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What are the Waves Modeling when Referring to the Atomic Orbitals
It is taught that the orbital shapes derive from wave functions with different numbers of nodes. For example, the "s" orbital comes from a wave that has one node. But what are the waves modeling? A light wave for example is showing a relationship between an electric field and magnetic field. What relationship does an atomic orbital wave show? In other words, if you were to label the axes, what would the labels be?
Thanks!

I believe the waves are the psi function representing the probability of finding an electron at a certain location, but it's been such a long time since I've had to do p-chem I can't remember any more details than that, and may have gotten something wrong. The psi function uses distance from the nucleus, and two angles as inputs, so it's a 3D polar function.

Thanks for your reply! So in the p-oribtal, I would  expect to find most electrons in a literal dumbbell shape around the nucleus?

yes. In fact, that may be 100% of the probability, but I can't really remember right now.

In a one-dimensional system (I choose this to keep things as simple as possible) the state of the system is described by the wave function $\psi = \psi(x, t)$, which depends on the coordinate $x$ of the particle at time $t$. Wave functions in general are complex functions of real variables. The symbol $\psi^{∗}(x, t)$ denotes the complex conjugate of $\psi(x, t)$. The quantity $P(x, t) = \psi^{∗}(x, t) \psi(x, t) \mathrm{d}x = | \psi^{∗}(x, t) |^{2} \mathrm{d}x$ gives the probability that at time $t$ the $x$ coordinate of the particle lies in the small interval $[x, x + \mathrm{d}x]$. The probability of the particle being in the interval $(a, b)$ on the $x$ axis is given by $\int_{a}^{b} | \psi^{∗}(x, t) |^{2} \mathrm{d}x$. The probabilistic interpretation of the wave function was proposed by Max Born. By analogy with the formula: mass = density $\times$ volume, the quantity $\psi^{∗}(x, t) \psi(x, t)$ is called the probability density that a particle at time $t$ has position $x$.
Edit: Atomic orbitals
For atoms you have a system that is spherically symmetric. Thus, it is most conveniently expressed in spherical coordinates instead of cartesian ones. The wave function solving the time-independent Schroedinger equation is then expressed in terms of the radial distance $r$, the polar angle $\theta$ and the azimuthal angle $\varphi$.
This wave function $\Psi(r, \theta, \varphi)$ can be split into a radial part $R_{n,\ell}(r)$ and an angular part $Y_{\ell,m} (\theta, \varphi )$, so that $\Psi(r, \theta, \varphi) = R_{n,\ell}(r) Y_{\ell,m} (\theta, \varphi )$, where $n$, $\ell$ and $m$ are the Principal, Azimuthal and Magnetic quantum number, respectively, and the functions $R_{n,\ell}(r)$ are the solutions of the radial Schroedinger equation
\begin{equation}
\bigg( \frac{ - \hbar^{2} }{ 2 m_{\mathrm{e}} } \frac{ \mathrm{d}^{2} }{ \mathrm{d} r^{2} } + \frac{ \hbar^{2} }{ 2 m_{\mathrm{e}} } \frac{ \ell (\ell + 1) }{ r^{2} }  - \frac{ Z e^{2} }{ 2 m_{\mathrm{e}} r } - E \bigg) r R_{n,\ell}(r) = 0
\end{equation}
with the nuclear charge $Z$ and the mass of an electron $m_{\mathrm{e}}$. Those radial wave functions are the part of the atomic wave functions $\Psi$ that determines their "extent" and their radial nodes, i.e. the nodes that grow with the Principal quantum number $n$: for example a $1\mathrm{s}$ orbital has no radial nodes (it is a simple sphere without any sign-change in the wavefunction) while the $2\mathrm{s}$ orbital has 1 radial node (it is a smaller sphere wrapped in a bigger spherical shell whereby the inner sphere has a different sign than the outer spherical shell such that you get one nodal plane between inner and outer part). The angular wave functions $Y_{\ell,m} (\theta, \varphi )$ are the so called Spherical Harmonics. They determine the angular momentum of the atomic wave function $\Psi$ and its characteristic shape, i.e. $\mathrm{s}$ orbitals are spheres while $\mathrm{p}$ orbitals are dumbbell-shaped (side note: the Spherical Harmonics are complex functions which are rather hard to visualize; the shapes that every chemist is used to comes from the usage of the so called real Spherical Harmonics, which are just linear combinations of the complex-valued Spherical Harmonics).


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Nitrobenzene reduction with (tin) Sn catalyst: Why is C-H bond cleavage preferred over O-H bond cleavage?
I am trying to understand this mechanism for nitrobenzene reduction with $\ce{Sn/HCl}$. I do not understand why it implies that you pull off the $\ce{C-H}$ hydrogen instead of the $\ce{O-H}$ hydrogen in the methanol.  Is it actually suggesting that the resulting $\ce{.CH2-OH}$ radical formed from the methanol eliminates to form formaldehyde or do you get a rearrangement where you form $\ce{(CH3-O)-}$? If the latter is the case, why not just pluck the proton from the alcohol group instead of breaking the $\ce{C-H}$ bond?
Also, it mentions that organic solvents are preferred for this reaction, why exactly is that over water?

I'm skeptical about the mechanism, but it's worth putting it out there that in radical reactions C–H bonds are easier to cleave than O–H bonds due to their lower bond dissociation energies.

I don't buy that mechanism.  Like you noted, breaking a C-H bond in methanol seems unlikely, especially since the reaction is run in the presence of acid, so a ready source of protons is available.
Most of these reactions that involve a metal surface proceed through what is known as a single electron transfer (SET) mechanism.  Reductions of nitro compounds, be they catalytic hydrogenations or metal surface - acid catalyzed reductions, usually proceed through nitroso and hydroxylamine intermediates.  This link provides some discussion of the reaction and presents the following (more reasonable) mechanism where electrons are transferred from the metal surface to the various intermediates and the acid provides a ready source of protons.  These reactions are usually run in an organic solvent in order to bring the organic nitro compound into solution.

No problem, though I have to say, I really dislike SET mechanisms. There is too much unknown about them, they seem to involve some kind of magic... (And I also would suggest, that the nitroso intermediate will be skipped in this.)

@Martin "the nitroso intermediate will be skipped" Why? See http://books.google.com/books?id=tBxxZclgKyMC&pg=PA398&lpg=PA398&dq=reduction+of+nitrobenzene,+nitroso+intermediate&source=bl&ots=_ZhVohGO6j&sig=hEYPU0ixloZpB6wNQfLcwSQ9USY&hl=en&sa=X&ei=WqT-U-37B9D6yATplYD4Cw&ved=0CDUQ6AEwAg#v=onepage&q=reduction%20of%20nitrobenzene%2C%20nitroso%20intermediate&f=false

I would expect the oxygen to be protonated all the way, at least to the major extent. In your scheme you go from $$\ce{R-N(OH)(OH2)+ ->[-\ce{H2O;\ - H+}] R-N=O ->[+\ce{H+}] R-N=OH+ },$$ but I would assume that it proceeds like this: $$\ce{R-N(OH)(OH2)+ ->[-\ce{H2O}] R-N=OH+ <=>[\pm\ce{H+}] R-N=O}$$ So you can trap the nitroso compound, but it is not actually part of the mechanism.

Is there any particular reason why you use zinc here? I assume it does not make a big difference, but the OP asked about tin.

@Martin No particular reason, it was an available drawing of the mechanism and zinc, tin , mercury amalgam, etc. should all proceed by the same mechanism.

Thanks for the nice discussion! I know that this has been studied on semiconductors as well such as CdS where the electron is photogenerated. It seems like the important part is the H+ and the electron.
@Martin- I think that the -NO product requires 2 electrons while -NHOH requires 4. If that is the case would't you form the -NO before the -NHOH if you are going to form it at all?

In the mechanism you are considering, there are two options for reduction of R-$\ce{N^{.}H}$ (and similar for other neutral radical species)

$\ce{CH3OH + R-N^{.}H -> RNH2 + C^{.}H_2OH }$
$\ce{CH3OH + R-N^{.}H -> RNH2 + CH3O^{.}} $

In the first option (1) nitrogen oxidizes carbon - rare, but possible, as carbon has lower electronegativity. In option (2) nitrogen oxidizes oxygen - an element that has higher electronegativity, and this is unlikely.
In addition, in option (1) free radical is stabilized by forming 3-electron 2-center bond $\pi$-bond C...O, while in option (2) such stabilization cannot occur.
Taking proton (i.e. hydrogen cation) from $\ce{OH}$ is not an option for reduction, and $\ce{R-N^{.}H}$ is a radical, and expects hydrogen atom, not ion. Taking proton will produce $\ce{R-N^{+.}H2}$ and $\ce{CH3O-}$ and the first particle is not much better than the original $\ce{R-N^{.}H}$ while heterolitic bond breaking is generally less favorable, than radical one.
Proton may be accepted by negatively charged radical anions, like $\ce{R-N(O)O^{-.}}$ but after that they will seek a hydrogen atom for reduction.
Of course, it IS theoretically possible, that the radicals are actually reduced by tin and all the reaction happens on metal surface, and this is likely the case if the reaction happens in water, but in case it happens in organic solvent, especially reducing one like methanol, it is theoretically possible that it does participate in reduction. Until proper proofs, like trapping of the intermediate species, are obtained, any mechanism is no more then speculation anyway.

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Bromination Pathways with alkane, alkene, and alkyne substrates
I'm confused about halogenation pathways. From what I understand, there are multiple halogenation pathways. I'll try summarizing them below; let me know if I do anything incorrectly.

Halogenation with the diatomic halogen - i.e. $\ce{Br2}$. This reaction proceeds through a halonium intermediate.
Halogenation with the diatomic halogen in the presence of light or heat. This reaction proceeds via radical intermediates. From what I understand, the light or heat cleaves the weak halogen-halogen bond and we get two radicals, which can initiate the radical reactions. Also the cool thing about this reaction is that it doesn't require an nucleophilic pi bond, so if we were starting from an alkane, we don't even need to first form an alkene or alkyne; we can free-radical halogenate that alkane and then eliminate a beta proton as to make the alkene. All the other reactions mentioned here require some sort of nucleophile to start with.
Halogenation in the presence of water - i.e. $\ce{Br2}$ with $\ce{H2O}$. Again, we go through a bromonium intermediate. The $\ce{Br-Br}$ bond is instantaneously polarized and at that point we can form the bridged halonium ion and that gives water an entrance - the ring is highly strained and has partial charges all over the place - perfect for water to do something.
Are these all the halogenation pathways I need to know for an elementary understanding of organic chemistry? Did I miss any? I think I finally clarified all of them to myself through typing up this post ...
Oops yes I missed a few. NBS for example is used for allylic (or benzylic) bromination, and this directly implies the need for a starting alkene ... so this isn't as synthetically useful as free-radical bromination in the presence of light.
Addition of $\ce{HX}$ - the pi bond acts as a simple nucleophile and attacks the proton. Then the anionic halogen adds to the resultant carbocation.
Addition of $\ce{HX}$ in the presence of peroxides - the pi bond again acts as a simple nucleophile but this reaction proceeds via a radical intermediate and as a result the reaction is overall anti-Markovnikov.

Radical bromination is not very selective and hardly controllable.

That's true but it also gives you at least something if you have a plain alkane.

Basically everything you've said here is correct. You will probably cover more halogenation reactions as you continue to study organic chemistry, but if you're limiting yourself to starting from alkanes/alkenes, then you've got it.

I wanted to offer some commentary on the points you've enumerated:

You mentioned halogenation of alkenes in the presence of $\ce{H2O}$. This reaction is known as halohydrin formation. It's worth mentioning that this reaction can also be performed in the presence of other nucleophiles, perhaps most commonly alcohols and salts of other halogens to give mixed dihalogenated products. (Other types of nucleophiles may also successfully react, but I can't recall seeing any specific examples.) The reaction isn't always efficacious. Frequently, addition of both halogen atoms (i.e., ordinary halogen addition) will occur; the proximity of the halide leaving group to the halonium bridge will often result in a rapid reaction before any other nucleophile can intercept.
Radical halogenation can be very difficult to control, even when dealing with substrates that should exhibit high regioselectivity. NBS (and its chlorinated counterpart, NCS) is actually an extremely useful and versatile reagent which mitigates some of the practical pitfalls of radical halogenation reactions. Under photocatalytic conditions, or with sufficient heating and a radical initiator, it can react by free radical pathways. It's often preferable to using the diatomic halogen directly (unless exhaustive halogenation is really the goal) because it generates lower concentrations of $\ce{Br2}$ in situ, so the reactions are somewhat more controllable and side-products from direct addition to $\pi$-bonds are minimized. It can also react analogously to $\ce{Br2}$ with alkenes to give a halonium ion, which makes it preferable for halohydrin formation (since there is no $\ce{Br-}$ leaving group present to immediately attack the bromonium bridge).

Overall, my opinion is that you have a good core knowledge base re: halogenation reactions within the confines of introductory organic chemistry. Of course, there are a multitude of other halogenation reactions operating on myriad functional groups, but I presume those aren't germane to this particular question.

Diatomic halogenation with alcohols give rise to ethers right? The general mechanism looks like the alcohol attacking the halonium ion and then the alcohol is deprotonated and boom, ether, right?

@Dissenter, yes, that's right.

Thanks; also I did not know about NCS. I'll have to remember that and add that to my arsenal.

@Dissenter, yes, there's also apparently an N-iodosuccinimide (NIS). NBS really is a tremendously useful reagent, and it has numerous specialty applications far beyond the scope of what I've described. I can also attest to the fact that it's much more pleasant and less hazardous to handle in the lab.

Reminds me; I once tried using H2S in a synthesis and my prof told me never to use that; it would likely kill me if I tried using that in real life.

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Designing a theoretical synthesis of methamphetamine from phenylpropyne

I am asked to synthesize "methamphetamine." Note that there is no configuration designated for this molecule, and methamphetamine refers to a very specific configuration. In any case, here are the steps I took:

Reduction with hydrogen gas and Lindlar's catalyst. This transforms the alkyne into a cis alkene.
Epoxide formation with m-CPBA.
Ring opening with sodium methylamide. This is a base, and it should attack the less sterically hindered side of the epoxide - the side without the phenyl substituent. This opens the ring and gives me almost what I want. Now I just have to remove the hydroxyl group.
Removal of hydroxyl group: I decided to use TsCl and heat. The tosylate group should leave upon heating and a secondary benzylic carbocation that'd form. Sodium hydride was added as to add to the carbocation formed when the TsCl leaves. However, a possible issue is that the hydride anion, being a good base, would give me the elimination product as well.

Alternate pathway below

Did you take the tosylation of the secondary amine as a possible side reaction into account? ;)

Supposed that the tosylation of ephedrine works as expected, there still the chance to end up with 1,3-dimethyl-2-phenylaziridine upon heating of the tosylate.

I like your second route better, but you still have to worry about E2 competing with SN2 in the last step. Greg E.'s suggestion below is likely the best, which is WW's favorite route.

That's quite an endorsement for Greg's pathway!

I think the synthetic route you've proposed is quite good for the most part. However, I think your use of $\ce{NaH}$ to remove the tosylate is incorrect. $\ce{NaH}$ is a very strong base, but not a good reductive hydride donor in most instances. Here, it will almost certainly just react by eliminating the tosylate to give a double bond. However, LAH (and various similar reagents) should achieve the desired effect. Alternatively, one of the more popular dehydroxylation reactions is the Barton-McCombie deoxygenation.
An alternative route might begin with hydroboration-oxidation of the alkyne directly. Generally, when the two ends of the alkyne are equally substituted, one obtains a mixture of isomers, and electronic factors are usually more decisive than sterics. That being said, in this case, I suspect that it might be possible to control the regiochemistry by selecting a sterically hindered borane, perhaps 9-BBN or catecholborane. I'm not certain, however. Assuming one can indeed install the ketone at the end of the alkyne further from the phenyl group, then reductive amination should give the desired product. Direct reductive amination should be possible in this case (e.g., with $\ce{NH2Me}$, catalytic acid, and $\ce{NaBH3CN}$), allowing for a one-pot reduction.
Hence, the proposed reaction scheme would be:
\begin{align}\ce{
Ph-C#C-CH3 &->[\mathrm{1)~9-BBN}][\mathrm{2)}~\ce{H2O2,~NaOH}] Ph-CH2-C(O)CH3\\\\
           &->[\ce{NH2Me,~[H+]}][\ce{NaBH3CN}] Ph-CH2-CH(CH3)NHMe
}\end{align}
An additional point in favor of this approach is that it can be made highly enantioselective by use of asymmetric catalysts (q.v., Noyori asymmetric hydrogenation).
Addendum
In response to your second reaction scheme, jerepierre's comment above is definitely correct about a competing elimination reaction in the last step being a problem. It's especially likely because the proton transfer is fast, $\ce{NaNHMe}$ is extremely basic, the elimination is highly entropically favored, and the resulting $\pi$-bond is conjugated to the ring, giving an especially stable product (the major elimination product by a large margin should be the Zaitsev one). In sum, I think both kinetic and thermodynamic factors are heavily against you here. If you really want to go the route of nucleophilic substitution to attach your amine, better approaches are probably the Delépine reaction, any of the azide reduction methods, or the Gabriel synthesis (approximately in that order, I think). All of those suffer from the fact that you would then need to methylate the resulting primary amine, and avoiding extensive polyalkylation is more or less impossible, as far as I know.
There's also an issue before that stage, in that hydroboration-oxidation will not predominantly install the alcohol at the correct position, giving mainly the benzylic alcohol instead, unless you use a suitably bulky borane to achieve the desired regioselectivity. Since you only want to replace the alcohol with a bromine, I think $\ce{HBr}$ in the presence of peroxide (or some other radical initiator) should work, with no intermediate step and a reasonably low proportion of unwanted side-products.

"Here, it will almost certainly just react by eliminating the tosylate to give a double bond." Why wouldn't H- be a good reductive hydride donor? Does this have to do with solvation?

@Dissenter, I suspect that it's because it's a very strong base, proton transfers are fast, and $\ce{H2}$ will bubble out of the reaction mixture (providing an entropic driving force for deprotonation). It's also so reactive that a very limited range of solvents are acceptable, so that one normally forms slurries/suspensions and reactions take place at the solvent system interface. I'm sure there are other factors (including the lack of coordination of the target group, as seen with LAH), and HSAB theory might be applicable as well.

@Dissenter, that explanation is mostly speculative, though. I've never actually seen more than a very superficial discussion of its (in)efficacy as a hydride donor with organic substrates. Generally, it's simply presented as a very strong, almost completely non-nucleophilic base.

@GregE. Assuming that Ph is larger than Me, don't you want a small borane?

@jerepierre, well, we want the boron to add to the carbon further from the phenyl ring during the hydroboration step prior to oxidative workup. The alkyl groups attached to the boron will be roughly on on side of the alkyne while the hydrogen will be on the other. With bulky groups, they should orient themselves opposite the larger group attached to the alkyne, hence my thought that we want a large borane.

@jerepierre, compare the yields in this schema, for example. The bulk of the isopropyl presumably repels the alkyl groups of the bulkier 9-BBN and leads to more selective addition of boron at the less hindered position, despite both carbons of the alkene being monosubstituted.

@jerepierre, a few more schemas I found: 1, 2 (page 2). Granted, these examples all involve alkenes, so there may be factors specific to alkynes that I'm overlooking. However, with a small borane, we will get predominantly addition at the wrong position (the one adjacent to the ring), see here.

@GregE. Ah, sorry! I had it totally reversed in my head. You're absolutely right.

@jerepierre, no problem, and thanks for the confirmation. I've searched a bit for literature on regioselective hydroboration of alkynes specifically, but thus far to no avail. In principle, I think the same reasoning should apply.

There has been explosive growth in the use of hydroamination reactions to synthesize organic, nitrogen-containing compounds over the past 10-15 years.  It seems that regioselective addition to produce either the Markovnikov or anti-Markovnikov product can be attained by proper selection of catalyst and/or reaction conditions.  Both terminal and internal alkynes can react, aliphatic or aromatic amines can react, and the reaction can be run in an inter- or intra-molecular (e.g. the amine is already part of the molecule containing the alkyne, so a ring is formed in the process) fashion.
Here are two model reactions that suggest this methodology can be used to produce your synthetic target in a one pot, two step reaction process.  The catalyst used in these reactions is bis(indenyl)dimethyltitanium(IV).  The first step in the reaction involves addition of the amine to the triple bond, and the

second step involves reduction of the resultant imine to the corresponding amine.  It appears that if you

were to run this reaction using your alkyne and methylamine you could expect overall high yield and regioselectivity to produce methamphetamine in a one pot process.  Here is a link to a review (see page 291-2 for more information on these specific examples) that includes more references to this reaction.


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Are there any negatively charged electrophiles, or simply neutral and positive?
When reading up on electrophilic aromatic substitution, most resources refer to electrophiles as positively-charged species, "in general" or "mostly".
What are non-positively charged electrophiles? Are they simply neutral atoms/molecules more willing to gain electrons due to having less grasp on their own electrons? Are there any, as strange as it sounds, negatively charged electrophiles?

You can always create a molecule where the negative charge (generally some deprotonated functional group) will sufficiently far from the electrophile group that its charge will have no influence on the electrphile's reactivity.

Yes, there are many neutral electrophiles.
One large class are the compounds with leaving groups. Some that you have likely encountered are: acids (HCl, HBr, sulfuric acid, etc) which are electrophilic at the hydrogen; unhindered alkyl halides/sulfonates (e.g., methyl iodide for SN2 reaction); dihalides (e.g., Br2 for electrophilic addition to an alkene).

A second large class are compounds with pi bonds to an electronegative atom. Most importantly this class includes the carbonyls (aldehydes, ketones, esters, etc). Michael acceptors fall into this class as well, which have an extended pi system containing an electronegative atom.

Since you are covering electrophilic aromatic substitution, soon you may learn about nucleophilic aromatic substitution, where a highly electron poor aromatic ring (e.g., having multiple nitro groups) bearing a leaving group can be attacked by a strong nucleophile.

In all of these cases, although the compounds are neutral, there is a partial positive charge at the electrophilic site explained by a polarized bond (acids, alkyl halides), resonance (carbonyl), or spontaneously (dihalides).
Neutral electrophiles are less reactive than their positively charged counterparts, which means that the nucleophiles are typically stronger for a reaction to occur.
As for negatively charged electrophiles, I can think of only a couple of cases that could be considered that way. For example, in the reduction of a carboxylic acid with lithium aluminum hydride, the first step is deprotonation of the acid to give a  carboxylate. LiAlH4 is very reactive and adds a hydride to the carboxylate. That could be seen as nucleophilic attack on a negatively charged electrophile, and many people will draw it that way. In reality though, there is something stabilizing the negative charge, either the lithium or aluminum. So it's not really a negatively charged carboxylate that's being attacked. I don't think that there are really any negatively charged electrophiles, but someone may have an example I haven't seen or thought of.

Well, if we take a broader look, strong neutral, and weaker negatively charged electrophiles are very common in inorganic chemistry - all Lewis acids are electrophiles, and even negatively charged complexes with number of coordination lower then optimal are still Lewis acids and therefore electrophiles.


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Homemade filter paper
I used normal paper from my notebook to perform filtration. The filtration is 100% perfect, but takes incredibly long to happen, because the paper is too dense.

I'm not planning to buy filtration papers. They're cheap, but I don't filter something every day, so they'd be laying around useless. I would prefer to be able to use some household item as a filtration paper.
One idea I had was to apply some substance on normal writing paper to make it softer.
I have tissues which are very soft, but these are too soft and will leave paper chunks in my filtrate.
Do you have any idea what should I use for filter paper, or how could I alter normal paper to suit my needs?

Apparently, the particles of your solid are either very small, or / and not crystalline.  This may complicate filtration, though this is one of the reasons filter papers used in the chem labs are sold in different grades of paper hardness.
A practical alternative to separate such a slurry from the liquid phase would be to substitute filter and funnel by a centrifuge.  With some luck you still find some with a manual crank (with conic tubes, about $\pu{20 mL}$ each). Based on own experience, this often is faster than setting up and then performing a suction filtration over a glass filter. On small scale, it is simpler to wash the slurry, and clean up the tools, too. Just keep the opposite tubes balanced and don't mind the surprised gaze of your colleagues around.

(credit)

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Nitrosation of benzene
How does benzene react with $\ce{NaNO2/HCl}$? Will $\ce{PhNO}$ be obtained? I read that pyridine can be produced under special conditions, but what will happen in some less exotic conditions?

Afaik, $\ce{NO^+}$-cation is too weak as electrophile to react with benzene, nitrosobenzene is obtained indirectly.

As you noted, the link you provided was to a reaction run under rather exotic conditions (low-temperature plasma). Under more everyday laboratory conditions, for example, in solution, activated aromatic rings can react with $\ce{NO^+}$ as shown below.

However if there is a ring-activating amino substituent present, then the $\ce{NO^+}$ will react with the nitrogen to produce the synthetically important diazonium ion.


text:
Cell Representation of a galvanic cell:
I know basic cell representations of galvanic cells and can identify the anode/cathode. But here's a weird representation that I encountered and couldn't understand it:
$$\ce{Pt | H2(g)}(\pu{1.0 atm})\ce{ | HX || KCl(1M) | Hg2Cl2(s) | Hg | Pt}$$
What does this actually translate into?

You will probably know that the notation for galvanic cell reads from left to right as anode || cathode, where || typically denotes a salt bridge.
Within each half cell, different phases are separated by |.
In your case, the anode is a platinum rod in an acidic solution under a hydrogen atmosphere. This is a Standard hydrogen electrode (SHE).
The cathode consists of a platinum rod which is contact with a layer of mercury and a layer of solid mercury(I) chloride. It is an aqueous cell with 1M KCl solution as the medium. This electrode is known as Saturated calomel electrode (SCE).
Both electrodes are used as reference electrodes.

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text:
Scary jelly forming on zinc anodes
I'm continuing with my electrolysis experiments. Over a long time, I seek a perfect electrodes. It seems, that when you pick the right anode and cathode, it might be a lot cheaper than universal super-electrode.
The cathodes are usually OK, since (+) ions do not hurt metals. However, $\ce{O-}$ and $\ce{Cl-}$ are both very agresive and will, sooner or later, destroy any anode I found so far.
My latest experiment was Zinc. Zinc is often used as non-corrosive layer added on iron via electro-plating. So I used it as an anode in $\ce{NaCL}$ aqueous solution. The anode I used was a piece of iron plated with zinc layer.
This scary jelly is forming on it. I also do not see any bubbles at all coming from the electrode - which mean that both chloride and oxygen ions are reacting.

The anode is consumed during the process - but what is produced? The jelly does not burn. When heated intensively, it bubbles and smells like chlorine. Does it decompose?

The orange colour is caused by the iron centre of the anode being oxidized. I tried other pieces of zinc I could find - they all produced this jelly.
Keep in mind, that solution does contain following additional compounds formed from salt and water:

$\ce{NaOH}$
$\ce{HCl}$
$\ce{Cl2}$

It looks like iron(III) hydroxide.

It forms also on zinc with no iron on it.

Where did you obtain the zinc anode from? I only ask since some metals, depending on what they were intended for, might have some sort of coating on them.

The zinc itself is coating. The electrode I used was iron coated with zinc - as the image and my question suggests. I said that zinc is often used as coating.

I also tried zinc from a battery. Same effect.

It wasn't actually clear whether it was galvanised iron or not, but thanks for clarifying.

http://en.wikipedia.org/wiki/Zinc_alloy_electroplating | here we see that zinc plating may contain other metal, say, iron. Zinc may form white gelatinous hydroxide, but it is white, the coloration is definitely from iron (III) (nickel would give greenish and cobalt probably would give pink or black). In case of zinc from alakaline battery, manganese oxides also may be brownish.

@permeakra In my original post, I said: "The orange colour is caused by the iron centre of the anode being oxidized." Thanks for proving me right. The zinc from battery formed mostly transparent jelly.

Judging from the color, this clearly looks like more or less hydrated $\ce{Fe(OH)3}$ or $\ce{FeO(OH)}$, and it could also contain dissolved $\ce{FeCl3}$ formed by the reaction of iron with $\ce{Cl2}$ evolved at the anode. The fact that it smells like chlorine when you heat the gelatinous precipitate would support the latter assertion, but the smell could also arise from residual $\ce{Cl2}$ which was still dissolved in the water of the jelly. Heating $\ce{Fe(OH)3}$ or $\ce{FeO(OH)}$ intensively above 200 °C will give you $\ce{Fe2O3}$ by thermal decomposition (dehydration).
\begin{aligned}
\ce{2Fe(OH)3 ->& Fe2O3 + 3H2O} \\
\ce{2FeO(OH) ->& Fe2O3 + H2O} \\
\end{aligned}
During electrolysis, the zinc layer on the electrode is degraded by pitting corrosion of $\ce{Cl-}$ and oxidation in the presence of oxygen. When the coating is completely removed by the corrosion in some places, the exposed iron reacts with oxygen and chlorine to the products mentioned above. On anodes made of pure zinc, $\ce{Zn(OH)2}$ will likely be the primary product, which will form a transparent to white, jelly-like precipitate. Zinc hydroxide formed by oxidative degradation of the $\ce{Zn}$ layer would also be a constituent of the precipitate on a zinc-coated iron anode.


text:
Distillation: Simple vs Fractional
Here is the excerpt of my lab manual:
"Simple distillation is more useful for purifying a liquid that contains either a non-volatile impurity or small amounts of higher or lower boiling impurities. Fractional distillation allows for several condensation-condensation cycles in a single operation. It can be used to separate liquids with comparable volatilities and to purify liquids that contain relatively large amounts of volatile impurities."
So, why does simple distillation put emphasis on small amounts and fractional distillation, on large volumes?

So, why does simple distillation put emphasis on small amounts and
  fractional distillation, on large volumes?

It's not small and large volumes of material being distilled, but rather small and large volumes of impurities present in the material to be distilled.  Both simple and fractional distillations can be used to distill large quantities of material.
The efficiency of a distillation column is measured in "height-equivalent theoretical plates" (HETP), the more plates the more efficient the column, the more efficient the column the more likely that it will be able to separate two closely boiling materials.  If you have a small amount of a lower boiling impurity in your solution and you want to separate it from a large amount of the desired material, then you can probably get away with using a simple distillation column and discarding some of the forerun (the first drops of liquid that distill over).  On the other hand, if there is a large amount of impurity and it boils close to the desired product, then you will need a column with a higher efficiency (fractional distillation column) to effect a good separation between the two fractions (or "cuts" as they're sometimes called) and get pure product.
To sum up, perhaps the original question would be better written as

So, why does simple distillation put emphasis on small amounts and
  fractional distillation, on large amounts of impurities?

and the answer would be that even an inefficient (fewer HETP) simple distillation column has enough power (enough theoretical plates) to separate a small amount of impurity from a large amount of product, especially if you are willing to discard the initial fore-run from the distillation.  If there is a larger amount of impurity, then we may need a more powerful (higher efficiency column) to produce the desired separation.
The higher the efficiency of a distillation column, the greater its ability to separate two things.  If we have to separate two materials that are both present in large quantity, or distill at boiling points close together, then we need a more efficient column than if they boiled further apart or one was present in only a small amount and we were willing to throw away the forerun (e.g. throw away some of the desired product in that early fraction that contained the impurity).  When we call a column a fractionation column, we're really saying that it has greater separating efficiency than a simple distillation column.  The more efficient the column, the better the job it will do on difficult separations.


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Deamination of Spermidine?
I was making a solution of spermidine, and noticed that Sigma Aldrich warns that spermidine solutions deaminate over time. I tried to think of how that would happen, and made a diagram:

Assuming some of the spermidine has the secondary amine in an uncharged state, the lone pair electrons can attack either charged terminal amine group. I don't know which is more likely, I have drawn the 4 membered ring, but the other amine would make a 5 membered ring, which would probably be more stable. Either way, this would kick off an ammonia, which would be stable enough in water. Then a water molecule would attack a carbon on the ring and open it, making an OH2+ charged group, which would give up a proton. Of course the protons would be balanced because the ammonia would pick up a proton.
Does this make sense? It reminds me of nitrogen mustard chemistry, but must not be as reactive because spermidine is used to condense DNA, and nitrogen mustards are used to damage DNA.
Additionally, spermidine seems hard to work with. I tried to make 10mL of a 10mM solution, and calculated what volume to use to get the right mass, because spermidine melts at 20-25C. However, when I used that volume, I had about 2.5 times as much mass as I wanted. I assume this is because the spermidine was not totally thawed out, and the density was higher than I assumed (925mg/mL at 25C). Has anyone worked with this compound and has suggestions for how to handle it better?

What solvent is the spermidine shipped in?  Was it packaged under an inert atmosphere?  Has your sample been exposed to air?

It's shipped as pure spermidine, it just has a melting point at approximate room temperature. I tried to keep it cold and weigh out a solid, but it was very hard to scoop any out, and what I managed to get just melted on the spatula. I assume it was packaged under nitrogen because the label says to store under inert atmosphere, and it was exposed to air when I opened it. It's a cheap compound, I don't need much, and I have 4 other bottles, so if I have to discard the opened the container I can.

Here are some links to the mechanism of the degradation of spermidine and some analogues in living systems.

reference 1
reference 2
reference 3
reference 4

Based on these reactions I would guess that an oxidative pathway is followed here as well, once the sample has been exposed to air.  Instead of a nice enzyme catalyzed oxidation, here we would have a free radical oxidation. Here is a general outline of such a process.  The first step would involve generation

of the more stable secondary nitrogen radical (as opposed to the primary terminal nitrogens) and formation of the nitrogen hydroperoxide.  Elimination of hydrogen peroxide could produce the two possible imines which could be further oxidized by oxygen to produce both possible diamines and both possible amino-aldehydes.  You could test your material for the presence of hydrogen peroxide, but the smell of putrescine ought to be a "dead" giveaway.

The spermidine itself smells bad enough. Not sure I'd be able to tell the two apart, but puns are always appreciated.


text:
Does changing the stoichiometric coefficients affect the Nernst equation?
If I take the electrodes as nickel and silver, I've according to the nernst equation that
$\ce{Ni(s) + 2Ag+(aq) -> Ni^2+(aq) + Ag(s)}$
Now $E=E_{0} - \frac{RT}{2F} \ln\left(\frac{[\ce{Ni^{2+}}]}{[\ce{Ag+}]^2} \right)$
But if I double the stoichiometric coefficients, the value inside the logarithm would square up. So if I write the same reaction but with different coefficients, I'll get different values of potential of the cell?

No, you won't get a different value. The number of electrons transfered in that case is 4, so it will be $E=E_{0} - \frac{RT}{4F} \ln\left(\frac{[\ce{Ni^{2+}}]^{2}}{[\ce{Ag+}]^{4}} \right)$.


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Boat conformation: axial hydrogens
In my exam, my teacher asked me if the following is true:
"The chair cyclohexane has more axial hydrogens"
I did not have time to think, so I just put false.
And apparently I was right.
But, at home, I thought to myself: Does this question even make sense? Axial hydrogens are by definition only seen on cyclohexanes according to what I know.

The two should have the same number of axial hydrogens.

Hmm, how is that?

Cyclohexane has 12 hydrogens and at any given time half are axial, and half are equatorial.

http://www.chem.ucla.edu/harding/IGOC/B/boat_conformation01.jpg Can you show me the axial and equatorial here?

The boat conformation only has pseudo axial and equatorial positions. What's the complete question?

It says fully:
TRUE OR FALSE:
Boat cyclohexane is a slightly higher energy conformer than chair cyclohexane because:
(a).... (a sub question) T OR F
(b).... (a sub question) T OR F
(c)chair cyclohexane has more axial hydrogens. T OR F

It says fully: TRUE OR FALSE: Boat cyclohexane is a slightly higher
  energy conformer than chair cyclohexane because: (a).... (a sub
  question) T OR F (b).... (a sub question) T OR F (c)chair cyclohexane
  has more axial hydrogens. T OR F

The answer to part (c) of that question is "False."  It turns out that the chair and boat conformations of cyclohexane both have 6 axial hydrogens - so the chair does not have more axial hydrogens.  Here's a picture of the chair (on the left) and boat (in the middle), you should be able to see the 6 axial hydrogens in both the chair and the boat.

Ron, I still can't see all the axial hydrogens on the boat. I only see 4, the ones that point down in this picture.

@yolo123 Good start, there are two more axial hydrogens that have special significance and one of them is labeled with an "f" in the middle drawing.  The "f" hydrogen is axial, but is often referred to as the "flagpole" hydrogen.  It causes steric problems with the other axial, flagpole hydrogen on C-4 (for reference, the original flagpole hydrogen is on C-1).  This flagpole-flagpole interaction is one of the factors that destabilizes the boat relative to the chair.

So, universally speaking, does the term "axial" hydrogen on boat conformations make sense? I always thought the term was restricted to chair conformations.

The term "axial" is widely used to describe the axial hydrogens in the boat.  Personally, I think it makes sense because if you look at a model, they look a lot like the axial hydrogens in the chair.

Are you a professional chemist :O?

Yes.  I need to use up 15 spaces.

If you want to be very precise (perhaps pedantic), then I would conclude that the question itself is a little dubious without additional clarification. Consider the IUPAC recommendation on Basic Terminology of Stereochemistry. It contains the following definition of axial with respect to ring conformations (emphasis mine):

In the chair form of cyclohexane ring bonds to ring atoms (and molecular entities attached to such bonds) are termed axial or equatorial according to whether the bonds make a relatively large or small angle, respectively, with the plane containing or passing closest to a majority of the ring atoms.

The point here is that the definition is contextually limited to chair conformations. Granted, this is only a recommendation, but I think it's a logical and authoritative one. Seen in that light, the question itself is malformed inasmuch as it admits either answer in the absence of an agreement on exact terms. To wit, "true" might be a valid answer, since by the strictest definition a boat conformation has zero axial hydrogens; on the other hand, if a similar definition (based on bond angles relative to a plane containing a majority of ring atoms) is applied to the boat conformation, one can identify six axial hydrogens roughly analogous to those in the chair conformation and the correct answer would be "false." Really, the question is murky unless you (and more importantly, your professor) are clear on the formal definition being used.

wrote, "The point here is that the definition is contextually limited to chair conformations."  But further down in the IUPAC link, the definition goes on to state, "The terms axial and equatorial have similarly been used in relation to the puckered conformation of cyclobutane, crown conformer of cyclooctane, etc."  In my understanding of these statements there is no limitation on the use of "axial" to chair conformations.

If we use apical not axial then all these problems can be circumvented.

@ron, in my opinion, the use of the term "axial" w/r/t other conformers and molecules is usually valid. This question, however, is itself proof of the potential for ambiguity with that usage. I can only assume there's a reason that the chair conformation was explicitly mentioned at the outset. While I'm definitely not claiming that IUPAC disapproves of related uses for the term "axial" in other conformers and molecules, that sentence you cite looks to me like a usage note, not an endorsement. I would argue that some explicit definition be given if the term is used other than for chairs.

In my exam, my teacher asked me if the following is true: "The chair cyclohexane has more axial hydrogens" I did not have time to think, so I just put false. And apparently I was right.

What's the complete question? The chair form has more axial hydrogens than what other form? If the test were referring to the other chair conformation, then you are right.
Cyclohexane has 12 hydrogens, and in its two chair conformations, six substituents are axial while the other are equatorial.

Basically, the question is does boat have more axial H's than
  cyclohexane?

The boat conformation doesn't have any true axial positions. I'm not sure what your professor considers to be axial or not, but if we take the flagpole hydrogens to be "axial," then both the chair and the boat should still have the same number of axial substituents.

TRUE OR FALSE: Boat cyclohexane is a slightly higher energy conformer
  than chair cyclohexane because: (a).... (a sub question) T OR F
  (b).... (a sub question) T OR F (c)chair cyclohexane has more axial
  hydrogens. T OR F

Regarding the question now in context the instability of the boat conformation likely has more to do with flagpole-eclipsing interactions rather than the number of hydrogens.

Dissenter, I may not have been clear. The teacher asked to compare boat and chair conformation. Basically, the question is does boat have more axial H's than cyclohexane?

Do you refer to flagpole H's as the ones attached to C4 and C1? In that case, there is only 4H's that are "AXIAL" according to the "hypothetical definition" of my teacher. Am I lost?

@Dissenter, your chair conformation next to the boat one, is incorrect.

C4 and C1 have flagpole hydrogens; the Hs on the other carbons may be considered to be axial.

Dissenter, so sorry for not having been clear from the beginning. I'll edit my question and I'll scan my exam. Please comment again for any thing you may want to add.

Ok, I uploaded the question so you can adjust your answer.


text:
What is the exothermic reaction of superglue with cardboard?
I glued a sheet of paper to a cardboard package using superglue (labelled with ethyl 2-cyanoacrylate).
To my surprise the paper became very warm.
What kind of reaction is it?

I believe this is due to the fact that cyanoacrylates polymerize by anionic (or sufficiently strong neutral nucleophile) addition. In particular, they react with hydroxyl ($\ce{-OH}$) groups; paper, being primarily cellulose, is abundant in hydroxyls. The bond formation involved in the polymerization reaction is inherently exothermic (generating a $\ce{C-O}$ bond on initiation and many subsequent $\ce{C-C}$ bonds as the polymerization proceeds), and with a large surface area and many reactive groups, I could certainly imagine the reaction being surprisingly violent. The Wikipedia page on cyanoacrylates notes its reactivity with cotton fibers (again, being chiefly composed of cellulose), and the MSDSs for various similar compounds warn of exothermic reaction with paper and similar materials (q.v., 1, 2, 3).


text:
What chemicals are used in receipt paper?
Most "invisible inks" turn brown when heated and they take a while to transform.  I would like to make (or obtain) an "ink" that responds to the heat of an iron and permanently turns black or a similarly dark color - instantly.  Thermal receipt paper does this where the paper is white until it is heated with almost instant response times AND the "ink" has a permanence of sorts (e.g. sticking a receipt in a freezer won't revert the reaction).  What chemical combinations in what amounts will do an equivalent transformation?
Bonus question:  What chemical combinations would create similar effects for standard palette colors?  That is, transparent/hidden until heat from an iron is applied, whereupon it instantly appears.
My goal is to use such an ink with paintbrushes and fountain pens on canvas.  Black is probably sufficient, but knowing how to make standard colors would dramatically expand my options.  The ink ideally shouldn't be toxic to touch after drying nor emit toxic fumes when heating with an iron at a "Cotton" setting.  It can be semi-toxic while wet but nothing that will melt nitrile gloves, fingers, or human lungs.  I will, of course, wear sufficient protection at all times as I don't really need any chemicals leeching through my skin.  Having a drawing/painting appear instantly in front of an audience from a seemingly blank canvas would be very dramatic and painting with invisible ink will be an interesting adventure.

Most paper used for receipts now are not thermal paper. If you scratch hard enough (even with a fingernail) you can leave a line.

Have a look at wikipedia article on Thermal Paper. The whole trick is in the three components, which are mixed as solid substances and applied as coating to the paper. When you heat the paper, the solid substances melt and the coloring reaction occurs.
For this you need fine pulverized chemicals, stirred in a form of suspension in some volatile solvent and apply them on the canvas. It should not be a big problem, see e.g. patent on Thermal Paper.
If all this effort is for an artistic purpose, I would suggest as a first try to approach the companies manufacturing the thermal paper (there should be lot of them) and ask them for a bottle of the prepared suspension, telling them your plans. The success rate would not be high, but hopefully non-zero and it saves you 99% of work.
Additionally, the painting itself could be quite fun, I would expect the components to be UV-active, so the black light could help you.


text:
Black Smoke Produced from Regular Lighter but Not Torch Lighter Applied to Copper
I have a piece of copper mesh from which I would like to remove all coatings/oils. The copper comes from a pot scrubber like this. In order to remove any residue, I held a standard bic lighter up to it. It gave off black smoke. I moved the lighter around the mesh, waiting for it to stop giving off any such smoke, but seemed to do so indefinitely. Even when I just held the lighter in one place, it would not stop giving off a steady stream of black smoke.
I then did the same thing with a torch lighter on a fresh piece of copper from the same source. This time it was hard to tell if there was any smoke at all-- and if there was, it stopped being produced very quickly.
With both the normal lighter and the torch lighter, the surface of the copper turned gray-black. I assume part of this is from the formation of copper oxides, but part of it may also come from the deposition of incompletely combusted carbon on the surface. In both cases I was able to restore the shiny surface by putting the mesh in $\ce{HCl}$-based toilet cleaner for a few minutes.
Why is the normal lighter producing black smoke where the torch lighter is not? What does this smoke consist of?

Common copper wire, especially used in electronic components like inductors and transformers, is usually covered with insulating varnish. This is the common source of black smoke, sometimes it may even burn on its own. I doubt, however, that kitchenware is coated with such varnish, there is no point in it.
There are two types of devices with flames: ones where the fuel is premixed with air, that usually produce blueish flame, and ones where fuel is not pre-mixed with air, but burns and mixes with it in same time, that may produce bright yellow flame.
For many organic compounds, especially ones that require large amount of oxygen per molecule to fully burn, in case the fuel is not premixed with enough air, the flame usually contains oxygen-deficient region where tiny particles of carbon and molecular hydrogen are formed. These carbon particles, since they are heated, give thermal light emission and are responsible for bright non-transparent orange-yellow organic flames. A typical example of such flame is a flame of a candle. It is a flame of heavy hydrocarbons or fatty acids (c20+), that require tens of molecules of oxygen per molecule of the fuel to fully burn.
In case gases from this yellow part of the flame are cooled without mixing with more air, this carbon particles may be obtained in form of deposits or black smoke. For this, a cold effective conductor of heat (i.e. metal, especially copper and silver) or a solid surface with high heat capacity introduced into yellow part of the flame may be used. In some cases, when too many of such particles are produced and the flame is not particularly hot, there is no need even to cool the flame. This is the usual case of hydrocarbons with big amount of carbon, like aromatics, and chlorinated hydrocarbons, like PVC (as long as they burn at all, which often is not the case even if there is no much chlorine in the compound).
Propane torches usually have a mean to regulate amount of air premixed with fuel. Torch lighters are small propane torches, usually without such means. When propane in such torch is premixed with excess of air, a very hot blueish short flame is formed. However, when air is not premixed with propane, the flame is organized in layers with more oxidized outer layers and more reduced inner layers. In inner layers, the dominating reactions are thermal decomposition
$\ce{C3H8 = 3C + 4H2}$
and incomplete combustion
$\ce{C3H8 + 2O2= 3C + 4H2O}$
(this is an oversimplification, as numerous other intermediates are formed, but that's advanced stuff)
In outer layers, propane combusts completely, just as if it was premixed with air
$\ce{C3H8 + 5O_2= 3CO2 + 4H2O}$
and carbon particles (and other reduced products) from inner layers also combust here.
Now, to get carbon particles from the flame in form of smoke the key is to cool them before they mix with more air. An ideal solution would be to suck them with tiny pipe and cool with water. However, it seems, that skein of copper mesh, at least until it is heated enough, works good for this purpose, both cooling the inner layers and hindering their mixing with air.
Since flames of fuels that are not pre-mixed with air have very efficient emitter of visible and and infrared light in it (many carbon particles), they tend to be cooler, especially the inner part. The flames with fuel pre-mixed with air (like in torch lighters with blueish flame like this), the flame is thin, quite short and much hotter.
Copper wire, when heated in air, also oxidizes. If done properly, it is a good way to produce pure black copper (II) oxide. However, it usually stay on the surface of the wire and indeed, can be washed with hydrochloric acid. If there are black particles that are not dissolved in hydrochloric acid, they are carbon particles.

Thanks for your answer. It has good information but my question is not really answered. Why do you think the normal lighter is producing the black smoke, whereas the torch lighter does not? Also, do you know what the chemistry of the varnish/smoke is?

@SeanMackesey because in case of torch lighter with blue flame (is that the case?) like this http://www.southsmoke.com/pd-shark-series-torch-lighter.cfm the fuel is already pre-mixed with air, so no oxygen-deficient region arises and the flame is a good deal hotter. I doubt that in case of kitchenware any varnish was used, so the black smoke is likely to be plane soot (mostly carbon). Do I have to add it to the answer?

OK, yes, please do add it to the answer-- could you also explain why the soot would make a stream of smoke? I understand the deposition of soot on the copper, but I don't understand why it would be put off as smoke. If it's mostly carbon, shouldn't it combust to make $\ce{CO_{2}}$ and $\ce{H_{2}O}$?

@SeanMackesey I edited the answer. As for this question, see bold font.


text:
What is the wavelength of an electron?
The kinetic energy of an electron is $1.67 \times 10^{-17}\mathrm{J}$. Calculate the wavelength ($\lambda$) of the electron.
I know the formula $\lambda= \frac{h}{mv}$; where h is Planck's constant.
How do I find $\lambda$ without $v$ (velocity) and $m$ (mass)?

The mass of electron is never given in question you should remember it. Mass of electron is $9.109 \times 10^{-31} \mathrm{kg}$.
Now read the question Kinetic Energy is given, so from that you can easily find velocity of electron.
$$K.E.=\frac{1}{2}mv^2$$
so You will get $v= 6 \times10^6 \mathrm{m/s}$
Now you can substitute value of m and v in formula $\lambda = \frac{h}{mv}$
You will get wave length of electron as $0.12 \times 10^{-9} \mathrm{m}= 0.12 \mathrm{nm}$

You do not even need to laboriously calculate the velocity of the electron. All you require is the mass (which should be in your knowledge), and the kinetic energy - already given.
See, momentum P = mv;
And kinetic energy = 1/2mv*v
Now, if you multiply and divide the KE expression by mass m, you get -
(1/2)(mv)(mv)*(1/m) = (1/2)P^2(1/m)
So, all from here, you can cross-multiply and find momentum directly, and hence substitute in de-Broglie's formula to find the wavelength of the electron.


text:
Graphene Vs Copper?
I was wondering about the amazing conductive properties of graphene, lets assume a large copper bar that is 10 kg, current of 1kA and probably more can flow in it, what about graphene? It certainly can since its a better conductor, but it would be much lighter? In the milligrams possibly?
Also, can graphene replace the heavy busbars that carry MW of power?
Making it much lighter?

Currently Aluminum is the next best conductor to copper, when it comes to bang for the buck; but aluminum is only 61% as efficient at conductivity as copper is, making it less desirable. The weight of aluminum and its low cost are big plusses when compared to copper. What do you suppose you'd have to do to augment aluminum with graphene in order to increase it's conductivity, while keeping its low-weight and flexibility. If you could do that, you could revolutionize the power transmission industry, saving us all a ton of money.

The trick with graphene is that a lot of its amazing properties only work when you have continuous perfect sheets of it, and making graphene like this is currently beyond us, for large scales anyways. It is true that graphene has very high electron mobility $\approx10^{5}~\mathrm{cm^2/Vs}$ at room temperature, which works out to on the order of $10~\mathrm{n\Omega\cdot m}$ in bulk (which assumes you can make a perfect multilayer structure that maintains the properties of a single sheet). An impressive figure, but only $\approx40\%$ better than copper.
Of course, copper is about 6 times more dense than graphene, so if mass is the main concern, graphene would be a pretty good improvement. Still, we're not talking about replacing $10~\mathrm{kg}$ of copper with a few milligrams. Given how much cheaper and easier it is to make metal wires, we're not at the point of using graphene for this kind of bulk application. There are also some more mundane problems to sort out, e.g. graphene is brittle like a ceramic, which might cause mechanical issues.

I hope this amazing materials get to be more available in the near future, there are some discoveries that helps its mass production. I hope it can replace copper quite soon! When it's improved of course.

These things are always hard to predict, but I'm guessing that we'll see smaller scale electronic applications first—replacing ITO in touchscreens and the like. When we can make long runs with multiple layers, then maybe we can have the power transmission stuff, along with space elevators and other cool stuff. I hope I'm around to see them!

Hopefully, you and I both!

Michael, to get an idea of how graphene is better than copper, if a copper conductor can carry 1kA continuously a graphene conductor could carry x1000 times that amount? I'm trying to grasp how good is graphene's conductivity vs copper and silicon.

As I said in my answer, if you do a back of the envelope calculation from the electron mobility, it's something like 40% more conductive than copper.

I found something interesting: http://arxiv.org/pdf/0906.4156.pdf , $10^8$ $A$/$cm^2$ that is huge...

It's certainly an impressive number, but you have to be careful using it in comparison with other materials. Because the failure mode when conductors carry current is generally a thermal effect, the current capacity has a lot to do with how well the conductor can lose heat. A small wire has a much larger surface area to volume ratio than a large one and takes more current to heat. I expect graphene would fare better than copper of the same shape, but be careful extrapolating these numbers to macroscopic wires. For something big like a bus-bar, the current capacity may well be 5 orders smaller.

If the failure mode is due to the thermal effect, graphene is an amazing thermal conductor as well, it can transport heat quite fast. So maybe if the issue of heat can be resolved in the future it could be able to carry currents up to $10^6$+, still it's mind boggling that such a light 1-atom thick material can have such properties.

Well, if we take the thermal conductivity number in the paper you linked of ~1000 W/m-K, it's only a bit more than double that of Cu (400 W/mK), though since it's carbon, graphene may remain intact above Cu's melting point of 1358 K, and thus be able to operate under even higher current.

I also found another theoretical material that could conductor electricity at 100% efficiency(I'm assuming no resistance) that is "Stanene". And it's not a superconductor, has the potential to surpass that of Graphene, imagine if they could make it... unlimited current. The future looks bright.

Stanene is strictly theoretical at this point as was the theoretical hardest substance C3N4 for many years ($\ce{C3N4}$ is extremely expensive and produced in very small quantities in research laboratories - a lot like antimatter). Graphene is 1.4 times as conductive as copper by volume, but if you factor in density, it is 5.8 times more conductive (if a given sample of copper can carry $\pu{1kA}$ with a $\pu{1V}$ drop in voltage, the same weight, and length, of graphene, could carry $\pu{5.8 kA}$ with with a $\pu{1V}$ drop in voltage - in theory). Of course, fabrication methods can degrade this. For electrical power transmission, which depends a lot on weight, $\ce{Al}$ is 2 times as conductive as copper BY WEIGHT and is, therefore, the current choice for power transmission. The fact that it costs less is a bonus. A graphene/$\ce{Al}$ alloy might give a conductor with the fracture resistance near $\ce{Al}$ but better conductivity. This is an active area of research currently.

On the graphene/Al alloy, what do you make of this recent announcement? https://www.graphene-info.com/india-based-nordische-technologies-and-cipet-announce-aluminium-graphene-battery


text:
Enzyme Assay - pectinase
I am working on Enzyme Assay.I assayed the enzyme in different buffers from $pH\ 1-12.5$ However,the enzyme has good activities starting from $pH\ 1-10.5$ Is that possible to have enzyme activity in such broad range ?


text:
Separating citric acid from sugar
I'm trying to get as pure citric acid as possible from a mandarin. It's not because I desperately need citric acid, but because I want to achieve this.

I started by grating the mandarine and filtering the juice. After filtering overnight, I was left with slightly yellow liquid. I heated it while avoiding boiling and I reduced the volume to 50%. When volume was reduced, the solution started to become saturated and crystals appeared. At this point, I stopped.
Now, I'm looking for a way to remove sugar from the solution, as this is the most annoying part. It's sticky.
When crystals appeared while I was boiling the solution, I came to think, that perhaps that could allow to separate sugar. Citric acid has solubility in water 1477.6 g/L (20 °C), while sugar has 2000 g/L (25 °C). Wikipedia unfortunately doesn't provide values for the same temperature.
From the above it seems that I can have more sugar dissolved than citric acid. I think that means that citric acid will crystallise first. Does it work like this? Are the crystals I've produced the citric acid crystals?
Is there another way to separate the compounds? Does something react with sugar to produce precipitate that can be filtered?

This would be very complicated as mandarin juice would contain thousands of compounds. Do you mind if the citric acid is potentially contaminated with a lot of those compounds- mainly organic compounds?

No I don't. But I want it to be acid, rather than sweet. As I said, this is an experiment for the sake of experiment. I will do as good as I can, but there's no cost for failure.

I don't think citric acid not sugar (which?) would be a major component in this juice that would single out from other components. Anyway, crystallisation has very limited value in separating such a diverse mixture. You can purify sugar with crystallisation, but cannot remove sugar from something else this way.

I would guess you need a lot of mandarins to isolate a tiny bit of citric acid. You could try fermenting the sugar to alcohol, that will take some time, and I am not sure how well it will work. Of course you won't get rid of any other contaminating components.

Can you make an alumina column to separate the citric acid? http://www.sciencelab.com/page/S/PVAR/SLA1906

First for the amount of citric acid, let's assume 0.01mol/L (http://www.answers.com/Q/Do_lemons_and_oranges_have_the_same_amount_of_citric_acid), ~ 2g /L. Sugars wildly guessing 100g/L. Without mentioning other compounds, citric acid is better considered impurity, rather than constituent.
If I would have to do it, I would probably turn the citric acid into calcium citrate, separate (again very impure) and turn back to free acid.

Citric acid is relatively soluble in ethanol (62g/100g) whereas sucrose is relatively insoluble in ethanol (1g/170ml). There are many different kinds of sugars and sucrose is probably not the primary one in this extract; branched sugars... oligosaccharides (which have generally higher solubilities than sucrose) are likely, and I'm sure there are a few other organic acids involved with a crude, fruit extract, but the different solubility characteristics should be relatively helpful for extracting citric acid from sugars and some other organic acids.
So, I would try extracting the citric acid from the dried fruit or dried water extraction with 99%+ ethanol. I would then allow the alcohol to evaporate to help concentrate the acid.
Dilute citric acid (from a concentrated extract) can be precipitated with calcium hydroxide as practically insoluble calcium citrate, which is then reacted with sulfuric acid to form (food grade) citric acid and calcium sulfate (gypsum) as a recoverable by-product.
To really purify the stuff, an ion exchange resin or some other chromatographic method would be the last step.

Try micro-filtration with gelatin https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=3&cad=rja&uact=8&ved=0ahUKEwj_gs_7hMHNAhUG6SYKHXfyAnwQFggsMAI&url=http%3A%2F%2Fwww.nytimes.com%2F2007%2F09%2F05%2Fdining%2F05curi.html&usg=AFQjCNHNgFO_oZ5o_AfUU9dazQgt7wrzWw&sig2=ham-sb2aV1PEYkMSe8qutw&bvm=bv.125221236,bs.1,d.dmo
search gelatin filtration on google

Welcome to Chemistry! Whilst this may theoretically answer the question, it would be preferable to include the essential parts of the answer here, and provide the link for reference.


text:
How to calculate the solubility product for PbS from galvanic potential?

Knowing that at 25 °C the following galvanic cell:
  $$\ce{Pb~|~Pb(NO_3)_2~1M~||~PbS~saturated~|~Pb}$$ shows an $\mathrm{EMF} =0.413~\mathrm{V}$, find the $K_\mathrm{sp}$ of $\ce{PbS}$.

My Approach
This is a concentration cell based on $\ce{Pb^2+}$. Since $\ce{Pb(NO3)2}$ dissociates completely, while $\ce{PbS}$ is a salt with a low solubility, the left semicell will be the cathode and the right one the anode.
So we have the following semireactions:
\begin{align}
 \ce{Pb^2+ + 2e- &-> Pb}  && \text{(cathode)} \\
 \ce{Pb  &-> Pb^2+ +2e- } && \text{(anode)}
\end{align}
And for the anode we also have $$\ce{PbS <=> Pb^2+ + S^2- }$$ where $\ce{[Pb^2+]} = \ce{[S^2- ]}= \sqrt{K_\mathrm{sp}}$.
So the semicell potentials are:
\begin{align}
E_\text{cathode} &= E^\circ\\
E_\text{anode}   &= E^\circ - \frac{0.059}{2} \log_{10}{[\ce{Pb^2+}]}\\
\end{align}
Thus:
$$0.413~\mathrm{V} = E_\text{cathode} - E_\text{anode} = \frac{0.059}{2} \log_{10}\ce{[Pb^2+]} \Rightarrow \ce{[Pb^2+]} = 10^{14} $$
And: $$K_\mathrm{sp} = \ce{[Pb^2+]}^2 = 10^{28}$$
I'm sure that I'm wrong for a sign but I don't understand where is the error.

The galvanic cell must be written like this:
$$\ce{Pb | PbS saturated || Pb(NO3)2 \pu{1 M} | Pb}$$
in order to have a positive  $\mathrm{EMF} = \pu{0.413 V}$.
Your approach is right with a slight error.
This is a concentration cell based on $\ce{Pb^2+}$. Since $\ce{Pb(NO3)2}$ dissociates completely, while $\ce{PbS}$ is a salt with a low solubility, the left semicell will be the anode and the right one the cathode.
So, we have the following semireactions:
$$
\begin{align}
 \ce{Pb^2+ + 2 e^- &-> Pb} &\quad &\text{(cathode)} \\
 \ce{Pb &-> Pb^2+ + 2 e-} &\quad &\text{(anode)}
\end{align}
$$
And for the anode we also have
$$\ce{PbS <=> Pb^2+ + S^2-}$$
where $[\ce{Pb^2+}] = [\ce{S^2-}] = \sqrt{K_\mathrm{sp}}$.
So the semicell potentials are:
$$
\begin{align}
E_\text{cathode} &= E^\circ\\
E_\text{anode}   &= E^\circ + \frac{0.059}{2}\log_{10}[\ce{Pb^2+}]\\
\end{align}
$$
Thus:
$$\pu{0.413 V} = E_\text{cathode} - E_\text{anode} = -\frac{0.059}{2}\log_{10}[\ce{Pb^2+}] \implies [\ce{Pb^2+}] = 10^{-14}$$
And
$$K_\mathrm{sp} = [\ce{Pb^2+}]^2 = 10^{-28}$$

Actually, I think the question has an error in the EMF of the cell given. It should be $\pu{-0.413 V}$ rather than $\pu{0.413 V}.$ So, actually you are right in saying that the left half cell would be cathode if the EMF of the cell has to be positive.
The mistake that you are doing is when you wrote the reduction potential of anode, you should have written
$$E = E^\circ - \frac{0.059}{2}\log\frac{1}{[\ce{Pb^2+}]},$$
as the reduction reaction would be
$$\ce{Pb^2+ + 2 e- -> Pb},$$
and the reduction potential is given by
$$E = E^\circ - \frac{0.059}{2}\log\frac{[\text{Products}]}{[\text{Reactants}]}.$$
$\ce{Pb^2+}$ is on the reactant side, so it should come in the denominator.

Welcome to Chemistry.SE! I took a liberty to format your answer a bit; if you are interested further in technical details, feel free to visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.

You swapped the cathode an anode. In a galvanic cell diagram, on the left salt of the bridge is the anode, and on the right side is the cathode.
Also, you should have written
$$E(\text{anode})= -E^\circ - 0.059/2 \ldots$$
and
$$\text{EMF} = E(\text{cathode}) + E(\text{anode}),$$
or
$$E(\text{anode}) = E^\circ + 0.059/2\ldots$$
Then
$$\text{EMF} = E(\text{cathode}) - E(\text{anode}).$$
There are two ways of writing the Nernst equation and you are mixing them, so you get something wrong. One way is by using the reaction coefficient $Q$ and the second by using $[\text{Ox}]/[\text{Red}].$
Make sure that you first understand these two ways of writing the Nernst equation, then try to derive the $\text{EMF}$ by each method. You need to make these things clear and then return to the problem again. The red-ox potential of the cathode must be higher than the one of the anode.

It should conventionally be. But in this problem, it is asked also to determine which of them is the cathode/anode (and the order in the diagram has no importance). My reasoning brought me to think that the left one is the cathode, even if I thought too it should be swapped. What to think so?

The order in the diagram has big importance, it is always like that, and by looking at it you can easily deduce which electrode is the cathode and which is the anode. The EMF would be negative in the opposite case which is thermodynamically unfavorable.

Also, as far as I know, it's written Ksp for solubility product.

Sorry, just forgot to translate it (in italian is Kps). Anyway I'd like to know a qualitative/quantitative method, if there's, to move in this kind of situations without using this kind of conventional rules.

You can deduce which electrode is the cathode/anode from the Nernst equation. The potential of the system where Pb is immersed in PbS, will be higher than the potential of the other electrode, as the concentration of the lead(II) ion is very small (because PbS has low solubility). Check this for yourself. The more positive electrode is the cathode

I'm sorry to be so insistent, but I don't completely understand and I want it really. Without consider any kind of concentration and without knowing anything about this cathode and anode, i can write from the semireactions that the semipotenatials have to be:
$E_{cathode} = E^o - \frac{0,059}{2} \log_{10}{\frac{1}{[Pb^{2+}]}} = E^o + \frac{0,059}{2} \log_{10}{[Pb^{2+}]}$
$E_{anode} = E^o - \frac{0,059}{2} \log_{10}{[Pb^{2+}]}$

Now how i can apply what you was talking about? @Marko

Sorry, I didn't notice this: You should have written E(anode)= -E(O)- ... and EMF=E(c)+E(a), or E(anode)=E(0)+ ... EMF=E(c)-E(a) There are two ways of writing the Nernst equation and you are mixing them, so you get something wrong. One way is by using the reaction coefficient Q and the second by using [Ox]/[Red]. What this means you can see here: http://en.wikipedia.org/wiki/Nernst_equation. Make sure that you first understand these two ways of writing the Nernst equation, then try to derive the EMF by each method. You need to make these things clear and then return to the problem again.

So bad! Now I think I really understood, thank you very much for your precious patience.

@Marko Could you expand your answer with the tips you gave in the comments. I guess anyone who is not an everyday-trained electrochemist will have trouble to understand the point you made with this wording.


text:
Show proposed mechanism is consistent with the experimental rate law
The mechanism for the reaction:
\begin{align}
\ce{NO_{(g)} + O_{2(g)}  &<=>[k_{-1}][k_{1}] NO_{3(g)}} &\text{(fast)}\\
\ce{NO_{3(g)} + NO_{(g)} &<=>[K_{2}] 2\,NO_{2(g)}}      &\text{(slow)}\\
\end{align}
The experimental rate law:
$$Rate=-\dfrac{\Delta\ce{[NO]}}{\Delta t}=k\ce{[NO]}^2\ce{[O_{2}]}$$
At equilibrium, the forward and reverse rates are equal to each other. We then have
$$k_{1}\ce{[NO]}\ce{[O_{2}]}=k_{-1}\ce{[NO_{3}]}$$
This can be re-arranged to
$$\ce{[NO_{3}]}=\dfrac{K_{1}}{K_{-1}}\ce{[NO]}\ce{[O_{2}]}$$
Since $2\,\ce{NO}$ disappear in the overall reaction for every NO that reacts in the second step, we multiply the overall reaction rate by two and thus have:
$$2k_{2}\ce{[NO_{3}]}\ce{[NO]}$$
I understand everything up to this point. What I do not understand is how we get to this equation from the previous one.
$$Rate=2k_{2}\dfrac{k_{1}}{k_{-1}}\ce{[NO]}^2\ce{[O_{2}]}$$

The final rate equation is generated from substituting $\ce{[NO3]}$ defined in the third to last equation:
$\ce{[NO3]} = \frac{k_1}{k_{–1}}\ce{[NO][O_2]}$
into the second to last equation:
$2k_2\ce{[NO_3][NO]}$
This will give you the final rate equation defined only in terms of the starting concentrations. The $\ce{[NO_3]}$ term is replaced, $\ce{[NO]}$ squared, and $\ce{[O_2]}$ remains. None of the equilibrium constants cancel out. The intermediate $\ce{[NO_3]}$ is not included in the final equation, as it is consumed in the overall reaction.


text:
Bond Angles and intermolecular interactions
It's fun to consider bond angles of molecules in isolation. For example, what is the number of different bond angles exhibited by the most polar form of $\ce{SIBrClF2}$?
However, can bond angles really be considered in isolation? The method I would employ to determine the number of different bond angles only considers the molecule in isolation; I just consider van der Waals repulsions between the component atoms. Molecules generally do not exist in isolation. We can study molecules in the gas phase as to minimize intermolecular forces, but still, intermolecular forces exist, as it is hard to isolate exactly one molecule.
In the case of $\ce{SIBrClF2}$, I would expect a number of rather strong dipole-dipole interactions among the $\ce{SIBrClF2}$ molecules.
Therefore my question is: are there any cases in which considering solely intramolecular forces leads one astray when determining the bond angles of a molecule? I.e. one might look at a molecule and expect a certain bond angle to be < 180 degrees but find out the angle is in fact > 180 degrees due to intermolecular forces prying the bond angle open?

You are asking so many questions on this site, that by now you should know that meaningful titles are very important. It would be a charm if there actually wasn't only a very general and broad topic in the title field.

For every bond angle larger than $\alpha > 180^\circ$, there is a bond angle $\alpha' < 180^\circ$, because three atoms always have to be in a plane, hence $\alpha' = 360^\circ -\alpha$.

When people talk about bond angles in molecules, they are usually talking about the equilibrium bond angle in isolation. This can be derived from the energy minimized structure of the molecule in isolation.
As you point out, the angles in a real, non-isolated molecule will fluctuate due to interactions with neighboring molecules. However, the actual angle formed by three successive atoms in a molecule will naturally fluctuate even if the molecule is isolated and has some kinetic energy. Everything's always jiggling around.

authors: Přâţīķ Mhâţřë PM


text:
During phase change in matter, why doesn't the temperature change?
I was working on something in school and came across the question:

Why does the temperature not change much during a phase change?

I'm really not sure why this happens in matter and I couldn't find an answer in my school resources. Does anyone here know?

From Changes of Phase (or State):
...
So, how could there be a change in heat during a state change without a change in temperature?
"During a change in state the heat energy is used to change the bonding between the molecules. In the case of melting, added energy is used to break the bonds between the molecules. In the case of freezing, energy is subtracted as the molecules bond to one another. These energy exchanges are not changes in kinetic energy. They are changes in bonding energy between the molecules.
"If heat is coming into a substance during a phase change, then this energy is used to break the bonds between the molecules of the substance. The example we will use here is ice melting into water. Immediately after the molecular bonds in the ice are broken the molecules are moving (vibrating) at the same average speed as before, so their average kinetic energy remains the same, and, thus, their Kelvin temperature remains the same."

Couldn't we say that during the change the molecules absorb the heat as kinetic energy, move faster so they can overcome the attractions (potential energy)? But the attraction are still there. So after a little time the molecules would be in a greater distance (higher potential energy) with same kinetic (conservation of energy) so the temperature will remain the same.

For a first-order phase transition, you need to add the enthalpy of the phase transition.  As an example, starting with ice below the melting point, you pump heat in, and raise the temperature.  When you hit the melting temperature, the heat you put in goes towards the enthalpy of melting, and starts converting ice (sold) to water (liquid).  Additional heat continues to melt more of the ice.  Once all of the ice is converted (still at the melt temperature), than more heat starts increasing the temperature of the water.
A second-order phase transition does not have an enthalpy associated with it.

Thanks so much for your answer.  This seems to tentatively make sense on an initial read, but unfortunately I am not familiar with the concept of enthalpy and how it effects the phase transitions. Would you mind explaining this in layman's terms inline with your succinct explanation?

@Ylluminarious - The concepts of enthalpy (H) and entropy (S) are fundamental to thermodynamics.  The Gibbs free energy $G = H - TS$.  A first order phase transition has a discontinuity in $G$, which is the enthalpy of the phase transition. An introductory thermodynamics textbook should handle this pretty early on.

@Ylluminarious perhaps a simple introductory video such as this could help clarify the concept of enthalpy.

It is because the heat energy is used to overcome the inter-molecular force of attraction but when this inter-molecular force is broken it changes it state and the temperature starts increasing.

Welcome to chemistry.SE! If you had any questions about the policies of our community, you can visit the help center or take a tour of the website.

Mix water and ice in a container. Keep stirring the mix while the ice is melting. Then the following three statements are true:

The temperature of the water is greater than or equal to 0.
The temperature of the ice is less than or equal to 0.
The water and the ice have roughly the same temperature.

It follows that the temperature of the mix is roughly 0.

Statements 1 and 2 are only true under certain conditions. See http://physics.stackexchange.com/q/60170/12613. Statement 3 may not be true. If I drop ice at -5C into an equal mass of water at +50C it will take an appreciable time before the temperatures of the water and ice approach a common temperature.


text:
Donation by inductive effect and withdrawal by conjugation
Is there a group that would attached to a benzene ring, donate electrons by inductive effect and withdraw them by conjugation?

Well, I guess a $\ce{-BH2}$ group attached to benzene ring would have this effect. It has an empty $\ce{p}$ orbital that would withdraw electrons via conjugation, but $\ce{B}$ is less electronegative than $\ce{C}$ and so it would have a positive inductive effect.

Thank you for the answer. Now, an additional question: Where would substitution happen on the aromatic ring?

Usually conjugation beats inductive effects. So substitution should happen in the meta position.

How probable is ipso substitution?

@Marko It's okay to ask a new question for follow-ups.  If you want to reference that it's a continuation of this question, you can link to this one.

I guess a $\ce{-BH2}$ group attached to a benzene ring would have this effect. It has an empty $\ce{p}$ orbital that would withdraw electrons via conjugation, but $\ce{B}$ is less electronegative than $\ce{C}$ and so it would have a positive inductive effect.


text:
Determining exponents in a rate law with different data
Here is the experimental data: Note all concentrations are the initial amounts
\begin{array}{cccc}
\text{Experiment} & \ce{[C2H4Br2]} &\ce{[I-]} & \text{Initial rate of formation of }\ce{I3-}\\\hline
1 & 0.127 & 0.102 & 6.45\times 10^{-5}\\
2 & 0.343 & 0.102 & 1.74\times 10^{-4}\\
3 & 0.203 & 0.125 & 1.26\times 10^{-4}\\\hline
\end{array}
$\mathrm{Rate}=k\ce{[C2H4Br2]}^m\ce{[I-]}^n$
We can determine the exponent $m$ easily because experiment one and two have the same initial $\ce{[I-]}$. However, when determining $n$, there is no data that has the same initial concentration. What the solution manual does is the following:
$$n=
\dfrac
{\ln \left(\dfrac{\mathrm{Rate}_{3}\cdot [\ce{C2H4Br2}]_2}
{\mathrm{Rate}_{2} \cdot [\ce{C2H4Br2}]_3}\right)}
{\ln \left(\dfrac{[\ce{I-}]_3}{[\ce{I-}]_2}\right)}=\dfrac{\ln \dfrac{(1.26\times 10^{-4})(0.343)}{(1.74\times 10^{-4})(0.203)}}{\ln \left(\dfrac{0.125}{0.102}\right)}=1$$
My question is, why are they multiplying the rate of the third experiment with the initial concentration from the second experiment and then multiply the rate from the second experiment with the initial concentration of the third experiment. I don't see any correlation with these numbers and how they can be used to determine the exponent.  Any thoughts?

If you've correctly determined exponent $m$, then to ascertain $n$ you can proceed by first dividing rate equation $(3)$ by rate equation $(2)$ (note that the rate constant $k$ cancels immediately and we have that $m = 1$):
$$
\dfrac{\text{rate}_3}{\text{rate}_2} = \dfrac{\ce{[C2H4Br2]}_3}{\ce{[C2H4Br2]}_2} \bigg(\dfrac{\ce{[I-]}_3}{\ce{[I-]}_2}\bigg)^n
$$
Rearranging this equation, one obtains:
$$
\dfrac{\text{rate}_3}{\text{rate}_2} \cdot \dfrac{\ce{[C2H4Br2]}_2}{\ce{[C2H4Br2]}_3} = \bigg(\dfrac{\ce{[I-]}_3}{\ce{[I-]}_2}\bigg)^n
$$
Now, to isolate $n$, take the logarithm (in any base of your choice, though the book arbitrarily uses the natural log, and I'll do likewise out of concern for consistency) of both sides of the equation, and remember that one of the basic properties of logarithms is that $\log_b x^n = n\log_b x$:
$$
\ln\dfrac{\text{rate}_3 \ce{[C2H4Br2]}_2}{\text{rate}_2 \ce{[C2H4Br2]}_3} = \ln\bigg(\dfrac{\ce{[I-]}_3}{\ce{[I-]}_2}\bigg)^n = n\ln\dfrac{\ce{[I^{-}]}_3}{\ce{[I^{-}]}_2}
$$
Solving for $n$, one finds:
$$
\frac{\ln\dfrac{\text{rate}_3 \ce{[C2H4Br2]}_2}{\text{rate}_2 \ce{[C2H4Br2]}_3}}{\ln\dfrac{\ce{[I-]}_3}{\ce{[I-]}_2}} = n
$$
Which is the final equation from your textbook's solutions manual.

I think that there is a simple method to solve this question. I did this question with these 6 steps and my answer is right (although I had to do some approximation in calculation)

Take values from experiment 1 and put them in rate equation. You will obatain an equation with $m$, $n$ and $k$ as varibales. Lets assume it equation 1.
Now take values from experiment 2 and put them in rate equation. You obtain another eqaution like equation 1. Lets assume it equation 2.
Divide equation 1 by 2 or 2 by 1 (whatever you like.) Solve it and you will easily get $m$ (equals to 1).
Now put the value of $m$ in equation 2. You will obtain an equation with $n$ and $k$ as variables. Lets assume it equation 3.
Now take values from experiment 3 and put them in rate equation along with value of $m$. You will get another equation like equation 3. Lets assume it equation 4.
Divide equation 3 by 4 or 4 by 3 (whatever you like.) Solve it and you will easily get n (equals to 1).

PS, this might not be a 'by-the-book' method but I will solve it this way when I don't have too much time to solve a question and also because I am lazy enough to solve natural logarithm and stuff.

From the equation : $\text{rate}=k[\ce{C2H4Br2}]^m[\ce{I-}]^n$
you can get by applying logarithms : $\ln\text{rate}=\ln k + m\ln[\ce{C2H4Br2}]+n\ln[\ce{I-}]$
That has the form : $\mathrm Y= w_0+w_1x_1+w_2x_2 $
where $y=\ln\text{rate}$,  $x_1=\ln[\ce{C2H4Br2}]$, $x_2=\ln[\ce{I-}]$ and $w_1=m$, $w_2=n$ and you can consider $x_0=1$ and $w_0=\ln k$
That becomes a linear regression problem, you can apply least squares or for better visualising it, you can draw the experimental values on logarithmic paper and fit a line. The result will be: $m=0.99884$ and $n = 0.98919$


text:
Is 4-nitrobenzaldehyde or 4-aminobenzaldehyde more likely to be hydrated?
Of 4-nitrobenzaldehyde and 4-aminobenzaldehyde, which has the higher $\ce{K_{eq}}$ of hydration?
 $\qquad \qquad$

Steric factors can likely be ignored in this problem since we are dealing with rather structurally similar compounds.
I suspect the problem comes down to recognizing that a more polar compound is more favorably solvated by water.
I know that the nitro group is always electron-withdrawing, and I drew a few resonance structures as well. The resonance structures for both, however, seem to suggest that both are rather polarized (given the distribution of formal charges).
Resonance forms for 4-nitrobenzaldehyde

I expect the ones with the oxygen possessing the incomplete octet are very minor resonance contributors.
Therefore it seems that the main thing about this molecule is that the nitro group is withdrawing electron density from the ring and the aldehyde group.

Resonance forms for 4-aminobenzaldehyde

The resonance structures here are all somewhat reasonable.

I think you need to consider the relative electrophilicities of the two compounds...

@jerepierre i would expect the p-nitrobenzylaldehye to be more electrophilic, but what does that have to do with hydration?

What is the reaction that you're trying to determine the Keq of? You might also want to write that out.

I think you're asking about the following equilibrium

This equilibrium will shift to the left as the aldehyde becomes more stabilized.  So the question becomes, which R group, $\ce{NO2}$ or $\ce{NH2}$, will tend to stabilize the aldehyde the most?  A substituent that further delocalizes the positive charge shown in the dipolar resonance form would tend to be stabilizing.  Just like in electrophilic aromatic substitution, additional stabilizing resonance structures can be drawn involving the amino substituent, but similar structures involving the nitro group are not as significant due to electrostatics (2 adjacent positive charges). Therefore, since the amino will stabilize the aldehyde better than the nitro substituent, the equilibrium involving the amino substituent should lie further to the left, and the aldehyde with the nitro substituent should be more hydrated.

How come in your hydrated picture you have two OH groups? Is hydration not the same as solvation?

No, hydration is not the same as solvation.  In the case of carbonyl compounds, when one speaks of hydration, and especially the equilibrium constant for this process, one is speaking about the addition of water across the carbonyl double bond to produce a gem-diol.  I thought perhaps you were really asking about solvation. That's why I wrote, "I think you're asking about the following equilibrium."

authors: พุด แก้วพะวง


text:
Significant figures used to imply stated error?
The question reads, " express each of the following quantities using significant figures to imply the stated error".
a)$2.3 \pm 0.001$
b)$1.989 \pm 0.0003$
I think the first one is 2.300 because the zero in the thousands place implies an error of $\pm 0.001$ but how do I use sig figs to imply an error of $\pm 0.0003$?

The use of significant figures always implies some error, that is not explicitly stated. It is therefore irrelevant how big this error is. I merely states that the last digit is rounded.
In the first case the quantity is given as $2.3\pm0.001$, which gives you the interval $(2.301;2.999)$, which is expressed as $2.300$ in significant figures.
The same applies to the second quantity, which is given as $1.989\pm0.0003$, which in turn gives you the interval $(1.9893;1.9887)$, which is expressed as $1.9890$.

Yes, In addition and subtraction we have to see digits after decimal fraction not significant figures. I messed up rules.


text:
What happens if you autoclave HEPES?
Many molecular biology protocols advice against sterilizing HEPES (4-(2-hydroxyethyl)-1-piperazineethanesulfonic acid) solutions by autoclave.
Supposedly, HEPES is heat labile, and the high temperatures used for sterilization somehow impair the function of the buffer, or perhaps generate unwanted chemical species (this is speculation on my part; I've often seen the advice to not autoclave HEPES, but never seen anything about why not). Typically autoclaves heat solutions to 121C and increase pressure to slightly above 2 atm, which is about a degree short of the boiling point of water. This environment is maintained for 20 minutes in order to destroy biological contaminants.
I have seen this claim often, but had trouble locating an actual source or reasoning for why HEPES should not be autoclaved. Has there been a publication that actually tried to heat a HEPES buffer to autoclave temperatures (121C) and see what happens? What exactly is the consequence, especially with regard to the biological uses of HEPES (ie. pH buffer)?

From here:

Some buffers (e.g., MOPS and HEPES) cannot be autoclaved because they degrade upon heating.

and from here:

[HEPES] Solutions may be autoclaved under standard conditions.

I guess nobody knows the answer.

So there isn't even a consensus about it. Interesting! Unfortunately the source for the second claim, Medzon and Gedies 1971, is paywalled and beyond my reach, so I'm a little stuck there.

Sorry, I can't access it either. It's likely that they autoclaved the HEPES solution and found out that it was still usable. It might be that some autoclaves have a slightly higher working temperature (like 125°C), so the HEPES is more likely to degrade in those autoclaves.

Today I tested a set of competent E. coli cells that were prepared using a HEPES-based calcium chloride buffer solution. The solution was autoclaved as I described in the answer.
I obtained a fair transformation efficiency from these cells, and viability was also not perturbed noticeably. There are several confounds in the preparation of my solution that could affect the transformation efficiency, but in conclusion, autoclaving the HEPES has not abrogated its ability to make cells chemically competent.
I will speculate further and say that autoclaving HEPES may degrade some of the HEPES molecules or produce reactions and generate unexpected subspecies - but both of these will be generated in small, insignificant amounts which should not affect most biological applications.

Sigma states that you can autoclave HEPES solutions:
https://www.sigmaaldrich.com/content/dam/sigma-aldrich/docs/Sigma/Product_Information_Sheet/2/h3784pis.pdf
Sigma cites the following reference:
Medzon, E. L. and Gedies, A., Substitution of 4-(2-hydroxyethyl)-1-piperazineethane sulfonic acid (HEPES) for bicarbonate in protein-free animal cell culture medium: application to vaccinia virus quantitation and fluorogenic acetylesterase assay in living LM cells. Can. J. Microbiol., 17(5), 651-653 (1971).
If this paper is behind a paywall I suggest you use a Scihub proxy. It bypasses paywalls for research articles.

I've read the paper of Medzon & Gedies. They autoclaved HEPES medium and see no statistical difference between it and bicarbonate solution in cell culturing. There is nothing said about impossibility of HEPES autoclaving.


text:
Implicit dependence of reaction rate through reduction potential and activation energy
On one hand, the larger the reduction potential of a chemical reaction and the smaller its activation energy, the larger the reaction reaction. on the other hand the rate of a reaction depends on thermodynamic conditions, such as temperature and pressure, and the concentration of the species involved.
Now given that both the reduction potential and the activation energy of the reaction depend on thermodynamic conditions, can we deduce that the whole thermodynamic dependency of the rate of the reaction lies inside the reduction potential and the activation energy?

For the sake of clarification, that I said the larger is the reduction potential and the smaller is the activation energy the greater will be the rate of the reaction is a conclusion based on the analogy of a redox reaction and an electric circuit, the reduction potential being in place of the voltage difference, the activation energy as the energy consumer being in place of the electric resistance and the electron transferred during the reaction being in place of the electric current. It is clear that the greater the electric current, the greater the rate of reaction should be.

What exactly do you mean by thermodynamics condition? What are the equations you draw your entering statement from: "On one hand, the larger is the reduction potential of the chemical reaction and the smaller is its activation energy the larger would be the rate of reaction, on the other hand the rate of reaction of a reaction depends on the thermodynamics condition upon which the reaction takes place."?

@Philipp, tried to clarify my point and also my background to the question. Please check if if the question is now clear and relevant. Thank you.

The rate of a reaction depends on the thermodynamics of its transition state.
It would be better to look at it in terms of Gibbs activation energy. Therefore there two factors affecting the rate of a reaction:

the activation enthalpy $\Delta^‡ H $, that corresponds to the Arrhenius activation energy
the activation entropy $\Delta^‡ S $ that takes into account the probability of a transition state.

The whole dependency of the kinetic constant on these parameters is expressed by the Eyring and Wynne-Jones equation:
$$ k = \frac{k_\text{B}T}h \operatorname{e}^{\frac {\Delta^‡ S} R }\operatorname{e}^{-\frac{\Delta^‡ H} {RT} }$$
Where $T$ is the temperature, $k_\text{B}$ is the Boltzman constant and $h$ is the Planck constant.
Sometimes a reflection coefficient $\kappa$ is added to the equation to take into account some other dependencies. It is near to 1 for most of the simple reactions.
$$ k = \kappa \frac{k_\text{B}T}h \operatorname{e}^{\frac {\Delta^‡ S} R }\operatorname{e}^{-\frac{\Delta^‡ H} {RT} }$$