text:
α,β-unsaturated carbonyl compounds and alkyl addition
In my book it is suggested that in order to add alkyl groups in α,β-unsaturated carbonyl compounds we use organometallic reagents such as organolithium reagents.
In the previous chapters it was suggested that the addition of alkyl groups in carbonyl compounds with saturated chains it is conducted by using secondary amines that produce enamines.
I would like to know if it is possible to add alkyl groups in α,β-unsaturated carbonyl compounds by using secondary amines.

Where is the alkyl group being added to the unsaturated carbonyl? In a 1,2- or a 1,4-sense?         Could you possibly [edit] your post to include the example from the book (you can upload photographs to imgur using the toolbar when editing)

The addition is 1,2 with organolithium reagents and 1,4 with organocopper reagents. The book is the 6th edition of ''Organic Chemistry" by Vollhardt and Schore. I'm sorry but I am not able to upload a photo.

Do you have a page reference? I have a copy of the book and can upload a picture of the relevant scheme for you

In chapter 18-4 (p. 934 of the Greek edition) the use of secondary amines is presented, while the use of organometallic compounds is presented in chapter 18-10 (p. 950).
Thank you very much for your help.

Enamines do not add alkyl groups to carbonyl compounds. They are enolate synthons ($d^2$). Are you confusing this with the Baylis-Hilman reaction?

Which position are you trying to alkylate at?

@KG3 I'm not saying that enamines add alkyl groups. What it is suggested is that by forming an enamine, one has a better reactant to then add an alkyl chain, instead of using a base straightforward since that would end up as an aldol reaction.

@orthocresol Either 1,2- or 1,4-. However, I think only 1,2- would be plausible.

If I understand the question correctly (I do not have the book), it goes like this:

(Ref. J. Am. Chem. Soc.  85, 2, 207-222)
In short I would say that it is not possible to perform the reaction in question in the same way with the Stork enamine alkylation simply because—due to the α,β unsaturation of the ketone—the enamine can not be produced with pyrrolidine. Also, I performed a Reaxys search and found no relevant results. (Of course, I might have missed something.) However, I did find a couple of interesting reactions between cyclohexenone and pyrrolidine which I submit below for completion:

And the following L-Proline organocatalytic reaction taken from: Tetrahedron Letters, 2010, vol. 51, #38, p. 5086 - 5090. Note that compounds 1 and 2 in the scheme were isolated.

However it must be noted that generally, using a different α, β unsaturated ketone and a different amine to the ones used above, the reaction in question is possible as demonstrated by user55119 in his answer here: Why enamine formation does not occur with α,β unsaturated ketones?

The pyrrolidine enamine of cyclohexanone reaction with methyl vinyl ketone does not stop at the Michael addition product but undergoes an aldol condensation (overall Robinson annulation) to form the pyrrolidine enamine of the enone. This species hydrolyzes with buffered aqueous acetic acid and not with water or aqueous mineral acid. [G. Stork, et al.; pubs.acs.org/doi/pdf/10.1021/ja00885a021]

@user55119 Well spotted, i have edited it accordingly but did not include all the details so it wont get unnecessary complicated for what the question is asking. Added your reference however so people can take a closer look if needed. Ta!

@AMM Thank you for your response. However, is there no way the last step of the enamine formation mechanism, the deprotonation, occurs at the CH3- side or would such a structure be impossible to exist?

@ΑντώνιοςΚελεσίδης I was thinking the same thing . The only reports of such compounds i found were: Chem. Commun. (London), 1967,0, 862-863 , and Journal of the Chemical Society, Perkin Transactions 1:  Organic and Bio-Organic Chemistry (1972-1999) (1977), (23), 2571-80;  really old articles which i am not sure i trust (for the search i kept the 2,3 and 5,6 unsaturated ring with NA2 on the 1 carbon with A stands for Any atom). Regardless, it is not common and if it does exist it looks like not accessible via the route above. This makes a good question by itself if u like to post, or shall i?

I see. You mean to post a different question regarding the existence of a "diene/amine" structure? If you'd like to, sure. :)

@ΑντώνιοςΚελεσίδης I was thinking something like: why the reaction you suggest is not observed and why the putative product is not commonly found in the literature. Maybe the computational people will jump in (kai ego Antonis parepiptontos).

People might redirect it back to this original question but I don't see why not try it. Go ahead and I'll keep an eye συνονάματε.

@ Αντώνιος Κελεσίδης: Could you be more explicit as to what you mean by the "CH3- side in your response to AMM. Perhaps I can help.

@user55119 I have posted the new question here: https://chemistry.stackexchange.com/questions/99987/why-enamine-formation-does-not-occur-for-α-β-unsaturated-ketones,  an organic chemist would be helpful here i suppose so let us know!

@AMM: The buffered hydrolysis conditions are for dienamines, not enamines. See, https://chemistry.stackexchange.com/questions/99987/why-enamine-formation-does-not-occur-with-%ce%b1-%ce%b2-unsaturated-ketones

Hi, may I ask, which software you used to plot these nice organic chemistry structures? Thanks!

@YangxinZhang It is called chemdraw.

I think that adding alkyl groups to $\alpha$ - $\beta$ unsaturated carbonyl compounds are possible through Enamine synthesis by Secondary amines. But the alkylation will be at the $\alpha$ and $\gamma$ positions as follows.  For simplicity, if we take the corresponding carbonyl compound as Cyclohex-$2$-en-$1$-one, the plausible pathways for alkylation are as follows,
Now, the first step of forming an enamine is already reported. The link for the JACS paper is https://pubs.acs.org/doi/pdf/10.1021/ja061104y .In this paper, preparation of enamine (by using S-Proline as a secondary amine)  of similar kind of $\alpha$ - $\beta$ unsaturated compound has been reported. One of the major reaction in this paper is the following,
Thus it is confirmed that such kind of doubly conjugated enamine formation is possible. The remaining two steps for $\alpha$ - alkylation is also reported earlier. For details, see the paper https://onlinelibrary.wiley.com/doi/pdf/10.1002/recl.19881070906
 Here they have done the following reaction related to this context,

Here $\gamma$ alkylation was disfavoured to preserve the aromaticity of the benzene ring. But in general, $\gamma$ alkylation forms again a conjugated carbonyl compound, and hence forms the thermodynamically more stable product. So,$\gamma$ -alkylation is more favoured at higher temperature, whereas the $\alpha$ -alkylated product is a kinetic product.  Thus, it is also possible to do alkylation of $\alpha$ - $\beta$ unsaturated compound using a secondary amine, but those alkylation will be specifically in $\alpha$ and $\gamma$ positions (not on $\beta$ positions).

authors: Αντώνιος Κελεσίδης


text:
Which holds the electrons stronger: 2p or 3s?
The s orbitals hold electrons stronger than p orbitals, but I do assume this is under the assumption that they are in the same energy levels.
Is there a rule for determining which is stronger between different energy levels, such as between 2p and 3s?
This question was inspired by electronegativity differences between sp2 and sp3 hybrid orbitals (sp2 more electronegative because higher s character)

My thoughts are that 2p holds harder as is on average closer to nucleus.  Regarding hybrid orbitals; sp2 has more of the character of an orbital closer  to nucleus than sp3 coz is likely to be 2s and 3p orbitals. Don't think is much to do with it being spherical.

The orbitals that hold the electrons stronger will populate earlier. Look at noble gases and ions of alkali metals. They don't have 3s electrons, but have 2p electrons. It means that p2 is lower than 3s.
Comparison of different orbitals and different energy levels gets tricky starting with transition metals. It gets even more difficult for actinoids.

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text:
Best way to mix essential oils with water?
I am using alcohol as an emulsifier. Of course, I can use a lot of alcohol to mix the oil and water but I do not like the color and the % of alcohol in the mix. The best mix I made was 10ml water, 0.5ml (10 drops) essential oil, 5ml alcohol because the mix looks super milky and when left alone the oil does no lt deposit. I repeated this formula again but the result is different (maybe the temperature was different, or the stirring, etc). My question is: what is the best way to stir the essential oil with the alcohol? Should I heat it? Should I stir it with an instrument other than just stirring in a beaker? All emulsifying tips are welcome!

It is not really clear what you would like to reach. Do you want to make a stable emulsion? I cannot compute what does "I don't like the % of alcohol in the mix" or color (isn't it white??) mean. Also, why do you use alcohol as emulsifier?

First of all, alcohol is not an emulisfier but a cosolvent. Please specify whether you want to obtain a homogenous solution, emulsion or dispersion. These are three different goals and require three different approaches.

I want to make a stable emulsion. If I put too much alcohol the mix turns yellow. If I put too little alcohol the oil does not mix with the water. If I put too much alcohol the mix becomes flamable. The perfect mix should turn it into a milky white, but I cannot accomplish it constantly. I use alcohol because it is cheap and easy to get anywhere.

I am not sure if what I need is defined as a solution or emulsion; I want to spray essential oils diluted in water, but need the best result to make the water and oil mix without having to shake the bottle every time. – Giancarlo 1 hour ago   delete

Any tips or recommendations?

A solution would be the best for spraying, and it would require dissolving the essential oil in an appropriate organic solvent. However, if you are concerned about the flammability, then a dispersion in water would be your best option. Basically, all you need is a stabilizing agent and a proper procedure. I am sure that you will find all the necessary information on this site. It has all the basics about emulsions, it lists various emulsifiers (stabilizing agents) and procedures. For your purpose, I think the best option would be the Bottle Method.

Those two html pages are not available.  Can you provide the information here?

To make an emulsion, a stable milky mixture of oil and water, you need an emulsifier. Alcohol is not an emulsifier, so the mixture will separate at some point.
The most common emulsifiers used in cosmetics can be found here.
If you are making this at home and do not have access to these kind of products, you can try with organic soap. This is normally oil hydrolyzed with caustic soda (NaOH) and is in essence what is listed as neutralized fatty acid soap.
Normal stirring is also many times not enough and a high shear mixer is generally used. So mix it as hard as you can.

The link you shared is not available.  Can you share the information here?

By alcohol, I assume you mean ethanol. A better choice might be isopropanol (rubbing alcohol). If either ethanol or isopropanol can be mixed with your essential oil in a 10:1 ratio to give a clear solution, then the alcohol is a solvent. The water is not a solvent for the essential oil, but the alcohol might be able to dissolve some water without expelling the oil.
I suggest you make a triaxial graph to plot the concentration ratios of ingredients that are clear (solution) or milky (stable (?) emulsion) or quickly separating (unstable emulsion). The advantage of a triaxial graph is that you can find a location on the equilateral triangle for any combination of three ingredients - and do some interpolations to find where you can be. This is a picture of a triaxial graph:

It might be possible to use another solvent, e.g., acetone, in whole or in part. One way to stabilize an emulsion is with surfactants; high shear is best for reducing the size of the micelles. Another way to stabilize an emulsion is to use a thickener that prevents the oil micelles from settling or creaming.

I'm trying to make a 1 to 2% solution of cedarwood oil with water/alcohol (rubbing) to spray for fleas inside a living space.  My plan is to mix my 1 ounce bottle of pure essential oil with 10 ounces of isopropyl.  That should look milky white.  Then I'm going mix that with 5 to 10 parts water to make the ultimate solution stable enough to shake once and spray for several minutes (1 to 2% cedarwood oil).  Should that work?

actually more 70% isopropyl will need to be mixed than 10 to 1 ratio because it's only 70% ??

@user12711: The 10 to 1 ratio is to determine whether the oil is soluble in pure alcohol (i.e., isopropanol). If so, maybe water can be added. But some oils are not soluble even in pure isopropanol, so the mixture will be cloudy and will probably separate. In this case, adding water will not help. The OP made one successful mixture, but could not replicate. Using a triaxial graph can map what compositions are useful. It may show how some compositions are milky, some settle, some are clear, depending on composition. Procedure (heating, stirring) needs other experiments.

You can use ultrasonication and no solvent (water + oil + sonicating = emulsion). The sonicator degasses the water which allows the oil to disperse homogeneously throughout the water.
I hope you can access a sonicator if not there are freezing techniques which also degas water.

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Band gap of fullerenes
From what I have read about fullerenes, the lower fullerenes like $\ce{C60}$ or $\ce{C70}$ have higher bandgaps around 3.5 eV or such, while the higher fullerenes have much smaller bandgaps of the order of 0.5 eV. Initially I thought that this could be explained using electron delocalization on the surface area of the sphere, but later I read that the fullerenes were not necessarily super-aromatic. What could be the reason behind such a dip of the HOMO-LUMO gap?

Well, sufficiently large fullerenes can be thought of as multiple graphene sheets stitched into a 3D shape, and since graphene sheets have zero bandgap, you know what the limit of the system must tend to.

@NicolauSakerNeto An interesting thing is that usually band gaps increase when layers are thinner. Monolayers of most semiconductors and insulators have a larger band gap than their bulk. So I am not sure how valid is that logic.

@QuantumMOCHACCINO same logic that I understand, the schrödinger wave pumps up the subsequent energy gaps when it is localized. Also, although I am new in this field, I am simply using my previous knowledge to tie the loose ends, graphene is superaromatic, which means that the $\pi$ and $\pi^*$ orbital waves are resonant over the entire exposed surface area. This is not true in the case of fullerenes, since the TPJs(Tri pentagonal junctions) act as a barrier to the pi orbital  delocalization.

related: http://chemistry.stackexchange.com/questions/37641/is-buckminsterfullerene-aromatic-or-not

@ubuntu_noob I have little knowledge about graphene and fullerenes. However, I do not see why we have to consider fullerenes as a graphene analog. If we consider the fullerenes as semiconductors or insulators, the trend makes sense. I think that graphene is a special case due to its aromaticity. But who knows I am not the guy studying fullerenes, you are.

@QuantumMOCHACCINO A decrement from 3.5eV to around 0.5eV is quite a huge jump considering that I'm just going from 60 atoms to 80. Even if we talk about the most common $CdSe$ QDs, the bandgap goes from 1.7eV to 2.3eV while it transitions from the bulk(theoretically infinite) atoms to the 4~5nm strong confinement regime. A jump from 3.5eV to 0.5eV, I'm almost dead convinced that something else is going on here, also the lack of super-aromaticity means that the change in the area of the wave localization is just a fraction of the area of the supposed sphere, doesn't seem that convincing.

@ubuntu_noob  I agree. It is quite interesting. I will try to look into it. From what journal did you get the results?

Its maybe a bit misleading to talk about band gaps in this context. The electronic structure of fullerenes is still better understood thinking of molecules (for which "jumps" are notorious). Hence HOMO-LUMO gap is more appropriate. Now look at benzene and cyclobutadiene, the former has a substantial gap the latter not (approaching even "zero" depending on the degree of geom. distortion). C${60}$ is non-aromatic while C${60}^{10+}$ is. The gap size really depends on the peculiarity of the electronic structure. (Multiple) Spherical Aromaticiy is maybe the simplest concept explaining it.

One of the explanation for answering this question can be provided as, moving from lower to higher Fullerenes the number of atoms in the structure increase, increasing the number of atoms in the system results in reduction of the discreteness of energy states, which results in decrease of the gap between HOMO and LUMO, thus the band gap is reduced.
You can also check for the particle in box model of the Quantum Physics for better understanding.

Fullerenes are not fully aromatic, there are certain junctions which prevent the delocalization of the gas and hence the box boundary is not necessarily extended up to he physical dimensions defined by the number of atoms...so


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Where does the energy required for excitation of electrons in hybridization come from?
We're commonly taught that when atoms want to hybridize in for instance methane, an electron is excited from $2s$ to $2p$ to allow for four unpaired electrons, which is necessary for bonding.
Where does the energy for this excitation come from?

If you are talking about the sp3 hybridization: there is no excitation involved...

Ab initio hybridisation is a concept taught in schools to help understand how carbon with its one s- and three p-orbitals can achieve a tetrahedral environment. As part of this process, it is required to formally excite an electron.
Formally is the key word here. The entire process is a formal process that has practically nothing to do with real-life atoms (especially because these real-life atoms rather rarely end up in a vacuum without anything around them as the theory assumes). Thus, in real life this formal energy is not needed.

In this process the energy of s orbital increases, but the energy of three p orbitals decrease.
This is like irl two hungry friends one of them having spam. The one with spam will share it with a friend and (if you look at him alone) his state worsens (say -1). On the other hand his friend benefits (say +2). Sharing still brings the benefit to the system.


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Adding chemical equations together
This is something that may seem small, but it bothers me … I mean really, really bothers me. The following issue will be expressed through the Brønsted-Lowry theory of acids and bases and specifically through the neutralistaion reaction of ammonia and hydrochloric acid.
When I dissolve hydrochloric acid in water, the following equilibrium reaction takes place:
$$\ce{HCl + H2O <=> Cl- + H3O+}\tag{1}$$
When I then add a base such as ammonia, $\ce{NH3}$:
$$\ce{NH3 + H3O+ <=> H2O + NH4+}\tag{2}$$
Adding these 2 equations should give:
$$\ce{NH3 + HCl <=> NH4Cl}\tag{3}$$
This implies that the first reaction produces hydroxonium ions which the second reaction uses up, hence the number of hydroxonium ions in the solution should remain the same as before the addition of the hydrochloric acid.
Now let's actually asses the implications of such a procedure. First of all, assume that the reaction has gone to completion. This implies that when I mix water with hydrochloric acid, I will get hydroxonium and chloride ions produced. Subsequently, when I add the same number of moles of ammonia as hydrochloric acid, this will react with all of the hydroxonium ions present in the solution, as is shown in reaction (2). One problem though; this isn't a full reaction. If I add the same number of moles of ammonia as hydrochloric acid, this time there will be more hydroxonium ions than there were in the solution before addition of the hydrogen chloride (this what what I guess will happen, and this is the center of my problem). So how do I become sure that the hydroxonium ions cancel out when the equations are added?

The first problem: you can't assume that "the reaction has gone to completion", because the ammonia clearly doesn't. Next, you're assuming that the balanced equation represents everything that is going on in the system. You know already that the system involves the reaction of species such as $\ce{H2O}$ and $\ce{H3O+}$, so don't expect an equation that doesn't include those species to necessarily be accurate. A more accurate equation would be something like $\ce{NH3 + HCl + H2O <=> $(1-x)$NH4+ + x NH3 + Cl- + x H3O+}$, however that's incredibly tedious and nobody likes writing that.

Why would the real equation look like that? I'm interested. Please do tell me through a full answer.

Simply because some of the $\ce{NH4+}$ ions will undergo hydrolysis: $\ce{NH4+ + H2O  <=> NH3 + H3O+}$. The issue is that without consulting a thermodynamic data table (and I am not about to do so right now) you don't know exactly how much of it will do that, hence the variable $x$. Sorry, don't have time for lots of explanation.

Also, I made the assumption that the reaction goes to completion just to shown you why I thought it's ok to cancel out some species in the case where the reaction is complete. However, since NO REACTION is ever complete, I also wanted to show that this whole cancelling thing which we learnt through Hess' law was actually wrong. I only noticed that it shouldn't work when I started studying acid- base equilibria (which I'm still on). Furthermore, since you just said they don't actually cancel, I'm doubting the truth of the Bronsted-Lowry theory.

The reaction equation you have written is a simplification of the real thing. Yes strictly speaking that means it is "wrong", however as was mentioned in my comment (as well as the answer) the equation is not meant to be rigorously correct, it is just a tool that captures the essence of the reaction (from a glance one can see it is acid-base) and also keeps track of stoichiometry. I don't see how there is a problem with Bronsted-Lowry theory, which merely says that an acid is a proton donor and a base a proton acceptor - nothing less, nothing more.

I don't need the value of x. Do you have time (later) to show me how you added the equation such that it took the form you proposed and NOT this: NH3 + HCl <=> NH4Cl

I'm studying A2 level (grade 13), so should I just assume that the equation is how I wrote it? Is there a later stage during my chemistry education when I will come to learn such things?

Simply take eq. 1: $\ce{NH3 + HCl <=> NH4+ + Cl-}$ and eq. 2: $\ce{NH4+ + H2O <=> NH3 + H3O+}$ and take a linear combination of the two equations: $(1) + [x \times (2)]$. There should be an $x$ in the $\ce{H2O}$ on the reactant side in my first comment. The original equation is, for all intents and purposes, "correct", in the sense that it is an appropriate (albeit incomplete) description of the reaction. Nobody at a higher level will use the thing I've written in my first comment.

What I meant is: is there a certain topic specified for learning how to write chemical equations? Did you self teach yourself this?

Sorry, but I don't think anybody teaches the art of writing chemical equations. They are just a tool to us, do not take them so seriously... The point of chemistry is not to write the most accurate equations, it is what you do with them that counts. In any setting, if you intend to account for the acidity of the ammonium ion, all you need to do is to write the two composite equations in my previous comment. Taken alone, neither of them are a complete representation of the actual reaction. Together, they are. And two easy equations is usually better than one complicated one.

But the equations which you have given don't represent the full situation. My original point was that we should add the equation of the reaction of ammonia with hydroxonium ions to the equation of the reaction of hydrochloric acid with water in order to give the equation of the reaction of HCl with NH3. My focus was on whether the hydroxonium ion and water molecule truely cancel out of the equation, or whether they remain, not on these 2 equations.

They fully cancel out in the equation, but not in real life.

That's what I wanted to hear. Also, I just need to make the assumption that the equations perfectly represent real life, right? And just ignore the extra water molecules and hydroxonium ions present in real life, right?

Yeah... somewhat. I don't think you have to assume that the equations are perfect, I'd say you should rather acknowledge that they are imperfect.

What's the evidence that this will happen?

When you write a chemical equation, you don't care whether the reaction has gone to completion or not. On a deeper level, none of them do. There is always an equilibrium, and there is always some starting compound left. To find out how much is that "some", you need to know the equilibrium constant and make certain calculations.
The written equation does not tell you that each and every molecule in the sample should react exactly like that. This is simply not what it is all about. It does, though, tell you another thing: if you got such-and-such amounts of products, you must have lost such-and-such amounts of the starting compounds. This is simply the conservation of mass, and there is no going around it.

(1) "H2O+HCl-> H3O$^+$ + Cl$^- has gone to completion" fair enough if the concentration of HCl is low.
(2) NH3+H2O <-> HN4OH never goes to completion. So, when you add 1 mole of HCl*H2O to 1 mole of NH3*H2O you actually add 1 mole of Cl$^-$+1mole of H3O$^+$+ x mole of OH$^-$+ x mole of Nh4$^+$ + 1-x moles of NH3*H2O. (0


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The effect of temperature on the caffeine intensity
Does the brewing temperature have an effect on the caffeine level?
For instance, coffee with 100*C have higher caffeine level compared to coffee with 50*C?

You may be interested by http://cooking.stackexchange.com/questions/1859/caffeine-content-of-cold-brewed-coffee-higher-or-lower-than-hot-brewed

When coffee is being brewed in a moka pot, the water keeps evaporating until it reaches the coffee powder, then it condenses and falls down again, bringing some extracts from the coffee powder: it's just like a small reflux condenser.
When the pressure is high enough, the extract reaches the valve, and it escapes, giving the brewed coffee.
When temperature is too high, the pressure raises too quickly, and water fails to make a good extraction (just think about a reflux condenser which is kept running only for a few cycles).
If the temperature is appropriate, the water performs many evaporization/condensation cycles before it comes out of the valve, hence performing a better extraction (which, I guess, implies an higher caffeine level).
That's why usually a low temperature and a slow brewing are recommended.

Are you sure that's how moka pots work ? This paper seems to say it is pure percolation, which makes much more sense to me.


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What would be the enthalpy change for a isothermal expansion?
I was doing a question on thermodynamics and came across this one stating

During isothermal expansion of gas of an ideal gas :
a) Internal energy decreases
b) Enthalpy increases
c) Enthalpy reduces to zero
d) Enthalpy remains unchanged

I cannot agree to any of these as $\Delta H = \Delta U + W$, $\Delta U = 0$ as $\Delta T = 0$ but $W \neq 0$ so $\Delta H = W$ which is $\neq 0$ and as $W \lt 0$ the enthalpy should decrease but how much how do I know and to add to the confusion the answer given is enthalpy remains unchanged in the light of the equation $\Delta H = \Delta U +\Delta n_\mathrm gRT$ and $\Delta U$ and $\Delta n = 0$  so $\Delta H = 0$ is it right.

@ChetMiller dPV or nRdT also defines work and nRTln(v2/v1) also defines work....then how work has zero and non zero values at the same time in the same process(isothermal expansion of ideal gas)

You have already mentioned that the change of internal energy is zero $(\Delta U=0)$ since, for an ideal gas, the internal energy $U$ only depends on amount of substance $n$ and temperature $T$, and  in a closed system $n$ is constant $(\Delta n=0)$ and during an isothermal process also $T$ remains constant $(\Delta T=0)$.
Enthalpy $H$ is defined as
$$H=U+pV$$
and the ideal gas law states that
$$pV=nRT$$
Thus
$$H=U+nRT$$
Since $n$ (closed system) and $T$ (isothermal process) are constant, the product $nRT$ is constant, and therefore, according to the ideal gas law, also the product $pV$ is constant.
Furthermore, since $U$ is constant during the given process, the sum $H=U+nRT$ as well as  $H=U+pV$ has to remain unchanged (answer d).

Note that your assumption
$$\Delta H = \Delta U + W$$
(where $W=p\Delta V$ is the pressure-volume work) only applies to processes under constant pressure $p$, since enthalpy is defined as
$$H=U+pV$$
and thus
$$\begin{align}
\mathrm dH&=\mathrm dU+\mathrm d(pV)\\
&=\mathrm dU+V\mathrm dp+p\mathrm dV
\end{align}$$
which simplifies to
$$\mathrm dH=\mathrm dU+p\mathrm dV$$
at constant pressure $(\mathrm dp=0)$.
However, the pressure does not remain constant during the process given in the question.

Yes.  The enthalpy of an ideal gas is a function only of temperature, no matter what kind of process is imposed.

@chester Miller if it were so then all isothermal process would be isoenthalpic which is not the case

@PrateekChauhan For a single phase ideal gas, of course it is.

@ChesterMiller If I have an ideal gas, then will the change in enthalpy be zero, for an isothermal path? Does it matter whether the process is carried out reversibly or not?

If the initial and final equilibrium temperatures are the same (irrespective of what happens along the path), the change in enthalpy of an ideal gas is zero.  The enthalpy and internal energy of an ideal gas are functions only of temperature.

@ChetMiller I don't understand if $\Delta H$ is 0 then which energy causes the isothermal work ?

@maverick It is caused by the heat that is absorbed (i.e., transferred to the gas from its surroundings).

@Chet Miller just one more thing, is it right that ,in isothermal process, if $Q$ heat is absorbed from surrounding , then all the heat is converted into useful work done by system (Gibb's energy) and it has no contribution in enthalpy change of system??

@ChetMiller And that useful work done by system is same as the  isothermal work which is either reversible or irreversible??

Yes.  That is correct.


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Separating glucose and fructose
If I have a mixture of glucose and fructose in aqueous solution, what is the simplest method to separate them? Heating to 110 °C to melt only the fructose would probably work theoretically, but this would cause a huge mess.

The most common industrial method is one of various types of liquid chromatographies, which is usually how glucose-fructose solutions are enriched in fructose.
I did see a couple other possible methods though:
This one from 1965—before good LC methods existed—relies on the difference in solubility in partially water-miscible solvents like n-butanol to perform a liquid-liquid extraction.
Unfortunately, the pdf seems to be incomplete, but basically the water-solvent-sugar system separates into two phases with each sugar partitioning more into a different phase.
They calculated that it would take around 40 cycles to get 90% fructose from this method, but it doesn't seem to be very efficient due to slow diffusion of the sugars (and since LC is the dominant method today, it doesn't sound too promising on any reasonable scale).
In the introduction, they also note that some of the other attempts of the era involved chemical means—specifically precipitation of the fructose as calcium fructosate, getting the fructose back out by reaction with carbon dioxide, oxalic acid, or sulfuric acid in what is known as Dubrunfaut's method.
Again, this did not result in an industrially useful process.
This one involves careful fractional crystallization of glucose-fructose solutions, as fructose is far more soluble in water than glucose.
The solution is cooled from room temperature to -20 °C fairly slowly (0.5 K/min) while being agitated and in the presence of small glucose seed crystals.
Some of the glucose crystallizes out of the solution, which is separated by centrifugation, and the process is repeated going to -35 °C.
The crystals they get are 89% glucose and 11% fructose, but a big red flag is they don't actually measure the concentrations in the liquid left behind (separating fructose is far more important industrially because pure glucose is easy to get by hydrolysis of starch).
They say that "a concentrated solution of fructose containing a small amount of glucose was obtained", but don't actually prove it, and apparently it only works properly if you can maintain a total sugar concentration near 50%, so I'm pretty sceptical of the method's utility.
You could probably do something similar at home if all you're interested in is getting some somewhat purified glucose.
Unfortunately, if there were a simple and effective way to do this, industrial production probably wouldn't bother with LC.
If you just want to play around with separations, these might be some things to try, but if you just want glucose and/or fructose, I have seen both at bulk food stores. (glucose is often labelled as dextrose and fructose is sold as a low glycemic index sweetener)

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Will Sodium Amide react with Epoxides in the same manner as Grignard Reagents?
I have been unable to find the reaction anywhere on the internet so I'm assuming the answer is no, although I would like an explanation as to why.  I can't seem to find any reason why it would not work, provided the reaction is carried out in two steps.  In the first step, H2N:- will act as a nucleophile, attacking the least sterically hindered carbon of the epoxide.  The second step would of course be an acidic workup.  As stated in the question, I would expect the same exact mechanism as the reaction of Grignard Reagents with epoxides (substituting sodium amide as the nucleophile) due to the high basicity of sodium amide and Grignard Reagents.  Like I said, I'm not expecting the reaction to be viable.  However, I would like to know exactly why not.  Im assuming it's something to do with the Hydrogen atoms of the amide anion, though I could be wrong.  Enlighten me.

It actually is possible but it works well only with simple epoxides so it is not wildly used. $NaNH_2$ is extremely basic (it can deprotonate alkynes) and too many functional groups can react with it: alkyl halides (elimination), aldehydes, ketones...
To work around that basicity, sodium azide $NaN_3$ is commonly used instead, followed by the reduction of the azido group into an amine.
For example, see this article.

Thank you!  Since I was unable to find any info on this, I was beginning to think I was crazy for even considering it!  And I was aware of the 1) NaN3 followed by 2) LiAlH4  method for forming primary amines, I was only considering this possibility as a possible more green (less chemicals used) and more efficient (less steps) method.  However, with all the side reactions that could occur with the amide route, the better option is clearly the azide/hydride route.  Id upvote you if I could but my rep is too low.

@Ki11akd0g Well maybe you are crazy but for another reason :-) For the reduction of azides, I usually preferred hydrogenation when possible, or the Staudinger reaction (triphenyhosphine with a little bit of water) when I had reducible moieties, though thriphenylphosphine oxide can be a pain to get rid of sometimes.

Well from what I have heard, Lithium ions can also be a pain to get rid of as well.  Guess you just have to pick between whatever the lesser of two evils is for a particular application X-)


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Caffeine concentration method!
I'm trying to measure the concentration of caffeine after the extraction process. If I dissolve the same amount of extracted caffeine into same amount of distilled water, will I be able to calculate the concentration of caffeine?

Which analysis method will you use?

I'm using the DCM (organic solvent) method

Measure mass of extract 2) Evaporate water or other solvents 3)Measure mass of precipitate 4) Concentration by mass= $\frac{precipitate}{mass of extract}$

what do you mean by the mass of extract? Is it the mass of water and coffee at the beginning of the experiment? or is it the mass of extract after filtering through separatory funnel?

@yjasz98 It depends. If you want to measure the the concentration of caffeine of "the mass of water and coffee at the beginning" then you can use "the mass of water and coffee at the beginning" as the denominator. If you want to get the concentration of caffeine "after filtering through separatory funnel" then use the mass of "after filtering through separatory funnel"

However, if I were you, I would use the mass after filtering through separatory funnel. There are lots of other things in coffee other than caffeine. So using the mass after the separatory funnel will probably be more accurate.

What is filtering through a separatory funnel? A separatory funnel isn't for filtering, it's for liquid-liquid extraction.

After boiling coffee and distilled water, I added sodium carbonate. Then filtered through a separatory funnel.

For analysis you need a method to tell the difference between caffeine and other chemicals. Let's say that you dissolve a 1 g mixture of caffeine and contaminant. You can be sure that you have no more than 1 g of caffeine dissolved, but that doesn't tell you much. You can measure concentration of contaminant, convert to weight and subtract it to get the amount of caffeine. This is impractical, because you don't know what the contaminant is.
You can use iodometric titration (http://www.hkasme.org/News/818/Report_6B_Christiam%2520Alliance%2520SC%2520Chan%2520Memorial%2520Secondary%2520School_Chemical_test_for_caffeine.pdf), but this gives you the sum of all oxidizable compounds including caffeine.
You need to tell more about the instruments you have, so that we can help to choose you the right ones.


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Producing ketones from secondary alcohols using hydrogen peroxide
Usually, to produce ketones from secondary alcohols (e.g. propanone from 2-propanol), people use a solution of potassium or sodium dichromate with a small amount of sulfuric acid. Can other oxidizers like hydrogen peroxide be used, and will the use of hydrogen peroxide produce organic peroxides?
$$\ce{C3H8O + H2O2 -> C3H6O + 2 H2O}$$
This is not for any particular purpose.

You can use acidified potassium dichromate.

@KennyLau The question is "can I use other oxidizers", since potassium dichromate is quite toxic and I don't have access to it.

Oh, sorry, I thought oxidizers other than hydrogen peroxide.

H2O2 oxidation of secondary alcohols has been reported in many articles but I suspect you won't have any luck with just H2O2 - most have some other stuff added in e.g. transition metal catalyst

Bleach, aka sodium hypochlorite. The one you can get from a corner store will work just fine. And is, probably, the greenest way you could find. Benzylic alcohols can also be oxydized with oxygen from air in an ultrasonic bath (although, that's not the most "home friendly" experiment, since, well, ultrasonic baths aren't easily available from a corner store :) )

I think it is possible to use hydrogen peroxide. This paper discusses it: http://greenchem.uoregon.edu/PDFs/ResourceID88.pdf
Other alternatives:

Pyridinium Chlorochromate (or PCC for short)
Dess-Martin Reagent
Swern Oxidation: oxalyl chloride in DMSO with triethyl amine

Fétizon Oxidation: silver(I) carbonate absorbed onto the surface of celite (Fétizon's reagent). This can be prepared by adding silver nitrate to aqueous sodium carbonate, in the presence of purified celite.

Oppeneaur Oxidation:  excess ketone reagent (such as acetone) and an aluminium triisopropoxide catalyst

Pfitzner–Moffatt oxidation: dimethyl sulfoxide (DMSO) activated with a carbodiimide, such as dicyclohexylcarbodiimide (DCC)

Corey-Kim Oxidation: Dimethyl sulfide is treated with N-chlorosuccinimide (NCS), and then addition of triethylamine

There are several moments:
(1) Chemists are mostly interested in routes yielding products that are otherwise difficult to obtain. Oxidation of alcohols to ketones is a generic operation. There must be zillions of undiscovered methods to produce ketones, but there is no motivation to discover those, because the traditional methods work just as well.
(2) Peroxides are reactive/corrosive/hard to work with compounds. There are copious synthetic methods that use peroxides, but this is done only if milder methods are not known. For the same reason you don't put sodium perchlorate in your soup to enhance taste, since table salt has very similar effect on taste.
(3) You definitely can turn alcohol to a ketone using a peroxide (based on delta G), but the yield might be low, other reaction groups (such as double bond) might not survive and there is zero incentive to come up with yet another way to turn an alcohol to a ketone.

http://www.organic-chemistry.org/synthesis/C2O/ketones/oxidationsalcohols.shtm This page has 40+ common oxidation reactions to prepare a ketone from an alcohol. Choose less toxic alternative if you want.
Bottom line: you will not make a Nature paper with yet another oxidation of alcohol to a ketone. Thus, academia doesn't care. H2O2 is more dangerous (especially considering explosion hazard) than CrO3. Industry doesn't care.

The motivation is to be able to use a less toxic and more easily accessible reagent.

@f'' CrO3 is much easier to work with than H2O2. Similar toxicity, while H2O2 is more corrosive and explosive. There is no demand on the market for a new oxidation of alcohols to ketones on the market.

The OP is not trying to make a Nature paper. They are trying to oxidize an alcohol using reagents that they can access at home. Hydrogen peroxide solutions are easily obtained in stores, while chromium trioxide is not. Chromium trioxide is quite dangerous in its own right and can ignite alcohols on contact. Dilute solutions of hydrogen peroxide, which are widely available, are much safer.

If you look at the work done by Shannon Stahl and others, it looks like industrially at least there is huge demand for new and environmentally benign oxidations (eg using oxygen from the atmosphere). I'd also echo f" with the comments about chromium and related oxidations.

Nature paper? I think it's a valid topic


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Solubility of NaOH in Organic solvent?
I am conducting organic reaction in Dimethoxyethane as solvent in the presence of base (NaOH), but the NaOH is not completely soluble ( Sparingly soluble) in Dimethoxyethane even after the addition of excess of solvent. Peoples says that for a reaction, the reactants must be completely soluble in solvent.
Can someone comments about solubility of Bases during organic reaction?

I actually want conduct similar kind of reaction, the paper mentioned that NaOH and organic compound stirred vigorously until to obtained clear solution. In my case my compound is completely soluble in DME but NaOH is poorly soluble.

I'd say it depends on the reaction you want to do. Can you tell what your reaction is?

@S.Chevalier please see my comment below.. thanks

Your best bet is to publish the procedure. Procedures from pier reviewed journals are generally reproducible (there are exceptions). So if they say "dissolve x grams in y mL" just do it.
NaOH is poorly soluble in dimethoxyethane. This might be sufficient if base is used as a catalyst or if reaction is slowly shifting to product.
Devil is in details. Heat, presence of water and other factors might impact soluility greatly.


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Calculating Electrode Potential for Electrolytic Cell at Non-Standard States
Recently, I came across an interesting question that left me rather baffled. I was asked to calculate the minimum voltage required for the electrolysis of 1.0 M of NaCl in neutral solution using the following equation:
$$\ce{2H2O + 2Cl- (1M) -> H2 (1 atm) + Cl2 (1 atm) + 2OH-} (\pu{1*10^{-7}~M})$$
I was left in a dilemma, if I were to use the Nernst Equation, I would get:
$$E = E_0 - \frac{RT}{nF \ln Q}$$
$$E = -1.36-(+0.83) - 0.012839\times\ln \pu{(1*10^{-7})^2}$$
$$E= \pu{-2.19 V} - (-0.4135) = \pu{-1.78 V}$$
So $+1.78$ V is needed to drive the reaction, theoretically and assuming ideal conditions.
However, I went online to search a bit, and came across a line written in Wikipedia that

... the standard potential of the water electrolysis cell is −1.23 V at 25 °C at pH 0 ([H+] = 1.0 M). At 25 °C with pH 7 ($\ce{[H+]} = \pu{1.0×10^−7 M}$), the potential is unchanged based on the Nernst equation.

Does that mean that regardless of the pH, electrode potential for water is still the same, or is there some rule or standard definitions that I'm missing out?
In this case, does this scenario apply only to water, what about other cases where other reagents are used at non-standard states, will my original method (using the Nernst Equation) work?


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Gaussian vs. Lorentz/Cauchy distributions for artificial broadening
When I am supposed to calculate an IR-, Raman-, or UV/Vis-spectrum, the calculations give me some x-y-values, e.g., Frequency-Intensity, or Energy-Oscillator Strength. Those values can either be printed as stick spectra or a sum of Gaussian-/Lorentz-distributions is being used to broaden the peaks to generate a more "natural" spectrum.
I've been asked a couple of times which broadening function to choose when, but actually never had a good answer/given explanation. The only thing that I wrote down quite some time ago, was that for IR- or Raman-spectra of solids a Gaussian broadening and for every other type of IR- or Raman-spectra a Lorentz broadening should be used.
Google searches for "gaussian or lorentz broadening" and "line broadening" did yield results that try to explain it, but I didn't find most of them really helpful.
So the question is: When do I choose the Gaussian or the Lorentz-functions for artificial broadening of IR-, Raman-, or UV-Vis stick spectra and why?

Shortly after posting I found this, which is at least related, if not the same http://chemistry.stackexchange.com/questions/324/when-simulating-spectral-line-broadening-which-convolution-is-preferred ... but the answer is not enough for my taste.

From the little bit I've learned, Lorentz would be the most accurate for a single broadening event but if you sample a population it would looks more like a Gaussian. That's what someone in my lab said at least...

I think that any of these functions, Gaussian, Lorenzian, Voight, will do, they do not differ by much but the Lorenzian has a slightly wider base and in overlapping spectra this may increase background too much. I assume that you will convolute your stick spectrum to get your 'real' spectrum, in this case you can zero pad the transform, it makes nicer looking spectra but adds no more information. Perhaps the most important point to check is  the spectral resolution of the instrument you are simulating spectra from, and adjust the width of you broadening function appropriately.


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Would a loaded cartridge ignite/explode when hit from the side or front by a bullet?
This question https://physics.stackexchange.com/a/193472/93888 explains what would happen after a cartridge is ignited: the bullet gets propelled by the gases in one direction and the cartridge (with the person, in the case of the question) in the opposite direction.
But would a cartridge explode/ignite if hit from the side (the rear part where the gunpowder is, not the side of the bullet) or the front by a bullet? Or is the rear the only side able to start the reaction?

The primer and gun powder are located at the rear of the cartridge. It's probably possible that a significant impact somewhere other than the rear of the cartridge can cause the reaction to take place, but I am no expert on firearms and ammunition. Have a look at the Wikipedia page on rimfire ammunition and the illustrations there in particular.

The cartridge has usually a significant amount of gunpowder. I meant a lateral impact on the rear part, where gunpowder is. I update the question.

Maybe
Unfortunately, the correct answer is “it depends.” Small arms cartridges struck by small arms projectiles generally do not ignite – but they do sometimes.
The issue is quite important in the armed forces. If you come under enemy fire, the last thing you want is for a near miss to cause your own ammunition to explode. There has been formal research on the subject for decades. However, it can be quite difficult to prevent.
There are formal test protocols to assess the safety of new ammunition types, and they can involve munitions struck by a range of different “stimuli” from various distances, points of impact and angles. Depending on the exact set-up, the response can vary from 100% explosion to barely scratched.

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Why Is This The Worst Beer Cup
My wife bought me a special cup for me to drink beer out of.  The problem is that, if I pour any beer in it, it goes crazy with foam.  I will get just a few ounces of beer in there and the rest of the 16 ounce cup will be full of foam.  The cup is hand spun from clay.  The texture is still pretty rough.  So, if there is any glazing, it is not really noticeable.  The cup also has little brown specks in it here and there.  Does anyone know the reaction that is taking place to produce all this foam?  Is there a way to fix it?  Let me know if I need to supply any more details.  You should assume you are talking to someone whose last chemistry class was high school.

Yep the rough texture is a physical catalyst for the evolution of carbon dioxide from dissolved carbonic acid. Pottery glaze fixes the foaming issue.

Exactly. Chemistry has little to do with it. I mean, sure, there is a chemical reaction going on, but it is the same as in any other cup. Rough surface is the key.

This could be a business idea. We could make a cup that is glazed except at the top.

I'm guessing these bumps and rough spots are nucleation sites?

@BenWelborn If you wish to add your comment as an answer, I can give you credit.

The rough texture is a physical catalyst (provides nucleation sites) for the evolution of carbon dioxide from dissolved carbonic acid. Pottery glaze fixes the foaming issue.


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Why are the electronegativities of elements in the middle of the transition groups quite high compared to surrounding elements?
Is it something to do with the d orbital being used, and if so, what effect does this have and why does it make the electronegativity higher?

Or maybe http://chemistry.stackexchange.com/questions/33611/why-are-there-peaks-in-electronegativities-in-d-block-elements is a better dupe?


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R and S Configuration Understanding
So I have been trying to understand how to draw R and S configurations. I feel like I have understood how to draw them, but every time I do a question, I get a wrong answer.
So here is what I have understood,

The lowest priority goes to the back of the compound meaning on the dash line.
The highest priority is on the top.
You can play around with the 3rd and 4rth, so that one is on the
'out of page' and the other is on the line, so that you can
differentiate between the R and S.

But then there is this picture for example from my book,

I have been going crazy trying to understand it. The things I do not understand are these,
1) I understand the diagrams on the right side. That I have been doing as well
2) But then when we look at the diagrams on the left. The configuration is completely opposite. Even though it says S, the configuration is like of a R. Plus the highest one is going into the page and not the hydrogen. WHY? It is driving me crazy because they do that for all the examples. When they S, they actually make an R and vice versa. Whereas on the internet it is a different story. So I really need someone to tell me what I should do.

You just need to rotate the structure on the left to make the H go back into the plane of the page. It takes a bit of getting used to at first, and making a model can be extremely helpful, but eventually you'll get there. Good luck!

You are confused because in the left diagrams, the low-priority atom (Hydrogen) stands up: it's not on the back of the molecule!
To explain the first model, you have CH2OH coming towards you, OH going away from you, and CHO and H at the same distance from you.
You should imagine yourself looking at the molecule from another perspective: if you look at it from the bottom, you have hydrogen away from you, and the other groups at the same distance from you, in a triangle shape (that's the right diagram!).
Remember: the low priority atom should always stay away from you.

Transcribing chiral structures is invariably going to cause misassignments. Here is my advice: https://chemistry.stackexchange.com/questions/99721/twisting-stereoisomers-with-rings-to-determine-r-s/99750#99750

This is difficult to explain, even face to face, but I'll give it a try as I know how frustrating this can be to grasp.  Looking at the wedge and dash model, pick one of the four atoms/groups that ISN'T the lowest priority and ISN'T in the dashed (going away from you) position.  Doesn't matter which as long as it meets the two aforementioned rules.  Personally, I prefer to pick the wedged (going towards you) group.  Now, "freeze" whichever atom/group you chose, and by this I mean this group/atom will remain in the exact same position it's already in.  In fact, go ahead and draw the central carbon atom with the dashed line, wedged line, and both normal lines, but omit the groups/atoms for now (and draw it the same way it is pictured above on the left, it makes it much easier to visualize).  So you should have a C with all 4 bonds drawn out from it.  Now, go ahead and write in the atom/group you decided to freeze in the SAME position as it was before.  Now, the objective is to get the H atom (or lowest priority group) to be at the dashed position.  And remember, the frozen atom/group cannot be moved.  There are only two other positions the H atom could assume, as it cannot occupy the frozen atom/groups location, and you don't want it in the position it's already in.  So in the case of the first molecule, it can either occupy the position of CHO or OH (if you chose the wedged group/atom to be stationary).  Well, you want the H atom to be in the dashed position, which is currently occupied by OH.  So on your new molecule, go ahead and draw in the H at the dashed position where the OH was in the original drawing.  Now, youve effectively rotated the H into the position that the OH was.  Now the final rule you must follow, is that no group EXCEPT the one you chose to freeze can remain in the same position as it was in the original drawing.  This leaves only one location for the OH to be moved to, and one location for the CHO to be moved to.  The hydrogen was already moved to a new location, so now you must move the OH group and the CHO group to new positions, and it cant be either of the already occupied positions (the positions occupied by hydrogen and the group you chose to freeze, which in this example is the wedged position).  The OH must occupy the old position of CHO, and the CHO group must occupy the old position of H (why must they occupy these exact positions?  Because there are only two unfilled positions, and remember the newest rule, all groups/atoms that arent the frozen group MUST occupy NEW positions).  So go ahead and draw them in.  Now your molecule should be ready to evaluate for R/S configuration.
Be consistent when redrawing the molecule with the original drawing.  Don't draw the wedged bond on the top of the molecule if it was on bottom in the original drawing. Hopefully this helped, this is a confusing subject though and like I said earlier it's difficult to explain even in person.  I would definitely recommend getting a set of ball and stick models as it makes visualization much easier.  Practice enough and you'll get to where you dont need them anymore.
By the way to clear up any confusion, this step-by-step is essentially rotating the molecule (you aren't changing the structure or stereochemistry).  I simply laid out detailed directions so that if you were having difficulties visualizing rotating the molecule in your head, you wouldn't have to think about it.  As long as you follow the directions and rules I have laid out, it should result in a new drawing of the original molecule from a different angle, the angle in which the lowest priority group is in the dashed (going away from you) position

But then when we look at the diagrams on the left. The configuration is completely opposite. Even though it says S, the configuration is like of a R.

Use the simulation linked below to go back and forth between the priority 4 substituent on the top (OPs left panels) and behind the plane of the display (OPs middle panel).
http://proteopedia.org/wiki/index.php/R/S_nomenclature
It is like looking at a clock from behind. If you look from the front, clocks go clockwise, but if you look from the back (and the clock is translucent), if goes counterclockwise. That is why it is important to correctly orient the priority 4 substituent before determining R vs. S.

The lines are on vertical positions of Fischer projection. CHO is at the bottom of the vertical line, and it seems to be on the right side.So, the dashed line is on the left side of the horizontal position and the solid wedge on the right side. Now, draw Fischer projection as such.
H is up CHO is down on vertical axis.
CH2O right and
OH is left on horizonatal line.
Select priorities.
Priorities are shown in the figure as numbers from 1 to 3.
The chiral carbon has S configuration.
This is according to my knowledge without being responsible if there is something wrong in my answer. Better check with your teacher.

The drawing in the question are not Fisher projections, so how is this relevant?


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Electronic configuration and the periodic table
I came across the following question:

If each orbital can hold a max. of 3 e–  what is the number of elements in the 4th period of the periodic table?

I was unable to even start the thought process of answering the question because I'm rather horrible at discerning the period number using the electronic configuration, the number of elements that can be accommodated in a period etc. If someone could please explain it to me, that would be so great, especially since I've googled this too many times already and haven't been able to wrap my head around it yet!
Thanks in advance :) Cheers!

In the 4th period of the periodic table there is 1 s orbital, 3 p orbitals, and 5 d orbitals. That's a total of 9 orbitals. $9 \times 3=27$
That's 27 elements.

Since when there's supposed to be 3 electrons per orbital?

It's a hypothetical question.

It's not hypothetical, but sarcastic - there can be only two...

Yes, and in the "there can only..." stream of thought, the double slit experiment, GR, and the quantum model of the atom were attempts at sarcasm.

https://en.wikipedia.org/wiki/Spin_quantum_number - seems you're somehow not getting my point. It's 9*2 = 18 as spin quantum number can have only two different values, orbitals can be occupied only by 2 electrons. Geez you can count elements in periodic table to see there are 18 elements in this period.

Everyone knows that there are two electrons in an orbital. The question is testing your understanding of how the number of electrons in an orbital affects the number of elements in a period. It isn't a real-world problem. When a physics problem says, "a ball rolls on a frictionless surface", you don't tear up the question paper and declare,"it's impossible to have zero friction". It's a pretty basic intro chem problem: its purpose is to test understanding, not to be realistic.

Oh, my I get it... Sorry!


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Is there a formula or way to determine whether an acid is a strong one or a weak one without conducting experiments in real?
I know the common strong and weak acids but if a mineral acid like $\ce{H3BO3}$ springs up, I can't actually tell whether it is strong or weak.
Also can I justify in chemical terms why $\ce{H2SO4}$ is a strong acid but $\ce{H2SO3}$ a weak one?
Why is $\ce{HNO3}$ a strong acid whereas $\ce{HNO2}$ a weak one ?
http://catalog.flatworldknowledge.com/bookhub/2273?e=ball-ch12_s04
The above link shares a list of strong and weak acids but fails to state why one acid is weak while the other strong.

http://chemistrytextbookcrawl.blogspot.com/2013/02/paulings-rules.html Apart from that 8-5p formula itself, you obviously have to know the molecular structure of the acid, i.e. you will have to know that H3BO3 is actually B(OH)3 which means that the approximate pKa as given by Pauling's rules is 8 - 5(0) = 8. On the other hand H3PO2, H3PO3 and H3PO4 all have very similar pKa's because p = 1 for all three.

Is there a formula or way to determine whether an acid is a strong one or a weak one without conducting experiments in real?

Yes, there are ways to determine whether an acid is strong or weak without going into the lab yourself. That being said, someone must still go into the lab.
The most straightforward way would be calculating ΔG for an acid-base reaction. Remember, a negative delta G means that a reaction is spontaneous. Since strong acids have a Ka > 1, their ΔG values should be negative.

A "strong" acid simply means that when added to water, the acid will dissociate completely, therefore a strong acid is strong because it dissociates completely in aqueous solution while a weak acid is weak because it does not dissociate completely in aqueous solution. There are only 7 common inorganic strong acids so I would simply memorize them.
As far as why they are strong acids i.e. why they dissociate completely while other acids do not is a little more complicated and to be honest I am probably not the person to ask. I haven't looked into inorganic acids too extensively, though organic acids you must look at the stability of the conjugate base. A more stable conjugate base means the acid will more easily donate a proton. I do know for a fact that the conjugate base of sulfuric acid exhibits resonance which is a major stabilizing factor.  However, based on the structure of sulfurous acid I would expect it to also exhibit resonance, though I could be wrong.
The acid dissociation constant Ka is determined experimentally.  Strong acids have a Ka that is >> 1 while weak acids have a Ka that is << 1.
Hopefully I answered your question, but if not I'm sure someone with a little more inorganic chemistry experience will fill in the holes.

That there's only 7 common strong acids mentioned in a textbook or somewhere doesn't mean that these are only ones; in the future put in answers things you're really sure, best if you have sources to back it up.

@Mithoron Sorry, that was what I was taught throughout college, and I even had re-googled it, finding several college websites that had lists of "the 7 strong acids".  I was unaware there were more.  But seeing as I am in my 4th year at an acredited university working towards a BS in Biochemistry and all they've expected me to know are 7 of them (HCl, HBr, HI, HNO3, H2SO4, HClO4, HClO3), Im sure this is all the asker will be expected to know.  I figured a grad student wouldn't be asking the definition of a strong acid.  My apologies though, I am far more interested in organic chemistry.

@Mithoron Correct me if I'm wrong but I do believe there are only 7 strong inorganic acids which I believe is all the asker was referring to.  I actually did know there were more strong acids, but they are all organic acids.  I have edited my post to omit the ambiguity.

I meant strong inorganic acids - it's impossible to tell how many of them exists - there may be millions of inorganic compounds and lots of them are strong acids, starting from H2SeO4 through halogenated carboranes to various complexes.

+1 to @Mithoron's comment. Tetrafluoroboric acid is also quite strong. But, it is less common, and I think the problem is that the list of 7 is really the list of 7 most common inorganic acids (that one might encounter in an intro chem class).

Short answer: In general, no; but oxyacids and polyprotic acids have an extensive literature. (See Wikipedia: Acid dissociation constant).
Your second question, which is a "why" question is best answered on a case-by-case basis. Your question implies to me that you are not familiar with the quantum mechanical aspects of the dissociation of a protic acid in aqueous solution. I'm pretty sure they would almost immediately require much more advanced knowledge of quantum mechanics (specifically, DFT (and if you don't know what DFT is, then that proves my point)) than you have. In general, modern Physics and Chemistry (the hard physical sciences) aren't so good at answering "why?" questions. I encourage you to read (at least twice) the Wikipedia article Acid dissociation constant. It mentions a variety of contributing factors which influence dissociation. You have not only electrostatic effects (which in all but the most simple cases are quantum mechanical in origin (meaning require a QM explanation for all but the least sophisticated student - the least sophisticated students will accept half-baked explanations (like Pauling's 2 rules)) but steric effects, resonance effects, etc. QM is the basis for chemistry. Add statistical mechanics to that, and a "sufficiently powerful" computer, and you could recreate modern chemistry. By which I mean that fundamental "why?" questions can "really" only be addressed by QM. I'm thinking that that is not what you're looking for. (but, there it is.)


text:
Physical properties of phenols
Order of increasing boiling point is p-nitrophenol > m-nitrophenol > o-nitrophenol. The difference between bp of p- and o-nitrophenol is that one form intermolecular hydrogen bonding while other form intramolecular hydrogen bonding. Then what is the reason for high bp of p- than m-nitrophenol as both form intermolecular hydrogen bonding?

Related: Intermolecular Hydrogen Bonding and Strength of types of hydrogen bonds?

In case of p-nitrophenol the lone pair on oxygen of OH takes part in an extended resonance upto NO2 , more delocalisation of lone pair results into more delta minus on H atom ,so stronger H bonding, whereas  in case of m-nitrophenol it cant have resonace structures upto NO2. Since less delocalisation results in comparitively less delta minus, H bonding is comparitively weak.


text:
What happens first? Dissolution or solvation?
Can somebody help with the phenomenon what happens first?
Dissolution or Solvation.
I am bit confused with it

They happen together.

"Solvation, also sometimes called dissolution, is the process of attraction and association of molecules of a solvent with molecules or ions of a solute. As ions dissolve in a solvent they spread out and become surrounded by solvent molecules."
Source: https://en.m.wikipedia.org/wiki/Solvation


text:
How to convert metre to ångström?
In my book it says

Convert $1.213\cdot10^{-11}$ metre into Angstrom.

The answer is $12.13\cdot10^{-2}$ Angstrom.
How is it possible? I thought of multiplying the metre with $10^{-10}$, but I didn't get the correct answer. The same applies to the conversion of centimetres to Angstrom.

Related: May I treat units (e.g. joules, grams, etc.) in equations as variables?

The ångström (symbol: Å) is a widely used non-SI unit of length. Its value in SI units is:
$$1\ \mathring{\mathrm{A}}=0.1\ \mathrm{nm}=100\ \mathrm{pm}=10^{-10}\ \mathrm m$$
Thus, the required conversion factor is given by
$$\frac{1\ \mathring{\mathrm{A}}}{10^{-10}\ \mathrm m}=1$$
or
$$\frac{10^{-10}\ \mathrm m}{1\ \mathring{\mathrm{A}}}=1$$
This conversion factor can be used to express the given length of $1.213\cdot10^{-11}\ \mathrm m$ in terms of the unit ångström:
$$1.213\cdot10^{-11}\ \mathrm m=1.213\cdot10^{-11}\ \mathrm m\cdot\frac{1\ \mathring{\mathrm{A}}}{10^{-10}\ \mathrm m}=0.1213\ \mathring{\mathrm{A}}$$
Note that $0.1213\ \mathring{\mathrm{A}}=12.13\cdot10^{-2}\ \mathring{\mathrm{A}}$, which is the answer given in your book.
Also note that you do not simply divide the given length by $10^{-10}$ since the units are part of the conversion factor. You actually divide the value by $10^{-10}\ \mathrm m/\mathring{\mathrm{A}}$.

You're correct in saying you need to do is multiply by $10^{-10}$. The answer you gave is equivalent to $1.123\cdot10^{-1}$ (just divide by $10$), which is exactly $(1.123\cdot10^{-11}) \cdot (10^{-10})$. If you're using a calculator, make sure you wrap each term in brackets so that the order of operations is correct.


text:
Silver electroplating
In most of silver electroplating (or electrorefining) methods, it is mentioned that silver anode is oxidized in aqueous solution of silver nitrate, and that silver is deposited at the cathode (regardless of cathode material, let's assume it's also silver for the sake of simplicity). What I don't get is why would silver anode be oxidized before $\ce{OH-}$? The reduction potentials are ~0.8 V for silver and ~0.4 V for $\ce{OH-}$, at standard conditions.

Hydroxide is bad for this process. You will have additional problems if hydroxide anion is present.
Silver cation reacts with hydroxide anion to form silver hydroxide, which spontaneously decomposes into silver oxide:
$$\begin{aligned}\ce{Ag+ + OH-}&\ce{ -> AgOH}\\
\ce{2AgOH}&\ce{ -> H2O +Ag2O}\end{aligned}$$
In addition to shutting down the oxidation of silver, hydroxide precipitates any silver cation that does dissolve.
For this reason, the electroplating process is probably run at a slightly acidic pH, and the competing oxidation reaction is not of hydroxide, but of water (reduction potential 1.229 V):
$$\ce{2H2O -> O2 + 4H+ + 4e-}\ E^\circ = +1.229\mathrm{\ V}$$
So long as you don't apply too much of an overpotential to the electroplating operating, silver will oxidize before water.


text:
Convergence of spectra calculated from molecular dynamics trajectories
How can we check whether or not the power spectra (vibrational density of states) calculated by taking the Fourier transform of the velocity auto-correlation function (VACF) from a molecular dynamics trajectory is converged? How to estimate the statistical error in it?
What is the meaning of the word "convergence" in this context?

There are multiple ways in which the VACF, along with other properties derived from molecular dynamics trajectories like the dipole ACF to give IR spectra, can be considered "converged".

The molecular dynamics trajectory itself must be converged by some time-dependent statistical metric, such as measuring fluctuations in the energy or temperature over a number of steps and ensuring the average is below some threshold. This will be related to the ensemble chosen, the thermostat and barostat and associated parameters (if relevant), the integrator used for time evolution, and the size of the simulation, among other things.
Molecular simulations are layers of approximations piled on top of each other. In the case of purely classical dynamics, this usually refers to the force field parameters for the solute and solvent. Force fields are designed for different systems and may or may not be transferable, but the goal is usually to accurately predict dynamics. In the case of the VACF, if the dynamics of a system are described correctly, then the resulting power spectrum should be as well. However, in the case of the dipole ACF, the force field parameters enter directly as the nuclear charges (rather than just indirectly through nuclear velocities), which places stricter requirements on the quality of the force field. One hopes that newer force fields are better, but improvement over time isn't necessarily systematic and there is no quality hierarchy.
As an extension of 2, if the simulation is quantum mechanical in any way (such as in QM/MM, Born-Oppenheimer MD, or Car–Parrinello MD), then there are approximations in how the electronic structure is treated. For spectroscopic properties, it is more clear how to systematically improve results with regard to electronic structure than force fields due to well-established quantum chemical method hierarchies. Benchmarks of this hierarchy have been performed on many properties, including geometric parameters, dipole moments, and vibrational frequencies. In theory, the error in both dynamics and spectra should decrease up going from Hartree-Fock to MP2 to CCSD to CCSD(T), or by using more advanced density functionals (say, BP86 to PBE0 to M06-2X to B2PLYP). In practice, this would require benchmarking to draw any conclusions.
As an extension of 3, there are approximations in QM/MM of how the classically-treated and QM-treated regions interact with each other. Rather than go into more detail, here is a review of different QM/MM methods covering different definitions of the boundary and how one region may interact with the other.

Regarding estimation of the statistical error in resulting spectra, I am only aware of direct comparison (absolute error) between experimental and computed spectra, because it is assumed that the MD trajectory is converged (point 1). Performing multiple independent MD simulations and calculating the VACF from each should give an error measure closer to true statistical error, though I am unsure of how this would be computed.
Here is good almost-review (with good examples) of using Born-Oppenheimer molecular dynamics to calculate power spectra, IR spectra, and Raman spectra. I am not aware of any comparisons between classical MD and BOMD for computing spectra.
On the surface, it may appear as though the introduction of quantum mechanics makes things more difficult; certainly things become more time-consuming, however it is much clearer how results will change by treating electronic structure explicitly and with less severe approximations, rather than with a force field.

whatever you are suggesting is related to the quality of MD trajectory, the way to check convergence is still not clear to me! As far as my understanding, we can check  whether VACF is converging to zero or not within the simulated time, since if the simulation time is small, it may not get converged! And, if we have long enough trajectory, we might consider spectra calculated from different portion of it, ie taking different initial starting point.

To your first comment, "whatever you are suggesting is related to the quality of MD trajectory, the way to check convergence is still not clear to me", you should read the peer-reviewed literature for systems related to yours in order to understand this. This is both fundamental and critical to running any kind of MD simulation.

To your second point, "we can check whether VACF is converging to zero or not within the simulated time", that might be valid, but what does the literature say? It sounds like you have some knowledge of this already.

To your third point, "if we have long enough trajectory, we might consider spectra calculated from different portion of it, ie taking different initial starting point", the simplest case would be to take the entirety of your production run and none of your equilibration run, assuming that your system is properly equilibrated. This would ensure your VACF captures all the points (hopefully unbiased) that your trajectory samples.


text:
What are APG and AEO7 and should I be concerned?
I'm considering importing some granite cleaner for my stone business to save some money and the following is the list of ingredients:
APG, AEO7, silicone oil, water, White Mineral Oil
My question is should I be concerned about any of these ingredients?
 May primary concern is "AGP" and "AE07" as I do not know what these are.

The company supplying the product are required (in the USA/UK) to provide a material safety data sheet (msds) which explains what chemicals are in the product and any known hazards. This should be available on their website or through rmailing them. The msds is often a little vague but should tell you about any associated hazards/risks and the appropriate cautions to take.

I'd also echo the MSDS - some searching doesn't turn up what "APG" or "AGP" or "AE07" are. They are probably some sort of abbreviation, but the company is obligated to provide safety information in the US and UK (and likely EU as well).

APG = alkyl polyglycoside and AEO7 =  aliphatic alcohol and ethylene oxide adduct. Both are non-ionic surfactants.


text:
Why is this not an isomer of pentane
Given was the fact there are 3 isomers of pentane, two were given: pentane and 2-methyl butane. Then I was asked to draw to other one. I drew up the star like structure and checked it to be ok (coincidental I might add).
Then I drew some more and came up with amore turnstile shaped molecule.
     H
     |
   H-C-H   H   H
     |     |   |
   H-C-----C---C-H
     |     |   |
   H-C-H   H   H
     |
     H

To me this also seems like $\ce{C5H12}$ molecule and all valences seem ok. Why isn't this a valid isomer?
Edit: this is 2-methyl butane according to the book:
        H
        |
   H  H-C-H   H   H
   |    |     |   |
H--C----C-----C---C-H
   |    |     |   |
   H    H     H   H

It is surely valid, and is known as 2-methyl butane.

That one has 4Cs adjecent to each other, doesn't look like this according to the book.. (see edit) and if they are the same, then why? Can you just rotate the methyl group and the hydrogen? Am I thinking too much in 2 dimensions?

A man with arms outstretched and with arms down does not look quite the same, but still we will recognize him instantly. Same thing with molecules. Yes, the real molecule is very much 3D. Yes, you kinda can rotate (unless there is a double bond, but that's another story).

Right, try using sticks and balls, bonds are 109.5° .

Ok, clear now! If you change it to/add an answer I can approve it :)

See also https://pqr.pitt.edu/mol/QWTDNUCVQCZILF-UHFFFAOYSA-N for a 3D view

To clarify Apeiron's answer. You have drawn 2-methyl butane, the same isomer that was given in the question. The rotation of a bond is probably the source of any confusion. A 3D model of your two structures would be the same.
To answer your question:
Neopentane is the third isomer of the pentane series.


text:
Reactivity and transport safety of substituted methyl benzoates
This was a practice question from an MCAT book whose answer I didn't really understand:

The normal reactivity of methyl benzoate is affected by the presence of certain substituents. Which of the following substituents will decrease methyl benzoate reactivity making it safer for transport?
a) $\ce{NO2}$
b) Hydrogen
c) $\ce{Br}$
d) $\ce{CH3}$

They give the answer as (a) because "in order to decrease the reactivity of  benzene molecule, an electron withdrawing deactivating substituent must be used." However, I thought the reactive part of the molecule would be the carboxylic acid, and so I would've expected (d) to decrease the reactivity of the molecule because it's electron donating.
I suppose if the question meant the reactivity of the ring itself, then the answer is (a) since electron withdrawing substituents on a benzene ring lower its reactivity.
So my question is: How would one reasonably infer that the reactive part of this molecule is the benzene ring itself instead of the carboxylic acid?

Methyl Benzoate is an ester, not a carboxylic acid.  As far as reasonably infering the reactive part of the molecule, you can assume they are talking of ring reactivity when they ask which "ring substituent" will decrease reactivity, although they didn't specify ring in the question.  I wouldn't think much of it as I've noticed a lot of ambiguity in my PCAT study book as well (it's a Kaplan).  But technically speaking, electron withdrawing groups will decrease the reactivity of the ring portion of the molecule, thereby decreasing the reactivity as a whole.

You can also make the inference by looking at the answer choices, NO2 is the only strongly deactivating group as Br is capable of both donating and withdrawing electron density, CH3 is activating, and H is neither (it's what EWG and EDG are measured against)

@Ki11akd0g Whoops, even wrote it down as an ester but changed it in my head. I'll assume the question will be much less ambiguous then on the actual exam. Appreciate your insight!

Methyl benzoate isn’t even something I would consider dangerous to transport, so the entire question is somewhat moot.

Toluene is dangerous to transport... let's add a nitro group to it to reduce the reactivity of the benzene ring! But why stop at one nitro group? Surely three nitro groups should be even better...

I’ve already addressed this in a comment, but now I went the lengths to check out the material safety data sheets of these compounds from Sigma Aldrich. My suspicions have proven correct: methyl benzoate is not a dangerous compound to transport so any recommendations to reduce its reactivity by adding a substituent must be moot.
In short, section 14 of the MSDSs as I access them contains transport information; and this entry for the compounds methyl benzoate (CAS 93-58-3), methyl 3-nitrobenzoate (CAS 618-95-1), methyl 2-methylbenzoate (CAS 89-71-4) and methyl 2-bromobenzoate (CAS 610-94-6) contains nothing but ‘not dangerous goods’ and the required iterations and variations of that theme to the different categories there.
Therefore, the question must be rejected as totally pointless: the starting compound is not considered a dangerous good for transportation and neither is any of the three possible substituted ones (one generic placeholder for each; I’m not going to check an additional six MSDS sheets).
Naturally, the only answer you can remove by deduction is b, because a hydrogen ‘substituent’ does not change the properties of the benzoate residue since hydrogen is already present there five times.

Considering we have found out that the compound is not dangerous to transport, how can we approach this question? Well, it’s hard. Concerning the reactivity of electrophilic aromatic substitutions — which is low anyway because of the deactivating ester substituent — of course an electron-withdrawing substituent would help; the nitro group suggests itself. On the other hand, concerning the reactivity of the ester group’s carbonyl function, only electron-donating substituents would help; the strongest being the inductive effect of the methyl group. So we are left with two possibilities and it is unclear which one we should follow.
Of course, neither the reaction with electrophiles nor the reaction with nucleophiles is inherently more dangerous than the other. We have again reached a vicious circle in which we cannot proceed at the risk of taking the wrong path.

Therefore, the best advice I can give is to ignore the question in the book. If it is given in an exam, challenge the exam for there is no reasonable answer to it.


text:
The Hund's Rule
According to hund's rule, electrons must occupy singularly first in the orbitals before pairing up. And if not, the atom would not be stable. Imagine if there are electrons that do not obey Hund's rule, how unstable would the electron-electron repulsion be, what would be the consequences and how are they even unstable other than electron-electron repulsion between them?

There are exceptions to Hund's Rule. In general Hund's Rule would apply to single atoms in free space. Once you form compounds or solid crystals, then violations of Hund's Rule are more common.


text:
Determining solubility thresholds
Are there any sources that could provide Ksp value thresholds for things that are "completely soluble", "partially soluble", and "not soluble"?

A $K_{sp}$ value is a parametric measurement but  "completely soluble", "partially soluble", and "not soluble" are non-parametric measurements. So it seems you're asking about apples and oranges. Please clarify. Also note that solubility is significantly dependent on temperature which is another  factor to consider.

I am looking for some rules of thumb for Ksp values such that if the Ksp is greater than x at 25C, it is refered to as "completely soluble". In my mind, anything with a Ksp value around that of NaCl I would say is "completely soluble".

Then you already have a rule of thumb - don't expect better rules because they are no strict rules.


text:
How do you calculate the dipole moment of larger molecules?
I know that the dipole moment is obtained by:
$$
\vec{\mu} = Q \cdot \vec{d}
$$
with the condition that charges of a pair of atoms are equal and opposite, i.e. $\ce{NaCl}$.
But, in order to obtain the molecular dipole moment of water, how is the $\ce{-OH}$ group handled? What about even larger systems such as $\ce{NH3}$?
For example, the water molecular dipole moment is 1.85 Debye. How was this value was obtained?
I have read 1, 2, 3, and chemistry textbooks.
I am interested in a more mathematical or physical description.

The total dipole of a molecule can be thought of as the sum of dipoles of individual functional groups:
$$
\vec{\mu}_{\text{total}} = \sum_{i}^{N_\text{groups}} \vec{\mu}_{i}
$$
Because each dipole is represented as a vector with both magnitude and direction, this amounts to vector addition.
In the case of water, consider the dipole of each $\ce{-OH}$ group. The above reference gives it as 1.53 Debye, and the $\ce{H-O-H}$ angle is $105^{\circ}$. Because coordinates aren't given, the angle has to be used instead:
$$
\vec{P} + \vec{Q} = \sqrt{\lVert \vec{P} \rVert^{2} + \lVert \vec{Q} \rVert^{2} + 2\lVert\vec{P}\rVert\lVert\vec{Q}\rVert\cos{\theta_{PQ}}}
$$
which gives an estimate of 1.64 Debye. Not very good compared to 1.85 Debye! The group addition approximation completely neglects how the electronic structure of the molecule changes upon going from (using water as an example)
$$
\ce{2OH^{.}} \rightarrow \ce{H2O}
$$
The issue is probably that the two individual $\ce{-OH}$ groups interact strongly and favorably when combined together in water, compared to when they are apart; that is, the electron density changes quite a bit due to the bonding.
I've also neglected to explain how these groups can be defined; there is no unique definition of how to partition a molecule. For example, take a carboxyl group:

Are all four atoms treated together as a single group, or are the $\ce{C=O}$ and $\ce{C-O}$ and $\ce{-OH}$ treated separately? Maybe this is a simple case but it becomes more unclear for something like a protein or a coordination complex.
The molecular dipole can also be calculated using electronic structure theory, which removes the ambiguity surrounding how to partition a molecule and incorporates those "non-additive" effects.
As Geoff mentioned in the comments, the definition of the dipole moment created by a set of charges can also be used:
$$
\vec{\mu} = \sum_{a}^{\text{charges}} \vec{r}_a \times q_a
$$
where $q_a$ is the charge, and
$$
\vec{r}_a = \vec{R}_a - \vec{O} = (R_{ax} - O_x, R_{ay} - O_y, R_{az} - O_z),
$$
where $\vec{R}$ is the position of the point charge, and $\vec{O}$ is a common origin, usually taken to be either $(0,0,0)$, the center of mass, or the center of nuclear charge. It doesn't matter what this point is for uncharged (neutral) systems, but there is an origin dependence for anything with a non-zero total charge. The $\{\vec{R}\}$ are taken to be atomic positions, and the charges $\{q\}$ can come from simpler Mulliken or Löwdin population analyses, or more complicated schemes of which many are designed to reproduce total dipole moments. Care must be taken to deal with units properly, specifically when converting: $0.393430307 \, ea_0 = 1 \, \mathrm{Debye}$.

+1 I think the only thing I'd add is that you could also, in principle add up atomic partial charges: q * v, where v is the atomic position relative to the center, but you'll need to be careful to convert from e*Å to Debye. Of course this requires that the atomic partial charges reproduce the dipole moment, but that's the goal of many partial charge schemes.

I did neglect to point that out, and it's mentioned here. This is also how one would calculate the dipole in molecular mechanics/from a force field.

in H2O the lone pairs of electrons are considered functional groups ? if yes then we should calculate the dipole moment from hydrogen atoms and lone pairs of electrons of the oxygene atom ?


text:
Rate equation variations
I am having a bit of trouble understanding variations to rate equation problems.
What I know:

For $\ce{A + B -> AB}$, if the rate is proportional to the concentration of A and B, the rate equation is:

$\quad$ rate=$\ce{k*[A]^1*[B]^1}$

For $\ce{A + A -> 2B + C}$, if the rate  proportional to $\ce{[A]}$:-

$\quad$ rate=$\ce{k*[A]}$ , as per  Question 1c and Answer 1c.

Question 1 : Why is this the case?
E.g. when it was $\ce{1A + 1B}$ and the rate was proportional to the concentration of $\ce{A}$ and $\ce{B}$ our rate equation contains $\ce{[A]*[B]}$. But for $\ce{1A + 1A}$ why does the rate equation contain $\ce{[A]}$ instead of $\ce{[A]*[A]}$?
Question 2: Do molecules change anything (compared to single atoms)
Is $\ce{A + A}$ equivalent to $\ce{A2}$
E.g. will all of these have the same rate equation (if the rate is proportional to $\ce{[A]}$)

$\ce{A + A -> 2B + C}$
$\ce{A2 -> 2B + C}$
$\ce{2A -> 2B + C}$

I can't see why they wouldn't all have the same rate equation as it would just be different ways to represent the same total $\ce{[A]}$.

The issue is that the rate equation is determined not by the reaction equation, but rather by the actual mechanism that lies beneath it. It turns out that reactions that appear fairly simple, such as the burning of methane in oxygen in the flame of a Bunsen burner, do not necessarily take place in a simple manner, I understand that the even the reaction for the Bunsen flame can in real life contain a very large number of steps. Depending on the reaction, different steps will require different reactants, and will take place at different speeds. Maybe for the reaction $\ce{A + B->AB}$, some step that takes a long time involves $\ce{A}$ decomposing on heating, the products of which go on to quickly react further to eventually form $\ce{AB}$. This step that takes a long time is called the rate-determining step, and decides the form of the rate equation. It may be that this causes only $[\ce{A}]$ to feature in the rate equation.
In this case however, the rate equation is given to you by the comments to the right of the equation. Just because we write it as $\ce{2A}$ or $\ce{A+A}$ doesn't make a difference, it's just how someone decided to write the equation, which does not necessarily reflect the steps of the mechanism. In fact, if a mechanism involved a slow step $\ce{A + A -> B}$, the reaction would be second order with respect to $\ce{A}$.
Hopefully that helps.

Thanks for the response. That is what I thought but for this 2A⟶2B+C , this link (question 1 c) http://www.chemguide.co.uk/physical/questions/q-reactionorders.pdf says the equation is 1st order not second order with respect to a (answere here, question 1 c : http://www.chemguide.co.uk/physical/questions/a-reactionorders.pdf)


text:
What is the difference between a ylide and a betaine?
I'm not sure what the exact difference between a ylide and a betaine is...
Are both zwitterions?
Something along the lines of: ALL ylides are betaines, but not all betains are ylides(according to wikipedia the cation/anion group can be adjacent or nonadjacent)?

The more definitive source for chemistry terminology is the IUPAC Gold Book.
Betaine

Originally, the compound betaine, $\ce{(CH3)3N+–CH2C(=O)O−}$ (N,N,N-trimethylammonioacetate), and similar zwitterionic compounds derived from other amino acids. By extension, neutral molecules having charge-separated forms with an onium atom which bears no hydrogen atoms and that is not adjacent to the anionic atom. Betaines cannot be represented without formal charges.

Ylides

Compounds in which an anionic site $\ce{Y−}$ (originally on carbon, but now including other atoms) is attached directly to a heteroatom $\ce{X+}$ (usually nitrogen, phosphorus or sulfur) carrying a formal positive charge. They are thus 1,2-dipolar species of the type $\ce{R_mX+–Y−R_n}$.

So, ylides are not betaines and betaines are not ylides. Primarily due to the atoms in a ylide that contain the charges must be adjacent, and in betaines they must not be adjacent.
An example of a ylid would be the phosphonium ylid used in Wittig chemistry:

Since it has adjacent positive and negative charges, it is classified as a ylid.
The addition of this ylid to a carbonyl compound would form a betaine (note that the mechanism of the Wittig reaction is now thought to be a concerted [2+2] cycloaddition leading to the oxaphosphetane immediately, without the intermediacy of the betaine):

Note that the ylid has an alternate resonance form (the one with a P=C double bond) which has no formal charges, whereas the betaine cannot be drawn without formal charges.

Quite nice answer. Welcome to Chemistry SE!

Great answer. It seems wiki has got it wrong.

@Nova Wiki is a great starting point, but not always technically correct.

You can format your posts using MathJaX: http://meta.chemistry.stackexchange.com/questions/86/how-can-i-format-math-chemistry-expressions-here


text:
Mechanism for Hydrogen Peroxide decomposition?
Someone asked this question earlier and made no attempt at understanding it on their own so the question was flagged.  However, it prompted me to attempt to figure it out.  This is what I have come up with, and the reason I'm asking about it is I found the mechanism quite interesting (if it is indeed the mechanism).
The first step is of course the breakage of the $\ce{O-O}$ bond in hydrogen peroxide to form 2 hydroxy radicals ($\ce{H - O•}$)
In the second step, a hydroxy radical reacts with an unsplit hydrogen peroxide molecule to form water and a peroxy radical ($\ce{H - O - O•}$).
In the third step, two peroxy radicals react to form a molecule of hydrogen peroxide and what google terms "triplet oxygen" with the structure $\ce{•O - O•}$.  Heres the interesting part, if I am indeed correct.  The triplet oxygen goes through an intramolecular free-radical termination reaction to form oxygen gas.  It may be more complicated than an intramolecular reaction.  I only assumed this is what happens based on my knowledge of intramolecular friedel-crafts acylation, which is much faster than intermolecular friedel-crafts acylation.  This is due to intramolecular acylation having no change in entropy (one reactant forms one product) while intermolecular acylation has a decrease in entropy (two reactants form one product).
This would be quite interesting if it is indeed the mechanism!
Edit: Regardless of whether or not the oxygen gas is formed by an intramolecular free-radical termination, I do believe this is the correct mechanism because the sum of all the elementary reactions produces the overall reaction of:
$$\ce{2 H2O2 -> 2 H2O + O2}$$
Elementary steps:
1)    $\ce{H2O2 -> 2 HO•}$
2)    $\ce{2 HO• + 2 H2O2 -> 2 H2O + 2 HOO•}$
3)    $\ce{2 HOO• -> H2O2 + •OO•}$
4)    $\ce{•OO• -> O2}$
Do correct me if I am wrong though

related http://chemistry.stackexchange.com/questions/42736/disproportionation-of-hydrogen-peroxide

also http://chemistry.stackexchange.com/questions/10477/what-exactly-happened-when-i-added-koh-to-hydrogen-peroxide http://chemistry.stackexchange.com/questions/8124/how-do-i-create-a-concentrated-solution-of-singlet-oxygen-in-the-dark

That's all great, but what's the question? Also the ground state of oxygen is already a triplet state

@orthocresol I was asking whether or not the mechanism was correct.


text:
The Definition of a Solution
Can there be more than one solute or solvent in a solution. Why? Why not? The largest component in the solution is called a solute and smallest a solvent. How is this possible when solutions contain more than 2 substances?

You can have as many solutes as it gets. Also there can be more of them in total then the amount of solvent present, so could you care less about semantics?

http://chemistry.stackexchange.com/questions/23519/can-there-be-more-solvent-in-a-solute

Well no, I have an exam on this. Solutes are defined as the smallest component in the solution, so if there are multiple solutes, what are they called?

Here is the solution defined by IUPAC:

A liquid or solid phase containing more than one substance, when for convenience one (or more) substance, which is called the solvent, is treated differently from the other substances, which are called solutes. When, as is often but not necessarily the case, the sum of the mole fractions of solutes is small compared with unity, the solution is called a dilute solution. A superscript attached to the  ∞ symbol for a property of a solution denotes the property in the limit of infinite dilution.

When the solution contains more than 3 components, one or more will be solvent(such as you solute NaOH in water and ethanol).
For example, You can regard glass as solution, because it contains SiO2, Na2CO3, and many other components.
I think which is solvent is choose by human, usually the most part of the "solution" will be regard as solvent.