text:
Conflict Between Organic and Physical Chemists--explaining the intensities of IR absorptions
I am currently a Chemistry undergrad and I love to fact check everything my O Chem professor says against my research advisor, a physical/computational chemist. Well, my O Chem professor, when discussing IR absorptions, said something which was a simplification for teaching purposes but which seemed rather unusual to me. He said that high intensity peaks are due to a large change in dipole (clealy true) which often corresponds to more electronegative bonds and then gave the example of carbonyl absorptions which are intense, or nitrile absorptions, or $\ce{CO}$ single bond absorptions, etc.
What he then said, and says he has no good explanation for, is the following:
The absorption of an $sp^3$ hybridized $\ce{CH}$ stretch is high-medium intensity.
The absorption of an $sp^2$ hybridized $\ce{CH}$ stretch is medium intensity.
The absorption of an $sp$ hybridized $\ce{CH}$ stretch is medium-low intensity.
Of course the hybridizations given refer to the hybridization state of the carbon to which the hydrogen is attached.
The reason he has no good explanation for this phenomenon is that in most other situations, an $sp$ hybridized carbon will behave as if it is more electronegative than an $sp^2$ or $sp^3$ hybridized carbon.
An example of this would be that $pK_a$ of an $sp < sp^2 < sp^3$. That is a characteristic of how electronegative they are in an acid-base sense.
Here, however, we see the opposite trend. That is, when the more electronegative stretch should have a higher intensity peak, it is weaker, and the less electronegative stretch has a high intensity peak.
For the record, the arbitrary assignment of more and less electronegative based on behavior isn't super satisfactory to me, and is probably why I'm digging deeper on this.
So, my advisor essentially said all that matters is this equation: $$f=\left\langle\psi_0\right| \mu\left| \psi_v\right\rangle^2$$
Basically, he said I can ignore the wavefunction details of the equation because the wavefunctions of the three stretches are not going to differ by much because they all have identical reduced masses and some other stuff I can't remember.
So, all that matters is the dipole moment change as the bond is displaced from its equilibrium position.
His physical explanation of the intensities described above is that there is a smaller dipole moment change for $sp^2$ and $sp$ hybrid carbons because the electron density is trapped in the double and triple bonds more so than in the single bond. That seems reasonable to me.
The Organic Chemist fires back, however. He says that he would expect intensity to increase with the double and triple bonds because the electron in pi bonds are in a higher energy state than the sigma bond electrons and are thus more easily displaced which ought to result in a larger dipole moment change. That also makes a lot of sense I think.
So, is my advisor correct or is there a more complicated explanation?
Sorry for being so long-winded, I just wanted to get past the simple explanations so that we can have a deeper, better answer.
Edit: A very good answer would also explain why $sp$ hybridized C-H stretches are sometimes very high intensity. Even as high as $sp^3$ stretches at times.

I have no idea what the correct answer is, but when your organic prof talks about the electrons in pi bonds being higher in energy, how is that relevant to the C-H bond itself (which of course isn't a pi bond)?

The idea is that the change in dipole has to do with the electron in that carbon-carbon bond, so having pi bonds ought to allow the electrons to be displaced easier as the hydrogen vibrates, so you  might expect a larger change in dipole and thus greater intensity, but that's not what happens.

@jheindel - Your adviser is correct. The IR spectrum intensity goes as the magnitude of the transition dipole moment, which in turn is reflected by the (induced) electronic dipole moment (your formula gives the intensity for the expectation value of the linear approximation of the transition dipole moment operator). I don't see how the hybridization argument holds water, especially when you're in the ground electronic state and are looking at roto-vibrational states.

Right I agree that he's right simply because that's what actually happens but I'm looking for a good explanation as to where the reasoning of my O chem Prof is wrong. Cause it seems to make sense to me when though it gives the wrong result.

The intensity is proportional to the transition moment but not quite as in your question. $\mu$ should be expanded as a series in x (displacement about equilibrium position) then $f=\mu\int \psi_0 \psi_vdx+(d\mu/ dx)_e\int\psi_0~x~\psi_vdx+ \cdot$. Due to wavefunction orthogonality the first term is zero so the intensity depends on the slope  the dipole has at the equilibrium geometry as the bond vibrates, not the absolute value. So this may cause the effect you are describing.

In considering the C-H stretch, let's consider that the C atom is attached to something else via an "immobile" bond(s).
For $sp^3$ hybridized the C would have 3 other bonds besides the immobile one.
For $sp^2$ hybridized the C would have 2 other bonds besides the immobile one.
For $sp^1$ hybridized the C would have 1 other bonds besides the immobile one.
For a non linear molecule there should be 3N-6 modes of vibration, but a linear on has 3N-5.
   Hybridization  # Atoms  Linear   Vibration Modes
       sp3            4       No         6
       sp2            3       No         3
       sp1            2       Yes        1

The more vibration nodes, the more microstates for the molecule in both its current state and its excited state. Thus having more microstates increases the probability that there are more transitions from the current state to some excited state.
For an analog on the microstate overlap, think about atomic spectra vs UV-VIS. Atomic spectra are line spectra, UV-VIS is band spectra.

Isn't it true, however, that many of those modes are vibrational modes involving the other atoms bonded to carbon and not just the hydrogen? Are you saying those vibrations increase the intensity of the C-H vibration because there is a change in dipole moment regardless of whether or not the C-H is the thing vibrating? I think I just need you to be a little more explicit in explaining the trend of intensities.

I'm not thinking dipole at all that explains the energy shift. // Think of a $\ce{=CH_2}$ group bonded to a wall. The various CH motions are shown here. https://en.wikipedia.org/wiki/Infrared_spectroscopy#Number_of_vibrational_modes But this is resonance stretching as if the molecule were in a continuous wave. A microstate would be the 3 atoms having some motion which would quickly damp out. In a real matrix the 3 atoms just aren't rock steady until a photon comes along. Does that make more sense?

Uhh... There's no damping out in molecular vibrations. The ground state of all vibrations has an energy greater than zero. Also, change in dipole most definitely has something to do with the absorption since that's what determines the intensity of a transition which is what IR shows... The microstate argument might have something to do with it, but I'm not convinced.

Also, the modes which you're referencing (bending, scissoring, etc.) are absorbed at a much lower energy than straight up C-H stretch. Bending modes are down below 1000 wavenumbers usually.

So if ethylene is in carbon tetrachloride then the ONLY movement of the atoms in the molecule has to be the "pure" modes like the ethylene is in a rarefied gas? // The point is that if the molecule was in a non-pure vibration mode then that motion damps out in a very short time. The absorption of a photon is must be fast. The photon isn't really a standing wave in which the atoms move in resonance. It is really more like a particle going 3E8 meters/sec. The gist is that in a solution the "allowed" vibrations last a time at least comparable to the time between intermolecular collisions.

Think of atomic spectra. "Forbidden" lines show up because atoms are not really individually isolated in free space. They are really bumping into one another creating all sorts of microstates in which the nucleus and electron cloud are recoiling from a collision.


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Convert from Pa to Pa to J/dm^3
Given the value 380 PaK^-1, how do I convert Pa to Jdm-3?
I know that 1 joule = 1 Nm, and that 1 pascal = 1 N/m^2.
But where does the dm^-3 (or L) come into this? I don't know where to start.

The conversion factor is 1.

Thank you, but where can I find workings to justify this?

Derived units are defined as products of powers of the base units. Some of the coherent derived units in the SI are given special names and symbols.
The joule (unit symbol: $\mathrm J$) and the pascal (unit symbol: $\mathrm{Pa}$) are such coherent derived units.
Expressed in terms of SI base units, one joule is
$$\mathrm{1\ J=1\ \frac{kg\ m^2}{s^2}}$$
and one pascal is
$$\mathrm{1\ Pa=1\ \frac{kg}{m\ s^2}}$$
(see: BIPM: The International System of Units (SI))
Hence,
$$\mathrm{1\ \frac J{m^3}=1\ \frac{\frac{kg\ m^2}{s^2}}{m^3}=1\ \frac{kg\ m^2}{m^3\ s^2}=1\ \frac{kg}{m\ s^2}=1\ Pa}$$
SI prefixes may be used in forming the decimal multiples and submultiples of units. Note that, when SI prefixes are used with SI units, the resulting units are no longer coherent, because a prefix effectively introduces a numerical factor.
$$\begin{align}
\mathrm{1\ m}&=\mathrm{10\ dm}\\[6pt]
\mathrm{1\ m^3}&=\mathrm{1000\ dm^3}\\[6pt]
\mathrm{\frac1{m^3}}&=\mathrm{\frac1{1000\ dm^3}}=\mathrm{0.001\ \frac1{dm^3}}\\[6pt]
\mathrm{1\ \frac J{m^3}}&=\mathrm{0.001\ \frac J{dm^3}}\\[6pt]
\mathrm{1\ Pa}&=\mathrm{0.001\ \frac J{dm^3}}\\[6pt]
\mathrm{1000\ Pa}&=\mathrm{1\ \frac J{dm^3}}
\end{align}$$

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I'm not sure I understand the stoichiometry of this mechanism
I was going through the experimental mechanism of nickelocene synthesis, and I looked through the notes and it said that there would be $\frac12$ $\ce{NiCl2}$.  I'm sort of confused by the $\frac12$.  It may be the fact I'm pretty sleep deprived and missing something.

It's just a matter of stoichiometry - since there are 2 Cp ligands for each Ni, you have to add 0.5 equivalents of Ni to the Cp ligand you already prepared. BTW, KCl2 doesn't exist, so you are probably pretty sleep deprived :p There's only one equivalent of Cl- in 0.5 NiCl2.

@orthocresol You are correct!   Yes, I am such an idiot! But it might be because I was very tired yesterday xD!  Thank you very much for the clarification!

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text:
Nucleophile substitution in nonclassical carbocation
In nucleophilic substitution of exo-2-norbornyl brosylate, carbocation formation is the rate determining step and the carbocation formed is: (in equilibrium)

After forming carbocation, the nucleophile will fill its electron on newly formed the vacant orbitals:

But this is not the product given in the book by FA Carey.
The given product is:

It's worth noting that the equilibrating structure shown in the first line was finally proved to be incorrect recently, after many years of debate, by successfully obtaining an x-ray crystal structure of the 2-norbonyl cation. See here for the paper.

What about the final product?


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What are some uses for an O2/N2 mixture that is more N2-rich than air?
What are some uses for an $\ce{O2}$/$\ce{N2}$ mixture that is more $\ce{N2}$-rich than air?
I need help with this, my professor did not do a good job explaining it, and I can't find a source to read about it.

Do they use it in scuba diving tanks?

They kinda do, only the other way around: they use oxygen-enriched mixture (before resorting to $\ce{He}$, but that's another story).

Nitrogen enriched air is used as an inerting gas in fuel tanks, (WARNING:  pdf), where the oxygen content is low enough that you don't get flammability.

Nitrogen is an inert gas. It does not react very easily. Conversely, oxygen is highly reactive. If there were more O2 than N2 we may expect to age more quickly, and also insects would be GIGANTIC. N2 is also responsible for forming NH3 with H2 in small traces. The Miller-Urey experiment proved that amino acids can form with sufficient energy in the presence H2, H2O, CH4, and NH3.

Some laboratories do use, and possibly some processes use "synthetic air", which contains 5% oxygen. This is to carry out oxidation reactions or processes but reduces the amount of excess oxygen to prevent unwanted combustion e.g. if flammable solvents are also being used. Many university research labs will use this if carrying out pressurised reactions.

I personally used custom $\ce{O2}$/$\ce{N2}$ mixtures in my Ph.D. work, both higher and lower in $\ce{N2}$ than air.  The application dealt with precisely controlled delivery of nitric oxide $\left(\ce{NO}\right)$ to aqueous solutions, to study how changing the $\ce{NO}$ concentration affects cell damage, cancer formation, etc.  In much of the body, the $\ce{O2}$ concentration is reduced below the air-equilibrated value, so it was important to design the system such that tests could be run with similar, lower-than-ambient $\ce{O2}$ concentrations.  I thus used nitrogen-enriched mixtures to hold the dissolved oxygen concentration steady at the target values.
For more detail, see my thesis here; particularly, Section 2.2.9 (p51) and Section 2.3.3 (pp60-65, including Table 2.3, Figure 2.6, and Figure A.5)

One potential use is in anything to do with fire suppression. Some low oxygen mixtures are breathable while suppressing common burning materials.
This may be a safer way to prevent harm from burning in places like laboratories and server rooms that some of the alternatives (freon fire suppression is dangerous to people, for example). Commercial systems to fill such locations with a low-oxygen atmosphere exist (see here). At oxygen levels below about 15%, things like candles and matches won't burn, but people can still breathe.


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Writing Mass Balance Equations
I was curious if there is a specific reason why when writing mass balance reactions we always leave a $\ce{H+}$ on the left side of the equation as the professor did not explain so, or is it just convention?
For example, write the mass balance equation of $\ce{H3PO4}$:
$$\ce{H2O <=> H+ +OH-}$$
$$\ce{H3PO4 <=> H+ +H_2PO4^{-}(aq)}$$
$$\ce{H2PO4- <=>H+ + HPO4^{2-}(aq)}$$
$$\ce{HPO4^2- <=> H+ + PO4^{3-}(aq)}$$

What do you mean by 'we always leave a $H^+$ on the left side of the equation'

It does not have to be done that way but that is how most people show it.

Those are not mass balance equations. The mass balance equation is $[\ce{H3PO4}] + [\ce{H2PO3-}] + [\ce{HPO4^2-}] + [\ce{PO4^3-}] = c_{\ce{H3PO4}}$.

I apologize I miserably attempted to write them in latex it did not work.  I think I still do not understand the general concept of mass balance, so it is something beyond what I would ask here.

Do you mean why the proton is "always" written as the first product? I am not aware of such a convention. As long as all products are there with the correct stoichiometric coefficient, it does not matter if the proton is written first, last, or somewhere in-between.

Well, I guess you can call these "dissociations of $\ce{H3PO4}$". Anyway, as Anders and Frisbee have said, there's absolutely no requirement that the proton be the first product.

When writing charge balance reactions we always leave an [$\ce{H+}$] on the left side of the equation as
$$[\ce{H+}]
= 3[\ce{PO4^3-}] + 2[\ce{HPO4^2-}] + [\ce{HPO4-}] + [\ce{OH-}] $$

There is no such convention explicitly telling what comes out first.
Here, for a traditional formula of phosphoric(V) acid $\ce{H3PO4}$ writing a proton first serves only a didactic purpose.
It's visually easier for students to keep a track on dissociation as the order of the elements both in formula and among the products is preserved.
On the other hand, shall one use a coordination formula of phosphoric(V) acid $\ce{[PO(OH)3]}$, it probably would make more sense to use a reversed order and put $\ce{H+}$ at the end:
$$
\begin{align}
\ce{[PO(OH)3] &<=> [PO2(OH)2]- + H+}\\
\ce{[PO2(OH)2]- &<=> [PO3(OH)]^2- + H+}\\
\ce{[PO3(OH)]^2- &<=> [PO4]^{3-} + H+}
\end{align}
$$
One more minor thing.
Equations are usually aligned about arrows, and a tabular array of products and reactants emerges when the same repeating compound ($\ce{H+}$) is shown first.
Note how easier it is to grasp the information when more compounds are aligned:
$$
\begin{align}
&\color{green}{\text{aligned}} & &\color{red}{\text{misaligned}}\\
\ce{H3PO4 &<=> H+ + H_2PO4^{-}(aq)} &\quad \ce{H3PO4 &<=> H_2PO4^{-}(aq) + H+} \\
\ce{H2PO4- &<=>H+ + HPO4^{2-}(aq)} &\quad \ce{H2PO4- &<=> HPO4^{2-}(aq) + H+} \\
\ce{HPO4^2- &<=> H+ + PO4^{3-}(aq)} &\quad \ce{HPO4^2- &<=> PO4^{3-}(aq) + H+}
\end{align}
$$

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Error in Titration
When doing a titration with a burette, the error of the burette is usually $\pm 0.05\ \mathrm {ml}$, as we have to read two times on the burette (initial/final vol), the error will be $\pm 0.07\ \mathrm{ml}$ ($\sqrt {(0.05)^2 + (0.05)^2}$)
When performing an titration, we usually do three determinations. The error of the titration is half the range of the three determinations.
However, What if I get values such as $14.2\ \mathrm{ml}$, $14.3\ \mathrm{ml}$, $14.2\ \mathrm{ml}$   or even three times the same volume. The error of the measure would be less than the error in the instrument. Which rule do I have to apply here?

Realted: Uncertainties after Averaging Trials

the error of the burette is usually ±0.05 ml

Well that depends on the quality of the burette and your technique. The "problem" with a burette is that the marking will be linear whereas the burette's diameter might not be constant.
I think that the burettes I last used were marked at the 0.05 ml mark, but you estimated between the marks. So you recorded to the 0.01 ml amount. It was easy to get at least ±0.02 ml on average for a reading.

I think the ±0.05 ml notion is from 20 drops to the ml. But you can dispense part of a drop by touching the inside of the flask to the tip of the burette.

if I get values such as 14.2 ml, 14.3 ml, 14.2 ml

Then you average and use the values calculated for the mean and the standard deviation of the measurements themselves. It is a "sanity check" to consider the precision of the titration procedure itself.
I suspect that your "problem" is that you're not carrying enough significant figures through your calculations. You only round on the the final results.
So start at 6.72 ml end at 20.92 yields 14.20 ml, not 14.2 ml.

If you get 14.20 ml, 14.20 ml, and 14.20 ml then I'd strongly suspect that your assertion that the precision of the titration is 0.07% is wrong.

The first titration I every did was with potassium permanganate to determine the equivalent weight of an unknown oxalate compound. My titration technique was horrible and I didn't have enough sample left to do more titrations. But I knew few oxalates were soluble, and the sample I had was obviously crystalline.  So I did some quick cation tests and found that the cation was NH4+ which has two waters. The prof wanted how I got the right answer to four significant figures with such a poor technique.

@user164612 -Well I added something about a triple reading that I don't will probable be useful to you. To get more specific I need to know how your burettes are marked and how many significant figures that you are recording.

I see you did add something but you are correct in asserting that it is the least helpful thing possible. It is entirely feasable to get the same result in titration three times (I had it happen to me back in the general chem lab course in the days) so just suspecting that some error is wrong does not help.

Both statements are unlikely. (1) I got three readings of 14.20, 14.20 and 14,20. (2) The error of the titration is +/- 0.07 ml. Remember that 14.20 is not equal to 14.2 in analytical chemistry.


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Why some compounds form different crystal structure at different temperature?
Lets take an example :-

Sodium superoxide has three different crystal structure, the marcasite
  structure at nearby 81 K, the pyrite structure between 196 K and 223 K
  and calcium carbide structure at room temperature.

There are many more compounds which have different structure at different temperature. If temperature is increased, bonds breaks, atoms scatters and decomposes into a different product (compound or element) . But this does not happen. Rather bond breaks, atoms rearranges to form a different structure of same compound.  Why does this happen?

You may just as well think it all did happen: bonds broke, atoms scattered, etc. After all, same compound with different structure is but a particular case of "different product".

Why do things melt or boil?

You are just entering into the field of Polymorphism. I couldnt find any source for Why polymorphism occurs but you can think about it as Energetically favorable structure at a particular temperature.
Related - Why polymorphs form?


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How to prevent fuel cell gas leak?
My research is fuel cell. I need to do a open system fuel cell, but in cathode side always have gas leak. Have any one have experience to tell me, how to prevent gas leak?

What gas is leaking?

Since it's the cathode where the leak is I presume it's the hydrogen you can't hold. Hydrogen is very hard to contain because of it's small molecules. I use plastic glass as its bubbler and a small hose to move it elsewhere. If it's your bubbler where the leakage is insulate it using silicon around the connections/openings.

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What is the difference between phase of matter and state of matter?
I am still confused on the differences between the phase of matter and state of matter


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Why do these two calculations give me different answers for the same acid-base titration?
In class, I performed an weak acid-strong base titration using commercial white vinegar and sodium hydroxide with the aim of finding the concentration of of ethanoic acid in the vinegar, and I used stoichiometric calculations as well as the $\mathrm{p}K_\mathrm{a}$ equation to find the concentration from the data I collected. However, these two methods are giving me different answers and I have no idea if I did something wrong in my calculations, or used the wrong method outright.
As a bit of background info, the label on the vinegar says it has a $\ce{CH3COOH}$ concentration of $5\%$, which according to my calculations (I measured the density of the vinegar to figure this out) is around $0.9\mathrm{~mol~dm^{-3}}$.

Here's the data from the titration:
Volume of $\ce{CH3COOH}$: $3.1\mathrm{~cm^3}$ (diluted with water to the $50\mathrm{~cm^3}$ mark for experimental purposes)
Concentration of $\ce{NaOH}$: $0.1 \mathrm{~M}$
Volume of $\ce{NaOH}$ added: $33.1\mathrm{~cm^3}$
Initial $\mathrm{pH}$ of $\ce{CH3COOH}$: $3.0$
Half-equivalence $\mathrm{pH}$: $4.6$

So, using the stoichiometric method, we have the formula
$$M_\mathrm{acid} V_\mathrm{acid} = M_\mathrm{base} V_\mathrm{base}$$
so,
$$M_\mathrm{acid}(0.0031) = (0.1)(0.0331)$$
$$M_\mathrm{acid} = \frac{(0.1)(0.0331)}{0.0031} = 1.067 \mathrm{~mol~dm^{-3}}$$
This answer is fairly close to the literature value, so I'm assuming it's accurate. (The percentage concentration works out to be about $6.1\%$.)

On the other hand, the $\mathrm{p}K_\mathrm{a}$ equation is $K_\mathrm{a} = \frac{[\ce{A-}][\ce{H+}]}{[\ce{HA}]}$
Since $\mathrm{p}K_\mathrm{a} = \mathrm{pH}$ at the half-equivalence point, the $K_\mathrm{a}$ is $10^{-4.6} = 2.5 \times 10^{-5}$; similarly the initial $\mathrm{pH}$ is $3$, so the initial $[\ce{H+}]$ is $10^{-3}$.
So,
$$2.5 \times 10^{-5} = \frac{(10^{-3})^2}{[\ce{HA}]-0.001}$$
Entering this into a calculator returns the value $[\ce{HA}] = 0.041 \mathrm{~mol~dm^{-3}}$.

So, why is the second answer so much smaller than the first? Have I used one of the equations incorrectly or made an error in calculation? Does one of the methods not work for the acid-base titration in question (and why would that be?)
I sincerely apologize for the messy formatting; I literally had to learn LaTeX from scratch just to post this question. Thank you so much in advance.

Not sure what your convention is but $dm^3$ and $cm^3$ are weird units to me. I would have used liters ($l$) and milliners ($ml$).

I reside outside of the U.S. where our education system is predisposed to the S.I. unit system rather than the metric system which uses liters and milliliters.

Part of the reason why there is such a large discrepancy is because in your second method of calculating $[\ce{HA}]$, you failed to take into account the fact that your vinegar was diluted from $3.1\mathrm{~cm^3}$ to $50\mathrm{~cm^3}$.
Effectively, if the original concentration of $\ce{HA}$ was $x$, then your dilution would have brought it to a concentration of $\frac{3.1}{50}x$. And you plugged the $\mathrm{pH}$ of this diluted solution into the the $K_\mathrm{a}$ equation. Let's adjust for that dilution:
$$\begin{align}
\frac{3.1}{50}x &= 0.041\mathrm{~mol~dm^{-3}} \\
x &= 0.661\mathrm{~mol~dm^{-3}}
\end{align}$$
As you can see, there is still some difference between the two values. However, you can calculate the theoretical $\mathrm{pH}$ of a $1.067\mathrm{~mol~dm^{-3}}$ solution of $\ce{CH3COOH}$, using the fact that the $\mathrm{p}K_\mathrm{a}$ is $4.6$. Here, I will make the simplifying assumptions that the final and initial concentrations of $\ce{HA}$ are the same, i.e. $[\ce{HA}] = c_{\ce{HA}}$, and also that $[\ce{H+}] = [\ce{A-}]$:
$$\begin{align}
\frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]} &= 10^{-4.6} \\
&= 2.511\times 10^{-5}\mathrm{~mol~dm^{-3}} \\
[\ce{H+}] &= \sqrt{(2.511\times 10^{-5})(1.067)}\mathrm{~mol~dm^{-3}} \\
&= 0.00518\mathrm{~mol~dm^{-3}} \\
\mathrm{pH} &= 2.28
\end{align}$$
You could go all the way and do mass-balance and charge-balance equations in order to avoid making that assumption, but I doubt it alters the result significantly. So, if you really had a concentration of $1.067\mathrm{~mol~dm^{-3}}$, your observed initial $\mathrm{pH}$ would not have been $3.0$. There is likely some experimental error present, or perhaps the $\mathrm{pH}$ meter that you used to measure the initial $\mathrm{pH}$ is not calibrated well.

For the first point: LOL derp - I can't believe I missed that! I see, that makes perfect sense now.

As for the pH meter being calibrated incorrectly - this is certainly a plausible reason for the discrepancy, but I also performed  the titration using phenolphthalein indicator and obtained a very close equivalence volume of $32 \mathrm{cm^3}$. Is this to be expected even if there is an experimental error with the pH meter?

Also, due to the fact that the vinegar was titrated about $16$ times, your equation gives me a theoretical pH of about $2.8$ to $2.9$. Although, yes I believe this could be due to a slight calibration error in the pH meter.

"Also, due to the fact that the vinegar was titrated about $16$ times, your equation gives me a theoretical pH of about $2.8$ to $2.9$." What do you mean? As for your earlier question, I don't know what else might have gone wrong. It might just worth repeating if you could do so. However I would guess the most likely case is the pH meter is off.

Sorry, I meant that I think in calculating the theoretical pH the concentration you should have used was $0.669$ (that's $1.067/16$), which corrects for the dilution, unless I'm very much mistaken there.
Oh, and I meant 'diluted', not 'titrated'. That was a typo.

How is $1.067/16 = 0.669$?

Well, I just want to be clear, I guess. 1) Your titration calculation is absolutely correct. The answer from there is 1.067 M. If you divide by 16 again, you are counting the dilution twice. 2) If the pH measurement was accurate, and you accounted for the dilution, you would have gotten 0.661 M. 3) The second part of my answer is just a way that I would recommend to help you whenever you receive wildly different, or unexpected results like this. You take one result, in this case, 1.067 M, and you do the second method "backwards" to see whether your two methods are logically consistent (...)

(...) So in this case, a concentration of 1.067 M means that your original solution would be pH 2.28. Since this is not the original pH you obtained, it means that something is wrong with your experimental values (i.e. the initial pH measured), and not your calculations. || Why are you counting the dilution twice? It's because your titration calculations are done using the volume of the concentrated solution (3.1 cm$^3$), so in effect, you are measuring the concentration of the concentrated solution. However, for the pH measurement, you measured the pH of the diluted solution.


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Difference in color between the chromate and permanganate ions
As the title states, I'm wondering why the $\ce{[CrO4]^2-}$ and $\ce{[MnO4]-}$ ions display different colors in solution. Both of the metal ions have a $\mathrm{d}^0$ electron configuration, have the same ligands, yet their respective absorption spectra max out about $100\mathrm{~nm}$ apart, with the chromium complex absorbing at the shorter wavelength of $420\mathrm{~nm}$, and the manganese absorbing at $526\mathrm{~nm}$.
What causes this difference?

http://chemistry.stackexchange.com/questions/39829/how-can-the-intense-color-of-potassium-permanganate-be-explained-with-molecular/39848 Put very simply, the energy gap $\Delta E (= h\nu)$ between the $e$ and $t_1$ orbitals is different. And this difference stems from the fact that the "$\ce{Cr^6+}$" ion has d orbitals that are higher in energy than those of "$\ce{Mn^7+}$".


text:
How to determine bond order for special cases?
In AP Chemistry, we learned two ways to determine bond order. The first method is: Bond order = (bonding electrons - antibonding electrons)/2. The second method is: number of bonds / number of bond locations.
Applying the first method for the bond order of CO gives an answer of (8-2)/2 = 3. Using the second method, CO has one triple bond which gives an answer of 3/1 = 3.
Both methods also yield the same answer for CO^2+. One important characteristic of this bond is that carbon and oxygen make a double bond instead of a triple bond because the CO molecule is electron deficient. Since there is one double bond and one bonding location, the answer is 2/1=2. The result is the same for the other method as well.
In CO^+, the methods give different results. The first method gives an answer of (7-2)/2 = 2.5. I am unsure of the exact structure of the CO^+. Using a triple bond and leaving either the oxygen electron deficient by one or the carbon electron deficient by one gives an answer of 3. I am not sure how else the molecule can be depicted, perhaps with a delocalized bond somehow? I do not know much about delocalized bonds, so I do not know their rules and limitations. Am I incorrectly drawing the structure and is there a way to draw it so that it can satisfy the second method? Feel free to make edits on my formatting, I am new to this.

Your second method is essentially valence bond theory. If you used a valence bond approach to describe the bond in $\ce{CO+}$ you would probably have to invoke a number of resonance structures, for example: http://i.imgur.com/2ZCkFOx.png There are more possibilities e.g. putting a 2+ charge on oxygen and a 1- charge on carbon but that's kind of pointless. So, if you just take an "average" of those two resonance structures, you get a "bond order" of 2.5.

Could you add some paragraph breaks to your post to make it easier to read? Also, you mention nitrogen somewhere but you are talking about carbon monoxide ions aren’t you?


text:
Which type of orbital, 4s or 3d, is closer to the nucleus of an atom?
In my textbook it is explained that 4s electrons show greater penetration than that of 3d electrons, and, thus, the energy of the 4s orbital is lower than that of the orbitals in a 3d energy sublevel. Here is what my textbook (General Chemistry Principles and Modern Applications, Tenth Edition) says exactly:

"For example, because of the extra penetration of a 4s electron over that of a 3d electron, the 4s energy level is below the 3d level despite its higher principal quantum number n..."

I took this to mean that 4s electrons are closer to the nucleus than that of 3d electrons. Is this true?
Furthermore, after doing research on why electrons are removed from the 4s orbital before the 3d orbital (a topic that still confuses me) I ran into a source that seemingly contradicts what my textbook says:

"When you talk about ionisation energies for these elements, you talk in terms of the 4s electrons as the outer electrons being shielded from the nucleus by the inner 3d levels."

From: http://www.chemguide.co.uk/atoms/properties/3d4sproblem.html
This sounds like it contradicts my textbook. Perhaps it does not. However, it does raise a few questions (one of which I posed earlier). Does the term "penetration" mean that the 4s electrons are closer to the nucleus than the 3d electrons? If not, what does it really mean? Also, what distinction is there between the aufbau process and ionization regarding differences in energies between 4s and 3d orbitals?

I found the answer to my question here: https://chemistry.stackexchange.com/questions/33297/why-do-3d-orbitals-have-lesser-energy-than-4s-orbitals-in-transition-metals/33301#33301


text:
Why does burning wood produce methane and other hydrocarbon chains?
This post kinda helps me understand and it sounds like it has to do with pyrolysis, but I'm having trouble putting the exact reactions together.

"Wood" is not a chemical. It is a mixture of thousands of chemicals. To detect all of the chemical products when wood burns would take lifetimes of work. Wood is generally stated to burn to water, carbon dioxide, and ash but that is a vast oversimplification.

Under the proper conditions, the hydrocarbons present in wood are completely oxidized to $\ce{CO2}$ and water upon combustion.  In fact, this happens every day in analytical chemistry laboratories for example: a sample of wood is dried, ground and placed in a combustion chamber of oxygen with a combustion catalyst at ~$\mathrm{950^oC}$. The evolved $\ce{CO2}$ and $\ce{H2O}$ (along with other gases and ash) from the combustion can then be analyzed to quantitatively determine how much carbon and hydrogen were originally in the wood.
Now lets consider a campfire.  Do you dry and grind your firewood, mix it with a catalyst and apply a heat source of $\mathrm{950^oC}$ under an $\ce{O2}$ atmosphere? Of course not, so why do we go through all these rigors to get a quantitative burn?  So that it all gets burned, no partial combustion allowed. And partial combustion is the key to most of the unburned hydrocarbons that you refer to. Parts of the wood are barely hot enough to burn at all, but still can thermally break down the hydrocarbons into shorter, volatile organics which are are carried away by wind or convection from the fire before the complete combustion to $\ce{CO2}$ and water can be completed.
You also mentioned pyrolysis in your question, and the answer given in your link is certainly relevant here, and it represents one of the processes that I'm talking about. For pyrolysis, you need an oxygen-poor environment.  Going back to the campfire scenario, this largely describes the situation down in the smouldering embers on a calm night.  It's very hot, there is plenty fuel, but no process bringing in sufficient oxygen for combustion. Under these conditions, thermal decomposition rather than just partial oxidation plays a major role in producing unburned hydrocarbons.  Think about methane for example; it is certainly not partially oxidized,


text:
Is readily available ammonia a substitute for ammonia water?
I am an artist attempting to duplicate a recipe for a painting medium.  The source says I can use powdered ammonia or ammonia water.  The recipe is from the 1940-50 and I am no chemist.
So, my question is can I use the readily available ammonia in place of "ammonia water" ... are they the same thing?

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. While it may seem strange at first we do not use salutations in our Questions and Answers. Please use the upvote button to say thanks, or leave a comment how the post can be improved.

Ammonia at standard temperature and pressure is a colourless gas. So when your recipe says "powdered ammonia" there must be some kind of mistake. It might refer to a salt, that can be produced with ammonia and hence serve as a source of it. One of such salts is ammonium chloride.
In household stores available ammonia is a solution of the gas in water. Depending on the product there might be other ingredients as well, for example surfactants if used as a cleaner. In any case there should be some information about the concentration of the solution. Often given in percent and it strength may vary, but I doubt it contains more than 10% in any case.
In conclusion, the available ammonia is ammonia water.
Lastly a word of warning. As an artist you are probably handling paint thinners and other chemicals all the time. Treat ammonia just as carefully as these. And keep your place well ventilated. You will probably do this anyway, because the smell is just horrible.


text:
Group 15 elements reactivity towards water
Why only tin in group 14 elements react with water and others do not?

Under what conditions? Steam oxidation is used in silicon integrated circuit production (well, probably not on leading-edge CMOS anymore).


text:
What is a Sodium wire?
I read in a paper:

[...] we prepared the $\ce{PhTeNa}$ in situ from $\ce{(PhTe)2}$ and sodium wire in THF.

Can anyone explain me this procedure (the mechanism) focusing on the definition of "sodium wire"?

Sodium is normally stored as cubes under petrol or similar solvent. As a metal it is very soft, you can easily cut it with knife. For the reaction you need fresh reactive surface of the metal, so the wire is one of best choices. For lab-scale preparation see e.g. https://www.youtube.com/watch?v=btvQdW4fXJE

In case one doesn't have the equipment required, it is quite easy to use a glass plate, a glass tube, a bit of vaseline oil and a knife to make curls of sodium foil that are just as good in many cases.

I have never used sodium foil as curls, if I want a large surface area of sodium I tend to make sodium sand. My normal method is to stir up sodium violently in boiling hot toluene and then turn off the heating while allowing it to stirr while cooling. This works nicely for me.


text:
Does storing a hygroscopic reagent in a bottle makes it prone to absorb humidity from the air trapped inside?
I've got a full bottle of 99,5% acetone, which I'd really like to keep it as dry as possible, though without using molecular sieves or other dessicants.
Once I open the bottle for the first time, use part of the reagent (say 25% volume) and then close the bottle again, will the remaining 75% acetone begin to absorb water from the 25% trapped air inside the bottle?
Also, does vapor pressure plays a role on this?

Due to acetones low boiling point it can easily be distilled again  prior to use. The significant  difference between the b.p of acetone and water makes them easy to seperate in this way and produces a pure distillate if done correctly.

Yes over time the acetone will absorb moisture from the air (humidity) inside the bottle.

Does storing a hygroscopic reagent in a bottle makes it prone to absorb humidity from the air trapped inside?

Yes, there is that tendency. I'll make a couple of points which I'll number for possible further discussion.
(1) 99.5% acetone is probably 0.5% water. So the "extra" water from the air above the acetone in the bottle (from one opening and closing of the cap) would be negligible compared to the amount of water already in the acetone. Obviously when using the acetone, uncap it and recap it ASAP to keep it as dry as possible. The "problem" is opening and closing the bottle many times.
(2) I looked quickly at Wikipedia, and do a quick a google search and didn't find any water-acetone azeotrope. So the distillate would be "pure".
(3) I'm not sure how pure a simple distillation in air of the acetone would be in removing water. I'd guess that the resulting purity would be like your 99.5% starting purity.
(4) Vapor Pressure would play a role in that the acetone would have a partial pressure in the bottle. So not all of the void in the bottle would be air. Acetone doesn't "boil" at room temperature though so the partial pressure won't be enough to keep some air from seeping in and out of the bottle. In other words the void in the bottle is mostly air with just some acetone.
(5) The !@#$%^&*( bottles "breath" as the air pressure changes if they are not absolutely sealed.

I did some more research and also got some numbers: density of water in air at room temp. and 100% relative humidity is around 20mg/L. That'd mean, for my example, that 250ml of air "contains" around 5mg of water, which is ~0.0008% in weight of the remaining acetone, totally negligible compared to the starting 99.5% purity. You could add this information to your first point for completeness. Thanks! ;)

That of course is a "one time exposure" as you open and close the bottle. The other thing is that the !@#$%^&*( bottles "breath" as the air pressure changes if they are not absolutely sealed.


text:
Order of citation of alkyl and halide prefixes in IUPAC names
I've been playing around with ChemDraw and found that it assigns the priorities of substituents in a way I wouldn't have expected, depending on the length of the alkyl substituent. I'm not a nomenclature buff, so I'd like to know if it's correct and if so, why.

$n = 1$: 1-ethyl-2-iodobenzene
$n = 2$: 1-iodo-2-propylbenzene
$n = 3$: 1-butyl-2-iodobenzene
$n = 4$: 1-iodo-2-pentylbenzene

What determines the priorities of the alkyl group and the iodo group?

I figured it out! It's according to the alphabet!

Related: http://chemistry.stackexchange.com/q/31709/7951

You correctly figured out that the prefixes that represent the substituent groups are arranged alphabetically. However, note that the order of citation and the numbering are laid down in two different rules.
With regard to the alphabetical order of citation of substituent prefixes, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.5 ALPHANUMERICAL ORDER
Alphanumerical order has been commonly called ‘alphabetical order’. As these ordering principles do involve ordering both letters and numbers, in a strict sense, it is best called ‘alphanumerical order’ in order to convey the message that both letters and numbers are involved
Alphanumerical order is used to establish the order of citation of detachable substituent prefixes (not the detachable saturation prefixes, hydro and dehydro), and the numbering of a chain, ring, or ring system when a choice is possible.
(…)
P-14.5.1 Simple prefixes (i.e., those describing atoms and unsubstituted substituents) are arranged alphabetically; multiplicative prefixes, if necessary, are then inserted and do not alter the alphabetical order already established.

Therefore, the alphanumerical order in your examples is correct:
x-ethyl-y-iodobenzene
x-iodo-y-propylbenzene
x-butyl-y-iodobenzene
x-iodo-y-pentylbenzene
The relevant rules concerning the numbering of locants for substituent prefixes are:

P-14.3.5 Lowest set of locants
The lowest set of locants is defined as the set that, when compared term by term with
other locant sets, each cited in order of increasing value, has the lowest term at the first
point of difference; (…)

and

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

Note that Rule (f) takes precedence over Rule (g).
In accordance with Rule (f), your first example could be named as 1-ethyl-2-iodobenzene as well as 2-ethyl-1-iodobenzene since both names correspond to the locant set ‘1,2’. However, according to Rule (g), this example is named as 1-ethyl-2-iodobenzene rather than 2-ethyl-1-iodobenzene since ethyl is cited first as a prefix in the name.
Hence, the numbering in your examples is correct:
1-ethyl-2-iodobenzene
1-iodo-2-propylbenzene
1-butyl-2-iodobenzene
1-iodo-2-pentylbenzene


text:
Why does the Reimer-Tiemann reaction give an ortho product?
On treating phenol with chloroform in the presence of sodium hydroxide, a $\ce{-CHO}$ group is introduced at the ortho position of the hydroxy group. Why is this so? Why is 4-hydroxybenzaldehyde not obtained?

I know this might seem unrelated, but do you know about the selectivity in the Kolbe-Schmitt reaction? Treatment of phenol with NaOH and carbon dioxide also gives the ortho product, salicylic acid, exclusively.

The Reimer-Tiemann reaction generally proceeds in poor yield.  In the case of phenol, both the ortho and para products are formed in the expected statistical ratio of 2:1 respectively (20% : 10% isolated yield). See this earlier question and the link provided in my answer.


text:
How many peaks would there be in H1 NMR of ethene?
Since the bonds can't rotate, would there be 2 peaks? Likewise, how many peaks would there be in trans-but-2-ene and cis-but-2-ene?

You need to consider whether the protons in ethene are chemically equivalent, magnetically equivalent, or both. The lewis structure of ethene:

Because of the symmetry of the planar ethene molecule, all H atoms are chemically and magnetically equivalent. The spin system is therefore A$_4$. The signal is not split because the environment is the same for all H atoms. The signal would therefore be a single peak. Compared to an "average" methyl group, the ethylene protons are less shielded, and would appear at a higher chemical shift than methyl groups.


text:
Implications of orthonormal wavefunctions
For orthonormal functions $\psi$ and $\phi$, by definition;
$$\langle \psi|\phi\rangle=0$$
However, when a wavefunction, $f$, is expressed as a linear combination of these orthonormal wavefunctions:
$$f=c_1\psi+c_2\phi$$ operating on this function with an arbitrary operator, $\hat A$ results in:
$$\langle f|\hat A|f\rangle=\langle c_1\psi+c_2\phi|c_1\hat A\psi+c_2\hat A\phi\rangle=c_1^2\langle\psi|\hat A|\psi\rangle+c_1c_2\langle\psi|\hat A|\phi\rangle+c_2c_1\langle\phi|\hat A|\psi\rangle+c_2^2\langle\phi|\hat A|\phi\rangle$$
My question is, do the crossed terms equal zero? (i.e does $\langle\psi|\hat A|\phi\rangle$ and $\langle\phi|\hat A|\psi\rangle$ equal zero?)
I initially thought that, yes, they do equal zero because, suppose that $\psi$ and $\phi$ are eigenfunctions of the operator $\hat A$ then it follows that: $$\langle\phi|\hat A|\psi\rangle=\langle\phi|a|\psi\rangle=a\langle\phi|\psi\rangle=a(0)=0$$
However, in a problem sheet I have just done, the question relies on these terms not being zero. Is my logic wrong or does this only equal zero in the case where the functions are eigenfunctions of the operator?

What's the exact problem? I think your logic is right, if $\hat{A}|\psi\rangle \neq a|\psi\rangle$ then the term is not equal to $0$.

The problem I was set isn't massively relevant I just want to know if $\langle\psi|\hat A|\phi\rangle=0$ or not. Are you saying that it is non zero if the function is not an eigenfunction?

It could be zero, but there's no reason it should be if the operator is arbitrary.

If the $|\phi|\rangle$ and $|\psi\rangle$ are eigenfunctions of $\hat A $ and are orthogonal, then you are right: $\langle\phi|\hat A|\psi\rangle=\langle\psi|\hat A|\phi\rangle = 0$.

And if they're not? Assuming that the operator is hermitian (if that makes a difference)

The eigenfunctions of a hermitian operator are always (or can always be chosen to be) orthogonal (even if there are degenerate eigenvalues).

Let's assume that $\phi$ and $\psi$ are eigenstates of some hermitian operator $\hat{B}$. This operator will generate a complete basis set of eigenfunctions, which includes $\phi$, $\psi$, and infinity more wavefunctions, which we'll call $|f_1\rangle$, $|f_2\rangle$ and so on. Now, if we denote $A|\psi\rangle = |g\rangle$, then you can expand $|g\rangle$ in the $\hat{B}$ basis: $|g\rangle = a|\phi\rangle + b|\psi\rangle + c_1|f_1\rangle + c_2|f_2\rangle + \cdots$ Only if the value of $a$ in this expansion is $0$, will $\langle\phi|\hat{A}|\psi\rangle$ be equal to $0$.

Now, unless the operators $\hat{A}$ and $\hat{B}$ commute, the eigenstates of $\hat{A}$ will not be the eigenstates of $\hat{B}$ (even if they do, there is an additional requirement that there is no degeneracy). Therefore, there's no reason to assume that $|g\rangle = A|\psi\rangle$ is an eigenstate of $B$, and therefore no reason to assume that $\langle\phi|\hat{A}|\psi\rangle = 0.$

Ah, @orthocresol has a good point. I thought that the wavefunctions are supposed to be eigenvectors of $| \hat{A} \rangle$. But if they are not then $\langle\phi|\hat A|\psi\rangle=0$ is not generally true.

How about making this discussion into an answer?

If you have two orthonormal states $|\psi\rangle$ and $|\phi\rangle$,
$$
    \langle \psi|\phi\rangle = 0,
$$
there is no guarantee that the term
$$
    \langle\psi|\hat{A}|\phi\rangle
$$
vanishes. All depends on the action of $\hat{A}$ on your state.
As you already noticed, if $|\phi\rangle$ is an eigenstate of $\hat{A}$ with eigenvalue $a_\phi$, the term is obviously zero:
$$
    \langle\psi|\hat{A}|\phi\rangle = \langle\psi|a_\phi|\phi\rangle =a_\phi \langle\psi|\phi\rangle = 0.
$$
But this is only a special case, since $|\phi\rangle$ is not necessarily an eigenstate of $\hat{A}$ in general.
One of the simplest situations where this term does not vanish can occour  when $|\psi\rangle$ and $|\phi\rangle$ are eigenstates of the harmonic oscillator, that we can denote $|n\rangle$ and $|n'\rangle$. As you probably know, these states are orthonormal:
$$
    \langle n|n'\rangle = \delta_{n,n'}.
$$
If you consider the specific case $|\psi\rangle=|n\rangle$ and $|\phi\rangle=|n-1\rangle$, it immediately follows that
$$
    \langle n|n-1\rangle = 0.
$$
Now if you take $\hat{A}= \hat{a}^\dagger$, for which you have $\hat{a}^\dagger|n\rangle=\sqrt{n+1}|n+1\rangle$ ($|n\rangle$ is not an eigenstate of $\hat{a}^\dagger$), you have ($n\geq1$)
$$
    \langle n|\hat{a}^\dagger|n-1\rangle = \langle n|\sqrt{n}|n\rangle = \sqrt{n} \langle n|n\rangle = \sqrt{n}\delta_{n,n} = \sqrt{n} \neq 0.
$$
As I already stated, all depends on the action of $\hat{A}$ on your state.

Nice explanation, but perhaps more importantly, nice example!


text:
Do free radicals re-arrange?
For example reaction of but-1-ene in the presence of $\ce{Br2}$ and light should give 1-bromobut-2-ene which is thermodynamically more stable than  3-bromobut-1-ene on the basis of number of alpha hydrogen.

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@MaxW You can edit incorrect naming, you know. Also the correct nomenclature is but-1-ene, not 1-butene.

@user22443, what actually is your question?

the reaction comes under substitution by free radical mechanism , in presence of peroxide + light or just light , halogens show this reaction in particular, with the exception of fluorine and iodine( since backward reaction is favorable). I chose bromine since it is more selective( w.r.t this reaction). Reaction is initiated by homolysis of halogen. halogen free radical takes away hydrogen of  sp3 carbon. after which the free radical substrate takes halogen free radical . Reaction is similar to that by N-bromo succinamide which takes place in allylic system.

@Jan My teacher says that free radical intermediates do not rearrange like carbocations do, but the reaction i mentioned above involves free radical rearrangement as it leads to a more stable product as compared to the product obtained by no re arrangemnt

LOL - I wonder if the nomenclature changed. I surely think it was 1-butene when I took organic chem.

If I am not wrong, for this reaction to be carried out, you have to keep the concentration of $\ce{Br2}$ extremely low to avoid forming the unwanted addition product. It's for that reason that NBS is often used: in combination with $\ce{HBr}$, it maintains a steady, but low, concentration of $\ce{Br.}$ Anyway, in carbocations, the mechanism of shifting is called a "hydride" shift: basically, the hydrogen is taking both of the electrons in the C-H bond together with it. This can't happen in radicals as it would result in a 9-electron carbon. I'm not sure if a hydrogen radical shift can occur.

It is a matter of timing. The "normal" reaction between a carbon double bond and Br2 would be to brominate both carbons. But if UV light breaks Br2 into 2 Br atoms (radicals) then the Br "plucks" a H from the hydrocarbon giving a free radical. The hydrocarbon free radical rearranges, then sometime later finds a Br2 molecule and forms Bromohydrocarbon and yielding another Br atom.

@MaxW since 1993, the IUPAC name is but-1-ene.

@ortho seal: Wagner-Meerwein rearrangement ;)

@user22443 Please add that relevant information to the question using the edit button.

@Jan Is that not a "normal" alkyl migration? Same as a hydride shift just that it's C-C instead of C-H. I mean alkyl migrations can certainly happen in carbocations too (Friedel-Crafts alkylation is the oft-cited example) but I ran out of space in that comment ._.

@Loong - I took organic chem in the summer of 1971 so you made me feel a bit better. I haven't done any chemistry in 20 years.

@orthocresol yes a hydride shift is not possible but is this kind of double bond shift allowed?

@user22443 Oh, you're right. I just realised it's not about migration of hydrogen... just the double bond shifting. Sorry.. And yes, of course it is possible. http://i.imgur.com/5iuQbxM.png See this question. http://chemistry.stackexchange.com/questions/16826/alllylic-bromination-of-1-butene


text:
Does carbonated water change over the course of several days, or was my container leaky?
I have a Klean Kanteen flip-top growler, allegedly leakproof at up to 35psi of pressure, that I filled with some freshly-carbonated water 4 days ago — just to see how long it would last. I used a Soda Stream for carbonation. The bottle was filled to the brim.
When fresh, the carbonated water fizzed and sparkled. You could see the bubbles constantly rising to the surface. It felt like pop rocks when you put it in your mouth.
Now, 4 days later, the water is still clearly carbonated, but it's a lot more subdued. You can taste the fizz when you drink it, but there's no more surface bubbling. When you pour it into a cup, you see bubbles for about a second before the water clears up.
The obvious theory is that some CO2 escaped over the course of the week, but I'm fairly certain there were no liquid leaks. I also couldn't hear any hissing. The bottle was stored on its side in the fridge.
Is it possible for a container to be liquid leak-proof, but not gas leak-proof?
If not, does the nature of carbonated water change over time? Does the water, perhaps, absorb more of the CO2 the longer it sits? If so, why does this not happen with soda cans and such?

Update 2015-11-3:
I repeated the experiment with a 20oz screw-top Klean Kanteen I had lying around, ensuring that the water was as close to the cap as possible and that the lid was screwed on as tightly as possible. The results after 4 days were more carbonated than my initial trial with the flip-top Kanteen, though still not as much as fresh. Currently repeating the experiment once again with the flip-top, this time making 100% sure that the water is touching the lid.

What temperature was the water when you carbonated it? What temperature was the water when you opened the container?

Going in, it was chilly, but not ice-cold. The bottle was stored in the fridge and was fridge-temperature when opened. I can measure the exact temperature if necessary.

No, I don't think that will be necessary. You say the container is filled the the brim, but was there any air space between the brim and the cap?

It's likely there was a small amount (half a centimeter?), but I tried to make it tight.

The difference is in what is leaking. Gas leaks out a lot easier than fluid. The container is rated not to leak liquid, not that it won't leak air. So if you turned it upside down with 35 psi inside (a carbonated drink) the fluid won't leak out.
So turn the Klean Kanteen flip-top growler upside down. ;-)

Hey, now that's an experiment! I agree that gas will leak much easier than the liquid, but I suppose if it has to also diffuse out of solution, that might make the difference. It's rather a trick to get a good leak tight seal for small gas molecules. You could also try to find an analytical balance and weigh before and after. Carefully. Several times. Just for kicks.


text:
How can we produce gasoline from electricity without fossil fuels?
The other day, I was wondering, if cheap and easy nuclear fusion were developed tomorrow and fossil fuels disappeared, would there be a way to fuel all the gas-powered cars out there? After all, we have the Haber-Bosch process for producing ammonia. But then gasoline is not a chemical but a mixture of different organic compounds. So, let me structure a question to get at the heart of my curiosity:
How might we produce gasoline from electricity on an industrial scale without the use of fossil fuels as input?
Assume the electrical power came from an unlimited source, such as nuclear fusion, solar arrays in space, or very specific magic.
You may also assume that the gasoline doesn't need to be replicated perfectly -- that which can fuel anything that uses gasoline, or which satisfies an industry definition of gasoline, is sufficient. Ethanol does not count, since most cars cannot use it solely.
Feel free to comment on the environmental friendliness of the solution, excepting the electricity generation itself.
This basic question was originally asked here, and where it was suggested to migrate to chemistry SE.

Hmm, it would be probably better if you waited for migration and edited later.

related http://chemistry.stackexchange.com/questions/6516/how-to-change-co%e2%82%82-to-a-less-polluting-gas

this pretty much answers it: http://chemistry.stackexchange.com/questions/35661/can-fuel-exhaust-be-recycled

Paper recently published on the topic at: https://www.cell.com/joule/fulltext/S2542-4351(18)30225-3
"The design captures ∼1 Mt-CO2/year in a continuous process using an aqueous KOH sorbent coupled to a calcium caustic recovery loop."

Here is an article about a paper using an iron-manganese-potassium catalyst to produce long-chain hydrocarbons: https://arstechnica.com/science/2020/12/with-the-right-catalyst-we-might-make-jet-fuel-from-co%e2%82%82/

Historically, this sort of process has been done with the Fischer–Tropsch process.
The core of this process is to use "syngas" (a mixture of $\ce{CO}$ and $\ce{H2}$) in the presence of a catalyst to make higher molecular weight hydrocarbons:
$$\ce{(2n + 1) H2 + n CO → C_{n}H_{2+2n} + n H2O}$$
Under normal conditions, this will actually make mostly straight-chained hydrocarbons, which are great for diesel fuel, but not so great for gasoline/petrol engines. However, there are standard catalytic cracking techniques in the petrochemical industry which can adjust the branchedness of hydrocarbons, making them more "gasoline like".
Now, the only question is where to get the $\ce{CO}$ and $\ce{H2}$ from? The $\ce{H2}$ is easy, as it is readily obtained from the hydrolysis of water.
The $\ce{CO}$ can be obtained from $\ce{H2}$ and $\ce{CO2}$ by the water-gas shift reaction (here technically run in reverse):
$$\ce{CO2 + H2 \rightleftharpoons CO + H2O  }$$
As I mentioned, the Fischer–Tropsch process was actually used during WWII by the Germans to produce motor fuel, as conventional sources were blocked off by the Allies. Their syngas didn't come from $\ce{CO2}$ and water, though, but from reformed coal, which they had a ready supply of.
This process is not done much because it's hugely inefficient, compared to just digging oil out of the ground. But if you have a "too cheap to meter" supply of electricity, the efficiency concerns go away. The main issue is getting a ready supply of $\ce{CO2}$, and there are a number of techniques being proposed at the moment for scrubbing $\ce{CO2}$ from the atmosphere - either by capturing it from point-source generators like power plants and cement manufacturers, or from "artificial trees" which capture $\ce{CO2}$ directly from standard atmospheric concentrations.
These "artificial trees" tend not to use normal photosynthesis for carbon capture, but instead often use something like a (strong) base in aqeous solution or $\ce{CaO}$. In these systems the base or the $\ce{CaO}$ reacts with dissolved $\ce{CO2}$ to form a carbonate salt, which either stays in solution or precipitates out. This carbonate salt formation is favorable, and happens spontaneously. The energy intensive portion of the process is the regeneration of the basic solution/$\ce{CaO}$. This normally happens by heating the carbonate salt, which causes $\ce{CO2}$ release. If done correctly, this can produce a stream of relatively pure carbon dioxide. But again, the regeneration step is quite energy intensive, and so would not be feasible without a cheap source of power.
Other environmental considerations would be minimal, aside from the land needed for the $\ce{CO2}$ devices (you would need a fair number to get a suitable steady volume of $\ce{CO2}$), and any disposal concerns about the process catalysts (which shouldn't be any more onerous than dealing with the catalysts now used in the petrochemical industry).

Thank you for the answer, R.M. I like especially how you go through the whole process. Might you be able to put some rough numbers on the efficiency? I'm wondering what order of magnitude we looking at -- for every 100J of electrical energy put in, are we getting something like 10J or 1J or 0.1J of chemical energy back? Or worse?

@einnocent I'm not a chemical engineer, so such efficiency estimates are not my area of expertise. A quick Google search estimates thermal efficiency of Fischer–Tropsch is ~60%, electrolysis is ~70%, and $CO2$ capture is ~$100/ton, which I think back-of-the-envelopes to the ~60% range. Combined,  you're probably running in the low double digits: so 10J of electricity to 1J of fuel is probably the right ballpark. All that depends massively on how you run your plants, though - there's opportunities for heat scavenging, etc. from a combined operation.

Thank you taking the time to look in to those numbers. This addresses another aspect of my question, which is the practicality of industrialized fusion to gasoline as replacement for current world consumption. I recognize fusion-to-gasoline is not an efficient process, but I wonder does that mean that it's not worth bothering until electricity is "too cheap to meter", or does it mean covering the surface area of the planet ten times over with fusion reactors? It sounds like the former.

How might we produce gasoline from electricity on an industrial scale without the use of fossil fuels as input? Assume the electrical power came from an unlimited source, such as nuclear fusion, solar arrays in space, or very specific magic.

I love this sort of question.
If we had magic and wanted gasoline, why would we make electricity?

To get serious...
You could make gasoline from the CO2 in the air and electricity, but it is very energy inefficient. So it is something like using 100 units of power in the form of electricity to make 1 unit of  power in the form of gasoline. So to say that electricity can't be used to make gasoline, really means that in this case that it isn't practical to do so.
The deeper rub is that even solar panels are not "free energy". It costs energy to make the solar panels, to ship them, to install them, and to maintain them.
But certainly converting energy types is a valid way to mull over the energy problem.

Thanks for the response. Remember, we have as much energy as we want, in the form of electricity, so efficiency isn't important, unless the waste heat of production causes environmental issues. What chemical reactions, as part of a larger industrial process, would produce the gasoline?

Fundamentally CO2 + H2O yields gasoline and O2.

You can think of gasoline as mixture of various $C_{x}H_{y}$ compounds.


text:
Number of grams of CO2 released from a carbonated water
The partial pressure of $\ce{CO2}$ gas in a bottle of carbonated water is $3.84\mathrm{~atm}$ at $25\mathrm{~^\circ C}$.  How much $\ce{CO2}$ gas (in g) will be released from $0.5\mathrm{~L}$ of the carbonated water when the partial pressure of $\ce{CO2}$ is lowered to $1.04\mathrm{~atm}$?  $25\mathrm{~^\circ C}$, the Henry’s law constant for $\ce{CO2}$ dissolved in water is $1.65 \times 10^3\mathrm{~atm}$, and the density of water is $1.0\mathrm{~g/cm^3}$.
First, I found the number of moles of $\ce{CO2}$ using this formula:
$$n = \frac{PV}{k}$$
So the answer that I found which is $0.000351$, can I multiply it by a molar mass of a $\ce{CO2}$ to get the number of grams inside the bottle? This approach seems too easy for me, but at the same time it makes sense.

@Ghost nice editing, but you missed out one CO2 :) Also, we normally add the units inside math mode, you can check out the current code to see how it's done.

This shows part of the problem.
Henry's law as given is
$K_{AW}^{px} = $1.65*10^3$ = \frac{p_{\ce{CO2(gas)atm}}}{[\ce{CO2(aq)}]}$
where:
$K_{AW}^{px}$ atm = aqueous-phase mixing ratio of $\ce{CO2}$ and the units are $atm$
$[\ce{CO2{(aq)}}]$ = the concentration of $\ce{CO2{(aq)}}$ in $mole/L$
$p_{\ce{CO2(gas)atm}}]$ = the partial pressure of $\ce{CO2{(gas)}}$ in $atm$
So rearranging...
$[\ce{CO2(aq)}] =\frac{p_{\ce{CO2(gas)atm}}}{K_{AW}}$
The moles of $\ce{CO2}$, $m_{\ce{CO2}}$, is given by concentration, $[\ce{CO2}]$, times the volume, $v$ in $L$.
$m_{\ce{CO2(aq)}} = [\ce{CO2(aq)}]v =  \frac{p_{\ce{CO2(gas)atm}}}{K_{AW}}v$

It does not seem right to me, if you use the formula Hpv and substitute all values in it which will be: 16500.51.04 = 858 grams. This number does not make sense to me, how can 858 grams of CO2 be in 0.5 L of bottle?

Could you double check the constant? Is it $1.65*10^{-3}$.

No, it says it is 1.65*10^3

This table https://en.wikipedia.org/wiki/Henry's_law#Values_of_Henry.27s_law_constants shows the problem. You didn't give any formula for the Henry's law and there were a number of them. I'm sorry. Instead of guessing one, I should have clarified that point. I'll fix the analysis.

In such cases a dimensional analysis is good to be sure that the right units are being used.

ok, I think I finally have this right.


text:
How are related the electrons magnetic dipole moments in atomic orbitals?
I wondering, has somebody ever try to take in account the electrons magnetic dipole moment for the shapes of the probability distribution in atom? Does this magnetic dipole moment play any role in the inner bounds of molecules?

Some thoughts of mine see here: https://www.academia.edu/18391675/About_the_distribution_of_the_electrons_magnetic_dipole_moments_in_the_atoms_Part_1

In my paper, mentioned above, I have had some doubts about the role of the electrons dipole moment distribution in chromium atom which is quite symmetrical if I use my octahedral model instead of s- and p-orbitals.
Now a have read on Wikipedia about the magnetic characteristics of pure chromium: "Chromium is remarkable for its magnetic properties: it is the only elemental solid which shows antiferromagnetic ordering at room temperature (and below). Above 38 °C, it transforms into a paramagnetic state." This matches perfect into my model abot octahedral structure of neon and argon shell electrons.


text:
How it was observed, that elements in columns 6 and 11 assume abnormal electron configuration?
Reading this question about why do elements in columns 6 and 11 assume abnormal electron configurations I wondering, how this was observed.

My guess is that it was observed in spectra of the of transition metal ions.

How it was observed, that elements in columns 6 and 11 assume abnormal electron configuration?

I assume you to mean

How it was first observed, that elements in columns 6 and 11 assume abnormal electron configurations?

Chemists (starting with alchemists) have were doing stoichiometric reactions for a very long time.
The point is that if there was only a single "normal electron configuration" for chromium, then there would only be one chromium chloride. However chemists have synthesized three.

Chromium(II) chloride, also known as chromous chloride.
Chromium(III) chloride, also known as chromic chloride or chromium
trichloride
Chromium(IV) chloride

Thus the fact that there must be "abnormal" electron configurations was first discovered by stoichiometry, not spectroscopy.
Other columns in the periodic chart do have elements with multiple electron configurations. So such an "abnormality" isn't unique.

If your question means

How are the bonding energies measured for the elements in columns 6 and 11 in various electron configurations?

Then ESCA would be one technique.

I don't see how 'abnomal configurations' follow from different chlorides. Iron has two chlorides ($\ce{FeCl2}$ and $\ce{FeCl3}$) but it does not have the same 'abnormal configurations'.'

It isn't an abnormality of only columns 6 and 11, the fact that different stoichiometries exists shows that an element can have more than one electron configuration. If no "abnormalities" existed, then the Lewis dot would work for everything.

Could you please explain what these "different electronic configurations" are, and how they lead to "different stoichiometries" observed in the chromium chlorides?

Max, I hope to get more answers. Thank you for yours. With ESCA I'm not fully satisfied because this is a technique introduced long after the knowledge about the aperiodical filling of the electron shells.


text:
What is the term for a compound where one element exists in two oxidation states in the same molecule?
Example nitrogen in $\ce{N2O}$ (−3 and +5). I'm pretty sure there's a term but I've forgotten what it is I thought it was disproportionation but this seems to only refer to a reaction

"Mixed oxide" is sometimes used to describe oxides where the central atom exists in more than one oxidation state. Another example is $\ce{Mn3O4}$, which exists in both the +2 and +3 states. See more at https://en.wikipedia.org/wiki/Mixed_oxide

When a species changes to two different oxidation states it's called "disproportionation" so could it be described as "disproportionate"?

What would you call a compound like $\ce{NH4NO3}$, where the nitrogen atoms are in differing oxidation states?

There's no such term and almost all organic compounds have oxidation state varying between atoms.


text:
what is the difference between a redox reaction and an ionic bond?
I am new into chemistry...
I'd like to know what the difference is between an ionic bond: which involves a bond with an electron transfert, and a redox reaction; which seems to be exactly the same to me.
Thanks

Redox reaction and$,\ldots$ ionic bond?!

I know that one is a reaction and one is a bond, if that is the point you are trying to make. But at the end they both seem to do the same: transfert an electron. So could we say that redox reactions creates ionic bonds?

A redox reaction is where one compound/molecule  is reduced and the other oxidized , one gains electrons and the other loses electrons . An ionic bond is the force of attraction between two oppositely charged ions in which electrons are shared between the ions and the net charge is balanced.

In short, redox reactions may or may not create ionic bonds. There are examples of redox reactions which actually destroy ionic bonds, and others that have no relation to anything ionic whatsoever.

I know that this is how an ionic bond is first presented to beginners, but you should not think of ionic bonds as bonds involving electron transfers. Ionic bonds are merely "electrostatic attractions between oppositely charged ions". Now in the case of $\ce{NaCl}$, these oppositely charged ions are $\ce{Na+}$ and $\ce{Cl-}$, and these are obtained from neutral $\ce{Na}$ and $\ce{Cl}$ by transfer of one electron from $\ce{Na}$ to $\ce{Cl}$. However, the electron transfer is not part of the formation of the ionic bond.

The ionic bond only has to deal with the positively and negatively charged species. Where those positively and negatively charged species come from is not relevant to the formation of the ionic bond. For example, I could mix $\ce{AgNO3}$ and $\ce{NaCl}$, which causes $\ce{AgCl}$ to precipitate out - you could say there is an ionic bond between $\ce{Ag+}$ and $\ce{Cl-}$. However, this ionic bond was not formed by electron transfer from $\ce{Ag}$ to $\ce{Cl}$.

@orthocresol Perfect explanation, that clears it up for me. /me wonders why we can't award bounties on comments

An ionic bond is formed by electrostatic attraction of electrons between one atom and the other whereas in redox reactions the electrons are transferred from one atom to the other. Hope this helps.

But... the electrons of both atoms are negatively charged, so surely they should electrostatically repel each other?

"$\ldots,$electrostatic attraction of electrons$,\ldots$" -- Huh? And BTW, this seems like an attempt at providing the definitions, not similarities or differences. Not much of an answer.


text:
Oxidation state of metals in complexes with bridging ligands
How do I find the oxidation number of $\ce{Co}$ in a coordination complex like the
tetraaminecobalt(2) $\mu$-amido $\mu$-oxo tetraamine cobalt(2) ion?
I searched all over the net but can't find it anywhere. Please help!

I meant oxidation number..sorry

Maybe this can help you. http://chemistry.stackexchange.com/questions/22307/how-do-i-find-the-oxidation-number-of-a-coordination-complex?rq=1

But what should I consider as the oxidation states of oxo and amido ? @Anders MB ?

It’s no different from complexes with non-briding ligands.


text:
Balancing a reaction with unknown products
I'm given this equation in word form:
$$\ce{sodium (s) + water (aq) -> unknown~products (s/g/aq)}$$
and I have to convert this to a chemical equation. I understand how to covert this normally, that equation can hence be represented by:
$$\ce{Na + H2O -> unknown~products\:(s/g/aq)}$$
This is indeed a single displacement reaction so it should become (according to what I know so far) $\ce{Na2O + H2}$ but after doing some research, I got that it should be:
$$\ce{Na + H2O -> NaOH + \frac{1}{2}H2}$$ or $$\ce{2Na + 2H2O -> 2NaOH + H2}$$
I'm lost now, where did $\ce{NaOH}$ come from? Especially the H since there is no reason for hydrogen to be present.
Please explain why and how to balance this reaction or convert this and what factors to look at when dealing with these types of equations.

The Na2O is unstable so it would very quickly convert to the more stable NaOH.

Please visit this page, this page and this one on how to format your posts better. Alternatively, visit this chatroom for further formatting guidance.

The hydrogen is created from the exothermic reaction between the highly reactive sodium metal and water.

@Joel But how can I identify when it does that? and where did H come from?

The H comes from the water, think about what happened to the water molecule that reacted with the sodium ion and left an oxygen atom , hence the speculation of sodium oxide formation, where did the hydrogens go?. Sorry identify when what does what?

If you did the reaction practically you can definetly observe the sodium reacting with the water, the pH of the water will turn alkaline as the sodium hydroxide is formed.

I don't really get it though :/... I thought single displacement would take the form of A + BC -> AB + C with no exceptions. Na+H2O... So Na bonds with oxygen right? And they form Na2O and O2 is left? @Joel

@Joel I think it's the same case with Potassium + Water but I don't know why

No, think if the sodium or potassium (as you just mentioned)  takes the oxygen from a water molecule (H2O) , hydrogen is left. Sorry I can't do the required format for any equations on my phone. Due to the fact that these oxides are so unstable , the product is just written as sodium hydroxide , using the thinking that there's no need to mention the intermediate sodium oxide .

Oh okay @Joel that makes a lot of sense. I have this question for a science lab, however we weren't taught this so should I keep the answer with OH? Thank you so much!!!

Sodium ion taking the oxygen atom or water molecule leaving oxygen atom ,  same thing ( sorry if it was confusing) .

It is correct both ways (to write the intermediate or not) but more conventionally written with the hydroxide byproduct as under normal conditions the oxide fully converts to the hydroxide so it is the end product.

related http://chemistry.stackexchange.com/questions/27393/why-does-hot-steam-produce-mgo-instead-of-mgoh2

Sodium(s)+water(aq)⟶Unknown Products(s/g/aq)
In this case you treat the hydrogen as a cation, or as a metal which is why it is displaced. Treat H2O as H(OH).
I know I have answered my own question but this is the answer I was looking for a long time ago.

This is unfortunately wrong. The reaction is not a displacement reaction. It is a redox reaction. One electron of sodium gets transferred to the hydrogen of water, which liberates a dihydrogen molecule.

Water is normally a mix of hydroxide and hydrogen, at very dilute levels.  For example, neutral water (pH of 7) is 10-7 of OH- and H3O+ ions.  Add sodium and it combines with the OH-.  The loss of electrons by sodium are available to the hydrogen ions, which combine to make hydrogen gas.  Sodium in metal form is so reactive, it will sustain this reaction even at the low hydroxide levels in pure water, and even in slightly acidic seawater.

There are some pretty significant errors in your answer (including but not limited to the first sentence, especially regarding hydrogen).


text:
Brewing tea with deionized water
I am wondering what efect brewing tea (let us assume for the sake of simplicity it is green tea, brewed at about 80 °C) in deionized water might have on the product.

Will the lack of ions in the water increase or decrease the solubility of alcaloids from the tea leaves. Will the effect be different for alkaloids in salt and alcaloids in amine form?
Will drinking moderate amounts of such tea (0.5–1 l per day) detrimentally affect one's electrolyte metabolism? I am guessing maybe it does not matter, since the tea fills the water with electrolytes anyway – but will these electrolytes be significally different from what you would get in tap/mineral water? and if so, will it matter?

I'd say there'd be just less electrolytes in it than a normal cup of tea. As for if it's healthy or not there's too many variables and factors that could lead to how it would effect your health  in particular that I wouldn't want to make an assumption as I don't know you. Sorry I don't know if it would be bad or not for you.

In lower Michigan there have been places where the water has $\ce{H2S}$ in the tap water. Using bottled or deionized water for tea in those places would make the tea drinkable. So unless you can taste a difference the more expensive water is pointless.

You might contemplate the difference between deionized and distilled water. Distilled water is corrosive because it really wants solutes in it. Deionized has merely exchanged certain ions for other ions. Now, I once worked in a lab where we always made coffee with distilled water, and it tasted just fine and I'm still alive.

And to add anecdotic evidence to @JonCuster’s: Yes, I was an intern in a group that used a samowar; the tea concentrate was brewed with deionised water while it was diluted with normal hot water. They did it for a reason which I can’t remember, but probably had something to do with easier cleaning.


text:
Conservation equation in an enzyme-catalysed reaction
Consider the reaction scheme:
$$\ce{S + E ->[k_1] C1} \qquad \ce{C1 ->[k_2] E + P} \qquad \ce{S + C1 <=>[k_3][k_4] C2}$$
where $\ce{S}$ is the substrate, $\ce{E}$ is the enzyme, $\ce{P}$ is the product, $\ce{C1}$ and $\ce{C2}$ are enzyme-substrate complexes. Let $[\ce{S}] = s$, $[\ce{E}] = e$, $[\ce{C1}] = c_1$, $[\ce{C2}] = c_2$ and $[\ce{P}] = p$ be the concentrations of each respective chemical. I have used the law of mass reaction to convert this reaction into a system of differential equations:
\begin{align}
\frac{\mathrm{d}s}{\mathrm{d}t} &= -k_1se-k_3sc_1+k_4c_2\\
\frac{\mathrm{d}e}{\mathrm{d}t} &= -k_1se+k_2c_1\\
\frac{\mathrm{d}c_1}{\mathrm{d}t} &= k_1se-k_2c_1-k_3sc_1+k_4c_2\\
\frac{\mathrm{d}c_2}{\mathrm{d}t} &= k_3sc_1-k_4c_2\\
\frac{\mathrm{d}p}{\mathrm{d}t} &= k_2c_1
\end{align}
I am asked to find a conservation equation and use it to simplify this system however I haven't been taught as to what a conservation equation is. Can anyone push me in the correct direction? (and check the above differential equations are correct).

The reaction scheme is:
$$\ce{S + E ->[k_1] C1} \tag{1} $$ $$\ce{C1 ->[k_2] E + P} \tag{2}$$ $$\ce{S + C1 <=>[k_3][k_4]C2} \tag{3} $$
Let's define a bit of nomenclature.

S = compound S in chemical reaction equation
s = $[S]_x$, the concentration of S at t=x.
$S^*$ = ds/dt

The whole set of differential equations has 5 equations with 5 unknowns.
\begin{align} \\
S^* = \frac{\mathrm{d}s}{\mathrm{d}t} &= -k_1se-k_3sc_1+k_4c_2  \tag{4} \\ E^* = \frac{\mathrm{d}e}{\mathrm{d}t} &= -k_1se+k_2c_1\tag{5} \\
C^*_1 = \frac{\mathrm{d}c_1}{\mathrm{d}t} &= k_1se-k_2c_1-k_3sc_1+k_4c_2 \tag{6}
\\ C^*_2 = \frac{\mathrm{d}c_2}{\mathrm{d}t} &= k_3sc_1-k_4c_2 \tag{7}\\ P^* = \frac{\mathrm{d}p}{\mathrm{d}t} &= k_2c_1 \tag{8}\end{align}
What is a conservation equation
In mathematical terms a conservation equation is some linear combination of the time derivatives which add up to zero on the right hand side.
The gist is that the various conservation equations allow some simplification in the various interactions in the system.
Rather than just search blindly, there are a couple of relationships that can be exploited to find time derivatives which add up to zero on the right hand side. First is a sort of mass balance. The initial concentration of the reactants has to be spread as the various intermediate species and the final product. A second line of attack is to exploit and equilibrium conditions such as equation (3).
First conservation equation from "mass balance"
The first conservation equation is found noting that at t=0 then all is [S], a constant. Noting that it takes 2 $\ce{S}$ to make one $\ce{C_2}$, at some later time t=x, thus:
$$[\ce{S}]_{t=0} = [\ce{S}]_{t=x} +  [\ce{C_1}]_{t=x} + 2[\ce{C_2}]_{t=x} + [\ce{P}]_{t=x}\tag{9} $$.
Now for some intermediate time we can take derivative of the mass balance equation with respect to time. All the dt's get cumbersome so lets use just $S^*$ from now on to mean ds/dt. From the mass equation then since the initial concentration is a constant we get
$0 = S^* + C^*_1 + 2C^*_2 + P^*\tag{10} $
Most likely solving for either $S^*$, or $P^*$ would give an interesting result.
A second conservation equation from mass balance
The second conservation equation comes from noting the enzyme participation in the reaction.
$$[\ce{E}]_{t=0} = [\ce{E}]_{t=x} +  [\ce{C_1}]_{t=x} + [\ce{C_2}]_{t=x}\tag{11}$$.
Taking the derivative we get:
$ 0 = E^* + C^*_1 + C^*_2 \tag{12}$
which can be rearranged to
$E^* = - C^*_1 - C^*_2\tag{13}$
Substituting equation 6 for $C^*_1$ and equation 7 for $C^*_2$ this gives us equation 5 again which isn't helpful.
A third conservation equation from the equilibrium condition
From the third reaction we can derive the equation.
$ \frac{k_3}{k_4}= \frac{[S]_x[C_1]_x}{[C_4]_x} = \frac{sc_1}{c_4}\tag{14}$
Taking the derivative:
$ 0 = \frac{c_1}{c_4}S^* + \frac{s}{c_4}C^*_1 - \frac{sc_1}{c_4^2}C^*_4\tag{15}$
rearranging
$  S^* = -\frac{s}{c_1}C^*_1 + \frac{s}{c_4}C^*_4\tag{16}$
Infinite time as a boundary condition
Obviously at $t=\infty$, all the S has been converted to all P which is a constant. This factoid isn't very useful though in predicting how long that an "effective infinite time would be. Rather it is just a boundary condition.
$$[\ce{P}]_{t=\infty} = [\ce{S}]_{t=0}\tag{17}$$

This is not correct because the time derivative of s +c1+c2+p is not equal to zero.

fixed equation because it takes 2S to make a C2.

As I said before, and as the differential equations indicate, the quantity that stays constant is $(s+e+2c_1+3c_2 +p)$, so $\Delta (s+e+2c_1+3c_2 +p)=0$

Ok, I believe you on what the "conservation equation" is. (Question 1) What I have is the "mass balance" equation right? (Question 2) Can't I take the derivative with respect to time on that equation?!?

I'm thinking about what has to be done to solve the problem for dP/dt, not just figuring out the "conservation equation" what is what the OP wanted.

All he's asked to do is to "simplify the system," not to solve for dP/dt. The original system consists of 5 coupled ODEs in 5 unknowns. By making use of the conservation equation, one can solve algebraically for any one of the unknowns in terms of the others, substitute that into the other four ODEs, and end up with 4 coupled ODEs in 4 unknowns. I call that "simplifying the system."

I've now "solved" the problem.

You've done something very interesting. You've shown that there is a second linear combination of the time derivatives that is equal to zero. Based on this, you've developed a second quantity that stays constant and a second linearly independent conservation equation: Δ(e+c1+c2)=0. So, e(t)=e(0)−c1(t)−c2(t). So we can reduce the system from 5 coupled ODEs in 5 unknowns to 3 coupled ODEs in 3 unknowns. –

I'm sort of confused. Is A "conversion equation" any combination summing to 0, or is it THE "conversion equation" meaning it must be for all (5 in this case) the species. For the "THE" case there could just be one case (plus integer multiples of  course which would be redundant). // I had an elementary school teacher who just loved A/THE true false questions.

I think you mean conservation equation.  It's any linear combination of species concentrations that stays constant as the reactions progress.  In this reaction scheme, there are two linearly independent combinations that remain constant.  So, for this sequence, there is not just one THE conservation equation.  There are two.  Some of the coefficients in the linear combination are allowed to be zero, so the conserved quantities do not have to involve all 5 species.  This all leads me to wonder whether there are other conserved combinations.  I doubt it, but maybe.

Yes, I'll plead a senior moment. I did mean conservation equation. So I'll edit the answer appropriately.

@MaxW Hi, I don't really understand what you have done when simplifying the system. I understand up until the point where you find $E$ in terms of $C_1$ and $C_2$ but you say there are three unknowns ($S$, $E$, and $C_1$) however how do you get $S$? Since it is not involved in the equation and how do you know $P$ and $C_2$ are known?

@Michael-Howlard - Ok I expanded out the part that was confusing you. Does that help?

You need to find a linear combination of the time derivatives such that they add up to zero on the right hand side.  This will tell you the corresponding linear combination of concentrations that is conserved.  For this problem $s+e+2c_1+3c_2+p$ is conserved.

ah... I'm guessing that the "conservation equation" gives dP/dt?

No.  The conservation equation says that the quantity $(s+e+2c_1+3c_2+p)$ is conserved during the chemical reaction because its time derivative $\frac{d(s+e+2c_1+3c_2+p)}{dt}$ is equal to zero.

Ok, I think I get it now. I gave equation for "mass balance" which yields one set of linear equations, the "conversation equation" yields another, and using them both you can get a simplified expression for dP/dt. // I need to correct my mass balance equation too. there are two S in C2.

@ChesterMiller How does this equation simplify the system?

The way that this simplifies the system is explained in the comments to MaxW's answer.  MaxW came up with a second conservation equation that also applies to this problem.  Basically, the conservation equations allow you to directly solve for s and e in terms of the concentrations of the other 3 species, and thereby allows you to reduce the number of differential equations from 5 to 3.

@Michael Howlard, see my comment above.

So how do we directly solve for $s$ and $e$? I can't see how you could without knowing what the other concentrations are.

@MaxW gave the equation $s=s(0)-c_1-2c_2-p$ in his answer, and also implied $e=e(0)-c_1-c_2$ in that same answer, where the (0)'s indicate the initial concentrations of these species.  Substituting these two algebraic relationships into the differential equations for $c_1$, $c_2$, and p reduces the system to 3 differential equations in the three unknowns $c_1$, $c_2$, and p.


text:
DNA's physical properties
If we took a bunch of DNA molecules and put them together what properties would it have?
I learned that long molecules tend to make a solid substance because they get tangled. A single DNA molecule is very very long so I guess it has to be solid as a substance. I also guess it would be resistant to pulling it apart. Is it true? Has anyone actually tried doing that?

Yes, I've cyrstallized DNA from peas before. We used PCR on the isolated DNA to bring the concentration up to a practical amount to work with then used a procedure involving absolute ethanol and a hook to retrieve the DNA as fibrous , geletinous strands (which looked like a little white clump ), although small you could easily observe the pure DNA.

Regarding first of two questions:
Assuming duplex DNA (dsDNA), the molecules should repel each other and thus stay in solution at low ionic strength, such as in Tris/EDTA buffer. This is due to the abundance of negative - repulsive - charges. Increasing the ionic strength of the solution, such as high concentration of sodium acetate should neutralize the charges and allow the dsDNA fibers to sediment. In fact, a controlled sedimentation condition may help crystallize the nucleic acids.


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Nucleophile attack on Epoxy rings
My professor had explained to us that a nucleophile always attacks on that carbon connected to an epoxy ring, which would make a stable carbocation, when the given compound with the epoxy ring is in an acidic solution(containing H+)
For example if this is the epoxy ring :
The Nucleophile would attack on carbon A, as if A forms a carbocation, it is more stable than B forming a carbocation, if it is in acidic solution. If not in acidic solution the nucleophile attacks on the less hindered carbon(which is B).
But my book has a contradiction :
In this reaction, C2H5- attacks on the less hindered carbon, not the carbon with more stable carbocation.
Is the book right. Also please explain are the above reactions(that my professor said) with epoxy rings always valid ?
Thanks.


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What's the difference between HOMO-LUMO and valence-conduction bands?
From Wikipedia:

Roughly, the HOMO level is to organic semiconductors what the valence
  band maximum is to inorganic semiconductors and quantum dots. The same
  analogy exists between the LUMO level and the conduction band minimum.

Why such distinction even exists? If it's the same physical concept, why use different names?

A band is a multitude of energy levels that are quasi-continuously distributed over a certain energy range. It comes about because of crystal (translational) symmetry. A HOMO/LUMO is just a single energy level. While thinking of the valence and conduction band edges as the "HOMO or LUMO of a crystal" is often quite fruitful, there are some effects present in solids that cannot be understood by this simple pov.

The properties of solids simply are not as much influenced by single energy levels as is the case for molecules. Here bands are the appropriate conceptual unit to think of.

@Philipp, so whenever "HOMO-LUMO" is used, the assumption is that energy levels are discrete?

Yes, unless you encounter some sloppy terminology.

@Philipp so why quantum dots are described with bands and not with HOMO/LUMO?

Have not worked in the area of quantum dots yet, so I can't really help you there. But I can recommend this book by Roald Hoffmann to you if you want to get a good understanding about the relations between molecular energy levels and crystal bands. It's a pretty short book but it really does a wonderful job at teaching you the links between solid state and molecular chemistry.

Because convention. Yes, that’s not the answer you wished for but it’s the way it is.
Coming from an organic perspective, one would think of concepts such as linear combination of atomic orbitals or others to arrive at molecular orbitals for entire molecules. Since organic chemistry always deals with rather small molecules and since these have discreet energy levels with discreet differences, it makes sense to speak of a single HOMO, i.e. one single highest occupied orbital (and likewise for a LUMO). The larger a molecule gets, the more orbitals it has and the less the difference between HOMO and LUMO might turn out to be, but from an organic perspective it is still a molecule and must begin and end somewhere.
(Note that this applies to a lesser extent to polymers and carbon nanotubes etc. This is where the shady grey area in which unicorns live begins.)
From an inorganic perspective, if you are dealing with solid materials, they will usually be crystalline in some way; hence they have infinite extents by definition. It no longer makes sense to think of discreet orbitals but rather one realises that certain levels contain a large number of orbitals — hence bands.
If organic molecules end up being sufficiently large, there is no longer a relevant physical difference between HOMO/LUMO designations or bands. But because nobody likes throwing their traditions and customs overboard, the two different names exist.

are all non-covalent arrays currently described in HOMO/LUMO? in other words, how closely packed small molecules need to be, to "justify" a band description? in particular, are supramolecular polymers described in bands or in HOMO/LUMO?

@Sparkler I don’t really work in the area, so I don’t really know, but since the underlying physics are the same I see no reason why there should be any rules on what can be called what.

I guess when energy levels are discrete, using "bands" is somewhat misleading. I'll use HOMO/LUMO when unsure, just to stay on the safe side... :)


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What defines incomplete combustion?
I know that complete combustion is a process where a fuel reacts with $\ce{O2}$ to become $\ce{CO2}$ and $\ce{H2O}$. But what is incomplete combustion?
According to the Internet, $\ce{CH4 + O2}$ and $\ce{C6H12O6 + O2}$ are examples of incomplete combustion. What makes them so?

Complete combustion is, as the name suggests, complete. That on its own doesn’t mean much, but I can tell you that the completeness implies all atoms that were put into combustion leave it fully oxidised (or fully reduced in the case of oxygen, fluorine, chlorine; or unchanged in the case of nitrogen).
An incomplete combustion is one where molecules remain that may still be oxidised, i.e. can react with further oxygen. Examples would be carbon monoxide or elemental carbon (colloquially ‘soot’). Incomplete combustion often results in a yellow flame stemming from carbon particles, while complete combustion typically yields hotter, colourless flames.
In your average Bunsen burner, you can choose between complete and incomplete combustion by regulating the amount of air (meaning: oxygen) that reaches the flame. Complete combustion would follow the equation $\ce{CH4 + 2 O2 -> CO2 + 2 H2O}$. There is no unique equation for incomplete combustion, but one possibility would be $\ce{CH4 + O2 -> C + 2 H2O}$.
By the way, the way living beings burn carbohydrates in cellular respiration is, by definition, a complete combustion since every carbon atom ends up in $\ce{CO2}$ and a set of two hydrogen atoms always ends up as $\ce{H2O}$. The overall equation is $\ce{C6H12O2 + 6O2 -> 6CO2 + 6H2O}$. If glucose is combusted with less than 6 equivalents of $\ce{O2}$, then incomplete combustion results.

More or less ok except that cellular respiration isn't a combustion.