What would the phosphorus-NMR of trifluorodiphenylphosphorane look like? I'm fairly new to NMR especially $\ce{^{19}F}$ and $\ce{^{31}P}$ NMR, we've not been formally taught NMR yet but we're expected to know $\ce{^{19}F}/\ce{^{31}P}$ NMR for our exam for basic molecules.

I know $\ce{PPh2F3}$ will have a trigonal bipyramidal geometry, I know in $\ce{PPh2F3}$ the phenyl groups will be equatorial due to the torsional angle and sterics. This leaves two fluorines axial and one fluorine equatorial, but  what I don't understand is why this gives a doublet of triplets signal (for $\ce{^{31}P}$). Any explanation to why you get this pattern would be helpful, taking into account my fundamental understanding of $\ce{^{31}P}$ NMR is not so great.

also what would the $\ce{^{19}F}$ NMR be for the same molecule?

Well, these are the fundamentals of NMR that are important:

Signals will be split if they couple to neighbouring atoms with a spin $\ne 0$ (though coupling to spin ½ nuclei is most common especially in organic NMR).
Every different environment of a coupling nucleus will produce a distinct coupling.
Depending on the number of nuclei in a given environment, the number of signals (coupling pattern) is given by the equation $2nI+ 1$, where $n$ is the number of nuclei and $I$ is that nucleus’ spin.

In your example you have correctly identified two different fluorine environments with one and two fluorines in them respectively.

The single equatorial fluorine will give a doublet because $2nI +1 = 2\ (n = 1, I = \frac{1}{2})$.

For the same reason, the axial fluorines will give a triplet: $2nI + 1 = 3\ (n =2, I = \frac{1}{2})$.

If one coupling gives a triplet and the other a doublet, the final result is either a doublet of triplets or a triplet of doublets. (The larger coupling constant i.e. the greater splitting is given first.)

The fluorine NMR would be more complex: They can couple to both phosphorus and the other fluorine. It is important to know that the axial fluorines are homotopic, i.e. the molecule can be rotated exactly exchanging their places — this means that the two give an identical signal. The equatorial one cannot be brought to overlap with the axial fluorines by any mean of symmetry; the two environments are heterotopic, i.e. they will produce different signals (and typically couple).

The axial fluorines will couple to the equatorial fluorine and phosphorus.
The equatorial fluorine will couple to the axial ones and phosphorus.

I’ll leave you to work out the coupling patterns. (Hint: The less bonds between two coupling nuclei, the larger the coupling constant usually is.)

A final note: It is due to electronic reasons that the phenyl groups occupy the equatorial positions rather than going axial. A four-electron three-centre bond is more stable for electronegative elements than electropositive ones.


Understanding the partial derivative of temperature with respect to thermodynamic beta (coldness) I am trying to prove that the specific heat is related to the fluctuations in the energy:

$$c_V = \frac{\langle E^2 \rangle - \langle E \rangle^2}{k_\mathrm BT^2}$$

Where:

$$\beta = \frac{1}{k_\mathrm BT}$$

$$\langle E \rangle = -\frac{ \partial \log(Z)}{\partial \beta}$$

I did:

$$\frac{\partial^2}{\partial \beta^2}\ln Z = \frac{\partial}{\partial \beta}\frac{1}{Z}\frac{\partial Z}{\partial \beta} = -\frac{\partial \langle E \rangle}{\partial \beta} = -\frac{\partial \langle E \rangle}{\partial T} \frac{\partial T}{\partial \beta}$$

My issue is that I do not understand why:

$$-\frac{\partial T}{\partial \beta} = k_\mathrm BT^2$$

You know that:

  $$\beta = \frac{1}{k_\mathrm BT}$$

Rearranging yields:
$$\color{red}{\beta} = \frac{1}{k_\mathrm B\color{blue}{T}}\implies \color{blue}{T} = \frac{1}{k_\mathrm B \color{red}{\beta}}$$

Thus:

$$\frac{\partial \color{blue}{T}}{\partial\beta} = \frac{\partial}{\partial\beta}\left(\frac{1}{k_\mathrm B \beta}\right) = \color{red}{-}\frac{1}{k_\mathrm B \beta^\color{red}{2}}$$

Substituting $\beta = \color{red}{\frac{1}{k_\mathrm BT}}$ yields:

$$\frac{\partial T}{\partial\beta} =  -\frac{1}{k_\mathrm B \color{red}{\left(\frac{1}{k_\mathrm BT}\right)}^2} = -\frac{k_\mathrm B^2T^2}{k_\mathrm B} = -k_\mathrm BT^2$$

$$-\frac{\partial T}{\partial\beta} = k_\mathrm BT^2$$
Q.E.D.


Using solubility guidelines to predict a reaction of two ionic, water-soluble salts Will a reaction occur in the following case? If so, write a net ionic equation for it:
$\ce{(NH4)2SO4 (aq) + ZnCl2 (aq) -> ?}$

The book says that all possible combinations of positive and negative ions lead to water soluble compounds, all of the ions remain in solution. No reactions occur.
Now I am a little confused. Why didn't $\ce{(NH4)2}$ join with $\ce{Cl2}$ and why didn't $\ce{SO4}$ join with $\ce{Zn}$? Or did they?

Of course $\ce{NH4+}$ can join $\ce{Cl-}$ and $\ce{Zn^2+}$ can join $\ce{SO4^2-}$, but the formed salts are completely soluble in water, and they completely dissociate in water to gives the corresponding ions.

On the other hand, the starting salts:  $\ce{(NH4)2SO4}$ and  $\ce{ZnCl2}$ are completely soluble in water, and they completely dissociate in water to gives the corresponding ions.

So, all of the ions remain in solution, as if no reaction occurs.


How to calculate molality, molarity and mole fraction given density (g/mole)? Calculate the molality, molarity and mole fraction of a $\ce{CuSO4}$
in a 17% m/m aqueous solution (density of the solution is 1.367 g/mole)

I did .17 * 1.367 to find grams of $\ce{CuSO4}$

and .83 * 1.367 to find grams of $\ce{H2O}$

Then used to them calculate:

molarity = 0.00146 M

molality = 1.287 m

Mole fraction = 0.023

I'm wondering if I'm doing it right?

First let me correct you, m/m means mass/mass % I think you assumed it to be mole/mole.

Assume you have 100g of solution in which Cu is 17g and solvent is 83 gm. 17 gm copper means 0.268 moles of Cu.

Now, density = Total weight/Total moles = 100/(0.268 + x) = 1.367 where x is moles of solvent, from here you will get x = 72.8849

Thus, mole fraction of Cu is 0.268/(0.268+72.8849) = 0.00366

Molality = Moles of solute/Weight of Solvent(in Kg) = 0.268/0.083 = 3.229 m

Since, density in terms of g/l is not given I don't think it's possible to calculate molarity.


Coupling of silyl imine and aldehyde LDA is a strong base and probably deprotonates somewhere which I'm not sure about. It I would to make a guess it would probably be beta to the imine. This is following by nucleophilic attack of the carboanion onto the use of the second reagent Pr-CHO to generate an alcohol. But I'm not sure the purpose of the 3rd reagent (oxalic acid).

This is a Peterson olefination, specifically this one is from the synthesis of Roseophilin published in J. Am. Chem. Soc. (2001) 123 8509. The LDA deprotonates next to the silicon as both it and the adjacent imine stabilise the anion.

The Wikipedia page gives a good summary of the mechanism and why an acidic work up is used in this case to give the required double bond geometry (Wikipedia).

(Image from www.chem.wisc.edu)


Major product of the dehydration of alcohols Well I was doing some questions and I have a bit of confusion on dehydration reaction. For instance, consider following reaction:

How do I decide which adjacent hydrogen gets attracted by $\ce{HSO4-}$?

now we have 3 possibilities:

either hydrogen 1,2, or 3 can leave resulting in 3 different products. So my question is how do I identify the major product in this case?

The mechanism of dehydration of 2° alcohol at high temperature (around 170°C-180°C) happens via $E_1$ mechanism. This means that the major product will be decided by the $\mathbf {Zaitsev}$ $\mathbf{rule}$, which says that major product will be the stables one. If you compare the three alkenes, you'll see that the middle compound you've drawn has the most hyperconjugative structures, and hence is the stablest. Therefore, that will be the major product.


Reaction of 1 equivalent of Mg in ether with 1,4-dibromobut-2-yne What will be the product formed the if the reaction of Mg in ether with 1,4-dibromobut-2-yne occurs? At first, I thought the answer would be cyclobutyne, but the correct answer is buta-1,2,3-triene.

Can anyone please explain how?

Cyclobutyne cannot exist.  There is no way to enforce 90° bends at both ends of a triple bond, but the carbanion formed by the Grignard reagent synthesis finds an easy way around this problem.

Let's say the magnesium reacts at position 1.  Then the negative charge would couple with the conjugated triple bond to spread to position 3:

$\ce{Br\overset{+}{Mg}\overset{-}{C}H2-C#C-CH2Br <-> Br\overset{+}{Mg}CH2=C=\overset{-}{C}-CH2Br}$

Then the carbon atom at position 3 can act as a nucleophile, displacing the remaining bromine as a bromide ion to give the product named in the book.


Acidity order of acetone, methyl acetate and N,N-dimethylacetamide Why is the acidity order true?
  $$\ce{CH3COCH3} > \ce{CH3CO2CH3} > \ce{CH3CON(CH3)2}$$

I deduced that in methyl acetate and N,N-dimethylacetamide there is a −I effect of the methoxy group, $\ce{N(CH3)2}$  and a +R effect, too so the attached keto-carbon develops a positive charge and the conjugate base becomes unstable due to the +R effect of methoxy and $\ce{N(CH3)2}$ groups.

Is my logic right?

The increased acidity of the protons on the alpha-carbon in these compounds, compared to the acidity of the protons in methane, is due to stabilization of the resultant negative charge through resonance with the adjacent carbonyl.  In the top line of the following figure we show a ketone where the proton has been removed.  We can draw 2 resonance structures to describe this carbanion.  In structure II the negative charge has been delocalized onto the electronegative oxygen atom - a very favorable and stabilizing situation.

To whatever degree we remove the carbonyl's ability to resonance stabilize that adjacent charge we will decrease the acidity of the alpha-protons.

In the case of both the ester (III) and amide (VI) we can draw a third resonance structure (V and VIII respectively).  To whatever extent these resonance structures contribute to the overall description of the ester and the amide, the carbonyl will be less available to stabilize the negative charge through structures IV and VII.

It turns out that resonance structures V and VIII are significant contributors, with the amide structure VIII having a positive charge on nitrogen being more important than resonance structure V which places the positive charge on the more electronegative oxygen (less favorable).  In fact, resonance structure VIII is such a significant contributor that restricted rotation occurs about the $\ce{C-N}$ bond and we can typically observe two different R group signals in the proton-nmr spectrum of amides.

So as stated earlier, as the contribution of the third resonance structure increases, the contribution of the second structure (the structure that stabilizes our carbanion) will decrease making our compound less acidic.  The ketone doesn't have a third structure so it will be the most acidic structure. Structure VIII is more important than structure V, so the ester will be more acidic than the amide.  Therefore, the overall order of decreasing acidity is
$$\ce{CH3COCH3 > CH3CO2CH3 > CH3CON(CH3)2}$$
just as you noted.


Which molecule has maximum bond angle amongst BF₃, BCl₃ and BBr₃? $\ce{Br}$ is the least electronegative from the given compounds. So, in $\ce{BBr3}$ the electrons will be the closest towards boron and thus due to electron-electron repulsions $\ce{BBr3}$ will have maximum bond angle.
Steric hindrance due to large size of bromine atom can be a factor too.

But the bond angles are same according to most of the websites.

With VSEPR theory you can predict the structures of $\ce{BX3}$. Since boron only has three valence electrons, all of which are used for σ-bonding to the halogen atoms, they arrange in a trigonal planar fashion. This means that all molecules $\ce{BF3}$, $\ce{BCl3}$, and $\ce{BBr3}$ have the same point group $D_\mathrm{3h}$. From this symmetry you can deduce that the bond angle is $\angle(\ce{X-B-X}) = 120^\circ$ for all of them.

Bent's rule won't have any effect on that.

There is also considerable π-backdonation in these molecules, but that only has an effect on bond lengths, not bond angles.


Is the following molecule aromatic or antiaromatic? This molecule has the same chemical formula $\ce{C20H10}$ as corannulene (which also is non-planar) and corannulene is aromatic, so I'm pretty sure that this molecule should be aromatic too.

I tried to analyse it using the MoCubed app on my phone (on my previous posts, I wrote "WebMo" instead of "MoCubed" because I have both on my phone and sometimes get confused) but it keeps adding unnecessary hydrogen atoms and I can't rotate the molecule to remove all of them so I have to ask here.

P.S. I do know Hückel's rules, but they don't apply to pericyclic rings.

It turns out this question has some subtleties.  The analysis presented below, based on Huckel MO theory, indicates that:

1.
The compound as presented by the OP is not expected to be aromatic and would be anti-aromatic by the Huckel model -- meaning in practice it would probably distort out of full conjugation to avoid this calamity.

2.
The dication, however, does appear to be aromatic.  Part of the reason is that while we ordinarily suppose that five-membered rings must be negatively charged in an aromatic system, a phenanthrene-like accumulation of three such rings can become aromatic with a positive charge instead.

Same formula, but ...

The OP suggests a comparison with the aromatic compound corannulene, which like the compound in question has the formula $\ce{C20H10}$ and twenty conjugated pi electrons.

Let's compare the Huckel energy levels for these two compounds, keeping in mind that an aromatic system will have (1) a stabilized highest occupied energy level and (2) a wide energy gap between this level and unoccupied states (figure made by the author):

Occupying the lowest available orbitals in corannulene on the left (including two degenerate pairs between $\beta$ and zero that are close together, thus appear as thick bars), we see the desired effect for aromaticity: stabilized highest occupied orbitals, wide gap to the unoccupied ones).  Not so on the right with the OP's compound: here there are nine bonding orbitals so putting in twenty pi electrons means having a mixture of unoccupied and occupied states at the nonbonding level where there is a degeneracy.  This is a classic antiaromatic situation (similar, actually, to triangulene which us invoked in another answer), and thus the answer to the original question is the compound we are examining will not be aromatic.  It will form a diradical or  find a way to get out of antiaromaticity instead.

Thrown for a loop

What happened?  In multicyclic systems aromatic coupling is not just a matter of having the right number of electrons.  Hidden in the energy calculations is a topological difference.  In corannulene five of the conjugated atoms are in an interior closed loop surrounded by the rest of the molecule, while the OP's compound lacks such an internal loop.  This topological change can alter aromatic coupling, which is sensitive to how loops of conjugated atoms are connected to each other.  For this particular combination of molecular formula and pi-electron count, corannulene has a good topology for forming an aromatic system.  The OP's compound does not.

Making lemonade out of lemons

Many Huckel-antiaromatic systems become aromatic if one places the proper charge on them, adjusting the number of pi electrons and thus occupied pi orbitals.  The classic case is cyclootatetraene which goes from an antiaromatic/distorted nonaromatic neutral molecule to an aromatic dianion when we combine it with potassium.  What can be done in this direction for the OP's compound?

According to the Huckel energy levels we ought to either fill the degenerate orbitals as in cyclooctatetraene, or empty them.  We see that in this case the latter is what works.  The dication, as opposed to the dianion, gas a stabilized highest orbital and a wide gap up to what are now unoccupied degenerate states.  In other words, the hallmarks of aromaticity.

But where can these positive charges reside?  We can make the seven-membered ring aromatic with one positive charge, but shouldn't the benzene rings be neutral and the five-membered rings negative like the cyclopentadienyl anion?

We crack open our topology textbook again and we discover that three separate rings are not connected in the same way as the accumulated five-membered rings in the OP's molecule/dication.  We'd better check the Huckel orbitals inherent to a cumulated five-membered ring system like the one in the OP's molecule, having the formula $\ce{C11H7}$ (this figure also by the author):

The seemingly similar phenanthrene system is a fourteen-pi-electron aromatic system.  Here, however, with fewer input atomic orbitals we see the widest gap, thus likely the greatest stabilization, occurs with only ten pi electrons.  This means the eleven-carbon system will have a positive overall charge.  The chain of three "cyclopentadienyl" rings generates an aromatic cation and thus can take on the second positive charge in the dicationic, predicted aromatic form of the OP's compound.


Bond energy and s-character While studying the basics of chemical bonding, I've often come across the following statements:
1. The bond energy increases with the increase in the s-character in a hybrid orbital, i.e sp3 sp2-sp2, and so on.

I can't seem to correlate the two. If a hybrid has a higher bond energy, shouldn't it also form stronger bonds?

Let us start with the following most important part of the IUPAC definition for the bond energy.

  bond energy (mean bond energy)

  The average value of the gas-phase bond dissociation energies (usually
  at a temperature of 298 K) for all bonds of the same type within the
  same chemical species.

I think it should be immediately clear now that the higher a bond energy is, the stronger is the bond, so that there is no any contradiction in a bond energy increase with the increase of the $\mathrm{s}$-character in hybrid $\mathrm{sp}^n$-orbitals.

Indeed, $\mathrm{2s}$-orbitals are lower in energy that the $\mathrm{2p}$ ones, and thus, the more $\mathrm{s}$-character a hybrid orbital has, the lower is its energy: $E_{\mathrm{sp}} < E_{\mathrm{sp^2}} < E_{\mathrm{sp^3}}$. The lower an atomic or hybrid orbital energy is, the lower is the energy of a molecular orbital in formation of which it participates, $E_{\sigma_\mathrm{sp}} < E_{\sigma_\mathrm{sp^2}} < E_{\sigma_\mathrm{sp^3}}$, and consequently, the higher is the corresponding bond dissociation energy, as well as, the average of them, i.e. the bond energy. So, the bond energy indeed increases as the s-character grows, say $E_{\ce{C(\mathrm{sp})-C(\mathrm{sp})}} > E_{\ce{C(\mathrm{sp^2})-C(\mathrm{sp^2})}} > E_{\ce{C(\mathrm{sp^3})-C(\mathrm{sp^3})}}$.

Below is the MO diagram for $\sigma$ $\ce{C-C}$ bond between two $\mathrm{sp}$-, two $\mathrm{sp^2}$-, and two $\mathrm{sp^3}$-hydridized $\ce{C}$ atoms that illustrates the logic above.


The chemistry of cutting I would like to know what is the chemistry of cutting. For example, when we use a knife to cut through bread what exactly happens on a molecular level? Are ions formed, do long organic compounds (e.g. in cheese) break, or are just intermolecular interaction that seize to exist?

Thanks a lot in advance.

It depends. Bread has moth small molecules (like 2-Acetyl-1-pyrroline). These molecules will just pushed aside by the blade. You have hydrogen bond network between water molecules. You break those. Bread has large polymers like starch. Knife breaks those.

A cleaner system to study is a polyethylene. You can freeze it in N2, break under inert atmosphere and study by EPR spectroscopy. You will find free radicals at the breaking point. This is because it costs less energy to break a bond then to move one molecule relative to others.


Finding out the value of rate constant on titration The question goes as follows

Consider the reaction $$\ce{X(aq) \longrightarrow Y(aq) + Z(aq)}$$

The above reaction is titrated with a reagent $A$ and all of $X, Y, Z$ reacts with $A$ with the ratio of 'n' factors of $X, Y, Z=1:2:3$

here n factor refers to change in oxidation number of substance per mole of the substance on titration with $A$

The following data was obtained.

The question is to find out the value of rate constant considering the titration reaction to be first order.

The value of rate constant is given by $\frac{0. 693}{t_{1/2}}$ So I have to find out half life of the reaction . I am facing trouble finding out since the rate of voume of A required depends on the amount of substance present at that instant. Any help shall be highly appreciated. Thanks.

From a mole balance, we find $C = n_x(t) + n_y(t) + n_z(t) = n_x(t) + 2n_y(t)$, where the second equality follows from the stoichiometry of the reaction. In particular, our boundary conditions are $C = n_x(0) = 2n_y(\infty).$

We assume that the reaction is first-order, so $n_x(t) = n_x(0)\exp(-kt) = C\exp(-kt).$ The two measurements give us the set of equations
\begin{align*}
80D &= n_x(5) + 5n_y(5) &&= C\exp(-5k) + \frac{5}{2}C(1-\exp(-5k)) \\
100D &= 5n_y(\infty) &&= \frac{5}{2}C,
\end{align*}
where $D$ is an undetermined proportionality constant for the titration, and the coefficient of 5 in $n_y$ accounts for the fact that 2 equivalents of $A$ react with $n_y$ and 3 equivalents with $n_z$. Dividing one equation by the other produces the final result $$\frac{4}{5} = \frac{2}{5}\exp(-5k)+1-\exp(-5k),$$ which can be solved for $k$ and hence $t_{1/2}$.


Percent of silver by mass in an unknown sample Question
  If all of the silver in a $7.563~\mathrm{g}$ sample of a mineral, which was dissolved by using nitric acid, is precipitated as silver thiocyanate ($\ce{AgSCN}$) by using $82.46~\mathrm{mL}$ of a $0.2500~\mathrm{M}$ potassium thiocyanate solution, what is the percent silver (by mass) in the sample?

My Attempt
I'm pretty sure the relevant chemical equation here is $\ce{Ag^+ + SCN- -> AgSCN}$.

Molarity is mols solute over L solution, so the number of mols of $\ce{KSCN}$ used is $0.25~\mathrm{M} \times 0.08246~\mathrm{L} = 0.020615~\mathrm{mol\ KSCN}$. According to the chemical equation above, the molar ratio is 1 to 1 so there is $0.020615~\mathrm{mol\ Ag}$  involved. That accounts for all of the silver in the original mineral sample, I believe. Therefore, there are $2.22370~\mathrm{g\ Ag}$, so the percent silver by mass is $2.22370~\mathrm{g\ Ag} \div 7.563 ~\mathrm{g} = 0.294$. Put into percent, that's $29.4\%~\mathrm{Ag}$ by mass.

I'm unsure about this because  I totally ignored the nitric acid.

Noah--

You're good to go. I was taught to use unit analysis to help me make sure I was working through these problems correctly:

82.46 mL KSCN solution * (0.25 M KSCN/1 L) * (1 M SCN- / 1 M KSCN) * (1 L / 1000 mL) = 0.020615 M SCN- (mL cancels, M KSCN cancels, L cancels)

0.020615 M SCN- * (1 M AgSCN / 1 M SCN-) * (1 M Ag+ / 1 M AgSCN) * (107.8682 g Ag / 1 M Ag) = 2.2237 g Ag ( M SCN- cancels, M AgSCN cancels, M Ag+ cancels)

2.2237 g Ag / 7.563 g sample = 0.294 g Ag per g sample * 100% = 29.4% Ag by mass

I found that it helps if you read your equations out loud you can work through the logic more easily. If you set the equation up as fractions, you can see the units that cancel each other. For the most past, if you can get the right units for your answer, you're doing it right.

And you get to ignore the nitric acid (as well as the potassium) because they don't factor into your analysis.


Why do electrons need to be paired? I’d read that it takes energy to place electrons in pairs, just less than it would take to place two electrons in different subshells. I get that. What I don’t get is, then, why would electrons want to fill their orbitals with electron pairs at all? Wouldn’t it better to just fill them up with single electrons, as per Hund’s rule, and then leave them like that? Would that not be more energetically favorable? Why do atoms insist of having full valence shells, then?

Wouldn’t it better to just fill them up with single electrons, as per
  Hund’s rule, and then leave them like that? Would that not be more
  energetically favorable?

You have to consider two different scenarios here.

Case 1. Orbitals of different energies.

Consider a two electron system, say, a $\ce{He}$ atom, and let us concentrate on $\mathrm{1s}$ and $\mathrm{2s}$ orbitals exclusively. Naively, one could think of the total energy of the system being simply the sum of orbital energies. Then for $\mathrm{1s^2}$ configuration we get $E_{\mathrm{1s^2}} = \varepsilon_{\mathrm{1s}} + \varepsilon_{\mathrm{1s}}$, while for $\mathrm{1s^1 2s^1}$ configuration we get $E_{\mathrm{1s^1 2s^1}} = \varepsilon_{\mathrm{1s}} + \varepsilon_{\mathrm{2s}}$. Now clearly, since $\varepsilon_{\mathrm{1s}} < \varepsilon_{\mathrm{2s}}$, we could expect that $E_{\mathrm{1s^2}} < E_{\mathrm{1s^1 2s^1}}$, i.e. that  $\mathrm{1s^2}$ configuration is indeed more energetically favourable than $\mathrm{1s^1 2s^1}$ one.

This is a bit naive way of looking at things since the total electronic energy is not equal to the sum of orbital energies. As you mentioned, pairing the electrons up takes some energy, so that $E_{\mathrm{1s^2}}$ is clearly not just the sum of $\varepsilon_{\mathrm{1s}}$ and $\varepsilon_{\mathrm{1s}}$, but this amount of energy is relatively small compared with the energy gain you get by having both electrons in the lower $\mathrm{1s}$ orbital. In other words, energy gain from placing both electrons in energetically more favourable $\mathrm{1s}$ orbital overcompensates for having to pair them up during the placement.

How do we know that? Well, we know it from the quantum theory, or, to be more precise, from its application to a chemical systems. Thus, a solid understanding of quantum mechanics & quantum chemistry is required for a quantitive discussion, but for now you can be certain that electrons fill the lowest available energy levels before filling higher levels. This is the principle known as the Aufbau principle which follows from quantum theory and generally holds.

Case 2. Orbitals of same energy.

Consider a $\ce{C}$ atom now and let us have a look at two electrons in  $\mathrm{2p}$ orbitals. You could populate the same, say, $\mathrm{2p_x}$ orbital by a pair of electrons, or you could let the electrons populate two different orbitals, say, $\mathrm{2p_x}$ and $\mathrm{2p_y}$. Naively, we could expect the total electronic energy for both configurations to be equal to each other, since $\varepsilon_{\mathrm{2p_x}} = \varepsilon_{\mathrm{2p_y}}$, so that,

$$
  E_{\mathrm{2p_x^2}}
  = \varepsilon_{\mathrm{2p_x}} + \varepsilon_{\mathrm{2p_x}}
  = \varepsilon_{\mathrm{2p_x}} + \varepsilon_{\mathrm{2p_y}}
  = E_{\mathrm{2p_x^1 2p_y^1}} \, ,
$$

but this is again "not quite right", since pairing the electrons up takes some energy. And this time there is no any gain to compensate for this effect since $\varepsilon_{\mathrm{2p_x}} = \varepsilon_{\mathrm{2p_y}}$, and as a result $\mathrm{2p_x^1 2p_y^1}$ configuration is energetically more favourable than $\mathrm{2p_x^2}$ one.


How many times can we recycle one sheet of paper? Its really interesting to see the process of recycling of paper (as shown in picture below) but at the same time I was thinking like for how many times can one sheet of paper can be recycled ?

Well apart from the fact that the answer depends on the type of the paper that is used. Secondly how much is wasted and also the process used. Below is an article that might be helpful in answering the question

Some industry sources estimate that an ordinary sheet of paper made from cellulose fibers derived from wood can survive only four to six trips through the recycling process. The Environmental Protection Agency puts the figure at five to seven times.
It is not surprising that the rigors of remanufacturing take a toll on the fibers.
Ideally, paper for recycling is separated into types, because paper with long fibers, like white office paper, offers the most flexibility for recycling, while newsprint, with its shorter fibers, is usually reserved for making more newsprint and other low-quality papers.
The paper is shredded and chopped, then subjected to a mixture of chemicals and water and heated as it is repulped. It is centrifuged and screened to remove impurities; de-inked with more chemicals; then sprayed onto a wire screen, drained, dried and squeezed through heated rollers.
With each step, the fibers become shorter, coarser and stiffer, so that eventually, recycled fiber needs to be mixed with virgin fiber to make paper of the desired quality.
According to the American Forest and Paper Association, 63.4 percent of the paper consumed in the United States was recovered for recycling in 2009. C. CLAIBORNE RAY

SOURCES: New York Times - http://www.nytimes.com/2010/12/21/science/21qna.html?_r=0


Less maintenance-dependent continuous pH testing I am creating an automated chemical component mixing system (DIY project). One of the features required is continuous monitoring of pH value in my main solution. I do not need it to be very precise, precision of 0.2 would be enough. Also, I want it to be as low maintenance as possible (cleaning the probe once a month would be great). pH of the solution will vary from $pH = 4$ to $pH = 7$.

So, the question is, which of this testing ways will require less maintenance of pH probe:

Constant submersion in the solution.
Once per hour dipping in the solution.

The most common approach is to measure periodically.

pH glass electrodes (the most common ones) need to be "wet" in order to be used, so if you are going to remove it from the solution every hour you'll need to keep it submerged in the appropriate solution. Not only that, you need to wash it (preferably with the most pure water you can get) between measurements. Read section 2.2 of this monograph to understand why it needs to be "wet" and what solution to store it in. There's information on a lot of types of electrodes. This explains how to wash it. Another problem you might face with continuous measuring is not the electrode, but the meter. From that monograph:

  The condition of the pH measuring assembly is subject to fluctuations over the course of time. Depending on the accuracy demands, a pH calibration should therefore be performed periodically, e.g. once a day.

You'd have to test and see what accuracy you get with your equipment.


How to experimentally measure number of protons or electrons in an element? How can one measure the numbers of protons or electrons (the atomic number) in a substance?

Like if the element I examine is carbon I expect the result to be 6.

Preferable a simple experiment which can be done in the kitchen with kids attending (cheap and safe). Any atoms will do. A mass spectrometry is not available.

Moseley measured the atomic number in 1913 with relatively modest equipment (by modern standards) but I doubt that you could reproduce his experiments in a typical kitchen.

Atomic number at Wikipedia

Expansion as suggested in comments.

Here is a quote from that article of what he did.

"Moseley measured the wavelengths of the innermost photon transitions (K and L lines) produced by the elements from aluminum (Z = 13) to gold (Z = 79) used as a series of movable anodic targets inside an x-ray tube."

It might be just possible at home if you really knew what you were doing but I doubt that it would be cheap or safe.  What is regarded as safe has probably changed a lot in the last 100 years.


Spacing between adjacent planes of a given set in a simple cubic crystal Today I learned about Miller indices in a cubic crystal, and I learned that adjacent planes $(hkl)$ in a simple cubic crystal are spaced a distance $d_{hkl}$ from each other, with $d_{hkl}$ given by:

$$d_{hkl} = \frac{a}{\sqrt{h^2 + k^2 + l^2}}$$

For example, take the origin any of the below vertices of the cube with z axis directed upwards, the $(0 0 2)$ plane cuts horizontally the middle of the cube, and for this set of planes, $d_{hkl} = \frac{a}{2}$.

May be a silly question, but if we followed the rule of spacing, we would get that the next $(0 0 2)$ plane is the very upper horizontal plane in the cube, which - i guess - is not true since the latter plane has $(0 0 1)$ and not in any case $(0 0 2)$ miller indices. However, if we considered that $(0 0 2)$ planes are only those passing horizontally through the middle of unit cells, then the distance between them will equal $a$ and not $\frac{a}{2}$ (using imagination and not rules).

I must be missing something very trivial here, but it is only today that i started learning these stuff and i am really confused. Thanks for help.

When Miller indices are different only in a common factor, the planes are in the same orientation. The higher order reflections share some planes with the lower order reflections. In the case mentioned, every second plane of the (2 0 0) planes coincides with the (1 0 0) planes, whose spacing is double that of the former.


How much salt (or any other substance one can find in a kitchen) do I need to add to make water boil at 104 °C? I've seen some formulas around in other questions and Google searches, but my chemistry is pretty much dead so I have no clue where to find the relevant values to calculate it myself. I just need to make an oil bath at 104 °C, but I would prefer to use a water solution instead since it is cheaper and easier to clean up. Doesn't need to be salt, can be anything commonly available at your average household.

According to [1, pp. 281–282], solution of sodium chloride $\ce{NaCl}$ prepared by dissolving 25 g of salt in 100 g of water has boiling point of 104.6 °C.
Additional data is available in the following table for the aqueous solutions of common salts and bases.
English transcription; column 1: Compound; columns 2–6: Concentration, g/100 g water — boiling points $(t_\mathrm{b.p.},~\pu{°C})$.

Note, however, that there is a significant drawback of using boiling salt solution due to the shift of boiling point upwards as the water boils off.
You either have to add water to the mark from time to time, or use a bath with another heat carrier and a thermal sensor.
References

V.A. Rabinovich and Z.Y. Havin. (Eds.) "Kratkii khimicheskii spravochnik" (Brief chemical handbook), 2nd ed., Khimiya, Leningrad, 1978. (in Russian)


Does tolerance to alcohol affect blood/breath alcohol levels? When blood / breath alcohol levels are taken to determine how intoxicated a person is, does it depend on their tolerance ?

i.e. If you had two people of identical age, weight, fitness, etc.

person A, a non drinker, drank 1 measure of alcohol over a period and felt quite "drunk"

while person B, a frequent drinker drank the same measure of alcohol over the same period, but because of their built up tolerance felt and acted considerably more sober.

would they give the same blood, or breath alcohol readings, despite the effects on the person being different ?

Sorry if this is a very poorly formatted question from a chemistry point of view, i know absolutely nothing, nor have interacted with any of the sciences since school.

I was also not sure whether this comes under chemistry or biology, if it needs changing to the biology stack exchange site please advise

Typical roadside 'breath-testing' measures metabolites produced in the metabolism of alcohol (primarily, they measure ethanal, oxidised ethanol). The level measured level of intoxication therefore definitely differs from person to person, and even day-to-day for everyone. Some people have higher metabolic rates, and will metabolise the alochol faster. Some people have greater amounts of alcohol dehydrogenase (enzyme that breaks down alcohol) than others. It follows, that a larger person will likely be able to consume more alcohol compared to a smaller person (compare the concentration of 1 g of salt in a small glass, compared to a large glass)

However, there isn't really any fudge room for arguing against the tests. And most police-forces will back up breath testing with far more rigorous blood tests.


Is it possible to use glycerol DNA/RNA drying during extraction? Is it possible to use glycerol with ethanol in order to avoid over-drying of DNA/RNA at drying step of extraction protocol?

if glycerol residues stay in DNA, it might interact with later PCR. Overdrying is an underestimated issue and often DNA resuspension is not only long (>3h) but also very partial even after one night resuspension. it is not possible to evaporate ethanol first but reaching the point where sample are still wet but ethanol mostly removed is the best option


Is the new equilibrium total pressure of this equilibrium reaction always more than its initial total pressure? In the case of the reaction $$\ce{2NO2 <=> N2O4}$$ at equilibrium and considering Le Chatelier's principle, if we add more $\ce{NO2}$, the pressure of $\ce{NO2}$ will increase acutely and then begin to decrease until the system reaches its new equilibrium. Why is the new equilibrium have a higher pressure of $\ce{NO2}$ than the original equilibrium? How can we tell this just by looking at it? Will it always be larger?

The higher pressure of $\ce{NO2}$ even after reaching equilibrium after adding some amount of $\ce{NO2}$ can be shown mathematically.

Suppose initially before adding $\ce{NO2}$,the pressure of $\ce{NO2}$ was $p^0_{\ce{NO2}}$, and that of $\ce{N2O4}$ is $p^0_{\ce{N2O4}}$.So, the equilibrium constant is $$K_\mathrm{p} = \frac{p^0_{\ce{N2O4}}}{(p^0_{\ce{NO2}})^2}$$ Now assume that the addition of $\ce{NO2}$ just increases the pressure instantaneously to $p_x$ but immediately after that due to dissociation to achieve equilibrium, the pressure decreases by $\ce{p_y}$ atm.

$$\ce{2NO2 <=> N2O4}$$
$$p^0_{\ce{NO2}} + p_x - p_y \ce{<=>} p^0_{\ce{N2O4}} + \frac{p_y}{2}$$

So, the increase of pressure $\ce{N2O4}$ will be $p_y/2$ atm. So, now a new equilibrium is achieved where the equilibrium constant $K_\mathrm{p}$ will be still the same assuming that during addition of the gases, temperature is not changed( or the change is very much negligible in real cases). Thus, equating $K_\mathrm{p}$ on both the sides we have,

$$\frac{p^0_{\ce{N2O4}}}{(p^0_{\ce{NO2}})^2}= \frac{p^0_{\ce{N2O4}} + \frac{p_y}{2}}{(p^0_{\ce{\ce{NO2}}} + p_x - p_y)^2}$$

Taking the squares on numerator and denominator and taking square root we will find:

$$1 + \frac{p_x -p_y}{p^0_{\ce{NO2}}}= \sqrt{1+\frac{p_y/2}{p^0_{\ce{N2O4}}}} > 1$$

So, we see that $1 + \frac{p_x -p_y}{p^0_{\ce{NO2}}} > 1$. Hence it is proved that $\ce{p_x - p_y > 0}$. So, the net pressure of $\ce{NO2}$ after achieving equilibrium is $p^0_\ce{NO2} + p_x - p_y > p^0_{\ce{NO2}}$. So, its pressure has increased than the previous conditions even after reaching equilibrium.


How does UV affect skin colors in dark-skinned people? Skin color is one of the things one would rather not ask anything about! Only in humans, it can vary from very dark brown to pale pink. In darker-skinned people, the color is mainly due to melanin, which is produced by melanocytes.
There are three types of melanin: eumelanin, pheomelanin, and neuromelanin. Eumelanin is the most common; so it must be the reason of the dark skin of the dark-skinned. And that's true, since the two major groups of eumelanin are black and brown species.
The first ray produced by sun that comes to mind when you think about affecting the attributes of skin is UVA, since it penetrates most into the skin (Even though its energy is lesser than UVB). So, when I attach these two together I reach the fact that if UV rays do not cause free radicals they will eventually change the skin's color. (This is what is mostly believed and its effect on natural selection is impossible to hide)
Now, I give up. How does UV do to the skin that makes it darker? This must be very easy, huh? What am I missing?

There is a journal article almost exactly about the title question, Mechanisms of Skin Tanning in Different Racial/Ethnic Groups in Response to Ultraviolet Radiation Journal of Investigative Dermatology (2005) 124, 1326–1332.

However, the body of the question assumes that eumelanin changes color.  Instead, the article finds that melanin gets closer to the surface of the skin, and the degree to which this effect occurs varies with race/ethnicity.


Hair styling gels and 'liquid gels' - effect of formulation / composition on viscosity and performance [Disclaimer: I am aware that this is a very 'first-world problem', 'light-hearted' kind of question, given the health crisis we are still going through. Just to avoid offending anyone, here's an easy solution: no need to continue to read if you disapprove :) ]
I have 3 brands of 'liquid gel', each with rather different characteristics.
Here are their compositions:
Brand 'A': Alcohol Denat., Aqua, Octylacrylamide/Acrylates/Butylaminoethyl Methacrylate Copolymer, Panthenol, Parfum, PEG-12 Dimethicone, Isoamyl p-Methoxycinnamate, Benzoic Acid, Linalool, Aminomethyl Propanol, Citric Acid.
Brand 'B': Alcohol Denat., Aqua/Water, Acrylates/t-Butylacrylamide Copolymer, Niacinamide, Hydroxyisohexyl 3-Cyclohexene Carboxyaldehyde, PEG/PPG-4/12 Dimethicone, Aminomethyl Propanol, Limonene, Panthenol, Benzyl Salicylate, Butylphenyl Methylpropional, Hexyl Cinnamal, Parfum, Fragrance.
Brand 'C': Aqua, Alcohol Denat., VP/VA Copolymer, Polysorbate 20, PEG-12 Dimethicone, Panthenol, Parfum, Linalool, Geraniol, Hexyl Cinnamal, Citronellol, Alpha-Isomethyl Ionone, Hydrogen Peroxide, Phenoxyethanol, Ethylhexylglycerin, Piroctone Olamine, Citric Acid.
My favourite brand is 'C'.
The shops where I used to buy it are no longer stocking it, they replaced it with the same stuff (same commercial name and 'strength'), but in 'gel' form, which I really don't like.
The composition of the gel is similar to the one of the liquid gel, up to a point:
Brand 'C-gel': Aqua, Alcohol Denat., VP/VA Copolymer, PVP, Propylene Glycol, Carbomer, Polysorbate 20, (PEG-12 Dimethicone - removed), Panthenol, Parfum, Linalool, Geraniol, Hexyl Cinnamal, Citronellol, Alpha-Isomethyl Ionone, Hydrogen Peroxide, Phenoxyethanol, Ethylhexylglycerin, Piroctone Olamine, Tetrasodium Iminodisuccinate, Sodium Hydroxide, Citric Acid.
Question 1: does anybody know what specific components, if any, are responsible for 'C' being liquid and 'C-gel' being indeed a rather thick gel ?
Question 2: could one turn the gel into a liquid that can be sprayed, and if so, how ?
This is no academic question: I would actually like to do that. I tried diluting the gel with water, but those formulation chemists did a very good job: it stayed quite gel-like and viscous, until I really diluted it a lot, like 1:3 or more, and even then it was still a viscous liquid that the spray bottle could not spray well at all.
Could a change of pH perhaps loosen it ? By which I mean a small change; not planning to turn this into battery acid or lye :)
Question 3
When using the liquid gels (applied on wet hair and dried either naturally or with a hairdryer), I noticed very different performances.
'A' results in a very unpleasant, 'doll-like' feel and look, and a lot of static electricity, when you comb your hair. And it doesn't smell nice.
'B' is marginally better, it sort of works in terms of disentangling your hair, but it tends to get a bit sticky, and they put something in it that smells absolutely repulsive (at least to me). I hope it's not the 'fragrance'.
'C', as I mentioned, is my favourite (despite being the cheapest): it smells good, it makes your hair very smooth and easy to comb, it does not stick, and even when dry your hair stays soft and tidy (unlike with B and especially A).
So: can you identify any components, or any formulation features, in A, B and C, that cause the performance I am describing ?
E.g. what might be responsible for 'A' turning your hair into individual doll-like plastic-y, electrostatic strands?
Or, do you know if any of the components of 'B' has a particularly offensive smell ?
Or, is there any specific component in 'C' that 'lubricates' hair and prevents it from sticking?
Thanks!

Question 1: Carbomer
Question 2: Not sure about the question, but anyway. Sure you can create a spray (theoretically), but the process is really hard (please, do not make it at your house) because you need a propellant. By the way, such a high concentration of Parfum.
Question 3:
Part1:
According to your observations, in my own opinion, the concentration of these components maybe are too high: Octylacrylamide/Acrylates/Butylaminoethyl Methacrylate Copolymer, Niacinamide, Hydroxyisohexyl 3-Cyclohexene Carboxaldehyde.
Part2:
Maybe because of the oxidation of hexyl cinnamaldehyde to Cinnamic acid you can feel a very unpleasant smell. But it is hard to tell because everything you describe is not in a detail (for example, the scent, what reminds and etc..)
Part3:
VP/VA Copolymer, PVP makes your hair shiny and not sticky.
Panthenol makes your hair not dry.
JUST A REMAINDER:
It does not mean, that those components which I have mentioned are very good or very toxic. It all depends on concentration and experimental data. We use poisons' as drugs, but lethal and effective concentrations differ.
I answered what could possibly be in my own subjective, but a trained opinion based on your poor objective/experimental data. Try to say more about what you smell, what you feel, and there many databases of components like (found some reliable databases for you):
Official Cosmetics ingredients review database: https://www.cir-safety.org/ingredients
Cosmetics and their function (easy to use database): https://www.cir-safety.org/ingredients https://www.ewg.org/skindeep/
Chemical compounds database: https://pubchem.ncbi.nlm.nih.gov/


Is nucleus at the centre of the atom? If so why? In postulates of Bohr's model of the atom, it is said that the nucleus is at the centre of the atom. In Rutherford’s α scattering experiment he said that out of 20,000 alpha particles only 1 particle is deviated by $180^\circ$. If the particles in the atom are moving continuously (like an electron), why is that nucleus at the centre of the atom? If that is so, then when we focus all the α particles at the nucleus (the point where an α particle got deviated by $180^\circ$) all should get deviated by $180^\circ$ right?

What I mean is that why is the nucleus at the centre of the atom?

Why is the nucleus the centre of the atom?

Definition. Whatever the electron does, chemists define the position of the nucleus to be the centre of the atom meaning that electrons can be polarised away.
Quantum mechanics. When studying an isolated atom, the only thing that influences electron behaviour is the nucleus. You can calculate the ‘path’ of an atom‡ relative to the position of the nucleus with quantum mechanics and your result will be that quite a few electrons will occupy a perfect sphere whose centre is the nucleus.

The second point may require elaboration. But it really is quite simple. You have a very lightweight, very fast particle with a charge of $-1$ (your electron) and you have a very heavy, very slow particle with a charge of $+1$ (your proton). By the time the proton has moved an inch, the electron has moved a mile so we can approximate that only the electron really moves. (This has the fancy name Born-Oppenheimer approximation.) Now remember that both are attracted to each other. The proton doesn’t move, but the electron feels the ‘desire’ to ‘move towards’ the proton. (Inverted commas because this is a macroscopic view that only gets us so far when dealing with quantum effects.) If you average this over time, you will find that the electron will have spent the same amount of time on one side of the nucleus as on the other and the mean electron position will be in the nucleus. Spherically outwards from the proton, the chance of locating the atom decreases monotonously with distance; the angle does not matter. We have a perfect sphere with the proton being the centre.


How to compare energy levels in hybridized sp orbitals? How to compare energy levels in $\ce{sp, sp^2, sp^3}$ orbitals?

Since a higher energy level implies lower stability, an $\ce{sp-sp}$ bond must have the lowest energy level, since it is formed by the overlap of one sigma and two pi bonds in total, more than those in $\ce{sp^2-sp^2}$ or $\ce{sp^3-sp^3}$. Hence, $\text{energy level(}\ce{sp < sp^2 < sp^3}\text{)}$.

Is my calculated trend correct? If so, then is this the intended reasoning? If not, then why?

How to compare energy levels in $\ce{sp, sp^2, sp^3}$ orbitals?

Background

Electrons in an atomic $\ce{2s}$ orbital will be lower in energy than electrons in an atomic $\ce{2p}$ orbital since the s orbital is closer to the nucleus.
When we describe a hybrid orbital as $\ce{sp^n}$, this is really a shorthand notation for $\ce{s^1p^n}$, which means that the orbital is made up from 1 part s orbital and n parts p orbital.

Analysis

As the "n" in $\ce{sp^n}$ grows larger we will have more p character and less s character in the hybrid orbital and, as explained above, it will consequently be higher in energy. So an $\ce{sp}$ hybrid orbital that has 50% s character will be lower in energy than an $\ce{sp^3}$ orbital which only contains 25% s character.

The energy ordering you proposed in your question is correct: $\ce{sp < sp^2 < sp^3}$.


Interesting ionization of sulfuric acid I just need an explanation of why it is a strong acid in its first ionization and a weak acid its second ionization.

It becomes $\ce{HSO_4-}$. From what I have learned about oxoacids, the bond strength determines the strength of the acid. So why does it become harder for the molecule to ionize the second time after it loses a hydrogen atom. Just can't think of how it will affect the strength of the bond connecting oxygen and hydrogen.

From what I have learned about oxoacids, the bond strength determines the strength of the acid.

Yes. But that is not the whole picture. You also have to consider what else is floating around in the solution.

The reaction of the species (I assume aqueous solution) is given below:
$$ \ce{H2SO4 <=> HSO4- + H+ <=> SO4^{2-} + 2H+} $$

As you can clearly see, the concentration of the protons is raised when going from left to right. There is a natural equilibrium that pushes backwards.

However, as you correctly point out, the bond stability (or "strength") directly correlates to the stability of the acid species. It might well be (and I'd even bet on it) that the bond strength of $\ce{H-SO4-}$ is higher than $\ce{H-HSO4}$ (note that the bonds indicated here are not the real bonds as they occur but merely a representation) but this requires an in-depth (possibly computational) analysis, as a quick search of the internet did not yield any useful results.

Edit: I did some small, very basic geometry optimization calculations with turbomole (on default settings, so def-SV(P) basis set for all atoms) to find out the bond length of the hydrogen to its oxygen. Turns out my hunch was right, the bond is shorter in $\ce{HSO4-}$ (95.07 pm) than in $\ce{H2SO4}$ (95.72 pm), indicating stronger bonding once the first proton went flying away.


What is this white layer on brass? I have an object made of what I assume to be brass, i.e. an alloy of copper and zinc.
The curved parts have both a black layer and a white layer (also note the verdigris (I guess) on the left top, but this is the only spot on the whole piece) :

If I understand correctly, the black layer is tarnish, or patina, and due to oxidation from sulfur dioxide in the air.
But I can't find reliable reports of white corrosion on brass.  Could it be wet storage stain, a.k.a. "white rust" ?
Hints on how to remove it are welcome.

Judging from the location and colour, it is most likely residue from liquid polish that was not properly rubbed/polished off. Quite a few types of liquid polish for brass turn white when they dry, and any folds and creases in the design is where they would be most likely to accumulate if the person doing the polishing is not very thorough.
Depending on how long its been there, you may be able to shift it with just warm water and a soft brush or cloth, leaving the patina mostly in place.  Otherwise you may need to use brass polish and polish the whole thing.
You could try asking advice from some of the better antique dealers or ideally someone who specialises in conserving antique brasses if the piece is a valuable antique.


How to calculate bond angle of simple molecules quantitatively? Hi I'm new to this forum so spare me if I've made mistakes.

Our high school chemistry teacher explained that bond angles of different molecules varies slightly, e.g. 107 degrees in ammonia and 104.5 degrees in water, which could be predicted by VSEPR theory qualitatively.

I have checked literature either through google or stack-exchange, but none point me to any formula/algorithm/computationally inexpensive methods to  find non-ideal bond angles by calculation.

This has a simple reason: To the best of my knowledge there is no quantitative way to predict bond angles without resorting to computational chemistry.

This is because the entire subject of predicting bond angles is a lot more tricky than it was displayed to you. Take phosphane and hydrogen sulphide, the heavy analogues to ammonia and water. Their bond angles are nowhere near $109^\circ$ but much closer to $90^\circ$.

If you dig deeper into the matter, you find that ammonia and water are more the exception than the rule, that the VSEPR model fails spectacularly on the long run, and that their geometries are actually finely balanced between an ideal $90^\circ$ angle and the hydrogen-hydrogen repulsion.

In fact, not even with computers can you calculate the bond angle directly with a certain set of axiomic assumptions. Rather, computational chemistry would empirically calculate energies for different angles and deem the angle that has the lowest energy to be ‘correct’ and ‘calculated’.

So no simple quantitative rule exists nor will it exist.

Note: There are, of course, semi-quantitative rules like the one mentioned by Ringo in a comment. But I don’t consider them quantitative, because they still contain approximations.


Mechanism of Hofmann Rearrangement Reaction In the Hofmann Rearrangement Reaction (in basic medium), when $\ce{RNH-}$ ion is produced, why does it not attack $\ce{Br2}$  or take $\ce{H+}$from the amide itself?
Also, why does it take $\ce{H+}$ from water only?

Water is present in large excess when compared to the amide, so it is statistically not likely that the proton would be removed from the amide rather then water. Even if it happens, proton exchange is reversible so it would be formed/unformed many times that eventually the removal of the proton by water would happen and proceed with the mechanism.

Regarding the attack to $\ce{Br2}$, I believe the reason it doesn't happen is that the ion would be surrounded by many many water molecules because: 1) the amide forms H-bonds with water, 2) the ion is solvated efficiently by water, so more molecules would be around to have their protons removed.


Why is a TTT diagram called isothermal transformation diagram? Why are TTT diagrams called isothermal temperature diagram?As far as I know the temperature variation is plotted with the variation of time in a TTT diagram. Then why do we call it "isothermal".What is so constant about the temperature?

They are two names for the same procedure.  A steel is heated to a temperature in the austenite range then quenched/quickly cooled to some specific temperature. Held at that constant ( isothermal) temperature ( T) for a specific time (T) then quenched to room temperature for examination of the transformation (T) products.  You may be thinking of CCT ; continuous cooling diagrams ; where a steel sample is austenitized  then continuously cooled at some specific rate.  Then examined for transformation products. The results can have similarities depending on alloy content , etc.


What are some materials that expand a lot proportionally to heat? There's never any source on the internet that can simply give me a list of materials that have the kinds of properties I'm looking for. I'm wondering what materials, like for a solid block of something, in general expand either a lot or at least very quickly when exposed to heat. I also can't remember what property this is exactly, but I think it's the coefficient of thermal expansion.

Am I looking for something with a high coefficient or low coefficient to achieve this?

You are looking for thermal expansion and its coefficient.  Of course such tables, and diagrams exist

(source)

as it is of relevance in daily live.  Thermometers based on liquid mercury, or alcohol, depend on this.  Bridges take this into account in expansion joints, too:

(source)

Joining two materials of different expansion coefficient equally is of relevance for the design of bimetallic stripes, found not only in thermometers and mechanical clocks:

(source)

So in reply to your "there is no source available" ...


organic hard surf wax? I made some organic surf wax with 4 parts beeswax and 1 part coconut oil. The wax turned out to be sticky but had no tack.

I'd like to make wax for warm water that forms a layer of hard bumps on the board when wax is applied.

Not sure how I can achieve this w all organic items.

Any suggestions?

Carnauba wax is a high-melting wax that should be suitable. In the Wikipedia article cited,
"Carnauba wax can produce a glossy finish... It was commonly used in its purest form as a coating on speedboat hulls in the early 1960s to enhance speed and aid in handling in salt water environments. It is also the main ingredient in surfboard wax, combined with coconut oil."


In a galvanic cell why does using a more reactive metal increase the voltage? As the title says, why does using magnesium produce a higher voltage compared to using zinc when reacting with iron? Im having trouble figuring out why the voltage changes.

magnesium+copper > zinc+copper in terms of voltage, why?

Any help is appreciated.

It is a good question because textbooks do not discuss it. Scientific explanations are of two types, how and why. The "why" part of the explanation is very difficult to answer. Often we don't know the answer at a fundamental level. A simple example is that of what is universe is made? What is the origin of mass or charge? In the same way the electrode potential can also be understood at various levels. The most advanced discussion of electrode potentials can be done in terms of Fermi levels, but this is not my area of expertise. The macroscopic level discussion is that one can consider the electrode potential as a (thermodynamic) tendency of a substance to gain electrons. If you look up modern electrode reduction tables, all entries are listed as reduction process (all electrochemists agreed to do that after a long academic fight to write it this way). Hydrogen has been given an arbitrary value of 0.0000 V for the sake of comparison. You always need a reference in electrochemistry so electrochemists decided to use H2 electrode as a reference.

a) If you look at the table below, look at lithium entry: It shows that Li+ has very low (or negative electrode potential) tendency to gain electrons. Read it the other way (right to left), that Li metal has a very high tendency to become a lithium ion.
b) Look at the silver entry: It shows that Ag+ has a high tendency to gain electrons (positive electrode potential). Read it the other way (right to left) that Ag has a very low tendency to become Ag+.

If I make a galvanic cell consisting of two beakers connected by a salt bridge, having a lithium electrode in 1 M Li+ solution, and a silver electrode dipped in 1 M Ag+ ions. Connect the leads to an voltmeter.

Now Ag+ has a tendency to accept electrons and become Ag and lithium metal has the tendency to lose electrons and become Li+, what do you think happen? Electrons would feel a "pressure" or better "potential difference" causing them flow from higher potential to a lower potential, which is given by

Ecell=  Ecathode - Eanode= 0.7996 - (-3.040) = + 3.8396 V, this should be your voltmeter readings. Why did we choose Ag+ /Ag as a cathode? Because Ag+ ions have a higher tendency to reduce than lithium ions.

In the same way you can choose Li and Be cell. What would be the cell voltage?

For example, electrode potential tables look like:

Li+(aq) + e−  ⇌  Li(s)      –3.040

Be2+(aq) + 2e−  ⇌  Be(s)    –1.99

Al3+(aq) + 3e−  ⇌  Al(s)    –1.676

2H+(aq) + 2e−  ⇌  H2(g)  0.00

Ag+(aq) + e−  ⇌  Ag(s)       0.7996

Ce4+(aq)+e−⇌Ce3+(aq)        +1.44

PbO2(s) + HSO4−(aq) + 3H+(aq) + 2e− ⇌ PbSO4(s) + 2H2O(l)    +1.690

H2O2(aq) + 2H+(aq) + 2e− ⇌ 2H2O(l) +1.763

F2(g) + 2e−⇌ 2F−(aq)         +2.87


Tandem mass spectrometry: a1 and immonium ions In tandem mass spectrometry of proteins one will usually end up with a mix of $b$ and $y$ fragments but also $a$ fragments and immonium fragments. My problem is that wouldn't an $a_1$ ion be the exact same as an immonium ion of the first amino acid in the peptide?

I don't know if you are still interested in this question. But yes, the a1 ion would have exact the same m/z as the immonium ion of all amino acids of the same type in the pepitde. Peptides typically fragment in low energy fragmentation methods such as CID by charg-directed mechanisms, which mainly produces y and b series ions. These ions can fragment further to lower b ions, a ions, or internal fragments and immonium ions. y and b ions are much easier to understand and we therefore mostly use these ions for interpretation of mass spectra.


Determine structure from NMR So I've come across this lovely spectrum of a compund where only its chemical formula is given with C5H9NO.
I have made the pictures of the spectrum available below.
obviously there is some pretty hefty coupling going on but I cant really figure out what it could be.
So far I think we definitely have a CH3 group that could be attached to either heteratom. On the other side, the single proton at 6.66 ppm could also be an H attached directly to the N-atom (maybe?).
Anyways, I hope you guys can help me here and maybe also explain me the couplings that are going on.

EDIT: to not get marked as homework, because it isnt, I suggesting that it should be some sort of pyrrole or pyrrolidine structure type. Further I think that the CH3 group must be shielded from other protons and is either near or is sitting at a heteroatom.
Next I have narrowed down two structural possibilities: either we have a proton sitting at the nitrogen and the other sitting on a double bound, or we have diastereotropic protons.
My best candidate so far is the picture below (please let me know if it would be possible):

This is an interesting question. However, disadvantage is having only $\ce{^1H}$-$\mathrm{NMR}$ to deal with. Yet, it has pretty good resolution (probably using $\pu{400 MHz}$ machine) so we can resolve the splitting pattern very easily and predict the structure.

First, OP has correctly assigned two degree of unsaturation. For molecular formula, $\ce{C5H9NO}$:
$$u=\frac{5\cdot2+2+1-9}{2}=2$$

However, OP's suggested structure was incorrect since chemical shift of -$\ce{OCH3}$ is usually greater than $\pu{3 ppm}$. Because of the chemical shift of -$\ce{CH3}$ group in the given $\ce{^1H}$-$\mathrm{NMR}$ is about $\pu{2.0 ppm}$, one can conclude that it is either olefinic or one $\alpha$ to a cabonyl group. Nonetheless, OP's suggestion of proton resonance around $\pu{6.66 ppm}$ is due to $-\ce{NH}-$ group is correct, but due to its deshielding nature ($\pu{6.66 ppm}$), one can suggest that it is an amido $-\ce{NH}-$ group. However, OP correctly assign it next to a $-\ce{CH2}-$ group.

Since I suspect non cyclic nature of the unknown, I have assigned splitting patterns and calculate relevant coupling constants assuming the spectrum is taken in $\pu{400 MHz}$ machine. All splitting patterns of remaining resonances are $ddt$ as depicted in fallowing diagrams:

Accordingly, these patterns and chemical shifts well match with N-allyl acetamide (see insert in first diagram).


Why is my sugar paste becoming a brick? I know this question may sound to be not inherent with chemistry, but since making the sugar wax paste involves acids and heating, I thought this would be the right place.

Preparing the paste is quite simple:
1 cup of sugar, 2 tbs of lemon juice and 2 tbs of water. Then the process expects to heat the compound until it becomes abra colored. The the mixture is moved away to cool down.The result would be a malleable and sticky paste at ambient temperature. My problem is that my paste becomes really hard and impossible to shape.

What could I do to avoid the compound to become a brick?

You don't have enough liquid in the recipe. Double the lemon and water to 4 tablespoons each, and you should have better luck. Adding more liquid will bring the sugar/liquid ratio more in line with what is needed to make the paste pliable. Add more liquid if you still have issues.


Stoichiometry of alkane-perchlorate reaction What are the products of an alkane and a perchlorate reaction?
e.g., making the alkane methane and the perchlorate anhydrous for simplicity:

My effort:
$$\ce{Mg(ClO4)2 + 4CH4 -> MgCl2 + 4CO2}$$
leaves 16H free so obviously incorrect, but shouldn't a salt and a gas be part of the products? I guess that the anhydrous form would make this explosive.

I think that there is some $\ce{H2O}$ as product:
$$\ce{2Mg(ClO4)2 + 4CH4 -> 2MgCl2 + 4CO2 + 8H2O}%edit$$


Calorific value of a gas in combustion The Wikipedia article on heat of combustion says:

  The calorific value [...] may be expressed with the quantities:

  energy/mole of fuel
  energy/mass of fuel
  energy/volume of the fuel

Is the energy obtained by the combustion mainly proportional to the mass of the gas, being then values in $\pu{J/mol}$ and $\pu{J/kg}$ a characteristic of the gas composition and basically independent of the pressure and temperature?

That is, for a specific gas, values in $\pu{J/mol}$ and $\pu{J/kg}$ will be mainly independent of the pressure/temperature of the gas in combustion, while value in $\pu{J/m^3}$ will be strongly dependent due to the relation between volume, pressure and mass.

In the latter case, given the heating value, $f_0$ (in $\pu{J/m^3}$) for a gas at $\pu{°C}$ and pressure $p_0 = \pu{101 kPa} = \pu{1 atm}$, we could (?) find new heating value $f$ (in $\pu{J/m^3}$) at some other temperature $t$ (in $\pu{°C}$) and pressure $p$ (in $\pu{Pa}$) applying the expression:
$$f = f' \frac{p}{p_0} \frac{273}{273+t}$$

Calorific values are usually quoted under standard conditions making the distinctions a matter of preference and convenience

The apparently different units used in quoting heat of combustion are not as different as they appear. We need to know heat of combustion for many reasons (eg how to bill domestic customers for the correct amount of heating or cooking their gas supply delivers). But the units used are all equivalent once we define a standard set of conditions when measuring them.

For example, a mole of pure methane at standard temperature and pressure (STP) occupies 22.4 L and will weigh about 16g and will burn in air releasing about 890kJ of energy. This means that, if we stick to the same conditions (STP in this case) we can interconvert the energy/mol, energy/mass and energy/volume with a simple calculation.

Which one we choose to use is a matter of convenience. Domestic gas meters might choose the energy/volume as they measure volume flow at a standard pressure and temperature. Liquefied gas ships may choose to measure by mass as the methane is mostly liquefied and the mass may be easier to measure than the volume, temperature or pressure (but, knowing the mass, we can work out the volume at STP to work out how much domestic gas can be produced). If a chemist is working on some reaction scheme, then they will probably prefer molar metrics as they make balancing the equations easier.

But the essential point is that a given amount of substance always has the same heat of combustion. How you measure the amount is a matter of convenience.


Does computational chemistry get the molecular orbitals of dioxygen wrong? The molecular orbitals of $\ce{O2}$ are typically shown as follows, with every orbital filled by a spin alpha (or up) and spin beta (or down) electron (except the homo levels which are singly occupied).

Calculations using for instance Hartree-Fock and density functional theory however produce alpha and beta orbitals with different energies - no degeneracy of alpha and beta orbitals. For instance, the B3LYP/6-311G** HOMO/LUMO levels (taken from the computational chemistry database) are as follows: alpha HOMO (-0.31186), alpha LUMO (0.19755), beta HOMO (-0.46536), beta LUMO (-0.11702). Clearly the HOMO levels are not degenerate, at odds with what we expect from the above molecular orbital diagram. Similar mismatch between the alpha and beta orbitals exists for orbitals below the HOMO, as well as using other levels of theory.

Is computational chemistry getting the orbitals of $\ce{O2}$ wrong? Am I missing something? Conceptually I can't grasp that alpha and beta orbitals have different energies since at least for the case of filled orbitals when the electrons pair up I expect them to have the same energy.

A triplet species, as your MO scheme shows, has two unpaired alpha spin electrons, which is why you don't have to look at the beta orbitals at all but you have to look, if the $\alpha$-HOMO and $\alpha$-HOMO-1 are equivalent or not.

After optimizing the geometry of triplet oxygen using the method you states, B3LYP 6-311G**, we end up with a bond length of $\pu{1.2058 Å}$ and are given the following energies:

 **********************************************************************

            Population analysis using the SCF density.

 **********************************************************************

 Orbital symmetries:
 Alpha Orbitals:
       Occupied  (SGG) (SGU) (SGG) (SGU) (PIU) (PIU) (SGG) (PIG)
                 (PIG)
       Virtual   (SGU) (PIU) (PIU) (SGU) (SGG) (SGG) (PIG) (PIG)
                 (SGU) (PIU) (PIU) (DLTG) (DLTG) (DLTU) (DLTU)
                 (SGG) (PIG) (PIG) (SGU) (PIU) (PIU) (SGG) (PIG)
                 (PIG) (SGU) (SGG) (SGU)
 Beta  Orbitals:
       Occupied  (SGG) (SGU) (SGG) (SGU) (SGG) (PIU) (PIU)
       Virtual   (PIG) (PIG) (SGU) (PIU) (PIU) (SGU) (SGG) (SGG)
                 (PIG) (PIG) (SGU) (PIU) (PIU) (DLTG) (DLTG) (SGG)
                 (DLTU) (DLTU) (PIG) (PIG) (SGU) (PIU) (PIU) (SGG)
                 (PIG) (PIG) (SGU) (SGG) (SGU)
 The electronic state is 3-SGG.
 Alpha  occ. eigenvalues --  -19.28009 -19.27992  -1.31462  -0.83929  -0.56862
 Alpha  occ. eigenvalues --   -0.56862  -0.55311  -0.31192  -0.31192   <------
 Alpha virt. eigenvalues --    0.19733   0.66242   0.66242   0.69937   0.71138
 Alpha virt. eigenvalues --    0.77370   0.78062   0.78062   1.23343   2.23120
 Alpha virt. eigenvalues --    2.23120   2.53472   2.53472   2.80188   2.80188
 Alpha virt. eigenvalues --    2.83167   3.40715   3.40715   3.75295   4.50288
 Alpha virt. eigenvalues --    4.50288   4.64528   4.80485   4.80485   5.97229
 Alpha virt. eigenvalues --   49.48209  49.67839
  Beta  occ. eigenvalues --  -19.24954 -19.24923  -1.25730  -0.75086  -0.51327
  Beta  occ. eigenvalues --   -0.46530  -0.46530
  Beta virt. eigenvalues --   -0.11708  -0.11708   0.24099   0.70896   0.70896
  Beta virt. eigenvalues --    0.71935   0.72389   0.79954   0.83108   0.83108
  Beta virt. eigenvalues --    1.27043   2.27711   2.27711   2.62147   2.62147
  Beta virt. eigenvalues --    2.87800   2.89802   2.89802   3.46348   3.46348
  Beta virt. eigenvalues --    3.78751   4.58054   4.58054   4.66543   4.88322
  Beta virt. eigenvalues --    4.88322   6.00448  49.51237  49.70844

You can clearly see, that the $\alpha$-HOMO and $\alpha$-HOMO-1 are degenerate with both an energy of $\pu{-0.31192 Eh}$. If you look at both MOs you can easily see why, because they are qualitatively the same combination of 2p AOs but with different orientation.

As symmetry was turned on, I guess, this needs to be the case, but the result stays the same, if it is turned off.

So, does computational chemistry get the molecular orbitals of dioxygen wrong? No, the MOs are right.

But actually you have a different question: Why are the energies of alpha and beta MOs in unrestricted open shell calculations not identical?

There are two methods to treat open shell systems. One is restricted open (RO*) and the other is unrestricted (U*). To not expand it too much, the main difference is, that as restricted closed shell methods, the restricted open shell methods force the alpha and beta orbitals to share the same MOs. Then, of course, they also would have the same energies.

But the energies that you show are from an unsrestricted open shell calculation. If you use this method, alpha and beta MOs are not forced to be in the same spatial MOs. And this freedom leads to different energies alpha and beta MOs.


What is the purpose of CaO in Oakwood reaction In the Oakwood reaction, i.e.
$$\ce{RCOOH ->[NaOH + CaO] RH + Na2CO3},$$
is there any role or importance of $\ce{CaO}$, because it doesn't appear to be anywhere in the product.

Sodium hydroxide plus calcium oxide forms soda lime. Its use is mainly down to handling covenience as it is far less hygroscopic than powdered sodium hydroxide.


Variation of compressiblity factor with temperature While I certainly understand the order of temperatures, I can't find a reason for the curves to intersect at one common point.

Why do the curves intersect at one point? or do they really intersect at a common point or is it a coincidence that they are intersecting in this diagram?

Remember the mathematical definition of $Z$:

$$ Z = \frac{pV}{nRT}$$

Given all the $Z$ graphs I've ever seen -- the single-point interaction of all these isotherms is just coincidence (and kind of misleading). There is nothing in thermodynamics that suggests for a given gas, there exists a $p$ for which $Z$ is constant, regardless of temperature. Even for constant $V$ and $n$, you have that $Z \propto \frac{p}{T}$, so changing $T$ can in no way leave $Z$ constant.

You can find graphs that look similar even for different gases at the same $T$:

Nonetheless, there is no reason for a given material to have the same $Z$ at a given $p$ regardless of temperature:

Generally speaking, if you zoom-out far enough, and have significantly different temperatures (like 100s of Kelvin apart), some $Z$ graphs may appear to have a uniform $(p,Z)$ point, but in reality the intersections are not uniform for any physical/mathematical reason. For this reason, graphs like that can be a little misleading.


How to save the force on a molecule in GROMACS? I am a new student of molecular dynamics. I have started using GROMACS a month ago.

I have a small box of 8192 molecules each with 3 atoms. I froze the last molecule and ran the simulation. I even converted the .trr file from binary to .gro and checked that the last molecule indeed stayed in the same place.

However, I want the .trr file to print the force on each molecule or each atom - in the x, y, and z-direction.
I even set the nstfout in the grompp.mdp file to a reasonable non zero number - 5000.

However, the .trr file - when converted to .gro - did not show the force. Can someone tell me what I am doing wrong and how to get GROMACS to save/reveal the force in the .trr file?

Gro output format does not support forces, so using it as an intermediate output format defeats your purpose.

Use

gmx traj -f your.trr -of

which runs the gmx wrapper binary, calling the traj module to do trajectory processing, names your input trajectory file after -f and triggers the output of forces with -of (optionally named after -of, defaulting to force.xvg), or similarly

gmx dump -f your.trr

to dump the entire contents of the file including forces) to the terminal.

See also

gmx help traj

or

gmx traj -h

or http://manual.gromacs.org/documentation/5.1/onlinehelp/gmx-traj.html

Note that I currently develop GROMACS.


How can methane absorb infrared light within wide range of frequency but CO2 can't? Methane is one of the most potent greenhouse gases in Earth's atmosphere.I saw a lot of answers related to it but lacks de facto exegesis to it.

Comparing the vibrational spectra of two different molecules is just like wondering why two people do not have the same fingerprints. The simplest answer is that these two molecules are different molecules with different atoms hence their vibration modes are supposed to be different.

IR Spectra of Methane and Carbon dioxide are available in NIST. This is how real spectroscopy looked like before the focus shifted from true science to fancier presentations and instruments became black-boxes. Anyway, all you have to note both methane and carbon dioxide absorb quite strongly in the typical range of infrared light 400 to 3800 $cm^{-1}$. The question of wide range does not arise as you can see from the spectra, their absorption band positions are different. Methane's spectra is more complex just because it has more active vibrations (dipole moment change is non-zero) and one can see the fine rotational structure of methane but carbon's dioxide's rotational lines are not resolved.


Instantly separate oil and water I have a solution of water and oil mixed to as much as possible, since they won't actually mix. The supply is coming down a pipe at a relatively fast rate and I need to separate it so that the water goes down a pipe and the oil goes through another pipe.
How would this be done, without waiting days for it to occur?

Oil-water separation is a big money industrial process, and there are a number of methods for successfully separating the two. So the answer to your question is best approached in two parts: a theoretical answer, and a practical answer.

Theoretically, yes. A number of systems exist and the best are described below. There are off-the-shelf applications, as well as major capital infrastructure for major water treatment plants. Whether these solutions are for you will depend on your application, as well as budget. Hence, the practical answer may be different if you are looking for a cheap, backyard-employable device to solve the problem created by Granny filling the radiator through the sump oil inlet again.

Wikipedia has a very nice summary of water treatment, and includes a lot of information about the API (American Petroleum Institute) oil-water separator, which has been in effective use for decades. It is a very effective device, albeit costly, and can treat upwards of 50000 litres of water per hour. You can read about it in detail here, but I don't doubt it is what you are after.

Something more in the lines of what you describe is the Vertical Gravity separator. It is a pressurized system which promotes separation through convection mixing as the liquid is pushed though a complex path. The differences in liquid densities encourages separation of the two components. VGS systems can treat up to 3000L per hour. A document describing the operation of this system can be found on a link to a pdf document on this page. There are some problems associated with this type of device, in that the 'filter' mesh can get blocked if the material is very dirty.

Of course, any treated water is not 100% pure, and may need additional treatment depending on your requirements.

There are a number of interesting web resources where you can read more about oil-water separation methods. Pielkenrood have an excellent page of various methods, past and present. This blog also has a number of useful pages, although it is a commercial site promoting their own preferred devices.


How does brewing time and brewing temperature affect the flavonoid concentration of green tea? I've searched a lot for this, but I still cannot find any good explanations for how these two factors affect flavonoid concentrations.

You might want to have a look at

J. Peterson et al., Major flavonoids in dry tea, Journal of Food Composition and Analysis, 2005, 18, 487--501, http://dx.doi.org/10.1016/j.jfca.2004.05.006

and the references therein, e.g.

Christiane Lakenbrink et al., Flavonoids and Other Polyphenols in Consumer Brews of Tea and Other Caffeinated Beverages, J. Agric. Food Chem., 2000, 48*, 2848--2852, http://dx.doi.org/10.1021/jf9908042

From the abstract of the latter:

  [...] At brew times up to 2 min the composition of the brew solids was for each product practically independent of brew time, with flavonoids again comprising the major proportion (86−88%) of the total phenolics. [...]

Longer extraction times (> 10 min) or continuous (Soxhlet) extraction lead to partial decomposition of the flavonoids, most likely due to phenolic oxidation, cf.

Won Jo Cheong et al., Determination of Catechin Compounds in Korean Green Tea Infusions under Various Extraction Conditions by High Performance Liquid Chromatography, Bull. Korean. Chem. Soc., 2005, 26, 747--754, http://dx.doi.org/10.5012/bkcs.2005.26.5.747

I hope that helps a bit.


Is there a name for the pair (c, R), where c is the stoichiometric coefficient of a reactant R in some chemical reaction? I am programming a simulation, and I need a good name for a data structure that stores pairs $(c, R)$, where $c$ is the stoichiometric coefficient of a reactant $R$ in some chemical reaction. Is there a name for these pairs $(c, R)$ in chemistry?

In chemistry, often quantities get denoted by an index $i$, which then stands for some compound ($i=\text{A, B, C, ...}$).

Examples:

The stoichiometric coefficient $\nu_i$
The concentration $c_i$
The amount of substance $n_i$

The notion of grouping together the stoichiometric coefficient with the substance it uses is a novel one for me, and will most likely not have a name.

I suggest you use a name such as "coefpair" or the like and document your code accordingly.


Plasticisers and other impurities from polyethylene laboratory bottles? Does polyethylene (PE), widely used for wash/squirt bottles in labs, contain soluble compounds such as plasticizers that can accumulate in organic solvents (e.g. alcohols, acetone) after prolonged storage? I noticed an odd, "plasticky" smell from old propanol/ethanol from such bottles and the bottles, once emptied, also give off a similar smell. This smell is not present in fresh, analytical grade solvents or new, unused bottles.
Can plastic bottles be a source of e.g. bisphenol A and get absorbed by the skin, when applying alcoholic hand rub from such bottles?
It has been shown that plastic labware can contaminate their content with various autofluorescing compounds.
PS.: I have the exact model, but I'm not sure if one is allowed to post links to specific products/manufacturers here...

Polyethylene (polythene, to some) may have plasticizers, such as adipates, terephthalates, or phthalates, but they are not inherently needed in all cases. Hopefully, the supplier of those alcohols does not bottle them in cntainers with plasticizers, but a wash bottle, purchased separately, might very well have them added to the plastic, particularly because a squeeze bottle must be very flexible.
It is also possible that the polymer itself degraded over time, and with exposure to light, yielding ethylene monomer and other scission products. Compare the odor of the alcohol with that of ethylene, a faint sweet and musky odor (or "odour" if it's ethene).
As to which plasticizer is used, if any, you'd need to consult the bottle manufacturer. You could substitute a glass wash bottle, e.g. a gas wash bottle with aspirator tube, or a PTFE pipette pump, if you're concerned about contamination, e.g. because your measuring phthalates.
However, if your concerned about personal health, this is not the correct forum.


Is it possible to save an insoluble snowflake? In The Big Bang Theory Season 3 Episode 1 "The Electric Can Opener Fluctuation"

Leonard gives Penny a snowflake that will never melt. He said some type of chemical gas name when he try to explain, but I don't understand what he said exactly so I can't do any research about that. You can watch it on this YouTube video as well.

I used Google for some searches like doesn't melt snowflake and insoluble snowflake but results gave me nothing.

So, is this possible to save a snowflake preserved in glass with a chemical element, gas or something?

With gas no absolutely. To preserve ice at room temperature using the pressure of a gas you should reach a pressure of one billion Pascals. Too dangerous for Penny safe...
In fact there are some imprecisions in the video Leonard cites two different methods to "preserve" snow flakes with an organic polymer. In fact these methods create a snow flakes replicas, they are called snow crystal fossils, because there is no more ice inside them. Both methods are describe here, have a look!
Formvar Replicas

Leonard: "it will last forever I preserve it in a 1% solution of
polyvinyl acetal resin"

This is the Schaefer and Day technique [1] the solution Leonard is talking about is made from 1% of polyvinyl acetal resin in ethylene dichloride.
Superglue Replicas

Leonard: "in fact is a pretty simple process, you see, Cyanoacrylate
are monomers which polymerize mmm... "

Cyanoacrylate are the main components of Super Glue  this is easy method propose by the chemist Tryggvi Emilsson that you can do at home.
You have to choose one of these methods, you can't do both of them as Leonard (it seems) did... Take in account that Leonard is a physicist he doesn't know much about this stuff...;-)
[1] V. J. Schaefer and J. A. Day, Peterson Field Guides: Atmosphere (Houghton Mifflin, 1981).


Why is barium used instead of bismuth as a contrast in gastrointestinal xrays? Why is barium used instead of bismuth as a contrast in gastrointestinal x-rays? Bismuth is a heavy metal that has a relatively low toxicity in comparison to barium. Is there a particular property of barium that makes it a better contrast? Is it less expensive?

Is it less expensive?

Yes, but not so much that it would be favored, i.e. you don't worry about saving \$5 when your doing a \$5000 procedure.

  Is there a particular property of barium that makes it a better contrast?

No both absorb x-rays as a result of their high atomic numbers, in fact bismuth has more electrons per atom and thus is a better contrast agent..

  Bismuth is a heavy metal that has a relatively low toxicity in comparison to barium.

Yes but this is for the soluble metal ion. Barium sulfate is inert under most conditions, importantly the acidic ones in the digestive tract. Because the sulfate binds the barium ions to the solid and thus barium cannot enter the blood stream to cause a physiological response. Contrast this to Bismuth (no pun intended) which is non-toxic but typically soluble in acid and if not soluble, the bismuth is at least leachable. Now typically bismuth ions are not an issue as they will harmlessly be excreted, but there is a potential it could interact with a patient or the drugs of a patient in a negative or uncontrolled way. Since doctors don't like to introduce variables when making diagnosis's nor do the like surprises, especially when it's caused by them, the fully inert barium sulfate is the favorite for radio contrast.


Why did my glass pot explode when I boiled sugar in it and added cold milk? I decided to try some cooking with sugar and put a glass bowl on the flame with half a cup of sugar in it and some water. After bubbles appeared I stirred the sugar for about a minute and then I decided to add milk, and as soon as I poured the milk, everything exploded and the glass pot shattered into little pieces rapidly.

What just happened? Why did my glass pot explode when I boiled sugar in it and added cold milk? And does this also happen with laboratory equipment?

Assuming initially your glassware contained a cup of sugar and a little of water, you basically prepared caramel.  Depending on the sugar employed, temperatures needed to trigger caramelization vary, but $\pu{160 ^\circ{}C}$ ($\pu{320 F}$) mentioned succrose, as example (which is the normal household sugar) is quite high. Most materials, including glass, dilate upon heating, which is fine as long as this is evenly done.

Now if you add cold milk, say of $\pu{20 ^\circ{}C}$, the glass aims to contract back to its initial state. Because of the quantity of milk added, and the large heat capacity of water as a major constituent of milk, milk served momentarily as considerable heat sink.  Now taking into consideration that the glass ware's walls are unevenly heated -- at the outside still heated by gas or stove to more than $\pu{160 ^\circ{}C}$, and the inner of $\pu{20 ^\circ{}C}$, this suddenly generates a lot of mechanical strain and stress on the material.  The more the walls of your glass ware are thick, the more easy these may then crack and shatter into pieces just by the sudden temperature change.  (For the same token, you place hot glass ware on a plank of wood to allow slow cooling to room temperature.)

In addition, if the glass ware were closed tightly just after addition of the cold milk, the hot ($\pu{160 ^\circ{}C}$) is able to boil off the water in the milk, too; generating steam that likes to expand, or -- if confined in volume -- will build up pressure.  Under normal circumstances, for each litre of (liquid) water, up to $\pu{1.7 m^3}$ of steam may be generated (at normal pressure). This represents an additional stress for the material, and standard kitchen glass ware is not designed to withstand such pressures.


Preferred IUPAC names for branched unsaturated hydrocarbons (polyenes)? What are the preferred IUPAC names for the following three compounds?

Compound 1 is 3-ethenylhexa-1,5-diene. 2 is 3-methylidenehexa-1,5-diene.
But I am still not sure about 3. I have narrowed it down to two choices: 3-ethylidenehexa-1,5-diene or 4-ethenylhexa-1,4-diene.

How do I decide which two of the three double bonds should be in the parent chain, considering that both choices of parent chain have the same number of complex and double bonds?
The information I used is from the following Q&As:
How is a side-group that contains a double bond named?
How do you name an alkene using IUPAC rules if the longest carbon chain in this alkene does not include the double bond?

According to the general methodology described in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), a double bond is expressed by changing the ending ‘ane’ of the name of the corresponding saturated parent structure to ‘ene’.

  P-31.1.1.1 The presence of one or more double or triple bonds in an otherwise saturated parent hydride (…) is denoted by changing the ending ‘ane’ of the name of a saturated parent hydride to ‘ene’ or ‘yne’. Locants as low as possible are given to multiple bonds as a set, even though this may at times give ‘yne’ endings lower locants than ‘ene’ endings. If a choice remains, preference for low locants is given to the double bonds. In names, the ending ‘ene’ always precedes ‘yne’, with elision of the final letter ‘e’ in ‘ene’. Only the lower locant for a multiple bond is cited, except when the numerical difference between the two locants is greater than one, in which case the higher locant is enclosed in parentheses.

  (…)

For example:

If the compound contains multiple double bonds, a multiplicative prefix is placed before the ending ‘ene’ and the letter ‘a’ is inserted between the root of the name of the parent structure and the multiplicative prefix.

  P-31.1.1.2 The multiplying prefixes ‘di’, ‘tri’, etc., are placed before endings denoting unsaturation to indicate the number of multiple bonds of each kind, as required, for example, ‘diene’ and ‘triyne’. For euphonic reasons, when the endings ‘ene’ and ‘yne’ are preceded by a multiplying prefix and a locant the letter ‘a’ is inserted. There is no elision of the final letter ‘a’ of a multiplying prefix before ‘ene’ or ‘yne’, for example, ‘tetraene’ and ‘pentayne’.

For example:

The names of substituent groups are derived from the respective parent hydrides. The presence of free valences formally derived from the loss of one, two, or three hydrogen atoms at the end of the longest chain of a parent hydride is denoted by replacing the ending ‘ane’ of the parent hydride name by ‘yl’, ‘ylidene’, and ‘ylidyne’, respectively.

  The suffixes ‘yl’, ‘ylidene’, and ‘ylidyne’ replace the ending ‘ane’ of the parent hydride name. The atom with the free valence terminates a chain and always has the locant ‘1’, which is omitted from the name. This method is recommended primarily for saturated acyclic and monocyclic hydrocarbon substituent groups (…). Substituent groups formed by this method are referred to as ‘alkyl-type substituent groups’;

For example:

methyl $(\ce{-CH3})$
methylidene $(\ce{=CH2})$
ethyl $(\ce{-CH2-CH3})$
ethenyl $(\ce{-CH=CH2})$
ethylidene $(\ce{=CH-CH3})$

Therefore, the name for the first compound given in the question is ‘3-ethenylhexa-1,5-diene’.

The name for the second compound is ‘3-methylidenehexa-1,5-diene’.

For the third compound, there is a choice among two different principal chains. In accordaqnce with the above-mentioned rules, the name could be either ‘4-ethenylhexa-1,4-diene’ or ‘3-ethylidenehexa-1,5-diene’.

The current IUPAC Recommendations stipulate various criteria for the selection of the principal chain. The first relevant criterion in this case provides that the principal chain has the greater number of skeletal atoms; i.e., the principal chain is the longest chain. If this criterion does not effect a choice of a senior parent structure, the following relevant criteria are applied successively until there are no alternatives remaining. According to Subsection P-44.4.1, the principal chain

  (a) has the greater number of multiple bonds (P-44.4.1.1);

  (b) has the greater number of double bonds (P-44.4.1.2);

  (…)

  (j) has the lower locant(s) for endings or prefixes that express changes in the level of hydrogenation, i.e., for ‘ene’ and ‘yne’ endings and ‘hydro/dehydro’ prefixes (P-44.4.1.10);

  (…)

Therefore, the correct name for the third compound is ‘4-ethenylhexa-1,4-diene’ (not ‘3-ethylidenehexa-1,5-diene’) since the locant set ‘1,4’ for the ‘ene’ ending is lower than ‘1,5’.


Polycyclic nitrogen network? So today I suddenly stricken with this structure and started to wonder if this polymeric structure of nitrogen is possible or not? If possible then what's its stability?

I searched CCDC for the infinite structures with the following building block:

where where each vertex contains any non-metal atom linked by any type of bond. The only known crystal structure that would meet these criteria is $\ce{[B12]2[CBC][C2]Mg_{1.42}}$ where $\ce{B12}$ icosahedra are linked via $\ce{B–C}$ and $\ce{B–B}$ bonds forming the moieties of the previously shown topology:

  Figure 2. Projection of the $\ce{B25C4Mg_{1.42}}$ structure onto the $0ac$ plane (the block of icosahedra of boron carbide structure is marked by shading).

Unfortunately, no similar structures are found for nitrogen yet. Also, the structural motif breaks on boron clusters, but this is the closest as I was able to get.

Reference

Ponomarev, V. I.; Konovalikhin, S. V.; Kovalev, I. D.; Vershinnikov, V. I.; Borovinskaya, I. P. Mendeleev Communications 2014, 24 (1), 15–16. DOI: 10.1016/j.mencom.2013.12.004.


Can fungi metabolize pluronic polyol f127? I am looking for a gelling agent that can be used in place of agar, where its carbon must not be bioavailable for fungi. I have read about pluronic polyol f127 as substitute for agar. Can fungi metabolize carbon from this polymer?

It is known that some species of fungi can use hydrocarbons as their carbon source. Polyethers like pluronic are less chemically inert than hydrocarbons. Indeed, degradation of polyethylene glycols by microorganisms is a well studied process (see DOI: 10.1007/s00253-001-0850-2 for example). Gloeophyllum trabeum is known to degrade poly(ethylene oxide) by a non-enzymatic, oxidative process (Proc. Natl. Acad. Sci. USA. 1998, 95(18):10373-7). Given those facts, there is a very good chance that some species of fungi can metabolize pluronics and similar compounds.


Preparing an amine from an azide Is there any mechanism available for reduction of alkyl azides to amines.
One of the reactions is conversion by $\ce{Na}$ in ethanol or by $\ce{LiAlH4}$.
But how do these proceed?
$\ce{LiAlH4}$ will produce a hydride ion which attacks where?

I googled it for a while and this was the best thing I found (Source):

  This is a tough question, that is probably not known with any certainty.
  Even Jerry March (Adv Organic Chem, 4th ed, page 1219) doesn't commit himself to a mechanism.

  You'll have to speculate.   Add a hydride on the N next to the alkyl group, then
  grab a proton on the terminal N.    Add a hydride on the terminal N, then grab a proton on the middle N.  After that, it gets a little sketchy for breaking the N-N bond.

  Try a SN2 attack by hydride on the middle N, giving hydrazine and  RNH-, which is a lousy SN2 leaving group, but we're stuck with it.
  Protonation of the RNH- gives the amine and hydrazine.   Or, you could do a SN2 on the N next to the alkyl group, and kick out a NH2NH- anion, which is also a lousy leaving group for the SN2.
  You can see why most books conveniently leave out a mechanism for this reduction.


Why is water a poor solvent for organic molecules? Few organic molecules dissolve in water but why is this the case?

Pure water can form two hydrogen bonds per molecule.  The enthalpy of these hydrogen bonds is about 23 kJ/mol of water.  You have to give water a very good reason to rip apart its hydrogen bond network, and the interactions water experiences with the organic compounds often aren't enough, even considering increased entropy, to make this favorable.