text:
Rate-limiting in transition state theory
Suppose an initial state $A$ can transition to either state $C$ or state $D$. Suppose further that both of these two processes are rate-limited by a transition to the same intermediate state $B$, as follows:

Note that $\Delta G_{AB}> \Delta G_{BC}>\Delta G_{BD}$.
According to transition state theory, rates are determined by the rate-limiting step. So the rate of each process will be the same,
\begin{align}
r_{A\to C}&=r_{A\to B} \\
r_{A\to D}&=r_{A\to B}
\end{align}
from which it follows that the number of transitions made during some large $\Delta t$ will also be the same,
\begin{align}
N_{A\to C}&=r_{A\to B}\Delta t \\
N_{A\to D}&=r_{A\to B}\Delta t
\end{align}
However, if we were to consider the two pathways collectively then we would predict more transitions to be made to state $D$ than $B$ since $\Delta G_{BD}<\Delta G_{BC}$.
How do we resolve this apparent discrepancy?

In transition state theory (TST, goldbook) one of the necessary assumptions is that reactants and products are in equilibrium. In principle this gives us
\begin{align}\ce{
A &<=> [AB]^{$\ddagger$} -> B\\
B &<=> [AB]^{$\ddagger$} -> A\\
A &<=> B,
}\end{align}
and in addition to this, also
\begin{align}\ce{
B &<=> [BC]^{$\ddagger$} -> C & B &<=> [BD]^{$\ddagger$} -> D\\
C &<=> [BC]^{$\ddagger$} -> B & D &<=> [BD]^{$\ddagger$} -> B\\
B &<=> C                      & B &<=> D,
}\end{align}
as well as
\begin{align}\ce{
C <=> [BC]^{$\ddagger$} -> & B <=> [BD]^{$\ddagger$} -> D\\
D <=> [BD]^{$\ddagger$} -> & B <=> [BC]^{$\ddagger$} -> C\\
C <=> & B <=> D,
}\end{align}
and
$$\ce{
C <=> A <=> D\\
[AB]^{$\ddagger$} <=> [BC]^{$\ddagger$} <=> [BD]^{$\ddagger$}
}.$$
You can use TST to estimate rate constants.
$$k = \frac{\mathcal{k}_\mathrm{B} T}{\mathrm{h}} \exp\left\{ −\frac{\Delta^\ddagger{}G^\circ}{\mathcal{R} T}\right\}$$
Since all components have to be in equilibrium, you can predict a ratio between them via Boltzmann statistics, hence:
$$\frac{N([BC]^{\ddagger})}{N([BD]^{\ddagger})} =  \exp\left\{ −\frac{\Delta^\ddagger{}G^\circ([BC]^{\ddagger})-\Delta^\ddagger{}G^\circ([BD]^{\ddagger})}{\mathcal{R} T}\right\}=\frac{N(C)}{N(D)} $$
The preceding equilibrium does not need to be considered to determine the ratio of the products. However, it is important to note, that there are many assumptions necessary for TST to work. For example, at higher temperatures, anharmonic corrections have to be considered. More crucial is the assumption, that the transition can be described as a translatory movement, i.e. it is treated with classical mechanics.
For a system with a preceding equilibrium, it is well possible, that the TST approximation breaks down completely.

Suggested edit:"one of the necessary assumptions is that reactants and products are in equilibrium"   products $\to$ activation complex

The use of the phrase "activation complex" or "activated complex" is not longer recommended. It should be used transition state as it marks the translatory movement which is connected to the transition from one state into another. Saying that the reactants and the TS are in equilibrium is only half the truth, since the reverse connection is as likely in TST. It is another requirement, that all products that arrive the TS from the reactants side will react to the product and vice versa. Hence the reactants are in equilibrium with the products, making the TS a reaction coordinate only.

Too bad then the the said Golden Book by IUPAC (that you linked) using always this term... It is even a separate entry: http://goldbook.iupac.org/A00092.html

If you do this calculation, and telling "everything is in equilibrium" especially the starting material and product then you fail to speak about the timeline when there is a net conversion from A to C and D (i.e. what kinetics is about). In equilibrium $$ \frac {d[A]}{dt} = 0$$ as equilibrium is a static state per definition. Any $$ \frac {d[A]}{dt} \neq 0$$ is nonequilibrium situation.

@Greg 1) You are absolutely correct about the phrasing. I was confusing it with "activated-complex theory", which is deprecated. I still do not like this phrase, as it suggests, that this state can be observed. There is also a much better definition: http://goldbook.iupac.org/T06468.html 2) I understand what you are saying. And it seems, that you have found one of the many problems that there are in TST. And there are plenty, and the one you pointed out is one of the lesser flaws. There are many examples, where TST fails, you can find some at the wikipedia page.

TST doesn't fail here.  TST is primarly a kinetic theory, it doesn't use the static assumption (i.e. the equilibrium between the activation-complex/transition-state to product), it doesn't need it. It assumes only a fast equilibrium between the initial materials to the activation complex.

@Greg Would you please be so kind and provide a reference for your claim, so that I can enlighten myself. There is also the possibility to downvote my answer, if you think my claim is wrong, you should use this unique opportunity.

Look, I really don't want to be confrontative, but your own references, both, contains it. Don't focus on the wording, scientist are hardly poets, but on the equations. Equilibrium is not reached between the product and activation complex, and the backward reaction is not considered at all, else you the derivation itself would be faulty and k would contain a correction dependent on the product concentration. Check the steps of the derivation. Instead of down voting, maybe you should edit your answer.

@Greg I think I made clear, that this is not going to happen. I am rather considering deleting it, because I already regret posting something in the first place. And if you do not want to be confrontative, then you should be wiser with your words.

Then I apologize, I should brush up my English.

On remark: just because activation energy high/low it doesn't necessarily mean that reaction is slow or fast. You cannot tell this without the pre-exponential factor.
The answer: Off course, the original statement should not applied as you do it.
"According to transition state theory, rates are determined by the rate-limiting step." doesn't mean automatically that all the reaction rates are equal.
A to B conversion is slow compared to other processes, is true, but still B to C and B to D are competing processes. The original statement only means that B will not build up in bigger quantities, so during the reaction all the final products (C and D together) are approximately equal with the used A after you consider stoichimetries.
Edited for clarification (sorry, pretty new to LaTex):
You should consider all the transformations as independent equations. Black magic is not involved, "rate determining step" and such only helps making approximations so one can solve the coupled differential equations on napkin.
Example 1.
Let us assume a reaction system with $A \to B \to C$. The kinetics of the system can be describe by two independent equations
$A \to B$ with a corresponding $k1$ and  $B \to C$ with a corresponding $k2$
To describe conversions we can get different equations:
$$ \frac{d[A]}{dt} = -k1 [A] $$
$$ \frac{d[B]}{dt} = k1 [A] - k2 [B]$$
$$ \frac{d[C]}{dt} = k2 [B] $$
We see that it is already pretty messy, so assumptions like $k1 << k2$ can help. The general way is that we say, this difference in the rate constants allows us to asume  $$ \frac{d[B]}{dt} \approx 0 $$. So we have  pipeline, and in first approximation have that much $B \to C$ conversion going as $A \to B$. I.e. the is no "clogging" in the middle. Not much magic here.
Example 2.
Now what if we have e.g. two slow reactions as second step instead of one? Noting special:
$A\to B$ , with $k1$,
$B\to C$ , with $k2$,
$B\to D$ , with $k3$
which translates to
$$ \frac{d[A]}{dt} = -k1 [A] $$
$$ \frac{d[B]}{dt} = k1 [A] -( k2 [B] + k3 [B])$$
$$ \frac{d[C]}{dt} = k2 [B] $$
$$ \frac{d[D]}{dt} = k3 [B] $$
Now we have the same assumption that the first step is much slower than the others than
$$ \frac{d[B]}{dt} = k1 [A] -( k2 [B] + k3 [B]) \approx 0$$
I think you can try it from here and see how the rates comes out.

Consider the following hourglass-analogy. The radius of mouth of each sphere indicates slowness in the sense, the slowest reaction has smallest radius.

Even though there is difference between the mouths at $B-C$ and $B-D$ junctions both are happening at same rate.This is what you thought.
But, the energy from $TS_1$has been* released in form of heat as an exotheromic reaction.Also when there is a difference in rates of r/d step this effect is diminished.
*Thanks to @Martin for pointing out an error
Also, the rate of B-D and B-C are different and the products must be proportional to the rates.Otherways, the product must increase in which reaction is fast. Consider this:

It is wrong to state that the activation energy is used in bonds. It will be released after the transition state. If $E(A)<E(B)$, the reaction would not take place. After the transition state there should always be more energy in the system.

it might be that my answer is wrong or maybe right, but please tell me why you consider that,I say, they need to gain energy again by collision/etc.

On the path towards the transition state kinetic energy is converted to potential energy. After the transition state the potential energy is converted back to kinetic energy, which could be lost to the system or to overcome the next transition state. The energy is not used up in bonds, but released to the system eventually (in one form or another). The reaction would not proceed if this was not the case.

@Martin Aha my bad, on second thought what must be done of a wrong answer(I know I must be knowing this beforehand)..

Your answer is not wrong per se, only this little statement. Just change " has been has been used up in bonds and " to " is released to the system. Additional ..." After that I think it will be a correct answer.


text:
Do certain cumulated dienes present geometric isomerism?
Let's take, for example: 2,3-pentadiene.
Can't this compound present geometrical isomers, like 2-butene?

Properly speaking, these sorts of allenes/cumulated dienes do not exhibit geometric isomerism in the way that alkenes can. Rather, when each of the two respective carbons bear two different substituents, the molecule is chiral and will exhibit enantiomerism. More specifically, these asymmetric allenes exhibit axial chirality. This is consequent to the two consecutive $\pi$-bonds being mutually perpendicular, which results in the molecule lacking any mirror plane of symmetry.
Generally, an even number of successive double bonds in a cumulene produces a potential axis of chirality (again, assuming each carbon at the end of the double-bond system bears two different substituents), while an odd number results in the substituents at the ends of the double-bond system being coplanar, which permits geometric isomerism akin to that seen in alkenes.

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sort: votes


text:
Why are oxides always of the form AyOx with A representing any atom except oxygen and fluorine
I notice that for quite a few elements, oxygen comes before it in alphabetical order. Here is one of the rules in inorganic chemistry for molecular formulas:
Put all the elements in alphabetical order.
So why do oxides always have what is being oxidized followed by oxygen?
That makes some molecular formulas like PO4 for phosphate violate the rule in inorganic chemistry that the elements are arranged in alphabetical order.
I guess all oxides have some element(except for oxygen or fluorine) followed by oxygen because:
1) it clarifies that what you are talking about is an oxide of that element.
and
2) the structure of an oxide is always some element with oxygen atoms around it.

Oxygen difluoride is OF2.  The convention seems to be less electronegative element first which means that O will always be second except with F.

I thought the convention was alphabetical order unless it is a hydrocarbon and if it is a hydrocarbon the C before the H and the H followed by everything else in alphabetical order.

Generally the least electronegative element is written first. And while naming also, +ide is attached as prefix to the more electronegative element.
For Ex- $ICl$ is written as Iodine Chloride not Chlorine Iodide. Similarly $OF_2$ is written as Oxygen Difluoride.
Coming back to your question, Oxygen is the second most electronegative element in the periodic table after Fluorine. Therefore it is generally written after all the other elements (except fluorine).

I have read that the convention is to use the alphabet rule when writing molecular formulas unless it is organic and then you have CxHy first and then the alphabet rule for everything else.

You have read correct, organic compound have a complete different nomenclature according to IUPAC, which has nothing to do with compounds like $OF_2$ AND $ICl$.

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text:
Is an SN1 mechanism feasible with allylic or benzylic halides as substrates?
My teacher stated that allylic carbocations are comparable in stability to tertiary carbocations. $\def\SN#1{\mathrm{S_N}#1}$
With this in mind I am confused why question 13 of this MIT practice test states that these two molecules are only expected to undergo $\SN 2$ reactions:

An $\SN 1$ reaction on the first molecule with the phenyl substituent would yield a primary carbocation, yes, but this primary carbocation is also benzylic, and can be stabilized through resonance.
An $\SN 1$ on the second molecule similarly yields a carbocation, but this is an allylic carbocation, and can easily be stabilized through resonance and delocalization of pi-electrons.
So, would $\SN 1$ be valid pathways of reactions for both of these molecules, or is $\SN 2$ simply the only pathway for these molecules to react?

I planned to add this as a comment but I don't have enough reputation ... A definite answer can only come by referring to the literature and/or from the lab. In the context of an undergraduate exam it is reasonable to accept both possible answers ($\mathrm{S_N1}$ and $\mathrm{S_N2}$ ) for allylic and benzylic halides. $\ce{EtOH}$ solvolysis of benzyl bromide is a known reaction and the hypothesis that it goes via $\mathrm{S_N1}$ is the most reasonable one. Is it possible that the solvolysis acually goes via $\mathrm{S_N2}$ instead? The answer lies on performing the appropriate mechanistic studies.
However, in this case, having checked the actual exam question, it specifies acetone as the solvent. This polar aprotic solvent favours $\mathrm{S_N2}$ conditions, which is aprotic (see this question also). Incidentally, aside from the use of $\ce{KI}$ over $\ce{NaI}$, this halide exchange is basically the Finkelstein reaction, which occurs via $\mathrm{S_N2}$. (Credit to Greg E. who wrote this in a now deleted comment.)


text:
Why does Co2+ have 7 electrons in the 3d orbital, and not 5 like Mn?
I'm taking general chemistry 2 this half of the summer. We are currently going over coordination complexes, ligands and transition metal ions. In particular, this question is in regards to the ligand field stabilization energy.
Why does $\mathrm{Co}^{2+}$ have 7 valence electrons in the $3d$ orbital? Doesn't the superscript $^{2+}$ denote that it is missing 2 electrons, and therefore isoelectric with $\mathrm{Mn}$?

Well, you can find some kind of explanation why the electronic configuration of $\ce{Co^2+}$ is $[\ce{Ar}] \, 3d^7$ here and there, but at the end of the day, you just need to * memorize it*.

When d-block elements form ions, the $4s$ electrons are lost first. Source

The $n+l$ rule tells you the order in which atomic orbitals are filled, and according to the rule the $4s$ orbital is occupied before the $3d$ orbital because it has lower energy.
Thus, the electron configuration of $\ce{Mn}$ is $[\ce{Ar}] \, 3d^5 4s^2$ while that of $\ce{Co}$ is $[\ce{Ar}] \, 3d^7 4s^2$.
But, the $n+l$ rule, as many other rules of old quantum theory, is not 100% working, and thus, sometimes gives wrong electron configuration.
For instance, in general, it is applicable only for neutral atoms in their ground state, and thus, if you apply it to cation $\ce{Co}^{2+}$ you will get the wrong electron configuration [\ce{Ar}] $3d^5 4s^2$.
As it is sometimes explained, the statement that $4s$ orbital is lower in energy than $3d$ orbital is true only when the orbitals are unoccupied. But while you fill $3d$ orbital with electrons it becomes lower and lower in energy and eventually ends up lower in energy than the $4s$ orbital. Thus, when electrons are lost from $\ce{Co}$ atom, they are lost from the 4$s$ orbital first because it is actually higher in energy when both $3d$ and $4s$ are filled with electrons.
P.S. And if I remember correctly this has nothing to do with stabilization in ligand fields.

Electronic configuration of $\ce{Co}$ is as follows: $1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 4s^2\ 3d^7$.
When Cobalt loses 2 electrons to become $\ce{Co^{2+}}$ it loses the electrons which are in $4s^2$, not the ones in $3d^7$ because the electrons in $4s^2$ have a high reactivity.
So $\ce{Co^{2+}}$ still contains 7 electrons in it's valence shell. The new configuration of $\ce{Co^{2+}}$ is
$1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 3d^7$.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$. I do also believe this is not quite correct.

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text:
How to use molality and molarity to determine the molar mass of a compound?

An aqueous solution of a certain organic compound has a
  density of $1.063~\mathrm{g/mL}$, an osmotic pressure of $12.16~\mathrm{atm}$ at
  $25.0~\mathrm{^\circ{}C}$, and a freezing point of $-1.03~\mathrm{^\circ{}C}$. The compound is known not to dissociate in water. What is the molar mass of
  the compound?

The molarity and molality can be determined with the given information.
$$m=\dfrac{\Delta T_{f}}{k_{f}}=0.554m$$ and
$$ M=\dfrac{II}{RT}=0.497M$$
$$\text{mass of }\ce{H2O}=\dfrac{1000~\mathrm{g}~\ce{H2O}}{0.554~\mathrm{mol}~\text{of solute}}\cdot 0.497~\mathrm{mol}$$
I don't understand how the last equation is used to determine the mass of $\ce{H2O}$. How are we justified in using the molarity and molality to determine the mass of water?

Try dividing the unit m by the unit M and see what you get.

It's impossible to find the mass of water. The problem is asking for the molar mass of the compound. Assume 1 mol of compound, use the definitions of molarity and molality, find the weight of the compound and as you assumed 1 mol of it, the number you found is going to be the molecular mass.

You can use 3 different approaches to solve that. One I've given in the comments, another one that I will use here that hopefully sheds some light to the equation which by itself out of context here doesn't make much sense.

In say 1000g of water you have 0.554 mol of compound (definition of molality)
In 1000 ml of solution you have 0.497 mol of compound (definition of molarity)
Thus in 1115 ml of solution you have 0.554 mol (linearity)
So you have 1000g of water and 0.554 mol of compound in 1115 ml of solution or 1115*1.063 = 1185 g of solution (using the density)
Obvioulsy enough you have 1185-1000 = 185g of compound. But we know already that there are 0.554 mol of the compound. So its molecular weight is 185/0.554 = 333.8 (so it's not sugar?)

Same result gives my (preferred) method in the comments

As:
$$m=molality\;=\;\frac{moles\; of\; solute}{mass\;of\;solvent}=\frac n{M_B}$$
$$M_B=\frac nm$$
And as :
$$M=molarity\;=\;\frac{moles\; of\; solute}{volume\;of\;solution}=\frac n{V}$$
$$n=MV$$
Now:
$$M_B=\frac {MV}m$$


text:
Byproducts of the neutralisation of hydrochloric acid with sodium hydroxide
Given a 1 M concentration of $\ce{HCl}$ and $\ce{NaOH}$ here are the pHs:
pH of $\ce{HCl}$: 1
pH of $\ce{NaOH}$: 13–14
and of course water has a pH of 7.
Here is a reaction that is most likely to occur:
$\ce{HCl + NaOH + H2O <=> H2O + NaCl(aq)}$
However I think that other reactions are theoretically possible.
Like for example theoretically sodium oxides could form from the $\ce{OH-}$ taking a proton from another hydroxide anion thus giving you $\ce{O^2-}$ which could react with the sodium to give you $\ce{Na2O}$ which is a sodium oxide.
I also think that theoretically $\ce{Cl-}$ could steal electrons from the $\ce{OH-}$ since $\ce{Cl}$ can exceed an octet of electrons. This would give you neutral $\ce{OH}$ which could then react with another chloride anion to give you $\ce{OHCl-}$(not sure if formula is written right or if it would be negatively charged) or in other words chlorine hydroxide. The proton could then be taken by another hydroxide anion giving you $\ce{OCl}$ also known as chlorine monoxide which could react with another chlorine to give you $\ce{OCl2}$ which is dichlorine monoxide.
Do these theoretical reactions that lead to oxyanions($\ce{O^2-}$ to be specific), chlorine oxides, chlorine hydroxides, and sodium oxides actually happen to some extent when you have $\ce{HCl}$ and $\ce{NaOH}$ in aqueous solution

Given a 1M concentration of $\ce{HCL}$ and $\ce{NaOH}$ here are the pHs:
pH of $\ce{HCL}$: 1
pH of $\ce{NaOH}$: 13-14

First, a 1 molar concentration of hydrochloric acid solution, $\ce{HCl}$, has a pH of 0.0 if we take ${-log[H_3O^+]=pH}$ to be true. (And it is mostly true for dilute solutions).

and of course water has a pH of 7.

Pure water at 25 degrees Celsius has a pH of 7.0. Boil the water and the water will no longer have a pH of 7.0. Cool the water and pH will similarly change. The only condition for neutrality is that the concentration of hydronium ion equals the concentration of hydroxide ion. It just so happens that water at normal lab conditions has a hydronium ion and hydroxide ion concentration of ${1.0*10^{-7}}$ M, which corresponds to a pH of 7.0. Your statement only holds at a specific temperature. At other temperatures, this reaction - the auto-ionization of water - which controls the pH of pure water - will go to varying extents:
$\ce{2H_2O <=> H_3O^+ + HO^-}$

Here is a reaction that is most likely to occur:
$\ce{HCl + NaOH + H2O <=> H2O + NaCl(aq)}$

Since we are talking about solutions of HCl and NaOH, your equation isn't reflective of what's actually going in on the system. HCl and NaOH both dissolve in solution (unless the solution is saturated with either). So you don't have HCl or NaOH in solution, but rather hydronium ion and chloride anions in the case of HCl and sodium ions and hydroxide ions in the case of NaOH solution. A more accurate equation would be:
$\ce{H_3O^+ + HO^- ->2H_2O}$
Regarding:

$\ce{NaCl(aq)}$

Sodium chloride has a very high solubility in water and dissociates unless the solution is saturated. So it would be more accurate to write simply sodium ions and chloride ions, instead of "aqueous" sodium chloride (unless you are referring to the ion-pair effect). Ion pairs often occur in solution; the name describes exactly what is implied - two oppositely charged ions are attracted to each other. For sodium chloride, ion pairing is likely minimal, but in other cases, such as between a carbocation and a halide anion, the ion pair effect has noticeable effects on chemical reactions (i.e. partial racemization rather than full racemization).

Like for example theoretically sodium oxides could form from the
  $\ce{OH-}$ taking a proton from another hydroxide anion thus giving
  you $\ce{O^2-}$ which could react with the sodium to give you
  $\ce{Na2O}$ which is a sodium oxide.

So hydroxide ion reacts as an acid and abstracts a proton from another hydroxide ion.
This is at least theoretically possible, but thermodynamically unfavorable. The oxide anion is very unstable in water solution to the point of being able to react with the solvent (water) itself to a large extent.

but the oxide anion could also react with the positive sodium ions and form sodium oxides and the chlorine could react with neutral OH to form chlorine hydroxide which can react with another hydroxide anion and chlorine oxides can form.

@caters, the reaction you suggest of $\ce{2NaOH -> Na2O + H2O}$ will not occur spontaneously under any reasonable conditions. Using Hess's law and standard thermodynamic data on enthalpies of formation, the reverse reaction is highly exothermic: $\Delta H^{\circ}_r \approx -151.9 \mathrm{\frac{kJ}{mol}}$. This is not even factoring in enthalpies of solvation, while the entropy change will be small either way. Intuitively, this should make sense. It should not be possible to spontaneously react two highly basic molecules to make an even stronger conjugate base and a stronger conjugate acid.

I didn't say it was major because that obviously is not true. However I think that $\ce{2 NaOH -> 2 Na^+ + 2 OH- -> Na2O + H2O}$ and $\ce{2 HCl -> OHCl-}$(chlorine hydroxide from chloride + neutral OH) $\ce{+ OH- -> ClO}$.(represents an unpaired electron) $\ce{+ Cl- -> Cl2O}$ happen to some extent when NaOH and HCl are in aqueous solution. And how are H2O, ClO, Cl2O,and Na2O stronger conjugate bases and acids than the HCl, H2O, and NaOH we started with?

Chlorine hydroxide is likely not a product. Especially not through the mechanism you propose (through a "neutral OH" or OH radical). The hydroxyl radical does not form spontaneously except under extremely high temperatures. Regarding the acid/base stuff - I would consult an introductory chemistry text.

Also, what is the point of all this speculation? Is this a homework problem? A thought problem that you came up with?

well chlorine hydroxide is an intermediate in forming chlorine oxides. It is a thought problem that I came up with as to "What other products can form from this besides NaCl?" and I came up with 2 theoretically possible products with some intermediates in it(Like the oxide anion for example being an intermediate in the formation of disodium monoxide.(At least that is what I think the name for Na2O is besides Na(x) oxide where x represents oxidation state.)

@caters, I'm referring to the fact that in your proposed reaction $\ce{2NaOH -> Na2O + H2O}$, you have hydroxide acting as an amphoteric molecule, one of the hydroxides a base, the second one an acid. The resulting oxide is a stronger base than hydroxide, and water is a stronger acid than hydroxide. I wasn't even referring to your hypothetical chlorine equilibria.

@caters, you can actually get an approximation of the (idealized) thermodynamic favorability of these reactions by calculating $\Delta G^{\circ}r$ from standard molar entropies and enthalpies of formation with Hess's law. NIST is a good source for the thermodynamic data. $K{eq} = e^{-\frac{\Delta G^{\circ}r}{RT}}$. A quick calculation on your proposed reaction gives $K{eq} \approx 1.28 \cdot 10^{-26}$, and that would be a generous estimate as it fails to account for solvation and aqueous equilibria.

but it isn't the best estimate since it doesn't account for those. Exothermic means that the reaction releases energy and thus the products are more thermodynamically stable. Thus the formation of sodium oxides based on this should be spontaneous since stable is favored over unstable(whether it is thermodynamics, kinetics, or charge).

@caters - you are clearly not understanding the fact that endothermic reactions can also be spontaneous.

but endothermic means the product is more thermodynamically unstable and so it is less likely to be spontaneous than an exothermic reaction is because again, stability is favored no matter which type.

@caters try looking into exergonic and endergonic reactions. Enthalpy change isn't the only measure of stability.

I know but if there is more stability in the products than the reactants than products will be favored. I think that this is the case because: 1) the chlorine can exceed an octet so it can have the equivalent of 7 bonds as it does in chlorine heptoxide. 2) O can ionically bond with several elements including sodium and magnesium. This is charge stability. There is then the fact that the products are more thermodynamically stable. That is 2 types of stability in the products vs 1 type in the reactants meaning NaCl,Na2O,ClO-,Cl2O, and H2O that forms should be favored over HCl and NaOH

If there is a high activation energy barrier, then the products won't be favored even if the products are thermodynamically more stable. Consider gasoline. Carbon dioxide and water are more thermodynamically stable than the hydrocarbons in gasoline. Yet gasoline doesn't spontaneously combust in your vehicle.

sure. and OH- is the conjugate acid of NaOH so it can get rid of it's 1 proton and form an oxide anion and water.

Let us continue this discussion in chat.

@caters Just to add as an example , thermodynamically graphite is more stable than diamond but it doesnt change into diamond just on hearing !! ( Though that would be quite helpful ) This is due to the kinetic restrictions .


text:
Mercury and 502 glue
I do some experiments with liquid mercury and I use a special bottle that was 3d printed (ABS plastic). To prevent leaking of mercury I use 502 glue layered around the bottle.
I want to know, if it is safe to store mercury in this bottle?
Does mercury interact with something in 502 glue?
502 glue: Methylene Chloride , Ethyl acetate, Toluene, Cyclohexane
I am not a professional at chemistry.

Mercury does not react with this compounds. However, I strongly encourage you to
1) use monolithic vial for mercury. 3d-printed pieces are very fragile in my experience. Glass vial should be fine, however I recommend double casing (keep sealed glass vial with mercury in a bigger plastic vial with sulfur dust.)
2) perform all operations with mercury over steel of plastic pan.
3) keep a piece of copper foil/plate and nitric acid at hand. Mercury drops stick to copper, freshly washed in nitric acid, which is used to collect mercury.
4) in case mercury is somewhere you don't have access to, drown it in concentrated iron (III) chloride solution. It will transform metallic mercury into mercury chloride, that does not give vapors.
5) Note, that mercury salts may react with some organic compounds, including aromatics (toluene). While risks are low, organomercury compounds accumulates in living tissues, so caution is required.

Mercury does not react with these products.
Consider investing in a mercury spill kit if you can afford it, otherwise find some powdered sulfur and keep it in a safe place.  See http://www.epa.gov/hg/spills/


text:
Is there a reason for the mathematical form of the equilibrium constant?
Why are the two molarities multiplied and not added, and why is each raised to the power of the coefficient rather than multiplied by it? What is the reasoning behind this form? Was it simply determined that the probability of reaction was proportionate to both and thus they are multiplied together? I'd like to know whether its form is purely empirical or has a theoretical backing.

After much research and work, I wrote a little explanation (not that the other answers weren't good, they just weren't well written for someone who didn't know about thermodynamics and other concepts …):
The below results were determined experimentally but this explanation gives some insight into why they are the way they are. Whilst many factors affect the probability of a reaction occurring, and thus the rate, the affect of concentration can be quite easily determined. Consider the following reaction:
$$\ce{a X + b Y -> c Z}$$
The probability of it occurring can be broken down into the probability of the particles reacting in a space, and the probability of them being in a space, thus:
$$P(reaction)=P(\ce{a X} \in \Delta V)\times P(\ce{b Y} \in \Delta V) \times P(\ce{X\bond{->}Y} \in \Delta V)$$
Wherein the probability $P$ of an atom being in certain volume $\Delta V$ is multiplied with the final probability of a collision (here displayed as $\ce{X\bond{->}Y}$) happening in said volume. The probability of a molecule being in a set space can be determined from its concentration:
$$C=\frac nV$$
$$P(\ce{1X} \in \Delta V)= C \times 6.02 \times 10^{23} \times 10^3 \frac{L}{m^3} \times \Delta V$$
This equation assumes that $\Delta V$ is in cubic meters and concentration is in $\mathrm{mol/L}$, however this may not be the case. The main concept is that the probability of one particle being present in some area is proportional to the concentration and some scaling factor:
$$P(\ce{X} \in \Delta V)=K_S×C(\ce{X})$$
Based on basic probability, it is known that the probability of a particles being in the volume is equal to the probability of one particle being in there to the power of $a$. Thus from this, assuming the scaling constant is $K_S$, and that the probability of a reaction occurring at the given temperature in the given space if particles are present is equal to $K_R$, the probability of a reaction occurring based on concentration is as shown below:
$$P(reaction)=K_R K_S^{a+b} [\ce{X}]^a [\ce{Y}]^b$$
If a reversible reaction is considered, the overall direction can be determined by finding the ratio between the rates of the reactions in each direction. Thus the ratio of backward reaction to forward reaction for the below reaction can be identified. (Note the constants on the top and bottoms are different)
$$\ce{a W + b X <=> c Y + d Z}$$
$$K_\mathrm{tot}= \frac{K_{R\ce{A}} K_S^{a+b} [\ce{W}]^a [\ce{X}]^b}{K_{R\ce{B}} K_S^{c+d} [\ce{Y}]^c [\ce{Z}]^d }$$
Dynamic equilibrium is the state for a reversible reaction, in which the rate of both the forward and backward direction is equal and thus the overall change is zero. In this case, the ratio would equal one.
$$1 = \frac{K_{R\ce{A}} K_S^{a+b} [\ce{W}]^a [\ce{X}]^b}{K_{R\ce{B}} K_S^{c+d} [\ce{Y}]^c [\ce{Z}]^d }$$
It can be easier however, to simplify repetitive calculations to just the concentrations, and create a new constant from all the constants shown:
$$\frac{K_{R\ce{B}} K_S^{c+d}}{K_{R\ce{A}} K_S^{a+b} }=\frac{[\ce{W}]^a [\ce{X}]^b}{[\ce{Y}]^c [\ce{Z}]^d }$$
This is the equilibrium constant, although it does change with temperature as the average kinetic energy of particles changes. The calculations made on the concentrations come up with a value called the concentration quotient ($Q$) which is equal to the equilibrium constant when a dynamic equilibrium is reached.

"Based on basic probability, it is known that the probability of a particles being in the volume is equal to the probability of one particle being in there to the power of a." - does it mean "to the power ot ifs molar quantity?" O_o

Here’s the link promised in the edit summary: http://meta.chemistry.stackexchange.com/a/3054

Equilibrium is also the state in which the forward and reverse reactions are proceeding at equal rate. Consider a hypothetical reaction
$$\ce{aA + bB} \overset{k_1}{\underset{k_{-1}}{\ce{<=>}}}\ce{ cC + dD}$$
If we assume that the forward and reverse processes are elementary reactions (a big assumption), then:
$$rate_1=k_1 [\ce{A}]^a [\ce{B}]^b$$
$$rate_{-1}=k_{-1} [\ce{C}]^c [\ce{D}]^d$$
At equilibrium:
$$rate_1=rate_{-1}$$
$$k_1 [\ce{A}]^a [\ce{B}]^b=k_{-1} [\ce{C}]^c [\ce{D}]^d$$
$$\dfrac{k_1}{k_{-1}}=\dfrac{[\ce{C}]^c [\ce{D}]^d}{[\ce{A}]^a [\ce{B}]^b}$$
The ratio of the rate constants is the equilibrium constant:
$$K=\dfrac{k_1}{k_{-1}}=\dfrac{[\ce{C}]^c [\ce{D}]^d}{[\ce{A}]^a [\ce{B}]^b}$$

The answer is both.  The formula for the equilibrium constant was first observed experimentally (that is, it's empirical) for many many reactions.  Then it was justified theoretically in terms of what is now called the Gibbs energy. The reason the formula is multiplicative is that (very roughly) the probability of finding a system in a certain state is proportional to the inverse exponential of that state's energy; and as you know when you take the exponential of a sum you get a product.
If you're curious to know more you should get a hold of Peter Rock's
Chemical Thermodynamics.  It rocks! :-)
If you like math and know calculus well you can read the derivation of the famous formula $K_{\text{equilibrium}}= \exp(-\Delta G^\circ/RT)$, where $\Delta G^\circ$ is the change in the Gibbs energy between reagents and products in their standard state, $R$ is the gas constant and $T$ is the temperature.

I will make "the proof" mentioned by Silvio Levy. Let's consider a chemical reaction, written as:
$$\sum_{i} \nu_i \ce{A}_i \longrightarrow 0$$
where $\nu_i$ are the stoichiometric coefficients of the chemical species $\ce{A}_i$. At equilibrium we have
$$\Delta_\mathrm{r} G = \sum_i \nu_i \mu_i = 0$$
and we also know that the chemical potential under non-standard conditions is given by
$$\mu_i = \mu_i^\circ + RT\ln\left(\frac{f_i}{f_i^\circ}\right)$$
where $f_i$ is the fugacity of the species $\ce{A}_i$. Then:
$$\begin{align}
\sum_i \nu_i \mu_i &= \sum_ i \nu_i \left[\mu_i^\circ + RT\ln \left(\frac{f}{f^\circ}\right)\right] \\
&= \sum_i \nu_i \mu_i^\circ + \sum_i \nu_i RT \ln \left(\frac{f}{f^\circ}\right) \\
&= \sum_i \nu_i \mu_i^\circ + RT \ln \left[ \prod_i \left(\frac{f}{f^\circ}\right)^{\nu_i} \right] = 0
\end{align}$$
Then
$$\frac{\sum_{i} \nu_i\mu_i^\circ}{RT} = -\ln\left[ \prod_i \left(\frac{f}{f^\circ}\right)^{\nu_i} \right]
\Longrightarrow
\frac{\Delta_\mathrm{r}G^\circ}{RT}= -\ln\left[ \prod_i \left(\frac{f}{f^\circ}\right)^{\nu_i} \right]$$
By definition, activity is define as :
$$a_i=\frac{f_i}{f_i^\circ}$$
Taking the exponential we obtain
$$\exp\left(\frac{-\Delta_\mathrm{r}G^\circ}{RT}\right) = \prod_i a_i^{\nu_i}$$
and we can conclude
$$K^\circ = \exp\left(\frac{-\Delta_\mathrm{r}G^{°}}{RT}\right)=\prod_{i}a_i^{\nu_i}$$
Here it's the exact result but we can make some approximations (obviously we do it, it's better for learn) and in the case of low concentrations, the activity of a solute can be approximated to the ratio of its concentration over the standard concentration: $a_i = [\ce{A}_i]/c^\circ$. That's why we write most of the time:
$$K^{°}=\prod_{i}[\ce{A}_i]^{\nu_i}$$
In this reaction, the stoichiometric coefficients of the reactants are negative and that's where the "fraction" comes from. It's exactly the same as you can find in all other answers.

EDIT : This is a misconsception to think that fugacities are only define for gaseous species. This is absolutely false and we can get them for liquid and solid too ! I really insist on that because this something use a lot to determine equilibrum


text:
What is the liquid used in instant shoe shiner?
What is the liquid used in instant shoe shiner products? Ive tried refilling it with WD-40 but I would like to know if there are any cheap alternatives to refill it?

It's silicone oil.
You can use a mineral oil. You can find it as baby oil and if you don't like the odor, also can be found as fleet enema products (eg. http://fleetlabs.com/fleet_enema_products.php).
For larger and cheaper amounts, you can buy it in computer hardware store, farm supply store, or even in vet clinic.


text:
What is the reason for the different solubility of silver halides in ammonia?
According to my knowledge, I know that $\ce{AgCl}$ dissolves in dilute ammonia, $\ce{AgBr}$ dissolves in concentrated ammonia and $\ce{AgI}$ does not dissolve even with concentrated ammonia.
What is the reason for this?

Multiply the Ksp of the salt by the Kf of the complex ion to get K for the reaction between ligand and salt.  The solubility trend of the salts is AgCl>AgBr>AgI so with the same ligand that's the solubility trend of the salts and ligand together.

Silver chloride is very thermodynamically stable in ammonia solution. The $K$ for the dissolution of silver chloride in ammonia solution is only $\pu{2.9E-3}.$ suggesting that the product — silver ammonia complex ion — is much less thermodynamically stable than the reactant, silver chloride.
$$\ce{AgCl + 2 NH3 <=> [Ag(NH3)2]+ + Cl-}$$
$$
\begin{align}
K &= K_\mathrm{sp}(\ce{AgCl})\times K_\mathrm{f}(\ce{[Ag(NH3)2]+})\\
  & = \pu{1.8E-10}\times\pu{1.6E7}\\
  & = \pu{2.9E-3}
\end{align}
$$
However, silver chloride will dissolve in an excess of concentrated ammonia solution due to Le Chatelier's principle. Despite the fact that silver chloride has a minuscule $K_\mathrm{sp}$ value, some free silver ions are present in solution. Addition of concentrated ammonia solution effectively consumes this free silver ion as ammonia, a good σ-electron pair donor ligand, combines with the silver ion.
Ions themselves are not precipitates, as all bulk matter must be approximately electrically neutral, and so the silver chloride dissolves when hit with excess, concentrated ammonia.
Also, note that the cyanide ion is also a good complexing agent with respect to silver. The cyanide ion has a surfeit of electrons on the carbon, and carbon is even less able to stabilize negative charge than nitrogen, making the cyanide ion reactive as a complexing agent. Hitting silver chloride and other silver salts with a source of cyanide ion will similarly cause the silver salt to dissolve into ions suspended in solution.

In fact you can predict exactly how much AgCl you can dissolve: so long as $\ce{[[Ag(NH_3)_2]^+]\cdot[ {Cl^{-}}]}<  0.0029\cdot\ce{ [NH_3]^2 }$ all the silver will be dissolved.  That's why people talk of "excess ammonia"; you need way more than a 1:1 molar ratio.  Because AgBr and AgI have even smaller solubility products than AgCl, the factor on the right is much less and the amount that you can dissolve even in pure ammonia is similarly much less.


text:
Can aromaticity arise from 3c-2e bonds
Electrons are delocalized between the three centers in a  3c-2e bond. If there was a molecule with 3c-2e bond in each direction, e.g. a boron cluster, would it be aromatic?

related: http://pubs.rsc.org/en/Content/ArticleLanding/2014/OB/C4OB01018C#!divAbstract

The cyclopropenyl (1) cation is believed to be aromatic as it demonstrates abnormal stability and NMR-shifts typical for aromatic compounds. Moreover, there are a good number of derived systems, where aromatic conjugation is maintained despite extra methylene insertions into the ring, such as is typical for the nornbornadienyl (2) cation. Some of these cations can be isolated in salts or observed via NMR in solution. Signs of aromaticity are found also for the cyclooctatetraene dianion and the cyclononatetraene dianion, both produced by reduction of the corresponding alkene with metallic sodium. Remember, please, that even though all systems noted above are 'aromatic' in some sense and are abnormally stable for their family of species, they are still very reactive, because they all belong to a very reactive family of species. While the neutral boron analog of the cyclopropenyl cation has not been isolated to my knowledege, a similar structure was found in organometalic compounds. Aromaticity of some ring systems may result in some weird properties of some compounds, like squaric acid (abnormally strong for ketoenol and a stable enol), high polarity and basicity of cyclononatrienol and cyclopropenone and so on.
(1) - cyclopropenyl cation

(2) nornbornadienyl cation

Good to know, but this still doesn't give an answer to my question.

Than please elaborate. You see, the only thing needed for aromaticity is to have 4n+2 electrons in conjugated pi-system in a ring. That's all. How orbitals and electrons are provided is irrelevant from theoretical point of view, though it does affect reactivity of the particle. I... don't understand what is hard to understand here.

That's according to the Huckel rule, but what about Mobius aromaticity? I thought here about some exotic type of aromaticity.

Hm. Well, there are rules for stability of metallic cluster particles and rules for extra stability of boron-based clusters. For example, $\ce{B12H12^{2-}}$ is a fully delocalized structure with all $\ce{B-B}$ distances equal. $\ce{B12}$ unit is extremely common in reduced boron chemistry. However, they are not widely recognized as 'aromatic' as they are not as widespread, and do not have so long history, this is a bit of esoteric knowledge of those involved in their respective field.

Could you list some of the rules?

@Marko http://en.wikipedia.org/wiki/Polyhedral_skeletal_electron_pair_theory

A complex with a borirene functioning as an aromatic ployhapto logand is known.  See here.  Supporting information contained in the link shows the structure of the borirene and its complex.

We may consider the pi electrons of the cyclopropenyl cation as a 3c-2e bond.  An even better example of a 3c-2e bond producing aromaticity is the trihydrogen cation, in which three protons are bound into a triangle with two electrons.  See here.


text:
Ring Formation and Alkenes
I was consulting Wade and I came across this problem and solution:

While I agree with the solution manual, I can't help but wonder if my proposed mechanism is correct.
I had the water attack the hydrogens attached to the alkene. I knew that alkenes are slightly electron-withdrawing and this makes the hydrogens an appealing target for a base.
Once the hydrogens were cleaved, I had the remaining pair of electrons attack the carbocation. Is this a valid pathway? Valid as in plausible? Would you accept this mechanism on a test?

The main problem I see with this mechanism is that yield is exceedingly low as sp2 carbons are not especially acidic, and water is a poor base. On the other hand the original problem does specify that this product is minor. But then, it appears that I chose the more minor of two minor pathways.

Not a definitive answer, but one problem I see with that mechanism is that it entails a concerted, bimolecular process, which should really be much less probable than any given unimolecular process of comparable energy and leading to the same product.

I'm also skeptical that the required configuration of molecular orbitals in the transition state is reasonable. I'm imagining your mechanism would require the electrons of the base flowing into a $\sigma^{*}_{CH}$ antibonding orbital, and I think this would really require the base (and the electrofugal proton) to be roughly coplanar with and on the same side as the newly forming $\ce{C-C}$ bond if it were all to happen concertedly. I could certainly be mistaken, but that doesn't seem at all plausible to me.

No, this is a serious mistake, look at the pKa of alkenes. Don't take it in a bad way, it's part of the learning process. What is the main product of the reaction? A pyran or just dehydration?

Acidity of the alkene was mentioned in my OP

Greg E.'s comments sum up the flaws with your version of the mechanism. While yours contains the same steps, it combines the nucleophilic attach and the proton transfer in a single step. Unless you have experimental evidence for a bimolecular step, the unimolecular step is much more likely. In this case, the unimolecular step is a 5-endo-dig cyclization, which is particularly fast. The alkene nucleophile is already present when the carbocation forms. The rate of the deprotonation is diffusion controlled, as we have to wait for a water molecule to find it.
Also, your logic relating the minority of the product to the unlikeliness of the mechanism is inherently flawed. A minor product does not occur at a lower proportion because the mechanism contains less valid steps. Instead, the distribution of products is related to the relative rates of different mechanisms and the relative energies of the products. That product is a minor product because some other product either forms faster or is more stable or both.
If this was on my exam, and I like the problem so I might use something like it, I would dock you a few points for combining steps that are not likely concerted into a concerted step. For example, if I decided this mechanism is worth 10 points, then you might receive 7 or 8.

Final note: Remember that experimental evidence trumps all else in science.

Regardless of how intuitively and aesthetically unusual your mechanism appears, if experiments support it, then it becomes a more valid mechanism. As an example of some of the non-intuitive things that are in "accepted" mechanisms, check out halonium ions and sigmatropic rearrangements, especially the one in the Baeyer-Villager Oxidation. Evidence in support of your mechanism might include a failure to observe or trap that final carbocation, a demonstration that the rate depends on the amount of water (since final deprotonation steps tend to be very fast), and others.


text:
On Heisenberg's uncertainty principle
Why can't we find the exact position and velocity of a particle? And why is it that if the uncertainty in position is very, very large then the velocity can be determined, and vice versa? Please explain to me.

Hi! Welcome to Chemistry.SE. The question you are asking is an extremely deep question that can be answered on a variety of difficulty levels. Could you please indicate your background and prior knowledge on the subject so the answers can be tuned to be understandable for you

There are probably already a lot of questions on [physics.SE] that address this, if you want another source of information.

If you imagine particles to be like billiard balls (the way you're taught in school) then this of course makes no sense. The only way you can understand it is to accept the quantum mechanical description of particles: that they are probability distributions that exhibit both wave-like behaviour and particle-like behaviour depending on how you measure them.
Once you accept that a particle is best imagined as a probability wave smeared out through space then the Heisenberg uncertainty principle is a rather simple mathematical result relating to Fourier transforms. The more concentrated some function $f(x)$ is, the more spread out its Fourier transform $f(k)$ is. And since position and momentum space are Fourier transforms of each other (see here) then this gives rise to the uncertainty principle.
In other words, if you measure the position very precisely (highly peaked $\psi(x)$) then the momentum representation $\psi(k)$ will become very broad (imprecise), and vice versa.

It's not that you cant measure it -- it's that particles just don't have well defined positions and velocities at the same time.
Actually, I believe the uncertainty principle actually refers to the position and momentum of a wave. But I don't want to split hairs.
Now, particles, until they interact with something, appear to exist as probabilities, (or I prefer to think them as "possibilities", it's more "magical" to me, but anyway...), and their probability functions, the areas where the particles might pop up.. for some reason or another they act like waves.
The think is, with a wave, how do you get the exact position? A wave can take on many forms. It can have just one peak, or it can have multiple peaks and troughs. Now, it it has just one peak, it's position is pretty well defined, at least, if you ask me. But if it has a bunch of peaks, it's position is more "spread out".
Now, if all you have is a glimpse in one instant, the more "spread out" the wave is, the easier to be to get an idea of where it's going. If the wave is all clumped up into one wavelet, it will be really hard to tell where it wants to go. But if it's spread out, like ripples in a pond, and you happened to get a glimpse, you can kind of tell where the wave is going, you will be able to tell that it's spreading in a certain direction, from the studying the ripples you will be able to guess it's speed, etc. But what is the exact position of the ripples? How do you define the position of .. "ripples"? Yeah, you can make up a suitable definition for the macroscopic world, but in the quantum case, the waves aren't actual, physical things. They are waves of "probability", and the actual particle, the actual physical thing we are interested in, well it's busy chilling out somewhere in these probability waves, at least until it decides to manifest itself.
So basically, the way it works out, the more spread out these probability waves are, the easier it is to guess where the particle wants to go. The more you know about where the particle wants to go, then the better you can define it's velocity.
However, if the probability wave is bottled up into a small spot, you have a good idea of where the particle is likely to be in that instant, but you will have no idea of where it wants to go. To know where it wants to go, the probability wave has to be smeared out. But then you lose the information of where it is, presently.
And that's just the way it is.

I like this, but "it's busy chilling out somewhere in these probability waves" makes the particle sound distinct from its probability wave which it's not.

Good point. I didn't quite mean to make it sound like that.

I believe that your best course is to search the web for sites aimed at explaining the principle. Hyperphysics does a good job of explanation, but might be a little technical. I haven't read this article on How Stuff Works, but it is aimed for the non-technical person.

The problem with that How Stuff Works article (and many other sources) is that it makes the uncertainty principle sound like a mere experimental limitation which is incorrect; a particle literally doesn't have a position, it's not just that we don't know where it is.

You ask "Why can't we find the exact position and velocity of a particle" - this can not be directly answered because there is no particle in this sense you use it.
The thing does not have an exact position and velocity - it just is not a particle in the "normal" sense.
The fact that things on this very small scale do not have a well defined location and velocity at the same time is just what we found out how the universe is like.
With this, the question becomes similar to "Why is there gravity?".
That's a very valid question too -  but a different kind of question.


text:
Hydrogen as a fuel really worth it?
Is hydrogen good to be used in cars? How much of traveling can a car do with medium sized balloon filled with hydrogen? Is it worth it?

Hydrogen escapes from vessels more readily than just about any other gas.   All of the schemes to store hydrogen seem unwieldy.  High pressure, low temperature, or hydrides.  I've seen NH3-BH3 proposed as a way to store hydrogen.  A good way to store hydrogen might be methanol.
Generating the hydrogen takes energy in the first place which factors in to the question of whether or not it is worth it.

TBH one of the best ways to store hydrogen is just CH4. That's 4 hydrogen atoms per molecule with just one extra non-toxic atom.

But methane is pretty powerful greenhouse gas, and you can't pretend it would never escape. Methanol is at least liquid and liquids are generally easier to handle and contain than gases. Not sure about Methanol's greenhouse gas properties. It is toxic, but we've put worse into the environment and it probably has a short half-life in the wild. If there were some way to electrolytically combine CO2 and H2 to make methanol, that would be cool. Could make fuel in a central location with clean energy and then ship it for use as fuel while stying carbon neutral.

The energy density of hydrogen at atmospheric pressure (which it roughly is inside a balloon) per unit volume is a mere 0.01005 MJ/L, which is roughly a factor 3600 less then that of gasoline (at 36 MJ/L).
A typical car does 15 km/L so where a gasoline filled balloon (say 3 L) does 45 km for you, a hydrogen filled balloon will get you only 12.5 meters, i.e. just out of the parking lot.
Therefore, any realistic application of hydrogen is to store it either at high pressure (e.g. 70 bar, where the energy density is about 6 MJ/L) or in so called metal hydrides which allow much higher energy densities. Another option which to me seems to most viable, is to use the Fischer-Tropsch process to make methane or even higher alkanes out of hydrogen and $\ce{CO}$. The big advantage of this route is that our current infrastructure of gas stations can still be used

Storing it under higher pressure would mean more energy and I guess that does not really make it a good alternative or could it?

@user7063: That becomes more of a physics question, but in short hydrogen is a small molecule (second only to Helium, and the smallest diatomic molecule). That unfortunately means it leaks, badly. At 70 bar the problem is much worse. And because it leaks so easily through many materials, it also has the annoying property of embrittling them.

@MSalters Do you have any reference about the leaking? I know that hydrogen molecules are tiny and will leak through many materials that other gas will not penetrate, but I also know that many chemical labs have steel containers with hydrogen at 200 bar and I have never heard of any leakage issues of those. Curious to know how safe those 200 bar steel containers are!

@Michiel: For lab use, the weight of the container is not a major concern, but for cars it is. Even so, leakage is still taken seriously in stationary situations. I know the physics building at my alma mater had a "shed" for the hydrogen storage, which was designed to direct any explosions away from the main building.

Hydrogen is very easy to leak. Any amount of hydrogen leak is dangerous. It's highly flammable, so if there's fire, naked light or electric spark nearby, it'll cause explosion and of course be dangerous for traffic.
By the way, hydrogen leak will eventually float up and deplete ozone layer.
Based on these facts, hydrogen isn't a promising fuel to be used in daily life.

This is assuming that you are going to use hydrogen in the gaseous form. A lot of research is focused on metal hydrides which don't have the leakage problem. Granted, they have other issues, but using leakage as the sole reason to say hydrogen is not a promising fuel is a little bit jumping to conclusions

Hydrogen destroys the ozone layer? And lasts long enough in the atmosphere to do so?


text:
Acid-base equilibrium of NH4CN
I'm having difficulties understanding this problem.
We have in the problem that $K_\mathrm a$ for $\ce{HCN} = 6.2\cdot10^{-10}$ and $K_\mathrm b$ for $\ce{NH3} = 1.8\cdot10^{-5}$.

Write chemical equations to represent the dissociation of $\ce{NH4CN}$ in water, and the acid-base equilibrium equation.
Will the salt react as an acid or a base?

I know this is part of the solution:
$$\ce{NH4+ + H2O <=> NH3 + H3O+}$$
$$\ce{CN- + H2O <=> HCN + OH-}$$
I understand Le Chatelier's principle and I know that the reaction will shift to the side of less energy.

Fun!  weak acid versus weak base!
Ammonium cyanide, $\ce{NH4CN}$, is a solid where the atoms are grouped into the same ions that are generated in solution: $\ce{NH4+}$ and $\ce{CN-}$.  Although it can be sublimed with very mild warming, it is fairly unstable. In particular it reacts with atmospheric water vapor, evolving ammonia and HCN.  Dangerous stuff!
The first part of the answer to 1. is the decomposition of $\ce{NH4CN}$ into ions in aqueous solution, which should be obvious; one subtlety is that you may want to add (s) and (aq) next to each species, especially if your professor makes a point of doing so.  So that's the dissociation. The data in the problem don't allow us to compute the equilibrium constant of the dissociation but it's safe to say that it is extremely high, that is, dissociation is complete: there is no $\ce{NH4CN}$ as such in solution.
The acid-base equilibrium equations can be the ones you wrote.  Again, you may want to add more information, like this:
$$\ce{NH4+ (aq) + H2O <=> NH3 (aq) + H3O+ (aq)}\quad K=5.5\times10^{-10}$$
Here the equilibrium constant $K$ is the acid dissociation constant of the acid $\ce{NH4+}$, the protonated form (aka conjugate acid) of ammonia.  I got its value by dividing $K_\mathrm w$, the dissociation constant of water ($1.0\times10^{-14}$), by the $K_\mathrm b$ of ammonia, given in the problem.
I encourage you to add similar information for the  equilibrium involving cyanide ion.  Hint: if you keep the reaction as you wrote it, the appropriate $K$ is not the $K_\mathrm a$ of $\ce{HCN}$ but the $K_\mathrm b$ of the deprotonated form (conjugate base), $\ce{CN-}$.
(But you can instead consider the alternate reaction $\ce{HCN + H2O <=> CN- + H3O+}$, and then the equilibrium constant is the $K_\mathrm a$ of $\ce{HCN}$.)
How about part 2?  The only correct answer to

Will the salt react as an acid or a base?

is "It can do either, depending on what it's reacting with".  But maybe what they mean is whether the solution of the salt is acidic or basic.  If so, one way to find out is to combine the two equilibrium equations and their $K$s.  I don't want to give problem away completely, but think of it this way:

What is the pH of a (say) 1 M solution of $\ce{HCN}$, based on the $K_\mathrm a$ of $\ce{HCN}$?
(I'm assuming the calculation of the pH of a solution of a pure weak acid or a pure weak base has been covered in the course before getting to the question you posted today.)

What is the pH of a 1 M solution of $\ce{NH3}$, based on the $K_\mathrm b$?

Is $\ce{HCN}$ a stronger acid than $\ce{NH3}$ is a base, or is $\ce{NH3}$ a stronger base than $\ce{HCN}$ is an acid?  Based on this, which would you expect to "win out" if you mix the two in equal proportions?

(The point here is that what you get by dissolving $\ce{NH4CN}$ is indistinguishable from what you'd get by mixing equal amounts of equimolar solutions of $\ce{HCN}$ and ammonia.  In other words, if you mix 1 L of a 2 M solution of $\ce{HCN}$ with 1 L of a 2 M solution of ammonia, the result is very nearly  2 L of a 1 M solution of $\ce{NH4CN}$.   Footnote: "very nearly" because volume is not always preserved exactly, unlike mass.)

I think this is a very good answer. The only thing I'd add (which may or not be apparent to the OP) is that combining the two acid-base reactions the OP has already written with the neutralization of hydronium and hydroxide (reversing the water autoionization reaction) gives what I think is probably the dominant equilibrium ($K_{eq} \approx 9.0 \cdot 10^{-1}$).

I'm fairly sure the calculation is right $\frac{K_w}{K_b} \cdot \frac{K_w}{K_a} \cdot \frac{1}{K_w} = \frac{1.0 \cdot 10^{-14}}{1.8\cdot10^{-5} \cdot 6.2\cdot10^{-10}}$. This agrees with the calculation of $10^{\Delta pKa}$ using reference values.

Yes, I realized my mistake in the meantime.  OP will appreciate the calculation though :-)

I think it is important to state, that $K_w=1\cdot10^{-14}$ only holds for $25~^\circ\mathrm{C};~1~\mathrm{atm}$.


text:
Cleavage of Ethers
Very confused by this video I stumbled across describing the cleavage of ethers with strong acids:

From what I can gather, the cleavage of ethers is run in aqueous strong acids such as HI or HBr.
Given that strong acids in water level to hydronium ion, why might HI or HBr be preferable to say HCl (another strong acid) in the clevage of ethers?
The video from which the above picture is taken says that HI and HBr are "stronger acids" and thus are better able to cleave the ether.
This sounds like garbage to me because if we're running this reaction in aqueous solution, there probably isn't a predominance of undissociated HI or HBr (very strong acids) reacting with the ethers. Instead, we'd only have hydronium ion and the conjugate bases of the strong acids.
So would the real reason HCl wouldn't work in cleaving ethers have to do with the nucleophilicity of the chloride anion in water solution (which is lower than that of either iodide or bromide ion) due to chloride anion's greater negative charge density and thus heavier solvation in water?

I agree with you, the higher nucleophilicity of bromide and iodide ions is what makes HBr and HI better reagents than HCl. But I leave the answer for anyone who has some references. A secondary reason might be because commercial HBr and HI come in higher concentrations than HCl (aqueous).

I'd say your on the right track. Protonation of the ether is the first step and shouldn't depend on which acid you use, so long as it is sufficiently strong. The reaction then proceeds by SN1 or SN2, both of which are obviously dependent on a nucleophile.

You are right that $\ce{HBr}$ and $\ce{HI}$ are preferred over $\ce{HCl}$ for this reaction because $\ce{Br-}$ and $\ce{I-}$ are better nucleophiles than $\ce{Cl-}$, for the reason you have provided. Actually, the hydrolysis proceeds significantly slower with hydrochloric acid, and reasonable rates are only achieved with hydrobromic and hydroiodic acid (source). And while all of the acids level to the hydronium ion in water, the $\ce{H3O+}$ concentration will be higher in aqueous solutions of $\ce{HBr}$ and $\ce{HI}$, as both are available in higher concentrations than hydrochloric acid (57 and 48 wt.-% versus 38 wt.-% for $\ce{HCl}$).


text:
Why should epoxides be more reactive at a position where positive charge is stabilised?

In the above reaction, I am told that the nucleophilic chloride anion attacks the trisubstituted carbon because it has the most partial positive character due to its ability to stabilize positive charge.
How does that statement make any sense?
I'm not following the train of "logic," which is as follows:

The nucleophile will want to attack the location with the greatest partial positive charge. Okay, Coulomb's law. I'm good with this statement.
The carbon with the greatest partial positive charge is the one that can best stabilize positive charge. (?)
Alkyl groups are slightly electron-donating; the tertiary carbon will be able to better stabilize partial positive charge.
Thus, the chloride anion attacks the tertiary rather than the secondary carbon. I don't see how this follows from the previous assertions. Doesn't stability correspond negatively with reactivity? If so, what is the real reason the chloride anion attacks the tertiary carbon rather than the secondary carbon?

Essentially the same question: Regioselectivity of acid-catalyzed ring-opening of epoxides

It is important to state, that in asymmetric epoxides, the $\ce{C-O}$ bond lengths are not equal. Upon protonation, this effect is enhanced, giving rise to a weak and a stronger bond that can be broken. This means the carbon that can better stabilise the charge is more weakly bonded to the oxygen. Thus this will be the carbon suffering the nucleophilic attack.

To the question mark in (2): this implication is simple:  if the charge were on a neighboring carbon you'd have a state of higher energy.  "Can better stabilize" is a synonym for "gives lower overall energy".
To the objection in (4): the stability ranking refers to various electronic distributions in the possible transition states. In the transition state of an SN2 reaction, and when there is a positively charged leaving group, the carbon undergoing nucleophilic substitution will bear a partial positive charge in the transition state. So, the degree to which this positive charge is stabilised settles the matter of which carbon will be attacked preferentially.
The reactivity ranking refers to the fitness of each atom in the intermediate as a site of attack by the nucleophile. The most likely target is the one which leads to the lower-energy transition state, i.e. has a lower activation energy barrier.
So there is no contradiction.  The carbon most able to stabilize a positive charge is also the one most likely to be the target of a nucleophilic attack (other things being equal; of course there could be factors, like sterics, changing the ranking).

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
How many different possible structures (isomers) are there for the complex, [Mn(OH2)3(NH3)3] 3+?
How many different possible structures (isomers) are there for the complex,
$\ce{[Mn(OH2)3(NH3)3]^{3+}}$?
I really don't know how to begin with this. I know the answer is 2, but i don't have the faintest idea how this is the case. Since $\ce{Mn}$ has a charge of $3+$, I know that its valence electron configuration is $3d^4$. I also know that $\ce{NH3}$ and $\ce{H2O}$ are monodentate, which means they contribute 1 lone pair.

Try sketching a few possible configurations.

Metal complexes with three each of two types of ligands $(\ce{MX3Y3})$ have two isomers: mer- and fac-isomers. For example, for $\ce{Co(NH3)3Cl3}$ there are two isomers:

Fac for facial, in which all three ligands of each type are clustered on a single face of the octahedron and the three other three ligands are on the opposite face.

Mer for meridional, in which the three ligands of each type are arranged meridianally in a row around the metal and the other three ligands are arranged likewise on the opposite side in a perpendicular meridian.

The compound you talked about is of the type $MA_3B_3$. I will illustrate a similar example here.
There shall be 2 geometrical isomers, 'FACIAL' and 'MERIDONAL'

To identify which is which, remember that in MERIDONAL maximum bond
  angles between the groups is 180 whereas in facial it is lesser
  than that. Also none of them is optically active as both have a
  Plane of symmetry (merdional has diagonal POS whereas facial has
  horizontal).

Therefore the compound has total 2 STRUCTURAL ISOMERS (2 GEOMETRICAL + 0 OPTICAL).


text:
Acid "Catalysis"

I don't follow this reaction. This is after all an acid catalyzed dehydration reaction; why then, is the final product the alkene and water? Water wasn't the original acid. Hydronium ion is the original acid. Shouldn't the final product be not water but hydronium ion (produced through deprotonation which in turn forms the alkene)?
In addition, this is the book's reaction mechanism, which clearly shows the formation of hydronium ion:

But this isn't a one-off sort of thing with the water as a "product" of acid-catalyzed dehydration.

So, is there a reason water is listed as the product in acid-catalyzed dehydration? Am I missing something?

Hi Dissenter: acid-catalyzed means hydronium is the catalyst, not a product.  Dehydration means water is the product.  Not clear what's tripping you.

Okay, I think I see it now. I was expecting the hydronium ion to be regenerated as it is a catalyst, but on the other hand, the problems don't show the hydronium ion as being consumed.

it is regenerated on the last step when $\ce{H2O}$ pulls a proton off the carbon.

I know, that's why I was expecting hydronium ion as a product, not water, but I see where the water is coming from now and since hydronium ion isn't being explicitly consumed, it isn't shown either as a product (explicitly).

Yes, I see the difficulty now.  The product water leaves at step 2.

Consider systematically:
The mechanism is an $\bf E_1$ reaction in which the substrate is a protonated alcohol.
Step I
In this step, an acid–base reaction, a proton is rapidly transferred from the acid to one of the unshared electron pairs of the alcohol. In dilute sulfuric acid the acid is a hydronium ion; in concentrated sulfuric acid the initial proton donor is sulfuric acid itself. The leaving group is a molecule of water.
Step II
The carbon–oxygen bond breaks heterolytically. The bonding electrons depart with the
water molecule and leave behind a carbocation. The carbocation is, of course, highly reactive because the central carbon atom has only six electrons in its valence level, not eight.
Step III
Finally, in step 3, a water molecule removes a proton from the $\beta$ carbon of the carbocation. The result is the formation of a hydronium ion and an
alkene.
In step 3, also an acid–base reaction, any one of the nine protons available at the three
methyl groups can be transferred to a molecule of water. The electron pair left behind when
a proton is removed becomes the second bond of the double bond of the alkene. Notice
that this step restores an octet of electrons to the central carbon atom.

Though water gives proton in step one and other molecule of water takes proton in last step, it is usually written as a produce of alkene and water until a mechanism is specifically mentioned.This is also reminding of the catalytic nature of water in this reaction.


text:
How can I reverse engineer a commercial cleaning solution's "other ingredients"?
I have a commercial lens cleaning solution. The manufacturer of the solution claims it is not only distilled water and that the formula is confidential. I suspected it was only or mostly distilled water after testing distilled water's cleaning abilities on different types of lenses with different coatings.
I have heard that you can ship a sample to a lab and they will conduct a quantitative analysis of the solution using a mass spectrometer, but there are so many different kinds of mass spectrometry that I am not sure what I would want to order.
What kind of mass spectrometry (or other lab analysis service) would be best suited to discover the formula for a non-volatile, non-corrosive, liquid sample of lens cleaner?
If that's not possible, I just want to know how much of the solution is essentially water, and would like to know what kind of service would be best to determine that.

This is not really a full answer, but if you'd be satisfied knowing that the mixture is principally water (to within some tolerable but not insignificant degree of error), you can determine its boiling point and density at home. You could also try freezing the mixture; if there's an organic solvent present in substantial amounts, it will typically remain liquid well below water's freezing point (at least for the solvents I could plausibly imagine being in a cleaning solution). Refractive index is also very simple and cheap to measure.

Having said all of that, most organic cleaning solvents are fairly volatile; they evaporate easily and usually have distinctive smells. I'd be fairly surprised if any combination of conventional solvents in non-negligible concentration(s) could be so inconspicuous.

@GregE To see if I could identify it as distilled water and isopropanol or something common, I compared evaporation to distilled water and it was roughly equivalent (very slight variation, hard to tell), there's no odor from it, and an online report from some unknown person said that they had tasted it and it had no distinctive taste. I have not tried determining the freezing or boiling point or density yet but those are my next steps. I didn't think about refractive index, so thanks for mentioning that.

Unlikely the the lab would just do mass spectrometry. More likely GC/MS. MS on an unknown mixture is fairly useless.

Reverse engineering, or deformulation, is a common industry practice. This is a very common type of request for commercial analytical laboratories, and is generally not cheap. You can discover just how widespread it is, and how many commercial providers there are with a quick Google search.
To get yourself a reasonable starting point, you should request a product MSDS from your suppplier. All suppliers should be able to provide a full MSDS, which will give an approximate ingredient listing and component concentration. Sometimes, you will find ingredients listed as "Trade secret" or "Proprietary", but many times you will get more information. It is very likley that your lens cleaner solution is, in fact, predominantly distilled water; perhaps up to 90-95%. However it is the other minor ingredients that are the active ingredients, and may include alcohols or glycols, anti-fog and anti-static agents, anti-baceterials and preservatives. Examples of lens cleaner MSDS can be found via a google search, and this may give you a useful bit of background information.
Note:
1. all suppliers are legally bound to providing MSDSs upon request in the local idiom (official language).
2. this document is safety-based which means it will (a) provide (useful) information on possibly hazardous materials, (b) assess the hazardous component(s), their type, level and correlate to concentration.
3. websites such as state Chemical authorities (ECHA, FDA, ...) publish ratings, studies and lists (of registered/authorized/forbidden) of chemicals for use in particular applications such as food contact, medical, pharmaceutical devices/products.
The next step is to determine what information you really are after, as this will determine the budget you will need to outlay. If you want to just determine whether the solution is predominantly water, you could try a few things yourself (if you are experiment-savvy). Try smelling it, as a first step; check for evaporation if you run a thin film on a glass slide, test density or melting point. These will all tell you much the same thing, which is, yes, the solution is predominantly water. A simple commercial 1H NMR analysis will show you this also, and confirm water concentration up to about 99.9%. This means minor ingredients will show up, but as long as they make up about 0.1% of the ingredient composition, and they contain hydrogen (sodium bicarbonate will not show up for instance). This method is commonly used when you have a reasonable idea about the identity of the minor ingredients, but can involve extensive analysis if you really are shooting in the dark.
Beyond this, if you actually want to determine the ingredients, and have no idea what they might be, you will need to head down the avenue of GC-MS and/or LC-MS; a combination of chromatography to separate all of the ingredients, and then Mass Spectrometery to try to determine the mass of each component. This type of analysis is generally charged by the hour, and the more variables you are trying to evaluate, the more expensive the procedure. Generally, when analysing a competitor's products, a reasonable knowledge of the types of raw materials exists, and fingerprinting chromatography and NMR traces against raw material libraries saves a lot of time (and money). Sometimes, other techniques such as FTIR, AAS, UV-Vis or ICP may also be used, but these would really only be if the MS methods failed to give you suitable leads.
Commercial analytical labs will be able to give you a very good summary of the best techniques for your sample, but be prepared; the very first question you are likely to be asked is "what is your budget?". Good luck.

This is a really informative answer and pointed out a lot of nice research points. I wasn't too far off in what I was going to ask of a lab, which is good to know. But at least now I know what kind of labs to look for and the types of services I can expect to see recommended. Thanks a lot!


text:
Inversion of Configuration - Opening of Epoxide
This doesn't look like a traditional inversion of configuration:

At first I dismissed the textbook author as making a typo. But then I realized that this "typo" was repeated throughout the solutions manual.
So I tried drawing a Sawhorse projection of the textbook's answer and my answer, and I found that both are equivalent.
So would my answer also be acceptable? In my answer I change the methyl from a wedge to dashes and the hydrogen from dashes to wedge and I place the SH group appropriately.

Do you mean adjacent to the OH? or opposite (like the drawing above)? It would really help to show what you have.

I have found that it can be difficult to recognize inversion when the leaving group and incoming nucleophile are both in the plane of the paper. The inversion in this case is really that the hydrogen and methyl are initially oriented "down" and are oriented "up" in the product.

Also, in your answer, without wedges and dashes, it's impossible to say if you are correct.

Well, the wedges and dashes aren't needed in a Sawhorse projection, because it's a pseudo-3D view of the molecule. Any substituent that looks like it's coming toward you is on a wedge in a flat, 2D representation.

Sorry, I see what you're doing there. I was looking at it as a "line-wedge" drawing.

It's slightly difficult for me to see clearly, but your drawing appears to be correct, and it corresponds to a $180^{\circ}$ rotation about the $\ce{C_1-C_2}$ bond axis relative to the textbook rendering ($\ce{C_1}$ being the carbon bearing the oxygen atom and $\ce{C_2}$ the sulfur atom, respectively). An unambiguous method of determining whether the two representations are equivalent is to assign the absolute configuration for the chiral carbon(s) in question. It may also help to use other projections (Fischer or Newman) if visualization of bond rotation and stereochemistry in the skeletal formula becomes difficult, as you did with the Sawhorse projection.


text:
Bohr Effect - related to H+ only?
Would I be correct in saying that the Bohr effect is ONLY related to the concentration of [H+] in the tissues.
Obviously the concentration of carbon-dioxide can contribute to this concentration by producing protons (through carbonic anhydrase) but it isn't directly involved in the effect?

I thin you may read this it explains...Bohr effect


text:
How can I determine the chemical composition of local rocks and minerals?
Specifically I live in the U.S. so I'm asking about an ordinary rock that you might find in your backyard or near a playground. Humans have discovered gun powder, petroleum, silica, iron, and many more types of useful materials by observing and experimenting with materials which lead to said discoveries.
Sometimes I'm walking outside and think to myself how people largely just ignore stuff outside like the rocks, dirt, grass and things like that, but I think if you knew how to separate the chemicals in these materials, then it could possibly lead to discoveries that might have otherwise been overlooked.
Especially rocks, because they can be cracked open and looked at with a microscope. Can't these rocks be used as raw materials to engineer something? If so, shouldn't they have inherent value? If not, why? If so, how?
What are the typical chemical elements that you will find in these types of materials, and how can you observe the chemical compositions by either isolating them or via some other physical process?

Rocks are made from minerals. Some rocks have a high abundance of important minerals that we already do extract - they're called ores.

@ozo are lands with ores available or are they owned buy big companies? what are the prospects of an individual finding some ores and extracting them himself then selling the purified minerals or using them to make something? just curious btw

You're more than welcome to ask questions about rocks and minerals in Earth Science SE!

Let's talk about rocks. The art and science of getting useful things from rocks is one of the oldest in human civilization. All of the example products you mention (gunpowder, petroleum, silica, and iron) were discovered or developed from experimentation on rocks.
In the ancient world, composition was determined by qualitative analysis, or a series of chemical tests and determination of important physical properties. For example, if a lump of shiny yellow stuff was soft, melted easily, did not lose its lustre in air or water, did not dissolve in vitriol (sulfuric acid) or aqua fortis (nitric acid), but did dissolve in aqua regia (made then by dissolving sal ammoniac (ammonium chloride) in aqua fortis), then we probably have gold.
In the modern age, the field of analytical chemistry was developed to answer the very questions you asked: What is this stuff? What are its constituents? How can we separate them? How do we determine that we successfully separated/purified them?
To continue the rock example, the modern technique of x-ray fluorescence is used to determine the elemental compositions of rocks (and other things, like ceramic fragments found at archaeological sites). In this technique, high frequency x-rays are used to excite the core electrons in the atoms of the sample. When these core electrons relax, they spit out x-rays of characteristic frequency/wavelength that is specific to the element. These emitted x-rays are collectively measured and converted into a spectrum like the one below.

This spectrum indicates that many elements will have more than one spectral line because there is more than one electronic transition that is possible. It also demonstrates the clear resolution of one element from another. After some maths (and better yet a calibration), you can use the heights of these peaks to determine the specific ratios of elements in your sample.
Not everyone is ignorant of the utility of natural objects. Most of us who become scientists do so to study the natural world and learn from it. For example, organic and medicinal chemists do not design new drug molecules from scratch. Rather, they take cues from nature. A folk legend about the curative properties of the leaves of a plant that grows along the riverside might turn in to an effort to extract and analyze the phytochemicals in those leaves, including throwing those compounds at bacteria, viruses, cancer, etc. Any that look promising are studied further. Taxol (paclitaxel) is a successful chemotherapeutic agent that was isolated from the bark of the pacific yew. It is now mass-produced by a semi-synthetic pathway which isolates a late-stage precursor from the needles of the European yew.
If you are specifically interested in rocks, then you should study geoscience.

are there cheap methods to crack open a rock in my yard and see what its made of? i don't own an x-ray machine  and don't see myself buying any expensive equipment in the near future, but i could maybe purchase a small kit for samples or something like that

@cortelldavis If you are that interested in rocks, you could contact your local mineral society. It's often possible to identify minerals without sophisticated analyses. And if you know the mineral you know its major elements.


text:
How to add two hydroxy groups to 1,5-diphenylpentan-3-ol with as high as possible yield?
From 1,5-diphenylpentan-3-ol (above), how to add the two hydroxy groups in red (below)?
Is there any proven successful synthesis route to 1,5-diphenylpentane-1,3,5-triol with higher yield (40 % or above)?

I would consider synthesis from acetophenone. Condensation of acetophenone enolate-anion with phosgene should give triketone, which in turn may be reduced by proper reagent into target product. Methods of introduction of substitute into benzilic position do exist, and some of them will work here, but I doubt you'll find anything working in a way you wish.

I've given an answer. That said, I think @permeakra is definitely suggesting a vastly superior approach, assuming you're not obligated to use the 1,5-diphenyl-3-pentanol as a starting material.

Two questions, 1) do you really need to start with the mono alcohol pictured above? 2) Your product contains 3 chiral centers resulting in the possible production of 4 diastereomers (2 chiral, 2 meso), is there one in particular you are interested in, or is a mixture acceptable?

A few possibilities for direct hydroxylation:

$\ce{SeO2}$ immediately comes to mind as one potential option; it is well known to selectively oxidize allylic positions to yield alcohols and carbonyl compounds, and there appear to be references suggesting it reacts at the benzylic position as well.
In Advanced Organic Chemistry, March notes that cerium(IV) triflate "converts benzylic arenes to benzylic alcohols, although the major product is the ketone when > 15% of water is present." Ref: Laali, K.K.; Herbert, M.; Cushnyr, B.; Bhatt, A.; Terrano, D. J. Chem. Soc., Perkin Trans. 1 2001, 578.
Intriguingly, March also reports a variety of oxidations accomplished enzymatically. I have no experience and very limited knowledge of these reactions, so I'll just give the relevant primary literature references:

Adam, W.; Lukacs, Z.; Saha-Moller, C.R.; Weckerle, B.; Schreier, P. Eur. J. Org. Chem. 2000, 2923.
Hamada, H.; Tanaka, T.; Furuya, T.; Takahata, H.; Nemoto, H. Tetrahedron Lett. 2001, 42, 909.
Adam, W.; Lukacs, Z.; Harmsen, D.; Saha-Moller, C.R.; Schreier, P. J. Org. Chem. 2000, 65, 878.

If those approaches aren't appealing, there are various less direct routes. Radical bromination (with, e.g., NBS under photochemical conditions), followed either by direct substitution with hydroxide, or alternatively elimination followed by subsequent acid-catalyzed hydration, would likely work well. If you need stereoselectivity, there are are numerous epoxidation reactions (that is, after bromination and subsequent elimination) that would likely be appropriate (Sharpless, Jacobsen, Shi, and certainly numerous others). Conveniently, your molecule would already have the allylic alcohol necessary for Sharpless asymmetric epoxidation, should you want to go that route. In any case, it should be possible to open the epoxides at the less hindered carbon with a suitable hydride donor (e.g., LAH) to yield the alcohols after work-up.

most oxidants will attack the alcohol group, radical bromination is statistical and acid will likely dehydrate the target molecule, especially if partially oxidated, into conjugated system. Because of it, most simple solutions targeting benzyl position will not work desired way.

@permeakra, well, oxidation of the alcohol isn't necessarily fatal; oxidation to a ketone can be reversed by hydride reduction easily, and none of the oxidants I've mentioned are aggressive enough to cleave $\ce{C-C}$ bonds. I agree with you about bromination. Even with the preference for the benzylic carbon, there will probably be roughly equal amounts of 1,1- and 1,5-dibrominated products, and they would be difficult to separate. In any case, I agree that the process would be comparatively painful and inefficient.

Actually, hell with that. I recall that organolithium compounds may be carefully oxidated at very low temperatures by different things, namely halogens, sulfur and - tadam! - molecular oxygen.
$\ce{BuLi}$ should selectively deprotonate each benzyl position once, and then the lithium compound may be very carefully oxidized with molecular oxygen at very low temperature followed by water hydrolysis into peroxide. Reduction of the peroxide should give the target compound, but it highly preferable to do it in basic conditions and avoid contact of target compound with acids as it looks to be able to dehydrate in acidic conditions. Another possible direction is to transform organolithium compound into boric acid and then oxidate it with hydrogen peroxide in basic conditions.

The idea of lithiation followed by oxidation to give the peroxide is really ingenious. It's not a reaction I see often, but this strikes me as an extremely elegant solution. Well done.

It's worth trying Greg's suggestion to do benzylic oxidation or bromination/hydrolysis. If it works it's cool, but a lot of times trying to change 2 functionalities of a molecule in one go has poor results with poor yields and formation of mixtures of products. Additionally you might find that you will need to protect your alcohol first (that's more possible with permeakra's enolate oxidation approach).
Another approach could be to start from acetone or an acetone equivalent (like its silyl enol ether) and do a double condensation with benzaldehyde. I'm sure you can find similar reactions in the literature.


text:
Why Does A Brønsted–Lowry Acid Accept Proton from Stronger Acid?

This is mechanism of Nitration of Benzene. In the first step, Nitric Acid (pKa = -1.4) accepts a proton from sulfuric acid which is stronger acid (pKa = -9). According to Brønsted-Lowry theory we consider acids dissociate increasing proton concentration and bases dissociate increasing hydroxy concentration. On the other hand bases also accept proton from acids. Therefore should we accept nitric acid as a base in this case ? If not, why does an acid accept a proton from another "stronger" acid ?

Methane can be converted to methyl lithium and CH5+ by strong enough bases and acids respectively.

The pKa of H2NO3+ (whatever protonated nitric acid is called) is actually just as relevant as the pKa of HNO3, since it's the conjugate acid of HNO3, and is what's going to be in your solution if HNO3 gets protonated.  Do you have H2NO3+'s pKa?

No I don't have but they are the values from book. I got no experimental or theoretical values.

Nitric acid is definitely a strong acid, but that does not exclude the possibility of protonating it with a stronger acid. It can be effectively protonated by a stronger acid such as sulfuric acid. Simply because something has "acid" in its name does not mean it has to dissociate into protons and its conjugate base, or protonate something else.
$$\ce{HNO_3 + H2SO4 <=>H2NO3^+ +HSO4-}$$
Note also that even strong acids in their pure forms - i.e. liquid nitric acid or liquid sulfuric acid - can and do auto-ionize to a fair extent. So if nitric acid can react with itself, of course it can react with a stronger acid.

I agree, but does stronger acid always protonate weaker acid in presence of both ? How do we know it ?

A stronger acid can protonate a weaker acid.

You can tell by calculating equilibrium constants for the particular reaction you have in mind.

@ordinarychemistrystudent Nope. A weak acid may be or not be basic enough. For example, $\ce{HCl}$ is not a base strong enough to be protonated by concentrated sulfuric acid, while nitric acid is.

It is still unclear that what rules we should follow when a stronger and weaker acid compounds present in solution. Is there a threshold of basicity of acids to be able to or be in a situation to accept protons from stronger acids ?

You're trying to find a clear-cut rule when there is none. The reality is that almost all molecules we consider acids and bases will react with themselves - i.e. they have both acidic and basic character. Even liquid sulfuric acid isn't pure; you'll find a bunch of hydrogen sulfate ions in the liquid as well as doubly protonated sulfuric acid molecules. Yes, it's a super-strong acid. But yes, it can also accept protons.

@ordinarychemistrystudent Of course, in order for the stronger acid to protonate the weaker acid, the weaker acid must be able to act as a proton acceptor as well. In this case, the nitric acid molecule has a lone pair which can form a dative covalent bond with a proton. If the nitric acid molecule did not have any lone pairs, it would then not be protonated.

Echoing Dissenter's comment, don't be hung up on the name "nitric acid" to mean that it can only release a proton. It's typical reactivity is as an acid, which is why it gained that name. Most compounds will change their reactivity under different reaction conditions. Take water for example. Under acidic conditions, such as when reacted with HCl, it will act as a base, becoming protonated to form hydronium. Under basic conditions, such as when reacted with butyl lithium, water will serve as an acid, being deprotonated to form hydroxide.
There also seems to be some confusion about acid-base concepts. Arrhenius theory considers compounds that increase hydrogen ion concentration to be acids and compounds that increase hydroxide concentration to be bases. That is a very limited definition. In the reaction between nitric acid and sulfuric acid, the Brønsted-Lowry concept is more useful. Brønsted-Lowry considers any compound that donates a proton (hydrogen ion) to be an acid (sulfuric acid here) and any compound that accepts a proton (nitric acid here) to be a base.

The first step of the mechanism would not, on its own, happen to a large extent - protonating nitric acid is unfavorable, so the equilibrium would lie far to the left. However, the second step of the mechanism is highly favorable (entropically), so it quickly consumes what little H2NO3+ is produced by the first step. This drives the overall reaction forward by Le Chatelier's principle


text:
How can you predict a double displacement reaction with no precipitate?
Is it possible to predict a double displacement reaction between two ions where the products remain in aqueous solution? I see how you can predict precipitation reactions if you know whether the product(s) is/are soluble, but that doesn't apply to all possible double displacement reactions.

If there is no precipitate, then there is nothing to drive the reaction. Both possible products would be soluble, and so there is no net reaction (it's just the reactant ions in solution).

Prediction, you say, is possible if you know the reactivity series like this for an elemtary level.

The reactivity series doesn't apply to double-displacement reactions; it's only relevant for redox (single-displacement) reactions.