Gaussian: Relaxed scan with modredundant optimization and dummy atoms I would like to perform a relaxed scan in Gaussian09 on my molecule using the modredundant optimization. Because I want to scan over two bond lengths simultaneously, I defined some dummy atoms which are close to the real atoms whose bond lenghts should be scanned. However, this approach does not work, because I can't properly index the dummy atoms. The Gaussian09 documentation says (emphasis mine):

  [Type] N1 [N2 [N3 [N4]]] S nsteps stepsize [[min] max]]

  N1, N2, N3 and N4 are atom numbers or wildcards (discussed below). Atom numbering begins at 1, and any dummy atoms are not counted.

How can I then perform a scan referencing a dummy atom? I don't want to do the scan using the z-matrix optimization because it does not converge.

Here is my input file:

%nprocshared=16
%mem=5gb
%chk=4MeAminoButanal
#P HF/6-31G(d) opt=(modredundant) scrf=(pcm) nosymm

4MeAminoButanal dummy scan

0 1
 C     0.000000     0.000000     0.000000
 N     0.000000     0.000000     1.451901
 C     1.325038     0.000000     2.055476
 C     1.991320     1.377897     2.032970
 C     1.229182     2.449098     2.809397
 C     1.309917     2.335179     4.306919
 X    -0.248373    -0.404008     1.611583
 X    -1.041414    -0.131882     1.066579
 H     1.218960    -0.323776     3.083767
 H     1.990264    -0.714465     1.564774
 H     2.993872     1.281489     2.436811
 H     2.106328     1.711600     1.006802
 H     1.602469     3.443240     2.566292
 H     0.178120     2.455789     2.536445
 H     0.708887     3.065770     4.854632
 H    -1.016528    -0.113770    -0.358506
 H     0.375985     0.943036    -0.379032
 H     0.606031    -0.798850    -0.431982
 X     1.641992     1.939406     4.607116
 X     1.311963     2.318294     5.471711
 O     1.974063     1.543637     4.907309
 H    -0.496745    -0.808017     1.771266

B 7 19 S 60 -0.05
B 22 7 F
B 2 7 F
B 6 19 F
B 21 19 F

When I run it, Gaussian09 obviously complains with

  The following ModRedundant input section has been read:
   Invalid atom number.
   B 7 19 S 60 -0.05

Disclaimer: This answer concerns the use of Gaussian 09. While the ModRedundant interface is retained and therefore the same restrictions also apply in newer versions, there is now a newer way of using internal coordinates. See my other answer for this option.

Like I posted earlier in the comments, you can indeed not use dummy atoms with a relaxed scan via the ModRedundant interface, because they are not existent in that coordinate system. Internal redundant coordinates are generated only from actual atoms. I tried to find an explanation of that, but failed. I believe the reason for that behaviour is, that dummy atoms do not have basis functions and are therefore not included in the wave function.

In this case you need alternative means to what you actually want to achieve. That of course depends on what you would like to model. From the input you provided, I can see, that you are trying to move the middle points of the carbon-oxygen and nitrogen-hydrogen bonds closer together, possibly to simulate the carbon nitrogen bond formation and proton transfer. While I seriously doubt this will happen in one step, this is at least one way to try to go about it.

If this method had worked, I believe you would have gotten a different than expected result. There are some more issues, that you might have not considered with your approach.
You might think that the following two lines keep the nitrogen hydrogen bond distance fixed. They do not.

B 22 7 F
B 2 7 F

Because the bond angle A 22 7 2 is still flexible, and the bond length B 22 2 is still flexible, this can possibly do whatever it wants. The same does hold for the carbon oxygen bond.

B 6 19 F
B 21 19 F

If I am correct with my assumption, then what you want look at is the carbon nitrogen bond formation. Why not take this internal coordinate and scan it? It is easy and it should pretty much do what you like. The following input should take you there. Note that I changed the method to pm6 to massively cut down computer time - and even with that, it took to long for my patience. (I did also not include any trailing lines about resources - see my notes below.)

#P pm6
opt=(modredundant)
scrf=(pcm)
nosymm

4MeAminoButanal scan

0 1
C        0.000000     0.000000     0.000000
N        0.000000     0.000000     1.451901
C        1.325038     0.000000     2.055476
C        1.991320     1.377897     2.032970
C        1.229182     2.449098     2.809397
C        1.309917     2.335179     4.306919
H        1.218960    -0.323776     3.083767
H        1.990264    -0.714465     1.564774
H        2.993872     1.281489     2.436811
H        2.106328     1.711600     1.006802
H        1.602469     3.443240     2.566292
H        0.178120     2.455789     2.536445
H        0.708887     3.065770     4.854632
H       -1.016528    -0.113770    -0.358506
H        0.375985     0.943036    -0.379032
H        0.606031    -0.798850    -0.431982
O        1.974063     1.543637     4.907309
H       -0.496745    -0.808017     1.771266

B 6 17 F
B 2 18 F
B 2  6 S 60 -0.05

Also note, that I did not check that the N-C bond length will become shorter than an Ångström, so the last steps take forever to converge and it will die in the last step running out of cycles. Run it with something more reasonable.
The calculation itself produced a somewhat reasonable result. The following animation cycles through all optimised structures. It was made with ChemCraft.

Some more notes on your input:

%nprocshared=16 is quite a number you are requesting for a Hartree-Fock calculation. Consider that parallelisation is not linear. While it became better throughout the years, I guess 4 threads is the most reasonable for this system size. You can go higher, I'd not recommend to go above ten though.
%mem=5gb is probably too much. I doubt that any of these processes will exceed 200MB of memory, probably a lot less.
%chk=4MeAminoButanal You should consider adding the .chk ending in all cases. Gaussian is known to mess things up when there are non-ordinary characters like _-. in a file name.
It is maybe a good idea to not write the whole route section in one line. Consider this as improving the readability of your input.  You can cluster similar commands (like pop statements) in one line, etc..
In previous versions, Gaussian might have had trouble reading in anything beyond the  80th char. I cannot reproduce this bug anymore and it's been a long time since I was told about it, so this should not be an issue. However, a well structured input file can help you finding possible syntax errors that result from typos.


Complexity in computing the integrals involving higher angular momentum I am quite familiar with Hartree Fock theory. However, writing it into program is an entirely different task, and that is where I have a couple of questions.

I wrote a Hartree-Fock routine in Python a couple of years ago. It involved a ultra-minimal basis: only one gaussian for two hydrogen atoms each. Kinetic energy term was pretty simple, so were the one-electron terms. Coulomb and exchange integrals were tricky, which involved evaluation of an erf. Now, my question is, if I included higher angular momentum basis functions (p,d,f,g type functions), what happens? Is it difficult to evaluate the J (Coulomb) and K (exchange) integrals because of the mathematical complexity? Or something else? Can someone give the math behind this and explain why it is difficult to evaluate?

It appears that the integrals that are difficult are the two-electron repulsion integrals. There exists a recursive relationship that can turn integrals with high-angular momentum functions into integrals with only s-type functions. Refer to this lecture for details.

See the following articles for details concerning these recurrence relations:

M. Head Gordon and J. Pople, J. Chem. Phys. 89, 5777 (1988)

S. Obara and A. Saika, J. Chem. Phys. 84, 3963 (1986)


Calculating the pH of a mixture of Na2HPO4 and Na3PO4? I know there are more questions about this on the forum, but I was just wondering: when we mix both solutions, would we have to consider the equilibria corresponding to $\mathrm{p}K_\mathrm{a2}$ and $\mathrm{p}K_\mathrm{a3}$, or just the latter?

I'll give a round about answer based on significant figures.

The whole truth is that any time that you add any phosphate ion into an aqueous solution, then you will have all four phosphate species ($\ce{H3PO4}$, $\ce{H2PO4^-}$, $\ce{HPO4^{2-}}$, and $\ce{PO4^{3-}}$) in solution. The relative amounts depend on the 3 equilibrium equations, and the total concentration of all of the phosphate species.

For phosphoric acid the three pKa's are different enough so that only two phosphate species will have a "significant" concentration at whatever pH the solution is at. Here is where the answer gets fuzzy. What is "significant"? To account for 99% of the species (2 significant figures) is typically good enough and at most two species would need to be considered. However if you want to account for 99.99999999999% of the species (12 significant figures), then you're going to have to consider all four phosphate species. So the gist is how many significant figures do you need to consider in the calculations?

What is typically done is to simply the four equilibrium equations to the two "significant" ones (maybe only 1 species at high or low pH's), and then calculate the concentrations of the last two species using the found concentrations of the first two.

In reality there is another consideration. If you look at the pKa values for phosphoric acid, Wikipedia lists pKa1 = 2.148, pKa2 = 7.198, and pKa3 = 12.319. There are only three significant figures in each of these equilibrium constants. (Only the mantissa counts, not the characteristic.) So you can only have three significant figures for any given phosphate species.

So, to three significant figures, for any sort of mixture of $\ce{Na2HPO4}$ and $\ce{Na3PO4}$ salts you'll need to consider both $pKa_2$ and $pKa_3$ and you'll end up with a quadratic equation to solve. The simplifying assumption is that $$\ce{[H3PO4] + [H2PO4^-] << [HPO4^{2-}] + [PO4^{3-}]}$$

An "exact" iterative solution, considering all four species, can easily be solved via a computer program, but it is really messy to do such a calculation by hand.

As a final check you should calculate the concentrations for $\ce{H3PO4}$ and $\ce{H2PO4^-}$ and verify that the assumption holds. (For phosphoric acid if the assumption doesn't hold to three significant figures, then normalizing the calculated values would yield a "good enough" result to 3 significant figures.)


What is a dl pair in stereochemistry? Is it two isomers with at least one stereogenic center each but the corresponding center on the other isomer will rotate the plane of polarized light in the opposite direction. Is this applicable to meso compounds which aren't chiral overall but have stereogenic centers?

d- l- Nomenclature as well as (+), (-) refer to a physical property of the molecule, its optical rotation. Optical rotation is a result of chirality, a topological property of the molecule. So d-, l- is something that you can generally measure with a polarimeter. If you have a 1:1 mixture of the 2 enantiomers then you will see zero rotation and this mixture is called racemic and named with the dl- prefix. A meso compound also has zero optical rotation since it is achiral but as it is not a (racemic) mixture the dl- notation doesn't make sense but you might see it used sometimes.

The problem now is if you are given a pair of enantiomeric molecules that are mirror images how do you name them so you can tell them apart every time? This is why the D-, L- and R-, S- notations were introduced. D- L- is the older and more naive approach that was later improved by the R-, S- (CIP) notation. Unfortunately as chemistry is a science with long history that is also used by other non chemistry-trained professionals, several times older naming conventions have survived. So you have to know all of them!

So very simply, d-, l- reflect a real physical property of the molecule and D-, L- (and R-, S-) is a way to distinguish between molecules that contain stereogenic centres (and might be or not be chiral). Actually even the R-, S- system has been criticized for not being rigorous enough. I saw a paper where a new naming system was suggested but it is not possible to read this without knowledge of advanced maths (group theory, algebraic topology)


How does the equilibrium shift when the products are solid and gas when increasing pressure? $$\ce{CaCO3(s) <=> CaO(s) + CO2(g)}$$
How will the amount of each element change if the pressure is increased?
My guess is that the equilibrium will shift to the left, because the right side has 1 gas molecule. I know my solution contradicts with Le Chatelier's principles, but there is no gas on the left. So how will the the amount of the elements on the right decrease and the solid on the left increase since pressure only has an effect on gases.

Consider the equilibrium expression for this reaction.
$$\ce{CaCO3(s) <=> CaO(s) + CO2(g)}\quad K_c^{'} = \frac{[\ce{CaO}] [\ce{CO2}]}{ [\ce{CaCO3}]}$$
The concentrations of solids and liquids are constant.  They are the molar densities. Since $\ce{[CaO]}$ and $\ce{[CaCO3]}$ don't change, they are moved to the left hand side and "folded into" the equilibrium constant.
\begin{align}
K_c^{'} &\cdot \frac{\ce{[CaCO3]}}{\ce{[CaO]}} = \ce{[CO2]} \\
K_c &= \ce{[CO2]}
\end{align}
Therefore, as long as solid $\ce{CaO}$ and solid $\ce{CaCO3}$ are present along with $\ce{CO2}$ gas there will be an equilibrium.  Only changes to the concentration of $\ce{CO2}$ will cause a shift in the equilibrium.
You asked how will the amounts change if the pressure is increased.  The pressure of $\ce{CO2}$ is increased by either adding more $\ce{CO2}$ or by reducing the volume of the container.  Adding more $\ce{CO2}$ will increase the concentration of $\ce{CO2}$ momentarily, which will shift the equilibrium to the left, using up some $\ce{CaO}$ and making $\ce{CaCO3}$. The pressure of $\ce{CO2}$ can also be increased by reducing the volume of the container.  Again, the concentration of $\ce{CO2}$ is increased, which increases the reaction with $\ce{CaO}$ to make additional $\ce{CaCO3}$.  The $\ce{CO2}$ pressure gets closer to what it was originally, as predicted by Le Chatelier's principle. Since the molar densities of CaO and $\ce{CaCO3}$ are constant, they don't appear in the equilibrium expression. That is why only changes to the pressure (concentration) of $\ce{CO2}$ affect the position of the equilibrium.
If the pressure in the container is increased by adding an inert or non-reacting gas, nothing happens to the amounts of $\ce{CO2}$, $\ce{CaO}$ or $\ce{CaCO3}$.  The added gas won't affect the partial pressure of $\ce{CO2}$.


How to check if azobisisobutyronitrile (AIBN) is still active? I will have some samples of azobisisobutyronitrile and I want to perform Barton-McCombie reaction. The problem is that the reagent is quite old and I need to know whether it has not decomposed.

Do you know any kind of test that I could perform? I was thinking if benzophenone in THF would not become blue after addition of fresh AIBN which would indicate the presence of the radicals. But I am not sure if such a test would be a good idea.

If azobisisobutyronitrile (AIBN) underwent homolytic cleavage over time, do you expect the resulting cyanoprop-2-yl radicals would survive as free radicals for a long period of time? If not, what would be the decomposed products?

Alkali metals reduce benzophenone in THF to form blue diphenylketyl radical anions. I'm not sure azobisisobutyronitrile (AIBN) or cyanoprop-2-yl radical can do that.

Maybe you can test the structural integrity of AIBN by 1H-NMR, IR or Raman spectroscopy?
The -N=N- stretching vibration of a symmetrical trans-azo compound like AIBN is "IR-forbidden" but absorbs in 1576cm-1 region of the Raman spectrum (1). If it decomposes, perhaps you can see the decomposed products in the mass spectrum.
Maybe you can try to activate the subject AIBN as a radical initiator?

a. Perhaps one can observe a minute amount of nitrogen gas upon homolytic cleavage?

b. AIBN can be used as radical initiator in radical polymerization. Once activated as cyanoprop-2-yl radical, it readily attacks olefinic double bond of vinyl monomers. Perhaps you can test such reaction and see if some polymerization occurs using mass spectroscopy?

Reference:

R.M. Silverstein and F.X. Webster: Spectrometric Identification of Organic Compounds (6th ed.)


Hard water filtration Why does hard water have to pass through the cation exchange resin first followed by the anion exchange resin? Why can't it be the other way around?

To answer your question, what if we did utilize the method you propose?
Before going into the details, it should be noted that the Ion-Exchange method is used to remove hardness of water at an industrial scale. The principal reason that this method is preferred over any other hardness removal process is that it has a unique advantage:
All exchanges take place in an aqueous medium in the form of ions alone.
Keeping that in mind, let's come back to the adjustment you suggested. In the Anion Exchanger, the usual reactions are
$$\ce{R'OH- +Cl^- -> R'Cl- + OH-}$$
$$\ce{2R'OH- + SO4^2- -> R'2 SO4^2- + 2OH-} $$
$$\ce{2R'OH- + CO3^2- -> R'2 CO3^2- + 2OH- }$$
Given that you still haven't removed the cations, the ensuing process would result as follows:
$$\ce{Ca^2+ + OH- <=>> Ca(OH)2} \   \\K_{-1}=\frac{1}{K_{spCa(OH)_2}}$$ and $$\ce{Mg^2+ + OH- <=>>Mg(OH)2}\\K_{-1}=\frac{1}{K_{spMg(OH)_2}}$$
Note that $K_{spMg(OH)_2}=1.8\cdot10^{-11}$ and  ${K_{spCa(OH)_2}}=6\cdot 10^{-6} $
. Further, due to the high concentration of $\ce{OH-}$ ions in the medium, the equilibrium shifts forward, meaning that a high amount of salt is precipitated and deposited within the exchanger, also called scaling. This is detrimental to the industry from any perspective.
Now, let's go back to the part where we discussed advantage of the catex. Do you get it now?
In simpler terms, the fundamental phenomenon behind this process (ION EXCHANGE) is challenged and its efficacy is vastly downgraded.
For future references:

Detailed exchange process
$K_{sp}$ values


How to derive this temperature-pressure-specific volume relationship? (for isentropic, adiabatic, ideal gas flow)
 $$\frac{T_x}{T_y} = \left(\frac{p_x}{p_y}\right)^{\frac{(\gamma-1)}{\gamma}} = \left(\frac{V_y}{V_x}\right)^{\gamma-1}$$ where $V$ is the specific volume, $x$ and $y$ are any 2 rocket nozzle axial sections (though I presume this relation holds true for many other things as well), and $\gamma$ is the specific heat capacity ratio $c_p/c_v$, numerically equivalent to the molar heat capacity ratio $(\bar{C}_p/\bar{C}_v)$.

Thanks!
p.s. I may have just learnt the hard way that it is probably not the best idea to write a markdown-dense blurb on a phone ;)

For an adiabatic system like a piston where $\delta Q = 0$, using the first law of thermodynamics gives you the following expression:

$$\mathrm dU = \delta Q + \delta W$$

$$\mathrm dU = - p\,\mathrm dV$$

This expression is pretty much useless however, in that you can't integrate it, since $T$, $V$, and $p$ are all constantly changing interdependently in the system. However if you assume the gas is ideal and making a few substitutions...

$$C_V\,\mathrm dT = \frac{-nRT}{V}\,\mathrm dV$$

$$C_V \frac{1}{T}\,\mathrm dT = -nR\frac{1}{V}\,\mathrm dV$$

$$C_V \int_{T_1}^{T_2} \frac{1}{T}\,\mathrm dT = -nR\int_{V_1}^{V_2} \frac{1}{V}\,\mathrm dV$$

$$C_V \cdot \ln\left(\frac{T_2}{T_1}\right) = -nR \cdot \ln\left(\frac{V_2}{V_1}\right)$$

$$C_V \cdot \ln\left(\frac{T_2}{T_1}\right) = -(C_p - C_V) \cdot \ln\left(\frac{V_2}{V_1}\right) = (C_p - C_V) \cdot \ln\left(\frac{V_1}{V_2}\right)$$

$$\ln\left(\frac{T_2}{T_1}\right) = \left(\frac{C_p}{C_V}-1\right) \cdot \ln\left(\frac{V_1}{V_2}\right)$$

$$\ln\left(\frac{T_2}{T_1}\right) = \ln\left(\frac{V_1}{V_2}\right)^{\gamma-1} ; \gamma = \frac{C_p}{C_V}$$

$$\frac{T_2}{T_1} = \frac{V_1}{V_2}^{\gamma-1}$$

From here it's just more cycling through variables; see if you can work the mathematics a bit so to get your second relationship.


Why is meta-nitrophenol less acidic than para-nitrophenol? My professor said while explaining the acidic nature of phenol that meta-nitrophenol is less acidic than para-nitrophenol, and gave the reason that resonance doesn't play on meta, but I didn't get it.

The acidity of nitrophenols (or any acid for that matter) is determined by the stability of the conjugate base. In the case of m-nitrophenol and p-nitrophenol, the relative stability can be determined by looking at the resonance structures.

You can see that p-nitrophenol has an additional resonance structure where the negative charge is delocalised onto the oxygens of the nitro group, which are highly electronegative and therefore stabilise the charge effectively. This stabilisation is not possible in the case of m-nitrophenol because there is no opportunity via resonance for delocalisation onto the nitro group.


Using Dalton's law of partial pressure to find percentages of composition A mixture of helium and neon gases is collected over water at $28.0 \rm~^\circ C$ and $745\rm~ mmHg$. If the partial pressure of helium is $368\rm~ mmHg$, what is the partial pressure of neon? (Vapor pressure of water at $28\rm~^\circ C =28.3~mmHg$). Calculate the weight percentage of He and Ne in the original mixture.

The question is based on Dalton's law of partial pressure. I got the first part right.In the second part, since we have to deal only with the original mixture, total pressure should be $$745-28.7=\pu{716.7 mmHg}$$ So we calculate the total moles in terms of "v" (total volume) and then moles of neon and helium in terms of "v" from which we calculate the weight of neon and helium and calculate the weight percentage. "v" will eventually cancel out and we get weight percentage of neon as $34.54\%$ and that of helium as $65.45\%$ but that is not the right answer.
In the solution to this question, they took the total pressure while dealing with the second part as $\pu{748 mmHg}$ only but why would we do that. We are only talking about the original mixture. Where am I wrong?
Answer in my textbook : Neon weight% = $82.58 \%$, helium weight% : $17.43\%$

First find the mole fraction of each gas in the original mixture using the formula, P(gas) = X * P(tot). For He:

$$X = \frac{368}{716} = 0.514$$

For Ne:

$$X = 1 - X(\ce{He}) = 0.486$$

Say you have 1 mol of mixture then you have 0.514 mol He and 0.486 mol Ne.
Converting these to weights using periodic table:

$$\rm 0.514~mol~He = 2.06~g~He \\\
0.486~mol~Ne = 9.81~g~Ne$$

The total mass of your mixture is the sum of these two masses, 11.9 g. Divide the individual masses by the total mass to get mass percent.

$$\rm\% w. He = 2.06/11.9 = 17.3 \%$$

Subtracting this from one for Ne (or you could apply the formula again):

$$\rm\% w. Ne = 1 - 17.3\% = 82.7\%$$

Note: I didn't carefully carry sig figs but you should!


How to find out unpaired electron in S2 molecule? Theoretically electronic configuration and number of unpaired electron can be found out by using molecular orbital theory. But, practically I have seen only up to 20. For example, for Ne2

But for S2 (32 electrons) how to find out the electronic configuration and number of unpaired electrons.

The molecular orbital energy-level diagram of $\ce{S2}$ is given below:-

(source)

We can see  how close the energy the 3s and 3p atomic orbitals are because their energy separation will determine whether the  $\ce{π_3_{p}_{x,y}}$,  or the $\ce{σ_{3}_{p}_{z}}$ molecular orbital is higher in energy. Because the ns–np energy gap increases as the nuclear charge increases , the $\ce{σ_{3}_{p}_{z}}$ molecular orbital will be lower in energy than the $\ce{π_3_{p}_{x,y}}$ pair.

Each sulfur atom contributes 6 valence electrons, for a total of 12 valence electrons. Ten valence electrons are used to fill the orbitals through $\ce{π_{3}_{p}_{z}}$and $\ce{π_3_{p}_{y}}$, leaving 2 electrons to occupy the degenerate $\ce{π^{⋆}_{3}_{p}_{z}}$ and $\ce{π^{⋆}_{3}_{p}_{y}}$ pair. From Hund’s rule, the remaining 2 electrons must occupy these orbitals separately with their spins aligned. With the numbers of electrons written as superscripts, the electron configuration of $\ce{S2}$ is $\ce{(σ_{3}_{s})^{2}(σ^{⋆}_{3}_{s})^{2}(σ_{3}_{p}_{z})^{2}(π_{3}_{p}_{z,y})^{4}(π_{3}_{p}^{⋆}_{x,y})^{2}}$ with 2 unpaired electrons at the antibonding $\ce{π^{⋆}}$ orbital and thus is paramagnetic. The bond order is (8 − 4) ÷ 2 = 2, so we predict an $\ce{S=S}$ double bond.


Self aldol condensation of butanal under basic conditions The reaction is such:

I've counted the number of carbons on each side: LHS=4, RHS=8
My working is such:

Conceptually, we can think of this reaction as occurring in three stages. I've provided diagrams for the self-condensation of acetaldehyde, however you should be able to draw an analogous mechanism for the self-condensation of butanal (your aldehyde).
Your mechanism is essentially correct, except for the final elimination step, which as you've drawn it gives the wrong product.
1. Enolisation of the aldehyde

Source: Organic Chemistry, 2 ed., Warren, Clayden, Wothers & Greeves

As you guessed, the hydroxide is basic enough to cause enolisation of the aldehyde. Using KOH/MeOH, we form the enolate under thermodynamic control, essentially meaning the enolate is in equilibrium with the aldehyde.
2. Addition of the enolate into another molecule of aldehyde

Source: Organic Chemistry, 2 ed., Warren, Clayden, Wothers & Greeves

The newly formed enolate may then attack an un-enolised molecule of the aldehyde (this is okay, since we've established that the aldehyde and the enolate are in equilibrium under these conditions). This nucleophilic attack leads to a beta-alkoxy aldehyde (the structure on the right), which will quickly pick up a proton to form a beta-hydroxy aldehyde.
Under these basic conditions (KOH, MeOH), we may be able to isolate the hydroxyaldehyde, however, they are prone to elimination (see below), which is what must happen to form the product you require.
3. Elimination of hydroxide

Source: Organic Chemistry, 2 ed., Warren, Clayden, Wothers & Greeves

The beta-hydroxyaldehyde formed (see above) is itself able to undergo enolisation, to form a new enolate. Rather than attacking another molecule of aldehyde intermolecularly, it may instead do an intramolecular E1cb type elimination, kicking out hydroxide (which is okay under these basic conditions), and giving you the product as drawn.
Your elimination step isn't the most favourable pathway to get rid of that alcohol, and infact the elimination only occurs under these conditions due to assistance form the carbonyl. The quote below puts this rather well.

You cannot normally eliminate water from an alcohol in basic solution as hydroxide is a bad leaving group. It is the carbonyl group that allows elimination here: these are E1cB reactions.
Source: Organic Chemistry, 2 ed., Warren, Clayden, Wothers & Greeves


What causes soda to fizz and how can it be stopped? When you go to a soda fountain and pour some soda into your cup, frothy fizz is generated at the top of the cup. Obviously, it has something to do with carbonation. However, when a cup of soda is just sitting and is not being filled or refilled, there is no fizz.

What causes the fizz? Is the fizz generated by the crashing of the soda into the cup?

I would also like to know how to stop the fizzing produced when the soda cup is being filled/refilled: Can I slow the rate at which the soda falls? Will spinning the liquid on the way down to the cup help attenuate the fizzing? What about changing the material from which the cup is made?

The soda is supersaturated with dissolved carbon dioxide since in the sealed container the soda is under pressure using carbon dioxide. A number of things can cause the formation of nucleation sites which degas the solution by forming bubbles which float to the surface. The foam is due to the surface tension of the liquid solution. The lay understanding is that the soda is "flat" when the soda is no longer supersaturated with carbon dioxide.

It should be noted that even in a saturated solution most of the carbon dioxide is in solution as a dissolved gas and not carbonic acid. Since soda solutions are acidic anyways, this also favors the dissolved gas as opposed to carbonic acid.

To minimize fizzing:

Use a clean glass. (Bubbles emanating from some sort on the
glass indicate that the glass is dirty.)
Use wet ice at the freezing temperature of water (not ice from
fridge below 32 F).
Cool the soda before pouring gently onto the side of the glass.
In other words do not just dump it into the center of the glass onto the
ice. Imagine trying to pour water into the glass without creating
any bubbles.


Which is a dye: Ferric Acetate or Ferrous Acetate? I’ve been trying to make Iron Acetate to dye leather and I cannot, for the life of me, figure this out experimentally.
I’ve been allowing steel wool to react with pure white vinegar in a sealed glass vessel with a pressure release for about a month. It turns dark black to green which I assume is due to the ferrous acetate ($\ce{Fe(C2H3O2)2}$). It also contains orange particulates that I assume are the ‘insoluble in water’ salt complex basic iron acetate, Ferric Acetate $\ce{[Fe3O(OAc)6(H2O)3]^+}$.
The solution has dyed leather black and acted as a mordant, and I assume it is a reducing agent. I’ve been trying to separate the iron acetates but I don’t know what properties they each have. Here’s the question:

Which is the dye? Which one turns leather and wood black? Any information anyone has on the properties of either one would be immensely helpful.

The gist is in the solubility products for the hydroxide precipitates of $\ce{Fe^{2+}}$ and $\ce{Fe^{3+}}$.

  $\ce{Fe(OH)3}$ doesn't really exist but is rather $\ce{FeO(OH)}\cdot x \ce{H2O}$. However the $K_\mathrm{sp}$ is calculated as $\ce{Fe(OH)3}$.

Now for the vinegar solution let's consider a somewhat "spent" solution where a lot of the acid has reacted. So consider a nominal pH of 5.5, so pOH = 8.5 and $\ce{[OH] = 3.16\times 10^{-9}}$. Let's calculate the equilibrium molarity of Fe species as "Mol(Fe) @ pH=5.5" in the table below.

\begin{array}{|c|c|c|}\hline
 & K_\mathrm{sp} & \text{Mol(Fe) @ pH=5.5}\\ \hline
\ce{Fe^{2+}} & 8.0 \times 10^{−16} & 80.12 \\ \hline
\ce{Fe^{3+}} & 2.79 \times10^{−39} & 8.84\times 10^{-14}\\ \hline
\end{array}

A $\ce{Fe^{2+}}$ solution can't be 80 molar of course, but what that does mean is that all the $\ce{Fe^{2+}}$ formed will stay in solution.

The calculations also show that a substantial amount, if not all, of the $\ce{Fe^{3+}}$ created will precipitate. The precipitation does lower the pH, so the equilibrium values are indeterminate without knowing the concentrations of the acetate anion and the nominal concentration of $\ce{Fe^{3+}}$ (amount dissolved plus amount in the precipitate). However looking at the overall reactions, it wouldn't seem that the pH could be lowered much.

\begin{array}{c c c c}
 & \ce{2Fe + 6H+} & \ce{<=>} & \ce{2Fe^{3+} + 3H2} \\
 & \ce{2Fe^{3+} + 6OH- } &\ce{<=>} & \ce{2Fe(OH)3} \\
 & \ce{6H2O} & \ce{<=>} & \ce{6H+ + 6OH-} \\ \hline
\text{Overall} &\ce{2Fe + 6H2O} & \ce{<=>} & \ce{2Fe(OH)3 + 3H2}\\
\end{array}

A further consideration here is that oxygen from the air, which dissolves into the solution, can oxidize $\ce{Fe^{2+}}$ to $\ce{Fe^{3+}}$.

Now I'd assume that it is the $\ce{Fe^{2+}}$ salt that you'd really want since that salt could diffuse into (soak into) whatever you're coating the solution with. The $\ce{Fe^{2+}}$ would then be oxidized to $\ce{Fe^{3+}}$ in the interior of the item.  The $\ce{FeO(OH)}\cdot x \ce{H2O}$ on the other hand would just sit on the surface of the item like a paint.


What is the chemical equation for cooking rice? What is the balanced chemical equation for the chemical reaction of cooking rice (including the formula for starch etc.)?

Essentially when rice is cooked all you are doing is adding water to it and heating it. No true chemical reaction occurs.  A way you could think of this addition of water in an equation is:
$$\ce{(C6H12O6)2 ->[\Delta][water] (C6H12O6)2.xH2O}$$
Not the most accurate representation as water was not added to a crystal system but I hope this helps you understand that no real reaction is occuring.


Why is grease from cooking hard to remove from wood? Cooking oil that jumps from a frying pan in use and over time accumulates around the frying pan on a wooden surface is sticky and hard to remove.

My question is why. Why is it hard to remove? What is the chemistry behind it? Does it bond with the wood somehow? If so, how is it different from oil used to give the wood a protective coating? Why is one sticky and the other not sticky?

This site explains the use of oil on wood fairly well:

  Oil is made of molecules small enough to seep down into the wood rather than merely sit on top. As a result, oil makes wood look richer and more translucent without adding a film on the surface. There are two different types of oils that woodworkers use: drying and non-drying oils. Drying oils will change from liquid to a solid film when exposed to oxygen in the air. Nut oils (boiled linseed, tung, etc.) are drying oils, but vegetable (peanut, olive) and mineral oils are non-drying. Edible mineral oil is popular on food contact items, like cutting boards. However, non-drying oils stay wet indefinitely, and they will wash off when the board is scrubbed with soap and water. Because they do not dry to a solid film, non-drying oils are considered a wood treatment, but not a finish.


If ions are floating around in a solution, how do we know they combine to form a compound in a solution? I apologize if my question sounds confusing I am not sure how to properly convey it here's an example:

For the reaction between lead nitrate and ammonium phosphate you get the reaction:

$$\ce{3Pb(NO3)2 (aq) + 2(NH4)3PO4(aq) -> Pb3(PO4)2(s) + 6NH4NO3(aq)}$$

but the ionic equation is
$$\ce{3Pb2+(aq) + 6NO3-(aq) +6NH4+(aq) + 2PO4^3-(aq) -> Pb3(PO3)2(s) + 6NH4+(aq) + 6NO3-(aq)}$$

so how do we know the compound $\ce{6NH4NO3}$ is formed if its just $\ce{6NH4+}$ and $\ce{6NO3-}$? Do the ions combine and then immidiately disassociate again? I'm very highly confused I would appreciate an explanation, and i apologize if this makes no sense.

User EdV comments above get to the heart of the matter. To summarize:

Your first equation is typically how a chemist would write the reaction.

$$\ce{3Pb(NO3)2 (aq) + 2(NH4)3PO4(aq) -> Pb3(PO4)2(s) + 6NH4NO3(aq)}$$

The equation is a "shorthand" explanation. What a chemist reads is:

On the left hand side is that that two solutions are mixed together.

Since $\ce{Pb(NO3)2}$ has an "(aq)" modifier it will be in solution, but since it is an ionic compound it will not be in solution as a molecule, but rather as the $\ce{Pb^{2+}}$ and $\ce{NO3-}$ ions.
Since $\ce{(NH4)3PO4}$ has an "(aq)" modifier it will be in solution, but since it is an ionic compound it will not be in solution as a molecule, but rather as the $\ce{NH4+}$ and $\ce{PO4^{3-}}$ ions.
Since the RHS shows a reaction the two salts can't be in one solution simultaneously. Therefore the salts must have been dissolved in separate solutions and the solutions were mixed.

On the right hand side $\ce{Pb3(PO4)2}$ precipitates and $\ce{6NH4NO3}$ remains in solution.

Also since $\ce{Pb3(PO4)2}$ has a tiny but finite solubility, a very small amount of $\ce{Pb3(PO4)2}$ will remain in solution.
Since $\ce{NH4NO3}$ has an "(aq)" modifier it will be in solution, but since it is an ionic compound it will not be in solution as a molecule, but rather as the $\ce{NH4+}$ and $\ce{NO3-}$ ions.

The "ionic equation" is harder to understand because of its verbosity. Did the $\ce{NO3-}$ ions come into the solution with the $\ce{Pb^{2+}}$ ions or the $\ce{NH4+}$ ions? Being chemists we can puzzle it out, but the equation is less succinct.
$$\ce{3Pb2+(aq) + 6NO3-(aq) +6NH4+(aq) + 2PO4^3-(aq) -> Pb3(PO3)2(s) + 6NH4+(aq) + 6NO3-(aq)}$$


Why does [Co(NO2)6]4- ion have 3 unpaired electron as opposed to 1? Why does $\ce{[Co(NO2)6]^4-}$ ion have 3 unpaired electron as opposed to 1?

The book says that it has 3 unpaired electrons. I thought $\ce{NO2-}$ was a strong field ligand, making $\ce{Co^2+}$ a $\ce{d^7}$ low spin ion. Six electrons go in the $\mathrm{t_{2g}}$ orbital and 1 in the $\mathrm{e_g}$ orbital.

Please explain the bonding in this compound (preferably with a suitable MO diagram to go with it).

According to Elliot, Hathaway & Slade (1966) cited by this paper, Co(II) with nitrite ligands is low spin.

So maybe it is just a typo. For d7 low spin, you would expect one unpaired electron.


Why are so many wave functions associated with hydrogen? According to Wikipedia, there's an infinite set of possible wavefunctions (orbitals) for the hydrogen atom:
$$\psi_{n\ell m}(r,\theta,\phi) = \sqrt {{\left (  \frac{2}{n a_0} \right )}^3\frac{(n-\ell-1)!}{2n[(n+\ell)!]} } e^{- r/na_0} \left(\frac{2r}{na_0}\right)^{\ell} L_{n-\ell-1}^{2\ell+1}\left(\frac{2r}{na_0}\right) \cdot Y_{\ell}^{m}(\theta, \phi )$$

Would an unperturbed electron ever go outside the lowest orbital?
At some point the perturbation (excitation) energy would exceed the ionization energy, so above that certain energy, no orbitals could ever be populated, so what's the point of having all these solutions?

"Probability densities for the first few hydrogen atom orbitals" (source)

1.Would an unperturbed electron ever go outside the lowest orbital?

First, there is no such thing as "outside the lowest orbital" because the lowest orbital has probability density at all points in space (no nodes).

But yes, at a given temperature electrons will be distributed amoung the energy levels in accordance with a Boltzmann distribution.

  2.At some point the perturbation (excitation) energy would exceed the ionization energy, so above that certain energy, no orbitals could ever be populated, so what's the point of having all these solutions?

There are infinite solutions none of which exceed the ionization energy (the ionization energy is only approached as n approaches infinity).

Think of a series like 0, 3/4, 8/9, 15/16, etc. where there are infinite members without reaching 1 (1 is approached in the limit of infinity).


Why does cotton dissolve in a dilute solution (1/2) of fuming sulfuric and nitric acids? A nitration experiment carried out showed that a solution of half diluted sulfuric and nitric acids would turn the submerged cottons into a gooey substance. The remnants of the cotton balls did not flash as nitrocellulose would have.

On the other hand, cotton balls submerged in a mixture of fuming nitric acid and fuming sulfuric acid remained intact and were nitrated, producing nitrocellulose.

What I'm really curious and confused about is how the dilution not only interferes with the nitration itself, but also affects its dissolution.

Any guidance is appreciated.

To address you second question first, in neither case did the cotton dissolve.  In the diluted acid mixture the cotton was just wetted, possibly also having enough oxidation of the cotton by the nitric acid to give it a different texture.

Regarding the effect of dilution on the nitration of the cotton, the reaction was basically just too dilute to proceed.  Also, according to this Wikipedia article, the sulfuric acid is present as a catalyst, producing the nitronium ion, $\ce{NO2+}$. The net reaction is given as:

$$\ce{3HNO3 + C6H10O5 → C6H7(NO2)3O5 + 3H2O}$$

Since water is a product of the reaction, by adding water to the reaction via dillution we further inhibit the formation of the products.


Why does acetone have a greater dipole moment than dimethyl ether? Below are the lewis structures for dimethyl ether and acetone. My chemistry textbook indicates that acetone has a larger dipole moment than dimethyl ether. I am familiar with dipole moments and the fact they are created by an unequal sharing of electron density. However I do not understand why acetone is more polar than dimethyl ether. My best guess is that having an extra carbon provides more electrons which allows the molecule to establish a larger difference in electron density between the oxygen and the rest of the molecule. I am unsure if this is correct and would greatly appreciate a more in depth explanation.

Dimethyl ether

The ether is fairly symmetrical though there is a bond angle of ~111°, "exposing" the oxygen on the side (and its electron cloud).  Acetone, however, has an oxygen off to the side, with its electron cloud offset ~120 picometers (pm), making it more polar.

Remember that oxygen is a strong electron acceptor, so that there will be a "lump" of electron probability surrounding it.

[Sorry about the hodgepodge illustration!]


Confusion about electron arrangement in elements My textbook says, "The first shell can hold only 2 electrons. The second and subsequent shells can hold 8 electrons to give a stable noble gas arrangement." Can someone explain me if it is MAXIMUM 8 electrons in the second and subsequent shells or is it possible to have more than that. I searched on the internet but it keeps showing this equation: 2n2. As I am a high school student, I have no idea about "quantum" stuff. Can someone explain? Thank you.

Basically, it depends on subshell. If you google something like "subshells", you'll see that s-subshell can hold 2 electrons, p - 6, d - 10, f - 14, g - 18. It's very unlikely that you'll need subshells f and g, since you're school student.
Look at this picture (about Aufbau principle).
It shows you the order of orbitals. Each s orbital holds up to 2 electrons, p - 6, so on. Fill them with all your electrons in given order.
For example, you have 12 electrons. The order (using Aufbau principle) is 1s-2s-2p-3s-3p-4s-3d...
1s holds 2 electrons (12-2 = 10), 2s holds 2 (10-2=8), 2p - 6 (8-6=2), 3 s - 2 (that's all). We see that 1st shell has 2, 2nd shell has 8, 3rd shell has 2, but 3rd is not full (since you still have empty 3p and 3d subshells).
So let's check. n is for number of shell. $2*n^2$ is for maximum number of electrons that shell can hold. For 1st, $2*1^2=2$, as we've seen. For 2nd, $2*2^2=8$, as we've seen. For 10th shell, the maximum is 200 electrons.
[edited]
Use Aufbau principle as well as info about subshells. Let's look at Y(39 electrons), which is in G3 P5. So we expect 5 shell to be outermost, and 3 valence electrons. Not "electrons in outer shell", but valence. For school course of chemistry, outer shell electrons are usually referred to as "valence", but these concepts are not always interchangeable.
Back to our example, Y has 1s-2electrons, 2s-2,2p-6,3s-2,3p-6,4s-2,3d-10,4p-6,5s-2,4d-1 (sum - 39). So yes, "maximum shell" is 5, as expected. But why do we have 1 electron on the outermost subshell, when we need to have 3? It's because valence electrons are those on 5s and 4d (2+1=3). Usually d subshell electrons are also considered "valence", when you have 1-10 of them (if total number of electrons, starting with 4d electrons is 18 or less). I highly recommend you to use periodic table with 18 groups, it's much more convenient for these purposes.


Can a reverse osmosis system remove radioactive elements in contaminated water? I am curious to know if a reverse osmosis system can clean water from radioactive elements like cesium-134 and cesium-137. If it can, how efficient can it be?

I don't know about cesium in particular, but according to the EPA's web site (see the section called "How will radionuclides be removed from my drinking water?"), both isotopes should be covered under the beta particle section.
I cannot quantify the efficiency part of your question, but presume it's acceptable to the EPA (which may or may not instill confidence to you in terms of whatever methodology they're using).

The following treatment method(s) have proven to be effective in removing radionuclides at levels below their MCLs:
Beta particle and Photon Radiation: ion exchange and reverse osmosis;
(Gross) Alpha Emitters: reverse osmosis;
Radium 226 and Radium 228 (Combined): ion exchange, reverse osmosis, lime softening;
Uranium: Ion exchange, reverse osmosis, lime softening,  coagulation/filtration.


Is there a element that shows this pattern of boiling point? Is there a compound that turns to gas at a certain temperature but if heated more turns back to liquid and if heated more than turn to gas?

Does it exist naturally? Has a such a substance be found? Can such a substance exist? if yes, then why?

Why couldn't at some temperature molecules interact in such a harmony that they actually cling together?

No. If you take a molecular system in the gas state, and you heat it up (at constant pressure), there is no reason for it to go into a denser, more organized state, such as liquid or solid. As you increase temperature, the interaction energy between molecules becomes smaller compared to thermal energy.


Are the bonds in the monomers at the end of a polymer filled or unfilled? In polymers the bonds between monomers are changed, usually from double to single1, to form a long-chain polymer with the now single bond joining another monomer. But in the monomers at the very ends of a polymer chain, there is an unfilled bond, a carbon atom with only three bonds filled. During the polymerization process are those unfilled bonds left unfilled, or is there some process that fills those bonds with, say, another hydrogen atom?

I say this because I don't know if there are any polymers made where the bond inside the original monomer is a triple bond. In all the commonly used plastics (those that have recycling symbols) the original monomers have double bonds between the carbon atoms.

You are asking about polymers made by radical polymerization.

  But in the monomers at the very ends of a polymer chain, there is an unfilled bond, a carbon atom with only three bonds filled.

During polymerization this is true, the radical will continue to react with monomer and propagate the polymerization, but the radical does not sit in the bulk. There are a number of ways which an additional atom amy bond with the carbon radical in a process called termination. Typically this is done by combination or inhibition.

Combination is the process where two radicals come together to form a bond leaving no free radicals to continue the reaction. This may occur when two radical chains combine or when a selected radical is added to cease the reaction. This radical could be the other half of the initiator which began the radical polymerization.

Combination of two polymerizing chains.

Combination of polymerizing chain with radical

For inhibition, a radical combines with a molecule that can form a stable radical such as oxygen or butylated hydroxytoluene (BHT).

 Radical chain combining with oxygen to form stable, non-ractive radical.

In these cases the radical will eventually find a stray radical or atom to bond with (like water), the unstable carbon radical will not exist permanently.

  I say this because I don't know if there are any polymers made where the bond inside the original monomer is a triple bond. In all the commonly used plastics (those that have recycling symbols) the original monomers have double bonds between the carbon atoms.

Polyacetylene uses a triply bonded monomer in its production.


Total number of isomers formed on monobromination of 2-methylbutane What is the total number of isomers formed on monobromination of 2-methylbutane (including configurational and stereoisomers)?

I tried working it out. However, I found only five:

According to the source, the answer is six. Which structure am I missing?

Here are the six structures that your source is probably referencing.


Entropy, reversibility and equilibrium Entropy is maximum at equilibrium, so if we have an equilbrium chemical reaction like:

$$\ce{HI <=> H+ + I-}$$

does it mean that the container of $\ce{HI}$ has maximum entropy or the universe?

Also, constant entropy implies reversibility but I don't understand how  the definition of reversibility applies to a system  of $\ce{HI }$ and $\ce{H+, I-}$ in a container. I mean, there's nothing to actually reverse, is there? By reversibility I mean that the system can be brought to its original state by an infinitesimal change(s) in it's condition.

Could anyone provide an explanation for this?

Note: I have edited my full question because previously it was very unclear.

Irreversibility is more a practical than a theoretical concept, in my opinion. 'In theory' all reactions are reversible. Take this highly schematised reaction:
$$\ce{A + B <=> C + D}$$
We define the equilibrium constant as:$$K=\frac{[C][D]}{[A][B]}$$
Using Nernst we can now establish a relation between $K$ and the left-to-right change in Gibbs Free Energy, $\Delta G^0$:
$$\Delta G^0=-RT\ln K$$
(Of course, as you know $\Delta G^0=\Delta H-T\Delta S$)
Evaluate this Nernst function to conclude that the more negative $\Delta G^0$ is, the higher the value of $K$:
$$K=\mathrm e^\frac{-\Delta G^0}{RT}$$
For very negative values of $\Delta G^0$, $K\to+\infty$ and by the equilibrium  constant equation, the concentration of the reagents at equilibrium is essentially $0$. Then we can write:
$$\ce{A + B -> C + D}$$
Such a reaction we would call irreversible. There is however no clear cut-off point and many reactions will have $K$ values that somewhat defy categorisation with respect to being 'reversible' or 'irreversible'.


PMMA cross-link temperature I premise, I’m a physicist and I’m working on a transfer procedure for graphene via CVD, which involves the use of PMMA as a resist layer. Studying this polymer, I still haven’t found what is the cross-link temperature (or window of temperatures) for it. Can someone give me a hand? or just a good reference?

UPDATE
Ok, maybe I’ve been misunderstood. My PMMA is diluted in anisole, and I apply it as a photoresist over the graphene sheet via spin coating. After this deposition NO Initiator is used, but it came across various heat treatment, depending in the recipe I use. My question is: does anybody know at what temperature the PMMA starts to seriously crosslink? or better: if I do more heat treatments, all at temperatures above 150 °C, am I going to have an highly crosslinked PMMA, so that a removal process in acetone solvent is impossible because it can’t get into the chains to districate them?

According to this thesis, the polymerization is determined by thermal initiator that is used in PMMA mixture. The polymerization (corsslinking) starts when the initator is added to the mixture, but it is really slow at low temperature. Higher temperature lowers hardening time (speeds up the crosslinking). Since the dustsheet does not provide the information about the initiator, I would suggest to determine the cross-linking temperature experimentally. Some material related to the polymerization process:http://www.chemistry2011.org/,

BULK POLYMERIZATION OF METHYL METHACRYLATE IN A KNEADER
REACTOR
P.-A. Fleury
LIST AG
CH-4422 Arisdorf,

High conversion polymerization of methyl methacrylate
in the presence of poly(methyl methacrylate) of different relative
molecular mass
J. BARTOŇ, V. PASTUŠÁKOVÁ, V. VAŠKOVÁ, and P. AMBROVIČ
Polymer Institute, Centre of Chemical Research, ChemicalPapers


Beer freezing after being opened I am watching the Stoke vs Swansea game, & to be honest, it's pretty boring. So much so, that I've started to wonder why it is that my beer has turned into a slushy after being taken out of the freezer.

I bought it this morning, it wasn't very cold, so just before the game I put it in the freezer to achieve optimal freshness. Started off pretty well, but rather than getting warmer as I drink, it's actually continued to freeze.

I am in France.

The windows are open.

It's 22 degrees Celsius.

I have attached a pic of the beer in question.

Have I discovered the holy grail of beer chilling?

When you opened the beer, carbon dioxide was lost to the atmosphere, increasing the mole fraction of water in the beer and raising its freezing point to a point above the temperature of the beer.  Not all of the water froze, leaving it a slush.

Edit:  There's also the possibility that the loss of CO2 from the solution is not the issue.  It could just be that when you opened the beer you disturbed a supercooled solution enough to start it freezing.

2nd edit:  Instead of continuing the comments I'll add here.  The freezing on the outside of the can shows that the temperature of the beer is below the freezing point of water.  Is the temperature of the beer lower than the freezing point of the beer?  If it were, the beer should freeze instantly upon being disturbed.  If the beer is above its freezing point, but the loss of CO2 raises the freezing point it should freeze gradually as it sits out.


Is HBrO2 a reductor? I'm working on a chemistry project and that features a reaction where $\ce{Br-}$ participates in a redox reaction as a reductor. Eventually, $\ce{HBrO2}$ competes with $\ce{Br-}$ to react with the same chemical as $\ce{Br-}$ reacts with.

Is is therefore true that $\ce{HBrO2}$ is also a reductor? If yes, what is the redox potential of $\ce{HBrO2}$?

If you need a good answer, you should not put an information embargo on the context of your question.

This is to be taken rather as an expanded comment than a full answer:
$\ce{HBrO2}$ oxidizes $\ce{Br-}$ to $\ce{Br2}$.
$$\ce{HBrO2 + 3 Br- + 3 H+ -> 2 Br2 + 2 H2O}$$
To tell if it may compete with $\ce{Br-}$ to react with the "chemical", we would have to know that chemical.
Regarding standard redox potentials, an exhaustive list is on Wikipedia, but $\ce{Br^{III}}$ species are not there.
But be aware $\ce{HBrO2}$ is unstable, strong oxidizer, while $\ce{Br-}$ is a weak reducer. I personally think the info may have been misinterpreted, having meaning $\ce{HBrO2}$ may compete with the chemical to react with $\ce{Br-}$.


What is Delta H for dissolution? An ionic compound has a solubility of $1\ \mathrm M$ in water at $25\ ^\circ \mathrm C$ and its solubility increases as the temperature is raised. What are the signs of $\Delta H^\circ$ and $\Delta S^\circ$ for the dissolving process?

Since the the solubility increases as the temperature is raised ($\Delta G^\circ$ becomes more negative), I know that $\Delta S^\circ$ is positive. However, I am unsure of how to determine the sign of $\Delta H^\circ$. The answer is $\Delta H^\circ > 0$.

I would solve this question rather simply.

Let's consider the general equation $\ce{ X-Y_{(s)} \rightarrow X^+_{(aq)} + Y^-_{(aq)}}$ with a $\Delta H = C$. Intuitively, if the reaction is endothermic ($\Delta H > 0$), an increase in temperature should help the dissolution (more heat to be absorbed by the salt while dissolving). You can rationalize this further using the Le Châtelier's principle:

$\ce{ X-Y_{(s)} + C \rightarrow X^+_{(aq)} + Y^-_{(aq)}}$

If you add more heat ($C$), the dissolution will be more favored.

More rigorously, you can use Van't Hoff equation (assuming a dissolution equilibrium): $\frac{d(ln(K))}{dT} = \frac{\Delta H}{RT^2}$. If $\Delta H > 0$, then $ln(K)$ increases with $T$, i.e. dissolution is more favored.


Why are s orbitals more “penetrating” than p orbitals? In polyelectronic atoms, the reduction in the net central force due to electron-electron repulsion is accounted for through an effective nuclear charge that depends on a “shielding effect” of inner electrons.
The reason for the splitting of energies of orbitals in the same quantum shell (s<p<d<f) is usually given that s orbitals are more "penetrating" i.e have a greater probability density near the nucleus, so experience less shielding and thus a greater effective nuclear charge.
However, I can't understand why a comparatively larger p. density is used to measure the penetrating power. the mean orbital radius actually decreases from s to f orbitals. Sure, s orbitals have a relatively higher p.d near the nucleus but the rest of the p.density is also relatively more further away than that in p,d or f orbitals. Using the mean distance as the appropriate measure, shouldn't the energy be in the order f<d<p<s?
the only explanantion that comes to mind is that the decrease in energy from being unusually close some of the time outweighs the decrease in energy due to being more farther away for the rest of the time (and thus on avg), but this is just a guess and a rather non-intuituve explanation

I found an old slide from when I was teaching.
These are plots of the 2s and 2p orbitals for hydrogen (assuming 1 electron). The radial wave function is on the left and the probability density as a function of distance from the nucleus is on the right ($4\pi r^{2}R(r)dr$ if you understand the calculus).
You can see that there's a blob of density that comes closer to the nucleus for the s orbital. There's your penetration.


How does 1 mole of a compound = the molecular weight of the compound? I am reviewing moles, and came across this sentence in my book:

  One mole of a compound has a mass in grams equal the molecular weight of the compound expressed in amu and contains $6.022\times10^{23}$ molecules of that compound.

Now, I perfectly understand the second part of the sentence. 1 mole of compound has $6.022\times10^{23}$ molecules. But how do $6.022\times10^{23}$ molecules have a mass same as the molecular weight of compound?
Ex. Molecular weight of $\ce{SOCl2}$ is $119\ \mathrm{amu}$. That is the mass of 4 atoms. How do $6.022\times10^{23}$ molecules have the same mass?

Well you can actually calculate it like this
$$1\ \mathrm{amu}=1.66\times10^{-27}\ \mathrm{kg}$$
Therefore
$$119\ \mathrm{amu}=119\times1.66\times10^{-27}\ \mathrm{kg}$$
But for 1 mole, multiplying by Avogadro number you get mass of one mole as $$\begin{align}
119\times1.66\times10^{-27}\times6.022\times10^{23}&=0.11899\ \mathrm{kg}\\
&=118.99\ \mathrm g\\
&\approx119\ \mathrm g
\end{align}$$


Why do disproportionation reactions take place? I want to know what are the conditions when disproportionation reaction is feasible. Is it just because of stability factors or it has other dimensions to it. As an example:
$$\ce{3HNO2 ->  HNO3 + 2NO + H2O}$$

Nitrous acid disproportionates to nitric acid and nitrogen monoxide. Here stability of +5 in nitric acid is understandable but how +2 in the monoxide is stable if the disproportionation takes place to gain stability.

Am I ignoring some other factors like pH or hydration enthalpy. I would be very thankful if someone gives me some general ideas on these reactions.

"Stability" is not really a driving force of a reaction. That would be the change in Gibbs Free Energy or ∆G. That takes into account both entropy and enthalpy, as well as concentrations of reactants and products, and temperature.

If you haven't yet discussed these in class, you should be able to read about these concepts by looking up "Gibbs Free Energy" in your textbook's index.


Strictly speaking, is air homogeneous? Why would air be considered homogeneous in its gaseous state (by Wikipedia and several other sources) even though its concentration keeps varying with time?

A homogeneous fluid as i think of it is as follows: If I take a small part of the fluid the number of particles of each kind is always the same.

Strictly speaking you might have a point there. But it is probably more difficult to judge.

I was unable to unearth an official recommendation by the IUPAC, so I unfortunately have resort to Wikipedia ([1] Kristi Lew (2009). Acids and Bases. Essential Chemistry. New York: Chelsea House Publishing.):

  Homogeneous is a term in physical chemistry and material science that refers to substances and mixtures which are in a single phase. This is in contrast to a substance that is heterogeneous.[1] The definition of homogeneous strongly depends on the context used. Generally it refers to smooth variation of properties with no discontinuities or jumps.

If we talk about all the air around the world, it is certainly not uniform. There are concentration differences across certain areas. When we consider a vector into space, the total concentration will decrease along this vector. As we know, we will pass certain areas with different chemical compositions, for example the ozone layer.
But all these changes will be gradually; there are no discontinuities. In this sense one would call air homogeneous, even though it does not fit in the definition of phase.

Additionally, if you take an arbitrary (volume) unit of air, all at the same temperature and pressure, and you allow it to establish an equilibrated state, then yes, this will not only be homogeneous, it will also form one phase.
The above linked article further considers air as a homogeneous mixture. It rambles on with considering it a gaseous solution of oxygen, carbon dioxide, etc. in nitrogen. This is technically incorrect, since the gaseous state of matter is specifically excluded from the IUPAC definition of solution. Also a solution can never be at the same time a mixture, as the former can be separated by mechanical means, while the latter cannot.

The problem here is obviously the lack of a unique and rigorous definition for homogeneity. For whatever purpose when it really matters, one should probably use a more robust definition like phase, which is essentially what your second statement describes.

Depending on the context you might even obtain a homogeneous mixture of water and oil - the question is then more or less how stable such a suspension is. After a while it will certainly separate in two phases.


Clarification on Oxidation Number and Ionic Charge I am confused on the differentiation between ionic charge and oxidation number (state).  For example and simplicity, $\ce{NaCl}$ is an ionic compound so it forms $\ce{Na+}$ and $\ce{Cl-}$.  Are these ($1+$ for $\ce{Na}$ and $1-$ for $\ce{Cl}$) both the oxidation states and the ionic charges?  Or is the net of the compound ($\ce{NaCl}$) the ionic charge?

Additionally, when you have something such as $\ce{C2O4^2-}$ in this case a covalent compound , carbon's is $3+$.  Is this carbon's oxidation state?  What then is the net charge of the compound ($2-$) called?

For simple mononuclear ions in ionic compounds, as you have in $\ce{NaCl, Al2O3, ZnS}$ or $\ce{GaN}$, the oxidation state of each element will always equal its ionic charge: $\ce{Na+}$ has $\mathrm{+I}$, $\ce{Al^3+}$ $\mathrm{+III}$, $\ce{Zn^2+}$ $\mathrm{+II}$, $\ce{Ga^3+}$ $\mathrm{+III}$, $\ce{Cl-}$ $\mathrm{-I}$, $\ce{O^2-}$ $\mathrm{-II}$, $\ce{S^2-}$ $\mathrm{-II}$ and $\ce{N^3-}$ $\mathrm{-III}$.

Something like the oxalate anion $\ce{C2O4^2-}$ is a multinuclear anion. Its ionic charge is $2-$, as evident by the superscript. However, you cannot always determine the constituent elements’ oxidation states a priori — the only thing you do know is that the sum of the oxidation states must equal the charge number. In oxalate, carbon has $\mathrm{+III}$ as you correctly mentioned and oxygen has $\mathrm{-II}$, as it should be of no surprise to you. Added up, this leaves us with:

$$2 \times (+3) + 4 \times (-2) = +6 + -8 = -2$$

And $2-$ is the ionic charge as we already know. This rule goes both ways, so you can use the (known) ionic charge of a multinuclear ion to determine an element’s oxidation state if the other oxidation states are known: in $\ce{SO3^2-}$ oxygen has $\mathrm{-II}$, so sulphur can only have $\mathrm{+IV}$.


Explanation for a basic decomposition of water experiment My son is trying to write a conclusion for a basic decomposition of water experiment using a 9 V battery (with a test tube placed over each terminal) placed inside a salt water solution.  The results were: the tube over the negative terminal nearly completely filled with presumably hydrogen gas while the tube over the positive terminal only had a minute amount of presumably oxygen gas.
His textbook didn't explain these results, other than to hint at the fact that the results were due to there being twice as many hydrogen atoms in the product.  He has several questions that I cannot answer.

If the hydrogen and oxygen molecules take up the same amount of space, then why did the tube over the negative terminal have much more than twice the gas as the other one?
The atomic mass of oxygen is nearly 16 times that of hydrogen, so then even if there was twice as much hydrogen in the product, wouldn't the oxygen still take up more space?

If the hydrogen and oxygen molecules take up the same amount of space, then why did the tube over the negative terminal have much more than twice the gas as the other one?

Good questions but keep in mind that you did not mention the concentration of sodium chloride. If you have too much salt in water, chlorine gas is produced (eventually bleach) instead of oxygen. This is why you did not see the expected 1:2 vol. ratios.
If possible, use the lowest possible salt concentration for the experiment or even better, can you use a small amount of Epsom salt? Then there is no issue of chlorine formation and only oxygen gas will form. You will certainly see 1:2 vol. ratio.

The atomic mass of oxygen is nearly 16 times that of hydrogen, so then even if there was twice as much hydrogen in the product, wouldn't the oxygen still take up more space?

This issue was tackled by scientists long time ago. The key point is that if you have 1 billion molecules of $\ce{H2},$ and 1 billion molecules of $\ce{O2},$ they will occupy the same volume at a given temperature and pressure. There is a named law behind it. The volume of molecules way too small as compared to the volume they occupy.
The radius of oxygen molecule is 152 picometer (symbol: $\pu{pm}$) and that of hydrogen is 120 picometer $(\pu{1 pm} = \pu{10^{-12} m}),$ so masses and volumes do not correlate exactly.


Reacting NaOH in solution with O3 I use common air to produce ozone. However the gas after the ozone generator does contain $\ce{NO2}$ gas, which I have to remove for the subsequent steps. I tried to remove the $\ce{NO2}$ with a stainless steel gas wash bottle filled with distilled water. This works fine, however the distilled water will subsequently get charged with $\ce{NO2}$ and turn to dilute nitric acid.

After a few days of operation, I have changed the water and tested its acidity. The $\mathrm{pH}$ was around 0.4 - it was possible to etch a copper pipe with it. So the formation of this acid is not very good for the stainless steel gas wash bottle. I have decided to add 2 more stainless steel gas wash bottles and fill the first one with $\ce{H2O + NaOH}$, in order to neutralize the $\ce{HNO3}$. The other two bottles are for capturing the remaining traces of $\ce{NO2}$.

Bottle 1: 5 liter $\ce{H2O}$ + $\pu{15g}$ $\ce{NaOH}$
Bottle 2: 5 liter $\ce{H2O}$
Bottle 3: 5 liter $\ce{H2O}$

Now I have done some ozone measurements with and without these gas wash bottles. The result was as follows:

It seemed as if $\ce{O3}$ would pass through pure $\ce{H2O}$ without any reaction once the solution is saturated and the saturation seems to happen almost immediately.
However the solution with the $\ce{NaOH}$ seems to react the first 2-3 minutes and then suddenly stop reacting, which is actually good, because it does not use up ozone, however I have no clue what is happening in detail.

Here are some charts from the measurement. Upper chart: test of the $\ce{H2O + NaOH}$ solution. Lower chart: test of the $\ce{H2O}$. Note: the lower the value in the chart, the higher the ozone concentration.

The duration of each test was about 5 minutes. I did measure the $\mathrm{pH}$ of the $\ce{NaOH}$ containing solution and it basically did not change due to the ozone treatment. (It was around 12.35 before and after ozone bubbling...)

Any idea what is happening there and why?

You could oxidize some hydroxide ions, to wit:

$\ce{OH^- + O3 -> HO2^- + O2}$

Such an oxygen transfer is favored by strongly basic solutions because $\ce{HO2^-}$ is a weaker base ($\mathrm{p}K_\mathrm{a}$ hydrogen peroxide = 11.75) than $\ce{OH^-}$, as well as ozone being a more powerful oxidizer than diatomic oxygen.


Why is H2S2O7 stronger than H2SO4? I've read that for an oxyacid, the oxidation number of the central atom shows the power of that acid, but here, the oxidation numbers of sulphur in sulphuric acid and pyrosulphuric acid are both the same, i.e. +6. That is confusing! How should I proceed now?

To answer this question at a more relatable level, typically the strength of oxoacids can be determined from the number of "lone" oxygen atoms in the compound--that is, the number of oxygen atoms that do not have a corresponding hydrogen. What usually happens is the central atom in an oxoacid will draw electrons from the bonded oxygen atoms, thereby making them more electronegative. This increases the polarity of the O-H bond, increasing acidity. In your example, H$_2$SO$_4$ has 2 lone oxygen atoms; while in H$_2$S$_2$O$_7$, 5 lone oxygens exist. We therefore suspect that the latter is more acidic. Although there are exceptions to this rule, for a general chemistry course it typically is valid.


Where does the extra electron come from in the Lewis Dot Electron structure of say for example nitrate ion? As most of you know, this is the diagram for nitrate ion:

To draw this, i have to make all bonds with electrons from nitrogen, nitrogen only has five valence electron so ONE of the bonds consists of one electron from nitrogen and the other from the MYSTERIOUS electron, that comes out of no where.

My book tells me that's if simply how you draw Lewis diagram for N, O, F, Cl. When the oxidation is positive, take away electron and when it is negative, add.

In any covalent compound, bonding atoms share their electron to complete octet.( yeh, there are exceptions)

In this case, N and O share their electron. But when I count the number of valence electrons, a negative charge means a extra valence electron. NOW, where does this extra valence electron come from?

Unfortunately your Lewis structure for the nitrate ion, $\ce{NO3-}$, is wrong. You did not include any charges and as it is an ion there is at least one. Let's first have a look at how an ion may be formed and then hopefully you see, that there is no mystery at all about it.

Take for example ordinary table salt, which has the chemical formula $\ce{NaCl}$. If you dissolve it in water the lattice will be broken and the ions will by hydrated and are now able to move more freely.
$$\ce{NaCl_{(s)} -> Na+{}_{(aq)} + Cl- {}_{(aq)}}$$

The same happens when you dissolve potassium nitrate, which releases the nitrate ion:
$$\ce{KNO3_{(s)} -> K+{}_{(aq)} + NO3- {}_{(aq)}}$$
Here you can see, that the nitrate ion has an extra electron which it has taken from the potassium. See ionic bonding for more information.

For the Lewis structure of the nitrate ion you first have to count all the valence electrons of the involved atoms:
\begin{array}{lclr}
1 &\times &\ce{N} & 5\\
3 &\times &\ce{O} & 6\\
1 &\times &\text{charge} & 1\\\hline
  &       &              & 24\\
\end{array}

In the Lewis structure you have to distribute 24 electrons in a fashion, that you do not violate the rule of two, which is better known as the octet rule. Therefore your structure may include a nitrogen oxygen double bond, a positive formal charge on nitrogen and a negative formal charge at each of the singly bound oxygens.
Another representation is that all oxygens are singly bound and therefore the nitrogen has a $+2$ formal charge and each of the oxygens has a negative formal charge. The problem with this structure is that nitrogen does not have a full octet. Nevertheless, it is a strong contributor to the actual electronic structure. The bonding situation is much more complicated than can be expressed through a Lewis structure. The concept of resonance structures has to at least be included. Better in this case would be molecular orbital theory and or valence bond theory. The molecule itself is a y-aromatic system, the charge is delocalised through the extensive $\pi$ system. The blue structure is an attempt of showing this feature.


Is the pressure of a gas proportional with the concentration of a gas? Is the pressure of a gas proportional with the concentration of a gas?

Because you can also use the pressure instead of the concentration in the  equilibrium condition.

Assuming your gas follows the ideal gas law, you have $PV=nRT$. Setting $c=\frac nV$ yields $P(c,T)=cRT$.

If the temperature is fixed, then $P(c)=(RT)c$ indeed tells us that the pressure of a gas is proportional to its concentration.


What materials for a boiler, if any, do not require a deaerator? I'm imagining a portable boiler/turbine unit that burns wood and creates steam for a turbine, generating electricity.

I was planning on using freshwater from a river or lake. (A simple filter will remove particulate matter and minerals.)

But then I came across something called a Deaerator.

Basically, the deaerator removes oxygen and carbon dioxide from the water. Those substances often dissolve in the water and, in a boiler, can cause metal corrosion.

So I'd like to choose a material that doesn't require this, if possible. Titanium? Bronze? Carbon-carbon? Iron-Nickle like they use in jet engines?

Edit: For the curious, this question arose from studying old steam trains. Pretty sure they did not de-aerate the feedwater there and just let boiler lifetime suffer. I'm wondering if some modern "miracle" material, maybe carbon composites, can eliminate the need for de-aeration.

There are many more common alloys in the Inconel , Hastelloy , and Incoloy families. And some odd balls like MP35N, 5o:5o Cr- Ni, Vitallium , and others. Titanum won't take the heat. For all of these materials; if you need to ask the price ,you can't afford them.


Order of C-O bond length Find the correct order of increasing $\ce{C-O}$ bond length in $\ce{CO}$, $\ce{CO2}$, $\ce{CO3^2-}$.

My solution: as bond length is inversely proportional to the bond order, the order would be $\ce{CO} < \ce{CO2} < \ce{CO3^2-}$. Am I correct?

The order is correct but this is my procedure.
Stronger the bond smaller the length


Is there a preferred way of naming the resonance hybrid in keto-enol tautomerism? While answering a question about keto-enol tautomerism the question arose, how can I refer to the resonance hybrid instead of on of the resonance forms. In the case of the deprotonated butane-2-one, I can name 1 3-oxobutan-2-ide and 2 but-2-en-olate (ignoring stereochemistry). I cannot come up with a name for the hybrid 3.

While this might be quite a simple case, there are many resonance stabilised structures out there, which have contributions from more than one form. It gets especially complicated for ionic substructures. I wondered if there is either already an official recommendation (preferred IUPAC name), or any other suggestion at all handling those cases. After all, we are treating the same bonding situation with two different names, there should be something that covers both (multiple) cases.

Nomenclature of Organic Chemistry, IUPAC Recommendations and Preferred names 2013 in the section P-76 DELOCALIZED RADICALS AND IONS mentions only "totally delocalized" ions like cyclopentadiene derived cyclopentadienyl radical, cyclopentadienylium cation, cyclopentadienide anion; benzo[7]annulenylium (azulenium?) and pentadienyl radical.

However, there is a draft having this chapter expanded, allowing more systematic naming of delocalized ions.

  P-76.1.2 Partial delocalization is denoted by the descriptor deloc preceded by the locants indicating the extent of delocalization. This descriptor and its locants are cited immediately before
  the appropriate suffix and enclosed in parentheses.

  cyclopenta-2,4-dien(1,2,3-deloc)ide
  ...

But I am not 100% sure how exactly would that apply to your 3 structure.. let's be brave and creative:

(2Z)-but-2-en-2-(2,3,O-deloc)olate or
(2Z)-but-2-en-2-ol(2,3,O-deloc)ate or if you don't need stereochemistry
3-oxobutan-2-(2,3,O-deloc)ide or maybe even just
butanon(2,3,O-deloc)ide ? .. or
butanon(O,2,3,-deloc)ide ?

However, that draft is from 2004, but this rule did not make it into the current official nomenclature (2013), so for now we can just speculate, or contribute to the upcoming version.


Melting ice with salt - how does it start? I know that salt reduces the melting point of ice. I can see how a little bit of salty water can melt ice to water which can dissolve even more salt. But how does it even start? There is ice, which is hard. There are grains of salt, which are hard. So how does the salt enter the ice, to begin with?

I believe that Ivan's comments are on the right track. Generally any melting phenomenon to do with ice begins with the fact that the surface of ice is disordered. By this I mean that ice on the whole is a periodic crystal like other solids, but at the surface there are imperfections, primarily due to the extra conformational freedom afforded to the surface molecules. This distortion then propagates down quite a few layers [1]. Ref. [1] discusses the properties of this surface layer which is liquid-like, and points out that this layer exists even down to quite low temperatures.

If these surface molecules "see" something charged (each ion that is) like $\ce{NaCl}$, then the surface molecules already have the conformational freedom to solvate the ions. Additionally, there is an energetic benefit to this because the hydrogen bonds between these ions and water are stronger than the water-water hydrogen bonds.

In general, melting phenomena are extremely complicated and the question of how melting "begins" is has not really been answered completely or satisfactorily (can provide refs. later if you want). This case is a bit different though because it is really a solvation phenomenon, which is adequately explained by the liquid-like layer at the surface.

[1]: Beaglehole, D., & Nason, D. (1980). Transition layer on the surface on ice. Surface science, 96(1-3), 357-363.


Putting sanitiser in PVC bottles can I put hand sanitiser in PVC bottles or must the bottles be PET.  With the shortage of bottles I am keen to know if a 70% ethanol or isapropyl based sanitiser will distort or soften a PVC bottle in the short term or long term?

A don't see major problems in storing it in a PVC bottle,   PVC is not soluble in ethanol, neither in isopropanol. However, in this site they rate isopropyl alcohol as A while ethanol as C (with possible swelling in the long term).


When mixing lye and citric acid, which citrate will form? I'm trying to combine precise amounts of citric acid and lye to get sodium citrate with minimal amount of reacted acid/base. I'm not so sure how stochiometry works, but I could sure learn that from some web article.

What's the real problem is, that I don't know which of three sodium citrates will form. Depending on the number of $\ce{Na}$ molecules, the amounts of $\ce{NaOH}$ greatly differs.

Which forms and why?

Citric acid has three pKs: 3.1, 4.8 and 5.4.  If you add 3 equivalents of lye, you will get essentially only trisodium citrate.  If you add a lesser amount, you will have a mixture of the various citrates (at least from microscopic point of view), as well as some remaining citric acid if you add less than one equivalent.

One "equivalent" being one mole of lye per mole of citric acid.


Stereo Isomerism - syn/anti I was studying the types of stereo-isomerism E-Z, Cis-Trans and then I come across syn-anti stereo-isomerism this one is giving me a bit of fight in understanding, especially where lone pairs are taken into consideration.How do classify the molecules with lone pairs in consideration as syn and anti Please help me out here!

Use of the terms "syn" and "anti" to describe the geometry about double bonds is no longer encouraged, but you'll still come across it, particularly in older articles.  Syn and anti are identical to Z(usammen) and E(ntgegen) and were often used to describe the geometry about carbon-nitrogen double bonds. In such cases, the lone pair of electrons is given the lowest priority and the sequence rule applied as usual.  For example, in the following N-methyl imines, the imine on the left has the highest priority group attached to the carbon in the double bond (ethyl) on the same side as the highest priority group attached to the nitrogen in the double bond  (the methyl has higher priority than the lone pair). Therefore the molecule is the syn isomer (but would now be called the Z isomer).

By the same reasoning, the molecule on the right is the anti of E isomer.  Again, the key point is that the lone pair has a lower priority than any atom.


Gas densities and the atmosphere The density of nitrogen is greater than the density of oxygen, hence the denser nitrogen would be present in the lower atmosphere. If so it would be deprived of oxygen as it would be pushed to the higher atmosphere due to its low density. But we live due to the intake of oxygen from the lower atmosphere. How is this possible?

Gas molecules in the air continuously collide with each, shoot off in different directions etc. This constant motion is like stirring a fluid. You wouldn't observe perfect density separation in a container with oil and water in it if you keep shaking/stirring it. The situation is similar here, the atmosphere is a fluid that is continuously being agitated. I believe this to be the reason, at least off the top of my head.


Calculating equilibrium constants: adding two or more equations If two or more reactions are added to give another, the equilibrium constant for the reaction is the product of the equilibrium constants of the equations added.

How is this true? What is the reason?

This is a direct result from the law of mass action.

When you have an equilibrium

$$\ce{A <=> B}$$

with the equilibrium constant

$$K_1=\frac {[\ce{B}]}{[\ce{A}]}\tag1$$

and a second equilibrium

$$\ce{B <=> C}$$

with the equilibrium constant

$$K_2=\frac {[\ce{C}]}{[\ce{B}]}\tag2$$

you can combine both chemical equations:

$$\begin{align}
\ce{A \;&<=> B}\\
\ce{B \;&<=> C}\\
\hline
\ce{A \;&<=> C}
\end{align}$$

Then you can rearrange Equation $\text(1)$

$$\begin{align}
K_1&=\frac {[\ce{B}]}{[\ce{A}]}\tag1\\[6pt]
\Leftrightarrow [\ce{B}]&=K_1\cdot[\ce{A}]
\end{align}$$

and substitute $[\ce{B}]$ in Equation $\text(2)$

$$\begin{align}
K_2&=\frac {[\ce{C}]}{[\ce{B}]}\tag2\\[6pt]
K_2&=\frac {[\ce{C}]}{K_1\cdot[\ce{A}]}\\[6pt]
\Leftrightarrow K_1\cdot K_2&=\frac {[\ce{C}]}{[\ce{A}]}
\end{align}$$

Thus the equilibrium constant for the combined equilibrium is the product $K_2\cdot K_1$.


Is there any way of representing a dative/coordinate covalent bond? I was wondering if there was any way of representing a dative/coordinate covalent bond on a chemical diagram.

A single bond can easily be represented as a single line, a double bond as a double line and a triple bond as a triple line.

Take the below structure of ethene, a compound with single and double non-coordinate covalent bonds: each bond is easily represented by a single and double line.

However, the ammonium ion does have a coordinate bond between the lone pair of electrons on the nitrogen and the H+ ion from whichever compound donated it (e.g. hydrochloric acid). Despite this, I am unable to see how a coordinate bond could be represented, as, to me, it just looks like 4 non-dative covalent bonds.

Is there any way of representing a coordinate/dative covalent bond on diagrams like these?

Generally, you wouldn't describe the bonds in an ammonium ion as dative bonds. They are usually considered to be proper covalent bonds. The term "dative bond" is typically used for bonds in (transition) metal complexes between ligands and a metal center. But for those complexes Lewis structures have lots of weaknesses and are rarely a good representation of the actual bonding. Yet they are used for the sake of convenience. Often an arrow from the ligand to the metal center is used to indicate a dative bond.

(source: uni-erlangen.de)
But just as often you see normal "covalent bonds" (straight lines) like in traditional Lewis structures being used.
Often both kinds of representations are even intermixed.
For a more detailed view on the subject of bonding in coordination complexes and its representation have a look at this answer of mine.


Comparison of acidic strength between nitric acid and nitrous acid These are the structures of nitric acid and nitrous acid respectively:

To compare acid strength, we compare the stability of the corresponding conjugate base. Here conjugate bases are nitrate and nitrite anions respectively.

Now, my point is that in $\ce {NO2-}$, the resonance effect is pronounced, but in the case of $\ce {NO3-}$, we have a case of cross resonance. Put simply, in  $\ce {NO2-}$ the negative charge de-localizes to double bonded oxygen atom. In the case of $\ce {NO3-}$, same thing happens, but we have another $\ce {O-}$ here due to which there is a cross of 2 resonance effects (If I am using the correct terminology). So, resonance effect of nitrate ion should be weaker than nitrite ion and therefore nitrite ion should be more stable.

If I am true then nitrous acid must be a stronger acid, but all my sources suggest otherwise. They insist that in $\ce {NO3-}$, the negative charge is distributed over 3 oxygen atoms, and hence its resonance is more powerful. My confusion is that there are also 2 negatives that need to be stabilized as opposed to one negative in $\ce {NO2-}$.

So what concept am I missing here?

The correct terminology is "cross conjugation" and "extended conjugation" (but the latter makes more sense in the context of more than two pi bonds, as in penta-1,3,5-triene).

Now, you have correctly identified that the nitrate anion has cross conjugation while the other has simple conjugation. However, you've incorrectly concluded one of them to be more stable. The conclusion is incorrect because there is a much dominant factor at play instead.

The three canonical resonating structures of the nitrate anion are "equivalent" i.e. wherever the negative charge may delocalise, the new resonance structure is exactly the same as the original one. Therefore, all the resonance structures are "equivalent" in their potential energy. This leads to a much rapid delocalisation of the lone pair, between the three oxygen atoms, leading to a rapid charge dispersal.

Both nitrate and nitrite anions have equivalent resonating structures. But, in the former, the negative charge is dispersed over three oxygen atoms (instead of only two in nitrite). From this, we conclude that the nitrate anion is much more stable than the nitrite anion.

Therefore, the next time you apply the fact that cross conjugation imparts lesser stability than extended conjugation, do look out for this additional dominant factor of "equivalent resonance"!


Making a 10L HCl solution with 8.77% HCl in a 100 mL solution from a 16 M stock solution How would you make 10L of HCl solution with 8.77% HCl in 100 mL from a 16M stock solution.

Here's what I have tried so far:

I put my percent HCl and got that down to match with the moles of HCl I had. Then I took 1.754% of 20 ml and converted that into liters. Then I found the molarity of the HCl I need to turn the 16 M solution into. Finally I did a dilution calculation to find the volume of water needed to make the 16 M stock solution into the starting percentage of HCl.

I am fairly certain that I am diluting with water, I don't know what else it would be with.

Is this the right way to go about this, or do I need to restart. Any help is much appreciated, and thanked for in advance!

I figured out the problem I had. I needed a way to convert grams to milliliters, which was density.

I found the density of HCl for the 8.77% concentration:
$$ HCl \ Density = 1.04 g/mL$$
I then found how much HCl would be in 100g of solution:
$$ 8.77\% \ of \ 100g, \ so \ 8.77g \ HCl \ and \ 91.23g \ solvent$$
I did some dimensional analysis to get to moles divided by liters, so that I could get the molarity:
$$ (\frac{8.77 g \ HCl}{1})*(\frac{1 \ mol}{36.461g})*(\frac{1}{100g \ solution})*(\frac{1.04g}{1 mL})*(\frac{1000 mL}{1 L}) = 2.5 M$$
Then, I did a nice dilution calculation:
$$ 10 L *2.5M = 16M *L_2$$
$$ \frac{25}{16} = 1.5625 L$$

To do the dilution I would place 1.5625 L HCl into a beaker, and then fill with solvent until I reached a total solution volume of 10 L.


Steric effects of a lone pair and piperidine I know that in piperidine, the lone pair on the nitrogen can occupy either an axial or a equatorial position in the two chair forms. I also read that the axial position for the lone pair is preferred by a small margin in the gas phase (so intermolecular forces are minimized).

Now, this was presented as an argument against the idea that lone pairs are "sterically demanding" or "bulky" or repel other entities of negative charge more strongly than do bonding pairs, etc. The conclusion was that a lone pair must occupy less space than a hydrogen atom since it prefers the axial position in piperidine!

This argument seems correct at first, but could there be other factors affecting why the axial lone pair conformation might be more stable than the equatorial lone pair conformation? Also, are there any other examples in which the lone pair behaves "unexpectedly," at least with respect to elementary theory?

ETA: On the other hand one could make the argument that this isn't a apples to apples comparison; we are comparing a lone pair of electrons with a bonding pair of electrons plus a hydrogen atom. So it may be that this is a steric rather than an electronic (i.e. repulsive) effect.

Or could there be some sort of hyperconjugative effect going on? Carbon withdraws electron density from hydrogen; hydrogen becomes slightly partially positive, is attracted to lone pair?

The energy difference between the two conformations in your example is extremely small.  This makes it difficult to say that "this" is the reason why things are the way they are. Nonetheless, it is a very interesting question. Perhaps, like in the "how many angels can you draw on the head of a pin" question, the discussion around the question might be as significant as the answer.

What we need is more data.  The phosphorus analogue of piperidine, phosphacyclohexane, has also been examined and the hydrogen resides in the axial position (no evidence for the equatorial isomer was observed).

There are important differences between these nitrogen and phosphorus analogues.  To start with, phosphorus-carbon and phosphorus-hydrogen bond lengths (180-85 pm and 145 pm respectively) are typically longer than the analogous nitrogen-carbon and nitrogen-hydrogen bonds lengths (150 pm and 102 pm respectively).  More important is the fact that the phosphorus is basically unhybridized, so the lone pair of electrons around phosphorous exists in a non-directional s-orbital.  This makes conformational analysis of phosphacyclohexane easier because we don't have competing (hydrogen and lone pair) substituent effects to consider and dissect.

The observation that the hydrogen in phosphacyclohexane exists in the axial position suggests that there is an attractive 1-3 synaxial H-H interaction and/or a sterically repulsive 1-2 H-H gauche interaction.  Given the long length of bonds to phosphorous, repulsive interactions seem unlikely and the attractive 1-3 synaxial H-H explanation was favored.

Another data point comes with N-methyl piperidine where the lone pair strongly prefers the axial position.  This was attributed to a repulsive 1-3 synaxial Me-H interaction.  The fact that the methyl group adopts the equatorial position also shows that the repulsive 1-2 Me-H gauche interaction must be of little consequence and any repulsion in the corresponding H-H case would be even smaller.

Piperidine itself appears to be somewhere in the middle, in between N-methyl piperidine and phosphacyclohexane.  Certainly the repulsive 1-3 synaxial Me-H interaction observed in N-methyl piperidine has been reduced, but not eliminated.

Whether just the repulsive 1-3 synaxial H-H interaction has been reduced or whether there is now also a contribution from the attractive 1-3 synaxial H-H interaction seen in phosphacyclohexane cannot be determined.


What does 0.12 N mean on a bottle of sulfuric acid? I just bought a bottle of sulfuric acid in order to simulate acid rain for an experiment. I was wondering what $0.12\:\mathrm{N}$ on the bottle means. I need a simplified version of the explanation as I'm currently a freshman in high school taking biology, I have never taken a chemistry course before. I've read that the $\mathrm{N}$ means normal, but the explanation included a ton of jargon that I couldn't understand.

The time in question: does the $0.12\:\mathrm{N}$ mean concentration, as in $12\%$ concentration?

How would you guys suggest making simulated acid rain with this? My guess is that I would keep titrating the sulfuric acid into the water until the $\mathrm{pH}$ gets to my preferred $\mathrm{pH}$.

0.12N is indeed 0.12 normal. In short, there are 0.12 moles of hydrogen ions per liter. Since this is sulfuric acid, it is 0.06 moles of sulfuric acid per liter.

0.12 moles of hydrogen ions per liter would give a pH just less than 1. To obtain a pH near 3, you would need to dilute 1 mL of this solution to 100 mL. For a pH of 4, dilute 1 mL to 1000 mL.


Is *rinsing* (after washing) with cold water is as efficient as rinsing with hot water? Let say I have, a dishwashing soap, plenty of cold water (5 °C) and only have 2 L of hot water (60 °C) to clean a t-shirt with oil stains. Should I use the 2 L to wash the t-shirt and then the cold water to rinse the shirt or should I wash the t-shirt with only 1 L of hot water and then use the second liter for a first rinse and then the cold water to finish rinsing?

If the soap has already "captured" the fat molecule (in the micelles), I shouldn't need the hot water to rinse the micelles, right?

Depends on carry-over & solubility

Generally, when you have two cleaning fluids, one of which is less effective than the other (the "rinse"), you need to use the good solvent to get the contaminant concentration below the solubility limit of the poor solvent, before switching solvents. Otherwise, you will need a heck of a lot of the poor solvent (albeit that in this case it is only cold water.) How do you do that with only 2 L of good solvent (soapy hot water)?

This sort of question is formally studied in both industrial cleaning plants, and decontamination processes. Determining the ideal process is not exactly rocket science but it is fairly complicated.

It depends on the solubility of the contaminant in the washing fluid -- usually, the dynamic solubility, since processes like these can't sit around waiting for a long soak. However, even if the contaminant is perfectly miscible in the cleaning fluid, it still also depends on "carry-over": the fraction of cleaning fluid that remains on or in the cleaned object between rinses.

In the case where the contaminant is miscible and the carry-over fraction is very small, then you actually get the best results (theoretically, at least) from a very large number of very small rinses. But that won't apply in your case: a wet t-shirt is capable of holding quite a large fraction of 2 litres.

For more practical size of problem like yours, the optimal solution is challenging to find analytically, but straightforward to find numerically. For example, you can set up a simple simulation in (say) MS Excel and run the Solver to get lowest residual concentration against an input parameter of "fraction of hot water used for each wash."

Oh, and it simplifies things considerably to know: you get best result when each titre is the same. That is, you don't need to consider using, say, 2/3 of the hot water on a first wash and 1/3 on a second. If you have n washes, just use 1/n each time.

Incidentally, for many practical parameters, the solution is so sensitive to carry-over fraction that often the most economical approach is to minimise carry-over; in your case, give the t-shirt a really thorough wringing out between rinses!


Lewis structures of H₂(C)xCH₂ molecules I need to calculate the angles between CCC, CCH, HCH in the following molecules:

ethene ($\ce{H2CCH2}$),
propa-1,2-diene ($\ce{H2CCCH2}$), and
buta-1,2,3-triene ($\ce{H2CCCCH2}$).

If I'm not mistaken, the Lewis structures of all of these molecules have the carbons in the middle, connected by double bonds to each other, and the hydrogen atoms connected to the carbons in the edge. For example:

H           H
 \         /
  C = C = C
 /         \
H           H

The carbons at the edge are $\mathrm{sp^2}$ and the middle carbon is always $\mathrm{sp}$ .

This means that the angle CCC is always 180°. But how can I determine the angles CCH, HCH?

In addition, how can I determine if the hydrogen atoms in the three molecules lie on the same plane?

What are the bond angles in a normal trigonal planar arrangement? Treat the center atom as in H C H as a trigonal planar arrangement with the middle (carbon) atom in the center with a third atom attached on the left over position (another carbon). What angle maximizes their separation?

Your other question about the plane of the molecule is a bit more interesting. The pi bonds between the carbons ensure that the molecule is not rotating in space. To maximize the distance between the terminal hydrogen atoms, one set should be aligned up and down, while the other two on the opposite side should be rotated 90° in space, otherwise they would ellipse one another. In other words, one pair is horizontal while the opposite is vertical for example, depending how you want to imagine it.