How to determine stable electron states in ionic and covalent bonds? I'm working on a program that needs to determine if a bond between two or more elements will result in a stable state.  I understand at a high-level how to fill electron subshells using the Aufbau principle, but I also read that in some cases, electrons will jump from a lower energy shell or orbital to a higher one in order to maintain a stable state (Filling Electron Shells).

For example, if I want to determine if Hydrogen and Nitrogen will form a stable bond I would fill the shells for Nitrogen like so:

When adding Hydrogen, would the two electrons in the 2s shell jump to the 2p shell because with the one electron from Hydrogen the HN would then have a full 2p shell? Like this:

Or is this not a stable state? Would it require a large amount of energy to excite or promote those two 2s electrons to the 2p shell?

I'm trying to understand if there are rules or heuristics I can use to estimate if two ore more elements will bond (on there own w/o adding a large amount of energy to the system) using their valence electron configuration like this, or if there are too many exceptions, making it not a simple programming task to estimate this.

I'm trying to understand if there are rules or heuristics I can use to estimate if two ore more elements will bond

You should try to find good textbook/course about Molecular Orbital theory. As I recall, good university-level textbooks on general/inorganic chemistry dive into this aspect and make analysis for dimers of second row elements. This heuristics, however, are of limited usability: they can be used to explain differences in family of compounds with similar structure, but are virtually useless for predicting geometry in complicated cases. And yet, this is the best you can get without diving into quantum chemistry.

Here is quickly found link on the subject. http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch8/mo.html#valence


Temperature Dependent Specific Heat of Nitrogen Calculate the change in enthalpy as 1kg of nitrogen is heated from 1000K to 1500K, assuming the nitrogen is an ideal gas at a constant pressure. The tempreature dependent specific heat of nitrogen is Cp = 39.06 - 512.79T^1.5 + 1072.7T^2 - 820.4T^3 where Cp is in kJ/kg-mol and T is in K

Answer is 697.27 kJ

This is the first time i've ever encountered something like "Temperature Dependent Specific Heat"

So I know

H = m*C_p*(T2-T1);
Cp = 39.06 - 512.79T^1.5 + 1072.7T^2 - 820.4T^3

So I try to make something like C_p1:

C_p1 = 39.06 - 512.79T^1.5 + 1072.7T^2 - 820.4T^3; where T = 1000
C_p1 = -8.1934*10^11

C_p2 = 39.06 - 512.79T^1.5 + 1072.7T^2 - 820.4T^3; where T = 1500
C_p2 = -2.766466215*10^12

Cp2 - Cp1 = -3.8585 * 10^12

So yeah... Its not looking too good. Am i missing a formula?

The function looks like its easy to differentiate or integrate. Should I be doing something involving calculus?

The function looks like its easy to differentiate or integrate. Should I be doing something involving calculus?

Yes, exactly!  The equation you should be using is the differential form of the enthalpy equation you are already using.

$$dH = m~C_p(T)~dT$$

$$\int_{H_1}^{H_2} dH = m \int_{T_1}^{T_2} C_p(T)~dT  $$

As you can verify, if heat capacity is independent of temperature, i.e. a constant, the equation reduces to

$$H_2 - H_1 = m~C_p~\left (T_2 - T_1 \right ) \implies \Delta H = m~C_p\Delta T$$

This is the equation you tried to use originally.  However, since we derived this equation assuming a heat capacity that is independent of temperature, and your question explicitly involves dependence on the temperature, we should not use the already integrated equation.  Instead you'll have to do the integration yourself by substitution of temperature-dependent equation.


sp² hybridized orbital I do not understand why and how an atom with $\ce{sp^2}$ hybridization has only one p orbital??

For example in pyridine why is the lone pair of nitrogen not counted as a pi electron.

The picture below from the UC Davis ChemWiki summarizes why that lone pair is not available. It also explains how pyridine can function as a base.

The lone pair in the $sp^2$ orbital is perpendicular to the $p$ orbitals and $\pi$ bonds in the molecule. the electron pair from the nitrogen that participates in aromaticity is the one in the $\ce{C=N}$ bond.
To answer your first question about hybridization, we have a conservation law that guarantees that when orbitals mix (for any reason), we need to get the same number of orbitals out as we put in. The law is the Conservation of Angular Momentum, which does apply in quantum systems.
Nitrogen, in its valence shell, has four orbitals: $2s,2p_x,2p_y,2p_z$. The $sp^2$ hybridization is achieved by mixing two $p$ orbitals and one $s$ orbital. Because of the conversation of angular momentum, four orbitals in means four orbitals out. Thus, the $p$ orbital that was not mixed is still there (and is the only one still there).
$$[s,p,p],p\implies[sp^2,sp^,sp^2],p$$
We cannot have two $p$ orbitals at an $sp^2$ hybridized atom because then we would have 5 orbitals (or we would not have $sp^2$).

Now, let's create a general rule for aromatic heterocycles.

Look at pyrrole (also from ChemWiki);

In the structure of pyrrole, the $\ce{N-H}$ bond uses the $sp^2$ orbital and the lone pair is in the $p$ orbital. In general, if there is a double bond drawn to the nitrogen atom (like in pyridine), then the $\pi$ bond uses the $p$ orbital and the lone pair is in the $sp^2$ orbital and does not count toward aromaticity. If there is no double bond drawn to the nitrogen atom, usually there is a $\ce{N-H}$ bond (or some other bond to the nitrogen). This $\ce{N-H}$ bond uses the $sp^2$ orbital and the lone pair uses the $p$ orbital and counts toward aromaticity.


Are there chiral compounds that don't rotate plane-polarized light? I know that meso compounds have chiral centers but don't rotate plane-polarized light, and I know that there can be non-traditionally chiral compounds (e.g. ones with large substituents that prohibit much rotation around single bonds) that can still rotate light, but I haven't been able to find any molecules that are chiral but don't rotate plane-polarized light (or rotate it impossibly little). I'm imagining something like an enantiomeric pair connected by one single bond with a slight modification to one that makes it have similar optical activity but not actually be identical (so the final molecule is indeed chiral), but I don't know exactly how to make it work.

Good question.
There's a phenomenon named cryptochirality[1] (meaning “hidden chirality”), when a compound, though chiral, has practically unmeasurable optical rotation activity.

It can happen to molecules with chiral center(s) bearing very similar substituents. (So, no tricks with bonded slightly modified enantiomeric pairs are needed.)

An example is 5-ethyl-5-propylundecane $\ce{CH3[CH2]5-\overset{$*$}{C}(CH2CH3)(CH2CH2CH3)-[CH2]3CH3}$,[2] don't call it “butyl(ethyl)hexyl(propyl)methane”, found e.g. in beans. Its specific rotation is $[\alpha] < 0.001$.

Another, more common example are fats, i.e. triglycerides, $\ce{RCOOCH2\overset{$*$}{C}H(OCOR')CH2OCOR''}$,[1] if containing e.g. only palmitic, oleic and similar long acyls, optical rotation is not demonstrable.[3]

(Related topic is chirality in polymers, see e.g. Q: Chirality on Carbon of PVC molecule.)

References:

Mislow K. & Bickart P.: An Epistemological Note on Chirality. Israel Journal of Chemistry 15, 1–6 (1976)
Wynberg H., Hekkert G.L., Houbiers J.P.M. & Bosch H.W.: The Optical Activity of Butylethylhexylpropylmethane. Journal of the American Chemical Society 87, 2635–2639 (1965)
Schlenk W.: Synthesis and analysis of optically active triglycerides. Journal of the American Oil Chemists’ Society 42, 945–957 (1965)


Why does sulfur, but not oxygen, catenate? Oxygen is a rather boring element. It has only two allotropes, dioxygen and ozone. Dioxygen has a double bond, and ozone has a delocalised cloud, giving rise to two "1.5 bonds".

On the other hand, sulfur has many stable allotropes, and a bunch of unstable ones as well. The variety of allotropes, is mainly due to the ability of sulfur to catenate.

But, sulfur does not have a stable diatomic allotrope at room temperature. I, personally would expect disulfur to be more stable than dioxygen, due to the possibility of $\mathrm{p}\pi\text{-}\mathrm{d}\pi$ back-bonding.

So, why do sulfur and oxygen have such opposite properties with respect to their ability to catenate?

First, a note: while oxygen has fewer allotropes than sulfur, it sure has more than two! These include $\ce{O}$, $\ce{O_2}$, $\ce{O_3}$, $\ce{O_4}$, $\ce{O_8}$, metallic $\ce{O}$ and four other solid phases. Many of these actually have a corresponding sulfur variant. However, you are right in a sense that sulfur has more tendency to catenate… let's try to see why!

Here are the values of the single and double bond enthalpies:
$$\begin{array}{ccc} \hline
\text{Bond} & \text{Dissociation energy / }\mathrm{kJ~mol^{-1}} \\ \hline
\ce {O-O} & 142 \\
\ce {S–S} & 268 \\
\ce {O=O} & 499 \\
\ce {S=S} & 352 \\ \hline
\end{array}$$
This means that $\ce{O=O}$ is stronger than $\ce{S=S}$, while $\ce{O–O}$ is weaker than $\ce{S–S}$. So, in sulfur, single bonds are favoured and catenation is easier than in oxygen compounds.

It seems that the reason for the weaker $\ce{S=S}$ double bonds has its roots in the size of the atom: it's harder for the two atoms to come at a small enough distance, so that the $\mathrm{3p}$ orbitals overlap is small and the $\pi$ bond is weak. This is attested by looking down the periodic table: $\ce{Se=Se}$ has an even weaker bond enthalpy of $\ce{272 kJ/mol}$. There is more in-depth discussion of the relative bond strengths in this question.

While not particularly stable, it's actually also possible for oxygen to form discrete molecules with the general formula $\ce{H-O_n-H}$; water and hydrogen peroxide are the first two members of this class, but $n$ goes up to at least $5$. These "hydrogen polyoxides" are described further in this question.


Crystallization issue For solid-liquid mixtures, cooling crystallization process can be used to separate the solute substances from each other. Assume that we have two salts that have very close solubility in a certain solvent which we are going to use. First, we dissolve the salts in the solvent while heating. After a point, we let the mixture cool down and observe the crystallization. My question is, how can we do this experiment accurately if they have really close crystallization point that they would nearly crystallize at the same time?

Under the right conditions, crystallization can be a very effective means of obtaining a pure solid form of the solute. Unfortunately, in your case one of those conditions is violated; you have a second solute with similar solubility characteristics.  You didn't quantify the solubilities of the solutes, but the phrase "very close solubility in a certain solvent" tells me that under the conditions of your system, crystallization will not be an effective means of obtaining a pure solid form of the solute.

One option, as mentioned in a comment to your question, is to choose a different solvent. Selecting the right solvent / solvent mix can be a bit of an art and will also require some research and experimentation on your part.  Because you didn't discuss the specific solvents or solutes used in your system, it is not really possible to give more specific advice on solvent selection.

Another option is to choose a different means of solute separation, like some type of chromatography. Again, not knowing anything about your compounds makes it difficult/impossible to give further advice regarding this technique.


Can I use calcium chloride instead of sodium sulfate as drying agent? In the synthesis of N-ethoxycarbonyl L-proline methyl ester from L-proline and ethyl chloroformate,
I dry the organic phase over sodium sulfate. Can I use calcium chloride instead of sodium sulfate?
I searched for information and I found that calcium chloride can contain basic impurities such as $\ce{Ca(OH)2}$ and $\ce{CaCl(OH)}$. Does someone know if it is actually the case?
If it would be correct I think that it cannot be used because the impurities can hydrolyse the ester.

Similar to @MathewMahindaratne's comment: if you know the drying agent suits your needs well enough, keep it.  In particular here, sodium sulfate is a neuter one (as is calcium sulfate, too) is a safe option because while drying the organic layer, the agent will neither protonate, nor deprotonate your product. Thus, if the protocol states $\ce{Na2SO4}$, stick to it.
This is in contrast to calcium chloride which not only will contain traces of calcium hydroxyde when taken from the jar, but upon contact with additional water (a.k.a. while drying your organic solution) will create new $\ce{Ca(OH)2}$ and thus is considered a basic (and potentially deprotonating) drying agent.  Note, $\ce{MgSO4}$ is considered as an acidic drying agent.  Using the wrong salt solutions or drying agents during the workup actually may cleave off (e.g., protecting) groups and thus complicate the isolation and purification of your product.


Why is ruthenium(III) intensely colored in water I was preparing a solution of ruthenium(III) chloride (aqueous) when I added some (nearly black) $\ce{RuCl3 . H2O}$ to the water. I was surprised that a tiny amount could dye the water so well, I'd say you can even compare this to $\ce{KMnO4}$.

The question is why is the color so intense?

Ruthenium(III) is a $\mathrm{d^5}$-system so the transitions should be spin-forbidden if we are talking about $\ce{d\bond{->}d}$ transitions. If this is the case we can consider a CT-transition. But if I look at the chemistry of manganese(II) it is slightly pink and doesn't dye water at all although it has the same ligands (chlorido and aqua).
And if we compare the colors of manganese(VII), chromium(VI) and vanadium(V) as well as manganese(VII) to renium(VII) with the same ligands, in this case oxido-ligands we see that the more stable states (vanadium(V) and renium(VII)) do not pull electrons that strong anymore and the color shifts further into something invisible to the human eye.

So ruthenium(III) should be even less colored than manganese(II) if we ignore the fact that it has 3 chlorido ligands for a second. So the only option I have left would be the π-donor ability of the chlorido-ligands.

I have read some time ago, that manganese is one of the only elements that form in as their aqueous $\ce{MCl2}$ a cis-dichlorido-coordinated species while all the other (3)d-metals form trans-coordinated ones. And I have already asked about the cis to trans difference in another post so I know now that it does indeed make a difference.

I'd like to compare this to the ruthenium(III) now but ruthenium(III) in water is somewhat complicated as a lot of possible compounds can form especially as the trihydrate will start to exchange chlorido-ligands for aqua ligands at a certain period of time and the trihydrate salt is said to be yellow-orange like iron(III) while I start with a monohydrate which is black.

I'm not sure whether this effect is now related to the rutheium only or if this is just the case because of some symmetry effect but I can't find a good answer on this anywhere.

As you said, in $\ce{RuCl3}$, the charge of the Ruthenium ion is +3, meaning d5. In water, the RuCl3 becomes $\ce{[Ru(H2O)6]^3+}$, an octahedral complex. The electron configuration of the Ruthenium ion can either be $\mathrm{t_{2g}{}^5\,e_g{}^0}$ (low spin), or $\mathrm{t_{2g}{}^3\,e_g{}^2}$ (high spin), as shown in the pictures below:

You also need to know about CFSE (crystal field stabilization energy), which is the difference in the energy of the free metal ion, and the metal ion in the complex.

According to the electron configuration diagrams above, the CFSE for each configuration is:

$$\text{Octahedral }\mathrm{d^5}\text{ — low spin}\\
\text{CFSE} = -0.4\Delta_\mathrm{O} \times 5 + 2P = 2P - 2 \Delta_\mathrm{O}\\[1.5em]
\text{Octahedral }\mathrm{d^5}\text{ — high spin}\\
\text{CFSE} = -0.4\Delta_\mathrm{O} \times 3 + 0.6\Delta_\mathrm{O} \times 3 = 1.2\Delta_\mathrm{O} - 1.2\Delta_\mathrm{O} = 0$$

Where $P$ is the mean pairing energy, the energy required to overcome the repulsion between the electrons and pair them together. $\Delta_\mathrm{O}$ is the octahedral splitting parameter, which I assume you already know.

As you can see, a high spin $\mathrm{d^5}$ octahedral complex will not lead to a change in energy compared to a free ion. Therefore, the spin is determined by the low spin configuration. If the pairing energy is bigger than the octahedral splitting parameter, the metal ion will gain energy, and will rather be high spin. If the pairing energy is smaller than the octahedral splitting parameter, the metal ion will lose energy and will rather be low spin.

The octahedral splitting parameter is mostly affected by the ligand type (sigma, pi acceptor, pi donor). The pairing energy is mostly affected by the size of the d orbitals — the bigger they are, the smaller it is (because the electrons have more room, the repulsion between them is smaller).

4d and 5d transition metals have large d orbitals, so that the mean pairing energy is usually smaller than the splitting parameter, which results in low spin complexes regardless to the ligand type.

Therefore, a $\ce{Ru^3+}$ ion, which is $\mathrm{4d^5}$, will form low spin octahedral complex, that has spin allowed transitions with intense colors.

About charge transfer. It's possible, but I personally don't know.


In the Favorskii rearrangement, why doesn't the base grab the alpha hydrogen of the carbon bearing the halogen? In the Favorskii rearrangement, the base (like hydroxide) always abstracts the alpha hydrogen of the carbon not bearing the halogen. But to me the hydrogens on the carbon directly connected to the halogen should be more acidic due to the electron pulling inductive effect of the halogen (which indeed does happen in the haloform reaction). This should result in an aldol condensation type reaction. Why is this not so?

Since it is much easier to discuss based on actual schemes, see scheme 1 for a general Favorskii rearrangement.

Scheme 1: General Favorskii rearrangement reaction conditions. Image taken from Wikipedia where a full list of authors is available.

And because it is also helpful, scheme 2 shows one of the most accepted mechanisms of the Favorskii rearrangement — the one used by both Wikipedia and Kürti-Czakó.[1] One should note, however, that a second mechanistic proposal[2] explains the reaction outcome well and is favoured by some calculations.[3] Because it does not require any deprotonation, it is outside of the scope of this answer.

Scheme 2: ‘Cyclopropanone’-mechanism of the Favorskii rearrangement. Image taken from Wikipedia where a full list of authors is available.

Your question asks why carbon 6 (from the usual numbering of 2-chlorocyclohexanone) is deprotonated to form the corresponding enolate and not carbon 2.

The most important notice here is the base used $\ce{NaOH}$ and the conditions employed (not mentioned in the image, but an example on organic-chemistry.org mentions room temperature). For comparison, table 1 shows a few $\mathrm{p}K_\mathrm{a}$ values from the Bordwell data — unfortunately, an α-chloroketone is not included in this data.

$$\textbf{Table 1: }\mathrm{p}K_\mathrm{a}\text{ values of different relevant acids in }\ce{H2O}\\
\begin{array}{ccc}\hline
\text{acid} & \mathrm{p}K_\mathrm{a}\left(\ce{H2O}\right) & \mathrm{p}K_\mathrm{a} \left(\ce{DMSO}\right)\\ \hline
\ce{Me2CO} & 19.3 & 26.5\\
\ce{PhCOMe} & 18.3 & 24.7\\
\ce{PhCO-CH2F} & & 21.7\\
\ce{Et2CO} & & 27.1\\
\text{3,3-dimethylcyclohexanone} & & 24.8\\
\ce{H2O} & 15.7^* & 31.4\\ \hline \end{array}\\
^*\text{ Other sources say 14.}$$

Obviously from this data, the choice of solvent is import when considering the acidity of a given substance — note the marked decrease in acidity of water in $\ce{DMSO}$. However generally, water is a weaker acid than carbonyl compounds; Professor Mayr taught us to estimate $\mathrm{p}K_\mathrm{a}$ values: alcohols $\approx 15$, ketones $\approx 20$. That means, the deprotonation is in no way complete; it is in equilibrium. This equilibrium will lead to a thermodynamic product distribution: carbon 2 will be deprotonated slightly more than carbon 6, but the most significant species will be the unenolised ketone.

Now as soon as carbon 6 has enolised, it is able to attack carbon 2 to form the bicyclo[3.1.0]hexan-6-one. This is an intramolecular attack and thus very fast. Once this has happened, the formation of the hydrate (6-hydroxybicyclo[3.1.0]hexanolate) is favoured due to the strained three-membered ring. And the subsequent intramolecular migration is also very fast. This leads to a stable product with little or no means for any back reaction; as soon as the initial attack has taken place, things are pretty much settled.

On the other hand, an aldol addition would not only require the enolate (doesn’t matter which one) to be around for some time; it is also inherently prone to retro-aldol dissociation. Usually in organic chemistry, aldol reactions are performed selectively by using kinetic deprotonation (e.g. $\ce{LDA}$; $\mathrm{p}K_\mathrm{a}(\ce{HN(iPr)2}) \approx 35$) ensuring that the forward reaction is strongly favoured. Hydroxide bases may be used in aldol reactions, but then generally the reaction is heated to reflux and the final product is the condensation rather than the addition.

Thus, to sum it up:

Both enolisations occur
The conditions do not favour aldol addition; they will revert readily
Once the Favorskii path has been entered, the rearrangement is basically irreversible.

Notes and References:

[1]: L. Kürti and B. Czako: Strategic Applications of Named Reactions in Organic Synthesis. Background and Detailed Mechanisms, Elsevier Academic Press, Burlington, MA, USA, 2005, page 164.

[2]: The alternative, semi-benzilic mechanism is initiated by the nucleophilic attack of a hydroxide ion on the carbonyl group. The thus formed 2-chloro-1-hydroxycyclohexanolate anion undergoes 1,2-migration with chloride displacement to form the product cyclopentyl carboxylic acid in a single step.

[3]: V. Moliner, R. Castillo, V. S. Safont, M. Oliva, S. Bohn, I. Tuñón, J. Andrés, J. Am. Chem. Soc. 1997, 119, 1941. DOI: 10.1021/ja962571q.


What happens to the t-butyl cation in the TFA deprotection of a t-butyl ester? I am a first-year chemistry graduate student and I am currently learning how to synthesize proteins. I have reached a stage in the process where I need to deprotect a carboxylic acid moiety protected by t-butyl using trifluoroacetic acid (TFA).
I went online trying to find the mechanism for this step, but in terms of step-wise mechanism with electron movement arrows, the only thing I was able to find was this:

(Taken from here, no references therein)

While this looks legit, I am wondering about the t-butyl cation and its fate.
My reaction is a bit simpler, and is presented below:

And my question is: is this truly where the deprotection ends? Does the t-butyl cation remains in solution as pictured?

(I know that a t-butyl cation is considered to be relatively stable on account of it being a tertiary carbocation, but still I was not sure that it can remain in solution like that.)

The reaction is, in principle, catalytic in TFA. If t-butyltrifluoroacetate is the by-product, then the catalyst is consumed. If only isobutylene is formed, catalytic is OK. Using excess TFA covers all contingencies. In the initial step, protonation is more likely on the C=O oxygen. Higher electron density.


Finding rate constants for parallel reactions A reactant $\ce{X}$ is converted into products $\ce{Y}$ and $\ce{Z}$
according to the following first order parallel reactions:

$\ce{X->[k_1]Y}$

$\ce{X->[k_2]Z}$

$\ce{[X]}$ has decreased from $\pu{1.00 M}$ at $t = 0$ to $\pu{0.549 M}$ after $\pu{10 min}$.
Calculate the rate constants $k_1$ and $k_2$, knowing that the yields of $\ce{Y}$ and $\ce{Z}$ obtained are $25.0$ and $\pu{75.0 mol\%}$, respectively.

So far, I have the ratio between $k_1$ and $k_2$ $=0.33$, which I found from the yield ratio but I'm unsure what to do with this ratio.

I also have derived the equation $[A] = [A]_0 \cdot e^{-(k_1+ k_2)t}$

By some simple mathematics, you can reach the following equations (the first of which you have reached):

$[X]=[X]_0\cdot e^{-(k_1+k_2)t}$

$[Y]=\frac{k_1}{k_1+k_2}[X]_0\cdot e^{-k_1 t}$

$[Z]=\frac{k_2}{k_1+k_2}[X]_0\cdot e^{-k_2 t}$

The $1:3$ ratio that you have arrived at is correct,  i.e. $\dfrac{k_1}{k_2}=\dfrac{1}{3}$.

Substituting this in the equation for $\ce{X}$, we get

$[X]=[X]_0\cdot e^{-4k_1 t}$

All that is left to do now is to substitute and solve, which leads to the answer:

$(k_1,k_2)\equiv(0.015\ (min)^{-1},0.045\ (min)^{-1})$

There you go!

PS: I've used $[X]_0$ for representing the initial concentration of $\ce{X}$.


Compound Polarity Lookup I know from the literature that the following compounds have decreasing polarity: Rosmarinic Acid -> TMF -> Sinensetin and I guess comparing compounds of similar structure it is manageable to compare bond types, groups etc etc by which to order their polarity, but what about compounds with very different structures, groups and bonds, especially when you want to rapidly order 10's-100's of compounds.
My questions:

Is there such a thing as an absolute measure of "polarity" for a compound by which other compounds can be compared/ordered?

Depending on 2) is there a resource whereby you can lookup your compound to obtain its polarity?

I am asking this question from a Chromatography perspective. I would like to know the order of separation for a group of compounds and how to know what "polarity" of mobile phase to start my experimentations with.

Polarity is an umbrella term which is popular among organic chemists. It is as vague as saying the weather is good today. "Good weather" may include, bright light, sunny, temperature, right humidity and so on.  Now read your own statement

Rosmarinic Acid -> TMF -> Sinensetin and I guess comparing compounds
of similar structure it manageable to compare bond types, groups etc
etc

Since you correctly use etc. etc. ...yes polarity includes et cetra et cetra. Its usage is discouraged by IUPAC.
The correct term is dipolarity which is a measure of the dipole moment (you can look them up in tables). Even that dipole moment is of no good because you have plenty of conformers and each conformer has a different dipole moment (at least in the gas phase).


Why is acetone less acidic than cyclohexanol? The $\mathrm{p}K_\mathrm{a}$ of acetone is 19, whereas the $\mathrm{p}K_\mathrm{a}$ of cyclohexanol is 16.

However, acetone is resonance stabilized and its major resonance structure displays the negative charge on an oxygen atom, like cyclohexanol. Is there any qualitative reason to explain why acetone is less acidic than cyclohexanol?

Thank you very much.

if you consider the structure of acetone, it has 2 $\ce{CH3}$ groups and a $\ce{C}$ double bond $\ce{O}$ ($\ce{C=O}$). Therefore, for the $\ce{H}$ to disassociate, it must be removed from the $\ce{CH3}$ group which leaves now a $\ce{C=O}$, $\ce{CH2}$, and a $\ce{CH3}$ group where the negative charge lies on $\ce{C}$ a part of the $\ce{CH2}$ group. Conversely, the cyclohexanol would have the negative charge lying on the $\ce{O}$ because the $\ce{H}$ a part of the $\ce{OH}$ group would be removed.
Just by simple electron negativity rules, $\ce{O}$ is much more favored to possess a (partial) negative charge than $\ce{C}$. Thus, cyclohexanol is a much more stable acid than acetone which is reflected in its $\mathrm{p}K_\mathrm{a}$ value.


About the way chemicals and molecules are represented The standard way of representing molecules with letters like $\ce{C}$ for Carbon, $\ce{O}$ for Oxygen etc.; and a lot of lines and double lines like a graph from Graph theory where each letter is a 'vertex', the lines like edges. I forget what this chemical notation system is called. Yet it is 2-dimensional. The formula for '$\ce{H2O}$' ,water doesn't indicate the 'angle' of 'separation' between the two Hydrogen atoms that are bonded with the Oxygen atom. Similarly a formula for some molecule like a protein might not indicate all the 'involved angles' in 3-dimensional space that 'allow' it to be folded in a precise way. So does the 2-dimensional notation system for molecules miss details of how a molecule is twisted in 3-D space?  Also can two distinct molecules 'look' the same in 2-D notation?

Below is 2 types of sugars; glucose and mannose.

If you know how to interpret the lines, then you can determine which one is which.


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

FCI of N2 with about 10^10 CSFs must be the largest such calculation...

Ref:
A full-configuration benchmark for the N2 molecule,
Elda Rossi, Gian Luigi Bendazzoli, Stefano Evangelisti, Daniel Maynau,
https://doi.org/10.1016/S0009-2614(99)00791-5


How can a sample that is in two different states be at a uniform temperature? In my chemistry course, the professor has discussed how a sample that contains both solid and liquid water would be at 0 degrees Celsius no matter what the proportion of solid to liquid water is. I find this concept to be confusing, because wouldn't the areas of the sample that had already changed from solid to liquid begin increasing in temperature even while the other parts of the sample continue to be in the process of fusion?

The other part of this idea that confuses me is that if the sample was at a uniform temperature, then why do bodies of water (ex. lakes that freeze in the wintertime) continue to have different temperatures of water within them?

For a system in equilibrium: Think of it as a closed system, with a certain amount of energy shared between the solid and liquid components (discounting the container).  At any instant, some of the faster-moving molecules of the liquid could hit the ice, melting a bit, and thereby losing some of their own energy and freezing.

It takes ~80 cal/g to change ice at 0 °C to water at 0 °C. This enthalpy of fusion, aka latent heat of fusion , "absorbs" any additional heat applied until all the ice has melted. So a gram of water at 0 °C has ~80 calories more heat than an equal mass of ice at 0 °C. If you have a 1 gram chunk of ice and apply 40 calories as heat, then about half will melt.

BTW, pure water does not necessarily freeze at 0 °C... it may supercool, which will change the calculations a bit. See this video or this one for a supercool experiment!

As @MaxW states, in real life most systems are not in equilibrium (they need to be shaken or stirred, despite Ian Fleming's direction). To view a clearly non-equilibrium demo, where there is water in three states at once, see Red Hot Nickel Ball on Ice.


Why are esters meta directors? Can't the carbonyl donate electrons into the ring? I'm having trouble understanding why an ester is a meta directing/deactivating group.

I'm not talking about an ester with its oxygen bound to the ring, like an acylated phenol. I mean the R group is bound to the ring and the alkoxy group is bound only to the carbonyl carbon.

It seems to me like the pi electrons in the carbonyl pi bond could be pushed down and delocalized into the ring. Why is this not possible?

As your concept oxygen gets positive charge and that leads to least stability which is as follows:


Does Suncream have any affect or help protect against radiation So I was just watching the HBO series Chenoybl and saw that people suffered various degreesvof burns from radiation form Uranium 235 and some were very minor and looked like a normal sun burn.

My question is, does Suncream protect against radiation burns?

The only radiation sunscreen protects you against is near UV light not radioactive emission

Sunscreens are designed to protect from the near UV radiation present in sunlight. Some block both UVA (315-400nm) and UVB (280-315nm) light but some don't work well for UVA. All UV light at ground level is harmful to skin to some extent and too much of it will cause skin damage that looks a little like a burn (though small amounts are good for encouraging vitamin D production).

Radioactivity can cause similar looking burns. But the cause of the burns is not near UV light but gamma rays (hard X-rays of much shorter wavelengths and much higher energy than UV light) and alpha and beta particles. All of these are much more dangerous than UV light (as the amount of energy deposited in tissues is far greater per photon or particle).

And sunscreen will offer no protection at all to radioactive emissions. Clothing can protect against alpha particles. But you will need several millimetres of aluminium to stop beta particles (and even then the secondary gamma rays from the interaction of those particles with the aluminium can be harmful). And gamma rays might need a thick chunk of lead to keep them out.


Stereoselectivity differences in Lindlar's catalyst and Na/NH3? Why does using the Lindlar's catalyst form cis-alkenes whereas $\ce{Na/NH3}$ forms trans-alkenes?

In Lindlar's catalyst, it is a matter of the physical way in which the molecules react. Firstly, $\ce{H2}$ dissociates into atomic hydrogen and binds to the catalyst, and the alkyne coordinates to the metal surface. Since the surface of the catalyst can be thought of as a vast plane, all carbon to metal bonds must be facing down onto the plane. The hydrogen atoms then get inserted in place of these carbon to metal bonds, and the result is syn-addition.

In the $\ce{Na/NH3}$ reduction, when the electrons are first added to the triple bond and the $\mathrm{sp}$ hybridization is broken, the groups bonds on either side the triple bond repel each other into the lowest energy conformation, the trans position. This is why the $\ce{H}$ atoms being donated by the $\ce{NH3}$ add through anti-addition.


Making Zn in school lab I want to make $\ce{N2O}$ in my school chem lab.

I'm thinking of making it from $\ce{Zn_{(s)}}$ in $\ce{HNO3_{(dil)}}$ :

$$\ce{
Zn_{(s)} + HNO3_{(dil)} -> N2O + Zn(NO3)2 + H2O
}$$

I'm not able to get $\ce{Zn_{(s)}}$, so I would like to make it on my own in lab.

Ahem, forget it.

Theoretically speaking, you can get $\ce{Zn}$ from alkaline batteries (see details here), but you have no chance to get pure $\ce{N2O}$ this way and reaction may easily go out of control.

Attention! Detonation hazard !!!

The usual and easy way to produce $\ce{N2O}$ is to heat very carefully widely available $\ce{NH4NO3}$, to $180-200~^\circ\mathrm{C}$. It is an absolute imperative to keep temperature of the reaction mixture below $240~^\circ\mathrm{C}$. Decomposition of ammonia nitrate ($\ce{NH4NO3}$) is exothermic and may proceed explosively. To be on the safe side, use sand bath to heat the reaction flask and put thermometer into the sand, so its temperature never risen above safe limits. Liquid bath with appropriate liquid may be better, but I'm not sure if a school lab would have proper liquid (silicone oil with high flash point). The produced gases should be cooled in Liebig condenser to separate water, and $\ce{N2O}$ may be further dried with $\ce{H2SO4}$. Of course, all joints must be sealed using proper oil. And keep a bath of cold water to cool the reaction mixture in case if temperature raises above safe limits.

It is advised to use lab grade, not fertilized grade ammonia nitrate.

Possible safer, but costlier route

It is reported that it is possible to produce $\ce{N2O}$ by heating mixture of   $5~\mathrm{mL}$ $6~\mathrm{M}~\ce{HNO3}$, $1~\mathrm{g}~\ce{NH4NO3}$ and $0.05~\mathrm{g}~\ce{NaCl}$ on water bath at $\approx 80~^\circ\mathrm{C}$. If this works, it is a safer route, as hot ammonia nitrate is diluted and autospeeding reaction cannot occur. Link


Transition state and free energy We have products $\ce{A + B}$ combining to form $\ce{C + D}$ through the transition state $\ce{X}$.

Are all reactions at least virtually reversible? Is the difference between a reversible reaction and an irreversible reaction simply that for the former, $\Delta G$ is small, where as the irreversible reaction will have a difference in free energy in products vs. reactants, that the reverse reaction simply doesn't occur, but theoretically could?
Why do concentrations affect the free energy? Is it because you sum the free energy for all reactant molecules, as well as for all the product molecules?

Following is from my textbook:

  [...] the transition state is too unstable to exist for long. It collapses to either substrate or product, but which of the two accumulates is determined only by the energy difference between the substrate and the product -- the $\Delta G$ of the reaction.

I would appreciate if someone could clarify this for me. Feel free to be very nitpicky, I don't want an understanding that is "approximately" correct or "functionally" good.

According to the principle of Microscopic Reversibility all elementary reactions, i.e. those that proceed in a single step, are reversible.
And you are right: For a reversible reaction $\Delta G$ is very small while for an irreversible reaction $\Delta G$ is large. This can be most easily described using a simple schematic energy curve diagram for your situation.

Each reaction is microscopically reversible. So either your reactants $\ce{A}$ and $\ce{B}$ or your products $\ce{C}$ and $\ce{D}$ can combine forming the transition state ($\ce{TS}$). As your text book says, the $\ce{TS}$, being an energy maximum, is very unstable and collapses very fast. Since the transition state represents a maximum of the energy curve the collapse may go in either the direction of the reactants or the direction of the products with equal probability, pretty much in the same way a ball placed at the top of hill may roll down on either side of it. So, once you got to the transition state you have a 50:50 situation.
The thing that determines which way your reaction goes is how easy it is to make the transition state or to "get the ball to the top". In order reach the transition state you have to overcome a thermodynamic barrier, which is the difference of the free energies of products and reactants $\Delta G$, and a kinetic barrier, which is the activation energy $E_{\mathrm{a}}$. If $\Delta G$ is a very large and negative then the reaction will proceed in the direction from $\ce{A}+\ce{B}$ to $\ce{C}+\ce{D}$ and will be irreversible because then it will take considerably more effort to "climb" the energy hill coming from the $\ce{C}+\ce{D}$ side than coming from the $\ce{A}+\ce{B}$ side. This means that in a given time a lot more $\ce{A}$ and $\ce{B}$ molecules are "elevated" to the $\ce{TS}$ than $\ce{C}$ and $\ce{D}$ molecules. So, even if $\ce{TS}$ will revert to $\ce{A} + \ce{B}$ 50 % of the time the reverted molecules will quickly be back "on top of the hill" in order to have another try while for $\ce{C}$ and $\ce{D}$ this takes much longer - the reaction rate for the forward reaction is much higher than the reaction rate for the backward reaction. So over time more and more $\ce{A} + \ce{B}$ is converted to $\ce{C}+\ce{D}$ until equilibrium is estabilished and the ratio of reactants/products is equal to the ratio of the reaction rates.
When you have a reversible reaction and $\Delta G$ is very small the "hill to climb to the transition state" is nearly the same height irrespective of whether you come from the product or from the reactant side. So, forward and backward reaction proceed nearly at the same rate.

Concerning your second question: I'm sorry, but I couldn't find a non-mathematical explanation as to why $G$ depends on the concentration. The Gibbs free energy is the thermodynamic potential of an ensemble at constant pressure and temperature. For variable particle numbers $N_{i}$ its fundamental relation is:

\begin{equation}
  dG = -S dT + pdV + \sum_{i} \mu_{i} dN_{i} \ .
\end{equation}

So, it is connected to the particle numbers (and thus to the concentrations) of the species in the system.


Polystyrene properties after dissolved by acetone Polystyrene plastic can be dissolved in acetone.
I would like to understand:

If the dissolved polystyrene will take-out of the acetone and becomes solid again (after all the acetone evaporates), will this solid polystyrene keep the same material properties as it was before the dissolving process?
How can I speed-up the solidification of the dissolved plastic?

What is happening is not actually the polystyrene melting but rather it dissolving in the acetone solvent. This is a physical change and not chemically altering the polymer. It should still retain its chemical properties as before.

However, other properties have changed. It is now a poorer insulator since the air pockets have been removed. It will also be more rigid for similar reasons.

The acetone ought to be able to be driven off by heating it in a ventilated space as acetone is quite volatile.


Is there a crystal structure of ammonia in ionic form? Like solid $\ce{PCl5}$, does solid $\ce{NH3}$ exist as $\ce{[NH4+][NH2-]}$ (i.e. ammonium amide)?

According to the work of Ninet et al. published in 2014 [1], ammonia has a β-phase at 150 GPa, where it transitions into ionic crystalline form, forming $\ce{[NH4+][NH2-]}.$

References

Ninet, S.; Datchi, F.; Dumas, P.; Mezouar, M.; Garbarino, G.; Mafety, A.; Pickard, C. J.; Needs, R. J.; Saitta, A. M. Experimental and Theoretical Evidence for an Ionic Crystal of Ammonia at High Pressure. Phys. Rev. B 2014, 89 (17), 174103. DOI: 10.1103/PhysRevB.89.174103.


Is NaCl crystal always going to have even number of atoms? I just read:

  In $\ce{NaCl}$, atoms don't aggregate so as to form discrete molecules but they are held together in a network structure. The ratio of the $\ce{Na}$ to $\ce{Cl}$ atoms in a sodium chloride crystal is $1:1$.

Here I have a $\ce{NaCl}$ crystal. Assume that I can break this crystal in all possible ways. So that I have its two pieces every time with a different mass ratio. Would these pieces always contain even number of atoms of $\ce{Na}$ and $\ce{Cl}$. Or, is this ratio going to be same for all conditions?

Suppose I have a $\ce{NaCl}$ crystal of mass $a$. I break it into two pieces of masses $x$ and $y$ such that $x,y<a$ and $x+y=a$. $a$ is a constant by its definition whereas $x,y$ are variables. Then, is the following true?

$$R(x)=R(y)=1:1$$

Where $R(x)$ and $R(y)$ denotes the ratio of $\ce{Na}$ to $\ce{Cl}$ atoms for all values of $x$ and $y$ respectively.

Tl;dr: Theoretically yes, practically no.

Needless to say this is a very mathematical question that almost calls for a philosophical answer. The key point to consider is usually the crystal’s surface. Here, there are most defects and the least ideal structure since there is at least one row of atoms which are lacking partners on one side. But I’ll keep that out of my reasoning for a second.

Consider an ideal crystal and a Subtle Knife. You can use the knife’s almost infinitely thin blade to cut exactly between two sheets of atoms. The initial crystal had a $1:1$ distribution, every single sheet has a $1:1$ distribution and thus so will your two bits.

Then consider cutting one of the fragments again, but three atom rows away from where you cut it before. If our ideal crystal is infinitely large, we still get a $1:1$ distribution. Now continue to consider cutting the fragments into smaller fragments until you arrive at a theoretical body of $3 \cdot 3$ size. This cube has to violate the $1:1$ condition. It will be charged depending on which atom type is present in the corners. (The same is true for any other cube of an odd side length when measured in atoms.) However, we have a charged particle (consisting of nine atoms) now. It will attrackt particles of the other charge and pull them towards it. In the long run, one of the following may happen:

An ion of the opposite charge is drawn towards the crystal and attaches to it or inserts into it.
An ion of the extraneous charge is removed.

Both will make us regain our $1:1$ crystal — but it is no longer ideal. However, see below for the evolution of a more realistic view.

Now let’s consider a slightly more realistic case. We no longer have an infinitely large ideal crystal but one with a finite size. Let’s have a cube of even atom length just for simplicity ($1:1$ ratio). And because we want to see our cube, let’s assume a macroscopic size, e.g. a cubic centimetre. Now cut it with your Subtle Knife so that the even side length is split into to cuboids with one odd length. One must have a $+1$ charge, the other must have $-1$. Thus, neither have $1:1$.

One could now say that maybe a chloride ion ‘jumps’ to the other side or something, again creating a non-ideal crystal. Or maybe moisture from the air supplies one side with a proton and the other with a hydroxide. But we could also say that we have a charge of $+1$ and one of $-1$ each delocalised over some $10^{20}$ atoms — no big deal if you think of it.

And let’s go a step further and consider a real crystal. It’s going to have defects, impurities, non-aligned domains and a surface. If it ever had a mathematically exact $1:1$ ratio, then that was purely by chance. We can still delocalise the effective charge over billions of billions of atoms. We would need an extensive charge in a very small area to cause a problem at all.

Even worse if our crystal is put into a saturated $\ce{NaCl}$ solution. (Saturated, because I don’t want it to dissolve.) All the time ions will be entering and leaving solvation in a non-coordinated manner. Both random in time and random in location. The numeric ratio of atoms will change all the time. If we consider a very sub-saturated solution, of course there will be more ions dissolving into solution than recrystallising. But also, those that end up in solution also dissociate from each other. Thus, there is nothing actually keeping an atom of a charge of one (or two, or three, see other metals) close to its counter-ion — charge dissociation is not as bad as it seems on the paper.

So real world crystals will only have the ideal stoichiometry in a short-lived transition state between being too positive and being too negative — and being philosophical I could say that that almost be like a person’s mood.


Solubility product when you add an acid to a solid? I ran into some trouble when I was resolving some exercises about solubility product principle.The exercise says:

  A concentrated, strong acid is added to a solid mixture of $0.015\ \mathrm{mol}$ samples of $\ce{Fe(OH)2}$ and $\ce{Cu(OH)2}$ placed in $1.0\ \mathrm{L}$ of water. At what values of pH will the dissolution of each hydroxide be complete?

So, I don't know how start when you have an acid and an insoluble salt when using the Ksp equation.

I know I can't to do this.

$\ce{Fe(OH)2 + HCl -> FeCl2 + H2O}$

$\ce{Cu(OH)2 + HCl -> CuCl2 + H2O}$

Not even do I have a common ion in order to establish a reaction. Do you know where I should begin? I tried using $K_\mathrm{sp}$, but I already have molar solubility, is $0.015\ \mathrm{M}$ for both. I can't ignore that. I don't find useful $K_\mathrm{sp}$ in this case. I know there's something that I'm not understanding. Hope you can help me to figure it out.

Are not you complicating the problem?

The solution is simple. You have $\pu{0.015M}$ of $\ce{Me(OH)2}$ and according to the reaction you will need twice as much acid (based on stoichiometry): $\pu{0.030M}$. Knowing that $\pu{pH = - \log [H^+]}$, the dissolution will be complete at $\pu{pH = 1.52}$.


Do plasma globes generate ozone? My kids and I were playing with a small plasma globe like this:

After touching it we all noted a faint smell like that created by other high voltage static discharges, which I always understood to be the smell of ozone.

So I guess this is actually two questions:

Is the smell associated with static electric discharge in fact ozone?
Do plasma globes like this create ozone?

(Presumably if the answer is "yes" then it would not be a good idea to leave one of these operating for an extended period in a closed space like a child's bedroom!)

Plasma globes do indeed produce small quantities of ozone when they are touched.
This is because the high-voltage within the globe is at very high frequency and this can drive high kinetic-energy electrons through the glass to produce ozone generating micro-discharges at your fingertips.
Nox generation is highly unlikely at the low energies used.


pH range outside conventional 0-14 Is a pH value outside 0 - 14 possible?
I asked my teacher who said: yes, it is, but very difficult to achieve.

Then on the internet, I found multiple answers, one saying it is but because of a fault in the pH glass meter we can't be sure, the other said that it is not possible because of the intrinsic value of water to ionize.
As you can see, I'm getting mixed answers from different sources and I'm really confused now. Could somebody here give a concise answer whether or not it is possible? A deep argument will be valuable.

Let me add a bit more to the answers already given. As has been said, $\mathrm{pH}$ is nothing but a measure of the activity of protons ($\ce{H+}$) in a solvent - $\displaystyle\mathrm{pH} = -\log_{10} \ce{\ a_{H_{\text{solvated}}^{+}}}$. In dilute solutions, a solute's activity is approximately equal to its concentration, and so you can get away with saying $\mathrm{pH} = -\log_{10} \ce{[H_{solvated}^{+}]}$.

The "normal" aqueous $\mathrm{pH}$ scale that goes from 0-14 delimits the region where both $\ce{[H^{+}]}$ and $\ce{[OH^{-}]}$ are lower than $\pu{1 mol L^{-1}}$, which is about the upper limit where concentrations and activities of solutes in solution are approximately the same and they can be used interchangeably without introducing too much error.

(The reason this is true is because a solute in a very dilute solution behaves as if it surrounded by an infinite amount of solvent; there are so few solute species that they interact very little with each other on average. This view breaks down at high concentrations because now the solute species aren't far enough apart on average, and so the intermolecular interactions aren't quite the same.)

Not only is it rarer to talk about pH for very highly concentrated solutions, because activity has a much more subtle definition than concentration, but also because you have to produce quite concentrated solutions to get unusual values of pH. For example, very approximately, to get a litre of an aqueous solution of $\mathrm{pH} = -1$ using hydrogen perchlorate ($\ce{HClO4}$, a very strong monoprotic acid) you would need to dissolve around $\pu{1 kg}$ of $\ce{HClO4}$ in a few hundred $\pu{mL}$ of water. A litre of $\mathrm{pH} = 15$ aqueous solution using potassium hydroxide ($\ce{KOH}$) would also require several hundred grams of the hydroxide in about as much water. It's so much acid/base in a small amount of solvent that many acids/bases can't even dissolve well enough to reach the required concentrations.

There is also another way to come across unusual values of pH. We almost always talk about acids in bases in water, but they also exist in other media. Notice that in the definition of pH, no direct reference is made to water. It just happens that, for water, we have:

$$\ce{H2O (l) <=> H^+ (aq) + OH^{-} (aq)} \\ K^{\pu{25^\circ C}}_{\text{autodissociation}}=k^{\pu{25^\circ C}}_\mathrm{w}=a_{\ce{H^+(aq)}} \times a_{\ce{OH^- (aq)} } \simeq \ce{[H^{+}]} \times \ce{[OH^{-}]}
 = 1.01\times 10^{-14}$$
$$\mathrm{pH}+\mathrm{pOH}=-\log\ k_\mathrm{w} \simeq 14 \ (25^\circ C)$$

However, for liquid ammonia, one has:

$$\ce{NH3{(l)} <=> H^+{(am)} + NH2^{-}{(am)}}
\\
K^{\pu{-50^\circ C}}_{\text{autodissociation}}=k^{\pu{-50^\circ C}}_\mathrm{am}=a_{\ce{H^+(am)}} \times a_{\ce{NH2^- (am)} } \simeq \ce{[H^{+}]} \times \ce{[NH2^{-}]} = 10^{-33}$$
$$\mathrm{pH}+\mathrm{p}\ce{NH2}=-\log\ k_{\mathrm{am}} \simeq 33 \ (-\pu{50^\circ C})$$

($\ce{(am)}$ stands for a substance solvated by ammonia). Hence, in liquid ammonia at $-50^\circ C$, the pH scale can easily go all the way from 0 to 33 (of course, it can go a little lower and higher still, but again now activities become important), and that neutral pH is actually 16.5.

For pure liquid hydrogen sulfate (some difficulties arise as $\ce{H2SO4}$ is a diprotic acid and tends to decompose itself when pure, but putting that aside and looking only at the first dissociation):

$$\ce{H2SO4{(l)} <=> H^+{(hs)} + HSO4^{-}{(hs)}}\\
K^{\pu{25^\circ C}}_{\text{autodissociation}}=k^{\pu{25^\circ C}}_\mathrm{hs}=a_{\ce{H^+(hs)}} \times a_{\ce{HSO4^- (hs)} } \simeq \ce{[H^{+}]} \times \ce{[HSO4-]} \simeq 10^{-3}$$
$$\mathrm{pH}+\mathrm{p}\ce{HSO4}=-\log\ k_{\mathrm{hs}} \simeq 3 \ (\pu{25^\circ C})$$

($\ce{(hs)}$ stands for a substance solvated by hydrogen sulfate). Thus, the pH in liquid hydrogen sulfate has a much smaller range than in water and ammonia, not going much further than interval from 0 to 3.

Notice also that not only does the range of $\mathrm{pH}$s change in each solvent, but there is no direct relationship between the values of pH between different solvents; formic acid in ammonia would behave as a strong acid and as such a $\pu{1 M}$ solution in ammonia would have a pH close to 0, but formic acid is actually a base in liquid hydrogen sulfate, and as such a $\pu{1 M}$ solution would have a pH above 1.5. An approximate comparison between the pH ranges and their relative positions in different solvents can be found in the figure below, from A Unified pH Scale for All Phases.


Not your ordinary hard water deposit We have a well that delivers hard water and we do not have a water softener. (There was a “green sand filter” that utilizes permanganate to oxidize other metals on the system, but it has been disconnected.) Analysis of the well water shows the presence of iron in noticeable amounts. No other metal ions were specifically identified. The system does have an ozonator to oxidize Fe(II) to Fe(III). Over the last couple of years, the porcelain toilet bowls have developed a blue ring at the air/water interface. When you scrub on the blue ring, it doesn’t feel like there is a deposit present (like normal hard water deposits). The ring is unresponsive to ammonia (household), strong acid (6 M HCl) or strong base (Drano), as well as to hydrogen peroxide (30 %).  (We haven’t tried a good reducing agent yet.) So, my question is, does anyone have an idea of what is causing the blue color? Is it possible that porcelain reacted chemically with salts, urea, uric acid, oxygen, etc?  Not an earth-shattering theoretical question, but I’m chemically curious as to what this could be if anyone has any ideas.

Toilet bowls are made of ceramics which can contain feldspar. Toilet bowls also often have a feldspar glaze. There is a study that disolved Cr(VI) can be removed by feldspar.

Studies on the removal of Cr(VI) from waste-water by feldspar
http://onlinelibrary.wiley.com/doi/10.1002/jctb.280530204/abstract

The IMA currently recognizes 90 Cr bearing minerals. 16 minerals if you only count those that contain also Si. Feldspar can be either K, Na or Ca-Feldspar. Of those 16 minerals 6 contain Na, 2 K and 5 Ca. Reading those papers might even narrow the choices down to a couple of minerals that can form in standard-toilet-bowl-conditions.

Mineral database that lets you select minerals by their chemistry:
http://rruff.info/ima/

I think that unless you are willing to donate your toilet bowl to science that the answer won't get any more precise than this.

Edit:
Cu would be the favorite choice for a blue color mineral (Azurite, ...). There is also a study about the removal of copper by feldspar:

Experimental studies of the interaction of aqueous metal cations with mineral substrates: Lead, cadmium, and copper with perthitic feldspar, muscovite, and biotite
http://www.sciencedirect.com/science/article/pii/S0016703797001178

Again there are 603 Cu-minerals. Cu+Si=32 minerals. With the common Feldspar cations your left with about 10 each. I don't think the minerals could be recognized visually. The favorite method would be x-ray diffraction of a powder sample of the blue stuff.


Why would breathing pure oxygen be a bad idea? My textbook mentions that SCUBA tanks often contain a mixture of oxygen and nitrogen along with a little helium which serves as a diluent.

Now as I remember it, divers take care not to surface too quickly because it results in 'the Bends', which involves the formation of nitrogen bubbles in the blood and is potentially fatal.

If that's the case, why not use pure oxygen gas in SCUBA tanks? It seems like a good idea since it would

a) Enable divers to stay underwater for longer periods of time (I keep hearing that ordinary SCUBA tanks only give divers a pathetic hour or so of time underwater.
b) Possibly eliminate the chances of developing 'the Bends' upon surfacing. Well, it seems plausible, that is if the diver were to take a 10 minute deep-breathing session with pure oxygen to flush out whatever nitrogen's there in his lungs before hooking up a cylinder of pure oxygen and going for a dive. So if there's no gaseous nitrogen in his lungs and blood, then he wouldn't have to worry about nitrogen bubbles developing in his system.

Now those two possible advantages aren't hard to overlook, but since no one fills SCUBA tanks with pure oxygen, there must be some reason that I've overlooked, that discourages divers from filling the tanks with pure oxygen. So what is it?

Also, I hear that the oxygen cylinders used in hospitals have very high concentrations of oxygen; heck, there's one method of treatment called the Hyperbaric Oxygen Therapy (HBOT) where they give patients 100% pure oxygen at elevated pressures.

Hence I doubt whether the increase in pressure associated with diving is the problem here. So I reiterate:

Why is it a bad idea for divers to breathe pure oxygen underwater?

I guess most of the recent answers have kinda missed a main point, so I'll rephrase the question:

Why is it a bad idea for divers to breathe pure oxygen underwater? If it is indeed due to pressure considerations as most sources claim, then why doesn't it seem to be a problem when patients are given 100% pure oxygen in cases like the HBOT (which is performed at elevated pressures) ?

The other answers here, describing oxygen toxicity are telling what can go wrong if you have too much oxygen, but they are not describing two important concepts that should appear with their descriptions. Also, there is a basic safety issue with handling pressure tanks of high oxygen fraction.

An important property of breathed oxygen is its partial pressure. At normal conditions at sea level, the partial pressure of oxygen is about 0.21 atm. This is compatible with the widely known estimate that the atmosphere is about 78% nitrogen, 21% oxygen, and 1% "other". Partial pressures are added to give total pressure; this is Dalton's Law. As long as you don't use toxic gasses, you can replace the nitrogen and "other" with other gasses, like Helium, as long as you keep the partial pressure of oxygen near 0.21, and breathe the resulting mixtures without adverse effects.

There are two hazards that can be understood by considering the partial pressure of oxygen. If the partial pressure drops below about 0.16 atm, a normal person experiences hypoxia. This can happen by entering a room where oxygen has been removed. For instance, entering a room which has a constant source of nitrogen constantly displacing the room air, lowering the concentration -- and partial pressure -- of oxygen. Another way is to go to the tops of tall mountains. The total atmospheric pressure is lowered and the partial pressure of oxygen can be as low as 0.07 atm (summit of Mt. Everest) which is why very high altitude climbing requires carrying additional oxygen. Yet a third way is "horsing around" with Helium tanks -- repeatedly inhaling helium to produce very high pitched voices deprives the body of oxygen and the partial pressure of dissolved oxygen in the body falls, perhaps leading to loss of consciousness.

Alternatively, if the partial pressure rises above about 1.4 atm, a normal person experiences hyperoxia which can lead to oxygen toxicity (described in the other answers). At 1.6 atm the risk of central nervous system oxygen toxicity is very high. So, don't regulate the pressure that high? There's a problem. If you were to make a 10-foot long snorkel and dive to the bottom of a swimming pool to use it, you would fail to inhale. The pressure of air at your mouth would be about 1 atm, because the 10-foot column of air in the snorkel doesn't weigh very much. The pressure of water trying to squeeze the air out of you (like a tube of toothpaste) is about 1.3 atm. Your diaphragm is not strong enough to overcome the squeezing and fill your lungs with air. Divers overcome this problem by using a regulator (specifically, a demand valve), which allows the gas pressure at the outlet to be very near that of the ambient pressure. The principle job of the regulator is to reduce the very high pressure inside the tank to a much lower pressure at the outlet. The demand valve tries to only supply gas when the diver inhales and tries to supply it at very nearly ambient pressure. Notice that at depth the ambient pressure can be much greater than 1 atm, increasing by about 1 atm per 10 m (or 33 feet). If the regulator were to supply normal air at 2 atm pressure, the partial pressure of oxygen would be 0.42 atm. If at 3 atm, 0.63 atm. So as a diver descends, the partial pressure of oxygen automatically increases as a consequence of having to increase the gas pressure to allow the diver to inflate their lungs. Around 65 m (220 ft), the partial pressure of oxygen in an "air mix" would be high enough to risk hyperoxia and other dangerous consequences.

Now imagine a gas cylinder containing 100% oxygen. If we breathe from it at the surface, the partial pressure of oxygen is 1 atm -- high, but not dangerous. At a depth of 10 m, the partial pressure of supplied oxygen is 2 atm -- exceeding acceptable exposure limits. This is a general pattern -- raising the oxygen fraction of diving gasses decreases the maximum diving depth.

And you can't lower the partial pressure much because the lower limit, 0.16 atm, isn't that much lower than the 0.21 atm of sea level atmosphere.

One general category of solutions is to change gas mixes at various depths. This is complicated, requires a great deal of planning, and is outside the scope of your question. But it is certainly not as straightforward as just simplifying the gas mixtures or just raising the partial pressure of oxygen.

Additionally, compressed oxygen is a relatively annoying gas to work with. It is not itself flammable, but it makes every nearby organic thing flammable. For instance using grease or oil on or near an oxygen fitting risks spontaneously igniting the grease or oil. Merely having grease on your hand while handling oxygen refilling gear (with a small leak) can burn your hand.


Bromination of hexene in presence of UV light or heat The reaction between hexene and bromine in presence of light gives 3-bromocyclohexene. Why is 1,2-dibromocyclohexane not formed instead?

Both products are formed although 3-bromocyclohexene is the major product.

Formation of major product: 3-bromocyclohexene

Under UV light, $\ce{Br2}$ undergoes homolytic splitting to generate $\ce{Br*}$ radicals:

$$\ce{Br2 ->[hv] 2Br*}$$

The formation of 3-bromocyclohexene is an example of substitution of alkanes, which require the free-radical mechanism:

In the first step of the upper mechanism, which is also the rate-determining step, a stable allyl radical is generated, which is stabilized by resonance:

As a result, the activation energy of the first step is significantly lowered.

Formation of minor product: 1,2-dibromocyclohexane

Individual bromine radicals are not electrophilic enough to attack the double bond in the cyclohexene, so the formation of 1,2-dibromocyclohexane requires the ions mechanism, typical for addition reactions (the lower mechanism in the following diagram).

The first step in this mechanism is the rate-determining step. In this step, bromine is ionized, which requires a moderate amount of activation energy, albeit still much higher than the rate-determining step of the upper mechanism.

Conclusion

Therefore, the upper mechanism occurs at a much faster rate than the lower mechanism, which makes the major product 3-bromocyclohexene and the minor product 1,2-dibromocyclohexane.

PS: Many people think that addition reaction is very fast. It is only true in water, where the bromine ion is stabilized by solvation in water.

Disclaimer: The mechanism I used for the addition reaction probably contradicts with your book. However, it does not matter. The point is that an ion is formed which makes the activation energy high.


How to calculate n factor for O2 and H2 in this reaction? How do I find n factor of $\ce{O2}$ and $\ce{H2}$ if I want to use equivalence principal on a reaction like $\ce{O_2 +2 H_2 -> 2 H2O}$?

Here is the unbalanced reaction
$H_2 +O_2----> H_20$
I hope you know how to calculate oxidation number
 On product side
In $H_20$ oxidation number of $H$ is  $+1$ and $O$ is $-2$
On reactant side
oxidation number of $H$ and $O$ is  $0$
So n-factor is magnitude in change of oxidation number
So n factor of $H$ is $1-0$ i.e $1$ and $O$ is $-2-0$ i.e $2$(magnitude)
Now to balance the reaction (for 2 reactant rex) you just interchange the n factors
You get
$1O_2 +2H_2 -----> 2H_20$
Now to calculate equivalent weight formula is
$E.w=molar mass/N.F$


Why doesn't this structure represent a substituted pentane? In the following ASCI structure (single bond under "$\ce{CH}$" containing $\ce{CH2,CH2, CH3}$)

$$
\begin{align*}
\ce{CH3-CH2-CH2-&CH-CH3}\\
&|\\
\ce{&CH2-CH2-CH3}
\end{align*}
$$

The correct way to name it is "4-methlyheptane".

But when I do it the way I have been taught I get the following "4-propylpentane".
Here are my systematic steps:

Firstly, I count the longest carbon chain. In this case its 5 carbons together so suffix -pentane. Then I look to see if there are any branched chains and number it from left to right. In this case the branched chain is on number 4. The branched chain contains $\ce{CH2-CH2-CH3}$, resp. $\ce{C3H7}$, aka propyl.

Ergo, 4-propylpentane, however, the correct answer as mentioned above is 4-methylheptane.

How is it 'heptane' ? The longest chain has FIVE carbons. I got the '4-' part correct, but how is $\ce{CH2-CH2-CH3}$ Methyl? Is this a mistake in my book?

You need to find the longest continuous carbon chain of:

$$
\begin{align*}
\ce{CH3-CH2-CH2-&CH-CH3}\\
&|\\
\ce{&CH2-CH2-CH3}
\end{align*}
$$

Stretch that out to $\ce{CH3CH2CH2CH(CH3)CH2CH2CH3}$ (with the $\ce{-CH3}$ coming off of the $\ce{CH}$) and you'll see that the book is correct.  The "side chain" of $\ce{-CH2CH2CH3}$ is actually part of the main carbon chain.  As far as the numbering, counting from either side give the 4- position due to the symmetry of the molecule.


How can I determine the chemical composition of local rocks and minerals? Specifically I live in the U.S. so I'm asking about an ordinary rock that you might find in your backyard or near a playground. Humans have discovered gun powder, petroleum, silica, iron, and many more types of useful materials by observing and experimenting with materials which lead to said discoveries.

Sometimes I'm walking outside and think to myself how people largely just ignore stuff outside like the rocks, dirt, grass and things like that, but I think if you knew how to separate the chemicals in these materials, then it could possibly lead to discoveries that might have otherwise been overlooked.

Especially rocks, because they can be cracked open and looked at with a microscope. Can't these rocks be used as raw materials to engineer something? If so, shouldn't they have inherent value? If not, why? If so, how?

What are the typical chemical elements that you will find in these types of materials, and how can you observe the chemical compositions by either isolating them or via some other physical process?

Let's talk about rocks. The art and science of getting useful things from rocks is one of the oldest in human civilization. All of the example products you mention (gunpowder, petroleum, silica, and iron) were discovered or developed from experimentation on rocks.

In the ancient world, composition was determined by qualitative analysis, or a series of chemical tests and determination of important physical properties. For example, if a lump of shiny yellow stuff was soft, melted easily, did not lose its lustre in air or water, did not dissolve in vitriol (sulfuric acid) or aqua fortis (nitric acid), but did dissolve in aqua regia (made then by dissolving sal ammoniac (ammonium chloride) in aqua fortis), then we probably have gold.

In the modern age, the field of analytical chemistry was developed to answer the very questions you asked: What is this stuff? What are its constituents? How can we separate them? How do we determine that we successfully separated/purified them?

To continue the rock example, the modern technique of x-ray fluorescence is used to determine the elemental compositions of rocks (and other things, like ceramic fragments found at archaeological sites). In this technique, high frequency x-rays are used to excite the core electrons in the atoms of the sample. When these core electrons relax, they spit out x-rays of characteristic frequency/wavelength that is specific to the element. These emitted x-rays are collectively measured and converted into a spectrum like the one below.

This spectrum indicates that many elements will have more than one spectral line because there is more than one electronic transition that is possible. It also demonstrates the clear resolution of one element from another. After some maths (and better yet a calibration), you can use the heights of these peaks to determine the specific ratios of elements in your sample.

Not everyone is ignorant of the utility of natural objects. Most of us who become scientists do so to study the natural world and learn from it. For example, organic and medicinal chemists do not design new drug molecules from scratch. Rather, they take cues from nature. A folk legend about the curative properties of the leaves of a plant that grows along the riverside might turn in to an effort to extract and analyze the phytochemicals in those leaves, including throwing those compounds at bacteria, viruses, cancer, etc. Any that look promising are studied further. Taxol (paclitaxel) is a successful chemotherapeutic agent that was isolated from the bark of the pacific yew. It is now mass-produced by a semi-synthetic pathway which isolates a late-stage precursor from the needles of the European yew.

If you are specifically interested in rocks, then you should study geoscience.


What is the geometrical structure of OF₂? Is it linear, angular, triangular or tetragonal? Well, since $\ce{O}$ has 2 electron matches of his own, I think it's tetragonal, but in wikipedia it is angular... Why? What about $\ce{CS2}$, is it linear?

Oxygen difluoride ($\ce{OF2}$) is analogous to water ($\ce{OH2}$) both in terms of structure and hybridization.

$\ce{OF2}$ has 4 ligands around the central oxygen, 2 $\ce{O-F}$ bonds and 2 lone pairs.  Just like in water these 4 ligands assume a roughly tetrahedral geometry around the central oxygen. Again like in water, the $\ce{F-O-F}$ angle is less than the perfect tetrahedral angle (109.5°), in this case being around 103°.

Carbon disulphide ($\ce{CS2}$) is linear.  It is analogous to the 3 carbons in allene, in both compounds the central carbon is $\ce{sp}$ hybridized making the $\ce{X=C=X}$ structure linear.


1,4 addition to α,β-unsaturated carbonyls? This is a numbering problem.

This is a 1,2-addition to an α,β-unsaturated carbonyl:

This is a 1,4-addition to an α,β-unsaturated carbonyl:

I don't see why the second reaction would be called 1,4. I'd call it rather 3,4, so obviously I am not getting the numbering rule. Help?

It is called 1,4 - addition because when you go through the mechanism of this reaction, then actually you will find that the carbonyl bond will break and oxygen will get negative charge.
Then nucleophilic will attack on beta carbon, i.e. carbon at 4th position and proton will migrate to negative oxygen which is at first position. Double bond will form between the carbonyl carbon and  carbon at 3rd position.

Thus, 1,4-addition takes place. But in the last step, keto-enol tautomerism takes place. And proton transfer takes place from oxygen to the carbon alpha to carbonyl. Therefore, final product is again saturated carbonyl compound.


Does 'Stoichiometry' refer to the mole ratio of atoms in compounds and the mole ratio of compounds involved in chemical reactions? For my intro to chemistry class, I was asked to answer whether 'stoichiometry' refers to the mole ratio of atoms in compounds and the mole ratio of compounds involved in chemical reactions?

To the best of my knowledge those ratios are an essential part of stoichiometry, but stoichiometry is a discipline, and not a ratio, so the answer would be false.

Is that correct?

That's right -- stoichiometry is a field which uses these numbers and the relationships between reactants and products, but not the numbers itself. The mole ratio is just the mole ratio, whereas stoichiometry is the entire discipline of using mole ratios to calculate relative quantities of reactants and products in reactions.


Average bond energy of H−F and H−O hydrogen bonds Which of the following hydrogen bonds (shown by the dotted line) has the highest average bond energy?
A) $\ce{F-H\bond{...}F}$
B) $\ce{F-H\bond{...}O}$

I am thinking the answer is $A$ since the covalently bonded F atom, being highly electronegative, would induce a high partially positive polarity on the H atom. But on the other side, the more electronegative atom F will form a stronger Hydrogen bond than the comparatively less electronegative O atom.
Any flaw in my reasoning?

So fluorine does have the ability to hydrogen bond. However, they are usually transient and depend significantly on the species that is donating the H-bond along with some other factors (i.e. what phase the molecules are in, temperature, etc...)
Fluorine is so electronegative that it holds its electrons much "tighter" then other atoms, making it a poor H-bond acceptor. Oxygen, being less electronegative, is a better H-bond acceptor since the electrons aren't held onto as tightly, allowing them to interact more with other atoms.
Since bond energy is "the average value of the gas-phase bond-dissociation energy for all bonds of the same type within the same chemical species. The larger the average bond energy, per electron-pair bond, of a molecule, the more stable and lower-energy the molecule" (IUPAC WEBSITE), I would suggest that the H-bond with oxygen has a higher bond energy, as it is the stronger interaction of the two.
EDIT
The OP requested some sources, in which I failed to do initially, so I apologize for that but here they are.

J. Am. Chem. Soc. 1983, 105, 10, 3206–3214

Tetrahedron, Vol. 52, No. 38, pp. 12613-12622, 1996

J. Am. Chem. Soc. 1962, 84, 16, 3221–3222

Here is a quote from reference 2.

Theoretical calculations variously estimate 5 the strength of a F...H bond to be between 2 to 3.2 kcal mo1-1.
This can be compared 6 to an O...H hydrogen bond which is typically between 5 - 10 kcal mo1-1. Consistent
with this the electrostatic influence of fluorine is approximately half that of oxygen 7. Thus the greater
electronegativity and lower polarisability of fluorine over oxygen, suppresses its electrostatic influence and
renders it a poorer hydrogen bond acceptor.

The last reference discusses hydrogen bonding in halides, not too sure how much it compares them to oxygen bonding but it is certainty worth taking a quick look.


Is Neutron radiation considered electromagnetic wave? I'm confused because gamma rays are classified as an electromagnetic wave. But why is neutron radiation not considered electromagnetic wave? I know Alpha and Beta, as well as positron and proton, is not considered electromagnetic wave because it can be deflected by magnetic or electric field (because they are + or -).

But a neutron is neutral, so why is it not electromagnetic wave? Can it be deflected too by electric and magnetic field? Is that why? Or is there another reason?

Electromagnetic radiation consists of electromagnetic waves (oscillations of the electric and magnetic field), whereas corpuscular radiation consists of actual particles. Certainly, wave–particle duality can make the distinction between waves and particles fuzzy. However, electromagnetic radiation (whether described as wave or as photons) is massless (its invariant mass is zero) and it travels at the speed of light. In contrast, the particles of corpuscular radiation have a non-zero mass and therefore a velocity below the speed of light. The particles of corpuscular radiation can be charged positively (e.g. α particles) or negatively (e.g. β− particles), or they can have no charge at all.

Neutron radiation is a corpuscular radiation because it consists of particles (i.e. neutrons). Neutrons are subatomic particles with no charge and a mass of $m=1.674\,927\,471(21) \times 10^{-27}\ \mathrm{kg}$ [source], which is slightly larger than the mass of a proton.

The kinetic energy of free neutrons can vary significantly. For example, the so-called fast neutrons, which are released during nuclear fission, have energies above $E_\text{kin} = 1\ \mathrm{MeV} \approx 1.6 \times 10^{-12}\ \mathrm{J}$. The thermal neutrons, which are used to initiate nuclear fission in moderated reactors, have energies below $E_\text{kin} = 0.1\ \mathrm{eV} \approx 1.6 \times 10^{-20}\ \mathrm{J}$.

Since, in classical mechanics, kinetic energy $E_\text{kin}$ is
$$E_\text{kin} = \tfrac{1}{2}mv^2$$
where $m$ is mass and $v$ is velocity,
the velocity of a neutron with a kinetic energy of $E_\text{kin} = 0.1\ \mathrm{eV} \approx 1.6 \times 10^{-20}\ \mathrm{J}$ may be estimated as
$$\begin{align}
v &= \sqrt {2\frac{E}{m}}\\[6pt]
&= \sqrt {2\frac{1.6 \times 10^{-20}\ \mathrm{J}}{1.674\,927\,471 \times 10^{-27}\ \mathrm{kg}}}\\[6pt]
&= 4.4\times 10^3\ \mathrm{m\ s^{-1}}
\end{align}$$
which is clearly less than the speed of light.


Reaction between halogens and concentrated nitric acid I have read that the reaction between $\ce{I2}$ and conc. $\ce{HNO3}$ gives $\ce{HIO3}$ and $\ce{NO2},$ but there is no mention of the other halogens' products. Do they all give halous acids and nitrogen dioxide? Specifically with chlorine, how does the reaction proceed?

This question can easily be addressed considering reduction potential of each compound. Concentrated nitric acid is a powerful oxidizing acid, which has a high reduction potential:

$$
\begin{align}
\ce{NO3- + 4H+ + 3e- &<=> NO + 2H2O} &\quad E^\circ &= \pu{0.957 V} \tag1\\
\ce{NO3- + 3H+ + 2e- &<=> HNO2 + H2O} &\quad E^\circ &= \pu{0.934 V} \tag2\\
\ce{2NO3- + 4H+ + 2e- &<=> N2O4 + 2H2O} &\quad E^\circ &= \pu{0.803 V} \tag3
\end{align}
$$

Now, let's look at reduction potential of halides:

$$
\begin{align}
\ce{F2 + 2e- &<=> 2F-} &\quad E^\circ &= \pu{2.866 V} \tag4\\
\ce{Cl2 + 2e- &<=> 2Cl-} &\quad E^\circ &= \pu{1.3583 V} \tag5\\
\ce{Br2 + 2e- &<=> 2Br-} &\quad E^\circ &= \pu{1.066 V} \tag6\\
\ce{I2 + 2e- &<=> 2I-} &\quad E^\circ &= \pu{0.5355 V} \tag7\\
\end{align}
$$
Accordingly, we can list oxidizing power at standard conditions in decreasing order:

$$\ce{F2} > \ce{Cl2} > \ce{Br2} > \ce{NO3-} > \ce{I2}$$

Therefore, at standard condition, concentrated $\ce{HNO3}$ acid (meaning $\ce{NO3-}$ in acidic medium) can easily oxidize only $\ce{I2}.$ One such example is documented in Ref.1. According to the authors, the oxidation of iodine by hyperazeotropic nitric acid ($\pu{16–21 M}$) was shown to proceed in at least two steps:

$$
\begin{align}
\ce{I2 + 3HNO3 &<=>[$k_1$][$k_2$] 2I+ + 2NO3- + 2HNO2 + H2O}\\
\ce{I+ + 6HNO3 &<=>[$k_3$][$k_1$] I^5+ + 4NO3- + 2HNO2 + 2H2O}
\end{align}
$$

Since reduction potentials of $\ce{HNO3}$ and $\ce{Br2}$ are in such close range that each can oxidize the other. In fact, the kinetic studies of this phenomenon has been performed in literature (Ref.2).

In different aspect, there is a publication (Ref.3) claiming that the presence of bromine in fuming nitric acid makes it a brominating reagent (electrophilic aromatic substitution):

This bromination has worked well with deactivated aromatic nucleus. In absence of bromine, the reagent nitrate the ring. Thus, it is safe to suggest that $\ce{HNO3}$ might promote formation of '$\ce{Br+}$', meaning oxidation.

References:

J. C. Mailen, T. O. Tiffany, "The reaction of iodine with concentrated nitric acid," Journal of Inorganic and Nuclear Chemistry 1975, 37(1), 127-132 (https://doi.org/10.1016/0022-1902(75)80139-4).
Istvan Lengyel, Istvan Nagy, Gyorgy Bazsa, "Kinetic study of the autocatalytic nitric acid-bromide reaction and its reverse, the nitrous acid-bromine reaction," J. Phys. Chem. 1989, 93(7), 2801–2807 (https://doi.org/10.1021/j100344a021).
Alexander M. Andrievsky, Vera I. Lomzakova, Mikhail K. Grachev, Mikhail V. Gorelik, "Aromatic Bromination in Concentrated Nitric Acid," Open Journal of Synthesis Theory and Applications 2014, 3(2), Article ID:44941, 5 pages (doi: 10.4236/ojsta.2014.32003).


Why is dimethylglyoximate a bidentate ligand and not a tetradentate ligand? Why can't the 2 oxygen atoms donate a lone pair and hence become binding sites? Is it because of the angle made by the oxygen and Nitrogen atoms?

You are right. I took this two pictures from Wikipedia to compare EDTA (a tetradentate) to dmg (a bidentate).

For EDTA you got this:

For dmg, on the other hand, you got this structure:

You can notice that for both of the cases, the metal complex is connected to the ligands through five-membered rings. This observation is general in chemistry: to have stable rings, you need five (sometimes six) members for those rings.

Now it is clear to see a general rule of thumb for multidentate ligands: the two donating atoms must be connected through 2 (sometimes 3) atoms. For dmg, you can see that the two nitrogens satisfy this rule, while the oxygen-nitrogen pairs do not.

For EDTA, you can see that going through the molecule, you can find an oxygen, a nitrogen, a second nitrogen and a second oxygen, all bridged by 2 atoms to each other (while the two oxygens of the carboxyl group are not connected by 2 bridge atoms, otherwise you might even have a hexadentate ligand).


Which is anode and which is cathode? A maybe (hopefully) simple question about the denotations of "anode" and "cathode". The below image is a schematic of a polymer solar cell (Source (WBM)). (The figure text is quoted as well for the sake of context).

My question is only about the naming of electrodes. Isn't it correct that

the cathode is the reducing electrode, which supplies electrons, while
the anode is the oxidizing electrode, which accepts electrons?

Why would electrons (black spheres) move towards the negatively charged cathode, while positive holes (white spheres) move towards the positively charged electron-absorbing anode? Wouldn't the opposite make sense?

Is it a mistake on the drawing or should I re-think my understanding of the terms "anode" and "cathode"?

  Figure 1. The working principle of the solar cell. Light enters the cell through the transparent anode, and is absorbed in the bulk heterojunction layer through generation of excitons (1). The excitons diffuse in the bulk heterojunction until they either recombine or reach a donor-acceptor interface, where they separate into electrons (black) and holes (white) (2). The electrons and holes will then move to the respective anode and cathode, through the donor and acceptor material phase (3).

Edit

If the figure is correct and I am the one being confused, then there are several other cases that seem to teach false. E.g. the following "PANIC"-remembering-rule to remember, which is cathode and which is anode (from this source, slide 8):

"*Negative is cathode", it says, and we are back to my question in the top: why do the electrons (black spheres) move towards the negative cathode in that schematic?

Many more diagrams agreeing with this latter one are found from a simple google search.

Regardless of the polarity, the electrode where oxidation takes place is the called the anode and therefore reduction must take place at the cathode. The electron flow from the anode to cathode is as shown in your top picture.

By way of an example, in an electrochemical cell, suppose that two beakers are connected by a salt bridge. In one beaker is a strip of zinc metal immersed in a $\ce{Zn(NO3)2}$ solution and in the other a strip of silver in $\ce{AgNO3}$ solution. The two metals are then connected by a wire and a current will flow. (The salt bridge supplies a return path so that the solutions remain electrically neutral). The redox of the Zn electrode is $-0.763$ V and that of the Ag $+0.799$ V. This means that electrons will flow from the zinc to the silver electrode. The zinc is oxidised to $\ce{Zn^{2+}}$ and the electrons are released into the metal and flow to the silver electrode through the wire. The zinc electrode is the anode and the silver the cathode.


Stereochemistry: synthesis of an epoxide This is a procedure to synthesize a propylene oxide.

I wonder if the inversion occurs or not after the reaction. That is, I want to know which product is correct in the diagram above.

I thought that this reaction is an SN2 reaction, so inversion occurs. However, I'm a little confused because the whole reaction is intramolecular.
Is it correct that this reaction is SN2, and the inversion occurs?

"Is it correct that this reaction is SN2, and the inversion occurs?" Yes, and yes, so the product is the one on the top.

"Inversion" in this context doesn't mean that every substituent has to be flipped. Maybe it's clearer if you take away the intramolecular aspect of it. This is a typical SN2 reaction as introduced in many textbooks:

Would you agree, here, that the stereochemistry at carbon has undergone inversion? It is nothing to do with the methyl group or the hydrogen, but rather it is because of the fact that the chloride (that was originally pointing down) was replaced with an oxygen (that is now pointing up).

Now to make it an intramolecular reaction, just replace the R with a single bond to the OH.


Number of pi-electrons in an energy state I am learning quantum chemistry and I am trying to understand the statement in a .pdf I'm reading which states
"The energy transition associated with the spectral transitions observed is:
$$\Delta E = E_{LUMO} - E_{HOMO} = \frac{(N+1)h^2}{8mL^2}$$
where $L$ is the length of the "box" (it's talking about particle in a box), $m$ is the mass of the electron, and $N$ is the number of $\pi$-electrons in the system. I know due to the solution of the Schrodinger equation that $$E = \frac{n^2 h^2}{8mL}$$
So, $$\Delta E = E_{HOMO} - E_{LUMO} = E_{k+1} - E_{k} = \frac{(2k+1)h^2}{8mL^2}$$
So, I want to figure out why $2k = N$?

Your calculation is indeed correct. However, because each electron in the $\pi$ system can have two spin states, there are two electrons per energy level of the “particle in a box” Hamiltonian. Hence, $N=2k$.


How to calculate molarity and the number of molecules for a mixture in a simulation box? I want to do a simulation of a mixture which consists a total of $512$ molecules of glycine, ammonia, and water molecules mixture in a cubic box of, say, length $\pu{2.6 nm}$; $\pu{2.6 nm}^3 = \pu{17.576 nm3}$ in terms of volume.

I want [1 glycine molecule + $x$ ammonia molecules + $y$ water molecules] which together comes to a total of $512$ molecules and in a box volume of $\pu{17.576 nm3}$ and molarity should be $\pu{15 M}$.

Actually here I am fixing the glycine molecule which is only one and varying only the number of ammonia and water molecules to get different molar concentrations, say, $\pu{5 M / 10 M / 15 M}$ and $\pu{18 M}$, where the total number of molecules is constant, i.e. $512$ molecules only (see the example below). Since a maximum of $\pu{18 M}$ solution can be prepared (since ammonia solubility is $\pu{31 g}$ in $\pu{100 mL}$), so, I am trying to study with varying the $\%$ composition of ammonia and water molecules.

In this case I want to get a concentration of $\pu{15 M}$.
How to calculate the number of $x$ ammonia and $y$ water molecules that need to be taken in order to get $\pu{15 M}$, considering the conversion of number of molecules to mass, amount of substance, etc..

For example:

$1$ molecule of glycine + $52$ molecules of ammonia + $459$ molecules of water is $10 : 90$ mixture (total $512$ molecules);
$1$ molecule of glycine + $104$ molecules of ammonia + $408$ molecules of water is $20 : 80$ mixture (total $512$ molecules).

Then what would be the molarity of the above two mixtures ($1 : 90$, $20 : 80$) considering the same volume ($\pu{17.576 nm}$)?

I would use a slightly different approach than andselisk, based on the volume on your box, and that I don't think you can neglect the volume of your ammonia molecules. Given $V(\text{sol}) = \pu{17.576E-27 m3}$ and $c(\ce{NH3}) = \pu{15.0E3 mol  m-3}$ we can calculate the number of ammonia molecules we need:
\begin{align}
  && c(\ce{NH3}) &= \frac{n(\ce{NH3})}{V(\text{sol})};\\
  && n(\ce{NH3}) &= \frac{N(\ce{NH3})}{N_\mathrm{A}};\\
&\to&
     c(\ce{NH3}) &= \frac{N(\ce{NH3})}{N_\mathrm{A}\cdot V(\text{sol})}\\
&\Longleftrightarrow&
     N(\ce{NH3}) &= c(\ce{NH3})\cdot N_\mathrm{A}\cdot V(\text{sol}) \approx 159
\end{align}

Since you want to limit the number of molecules to $512$, we know that all other molecules have to be water:
$$N(\ce{H2O}) = 512 - N(\ce{Gly}) - N(\ce{NH3}) = 352$$

Lastly, let's check whether the density is anywhere close to what we would expect, we take $M_\mathrm{m}(\ce{Gly}) = \pu{75 u}$, $M_\mathrm{m}(\ce{H2O}) = \pu{18 u}$, and $M_\mathrm{m}(\ce{NH3})= \pu{17 u}$.
\begin{align}
  \rho(\text{sol}) &= \frac{m(\text{sol})}{V(\text{sol})}\\
  &= \frac{\left(
       N(\ce{Gly})M_\mathrm{m}(\ce{Gly})
     + N(\ce{H2O})M_\mathrm{m}(\ce{H2O})
     + N(\ce{NH3})M_\mathrm{m}(\ce{NH3})\right)\frac{m_\mathrm{u}}{\pu{u}}
     }{V(\text{sol})}\\
  \rho(\text{sol}) &\approx \pu{861 kg m-3} = \pu{0.861 g cm-3}
\end{align}

Citing Wikipedia's entry of ammonia I would say you are quite close:

  Solvent properties
  Ammonia is miscible with water. [...] The maximum concentration of ammonia in water (a saturated solution) has a density of $\pu{0.880 g cm-3}$ and is often known as '.880 ammonia'. [...]

The Handbook of Chemistry and Physics has a concentration table (91st Edition p. 8-53; currently 98th Edition, also online), and I'll reproduce a couple of values for comparison, at $\pu{20 ^\circ C}$.

\begin{array}{rrr}
  \text{Mass\%} & c / \pu{mol L-1} & \rho / \pu{g cm-3}\\\hline
    5.0   &   2.868    &   0.9770\\
    9.0   &   5.080    &   0.9613\\
   12.0   &   6.695    &   0.9502\\
   18.0   &   9.823    &   0.9294\\
   24.0   &  12.826    &   0.9102\\
   28.0   &  14.764    &   0.8920\\\hline
\end{array}

In conclusion, you are a few water molecules short, or your box is slightly too large, or your solution is not dense enough. I would expect some trade-offs when doing such a study and keeping the named constraints, so I guess you should have a close look at everything and settle for the optimum you can afford.


d-orbital splittings in WS2 monolayer The formerly degenerate $d$ orbitals of the tungsten atoms in the $\ce{WS2}$ monolayer are split into three groups: (1) $\mathrm{d}_{z^2}$, (2) $\mathrm{d}_{x^2-y^2}, \mathrm{d}_{xy}$ and (3) $\mathrm{d}_{xz}, \mathrm{d}_{yz}$ in the order of increasing energy with the large band gap between the first and the second groups.

Now the valence band maximum consists dominantly of $\mathrm{d}_{x^2-y^2}, \mathrm{d}_{xy}$ orbitals, whereas the conduction band minimum is composed mainly of $\mathrm{d}_{z^2}$ metal orbital.

So, the order of $\mathrm{d}$ orbitals in the first two splitting groups now is reversed, since the valence band is lower in energy than the conduction band. Where is the correspondence between these two facts?

One vague explanation I came up with is the following.

A prismatic arrangement of atoms (first figure) does not capture fully all symmetries of a hexagonal arrangement in a crystal lattice (which is the case of the second figure). However, the most overlap (in terms of the number of symmetries) between the two occurs at the $\Gamma$ point, since it is the point of the highest symmetry. Thus, the first figure resembles mostly the orbitals at the $\Gamma$ point in the first Brillouin zone of the crystal lattice.
Please, if I am thinking in a right direction, refine my answer to make it more strict and rigid.


Reproducibility and Standard Deviation Would a set of chemical data with a relative standard deviation of 0.6% or 20% be more reproducible?

My reasoning is that because it's easier to make mistakes and have a wide range of data, the 20% would be more reproducible, even though it is not accurate.

A RSD of 0.6% would be extremely precise and therefore harder to reproduce.

At least this is my thoughts.  Thank you for your help!

Both accuracy and precision are desired for proper data analysis. They are not mutually exclusive. This is heavily dependent on the person and the measurement tool as well. You must have heard of the idiom "A bad workman quarrels with his tools"... this is true in science as well. Even if you have million dollar instrument, and the operator is unskilled, the data will be garbage. The premise that it is easier to make mistakes is not valid and generally RSD of 1-5 % is routine in scientific measurements. An RSD of 20% would be considered too high.


NMR coupling pattern for hydrocarbon What does the 1H NMR spectroscopy coupling pattern look like for the hydrogen attached to the carbon that the blue arrow is pointing to? There is a $\ce{CH2}$ attached to ether side of the carbon, but each of those carbons is in a different chemical environment from the other. Does that mean that it is a doublet of doublets? Or just a doublet?

The two $\ce{CH2}$ groups are, as you mentioned, in chemically different environments. Thus, each one will couple separately. But within each of the two groups, there are two protons. So you will not get a doublet of doublets, but a triplet of triplets.

Note that for each $\ce{CH2}$ group the hydrogens are chemically equivalent but not magnetically. Therefore, if you include a chiral environment, the pattern will further complexify itself into a higher-order signal that cannot easily be analysed. But since you’re probably measuring in chloroform you need not worry — there is no chiral information present and the couplings will be degenerate.

If I had to label a spin system here, I would probably choose $\ce{A3BB'CC'MM'X3}$.

By the way, it can be assumed that both coupling constants to the two different $\ce{CH2}$ groups be reasonably similar that the overall pattern looks like a quintet without actually being one.


Which is the bigger ion, F- or O-? Well, according to the proton-electron ratio $\ce{O-}$ should be bigger than $\ce{F-}$

What about the charge/electron density in $\ce{F}$? Will it not affect the size of the atom of $\ce{F-}$?

definitely $\ce{O-}$

In $\ce{O-}$ 7 outer electrons are tied to nuclei with 6 positive charges (number 8 minus 2 electons of inner shell), while in $\ce{F-}$ 8 outer electrons are tied to nuclei with 7 charges. The difference, however, should be quite small.


What is the use of chloroform in the procedure of determining peroxide value? In this procedure of determining peroxide value, the oil sample is dissolved in acetic acid chloroform solution. I understand that the acid is added so that the following reaction occurs

$\ce{2ROOH + 2H^+ + 2I^- -> 2ROH + I2 + H2O}$.

But what is the point of the added chloroform?

Acetic acid is required for this method for exactly the reason you gave.  It also has the fortunate property of being miscible with many polar and non-polar solvents.  Because this is a method for the analysis of oil samples, it is necessary to chose a non-polar solvent capable of dissolving the sample that is also miscible with the acetic acid.  Chloroform is one solvent that fits that description nicely.

As a quick side-note, the use of chloroform in this test procedure is actually now discouraged by the Standard Methods for the Analysis of Oils, Fats and Derivatives.  Isooctane seems to be the preferred solvent, apparently for health and safety reasons.