text:
Relative magnitudes of 2- and 3-bond coupling constants
Why is geminal coupling weaker than 3-bond coupling in complex coupling? It would make more sense that atoms that are closer to each other influence each other more. Take methyl acrylate for example. The $\mathrm{H_a}$ is causing the bigger split, and the $\mathrm{H_b}$ the smaller ones. Why?

https://www.chem.wisc.edu/areas/reich/nmr/05-hmr-04-2j.htm

J coupling is mediated by the interaction of spin-polarised molecular orbitals. 120°  bond angle geminal orbitals just don't interact well, as opposed to the situation at an sp3-hybridised CH2 group.

What do you mean by weaker (and stronger)? You can’t weakly couple with pregnancy (i.e. can’t be a little bit pregnant). Unless you are talking about the magnitude of coupling?

@Jan I'm pretty sure he's talking about the magnitude of the coupling constant. Perhaps the OP can confirm.

@DavePhD - Hooray! Someone else is onto Hans Reich's page as a great resource.

@Jan  That is sort of like saying being 8 months pregnant is identical to being 2 months pregnant-- which is true in only the most trivial of ways.  Magnitude is obviously implied from 'weaker' and 'stronger' in this case.

@Jan - Take care when using the terms weaker and stronger when referring to coupling in NMR. A weakly coupled system is one where the distance between resonances is much greater than the coupling constant. Strongly coupled nuclei are closer together, and give rise to roofing/tenting in spectra. An AB spin system is a strongly coupled system.

@long Point taken. Looking at bon’s comment, I wish I had added constant to my previous comment.

Be careful when generalizing from a single example.  There are many situations where the geminal coupling is much larger than vicinal coupling.  Pick pretty much any molecule that has an aliphatic CH2 where the two H's are diastereotopic and you'll find that the geminal coupling for those two H's can be 12-16 Hz vs. the typical ~7 Hz vicinal couplings.  However, if you're familiar with 6-membered rings in chair conformations, you've also seen vicinal coupling that range from ~2-15 Hz which leads us to the real answer to your question: the Karplus relationship.
Martin Karplus (who got a nobel prize for completely different work) is most well known in the NMR community for his theoretical work relating the values of J coupling to bond angles, bond lengths, electronegativity of substituents, etc.  In other words, there's no reason that 2 bond coupling is necessarily larger (or smaller) than a 3 bond coupling – the exact values will depend on the functional group and local geometry of the molecule.  Most organic chemistry texts describe a highly simplified Karplus relationship (ignoring all effects except for bond angles) for 2-bond and 3-bond J values.  I've attached two common examples.

You can see that for a typical aliphatic CH2, a bond angle of ~109° would predict a 2J ~ 14 Hz.  Whereas, a terminal alkene with a bond angle of ~120° would predict a 2J ~ 3 Hz.  Comparing this with the 3-bond Karplus relationship, the vicinal coupling for alkenes should be 8 Hz or 10 Hz depending on the cis/trans relationship.  For an aliphatic system with free rotation about a C-C bond, you have to average the Karplus relationship over all angles and you end up predicting ~ 7 Hz.  For a cyclohexane derivative in a chair conformation, you can see that an axial-axial coupling (180°) is predicted to be ~10 Hz, while an axial-equatorial coupling (~60°) is predicted to be < 2 Hz!
Qualitatively, this matches what we observe quite well.  Quantitatively, we quickly run into a lot of problems.  Many axial-axial couplings are 12-15 Hz and the J coupling observed in alkenes typically deviates quite a bit from the predictions given by these graphs.  Again, these graphs are grossly oversimplified - the real theory would fold in information about nearby substituants and steric effects, etc.  However, you'd then need run a bunch electron structure calculations and average these over many different conformations, etc.  So most people just live with the qualitative nature of these graphs to rationalize the patterns that they see.

Sorry about the large images, I'm not sure how to shrink them using the markup commands...


text:
19F MAS NMR chemical shift tensor and the crystal structure
$\ce{^{19}F}$ MAS NMR is measured on fluorine-doped $\ce{TiO2}$. $\ce{F}$ is to substitute $\ce{O}$ and the crystal structure is anatase, which has a tetragonal crystal system with a space group $\ce{I 41/a m d S(141)}$. Using the observed isotropic chemical shift value, a chemical shift anisotropy (CSA) analysis is carried out using Simpson and gives a uniaxial chemical shift tensor ($\ce{δ11 = δ22 ≠ δ33, etha = 0}$). This result dictates a three-fold symmetry at the fluorine site. In anatase with a tetragonal system, however, the oxygen ions occupy position 8e with a point group of ${mm}$ and no three-fold axis.
How can the uniaxial chemical shift tensor, indicating three-fold symmetry, correlate to the crystal structure of anatase which does not possess any three-fold axis?

You had the isotropic chemical shift and from that somehow back-calculated the CSA using Simpon? I wouldn't have thought that it is possible... Can you post the spin system section of your Simpson input file?

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sort: votes


text:
Nomenclature of an alkyne with bromo and hydroxy substituents

Is the IUPAC name of above compound 7,7-dibromo oct-5-yn-4-ol or 2,2 dibromo oct-3-yn-5-ol ?
I understand that the former name is given keeping in mind -OH as the functional group but by giving the latter name we will have the least sum of locants (ie, following lowest sum rule) .
When should the lowest sum rule be dropped ?
Thank You in advance.

Daily dose of pedantry from me: Alkynes shouldn't be depicted as being bent like that.

The smallest sum of locants rule does not exist in the IUPAC recommendations. The application of this ‘rule’ can lead to wrong results in many instances.

In this case, the relevant criteria for the numbering are:

lower locants for suffixes (here: ‘ol’)
lower locants for multiple bonds (here: ‘yne’)
lower locants for prefixes (here: ‘bromo’)

The corresponding subsection of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(e) saturation/unsaturation:
  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(…)

Therefore, the correct name for the compound given in the question is ‘7,7-dibromooct-5-yn-4-ol’ (not ‘2,2-dibromooct-3-yn-5-ol’) since the locant ‘4’ for the suffix ‘ol’ is lower than ‘5’.

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text:
Why isn't ethane used for cooking?
We commonly use methane and propane for cooking (and home heating), but not ethane. I would expect ethane to be suitable for this, being in between the two, but I've never heard of anyone using it for this purpose. Why is that?
On a related note, why is butane used for cigarette lighters and basically nothing else (in ordinary life, I mean)?

https://en.wikipedia.org/wiki/Liquefied_petroleum_gas - LPG is generrally  a mix of propane and butane isomers, so your second question may be unjustified.

Probably the biggest drivers behind using methane as a fuel is that it is abundant in natural gas and is (currently) mostly useless as a chemical feedstock.
Ethane makes up a few percent of natural gas and can also be obtained as byproducts of petroleum refining, but the big difference from methane is that ethane is extremely useful in chemical synthesis (mostly to make polyethylene).
In fact, household natural gas often contains a bit of ethane, which may vary depending on the current demand for ethane as a feedstock.
In short: it burns just as well, but ethane has other important uses.
Propane has some use as a feedstock, but it is an attractive fuel because it is easily stored as a liquid without requiring huge pressures.

The reason more methane than ethane is used as a fuel is simply its greater abundance in natural gas.  Most polyethylene is not made from ethane.  Polyethylene is made from ethylene, which is mostly made by cracking longer hydrocarbons. This is a rather imprecise process, so the products are separated and used as follows: Hydrogen and methane: burnt in the cracker furnace; ethylene -> polyethylene;  propylene -> dimerization to branched high octane C6 gasoline component and / or polymerization to polypropylene; C4+ -> recycled. https://en.wikipedia.org/wiki/Ethylene is a good starting point.

Ethane is one of the commonly-used feedstocks, along with LPG, propane, butane, etc. used for steam cracking when lighter products like ethylene are desired, whereas steam cracking of heavier feedstocks like naptha tend to produce bigger products like longer alkenes and aromatic compounds. Ethylene is made starting from many feedstocks, and I don't know how much ethylene is made from ethane, but ethylene production seems to be the primary use of ethane.

Given that the C2H4 production process basically consists of smashing hydrocarbons up with high temperatures, then quenching & separating the results, it doesn't care that much what the feedstock is. C2H6 isn't often deliberately separated from CH4, due to expense of low temperature distillation. It's becoming more available, because where liquefied natural gas is made, it's desirable to have pure CH4 for a single boiling point. But C2H6 isn't "extremely useful" in C2H4 synthesis, it's just not much use for anything else. Much of the C2H6 used in C2H4 synthesis comes from the recycle stream.

@LevelRiverSt, I think you're understating the role of ethane in ethylene production.  There are world scale crackers that run almost exclusively on ethane.  Others use an ethane/propane mix to get a little more propylene and heavier products.  Most new crackers being built in the US are made for light and not heavy feedstocks.  Unlike heavy feedstocks with their myriad products, ethane cracks to very few products (methane, H2, ethylene) with good yields.

@ByronWall it seems US used 51% ethane in 2005, while the world used 25.9%, a significant geographical difference.   http://images.pennwellnet.com/ogj/images/ogj3/9613jch02.gif  Old style plants get 30-40% C2H4 from heavy feed. That's OK as all other products are saleable. Post-cracking separation cost is high, but not as high as separating out C2H6 feed from mixed gases, or they'd have done it. New plants use more C2H6, because LNG has given high C2H6 availability, effectively separating C2H6 from CH4 "for free" so better yield but less saleable byproduct. C2H6 is nice to have, not essential.

@LevelRiverSt, thanks for the data.  I couldn't find a source with a quick search yesterday.  Looks like 2010 feedstock split for ethylene (see page 13) for the US was ~65% ethane.  I suspect that has only increased since then.  Maybe a US-centric view (I worked in the US ethylene industry making polyethylene), but ethane is critical to the production of ethylene here.

Ethane on the other hand could be produced from biomass ethanol through electrocatalysis using renewable energy producing both hydrogen gas and ethane, which could be used to power fuel cells (hydrogen) and for heating (ethane). Maybe it will be the most common fuel in the future.

In both cases, there appears to be a confusion of terminology between common and technical uses.

We commonly use methane and propane for cooking (and home heating), but not ethane. I would expect ethane to be suitable for this, being in between the two, but I've never heard of anyone using it for this purpose. Why is that?

In reality, anyone using natural gas as a cooking fuel likely is cooking both with $\ce{CH4}$ and $\ce{C2H6}$. From the above-linked Wikipedia page (emphasis added):

Natural gas is a naturally occurring hydrocarbon gas mixture consisting primarily of methane, but commonly including varying amounts of other higher alkanes, and sometimes a small percentage of carbon dioxide, nitrogen, hydrogen sulfide, or helium.

EngineeringToolbox.com reports the following representative composition ranges (probably in percent by volume?) of natural gas:
$$
\text{Composition (%)} \\
\begin{array}{cccccccccc}
\hline
& \ce{CO2} & \ce{CO} & \ce{CH4} & \ce{C2H6} & \ce{H2} & \ce{H2S} &
   \ce{O2} & \ce{N2} \\
\hline
\text{Min} & 0 & 0 & 82 & 0 & 0 & 0 & 0 & 0.5 \\
\text{Max} & 0.8 & 0.45 & 93 & 15.8 & 1.8 & 0.18 & 0.35 & 8.4\\
\hline
\end{array}
$$
Given that $\ce{CH4}$ is by far the major constituent of natural gas, it is sensible that it is referred to commonly by the term methane, even if it is often actually a mixture of methane, ethane, and trace higher hydrocarbons.

On a related note, why is butane used for cigarette lighters and basically nothing else (in ordinary life, I mean)?

Per the Wikipedia page for liquefied petroleum gas, linked in Mithoron's comment, most of what is commonly referred to as propane or butane is actually a mix of $\ce{C3H8}$ and $\ce{C4H10}$ in varying ratios (emphasis added):

Liquefied petroleum gas or liquid petroleum gas (LPG or LP gas), also referred to as simply propane or butane, are flammable mixtures of hydrocarbon gases used as fuel in heating appliances, cooking equipment, and vehicles. ... Varieties of LPG bought and sold include mixes that are mostly propane ($\ce{C3H8}$), mostly butane ($\ce{C4H10}$) and, most commonly, mixes including both propane and butane. In the northern hemisphere winter, the mixes contain more propane, while in summer, they contain more butane.

So, Mithoron is right: $\ce{C4H10}$ is used in much more than just cigarette lighters, it's just that common usage happens to apply the term butane for this context.

As a further note, I would guess the primary rationale for using different mixes of $\ce{C3H8}$/$\ce{C4H10}$ deals with the vapor pressures of the two gases.  The energy densities $\eta$ of the liquefied gases, approximated as $-\Delta H_c^\circ\rho \over \mathrm{MW}$, are nearly equal:
$$
\begin{array}{ccccc}
\hline
\text{Quantity} & \text{Units} & \ce{C3H8} & n\text{-}\ce{C4H10} & iso\text{-}\ce{C4H10} \\
\hline
\Delta H_c^\circ & \mathrm{kJ\over mol} & -2202^1 & -2878^2 & -2869^3\\
\mathrm{MW} & \mathrm{g\over mol} & 44 & 58 & 58 \\
\rho & \mathrm{g\over mL} & 0.58^4 & 0.604^5 & 0.56^6 \\
\hline
\eta & \mathrm{MJ\over L} & 29.0 & 30.0 & 27.7 \\
\hline
\end{array}
$$
Thus, roughly comparable energy value is obtained per volume of each, and there is little reason to favor one or the other on this basis.
Practically, the lower limit of acceptable vapor pressure is that which provides sufficient flow of gaseous hydrocarbon to the point of combustion.  The upper limit is more or less defined by the strength of the container and plumbing.  Consider the following vapor pressure data, calculated from fitted equations (sources: propane | n-butane | iso-butane):
$$
\text{Vapor Pressure (atm)} \\
\begin{array}{ccc}
\hline
 & 0~^\circ\mathrm C& 25~^\circ\mathrm C & 38~^\circ\mathrm C \\
\hline
\ce{C3H8} & 4.7 & 9.3 & 12.8 \\
n\text{-}\ce{C4H10} & 1.0 & 2.4 & 3.5 \\
iso\text{-}\ce{C4H10} & 1.5 & 3.4 & 4.9 \\
\hline
\end{array}
$$
Cigarette lighters (especially disposable plastic ones) presumably do actually use butane-rich fuel mixes, so as not to approach or exceed the mechanical limits of the lightweight, portable containers.  As well, the temperature at point-of-use is somewhat better controlled, as even on cold days the heat from the user's hand is likely to maintain the butane vapor pressure high enough to provide sufficient gas flow. Finally, as noted in a comment by A.K., lighters are generally charged with iso-butane, which is sensible as it is the isomer exhibiting modestly higher vapor pressures.
For applications where metal-walled containers are feasible (grilling, automotive fuel, etc.), however, structural considerations are less important and the higher deliverable pressure from propane becomes advantageous.  In hot summer months, though, I would assume the higher fraction of butane is used so as to mitigate the fairly dramatic increase in vapor pressure of pure propane with increasing temperature.

$^1$ Wikipedia, "Propane (data page)"
$^2$ Wikipedia, "Butane (data page)"
$^3$ Wikipedia, "Isobutane (data page)"
$^4$ Engineering Toolbox, "Chemical, Physical and Thermal Properties of Propane Gas - $\ce{C3H8}$
$^5$ Engineering Toolbox, "Chemical, Physical and Thermal Properties of n-Butane"
$^6$ AeroPres, "Physical Properties" datasheet (PDF link)

cigarette lighters use isobutane which has different physical properties than n-butane.

Methane and ethane are very hard to be liqufied with respect to propane and butane. To make methane liquid requires highee pressures and this means more cost and danger.
But especially, n- butane may be easily liqufied and liquid phase covers less volume. This makes it more suitable for lighthers.

This doesn't really add any information beyond the previous answers.

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text:
What is the difference of molar volume between dissolved 35Cl- and 37Cl-?
Im am geochemist, dealing with isotopes such as 35Cl and 37Cl.
For some model I'm trying to make I need to be able to estimate what is the difference of partial molar volume between dissolved 35Cl and 37Cl in water (as chloride Cl-).
Huge literature exists concerning the partial molar volume of Cl- and other ions, but I can't find anything on specific studies on variations of volume on different isotopes. As in isotopes geochemistry we're often looking a permil variations of ratios, any tiny difference in molar volume of 35Cl and 37Cl would have an importance for me.

I suspect the difference will be quite small, likely immeasurably so, and that there are many other larger uncertainties to take into account before this effect.

While there are some large chemical differences between deuterium and protium, these are mainly related to vibrational mode differences given the (relatively large) mass differences. By the time one gets to chlorine those are pretty much gone away. I see no reason to think that there would be a large difference in molar volume between the two isotopes.

Again, I'm measuring some 0.1 permil variations of chlorine isotopes ratios. So any percent/ permil variation of the molar volume would be of importance

As I said, I expect it to be immeasurable, and attempting to find the difference may lead to cases of false precision. I would be surprised if the difference was higher than ~1 ppm (absolute guess here, though). The best case to investigate would be with either hydrogen or lithium cations, as both are light and thus differences between isotopes are enhanced. Unfortunately there are probably many complicating factors involved with $\ce{H^+}$, so looking into a difference between $\ce{^{6}Li^+}$ and $\ce{^{7}Li^+}$ may be your best bet.

I don't have an exact answer for chloride specifically, but consider the theoretical equations in the following two articles:
Isotope Effects on Molar Volume and Surface Tension: Simple Theoretical Model
and Experimental Data for Hydrocarbons J. Chem. Phys. 44, 457 (1966)

We assume the molecules are held together in a
  condensed phase by van der Waals "bonds" or covalent
  bonds represented by 6-12 or Morse potential-energy
  functions, V(x). For monatomic molecules x is taken
  to represent the distance between molecular centers.

and
Lattice Constants of Separated Lithium Isotopes J. Chem. Phys. 27, 1030 (1957)
which finds atoms of Lithium-6 $4 \times 10^{-4}$ times more separated than those of Lithium-7 and also cited to a reference which considers lithium-6 versus lithium-7 fluoride-19.


text:
How to calculate minimum temperature of a reaction?
For burning ethanol (its reaction with oxygen), we need a minimum activation energy (a minimum temperature) to start the reaction.
Is it possible to calculate what are the minimum temperatures required for burning ethanol and methanol to compare them?

You want flash point or autoignition temperature?

If you mean this on a conceptual level: There are tables with bond energies; compare the intermediate form with the educts and calculate. If that is actually your question, I'll elaborate further.


text:
Are there any strategies to determine bond order and hybridization rankings without drawing the Lewis Structure?
I'm not Chemist. I'm taking a chemistry class in college, and the test review contains many questions where I'm asked to rank from least to greatest based on bond order, polarity, and bond angles. I feel like I can rate on electronegativity.
The problem is that I don't know a way to rationalize this in my mind without drawing the Lewis Structures. Many times each possible choice has like 5 or so, so it takes a lot of time. We are only given 50 minutes for the test and there are many more problems that require large amounts of time.
As a chemist or someone who is more knowledgeable in this field than I am, are there any techniques that would help you now or when you were first learning?

Although it is often taught otherwise, hybridisation is a result from molecular structure. Therefore you must know (or be able to predict) the structure. On a basic level, the easiest way to do that are Lewis structures.
A different way is using the VSEPR theory, a rudimentary model to predict shapes and molecular geometries. Be warned though that this model will fail when it comes to more complex molecules. You need a good imagination of the three dimensions to be able to do it without drawing it.
However you want to spin it, you will have to know which atoms are bonded to other atoms. If you can imagine that, that is good, if not you will have to draw it out. The longer you study, the more present some common molecules are and you will be able to tell certain properties by what we like to call chemical intuition. But that takes time.
Bond order is a bit of a tricky topic, because you have to know the bonding situation. This would include all possible resonance structures for some molecules. While knowing or judging this also comes with experience, it is probably best to draw such structures for learning purposes.
I personally do not like exam questions like these, because they usually just test if a student is good at cramming facts, rather than asking about if these things are understood. But things will get better, if you don't lose interest in the mean time.

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text:
When do I use the hydronium ion in an acid-base equation?
I have a question about when to use H3O+ and when to use the actual compound. I'm not really sure how to word this question so I'll show you an example of what I mean.

Example of using H3O:
"A titration for a sample of 25mL of 0.1 M NH3 by 0.1M HCl is conducted. Calculate the pH after the addition of 12 mL of 0.1M HCl"
$\ce{NH3 + H3O -> NH3 + H2O} $
(etc)

Example using the compound:
"Calculate the pH of 75 mL of 0.1M HC2H3O2 and .2M of NaC2H3O2 to which 9.5mL of 0.1M HCL is added"
$\ce{C2H3O2- + HCl -> HC2H3O2 + Cl-}$
(etc)

My question is, when are we supposed to use HCl in the equation and when are we supposed to use H3O+? Is it only when you are using a buffer that you are to use HCl? If so, why is this?

In both your cases, only H+ actually participates in the reaction. Remember that HCl is a strong acid, and dissociates completely into H+ and Cl-. For example, consider the net ionic form of your second equation:
$\ce{C2H3O2- + HCL -> HC2H3O2 + Cl-}$
$\ce{C2H3O2- + (H+) + Cl- -> HC2H3O2 + Cl-}$
$\ce{C2H3O2- + H+ -> HC2H3O2}$
For your first equation, also note that, similarly, although the entire H30+ molecule participates in the reaction, only the H+ is transferred. (Also, I think you meant NH4+ instead of NH3 as a product.) So:
$\ce{NH3 + (H3O)+ -> NH4+ H2O}$
$\ce{NH3 + H+ -> NH4+}$
So basically, they both use the same compound (H+). Usually water is the main producer of H+ (through self-ionization). You just need to consider factors such as the common ion effect that would make other compounds such as HCl the main producer of H+ instead of water.

This link may be useful in order to learn how to format chemical equations on this site.


text:
How to interpret TLC data regarding the purity of one’s sample
The picture provided is a sketch of my TLC data from the lab along with the $R_f$ values for $i$ (initial, un-recrystallized aspirin) $= 0.429$, $a$ (acetylsalicylic acid) $= 0.429$, and $s$ (salicylic acid) $= 0.571$.
One of the questions in my post lab was:

Based on the TLC data, discuss the purity of the impure aspirin sample, explain your reasoning.

So far, I have said that based on the data, the aspirin sample would be considered pure because there were no multiple spots in any lane. I don't know how else to elaborate on this. What can I say about the $R_f$ values?

If that is an accurate rendering of the TLC slide, I agree with your conclusion.

If your picture is in any way accurate then no way can the solvent front be 3.5 cm while salicylic acid is 2.0 cm.

TLC is a great and poor method at the same time. It is great because it is quick and accurate, and it is poor because it can only show substances that can be coloured in some way and are sufficiently non-volatile to not evaporate away.
You can be rather sure of the following points:

Your impure sample does not contain any more free acid since a spot of the same $R_f$ value is missing.
Your impure sample contains a probably very high amount of aspirin, since the spot has the same $R_f$ value.

However, you cannot be sure about:

Impurities that evaporated away during spotting, development or colouring of the TLC plate
Impurities that simply don’t show up in your method of colouring (e.g. due to the lack of a benzene ring for UV fluorescence).
Solvent impurities that will never show up due to both points above.

To prove purity you would need to resort to different analytic methods such as melting point (historically), HPLC-MS or GC-MS, NMR, etc.

If "i" is the "impure" aspirin sample, then yes. Both the aspirin and the acetylsalicylic acid have the same retention factor of 0.429. This shows that the impure aspirin sample contains acetylsalicylic acid. As you stated, the absence of any other spots in the TLC shows that the impure aspirin sample only contains acetylsalicylic acid. Therefore, it is pure.
In the real world, however, it is more complicated. Retention factors are not a definitive indicator. Many compounds can have similar (and even identical) retention factors.  Hence, it is theoretically possible that a compound with Rf value of 0.429 that is not aspirin is mixed with it. Furthermore, the retention factors change depending on the conditions (choice of mobile phase, stationary phase, temperature, etc). Thus, it is necessary to further analyze it with mass spectrometers, etc. And also to carefully adjust the conditions.
But for a chemistry lab, your reasoning should be enough.

An old school method is to make up known concentrations of the reference material and standardise the spotting method. The spot size was then used to approximate a concentration and give a purity range


text:
Etching ITO on PET without damage
I am planning to use some of this to construct a transparent touch panel for a prototype of mine. It is PET plastic coated in Indium Tin Oxide.
My plan is as follows:

Use some of this UV-cure dry film, a transparency, and a fluorescent light fixture placed very close to it for a while to cure it (not sure if enough UV there at the right wavelength...I may get one of those UV LED nail spa cure things).
Develop the film, remove excess with water.
Use some of this 32% HCl to etch the ITO.
Remove the film.
Enjoy my completed touch screen.

The problem is, I'm not really a chemistry person. My background is electrical engineering and my experience is limited to one semester in a cleanroom class where we did a lot of SiO2 etching (BOE) and some aluminum and nickel etching.
My questions are as follows:

Will dry film work well for this application? I worry that since it is designed for use with PCBs, made of the relatively stable FR-4, the etchant and remover provided with the kit I linked to will be too harsh and damage the PET or ITO
This paper seems to note that $\ce{HCl}$ is quite slow at $20^{\circ}$C for etching ITO. I estimate the molarity of the solution above is probably between 8-12M, giving me an etch rate of ~2-4angstroms/sec. I'm not sure what the thickness of the film is on the plastic, so I could be there a while. How can I determine the thickness of the ITO given its sheet resistance of 50 ohms? (What I'm really asking is how to determine the resistivity so I can solve for the thickness of the film.) I don't have access to anything fancy like a 4-point probe.
This chemical information sheet states that PET plastic does great with $\ce{HCl}$ at 10%, but terribly when "concentrated" (whatever that means). Am I going to destroy the PET during this process?
Is my plan fundamentally flawed? Is it likely not going to work? Do I need to use a different photoresist or etchant (aqua regia is terrible to PET according to that info sheet)?

I'm not too concerned with overetching or overdeveloping since the pattern doesn't need to be incredibly precise (this is a one-off thing and the contacts are around a half-inch square each).
Assume that I don't have access to a laser for ablation, fancy deposition equipment, or any complex measurement equipment (I've got a multimeter and oscilloscope). I can acquire basic glassware through amazon and such AFAIK.

My lab has a lot of experience with ITO-coated PET, but unfortunately I don't think anyone's tried etching it.
We typically etch ITO-coated glass with a mixture of hydrochloric acid/water/nitric acid (4:2:1 by volume) which is fairly fast at room temperature, going through a few µm in about 10 minutes and mask with a positive photoresist.
To your questions:

Will dry film work well for this application?
Because the resolution you need isn't great, it should work fine, so long as the film adheres properly to the ITO.
Underetching is a possibility, but if the features are big enough it should be okay.
The big issue here is trying to use a normal fluorescent light to expose the film.
This is probably hopeless; you're better off finding a proper UV lamp or even using the sun.
How can I determine the thickness of the ITO given its sheet resistance of 50 ohms?
The mixture I mention above is fairly quick, but either way, I wouldn't worry about the precise film thickness (which may actually vary somewhat depending on the batch/supplier).
The easy way to deal with this is to simply pull the sample out of the etchant occasionally and test the resistance with a multimeter in a region where the ITO should be etched away.
On PET, the conductivity should be negligible when the ITO is gone.
Am I going to destroy the PET during this process?
The datasheet you linked refers to the use of PET containers for storage of chemicals—it may actually last a few hours without significant damage, but I haven't tried it.
Is my plan fundamentally flawed?
I think it's worth a shot.
You will need to handle the ITO sheets fairly delicately—the ITO can crack if the sheet is flexed too much, but I don't see this method as being totally unworkable, though you may have issues with the quality of etching you'll get.

As a final note, I wouldn't worry about getting special glassware.
For measuring simple ratios and etching/developing baths, kitchen quality Pyrex dishes and measuring cups work just fine for this kind of thing.
We keep stacks of them in the cleanroom because they're nice and cheap.


text:
Why can't oxygen in furan be sp-hybridized?
From the Chem LibreTexts article on Aromaticity and the Huckel $4n + 2$ Rule:

So far, you have encountered many carbon homocyclic rings, but compounds with elements other than carbon in the ring can also be aromatic, as long as they fulfill the criteria for aromaticity. These molecules are called heterocyclic compounds because they contain one or more different atoms other than carbon in the ring. A common example is furan, which contains an oxygen atom. We know that all carbons in furan are $\mathrm{sp^2}$ hybridized. But is the oxygen atom $\mathrm{sp^2}$ hybridized? The oxygen has at least one lone electron pair and is attached to an $\mathrm{sp^2}$ hybridized atom, so it is $\mathrm{sp^2}$ hybridized as well.

Is this a general pattern - if an adjacent atom is $\mathrm{sp^2}$-hybridized, and if there's at least one electron pair, then our atom it $\mathrm{sp^2}$-hybridized as well?
Why can't oxygen in furan be $\mathrm{sp}$-hybridized? Is this only because the overall structure will not be consistent, because sp-hybridization will make the $\ce{C-O-C}$ line straight, and thus the $\ce{O-C-C}$ angles would have to be lesser than the carbons' $\mathrm{sp^2}$-hybridization would allow?

Is this only because the overall structure will not be consistent, because sp-hybridization will make the COC line straight, and thus the OCC angles would have to be lesser than the carbons' sp2-hybridization would allow?

Yes, this is the correct answer. A very common misconception is that hybridisation can be used to predict the geometry, or that hybridisation somehow involves an energy cost associated with 'promoting' electrons into the hybrid orbitals. This is entirely wrong.
Hybridisation is always determined by geometry. You can only assign hybridisation states to an atom if you already know its geometry, based on some experimental or theoretical evidence. The geometry of the oxygen in furan is trigonal planar and therefore the hybridisation must be $sp^2$.

I think bon's answer is excellent, but I wanted to comment about one piece of your question:

Is this a general pattern - if an adjacent atom is $\mathrm{sp^2}$-hybridized, and if there's at least one electron pair, then our atom it $\mathrm{sp^2}$-hybridized as well?

Whenever students ask "general pattern" kinds of questions, I hesitate, because chemists love to find counter-examples to general patterns.
The point of Hückel's rule (article) is that even heteroatoms can participate in aromatic rings. The specific rule is that if you have an $\mathrm{sp^2}$ conjugated system, the lone pair will be involved if it makes the system more stable. In this case, conferring Hückel $4n+2$ aromaticity.
Consider furan - in the Lewis diagram, you have two lone pairs on the oxygen atom. If we count electrons from the carbon atoms, we have 4 (one per carbon). So adding two electrons from one of the lone pairs will give 6 = 4(1) + 2, so Hückel aromatic.
If we look at thiazole:

Now there are two heteroatoms. The sulfur will act like the oxygen in furan (i.e., as a two-electron donor). So how many electrons does the nitrogen contribute? Do you think the nitrogen is $\mathrm{sp^2}$? Yes, clearly this is also aromatic.

Hybridization is driven by geometry, so ask whether a geometry that drives $sp$ hybridization on the oxygen, including a bond angle of 180°, is plausible in this small ring.  I'm guessing you might find such a geometry would be a bit destabilized by angle strain.


text:
Number of stereoisomers of 1-(2-methylcyclohexyl)ethan-1-ol

How many stereoisomers are possible for the compound in the picture? I tried to draw a rough picture and indicated the chiral carbons. I'm getting 8 stereoisomers are possible but my book gives the answer as 2. Can someone explain where I'm going wrong?

I'm wondering this as well - the only other interpretation I can see is if you had to specify how many types of stereoisomers there are. That would give 2, because you'd have diastereomers and enantiomers but not meso compounds.

Can't there be geometrical isomerism ?

Geometrical Isomerism doesn't apply in this case since there are no double bonds.

But one could be above the plane and one below...there need not be a double bond

True, but that would not change the name of the compound regardless of whether the particular group in question takes an axial or Equatorial position.

The comments on the question are incorrect [edit: the offending comments seem to be gone now], as is the answer given in the book. In general, answers printed at the back of a book should never be taken as the absolute truth. There are typos in all books. Even well-known and well-written textbooks have many errata (usually corrected in subsequent printings).
As you said, there are three chiral centres and therefore 8 stereoisomers.
There are no other elements of symmetry present in this molecule which render any of those 8 stereoisomers equivalent.


text:
What chemical properties that allow for colour exist in the dark?

Of course if there's no light around, there's no colour that you can see. On the other hand, the wall must have some property that makes it be blue. That property is still there in the dark.
— Mike W.

Is Mike referring to the property of excited states, or the property that the states are excitable (via split d-orbitals or conjugated orbitals)?
I understand that anything that we see as coloured but not glowing is transmitting unabsorbed light — the complementary colour of the absorbed light — which is only possible when electrons jumping from ground to excited states.
What I'd like some clarity on is whether those excited electrons fall back to their ground state in the absence of light (another potential reason why colour does not exist in the absence of light). If the electrons do, then I guess the property Mike's referring to is the excitability of the electrons — which makes sense.

Yes, excited states will eventually return to thermal equilibrium. Phosphorescence is a process where you can see the relaxation in the dark, as if the object is glowing.

http://chemistry.stackexchange.com/questions/16633/why-is-gold-golden http://chemistry.stackexchange.com/questions/26749/why-is-snow-white http://chemistry.stackexchange.com/questions/32484/what-causes-the-iridescent-colour-in-laboratory-grown-bismuth

There might be some useful things to note about when electrons will go back to the ground state and at what rate. VTR'ed.

The chemical property that creates colour is the ability to absorb light of a specific visible wavelength. There is more than one way to do this.
Mostly colour is caused by the existence of electronic transitions in substances that match the energy of some wavelength of light so when light hits the substance, some is absorbed by exciting electrons from a lower to a higher energy level. The higher energy levels will not be populated unless there is some energy input to excite the electrons. But the structure that enables this absorption to exist is present whether or not there is any light.
There are other ways a substance can create colour. For example, some compounds do it by interference: the physical structure contains features that have the same scale as wavelengths of some colour of light. When light hits the substance the reflected light creates interference patterns that can selectively eliminate some wavelengths. This is why oil films on water show rainbow colours and why the mineral opal shows beautiful iridescence. again, though, the cause of colour is present whether there is light or not.
In short, the cause of colour is sometimes a chemical thing (involving electrons being excited) and sometimes a physical thing (involving interference) but, either way, the property that creates it exists even when there is no light.


text:
What causes metallic objects to ring?
If I use a iron rod to hit a 3" × 12" iron plate there is the distinct sound of a metallic ring.
What is the cause of this distinct sound? Is it the crystal structure of iron or that iron is a metal and most metallic objects will ring when struck?

This is a physics question, and has to do with mechanical properties of materials. Also, it is not necessarily associated with metals; for example, lead certainly will not ring, while glass probably will.

see this:-http://physics.stackexchange.com/questions/121879/can-i-compute-the-mass-of-a-coin-based-on-the-sound-of-its-fall

Whether something rings when struck depends on the mechanical properties of the bulk material. Ringing depends on the ability of the bulk material to recover from a mechanical impact elastically with little loss of energy. That way the block of material can vibrate for a significant period of time and we hear this because of the interaction of these vibrations with air, creating sound.
The bulk properties of a material depend on both the nature of the material and the crystal structure of the bulk solid. Ringing doesn't happen if the material has ways of dissipating mechanical energy quickly rather than by vibrating elastically. This can be because of a low elasticity but also because the crystal boundaries may create ways to dissipate energy.
So some metals don't ring. Lead makes very bad bells (as does mercury when solid as demonstrated in this amusing video). Many polymers don't ring as they have many internal ways to dissipate vibrational energy quickly. But many elastic materials like brass, iron or glass do ring because they don't have any easy internal ways to absorb vibrational energy.

Internal ways of absorption as in co2 or h20 within its structure? Basically gaps of air?

@user5434678 There are lots of ways for structures to dissipate energy internally. Polymer chains in plastics can reorganise; crystal boundaries can slip or deform; air gaps can reduce material stiffness... And many materials are just not stiff enough so they deform rather than vibrating.

So no other particles in nature can form exactly in the crystal structure of iron or brass?

@user5434678 I wouldn't say that. Plenty of substances have the right mechanical properties but we know about the ones that are cheap and easy to make. BTW the shape of the object matters and the ease of making the shape from the substance will, therefore, matter. It is easy to cast bronze, for example.

How about ringing rocks? They are  uncommon and rare. They ring like metal, what are the mechanics behind these stones?

@user5434678 If rocks have the right shape and the right mechanical properties then there is no reason why they can't ring.

@user5434678 I'm not sure what you mean by "electro scope image".


text:
Acrylic container and electrolysis
Normally when you do electrolysis, you would use a glass container. Would it be wise to use a acrylic container? I have one spare but not sure if this would affect the electrolysis process. Is there any reaction between the acrylic container in the electrolysis process?

Are you performing an electrolysis of water?

yes. doing with other liquid probably is dangerous.

Is my question still on hold?

It isn't now. ;)

Poly(methyl methacrylate), PMMA withstands acids and bases of low concentration. So it should be safe for electrolysis of water.

I would be very cautious with what to electrolyse in an acrylic container. Anything too basic or too acidic will not do any good, neither aggressive oxidizers. This rules out any chlorides for example and anything with metals deposition (as it would leave acidic solution). Sulfates should be fine, though, and hydrocarbons probably too.
I would look at polyethylene and/or polypropylene container if glass is unavailable. These materials are more durable - they withstand most acids and bases in water with no problems and few oxidizers in water can seriously affect them (though chlorine and bromine potentially can - but not right away, it would take some time)


text:
Compute minimum energy paths from arbitrary positions on the potential energy surface
The minimum energy path is a useful concept in understanding chemical reaction paths. As I understand it, such MEP's are most often computed to verify that a transition state structure connects two minima of interest. Several algorithms exist for computing MEP's, but I suppose all have in common that they are able to follow the gradient.
I am in a situation where I think it would be useful to follow the gradient, but not from a TS structure. Can this be done? I think the IRC functionality of software expects to follow the gradient in two directions, and demand the reaction vector.
Is there a simple and common way of computing MEP's from arbitrary positions on the PES? (I will use Molcas). What would happen if I just do a geometry optimization from the arbitrary position? It is my understanding that geometry optimization algorithms displace the atoms in the direction of the force, and in that way a lower energy is guaranteed for each successive step. But will this actually follow the gradient?

http://theory.cm.utexas.edu/vtsttools/neb.html

What would happen if I just do a geometry optimization from the arbitrary position? || More or less. Usually some extrapolation techniques are used to improve convergence speed.

I am not sure, but have a look at technique like metadynamics? You can map entire potential energy surface on it. But I am not sure about MOLCAS

Not all optimization follows the gradient! Actually you should be able to do this with IRC.

There are some possibilities here:

The negative of the gradient of a function always points to the direction of optimal descent (for a discussion, see here).
A method that simply follows the gradient is called first order.
As permeakra pointed out, extrapolation techniques are commonly used for optimizations.
This is because we take finite steps in the PES and this introduces the risk of "jumping" over the minima.
Better approximations are either second order in nature (e.g., the Hessian matrix is calculated in each step) or quasi-second order (the Hessian matrix is approximated or approximately updated).
In order to simply follow the gradient you want the steepest descent-like, first order method with probably a very small maximum step and this can be accomplished by setting the Hessian or the Hessian approximation equal to the identity matrix.
In ORCA (I have never used Molcas, sorry) for example you may use (source):
! Opt
%geom
  inhess unit
  update noupdate
  step qn
end

permeakra also cited the Nudged Elastic Band, which is used for finding minimum energy paths connecting two structures.
This might be very useful and is available for instance in VASP and in NWChem (free).
You may also want to take a direction from one Hessian eigenvector.
This is commonly associated with a molecular vibrational degree of freedom.
This may be accomplished with NWChem.

If you just want to go down the PES from an arbitrary geometry without any jumping, solution 1. is probably the one you want.


text:
Thermodynamic quantities for the freezing of water
I am trying to do the following problem:

Freezing of water at $273~\mathrm K$ and $1~\mathrm{atm}$
Which of the following is true for the above thermodynamics process?
p) $q=0$
q) $w=0$
r) $\Delta S_\mathrm{sys}<0$
s) $\Delta U=0$
t) $\Delta G=0$

My answer was r, s, and t:

Since the reaction happens at constant temperature, internal energy is constant (yes for s)
Since the reaction is open, it is isobaric in nature; and since the volume changes due to the phase change the work done is not zero (no for q)
Heat must be exchanged (first law of thermodynamics) (no for p)
Randomness decreases so $\Delta S_\mathrm{sys} < 0$ (yes for r)

But, the answer given is q, r, and t. I don't understand why q is correct but s is not.

I recommend against closure; IMO sufficient explanation was given of the effort made on the problem.

Eh, there is plenty of justification in the question. No need to close as homework. (@Brian you beat me by a full nineteen seconds..)

@orthocresol

For option s, see this answer: http://chemistry.stackexchange.com/a/39719

Strictly speaking, you are correct that (q) is false, since the densities of ice and water differ at $0~\mathrm{^\circ C}$:

The density of ice is $0.9167~\mathrm{g\over cm^3}$ at $0~\mathrm{^\circ C}$, whereas water has a density of $0.9998~\mathrm{g\over cm^3}$ at the same temperature.
(source)

As the water freezes, the volume of the system expands slightly and thus $w = -P\Delta V \neq 0$ in a strict sense. (I'm using here the convention that $w$ is work performed on the system, not by the system.) However, for the purposes of the problem, the answer was probably written considering that this expansion is small enough to be neglected $(w\approx 0)$.
As for (s), $\Delta U = q + w \neq 0$ even though the process is isothermal. As you indicated, $q\neq 0$ is absolutely true since heat is being transferred. Since $w$ is very small, it has to be the case that there is a change in internal energy of the water/ice. This change is closely related to the enthalpy of fusion of water.

thank you very much sir for replying,thought the volume change is small,how can it be neglected if that is the only one which is non zero? though we can explain change in internal energy being non zero through first law of thermodynamics doesn't it violate the fact that internal energy is dependent only on temperature?isn't enthalpy of fusion  not the heat transferred during the reaction,please explain how is internal energy related to it.

@AjaySabarish You're interpreting the phrase internal energy is dependent only on temperature too strictly.  There absolutely is a change in internal energy as a substance undergoes a phase change.

@AjaySabarish In practical application of the sciences, one often must decide whether a phenomenon is of sufficient magnitude to actually affect the system of interest.  I would guess the $PV$ work done by freezing water to be at most a few Joules per kilogram of water, which is far less than the heat transfer required to freeze it. Thus, while rigorously it is the case that $w\neq 0$, practically it might as well be the case that $w=0$.

sir then when can we use the expression internal energy is dependent only on temperature?

@AjaySabarish Let's continue this discussion in chat: http://chat.stackexchange.com/rooms/39018/discussion-of-a-problem-from-thermodynamics

s is not correct because U is not just a function of temperature but also a function of chemical composition.  The internal energy of liquid water is higher than the internal energy of ice.  If you want to think of it in terms of temperature, then just view the heat capacity as being infinite at the phase transition (Dirac delta function of T), such that the integral over the phase change is equal to the heat of fusion.  As far as q is concerned, I guess they are assuming that the change in volume from solid to liquid is negligible.

sir please explain your answer in simple terms iam just a high school student and i don't know what is dirac delta function.and i have read in my books that internal energy is dependent only on temperature.is it also depended on chemical composition?please explain it sir.


text:
Why does this flame travel (in a circular path, multiple times)
We filled a graduated cylinder with Ever Clear and let it overflow. Then we lit the liquid on fire. To our surprise the flame started running around the base of the cylinder. Why does this happen?
http://webmshare.com/m7vb8

Very curious. Too bad the video isn't embedded, but it's cool.

Cross-posted to Physics SE as well.

I have removed it from Physics SE

I can't decide if I can see the cylinder rocking slightly or not? Possibly the unstable cylinder was set in motion by the combustion and the bright flame is on the lifted edge where more oxygen can get to the fuel.

A flame is a dynamic phenomenon, using up fuel and oxygen in places and driven by convection and diffusion. Flames are sensitive to factors such as external sound, and can be used to measure frequency and waveform with a Rubens tube.
Under micro-gravity, flames burn quite differently, due to the lack of convection.
So though not a specific answer to your questions, I assume that oxygen is being used up rapidly in one spot, and the flame front moves along as convection brings in fresh oxygen. You could test this hypothesis by piping in just a little air in one spot on the "ring of fire" to see if it stabilizes the flame there.
Thanks for sharing that interesting video!


text:
Why is the lone pair of pyridine's nitrogen atom not part of the aromatic ring?
In pyrrole, the lone pair of electrons belonging to the nitrogen is part of the aromatic ring. However, in pyridine it is part of an sp2-hybridized orbital. Why can't it be in the p-orbital and take part in the aromatic ring?
I mean, why can't the lone pair in pyrrole be in the p-orbital, with the sp2-hybridized orbital (which is not bonded to anything) sporting a single electron?
I'm not talking about the nitrogen atom having two p-orbitals participating in the aromatic ring. That would be strange. I wonder why the existing p-orbital of N cannot carry two electrons, the way it does in pyrrole.

P.S. A related question, with a very interesting answer: sp2-hybpidization in pyridine and pyrrole

Because there is already a $p$ orbital of nitrogen which takes part in the aromatic ring. You can't have two $p$ orbitals on one atom pointing in the same direction. And you can't have orbital take part in the ring if it is oriented differently.

Huckel's rule demands that there be 4n+2 pi electrons in the ring. Each carbon contributes one, making for a total of 5 from the carbons. If your nitrogen lone pair is in the ring, then you will have seven pi electrons. $7 \neq 4n +2$ for $n \in \mathbb{Z}$

@orthocresol - so it's solely to satisfy Huckel's rule? And the same applies to pyrrole - it cannot stick out the lone pair, because that way the aromatic ring will be disrupted, so it has to use the lone pair in the aromatic ring?

If your lone pair is pointing upwards, where would be your existing p orbital on N? That's my point, and it comes before counting electrons for Huckel.

@IvanNeretin - I worded my question badly in the beginning. My whole point was, why can't the lone pair be instead in that p-orbital poining upwards - instead of its real position, which is i the sp2-orbital.

Yes, the fact that you put one electron in the p orbital and two in the sp2 orbital is solely to satisfy Huckel's rule. For pyrrole, there are only four other carbons which contribute one electron each, and the nitrogen therefore has to give two of its own in order for it to reach $4n+2$. If the lone pair sticks out, the aromatic ring would not be "disrupted". The ring would just not be aromatic. Also, the sp2 orbital in pyrrole is bonded to a hydrogen 1s.

Well, then it's Huckel's rule alright. Also, if you'd have a pair on $p$, you'd have an unpaired electron on that $sp^2$, which is not nice.

Might find useful? http://www.chem.ucla.edu/~harding/tutorials/lone_pair.pdf

@orthocresol can you please explain this question to me ? The OP asks why can't the lone pair in pyrrole be in the p orbital.... But it already is! The lone pair is in the p orbital taking part in resonance.

@Ali I don't know what you're saying. The question is clearly about pyridine.

@orthocresol In the description, the OP says "I mean, why can't the lone pair in pyrrole be in the p-orbital, with the sp2-hybridized orbital (which is not bonded to anything) sporting a single electron"

@Ali I assume that's a typo and that the OP meant pyridine. It fits in with the general question.

@orthocresol I thought moderators can edit a question? Also ,no one bothered to correct a question having 12 upvotes ? So people don't actually read the description and just answer with whatever general view they have about the question ?

@Ali Anybody can edit a question, including you and I. I don't really understand why you're asking me about this, or what you hope to accomplish by this, because I can't control how other people behave on this website. How am I supposed to tell you what runs through the minds of every user here and why they didn't edit it or whether they read it or not?

@orthocresol I am asking you about this because -1. Because you are a moderator. I thought moderators had special edit privileges . My edits would be peer reviewed , yours wouldn't be. 2. Because you were involved with the question a long time ago 3. Because I was surprised that a question with so many upvotes has an error , although a trivial one and nobody pointed it out even in the comment section. That is unlike stackexchange. Major reason is point 3 ,not because I wanted to argue over some typo.

As follows from our discussion in the comments to the question, pyridine uses only a single electron in the p-orbital in order to comply with Huckel's rule. Kudos to Orthocresol:

Huckel's rule demands that there be 4n+2 pi electrons in the ring. Each carbon contributes one, making for a total of 5 from the carbons. If your nitrogen lone pair is in the ring, then you will have seven pi electrons. $7≠4n+27≠4n+2$ for $n ∈Z$

Of course, electrons know nothing of Huckel and his rule, but apparently they position themselves into the most energetically suitable configuration. I am not yet qualified to explain why exactly it is the most suitable one.

The key point is mentioned here! Electrons position themselves in the most suitable configuration, and if you look at the MO Diagram of Pyridine it becomes clear why the line pair will sit in Nitrogen rather than in the p Orbital

@ITTsoi - that may well be so, and I've been shown an MO diagram in the chat, but I haven't yet mastered the MO theory. If I do that, I'll expand my answer. Or maybe someone else will post an additional answer with an MO-based explanation.

Huckels rule is irrelevant and the orbitals are not "single" electrons.

It has the wrong geometry
The other P orbital can't participate in the aromatic ring because it has the wrong geometry: it points in a direction orthogonal to the orbitals involved in the aromatic part of the ring. The picture in the question shows this clearly.
There is already a P-orbital on nitrogen involved in the aromatic system of the ring (the one that points "up"). The orbital coloured yellow-ish in the picture doesn't overlap with the aromatic system at all as it points "out" from the ring and (if you did calculations) would have no overlap with the other orbitals in the the aromatic system. No overlap means it can't be involved in aromaticity. It is, ultimately, simple geometry and no other explanations are required.


text:
How to convert from w/w% to molarity?
After reading a number of explanations about it after googling, I'm a bit confused on this. Some explanations are doing it using a solution density, others aren't, though most are. On one site it's showing different densities for a solution of a compound at different w/w% values. I'm now very lost as to how you convert from a w/w% to molarity. Can someone clarify?

Weight percent is essentially the same as (I arbitrarily picked grams as the mass unit):
$$
\%~\mathrm{w/w~X} = \mathrm{g~X\over g~solution}
$$
In order to get to the $\mathrm{mol\over L}$ units of molarity, you have to convert by multiplying by the solution density, $\rho$, and dividing by the molecular weight of $\mathrm X$, $M_\mathrm{X}$:
$$
\mathrm{g~X\over g~solution} \times \stackrel{\left(1\over M_\mathrm X\right)}{\mathrm{mol~X\over g~X}}
 \times \stackrel{\left(\rho\right)}{\mathrm{g~solution\over L~solution}}
 = \mathrm{mol~X\over L~solution}
$$
One key thing to remember is that $\rho$ is the density of the solution, not of the pure solvent. If you're working with dilute solutions (below $\sim\!100~\mathrm{mM}$ for aqueous) these two densities will be similar, but in concentrated solutions they will diverge.

I'd rather write mass fraction $w(\ce{X}) = m(\ce{X})/m(\text{solution})$, which is obviously not in percent. I also would call it (amount) concentration, $c(\ce{X}) = n(\ce{X})/V(\text{solution})$ for clarity. Then you can also easily see that the density of the solution is $\rho(\text{solution}) = m(\text{solution})/V(\text{solution})$. Your dimensional analysis works fine though. (In your first equation you need to multiply the right hand side with $100%$, otherwise you're using mass fraction.)

Use the density of the solution to get the total volume of the solution.
Then use the weight percent of solute to determine the amount of substance of the solute.
Use the amount of substance of the solute divided by the volume to get molarity.

Molarity is relevant to our concept of chemistry and kinetics of reactions.  However wt/wt% is frequently easier to use on an industrial scale.

W/W% is just the same as a any percentage but now the difference is that let say that a concentrated solution contains  67% w/w ammonia that means that there is 67g of ammonia in 100g of the solution;
now let us consider a question

A concentrated solution of aqueous ammonia is $28.0\%\text{w/w }\ce{NH3}$ and has a density of $\pu{0.899 g/mL}$. What is the amount concentration of $\ce{NH3}$ in this solution?

Converting $28.0\%\text{w/w}$ to ratio form gives us $28/100$ then you multiply it with the density $\pu{0.899 g/ml}$ for ammonia and $\pu{17.04 g}$ is an equivalent of one mole.
$$28/100 \times \frac{\pu{0.899 g}}{\pu{1 ml}} \times \frac{\pu{1 mol}}{\pu{17.04 g}} \times \frac{\pu{1000 ml}}{\pu{1ml}}$$


text:
How to dissolve iron powder?
I know iron can react with many solutions, but when I tried to mix iron powder together with permanganate or dilute sulphuric acid, the iron powder would not react and instead, it just float on the surface of the solution. I tried to stir it for several minutes but it seems doesn't work.
It is pure iron powder, can be attracted by magnet
How can I dissolve iron powder and let it become $\ce{Fe^2+}$ or $\ce{Fe^3+}$ ? (Just dissolve it is ok)

To float on the surface of a water solution (density slightly above 1) is quite a peculiar thing to do for iron (density 7.8). Isn't the powder stained with some kind of oil, by any chance?

@IvanNeretin I've seen iron filings do this, and stirring (with a magnetic stir bar, at least) isn't exactly the best way to break surface tension at the top of water.

Then we have to lower the surface tension. I think a little drop of acetone might help.

As far as the question goes, I don't think you're going to get it to dissolve without making $\ce{Fe^{2+}}$ or $\ce{Fe^{3+}}$. There may be some wierdo $\ce{Fe^0}$  complexes that would do it, but those would be the exception. I'm not even sure it would work at all.

what is the iron powder made of? just pure elemental iron in its various oxidation states? I don't know how iron bonds so I'm not sure if it exists elemental or not :/, but if I can get a little bit of information about this I think I can help with the dissolving problem

I've tried to use magnetic stirrer to mix them, but the result would be the magnetic stirrer attracts most of the iron powder, while the remaining iron powder are just mixed with the solution physically

The last time I had to disolve iron powder, I placed the iron powder in a beaker, put the beaker down on a couple of strong magnets (Fe-Nd-B), then added the reactant to the beaker. The magnets hold the iron down and are under the beaker, so they are not corroded, etc. Alternatively, put a PTFE-coated magnetic stirring bar in the beaker with the iron powder. Then add reactant solution. This obviously would facilitate stirring in the customary way.

I did some brief research and I saw that sulfuric acid can dissolve iron:
Fe(s) + H$_2$SO$_4$(aq) → Fe$^{2+}$(aq) + SO$_4$$^{2-}$(aq) + H$_2$(g)
Of course, this would leave sulfate ions and perhaps some precautions would need to be taken because of hydrogen gas production. If you absolutely cant have any sulfate ions, you could precipitate them out by adding barium or silver ions to the solution, but that would require extra work and introduce new ions to the solution. Also note, the iron ions should be aqueous in +2 state but I read that if there is oxygen in the system (which there naturally would be), some iron(II) can oxidize to iron(III).

dissolution of iron in diluted acid is painfully slow.

The reaction of metallic iron powder with sulfuric acid solution is rather slow. But it produces at least some Hydrogen bubbles that are adsorbed against the iron powder. This may be sufficient for maintaining the iron powder at the surface of the solution.

(+1) Still, back in 1783, Jacques Charles and associates filled balloons with hydrogen that they generated via the reaction of iron and sulfuric acid. Brave folks!

@Ed V, Right. But Charles and Gay-Lussac used rather concentrated sulfuric acid : 70% if my memory is correct. With 2 molar solutions, as used in high school laboratories, the reaction is desperately slow.

I fully agree about the slowness! I first did this reaction many years ago and it was pathetically slow. And impurities in the iron powder I used made a nasty smell. Whenever I needed hydrogen gas in my basement chemistry lab, I used either zinc or manganese with non-concentrated sulfuric acid. Or I used aluminum granules with NaOH (fairly concentrated). Lucky I never had a major incident! ;)

@Ed V. I agree with you. And even pure zinc does not react as quickly as desired. It is better to add a few drops of cobalt chloride solution. This produce an electrical cell effect, and the reaction becomes tumultuous. I give you 1 point.

Ah, many thanks for the cobalt chloride tip!


text:
The potential range of the cyclic voltammetry of CoCat in KPi made of D2O
I have a phosphate buffer adjusted to pH 7, and another phosphate buffer made of deuterium oxide (D2O) adjusted to pH 6.6 (pD 7). I want to conduct cyclic voltammetry of cobalt oxide catalyst in these buffers.
I want to apply the following potential range for the buffer made of normal water: 0.9 to 1.25 V vs. NHE. So, my question is here: shall I apply the same potential range for the buffer made of D2O? or I should take the pH difference into account*?
*The pH difference would be: (7-6.6)*0.059 = 0.0236 V.
Then the potential range for buffer made of deuterium oxide would be: 0.9236 to 1.2736 V vs. NHE.


text:
How does spin flipping of triplet carbenes occur?
Below is presented a page from Clayden Organic Chemistry 2ed.

How and why does the spin flipping during the collision with solvent molecules actually occur?

"How" does not have a proper answer as in quantum physics there is no good way to describe interaction of two systems, but only one system. And for one system the question has no answer.  "Why" does have an answer: it is because spin flip requires consumption/release of energy that must go to/from somewhere.

The same factors that relax forbidden singlet-triplet spectroscopic transitions apply here.  See this earlier answer for some discussion of this topic.

Could you write an answer?

In these reactions you are not observing optical transitions between singlet and triplets, but the interconversion of a triplet to singlet state in competition with rotation of part of the molecule. Only in the singlet can a bond be formed as now each electron has a unique set of quantum numbers. As no photons are involved it is necessary for the triplet and singlet levels to become effectively equal in energy before a transition can occur. Thus there has to some kind of perturbation for this to happen and this has to accommodate one unit of angular momentum change, the difference between singlets and triplets. It can be useful to visualise singlet & triplets with a vector model. This is shown below. The arrows show the spin vectors of the two electrons in a triplet and singlet state.

The triplet state corresponds to the three images on the left and the singlet to that on the right. If you look at the triplet with $m_s=0$ you will see that to convert it to a singlet all that is needed is a re-phasing of the spins from parallel to anti-parallel.  In the diradicals you study, the electrons are rather distant from one another thus their interaction is weak, and the difference in energy between singlet and triplet small. This means that only a small perturbation can cause a crossing between levels. This can, for example, be produced by an external magnetic field which raises and lowers different triplet energy sub-levels and so reach the singlet's energy. A nearby solvent molecule can produce this energy perturbation as it also has electrons and it will be diffusing by rotation and translational motion, i.e. producing a time and spatially varying magnetic field. Similarly the diradical is rotating with whole body motion as are segments of it and in doing so presents different orientation of the electrons to the earth's magnetic field which may also induce changes in energy levels and triplet-singlet crossing. These interactions will also have to involve orbital angular momentum.The time scale for rotation will be sub-nanosecond (in most normal mobile solvents) but only occasionally will a perturbation occur at the same time as the groups are proximal, because segmental rotational diffusion will mainly keep them apart, and the low frequency of perturbations of sufficient magnitude and the geometric requirement just mentioned are likely to make the reactions slow.

How does this answer the question about the solvent molecules?

as I mention as a solvent molecule contains electrons then it has a weak magnetic field as it is moving and this can cause a spin flip.

also I forgot to mention that (a) heavy atoms in solvent, such as Xe gas,  I$^-$ or as an iodo solvent will cause spin-orbit interaction. Chlorinated solvents will also do so but to a lesser extent. (b) The solvent also mediates the interaction reciprocally due to its dielectric constant, the larger this is the smaller the interaction. (c) As the carbene tumbles in solution it cuts through the earth's magnetic field, thus it effectively experiences a random field, this also will induce transitions.


text:
How should I construct the orbitals of dilithium using only pen and paper?
For a week I have lessons about quantum chemistry and I have a trouble understanding how to arrive at the good answer for the drawing of the orbital $(\ce{III})$ for $\ce{Li2}$ (see the second picture).
In the first picture I did an example of my reasoning with a basic molecule such as $\ce{LiH}$ just to show you how I figure it must work.
And in case someone would like to mention it, yes I haven't drawn the $\mathrm{p}_x$ and $\mathrm{p}_y$ orbitals because it is not the point here.

With LiH

With $\ce{Li}_2$

The question is on this last picture, if you have trouble to read tell me.

Why are you using p-orbitals at all? $\ce{Li2}$ can be perfectly explained with only s-orbitals and the s-p energy difference is significant …

Ugh... I'm just asking how I should make my reasoning to draw the orbitals when there are interaction with four orbitals and why what I did doesn't match with what I must find. Answer me with $\ce{N}_2$ if you prefer or whatever you want...

Ah, that is why.

I have not the energies of these orbitals with me. The teacher said it works from Li to N so I chose the most light of them it is simpler to carry my sheet haha.

When combining orbitals, your first approximation should be to relate similar ones with similar ones. In this case, on each lithium (or nitrogen or fluorine) atom we have:

1 s-orbital
1 p-orbital in the direction of bonding
2 p-orbitals perpendicular to the bonding axis.

We should treat these separately initially, see what the results give us, and then think if we should combine them.
Realise that the point group of the molecule is $D_\mathrm{\infty h}$. However, that is a bad point group to work with. Instead, assume $D_\mathrm{2h}$ and correlate your results back to $D_\mathrm{\infty h}$ according to a table originally presented here:
$$\begin{array}{cc}
\hline
{\text{representation in }D_\mathrm{2h}} & {\text{representation in }D_\mathrm{\infty h}}\\
\hline
A_\mathrm{g} & \Sigma_\mathrm{g}^+\\
B_\mathrm{1g} & \Sigma_\mathrm{g}^-\\
B_\mathrm{2g} + B_\mathrm{3g} & \Pi_\mathrm{g}\\
A_\mathrm{g} + B_\mathrm{1g} & \Delta_\mathrm{g}\\
B_\mathrm{1u} & \Sigma_\mathrm{u}^+\\
A_\mathrm{u} & \Sigma_\mathrm{u}^-\\
B_\mathrm{2u} + B_\mathrm{3u} & \Pi_\mathrm{u}\\
A_\mathrm{u} + B_\mathrm{1u} & \Delta_\mathrm{u}\\
\hline\end{array}$$
Ignore the difference between $\Sigma^+$ and $\Sigma^-$ for now. What is relevant is that both the two s-orbitals and the two p-orbitals along the bond axis transform as $A_\mathrm{g} + B_\mathrm{1u}$ in $D_\mathrm{2h}$, and thus as $\Sigma_\mathrm{g}^+ + \Sigma_\mathrm{u}^+$ in $D_\mathrm{\infty h}$. Which means that we expect a bonding and an antibonding orbital from both — no surprises here. So initially, assume (as it looks you tried to do) a pair of $\unicode[Times]{x3c3}$ and $\unicode[Times]{x3c3}^*$ orbitals for both.
Since the energy difference between s- and p-orbitals is large enough (you can and should always assume that) the antibonding $\unicode[Times]{x3c3}_\mathrm{s}$ orbital should still be lower than the bonding $\unicode[Times]{x3c3}_\mathrm{p}$ orbital. These two now have a different symmetry and thus cannot mix productively.
Fill in your electrons. You have two, which will end up in the lowest $\unicode[Times]{x3c3}_\mathrm{s}$ orbital.

What can still mix is $\unicode[Times]{x3c3}_\mathrm{s}$ with $\unicode[Times]{x3c3}_\mathrm{p}$ and their corresponding antibonding orbitals. This will lead to the fidgety task of determining how much stabilising interaction can still be found. Assume the $\unicode[Times]{x3c3}_\mathrm{s}$ to be stabilised slightly by $\unicode[Times]{x3c3}_\mathrm{p}$, because the former is populated while the latter is empty. That should give the $\unicode[Times]{x3c3}_\mathrm{s}$ orbital a slight look as if it derived from sp hybrid orbitals. (Only weakly though!) Note however, that this secondary mixing can be ignored at first approximation.
Also note that any of this mixing must result in symmetric or fully antisymmetric orbitals with respect to the plane of symmetry perpendicular to the bond axis ($\sigma_\mathrm{h}$).

Taking all this and reviewing your picture:

Your lowest orbital looks good. A fully bonding orbital of the s-type with very slight hints of a p-orbital mixing in.
Orbital 2 should look much more like an antibonding $\ce{H2}$ orbital. Since both antibonding orbitals are unpopulated, mixing is unproductive. If you must, very slightly add a bit of p-contribution on the inner side of the atoms.
Orbital 3 should look almost like orbital 1, but clearly with a much larger p-look; so have two nodal planes, one through each atom. Once you get to and past diboron, this is the primary bonding orbital.
Orbital 4 is technically good. However, as per the initial approximation both lobes on each atom should be equal, and only due to the secondary mixing should the inner lobes be slightly smaller. Draw it, as you may be accustomed to $\mathrm{(sp^3)^*}$ hybrid orbitals.

If you want a shortcut, remember to consider s and p separately. Two s-orbitals mix as you know it from hydrogen, creating a very big spherish blurb and an antibonding double-sphere. Two p-orbitals mix to give something like your orbitals I and II (bonding and antibonding). The two s should be lower than both the p.
In a second step, you can mix the two bonding orbitals (and the two antibonding orbitals; but not bonding with antibonding) to add a little p-contribution to your s-type orbitals. However, the lower should clearly always remain an s-type (with a little p mixed in).

I have not learnt anything about point group yet and I don't understand why your answer is so complex, what I was asking is (maybe it wasn't enough clear) is for the orbital $\ce{III}$ what are the steps to get the right answer and not what I did? I guess your answer must be clear for someone who has a lot of knowlegde about this but for me it is really hard sorry. For the moment I don't want to know exactly what all of them should look like just how step by step can I find the good shape of the third orbital :-) after I'll do more complex things ^^

So to make it more clear it is why for the third MO, the two s-orbital are not in opposition of phase like I did in my reasoning because their energy will be higher than at the beginning?

It’s complex because I love being complex O:-) But did you follow the link to the other question (and answer)?

Yes I followed them but as I said we still have not do it in class and so we must be able to do it without them. And I don't understand all your shortcut. At least I would like your answer to be only this short cut but with more details or images too because I have some troubles to understand english sometimes.

I found how to do but for me your "answer" is not the answer of my question about how should I mix (and which ones) to obtain the diagram.


text:
What will be the reaction between tin (II) chloride and copper (II) chloride?
I just tried conducting an experiment during lunch today. Now, I calculated this reaction 4 days ago dealing with electrochemistry,
$$\ce{SnCl2 + CuCl2 -> Cu(s) + SnCl4}$$
Theoretically, I got these 1/2 cell reactions from a couple different sources (one of which are posted below):
$$\ce{Sn^2+ -> Sn^4+ + 2e-} \quad E° = -0.15~\mathrm{V}$$
$$\ce{Cu_{(s)} -> Cu^2+ + 2e-} \quad E° = -0.34~\mathrm{V}$$
Using basic electrochemistry, we know that Copper prefers to be copper metal. Due to its overbearing power over the tin(II) ion, the copper(II) ion forces the tin(II) ion to become tin(IV).
I did not check the reduction or oxidation potential of the chloride ion (since the chloride ion is the most stable form of chlorine due to its (relative to the other compounds in this experiment) high ionization potential. I concluded that chloride would be a spectator ion.) Finally resulting in this net ionic reaction.
$$\ce{Sn^2+ + Cu^2+ -> Cu_{(s)} + Sn^4+} \quad E^\circ_{\rm cell} = 0.19~\mathrm{V}$$
I mixed the compounds in $150~\mathrm{ml}$ of deionised water. Later combining them only to end up with a milky-green colored solution. I saw some brown-red precipitate at the bottom of the beaker. however, only a small amount of what I think is copper precipitated. Tomorrow I will place the beaker with the solution onto a hot plate in hopes of speeding up the reaction.
I was wondering if a smaller difference in electrical potential between the two half cells results in a slower reaction. The other thing I was wondering concerns any other ideas or misconceptions I may have faced while conducting this experiment.

The reactions taking place are probably quite complex. First there's also copper(I) involved ($\ce{Cu^{2+} + e- -> Cu+}, E° = +0.16~\mathrm{V}$.) Second the formation of complexes and precipitates that depends on pH and anions present.

Tin 2 in solutions is oxidized by air, and fine copper powder too.

This is indeed a redox reaction between copper and tin, but you should instead use reduction potentials for the $+2$ ions to calculate the potential.
This might be quite an exothermic reaction. http://www.periodicvideos.com/videos/mv_copper_chloride.htm


text:
Effect of His-tag on enzyme activity
In my biochemistry laboratory class, I designed an experiment to study the effect of a His-tag on enzyme activity. First, I measured the activity of the His-tagged enzyme. Then, I cut off the His-tag (using enterokinase) and measured the activity again.
After data analysis, I was able to calculate the $K_\mathrm{M}$ values and $k_\mathrm{cat}$ values for both enzymes. How should I interpret the data? I hypothesize that the His-tag could interfere with kinetics. But, how do I use the data to show this? Should I compare $K_\mathrm{M}$ values or $k_\mathrm{cat}$ values?

People would usually compare three numbers: $K_M$, $k_{cat}$, and $k_{cat}/K_M$.  If any of those numbers are significantly different for the wild-type enzyme compared to the His-tagged enzyme, then you could conclude that the His tag interferes, at least somewhat.

I assume the excised-off His-tag as well as the enterokinase protease are removed from the system.
I would use the catalytic efficiency, i.e. the ratio of kcat/KM as the metric to compare the results of two conditions.


text:
Possible E2 products of S-2-bromo-2-methyl-1-phenylbutane
I have been asked to predict wether the following reaction will proceed by elimination or substitution.
$\hspace{2.4cm}$
I believe that the reaction would proceed by the $\mathrm{E2}$ mechanism. I am confused, however, as to what products what would be formed from this reaction. Currently I believe that there would be a total of five products: 2 sets of E/Z isomers, and a product with a double bond with no carbon substituents on one end (i.e along the dashed bond). Is this the case?

The major products will be (E)-2-methyl-1-phenyl-1-butene and (Z)-2-methyl-1-phenyl-1-butene, formed in a 1:1 ratio. There is are reasons both for why elimination occurs specifically across the first and second positions, and as for why the product is formed in a 1:1 ratio.
The hydrogen on position 1 are called benzylic hydrogen, and are much more acidic than aliphatic hydrogen due to the resonance stabilization of the benzyl anion. This will cause these hydrogen to be removed much more quickly than the others, resulting in a majority of 1,2-unsaturated products. This is of course not to say that the other products are not formed, but they are to a lesser extent.
The products are formed in a 1:1 ratio because there are two ways for the $\ce{H}$ and $\ce{Br}$ to be anti-periplanar. This is easily seen when constructing the Newman projections of the starting material:
$\hspace{3.6cm}$
Upon extraction of the hydrogen, the electron pair forms a $\pi$-bond with the adjacent carbon, ejecting the $\ce{Br-}$, and locking in the stereochemistry.

Actually, with tert-butoxide as  a base, the less substituted Hofmann product will be favored.


text:
Why is pyrimidine less basic than pyridine?
From Wikipedia's article on pyrimidine:

Because of the decreased basicity compared to pyridine, electrophilic substitution of pyrimidine is less facile.

But why is pyrimidine less basic than pyridine? Pyrimidine has two $\mathrm{sp^2}$-hybridised lone pairs available for protonation, compared with pyridine's one.

It is not the number of lone pairs that in any way explains basicity. Take a random sugar and it will have ten times the number of lone pairs (albeit on oxygen, not on nitrogen) without being significantly basic.
The problem is electronics. Nitrogen, being a rather electronegative atom, is able to draw the π electrons towards it well — to the extent that pyridine derivatives are termed extremely electron poor and thus very slow in electrophilic substitution reactions.
Going from pyridine to pyrimidine we need to double that. Thus, there is a perceived severe shortage of electrons in that ring since both nitrogens are trying to draw them towards themselves.
Now assume you protonate that species. We now have two nitrogens of which one is positively charged. Immediately, that positively charged nitrogen multiplies its electron-withdrawing force. Equally immediately, the other nitrogen feels electron density leaving its vicinity. The whole system is much less favourable than the pyridine system with only a single nitrogen.
Rephrasing that argument: Basicity is increased when electron-donating neighbours increase the electron density (that equals a partial negative charge) on an atom. In pyrimidine’s case, we have an electron-withdrawing neighbour that reduces the electron density (giving a partial positive charge) and making protonation (i.e. more positive charge) less favourable.

One more thing worth noting, the lone pairs on nitrogen atoms cannot be used to increase the electron density of the ring, as no resonance structure will accept the lone pair on any of the nitrogens.

Come to think of it, there are certain compounds in which several lone pairs cooperate nicely to enhance basicity (guanidine comes to mind). Not the case here, though.

@IvanNeretin Or proton sponge, a.k.a. 1,8-bis(dimethylamino)naphthalene as another prototypical example.

As Ivan mentioned, guanidine is a very strong base. When is this reasoning applicable? Maybe this is how chemistry is and I should probably just memorise it up.

@user600016 The conjugate acid is a highly symmetric structure. The lone pairs and the double bond resonate in what Martin has called Y-aromaticity. Overall, it’s a good structure for a protonated species, so it is protonated easily.

Guanidine/pyridine = apples/olives.

In pyridine, the hybridisation of nitrogen atom is $\mathrm{sp}^2$. The electron pair on nitrohen lies outside the ring on an $\mathrm{sp}^2$ hybrid orbital and is available for protonation, making pyridine a basic heterocycle.
The $\mathrm pK_\mathrm{a}$ of the conjugate acid of pyridine is $5.25$

In pyrimidine, the nitrogen atoms are equivalent and $\mathrm{sp}^2$ hybridized. Both electron pairs lie outside the aromatic ring on $\mathrm{sp}^2$ hybrid orbitals. Both N are slightly basic. Pyrimidine is less basic than pyridine because of the inductive, electron-withdrawing effect of the second N atom.
The $\mathrm pK_\mathrm{a}$ of the conjugate acid of pyrimidine is $1.3$


text:
Application of limiting reagents and acid-base stoichiometry
I'm having a bit of trouble with the latter questions of this problem.

$\pu{15.0 mL}$ of $\pu{1.4 M}$ $\ce{HCl}$ was mixed with $\pu{1.00 g}$ of $\ce{CaCO3}$ until all the solid had dissolved. The solution was then transferred to a conical flask and made up to $\pu{200 mL}$ with water. A $\pu{20.0 mL}$ portion of this solution was then neutralized by $\pu{8.50 mL}$ of a $\pu{0.100 M}$ $\ce{NaOH}$ solution.
  Calculate:

Amount of substance in excess of $\ce{HCl}$ in the $\pu{20.0 mL}$ portion
Amount of substance in excess of $\ce{HCl}$ in the $\pu{200.0 mL}$ portion
Amount of substance in excess of $\ce{HCl}$ which reacted with $\ce{CaCO3}$

My approach
$$\ce{2HCl(aq) + CaCO3(s) -> CaCl2(aq) + H2O(l) + CO2(g)}$$
Finding the amount of substance of each reactant and hence the limiting reactant,
For $\ce{HCl}$:
\begin{align}
M[\ce{HCl}] &= \frac{\mathrm{mol}}{\mathrm{L}}\\
\mathrm{mol}[\ce{HCl}] &= \mathrm{M}\times \mathrm{L}\\
\therefore \mathrm{mol}[\ce{HCl}] &= 1.4\times (15.0\times 10^{-3}) = 0.021
\end{align}
For $\ce{CaCO_3}$:
\begin{align}
\mathrm{mol}[\ce{CaCO_3}] &= \frac{\mathrm{m}}{\mathrm{mm}}\\
\mathrm{mol}[\ce{CaCO_3}] &= \frac{1.00}{100.0869 }\\
\therefore \mathrm{mol}[\ce{CaCO_3}] &= 9.9913\times10^{-3}
\end{align}
Therefore (after dividing by stoichiometric coefficients) $\ce{CaCO_3}$ is limiting.
From here I'm not sure if what I did next is correct. I found the amount of substance of $\ce{CaCl_2}$ produced the remaining $\ce{HCl}$ in excess. Which were,
\begin{align}
\mathrm{mol}[\ce{CaCl_2}] &= 9.9913\times10^{-3} \text{ and,}\\
\mathrm{mol}[\ce{HCl}] &= 1.1009\times10^{-2}
\end{align}
(Meaning that we have $9.9913\times10^{-3}\mathrm{mol}$ of $\ce{CaCl_2}$ and $1.1009\times10^{-2}\mathrm{mol}$ of $\ce{HCl}$ in in $\pu{15 mL}$ of solution?)
From here on I'm stumped.

I understand the procedure but not the questions. So there are two reactions:
$$\ce{2HCl(aq) + CaCO3(s) -> H2CO3(aq) + 2 Cl-(aq) + Ca2+(aq)}$$
$$\ce{HCl(aq) + NaOH(aq) -> H2O(l) + Cl-(aq) + Na+(aq)}$$
The first reaction has reactants in non-stoichiometric amounts. The second reaction, a titration, is done with stoichiometric amounts, yielding a neutral reaction mixture.
The strange thing is that the problem is over-determined: I can calculate the amount of HCl not neutralized by calcium carbonate either by looking at the amounts of the two (assuming the reaction goes to completion) or by the titration (assuming that neutralizing means to add sufficient NaOH to be at stoichiometry ratio with the remaining HCl).
There is also a technical detail: it would be better to boil off the carbonic acid made by protonating the carbonate. Otherwise, the analyte will contain strong and weak acid, and buffer will form during the titration (with the endpoint of the titration ill-defined: is it when the HCl is depleted, or when all acid is depleted?).
However, it seems like the question wants us to make an argument about amounts starting with the titration. We will assume that the titration and the reaction between $\ce{CaCO3 and HCl}$ go to completion, and chalk up (get it?) any inconsistencies to impure $\ce{CaCO3}$.
Amount of NaOH
$$c_{\mathrm{\ce{NaOH}}} = 0.100\  \mathrm{M}$$
$$n_{\mathrm{\ce{NaOH}}} = c_{\mathrm{\ce{NaOH}}} \cdot V_{\mathrm{\ce{NaOH}}}$$
$$\ \ \ =0.100\  \mathrm{M} \cdot 8.50\  \mathrm{mL}$$
$$\ \ \ =8.50\times 10^{-4}\  \mathrm{mol}$$
Amount of HCl
The amount of HCl in the 20 mL sample was $\pu{8.50e-4 mol}$ (from the titration). So the amount in the entire 200 mL sample after adding carbonate was $\pu{8.50e-3 mol}$. The amount used up by reacting with carbonate is the difference between the original amount, $\pu{0.021 mol}$ or $\pu{21e-3 mol}$, and what is left after the first reaction.
$$n_\mathrm{reacted} = \pu{21e-3 mol}  - \pu{8.50e-3 mol} = \pu{12.5e-3 mol}$$
Amount of $\ce{CaCO3}$
As the OP writes, there is a 2:1 stoichiometry (HCl vs $\ce{CaCO3}$), so
$$n_\ce{CaCO3} = \pu{12.5e-3 mol} / 2 = \pu{6.2e-3 mol} $$
On the other hand, if the sample were 100% and everything has 100% yield and we had boiled off the carbonic acid, we would expect a higher amount in 1.00 g of calcium carbonate, i.e. $\pu{9.99e-3 mol}$ as correctly pointed out by the OP.
Why the discrepancy?
Without more information, there is no way of knowing.

This problem looks like a typical job, that I have often given in the lab sessions to my high school classes. The student has to analyze a sample of chalk, which is made of CaCO3, plus some insoluble impurities like alumina $\ce{Al2O3}$, silica $\ce{SiO2}$ or gypsum $\ce{CaSO4·2H2O}$. So the student adds an excess of $\ce{HCl}$, waits until no effervescence is observed, and then titrate the excess of $\ce{HCl}$ with a standard $\ce{NaOH}$ solution. This is what was done here.
Here the titration shows that the sample contains $0.625$ g pure $\ce{CaCO3}$. So it is a rather impure sample, with only $62.5$ % $\ce{CaCO3}$. This may happen.


text:
How do I correctly typeset metastable radionuclide symbols?
Does anyone know if there is an officially sanctioned way to typeset symbols like technetium-99m (99Tcm or 99mTc)? I have seen both, although in more recent publications, I think, the latter predominates.
For my part, it looks somewhat wrong to place the symbol for metastable in the position you'd normally expect the charge. How would you write the ion, 99Tcm+? That looks quite clumsy.
So does anyone know if there's some IUPAC (or similarly authoritative) guidance on this?

According to the international standard ISO 80000 Quantities and units – Part 9: Physical chemistry and molecular physics, Amendment 1, a metastable nuclide is indicated by adding the letter m (in roman type) to the mass number of the nuclide, as in the following examples:
$$\mathrm{^{133m}Xe}$$
$$\mathrm{^{99m}Tc^+}$$

I'm quite confused since BS ISO 80000-9:2009 (Annex B, p. 34) suggests otherwise: "State of nuclear excitation: $\ce{^110Ag^*}$ or $\ce{^110Ag^\mathrm{m}}$". Are you citing a more recent ISO, maybe?

@andselisk Yes indeed. The first edition of ISO 80000-9 (2009-04-01) had in Annex B: “$\ce{^110Ag^*}$ or $\ce{^110Ag^\mathrm{m}}$”. This mistake was corrected in Amendment 1 (2011-06-01).

According to IUPAP recommendations (Symbols, Units, Nomenclature and Fundamental Constants in Physics), part 2 SYMBOLS FOR ELEMENTS, PARTICLES, STATES AND TRANSITIONS, 2.1 Chemical elements, both placements are valid [1, p. 9]:

The right superscript position should be used, if required, to indicate a state of ionization (e.g., $\ce{Ca^2+}$, $\ce{PO4^3-}$) or an excited atomic state (e.g., $\ce{He^∗}$). A metastable nuclear state, however, often is
treated as a distinct nuclide: e.g., either $\ce{^118Ag^\mathrm{m}}$ or $\ce{^\mathrm{118m}Ag}$.

On the other hand, AIP Style Guide in Part III. General style, A. GRAMMAR AND PUNCTUATION, 12. Symbols for nuclides suggests that roman "m" is placed as superscript on the right [2, p. 16]:

Journals of AIP and its Member Societies follow the recommendations of the Symbols, Units, and Nomenclature (S.U.N.) Commission of the International Union of Pure and Applied Physics on the symbols to be used for nuclides and their states. [...]
A posterior superscript can indicate either a state of ionization:
$$\ce{Ca^2+}$$
or an excited state:
$$\ce{^110Ag^\mathrm{m}},~\ce{^14N^*}$$

References

Cohen, R. E.; Giacomo, P. IUPAP “Red Book” Symbols, Units, Nomenclature and Fundamental Constants in Physics (2010 Reprint); IUPAP recommendations; Commission C2 – Sunamco, 1987.
AIP Style Manual, 4th ed.; American Institute of Physics, AIP Publication Board, Eds.; American Institute of Physics: New York, N.Y., 1990. ISBN 978-0-88318-642-8.


text:
Conductometric titration of Strong Acid And Strong base
Consider conductometric titration of Strong Acid with strong base say HCl + NaOH.
My question is why initially the conductance decreases. If H+ of HCl is consumed by OH- of NaOH, forming water and hence conductance decreases.  I agree to these.
But On other hand NaCl is also forming in same amount which can give rise to Conductance. So in my point of view Conductance must remain same or differ slightly. Instead of decreases considerably.

You correctly identify that the number of cations do not change in solution, so from a charge perspective it appears that there should be no significant change in the conductivity.
However, consider two things:

Sodium ions are much heavier and larger than free protons and so should travel through water more slowly. Charge moving more slowly
means lower conductivity.
In addition to their smaller size and lightness, Protons can also "hop" from water molecule to water molecule and so can move through the solution much more quickly and easily. Other water molecules do not act as obstacles

Point 2 is called the Grotthuss Mechanism.

H+ -1091
Na+ -406

These are values of Hydration enthalpy of H+ and Na+. For H+ it is more means it is highly solvated by water molecules compared to Na+ hence it can roam freely in water compare to Na+. So By applying this concept Na+ must have higher conductance.

Is my concept is true?

@JayPatel It's very wrong as you're using thermodynamics instead of kinetics. The answer is ok +1

simple..proton is light than sodium so the mobility of proton is greater than sodium ion thats why conductance decrease

Please at least expand your answer to a paragraph...


text:
Stereoselectivity of hydroboration of dihydrofuran

The above compound undergoes hydroboration. I originally thought that the $\ce{OH}$ group can attach to any of the 2 carbons plus there are 2 possible enantiomers at each position. So there should be 4 different products formed.
However the answer states that only below compound would be produced:

Why is this the only compound that is produced? Also if the $\ce{OH}$ group had to be attached to one of the carbons wouldn't it be more likely to attach to the carbon next to the oxygen since it will be more electrophilic?

Generally, the regioselectivity (and stereoselectivity, depending on the situation) of hydroboration reactions is determined by steric factors. This is a prime example of when it isn't.
Regioselectivity
Borane, $\ce{BH3}$, is electrophilic on boron because of the vacant p orbital (low-lying LUMO). It will therefore preferentially add to the alkene carbon that is more nucleophilic. The enol ether you have drawn is clearly nucleophilic on the carbon β to oxygen:

and consequently it is this carbon that the boron atom becomes attached to:

Reading this:

Also if the OH group had to be attached to one of the carbons wouldn't it be more likely to attach to the carbon next to the oxygen since it will be more electrophilic?

I suspect you need to check the mechanism of hydroboration-oxidation again. There is no attack of $\ce{OH-}$, or indeed any oxygen-based nucleophile, on an electrophilic carbon. The C-O bond is formed by migration of carbon from boron to oxygen. Which carbon is more electrophilic is inconsequential.
Stereoselectivity
It is very common practice in organic chemistry to only draw one enantiomer when a racemic mixture is implied. The purpose of this is to save time: if you were discussing the diastereoselectivity of a particular reaction, you don't want to have to keep on drawing two enantiomers at every single step of the reaction.
Since attack from the top and bottom faces is equally likely to happen, an equal mixture of enantiomers is formed. As described above, the single enantiomer drawn therefore refers to a racemic mixture:

There is no diastereomerism present in this compound, but I would guess that the other reason why stereochemistry is indicated is to alert the reader to the fact that the product is chiral. Sometimes, this can be considered to be misleading. The most unambiguous way to indicate "undefined stereochemistry" is therefore using the wavy line as I did above.

Thanks for the answer. Yeah, I was confusing myself with some other reaction. So basically both enantiomers are equally possible then?

@Nanoputian Yes both enantiomers are formed

Older chemists (such as myself) also use an unbolded line to indicate unknown or ambiguous stereochemistry

According to the Graphical Representation of Stereochemical Configuration (IUPAC Recommendations 2006), it is preferred to use a plain bond if the configuration is unknown. A wavy bond may be used to place special emphasis on the unknown nature of the configuration (usually accompanied by explanatory text). If hashed wedged and/or solid wedged bonds are present (at all stereogenic centers), it will represent a single stereoisomer. Mixtures should be represented with explanatory text (a plus sign should not be used to replace the word “and”).