 6.4"

"The pH of [magnesium sulfate] hydrates is average 6.0 (5.5 to 6.5)."

"pH: Neutral to litmus"

Why not perform the experiment in class, using an accurate pH meter and/or indicators (see the pH indicator chart in Wikipedia for suggestions).


Comparison of atomic radii My teacher told me that Neon has a larger atomic radius than Fluorine.I am of the understanding that it is merely a consequence of the way we define the atomic radius and that we use Van der walls radius for neon which is larger that the covalent radius of fluorine.

He also said that the radius of a fluorine 1- anion is more than a neon atom by using the fact that they are iso-electronic species and that since neon has a higher nuclear charge, it will pull on the 10 electrons more than fluorine.

But what definition of atomic radius am i using for neon now? Aren't we always supposed to use the Van der walls definition for noble gases? In case we do use the Van der walls radius of the neon atom and the ionic radius of the fluorine anion, how am i supposed to compare these two and come to the conclusion that the fluorine anion is larger?

Atomic radii cannot really be uniquely nor accurately defined.  However, there is generally a very big difference between covalent, van der Waals, and ionic radii.

So, by most any definition of the respective terms the covalent radius of fluorine will be smaller than the van der Waals radius of neon.  Similarly, the ionic radius of fluoride ion will be larger than the van der Waals radius of neon. The reason for the latter is the higher nuclear charge on neon.

However, most sources list a slightly smaller van der Waals radius for fluorine compared to neon, but a slightly larger covalent radius for fluorine compared neon. So when comparing radii you have to carefully specify what kind of radii you are talking about.