No buffer region in the titration of strong acids or bases? So recently I have conducted a acid-base titration experiment. I encountered that there is no buffer region in strong acid-strong base reaction. However, when take a weak acid or base buffer region appears.
WHY THIS IS LIKE THAT? What I mean is there any rational explanation for this?

Strong acids and strong bases essentially completely dissociate. So if you're adding the strong base to the strong acid, the solution stays very acidic, then neutralizes, then goes basic with just a small amount of the strong base.

With a weak-acid at the pK value you have as much $\ce{H+}$ as $\ce{A-}$. Thus a small amount of strong base or strong acid won't change the pH value much. That is the very notion of a "buffer solution."

So yes there is a rational chemistry explanation. :-)


How can I determine the quantum number of a electron lost during ion formation? How do quantum numbers change for ions?

  A sample question: Give the set of four quantum numbers that could represent the electron lost to form the Rb ION from the Rb atom.

The answer given is $n=5; l=0; m_l=0; m_s= +\frac12$. I chose the answer: $n=4; l=1; m_l=1; m_s= -\frac12$.

I am looking for an explanation. I am confused because I thought the question asked me to describe $\ce{Rb+}$ and the answer given is the same that I would choose for $\ce{Rb}$ at ground state. If it is asking me to describe the electron, how is that technically possible if I don't know where the electron went? Sounds silly and I know I'm over thinking it. Some insight would be appreciated.

The correct answer has been given in the comments section of this post. I am providing the answer here in accordance with a meta post discussion.

The question you were asked was in reference to the quantum numbers that correspond to the electron lost when $\ce{Rb}$ goes to $\ce{Rb+}$. The electronic configuration of $\ce{Rb}$ is $[\ce{Kr}]\ 5\mathrm{s}^{1}$; thus, you need the set of quantum numbers for a $5\mathrm{s}^{1}$ electron. That set of quantum numbers is:

$$n = 5;\ l= 0;\ m_{l} = 0;\ m_{s} = +{1\over2}$$


How to calculate Lennard-Jones potential with quantum mechanical methods I want to know the procedure to calculate the Lennard-Jones potential for a metal-halogen pair (specifically vanadium-chlorine). Is it possible to calculate using any QM packages like Mopac, NWChem, or Gaussian?

I am specifically looking for values of A and B in the 12-6 potential.

Yes, this is technically possible. A basic tutorial for this is in the excellent Psi4Numpy project, which I'll reproduce here with minor modifications. Their example fits the counterpoise-corrected MP2/aug-cc-pVDZ total interaction energy of the helium dimer.

from __future__ import print_function

import psi4

import numpy as np

import matplotlib as mpl
mpl.use('Agg')
import matplotlib.pyplot as plt

he_dimer = """
He
--
He 1 **R**
"""

distances = [2.875, 3.0, 3.125, 3.25, 3.375, 3.5, 3.75, 4.0, 4.5, 5.0, 6.0, 7.0]
energies = []
for d in distances:
    # Build a new molecule at each separation
    mol = psi4.geometry(he_dimer.replace('**R**', str(d)))

    # Compute the Counterpoise-Corrected interaction energy
    en = psi4.energy('MP2/aug-cc-pVDZ', molecule=mol, bsse_type='cp')

    # Place in a reasonable unit, Wavenumbers in this case
    en *= 219474.6

    # Append the value to our list
    energies.append(en)

print("Finished computing the potential!")

# Fit data in least-squares way to a -12, -6 polynomial
powers = [-12, -6]
x = np.power(np.array(distances).reshape(-1, 1), powers)
coeffs = np.linalg.lstsq(x, energies)[0]

# Build list of points
fpoints = np.linspace(2, 7, 50).reshape(-1, 1)
fdata = np.power(fpoints, powers)

fit_energies = np.dot(fdata, coeffs)

fig, ax = plt.subplots()
ax.set_xlim((2, 7))  # X limits
ax.set_ylim((-7, 2))  # Y limits
ax.scatter(distances, energies)  # Scatter plot of the distances/energies
ax.plot(fpoints, fit_energies)  # Fit data
ax.plot([0,10], [0,0], 'k-')  # Make a line at 0
ax.set_xlabel(r'intertomic separation ($\AA{}$)')
ax.set_ylabel(r'interaction energy ($\mathrm{cm^{-1}}$)')
fig.savefig('1b_molecule.pdf', bbox_inches='tight')

Once the set of energies is calculated, a least-squares polynomial fit is performed, giving

$$
f(x) = 6677721.45419193x^{-12} - 11394.79882998x^{-6}.
$$

This isn't a good (efficient or general) way of evaluating the fit function, but illustrates the placement of correct signs in the above code:

def f(z):
    return coeffs[0]*z**powers[0] + coeffs[1]*z**powers[1]

From the above we can see that the first and second coefficients in the polynomial fit correspond directly to $A$ and $B$, respectively. In this example, the energies are in wavenumbers ($\pu{cm^{-1}}$). Most quantum packages print in atomic units (Hartrees) and dynamics packages print in kJ/mol, which is fine, but be aware that in the above fit the units are incorporated into $A$ and $B$. Here is the resulting fit compared against the individual interaction energy calculations:

This was not in the question, but I think it's important to ask if this is the correct approach and investigate further. Notice how the fit is qualitatively ok near the minimum, but the shape in the dissociation region is qualitatively wrong, and the potential simply isn't attractive enough in the limit of infinite separation. This could arise in three ways:

The curve fitting isn't working properly, or there's a bug in the code.
The functional form of the chosen interatomic potential is incorrect.
The (Boys-Bernardi) counterpoise correction, which is known to overcompensate for BSSE, is giving strange results at large separation.

Assume that point 1 isn't an issue (if there's a bug anywhere, please comment). Point 2 can be investigated by also fitting against the Buckingham and shifted Morse potentials, which should display better short-range and long-range behavior:

\begin{align}
V_{\text{Buckingham}}(r;A,B,C) &= A e^{-Br} - \frac{C}{r^{6}} \\
V_{\text{Morse}}(r;A,B,C,D) &= A \left( 1 - e^{-B(r-C)} \right)^{2} + D
\end{align}

Note that for these energies, an unshifted Morse potential (no $D$ parameter) fails miserably. See here for more information about fitting Morse-type potentials. The Buckingham and Morse fits require a more general (arbitrary) curve fitting routine, and the script that made this plot can be found as an HTML comment in the post source code below.

I find it interesting that the Buckingham potential gives a worse fit than the Lennard-Jones one, even though it is supposed to reproduce the repulsive wall better with $e^{-r}$ rather than $\frac{1}{r^{6}}$. The Morse fit is remarkably good, which makes sense considering there are 4 free parameters rather than 3 (Buckingham) or 2 (Lennard-Jones). Before going any further, a check on point 3 by setting bsse_type=nocp:

This reveals that the counterpoise correction is definitely interfering with the quality of the fit, and the functional forms of the Lennard-Jones and Buckingham potential appear to not describe dissociation properly, though extended to infinite separation, the Morse potential is the one that is qualitatively incorrect. This can be attributed to using a single functional form to describe the total interaction or binding energy, and not the different components. I suspect that the apparent poor fit is due to a non-zero Coulomb interaction, rather than van der Waals-type interactions which at least the Lennard-Jones potential is meant to be used for.

To test this, here are interaction energy calculations at two levels of sophistication, both based on symmetry-adapted perturbation theory (SAPT). This is still the helium dimer, at 7 angstroms separation, with the aug-cc-pVDZ basis set and the monomer-centered basis approximation. All units are kcal/mol.

$$
\small
\begin{array}{lrr}
\hline
\text{Component}  & \text{SAPT0} & \text{SAPT2+3(CCD)}\delta_{\text{MP2}} \\
\hline
\text{Electrostatics} & -0.01796877 & -0.01777334 \\
\text{Exchange}       &  0.00000000 &  0.00000000 \\
\text{Induction}      &  0.01587019 &  0.01508578 \\
\text{Dispersion}     & -0.00013235 & -0.00016773 \\ \hline
\text{Electrostatics + Induction} & -0.00209858 & -0.00268756 \\ \hline
\text{Total}          & -0.00223093 & -0.00285529 \\
\hline
\end{array}
$$

At both levels of SAPT, the non-dispersive part of the interaction energy accounts for 94% of the total interaction! Is this true at 3 angstroms separation?

$$
\small
\begin{array}{lrr}
\hline
\text{Component}  & \text{SAPT0} & \text{SAPT2+3(CCD)}\delta_{\text{MP2}} \\
\hline
\text{Electrostatics} & -0.04852183 & -0.04836334 \\
\text{Exchange}       & 0.02045965 & 0.02219227 \\
\text{Induction}      & 0.04202349 & 0.04081412 \\
\text{Dispersion}     & -0.02228767 & -0.02843936 \\ \hline
\text{Electrostatics + Induction} & -0.00649834 & -0.00754922 \\ \hline
\text{Total}          & -0.00832637 & -0.01379630 \\
\hline
\end{array}
$$

The non-vdW interactions now account for 78% and 55% of the total interaction energy for each SAPT flavor, respectively. Although induction (also called polarization) is repulsive, I am including it in the Coulomb-type interaction since it is not purely quantum mechanical in nature like the exchange term. It is the exchange term, not induction, that is closer to the charge cloud penetration picture. Even so, the interaction between two helium atoms is not purely based on dispersion, and requires fitting another nonbonded term for electrostatics, the Coulomb term:

$$
V_{\text{Coulomb}}(\vec{r}_{i},\vec{r}_{j},q_{i},q_{j}) = \frac{1}{4\pi\epsilon_0} \frac{q_{i}q_{j}}{|\vec{r}_{i} - \vec{r}_{j}|}
$$

Our goal is to separate out the Coulomb-like terms from the interaction energy, so that $A$ will describe only the exchange contribution and $B$ will describe only the dispersion contribution; electrostatics and induction will be handled separately in the Coulomb term. This will be done by fitting only to the combination of exchange and dispersion, which are taken from SAPT calculations. MP2 binding energies are not CP-corrected.

As you can see, using SAPT energies provides a much better fit than naively using the total interaction or binding energies. I'm not sure why the total SAPT energy works so well, but it may be due to the absolute magnitudes at each point being so small.

The importance of only fitting exchange and dispersion is obvious for $\ce{Na^+---Cl^-}$ (using def2-SVP for the basis).

This is an unfair example, since the dispersion interaction is dwarfed by exchange, and it is an ionically-bound molecule. As a final example, consider $\ce{V(III)---Cl^-}$ (using def2-SV(P) for the basis, vanadium as a quintet).

I had some trouble converging many of the calculations, presumably due to a non-optimal spin state at long range. For this reason (spin-state crossing), Lennard-Jones type models are poor for metal-x interactions. The problem appears to be similar to the sodium chloride, but the exchange plus dispersion fit looks good.

If you want to reproduce any of the work, rather than fish through the source for inputs and scripts, everything is in a GitHub repository. More references for SAPT can be found here, along with the Psi4 manual. Note that I didn't do any literature searching here, but I expect people have used energy decomposition approaches in the past as part of force field design. Hopefully this serves as a good starting point.


Relative strength of desiccants Is it meaningful to characterize the relative strength of desiccants?  For example, is there a measure of hygroscopy, or an ordering of desiccants, such that higher ones will always dry lower ones?

Let us restrict the question to "practical" desiccants, which are hygroscopic substances that can be "recharged" (i.e., dried out) by heating.  I'm not certain I know all possible mechanisms of water sorption within this class of chemicals.  But is it safe to say that for all such chemicals, at standard temperature and pressure, there is some water vapor pressure around a desiccant that has sorbed any water?

I am trying to imagine a series of tests in which two different desiccants are put in a humid chamber in adequate quantities that either could completely sorb all the water.  At hydrous equilibrium will the "stronger" desiccant contain virtually all of the water?  Or is the equilibrium distribution of water a function of the desiccants' relative "hygroscopy," rates of sorption, or some other factor(s)?

This is ultimately a practical question.  One example: I have a hygroscopic, hydrated salt X.  Can I dry it out by sticking it in a sealed box at STP with a "more powerful" desiccant?  If not what determines the limit of dehydration that would be achieved by exposure to other desiccants?

I'd say no, because there is also a reactivity dimension to desiccants. Some drying agents react with some solvents and therefore can't be used as a desiccant. This happens for many organic solvent and desiccant pairs. Some desiccants also react as desiccants and this changes their drying power. This is the case for phosphorous pentoxide, for example, which gradually forms a viscous layer of phosphoric acid which stops any drying. So any linear ordering you are attempting is likely to be overwhelmed by the reactivity side effects of the many desiccants.

On the other hand, if you are only interested in some theoretical point in time estimate of desiccant efficiency, this might be possible but would have virtually no practical import. A desiccant has to react over time and the side effects cannot be ignored.


If gases turn into liquids under pressure, what does carbon dioxide turn into? Since there is no liquid state of carbon dioxide, what will carbon dioxide turn into when exposed to high pressure?

There is a liquid state for carbon dioxide. Borrowing the $\ce{CO2}$ phase diagram from Wikipedia, we can see that $\ce{CO2}$ will condense at a few atmospheres, dependent on temperature. At still higher pressures, the liquid will solidify. Below the triple point temperature, the gas will transition directly to solid.


Does attachment of phenyl to an alkene increases its reactivity towards electrophilic addition reaction? According to me phenyl should decrease the reactivity as its electron withdrawing nature destabilises the carbocation.

Not always, and it can depend upon the groups attached to a phenyl ring, like $\ce{-OMe, -COOH, -NH2, -NO2, -X}$, etc.

The attachment of a phenyl ring to a carbon-containing double bond can affect both the stability of carbocation and the possibility of attacking with the help of the compound's $\pi$-electron cloud.

If you have only the phenyl ring with no substitution, first of all the $\pi$-electron cloud of the double bond will be in high conjugation with the phenyl ring, and thus the $\pi$-electron density will be reduced on the $\ce{C=C}$ double bond, which can reduce the possibility of attack of that double bond to a electrophile. Though the carbocation formed can be stabilised through resonance with the phenyl ring (as it is a benzyl carbocation), the step of carbocation formation can have a very high activation energy ($E_\mathrm a$) if you don't have enough stabilisation of the carbocation. But, if you have the carbocation which can stabilise through resonance with multiple benzene rings, the actual potential energy of the carbocation can be reduced to a great extent. In that case, the rate of reaction will increase.

So, actually this situation can be thought as a competition between the stability of carbocation and the decreased electron density. Generally, the stability of carbocation is the dominating effect if it is stabilised to a large extent.

The rate will be decreased even more if an electron withdrawing group is attached to the phenyl ring due to even more delocalisation of $\pi$-electron cloud, and less stabilisation of the carbocation.

If you have a strong electron donating group attached to a phenyl ring and the double bonded carbon is ortho/para w.r.t that group, that can itself exhibit the $+R$ effect, and increase electron density near the double bond through delocalisation of its own electron density through the $\pi$-bonds. In that case, the attacking of $\ce{C=C}$ double bond to a electrophile will be much easier, and the carbocation formed will be even more stable. Then, the rate of reactivity of the double bond in the electrophilic addition can increase for attachment of a substituted phenyl ring.


Is there any empirical upper limit for order of reaction? I learned from one of my teachers that empirically reaction with order greater that 3 are never found. But, I am suspicious of the truth value of his claim.

To add to my suspicion there is this paper from 1932 referring to a fourth-order reaction

It seems to me theoretically there should be no upper limit for a  reaction as it is possible for 4,5,6... chemical species to come together with the right activation energy in the right direction.

However, empirically making 4 or 5 chemical species to form a transition species is very hard. So, there should be a empirical limit to order of reaction in the light of chemistry we know today.

What is this limit?

The order of a reaction is an experimentally determined quantity and can be  positive, negative or fractional. The order need not be related to the stoichiometric coefficients, although sometimes it is. In some reactions it is not possible to define an order. There seems to be no reason therefore why the order cannot be greater than three; the iodate-iodine reduction (Dushman) reaction $\ce{IO3^- + 5I^- + 6H^+ -> 3I2 + 3H2O}$ has the rate expression $r \approx \ce{[I^-][IO_3^-][H^+]^2  }$.

If the reaction is $aA+bB \rightarrow cC+dD$ then the rate is

$$ r= - \frac{1}{a}\frac{d[A]}{dt} = \frac{1}{c}\frac{d[C]}{dt}= \cdots =k[A]^\alpha[B]^\beta[C]^\gamma \cdots$$

where the order n is $n=\alpha+\beta+\gamma$ and these need not be the same as a, b and c.

If the rate expression is not of this form then it is hard to define an order; one familiar example is the chain reaction $\ce{H2 + Br2}$ where the experimentally determined rate law for the production of HBr has the form $\displaystyle r=\frac{k_2[\ce{H_2}][\ce{Br_2}]^{1/2}}{1+k_2'[\ce{HBr}]/[\ce{Br}]}$ where an overall order cannot be defined.

The order should not be confused with molecularity which describes the number of species reacting in a  postulated elementary reaction and which is always a small positive number. Effectively this means 1 or 2. A molecularity of 3 is not impossible but the chance of three  reactive species colliding at the same time is vanishingly small. In practice when a molecularity of 3 is suspected it is often found that in  subsequent experiments an intermediate is formed between of the two species which then reacts with the third at a slightly later time.


Longest chain IUPAC rule I wonder, what are the exact IUPAC organic chemistry nomenclature rules that give the preference to the chain length over number of unsaturated bonds, i.e. that justify, or prefer following names:

4-ethenyl-4-ethynylheptane over
3,3-dipropylpent-1-en-4-yne

3,5-dimethylideneheptane over
2,4-diethylpenta-1,4-diene

?

In Chapter P-4 "Rules for Name Construction", section P-44 "Seniority Order for Parent Structures", of Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013, it is stated that (p 490):

  P-44.3.2 The principal chain has the greater number of skeletal atoms [criterion (b) in P-44.3].

The Blue Book lists several helpful examples, one of which is that the principal chain "octane" is senior to "hept-1-ene". Therefore, the following compound

is named 3-methylideneoctane instead of 2-ethylhept-1-ene.

The authors mention that this preference is in contrast to previous recommendations (e.g. 1979):

  In acyclic parent structures the order of seniority between unsaturation and length of chain given in earlier recommendations is reversed. Thus, the first criterion to be considered in choosing a preferred parent acyclic chain is the length of the chain; unsaturation is now the second.


Why should I acidify twice in the procedure for qualitative analysis of chloride anions? Separation of the Chloride ($\ce{Cl-}$); Confirmation of Chloride.

  Chloride ion forms an insoluble silver compound. Silver chloride is a white solid.
  $$\ce{Cl^-(aq) + Ag+(aq) -> AgCl(s)}$$
  Silver chloride dissolves in $\pu{6 M}$ ammonia, $\ce{NH3}$, forming the colorless ion $\ce{Ag(NH3)2+}$. If nitric acid, $\ce{HNO3}$, is added to a solution containing this ion, the ammonia in the complex reacts with hydrogen ions to form ammonium ions, and the silver recombines with the chloride ions that are still present in solution.
  \begin{align}
\ce{AgCl(s) + 2NH3(aq) &-> Ag(NH3)2+(aq) + Cl^-(aq)} \\
\ce{Ag(NH3)2+(aq) + Cl^-(aq) + 2H+(aq) &-> AgCl(s) + 2NH4+(aq)}
\end{align}

  a. Place 10 drops of the original test solution (or unknown solution) in a test tube. Test to see if the solution is acidic. If it is not, add $\pu{6 M}$ acetic acid, $\ce{CH3COOH}$, dropwise with stirring until the solution is acidic.

  b. Add 10 drops of $\pu{0.1 M}$ silver nitrate, $\ce{AgNO3}$. A precipitate of $\ce{AgCl}$ will form.

  c. Centrifuge and pour off the supernatant liquid.

  d. Wash the solid with $\pu{0.5 mL}$ distilled water, centrifuge and discard the wash water.

  e. Add $\pu{0.5 mL}$ $\pu{6 M}$ ammonia, $\ce{NH3}$, to the precipitate. Stir to dissolve any $\ce{AgCl}$.

  f. Centrifuge, and pour the supernatant liquid into another test tube to test for chloride ion.

  g. Add $\pu{1 mL}$ $\pu{6 M}$ nitric acid, $\ce{HNO3}$, to the solution containing the dissolved silver chloride. The solution will get hot and smoke from the reaction with the excess ammonia whether or not silver chloride is present.

  h. Test with litmus or pH paper to see if the solution is acidic. If it is not, add more $\ce{HNO3}$ until the solution is acidic. The appearance of the white precipitate of $\ce{AgCl}$ in the acidic solution confirms the presence of chloride.

From the procedure above, if you all could take a loot at step 1a. It makes no sense to me. I do not understand why adding $\ce{CH3COOH}$ has any effect on the solution whatsoever. What is the point of adding it when we later add $\ce{HNO3}$?

Playing around with pH values, going from acidic to basic and back to acidic is a major element of the separation methods used in traditional analytic inorganic chemistry. It takes advantage of the facts that both acids and bases were easy to acquire even in the early days of chemistry and that the pH can be monitered well by indicators which had also been known for a long time (even though in the past they might have used different indicators than nowadays). Making sure the pH is at a correct value at a certain time is also important in this cycle.
(Because the Markdown syntax does not allow calling enumerated lists with lowercase letters, I’ll use numbers instead to refer to the different steps.)

You have a solution of potentially many different ions. For the sake of this answer, let’s assume chloride, bromide, nitrate and carbonate. If you measure the pH of this solution, you will notice it is slightly basic due to the carbonate. If you were to add silver nitrate immediately, silver carbonate $\ce{Ag2CO3}$ would also precipitate, complicating later analyses.
Therefore, you add an acid (whether it is acetic or nitric acid does not matter. It should not be hydrochloric acid for obvious reasons and it should not be too strong.) until the pH of the solution is acidic (i.e. at least 5).

You can now add your silver nitrate. Only chloride and bromide will precipitate as $\ce{AgCl}$ and $\ce{AgBr}$, respectively.

By centrifuging, you now have only the silver chloride and bromide precipitates remaining since you are discarding the supernatant. Due to residual amounts of whichever acid you used (acetic acid in the protocol), the damp precipitate is still acidic. However, most of the original acid will have remained in the supernatant and has been poured away.

Acidic residues remain even after washing albeit to a lesser extent. We can basically consider the precipitate neutral now. This is important for step 5.

Adding diluted ammonia (not $25~\%$ concentrated) will dissolve $\ce{AgCl}$ as $\ce{[Ag(NH3)2]Cl}$ while leaving the $\ce{AgBr}$ untouched. Due to the ammonia you are adding, the pH of the resulting solution is basic. Any remaining residues of acid from step 1 have been neutralised now at the latest.

Centrifuging makes sure to remove any remaining dispersed silver bromide. The supernatant should contain $\ce{[Ag(NH3)2]Cl}$ in solution.

Now we want to reprecipitate the silver chloride to confirm we actually had it in the precipitate. (Note we cannot be certain that silver chloride was present if silver bromide has precipitated out.) We do this by reacidifying the solution until all ammonia is protonated to give ammonium and the ammine ligands therefore dissociate from the $\ce{[Ag(NH3)2]+}$ complex. This causes silver chloride to reprecipitate.

In case you did not add enough acid you are here reminded to do so. Failing to add sufficient amount of acid would lead to the silver ions remaning complexed in solution and you incorrectly assuming there be no chloride in your unknown sample.


Materials for growing large quality sugar crystals I am trying to grow large, solid, quality crystals on a course fabric, and I was wondering what other than sugar I would need to put in my solution, as well as how long it may take, and if food coloring/another dye would change the color of the crystals without destroying them. Any tips or advice would be great.

The Exploratorium has directions for making rock candy, the common name for sucrose crystals. In short:

Make a saturated solution of sucrose -- 4 cups sugar dissolved in just 2 cups of boiling water.
Immerse the fabric in the solution, and let it cool slowly (to make larger crystals). You might use a covered styrofoam container to slow the cooling. If crystallization doesn't start in a few days, you might need to drop one seed crystal of sugar into the container.
You can experiment with different types of food color in the sugar solution to color the crystals, but in the process of crystallization, foreign molecules of dye may be excluded from the crystals. The larger the crystal, the less colored it may appear.

Enjoy! Be sure to use sanitary equipment and safe colors, if anyone is to eat this. BTW, ordinary cane sugar is one of the purest substances sold in quantity, better than 99.9% sucrose.


How do I write orders of magnitude in scientific notation when margin of error is involved? My TA has me confused about how to use scientific notation when margin of error and orders of magnitude are involved.

The number I wrote: 3.78 ± 0.02 * 10^-3

My TA told me that this means: 3.78 ± 0.00002

I thought that this meant: 0.00378 ± 0.00002

Which one is correct? I've been looking online to find something that explains the proper notation when writing out margin of error with orders of magnitude but I haven't been able to find anything (I'm at a loss of what terms to search for). If what I wrote isn't the correct way to say 0.00378 ± 0.00002, then what is? Writing it as 3.78 * 10^-3 ± 0.02 or  3.78 * 10^-3 ± 0.02 * 10^-3 seems kind of foolish.

Sometimes you need to go the extra mile to be entirely unambiguous. Your original formulation (3.78 ± 0.02 * 10^-3) could be interpreted in two ways, the way you intended and the way your TA suggested. The majority of people, I expect, would use context to interpret it correct as you intended, but some (as your TA demonstrates) will not.

There is no clear cut widely agreed standard that will resolve this. But you can apply standard mathematical notation (which will look superfluous to some) to make what you write entirely unambiguous. As Ivan suggested in a comment the simple addition of brackets giving (3.78 ± 0.02) * 10^-3 makes your statement unambiguous in its interpretation. This is more compact than 3.78 * 10^-3 ± 0.02 * 10^-3 but just as unambiguous.

If you were using properly typeset formulae, you might well be able to avoid some of the apparent redundancy of the last formulation, but that would take more effort. Better to accept that redundancy is sometimes required to make things clear.


Oxidation state of sulfur in H2S2O5 How can I find the oxidation state of sulfur in $\ce{H2S2O5}$ using its structure?

This is the structure of $\ce{H2S2O5}$:

Use the states of oxygen and hydrogen to know the oxidation state of sulphur.


How the plate height is assumed to be the standard deviation? Fundamentals of Analytical Chemistry book states that the  plate  height  can  be  thought  of  as  the  length  of  column  that  contains  a  fraction  of  the  analyte  that  lies  between  $L$  and  $L - \sigma$ [1, p. 870].
But what I know is that $H = \sigma^2 / L$ (At the retention time), so, how can plate height be equal to $L - (L - \sigma)$?
$H$ is the plate height. $L$ is the column length. $\sigma$ is the standard deviation.
Reference

Skoog, D. A.; West, D. M.; Holler, F. J.; Crouch, S. R. Fundamentals of Analytical Chemistry, 9th ed.; Brooks Cole: Belmont, CA, 2013. ISBN 978-0-495-55828-6.

I feel that you misunderstood what the authors were trying to say in that textbook. Forget about everything for the time being.
Imagine if you injected a single analyte band into an HPLC column of length L at a constant flow rate. You waited for some time, and now the band is exiting the column. For simplicity, assume that the band is rectangular.
Ask yourself, how much average distance did the analyte band travel? Indeed, it must have traveled a distance L if it has to exit the column. Most modern HPLC columns are 10 cm long; assume analyte traveled an average of  L=10 cm.
Now a band cannot be infinitely thin inside a chromatographic column. The analyte spreads in space, i.e., its width increases as it passes through the column due to diffusion and other processes.
So let us say the band is now a 0.05 cm wide rectangle when it reached the end of the column. Half of it is inside the column, and half of it has exited the column. What width of the band is inside the column, and what width is outside the column? You would say 0.05/2 cm of the band is inside the column, and 0.05/2 cm of the band is outside the column.

Total distance traveled by the portion of the band which has exited = 10 + 0.05/2 cm
Total distance traveled by the portion of the band which is inside = 10 - 0.05/2 cm

Now in chromatography, as you have seen, peaks are not rectangles, but they are Gaussians. Furthermore, Gaussians are characterized by their "standard deviation," which is a measure of their spread. If you apply the sample logic

The mean (=center of the Gaussian) has traveled a distance of 10 cm or L
The portion of the Gaussian which is still inside traveled a distance L-1*standard deviation
The portion of the Gaussian which exited the column traveled a distance L+1*standard deviation.

Note that conventionally a full Gaussian band at the base is considered to have a width of 4xstd. dev.


Questions about van't Hoff factor in colligative property I learned from many websites that $i$ of strong acids and bases and some weak bases could be determined by how many ions are dissociated. But, I realized that I need to calculate the $n$ in $i=1+(n-1)a$ ($a$ is the degree of dissociation), not the $i$ in my problems.

Is there any rule to calculate the $n$ for strong acids and bases and weak acids and bases? Do I also need to check solubility table to solve colligative property problems that using van't Hoff factor?

In general, 'n' is the number of species the compound dissociates into, and such formulas are applicable only when the acid itself is monobasic, i.e.only one dissociaton is involved.For sulphuric acid,this is not applicable as there are two dissociations with two dissociation constants.The best way to find the van't hoff factor is to find the ratio of the final number of particles to the initial number of particles.


Is there an algorithm that counts Joback groups in a structure? So I was wondering, there are a lot of cheminformatics tools (eg Babel) out there, to analyze and convert representations. I also saw some codes which counts certain groups, but ignores the carbon chain, which is obviously not suited for a Joback script. My google research only turned me a "homemade" matlab code found at Reddit, but it is no way general, as the first step of it is to exclude most of the rings.

So is there any open cheminformatic tool for this problem?

So it looks like 'Jobeck groups' are some set of functional groups that you want to find in molecules.

Sounds a lot like substructure search, for which there are many cheminformatics tools (like SMSD, for example) however, since you mention OpenBabel:

http://open-babel.readthedocs.org/en/latest/WritePlugins/AddNewDescriptor.html

talks specifically about 'group contribution methods' and gives a simple example of how to get the counts for a set of functional groups.


What material most closely resembles/emulates dog's excrements? Please get me right - this is no joke!

Our engineering team has a terrible contract - we need to find a way to get height from which some dog excrements were dropped judging by what was found on earth when it hit the ground. We need to find some invariant in excrements form, scattering of droplets/material, I don't know whatever else...

We did not start math modelling yet, because it makes no sence - we need to get somewhere physical propeties of "material", besides it is clear that the best way to start with is real-life experiments - but throwing such stuff from different heights over and over again is definitely a bad idea.

The best solution would be to find some material which is maximally close to what our target is - like ballistic gel instead of a human body.

Sorry to ask, but what material most closely resembles/emulates dog's excrements?

So far we found nothing in literature, maybe at list please a hint where to look for...

Sorry once again - this is absolutely no joke, this is for real. The height is within 70 meters limit (this is the max height), the surface is hard - it is pavement (just where people walk). Actually we need to create methodology allowing to calculate from what floor of a building it was dropped (or proove that required resolution of 2-3 meters cannot be reached) by teenage pranksters who do it regularly in different locations (they play it from their rooms and they are networked through some kind of viral social network). This is a real-life problem and we understand why it arose, but it is terrible from the technical point of view.

You'll never be able to get to 2-3m accuracy with this. With a reproducible sample, you could make a few dozen tests and create a convincing statistics.

But dog excrements? Even if they always used the same dog, which they kept on a constant diet, and which you had impounded, it'd be impossible. The sample is much, much too complex, and, unfortunately, very much alive. (I don't mean the dog.)

I say stick to old-school policework: Keep interrogating your pranksters until one of them sings. ;-)


Reaction of SeCl4 with alkene and possible consecutive reaction? There is a reaction that confused me a bit:

According to my research on $\ce{SeCl4}$, it can be used as a dichlorinating agent for alkenes. However, I am totally confused about the mechanism. I've read about several mechanisms using $\ce{PhSe-SePh}$ as a precatalyst, but in this reaction, there is no other information, apart from the reagent $\ce{SeCl4}$.
If I assume that there is a dichlorination with $\ce{SeCl4}$, I think, the syn-specific dichlorination happens on "top" as the methylene group is sterically less demanding.

Now, if this happens, could there be a subsequent $\mathrm{S_N2}$ reaction with the $\ce{OH}$-group? It could form a 5-membered cycle through a back-side attack on the carbon with the chlorine substituent.

Can you confirm or correct my thoughts about this reaction? Thank you a lot in advance!

We humans are still at best mediocre at predicting the outcome of chemical reactions. Even computers are at best a little bit better. If there is no literature on this compound (and Orthocresol commented that he found none), no experiment has likely been performed and we cannot be certain.
You could find a computational chemist of your choice with free time and the appropriate software and let them simulate this reaction. This would allow a slightly better prediction as it would answer the question ‘is there a reasonable reaction pathway that this compound could follow?’ However, even when simulating ‘obvious’ reactions we chemists might overlook problems with a simple-looking pathway that make this pathway non-accessible in the first place. Thus, even if the simulation looks promising, the only true answer is to go into the lab and try it out.
On paper, your scheme seems promising. You may require additional base (maybe a weak base like carbonate will do the trick, maybe you need selective oxygen deprotonation e.g. by sodium hydride) or it may happen during the chlorination reaction.
There may be other side reactions. Without knowing whether it is applicable in your case, I would attempt to reseach potential non-classical cations and how they might interfer, in case this reaction would follow a more $\mathrm{S_N1}$-style pathway.
In addition, I wonder if an elimination reaction to a chloro-norbonene might be a competing side-reaction.
I don’t think that a nucleophilic substitution on the other chloride would pose much of a problem as five-membered rings are typically kinetically preferred over six-membered rings.


(At what temperature) can salt evaporate? Related to my previous question, can sodium chloride evaporate, and if so, at what temperature? Since salt is a solid, and apparently all substances can at various temperatures exist as liquids and vapors as well as solids, salt should also have a melting and a vaporization points. At what temperatures are these points if they occur, and if they don't occur, why? (This is assuming the pressure is 1000 mb)

All substances may have a melting and boiling point. I say "may" because in some instances there can be decomposition, for example, before reaching either temperature.

Sodium chloride has a melting and boiling point. Check Wikipedia:

https://en.wikipedia.org/wiki/Sodium_chloride

The sidebar lists the melting point at 801 °C (1,474 °F; 1,074 K) and the boiling point at 1,413 °C (2,575 °F; 1,686 K). The values are relatively high being consistent with the ionic nature of NaCl.


R and T conformations in relation to Activators and Inhibitors; Km of reaction in relation to uncompetitive and non-competitive inhibitor Do

-Activators

-Competitive Inhibitors

-Non-Competitive Inhibitors

-Uncompetitive Inhibitors

all change the conformation of the targeted enzyme from T to R (activator), and from R to T (the rest) respectively?

Also, why does the Km of a reaction decrease with an uncompetitive inhibitor, while remaining constant with a non-competitive inhibitor?

I have tried to find a coherent answer to this all across the internet, but to no avail. Many thanks!

Allosteric inhibitors changes the conformation of enzyme enzyme for preventing the substrate binding, and allosteric activators modify the enzyme conformation so that  substrate binding can take place.

Competitive inhibitors don't change the enzyme conformation, as these inhibitors compete with the substrate to bind at enzyme's active site.

Non-competitive inhibitors change the enzyme conformation, as they inhibit enzyme activity by binding to site other than active site and prevent the substrate binding by changing the active site's structure so that substrate can not bind to enzyme.

Uncompetitive inhibitors inhibit enzyme activity by binding to enzyme substrate complex and these inhibitors change the conformation of the enzyme which reduces the affinity of the substrate for the enzyme's active site.

Km decreases in uncompetitive inhibition because inhibitor binds to the enzyme-substrate complex and it becomes incredibly difficult for the substrate to leave the enzyme. This results in increased affinity and Km represents an increase in affinity.

(Via: https://courses.lumenlearning.com/boundless-microbiology/chapter/enzymes/

https://en.wikibooks.org/wiki/Structural_Biochemistry/Enzyme/Reversible_Inhibitors)


Detection of sulphide in solution also containing sulphate Question:

The reagent(s) that can selectively precipitate $\ce{S^2-}$ from a mixture of $\ce{S^2-}$ and $\ce{SO4^2-}$ in aqueous solution is(are):

(A) $\ce{CuCl2}$
(B) $\ce{BaCl2}$
(C) $\ce{Pb(OOCCH3)2}$
(D) $\ce{Na2[Fe(CN)5NO]}$

Answer:

 (A) or (A) and (C)

Why (C) is considered as correct option even though lead sulphide forms black precipitate and lead sulphate forms white precipitate?
Question source: JEE Advance 2016

I will add one thing that may be the reason behind why (C) is an acceptable answer choice. Yes, both $\ce{PbS}$ and $\ce{PbSO_4}$ are insoluble. However, if you look at the $\ce{K_{sp}}$ for both compounds, $\ce{PbS}$ is about $\pu{3.2 x10^-28}$ while the $\ce{K_{sp}}$ for $\ce{PbSO_4}$ is $\pu{1.3 x 10^-8}$. When you have a mixture of ions, the resulting molecule with the lowest $\ce{K_{sp}}$ will precipitate first. In this case, since $\ce{PbS}$ has a lower  $\ce{K_{sp}}$ than $\ce{PbSO_4}$, it will precipitate first.
This technique is called fractional precipitation where you separate two ions, in our case $\ce{S^{2-}}$ and $\ce{SO_4^{2-}}$, by precipitating one of the ions selectively through taking advantage of their differing $\ce{K_{sp}}$ values.
This is the likely reason behind why choosing (A) itself is acceptable or choosing (A) and (C) is also acceptable.


What bases are fit for human consumption and can be used in foods? I know that are several acids that can be found in the kitchen, such as citric acid and vinegar.

However, I never heard of a base that can be used in the kitchen to raise the pH levels of any food. I am well aware that this mix would probably make a salt as a result. Something like hydrogen peroxide doesn't sound safe for food. So, my question is:

What are (if any) the bases that can be used on those acids and still be human-consumable?

Unfortunately Googling for "alkaline food" these days is a bad idea; results are dominated by nonsense sites espousing the "alkaline diet" nonsense / woo.

There aren't many foods that are naturally alkaline.  The two exceptions I know of are:

baking soda
egg white

Sometimes when I'm cooking and accidentally add too much acid (e.g. lemon juice or vinegar) to a broth or sauce, I can partially correct the mistake by adding small amounts of baking soda.  This converts the excess acid to the sodium salt and $\ce{CO2}$ is evolved.  For the example of vinegar (acetic acid):

$\ce{CH3COOH + NaHCO3 -> CO2(g) + CH3COONa + H2O}$

So sodium acetate salt is left behind, which limits the degree to which you can correct overaddition of acid.  (You don't want your food to be too salty.)$%edit$


What does "parts by weight" or "parts by mass" specifically mean? Equivalent mass may be defined as the number of parts by mass of a substance which combines with or displaces directly or indirectly 1.008 parts by mass of hydrogen or 8 parts by mass of oxygen or 35.5 parts by mass of chlorine.

Why isn't it defined like "that mass of a substance which combines/displaces 1 gram of hydrogen"

Why does it use the term "parts by mass" ?

It seems that the unusual wording comes from the fact that we are used to having all of the atomic masses available to us simply because we have measured all of them. In the early days of chemistry, however, things were not so simple. Thus, the equivalent weight (technically mass since it has units of $\mathrm{kg}$ rather than $\mathrm{N}$) was used as a method for determining a weight relative to some predefined standard. In this case, hydrogen.

Some of the simplest reactions are acid/base reactions, and this explains why the scale is measured relative to equivalents (where one equivalent is one mole) of hydrogen. That is, in your quote from the wikipedia page, one mole of hydrogen, which has a mass of 1.008 grams, can react with 8 grams of oxygen or 35.5 grams of chlorine to form $\ce{H2O}$ and $\ce{HCl}$ respectively. Another way of saying this is that every hydrogen atom reacted with two oxygen atoms and every hydrogen atom reacted with one chlorine atom.$^1$ Thus, we see what is meant by the term "parts by mass". Notice this is not the same as the phrase "parts per mass". "Parts by mass" could be stated more clearly as the "parts of X measured by mass". That is, we are indirectly measuring how many parts of oxygen, chlorine, or whatever other X goes into a reaction by measuring the final mass of a substance which reacts with a known mass of 1.008 grams of hydrogen.

The wikipedia article notes that not everything reacts with hydrogen, so in order to find equivalent weights for some materials, you must measure its equivalent weight to a known mass of oxygen and then use the 8:1 conversion to hydrogen equivalent weights.

I hope this clarifies the idea. Don't get worried though, this does not contain any more information than simple atomic weights.

$^1$This is kind of cheating though because we know the weights of oxygen and chlorine, while chemists in the 1800s may not have known this.


Is pyrite (FeS₂) an ionic or a covalent compound? I have searched all over the web and found a lot of diverse explanations, but none of them are concluding exactly whether $\ce{FeS2}$ (solid - pyrite) is a covalent or an ionic compound.

From electronegativity, it should be covalent as the $\Delta\chi=0.7$ which is less than $1.5$ and thus said to make covalent bonds and therefore be a covalent compound.

From the definition of ionic bonds, which are bonds between a metal and a non-metal element (whereas covalent bonds are bonds between non-metal elements), it should be an ionic compound.

Does someone know which of those is 'true', or better, if there is another, more detailed explanation?

You seem to have fallen into the trap of thinking that ionic and covalent bonds are fundamentally different. They are not - they are just two ends of a spectrum, which has an arbitrary division somewhere in the middle into an ionic and covalent regime. This is explained in the answers to this question.

In the case of pyrite we have a relatively hard cation, with a small ionic radius and charge of +2, and a rather large anion, with a charge of -2. Therefore there will be a significant degree of covalent character in the Fe-S bonds due to the polarising effect of the cation on the anion. This is confirmed by experimental results and theoretical calculations which suggest that the charge on Fe is about +2/3 and the charge on S is about -1/3. This is significantly less than the expected charges of +2 and -1 from a purely ionic model and so indicates that there is significant sharing of electrons. This is supported by measurements and calculations of the electron density, which show significant electron density between the atoms.

Reference: http://pubs.rsc.org/en/content/articlepdf/2014/sc/c3sc52977k


What kind of chemicals would exist at the heat death of the universe? As the universe reaches the maximum entropy, presumably only the most stable chemicals would exist.

Assume that all the elements haven't turn into iron yet, what kind of chemicals would dominate at that point?

There is a lot of uncertainty regarding the far future of our Universe, but it seems that chemistry as we know it will be gone long before the end. Both free and bound protons (and neutrons) are expected predicted to decay through at least one of several mechanisms, with a half-life somewhere between the range of $10^{35}$-$10^{200}$ yr (far shorter than the timescale for nuclear reactions to convert all matter into iron, ~$10^{1500}$ yr). Therefore, at some point the statistically expected number of atoms inside our observable bubble could potentially dip below 1.

Should proton decay happen relatively quickly, perhaps the last elements in the Universe will be present near the end of the black hole era (~$10^{100}$ yr), when the last supermassive black holes decay in a flash of subatomic particles, possibly creating small amounts of new protons/neutrons. These will decay in a comparative blink of an eye at this timescale, however, and the Universe will soon become a barren place once more.

The last structures with any semblance to matter in the Universe will likely be positronium "atoms", where an electron and a positron (anti-electron) orbit each other in a pseudo-hydrogen atom. Though ground-state positronium decays in a matter of nanoseconds or less, it will be very long-lived in the future due to the extreme sparsity of subatomic particles in the hugely expanded observable bubble, with a mean distance between a bound positron and electron measured in trillions of trillions of trillions of trillions of... billions of lightyears or more. Even in these conditions, positronium too shall eventually decay, in a timescale of anywhere between ~$10^{140}$ yr and ~$10^{800}$ yr.

Before all of these nuclear processes take place, there may also be a significant amount of unusual chemical reactions happening solely due to tunnelling effects, even at arbitrarily low temperatures, since the kinetic barriers for chemical reactions are far smaller than that for nuclear reactions. For example, most molecules containing nitrogen atoms are thermodynamically unstable with respect to release of $\ce{N2}$, and so might spontaneously break up even at $\mathrm{0\ K}$ given enough time. I don't know what the timescale for these processes is, however.


Why does increasing temperature for an endothermic reaction always increase equilibrium constant? According to Le Chatelier's Principle, increasing temperature for an endothermic reaction shifts the reaction towards the products.

However, unless there are things that I am missing, I believe there are three ways to imagine this phenomenon, albeit they all arise from the same equation/concept.

Classic $\Delta G = \Delta H - T \Delta S$
$\Delta S_{\text{sys}} + \Delta S_{\text{sur}} = \Delta S_{tot\text{}}$
Van't Hoff Isochore: $\ln\left(\frac{K_2}{K_1}\right) = \frac{-\Delta H}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$

In the second case, it makes sense that an endothermic reaction would favor an increase in temperature because $\Delta S_{\text{sur}} = \frac{-\Delta H_{\text{sys}}}{T}$ assuming constant pressure. Thus an increase in temperature would mean an increase in total entropy change, indicating a more favorable change.

Van't Hoff's isochore also predicts that an increase in temperature favors the forward reaction of an endothermic reaction as well if you do out the math.

However, in the first case of the classic equation, the equation suggests that there is actually an entropy dependence rather than an enthalpy dependence of the gibbs free energy, which is related to the equilibrium constant by $-RT \ln(K) = \Delta G$.

It further confused me when considering that case 1 derives from case 2, and case 3 derives from case 1. I need help identifying my mistake or rectifying the confusion on why equilibrium is not entropy dependent.

The first equation, the corresponding change in G,  and the equilibrium constant is supposed to be related to the following two initial and final states:

State 1:  Pure reactants in separate containers at temperature T and pressure 1 bar

State 2:  Pure products in separate containers at temperature T and pressure 1 bar

The $\Delta G$ you are talking about refers to the change between these two thermodynamic equilibrium states.  But please note that these  thermodynamic equilibrium states that are not in chemical reaction equilibrium with each other.  So, to get the equilibrium constant for the reaction, you are using the $\Delta G$ for these two states which are not in chemical equilibrium.  To understand how this plays out, you need to review the derivation of the relationship between this $\Delta G$ (actually denoted $\Delta G^0$ to distinguish the specific states 1 and 2) and the equilibrium constant K.

Now, when you change the temperature, all three quantities in the equation for $\Delta G^0$ change:  T, $\Delta H^0$, and $\Delta S^0$, not just T.  This is why the Van't Hopf equation gives the correct result for the effect of temperature on the equilibrium constant, and the other approach you described does not.


Chemistry behind lip balm coloring based on skin pH I stumbled across an ad for a clear, colorless lip balm that purportedly tints your lips a shade of pink determined by your skin's pH.

  This balm stain applies on clear then reacts with your skin's pH levels, uniquely creating the perfect shade of pink lips just for you.

Out of curiosity (and the fact that I'm an easy target for marketing), I read further. The product contains the following ingredients:

Hydrogenated Polyisobutene
C12-15 Alkyl Benzoate
Diisostearyl Malate
Octyldodecanol
Dibutyl Iauroylflutamide
Dibutyl Ethylhexanoyl
Thylhexanoyl Glutamide
Isononyl Isononanoate
CI 45410
Chrysanthemum (a whole flower added to the bottom of the tube, presumably just to make it look nice.)

Other than CI 45410, I can't find anything on the above list that mentions anything about color -- and there's nothing listed that I think could change color based on pH.

Note: Some of the ingredients are very hard to find data for, but most seem to be thickening agents.

Does this clear lip gloss actually have the ability to change color based on pH or does the color come from the CI 45410? If it's from the CI 45410, what causes it to go from clear to pink?

CI 45410 (A.K.A. Phloxine B or acid red 92) is a pH dependent dye which is clear below about pH=4.1 and pink at pH>4. The formulation is mildly acidic and is neutralised by saliva upon application to become pink.

(Thereby "uniquely creating the perfect shade of pink lips just for you.")


Is the scientific explanation behind Smartkleen legit? I'm trying to have a conversation/debate about the SmartKleen laundry detergent alternative, but lack the knowledge to understand if their explanation is science or pseudoscience.

Far Infrared emission at cold water temp, smaller water molecule clusters, magnets that increase levels of dissolved oxygen, etc., can someone offer a solid critique?

I think the idea of "smaller water clusters" penetrating the cloth more efficiently is actually quite smart as the same idea is utilised in steam cleaners. But the rest of the article is a bit too psychotic. How can they say that colours are preserved while the Ball constantly generates H$_2$O$_2$ (in every second)? And why do they recommend the Stain Remover for stains if the ball was such a miracle...


Over potential -what and why ? Value of standard electrode potential for the oxidation of Cl– ions is more positive than that of water, even then in the electrolysis of aqueous sodium chloride, why is Cl– oxidised at anode instead of water?

Standard electrode potentials describe the tendency of getting reduced.Positive value indicates that H2 can reduce that element or reduced form of the element is more stable than proton.
The answer to the question above is over potential
Please explain how it came into the story.My book explained it under "products of electrolysis",but the explanation is not lucid.
They explained it as :
Some electrochemical processes although feasible, are so slow that at  lower voltage these do not seem to take place and extra potential called over potential has to be applied which makes such process more difficult to occur.

When I took P-Chem, decades ago, one experiment we had was using a dropping mercury electrode, DME. I don't recall the details, but the set-up was fairly elaborate, especially compared to a Ag/AgCl electrode. The DME is fairly common (or was, before we decided that Hg was probably better avoided by the undergrad). The reason WHY we used it was it provided a CLEAN surface. Think about it. We wouldn't go to such great lengths if the problem of contamination wasn't significant.

All materials absorb contaminants, may chemically react with near-by chemicals (solvent, ions, gasses), and will form a charge polarization layer, especially in polar solvents like H2O. In fact some electrochemical methods rely on it. But for getting the "pure" ideal thermodynamic (reversible) numbers the large tendency of electrode surfaces to change is a real problem.

Overpotential is the term we use to describe the additional energy (voltage) we have to put in to these non-ideal systems (either at the anode or the cathode) to overcome these "side reactions". This energy is ultimately converted (mostly) to heat. Although maybe not an overpotential issue, surface area can be important, so smoothness does matter.

I can think of no area in chemistry (short of biochem) where you have to get into the weeds (details) so quickly as you do with electrochemistry. I wonder if its because chemistry is actually about electric interactions between atoms, molecules, ions, etc.? I should also mention that on other nice thing about the DME was that the surface was smooth - although I can't recall whether the surface was smooth at the atomic level, most solids are far from smooth, another problem in reproducibility.


Why does lattice enthalpy decrease with increasing ionic size? Lattice enthalpy decreases as ions get larger, but I have found two explanations:

The charge density is greater in smaller ions, so greater attraction
The ions are themselves able to get closer together, so greater attraction

Do these two explanations relate in any way? What would be the best explanation?

Think about Coulomb's inverse-square law:
$$F = k\frac{q_1q_2}{r^2},$$
where $q$ is the magnitude of the charge, $r$ is the distance, and $k$ is the Coulomb’s constant.
Mathematically speaking, as $r$ increases, the magnitude of $F$ becomes smaller. As far as the charge density, as it increases, the magnitude of $F$ increases.
So both explanations are saying the same thing from different point of views. As ionic radii increases, charges are further apart so attraction decreases. As charge density increases, the attraction increases since the charges are increasing in magnitude.
Just as $\ce{MgO}$ is stronger then $\ce{NaCl}$ since $\ce{Mg}$ and $\ce{O}$ have smaller ionic radii and they both bare a $2+$ and $2-$ charge, respectively, making the lattice enthalpy greater.


Reason for the stronger acidic property of phenol than alcohol In phenol, pulling the $\mathrm{p}_z$ electrons from the oxygen atom into the ring causes the hydrogen atom to be more partially positive than it is in aliphatic alcohols. This means it is much more easily lost from phenol than it is from aliphatic alcohols, so phenol has a stronger acidic property than ethanol.

Could someone please explain the link between the $\mathrm{p}_z$ electrons of oxygen overlapping with the cloud of delocalised electrons of the phenol ring and the increase in positivity of the bonded hydrogen … which in turns allows phenol to lose a proton more easily?

In fact I find a more simple reasoning with resonance structure. When phenol loses the $\ce{H+}$ the phenolate ion is stabilized due to the resonance effect, as shown below:

The energy of the dissociated form is lower and so phenol has more chance to be in the solution dissociated with the phenolate ion. Aliphatic alcohols are not stabilized by resonance so they are not very prone to dissociation.


Reaction speed with solids It seems that reaction rate is defined as change of concentration over time
and the speed doesn’t change if you add more solid. But imagine we have a cube (1 kg) of $\ce{CaCO3}$ that yields 1 mole $\ce{CO2}$ in $t$ seconds and we have another cube (5 kg) of $\ce{CaCO3}$. Does this one also yield 1 mole $\ce{CO2}$ in $t$ seconds? (my teacher said yes)

But if we think of second cube as 5 separate 1 kg cubes, then each one again yields 1 mole per $t$ seconds, making it 5 moles of $\ce{CO2}$ in $t$ seconds.

So which one actually happens and is the speed the same?

Reaction rates are in most cases influenced by concentration for liquids/dissolved solids and surface area for solids.

Surface area

The surface area of the calcium carbonate is one factor in the rate of the reaction. The more surface area, the faster the reaction. In this case, certainly, the five $\pu{1kg}$ cubes will react much faster. The reason large cubes are used to demonstrate this instead of powder is because the surface area is more or less constant throughout the reaction.

The rate law
$$\text {Rate} = k[\text A]^\text m[\text B]^\text n $$ where $\text A$ and $\text {B}$ are the reactant concentration, $k$ is the rate constant (does not change in the same reaction), and $m$ and $n$ are the rate orders (which also does not change in a reaction)

This means that unless m and n are 0 making it a zero order reaction which is independent of concentration, in most cases concentration does have an effect on rate.

In this case, with $k$, m and n constant and either  m or n non-zero, and increase in $[\text A] \text{ or } [\text B]$ would increase the rate.

If $[\text A]$ is calcium carbonate and $[\text B]$ is hydrochloric acid, it shows that most likely an increase in concentration of $\text B$ will increase rate. However, calcium carbonate is a solid and has no concentration. We can imagine $[\text A]$ as the surface area of $\ce{CaCO3}$ instead, and m as 1.

We could also imagine the $\ce{CaCO3}$ dissolved in a solvent, like sugar in water. Now, the reaction rate will be dependent on the amount of
$\ce{CaCO3}$
(Note: the rate orders can only be determined experimentally)


Is m-cresol or p-cresol more reactive towards electrophilic substitution? Which is more reactive towards electrophilic substitution, meta-cresol or para-cresol?

Both oxygen and methyl are activating groups. I don't know on what basis do I compare their reactivity. There's just a change in the position of one of the group.

My book says meta-cresol is more reactive and the reason given was

  If two groups are working for the same effect at the same time, then their effect decreases.

I don't understand the reasoning and hence the answer. Any help would be appreciated.

To start, it is important that you understand that the methyl and hydroxy groups are activating only in the ortho and para positions. For example, if I were to substitute plain phenol, which contains an activating hydroxy group, which position would my electrophile (in this a bromine electrophile) add to? You can see that the intermediary sigma complex involves a positive charge. When the bromine attacks at the ortho or para (I did not include a diagram for the para attack, but it is similar to the ortho attack) positions, it is stabilized by an additional resonance structure provided by the oxygen lone pair. In the case of the meta attack, as you can see in the diagram below, this does not happen, and the intermediate is left with one less resonance structure.

Most activating groups will have the same directing effects as shown with the activating hydroxyl group above (para-ortho directors), so a methyl group, which contains no lone pair (and thus no additional resonance structures) will still stabilize the intermediate by donating electron density in much the same way as if an additional resonance structure was present.

Now getting back the para and meta cresol, the meta cresol contains two activating groups

Both of these activating groups activate the same positions, so this combination of activating effect will make the meta cresol more reactive than para-cresol. In para-cresol, the two groups do not work to activate the same positions, so, no matter which carbon is substituted, it will not be activated by both groups simultaneously like in meta-cresol.


Oxygen manufacturing on Mars I was reading about the Perseverance lander on Mars and was surprised to find that it has an oxygen manufacturing instrument on board. The oxygen is needed as a part of a binary rocket fuel to get back from Mars. It makes oxygen from carbon dioxide, a technology that would be very useful for carbon capture here on Earth. I can not find the chemistry or the technology that makes this work. Does anybody know the chemistry of this instrument?
The following is the text from nasa.gov explaining the mechanism of MOXIE:

Mars’ atmosphere is 96% carbon dioxide. MOXIE works by separating
oxygen atoms from carbon dioxide molecules, which are made up of one
carbon atom and two oxygen atoms. A waste product, carbon monoxide, is
emitted into the Martian atmosphere.

On this site, Nasa give some closer information on the MOXIE system.
It splits electrochemically $\ce{2 CO2}$ into $\ce{2 CO}$ and $\ce{O2}$.


Question on d block contraction For size trend down a group, it is known that:$$Al\gt Ga$$
Due, to d block contraction of gallium.
Why isn't this the case when it comes to silicon and germanium?
Won't the poor shielding effect from the d electrons have an effect here?

The Answer on the SE for a similar question does not really answer the question.

If germanium is supposed to be bigger than aluminum, in fact, its empirical values correspond to $\pu{125pm}$ for each one respectively, but germanium being lower in the periodic table is bigger at the same time which is why we say that $\ce{Al < Ga}$. Now yes, the atomic radius of silicon is $\pu{118pm}$ while as we said before, the atomic radius of germanium is $\pu{125pm}$, in the same way the size is deduced based on its position in the periodic table, so $\ce{Si < Al < Ga}$.
This occurs because there is a relationship between the effective nucleus charge and the electronic levels of the atom, since the positive nucleus attracts the electrons, but in turn, the electrons repel each other (except those of the first level), it is by This means that the forces and counter forces end up having an effect on the valence electrons, and these determine the atomic radius of the atom. This difference of attraction by the nucleus and repulsion by the electrons increases as there are more electronic levels, that is, if I have a compound with Z = 11 and e = 11 that we will call "A", it will be greater than a compound "B" whose Z = 10 e = 10, leaving B <A since although both have the same number of electrons as protons, compound A has an electronic configuration $\ce{1s^2 2s^2 2p^6 3s^1}$ with 11 protons while compound B has an electronic configuration of $\ce{1s^2 2s^2 2p^6}$ with 10 protons, that is, compound A has a higher electronic level than compound B, which gives a larger atomic radius.
Reference
1.https://es.wikipedia.org/wiki/Anexo:Radio_at%C3%B3mico_de_los_elementos_qu%C3%ADmicos
Ah, I just realized that it was Galio and not Germanio, sorry.


How can I get analytical form of GTO primitives for a atom? As I know, for Hydrogen atom a $1s$ Slater Type Orbital (STO) equation is (I can get it from here):
\begin{equation}\label{STO}
    \mathrm{STO} = \sqrt{\frac{\zeta^3}{\pi}} e^{-\zeta r}.
\end{equation}
With remark from Szabo A., Ostlund N.S. Modern Quantum Chemistry: Intro to Advanced Electronic Structure Theory, Dover, 1996:

For example, the standard exponent  for the $1s$ basis function of hydrogen is , $\zeta = 1.24$. This is larger than the $\zeta = 1.00$ exponent of the hydrogen atom, since the hydrogen $1s$ orbital in average molecules is known to be "smaller" or "denser" than in the atom.

Gaussian Type Orbital (GTO) equation for $1s$-orbital (where did this coefficient $\left(\frac{2\alpha}{\pi}\right)^{3/4}$ come from?):
\begin{equation}
    \mathrm{GTO} = \left(\frac{2\alpha}{\pi}\right)^{3/4} e^{-\alpha r^2}
\end{equation}
So, I have good analitical form of Hydrogen STO and GTO, and using formula
\begin{equation}\label{}
    \sqrt{\frac{\zeta^3}{\pi}}e^{-\zeta r} \approx C_1 \left(\frac{2\alpha_1}{\pi}\right)^{3/4}e^{-\alpha_1 r^2} +  C_2 \left(\frac{2\alpha_2}{\pi}\right)^{3/4}e^{-\alpha_2 r^2} +  C_3 \left(\frac{2\alpha_3}{\pi}\right)^{3/4}e^{-\alpha_2 r^2}.
\end{equation}
and copyng $\alpha_i$'s and $C_i$ from basissetexchange.org
#-----------------------------------------------------
# Basis Set Exchange
# Version v0.8.13
# https://www.basissetexchange.org
#-----------------------------------------------------
#   Basis set: STO-3G
# Description: STO-3G Minimal Basis (3 functions/AO)
#        Role: orbital
#     Version: 1  (Data from Gaussian09)
#-----------------------------------------------------

BASIS "ao basis" PRINT
#BASIS SET: (3s) -> [1s]
H    S
      0.3425250914E+01       0.1543289673E+00
      0.6239137298E+00       0.5353281423E+00
      0.1688554040E+00       0.4446345422E+00
END

I can plot Hydrogen orbital

Now I want to do the same for a Lithium atom using data
BASIS "ao basis" PRINT
#BASIS SET: (6s,3p) -> [2s,1p]
Li    S
      0.1611957475E+02       0.1543289673E+00
      0.2936200663E+01       0.5353281423E+00
      0.7946504870E+00       0.4446345422E+00
Li    SP
      0.6362897469E+00      -0.9996722919E-01       0.1559162750E+00
      0.1478600533E+00       0.3995128261E+00       0.6076837186E+00
      0.4808867840E-01       0.7001154689E+00       0.3919573931E+00
END

But, I do not understand where I can get a analitical formula for GTO's primitives. All of references,  I saw:
\begin{equation}\label{GTO}
\mathrm{GTO}(x,y,z;\alpha,\ell_1,\ell_2,\ell_3) = N  x^{\ell_1} y^{\ell_2} z^{\ell_3} e^{-\alpha r^2}
\end{equation}
with  unknown normalizing factor $N$. Sometimes I saw for $1p$-orbital
\begin{equation}
\mathrm{GTO}(p_x) = \left(\frac{128\alpha^5}{\pi^3}\right)^{1/4} x \exp(-\alpha r^2)
\end{equation}
but I don't understand where the factor $\left(\frac{128\alpha^5}{\pi^3}\right) $ comes from.
So, my general question: where can I see the analytical view of the GTO primitive (like for STO's), and how can you find $\zeta$ for atom obital? (how to read such data?)

Main differenece betweet STO's and GTO's is a $r$-expotetnt. Normalizing factor of GTO one can get from common procedure:
\begin{equation}
N^2 \int_0^\infty \left(r^{n-1}e^{-\zeta r^2}\right)^2 r^2 dr =1
\end{equation}
Of course, computation is not so simple for $n > 1$, but we can use online WolframAlpha (clickable calculation) and we get the formulas from the question post.
For a Lithium atom we have contraction (6s,3p) -> [2s,1p], which means
the first 3s of 6s GTO's cotract to one s "1s"-STO
$$\mathrm{STO}(1s) = \sqrt{\frac{\zeta^3}{\pi}} e^{-\zeta_1 r},$$
second 3s of 6s GTO's contract to another s "2s"-STO
$$\mathrm{STO}(2s) = \sqrt{\frac{\zeta^5}{3\pi}} r e^{-\zeta_2 r},$$
3p GTO's contract to one p - "2p"-STO
$$\mathrm{STO}(2p) = \sqrt{\frac{\zeta_2^5}{\pi}} x e^{-\zeta_2 r}$$
(notation in orbitals in quotation marks "1s", "2s" means the same as for the hydrogen atom). So, as we can see, every atomic STO contract by only 3 GTO's (which justifies their name).

For a Litium atom we have for "1s" atomic orbital
\begin{equation}
\sqrt{\frac{\zeta_1^3}{\pi}} e^{-\zeta_1 r} \approx
C_1 \left(\frac{2\alpha_1}{\pi}\right)^{3/4}e^{-\alpha_1 r^2} +
C_2 \left(\frac{2\alpha_2}{\pi}\right)^{3/4}e^{-\alpha_2 r^2} +
C_3 \left(\frac{2\alpha_3}{\pi}\right)^{3/4}e^{-\alpha_3 r^2}
\end{equation}
where
\begin{align}
\alpha_1 = 0.1611957475E+02; C_1 = 0.1543289673E+00;\\
\alpha_2 = 0.2936200663E+01; C_2 = 0.5353281423E+00;\\
\alpha_3 = 0.7946504870E+00; C_3 = 0.4446345422E+00;
\end{align}

For "2s" atomic orbital
\begin{equation}
\sqrt{\frac{\zeta_2^5}{3\pi}} r e^{-\zeta_2 r} \approx
C_4 \left(\frac{2\alpha_1}{\pi}\right)^{3/4}e^{-\alpha_4 r^2} +
C_5 \left(\frac{2\alpha_2}{\pi}\right)^{3/4}e^{-\alpha_5 r^2} +
C_6 \left(\frac{2\alpha_3}{\pi}\right)^{3/4}e^{-\alpha_6 r^2}
\end{equation}
where
\begin{align}
\alpha_4 = 0.6362897469E+00; C_4 = -0.9996722919E-01;\\
\alpha_5 = 0.1478600533E+00; C_5 = 0.3995128261E+00; \\
\alpha_6 = 0.4808867840E-01; C_6 = 0.7001154689E+00;
\end{align}

For "2p_x" atomic orbital
\begin{equation}
\sqrt{\frac{\zeta_2^5}{\pi}} x e^{-\zeta_2 r} \approx
D_1 \left(\frac{128\alpha_4^5}{\pi^3}\right)^{1/4} xe^{-\alpha_4 r^2} +
D_2 \left(\frac{128\alpha_5^5}{\pi^3}\right)^{1/4} xe^{-\alpha_5 r^2} +
D_3 \left(\frac{128\alpha_6^5}{\pi^3}\right)^{1/4} xe^{-\alpha_6 r^2}
\end{equation}
where
\begin{align}
\alpha_4 = 0.6362897469E+00; D_1 = 0.1559162750E+00;;\\
\alpha_5 = 0.1478600533E+00; D_2 = 0.6076837186E+00; \\
\alpha_6 = 0.4808867840E-01; D_3 = 0 0.3919573931E+00;
\end{align}

(For $p_y$ and $p_z$ --- do the same)
Also, for STO's best fit $\zeta_1 = 2.69$ and $\zeta_2 = 0.81$ are consistent with those given in the article Hehre, W. J., Ditchfield, R., Stewart, R. F., & Pople, J. A. (1970). Self‐Consistent Molecular Orbital Methods. IV. Use of Gaussian Expansions of Slater‐Type Orbitals. Extension to Second‐Row Molecules. The Journal of Chemical Physics, 52(5), 2769–2773. doi:10.1063/1.1673374


Formation of methyl esters Can't we perform Fischer esterification using methanol to make a methyl ester from a carboxylic acid?

If so, why might we use diazomethane? Would the release of nitrogen gas confer an entropic advantage to the reaction? As I understand, the Fischer esterification process is an equilibrium process and at equilibrium there is often significant amounts of both product and reactant. So we might have to add excess alcohol to drive the reaction to the right.

I ask because my book just presents the use of diazomethane as a way of making methyl esters without any discussion of why might someone do this rather than just use a few gallons of methanol.

On the other hand I know from working in a research lab that methanol is poisonous and protocol calls for its use in a fume hood ... could that be another reason? On the other hand diazomethane is also toxic (and explosive to boot) ... so I would definitely be inclined to use that in a fume hood (perhaps more so than when using methanol).

Both methods are used in practice (and a a lot more exist). Diazomethane is a last resort method when everything else fails. For example you might have a molecule that is acid sensitive or heat sensitive and then $\ce {MeOH/H^+}$ is out of the question. Although diazomethane is an excellent reagent to do the job, as it works selectively with almost every substrate, to give good yields under mild conditions, generally no extensive work-up or purification are required, it has some major disadvantages: It is not commercially available, it has to be prepared fresh before the reaction using a (somewhat) specialised equipment and a toxic starting material (Diazald). It is explosive and very toxic and obviously can not be stored. An alternative to using diazomethane is TMS diazomethane that is stable enough to be stored as a solution and it is commercially available however it is not innocent at all as this article  shows.


How can change in Gibbs free energy ever be positive? I have a question regarding this formula for Gibbs free energy of a system: $$\Delta G=\Delta H-T\,\Delta S$$
According to the second law of thermodynamics: $T\,\Delta S\ge\Delta Q$ and to my understanding, $\Delta H$ is just the same as $\Delta Q$.
Now, if $\Delta Q$ will be always smaller than (or equal to) $T\,\Delta S$ and $\Delta H=\Delta Q$, then looking at the first formula again it seems that $\Delta G$ can only be smaller than (or equal to) zero.
What am I doing wrong here?

You didn't do anything wrong here. This was a very good observation. This is one of the criteria of spontaneity, which I believe you haven't learnt yet.
As you correctly wrote, $\Delta G$ can only be negative or 0 for any reaction. This is actually a consequence of the second law of thermodynamics, which also goes on to say that in any spontaneous reaction the entropy change of the universe is always positive.
Essentially, the answer to your question heading is: in any spontaneous reaction, the Gibb's free energy change can never be positive. Since you took the conditions of the second law, which essentially explains spontaneity, you arrived at this result.


Why do polarizable nucleophiles favor SN2 over E2? Weakly basic ions and polarizable bases favor $\mathrm{S_{N}2}$ over $\mathrm{E2}$. Why? The argument used that polarizable nucleophiles form bonds earlier, stabilizing the transition state, can be applied to either $\mathrm{S_{N}2}$ or $\mathrm{E2}$.

Polarisable Nucleophiles increase the rate of an Sn2-type reaction but not an E2 reaction because the more polarisable a base is, lower the electronegativity and charge density (conversely, higher ionic radius) it has.

Now, we know that the basic strength depends on these factors( in a period, we check the electronegativity and in a group, the ionic radius).

As, the Rate Determining Step of an E2 reaction involves the attack of a base, lower basic strength would mean that the reaction would be slower. Hence, more polatisable bases means that E2 would be slower.


Why doesn't the bromide anion attack during electrophilic addition of bromine in methanol solvent? It is known that addition of bromine (in methanol) to ethene will yield 1-bromo-2-methoxyethane. The reaction proceeds with the formation of a bromonium intermediate in the first step. In the second step, methanol attacks either carbon through the lone pair on its oxygen atom. In the last step, the $\ce{CH3OH+}$ loses a proton, which combines with the bromide anion obtained in the first step, to release $\ce{HBr}$ as a by-product.

My question: Why doesn't the $\ce{Br-}$ (obtained in the first step) attack as a nucleophile in the second step?

We know that nucleophilicity is directly proportional to the atom's charge and size. Bromide anion has a unit negative charge as opposed to the neutral oxygen atom in methanol. Bromide anion is also much larger in size. Both these factors favor bromide anion being a much better nucleophile than the neutral oxygen atom in methanol. Yet, the latter only attacks in the second step. Why?

There's a very simple explanation for this. Your reasoning regarding $\ce{Br-}$ being a better nucleophile is absolutely correct. The reason as to why methanol attacks instead of $\ce{Br-}$ is because it is the $\mathsf{\color{red}{solvent}}$. The solvent concentration is simply too high. Only one $\ce{Br-}$ is generated.

More or less the answer is this paragraph from J. Clayden - section "Unsymmetrical bromonium ions open regioselectively" in chapter 20 "Electrophilic addition to alkenes" :

  ...when a bromination is done in a nucleophilic solvent—water or methanol, for example—solvent molecules compete with the bromide to open the bromonium ion. As
  you know, alcohols are much worse nucleophiles than bromide but, because the concentration of solvent is so high (remember—the concentration of water in water is $\pu{55M}$), the solvent gets there first most of the time. This is what happens when isobutene is treated with bromine in methanol. An ether is formed by attack of methanol only at the more substituted end of the bromonium ion.

Note: Concentration of pure methanol is around $\pu{24.75 M}$, calculated from the density given on ChemSpider


Trial by fire: Safely heating sand without it agglomerating? Wikipidia quotes a melting point of 1713C for sand.
I went on the beach to get some sand. I assume it's a mix of all kinds of stuff.
The finality is to prepare a big pile of clean sand to use as filtering aid above paper filter (among other uses).
I'd like to get rid of plastics & organics by heating it up.
What temperature can I bring it up to before the smallest particles start to agglomerate? What temperature would you recommend for pyrolysing as much of everything in there that is not SiO2 such that it either degrades or later more easily disssolves in water?

Sand is not available as a reagent grade chemical, but rather is mineral origin particles of a size between 0.074 and 4.75 millimeters. Gravel is bigger, silt is smaller.
Pure silica (SiO$_2$) may melt at 1713$^o$, but the melting point of real sand is given as 1500 to 1610, and you can probably find some numbers outside that range.
Perhaps more of an issue than the exact melting point is how much are you preparing to treat? And with what heat source? And what sort of vessel are you going to contain the sand in while you heat it up and cool it down? And for what purpose?
Thin layers of sand (small overall quantity) may burn off organics relatively easily, say in a stainless steel measuring cup, with a propane torch  - although, the handle will get hot, so arrange some sort of apparatus to hold the vessel.
Different materials will need different temperatures and times to burn off completely; perhaps you can heat till smoke stops coming off and the sand looks like it is a bright red or orange color. I doubt that you would have to get to a yellow.


why CO2 is not organic this might be very childish question but i am comfused because they say that organic molecules are those that contain H and C if that is so how can you call carbontetrachloride organic, even it hasn't any H atoms

Carbon dioxide
"is" neither organic nor inorganic, it's just carbon dioxide.
This distinction is nothing but a convention to separate things
between the textbooks and professors.
Carbon dioxide forms carbonic acid in water and the usual salts.
This was/is reason to think of it as inorganic.
On the other hand, urea, the diamide of carbonic acid was thought
to be organic when Wöhler synthesized urea for the first time
in lab, today it is categorized as inorganic.


What is this Boron Compound called? What would this aromatic boron compound be called?

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), heteromonocyclic compounds with up to and including ten ring members are named by the extended Hantzsch–Widman system. According to P-68.1.1.3.2, this also applies to heterocyclic parent hydrides containing Group 13 (boron group) atoms.

The Hantzsch-Widman system stem for an unsatured 7-membered ring is “epine”. Therefore, the preferred name for the compound that is given in the question is 1H-borepine.
The name 1H-1-boracyclohepta-2,4,6-triene is not recommended.


Effect of relative strength on nature of solution If a solution has equal no. of gram equivalents of a week acid (say acetic acid) and a strong base say NaOH , will the soltution be neutral or the relative strength has some role making solution slightly basic?

The conjugate base of a weak acid is basic, whereas the conjugate acid of a strong base is not acidic. When a weak acid and strong base neutralize each other, the basicity of the conjugate base formed makes the solution basic.


Why does benzene have 3 π–π* transitions in the UV range, but ethene, butadiene, and hexatriene each have only 1? Are the π and π* orbitals of benzene much closer together in energy which allows for multiple transitions? Why is this not the case for hexatriene if they both contain the same number of pi bonds?

It seems that here we only need to consider the lowest energy transitions, i.e. HOMO–LUMO transitions. The difference between benzene and the linear polyenes is that both the HOMO and LUMO of benzene are doubly degenerate, whereas the linear polyenes do not have degeneracy.

If we think about this HOMO–LUMO excitation in the case of benzene, we can schematically represent it like this:

The ground state has overall symmetry $\mathrm{A_{1g}}$ because all electrons are paired, but the first excited electronic configuration above $(\mathrm{e_{1g}})^3(\mathrm{e_{2u}})^1$ corresponds to three different electronic states, each with a slightly different spatial distribution of electrons. The symmetries of these three states are obtained by taking a direct product of the relevant irreps:

$$\mathrm{E_{1g} \times E_{2u} = B_{1u} \oplus B_{2u} \oplus E_{1u}}$$

Because these three states arise from the same electronic configuration, their energies are going to be fairly close to each other, and all three electronic transitions from the ground state

$$\mathrm{A_{1g} \to B_{1u}, \quad A_{1g} \to B_{2u}, \quad A_{1g} \to E_{1u}}$$

should fall within the UV range.

For any linear polyene (including ethene, butadiene, and hexatriene), both HOMO and LUMO are nondegenerate, so the $(\pi)^3(\pi^*)^1$ configuration will only correspond to one electronic state.

This is a rather simplified analysis, because we are only considering transitions which conserve spin, i.e. singlet–singlet transitions. These are so-called "spin-allowed" transitions. The singlet–triplet transitions also occur within the UV range, but are technically "forbidden". On top of that, not all of the three singlet–singlet transitions in benzene are equal: two of them are also "forbidden", as the transition dipole moment $\langle i | \hat{\mu} | j \rangle$ vanishes. But that is a story for another day. The keen reader may wish to check out J. Mol. Struct. 1973, 19, 143–166 for an assignment of the electronic transitions of benzene.


Why does magnesium oxide not react with water? There is a question in my textbook which asks "which of the reactions are not possible to occur". The right answer is $\ce{MgO + H2O}$ and my chemistry teacher confirmed it. Why is that so? I thought that it would form $\ce{Mg(OH)2}$.

Does the reactivity also depend on the temperature?

Like a lot of things in chemistry, "it depends".

Magnesium oxide is commonly produced by calcination of magnesium hydroxide or carbonate. The magnesium hydroxide or carbonate decomposes at a fairly low temperature whereas magnesium oxide is itself stable to both decomposition and melting  beyond 2000°C. So, we can choose from a wide range of calcining temperature, and as explained here we can use this to dial in the reactivity we want.

At a low calcining temperature the magnesia is quite reactive, even hygroscopic (the "magnesium oxide" film formed on magnesium in ambient air is thus usually magnesium hydroxide); this "light-burned magnesia" is used where easy reactivity is required as in water or soil treatment.  And, yes, the reaction with water gives magnesium hydroxide.

Calcination at higher temperature gives less reactive "hard-burned" magnesia, or in the extreme case the totally unreactive "dead-burned magnesia" used in furnace refractories.


Is pigment grade titanium dioxide likely to contain any impurities that are toxic? I have a bag of titanium dioxide of pigment quality, unknown source. I am curious to know if the manufacturing process of titanium dioxide makes it likely that it contains any toxic by products / impurities.

Would it for example be safe to use pigment grade titanium dioxide to manufacture sunblock?

According to the MSDS for titanium dioxide, the substance itself is suspected of causing cancer. See, for example, this link.

My opinion is that, in general, it's not a good idea to use reagents from unknown sources for any purpose, especially something like sunblock.


How would I measure the Carbon Dioxide content in Coca Cola over a period of time? I'm sorry if this is a broad question, but I am trying to plan a simple experiment. I am wondering is there a somewhat simple way of continuously measuring $\ce{CO2}$. I already know the initial carbon dioxide content is around $\pu{2.2g}$ from research, but I don't know how this result is reached.

Measure the change in mass over time of the remaining liquid. Though some water will also evaporate, you can control for that by keeping the humidity near 100%.

If you have to be precise, collect the outgassed $\ce{CO2}$ in a liquid-nitrogen cold trap. Check the mass of the condensate, which should equal that lost from the soda. Check purity, to be really finicky.


Heat of combustion for octane I'm a Mechanical Engineering student and I have been working on a school project but I want to go a bit beyond that and it has led me here. I'm designing an engine cylinder, and I figured I should be able to calculate the heat of combustion of the working fluid.

Please remember that I have no chemistry textbook and all I have been doing up to now is using the internet and some vague memory from high school.

So I wrote the balanced equation,

$$\ce{C8H18 + 12.5\left(O2 + 3.76 N2\right) -> 8 CO2 + 9 H2O + 23.5 N2}$$

using that I was able to compute the air-to-fuel ratio, individual masses for the $1.8\ \mathrm L$ volume, etc. I settled on a compression ratio of 8 which after some thermodynamic computation (if neccessary I'll present them) I got to the following state,

$$
T_2 = 650.9 \; \text{K} \\
P_2 = 1779 \; \text{kPa}
$$

basically those are the conditions prior to the spark. I'm trying to compute the heat of combustion using those properties. Using Mathematica I found out that for Octane the heat of combustion is somewhere along,

$$
\Delta H=5474\;\frac{\text{kJ}}{\text{mol}}
$$

but I assume this is for STP, my interest is to calculate for my conditions at state 2. After some bit of research I noticed one has to use the specific heats of the participants, but the website stated 'find these at the end of the book' – which I don't have one. To be honest, it will probably take me days to figure out this problem on my own using the internet. So does anyone know of a shortcut such as an online database or some public tables that display specific heats (maybe as a function of temp) for octane. Or how would you go about solving my problem?

Remember that I don't have much knowledge regarding advanced chemistry, and most of my interest lies in the field of engineering.

Enthalpy of a system is a state function, meaning, it does not depend upon the path you take to reach the end state. What that means in this case is that the change in the enthalpy of combustion will be the heat required to raise the temperature of the combustion material to the new temperature and to increase the pressure. Essentially, think of it as if the combustion is occurring at STP, and you must construct a path from the beginning state to the end state which will pass that combustion at STP. In this case, that would be:

Initial state of reactants -> STP reactants
Heat of combustion at STP
STP products -> final state of products

Your final state will be the same as your initial state, 650.9 K/1779 kPa. The important point is that the heat in the change of the reactants to STP will be different from the change of the products back to the initial state. This difference would be reflected if you found a heat of combustion at some condition other than STP.

Nitrogen is a spectator to the reaction, so you can completely ignore it for the sake of reaction. However, you do need to account for raising it to the final pressure/temperature if either is changing. I'll assume you've got that under control.

Now, the issue comes in at the fact these are somewhat extreme conditions, at least in pressure. To calculate this properly would take a great deal of time and effort. That said, there may be corners you are capable of cutting in the process. The general process you need to follow is:

Take octane and $\ce{O2}$ from 1779 kPa to 1 atm, 650.9 K to $T_{1,\mathrm i}$.
Take octane and $\ce{O2}$ from $T_\mathrm i$ to 25 °C at constant pressure (using $C_p$).
Use the heat of combustion.
Take $\ce{H2O}$ and $\ce{CO2}$ from 1 atm to 1779 kPa, 25 °C to $T_{2,\mathrm i}$.
Take $\ce{H2O}$ and $\ce{CO2}$ from $T_{2,\mathrm i}$ to 650.9 K.

The overall enthalpy change of this process will be the heat of combustion at your conditions. Now, you'll have to decide for yourself about steps 1 and 4, as you need to be considering intermolecular interactions, and thus need to pick an equation of state (not ideal gas). I feel like I am not necessarily knowledgeable enough in this specific field to offer suggestions in that regard. I personally would just pick something like VDW, but there might be a specific standard for combustion processes. Further, only one book that I have access to right now that has the proper temperature-dependent-equation form for $C_p$, and it doesn't have one for octane. These are the ones for the other gases involved:

$\ce{O2}$: $C_p = 25.46 + 1.519 10^{-2} T - 0.715 10^{-5} T^2 + 1.311 10^{-9} T^3$

$\ce{CO2}$: $C_p = 22.243 + 5.977 10^{-2} T - 3.499 10^{-5} T^2 + 7.464 10^{-9} T^3$

$\ce{H2O}$: $C_p = 32.218 + 0.192 10^{-2} T + 1.055 10^{-5} T^2 - 3.953 10^{-9} T^3$

From Chemical and Engineering Thermodynamics (Sandler).

I wish I could help more. Hopefully this explanation at least points you in the right direction.


Is there any diprotoic acid where both pka's are less than 0? The question states it all.

I'm working out special cases for a different problem and such an acid would create another special case.

Sulfuric Acid is the strongest that I remember. pKa's −3, 1.99

Ethanedisulfonic acid does indeed have $pK_{a1}=-2.06, pK_{a2}=-1.46$.
Using polycyano substutution can also work.  This paper reports the table below, note that the tetracyano hydroquinone "makes it" (in the excited state produced by irradiation).


Reduction with NH2-NH2 , OH-, and glycol this question was asked in my exam and I have done it properly until the very third step but unfortunately, although the last step looked like wolfkishner reduction to me I could not do it.
My approach:
At first, via ozonolysis aldehyde groups will come and then aldol reaction would take place and by heating alpha-beta unsaturated aldehyde product will be formed.
But then what if I reduce it in the last step my answer is not matching with any of the options given. Any help will be highly appreciated.

If you look carefully, the reagent for the first step is $\ce{O3/H2O}$, which is used for Oxidative ozonolysis. If the mechanism of ozonolysis is drawn, you will realise that a molecule of Hydrogen peroxide($\ce{H2O2}$) is released by the oxidation of the alkene to aldehyde and if that $\ce{H2O2}$ is not removed from the reaction mixture, it will oxidize the aldehyde ultimately to acid. That's why during reductive ozonolysis, people use reducing agents like $\ce{Zn}$ or $\ce{Me2S}$ or $\ce{PPh3}$ which can get oxidised themselves by the $\ce{H2O2}$  produced  to $\ce{ZnO}$, $\ce{Me2S=O}$ and $\ce{Ph3P=O}$ respectively and your aldehyde is not oxidized to acid any more. But in this case no such reducing agents is used and thus the product of the first step in the diacid not di-aldehyde. After that heating with $\ce{Ca(OH)2}$ doesn't perform aldol reaction here but eliminates one molecule of $\ce{CaCO3}$ from the compound i.e. essentially decarboxylation and you are left with one carbon less i.e. $\text{Cyclopentanone}$. Then as you recognised correctly is the Wolff-Kisher reduction.

So your final product is $\text{Cyclopentane (B)}$. Here is the schematic of Oxidative ozonolysis releasing $\ce{H2O2}$ responsible for oxidizing the di-aldehyde to di-acid.


Which ions determine ionic conductivity of a salt solution? Let's take $\ce{ZnSO4}$ salt solution in water for example. So the ions inside my solution will be $\ce{H+}$, $\ce{OH-}$, $\ce{Zn^2+}$ and $\ce{SO4^2-}$.

My questions are:

Is the ionic conductivity measured for the solution gives an idea of mobility of zinc ions and sulfate, or is it a result of all four ions combined?
When we change the salt concentration of zinc sulfate, do we change the ionic conductivity of zinc and sulfate ions only, or all four ions?

Also, any undergrad university level source to understand this concept will be helpful.

Di water H2O has very poor conductivity. Because H+ and OH- only driven by the ionic product.

2 H2O <=> H3O+ +OH-

If any other ions will be add the condictivity rises . So that can be for example by acids (H+), alcaline  solutions (OH-) and salts (Cations and anions). The conductivity of the salts  >>>>>> as of the water. So mainly in your case zinc and sulphate drive it. But to calculate correctly all 4 ions participate.


Attack on cyclic intermediate during NGP I have studied that when an epoxy ring opening takes place in acidic medium, the ring opens in a way that the most stable carbocation forms. This is due to the positive charge on the oxygen, which gives partial positive character to the carbon atoms.

A very similar intermediate is formed in NGP (neighboring group participation) reactions with sulphur as the neighboring atom (source). How is the attack carried out then? Is it decided sterically as a regular SN2, or does the logic with oxygen extend to sulphur?

Inductive reasoning in organic chemistry can be somewhat hard at times, given all of the exceptions that can pop up, for example the SN2-like mechanism of opening an epoxide in basic conditions versus the acidic conditions.

For the epoxide opening in base; as you know sterics wins, and in acid the stability of the reactive transition state is our reasoning for the product observed. I believe for the NGP reactivity that stabilization of transition states is an acceptable platform to reason the important question as to exactly how this reaction happens in the first place.

lets look at the NGP example you've pulled from Wikipedia
(Awesome btw! reading organic chem on wiki for fun?!)

I have a hard time fully following your question but here's the best answer I could come up with:

I see a symmetrical transition state where the reactive-intermediate charged cyclopropyl-S-heterocycle can probably just be attacked from either side in a SN2 like fashion. If you prefer charge polarization to be our main reasoning in the S-hetero cycle, given that it is a symmetrical substrate I believe the same product will form either way due to symmetry.

Did you read the part of that wiki article below that depicts NGP reactivity by an Alkene? The driving factor there appears to be pi-stabilization of the carbocation in the transition state.

This is a physical organic chemistry question that is not trivial. I applaud your curiosity.