The influence of diffuse functions on electron density calculations I am not a quantum chemist, but I am (starting to be) a user of electronic structure software. In the picture (screenshot) is the electron density of the same molecule. On the left I calculated the density using B3LYP/6-311++G*, on the right I used B3LYP/6-311G*.

I have visualized the densities using Avogadro. In Avogadro, I can visualize a few of the HOMO/LUMO orbitals. The screenshot is of the first LUMO for both calculations.

In the images, red indicates electron rich areas, blue indicates electron poor areas. The molecule (caffeine) was in the presence of point charges which created an electric field, explicitly polarizing the electron density.

My Question: Why does the presence of diffuse function (the plus's in 6-31++G*), make the LUMO orbital opposite in electron density? I understand allowing diffusion would change the electron density, but it looks nearly exactly opposite now (however the total density looks pretty much the same - see last image).

Note that these are screenshots of just two orbitals, the overall density plots shown below, look very similar.

I am not a quantum chemist, so please use small words...

The colours you see in the first two pictures are not electron densities, but phases of the orbitals.
The colours of a single molecule could matter if you were calculating the interactions between different molecules (in which in-phase or out-of-phase interactions could matter) but they make no difference between the molecule(s) within the same calculation.
The electron density arises from the square power of a function which is, say, positive in the red lobe and negative in the blue one.
The square power means that there is no difference between blue and red, or + and - (See Born's law). They are just with inverse phase in respect to each other, but they have no other influence on the system otherwise.
You might find in some software packages a "switch phases" button, which inverts the colours.


Is the Henderson–Hasselbalch equation volume independent? Looking at this equation $\mathrm{pH} = \mathrm{p}K_\text{a} +\log\frac{[\ce{CB}]}{[\ce{A}]}$ makes me think that it is concentration independent.

What I mean is, when we look at $\frac{[\ce{CB}]}{[\ce{A}]}$, the volumes cancels out, so we are only left with the molar ratio of the conjugate acid and conjugate base. So does this mean that the pH is only dependent on the molar ratios? Obviously not, because $0.003\ \mathrm{mol/L}$ of acid $\ce{A}$ and $0.003\ \mathrm{mol/L}$ of its conjugate base is going to have a different $\mathrm{pH}$ from $0.3\ \mathrm{mol/L}$ of acid $\ce{A}$ and $0.3\ \mathrm{mol/L}$ of its conjugate base right?

What am I missing here? This is seriously confusing me. Please help, I want to use the Henderson–Hasselbalch equation to calculate the $\mathrm{pH}$ of a weak acid as it’s being titrated with a strong base, but this is making me reluctant about the results.

For a much shorter and clearer explanation of this problem, look here.
The Henderson-Hasselbalch (HH) equation is not to blame here. It is an approximate equation, with a certain region of validity. By its nature, it does not take into account the self-dissociation of water, which becomes increasingly important in dilute solutions. When concentrations reach somewhere around $\pu{10^-5 mol L-1}$ or lower, the true $\mathrm{pH}$ will deviate significantly from the value predicted by the HH equation. However, this is not the problem most of the time.
The reason the HH equation might produce poor predictions when calculating buffer $\mathrm{pH}$ is really because of an oft-made assumption which has nothing to do with the HH equation itself: the weak acid (or weak base) is assumed to be so weak that its ionization contributes almost no conjugate base (or conjugate acid) in comparison to the dissolution of the buffer salt. In other words, we assume the formal concentrations $C_X$ of species in the buffer are equal to their actual concentrations $[X]$; we replace
$$
\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \frac{[\ce{A^-}]}{[\ce{HA}]}
$$
for the worse approximation
$$
\mathrm{pH}
  = \mathrm{p}K_\mathrm{a} + \log \frac{C_{\ce{A^-}}}{C_{\ce{HA}}}.$$
This assumption is incorrect, but it is used to make the $\mathrm{pH}$ calculation much easier at the cost of accuracy. The resulting approximation breaks down in sufficiently dilute solution, and is already quite noticeable far before reaching water self-dissociation issues. It is also a noticeably poor approximation for weak acids/bases with relatively high ionization constants (say, $K > 10^{-2}$).
So how can we visualize the effect of diluting a buffer without approximations, and where do the approximations start to break down? Let us consider the simple case of solutions of a weak monoprotic acid $\ce{HA}$ with acid dissociation constant $K_\mathrm{a}$ where the concentration of acid and conjugate base are formally equal. This can be done by titrating half of the weak acid with a strong monoprotic base (e.g. $\ce{KOH}$). This problem can be solved exactly (assuming all activity coefficients are equal to $1$, which is generally a good approximation for solutions below about $\pu{1 mol L-1}$), and I have previously outlined the solution here. The equation relevant to this problem is the following:
\begin{multline}
0 =
  [\ce{H^+}]^3
    + \left(K_\mathrm{a}
      + \frac{C^\circ_\ce{B}V_\ce{B}}{V_\ce{A} + V_\ce{B}}
    \right)[\ce{H^+}]^2 \\
    + \left(
      \frac{C^\circ_\ce{B} V_\ce{B}}{V_\ce{A} + V_\ce{B}} K_\mathrm{a}
      - \frac{C^\circ_\ce{A} V_\ce{A}}{V_\ce{A} +  V_\ce{B}}
      K_\mathrm{a} - k_\mathrm{w}
    \right)[\ce{H^+}]
    - K_\mathrm{a}  k_\mathrm{w}
\end{multline}
$V_\ce{A}$ is the volume of the weak acid solution being titrated and $C^\circ_\ce{A}$ is its formal initial concentration (before mixing and reacting with the base), while $V_\ce{B}$ is a variable volume of strong base solution added with formal initial concentration $C^\circ_\ce{B}$ (before mixing and reacting with the acid). The presence of the self-dissociation constant of water $k_\mathrm{w}$ shows that it is being taken into consideration.
We can now substitute values as desired to obtain a polynomial in $[\ce{H^+}]$. I'll set up the volumes and concentrations arbitrarily so we can reach the acid and conjugate base formal concentrations of $\pu{0.3 mol L-1}$ as mentioned in the question. We can start off with $\pu{100 mL}$ of weak acid solution at a concentration of $\pu{0.9 mol L-1}$ with a dissociation constant $K_\mathrm{a} = 10^{-3}$ (this constant can be changed at will). We can reach the target concentrations of acid and conjugate base by adding $\pu{50 mL}$ of strong base at the same concentration $\pu{0.9 mol L-1}$ (neutralizing half the original acid). You can check the resulting concentrations after mixing the solutions, if you wish. Solving this polynomial, the resulting buffer solution containing formal concentrations $C_{\ce{HA}} = \pu{0.3 mol L-1}$ and $C_{\ce{A^-}} = \pu{0.3 mol L-1}$ has $[\ce{H^+}] = \pu{9.93399E-4 mol L-1}$. This results in $\mathrm{pH} = 3.00288$, which is very close to the value predicted by the approximations made at the start ($\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log 1 = \mathrm{p}K_\mathrm{a} = 3$). For now, they work.
After looking at this first case, we shall investigate the effects of dilution by factors of 10. This can be performed simply by diluting both the initial weak acid and strong base concentrations by 10 (i.e., $\pu{100 mL}$ of $\pu{0.09 mol L-1}$ $\ce{HA}$ plus $\pu{50 mL}$ of $\pu{0.09 mol L-1}$ $\ce{KOH}$ result in a buffer with acid/conjugate base concentrations of $\pu{0.03 mol L-1}$, and so on). Before going into the calculations, it's quite simple to see an issue with the approximation in the second paragraph. Independent of the level of dilution, so long as the concentration of acid and conjugate base remain equal, then according to the approximation, $\mathrm{pH} = \mathrm{p}K_\mathrm{a} = 3$. However, consider the case when starting with a solution of $\ce{HA}$ with initial concentration equal to $\pu{9E-4 mol L-1}$. This is a weak monoprotic acid, so the initial solution must have $[\ce{H^+}] < \pu{9E-4 mol L-1}$ and thus $\mathrm{pH} > 3$. Clearly it's impossible to add any amount of strong base to this acid and get a buffer solution with $\mathrm{pH} = 3$! The approximations at the start have broken down at this point.
Now, for the calculations. To get the proton concentration, I replaced every value of $0.9$ in the previous Wolfram Alpha link with $0.09$, then $0.009$, and so on. With the proton concentration determined, I used equations shown in my other linked answer to calculate the actual concentrations $[\ce{HA}]$ and $[\ce{A^-}]$ for comparison with the formal concentrations $C_\ce{HA}$ and $C_{\ce{A^-}}$. Note that they are almost equal at high concentration, but the error increases quite suddenly at lower concentrations. While the ratio $\frac{C_\ce{A^-}}{C_\ce{HA}}$ is always equal to $1$, the ratio $\frac{[\ce{A^-}]}{[\ce{HA}]}$ rockets away.
$$ \scriptsize
\begin{array}{c|ccc|cc}
\hline
C_\ce{HA} = C_\ce{A^-} / (\pu{mol L^-1})  &
[\ce{HA}] / (\pu{mol L^-1}) &
[\ce{A^-}] / (\pu{mol L^-1}) &
\frac{[\ce{A^-}]}{[\ce{HA}]} &
[\ce{H^+}] / (\pu{mol L^-1}) &
\mathrm{pH} \\\hline
0.3 & 0.29901 & 0.30099 & 1.00664 & 9.93399\times 10^{-4} & 3.00288 \\
0.03 & 0.029061 & 0.030939 & 1.06464 & 9.39282 \times 10^{-4} & 3.02720 \\
0.003 & 0.00235421 & 0.0036458 & 1.54858 & 6.45751 \times 10^{-4} & 3.18993\\
3\times 10^{-4} & 1\times 10^{-4} & 5\times 10^{-4} & 5 & 2 \times 10^{-4} & 3.69897 \\
3\times 10^{-5} & 1.6539\times 10^{-6} & 5.83461\times 10^{-5} & 35.2778
 & 2.83464 \times 10^{-5} & 4.54750 \\
3\times 10^{-6} & 1.78596\times 10^{-8} & 5.98214\times 10^{-6} & 334.953
 & 2.98549×10^{-6} & 5.52498 \\
3\times 10^{-7} & 1.97992\times 10^{-10} & 5.99802\times 10^{-7} & 3029.42
 & 3.30096×10^{-7} & 6.48136 \\
3\times 10^{-8} & 6.96609\times 10^{-12} & 5.99939\times 10^{-8}  & 8612.15
 & 1.16115×10^{-7} & 6.93511 \\
3\times 10^{-9} & 6.09004\times 10^{-13} & 5.99939\times 10^{-9} & 9851.15
 & 1.01511×10^{-7} & 6.99349 \\
\hline
\end{array}
$$


Why is this compound named resorcinol? Substituted phenols or polyhydroxy phenols have their general name apart form IUPAC names. The general names comes from the source it is found like:-

Catechol (1,2-dihydroxybenzene)- obtained from catechu
Cresol (methylphenol) - obtained from creosote, a component of coal tar
Gallol (1,2,3-Trihydroxybenzene) - synthesized from gallic acid which is obtained from  oak galls, not a compound of gallium which many people get confused.

But what about resorcinol? Why is it named so? Is it resourceful? It is primarily obtained from argan oil. So, it should be named as arganol. But it is not so.

Same thing happens with phloroglucinol. Various sources are there but none of them matches with the general name.

For resorcinol, the naming is covered in a footnote to the Wikipedia article:

  "Wir nennen den neuen Körper, da wir gefunden haben, dass er auch aus dem Ammoniakgummiharz erhalten werden kann, Resorcin, um an seine Entstehung aus Harzen und seine Beziehung zum Orcin zu erinnern." (We name the new substance, since we've found that it can be obtained from ammoniated resin gum, resorcin, in order to recall its creation from resin and its relation to orcin [i.e., orcinol].)

Which is itself a quote from  H. Hlasiwetz and L. Barth (1864) "Ueber einen neuen, dem Orcin homologen Körper", Annalen der Chemie, 130 (3) : 354-359.

At some point the ending -ol got added, to reflect the alcohol moieties in the compound.

Then name orcinol itself is derived from its association with orcein, a dye substance derived from orchil lichens.


Halflife of hydrogen bonds Related: Covalent Bond/Hydrogen Bond

What is the half-life of hydrogen bonds in a glass of water at room temperature, say 18 degrees Celsius? I guess values may be different for distilled water and for tab water, but both may be intersting.

My attempts to get an answer involve searching with google, google scholar, and google books. Complications were that I cannot get access to all articles using my universities database.

From this page we have

  The half-life of a hydrogen bond is about 10 seconds

Is this correct?

I'm not sure whether the term "half-life" means the same as "lifetime" but I found a very interesting website about hydrogen bonding in water (with a lot of references) where it is stated that the lifetime of a hydrogen bond is in the range of 1 - 20 ps (it gives the following reference for this: F. N. Keutsch and R. J. Saykally, Water clusters: Untangling the mysteries of the liquid, one molecule at a time, Proc. Natl. Acad. Sci. U.S.A. 98 (2001) 10533-10540). That is quite a big difference to the 10 seconds from your reference. Thinking of the very dynamic nature of a hydrogen bond network and of the typical time scale for molecular processes in general I would tend to think, that the breakage and formation of hydrogen bonds on a molecular level should take much less time than 10 seconds. So if you are looking for the lifetime of a hydrogen bond on a molecular level a few picoseconds sound reasonable.

Edit: Here is another website, which also states a hydrogen bond lifetime of about a few picoseconds.


Why is the ester carbonyl peak in benzocaine IR spectrum so low? Having looked at multiple IR spectra for benzocaine, I am struggling to understand why the carbonyl C=O peak is so low in the spectra compared to where an ester carbonyl group usually is on other IR spectra

Let's first have a look at the structure of benzocaine, to see how it is different from an usual aliphatic ester:

One thing that should be immediately obvious is that the ester is attached to an aromatic ring (which has an amine group). The aromatic ring would participate in conjugation with the ester oxygen, particularly because the amine group is strongly electron-donating (via conjugation).
This is the IR spectrum of benzocaine (from Chemical book):

The peak at $\mathrm{1682\;cm^{-1}}$ can be assigned to the ester C=O stretch. Usually aliphatic esters have IR stretches around $\mathrm{1730-1750\;cm^{-1}}$. Esters attached to unsaturated systems or aromatic groups have lower stretches, as the conjugation weakens the C=O bond.
Compare the $\ce{C=O}$ IR stretches of some esters:

Ethyl acetate  : $\mathrm{1752 cm^{-1}}$
Ethyl benzoate  : $\mathrm{1726 cm^{-1}}$

The presence of the benzene group has reduced the stretch wavenumber by $\mathrm{\sim30 cm^{-1}}$.

methylparaben  : $\mathrm{1682 cm^{-1}}$

Here the $\ce{-OH}$ group is quite strongly electron-donating, and it brings down the ester C=O stretch wavenumber even further, and now the IR stretch resembles that of benzocaine. $\ce{-NH2}$ is also a stronlgy electron-donating group like $\ce{-OH}$.


Flame Test Spectrograms Not Lining Up With Reality In my AS Level chemistry class we had tried some DIY spectroscopy of flame tests with the aim of identifying anions in certain salts. We got as far as building the spectrometer inspired by Paper Craft Spectrometer (though we built a much bigger version such that a DSLR could be used) and performing the experiment and collecting the data. The problem was that I couldn't manage to get many of the spectrograms to line up with those sourced from the web. Many of our spectrograms, while plenty bright and with lines covering the whole visible range, had too few lines and those that were there didn't align with any real data we could find.

For instance with the following recorded spectrogram:

Some processing was done on the above image to produce the spectrum below it. The scale is calibrated from an LED which was also recorded and then confirmed my matching it against the emission of sodium which also roughly lined up to this scale +-10nm. From the colour of the flame we assumed it was calcium, given that all salts tested were chlorine salts, and the metal is one of: Ca, Cu, Cs, K, Li, Na, Sr. I don't know if there was an experimental error or if we were not looking at the correct reference spectrograms but nothing I could find would line up with that,does anyone have any insight into this?

The first step you can take to make the spectral bands narrow is to reduce the slit width. It is slightly wide at this moment. We are not sure where you were looking for wavelengths? In the compiled tables in handbooks or schematic diagrams on the web, the emission wavelengths are listed for the excited atoms using very high temperature flames using air acetylene, or nitrous oxide acetylene flames. The ordinary flame temperatures in a laboratory are too low to electronically excite the atoms of any other element other than alkali metals. For Li, Na, K, Cs, you would see a line spectrum for each element (one or two lines). For potassium is extremely difficult to see the far red wavelengths (the resonance wavelengths listed in tables) until and unless your eyes are sensitive to red. The flame colors Ca, Ba, Sr, Cu are usually due to molecular compounds of these metals in the ordinary flames.

The key point is that the flame temperature is too low to see the line spectra you were expecting. However, if you look at discharge tubes, then you can do a good comparison.


Pros and cons of Cartesian vs. Z-matrix representations of molecules? Over the course of my studies, I have switched largely from using Z-matrix representations of molecular geometries in calculations to Cartesian representations.

The software that I use now makes it easy to add the sorts of constraints/restraints/transits that I would have previously used Z-matrices for, and I know that Z-matrix geometries can be problematic in large molecules* where minute changes in a bond angle or dihedral (due, for instance, to rounding errors/low-quality gradients) can result in large movements in peripheral atoms.

What pros or cons exist for either geometry definition that I don't know about? What circumstances recommend one representation over another?

*Or small molecules with silly Z-matrices.

Cartesian Space

In Cartesian space, three variables (XYZ) are used to describe the position of a point in space, typically an atomic nucleus or a basis function.  To describe the locations of two atomic nuclei, a total of 6 variables must be written down and kept track of.  The general ruling is that for Cartesian space, 3N variables must be accounted for (where N is the number of points in space you wish to index).

Internal Coordinates

Z-matrices use a different approach.  When dealing with Z-matrices, we keep track of the relative positions of points in space.  Cartesian space is 'absolute' so to speak.  A point located at (0,0,1) is an absolute location for a coordinate space that extends to infinity.  However, consider a two atom system.  The translation of the molecule through space (assuming a vacuum) will have no affect on the properties of the molecule.  An H2 molecule centered around the origin (0,0,0) is no different from the same H2 molecule being centered around (1,1,1).  However, say we increase the distance between the hydrogen atoms.  We now have altered the molecule in such a way that the properties of that molecule has changed.  What did we change?  We simply changed the bond length, one variable.  We increased the distance between the two atoms by some length R.  With Z-matrices, we keep tabs on internal coordinates: bond length (R), bond angle (A), and torsional/dihedral angle (T/D).  Using internal coordinates reduces our 3N requirement set by the Cartesian space down to a 3N-6 requirement (for non-linear molecules).  For linear molecules we keep tabs on 3N-5 coordinates.  When performing complex computations, the less you have to keep track of, the less expensive the computation.

Symmetry

Consider the following molecule, H2O.  We know from experience that this molecule has C2V symmetry. The OH bond lengths should be equivalent.  When using some sort of optimizing routine, you may want to specify symmetry in your system.  With a Z-matrix, the process is very straightforward.  You would construct your Z-matrix to define the OH(1) bond as being equivalent to the OH(2) bond.  Whatever program you use should automatically recognize the constraint and will optimize your molecule accordingly giving you an answer based off a structure that is constrained to C2v symmetry.  With Cartesian space this is not guaranteed.  Rounding errors can cause your program to break symmetry, or your program may not be very good at guessing the point group of your molecule based on the Cartesian coordinates alone.

Picking the Right One

As a preface, programs like Gaussian convert your Cartesian coordinate space (or your pre-defined Z-matrix) into redundant internal coordinates before proceeding with an optimization routine unless you specify it to stick with Cartesians or your Z-matrix.  I warn you that specifying your program to optimize using Cartesian coordinates makes your calculation much more expensive.  I find that I will explicitly specify 'Z-matrix' when I know I'm dealing with high symmetry and when I know my Z-matrix is perfect.

You will want to use Z-matrices on systems that are rather small.  If dealing with systems with high symmetry, Z-matrices are almost essential.  They can be rather tricky to implement and you will likely spend some time figuring out the proper form of your Z-matrix through trial-and-error.  If you wish to scan a particular coordinate, Z-matrices are also very helpful as you can tell a program to scan across a bond length, angle or torsion with ease (as long as you've properly defined that coordinate in your Z-matrix).

I use Cartesian coordinates for large systems, systems with very little or no symmetry, or when I'm in a hurry.


Do lithium and beryllium want to lose electrons to be like helium? I can't seem to remember. I know that they all want to fill their valence shells, but, aren't these different?

Yes, lithium wants to lose electrons to be like helium because full valence shells are more stable states and all noble gases have full valence shells. So alkali metals lose one electron to achieve nearest noble gas configuration.

But beryllium is exceptional because of its small atomic and ionic radii. Due to this, its ionization potential is very high. So it mostly forms covalent bonds.


Why the optimization in gaussian shows no bond with some atoms in diborane? I have optimized the structure of diborane with Gaussian at the B3LYP/6-31+g(d,p) level and when check it on Gaussview I found that there are two hydrogens with no bond to boron atoms (like free atoms) and the two B atoms are bonded. Frequency calculations gives zero imaginary frequencies and when I compute properties (like ionization potential) the values are good approximations.

I wonder if the problem is with the calculation or only the graphical interface of Gaussview doesn't recognize it as an actual bond.

Also, the bond distances are in accordance with the literature (~1.3 Angstroms for H shared between the borons and ~1.19 Angstroms for the others).

I wouldn't put too much faith in the representation of bonding in GaussView, it's simply determined by bond distances and often it doesn't show the bond at all for some unknown reason. If your geometry is good, then you can use your knowledge and intuition of bonding to recall the true structure.


Ionic reaction between lithium and chlorine I got a weird question in my chemistry book. It says Li and Cl react to form LiCl an ionic compound. Cl has .... electrons in its outermost occupied shell and the answer of the book is 6 electrons . How?

Maybe your book is wrong. It can't be. There has to be 8 electrons in the outermost occupied shell and to complete its octet. I searched over the google about it and found three links to cover my answer.

https://in.answers.yahoo.com/question/index?qid=20111018173152AAn9DHz
https://quizlet.com/38997670/chem-1305-exam-2-study-guide-flash-cards/ (question number 3)
http://slidegur.com/doc/1661637/ionic-bonds (slide number 18)


Why is the nitrile group not hydrolyzed as well during hydrolysis of imine in Thorpe reaction? while imine is getting hydrolysed, why does not nitrile group hydrolyse to give carboxylic acid?

It has to do with hybridization. The nitrogen in a nitrile is sp hybridized; imine sp2 and ammonia sp3. The more s-character in hybrid orbitals, the less basic the electron pair (protonated nitrile pKa ~ -10; pKb ~24). Ammonia has the lowest percent s-character (25%, NH4+ pKa 9.25). Imines lie in between although most examples you will find are measured in DMSO and are loaded with biasing phenyl groups.
Compare the carbon series: acetylene (sp, pKa ~25), ethylene (sp2, ~44), methane (sp3, pKa ~50). The closer the electron density is to carbon, the more acidic the C-H bond. If electrons are held closer to carbon, then it is more difficult for homolytic cleavage of a C-H bond. That is, acetylene is willing to give up a proton but not a hydrogen atom. Thus, Bond Dissociation Energy runs in the opposite order as acidity.

The harder it is to protonate, the harder it is to hydrolyze.

There were no conditions provided by user 481311 for the hydrolysis. Imines, such as the acetaldehyde methylimine (CH3CH=NCH3), will hydrolyze in water. As the User illustrates, imines hydrolyze readily in acid medium. These hydrolyses may be conducted at ambient temperature or at mild temperatures. Nitriles, owing to their low basicity, require more vigorous conditions like refluxing aqueous acid. Its all about conditions. I hope this helps.


Electrical conductance of NaCl solution I'd like to ask whether EC of a salt solution of iodized table salt differs from that of uniodized and to what extent?

Is salt an electrolyte?
Common table salt (NaCl) is an electrolyte, and when this is dissolved in water to form salt water, it becomes sodium ions (Na+) and chloride ions (Cl-), each of which is a corpuscle that conducts electricity.
Let's go back to conductivity. Conductivity is an index of how easy it is for electricity to flow. In water, it is the ions that pass electricity from one to the next. This means that the more Na+ and Cl- contained in water the more electricity is carried, and the higher the conductivity.
Here's a chart:

(Don't mind the sugar water side)
Source linked here.


Aflatoxin versus Benzene, permissible levels Alfatoxin and benzene are carconogenic, and this is the major concern for low, chronic doses. They are regulated to the few parts per billion level in food. However, according to http://toxnet.nlm.nih.gov/cpdb/pdfs/ChemicalTable.pdf,
aflatoxin is three to five orders of magnitude more carcinogenic, as evident from the lower doses needed to induce tumors in rodents.

Why are the permissible thresholds similar?

It is usually assumed that there is no finite threshold below which carcinogens will not exert an effect. Therefore, there may be no safe exposure level below which there is no health risk.

Accordingly, a target value that is based solely on possible health risks (considering exposure over a lifetime, possibly even demanding an adequate margin of safety) would be zero.

Nevertheless, permissible limit values for carcinogens should not be set as low as possible since the exposure should not be reduced as much as possible at all costs. Instead, the potential exposure and limit values should be as low as reasonably achievable taking into account economic factors, other benefits and disadvantages, and the ability to detect and remove contaminants.

Hence, there may be similar limit values for various substances with different carcinogenic potentials. On the other hand, there may be different limit values for one substance depending on the exposure pathway (e.g. for different foodstuffs).


What is chemical decomposition in the context of a crystal or amorphous solid? The IUPAC Gold Book definition of chemical decomposition is:

  The breakdown of a single entity (normal molecule, reaction intermediate, etc.) into two or more fragments.

That is a very general definition, but I am not sure if it is supposed to be this inclusive. For example, think of a diamond crystal. It is a single entity. Now if chemical decomposition occurs, it can break down into multiple fragments. These fragments are not defined more closely, so it could be that the fragments are single C atoms, polyatomic C molecules... or even diamond fragments again.

In this sense, simply cutting or polishing a diamond would be a special case of chemical decomposition. This seems quite absurd to me. So the question is: Is chemical decomposition really that broad, or is there some better definition that is more commonly used in chemistry? It seems to me that the intuitive definition of chemical decomposition would be that a molecule is broken into smaller fragments, while a crystal or an amorphous solid would need to be broken into non-crystalline (non-amorphous) fragments in order for it to be considered chemical decomposition. Otherwise it's just cutting!

The question is relevant to me because I am writing my PhD thesis on laser-induced breakdown spectroscopy. The "breakdown" in the name of the technique refers to a chemical breakdown, which is the same as chemical decomposition. Usually, solid samples are analyzed, in which case some material from the sample surface is removed by laser ablation. Laser ablation is basically the same as laser-induced breakdown, if you look at the definitions. So there is some confusion on the terminology in general, and I want to unravel it as good as possible in my theory chapter.

Thank you very much for your input!

(Note: By the way, some non-IUPAC definitions I have found for chemical decomposition also confusingly define it as relating to compounds. As far as I know, the definition of a compound excludes crystals and amorphous solids, but it also excludes homonuclear molecules. So this definition seems way too exclusive, because it makes sense that decomposition can also occur for homonuclear molecules.)

When it comes to non-molecular solids such as glass, metal or diamond, "a single entity" is in fact the formula unit, the concept identical to a "molecule" for, well, molecular compounds. Both formula unit and molecule are the smallest portion of a substance that preserves its chemical properties. As long as the formula unit remains the same, there is no chemical decomposition to speak of.

However, this doesn't mean that there is no chemical processes that can be initiated mechanically. An entire branch, mechanochemistry deals with them. As for laser ablation, I think you are basically introducing a secondary phase, a thin film of another substance (for example, metal oxide) on the surface.

One more practical example I can think of: for a single-crystal diffraction experiment one has to use a 3D crystal (in a sense that it shouldn't be a 2D plate for the proper absorpton correction) with linear dimensions not exceeding fractions of a millimeter. For the probe preparation a crystallographer often has to literally cut the crystals with a special scalpel under the microscope. However, the resulting crystal structure doesn't depend on how many cuts and in what direction the crystallographer did. The x-rays though, can alter the structure of some sensitive molecular crystals (often biological samples, proteins), but this is another story.


How would I determine the relative acidity of substances? Which of the following salt will produce a neutral solution (If any)? A) $\ce{KI}$ (Potassium Iodine) or B) $\ce{SrS}$ (Strontium Sulfur)

I tried to solve this problem by using the "Relative Strengths of Brønsted-Lowry Acids and Base in aq solution at room temp table". From what I know each of those two substance will dissociate into ions. Having the ions $\ce{K^+, I^-,Sr^{2+}, S^{2-}}$

Since K does not hydrolyze and I is a product of HI which is a strong acid it won't hydrolyze either. I also know that the only cations that hydrolyze are Al,Fe and Cr hence Sr will not hydrolyze. However how do I know weather S will hydrolyze? The other question I had difficulties was "Order the following from most acidic to least. SO3, CO3 and CO2".

In my homework one of the question was which of the following salt will produce a neutral solution(If any)? A) KI (Potassium Iodine) or B) SrS (Strontium Sulfur)

For (A) both K+ and I- will essentially stay as the ions. So assuming the solution is neutral to start, it will remain neutral.

But for (B), in a neutral solution, $\ce{Sr^{2+}}$ will stay as the ion,  $\ce{S^{2-}}$ will react as:

$\ce{S^{2-} + H2O <=> HS^- + OH-}$

so the solution will become slightly more basic.

"Order the following from most acidic to least. $\ce{SO3}$, $\ce{CO3}$ and $\ce{CO2}$".

I can't really make sense of this question since $\ce{CO3}$ is highly unstable, and there are three isomers.

Bubbling  $\ce{SO3}$ and $\ce{CO2}$ gases into water would produce acidic solutioons via reactions

$\ce{SO3 + H2O -> H2SO4}$

$\ce{CO2 + H2O -> H2CO3}$


Why is melting point of Vanadium higher than Iron? The iron is found to have four unpaired electrons while vanadium has 3 of them. we say the binding energy is directly proportional to number of unpaired electrons and hence number of  metallic bonds in the structure. By this logic, iron should have high melting point than vanadium but no. Iron has a m.p of 1535 degree celsius while vanadium has 1917.

@Jon Custer makes a good point in his comment so lets look beyond the electronic shell configuration of V and Fe to help understand the different in melting points.
Consider the following:
Pauling’s Electronegativity Fe > V  1.8 vs 1.6
Vander walls radius Fe < V  0.126 nm vs 0.134 nm
And consistent with Fe having a higher electronegativity the first ionization energy of Fe [761 kJ per mol] is greater than that of V [ 649.1 kJ per mol]
Given that we think of melting as the breaking of lattice forces holding atoms together, Fe being a smaller and more electronegative atom [vs V] is consistent with the relative melting points of these two elements.
Ref:
https://www.lenntech.com/periodic/elements/fe.htm
https://www.lenntech.com/periodic/elements/v.htm


How to find the vitamin C content of orange juice? I'm an IB chemistry SL student doing an IA on the vitamin C concentration of orange juice. We haven't learned about the titration process, so here is what I have done so far:

$\ce{5KI + KIO3 + 3H2SO4 -> 3I2 + 3K2SO4 + 3H2O}$

This is the equation for producing the iodine solution. Now using a titration process, I added the iodine solution to the orange juice. I measured the volume of Iodine solution used.

Here is the formula for the reaction between the vitamin C and the Iodine solution.

$\ce{C6H8O6 + I3- + H2O -> C6H6O6 + 3I- + 2H+}$

I'm not sure where the Triiodide comes from. Can someone explain this to me? I'm not a very good chemistry student.

Iodine and iodide will form triiodide in aqeous solution. This is a common way to increase the solubility of $\ce{I2}$ in water. For a titration, one will use excess $\ce{KI}$ and control the formation of iodide by adding $\ce{KIO3}$ using a burette.

In response to the comment:

$\ce{KI}$ contains iodide anions. I strongly suspect that an excess of $\ce{KI}$ was used (more than fivefold compared to $\ce{KIO3}$), leaving enough iodide anions to form the triiodide. Also note that effectively a $\ce{I-}$ can be subtracted from either side of the second equation of the question.


What is the definition of liquid phase? Why is a supercritical fluid not a liquid? My book says that because in a container above the critical temperature and critical pressure, a fluid of density similar to liquid exists uniformly in the container and there is no surface or meniscus that separates the liquid from the gas. Does that mean that a liquid always has a meniscus which separates it from the liquid's vapour?

Does that mean that in free space without no gravity, a gas/vapour will try to expand and the liquid will try to contract and form spherical blobs due to surface tension? Can I draw the conclusion from this that a liquid is the phase when attractive forces are dominant and repulsive forces are dominant in gas/vapour phase?

Why is a supercritical fluid not a liquid?

Beyond the elementary school list of solid liquid and gas, there are many states of matter which exhibit peculiarly different properties. Wikipedia gives a list of the states of matter.

  My book says that because in a container above the critical temperature and critical pressure, a fluid of density similar to liquid exists uniformly in the container and there is no surface or meniscus that separates the liquid from the gas.

That is true.

  Does that mean that a liquid always has a meniscus which separates it from the liquid's vapour?

No, if the container is absolutely full of liquid then there is no gas phase.

  Does that mean that in free space without no gravity, a gas/vapour will try to expand and the liquid will try to contract and form spherical blobs due to surface tension?

Yes, but for the liquid there will also be evaporation. The rate of evaporation depends on the temperature of the liquid. As evaporation occurs the liquid will also cool. If there is enough liquid in the blob, then the matter can get cold enough to form a solid which will then sublime.

  Can I draw the conclusion from this that a liquid is the phase when attractive forces are dominant and repulsive forces are dominant in gas/vapour phase?

This is more or less true. I'd quibble about the nature of the "repulsive forces." I don't think that "repulsive forces" is the best way to model water in the form of steam. Rather the water molecules have too much kinetic energy to stick together.


How to evolve hydrogen from aluminum when the household drain cleaner is safe to use on aluminum? I live in Japan. I would like to evolve hydrogen to demonstrate to my son buoyant balloons (we have helium aerosol cans here which barely fill one balloon). There are drain cleaners here with low concentration of sodium such as up to 4% and say safe for aluminum (probably the calcium hydroxide also contained). I have obtained some wire and batteries and I think electrolysis could help- what should I do  for the best chance of evolving hydrogen?

First, any drain cleaner that is at all effective is probably not safe for a child to use, even under adult supervision.
Second, you can easily produce enough hydrogen to inflate a plastic bag, perhaps one to four liters, using electrolysis. Unlike an elastic balloon, the bag will not resist inflation appreciably until full. Try carbon electrodes (e.g., from zinc-carbon cells) with $\ce{MgSO4}$ (Epsom salt) in water as the electrolyte.
You can also capture oxygen generated and test it with a glowing wooden splint.
It is an exercise for the experimenter to calculate how many ampere-hours it would take to inflate the specific bag.
BTW, you might want to perform this outdoors, in case a spark sets off the hydrogen!


Are there any poisons with an activation threshold? Are there any poisons that lay dormant unless a significant quantity of one has been added, i.e. poison X might as well be water in concentrations less than 5M, but at 6M, all of a sudden a threshold has been reached and the poison has activated.

Yes, there are certainly some chemicals that are harmless in low quantities but really dangerous in higher doses.

The common painkiller paracetamol (or acetaminophen) is one such candidate. In low doses it is a mild analgesic but in higher does it is deadly. The reason is that there is a metabolic side-reaction in the liver that produces a small amount of a toxic compound. Under normal circumstances this is mopped up by other pathways in the liver and causes no damage. But when large amounts are taken the liver's capacity to mop up the toxin is exhausted, the toxin builds up to dangerous levels and it causes severe liver damage often resulting in death.


IUPAC Naming. Which one is correct? I have named this according to the numbering shown as
3-ethyl-6-(1-methylbutyl) oct-2,4-diene-7-ynenitrile.
But my book's answer key says
3-ethyl-6-ethynyl-7-methyl deca-2,4-dienenitrile
This means it has taken the triple bond as a side chain. But IUPAC rule says that the chain with maximum number of double and triple bond is taken even if it is shorter than a chain with lesser double or triple bonds. Which answer is correct and why, for the given example?

As Loong said, the up-to-date IUPAC recommendation stipulates that the chain with the longest continuous chain of carbon atoms should be the parent chain. For this reason, I would agree with your textbook's answer.


Reaction of substituted furan with methyl vinyl ketone with subsequent rearomatisation I am supposed to provide a mechanism for the above transformation. I observed that the order of substituents on the product is same as that on furan, and since furan undergoes Diels–Alder reactions, I came up with the following mechanism:

Is my Diels–Alder product correct? After this, however, I have no idea how aromaticity is restored.

Your Diels–Alder adduct is pretty much correct. Without any further evidence we cannot definitely say whether the exo or the endo adduct is formed (you have drawn the exo adduct, which generally arises via thermodynamic control). However, the stereochemistry of this adduct does not affect the final product, so it does not matter which one is formed.

By the way, you also should think about the regioselectivity of the Diels-Alder reaction, i.e. why do the $\ce{NH2}$ and $\ce{COMe}$ end up ortho to each other and not meta. You need to have a better reason than simply "that's the way it was drawn in the product". Generally, this is rationalised by looking for the largest coefficients in the HOMO of the diene and the LUMO of the dienophile. The HOMO of 2-aminofuran is likely to have a large coefficient on C–5 (opposite the amino group), whereas the LUMO of methyl vinyl ketone has a large coefficient β to the carbonyl group.

The subsequent steps are not anything fancy. Generally, $\mathrm{sp^3}$ carbons that are bonded to two heteroatoms tend to collapse (e.g. tetrahedral intermediates in nucleophilic acyl substitution; hydrates; etc.) In this case the nitrogen kicks out the oxygen. Note that the other option of oxygen kicking out nitrogen is not possible, as that would lead to a bridgehead double bond.

Ordinarily you need some sort of acid or base catalyst, but presumably here, the thermodynamic driving force for the collapse is large (relief of ring strain + product is aromatic), and the high temperature provides the energy required to overcome the activation barrier. The last elimination step probably occurs via an E1 mechanism, since there's no good base present.


What are the products of the reaction of NH4Br (aq) + Hg2(NO3)2 (aq)? Predict the products of the following reaction based upon solubility rules. (Separate substances in a list with a comma. Include states-of-matter under the given conditions in your answer). The reaction is:

$$\ce{NH4Br (aq) + Hg2(NO3)2 (aq) -> ?}$$

I can not figure out what the products of this reaction are. I would think $\ce{NH4(NO3)(aq)}$, $\ce{Hg2Br2(aq)}$, but this is not correct.

According to this table, mercury (I) bromide ($\ce{Hg_2Br_2}$) is somewhat insoluble (among other mercury (I) salts), so it gets precipitated as $\ce{Hg_2Br_2(s)}$ instead of $\ce{Hg_2Br_2(aq)}$, making the overall equation this:

$$\ce{2NH4Br(aq) + Hg2(NO3)2(aq) -> 2NH4NO3(aq) + Hg2Br2(s)}$$

Mercury (I) ions are insoluble because every orbital is full, making it unable to form ligands with water.


Are there ionic solids that conduct electricity? We are taught in school that ionic substances don't conduct electricity, and when they do, it is either because they are in a molten state or because they are in solution. I understand these concepts.

Are there any exceptions? By exceptions, I am looking for ionic compounds that may not conduct electricity well, but they conduct electricity better in comparison to other ionic compounds. Additionally, what makes such a compound a better conductor of electricity?

Yes.

Another example is VO2. Over 67 °C its crystal structure changes to rutile with a band gap of ~0.1 eV. Dopants of other substances can reduce the temperature at which it becomes a conductor.


Hydrolysis of ammonium cloride As we all know ammonium chloride is a acid. And what explains this is its salt hydrolysis. It breaks down into ammonium and chloride ions. Ammonium react with water to produce hydronium or hydrogen ions.

But then why can't chlorine ions react with water to produce hydroxide ions?
Is there any way other than using an indicator to know, whether a compound is acidic or basic? We know the second reaction can't happen, because we can tell the compound is acidic by using an indicator.

What really stops the second reaction from happening?

Chloride ion does not react with the water to any significant extent. To understand this, we must first accept that all chemical reactions are equilibriums. This means that all chemical reactions are reversible to some extent. Some reactions aren't very reversible at all (at least not in the short or medium term) - such as the combustion reaction in your car's gasoline engine. But if you give it a few million years, the carbon dioxide released by your car's combustion process will be fixed (biological term) by plants and these plants might eventually turn into fossil fuels again.

So we've established that all chemical reactions are equilibriums. Now we can consider the case of the chloride ion and why this reaction does not happen, as you say, or does not happen to a significant extent, as I say:

$\ce{Cl^- + H_3^+O -> HCl + HO^-}$

The reason is that if the above reaction happened to any significant extent, then this would make HCl a weak acid. Strong acids are defined to dissociate nearly completely in water. HCl is a strong acid by any account. Think of it as a waterfall. This reaction represents going down a waterfall (this reaction is large extent; heck, it's practically 100%):

$\ce{HCl + H_2O -> Cl^- + H_3^+O}$

So if the above reaction is the going down part of the water fall, its reverse reaction must represent the going up part of the waterfall. Going up the waterfall is hard no matter how you slice it; the only difference is whether you're trying to go up a small waterfall or Niagara falls. The reaction of hydrochloric acid with water is akin to going down Niagara falls - this is a favorable reaction. This is basically a one-way street.

Therefore, chloride ion can't affect solution pH; if chloride ion could somehow react again to form more hydrochloric acid then we'd violate the definition of a strong acid - nearly complete dissociation. This is what makes hydrochloric acid a strong acid!

And this is empirically true. Chloride ion can't even be protonated effectively by the strongest acid in a solution of hydrochloric acid, which is hydronium ion, so there is no way that you can expect chloride ion to be protonated by water (which is only about a trillion times weaker than hydronium ion as an acid). Thus, your solution of ammonium chloride has a pH controlled not by the chloride ion but by the ammonium ion.


How to determine net charge of a polycation using Wade's rules? In order to determine the net charge of polycations, which adopt a Wade-like structure (octahedral, trigonal bipyramidal, ...), Wade's rules can be used.

For example, the charge $x$ of the octahedral cluster $\ce{[Bi6]^x+}$ should be determined:

It is known that a Wade-cluster overall "needs" $2n+2$ electrons (to fill the $n+1$ energetically favoured, bonding MOs).
Bi atoms contain three valence p-electrons (the two s-electrons do not take part in the cluster-bonding). So in a $\ce{Bi_n}$ cluster, there are $3n$ electrons already available for bonding.
But only $2n+2$ electrons are needed: We need to remove $2n+2-3n=-n+2=-(n-2)$ electrons to obtain the favoured $2n+2$ electron count in the cluster.
So for example the $\ce{[Bi6]^x+}$ should have $-(6-2)=-4$ electrons "removed". We would arrive at a $\ce{[Bi6]^4+}$ net charge... ...but in reality, there is "only" a $\ce{[Bi6]^2+}$ - with two electrons more than determined according to the Wade's rules.

EDIT:  In contrast to that, the rules aboves apply to $\ce{[Bi5]^3+}$ (Thanks to @orthocresol).

Is there a misconception in my above thoughts? I can not find an illogical step - Maybe Wade's rules can not be applied to the polyatomic clusters (my textbooks suggests this nevertheless)? Or I should have take the s-electrons into account?

As far as I can tell, your calculations are correct. A closo structure would require $2n+2$ cluster electrons (or $4n+2$ total electrons), which is satisfied by the formula $[\ce{Bi}_n]^{(n-2)+}$. As such, for $n  = 6$, a closo octahedral structure would correspond to the ion $\ce{[Bi_6]^4+}$ as you have outlined.

The contradiction seems to be that the ion $\ce{[Bi_6]^2+}$ is not closo, but rather nido, according to e.g. Hampel & Ruck:[1]

  The $\ce{[Bi_6]^2+}$ cation, a nido cluster with 16 skeletal electrons, has the shape of a distorted octahedron with an opened edge.

These require $2n$ cluster electrons for complete population of the bonding molecular orbitals, consistent with the formula $\ce{[Bi_6]^2+}$. I do not currently have access to Holleman-Wiberg (nor would I understand the German version!), so I cannot check what they have written, but it is possible that it is a simple typo.

Interestingly, some of the higher bismuth polycations also display a tendency to have "extra" electrons. For example, Greenwood and Earnshaw[2] describe the following cations:

$$\begin{array}{cccc}\hline
\text{Cation} & \text{Formal oxidation state} & \text{Cluster structure} & \text{Point group symmetry} \\ \hline
\ce{Bi+} & 1.00 & - & - \\
\ce{Bi3+} & 0.33 & \text{Triangle} & D_\mathrm{3h} \\
\ce{Bi5^3+} & 0.60 & \text{Trigonal bipyramid} & D_\mathrm{3h} \\
\ce{Bi8^2+} & 0.25 & \text{Square antiprism} & D_\mathrm{4h} \\
\ce{Bi9^5+} & 0.56 & \text{Tricapped trigonal prism} & C_\mathrm{3h} (\sim\!D_\mathrm{3h}) \\ \hline
\end{array}$$

The rationalisation which they provide for this is

  [The application of Wade's rules] would account for the stoichiometries $\ce{Bi3+}$ and $\ce{Bi5^3+}$ but would also lead one to expect $\ce{Bi8^6+}$ and $\ce{Bi9^7+}$ for the larger clusters. However, these charges are very large and it seems likely that the lowest-lying nonbonding orbital would also be occupied in $\ce{(Bi+)_n^2-}$. For $\ce{(Bi+)8^2-}$ this is an $\mathrm e_1$ orbital which can accommodate 4 electrons, thereby reducing the charge from $\ce{Bi8^6+}$ to $\ce{Bi8^2+}$ as observed. In $\ce{(Bi+)9^2-}$ the lowest nonbonding orbital is $\mathrm a_2''$ which can accommodate 2 electrons, thus reducing the charge from $\ce{Bi9^7+}$ to $\ce{Bi9^5+}$ as observed.

References:

Hampel, S.; Ruck, M. Synthesen, Eigenschaften und Kristallstrukturen der Cluster-Salze Bi6[PtBi6Cl12] und Bi2/3[PtBi6Cl12]. Z. Anorg. Allg. Chem. 2006, 632 (7), 1150–1156. DOI: 10.1002/zaac.200500379.
Greenwood, N. N.; Earnshaw, A. Chemistry of the Elements, 2nd ed.; Butterworth–Heinemann: Oxford, U.K., 1997, pp 590–591.


Is ammonium a strong or weak conjugate acid? Ammonium, $\ce{NH_4^+}$, is the conjugate acid of ammonia, $\ce{NH_3}$. I have searched the Internet and so many different answers pop up. The rule of thumb that I read was that a strong acid has a weak conjugate counterpart.

For example, a strong acid has a weak conjugate base. Is ammonium a strong or weak acid? Is ammonia a strong or weak base? Is there something to the rule of thumb that I mentioned?

Both ammonia is a weak base and ammonium ion is a weak acid.  Many, even most, acid/base conjugate pairs are like that.  We should be using comparative instead of absolute adjectives in the rule about conjugate acid-base strengths:

A weaker acid has a stronger conjugate base, not necessarily a totally strong one.

A weaker base has a stronger conjugate acid, not necessarily a totally strong one.


Why are Tyrosine and Tryptophan considered hydrophobic? Since Tyrosine and Tryptophan are amino acids, their polarity is determined on their side chains or R groups. If their R groups are polar, the amino acid is polar. Both Tyrosine and Tryptophan are listed as non-polar molecules. However, Tryptophan has nitrogen in its side chain and Tyrosine has oxygen, both of which have a high electronegativity compared to the carbon that makes up the bulk of both molecules, which is something commonly seen in polar amino acids like Serine, which has oxygen in its side chain. This should mean that Tyrosine and Tryptophan are polar, so why are they non-polar, even with an electronegative atom in their side chains?

Unlike that in organic chemistry, polar and non-polar compounds determination is done differently in biochemistry, probably because most biochemical reactions are done in aqueous medium. Therefore, the classification of them as polar and non-polar is somewhat different from source to source. In general, amino acids, which can make hydrogen bonding in aqueous solutions are called polar, and ones repel from water are non-polar. According to Ref.1:

All polar amino acids have either an $\ce{OH}$ or $\ce{NH2}$ group (when in aqueous environment), and can therefore make hydrogen bonds with other suitable groups.

Following image shows 20 amino acids categorized to polar (10) and non-polar groups (10):

All 10 non-polar amino acids are hydrophobic and 10 polar amino acids are hydrophilic. Polar amino acid residues have a tendency to be on the outside of a protein, due to the hydrophilic properties of the side chain (Ref.1). Keep in mind that tryptophan has indole function, but its lone pair of nitrogen is involved in the aromatic system. Thus, it make only weal H-bonding, which are not good enough to categorize as "polar."
References:

Bruce Alberts, Alexander Johnson, Julian Lewis, Martin Raff, Keith Roberts, Peter Walter, In Molecular Biology of the Cell; 5th edition, Garland Science: New York, NY, 2008, p. 125-129 (ISBN-13: 978-0815341055).


Molecular orbitals symmetry - states notation I'm reading some papers and I'm repeatedly seeing the following notation of system states:

Could you, please, explain to me the meaning of it?

E.g. the state $B^2 \Sigma_u^+$ - I'm aware, that $\Sigma$ is used as a substitute for $A$ in both $C_{\infty h}$ and $D_{\infty h}$. I'm also aware, that $u$ stands for "ungerade", i.e. it tells us, that the state is anti-symmetric with respect to the inversion.

But what does the $B^2$ before $\Sigma$ stand for? And what does $\Sigma^+$ mean?

I'm aware of the Mulliken notation, but it doesn't seem to contain these symbols.

The symbol $B$ is explained in greater detail here: Letter codes in molecular term symbols. In this context, it indicates that it is the second excited state of the same multiplicity as the ground state. The ground state, labelled with $X$, is a doublet; the first doublet excited state is labelled with $A$, and the second doublet excited state is labelled with $B$.

The superscript $2$ denotes the multiplicity, i.e. this is a doublet state, with total spin $S = 1/2$ (technically the spin quantum number) such that $2S + 1 = 2$. The dinitrogen cation $\ce{N2+}$ will have one unpaired electron in its low-lying electronic states, so it's not too surprising that the total spin is $1/2$.

Finally, there is a bit more to the symbol $\Sigma$ than just being a substitute or a generalisation of $\mathrm{A}$. In the linear point groups, this denotes the total orbital angular momentum about the internuclear axis. In this state, there is no orbital angular momentum about this axis, and the relevant quantum number is $\Lambda = 0$, which is denoted by the letter $\Sigma$. ($\Lambda = 1$ is a $\Pi$ state, $\Lambda = 2$ a $\Delta$ state, and so on.)

If you think about the ground state electronic configuration of $\ce{N2+}$, it is essentially the same as $\ce{N2}$, but with one unpaired electron in the $\sigma_\mathrm{g}$ orbital. In order to generate any kind of orbital angular momentum, you need to have an asymmetric population of $\pi$-type MOs. Since this isn't the case, the ground state of $\ce{N2+}$ has no orbital angular momentum, and is consequently a $\Sigma$ state. A similar argument can be applied to the second excited state, which seems to arise from a $\sigma\to\sigma$ excitation.


At what amount is digesting soap unhealthy? Several of my friends, when they do the dishes manually, do not rinse the dishes after cleaning but merely dip them in the dishwater and then dry them (using a towel or a dish rack). I myself on the other hand always rinse the dishes with fresh (hot) water from the tap before drying them, mostly because the idea of having a dose of soap with my next meal freaks me out.

But is there some guideline as to how much of the soap you can ingest until something goes wrong, be it the taste of the next meal or some poisonous reaction? The soap used for doing the dishes manually doesn't have any warning labels on it... I've found this describing poisonous effects of automatic dishwater soap, and this pertaining to a dish soap that is used for manual dishwashing.

I'm less familiar with safety regulations elsewhere in the world, but in the US, "chemical manufacturers" (which includes soap, sugar, salt, and just about everything else you can imagine) have to produce Materials Safety Data Sheets (MSDS) and make them publicly available.

Out of curiosity, I did some digging for soaps. Proctor and Gamble is a huge consumer producer, including dish detergent. So there are MSDS sheets for soap.

The MSDS gives oral LD50 values. That is, the lethal dose required to kill half the population.

The LD50 for soaps seem to vary by animal, but judging from a few MSDS I pulled up, they're ~2 g/kg or so. (That value was for rats.)

So if I had to guess, given, say a ~70 kg adult, we'd be talking about 140 g of soap, minimum to kill a person.

Now.. that's acute doses. I have no ideal if anyone has established long-term toxicity of soaps, and I can't find anything obvious online.


Detail from Integral Evaluation in Szabo (Appendix) This is a step from Appendix A in Szabo and Ostlund's Quantum Chemistry. In short, the question is how to get from \ref{A.20} to \ref{A.21}. We can factor out some constants $K$ in the expressions to get:

\begin{align}
  (A| -Z_C/r_{1C} |B)
  &=
    K \left(\frac{1}{2\pi^2}\right)
    \int \mathrm{d}\mathbf{k} \; e^{-k^2/4p} k^{-2}
    \exp\big(-i\,\mathbf{k}\cdot(\mathbf{R}_P - \mathbf{R}_C)\big)
    \tag{A.20}\label{A.20}\\
  &=
    K \left(\frac{2}{\pi|\mathbf{R}_P-\mathbf{R}_C|}\right)
    \int\limits_0^\infty \mathrm{d}k \; e^{-k^2/4p} k^{-1}
    \sin(k|\mathbf{R}_P-\mathbf{R}_C|)
    \tag{A.21}\label{A.21}
\end{align}

Part of the note between these steps is that (paraphrasing) we can let $\mathbf{R}_P-\mathbf{R}_C$ lie on the $z$-axis and use that $k\cdot (\mathbf{R}_P-\mathbf{R}_C)= k|\mathbf{R}_P-\mathbf{R}_C|\cos\theta$, "then we can easily perform the angular part of the integration over $\boldsymbol{k}$ to obtain [\ref{A.21}]."

What I have done: It seems that Euler's formula must be applied to the second exponent in \ref{A.20} but using the hint we'd have something like $$\cos(k|\mathbf{R}_P-\mathbf{R}_C|\cos\theta) + i\sin \dots \text{etc}. \tag{???}$$
 I guess that the sine in \ref{A.21} follows integration of the (real) cosine in Euler's expansion. In line with this, \ref{A.21} is multiplied by a factor of $2\pi$.

I will keep at this but I think if it puzzles me, then others may have stumbled over it so I am posting.
Hopefully removal of the common factors ($K$) makes this easier to read but the problem is on pages 413-414 of Szabo and Ostlunds book.

As commented, do note that the integral (A.20) is taken over all $\mathbf{k}$-space, whereas the integral (A.21) is one-dimensional. The explicit calculation of this integral follows as

$$\begin{align} \int\text{d}\mathbf{k}\,\Psi(k)e^{-i\mathbf{k\cdot r}} &=\int_0^\infty\text{d}k\,k^2\int_0^\pi\text{d}\theta\int_0^{2\pi}\text{d}\varphi\,\sin\theta\Psi(k)e^{-ikr\cos\theta}\\
&= 2\pi\int_0^\infty\text{d}k\,k^2\Psi(k)\int_0^\pi\text{d}\theta\,\sin\theta e^{-ikr\cos\theta}\\
&= 2\pi\int_0^\infty\text{d}k\,k^2\Psi(k)\left[\frac{e^{ikru}}{ikr}\right]_{u=-1}^1\\
&= \frac{4\pi}{r}\int_0^\infty\text{d}k\,k\Psi(k)\sin(kr).\end{align}$$

In the first equality we convert this integral from Cartesian to spherical coordinates, in the second equality we evaluate the integral over $\varphi$, in the third equality we substitute $u = -\cos\theta$, and in the fourth equality we make use of Euler's formula to get the exponential representation of $\sin(kr)$.


Are the triply-bonded carbons in pyridyne (and benzyne) sp-hybridised? I was given this molecule (3,4-pyridyne) in a test:

I had to find out if it had any atoms which had a hybridization state other than sp2. The solution said that it doesn't have any. But aren't the triply-bonded carbon atoms sp-hybridized?

Edit:
Benzyne also has a similar structure, with a distorted triple bond -

TL;DR It is sp-hybridised, but maybe not in the way we normally think of alkynes as being.

There are a couple of things to say before going into the actual answer.

Firstly, hybridisation is derived from geometry, not from the bonding pattern (in other words, how many σ- and π-bonds there are). In most cases, the bonding pattern dictates the geometry: for example, most saturated carbons are tetrahedral, and a tetrahedral geometry usually signals sp3 hybridisation, so saturated carbons are usually described as being sp3-hybridised. So, effectively, the bonding pattern does dictate hybridisation. This is all fine, but only up to a certain point, because it can break down in exceptional cases such as cyclopropane. Because of the geometrical constraints of a three-membered ring, there is not quite enough room for the carbon to be perfectly tetrahedral, and so it is not correct to label the carbons as being sp3-hybridised in the usual sense.

Secondly, hybridisation is usually more nuanced than "sp", "sp2" or "sp3". It is individual orbitals that are hybridised, not atoms; so for example, we can talk about an sp orbital, which is a linear combination (a sum) of one part s-orbital and one part p-orbital. When we say that "a carbon atom is sp-hybridised", what we really mean is that the bonding of that carbon atom can be described using four valence orbitals: two sp-hybridised orbitals, and two unhybridised p-orbital. Likewise, an "sp2 carbon atom" uses three sp2-hybridised orbitals and one unhybridised p-orbital; and an "sp3 carbon atom" has four sp3 orbitals.

Again, in the vast majority of cases, these three categories are sufficient to accurately classify all carbon atoms in a molecule. However, as you may guess, this breaks down in exceptional cases: cyclopropane is the most well-known example. Here, each carbon has two sp5(ish) orbitals and two sp2(ish) orbitals. It uses the sp5 orbitals to form σ-bonds to the carbons next to it, and the sp2 orbitals to form σ-bonds to the two hydrogens on it. "How is it possible to have five p orbitals?" sp5 does not literally mean one s orbital and five p orbitals; it just means the ratio of s-orbital and p-orbital contribution to this particular hybrid orbital is 1:5.

With that out of the way, here is how benzyne works (PBE0/cc-pVTZ gas phase optimisation + NBO analysis). We're going to look at the orbitals that the triply bonded carbon (C1) uses for bonding to its neighbours (C2 and C6, see structure below):†

One sp orbital (50.02% s, 49.75% p) for forming a σ-bond to C2;
One p orbital (0.39% s, 99.38% p) for forming the π-bond to C2 in the plane of the molecule;
One p orbital (0.00% s, 99.84% p) for forming the π-bond to C2 perpendicular to the plane of the molecule;
One sp orbital (49.64% s, 50.17% p) for forming a σ-bond to C6.

On their own, the numbers don't tell you which direction the p-orbitals are pointing in, but you can use visualisation software to figure this out. From left to right, these are the (σ- or π-, not the individual hybrid) orbitals, in the same order as listed above.

Based on the above, the orbitals can be crudely illustrated as such:

This is, in fact, pretty much what an organic chemist would describe as an "sp-hybridised carbon atom". However, it is not sp-hybridised in the same sense as an ordinary alkyne. In an ordinary alkyne, the σ-bond between the two triply bonded carbons comes from overlap of sp-orbitals. This is also the case here, but there is a distinction. In an ordinary alkyne, the sp-orbitals point directly towards each other. Here, they are not quite aligned in the same way; they instead form what is sometimes called a "bent bond". The same can be said of the C1–C2 π-bond.

This is fundamentally because of the geometry of the ring. Since we have already allocated two p-orbitals to two other bonds, the two sp-orbitals must be derived from the same p-orbital. As a consequence, the two sp-orbitals must point in diametrically opposite directions, 180° away from each other. If C1 were to use an sp-orbital pointing directly towards C2, then the remaining sp-orbital would point rather far away from C6, significantly weakening the corresponding bond with C6. The orbital geometry that you see is a compromise.

The triply bonded carbons in pyridyne, for all intents and purposes, can be described in the same way as those in benzyne. The effect of nitrogen on the orbitals above is tiny; qualitatively they look the same (the colours are flipped for the first and last ones relative to the benzyne orbitals earlier, but this is of no physical significance).

† The numbers don't add up to exactly 100% because the tiny d- and f-orbital contributions are neglected.


Removal of octadecanoic acid from paraffin wax I have some very plain candles. I don't think they contain anything but paraffin wax and octadecanoic (stearic) acid, since they are odorless and very hard to bend, tending to crack instead (even at around 45 degrees Celsius). I want to make some moldable wax plugs for bottles without heating over about 50 degrees Celsius (it'll burn my hands, I have no other tools to use). How can I get rid of the acid? I don't particularly care if the acid is destroyed; but it would be nice if it wasn't so I could make soap or something.

Some possible ways I've thought of:

Distillation. I can't do this because it stinks and clogs up my glassware. Also, stearic acid decomposes around the same temperature as paraffin wax boils.
Dissolution of octadecanoic acid in some alcohol. Wax won't be dissolved because alcohols are polar and wax is not. However, I don't know how to actually do this as the wax may block access to most of the acid.
Neutralization of the acid using aqueous solutions of sodium hydrogen carbonate, sodium carbonate, or sodium hydroxide. This seems like the fastest way of removing acid; however the base may not come into contact with the octadecanoic acid due to separation of the layers.

What seems like the best or most effective method of removing the acid?

It should be noted that typically, candles are about 10% stearic acid and 90% paraffin. Also, there are many different grades of paraffin waxes which are generally identified by their melting points (often given as ranges); candles are usually made from paraffin graded from 53.3 °C (128 °F) to 65.6 °C (150°F), while paraffin of 76.6 °C (170 °F) is more common for molding and sculpting.

So I'm pretty sure that you didn't know what grade of paraffin you are after, and I would guess that you are looking for something more like cheese wax. While cheese wax is significanly softer, it actually has a generally higher melting range of perhaps 72 - 82 °C (161 - 179 °F).

Cheese wax is a blend of paraffins:

<30%  paraffin wax fume (Cas# 8002-74-2)
<30%  paraffin waxes and hydrocarbon waxes, microcryst. (Cas#
63231-60-7)
<20%  petrolatum (Cas# 8009-03-8)
30% Slack wax, petroleum, hydrotreated (Cas# 92062-09-4)

However, there may be multiple grades of paraffin used for your candles or the grade used may be of a low enough melting point (MP) to be useful for making the product you desire (in conjunction with other materials to make it softer for sculpting or molding).

The "analysis" and the separation procedure would be based on the melting point. So, (I would) try to use the melting points to separate the stearic acid from the lower temperature fractions.

Stearic acid's MP is 69.3 °C (156.7 °F)

Paraffin's MP may be as low as 37 °C (98.6 °F), but finding that grade of paraffin in your candles is unlikely. With a bit of luck, a low temperature grade of paraffin (48.3 °C (119 °F)) might show up; but more likely would be grades having a melting point of 53.3 °C (128 °F) and/or 61.1 °C (142 °F), which are common for candle making.

Step 1. Melt the candles in a pot and remove the strings. If possible, you would want to use a hot plate with accurate, good-quality temperature control.

Step 2. Insert a (candy) thermometer and a spoon or ladle in the pot (maybe a screen-type of ladle, like what is used for noodles and dumplings). You need the ladle to be warm, so that the wax doesn't freeze and stick to it when you use it to scoop out the stearic acid.

Step 3. Reduce the heat until precipitates begin to form. This should begin to happen somewhere around 70 °C. With an open pot, the cooler surface will probably cause the top to skim over. The higher melting point fractions might be what solidify first on the top, like slag on molten steel. The first precipitates or slag should be higher in octadecanoic (stearic) acid concentration. Use the ladle to scoop/filter out the precipitates (put the stearic acid in a in another pot or a bowl). You should keep collecting precipitates until it reached about 55 °C or until it all becomes solid. Be sure to make note of the final temperature. If the temperature at which it all solidifies is at or above 60 °C, then the kind of paraffin that you have is probably not what you wanted; and it will be quite difficult to separate, and this procedure is not very useful for separating the stearic acid from the paraffin.

Step 4. Even if the paraffin has a low MP, this will not be a 100% perfect procedure (far from it); you will probably need to repeat the (enrichment) procedure a couple, or maybe several, times. Also, you could probably recover some (low MP) paraffin from the precipitated/enriched stearic acid filtride. It might be favorable to have a few bowls or pots which represent collections at different temperature ranges... like, for example, the first bowl would contain solids from 70 - 65 °C; the second bowl contains solids from 64 - 58 °C; and the third bowl contains solids from 57 - 52 °C.

Step 5. Measure the new MPs. A good anylysis of the cleanliness of this procedure will be to compare the MPs. A pure solution of stearic acid will melt/solidify at exactly 69.3 °C, but obtaining a pure solution in this manner, is a pretty lofty dream. So, after the stearic acid fraction has melted, let it cool and  mark the temperature at which it begins to solidify and the point at which it is completely solid. If the two fractions that have been separated have distinctly different melting points, that's pretty good. Specifically regarding the paraffin, if it doesn't begin to solidify until it's below 50 °C then, great job, it's pretty clean and you obtained the low temperature paraffin that you desired!

Step 6. Depending on the cleanliness (melting point) and amount of the stearic acid that you have, you may want to try to salvage some of the paraffin from the stearic acid (filtride). I believe that this will be more difficult, and I would advise using a third pot (or second bowl) to collect the salvaged paraffin. And then work on cleaning the stearic acid out of the salvage, checking the cleanliness as in step 5, before combining the salvaged paraffin with the first "batch".

Step 7. Further chemical reactions for cleaning up the stearic acid (or converting it to soap) could be performed with better results from the enriched fractions.


Is there safe chemicals for making voice pitch or bass? I'm looking for safe experiment that can pitch and bass my voice.finally I find a video that do it but he said "not try in home,ok..?".Is there safe tool or chemicals for Manipulation of the larynx is used to generate a source sound with a particular fundamental frequency.

other my question is about below viedo link:How these chemicals make voice pitch or bass!(Helium and Sulfur Hexafluoride)

Pitch is a perceptual property of sounds that allows their ordering on a frequency-related scale,1 or more commonly, pitch is the quality that makes it possible to judge sounds as "higher" and "lower" in the sense associated with musical melodies.

video link

Assuming that helium and sulfur hexafluoride (as shown in the video) will give you the correct sounding "pitch" and "base", it would be fine if done correctly and safely.

Helium is incredibly inert and so will not cause any sort of toxicity. It is also a very light gas, so it will exit your lungs quickly. Just do not breath only helium, or for a long time or you could suffocate. Make sure to get lots of normal air.

Sulfur Hexafluoride is also rather unreactive, but more dangerous. This is because it is heavy and will tend to stay in your lungs. Only use it for a short period of time, or you can asphyxiate. While the same goes for helium, it tends to exit your lungs more easily.

The actual mechanism that causes them to work is based of how dense they are. It does not change the pitch, however. Instead, it changes something called timbre. This is basically the fact the helium allows sound to travel much faster through it, so it is a similar affect to playing a video at a faster speed.

Sulfur hexafluoride does the exact same thing, except it is much denser and does not allow sound to travel through it very quickly.

If you want to do this, be careful! You can suffocate, especially with sulfur hexafluoride.


Definition of “solution” In my textbook “solutions” are defined as follows:

  Homogenous mixtures of two or more substances are known as solutions.

Should the two substances always be non-reacting?

The definition has no indication of this, buts it’s just something that came to my mind.
Won’t the solution still be homogeneous so shouldn’t we still call it a solution?

This is what Encyclopaedia Britannica definition of a solution in chemistry:

  Solution, in chemistry, a homogeneous mixture of two or more substances in relative amounts that can be varied continuously up to what is called the limit of solubility. The term solution is commonly applied to the liquid state of matter, but solutions of gases and solids are possible. Air, for example, is a solution consisting chiefly of oxygen and nitrogen with trace amounts of several other gases, and brass is a solution composed of copper and zinc.

A solution consists of solutes (at least one) and a solvent. The solute is define as the substance that is dissolved in the solvent. In other words, for solutions with components in the same phase, the substance(s) present in lower concentration are solutes while the major substance present in highest abundance is the solvent. For example, 190-proof alcohol is a liquid-liquid solution, which is the mixture of 95% ethanol and 5% water by volume. The major substance in this mixture is ethanol. Therefore, it is the solvent. The 5% water is the solute. Brass is a solid-solid solution of copper and zinc metals. General composition of brass is 65% copper and 35% zinc by weight. Hence, in brass, copper is the solvent and zinc is the solute. Air is a solution of multi-gases. In air, 72% nitrogen and 20% oxygen, and rest is other gases including $\ce{CO2}$ and $\ce{H2}$. Thus, the solvent of air is $\ce{N2}$ while $\ce{O2}$ is one of its solutes. Saline solution, on the other hand, is a solid-liquid solution. In a saline solution, solid salt ($\ce{NaCl}$) is the solute dissolved in water, which is the solvent (liquid).


Can I express the rate of a reaction as percent per second (%/s)? Rate is calculated as change in concentration over change in time, so usually $\pu{M/s}$ $(\pu{mol L-1 s-1}).$

Say, I'm calculating it for $20\%$ solution of ammonium hydroxide, so my final rate is in $\pu{\%/s}.$ Is this a proper format, or should I calculate for $\pu{mol L-1}?$

According to IUPAC Recommendations for chemical kinetics [1], rates $\nu$ are defined for reactant $\ce{B}$ and product $\ce{Y}$ as a time $(t)$ derivative of the amount $n$ (in general) or per unit volume $V$ (for the closed system):

$$
\begin{array}{lll}
\hline
\text{System} & \text{Consumption} & \text{Formation} \\
\hline
\text{Open} & \displaystyle\nu(n_\ce{B}) = -\frac{\mathrm dn_\ce{B}}{\mathrm dt} & \displaystyle\nu(n_\ce{Y}) = \frac{\mathrm dn_\ce{Y}}{\mathrm dt} \\
\hline
\text{Closed} & \displaystyle\nu(c_\ce{B}) = -\frac{1}{V}\frac{\mathrm dn_\ce{B}}{\mathrm dt} & \displaystyle\nu(c_\ce{Y}) = \frac{1}{V}\frac{\mathrm dn_\ce{Y}}{\mathrm dt} \\
  & \displaystyle\nu(c_\ce{B})|_{V = \mathrm{const}} = -\frac{\mathrm d[\ce{B}]}{\mathrm dt} & \displaystyle\nu(c_\ce{Y})|_{V = \mathrm{const}} = \frac{\mathrm d[\ce{Y}]}{\mathrm dt} \\
\hline
\end{array}
$$

Rates of consumption and formation that are measured in $\pu{mol L-1 s-1}$ are defined for the closed system at constant volume, and molar concentration arises from its definition.
Whether you can or cannot substitute molar concentration $c_i$ with fractions can be answered by looking at the relationship between the two.

For example, molar concentration $c_i$ expressed in terms of mass fraction $\omega_i$ via the density of solution $\rho$ and the molar mass $M_i$ of the $i$th component:

$$c_i = \frac{\omega_i \rho}{M_i}\tag{1}$$

Similarly, molar concentration $c_i$ expressed in terms of mole fraction $x_i$ via the density of solution $\rho$ and the average molar mass $\bar{M}$ of the solution:

$$c_i = \frac{x_i\rho}{\bar{M}}\tag{2}$$

As you can see, there is a problem: as the reaction progresses, not only the amounts (or fractions) would change, but also the density and the molar mass of the solution may and will change in most cases, which leads to inapplicability of percent fractions as time derivatives.
All in all, in chemistry it's just easier to manipulate with the quantities based on amount of substance.

References

Laidler, K. J. A Glossary of Terms Used in Chemical Kinetics, Including Reaction Dynamics (IUPAC Recommendations 1996). Pure and Applied Chemistry 2009, 68 (1), 149–192. DOI: 10.1351/pac199668010149. (Free Access)


Dependance of heat capacity at constant volume on temperature for solids and liquids In thermodynamics we have an equation for the change in internal energy:
$$\mathrm dU = nC_V\,\mathrm dT\tag{1}$$
$$\int{\mathrm dU} = \int{nC_V\,\mathrm dT}\tag{2}$$
Since $C_V$ is independent of temperature, for an ideal gas
$$ΔU = nC_VΔT\tag{3}$$
Why is $C_V$ independent of temperature for an ideal gas? On the other hand, in case of solids and liquids, $C_V$ may or may not depend on temperature. Why?

The molar heat capacity of classical ideal gases at constant volume is temperature independent, because there are not considered quantization steps of electronic, vibrational and rotational energy.
The molar heat capacity ideal gases in the context of quantum-mechanic-aware gas theory is not temperature independent anymore. Because the electronic, vibrational and rotational energy contribution depends on ratio of mean energy per degree of freedom $E=\frac{1}{2} \cdot k_\mathrm{B} \cdot T$ versus the energy of quantization steps.
The capacity raises as rotation energy contribution kicks in, and later vibration energy as well, and finally electronic energy, as their quantization steps grows in this order.
There is a similar temperature dependent effect for liquids and solids, where quantized vibration is essentional, especially for solids.

The classical model of ideal gas does not take into account quantum effects. By warming up of an ideal gas, the mechanical energy gets evenly distributed between energy of linear motion, bond vibration and molecular rotation, proportionally to absolute temperature.
Quantum effects, taking part in behaviour of real gases, liquids and solids affects energy distribution. Vibration, rotation or electron energy are allowed by laws of quantum mechanics to have just some energy levels.
Imagine you throw small stones upwards to a steep hill. Stronger you throw it, the higher it would stay. This is analogous to an ideal gas.
Now you try to throw such stones to open windows of a tall building. If you make weak throwing, it will not flow even to the nearest window and will fall back. Stronger and stronger throwing ( = higher temperature ) would eventually cause some stones would reach at least the 2nd floor. Later even 3rd one etc.  This is like quantization of possible energies for real gases and condensed phases.


Is there always a limiting reagent in every equation? Is there always a limiting reagent in every equation? For instance, is there a limiting reagent in $\ce{CaCl2 + Na2CO3 -> 2NaCl  + CaCO3}$ where the same number of moles are reacting with each other?

An interesting philosophical question. We are given a chemical equation: $$\ce{CaCl2 + Na2CO3 -> 2NaCl  + CaCO3}$$ and that the same number of moles of the reactants are used. Now the OP poses the question "Is there a limiting reagent?"

As chemists we can say that 1.00 moles of $\ce{CaCl2}$ were reacted with 1.00 moles of $\ce{Na2CO3}$. We'd pragmatically say that the moles were equal and that there was no limiting regent.

However this is a significant figure argument, not reality. Avogadro's number is about $6.022\times10^{23}$. There is no lab in the world that can weight out precisely one mole of $\ce{CaCl2}$ accurate to 23 decimal places. We don't even know Avogadro's number accurate to 23 decimal places! The best value is $6.022140857(74)\times10^{23}$.

So in reality, despite our best efforts as chemists, one reagent or the other will be in some small excess which we'd consider to be insignificant given the overall precision of the experiment.


Water reacts or doesn't react? I know that when you have $\ce{HCl}$, $\ce{H2O}$, and $\ce{NaOH}$ that you produce $\ce{H2O}$ and $\ce{NaCl}$.

However some people say water does react and so would think that this 2 step reaction is correct and the 1 step reaction is wrong.

$\ce{HCl + H2O + NaOH -> H3O+ + Cl- + NaOH -> 2 H2O + NaCl}$

However others say that water does not react and think that the 2 step reaction is wrong and the 1 step reaction is correct.

$\ce{HCl + NaOH -> H2O + NaCl}$

I personally think that both of these happen and so water consumed = water produced and thus there is no change in volume. So does water react with the $\ce{HCl}$ via the 2 step reaction or is it just a spectator molecule waiting around until there is more water and salt?

The answer to this depends on how the reagents are mixed. For one, you can have the reaction without water, as pointed out by @Klaus in the comments. If you blow dry HCl on dry NaOH, they will react without involving water in the reactants.

$$\ce{HCl + NaOH -> H2O + NaCl}$$

This is probably not the case though, and if you're dealing with solutions, then yes, water will react. If you start by adding the HCl to pure water, you'll have

$$\ce{HCl + H2O -> H3O+ + Cl-}$$

This representation is prefered over the simple dissociation $\ce{HCl -> H+ + Cl-}$ because it shows that you don't have "free" protons in solution. You could actually even say that this reaction has already happened since, considering you're using hydrochloric acid it's already dissolved, and this step would be just a dilution.

If you start with the NaOH solution, you could argue that water does not react since

$$\ce{NaOH -> Na+ + OH-}$$

but that only happens in the liquid state (usually in solution). Also, you can't ignore the ions solvation (which is one, if no the most important factor which affects the solubility of the sodium hydroxide salt) and equilibrium

$$\ce{H2O + OH- <=> OH- + H2O}$$

And what this means is that you don't have fixed hydroxil ions and water molecules, so you can't simply ignore the solvent and assume it's just a spectator molecule waiting around.

To write the two step reaction, it's not a good idea to do it in one line, because as you can see on the first step you have NaOH on both sides, which is something that doesn't make much sense. It's better to do it like this:

$$\ce{HCl + H2O -> H3O+ + Cl-}$$

$$\ce{H3O+ + Cl- + NaOH -> 2H2O + NaCl}$$

It's usually a good idea to represent salts in their ionic form when dissolved, which would be

$$\ce{H3O+ + Cl- + NaOH -> 2H2O + Na+_{(aq)} + Cl^{-}_{(aq})}$$


Why does benzaldehyde not respond to Fehling's test? All aldehydes are supposed to respond to Fehling's test. So why doesn't benzaldehyde respond to it?

Does it have something to do with the fact that benzaldehyde is an aromatic compound? If so, then how?

The rate-limiting step of the Fehling’s test reaction with aldehydes is the formation of the corresponding enolate:

The subsequent reaction of the enolate with copper(II) proceeds through a single electron transfer mechanism.

Aldehydes that lack alpha hydrogens, such as benzaldehyde or pivalaldehyde (2,2-dimethylpropanal) cannot form an enolate and thus do not give a positive Fehling’s test result under usual conditions.


How or Why does diesel/kerosene have a much higher 'flash point' but lower auto/self-ignition point than gasoline/petrol? This is a very common question, but I have yet to find a good theoretical answer to both questions (diesel has higher flash point BUT lower spontaneous ignition temp.)

Flash point and autoignition temperature measure different things: gasoline and diesel are optimised for different ignition conditions
First, the standard definitions.
Flash point
Flash point is the lowest temperature at which a liquid can gives off vapor to form an ignitable mixture in air near the surface of the liquid.
Autoignition point
The autoignition temperature or kindling point of a substance is the lowest temperature in which it spontaneously ignites in a normal atmosphere without an external source of ignition...
The first point to make on the differences between diesel and gasoline is that the liquids are designed for different properties in engines.
Petrol/gasoline engines need fuels that vaporise easily to give a mixture that is ignited by a spark. So they want a mixture that is easy to ignite but that does not explode spontaneously in the engine (which is bad for the engine and upsets the control of when the exploding fuel happens). So a low flash point is good but a low autoignition point is bad.
Diesel engines work differently. The explosion is not triggered by a spark but by compression of the fuel mixture which heats it and leads to a spontaneous explosion in the mixture driven by the heat of compression. So a diesel engine wants a specific autoignition temperature and one that is low enough to create reliable fuel ignition.
So the properties of gasoline and diesel are there by design as that is how the engines they feed are meant to work.
The chemical properties that create these properties are, broadly, different mixtures of hydrocarbons. Gasoline tends to be more volatile and contains far more unsaturation (aromatics and branched hydrocarbons). Volatility is necessary so the mixture can be vaporized easily to make a flammable mixture. Diesel tends to have many more straight chain hydrocarbons which are more susceptible to attack by oxygens in the air. Diesel also tends to be much less volatile than gasoline as volatility is much less important to creating an effective mixture in the engine. This has the side effect of increasing the flash point for diesel.
So design intent and chemical properties make the metrics different.


What exactly is temperature? I've read at many places that temperature is the average kinetic energy of particles present in an object. I just don't intuitively get how kinetic energy is connected with temperature. And how is heat connected with temperature then? Then what exactly is temperature? All of the descriptions given online are very confusing.

Heat is the transfer of energy to or from the body in forms other than matter flow or work (organized energy transfer, such as pushing).

Temperature is only a well-defined property for a collective body (you wouldn't be able to tell me the temperature of a single atom, for example). Like you said, it's the property of matter describing the amount of kinetic energy of the particles in the body. As to why this is, I'd ask: what happens at absolute 0?

   At absolute 0, heat has transferred out of the system so much so that you cannot lower the energy of the system any more.

(Note for the knowledgeable reader: there is indeed still a quantum mechanical phenomena -- zero point energy-- that prevents some energy from leaving the molecule, but that's a conversation for another time).

From a thermodynamic definition, temperature is the description of how the internal energy changes with entropy for a closed (no matter flows in or out) system of constant volume:

$T=\left(\frac{\partial{U}}{\partial{S}}\right)_{N,V}$

By increasing the entropy of the system by a fixed amount, the temperature of the system tells me by how much the internal energy will increase. Now, this is not a very useful form, as you cannot directly increase the entropy of a body (you must add energy and then let the entropy indirectly increase). It's much more useful to consider the inverse temperature:

$\frac{1}{T}=\left(\frac{\partial{S}}{\partial{U}}\right)_{N,V}$

At absolute zero, everything is in the lowest energy state. Any small transfer of energy to the system will result in a large increase in entropy. But this was only a small change, so the internal energy doesn't increase by much. Compare to a system at room temperature, where I must transfer much more energy to the system to achieve the same magnitude of increase in entropy.


Electrical conductivity of organic solvents Do organic solvents such as pure ethanol and pure methanol conduct electricity? why is this so?

I have done some research and have not been able to find anything conclusive. I would think it would not conduct because I cannot see any medium for the passage of charge.

Ethanol and methanol dissociate into ions, just as water does.  For water:

$$\ce{2 H2O <=> H3O+ + OH-}$$

For an alcohol $\ce{ROH}$:

$$\ce{2 ROH <=> ROH2+ + ^-OR}$$

For water the self-ionization constant is $K_\mathrm{w} \approx 10^{-14}$.  For ethanol or methanol, I was unable to find data on the self-ionization constant, which is essentially the same as the $\mathrm{p}K_\mathrm{a}$ of the alcohol in a solution of itself.  The $\mathrm{p}K_\mathrm{a}$ of ethanol in water is slightly higher than the $\mathrm{p}K_\mathrm{a}$ of water in water.  Also, the lower dielectric constant in alcohols will further disfavor self-ionization relative to water.  So for methanol and ethanol the self-dissociation into ions will be weaker than in water.

These ions will still be present, though, and will conduct electricity even in pure ethanol and methanol solutions, but the conductivities will be very low.  Very pure water has a resistivity of $\pu{18.2 MΩ cm}$.  If they self-ionize to a lesser extent than water does, the resistivities of methanol and ethanol will be even higher than water's.


A titration mixture having pottasium oxalate and pottasium hydrogen oxalate! I came across this question, this is how it goes-

A solution is prepared by dissolving a solid mixture of $\ce{K2C2O4}$ and $\ce{KHC2O4}$. A $\pu{10 ml}$ portion of this solution required $\pu{10 ml, 0.05 M \ce{KOH}}$ solution for titration reaction. In a seperate analysis $\pu{10 ml}$ of the same stock solution required $\pu{10 ml, 0.06 M}$ acidified $\ce{KMnO4}$ solution for titration.

I  have  to  find the ratio of $\ce{K2C2O4}$ to $\ce{KHC2O4}$ in the initial reaction mixture.

I know that(I think), 1 mole of $\ce{KHC2O4}$ would require 1 mole of of $\ce{KOH}$ for neutralization. While $\ce{K2C2O4}$ would not react with $\ce{KOH}$ (please correct me if I am wrong).

Also, $\pu{31.8 g}$ of $\ce{KMnO4}$ would oxidise $\pu{63.5 g}$ of $\ce{K2C2O4}$. But when I don't know how (differently) it shall oxidise $\ce{KHC2O4}$.

You are right, $1$ mole of $\ce{KHC2O4}$ reacts with $1$ mole of $\ce{KOH}$. From that first piece of information in your exercise, you can find out that, in $\pu{10mL}$ of solution, you have $\pu{0.05 * 0.01 = 5 * 10^{-4}}$ moles of $\ce{KHC2O4}$. The next piece of information is meant to let you how much substance there is in the solution. $10$ moles of both substances react with $4$ moles of acidified $\ce{KMnO4}$. Which means the sum of the quantities of both substances in that solution is $\pu{(0.06 * 0.01 * 10) / 4 = 15 * 10^-4}$. So you got $\pu{a = 5 * 10^{-4}}$ and $\pu{a + b = 15 * 10^{-4}}$. Do the math.

Please tell me if I was clear enough.


Why is a buffer solution most effective when concentrations of salt and acid are the same? Why does a buffer work most effectively when the concentration of salt and acid is 1:1?

Using the Henderson-Hasselbalch-Equation:
$$\mathrm{p}\ce{H}=\mathrm{p}{K_a}+\log\frac{\left[\ce{salt}\right]}{\left[\ce{acid}\right]}$$
and buffer capacity:
$$\phi=\frac{\text{number of moles of acid/base added to 1L solution}}{\Delta\mathrm{p}\ce{H}}$$
where:
$$\Delta \mathrm{p}\ce{H} = \left|\log\frac{\left[\ce{salt}\right]}{\left[\ce{acid}\right]}\right|$$
it is minimum when $$[\ce{salt}]=[\ce{acid}]$$


Is tea powder better dissolved when put in hot water? I usually make tea by adding two spoons of tea powder to water at room temp. and start to heat it. My friend said it will be better to add the powder once the water starts boiling. She said that it will dissolve tea more easily because of the sudden temperature difference experienced by tea powder.

Does it make a difference? Why?

I would say that your method is actually quicker (if that's what you mean by 'better'). My reasoning is as follows: the solubility of the powder in water increases with increasing temperature, as does the dissolution rate but the dissolution rate remains finite all the time.

1) If you add the powder and start heating at $t=0$ the concentration of dissolved powder will start to increase from $c=0$ to $c>0$ immediately. Once the water reaches the boiling point at $t=t_b$ already a certain amount (maybe even all) of the powder has dissolved i.e. some concentration of dissolved powder is present $c=c_1$.

2) If you boil the water first and then add the powder you start with $c=0$ at $t=t_b$, which is a lower concentration than in the powder-first case.

Because there is already more powder dissolved at the boiling point ($c=c_1$ vs. $c=0$ at $t=t_b$), the moment at which all powder is dissolved comes sooner for the powder-first case (situation 1).

Note that I approach this question purely from the physical point of view, I have no clue whether the two methods might result in differently tasting tea.

P.S. I ignored any possible changes in the boiling point due to the already dissolved powder. For low concentrations, as in the case of tea, this is no problem. If the concentrations become much higher then the change in boiling point might become dominant and the story changes


What is the reaction in a child's 'add water and watch the egg hatch' toy? My son and daughter got given gifts where you place a plastic egg in water, and the (rubber?) toy inside expands (over three days) and breaks the shell of the egg so it 'hatches'.

I also observed that three weeks later the toys shrank again, but grew once placed in water.

I found the chemical reaction fascinating.

My question is: What is the reaction in a child's 'add water and watch the egg hatch' toy?

The egg hatch toys are composed of a superabsorbent polymer. These polymers are able to absorb up to 30 to 60 time their volume in water by forming hydrogen bonds with the water molecules.

Sodium polyacrylate ($\ce{[-CH2-CH(CO2Na){}-]}_n%edit$) is the most common superabsorbent polymer and according to its wikipedia page, is the one used in the egg hatch toys.


Enantiotopic hydrogens in cyclopropanone Are the hydrogens that on the same carbon and those that are cis enantiotopic? And are they isogamous or anisogamous?

And will they anisochronous in a chiral environment NMR?

Isochronous - hydrogens that are in the same chemical environment leading to the same shift and splitting in NMR. If the hydrogens are enantiotopic they will be isochronous in an achiral nmr environment but anisochronous in a chiral environment.

I can see there is there is a C2 and two planes of symmetry but the TA said they will be isochronous in a chiral NMR resulting in a single singlet but as i did the replacement test I got two enantiomers which means that they will be anisochronous in chiral NMR, but I'm not sure.

Cyclopropanone is a reactive species, as such there aren't many experimental reports in which it is characterised, however some (possibly questionable data) does exist:

  Cyclopropanone is a highly reactive compound. It is stable for a few days at liquid nitrogen temperature but rapid polymerisation takes place above o or even below if traces of water are present. The reaction is highly exothermic.

  Ref: Recueil des Travaux Chimiques des Pays-Bas, 1966, 11, 1170. DOI:10.1002/recl.19660851113

.

  The NMR spectrum  (neat) shows a singlet  at 1.72 p.p.m.  relative  to  tetramethylsilane.

  Ref: Recueil des Travaux Chimiques des Pays-Bas, 1966, 11, 1170. DOI:10.1002/recl.196608511

It appears that in the absence of a chiral environment, all of the protons are in a single environment at 1.72 ppm (Hans Reich quotes a single peak at 1.65 ppm) which is what we'd expect (the molecule is essentially planar, and all of the protons are equivalent).

As you point out in your question, replacing any one of the protons (hypothetically) would yield two enantiomers, which, in the presence of a chiral environment, should be distinguishable. It can therefore be concluded that in a chiral environment, the protons should appear to be enantiotopic.*

* there are two slight caveats to this. Firstly, the compound would still be unstable and prone to polymerising, therefore there is no way of actually testing this. Secondly, the chiral environment must be able to interact with the molecule, for things like chiral alcohols, lanthanide shift reagents are (/used to be) commonly used, but I'm less convinced how you put a ketone into a chiral environment.