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Why is an S-S bond stronger than an O-O bond?
I'm wondering why exactly the single bond between two sulfur atoms is stronger than that of two oxygen atoms. According to this page, an $\ce{O-O}$ bond has an enthalpy of $142~\mathrm{kJ~mol^{-1}}$, and a $\ce{S-S}$ bond in $\ce{S8}$ an enthalpy of $226~\mathrm{kJ~mol^{-1}}$. This one reports the $\ce{S-S}$ bond enthalpy to be $268~\mathrm{kJ~mol^{-1}}$, but I'm not sure which molecule they mean, or how they measured it. Anyway, it's still higher than that of $\ce{O-O}$.
Searching the Net, the only justification I could find was something similar to concepts they apply in VSEPR, like in this Yahoo Answers thread with such remarkable grammar. Quoting the answer, which might have borrowed some stuff from a high school textbook,

due to small size the lone pair of electrons on oxygen atoms repel the bond pair of O-O bond to a greater extent than the lone pair of electrons on the sulfur atoms in S-S bond....as a result S-S bond (bond energy=213 kj/mole)is much more stronger than O-O(bond energy = 138 kj/mole) bond $\ldots$

Other variations of the same argument can be seen here, but it doesn't make sense, since one couldn't apply the same argument to $\ce{O=O}$ and $\ce{S=S}$. The first reference documents the $\ce{S=S}$ and $\ce{O=O}$ bond enthalpies to be $425$ and $494~\mathrm{kJ~mol^{-1}}$, respectively.
It's a bit shaky, and I'm looking for a solid explanation using MO or VB, or anything else that actually works. So, why is an $\bf\ce{S-S}$ single bond stronger than $\bf\ce{O-O}$, despite $\bf\ce{O=O}$ being obviously stronger than $\bf\ce{S=S}$?

questions to consider: 1) is dioxygen (and disulfur) paramagnetic or diamagnetics? 2) assuming that there are lone pairs in both H2O2(H2S2) and O2 (S2) which are angles O-O-:  (S-S-:) ?

Just making a wild guess, it has something to do with catenation. Sulfur has more catenating power than oxygen so, $\ce{S-S}$ bond enthalpy is higher than $\ce{O-O}$. The readings (bond enthapy) is given here - http://chemistry.stackexchange.com/questions/431/why-does-sulphur-but-not-oxygen-catenate/432#432

@Nilay there are no such things as "catenation power". That stuff is actually derived from the problems of this question.

I meant "catenation power" to be "tendency to catenate".

Yes @Nilay, but that tendency is something you conclude from this data, not the vice versa. It's the result, not the reasoning.

TL;DR: The $\ce{O-O}$ and $\ce{S-S}$ bonds, such as those in $\ce{O2^2-}$ and $\ce{S2^2-}$, are derived from $\sigma$-type overlap. However, because the $\pi$ and $\pi^*$ MOs are also filled, the $\pi$-type overlap also affects the strength of the bond, although the bond order is unaffected. Bond strengths normally decrease down the group due to poorer $\sigma$ overlap. The first member of each group is an anomaly because for these elements, the $\pi^*$ orbital is strongly antibonding and population of this orbital weakens the bond.

Setting the stage
The simplest species with an $\ce{O-O}$ bond would be the peroxide anion, $\ce{O2^2-}$, for which we can easily construct an MO diagram. The $\mathrm{1s}$ and $\mathrm{2s}$ orbitals do not contribute to the discussion so they have been neglected. For $\ce{S2^2-}$, the diagram is qualitatively the same, except that $\mathrm{2p}$ needs to be changed to a $\mathrm{3p}$.

The main bonding contribution comes from, of course, the $\sigma$ MO. The greater the $\sigma$ MO is lowered in energy from the constituent $\mathrm{2p}$ AOs, the more the electrons are stabilised, and hence the stronger the bond.
However, even though the $\pi$ bond order is zero, the population of both $\pi$ and $\pi^*$ orbitals does also affect the bond strength. This is because the $\pi^*$ orbital is more antibonding than the $\pi$ orbital is bonding. (See these questions for more details: 1, 2.) So, when both $\pi$ and $\pi^*$ orbitals are fully occupied, there is a net antibonding effect. This doesn't reduce the bond order; the bond order is still 1. The only effect is to just weaken the bond a little.
Comparing the $\sigma$-type overlap
The two AOs that overlap to form the $\sigma$ bond are the two $\mathrm{p}_z$ orbitals. The extent to which the $\sigma$ MO is stabilised depends on an integral, called the overlap, between the two $n\mathrm{p}_z$ orbitals ($n = 2,3$). Formally, this is defined as
$$S^{(\sigma)}_{n\mathrm{p}n\mathrm{p}} = \left\langle n\mathrm{p}_{z,\ce{A}}\middle| n\mathrm{p}_{z,\ce{B}}\right\rangle = \int (\phi_{n\mathrm{p}_{z,\ce{A}}})^*(\phi_{n\mathrm{p}_{z,\ce{B}}})\,\mathrm{d}\tau$$
It turns out that, going down the group, this quantity decreases. This has to do with the $n\mathrm{p}$ orbitals becoming more diffuse down the group, which reduces their overlap.
Therefore, going down the group, the stabilisation of the $\sigma$ MO decreases, and one would expect the $\ce{X-X}$ bond to become weaker. That is indeed observed for the Group 14 elements. However, it certainly doesn't seem to work here. That's because we ignored the other two important orbitals.
Comparing the $\pi$-type overlap
The answer for our question lies in these two orbitals. The larger the splitting of the $\pi$ and $\pi^*$ MOs, the larger the net antibonding effect will be. Conversely, if there is zero splitting, then there will be no net antibonding effect.
The magnitude of splitting of the $\pi$ and $\pi^*$ MOs again depends on the overlap integral between the two $n\mathrm{p}$ AOs, but this time they are $\mathrm{p}_x$ and $\mathrm{p}_y$ orbitals. And as we found out earlier, this quantity decreases down the group; meaning that the net $\pi$-type antibonding effect also weakens going down the group.
Putting it all together
Actually, to look solely at oxygen and sulfur would be doing ourselves a disservice. So let's look at how the trend continues.
$$\begin{array}{|c|c|c|c|}
\hline
\mathbf{X} & \mathbf{BDE(X-X)\ /\ kJ\ mol^{-1}} & \mathbf{X} & \mathbf{BDE(X-X)\ /\ kJ\ mol^{-1}} \\
\hline
\ce{O} & 144 & \ce{F} & 158 \\
\ce{S} & 266 & \ce{Cl} & 243 \\
\ce{Se} & 192 & \ce{Br} & 193 \\
\ce{Te} & 126 & \ce{I} & 151 \\
\hline
\end{array}$$
(Source: Prof. Dermot O'Hare's web page.)
You can see that the trend goes this way: there is an overall decrease going from the second member of each group downwards. However, the first member has an exceptionally weak single bond.
The rationalisation, based on the two factors discussed earlier, is straightforward. The general decrease in bond strength arises due to weakening $\sigma$-type overlap. However, in the first member of each group, the strong $\pi$-type overlap serves to weaken the bond.
I also added the Group 17 elements in the table above. That's because the trend is exactly the same, and it's not a fluke! The MO diagram of $\ce{F2}$ is practically the same as that of $\ce{O2^2-}$, so all of the arguments above also apply to the halogens.
How about the double bonds?
In order to look at the double bond, we want to find a species that has an $\ce{O-O}$ bond order of $2$. That's not hard at all. It's called dioxygen, $\ce{O2}$, and its MO scheme is exactly the same as above except that there are two fewer electrons in the $\pi^*$ orbitals.
Since there are only two electrons in the $\pi^*$ MOs as compared to four in the $\pi$ MOs, overall the $\pi$ and $\pi^*$ orbitals generate a net bonding effect. (After all, this is where the second "bond" comes from.) Since the $\pi$-$\pi^*$ splitting is much larger in $\ce{O2}$ than in $\ce{S2}$, the $\pi$ bond in $\ce{O2}$ is much stronger than the $\pi$ bond in $\ce{S2}$.
So, in this case, both the $\sigma$ and the $\pi$ bonds in $\ce{O2}$ are stronger than in $\ce{S2}$. There should be absolutely no question now as to which of the $\ce{O=O}$ or the $\ce{S=S}$ bonds is stronger!

How do we physically explain the greater antibonding nature of antibonding pi orbitals in the O-O bond? Is it due to the inter-electronic repulsions between the two dense electron clouds of the two oxygen atoms?

It's tempting to think that bonding character arises from electron-nuclear attraction and antibonding character from electron-electron repulsion, but it's not so simple. All factors (kinetic energy, potential energy) play a role in orbital energies. Even a physical explanation of why H2 is stable (i.e. why is a covalent bond stable?) isn't easy (see e.g. J. Phys. Chem. A 2013, 117, 7946), so I don't imagine it's any simpler in O2. My direct answer to your question, therefore, is "I don't know".

@orthocresol Can the same argument be used for weaker C-O single bond (~ 350KJ/mol) compared with Si-O single bond (~ 450KJ/mol)? (if you are still here, otherwise I will post a new question.)

@DeepakArya No. To get to the same MO diagram as above, you’d need to put so many more electrons onto carbon (or silicon), which doesn’t make sense chemically; no real compound has a carbon with a 3- charge. You need some other reason for that, and let me warn you against using “silicon d-orbitals” as the reason.

@orthocresol I was just wondering about typical Si-O single bond enthalpy in any compound with typical C-O bond enthalpy. I was not talking about diatomic species but single bonds as we encounter them in alcohols or Me3SiCN rxn with ketones. There the reasoin for forward rxn was given as a strong Si-O bond. Then I looked at this page (http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html) and it shows Si-O bond energy around 100KJ/mol higher than C-O bonds. Which seems weird. Either the page is wrong Or should I envoke oxygen lone pair going to empty d-orbitals of Si ;).

Sulfur atoms are larger than oxygen atoms.
Pi bonds are formed by overlapping of two parallel p orbitals. The further the distance between atoms, the lesser the overlapping and weaker the bond.
But sigma bonds in case of Oxygen and Nitrogen are not strong enough because you are bringing two very small atoms (with  large no. of electrons in the outer shell) too close which makes the sigma bond comparatively unstable  than that of S-S bond where sigma bond is more stable due to lesser electro static repulsion of non-bonding electrons.


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How to determine which of the given redox reactions will occur?

Which one of these reactions will occur?

$$\ce{KMnO4 -> K2O + MnO2 + O2}$$
$$\ce{KMnO4 + HCl -> KCl + MnCl2 + Cl2 + H2O}$$
$$\ce{KMnO4 + CO2 \longrightarrow K2CO3 + Mn2O7}$$
$$\ce{KMnO4 + CO2 -> K2CO3 + MnO2}$$
$$\ce{KMnO4 + SO2 -> K2SO4 + Mn2O3 + SO3}$$

And how am I supposed to know which one it is? It's not just this question, I struggle with these types of questions every time. There seem to be a lot of things to take into consideration.
I'm thinking whether it's under 3 or under 4. The oxidation numbers for the reaction 3 don't change at all. But in reaction 4, only the oxidation number of Mn changes (from +7 to +4) and the other oxidation numbers stay the same (if I'm right). So, I think reactions 1, 2 and 5 definitely happen, but I don't know about 3 and 4.

Indeed, you have to consider many things which together constitute a huge body of knowledge known as chemistry. First of all, check the oxidation states and determine whether a redox reaction is taking place (you must have an oxidizing agent and a reducing agent). See where this gets you.

Both 3 and 4 are impossible. $\ce{CO2}$ does not react with the $\ce{KMnO4}$, because $\ce{KMnO4}$ is strongly oxidizing. $\ce{C}$ is in the highest oxidation state and can not be oxidized, hence, it is impossible to react.

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Why do we need sulfuric acid in reducing nitric acid?
From Wikipedia:

Why do we need sulfuric acid in the left part of the formula? Wouldn't FeSO4 reduce nitric acid without the sulfuric acid taking part?

P.S.
I was answered that a milder form of the reduction would still take place:

Is that right?
So we need the acid to make the reduction more advanced, all the way to N(2+) from N(5+)? How does it work?

Sounds about right. Now, this is essentially the same reaction, but with different consequences. If you have excess $\ce{HNO3}$, it will (sort of) bring your $\ce{NO}$ back to $\ce{NO2}$. If you don't, it won't.

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Protein Data Bank: Asymmetric Unit/Biological Assembly
For a schoolproject I have to interpret the 3D structure of an imatinib-BCR ABLC kinase complex. When I visited the PDB website I downloaded the PDB file corresponding to this complex. Once I opened it on my 3D modelling program it struck me that the protein complex was portrayed twice, looking almost the same but not completely. So after that I stumbled upon something called the Asymmetric Unit/Biological Assembly on the PDB site. I read the description  down this post here, but didn't fully understand it.
Could anybody explain me what this is and how I'm able to import the right complex in my program?
http://pdb101.rcsb.org/learn/guide-to-understanding-pdb-data/biological-assemblies

Just to make sure: this is for a "school project" as in high school and not university/college? In general, you should look at the biological assembly. Also could you provide the reference for the structure in question (especially the 4 character code - e.g. 2HYY)

Well university is kind of called school in my country, so sorry if I confused you there. The code for the complex is 1iep.

Most of the structures in the Protein Databank are determined with X-ray crystallography. Like inorganic compounds and small organic molecules, under the right conditions proteins can form crystals - ordered, repeating patterns of protein molecules, filling a 3D space. That is, you can take a basic building block and rotate and translate this building block to completely fill the three-dimensional space. (2D example).
So what's being modeled in an X-ray crystallography experiment is that basic building block, the one which undergoes symmetry operations (rotation and translation) in order to fill out 3D space. This basic building block is called the asymmetric unit, because it is without the crystallographic symmetries which result in filling space.
But even in the asymmetric unit there can exist noncrystallographic symmetry. These are symmetries (rotations and translations) which relate portions of the asymmetric unit, but aren't lined up quite right to fill space. Because they don't contribute to the overall symmetry of the crystal, normally they're not exactly symmetrical. Usually there are slight differences, based on how the noncrystallographic symmetry fits into the crystallographic one.
For a 2D example, think of a triangle which is tiled on a square grid. The square grid is the crystallographic symmetry. The triangle has noncrystallographic symmetry, because the three-fold rotational axis of the triangle does not contribute to tiling the plane. This means the corners of the triangle are not exactly equivalent, as each corner is in a different position with respect to the neighbors in the next square.
So when an X-ray cryptographer goes to solve the structure of a protein crystal, the primary unit of interest is the asymmetric unit - even if there's noncrystallographic symmetry in the asymmetric unit. So that's what the PDB records as its primary entry - the contents of the asymmetric unit, including the slight differences between any repeated units.
In your case (1iep), that's what happened. Because of the specific way the protein crystallized, there just happens to be two copies of the protein in the asymmetric unit, almost - but not quite - identical to each other. This representation of the asymmetric unit - including both proteins - is what you've downloaded.
Because this might not be what you're interested in, the PDB also provides one or more "Biological Assembly" files. These are processed files which attempt to model what the experimentalists believe* is the actual biologically relevant unit. In your case, the relevant biological unit is a monomer, so they provide two biological assembly files - one for each (distinct) monomer in the asymmetric unit. For multimeric proteins, it sometimes happens that one of the symmetry axes for the multimer lines up with the crystallographic axis. In this case, you'd only get part of the protein in the asymmetric unit download, but the biological assembly file would contain a reconstructed representation of the multimer.
If you want further, extensive detail, I would second the recommendation made here to look at the textbook Crystallography Made Crystal Clear, by Gale Rhodes.
(*The biological unit provided by the PDB is a guess. Sometimes you'll see proteins which should be multimeric represented as monomers, or monomeric proteins represented as multimers. It depends in part on how accurately the depositors annotated what the relevant biological unit was.)

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What are unit cells, and how do you determine the number of atoms in one of them?
I've read several websites to try to figure out what exactly a unit cell is, but I can't seem to understand it. Could someone explain what a unit cell is, and the different types? Thanks!

The unit cell is the most primitive 3D structure that contains the arrangement of atoms which form the crystal of a compound. The unit cell is probably most meaningful in the study of minerals, so it is more of an inorganic analytical technique rather than an organic analytical technique. There are 7 3D crystal lattice systems which form 14 primitive 3D forms called the Bravais lattices. The 14 Bravais lattices can be further subdivided into 32 point groups.

Thank you MaxW, in bcc can we say two atoms at$000$ and $1/2 1/2 1/2$ form a unit cell? Their repetition will lead to generation of entire lattice.


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What causes soda to fizz and how can it be stopped?
When you go to a soda fountain and pour some soda into your cup, frothy fizz is generated at the top of the cup. Obviously, it has something to do with carbonation. However, when a cup of soda is just sitting and is not being filled or refilled, there is no fizz.
What causes the fizz? Is the fizz generated by the crashing of the soda into the cup?
I would also like to know how to stop the fizzing produced when the soda cup is being filled/refilled: Can I slow the rate at which the soda falls? Will spinning the liquid on the way down to the cup help attenuate the fizzing? What about changing the material from which the cup is made?

Please don't double-post a question; it's a chemistry question; so I would suggest you to delete the Phys SE question: http://physics.stackexchange.com/questions/240253/what-causes-fizz

I'm seeing there exists an answer there; so I would suggest you to ask the mods there to migrate the question here.

They told me to ask here so I did. Sorry I don't know how to migrate.

Is this really just a chemistry question? I think it would fit in either place. One or the other, though.

Also see this - http://chemistry.stackexchange.com/questions/37176/why-does-pouring-soda-slowly-and-smoothly-seem-to-keep-more-of-the-carbonation

It think this link will help you: http://www.adichemistry.com/inorganic/p-block/group-14/carbonates/carbonates-bicarbonates-1.html

The soda is supersaturated with dissolved carbon dioxide since in the sealed container the soda is under pressure using carbon dioxide. A number of things can cause the formation of nucleation sites which degas the solution by forming bubbles which float to the surface. The foam is due to the surface tension of the liquid solution. The lay understanding is that the soda is "flat" when the soda is no longer supersaturated with carbon dioxide.
It should be noted that even in a saturated solution most of the carbon dioxide is in solution as a dissolved gas and not carbonic acid. Since soda solutions are acidic anyways, this also favors the dissolved gas as opposed to carbonic acid.
To minimize fizzing:

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

Thank you for the link to Foam Wiki, I didn't image this would be available. I don't think this answers the question completely. Is surface tension the only reason? If I dump soda into a surface-tension-free glass, there would be no fizz? I am hoping you could delineate on "formation of nucleation sites"

?!? Surface tension is a property of the liquid, not the glass itself. // "Nucleation sites" are essentially "irregularities" in the solution which would cause the dissolved gas to form bubbles at that site. So particulates floating in the solution, sharp edges on the ice cubes, dirt on the surface of the glass, a chip or crack in the glass and so on. // Added wikipedia links for "nucleation sites" and "surface tension" rather than trying to explain those concepts.

I'll point out in addition that "fizz" and "foam" are two different things. I was thinking of fizz as the bubbles floating to the surface. The "foam" results because not all bubbles pop immediately upon reaching the surface.

Thanks this is super helpful. Please tell me if I misunderstand at all. So the bubbles don't pop because of surface tension which is a characteristic of the soda that we won't be able to change unless we add a different liquid in? The only way to stop the foam is to create less bubbles. Which means creating less nucleation sites. The only ways to do this are decrease impurities, keep substances at the same temperature, and pouring more gently. What property is pouring gently fixing? Velocity, acceleration, angular momentum, force ? Generation of Nucleation sites are directly proportional to?

RE: "The only way to stop the foam is to create less bubbles." You could add various chemicals which would "kill" the foam. You wouldn't be getting less bubbles, it would just be that the bubbles would pop when they hit the surface. For beer drinkers having a nice  "head" (layer of foam) is desirable since the bubbles busting release various smells which are detected by the nose.

RE: "pouring gently" - As I tried to indicate if you just dump liquid into a glass then you get bubbles from the air which form nucleation sites for the dissolution of the carbon dioxide.

is that due to a rise in kinetic energy? Or something else


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Nucleophilicity with respect to the solvent
I am currently studying nucleophilicity of molecules and am getting a set of conflicting information, so I wanted to clarify nucleophilicity trends with respect to solvent.
Master Organic Chemistry discusses how nucleophilicity increases down the periodic table with protic solvents and decreases down the periodic table with aprotic solvents; the main argument is that a solvating shell will form with the protic solvent, making the smaller atoms less likely to form a bond. When this effect is removed, the trend is reversed.
However, during my lectures, my professor said that there is no such change in the trend: the larger atoms such as $\ce{I-}$ will always be more nucleophilic than $\ce{F-}$ due to the fact that it is more polarizable and likely to donate an electron pair. The textbook that I am using, Organic Chemistry by Wade, also does not discuss this trend reversal depending on the solvent. The only discussion talks about how $\ce{F-}$ would be a poor nucleophile as it is small and holds its electrons "tightly."
A practice problem provided during class also asked to compare the nucleophilic strength between $\ce{H2O}$ and $\ce{H2S}$ in polar aprotic solvent; similarly, the correct answer provided was that $\ce{H2S}$ would be a stronger nucleophile as it is larger and more polarizable.
I would appreciate it if someone could clarify this matter.

Actually nucleophilicity is inversely proportional to electronegativity when comparing compounds across the period, and inversely proportional to size when comparing in the same group. This condition is meant in aprotic solvent.
In protic solvents  the more electronegative ions gets easily solvated to a larger extent (i.e., like a ball with the ion in the centre and the water molecules around it) so the order across the period remains same since electronegativity is not effected but the order about the group changes since the now smaller ions become the larger ones.
For example, the trend for nucleophilicity is  $\ce{F- > Cl- > Br- > I-}$ in aprotic solvent but in protic solvent it is
$\ce{I- > Br- > Cl- > F-}$    ($\ce{F-}$ is now larger than others in size because it is solvated.)

Hi. Can you please clarify the case of H2O vs H2S nucleophilicity in APROTIC solvent?

You seem to be describing the trends correctly.
Perhaps what is missing is the fact that only protic solvents form a meaningful shell. In aprotic solvents the nucleophile is "free" without a shell.
As a result of this the solvent identity is often the important factor b/c a protic solvent will "lock up" what is normally a very strong nucleohphile.
So in protic solvents like MeOH you'll usually see the neutral form be the nucleophile (H20) vs the charged, more nucleophilic form (-OH)
There are several competing factors here, solvent, nucleophilicity, charge density, and others. In an academic learning setting you won't be asked to distinguish between examples that might fall into a gray area. In a practical setting you'd just test the examples that might not be predicted well from trends.
These trends are also useful in the context of predicting whether SN1/SN2/E1/E2 will likely occur given a solvent/nucleophile combination. Sometimes this is straightforward. Sometimes it's a mixture of the reactions.

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Electronic configuration of excited states of iron
The electron configuration of the ground state of $\ce{Fe}$ is $\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6}$. What are the configurations of its first and third excited states? The first excited states seems to be $\mathrm{[Ar]4s^1 3d^7}$, but isn't the angular momentum not conserved because $l=0$ for $\mathrm{s}$ and $l=2$ for $\mathrm{d}$?

I am somewhat puzzled at the “isn't the angular momentum not conserved” part of the question… Among excited states, not all will have the same principal quantum numbers (such as angular momentum).

The electronic configurations of the excited states of neutral iron are listed on this NIST Database page:

3d64s2 — a 5D — ground state
3d7(4F)4s1 — a 5F — 6928.268 cm–1
3d7(4F)4s1 — a 3F — 11976.238 cm–1
3d7(4P)4s1 — a 5P — 17550.180 cm–1
3d64s2 — a 3P — 18378.185 cm–1
3d6(5D)4s14p1(3P) — z 7D0 — 19350.892 cm–1

The reference give for this data is: G. Nave, S. Johansson, R. C. M. Learner, A. P. Thorne, and J. W. Brault, Astrophys. J. Suppl. Ser. 94, 221 (1994).

So, while in general predicting electronic configurations for excited states is a tricky business, you are right: the first excited electronic configuration is to “promote” one 4s electron into the 3d orbitals. And the second excited state configuration (regardless of the different spectroscopic terms and sublevels) is to promote the 4s electron into 4p.


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Coupling constant in 1H NMR with phosphorus
I observed a doublet with peaks at 5.6336 and 5.5886 ppm with the NMR instrument running at 300 MHz, and got a $J_\ce{PH}= \pu{13.5 Hz}$. Is this correct for this molecule (ethoxycarbonylmethyl(triphenyl)phosphonium bromide)? This value is larger than the hydrogen coupling constants for the C-C bonds. Is this larger coupling constant a consequence of the phosphorous?

Yes, this is the 2JP-H coupling. You could verify this (if you had access to a NMR spectrometer) by running a 1H{31P} experiment; that is, a proton with 31P decoupling. This doublet would collapse back to a singlet. You would also notice some changes to the phenyl protons, as they all have some coupling to the 31P also. A 2J coupling of 13.5Hz is very typical for this type of 1H-31P coupling.
The magnitude of 1H-31P couplings varies enormously, depending on bond order, oxidation state of 31P and bonding arrangement.

1JPH of PH3 = ~175Hz.
1JPH of HP(O)(OMe)2 = ~1200Hz
2JPH P(CHx)y are typically 1-20Hz

It is sometimes observed in 31P-1H (and 31P-13C) that high bond order couplings are larger than lower bond order couplings (ie 3JPH > 2JPH)


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Balance chemical reaction
I want to balance reaction between glutamic acid and sodium carbonate. Can you check my answer whether it's true or false?? thank you.
C5H9NO4 + Na2CO3 = C5H7Na2NO4 + H2O + CO2

I think you're wrong.
I suspect This reaction forms Monosodium glutamate (NaC5H7NO4) which would mean you would have to balance by reacting 2 HC5H8NO4 for each Na2CO3, and result in double the H2O and CO2. (Decomposed from H2CO3)
But aren't questions like this (without a real question) Not Allowed?

thank u so much for your answer. Sorry, I'm new to this site.

Questions like this are right on the border @adam, it's always better to explain your work as to how you came to the conclusion rather than just asking if it's correct.  Welcome :)


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How can I pass this concentration to ppm?
So the exercise gave us the concentration of a solution (1,5 x 10^-3 M) and the volume (10 mL).
The 1st part of the exercise was to calculate the amount of volume we need from a more concentrated solution (0,010 M). I calculated the volume which is 1,5 mL.
The solution is from CoCl2.6H2O.
Now they want us to pass the concentration of our prepared solution to ppm. How can I do that? Thanks!

You make a stated assumption that 1.5 milliliters of solution has a mass of 1.5 grams. For such a dilute solution the assumption is surely good to 2 significant figures. // You do understand what ppm is don't you?!?

I'm not making any assumption! ppm is mass of solute/mass of solution x 10^6... Now how can I calculate it?

Is this a homework question?

BTW, I believe @MaxW didn't mean that you made an assumption, but rather that you should make one - since you want to solve your problem


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Boiling and Exploding Beaker?
A few days ago, I was creating a solution which involved boiling water. I had two 4L beakers on identical hot plates. After around 10 minutes, the first one boiled. Waiting another 30 minutes, the second one had still not started boiling, although small bubbles had formed at the bottom. Shortly afterwards, I heard a ping and the second beaker cracked and leaked water.
Is there any possible correlation between the long boiling time and the beaker breakage? Both beakers should have been heat-safe, although they were from different brands. I'm not convinced of the cleanliness of the beakers, so if this could've been caused by a contaminant, that's a definite possibility.

Was the beaker bigger than the hotplate?  A famous experiment was done with 10" glass frying pans. The pans were put on small stove eyes. Several pans exploded while others did not.  This is why 10" glass pans are not sold.

I'm inclined to suspect that your identical hotplates were less identical than you think.  They can have substantial differences in heating rate as well as the actual temperature they reach, particularly if they have been around for a while.  You could try boiling another pair of similar beakers and check the time they take to boil to confirm.
Usually lab glass breaks because of thermal shock - either hot glass with cold water or cold glass with hot water.  However, if you heated the beaker steadily for 40 minutes, it definitely wasn't shocked into breaking.  An exception might be if the hotplate is malfunctioning and get extremely hot extremely quickly, but that doesn't seem likely.  It is definitely possible that there was already a small crack in the beaker and that heating caused it to open further.
A contaminant is very unlikely to have been the source of the problem.  You should be able to boil dirty beakers all day long without them popping on you.
Unfortunately it's pretty much impossible to tell what happened for sure unless it winds up being reproducible.

Uneven heating on the hotplate surface could be another explanation. Since the second hotplate seemed to take much longer to boil, it seems that it may not have been heating correctly.

Questionable hot plates are definitely a possibility -- to the best of my knowledge, I'm younger than either. I can check to see if anyone else has encountered any issues with them on Monday, but I haven't heard anything recently.

Generally you should put boiling stones in a liquid to reduced "bumping." Basically the vessel becomes hotter than the boiling point and all of a sudden a large bubble of the vaporized substance forms. As you have verified the bumping can be violent enough to crack the vessel.

Wow, seriously? That's nuts. If this is the case, boiling should have occurred as the beaker brokem. @tikiking1, did the water in the second beaker boil violently once the glass broke? Based on your description, it doesn't sound like it bumped at all.

Though I agree that boiling stones should be used, if small bubbles had formed, the water didn't bump.

I was thinking that "exploding" in the title meant what I was calling bumping. If the beaker "just cracked" then it was probably uneven heating causing uneven thermal expansion of the glass.

I didn't see anything like a big bubble, but I wasn't watching the beaker until I heard the ping noise. It still wasn't at a rolling boil when the heat was removed. The exploding was what happened when attempting to move the beaker onto the table; the bottom fell clean off.


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Why does calculating formal charges help us draw correct Lewis structures?
Reading about how to draw Lewis structures, all I know is that formal charge is the charge assigned to an atom in a molecule assuming that electrons are shared equally, regardless of differing electronegativity between atoms. It is just a concept used to arrive at the correct Lewis structure and does not represent actual charges on atoms. But what I don't understand is, why does using this concept lead us to the correct structure?
For example, right now I am looking at the example of the sulfate ion. I get to this point before formal charges need to be calculated:

All the atoms have an octet around them (though I'm aware sulfur can have more than eight electrons around it), but it is not the correct structure. Why does considering the formal charge give the right structure, if as stated above it doesn't reflect any actual charges? It also says on Wikipedia "Formal charge is a test to determine the efficiency of electron distribution of a molecule. This is significant when drawing structures.", but I am not sure what it means by this.

It's mostly for electron bookeeping, especially useful when drawing resonance structure.

Formal charges are helpful for two reasons:

at lower levels, you use them to show your instructor that you can properly count electrons and ‘keep the books’. You would need to realise that each oxygen has an additional electron and therefore a formal negative charge while sulfur is lacking two electrons and therefore has a formal positive charge. Adding up the formal charges should give the charge of the molecule.
Since all lone pairs are typically drawn at this level, lacking formal charges can easily be rederived.

at higher levels, lone pairs are typically not drawn unless they are important for some reason (e.g. if they take part in a resonance mechanism discussed at this very moment). Therefore, the only information on the number of electrons around an atom we can extract from the formula is the formal charge.
If at higher levels, you draw a tetravalent sulfur atom without formal charges, we must assume that it is actually a formally neutral sulfur and that there is an additional lone pair. This, of course, makes a big difference; compare the electronic states of sulfur in $\ce{SO4^2-}$ and $\ce{SF4}$.


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Isn't the mechanism of aqueous sodium hydroxide and 1-bromobutane involves both heterolytic bond fission and attacking by a nucleophile?
What is involved in the mechanism of the reaction between aqueous sodium hydroxide and 1-bromobutane?
A. attack by a nucleophile on a carbon atom with a partial positive charge
B. heterolytic bond fission and attack by a nucleophile on a carbocation
C. homolytic bond fission and attack by an electrophile on a carbanion
D. homolytic bond fission and attack by a nucleophile on a carbocation
I think that A and B would be correct answers, but the correct answer is only A. Bromine accepts both two electrons from the carbon atom and the nucleophile attacks on that carbocation of the carbon atom. Why choice (B) is incorrect? I also searched the internet and found that this reaction is SN1, but why it occurs with the primary carbon, which is usually a reactant in SN2 reaction.

Have you got a link to the internet site claiming this to be SN1 ? I really dislike these questions, they give the impression to the student that chemistry is applying some textbook rules when generally in reality you either form some hypothesis and execute a suitable experiment (or experiments) to disprove/prove it or you search the literature for someone that has already done this. Maybe it is better to ask the questions as "What is the most likely mechanism/product to form etc ..."

Sorry for the rant above, I forgot to mention that the textbook answer is that it should be SN2 as a primary halide.

So sorry, I misread it.  Actually, it is as what you said that it should be SN2. At first, I misconception about the strongly charged nucleophile. @K_P

Well your thinking is absolutely right but only for tertiary halogenoalkane as this is a prinary halogenoalkane it will follow SN2 mechamism. In which necleophile attacks on a partial positive carbon


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Reaction between chlorobenzene and methyl chloride
What will be the product of the reaction between chlorobenzene and methyl chloride in the presence of anhydrous $\ce{AlCl3}$?
I think we should get 1-chloro-4-methylbenzene as the major product, as we'll have Friedel–Crafts alkylation. Is this correct? Or will we get toluene?

The answer is there in the mechanism of reaction itself.Friedel craft reaction is an electrophilic substitution reaction.Though,your doubt is common when the concept of electrophilic substitution and nucleophilic substitution  is not clear.Think what is the purpose of anh AlCl3 as catalyst.Once you find it you will have the answer (If this still doesn't address your problem please notify.thanks).

@AbhishekPallipparagopakumar Actually I was quite sure of the product. It's the book which confused me. Thanks anyways :)

You will actually get a mixture of different products, with the product with the highest yield being p-chlorotoluene, or 1-chloro-4-methylbenzene. The other products resulting from the reaction include o-chlorotoluene and m-chlorotoluene, with a significantly higher amount of the former being produced*. In fact, there would also be products with more than one methyl substituent being produced in various proportions depending on the precise conditions being used.
Yes, the reaction will proceed by Friedel-Crafts alkylation: Firstly, the Lewis acid catalyst abstracts the chloride ion from methyl chloride, giving a reactive, electrophilic methenium ion. Then, electrons from the $\pi$ system of the benzene ring of chlorobenzene then attack the electrophilic ion. Finally, a proton is lost to restore aromaticity of the ring, giving the product.
*Refer to the following post to understand why the chlorine atom as a substituent would be ortho/para-directing: Why are halogens ortho para directing even though deactivating?

1-chloro 4-methyl benzene because chlorine is an ortho para directing group so methyl group eithermust attack at ortho position or para position if chlorine. But chlorine shows negative inductive affect.so in order to increase stability methyl group must be attached at para position of chlorine

A para-substituted aromat is the product because the previous substituent is not para-directing? And I can’t understand your reasoning for why para should dominate over ortho …


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Why NOx will break the ozone sphere
I've found that a reason for people to oppose supersonic aeroplanes was the $\ce{NO_{$x$}}$ gases it released would do harm to the ozone layer. So what is the mechanism of this?

The mechanism is rather complex, I discussed some of it already for chlorine. In short, it basically removes the oxygen radicals from the ozone cycle: $$\ce{O2 ->[h\nu] O + O} \
\ce{O + O2 ->[] O3}\
\ce{O3 ->[h\nu] O + O2}$$

Don't UV and cosmic radiation also produce NOx?

According to Introduction to Atmospheric Chemistry, by Daniel J. Jacob, Princeton University Press, 1999,Chapter 10,

An important component of aircraft exhaust is nitric oxide ($\ce{NO}$) formed by oxidation of atmospheric $\ce{N2}$ at the high temperatures of the aircraft engine. In the stratosphere $\ce{NO}$ reacts rapidly with $\ce{O3}$ to produce $\ce{NO2}$, which then photolyzes:
$$\ce{NO + O3 -> NO2 + O2}$$
$$\ce{NO2 + $h\nu$ -> NO + O}$$

Note that there is no net consumption of $\ce{NO_x}$, thus it is able to act as a catalyst such that a single $\ce{NO_x}$ molecule can destroy many molecules of ozone by these reactions.


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Why does bisulfite (HSO3-) act as a base in this reaction?

I don't understand why the answer is (A) since $\ce{HSO_3^-}$ gives its proton out and becomes $\ce{SO_2}$ or $\ce{(SO_3)^{2-}}$ as show in both equations above, it'd rather be an acid. There is also no change in oxidation numbers of any of these atoms, too.

$\ce{HSO3-}$ is acting as an acid and a base in these equilibria. All the other answers are wrong so it must be A.

A screenshot or picture of an exercise is not searchable. Please consider rewriting it, so that it can be of help for future visitors.

I think HSO3 acts as a base because it donates a pair of electrons and this is another definition when you think of acids and bases. An acid is a substance that accepts a pair of electrons whereas a base is a substance that donates a pair of electrons.


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Numerical value of mole
Why is the numerical value of a mole of atoms equal to the atomic weight? For example, hydrogen has an atomic weight of 1.008, so 1 mole of hydrogen atoms has a molar mass of 1.008 g/mol. Why?

please clarify what you meant by "Why is the numerical value of a mole of atoms equal to the atomic weight? ".A mole of any substance has a mass equal to its atomic weight.The numerical value of mole is fixed for any substance and is not equal to its atomic weight.

Because we decided that 1 mole of C12 weighed 12 grams. The amu is a unit derived from the gram and the number of atoms in a mole.

http://chemistry.stackexchange.com/questions/41273/is-this-definition-of-mole-correct http://chemistry.stackexchange.com/questions/11169/why-is-the-definition-of-the-mole-as-it-is?s=3|2.8677 http://chemistry.stackexchange.com/questions/5321/why-is-the-mole-a-unit-of-measurement?s=4|2.8031

If i assumed your question right here is the answer.
1 mole=6.022*10^23 particles    (its not any unit of mass)
atomic mass of atoms are often expressed in a unit called amu.
1 amu=1.6605*10^(-24) gram      (more significant figures are there)
according to your question
mass of 1 hydrogen atom is 1.008 amu
= 1.008*1.6605*10^(-24) = 1.6737*10^(-24) gram
for 1 mole particles
mass =1.6737*10^(-24) * 6.022*10^(23) = 1.0079 g/mol (approximately 1.008)

@user27301 what is that you expect by evidence?(sometimes it may not be upto my level of knowledge to explain in depth)

maybe sources, website if you have any

@user27301 what proof do you want?I mean about which thing?(that is which part of my answer requires proof )

@user27301 just multiply those values in calculator and see for yourself.   !https://en.wikipedia.org/wiki/Mole_(unit)  Mole is a unit (just like other units)

The rest of the story - The atomic weight in the periodic table isn't for a single isotope, but for the mix of isotopes typical for that element. So chlorine is 75.77% $\ce{^{35}Cl}$ and 24.23% $\ce{^{37}Cl}$. Overall the mixture of isotopes still has $6.022*10^{23}$ atoms.


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why do we call lone *pairs*?
According to VSEPR theory, molecules adjust their shape to minimize the effect of repulsion.
Suppose I take the structure of $\ce{ClF3}$. It has three bond pairs and two lone pairs.
So to minimize the repulsion it gets T-shape (the two lone pairs are at corners of a trigonal bipyramidal shape).
But my doubt is why do we call two lone electrons a lone pair? I mean, if we keep a lone pair at some point in a molecule, how can two lone electrons be so near? Why don't we take electrons individually but always lone pairs?

What do you mean by "take individually" ?

We call them "pairs" because there are two electrons in each grouping. Why are there two instead of three or just one?
Spin and orbital energies.
One electron up and one electron down, in order to have all electrons in the lowest energy orbital available. Having them separate is a vastly higher energy state for the particles as a whole.
If you have not yet studied orbital theory - your question will be answered in much more depth when you do.

If you are having difficulty with why electrons pair, I am not surprised that you find the rest of orbital theory strange or illogical. Regardless, the topic is no less backed by experiment and evidence than any other in chemistry.


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Why does more acetic acid require more sodium hydroxide in order to be neutralized
I was recently titrating sodium hydroxide against vinegar and based off my results i realised that the vinegars with a higher concentration  of acetic acid required more sodium hydroxide in order to be neutralized. Why is this so? Does this happen for all acids regardless of pH?

Sounds quite obvious : more acetic acid needs more NaOH to be neutralised.

A higher concentration of acetic acid means more molecules (or moles) of acetic acid in the same volume that you used. So, it's pretty clear that you need to use more $\ce{NaOH}$ to neutralize the acetic acid.
1 mol of acetic acid reacts completely with 1 mol of sodium hydroxide.
$$\ce{CH3COOH + NaOH -> CH3COONa + H2O}$$
So obviously, increasing the number of moles of acetic acid (i.e. increasing the concentration) will necessitate a corresponding increase in the number of moles of $\ce{NaOH}$ (i.e. by using more $\ce{NaOH}$).
And yes, this should happen for all other acids too.


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How are ionic formulas written?

Why is $\ce{Ba(OH)2}$ Barium hydroxide not Barium dihydroxide?
Why is Lithium carbonate $\ce{Li2CO3}$ instead of $\ce{LiCO3}$?

Is  there a formula or is just the case for these two?

Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center. Could you tell me how you would deduce the formula of an ionic compound with hydrogen anions and calcium cations be?

Related: Why is CaCl2 called calcium chloride?

There is no other Ba(OH)x so there is no need to note a difference between several forms.
This happens with any metal ion that only attains precisely one positive charge, whether it is +1 or +10.
Lithium ions have +1 charge. Carbonate ions have -2 charge. Therefore to balance the charge in the formula we need 2 lithium to 1 carbonate.
This happens with all formulae, and it is a simple mathematical check to decide what the correct ratios should be.

In a different method that my teacher used to teach me, Lithium carry valence of one and Carbonate carry valence of two, and when they bond together, they interchange their valence which form Li2CO3 ( since we never wrote a valence of one as the subscript).

This is the general process to write the formula for an ionic name.

Name: Metal [Charge] (Ani)de

Ex.  Barium [Charge not needed] Hydroxide
Manganese (IV) Oxide
Lithium [Charge not needed] Carbonate

Charges:

if charge is given in parentheses, that is the charge for cation; otherwise, it is always the same. (Ex. $\ce {Ba}^{2+}$, $\ce {Mn}^{4+}$, $\ce {Li}^{1+}$)
Charges for anions are always the same (Ex. $\ce {OH}^{1-}$, $\ce {O}^{2-}$, $\ce {CO}_3^{2-}$)

Formula

The overall formula is the smallest amount of cations/anions in order to be neutral, i.e. the subscript of the cation is the charge of the anion, the subscript of the anion is the charge of the cation, then reduce (Criss-cross charges/subscripts of anions/cations, then reduce).
Ex: $\ce {Ba}_1^{2+}{(OH)}_2^{1-}$
Ex2: $\ce {Mn}_1^{4+}{O}_2^{2-}$
Ex3: $\ce {Li}_2^{1+}{({CO}_3)_1}^{2-}$

"1"s and "charges" are removed since they are implied.

Ex: $\ce {Ba}{(OH)}_2$
Ex2: $\ce {Mn}{O}_2$
Ex3: $\ce {Li}_2{({CO}_3)}$

The reducing subscripts and straightforward nature (only one right way to do it) are due to how these ionic formulas are "formula units" written to be neutral, as in the actual form, the ions form crystal lattices which can be described as large collections of these ratios.

It all depends on the valency of the atoms involved.
If you are have 1st group elements they all have a valency of +1 , and p-block elements have according to their periods*(it's better to remember groups of p-block elements - these start from the boron group to halogen group (fluorine.)*
Chlorine has valency -1 and sodium has valency +1 so multiply them. The valency of chlorine will be the subscript for $\ce{Na}$ and valency of $\ce{Na}$ will be the subscript for $\ce{Cl}$.
Similarly, Mg has valency 2 and oxygen also 2 $\ce{Mg_2O_2}$ will be the compound since both subscripts are common, we generally don't write them and we write $\ce{MgO}$ as compound.
There are still lot of exceptions.
Hope this helped.

So basically if both charges cancel out, I don't write any subscript? But if it needs balancing I put the charge on the bottom of the unbalanced element? Thanks for replying!

yes! that's correct


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Why does only one reaction occur when so many are possible?
Why does only one reaction take place, although so many would be possible? I'm more of a physics and math guy, so simply "accepting" it without a proper explanation is kind of hard for me. For example:
$$\ce{Na2CO3 + HCl -> NaCl + H2O + CO2}.$$
Why can't $\ce{H+}$ combine with $\ce{CO3^2-}$ to form a compound?
I'm going to highschool, and I am great at physics and mathematics, and I dislike chemistry since it's mostly memorizing. Please describe it in terms I can understand. Also, this is something that I genuinely want to understand rather than memorize, it is not related to school.

$\ce{H+}$ not only can, but actually does combine with $\ce{CO3^2-}$ to form a compound. The problem is what happens next. Also, welcome to Chem.SE.

@IvanNeretin Uh..... Then why is it not mentioned? Also, if it does react, what about the other possible combinations? Do they all occur? If so, shouldn't the products also react? If we perform an experiment in a lab, shouldn't that throw off everything?(So many questions, I know..) And thank you for the welcome.. :D

@IvanNeretin OK.. Never expected that'd be possible, but thanks. Also, can you post that as an answer? I'd like to mark it as one.

Your example is not quite the case when multiple reactions are possible. Also, $\ce{H+}$ not only can, but actually does combine with $\ce{CO3^2-}$ to form a new particle: $\ce{HCO3-}$, the hydrocarbonate anion. This is where it would end if you'd use just one tiny droplet of $\ce{HCl}$. Otherwise, that anion would attract another $\ce{H+}$ and form carbonic acid $\ce{H2CO3}$, which is unstable and decomposes into $\ce{H2O}$ and $\ce{CO2}$, to produce the summary equation which you already know.
As for your question in general, in most cases when so many reactions are possible, they indeed occur all at once. It just so happens that they go at different rates, and one of them outruns the others by a wide margin, so you may consider it "the only one" and safely ignore the rest. Or it does not, and you end up with a bucketful of brown goo. Such examples rarely make it to the textbooks, though.
All this is especially big deal in the field of organic chemistry, where all reactions produce more or less "side products". Read the details on just about any reaction, and you'll see these yield figures to the right: 96%, 92%... or maybe less if the authors were less lucky. That's how much of our reactants went down the desired pathway. The rest went someplace else.

Well, I don't think this quite explains the gist. // Regardless of pH, there is an equilibrium between $\ce{H+}$, $\ce{CO3^2-}$, $\ce{HCO3-}$, $\ce{H2CO3(aq)}$, $\ce{CO2(aq)}$, and $\ce{CO2(g)}$ where $\ce{CO2(g)}$ is the concentration of $\ce{CO2}$ in the atmosphere. The point is that the atmosphere is such a large sink for $\ce{CO2}$ that the reaction is driven towards pushing the carbonate out of solution. So mass balance wise 99.9+% of the carbonate is converted to $\ce{CO2}$ in the atmosphere. But there is still some minuscule amount of the other carbonate species in solution.

OK, point taken. So even in a reaction that simple, the yield is not quite 100%. Wait, it is never 100%, because the equilibrium constant is never $\infty$.


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Space in naming Acetophenone
The IUPAC name for Acetophenone is : 1-Phenylethanone.
An other name is : Methyl phenyl ketone.
My question here why there's a space in "Methyl phenyl ketone" although there's no space in "1-Phenylethanone"?

The two mentioned systematic names correspond to different types of nomenclature. The name “1-phenylethan-1-one” is formed according to substitutive nomenclature, and the name “methyl phenyl ketone” is formed according to functional class nomeclature. Both systematic names are in accordance with current IUPAC recommendations. However, the preferred IUPAC name (PIN) for such acyclic ketones is generated using substitutive nomenclature. Note that the omission of the locant “1” for the suffix “one” in “1-phenylethanone”, while permissible in general usage, is not allowed in PINs, thus the name “1-phenylethan-1-one” is the PIN. Furthermore, the name “acetophenone” is retained, but only for general nomenclature.

The corresponding subsection in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-64.2.2.1 Acyclic ketones
Unsubstituted acyclic ketones are systematically named in two ways:
(1) substitutively, using the suffix ‘one’ and the prefix ‘oxo’; the presence of several ‘one’ characteristic groups is denoted by the numerical multiplying prefixes ‘di’, ‘tri’, etc.; the final letter ‘a’ of a numerical multiplying prefix is elided before the suffix ‘-one’, for example, ‘tetrone’;
(2) by functional class nomenclature using the class names ‘ketone’, ‘diketone’ etc.; substituent groups are placed, as separate words, in alphanumerical order before the class name.
Method (1) generates preferred IUPAC names.

Note that the rule for functional class nomeclature (method 2) stipulates that the groups shall be written as separate words.


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Mixing bleach and ammonia?
I know that the two should not be mixed, but i want to know that if the two become mixed in a contained unit, does it build pressure? Or can you seal the container safely untill it can be disposed of properly?

No, the reaction forms a variety of gasses, including chlorine - it will build pressure.

A mixture of the two creates chlorine gas. It would be dangerous to try and seal the mixture in a container since the reaction will build up pressure. // The gist is that if you make this mistake then you will smell a strong chlorine smell. Ventilate the area and leave immediately. Your sense of smell diminishes even though the chlorine may be increasing in concentration. Although not immediately deadly like HCN gas, the chlorine gas is a strong corrosive and will damage your lungs.  Significant exposure will be fatal.

Bleach is a generic everyday term for a chemical substance that usually removes colour and disinfects. Whilst chlorine is common to many bleaches, some bleaches, such as hydrogen peroxide, do not contain chlorine. The reactivity will depend on the nature of your "bleach" (although will still probably be extremely hazardous). @Jacob, Can you be specific about what your bleach is?

In fact, here is just one of many recipes for "chemical free" bleach!! Note that the first ingredient is peroxide.........

Here is the usual citation for the toxic products and symptoms created upon mixing NH3 and NaOCl (not in lab conditions, which involves employing very cold dilute aqueous solutions):

When bleach is mixed with ammonia, toxic gases called chloramines are produced. Exposure to chloramine gases can cause the following symptoms:
Coughing.
Nausea.
Shortness of breath.
Watery eyes.
Chest pain.
Irritation to the throat, nose, and eyes.
Wheezing.
Pneumonia and fluid in the lungs.

The chemistry proceeds along the lines:
$\ce{NaOCl ⇌ Na+ + OH- + HOCl}$
$\ce{NH3 + HOCl ⇌ NH2Cl (monochloramine) + H2O}$
$\ce{NH2Cl + HOCl ⇌ NHCl2 (dichloramine) + H2O}$
$\ce{NHCl2 + HOCl ⇌ NCl3 (trichloramine) + H2O}$
Wikipedia also cites possible disproportion paths in acidic conditions starting with monochloramine, to quote:

In an acidic medium at pH values of around 4, chloramine disproportionates to form dichloramine, which in turn disproportionates again at pH values below 3 to form nitrogen trichloride:
$\ce{2 NH2Cl + H+ ⇌ NHCl2 + NH4+}$
$\ce{3 NHCl2 + H+ ⇌ 2 NCl3 + NH4+}$
At low pH values, nitrogen trichloride dominates and at pH 3–5 dichloramine dominates. These equilibria are disturbed by the irreversible decomposition of both compounds:
$\ce{NHCl2 + NCl3 + 2 H2O → N2 + 3 HCl + 2 HOCl}$

Note, with respect to any formed nitrogen trichloride (appears as a yellow oily liquid), extreme caution as it is a very unstable explosive compound upon exposure to shock, heating, sunlight and organics, even while in aqueous solutions. Because of its dangerous sensitivity, $\ce{NCl3}$ is best avoided.
Also, with respect to exposure of swimmers where chloramine can be formed in non-chloramine pools from free chlorine interacting with amine groups present in organic substances, or chloramine treated pools employing NH2Cl as a disinfectant, especially in the case of indoor pools, per the same Wikipedia source:

There is also evidence that exposure to chloramine, can contribute to respiratory problems, including asthma, among swimmers.[11] Respiratory problems related to chloramine exposure are common and prevalent among competitive swimmers.[12]

On the question, can one safely disposed of aqueous NH2Cl? Be careful, as on just adding it to a water source may increase the solubility of heavy metals. This can result, for example, in Lead poisoning for animals and humans (see 'Chloramine Catch: Water Disinfectant Can Raise Lead Exposure).

Yes, it will build pressure, and if you seal it, it becomes a potential bomb.  Unless you're doing this in a controlled environment, intentionally, with a need for the toxic compounds you'll produce (chloramine, hydrochloric acid vapors, and/or hydrazine, all in a mixture of unreacted ammonia and bleach) you should not do this, period.
There is a reason that everyone is taught not to mix these compounds.


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Trends in conductivity in groups 1 and 2
I have read that the trend for conductivity in groups 1 and 2 is a decrease down the groups (generally). Apparently this is because the atoms of each element get larger, so while the number of electrons per atom stays the same, the number of electrons per unit volume decreases. But the amount of shielding between the outer electrons and the nucleus increases, so surely the outer electrons become easier to donate to the delocalised sea, and will experience less attraction to the positive ions around them. Which I would have thought made them more mobile and therefore made the metal a better conductor. Obviously I'm wrong but can someone please explain why?

The conductivity of metals depends quite fundamentally upon the temperature at which it is measured - and without clarifying what temperature(s) you are interested in this question is difficult to answer. You probably mean low temperatures as you are ignoring the contribution to the resistivity from lattice vibrations, but now I am guessing.

the alkali  metal ions in aqueous solution have high electrical conductance. the electrical conductance of Li+ ions expected to be highest, which is smallest ions in size, as ionic size increases down the group.
               however, we find experimently that electric conductance increases down the group and Cs+ has the highest conductance while Li+ has lowest.

this is evidently the reverse of what we expect. This anomaly arises from the fact that alkali metal ions are hydrated to diffrent extents.The Li+ ion is hydrated to the maximum extent due to its  high charge/size ratio compared to other alkali metals the extent of hydration decrease down the group, thus Cs+ is least hydrated Li+ ion due to high hydration is the biggest while Cs+ ion due to smaller hydration is the smallest. thus, the ionic size in aqueous solutions, instead of increasing actually decreases on moving from Li+ to Cs+
as a result of which ionic mobility increases as we move from Li+ to Cs+. Hence, the ionic conductance, instead of decreasing, actually increases on moving from Li+ to Cs+

The question is about the elemental metals, not the ions in (aqueous) solution


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Why are X-rays used in crystallography?
Can any other wavelength of electromagnetic radiation be used?
Why can't gamma radiation be used?

There are a few reasons, but the most direct answer is that the wavelength of X-ray photons is on the order of the distance between atomic nuclei in solids, e.g. ~ 4 ångströms (bonds are roughly 1.5-2.5 Å). You can think of it like the waves fit nice and snugly between the atoms and "fill" the crystal and thereby give us information about where the "cavities" are, and thus by reciprocal analysis (inverse Fourier transform) we can determine where the "rigid" stuff is (the atoms).
Other kinds of incident radiation are used e.g. neutron beams (for crystallography) and even electron beams (for surface spectroscopy), but X-rays are a more readily available source of radiation. In other words, it's easier/cheaper to produce X-rays than a beam of pure neutrons.
Gamma rays are about 100 times smaller in wavelength than X-rays (so instead of 4 Å, something like 0.04 Å, and so they generally penetrate through any solid material and don't produce a clear diffraction pattern like X-rays. Also if I'm not mistaken, it is much more difficult to produce gamma rays.
It's important to realize that diffraction is not the same as reflection and refraction. The latter two are surface phenomena - diffraction is a bulk phenomena completely unrelated to surfaces.
Electrons interact strongly with the surface of solids not only because of their charge but their energetic wavelength - gamma rays interact very weakly with a bulk crystal and go right through it. X-rays and neutrons (bearing no charge) are in the "happy medium" and produce a sufficient interaction with crystals as to give information about the arrangement of particles.
Picture taken from wikipedia's X-ray article:

More useful info in Economou's book "The Physics of Solids" can be found.

To be more specific, the most common X-ray wavelengths used in crystallography are 0.7 and 1.5Å.

Indeed -- those are the wavelengths that correspond to atomic bonds.  X-rays are categorically described in the range 0.1 - 100A, so crystallography is a pretty specific application of x-ray energy.


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Pressure increase on rate of reaction
For reactions involving gases, an increase in pressure is no different from increasing the concentration of reactant particles, thus increasing the rate of reaction. But for reactions involves different phases, such as both solid and liquid, what will the pressure increase cause? Will it increase the rate of reaction? Or, because the rate also depends on other reactants and in reactions involving different phases, will pressure not change significantly even if gas is present?

I'm afraid this question can't be addressed generally

There is probably going to be a separate reaction rate law for each phase, assuming that the reaction even occurs in other phases.  Of course, in liquid and solid phases, the pressure is going to have much less of an effect than in a gas phase.  Another consideration is that there may be interphase mass transport between the various phases (at their interfaces), and this may have to be taken into account in the chemical species mass balances for the individual phases.


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Smell of the methanol
When I first time smell the methanol, for me it was similar to the benzene (completely different from the ethanol). But almost everywhere information says, that it should me very similar to the ethanol. (I am talking about almost dry methanol, not dissolved in water or mixed with ethanol.)
I understand, that smelling chemicals is not healthy, but kills the dosage, not the poison. The only breath won't kill anybody. (Off-course, if you are working with methanol or other dangerous compounds you must wear gas-mask and/or work with good ventilation!)
How would you describe the smell of the methanol?

Its a qualitative answer, so you'll get a lot of odd things in these responses.
I find methanol to be like ethanol but shimmed by a plastic/wood-life taste.
More importantly, having a higher volatility, its taste will be more intense because there are more vapors to stimulate your odor receptors which frequently change or augment tastes.

Yes, the smell is much more intense, than the ethanol.


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What is the name for a reaction with a solid product rising from its container?
The reaction between sugar and sulfuric acid is well known to produce a solid column of rising carbon which may leave its container.  A very similar rising column may also be produced from high temperature pyrolysis of certain organic substrates.
Is there a name exclusive to the type of reaction that produces a rising column?

One of my buddies poured himself a beer.  Beer going into the glass caused a frothing reaction.  Has anyone heard of solid frothing?

In pyrotechnic chemistry, they might call it a pharoh's snake (or serpent) reaction. To a degree, (or rather 330 °C) pyrolysis of the sugar is occurring (starting at 170 °C).
In cooking, they would call it caramelization, although the term is also applicable in general chemistry. Less specifically, you can call it a "browning reaction"... note- caramelization (non-enzymatic) should not be confused with a Maillard reaction (enzymatic). More specifically, you could call it dehydration or decomposition of sugar with sulfuric acid. Somewhere in-between you might find the term carbonization.
You may also find the terms carbon soufflé and carbon foam interesting.

Ok, that is the detailed Dad answer. Mom's more to point answer is - No. There is no specific name for such a reaction.

@MaxW Ha, you got me!

Plus there are also a number of names referring to the fact that it is something cylindric and it is rising.

While doing biochar research, sometimes this reaction happens instead of char formation.  I was hoping for something more scientific sounding than a d*** mess, which is what I was calling it. This answer works.

Concentrated sulfuric acid dehydrates almost any hydroxyl containing organic molecule. This is the specific type of reaction that is occurring.
The reason the carbon-column forms  is the intense heat that is released from the reaction. The water vaporizes and rises, carrying the carbon with it.

Dehydration does not always form a rising column (example  https://www.youtube.com/watch?v=IhE6CM6Qbws.)  Certain pyrolytic reactions do form a rising column.  I am looking for a name describing a reaction which produces this phenomena.

That is a completely different type of dehydration. The dehydration reaction that I was talking about specifically involves the removal of water from an organic compound, not the driving off of water from a hydrated salt.

Is there a name for this specific form of dehydration?

I believe it would be called acid catalyzed dehydration.

I changed the language in the question for clarity.

How about "Charring"?


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Can fumes from using bleach seep through plastic packaging and glass jars
A bulging can leaked inside our kitchen cabinet. I googled how to handle it. I followed the instructions I found online. I mixed 1/4 bleach and 2 cups water. Applied it on all surfaces. Then applied dish soap diluted with water.
Does this produce toxic fumes? And can the toxic fumes seep through food even if they are covered and placed about 3 meters away? The cabinet still smells like bleach. What should I do about it?

A quarter cup of bleach?

Yes, I used a quarter cup of bleach

Just wanted to clarify.

Sorry, was typing too fast I didnt notice I didnt type proper measurement.

Disclaimer: This answer is provided for information only. Use it at your own risk.

I mixed 1/4 bleach and 2 cups water. Applied it on all surfaces. Then
  applied dish soap diluted with water.

This was not smart. Bleach should never be mixed with other chemicals, cleaners, or detergents. Rinsing with plain water first would have been less risky.
With bleach it is important to ventilate the area. Bleach should be thoroughly rinsed if it gets on anyone. Gloves are highly recommended for this type of cleaning.

Does this produce toxic fumes?

Possibly.
Bleach attacks double and triple bonds in organic molecules. This action not only inactivates most disease causing organic particles, but also destroys the chromophores which give organic material their color. This reaction can sometimes liberate malodorous organic fumes which is best not to breathe.
Most inexpensive commercial bleach is a diluted form of sodium hypochlorite. Sodium hypochlorite can release chlorine gas; however, under normal usage bleach, nor its smell is likely to be harmful. The odor threshold for common bleach is 0.002 parts per million (ppm) which is well below the concentration it will cause any harm.
If on the off chance the smell is extremely irritating or if coughing occurs then get people and pets out of the area and ventilate the gas. Wait till the odor ceases to be irritating. If the odor is just annoying then it is likely safe.

And can the toxic fumes seep through food even if they are covered and
  placed about 3 meters away?

If ventilation is good over a relatively short period of time then seepage is unlikely. If ventilation is poor over several days then seepage could turn into a problem. If the food smells or tastes either odd or like bleach, it may be best to throw it out.

The cabinet still smells like bleach. What should I do about it?

Ventilating the cabinet till the smell goes away is best. Rinsing thoroughly with plain water, getting the water up as best as possible, and then allowing the surfaces to dry may work. Only after thoroughly rinsing and ventilating would it be worth it to clean with dilute vinegar help with the smell. Do not allow vinegar to mix with bleach!
A better disinfectant choice for just one bulged can may have been to use 70% isopropyl rubbing alcohol soaked into a dry rag. No chlorine fumes, much less likely to discolor, while still effective with germs.  This should go without saying...  Do not allow alcohol to mix with bleach!


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Where does the rate equation come from? (r = k [A]^x [B]^y)
Equations are great, but they're much better when you can understand them. I don't quite get why this works. Could someone explain it to me?r = k [A]^x [B]^y

Is this answer helpful?

A to the power of x or B to the power of y are telling you how the concentration of that reactant will affect the rate of reaction. For example, if you find that doubling the concentration of reactant A quadrupled the reaction rate than A could be raised to the power of 2. If doubling the concentration of B doubled the rate of reaction than you could raise B to the power of 1. K is the Arrhenius equation and its value is taking into account the frequency that the particles are colliding, the amount that are colliding with the right orientation and the energy required for the reaction to take place. K = Ae^(-Ea/RT) where A is the product of the frequency of the colliding particles and the amount that have the correct orientation. e is 2.718281828 and is raised to the power of negative the activation energy divided by RT where R is the ideal gas constant with a value of 8.314 and T is the temperature. You could use algebra to solve for A if you have the other values.

Agreed. Rate constants are found experimentally, and only indirectly reflect the mechanisms occurring at the molecular level. In fact: many reaction mechanisms are much more complex than you might imagine!


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Making a 10L HCl solution with 8.77% HCl in a 100 mL solution from a 16 M stock solution
How would you make 10L of HCl solution with 8.77% HCl in 100 mL from a 16M stock solution.
Here's what I have tried so far:

I put my percent HCl and got that down to match with the moles of HCl I had. Then I took 1.754% of 20 ml and converted that into liters. Then I found the molarity of the HCl I need to turn the 16 M solution into. Finally I did a dilution calculation to find the volume of water needed to make the 16 M stock solution into the starting percentage of HCl.
I am fairly certain that I am diluting with water, I don't know what else it would be with.
Is this the right way to go about this, or do I need to restart. Any help is much appreciated, and thanked for in advance!

I figured out the problem I had. I needed a way to convert grams to milliliters, which was density.
I found the density of HCl for the 8.77% concentration:
$$ HCl \ Density = 1.04 g/mL$$
I then found how much HCl would be in 100g of solution:
$$ 8.77\% \ of \ 100g, \ so \ 8.77g \ HCl \ and \ 91.23g \ solvent$$
I did some dimensional analysis to get to moles divided by liters, so that I could get the molarity:
$$ (\frac{8.77 g \ HCl}{1})*(\frac{1 \ mol}{36.461g})*(\frac{1}{100g \ solution})*(\frac{1.04g}{1 mL})*(\frac{1000 mL}{1 L}) = 2.5 M$$
Then, I did a nice dilution calculation:
$$ 10 L *2.5M = 16M *L_2$$
$$ \frac{25}{16} = 1.5625 L$$
To do the dilution I would place 1.5625 L HCl into a beaker, and then fill with solvent until I reached a total solution volume of 10 L.


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Using a calcium metal anode in aqueous solution?
In school, I am doing an electrochemistry lab in which I need to create a galvanic cell battery with the highest voltage possible using two beakers, a salt bridge, and electrodes.
To maximize the voltage, I have decided to use calcium metal for the oxidation half-reaction and a sodium persulfate solution for the reduction half-reaction. I understand that for the reduction reaction in the cathode, I would merely dissolve the persulfate in water. However, calcium metal reacts with water, so how would I go about connecting the calcium to the salt bridge (which is a requirement in the lab)?
The salt bridge has an aqueous salt and my chemistry teacher said that the redox reaction that would result from letting the calcium touch the water would prevent it from being effective in the battery. Also, my class vaguely touched on the Nernst equation, but are there any other revisions I can make to the battery to increase the voltage, either by adding $\ce{OH-}$, $\ce{H+}$, or increasing/decreasing concentration of the persulfate solution?

Basically, the reaction of calcium with water short-circuits your battery; the electrons you wanted to go through the external circuit end up going directly into the electrolyte instead.
Magnesium will work better for your anode; it reacts little or none with water when cold and it's anodic potential still packs a pretty good punch.

Okay, that makes sense, thanks! I think other people will be using magnesium so we wanted to go with calcium since the standard reduction potential of calcium ions is lower than that of magnesium. Do you know of any ways to still use calcium and prevent the short-circuiting (use something in between the calcium and salt bridge so they're not in contact)?

Calcium does react with water, but it does so slowly especially in cold water.  You should be able to use it as you would a typical metal as some water will touch the metal. Just be aware that it will react and your voltage may be less than predicted.  This battery of course would be of no practically use as its shelf life would be on the order of less than an hour, but should last long enough to obtain a measurement and complete the experiment. You should also expect to see a change in voltage over time as $\ce{Ca(OH)2}$ is added to solution reducing the potential.

Thank you! Is there any way to make the battery work without having the calcium in direct contact with water? In the anode, I would be using a graphite rod as the electrode and touch it to the calcium metal. My main problem is the salt bridge. Is there a method of connecting the calcium to the bridge to maintain charge balance without sacrificing voltage from a potential reaction with water?

typically batteries use propylene carbonate for such reactive metals. access to such may be an issue.


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Colorimetric test for methanol
Will adding dilute potassium permanganate to impure methanol in alkaline conditions work as a good colorimetric method? I am planning to produce salicylic acid from methyl salicylate (by hydrolysing the ester and producing methanol), and was wondering how I can show that the reaction goes to completion under alkaline conditions. If not, what would be a good method to show this?

Usually, methanol is used in excess, so checking for methanol isn't terribly useful. You can check for salicylic acid with $\ce{FeCl3}$, which forms a purple complex with salicylic acid. See here. Permanganate may work for methanol, but it is such a strong oxidant, I'd be worried about it reacting indiscriminately.

Thanks for the quick response. However I think you've got it back to front about the reaction, I'm hydrolysing the ester, so producing methanol. Also iron(III) chloride will form a purple complex with both the reactant and product since they both have a phenol group. However, you may be right about other reactions occurring due to its oxidising ability and I hadn't thought about that so I'll have a think about that, thank you!

Oh yeah, I misread. My bad. Are planning on titrating or just doing a qualitative test for methanol? Because just qualitatively testing for the presence of methanol won't tell you anything about the completeness of the reaction. Also, I thought it was the chelating effect of the carboxylate and the phenol that caused the purple color.

Nevermind, you're right, ferric chloride tests positive for all phenols with a purple color.

Some form of quantitative test - I was thinking of trying colorimetry with the potassium permanganate reaction. Iron(III) chloride will turn purple with a phenol without a carboxylic acid (I've also tried it with both reactant and product; both go purple), but I might be able to use a ligand exchange to differentiate between the 2 compounds as methyl salicylate will only be a mono-dentate ligand, then use colorimetry.


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Is Van der Waal's force the real source of friction?
I'm currently taking a physics class, and my teacher explained that friction is the result of intermolecular interactions, and that energy must be applied in order to break these bonds.
This got be thinking about what kinds of things have a high coefficient of friction, and immediately my mind went to sandpaper. The atomic surface of these mineral crystals would be really smooth and not easily polarizable, as they are held tightly in place by the other ions. So does surface area have more to do with friction than actual Van der Waals forces? This would also seem to explain how geckos are so adept and clinging to things like glass, as the tips of their fingers are so highly ridged.

There are interactions other than van der Waal's forces, so in general, no.

Most of the forces that you encounter day-to-day (apart from gravity) are Coulombic repulsions of outer-shell electrons on the surfaces of macroscopic objects (walls, floors, balls, keyboards, etc.) While you are right about the stickiness of gecko feet being attributable to VDW, the friction from sandpaper is most likely due to its uneven surface which makes it prone to scrape against other surfaces via Coulomb. One might compare this kind of process to sliding strips of velcro against each other.

While van-der Wall forces contribute to the frictional force, they are not the only reasons. Other secondary inter-molecular forces like electrostatic attractions, polar interactions, hydrogen bonding and more also play a significant part. Apart from this features on the surface of materials, present as abrasions are also responsible for locking together and resisting change.
The case of a gecko sticking to wall is not just about these forces and interlocking, but more about how the wrinkles on feet of a gecko act as several small vaccum cups which provide it the stability of sticking upside down as well(think how friction plays almost no role here)

Except geckos do stick due to Van der Waals: http://www.sciencemag.org/news/2002/08/how-geckos-stick-der-waals.


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Why does final pH value of buffer solution after addition of acid not depend on amount added?
I'm brushing up and the example in my book is against my intuition.

A buffer solution contains 1.0 mol/L acetic acid and 1.0 mol/L
  sodium acetate. The solution is at pH = pKa = 4.75. What's the pH
  after addition of 0.01 mol/L HCl?

They don't write how much of the acid is added.
They go on stating that addition of x = 0.01 mol/L H+ results in pH = 4.75 - log[(1.01 mol/L) / (0.99 mol/L)] = 4.733.
That's all fine from the formalism. What's against my intuition is:
Why is it not required to state how much of the acid was added? Given the buffer solution, is the outcome of adding acid the same independent if I add a Liter of if I add a mL?

maybe they meant "addition of 0.01 mol" and the "per liter" is a mistake? at least that would make sense.

The concept of a buffer is that a small addition of acid or base won't change the pH significantly. The pH must change at least a small amount.

@MaxW I don't understand your comment. My question is not about what a buffer does. It's about why the amount of acid is not required in the question to calculate the change in pH. Will the pH change be the same no matter how much acid is added?

When more acid is added, then the pH must fall - at least a little. But a drop of conc. HCl doesn't change pH of 100 ml of pH 7.0 buffer much, but it changes pH of 100 ml of pure water a lot.


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

There really isn't a chemical formula for cooking rice.

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

There might not be a simple equation but there is more to it than simple hydration. Surely there is some breakdown of the starchy component (starch being a long polymer of sugars)?


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What wavelength of light is released when an electron drops an energy level in complexes?
I know that the color of transition metals is related to the splitting of the d-orbitals. Only a certain wavelength is absorbed in the promotion of electron to a higher d-orbital and therefore the remaining colors are interpreted as a certain color. For example in copper(II) sulphate the red wavelength is absorbed and therefore it the complex appears to be cyan. But shouldn't the same amount of energy be released once the electron drops back to its previous state and releases the energy is gained? What wavelength would this drop correspond to?

Not necessarily. The drop may occur gradually so that the colour is "blinded out" by the nonadsorbed colours. The drop may also occur in stages, so that several low energy emissions occur. Some are outside visible range and others are too weak or go unnoticed.

But if it is due to the d-orbital splitting, shouldn't there be only one possibility of a drop? Where else could the electron go, besides the lower d-orbital?

Are you asking about a specific case of splitting or in general? There are a lot of things that affect the energy of an orbital, and there are multiple splitting combinations amongst the metals.

I am asking in terms of a complex that is formed with a ligand. I was taught that the reason why complexes appear to be colored is that their d-orbital splits into a higher and lower energy, and when light is shined through the complex, a certain wavelength is absorbed. But I wonder what happens to the excited electron. If it were to drop to a lower d-orbital it would release the same wavelength as it absorbed...

Then the energy levels available are dependent on what metals and what ligands are involved. But there's never just two of them. What would happen if an electron in the lowest level jumped to level 6, then all electrons dropped by one level each? No transition is repeated and yet we absorbed then emitted energy.


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Glycine in magnesium glycinate
To be very specific I'm looking at a supplement that contains 2,000 mg of magnesium glycinate. In that serving it sates there is 2oo mg of elemental magnesium. How much glycine is there?

If there are no other ingredients like citric acid- 1800mg. It is probably magnesium BISglycinate. For each Mg ion, there are two glycine molecules.
After a calculation, if it's MagBisglycinate, for 200mg Mg you should have around 1235.5mg glycine. It probably means there are other ingredients or simply glycine that left unreacted. If it's simple Magnesium glycinate than you should have 617.7mg glycine for 200mg Mg.
Correct me if I'm wrong.


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What causes rapid nitrogen inversion?
When I asked my teacher whether a nitrogen bonded to 3 unique groups counted as a chiral center, they said that it did not because nitrogen undergoes rapid inversion at room temperature. What causes this? And what other atoms are capable of doing this?

There might be quantum tunnelling effects as well. It's possible that the smaller mass of N as compared to P etc. makes the tunnelling more likely

@orthocresol Along the same lines is the higher electronegativity of N vs P, which makes the quantum tunnelling more favorable. To be honest, though, I don't even know what that means. I'm just regurgitating the Anslyn and Dougherty book's ideas.

related http://chemistry.stackexchange.com/questions/464/why-dont-trigonal-s-and-p-compounds-undergo-inversion-at-room-temperature

In the second row of the periodic table, elements have relatively small differences between the size their $\mathrm s$- and $\mathrm p$-orbitals. Therefore, the orbitals of $\ce{NR3}$ can go from $\mathrm {sp}^3$ to $\mathrm {sp}^2$ with relatively little energy increase, so an amine can become planar and then reorient with the inverted stereochemistry. The same occurs with carbanions. I'm not certain, but I'm guessing oxonium cations do this, as well, and I would even guess that it would happen more quickly due to oxygen's smaller size.
The reason we don't see this with $\ce{NF3}$ is because orbitals with more $\mathrm p$ character are more affected by induction, so the highly electronegative fluorine atoms would lose electron density during inversion, which would be energetically unfavorable (to the point that it doesn't occur readily).
Chiral phosphines do not invert because the size difference between $\mathrm s$- and $\mathrm p$-orbitals for larger atoms is greater, so converting $\ce{PR3}$ $\mathrm{sp}^3$ orbitals to $\mathrm {sp}^2$ would require much contraction and, therefore, more energy than for the analogous amine.


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optical rotation of styrene oxide
Which optical isomer of Styrene Oxide  would be produced if it is synthesized via epoxidation of the halohydrin. i.e.

Is there a way to deduce this? I assume it would depend on which isomer the halohydrin itself is since that too seems to have an assymetric Carbon atom?
Well, the halohydrin itself is produced from styrene which is optically inactive by the action of hypochlorous acid. Does this enable us to deduce theoritically the optical rotation of the final product i.e. Styrene Oxide?

Thanks! Would there be a practical way to separate the optical isomers at scale for a tentative production process? Or would one rather choose an entirely different synthesis route;a route that starts from an optically pure reagent and utilizes chirality preserving reactions?

Styrene and HClO are going to make a racemic mixture, in which case the epoxide will also be racemic, i.e. there should be no optical rotation.
An alternative would be to use the Jacobsen epoxidation, the Shi epoxidation or some other enantioselective epoxidation.


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How can I determine the activation energy of a reaction if the reaction rate constant is unknown?
I am trying to conduct an experiment to determine the activation energy of the following chemical reaction:

$$\ce{IO3- (aq) + 3HSO3- (aq) -> I- (aq) + 3SO4^2- (aq) + 3H+ (aq)}$$
$$\ce{5I- (aq) + 6H+ (aq) + IO3- (aq) -> 3I2 (aq) + 3H2O (l)}$$

The Arrhenius Equation can be rewritten as $$\ln k = \frac{-Ea}{{\rm R}T} + \ln A$$ Therefore, if ln k and 1/T were plotted on a graph, the slope of the line should be equal to -Ea/R.
I can use temperature as my independent variable and the reaction time as my dependent variable. However, I am unsure as to how to determine the reaction rate constant (k) of the reaction experimentally from the reaction time.

You have two reactions there.  One will be rate limiting (I don't know which), so your plots will give you the activation energy for that part only.  As for how, this is standard rate law problems, although this one is a smidge messier than normal.  You might have to assume steady state Iodide and Hydrogen Cation


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Does the addition or multiplication rule have precedence deciding the significant figures in 5×5.364?

How many significant figures should be present in the answer of
  $5\times 5.364$?

Addition rule: The result cannot have more digits to the right of the decimal point than either of the original numbers.
Multiplication rule: The result must be reported with no more significant figures as there are in the measurement with the few significant figures.

Using rule 1,
$$5\times 5.364 = 5.364 + 5.364 + 5.364 + 5.364 + 5.364 = 26.82,$$
one would use 4 significant figures.
Using rule 2,
$$5 \times 5.364 = 26.82 = 30,$$
one would use 1 significant figure.
Which one is correct?

If this is the question in its entirety, then you should not try to 'guess' what the origin, and therefore the accuracy, of the measurement 5 is. You should take it on face value, and complete the calculation using the multiplication rule, as the equation is requiring a multiplication operation. In this case, you should have one significant figure only.
The only ambiguity you need to consider in calculating your significant figures is whether the '5' is an exact number or not. Exact numbers come about either by definition (1 hour = 3600 seconds) or by counting (5 measurements were made) and have an infinite number of significant figures. Hence, they do not affect the accuracy of your measurements. Exact numbers are usually integers (speed of light and number of centimetres in an inch are a couple of exceptions). My guess is that your question is simply a case of going through significant figures calculations, using the information provided. To avoid ambiguity in the case of a single significant figure measurement, one could consider writing 5 as 5E0, or 5x100.
If you know 5 to be an exact number (ie the question gives that information), then you would still complete the calculation using the multiplication rule, but assume infinite number of significant figures for 5. In this case you would have 4 significant figures.

Using the addition rule doesn't make any sense because then you would always use the addition rule for multiplication.
You would use the multiplication rule, and if this question was given as-is, 30 would be correct. If you're told the 5 is exactly 5 (like in many/most conversions), then you would consider it infinite significant figures and answer with 4 significant digits.

If the measured width of a rectangle is 5 meters and the length is 5.364 meters, then the area would only have one significant figure, so 30 square meters by the multiplication rule.
If you measure the width of 5 bricks and the measurement of each brick is 5.364 cm, then the number "5" is not a measured value, it is an exact value.  Four significant digits would be appropriate, consistent with the multiplication rule.
Contrary to the OP, literal application of the addition rule "cannot have more digits to the right of the decimal point than either of the original numbers", would mean 5 significant digits not 4.
However, when measurements are being added, the uncertainty in the square root of the sum of the squares of the uncertainties in each measurement, so the addition rule increasingly fails as more values are summed.

It really depends on what you're trying to signify, but I'd say that neither of the answers that you have given is entirely satisfactory. The "rules" you cite are good for sort of rule of thumb, but don't really work well when you're working with a single digit. 9 +/- 0.5 is about a 6% error, but 1 +/- 0.5 is a 50% error.
If you were working with a number which has 3 significant figures by a number which has 4 significant figures then the rules you cite work well enough.
The quirk here is that you can assume that 5 is +/- 0.5 so that there is an error of 10%. So 27 +/- 3 would really be a better answer.
Another thing to consider is that 30 has two significant figures. If you want to denote just one significant figure then you'd write the answer in scientific notation as 3E1 to clearly denote just one significant figure.
Of course if the 5 was an integer value then it has infinite precision. Say you're working with oxygen in the following chemical equation:
$$\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}$$

Perhaps I misunderstand you, but 30 has 1 significant figure.

@long - It depends on if you know it is 30 +/- 10 (which has 1 significant figure), or if you think it is 30 +/- 0.5 (which has two significant figures). I pass a speed limit sign that says 30 mph. I don't think the law means +/- 10 mph.