text:
Is there a relation between transition density and density differences?
When I have an excited state, is there a relation between the transition density associated with this electronic transition and the density difference between the excited state density and the ground state density? Can the transition density be equal to the difference density?

By definition, the transition density is not the same as the difference density. In the following derivation, the transition density is $\mathbf{T}^{\mathrm{CIS}}$ and the difference density is $\mathbf{P}^{\Delta}$. Although it may seem like they can be formed from the same quantities, the true difference density contains orbital relaxation terms and can be used to build a relaxed density, which is the correct excited state density. Neglecting those terms (the $P_{ia}^{\Delta}$) will give an unrelaxed density. I will also try to show that the unrelaxed difference density is still not equivalent to the transition density.
Derivation of individual CIS densities
Here is the derivation of the transition density for doubly-occupied spin-restricted MOs, with quoted blocks and equations directly taken from the original CIS paper.
The CIS approximation is identical to the TDA approximation in TD-DFT. Full TD-HF or TD-DFT follows the RPA equations, which include deexcitation terms in addition to excitation terms. The equations would be different for RPA, but conclusion would be identical.
The standard indexing convention is followed: $\mu, \nu, \lambda, \sigma$ for AOs, $i,j,k,l$ for occupied MOs, $a,b,c,d$ for virtual MOs, and $p,q,r,s$ for any MO. Define the Hartree-Fock (or SCF-type, valid for DFT as well) density as a sum over occupied MOs:
$$
P_{\mu\nu}^{\mathrm{HF}} = \sum_{i=1}^{n} c_{\mu i} c_{\nu i}
$$

The two-particle CI-singles density matrix, $\mathbf{\Gamma}^{\mathrm{CIS}}$, can be written in terms of the HF ground-state density matrix and the ground-to-excited-state transition density matrix, $\mathbf{T}^{\mathrm{CIS}}$:
$$
\Gamma_{\mu\nu\lambda\sigma}^{\mathrm{CIS}} = \frac{1}{2} \left[ P_{\mu\nu}^{\mathrm{HF}} P_{\lambda\sigma}^{\mathrm{HF}} + 2 T_{\mu\nu}^{\mathrm{CIS}} T_{\lambda\sigma}^{\mathrm{CIS}} - P_{\mu\sigma}^{\mathrm{HF}} P_{\lambda\nu}^{\mathrm{HF}} - 2 T_{\mu\sigma}^{\mathrm{CIS}} T_{\lambda\nu}^{\mathrm{CIS}} \right]
$$
where $\mathbf{P}^{\mathrm{HF}}$ is given (above) while $\mathbf{T}^{\mathrm{CIS}}$
$$
T_{\mu\nu}^{\mathrm{CIS}} = \sum_{ia} a_{ia} c_{\mu i} c_{\nu a}
$$

I can't find the definition of $\{a\}$ in the paper, but it is the set of coefficients from a converged CIS calculation that describe the single-particle excitation from every occupied orbital to every unoccupied orbital. As an aside, they are the elements of the $\mathbf{X}$ vector in the RPA eigenvalue problem:
$$
\begin{pmatrix}
\mathbf{A} & \mathbf{B} \\
\mathbf{B} & \mathbf{A}
\end{pmatrix}
\begin{pmatrix}
\mathbf{X} \\
\mathbf{Y}
\end{pmatrix}
=
\omega
\begin{pmatrix}
\mathbf{1} & \mathbf{0} \\
\mathbf{0} & \mathbf{-1}
\end{pmatrix}
\begin{pmatrix}
\mathbf{X} \\
\mathbf{Y}
\end{pmatrix}
$$
$\mathbf{A}$ and $\mathbf{B}$ are parts of the orbital Hessian. If you diagonalize the whole thing, you get the set $\{\omega\}$ of all excitation energies back as eigenvalues. As mentioned above, setting $\mathbf{B} = \mathbf{0}$ gives the TDA or CIS approximation, so there is no longer a deexcitation vector $\mathbf{Y}$.
Importantly, we always start from the set of ground-state MO coefficients $\{c\}$ and describe excitations on top of them.

The CI-singles excited-state density matrix, $\mathbf{P}^{\mathrm{CIS}}$, is also constructed as a sum of HF and excited-state terms:
$$
P_{\mu\nu}^{\mathrm{CIS}} = P_{\mu\nu}^{\mathrm{HF}} + P_{\mu\nu}^{\Delta}
$$
Here we have introduced $\mathbf{P}^{\Delta}$, the CI-singles $\Delta$ density matrix. This can also be called a "difference density matrix", since it represents the changes in electronic distribution upon excitation. It is not, however, the same as the transition density matrix $\mathbf{T}^{\mathrm{CIS}}$ defined above. As we shall demonstrate, it is the use of the true CI-singles density matrix required by (the CIS gradient expression that isn't important here) and not the simple one-particle density matrix (1PDM) which allows the realistic computation of charge distributions, orbital populations, and electronic moments of the excited state. To see this distinction, first consider the $\Delta$ density matrix which would be added to the HF density matrix in order to generate the 1PDM for an excited state. In the MO basis, it is a symmetric matrix with both occupied-occupied (OO) and virtual-virtual contributions
$$
\begin{align}
P_{ij}^{\Delta} &= -\sum_{ab} a_{ia} a_{jb} \\
P_{ab}^{\Delta} &= +\sum_{ij} a_{ia} a_{jb}
\end{align}
$$
with the occupied-virtual (OV) elements all zero. The true CI-singles density matrix required in (the CIS gradient expression) will have exactly the same OO and VV contributions, but the OV terms are not all zero. The appearance of these off-diagonal block elements in the excited-state density matrix can be interpreted as orbital relaxation following the initial gross charge rearrangement due to excitation. That is to say, the CI coefficients will by themselves describe some of the gross features of charge redistribution in the excited state, but the response of the wave function to an external perturbation will account for further refinement in electronic properties. These OV terms can be found by solving a single set of CPHF equations:
$$
L_{ai} = \sum_{bj} \left[ (ij||ab) - (ib||ja) \right] P_{bj}^{\Delta} + (\epsilon_a - \epsilon_i) P_{ai}^{\Delta}
$$
where the $\mathbf{L}$ vector is the CI-singles Lagrangian:
$$
\begin{align}
L_{ai} &= C1_{ai} - C2_{ai} + \sum_{kl} P_{kl}^{\Delta} (al||ik) + \sum_{bc} P_{bc}^{\Delta} (ab||ic) \\
C1_{ci} &= -2 \sum_{jab} a_{ia} a_{jb} (cb||ja) \\
C2_{bk} &= -2 \sum_{ija} a_{ia} a_{jb} (ik||ja)
\end{align}
$$
The total CI-singles $\Delta$ density matrix required for $\mathbf{P}^{\mathrm{CIS}}$ can be generated by backtransforming the entire MO basis $\Delta$ density matrix defined by $P_{ij}^{\Delta}$, $P_{ab}^{\Delta}$, and $P_{ia}^{\Delta}$:
$$
P_{\mu\nu}^{\Delta} = \sum_{pq} P_{pq}^{\Delta} c_{\mu p} c_{\nu q}
$$

Show that the transition density and unrelaxed difference density are not equivalent
We want to compare $T_{\mu\mu}^\mathrm{CIS}$ and $P_{pq}^{\Delta}$, but they are not in the same basis, so one must be transformed. Because I don't want to mess with the block-diagonal structure of $\mathbf{P}_{\mathrm{MO}}^{\Delta}$, I will transform $\mathbf{T}$ to the MO basis:
$$
\begin{align}
T_{pq}^{\mathrm{CIS}} &= \sum_{\mu\nu} c_{\mu p} T_{\mu\nu}^{\mathrm{CIS}} c_{\nu q} \\
&= \sum_{\mu\nu} \sum_{ia} c_{\mu p} c_{\mu i} a_{ia} c_{\nu a} c_{\nu q}.
\end{align}
$$
Now consider the unrelaxed difference density, meaning the ov/vo blocks are zero due to neglecting orbital relaxation effects:
$$
P_{pq}^{\Delta} =
\left(
\begin{array}{c|c}
-\sum_{ab} a_{ia} a_{jb} & \mathbf{0} \\
\hline
\mathbf{0} & +\sum_{ij} a_{ia} a_{jb}
\end{array}
\right)_{pq}.
$$
Here is a slightly hand-waving proof. I am not sure if the MO coefficients in the double sum can be simplified, but it doesn't matter; assume they are unity. $T_{pq}^{\mathrm{CIS}}$ contains terms with $a$, while $P_{pq}^{\Delta}$ contains terms with $|a|^2$. Unless there is an extra transformation that can be done, I don't see how they can be equivalent.
If you want a little test, generate some fake matrices and see for yourself. $\texttt{x} \equiv X_{ia} \equiv a_{ia}$.
#!/usr/bin/env python

from __future__ import print_function

import numpy as np

np.random.seed(42)

nocc = 2
nvirt = 2
norb = nocc + nvirt
nov = nocc * nvirt
nbasis = norb

x = np.random.rand(nocc, nvirt)
c = np.random.rand(nbasis, norb)
cocc = c[:, :nocc]
cvirt = c[:, nocc:]

p_delta_oo = -np.einsum('ia,jb->ij', x, x)
p_delta_vv = np.einsum('ia,jb->ab', x, x)

p_delta_ov = np.zeros(shape=(nocc, nvirt))
p_delta_vo = p_delta_ov.T

p_delta = np.asarray(np.bmat([[p_delta_oo, p_delta_ov],
                              [p_delta_vo, p_delta_vv]]))

# same as np.dot(cocc, np.dot(x, cvirt.T))
t_ao = np.einsum('ia,mi,na->mn', x, cocc, cvirt)
# same as np.dot(c.T, np.dot(t_ao, c))
t_mo = np.einsum('mp,mi,ia,na,nq->pq', c, cocc, x, cvirt, c)

print(t_mo)
print(p_delta)

which gives
[[ 1.82856579  1.89892758  0.59008502  3.09931192]
 [ 1.48134568  1.53535055  0.49193325  2.47904162]
 [ 0.54242862  0.56250364  0.17874155  0.9109361 ]
 [ 1.79242654  1.85791848  0.59456093  3.00118592]]
[[-1.75629929 -1.76345302  0.          0.        ]
 [-1.76345302 -1.77063588  0.          0.        ]
 [ 0.          0.          1.22441763  1.71443377]
 [ 0.          0.          1.71443377  2.40055604]]

Final comment
My original understanding of the term "difference density" comes from a coworker who was taking the real-space representation (cube files) of two state densities, say the ground state $0$ and an excited state $n$, and subtracting them.
$$
\Delta \mathbf{P}^{0\rightarrow n} = \mathbf{P}^{(n)} - \mathbf{P}^{(0)}
$$
Is this equivalent to the definition of the difference density from above?
$$
\begin{align}
\mathbf{P}^{(n)} - \mathbf{P}^{(0)} &= \mathbf{P}^{\mathrm{CIS}(n)} - \mathbf{P}^{\mathrm{HF}} \\
&= \left( \mathbf{P}^{\mathrm{HF}} + \mathbf{P}^{\Delta(n)} \right) - \mathbf{P}^{\mathrm{HF}} \\
&= \mathbf{P}^{\Delta(n)}
\end{align}
$$
Looks good!
References

James B. Foresman, Martin Head-Gordon, John A. Pople, and Michael J. Frisch. Toward a systematic molecular orbital theory for excited states. The Journal of Physical Chemistry 1992 96 (1), 135-149, DOI: 10.1021/j100180a030
Frank Neese. Prediction of molecular properties and molecular spectroscopy with density functional theory: From fundamental theory to exchange-coupling. Coordination Chemistry Reviews 2009 253 (5-6), 526-563, DOI: 10.1016/j.ccr.2008.05.014

You seem to be saying at the top that they're not identical, but your "Final Comment" makes it seem like they are identical. I'm guessing the "Final Comment" part is a full aside, and that your prior (incorrect?) understanding of the difference density is identical to... something?

No. My initial understanding of the difference density is identical to the difference density from the derivation.

This is actually a different question from the one you asked, and you should edit this question to reflect that. I would add the entire comment to the question. I am unfamiliar with Delta SCF so I'll need to read up on it. What program package are you using?

By the way, if you think this was useful, consider accepting it. I can't figure out the Delta SCF problem if you don't modify the question or ask another one.

I'll ask another question related to what I posted after your comment. Thanks @pentavalentcarbon

Thank you for coming back! This helped me clarify things about excited state densities myself and I look forward to attempting an answer for your next question.


text:
What is the effect of Helium-based surroundings (99% Helium) on electrochemical gas sensors?
I am currently working on a project measuring gas composition of car exhaust. In order to dilute the gas concentration down to something electrochemical sensors can read, I diluted the sample with pure helium.
The following is true for these sensors:

Importance of Oxygen. The reactions at the sensing electrode (anode) for some gases are as follows:
  \begin{align}
\ce{CO + H2O &-> CO2 + 2H+ + 2e-} \\
\ce{H2S + 4H2O &-> H2SO4 + 8H+ + 8e-} \\
\ce{NO + 2H2O &-> HNO3 + 3H+ + 3e-} \\
\ce{H2 &-> 2H+ + 2e-} \\
\ce{2HCN + Au &-> HAu(CN)2 + H+ + e-}
\end{align}
  Simultaneously, the reactions at the counter electrode (cathode) need oxygen molecules to complete the process:
  $$\ce{O2 + 4H+ + 4e- -> 2H2O}$$
  An inadequate supply of oxygen to complete the reaction will shorten the life of the sensors, hence the sensors will not operate properly.

Does this mean that doing what I intend is going to affect their performance and damage them?

Aiming for dilution of the exhaust gases, you need a gas that is inert, i.e. does not react at these conditions with neither the gases present, nor with the metals present in in the catalytic converter (if there is such device installed) and electrodes of the sensors.
Under this restraint, using a nobel gas was a good idea.  Yet instead of helium, that you still may use to spot leaks in the tubings, I would opt for argon, which is much cheaper per litre.  Note that at ambient conditions, its density as gas is higher than the one of helium and of air and has a slightly lesser thermic conductance.

Yet still, based on the spec sheet data above, that won't help with the bit about needing oxygen on the cathode. What is wrong with using dry air to dilute the effluent?

Oxygen may be either still present (Diesel engine, p. 6 at http://www.volkspage.net/technik/ssp/ssp/SSP_230.pdf) or may be added intentionally (and then monitored) by the design of the test engine.  At the other side, using dry air may to dilute the exhaust gases may render the analysis more complex: i) the then added oxygen may offer some combustion of previously not combusted material, ii) nitrogen may react at elevated temperatures to yield $\ce{NOx}$.

It would seem that a deeper reading of the spec sheet (and a call to their application engineer) might be in order for the OP.

@JonCuster I have read the sensor datasheets and they do not seem to provide any informaiton. Even the extract I added wasn't from a datasheet, but from a book. An when it comes to contacting them, I do not think that is possible since most of them are chinese copies.

@Buttonwood Do you think that there is enough Oxygen for these sensors to normally operate in the specified surroundings (95% Helium, 5% car exhaust) for both gasoline and diesel based engines?

@Erol From the spare information provided (OP), and as you say, the data sheet available, the experiment may provide an answer.  Do you aim for $\ce{O2}$? If so, modern cars perform oxygen monitoring regardless if based on a Diesel, or Otto engine; which may be seen as "there are sensors that by design may used on both".  A statement "The [$\lambda{}$] sensor does not actually measure oxygen concentration, but rather the difference between the amount of oxygen in the exhaust gas and the amount of oxygen in air." at https://en.wikipedia.org/wiki/Oxygen_sensor for some of them supports this.

@Buttonwood What do you mean? Are you impling on testing the sensor in various concetrations of oxygen found in the car exhaust ie. putting the sensor directly in the exhaust and seeing how it behaves?                                                Nevertheless, I think I have found a way for keeping the sensors working fine and preventing any potential damage. After the sensors give me the measurments, I empty the sample mixture out of the chamber in which they are located and fill it with air for a brief amount of time. Will this be sufficient for the cathode reactions and their completion?

@Erol is the sensor you are using of the electrochemical type (e.g. two solid electrolytes with an air gap) as described on Figaro's web site? It's frustrating when trying to read the data sheets for the current supply of the low-cost versions of old designs, and some of those (just for example MG-811) contain minimal and potentially wrong or incomplete information. Reading text books and older publications (the way you describe) is the right way to go.

@Erol slightly related although these are mostly MOS and not electrochemical sensors.

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text:
What chemicals are in hard wood ash? (Making lye)
I am trying to make lye or potassium hydroxide. So far I have been leaching ashes in a barrel for a few days. I decided to do some tests on the water to see if it was working. My pH paper said it was basic, so then I reacted it with vinegar to see what would happen. The equation of acetic acid + potassium hydroxide would be this I believe:
$$\ce{KOH + CH3COOH -> CH3COOK + H2O}$$
However, when I mixed the two, it began bubbling after a few seconds. The gas formed smelled like boiled eggs and was white. I assume it is some sulfur compound but I may be wrong. The ashes are mostly from locust wood. Is there anyway to find out what other chemicals are in the ashes and how to separate the potassium hydroxide out?

Interesting question. I'm unsure on what smells like "boiled eggs", but you can separate the $\ce{KOH}$ out by precipitating it by adding sodium tetraphenylborate.

Thanks for the tip but I'm trying to make potassium hydroxide from wood ash because I don't have easy access to manufactured chemicals. I'm using it as an electrolyte in electrolysis so I hope the other compounds won't react during that process.

The white you're seeing is probably just condensing water vapor (i.e. steam), especially since you said it was bubbling, which indicates some sort of heating. The smell is probably caused by sulfur compounds like you said, but it's probably not possible to nail down exactly which compounds without some advanced tools.

I thought lye was sodium hydroxide.  Sodium is much more prevalent in nature than potassium.  What makes you think locust wood would provide potassium hydroxide?

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text:
How do I calculate the standard enthalpy of formation for liquid ethanethiol? Do I use Hess's law?
Calculate a value for the standard enthalpy of formation for liquid ethanethiol, $\ce{C2H5SH}$. Use the equation given below and enthalpy of combustion data from the following table.
Just assume values and comment the methods to solve the problem, I'll try solving it.
Edit: I'm not sure whether you're supposed to use
$$\Delta H^\circ = \sum{\Delta H_f(\text{product})} - \sum{\Delta H_f(\text{reactant})}$$
because the given is enthalpy of combustion data, not heat of formation. I'm not quite sure what an enthalpy of combustion is.

Not going to leave a full answer since there are piece I don't remember...generally for a hydrocarbon X, the enthalpy of combustion is the enthalpy of $\ce{X + O2 -> CO2 + H2O}$, suitably balanced. The problem is I don't recall what product you need on the right for sulfur-containing compounds...I think it might be $\ce{SO2}$, but I really have no clue.

Welcome to Chem SE. It's good to see you trying to solve your own question. But please show it to us. Pinpoint where your stuck exactly. That can help us give you a specific and relevant answer. By the way, have a [tour] of this site. Visit the [help] to learn how to ask and answer questions here. Good luck!

Which table???Which equation???  Please consider adding the table and the equation (because knowing the substances whose heats of combustion are given is essential for solving this problem!) to your question. "Just assume values...", its like you are ordering us to solve your problem!


text:
Any way that I can extract sugar from drinks? Without boiling the drinks
Any method? Materials, engineering, any technology.
I am a product design student, so any possible method is useful to me.
I am thinking to add this kind of technology into my concept.

Pragmatically, it isn't possible...

If you have the money for a reverse osmosis machine, maybe.

What is a reverse osmosis machine @@?

We're gonna need some more details about what the goal is here. Is it okay if the sugar is poisonous to people after extraction (I can think of two techniques that do this)? Does it need to be dried later on? What if the process costs billions of dollars to work on a single can of soda? Is that acceptable? What if it's relatively cheap but takes a few months?

@MichaelYan You take a filter that lets water through, but not salt or sugar, then apply pressure to squeeze all the water from one side to the other. Salts, sugars, and other things get left on the first side. The problem is that this is very expensive and you still have to do a bit more extraction after it's finished.

@chipbuster ya thats what a want but is that possible that I can block the melted sugar in the water by using a filter? Do you have any related source and link that I can take a look :)

If by boiling you mean heating, then you could apply vacuum until the water was gone, or let it air dry.

It's possible to freeze out the water and get the crystals of sugar that separate out

Of course there is a way: the millennia-old nanotechnology known as yeast fermentation. It has a side effect of producing alcohol, though.

Already asked https://chemistry.stackexchange.com/questions/75160/methods-to-extract-sucrose-from-tea-after-its-been-stirred-in


text:
Diffusion/Streaky bands on TLC plate
These days I am working on a research project where I am supposed to run TLC for samples which are wax extracts from coconut butten nuts using chloroform as the solvent. All the samples have been concentrated to 10% from its original volume. I recently run a TLC for a sample using hexane(90%):diethyl ether(7.5%):acetic acid(1%) as the solvent system (this solvent system has been used to run similar wax extracts). But my TLC run resulted me very diffused streaks  having some curves towards both the sides  of the tlc plate. Can anyone tell me why this happens??

Welcome to Chem SE. Take a [tour] of our site. Visit the [help] for more information on how SE works

Curvature: Maybe it is a defective TLC plate? Or you did not allow enough time for the solvent to saturate the TLC chamber. Band:  If other TLC do not have the dark band: it is possible that there is an impurity in the solvent.

Since you said you spotted but got a band, the spot was always above the liquid level, right? Just asking to be absolutely sure.

Yeah, spot was few millimeters above the liquid level. Today I run phenolic extract (by methol) on a portion of the same TLC plate, it didn't make that problem. No streaky bands seen except the solvent front. Few spots seen.

Several things that may help your situation, listed roughly in the order in which I'd try them:
1. Application of the sample to the TLC plate
TLC is most usually ran by applying a small spot of the sample to the plate, and allowing it to run up. Analytically, this gives the greatest resolution, and as such for things such as monitoring reactions or identifying components in  column chromatography fractions, spots are the most widely used.

In some cases, the TLC plate is used to purify components of the sample, this is called preparative layer chromatography, or PLC for short. In PLC, bands are applied to the plate, to allow greater quantities of the sample to be applied and hence separated.
When applying bands to the PLC plate, it's crucial to ensure that the band doesn't get too close to the edges (as it has in the images you've uploaded), as a rule of thumb, at least 1/2 cm from each side should be left clean.
The reason for this is that the solvent doesn't run perfectly level up the plate, this is most evident at the edges of the plate. If you re-run your TLC plates with a shorter band, you should see a decrease in the curving effect that you're observing.
2. Quantity of sample loaded onto the TLC plate
Looking at the images you've provided, you may also have loaded too much sample onto the TLC plate, saturating the silica and reducing the separation.
You could try either increasing the size of the plate, increasing the thickness of the silica layer (special PLC plates are available with thicker layers of silica, allowing more material to be loaded), or simply just diluting your sample and loading less on.
3. Solvent system
Although you say that "this solvent system has been used to run similar wax extracts", its always worth trying different eluents with your chromatography, unless your sample is identical to the prep you're following.  A different modifier (formic acid or trifluoroacetic acid often work well), or different solvent mixtures (methanol in dichloromethane is often a lifesaver) might help tighten up the bands.
Finally, and possibly most of all, make sure you completely dry the plate out before visualising it (leave it sat on the side for a few minutes or use a hot air gun).

Thank you very much for your valuable comment. I just applied just a spot not a band. Eventhough I applied a spot finally I received some bandsas shown in pictures. As you mentioned the sample was a concentrated one. I will try to dilute it first.


text:
Do secondary alcohols prefer to undergo SN1 or SN2 reactions when reacting with a strong acid?
For example, let us say we have 2.2-dimethycyclohexanol reacting with $\ce{HBr}$. We know that the oxygen gets protonated and then leaves as water, but the question does it leave on its own or does a backside attack happen? If its $\mathrm{S_N1}$ we have 1-bromo-1,2-dimethylcyclohexane as the major product. If its $\mathrm{S_N2}$, we have 1-bromo-2,2-dimethylcyclohexane as the major product. This is assuming we are at 0℃.
I believe that its $\mathrm{S_N1}$ because bromide is a weak base.

Actually, secondary alcohols undergoes $\mathrm{S_N1}$ or $\mathrm{S_N2}$ depending upon the substrate you are using. For e.g. In your case(as provided above) the substrate has steric hindrance nearby alcoholic group and the carbocation will be stable due to the migratory aptitude of the methyl group. Hence, here should be $\mathrm{S_N1}$ and you have to see the structure of substrate to decide $\mathrm{S_N1}$ or $\mathrm{S_N2}$ for secondary alcohols.
Also, alcohols are not good leaving groups also! But after protonation the water is a very good leaving group.

Yes, here it'll be $\mathrm{S_N1}$ preferably. Why not $\mathrm{S_N2}$?, one cause is steric crowding in the adjacent carbon and for the attacking nucleophile and leaving group nature comparison see here, https://chemistry.stackexchange.com/questions/75356/how-can-aqueous-koh-give-alcohol-by-sn2-of-alkylhalide


text:
Electrophilic substitution on disubstituted benzene rings
If I have a strong deactivating group (like nitro) and a weak activating group (like methyl) attached to the benzene ring, which group's preference will direct the next substitution on the benzene ring? For example, if a chloro group is attached to meta-nitrotoluene, where will it substitute?
If I have both strong activating and also deactivating groups in the benzene ring, which position will be preferred for the next electrophilic substitution? For example, if I want to attach a chloro group to the meta-nitroaniline, where will it substitute?

Im not sure if I understood correctly, but I vaguely remember answering a similar question, not sure if it addreses your question but its found here: Rules for benzene halogenation

if I want to attach a chlorine group to the meta-nitrotoluene, where will be chlorine attached ? @xavier_fakerat

For that you may refer to this site: Electrophilic Substitution of Disubstituted Benzene Rings but if you wish to have that summarised, I could make an answer for it. Hope it helps

Thanks for the link.  It helps a lot. But this link does not deal with strong deactivating and weak activating group combined  at benzene ring like meta-nitrotoluene. Would you please clarify  the question :
if I want to attach a chlorine group to the meta-nitrotoluene, where will be chlorine attached ?" . Thanks in advance . @xavier_fakerat

To prevent a homework-answer scenario, I had to explain the concepts so you understand and be able to apply. Generally. I dislike this approach (question-direct answer) because it has long term negative consequences. So that is why I give Conceptual answers, once you get the concept you will apply it anywhere and still pass, suceed etc.. It may be a bit of content but trust me it helps :).

Yes , after observing your answer style, I understand that you don't want to answer directly. To give conceptual answer is a good approach. :). But some time,some students may not get it as the concept is written. Thank you sir

This is a very good question and a common one among students of organic chemistry. I removed the sentence asking about the directing effect of diazonium, because that really doesn't fit with the the rest of what you're asking. If you would like, you can submit another question about that.

Thank you for correcting my question style. @jerepierre .

Electrophilic Substitution of Disubstituted Benzene Rings
When a benzene ring has two substituents, each substituent exerts an effect on subsequent substitution reactions.

Electrophilic substitution of a disubstituted benzene ring is governed
  by the same resonance and inductive effects that affect
  monosubstituted rings. The only difference is that it’s necessary to
  consider the additive effects of two different groups.

If the directing effects of the two groups reinforce each other, the situation is straightforward. In p-nitrotoluene, for example, both the methyl and the nitro group direct further substitution to the same position (ortho to the methyl and meta to the nitro). A single product is thus formed on electrophilic substitution.

From above, the methyl and the nitro group direct further substitution to the same position (ortho to the methyl and meta to the nitro). This is similar case with chlorine.

If the directing effects of the two groups oppose each other, the more powerful activating group has the dominant influence, but mixtures of products are often formed.
Further substitution rarely occurs between the two groups in a meta-disubstituted compound because this site is too hindered. Aromatic rings with three adjacent substituents must therefore be prepared by some other route, such as by substitution of an ortho-disubstituted compound.

You may also wish to review this table showing positions where common substituents direct:

Basing on the above explanations,  it is easy to predict the products of  electrophilic substitution of disubstituted benzene ring, the only thing you have to know is the effect of each substituent whether it is activating or deactivating. In some cases, observe the positions of the substituents as point 3 mentions steric effects may come into play and substitution may not occur.
On the second note, concerning diazonium ion, I think you have to look at measured dipole moments, experimental data etc I doubt if there is a way of telling whether a substituent is strongly deactivating by merely looking at it

After explaining concepts now lets see what the answer for the question in comments will look like:

Reasons

the directing effects are opposing (nitro is meta while methyl is both ortho and para directing) but the methyl is more activating so we have a mixture:

meta to nitro and methyl
ortho to the methyl but para to the nitro

In this scenario since position 3 is not deactivated by nitro group hence there is also a possibility of a mixture here, also take note of how the nitro group is strongly deactivating at position 4 (more additive effect)
Hope this helps
References

McMurry Organic chemistry
Electrophilic Substitution of Disubstituted Benzene Rings

Thanks. So between activating and deactivating groups, always activating groups are preferred (even the activating group is weak).right ? Except Stearic Hindrance effect.

I think this had to do with charge stabilisation. When one substituent has a pair of non-bonding electrons available for adjacent charge stabilization (e.g some activators like -OH), it will normally exert the product determining influence, even though it may be overall deactivating

Thank you, The group which has lone pair electrons and if it can stabilize the adjacent atom, will be preferred. So if OH and nitro are present, OH is preferred. Right,sir ?

That only applies to case 2 :) otherwise you just look at all possible outcomes (the three points)

And, for your last reaction, I want to know about meta-nitrotoluene but you explain about ortho-nitrotoluene.. meta-nitrotoluene has a problem as two group oppose. neither toluene or nitro has lone pair. So which group will be preferred ? Sir I understand many things but the answer of this question is still in confusion. :-(

Thanks let me remove that specific reaction, and place the nitro on meta ;)

Fixed! There you go Like I said before its similar, you just have to apply concepts :)

sir,as far as I know,is not it also called ortho-nitrotoluene? as NO2 is attached to the adjacent  to the methyl? Thank you for noticing... This is from wiki:  https://en.m.wikipedia.org/wiki/3-Nitrotoluene

You are very right :)

Thank you sir for  all of these. :-) .  So as there is methyl and nitro group, we now get mixture. But if there is amino or hydroxy group instead of methyl , we may have get most of the products preferred by hydroxy or amino group.

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text:
High specific energy solid material for cryogenic work
I would like to know a few (solid) materials that have a very high heat capacity (specific heat?) that can absorb cold energy very well, and hold that cold for quite a while from 77K to 90K (at least 5 hours)
I am using a high tech stirling heat pump to cool down superconductors (along with very good insulation around the superconductors) to 77K. But I want to put that material around/on the superconductors, to maximise the time they stay cold. I would hold the heat pump against the material for perhaps 30 mins, and cool-down that material. Having an ability hold "cold energy" very well, and not release it too easily, would be needed. The material would need to handle cryogenic temps very well and not expand/contract too much (hence desire for a solid). That material would probably be inside a 2mm walled copper box (with superconductors strapped to the outside of that box)

The amount of heat it loses depends on the value of the exposed area of the solid. You can also protect the heat by polishing the surface, this reduces it's emissivity. By the way welcome to Chem SE.

cold energy????

You are looking for a thing which is simply not there. Heat capacity of anything is pitifully low when compared to the heat of phase transitions, especially boiling. This is the reason why we use liquid nitrogen for cooling in the first place.


text:
Boiling points of Hydroxylamine and water
Why does $\ce{NH2OH}$ have a lower boiling point than water? According to Wikipedia, hydroxyamine has a boiling point of 58 °C. Could this be due to the extent of hydrogen bonding among both molecules?

Before you ask "why", ask "if". Did you just ignore those "22 mm Hg" as meaningless?

I don't understand why this question was downvoted. Agreed, it's quite small but I don't see what else we should expect from a asker. The question is quite genuine too.

@PrittBalagopal Based on Ivan's comment, the question was downvoted (not necessarily by him) because the statement it makes is incorrect. Hydroxylamine would have a higher boiling point than water at 1 atm, if it did not decompose from heating. I don't mind this so much, since it's an honest mistake, and there's something to be learned from it.

Think about ammonia (NH$_3$) it has a bp. of -33.4°C , that tells about the possible intermolecular interactions between NH$_3$ molecules. They are much worse than those in water. In H$_2$NOH you are about in between. Very roughly speaking. It doesn't work perfectly, (averaging between H$_2$O$_2$ and hydrazine would give you a very bad estimate (actually above 100°C). That shows that the business is not that easy ...

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text:
How can we determine the Acidity or Base only from information molecular formula provides?No concentration given
How can we determine the Acidity or the base of a solution just from the information the molecular formula provides, like $\ce{KCl}$, $\ce{NH4Cl}$, $\ce{MgSO4}$, $\ce{NaCO3}$, like which ones have a pH of less or more than 7, if I was given the concentration , I would have used $-\log{\ce{[H3O+]}}$ formula, but without it, how can I determine the pH levels?

You cannot determine exact pH levels without knowing specific equilibrium constants (or equivalent values) and concentration of the substances.
Determining whether something is acidic is relatively straight forward. Inorganic compounds like the one you've presented can be broken down into cations and anions.
For the anions, you should consider:

Is this anion acidic ($\ce{H2PO4-}$)
Is this anion basic ($\ce{OH-}, \ce{HCO3-}$)

For the cations, you should consider:

Is this cation the hydrogen ion ($\ce{H+}$)
Is this cation a Brønsted-Lowry acid ($\ce{NH4+}$)
Is this cation Lewis acidic ($\ce{Al^{3+}}$)

For compounds where ions are both acidic and basic, the exact balance matters. Then, you're crossing into the complicated regime where things are interesting.


text:
Boron nitride polymerization with silicon based molecules
Since boron nitride is similar to carbon hexagonal structures and silicon is a main component of inorganic polymers. Is it possible to make it from a polymer under certain conditions, if so what catalysts are needed?


text:
Largest system studied by full CI
What is the largest system (organic molecule) which total energy was calculated by full configuration interaction? Lets say with DZV or similar basis set. How far can we get with CI in larger basis sets?

Lets do a competition: He-dimer He$_2$, Basis: [11s]4p3d1f+1p2d1f.
(1995) dx.doi.org/10.1063/1.469060 ,

Another hint: "deterministic FCI calculations for 6e^- have not exceeded the level of TZ basis sets" dx.doi.org/10.15278/isms.2016.TK07 (2016). While they themselves do Li$_2$ with a cbs extrapolation from AUG-CC-PCV5Z and AUG-CC-PCV6Z.

related: https://mattermodeling.stackexchange.com/q/9495/5

FCI of N2 with about 10^10 CSFs must be the largest such calculation...
Ref:
A full-configuration benchmark for the N2 molecule,
Elda Rossi, Gian Luigi Bendazzoli, Stefano Evangelisti, Daniel Maynau,
https://doi.org/10.1016/S0009-2614(99)00791-5

J. Chem. Phys. 125, 124311 (2006) reports full configuration interaction (FCI) calculations on BN, BP, AlN, and AlP in the cc-pVTZ basis set:

The full CI calculations consisted of about 45x10^9 determinants for the singlets and about 35x10^9 determinants for the triplets.

J. Chem. Phys. 147, 184111 (2017) reports FCI calculations for the chromium trimer with a (20e,20o) active space and for pentacene with a (22e,22o) active space. They also did a single CI iteration (they did not converge the FCI calculation!) with an active space of (24e,24o) for the chromium tetramer with 914 058 513 424 determinants.


text:
Why does acetic acid combined with dirt turn blue?
Around my faucet was some mysterious dirt from a previous tenant. I poured acetic acid with a concentration of 25 percent over it. I came back after a few hours and noticed some blue stains which hadn't been there before. What is this blue substance? Is it poisonous? What must have been in the dirt to produce this?
In Wikipedia I came across a statement about acetic acids and blue color:

A more sensitive test uses lanthanum nitrate with iodine and ammonia to give a blue solution.

Could the mechanism described there be what I'm looking at? Is it really possible that this dirt contains lanthanum nitrate?

Welcome to Chemistry! Take the tour to get familiar with this site. If you receive useful answers, consider accepting one.

The intersection of plumbing and blue is copper. Copper oxide is black. It will react with acetic acid to make copper acetate which is blue when hydrated.

@BenNorris It's definitely more likely then lanthanium nitrate, still unless OP finds out copper out there, it still may be like anything one could dump into sewage.

I understand that the information given might not be enough to provide a definite answer but I'm afraid I simply haven't got more information about my problem myself...


text:
Methanol vs Ethanol Substrate cleaning
We have to clean substrates, often made of STO or MgO, before PLD for thin film growth. Previously we would use ethanol for sonication to clean the substrates. We recently ran out of ethanol but have a lot a methanol. In looking on the web I have only found conflicting viewpoints on which to use, or how corrosive methanol is. Are the two interchangeable for cleaning substrates? From where do you know?

Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. If you receive useful answers, consider accepting one.

What is STO (strontium titanate?) and PLD? What kind of films are you growing on the substrates?

Yes strontium titantate. Pulsed Laser Deposition to grow doped STO or ScF3.

Both $\ce{SrTiO3}$ and $\ce{MgO}$ can safely be washed with methanol. These compounds are insoluble in most any solvent other than certain acids. I suspect that any online reference you've found saying methanol is corrosive is with respect to organic compounds, nothing like the composition of your substrates.
One (somewhat handwaving) way to convince yourself that this is the case is that we know the compounds are insoluble in water (see their Wikipedia and PubChem pages) and ethanol (from your experience, also confirmed for $\ce{MgO}$ in it's Wikipedia and PubChem pages). In terms of solvent polarity, methanol lies between water and ethanol. It would be extremely unlikely for an ionic solid to be completely insoluble in water and ethanol yet have some significant solubility in methanol.
Although we can be confident that the substrate materials will not be destroyed by methanol, only trial will let you know how effective it is for cleaning whatever contamination may be present in your situation. Methanol is fairly, though not completely, similar to ethanol in it's solvent properties, and there is a good chance that it will be as effective as ethanol. Again though, only trial will verify this in your situation.

Generally in semiconductor and thin film work, the preferred final rinse before drying is methanol, not ethanol (although much has to do with how much easier it is in the US to have reagent grade methanol in the lab vs ethanol).

Upvoted. Seems to be no theoretical reason for it not to work, just test it and see. The only drawback is while ethanol can be safely added to a coffee cup to make the laboratory day go a bit faster and smoother, methanol must not. ;-)


text:
Where can I find a downloadable database of basic physical and chemical properties?
The web has many online databases of properties of compounds locked behind search interfaces, but I can't find anything you can actually download.
I want a corpus of data that might allow me to, say, sort compounds by enthalpy of formation, or melting point, or other properties you'd encounter in a high school chemistry class. I don't need environmental impact properties, or toxicology data, or even molecular structure, and I don't need to know about exotic stuff hardly anyone's ever heard of. Just the basic simple things you would expect to be easily available like just about everything else on the web.
There are handbooks of general properties of compounds. Collections like these are perfect. Surely someone has put information that exists in a book into digital format.
I don't mind paying for it. But I can't even find it.

https://en.wikipedia.org/wiki/CRC_Handbook_of_Chemistry_and_Physics ? Theres a downloadable version, you'd have to script something to mine the data from it into a spreadsheet, however

@NotEvens -- Thanks. You mean the ebook? I dunno, it probably would be compressed or otherwise DRMed. I'm pretty sure even a PDF would be cryptic, right?

The eBook was what I had in mind, its fairly easy to extract data from a PDF, especially if you're on;ly interested in certain tables

@NotEvens -- Thanks, but I'd need to be certain the ebook data is raw enough before I spend money on it.

Regarding melting points, https://datahub.io/dataset/open-melting-point-data backed by Alfa Aesar or https://figshare.com/articles/Melt-ing_Point_and_Pyrolysis_Point_Data_for_Tens_of_Thousands_of_Chemicals/2007426 may be examples; for high school, however, likely an overkill.

Search specifically for MSDS sheets, these will be the most helpul. For molecular structures, use SMILEY or ChemSpider.

The NIST database has been succesfully crawled before. It would take some scripting as well.

http://blog.sciencenet.cn/upload/blog/file/2008/12/2008121510347360620.pdf

@DrMoishePippik -- This is very helpful. Thank you.


text:
Nucleophilic acyl substitution: conversion of butanoic acid to butanenitrile

How can I convert the carboxylic acid functional group into a nitrile?

Butanenitrile has 4 carbons not 5, and if you meant pentanenitrile, why are you trying to remove a carbon? Surely going from 4 to 5 is adding a carbon.

it has been edited, could you answer this question?

Convert to amide then dehydrate to nitrile, should work

That's not the formula for butanenitrile. It's butylamine... Which one do you want?

The direct conversion of a carboxylic acid to a nitrile is not a well defined reaction, although examples do exist, the reaction is not general to a wide range of carboxylic acids.
By far the most common method is to convert the carboxylic acid into the primary amide, followed by some method of dehydration. Strongly acidic reagents such as  $\ce{P2O5, POCl3, SOCl2, TiCl4, NaBH4}$ have all been used to achieve this, however milder methods such as $\ce{(COCl)2}$ or cyanuric chloride are also common, along with more modern catalytic methods (Matthias Beller has a very nice iron catalysed reaction).

A method that I've seen used extensively is to make the t-butyl amide and dehydrate that with PCl5

Carboxylic acid can be converted to nitriles using various methods:

Direct Preparation of Nitriles from Carboxylic Acids in Continuous Flow

A continuous-flow protocol for the preparation of organic nitriles from carboxylic acids has been developed. The method is based on the
acid–nitrile exchange reaction with acetonitrile, used as the solvent,
and takes place without any catalyst or additives under the
high-temperature/high-pressure conditions employed. At 350 °C and 65
bar, where acetonitrile is in its supercritical state, the
transformation of benzoic acid to benzonitrile requires 25 min. The
protocol has been tested for a variety of nitriles, including aromatic
and aliphatic substrates.

Preparation of nitriles from carboxylic acids: A new, synthetically useful example of the Smiles rearrangement

Reaction of 2,4-dinitrobenzenesulfonamide with acyl chlorides in the
presence of excess triethylamine produces the corresponding nitrile in
good to fair yields. Mechanistic studies indicate that the reaction
proceeds via a Smiles rearrangement of the initially
formed N-(2,4-dinitrobenzenesulfonyl)amide to form the nitrile,
2,4-dinitrophenol and sulfur dioxide.
The reaction of acyl chlorides with 2,4-dinitobenzenesulfonamide in
the presence of TEA gives the corresponding nitrile, 2,4-dinitrophenol
and sulfur dioxide.

Letts nitrile synthesis(for making aromatic nitrile)

The Letts nitrile synthesis is a chemical
reaction of aromaticcarboxylic acids with metal thiocyanates to
form nitriles. The reaction includes the loss of carbon
dioxide and potassium hydrosulfide. The polar basic substitution
reaction was discovered in 1872 by Edmund A. Letts.

A SIMPLE ‘ONE-POT’ CONVERSION OF CARBOXYLIC ACIDS INTO NITRILES

Carboxylic acid is first converted into mixed sulfonic anhydride using
methanesulfonyl chloride(MSC) in dry pyridine ar 0℃. This anhydride is
treated with ammonia to form an intermediate which on dehydrating
gives amide. The amide is converted to nitrile using MSC.


text:
Theoretical synthesis of dimetinden
I tried to find a way to synthesize Dimetinden (N,N-dimethyl-2-[3-(1-phenylethyl)-1H-inden-2-yl]ethan-1-amine) with an Indensynthesis
https://www.thieme-connect.com/products/ejournals/abstract/10.1055/s-0030-1261196
First of all I would like to know if there is anything wrong with the whole synthesis and how about the yield?
Second question would be the mechanism of the inden synthesis, because it looks kind of wrong.

Are you asking if the linked literature procedure is correct?

I am asking if my mechanism for the linked procedure makes sense and are the other parts of the synthesis possible.

A couple of things, first, inden-1-ol tends to tautomerize into the corresponding 1,3-dicarbonyl.  Second, when adding TsCl to an alcohol, the group formed is -OTs not -OTMS, the latter is a silyl ether.  The reaction with addition of TsCl requires a base such as pyridine, and because of the presence of a 1,3-dicarbonyl, rather than being added to the alcohol, it will be added to the carbon alpha to both the carbonyls.  Third, the reaction of Grignard reagent should be avoided due to loss of yield because of interaction with the carbonyl.  Fourth, the inden-1-ol formation is done in the presence of a base (to form enolates) which is then followed by an acid wash(to form enols which tautomerize to ketones), your mechanism seems okay if enolization occurs instead of tautomerization.  Finally, the correct structure of Dimetindene is below:
https://en.wikipedia.org/wiki/File:Dimetindene.svg
I will briefly explain how I would do it from the indenol you described.  First, would be a Wittig reaction using 1-(pyridin-2-yl)ethylenetriphenylphosphorane, (which can be prepared from 2-acetylpyridine) and the indenol tautomer.  This will yield a beta-gamma unsaturated ketone.  This product can undergo a photochemically induced [1,3] hydride shift to form an alpha-beta unsaturated ketone.  This can be subject to Wolff-Kishner conditions to form Dimetindene.


text:
If polyethylene glycol (PEG) is hygroscopic, how does it protect wood?
Archaeologists and conservators use polyethylene glycol (PEG) to replace water inside wood to protect it from shrinking and expanding with changes in humidity. But it is hygroscopic—even more so than wood, it sounds like. So how does it work? Is it like the polymer in a diaper that is able to hold on to moisture and therefore not allow any additional moisture to expand the wood further? Or is it simply filling in all the holes and its hygroscopic character is inconsequential?

Good question! Thanks for asking it here at Chem.SE. For more information about the site and its policies, check out the [tour] and [help]. In particular, the page on how to format posts using Markdown/HTML may be helpful for future questions. Finally, since your question has earned you 20 rep, feel free to drop by chat and say hello! Enjoy!

The main purpose of PEG is to provide osmotic pressure on lignin layers which would otherwise collapse under the absence of water.

The reason why PEG is suitable for this purpose has many reasons including the wood chemistry and degradation of wood in water.
Degraded wood usually very characteristic chemical losses. The cellulose of the wood is usually lost due to bacterial action and/or due to the action of the water itself. Over long period of time wood looses water-soluble components like starch and complex sugar like molecules to water. What remains in there is the water-insoluble component in the cell walls and wood tissues primarily consisting of lignin.
If the wood is in water, lignin provides structural stability to the wood. But wood becomes spongy and easy to break. If this wooden piece is exposed to air, water evaporated and wood becomes dry and shrinks. Most of the spaces between lignin layers is nor air, which makes it easy to break.
Here, comes in PEG. PEG is remarkable because it can dissolve in water.
PEG solution has very high osmotic pressure and this is one of the reason why PEG solution treatment works. PEG seeps in the tissues and provides the required osmotic pressure to keep lignin (and wood tissue) stable. This prevents the wood from expanding and shrinkage. While treatment, it is ensured that PEG penetrates the wood tissue which takes considerable times, and water is removed from the wood. In the end a large part of the original water in wood is replaced by PEG in the tissues providing structural stability.
Coming to your point on the hygroscopic property (also known as deliquescence). Higher molecular weight PEG has lower hygroscopic property because of more alkyl chained part. So it depends on the type of PEG used. But a balance has to be made since, lower MW chains are easier to penetrate tissue.
In case the relative humidity changes, for example increases, the effect due to absorption due to PEG inside tissue is negligible. Only the PEG which is outside tissues (excess, which was not removed during PEG treatment) might cause additional increase in weight. The removal of excess PEG is usually taken care of by experts.
Still however, it is the norm that >80 % relative humidity is not good for PEG treated wood.
Reference

D.W. Grattan, "A Practical Comparative Study of Treatments for
Waterlogged Wood, Part II, The Effect of Humidity on Treated Wood",
(Ottawa: ICOM Waterlogged Wood Working Group, 1981), pp. 243-252.
Taduteru Nishiura and Setsuo Imazu, "Experimental study of the
dimensional change of highly degraded waterloged wood according to
ambient humidity after preservation treatement"Bremerhaven: ICOM
Waterlogged Wood Working Group, 1990) pp. 107-118.

Thank you so much! I am creating an exhibit that includes the science behind conservation methods, and I was really struggling on how to model this for younger visitors. Your explanation makes it much more understandable and easy to adapt to a visual model.


text:
Does the methyl rotation in propene have a symmetric or asymmetric potential energy surface?
Depending on how one imagines the methyl rotation of propene to occur, it seems like it could exhibit either a symmetric or an asymmetric potential energy surface.  If the $\ce{H2C=CH\! -}$ portion were held fixed, the rotating methyl would seem to be symmetric due to the local $\mathrm C_3$ axis:

On the other hand, if the methyl group were held fixed, the rotating $\ce{C=C}$ portion would seem to be highly asymmetric:

Obviously this rotation can't be both symmetric and asymmetric. So which is it, and why?
Intuitively, given the logical contradiction of the above, my inclination is to assume that the rotor is asymmetrical. My best guess as to a more rigorous argument for it being asymmetrical is that it's not valid to imagine any part of the molecule as being "held fixed", since it's not actually anchored to anything.  Thus, both of the above images are probably subtly incorrect.

It does not matter which part rotates. The methyl group will experience a sinusoidal like potential as it rotates a full circle, with three equal main humps spaced at 120 degrees.  These barriers will be due to a small (steric) interaction with the  closest H atom on the $\ce{CH2}$. There may also be smaller three fold barriers displaced by 60 degrees from the bigger ones and caused by the H atom next to the methyl .

I don't mean symmetry in the PES -- I mean symmetry in the inertial properties of the rotation. Perhaps I should specify....

Yes I understand now. There is whole body rotation and internal rotation; so you can choose which bit is to be internal. As the total angular momentum is constant (in isolated molecule) it works out that it does not matter which bit you consider to be internal motion, its just corresponds to a change of reference /coordinate origin. The rotational spectrum (hence energy levels) cannot exhibit two different forms depending on which bit we consider to rotate.

@porphyrin , I was thinking hazily along those lines. By that logic, it has to be an asymmetric internal rotor, right?  How to prove it mathematically, though?  Is it because the axis of the internal rotation doesn't include the center of mass of the system?

I'm not sure what you mean by 'asymmetric', the rotor moves in a 3-fold potential (or 6 fold if you add the small other H atom interaction)  so is symmetric in this sense.  If the energy is large enough at room temp the barriers will be crossed, if not it will oscillate in the lower part of the potential. As an example see J. Chem. Educ., 2011, 88 (7), pp 929–931.

@porphyrin I think I was actually insufficiently clear in stating my question. I've edited it to be consistent with the majority interpretation in the comments and answers.  I'll be posting a new, clarified question shortly....

@porphyrin New question

These rotations are the exact same physical processes for a molecule because we cannot fix either end in place. In order for rotation to mean anything, we must choose an origin. Mathematically, I must be able to draw a vector from all of the nuclei to some arbitrary point, but be consistent about this arbitrary point. Or, I must determine positions of nuclei by some measurement relative to other nuclei.
Imagine I pick the carbon of the methyl rotor as the origin. If you were standing at this origin and this rotation you describe were happening, you would see the so-called "asymmetric rotation". If, however, we put the origin on C1, we would see the "symmetric rotation." We could also define internal coordinates so that we simply changed a dihedral angle and observed the same rotation. In this case it's almost like both motions are happening. Clearly only one physical process is happening, so we might be picturing this wrong in our heads.
I believe the problem you're encountering is that if you had a macroscopic object shaped like this, you could imagine grabbing either end and you would have two separate rotations which we could choose to call symmetric and asymmetric if we so desired (though I'm not convinced of this naming). We know the motions really are different because we are spinning around different total masses. On a large scale where we hold half of this molecule in place, it would seem easier to do the symmetric rotation than the asymmetric rotation. Stated another way, for an equivalent impulse to each of the possible rotors, the initial rotational frequency of the methyl rotor would be larger than that of the $\ce{C=C}$ rotor. Note that we would still be resisting torque in the hand with which we are holding the object and the size of this torque would change depending on the rotation.
Molecules, however, cannot be thought of as fixed in this kind of way, and the resistance to rotation does not come from friction as it probably would for a macroscopic object. The "resistance" to rotation for the molecule comes from how much energy it takes to rearrange electrons for a change in nuclear positions. You can measure these nuclear positions relative to each other using whatever method you want, three of which we already discussed, and this will not change the minimum energy of the electrons for the set of nuclear positions. This minimum energy path (dare I call it a gradient?) is the path along which the rotation would happen. Whichever actual rotation you picture in your head does not matter at all. Because the molecule is free during this rotation, you can get to any of the many rotations by imagining yourself on a different atom, but this will not change the electronic energy because the relative nuclear coordinates will remain unchanged when you change the origin. In the language of one of your comments, because no part of the molecule is fixed, the inertial properties of the rotation are determined by the PES, so whatever symmetry you're thinking of is the same as the symmetry of the PES.
I think I addressed the potential pitfall here. It is indeed quite weird to imagine what the "actual" internal rotation of the molecule looks like, and maybe someone has a better intuition for what that means, but I think describing the actual motion is fairly arbitrary.

I've edited the question to be more in line with how commenters and answerer(s), including yourself, have interpreted it. I'll be posting another, clarified, question shortly.... Thanks for the answer!

New question


text:
Why is Cr2O3 amphoteric but CrO not?
I faced a question where I had to find the amphoteric oxides among some given oxides. The answer key says $\ce{Cr2O3} $ is amphoteric but $\ce{CrO}$ isn't. But why is that?

Basic oxide are those oxides that dissolve in water to form soluble hydroxide. Basic oxides are very soluble for this reason. Acidic oxides are that oxides that dissolve in water to form strong acids. Generally, covalent oxides form acidic oxides as the element to which oxygen is bonded is electronegative. As charge increases, electronegativity increase and thus oxide become acidic. This also applies for elements having multiple oxides. As oxidation no. increases, charge increases and thus oxides become acidic. This link shares a good information on this topic:

Since the acidity of a cation rises rapidly with charge, d-block
  elements which exhibit a wide variety of oxidation numbers may have
  one or more oxides that exhibit only basic properties and one or more
  oxides that exhibit only acidic properties. The higher the oxidation
  number the more acidic the corresponding oxide. Chromium is an example
  of such an element.
$$
\begin{array}{|c|c|c|}\hline
\text{Oxide}&\text{Oxidation number}&\text{Category}\\\hline
\ce{CrO}&\ce{Cr^2+}&\text{basic}\\\hline
\ce{Cr2O3}&\ce{Cr^3+}&\text{amphoteric}\\\hline
\ce{CrO3}&\ce{Cr^6+}&\text{acidic}\\\hline
\end{array}
$$

The amphoteric character of chromium(III) is stated in its Wikipedia article:

Chromium(III) oxide is amphoteric. Although insoluble in water, it
  dissolves in acid to produce hydrated chromium
  ions, $\ce{[Cr(H2O)6]^3+}$ which react with base to give salts
  of $\ce{[Cr(OH)6]^{3−}}$. It dissolves in concentrated alkali to yield
  chromite ions $(\ce{[CrO2]^{-}})$.

So, the rule of thumb is as oxidation number increases, charge on metal increases, acidic character increases. Also, basic oxide is soluble in water, acidic oxide is insoluble in water. This can be used to differentiate between acidic and basic oxide.

I understand the explanation, but find it incomplete. Apparently, it is just circumventing the main underlying logic. You say that increasing oxidation number means increasing acidity. But what is the reason for this? Why does acidity increase with increasing oxidation number?

Increasing oxidation number on the metal atom leaves more empty orbitals to accept electron density, so shouldn't that increase the Lewis acidity contribution to the behavior of the metal oxide?
I agree, if you can still say, "but why?", then the underlying logic (if any) is not sufficient.


text:
Wouldn't radiolabelled phosphorus in DNA break it apart as it disintegrates?
The Hershey-Chase experiment was designed to prove that DNA is the genetic material in organisms. In this experiment, two batches of viruses were grown in two separate media A and B, with A having an abundance of nutrients involving $\ce{^32_15P}$, and B having a lot of $\ce{^35_16S}$.
Leaving solution B aside, we have $\ce{^32_15P}$ undergo a beta decay:
$$\ce{^32_15P -> ^32_16S + e- + \bar{\nu_e}}$$
This process releases $\pu{1.709 MeV}$ of energy, which is quite sufficient to break apart the phosphodiester linkages.

With all the phosphorus in the viral DNA being radiolabelled, this would very well break apart all the phosphodiester linkages. and give us a soup of sulfur-based nucleosides. In such a case, how come this hasn't affected the virus' mechanism to transfer the DNA to a host cell?

The decay phenomena of Phosphorous would be , most likely, included in the experimental design since 14 days half life means, that if experiment is finished in 3 -4 days the loss is less than 20 %.

@ankit7540 Agreed, but even if a few phosphorus disintegrated, it could create highly unstable species that form as the sulphur formed recoils out.

Of course it would break, just like you said; also, a high-energy $\beta$ particle would kill quite a lot of bystander molecules. Also, if not for other reason, the resulting molecule would no longer be DNA , since the decayed atom would no longer be $\rm P$. Also, the product would no longer be radioactive, so we wouldn't be able to detect it anyway.
The same applies to any other case of radiolabeling, in biochemistry or elsewhere, and you seem to be missing the very point of it: All this is not important.
Say, we have some $\ce{A + B -> C + D}$, and we label some atom in $\ce{A}$ so as to find out whether it will go to $\ce{C}$ or to $\ce{D}$. Then we run the reaction as usual. Now, radioactive atoms do not decay all at once; some decay earlier and some later. Those which decay earlier (that is, before and during the reaction) are lost to us, and we don't care about them one bit. Then we separate $\ce{C}$ from $\ce{D}$ and measure the radioactivity of each, and it is those atoms which decay now that matter.
Oh, and it is not like we label all atoms in $\ce{A}$. No, a very tiny fraction would suffice. Those viruses would probably get but a few radioactive atoms each. Some will be lucky enough to reproduce. Some will be killed by an early decay and remembered as martyrs of science.
So it goes.

Thanks for your answer. I have a few doubts. If we radiolabel $\ce{A}$, and it disintegrates as the reaction is run, how come the unstable species produced as a result of recoil of radioatoms not interfere with the reaction? Couldnt such a random unstable intermediate combine with $\ce{B}$ and form some other product $\ce{E}$ instead?

Sure, they would form all kinds of side products. It is just that we don't care, because we have extremely few of those. Again, the fraction of labeled molecules in $\rm A$ is so small that from chemical point of view they don't exist at all. Only the decay event makes them visible.

Why the disintegration of $\ce{^32_15P}$ does not affect the experiment. Following reasons:

As mentioned since the experiment is relatively short relative to the half-life of the radioactive element under question. From the original paper (https://www.ncbi.nlm.nih.gov/pmc/articles/PMC2147348/) it appears that 1-2 days would be maximum for needed for a set of experiment. And this work involved several sets of experiments. In such case, the theoretical amount of remaining radio-active element is > 88%.
The typical number of cells in this experiment was $\ce{10^9}$ cells per $\ce{mL}$. And starting cell culture broth usually involves several $\ce{mL}$s. Hence, the total phages would be very large in number. This means that of the < 12% cells can have $\ce{^32_15P}$ disintegration (assuming the maximal time of experiment). Even if going by this maximal approach this experiment would be accurate to give correct interpretation since the radio-active assay used has $\ce{\pm}20$ % error (check paper). Again, as mentioned by Ivan Neretin, the disintegration is random and and can happen in any cell absolutely randomly, so considering one phospho-diester breakage in a cell would not cause the phage to die. I am not sure but there might be correction mechanisms too.

Why this disintegration not affects the virus' mechanism to transfer the DNA to a host cell ?
Consider a phage which has undergone such disintegration.
If the phage(virus) survives and somehow can still replicate its genetic material  it can continue to act on bacteria. But nobody knows what happens !
If it dies, then when centrifugation is done it either be sticking to the bacteria or be in the broth(culture solution). Then either it can go with bacteria or be separated as lower weight fraction. This might skew the result a little though !

In this work, before releasing phage on bacteria initial separation of phages for labelling  was done.
The $\ce{^35_16S}$ is a stable species and can only affect the numerical accuracy of this experiment by < 12% by the maximal disintegration as discussed above.


text:
Does this wire gauze contain asbestos? Can I tell visually?
I bought a wire gauze which claimed to be made of ceramic, and asbestos-free. However, I bought it from Amazon (not a trustworthy source), and I see fibers around the edge of the center. Is there a way to tell whether this is asbestos?

Just wondering why you bought something from an 'untrusted' source that you might worry about the health impacts of. Seems, well, odd.

@porphyrin Part of my question is whether it's possible to tell visually or by its characteristics. If it's not, that is an answer.

that was my point, you cannot tell just from the photo what its made from.

@porphyrin Thank you. Then I'm assuming the fiber at least could be fiberglass. I'm not familiar with how wire gauzes are made.

This question should be reopened as it does not ask about safety, only identification.  If the OP had asked whether the wire was stainless or not I do not see a reason why the question would have been closed. Though I do acknowledge that there are two questions in this question.

@A.K. I'll remove the 2nd question and keep it just about identification.

Probabilities are high that it is 316 SS. It is often used for small stuff like this.

Unless the component is made to be used with a gas burner, in which case it is likely an Inconel.


text:
Number of isomers in a 3-sugar oligosaccharide
I recently read that 12288 different oligosaccharides can be formed from the D-isomers of glucose, mannose and galactose. I wanted to try to calculate this, but I keep falling very short of this number.

Each molecule has the potential to form 6 unique bonds:

The disaccharide can form 13 unique bonds:
(Note, bonds formed by anomeric carbons shown with an arrow)

So, if a monosaccharide can form 6 different bonds, this means that:
6 possible bonds x 5 bonds on 2nd sugar x 2 anomers for seconds sugar = 60 possible disaccharides
Each of these disaccharides has 13 possible bonds it can make to a third saccharide:
13 bonds x 5 bonds of 3rd molecule x 2 anomers = 130 combinations for a single dissacharide.
130 combinations per disaccharide x 60 possible disaccharides =7800 isomers.
This is probably a counting error. But maybe I am missing a stereocenter?

Do you have a reference for the 12,288 figure? I get twice that. Would like to look at work before posting.

Found it! Google stryer 12288 isomers.

This question is interesting and neglected. Bob made a sensible attempt at solving the problem. I took a different approach. [The number of isomers 12,288 was found in Stryer’s Biochemistry (Google: stryer 12288 isomers)]. Of course, oligo- means a few. I presume this means three in this case. But can D-glucose, D-galactose and D-mannose be used as many times as you want? That would mean 27 (33 = 27) permutations which would eventually make the total much greater than 12,288. However, if each hexose is to be used once, then six (6! = 6 factorial) permutations are possible for the location of the hexoses in the oligosaccharide. Division of 12,288 by 6 removes this issue from the question and leaves 2,048 as the number to justify. Now 2048 = 211. The problem reduces to finding 11 stereocenters, if possible!
Let us assume as a point of departure that diagramatic structure ABC (top of the illustration; The connectivity C1-C4 is arbitrary.) shows where bonds are formed between the C1 anomeric carbons and the hydroxyls at C2, C3, C4 and C6 of another hexose. Another pair of assumptions is that all rings are 6-membered (pyranose) and that all hexoses, as defined, are in the D-series. Finally, assume for the time being that there are no C1-O-C1 connections as typified  in figs. D and E.
Rather than define C1 of hexose A as having two stereocenters (α and β anomers), this assignment will be considered in a different way. Hexose A is given a value of zero because no attachments may be made at its hydroxyl groups at C2, C3, C4 and C6. Hexose B has four hydroxyl groups (blue dots) that can form two anomeric, glycosidic bonds at C1 with hexose A. The AB connection creates 24 (16) stereoisomers. In a similar fashion the hydroxyls at C2, C3, C4 and C6 of hexose C (red dots) form another 16 stereoisomers with C1 of hexose B. The total count thus far is 3! x 28 = 6 x 256 = 1536. This result is only one-eightth of the goal of 12,288 or short by a factor of 23. To this point the eight hydroxyl groups considered can only form two configurations each.
Now the anomeric hydroxyl of hexose C when bonded to the anomeric carbon of hexose B will give not two stereoisomers but rather four (αα, αβ, βα, ββ)(see fig. E). Similarly, the same relationship exists between hexoses B and A (see fig. D). These configurations add a factor of 16 (24) to the total. This reasoning leads to a value of (3!) x (212) = 6 x 4096 = 24,576.
If the original analysis considered a structure such as ABC having a mutarotating C1-OH, then the total might be 12,288. However, ABC is a connectivity map, not a chemical structure.

What may be concluded from all of this? The premises set down here may not be those applied by the proposers of 12,288. There may be a flaw in this, or their, analysis. The answer may be 12,288, 24,576 or something else.


text:
Addition of a noble gas - Le Chatlier
So, regarding a generic reaction (for the sake of argument, let us assume that it is completely gas phase), we have Le Chatelier's Principle. Let us use this example:
$$\ce{2SO2(g) + O2(g) <=> 2SO3(g)} \quad\quad \Delta H < 0$$
Now, we have the questions: oxygen gas is added, heat is added, a noble gas is added, etc.
For the addition of oxygen gas, we obviously shift right, and for heat, we shift left.
For a noble gas, I would expect no shift to happen. Indeed, this is what the SAT's seem to say, and I completely agree. However, my teacher seems to have the conception that when one adds a noble gas to a reaction such as above, you decreases molecular space, and the reaction does shift/speed up.
I have a test tomorrow, and if she gives me a question like this, I am planning to write that there is no shift or rate change. Now, if she marks this wrong, I need a formal argument to support my claim.
Am I correct in saying that with the addition of a noble gas, there is no change in either equilibrium shift or rate change? Why?

Do you mean "ideal gas" (= a gas that obeys the equation of state $pV = nRT$), or "noble gas" (= He, Ne, Ar, Kr, Xe)?

Noble gas, for example, many problems use Ar.

Oh, ok, I see. I guess the point is that the gas does not react i.e. it is inert.

@Equinox Congradulations on your Yearling badge!

For the record, SO3 is described by Wikipedia as "Colorless to white crystalline solid which will fume in air.[1] Colorless liquid and gas.[2]" And, "Melting point 16.9 °C (62.4 °F; 290.0 K)" and "Boiling point 45 °C (113 °F; 318 K)". So why does this equilibrium not specify a temperature as the SO3 could well be NOT in a gaseous state?  Temperature permitting,  presence could be as a particle suspension in air, so increasing pressure is expected to benefit SO3 formation.

See, in an equilibrium reaction in gas phase at constant temperature, noble gases can be added in two ways―

At constant volume in the system
At constant pressure in the system.

Effect of adding noble gases at constant volume
At constant temperature and volume, if noble gas is added, total number of molecules,i.e., amount of the element increases (or you can say, 'number of moles' increases, though it's not correct, have you ever heard of number of kilograms?) in the system. As a result, also total pressure of the system increases, but there is no change in partial pressure of the reactants and products (because total pressure increases and mole fraction decreases). So, there is no effect on equilibrium by adding noble gases in a system at constant temperature and volume.
Effect of adding noble gases at constant pressure
At constant temperature and pressure, addition of noble gas to the system leads to increase in volume (cause, amount of the element increases). As a result, partial pressure of the reactants and products decreases (because total pressure constant but mole fraction decreases) as well as the sum of partial pressure of the reactants and products also decreases. Here, according to Le Chatelier's Principle, the equilibrium would be shifted in that direction where number of molecules or volume increases.

@chail10...So the backward reaction is favoured. Interesting!

@Eashaan Godbole In this case, yes! As total number of gaseous molecules in the left (3) > total no. of gaseous molecules in the right (2)


text:
What would you call the polymerised sheet-like polymer arising from Fittig reaction of 1,3,5-trichlorobenzene?
If we dip a sample of chloroalkane in an ethereal solution, radicals form at the sites where chlorines are attached, which subsequently pair with other radicals.
Now, consider 1,3,5-trichlorobenzene:

If I attempted to perform a Wurtz-Fittig reaction with this, I suspect the product would be a two dimensional polymerised product. I drew a portion of it:

This polymer is quite interesting. Has this been observed before? Does it have a name?
While looking for the name of this molecule is my primary concern, it would be great if I could also get some information about its properties. This polymer looks like a "hollow" version of graphene, so I believe this would have some properties of graphene, although quite lighter. I think this could conduct electricity, due to conjugated $\pi$ systems of benzene rings. I also believe it would be quite porous, and probably allow small molecules like $\ce{H2O}$ to pass through.

Fittig isn't particularly good coupling reaction. Ultrasounds are needed even for biphenyls.

@Mithoron thanks for your insight, but my main concern is with the polymer itself. If Fittig isn't good enough, how about Kolbe electrolysis of trisodium 1,3,5-benzenetricarboxylic acid? Anything that can give me my polymer.

I think it simply can't be done with state-of-art methods. Interesting idea though. Problem is that even if you used Suzuki (no Kolbe wouldn't do) there would be nothing to force planarity and order.

@Mithoron I heard that benzene allows certain "aromatic ring currents", could this be used to align all the benzene rings perpendicular to a normal magnetic field?

Not "allows", rings create ring current, and no, no, no.

The first coupling will already be so far out of planarity, it is a biphenyl after all, that there would never be a 2D-structure. I believe with the third molecule the whole thing will start to coil. My best guess would be that there might be string-like oligomers.

Ignoring the complex challenges it would take to make such a structure, the name is rather unclever with the literature referring to it as "holey graphene". You could probably take this a step further with your hypothetical material and call it :

well-ordered holey graphene
or better yet, well-ordered monodisperse-C6-holey graphene.

with C6 alluding to the missing $\ce{C6}$ ring that would be present in perfect graphene. Materials science has very unrigid naming conventions. If you make it first, you can greatly influence the literature terms.
As far as conductivity goes, I agreed that it would likely be conductive or semi-conductive at a minimum similar to poly(9,10-anthracene diylidene).

I actually couldn't find any evidence that holey graphene would be saturated with hydrogen at the edges of the holes. Sometimes functional groups though. Anyway, I did not really look deep into the matter.

I think that the IUPAC nomenclature is not yet ready to name such regular periodic network structures. The IUPAC polymer nomenclature could be used to name it, but the meaning of the following names (source- and structure-based) are more or less vague, respectively:

net-poly(benzene)
net-poly(benzene-1,3,5-triyl)

(The latter name might be exact, maybe even just “poly(benzene-1,3,5-triyl)”, but the polymer nomenclature for network polymers does not expect such regular branching with the branch size of the monomer unit, or say regular trivalent monomer based mesh networks. IUPAC ‘Purple Book’ does not mention such regular networks, there are some polymers based on benzene-1,2,4,5-tetrayl unit, but these are linear strands; and also benzene-1,3,5-triyl unit, but as a center branching point of three-star polymer, where the rest are three polystyrene chains.
And there is also seemingly appropriate comb- prefix available, but it's something different, is not meant for such networks.)


text:
Effect of increasing volume on gas-phase equilibrium
$$\ce{P4(s) + 6F2(g) <=> 4PF3(g)}$$
If the volume of the system was increased and has now re-established equilibrium which of the following would NOT be correct?
a) The mass of $\ce{P4}$ would not have changed.
b) The concentration of $\ce{F2}$ would have decreased.
c) The number of moles of $\ce{PF3}$ would have decreased.
d) The number of moles of $\ce{F2}$ would have increased.
The answers say c; please help explain how they got this.

it depends, decreased from when? Is it referring to the decrease immediately after volume change or from the previous equilibrium?

look again, it should be B in the latter case. None are right in the former case

Why is no one closing this question as "off-topic" as the problem poster as shown no attempt in providing his own insight? Just curious because it seems to me that's usually what happens to this kind of problems.

Simply apply Le Chatelier's Principle, which states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change.
In this case, when the volume of the system is increased, the pressure of the system decreases. To increase the pressure of the system, the system would have to favour the side of the equation with more gaseous molecules. Thus, the position of equilibrium shifts to the left, increasing the number of gaseous molecules of fluorine and decreasing the number of phosphorous trifluoride molecules, resulting in an overall increase in gaseous molecules in the system, increasing the pressure of the system, to counteract the decrease in pressure. As a result, more phosphorus molecules would be produced.
Thus, A is incorrect. However, B is also incorrect as a bigger amount of fluorine molecules in the system would naturally lead to an increase in its concentration, as the other gas phosphorus trifluoride decreases.
I conclude that your question is problematic.


text:
Conductivity of water and chlorine?
I am a novice in chemistry and i have a question: I know that water is not a good conductive material (since there is a perfect bond in water I think) but I am thinking that if we add chlorine which has 7 valence electrons this means that there is a hole and this increases conductivity... am I right?

No, it doesn't work like that. There are ways to make water conductive, but they don't have anything to do with holes.

but chlorine does increase conductivity (or adding impurities in general)

Some impurities (known as electrolytes) do indeed increase conductivity, some don't.

If so can you explain me how does it work? thanks a lot

Yes, it indeed does. Impurities help in dissociating the ions in water, but as Ivan Neretin said, 'they don't have anything to do with holes', here you are not doping the Cl (or, other impurities) in water.

Oh i see (my background is in elettronics so i thought it was like semiconductors); so are the impurities that are conducting

In a nutshell, water is very much not like semiconductors. Its conductivity is due to altogether different mechanism, and follows different trends.
Typical semiconductor is a rigid network of atoms held together with bonds. Nearly any "wrong" atom would provide either extra electrons or holes (that is, missing electrons). Holes and electrons are the charge carriers here. Being very small, they move fast, so you only need a tiny amount of those impurity atoms (say, a few ppm) to create significant conductivity.
Water is quite different. You can't have free electrons in water, to begin with; they will instantly react with something and become ions. Ditto for holes. Ions are the only allowed charge carriers in water. Ions are as big as atoms, and each carries a thick coat of water molecules stuck to it, so effectively they are even bigger than atoms. An ion would move in an electric field, but with great difficulty, much like a man pushing his way through a busy crowd. (Compared to it, the electrons would be like little sparrows darting overhead.) Therefore you need a lot of ions to create some real conductivity.
Things that produce ions in water (dissociate, as we say) are called electrolytes, and there is plenty of them around. For example, table salt dissociates into $\ce{Na+}$ and $\ce{Cl-}$; sulphuric acid dissociates into $\ce{H+}$ and $\ce{SO4^2-}$. Elemental chlorine ($\ce{Cl2}$), being a neutral molecule, does not dissociate; also, you can't add much of it, for it is a gas, and its solubility is not that great. Over time, it would react with water to produce certain compounds which do dissociate, so you'll have some conductivity after all, but that's a weird and inefficient way of achieving it. Using any soluble salt is much more straightforward.
The choice of particular electrolyte depends on your ultimate purpose.

Thanks a lot man and sorry for saying dumb things like the holes one :D


text:
Why are the stoichiometric coefficients the powers in the rate law?
Generally, for a reaction of the form
$$\ce{$n$_A A + $n$_B B + \dots -> products},$$
the rate law is given by the following:
$$\text{rate} = k [\ce{A}]^a [\ce{B}]^b.$$
Do the stoichiometric coefficients equal the powers in the rate law?
\begin{align}
a &= n_{\ce{A}},&
b &= n_{\ce{B}}
\end{align}
If that is true, why is it so?

There's a fundamental word missing there: what you wrote is true for elementary reactions, that is, reactions that take place in a single step with a single transition state.
In an elementary reaction, then, what actually takes place at a microscopic levels fits the reaction formula we write (if we express the reaction as $\ce{A + B -> C}$, a molecule of $\ce{A}$ and a molecule of $\ce{B}$ actually collide and produce a molecule of $\ce{C}$ as a single step). But, in general, that's not the case; reactions in general have mechanisms comprising one or more elementary reaction steps.
A related concept is the molecularity of a reaction step, that is, the number of molecules that intervene in a certain step. Since elementary reactions match their chemical expression, if a reaction is elementary, the order of a reagent will match its stoichiometric coefficient (which is your question: that means that $a=n_a$ and $b=n_b$). This is not the case for multi-step mechanisms - kinetic order is not related to stoichiometry in general (and you can have the reaction rate depend on concentrations that don't even appear in the formula, such as in homogeneous catalysis).
Note also that molecularity is always low - the vast majority of reactions have molecularity 1 or 2 (meaning 1 or 2 molecules are involved). Molecularity 3 involves the simultaneous and spatially productive collision of 3 molecules so it's very, very rare - and molecularity 3 reactions are only observed for a few systems. As far as I know, no reactions with molecularity 4 have been observed at the macroscopic scale.
Note also that there's no way to know a priori if a reaction is elementary. Some reactions are clearly multi-step (for instance if the formula involves more than three molecules), but finding out if a simple reaction is elementary or complex requires empirically measuring its order, proposing mechanisms, testing for intermediates and transition states, and is always open to reassessment.


text:
Why doesn't a neutral solution at any temperature have pH 7?
I am given 4 statements to choose from and find which is correct about a neutral solution:

$K_\mathrm{w} = \pu{1.00\times10^-14 mol^2 dm^{-6}}$
The solution contains only $\ce{H2O}$
$\ce{[H+] = [OH-]}$
$\ce{pH =7}$

I can see that the first statement is wrong as that is only the case at $\pu{25 ^\circ C}$.
The second statement is also incorrect - it can be when $\ce{H+}$ ions and $\ce{OH-}$ are equal in concentration.
So I can see that option 3 is correct.
However, I am confused why option 4 is wrong.
Why is option 4 wrong?

What happens to [H+] if Kw =/= 10^-14? Say if it's 10^-12?

As the temperature changes the value of $K_\mathrm{w}$ changes(as you wrote) and then the concentration of $\ce{H^+}$ and $\ce{OH^-}$ also changes and now the concentration of $\ce{H^+}$ will not remain $10^{-7}$ and hence the neutral point shifts. Therefore, at any temperature the $\ce{pH}$ of neutral point is not 7.

Always welcome. Friend!


text:
Are benzyl alcohol and o-cresol functional isomers or position isomers?
My textbook categorizes them as functional isomers (phenol and aromatic alcohol) whereas

The above mentioned isomers are given as position isomers on 'chemguide.co.uk'

Welcome to Chem SE. I can understand your confusions, such classifications are vague and often overlap. You can call it as both positional and functional. By the way, have a [tour] of this site.

The chloride example is interesting because it highlights a fine line separating both.
The distinction between functional and position isomers isn't as simple as the typical examples make it to be - sometimes, what a substituent is attached to is critical to its functional group, so the same sequence of atoms with the same connectivity can behave, chemically speaking, very differently depending on what they are bonded to. And, conversely, position isomers in general don't have exactly the same chemistry either - you know, for instance, that n-butanol, 2-butanol and t-butanol will behave similarly but not identically.
So where do we draw the line? Technically, it's true that both are position isomers. In both cases the same substituent is attached to different positions (unlike, say, acetone and propenol, where the connectivity is clearly different). But it's also true that their chemistry is distinct - aryl compounds, in general, have different reactivity than their equivalent alkyl compounds, which is why we talk about the reactivity of "phenols" and "alcohols" as separate ideas.
So, from your examples, I'd say benzyl alcohol and o-cresol are clearly functional isomers, as their chemistry is quite different. For instance, compare pKa values:
o-cresol: $\mathrm pK_\mathrm a=10.3$
benzyl alcohol: $\mathrm pK_\mathrm a=15.4$
2-ethylphenol (similar to o-cresol): $\mathrm pK_\mathrm a=10.2$
2-phenylethanol (similar to benzyl alcohol): $\mathrm pK_\mathrm a=15.9$
Chloride, on the other hand, is less sensitive to the substrate; but I'd still classify chlorotoluenes and benzyl chloride as functional isomers, since I think they are different enough to merit that distinction, but it's in the twilight zone of vague definitions.


text:
Why can't the pH of a solution of a weak acid be calculated without the dissociation constant?
Why can't the pH of a solution of a  weak acid be calculated without the dissociation constant? How can pH of weak base be determined?

As much as I know the dissociation constant decides the degree of dissociation at that temperature and hence if you don't know how much the compound dissociates then how can you determine the concentration of it's component(which here is an acid having hydrogen atom as its component).


text:
How to convert ppm to g/m3?
I'm not very comfortable with chemistry and I need to convert $\ce{CH4}$ concentration values in $\pu{ppm}$ to $\pu{g/m3}$. Is that possible?
I've already researched a bit and realised that for water you can assume $\pu{1 ppm}$ equals $\pu{1 mg/L} = \pu{1 g/m3}$. But since I'm measuring concentrations in the air this might not be correct.
I really appreciate any help. Thanks!

1 ppm is like you have 1 part , here 1 molecule of $\ce {CH_4} $ in a million molecules of air. If we assume air to be ideal gas then you may use the Ideal gas equation for knowing the volume of total air and then take out value in $g/m^3$ and don't forget the $g $ represent weight of methane

@Physicsapproval Thanks for your help! I have estimated the volume using the Ideal Gas Law (assuming 1 mol of $CH_{4}$) but I'm unsure as to what to do next. Should I divide said volume per the molecular weight of $CH_{4}$?

I have tried a different approach. Knowing that: $ 1 ppm = 1 \frac{\mu g}{g}$; first, I multiplied the ppm values per the density of (in this case) methane ($656 g/m^{3}$) and them multiplied again by the factor $(10^{-6})$. Here's the units calculation: $\frac{\mu g}{g} \times \frac{g}{m^{3}} = \frac{\mu g}{m^{3}} \times (10^{-6}) = \frac{g}{m^{3}}$. What do you think?

okay is the methane gas in  a mixture I believe assuming air then how have you calculated the density ? Again have you used ideal gas law here to find density ?

I am trying to understand ppm, too. As far as I have understood there are different kinds of ppm, which is basically a ratio: it can be a ratio of amount of substance, masses, or volumes.
Assuming your ppm is a molar ratio I did this reasoning:
Indicating with $n$ the amount of substance, with $M$ the molar mass and with $V$ the volume, the concentration of your gas is:
$$c = \frac{n_\mathrm{gas} \cdot M}{V},$$
and defining the $\mathrm{ppm}$ as:
$$\mathrm{ppm} =\frac{n_\mathrm{gas}}{n_\mathrm{total}}\cdot 10^6.$$
Using the gas law:
$$n_\mathrm{tot} = \frac{p\cdot V}{R \cdot T}, $$
where $T$ is the temperature in kelvin and $p$ the pressure in pascal, and substituting, you get:
$$c = \frac{\mathrm{ppm} \cdot M \cdot p}{R\cdot T}\left[\frac{\mu \pu{g}}{\pu{m^3}}\right].$$

You really don't need to overcomplicate things for this answer.
The key things worth knowing is that in an ideal gas (a good approximation for most at standard conditions (0°C and standard atmospheric pressure)) one mole of the gas will occupy 22.4 L of volume. A mixture of gases isn't any different and to know the weight of the gas you want you just need to multiply the molar mass of the gas with the proportion in the mixture (ppm is the proportion here).
So each ppm of methane will contribute about 16 / 1,000,000 g to each 22.4L of the gas mixture. Or (adjusting for the volume conversion to cubic metres which contain 1,000 L) 44.7 * 16 /1,000,000 g/cubic metre.
By this formula, a cubic metre of pure methane would weigh ~715g at STP so you could just work with that by multiplying by the ppm value.
It only gets more complicated if you need proportions by mass in the mixture: then you have to know the molar masses of all the other components. But, if you stick with volumes, gas laws keep things really simple.
PS if your conditions (pressure or temperature are different) the only thing you need to adjust is the volume an ideal gas under those conditions (molar volume is closer to 24.8 L at 25 °C, for example).


text:
Why isn't there a third buffering zone in the pH curve of this titration?
I'm in chem 30 right now and we were learning about pH curves and buffering zones. I was looking at a question on the titration of $\ce{H3PO4}$ with $\ce{NaOH}$. The curve showed two qualitative reactions occurring as seen in the two sudden changes in pH. It also showed two buffer zones. Shouldn't there be 3 buffer zones not 2? My understanding is that once the reaction of $\ce{HPO4^{-2}}$ and $\ce{PO4^{-3}}$ is occurring, a third buffer zone would occur.

According to this, we have three pKas, as you observed: 2.12, 7.21 and 12.67 (at 25°C).
Your values may be different due to temperature.
The third value (12.67) can't show up in your graph because it doesn't go beyond pH 12.