text:
Keto Enol Equilibrium for Claisen Condensation Products
Do Claisen Condensation products prefer the di-keto form, or a keto/enol form due to increased conjugation and hydrogen bonding?

The Claisen reaction yields a product with an acidified proton in $\alpha$-position of carbonyl and ester, right at the same carbon atom of R depicted in blue.
In presence of a base, yes, $\beta$-dicarbonyl compounds undergo deprotonation at their methylene unit to yield a keto-enolate, stabilized by mesomerism with the secound carbonyl group in conjugation. Compare the $pK_a$-values of $\beta{}$-dicarbonyl compounds with ketones, for example (one reference), to estimate if the equilibrium is on the left, or on the right.
.
I think cations of the right size and charge equally favour the formation of the keto-enolate.  If you synthesize $\beta$-dicarbonyl compounds, the risk to loose product due to chelation  is lowered if initial work-up is concluded by drying the washed organic phase with sodium sulphate ($\ce{Na2SO4}$), rather than by magnesium sulphate ($\ce{MgSO4}$).

Generally, diketo compounds can react to form an enol-keto form and back again. This tautomery can not be considered a [1,3] proton shift due to orbital chemistry and has to be thought of as a combination of deprotonation/reprotonation or protonation/deprotonation steps — ask another question if you’re interested in details (assuming they are not already posted somewhere on the site).

Notice, that in its enol-keto form, the compound can form an intramolecular hydrogen bond. If desired, the $\mathrm{p}K_{\mathrm{a}}$ of the molecule itself is enough to protonoate another one, so the process can be thought of as self-catalysed. Thus, there is a tendency of these compounds to equilibrate to a desired ratio. Only minor differences need to be considered, if $\mathrm{R} = \ce{OR''}$
But what is that ratio going to be? It depends on the surroundings. If you dissolve it in solvents that are capable of donating hydrogen bonds, then both keto oxygens will accept H-bonds, and the diketo tautomer will be preferred. On the other hand, if the solvent cannot form any hydrogen bonds, the enol tautomer, which is capable of forming an intramolecular H-bond, is less disfavoured for that exact reason. Further parameters like other molecules dissolved, their $\mathrm{p}K_{\mathrm{a}}$’s, the dryness of the solvent, the pH-value etc. also factor in.
Which tautomer dominates in which solvent and what the ratio between the two is (it is extremely unlikely for one form to be the only one present) cannot be predicted exactly and needs to be measured experimentally.

The question is about $\beta$ keto-esters not 1,3 diketones though.

@bon See last line of second paragraph. Does it change much?

Missed that. I'm not sure how much it changes though so have to let someone else answer that.


text:
Born–Oppenheimer adiabaticity
The Born–Oppenheimer aproximation is said to be adiabatic.

What does "adiabatic" mean?
What would be a non-adiabatic approximation to the Schrödinger equation?

Note that the Born-Oppenheimer approximation is just further approximation of the adiabatic one, and as such, it is also adiabatic.

Also, if you want a less mathematical but more intuitive answer. You can have a look at here:
http://www.rpi.edu/dept/phys/Dept2/Courses/phys410/lct10.pdf

$\newcommand{\conj}[1]{\overline{#1}{}}
\newcommand{\braket}[2]{\langle{#1}\,|\,{#2}\rangle}
\newcommand{\bracket}[3]{\langle{#1}\,|\,{#2}\,|\,{#3}\rangle}
\newcommand{\mat}[1]{\mathbf{#1}}
\newcommand{\rel}{\vec{r}_{\mathrm{e}}}
\newcommand{\rnuc}{\vec{r}_{\mathrm{n}}}
\newcommand{\linop}[1]{\hat{#1}}
\newcommand{\Tnuc}{\linop{T}_{\mathrm{n}}}
\newcommand{\Tel}{\linop{T}_{\mathrm{e}}}
\newcommand{\Velel}{\linop{V}_{\mathrm{ee}}}
\newcommand{\Velnuc}{\linop{V}_{\mathrm{en}}}
\newcommand{\Vnucnuc}{\linop{V}_{\mathrm{nn}}}
\newcommand{\Hel}{\linop{H}_{\mathrm{e}}}
\newcommand{\psiel}{\psi_{\mathrm{e}}}
\newcommand{\psieli}{\psi_{\mathrm{e},i}}
\newcommand{\psielj}{\psi_{\mathrm{e},j}}
\newcommand{\Eel}{E_{\mathrm{e}}}
\newcommand{\Eeli}{E_{\mathrm{e},i}}
\newcommand{\Eelj}{E_{\mathrm{e},j}}
\newcommand{\psinuc}{\psi_{\mathrm{n}}}
\newcommand{\psinuci}{\psi_{\mathrm{n},i}}
\newcommand{\psinucj}{\psi_{\mathrm{n},j}}$
Setup
In short, our goal is to separate the coordinates of all electrons $\rel = \{ \vec{r}_{i} \}_{i=1}^{n}$ from that of the nuclei $\rnuc = \{ \vec{r}_{\alpha} \}_{\alpha=1}^{\nu}$. Thus, we write the time-independent Schrödinger equation in the following form
\begin{equation}
    \linop{H} \psi(\rel, \rnuc) = E \psi(\rel, \rnuc) \, , \tag{1}
\end{equation}
where the Hamiltonian operator is given as follows
\begin{equation}
    \linop{H} = \Tnuc + \Tel + \Velnuc + \Vnucnuc + \Velel \, ,
\end{equation}
or, explicitly in atomic units,
\begin{equation}
    \linop{H}
    = - \sum\limits_{\alpha=1}^{\nu} \frac{1}{2 m_{\alpha}} \nabla_{\alpha}^2
    - \sum\limits_{i=1}^{n} \frac{1}{2} \nabla_{i}^2
    - \sum\limits_{\alpha=1}^{\nu} \sum\limits_{i=1}^{n} \frac{Z_{\alpha}}{r_{\alpha i}}
    + \sum\limits_{\alpha=1}^{\nu} \sum\limits_{\beta > \alpha} \frac{Z_{\alpha} Z_{\beta}}{r_{\alpha \beta}}
    + \sum\limits_{i=1}^{n} \sum\limits_{j > i}^{n} \frac{1}{r_{ij}} \, ,
\end{equation}
where $m_{\alpha}$ is the rest mass of nucleus $\alpha$ and $Z_{\alpha}$ is its atomic number, $r_{\alpha i} = |\vec{r}_{\alpha} - \vec{r}_{i}|$ is the distance between nucleus $\alpha$ and electron $i$, $r_{\alpha \beta} = |\vec{r}_{\alpha} - \vec{r}_{\beta}|$ is the distance between two nuclei $\alpha$ and $\beta$, and $r_{ij} = |\vec{r}_{i} - \vec{r}_{j}|$ is the distance between two electrons $i$ and $j$.
Here for simplicity we ignore magnetic interactions and use the non-relativistic treatment.
Exact separation
We know that nuclei are much heavier as compared to electrons, $m_{\mathrm{n}} \gg m_{\mathrm{e}}$, and assuming that at average the kinetic energy of electrons and nuclei is of the same order of magnitude (see the equipartition theorem)
\begin{equation}
    \frac{m_{\mathrm{n}} v_{\mathrm{n}}^2}{2}
    \sim
    \frac{m_{\mathrm{e}} v_{\mathrm{e}}^2}{2} \, ,
\end{equation}
we conclude that even for case of the lightest nucleus (a single proton)
\begin{equation}
    \frac{v_{\mathrm{e}}}{v_{\mathrm{n}}}
    \sim
    \sqrt{\frac{m_{\mathrm{n}}}{m_{\mathrm{e}}}}
    \sim \sqrt{1836} \sim 43 \, ,
\end{equation}
i.e., nuclei at average are moving considerably slower than electrons (talking in terms of classical mechanics).
For a coupled motion of two subsystems with significantly different speeds it is often assumed that one could treat the motion of a relatively fast moving subsystem (electrons) as if it depends only on the static configuration of a relatively slow moving one (nuclei).
This assumption results in a procedure with two major steps, described below.
First step
First, we solve an equation describing the motion of electrons for a number of static configurations of the nuclei.
The equation is known as the electronic Schrödinger equation and it can be written as follows
\begin{equation}
    \Hel \psieli(\rel, [\rnuc]) = \Eeli([\rnuc]) \psieli(\rel, [\rnuc]) \quad \text{for} \quad i = 1, 2, \dotsc \, .
\end{equation}
where $\Hel = \Tel + \Velnuc + \Velel$ is the electronic Hamiltonian, $\psieli$ is the electronic wave function describing $i$-th electronic state, and $\Eeli$ is the electronic energy of the $i$-th electronic state --- the total energy of a molecule with nuclei being fixed at particular coordinates less the nuclear repulsion potential energy $\Eel = U - V_{\mathrm{nn}}$.
Note that the electronic wave function is a function of electronic coordinates only, though parametrically it still depends on nuclear coordinates. The parametric dependence of $\psieli$ and $\Eeli$ on $\rnuc$ means that for different sets of fixed nuclear coordinates $\rnuc$ we have different electronic Schrödinger equations with different solutions $\psieli$ and $\Eeli$, or, in other words, that the electronic Schrödinger equation written above is a family of equations parametrized by $\rnuc$, rather then one equation.
Second step
Once solutions of the electronic Schrödinger equation $\{ \psieli(\rel) \}_{i=1}^{\infty}$ and $\{ \Eeli \}_{i=1}^{\infty}$ are found for a number of different fixed nuclear configurations $\rnuc$, we took the second step: the task is to solve an equation of motion for nuclei which is obtained as follows.
Note first that since $\Hel$ is self-adjoint, its eigenfunctions $\{ \psieli(\rel, [\rnuc]) \}_{i=1}^{\infty}$ form a complete orthonormal set.
Now we use the assertion that a function of two independent variables $\psi(\rel, \rnuc)$ can be expanded over a complete set of functions of one variable $\{ \psieli(\rel, [\rnuc]) \}_{i=1}^{\infty}$ in the following way
\begin{equation}
    \psi(\rel, \rnuc) = \sum\limits_{i=1}^{\infty} \psieli(\rel, [\rnuc]) \psinuci(\rnuc) \, . \tag{2}
\end{equation}
Substitution of (2) in the Schrödinger equation (1), multiplication of both sides by $\conj{\psi}_{\mathrm{e},j}$, and integration over electronic coordinates $\rel$ only, leads to
\begin{equation}
    \Big( \Tnuc + \Eelj + \Vnucnuc - E \Big) \psinucj
    -
    \sum\limits_{i=1}^{\infty} \linop{\Lambda}_{ji} \psinuci
    = 0 \quad \text{for } j = 1, 2, \dotsc \, , \tag{3}
\end{equation}
where operator $\linop{\Lambda}_{ji}$ is defined as follows
\begin{equation}
    \linop{\Lambda}_{ji} =
    \sum\limits_{\alpha=1}^{\nu} \frac{1}{2 m_{\alpha}}
    \big(
        2 \bracket{ \psielj }{ \nabla_{\alpha} }{ \psieli } \nabla_{\alpha}
        + \bracket{ \psielj }{ \nabla_{\alpha}^2 }{ \psieli }
    \big) \, .
\end{equation}
and arguments $\rel$ and $\rnuc$ were dropped for brevity.
The last equation describes a system of differential equations (one equation for each value of $j$), and once the system is solved and a set of functions $\psinuci$ is obtained, one could write down the exact solution of the Schrödinger equation (1) in the form we introduced above (2).
Approximations
The task to solve the system of equations (3) is extremely difficult and the reason is that equations in the system are coupled in a sense that the solution of any $i$-th equation, $\psinuci(\rnuc)$, enters all other equations as the $\linop{\Lambda}_{ji} \psinuci$ term. One obvious way to decouple equations in the system (3) is by assuming that $\linop{\Lambda}_{ji} = 0$ either for all $i \neq j$, or for all $i$ without any exception. Operators $\linop{\Lambda}_{ji}$ may be regarded as the elements of a square matrix $\mat{\Lambda}$ and of course the fact that we equate these elements with zero is an approximation: in the first case, when we equate with zero all off-diagonal elements $\linop{\Lambda}_{ji}$, we introduce the adiabatic approximation, while in the second case, when we equate with zero all elements $\linop{\Lambda}_{ji}$, we introduce the Born–Oppenheimer approximation.
Both these approximations lead to a system of the so-called nuclear Schrödinger equations
\begin{equation}
    \linop{H}_{\mathrm{n}} \psinucj(\rnuc) = E \psinucj(\rnuc) \quad \text{for } j = 1, 2, \dotsc \, ,
\end{equation}
where $\psinucj(\rnuc)$ is known as the nuclear wave function and $\linop{H}_{\mathrm{n}}$ as the nuclear Hamiltonian, with the only difference being that in the first case the nuclear Hamiltonian takes the following form
\begin{equation}
    \linop{H}_{\mathrm{n}} = \Tnuc + \Eelj([\rnuc]) + \Vnucnuc - \linop{\Lambda}_{jj} \, ,
\end{equation}
while in the second case the last term $\linop{\Lambda}_{jj}$ is absent
\begin{equation}
    \linop{H}_{\mathrm{n}} = \Tnuc + \Eelj([\rnuc]) + \Vnucnuc \, .
\end{equation}
The first thing to notice is that in both the above mentioned approximations the only terms in the nuclear Schrödinger equation for $j$-th nuclear state $\psinucj$ that depend on electronic states, $\Eelj$ and also $\linop{\Lambda}_{jj}$, are related to the same and the only electronic state $\psielj$.
This means that in both these approximations each nuclear state $\psinucj$ depends only on one electronic state $\psielj$, while in the accurate physical picture (3), electronic states are coupled through off-diagonal $\linop{\Lambda}_{ji}$ terms, known as non-adiabatic coupling terms (NACT).
Note also that the first term in the nuclear Hamiltonian in both adiabatic and Born–Oppenheimer approximations represents the kinetic energy of nuclear motion, and consequently, the remaining terms collectively should stand for the potential energy of nuclear motion.
It is said thus that nuclei move on the so-called potential energy surface (PES) defined by the sum of these remaining terms: $\Eelj([\rnuc]) + \Vnucnuc - \linop{\Lambda}_{jj}$ in the adiabatic approximation and just $\Eelj([\rnuc]) + \Vnucnuc$ in the Born–Oppenheimer one.
Non-adiabatic coupling terms usually (but not always) can indeed be neglected, especially around the equilibrium nuclear configuration.
Besides, often the errors due to neglecting these terms are less than the errors resulting from some further approximation introduced when solving electronic and nuclear equations.
However, when two potential energy surfaces come close to each other or even cross, non-adiabatic coupling terms becomes large and are no longer negligible.
Answers

What adiabatic means?

The term "adiabatic" from Ancient Greek αδιαβατος (α - "not", δια - "through", βατος - "passable") literally means the situation when something "is not passing through" something else.
In thermodynamics adiabatic process is a process occurring without exchange of heat of a system with its environment, i.e. a process in which heat is not passing through system enclosure.
In quantum mechanics adiabatic refers to a process in which no abrupt transition from one state to another with respect to continuous changes of some parameters happens, and thus, energy of a system changes continuously with respect to continuous changes of these parameters.
And the reason why we call the above mentioned approximation adiabatic lies in the fact that in this approximation when solving the electronic Schrödinger equation for different fixed nuclear configurations by continuously varying the coordinates of the nuclei we indeed assume that electronic state of the system continues to be the same. In other words, we assume that we are "staying" at the same PES. This could be the case, of course, only when potential energy surfaces for different electronic states are well separated, so, in the regions of nuclear coordinates where potential energy surfaces come close to or even cross each other adiabatic approximation is inadequate.

What would be the case of a non adiabatic appox. to Schrodinger eq.?

Non-adiabatic solution refers to the solution of the system of coupled equations (3).

This answer is clearer than textbooks I have read on the BO approximation, and on adiabatic approximations in general!

Wildcat, thanks for your answer, it was very instructive. I just want to make sure I understood you properly. Regarding the part on nuclear Schrodinger equations, this is a system of Schrodinger equations, as each electronic state $\psi_{\mathrm{e},j}$ leads to a different Hamiltonian $\hat{H}{\mathrm{n},j}$, and each of these Schrodinger equations generates a set of nuclear wavefunctions ${\psi{\mathrm{n},k}}$ which only depend on one electronic state $\psi_{\mathrm{e},j}$ - did I get that right?

@orthocresol, yep. Each $j$-th nuclear SE has a set of solutions, so for each and every electronic state $\psi_{\mathrm{e},j}$ you indeed get a bunch of nuclear states ${\psi_{\mathrm{n},k}^{j}}$ where $k=1, 2, \dotsc$. These states can be further (approximately) subdivided into vibrational, translational, and rotational.

Firstly, may I ask, why is the assumption that the nuclei and electrons have approximately the same momentum is valid? Secondly, in the Gaussian DFT procedure to determine molecular vibrational frequencies, is the nuclear SE solved? I understand that DFT essentially solves the electronic SE and the ground state energy (PES) is sum of the internuclei repulsion energy and the electronic energy. And that the DFT procedure solves for the electronic energy (EE) first, then determines the PES later. I know the procedure for determining EE well, but not so sure as to how DFT gives PES.

@Tan Yong Boon Regarding the assertion that 'the nuclei and electrons have approximately the same momentum', I don't think it's true. In classical terms, if you have a gas of two particles with different masses then the equipartition theorem will lead to each particle having approximately the same energy, not momentum. See, e.g., https://arxiv.org/pdf/physics/0610189.pdf . As a consequence, the ratios of the velocities are not proportional to m1/m2−−−−−−√ . So, in classical terms, the nuclei do move slower than the electrons, but by a factor rather closer to 40 than to 2000.

@Lorents many thanks for for pointing that mistake out. Answer corrected.

@Wildcat Isn't the equipartition theorem applied to atoms or molecules and not electrons or protons? The wiki link for equiparition theorem doesn't state anything about the average kinetic energies for electrons and nuclei. So how can we justify that nuclei move slower than electrons?

Can you please elaborate on the "exact separation"? If the separation of motion is exact why then you say "this assumption results in a procedure with two major steps"? The fact we can expand the total (exact) wavefunction in the parametric electronic eigenfuctions has nothing to do with the time scale of motions. The latter becomes significant during "approximations", based on the adiabatic theorem. For adiabatic theorem to hold we need time scale separation of motion and energy gap between the potential energy surfaces.

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text:
Synthesis of bromobutane
What will be the cost efficient and high yielding method of synthesis for 1-bromobutane?
I can react butanol with $\ce{HBr}$ but I am worried butanol is not present at the moment in lab. But I need to hurry and make bromobutane.

Well, clearly what you do have in the lab will the rate determining step. What potential candidates have you got in the inventory?

Current prices are about 34 € (500 g) for 1-bromobutane and 32 € (500 mL) for 1-butanol in decent lab quality.
The most cost efficient method is ordering 1-bromobutane from a chemical supplier of your choice!
On a technical scale, one would probably react 1-butene with anhydrous hydrogen bromide under radical conditions, using peroxides or UV light to initiate homolytic cleavage of the $\ce{H-Br}$ bond.

Bromine radical is selective. You cannot expect to obtain a high yield of 1-bromobutane from such a reaction. Besides, such reactions are a mess to purify and this will be very tedious and difficult to achieve.

Start with butanal or butanoic acid, reduce them with LiAlH4 and aqueous work-up and you have 1-butanol.
If you have 1-butene, you do hydroboration/oxidation then phosphorus tribromide.
Are you trying to make a Grignard? or a substitution reaction?

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text:
What are some books, websites, etc. where I can find the detailed protocol for standardization of sodium thiosulfate with dichromate?
I am looking for a book or website such as "Vogel's Textbook of Quantitative Chemical Analysis" where I can find methods for analyzing/standardizing a particular species easily.
I am not looking for an exact sub area such as water analysis or drug analysis; rather I want a resource like Vogel's which indexes generic analytical methods, both classical and modern.
Edit:
Actually I am looking for the method of standardizing a sodium thiosulfate solution using a potassium dichromate solution. I can't find the method in "Vogel's Textbook of Quantitative Chemical Analysis 6th Edition".

Finally I found the book I was looking for. The method is listed in "Vogel's Textbook of Chemical Analysis 5th Edition" (page 392). They have removed it from the 6th edition.

My go-to for this is typically Kolthoff & Sandell's Textbook of Quantitative Inorganic Analysis.  They unfortunately do not report a detailed protocol for dichromate standardization of thiosulfate solutions therein, but instead cite Volumetric Analysis, Vol III by Kolthoff and Stenger (Amazon and Google Books).  I don't have access to this volume; WorldCat shows numerous copies in university libraries in my area of the U.S.; more than likely there's a library where you are that has a copy.

@chemkatku Sure! I was rather surprised that K&S didn't have it in detail. It must be that authors are either considering it an obsolete method, or so commonplace that it's not needed to include any more.

Book:
I can find one book which should have analytical methods for most of the species.
Encyclopedia of Separation Science
It is ten volume set.
websites:
There are too many websites on internet but I will choose following two:

Chemistry Stack Exchange
Wikipedia

Thanks. Encyclopedia of Separation Science is great. But I want some resource that indexes analytical methods.


text:
change in state (sublimation and deposition)
I am slightly confused as to how I should think of changes in state (especially sublimation and deposition). Should I think of sublimation and deposition as skipping the liquid phase (fig 1) or should I think of it as a circular pattern (fig 2).

fig 1

fig 2
Why during sublimation and deposition is the liquid phase skipped and how is it skipped, would not all matter have to be a liquid in between solid and gas?
Is it due to the fact that some substances are unable to be liquids? if so what prevents it from being a liquid?

Please have a look at http://chemistry.stackexchange.com/questions/6318/is-sublimation-caused-by-an-environmental-difference-or-an-elements-composition and http://chemistry.stackexchange.com/questions/15028/is-sublimation-a-property-of-all-substances and http://chemistry.stackexchange.com/questions/11104/why-doesnt-a-table-sublimate-while-ice-does.

Both are correct.

Firstly, in regards to the diagrams, as mentioned in the comments, both diagrams are correct. One thing that both diagrams imply is that sublimation/deposition is equivalent to the combination of melting/freezing and evaporation/condensation - another way to look at this is the following diagram from the UC Davis ChemWiki page Heat of Sublimation:

An explanation from the webpage is that:

Though in sublimation a solid does not pass through the liquid phase on its way to the gas phase, it takes the same amount of energy that it would to first melt (fuse) and then vaporize.

In regards to your related questions:

Is it due to the fact that some substances are unable to be liquids? if so what prevents it from being a liquid?

Not quite, liquids can occur in the right conditions, the explanation is below.

Why during sublimation and deposition is the liquid phase skipped and how is it skipped, would not all matter have to be a liquid in between solid and gas?

To understand why sublimation occurs requires an understanding of phase diagrams. A generalised phase diagram from UC Davis Chemwiki page Phase Diagrams is shown below:

The sublimation/deposition curve is the line from the origin to the triple point, represented as the red line in the diagram above. A generalisation that can be made is that a substance undergoes sublimation/deposition when either (or both) the temperature and pressure are lower than at the triple point. Consider the following examples comparing the phase diagrams of carbon dioxide and water (from the UC Davis Chemwiki page linked before):
First, carbon dioxide - at 'normal' atmospheric pressure (such as in the typical laboratory - 1 atm), it is lower than the 5.2 atm at the triple point (labelled 'D'). Assuming minimal pressure changes from standard laboratory conditions, as the $\ce{CO2}$ is cooled, it will eventually undergo deposition to the solid phase.

Now, for comparison, consider the phase diagram for water, the triple state (also labelled 'D') occurs at a very low pressure 0.006 atm, pressures lower than that would result in water subliming. Note for the standard laboratory conditions, water maintains the familiar solid, liquid and gas characteristics (depending on temperature).

Note, both do have a liquid phase.


text:
May I treat units (e.g. joules, grams, etc.) in equations as variables?
Suppose I have two units, Tonks, $T$, and Borks, $B$, as well as the equation $\frac{50T}{20B}=x$. May I manipulate the equation as though $T$ and $B$ were the kinds of symbols that we learned to manipulate in math class? For example:
\begin{align}
\frac{50T}{20B}&=x\\
50T&=20Bx\\
\frac{50T}{20}&=Bx\\
\end{align}

Note that if tonk and bork were actual units, their unit symbols would be printed in roman (upright) type. They would be printed in lower-case letters unless they were derived from a proper name. Thus, conforming unit symbols would be $\mathrm t$ and $\mathrm b$ (not $T$ and $B$). (However, $\mathrm t$ is actually used for the tonne and $\mathrm b$ is actually used for the barn.) Furthermore, a space is always used to separate the unit from the number. Therefore, your equation should probably read $\frac{50\ \mathrm t}{20\ \mathrm b}=x$.

Yes, you may. It is quite common to convert units into each other. The simplest conversion might be the prefixing of units, e.g.
$$\mathrm{\frac{km}{m}}=1000 \Longleftrightarrow \mathrm{1~km = 1000~m}.$$
Another example is the interconversion of units of energy. In some parts of chemistry, it is still quite common to use calories, in others Joule, as a SI unit, has taken its place.
$$\mathrm{\frac{cal}{J}}=4.184 \Longleftrightarrow \mathrm{1~cal = 4.184~J}.$$
In principle, all units are defined to be constants, special numbers in a wider sense. You can treat them as such.

It may be worth explicitly discussing units like F and C which don't have "natural" zero points. Can you still treat them as constants?

If you mean Fahrenheit and Celsius, I agree and will try to cover that as soon as i get the chance.

@R.M. Is it even allowed to use F and C in equations? I mean, what would 210°C be? How about 20°C? Easy to run into contradictions

Fahrenheit and Celsius can be used in equations just like any other physical unit: $F = \frac{9}{5} C + 32$. They can also be used in any equation where the temperature is assumed to be in those units (as for older empirical formulae). Celsius units can be used in any equation requiring either Celsius or Kelvins when it's the difference in temperatures that's important, since they share the same scaling relation (a change of 1 K is the same as a change of 1 degree Celsius). Of course, an absolute scale is required for any physical equation.

@jvriesem: Watch out, those symbols you wrote there are not the units, but represent the numeric values of a value given a specific unit. It's The temperature in degrees Fahrenheit is 9/5 times the temperature in degrees Celsius plus 32, not 1 degree Fahrenheit is .... You correctly wrote them in italics.

You would probably have more success working with Kelvins and degrees Rankine.

Treating units as if they were algebra is called “dimensional analysis”.
One example given in that article is the question of how many seconds are there in two years.
$$2\ \mathrm{yr} \times 365\ \mathrm{day}\ \mathrm{yr}^{-1} \times 24\ \mathrm{hr}\ \mathrm{day}^{-1} \times 60\ \mathrm{min}\ \mathrm{hr}^{-1} \times 60\ \mathrm{s}\ \mathrm{min}^{-1} = 63072000\ \mathrm{s} $$
Another example question is,

How many atoms of hydrogen can be found in 45 g of ammonia, $\ce{NH3}$?

This is how I would approach this:
$$45\ \mathrm{g}\ \ce{NH3}\ /\ 17\ \mathrm{g}\ \mathrm{mol}^{-1} \approx 2.647\ \ce{NH3}\ \mathrm{mol}$$
$$2.647\ \ce{NH3}\ \mathrm{mol}\ \times\ 6.02 \times 10^{23}\ \mathrm{molecules}\ \mathrm{mol}^{-1} \approx 1.594 \times 10^{24}\ \ce{NH3}\ \mathrm{molecules}$$
$$1.594 \times 10^{24}\ \ce{NH3}\ \mathrm{molecules} \times 3\ \mathrm{(H\ atoms)}\ \mathrm{(\ce{NH3}\ molecules)}^{-1} \approx 4.781 \times 10^{24}\ \ce{H}\ \mathrm{atoms}$$
The article works this out differently, as a single step. I myself find the multi-step approach easier to follow, but everyone’s different. The article renders this as,

$$45\ \mathrm{g}\ \ce{NH3} \times \frac{1\ \mathrm{mol}\ \ce{NH3}}{17\ \mathrm{g}\ \ce{NH3}} \times \frac{6.02 \times 10^{23}\ \mathrm{molecules}\ \ce{NH3}}{1\ \mathrm{mol}\ \ce{NH3}} \times \frac{3\ \mathrm{atoms}\ \ce{H}}{1\ \mathrm{molecule}\ \ce{NH3}}$$
$$ = 4.8 \times 10^{24}\ \mathrm{atoms}\ \ce{H}$$
(to 2 significant figures)

If the units cancel out correctly, that is further indication that you have the right formula. (And if they don’t cancel out correctly, that is a certain indication that you don’t have the right formula.)

One cannot stress enough how important dimensional analysis is and the realization that dimensions can be treated algebraically in the same manner as numbers. As TRiG says If you can master dimensions you're likely never to make mistakes again in your computations. And there is much more ... check out Buckingham's PI theorem to see where dimensions can be used to infer mathematical relationships of physical systems.

Of course, depending on what genre of music you prefer, you may be able to simplify this further by using either 86400 s day^-1 or 525600 min yr^-1 - bearing in mind that for the latter you are embedding an assumption about non-leap years into how you measure a year.

@docscience And the units themselves, not just the dimensions, can be treated as algebraic variables. In my example, there's just the one dimension  (I may add more examples when I'm not on a train), and more of the values are simple conversions.

Not only may you, but you absolutely should. As a physics professor, watching students toss away the information contained in units is incredibly frustrating. Typically students drop the units, "do the math," and then put back on whatever final units they think should be there (if they bother with the last step at all). This leads to a shockingly large number of mistakes (even multi-million dollar ones: http://www.cnn.com/TECH/space/9909/30/mars.metric/)
Whenever you write a physical value you should write both the number and the unit and "do the math" on both, exactly as you propose. If the units don't work out to a unit that you expect then you know that you've done something wrong.

You not only can, but also must treat symbols for units by the ordinary rules of algebra, since unit symbols are mathematical entities and not abbreviations.
The value of a quantity is expressed as the product of a number and a unit. That number is called the numerical value of the quantity expressed in this unit.
This relation may be expressed in the form
$$Q = \left\{ Q \right\} \cdot \left[ Q \right]$$
where $Q$ is the symbol for the quantity, $\left[ Q \right]$ is the symbol for the unit, and $\left\{ Q \right\}$ is the symbol for the numerical value of the quantity $Q$ expressed in the unit $\left[ Q \right]$.
For example, the mass of a sample is
$$m = 100\ \mathrm g$$
Here, $m$ is the symbol for the quantity mass, $\mathrm g$ is the symbol for the unit gram (a unit of mass), and $100$ is the numerical value of the mass expressed in grams. Thus, the value of the mass is $100\ \mathrm g$.
It is important to distinguish between the quantity $Q$ itself and the numerical value $\left\{ Q \right\}$ of the quantity expressed in a particular unit $\left[ Q \right]$. The value of a particular quantity $Q$ is independent of the choice of unit $\left[ Q \right]$, although the numerical value $\left\{ Q \right\}$ will be different for different units.
For example, changing the unit for the mass in the previous example from the gram to the kilogram, which is $10^3$ times the gram, leads to a numerical value which is $10^{-3}$ the numerical value of the mass expressed in grams, whereas the value of the mass stays the same.
$$m = 100\ \mathrm g = 0.100\ \mathrm{kg}$$
Since symbols for units are mathematical entities, both the numerical value and the unit may be treated by the ordinary rules of algebra. For example, the equation $m = 100\ \mathrm g$ may equally be written
$$m/\mathrm g = 100$$
It is often convenient to label the axes of a graph in this way, so that the tick marks are labelled only with numbers. The quotient of a quantity and a unit may also be used in this way for the heading of a column in a table, so that the entries in the table are all simply numbers.
Performing the mathematical operations of quantities is called quantity calculus. Quantities are multiplied and divided by one another according to the rules of algebra, resulting in new quantities.
The quotient of two quantities, $Q_1$ and $Q_2$, satisfies the relation
$$\begin{align}
\frac{Q_1}{Q_2} &= \frac{ \left\{ Q_1 \right\} \cdot \left[ Q_1 \right] }{ \left\{ Q_2 \right\} \cdot \left[ Q_2 \right] } \\[6pt]
&= \frac{ \left\{ Q_1 \right\} }{ \left\{ Q_2 \right\} } \cdot \frac{ \left[ Q_1 \right] }{ \left[ Q_2 \right] }
\end{align}$$
Thus, the quotient $\left\{ Q_1 \right\}/\left\{ Q_2 \right\}$ is the numerical value $\left\{ Q_1/Q_2 \right\}$ of the quantity $Q_1/Q_2$, and the quotient $\left[ Q_1 \right]/\left[ Q_2 \right]$ is the unit $\left[ Q_1/Q_2 \right]$ of the quantity $Q_1/Q_2$.
For example, assuming a volume of $V = 0.127\ \mathrm{l}$, the density $\rho$ of the above-mentioned sample is
$$\begin{align}
\rho &= \frac{m}{V} \\[6pt]
&= \frac{ 0.100\ \mathrm{kg} }{ 0.127\ \mathrm{l} } \\[6pt]
&= \frac{ 0.100 }{ 0.127 } \cdot \frac{ \mathrm{kg} }{ \mathrm{l} } \\[6pt]
&= 0.79\ \mathrm{kg/l}
\end{align}$$
Similarly, the product of two quantities, $Q_1$ and $Q_2$, satisfies the relation
$$\begin{align}
Q_1 \cdot Q_2 &= \left( \left\{ Q_1 \right\} \cdot \left[ Q_1 \right] \right) \cdot \left( \left\{ Q_2 \right\} \cdot \left[ Q_2 \right] \right) \\[6pt]
&= \left\{ Q_1 \right\}\left\{ Q_2 \right\} \cdot \left[ Q_1 \right] \left[ Q_2 \right]
\end{align}$$
Thus, the product $\left\{ Q_1 \right\}\left\{ Q_2 \right\}$ is the numerical value $\left\{ Q_1Q_2 \right\}$ of the quantity $Q_1Q_2$, and the product $\left[ Q_1 \right]\left[ Q_2 \right]$ is the unit $\left[ Q_1Q_2 \right]$ of the quantity $Q_1Q_2$.
For example, considering the standard acceleration of free fall $g_\mathrm n = 9.80665\ \mathrm{m/s^2}$, the weight $F_\mathrm g$ of the above-mentioned sample is
$$\begin{align}
F_\mathrm g &= m \cdot g_\mathrm n \\[6pt]
&= 0.100\ \mathrm{kg} \times 9.80665\ \frac{\mathrm m}{\mathrm{s^2}} \\[6pt]
&= 0.100 \times 9.80665 \times \mathrm{kg} \cdot \frac{\mathrm m}{\mathrm{s^2}} \\[6pt]
&= 0.98\ \frac{\mathrm {kg\ m}}{\mathrm{s^2}} \\[6pt]
&= 0.98\ \mathrm{N}
\end{align}$$
In forming products and quotients of unit symbols, the normal rules of algebraic multiplication or division apply.
For example, the expansion work $W$ at constant pressure $p = 100\,000\ \mathrm{Pa} = 100\,000\ \mathrm{kg\ m^{-1}\ s^{-2}}$ associated with a volume change of $\Delta V = 0.5\ \mathrm{m^3}$ is
$$\begin{align}
W &= p \cdot \Delta V \\[6pt]
&= 100\,000\ \frac{\mathrm{kg}}{\mathrm{m\ s^{2}}} \times 0.5\ \mathrm{m^3}\\[6pt]
&= 50\,000\ \frac{\mathrm{kg}}{\mathrm{m\ s^{2}}}\cdot\mathrm{m^3}\\[6pt]
&= 50\,000\ \frac{\mathrm{kg\ m^2}}{\mathrm{s^{2}}}\\[6pt]
&= 50\,000\ \mathrm J
\end{align}$$
Two or more quantities cannot be added or subtracted unless they belong to the same kind. The expression shall be written as the sum or difference of expressions for the quantities
$$l=12\ \mathrm m-7\ \mathrm m$$
or parentheses shall be used to combine the numerical values, placing the common unit symbol after the complete numerical value
$$l=\left(12-7\right)\ \mathrm m$$
but it is not permissible to write
$$l=12-7\ \mathrm m\quad\color{red}{\small\text{(wrong!)}}$$
For the same reason, quantities on each side of an equal sign in an equation must be of the same kind
$$\begin{align}
m_\text{total} &= m_1+m_2 \\
1.8\ \mathrm{kg} &= 1.5\ \mathrm{kg}+0.3\ \mathrm{kg}
\end{align}$$
However, quantities of the same kind do not necessarily have the same unit.
$$250\ \mathrm g = 0.250\ \mathrm{kg}$$
$$10\ \mathrm{m/s} = 36\ \mathrm{km/h}$$
Anyway, quantities on each side of an equal sign in an equation must not be of different kinds.
$$1\ \mathrm{mol} = 22.414\ \mathrm l\quad\color{red}{\small\text{(wrong!)}}$$

May I manipulate the equation as though $T$ and $B$ were the kinds of symbols that we learned to manipulate in math class?

Yes, you can and should (that is, never leave out the units when you plug values into an equation: a quantity is always the product of a numerical value and a unit, and leaving out units is asking for disaster). The set of rules that govern the algebraic manipulation of quantities is known as quantity calculus.
Among the several references that can be found about quantity calculus and quantity equations (others can be found in this answer on Academia.SE), I suggest you in particular this guide, §7.11, and the following paper, which is geared toward chemists:

M J ten Hoor, "Quantity calculus for chemists", Chemistry in action n. 57, 1999. Online

However, units are not variables (we strongly believe that units should be immutable quantities: whether they are actually so is a matter of scientific investigation), and to highlight this fact it is recommended to use an upright typeface to typeset the units, instead of the italic typeface reserved to variable quantities. So your quantity equations would be better written as
$$\frac{50\,\mathrm{T}}{20\,\mathrm{B}} = x.$$

Absolutely!
Having taught physics and planetary science for years now, this is how I've seen unit conversion taught most effectively.
Example
Suppose we wish to convert  423 feet to kilometers, where we know $3280.4 \; \text{ft} = 1 \; \text{km}$, or $\frac{1 \; \text{km} }{3280.4 \; \text{ft}} =1$:
$$\require{cancel} 423 \; \text{ft} = \left( 423 \; \text{ft} \right) \times 1 = \left( 423 \; \text{ft} \right) \times \left( \frac{1 \; \text{km} }{3280.4 \; \text{ft}} \right) = \frac{423 \; \text{ft} \; \text{km}}{3280.4 \; \text{ft}} = \frac{423}{3280.4}  \frac{\cancel{\text{ft}} \; \text{km}}{\cancel{\text{ft}}} \approx 0.129 \; \text{km} $$
There are a few catches, however...
First, radians are a pseudo-unit. Angular frequency ($\omega$) and regular frequency ($f$) are two closely related quantities: $\omega=2\pi f$, but they have different units. The units of angular frequency are radians per unit time (e.g. rad/sec), whereas the units of regular frequency are "per unit time" (e.g. [cycles, counts, etc.]/sec or Hz). Many scientists disregard radians as a unit and give angular and regular frequency the same units. This is acceptable, but you should be aware of the context, as an angular frequency of 20 Hz is not the same as an ordinary frequency of 20 Hz.
Second, some (particularly chemists) have an unfortunate tendency to be pretty loose with units in some situations—especially when it comes to calculating equilibrium constants and dimensional laws. For example, it is not permitted for the argument of a natural logarithm or exponential to have units. Many people calculate $\ln(P) = ...$ (as for the Clausius-Clapeyron equation), while it ought to be something like $\ln(\frac{P}{P_0}) = ...$, where $P_0$ is a standard pressure scale used to scale all pressures in an equation. In most cases, this scale factor cancels out in the equation, but if you are keeping track of units, this is essential. Because this factor cancels out, calculations can usually be done correctly without thinking about it, and many chemists simply choose to ignore it or are even unaware of it (including some professors I've known!). Some old textbooks still do this, too, so this sloppiness is prevalent. The bottom line is that understanding the context is important!
Third, this effectively only works for units that are directly proportional to each other. One example of this not working is when converting between degrees Fahrenheit and degrees Celsius and Kelvins. The units can (and should) still be treated as variables, but because these unit systems are not directly proportional to each other, special care must be taken with them. The conversion formulae must be used.

Radians should technically not be included as a unit because it is a ratio not a true unit.

@bon SI officially recognizes both the radian and steradian as "dimensionless derived units", but does effectively call them optional.

@bon: regardless of what one thinks "should be" a unit, nearly anything CAN be used as units, just like "counts" or "cycles" or "gross" or "dozen". They can be manipulated algebraically in the same way as ordinary units. That's what the original poster (OP) was asking about.

Another catch: degrees Celsius  & Fahrenheit behave oddly.

@bon: Using radians as a unit can help differentiate certain quantities. The one that comes to mind is angular frequency (often $\omega$), which has units of rad/sec, vs. ordinary frequency  (often $f$ or $\nu$), which has units of cycles or events (or some countable quantity) per second, though it is usually abbreviated as "per second", $\pu{s^{-1}}$, or Hz. Since they are related by $\omega = 2 \pi f$, the units (or context) help to distinguish which variable is given.


text:
How to determine the purity of lactic acid with titration?

$\pu{25 mL}$ of $\pu{0.5 \%(w/v)}$ lactic acid ($\ce{C3H6O3}$, molecular mass $\pu{90.1 g mol-1}$, $\mathrm{p}K_{\mathrm{a}} = 3.86$ was neutralised by $\pu{13.1 mL}$ of $\pu{0.1010 M}$ monobasic base. Estimate the end point $\mathrm{pH}$ and calculate the amount of $\ce{C3H6O3}$ in % of the initial one.

This question was part of my final paper. I can't remember the exact words used in the paper for the % part, but I think it is asking for the % purity. I am not sure how to do it.
I calculated the $\mathrm{pH}$ using the formula $\mathrm{pH} = 0.5 (\mathrm{p}K_\mathrm{a} - \log M)$ but I think it is wrong. How is this doen correctly?

At the end point, you have a solution of sodium lactate (weak base). The concentration of sodium lactate is given by the equation:
$$
C' =
\frac{C_\mathrm{b}\, V_\mathrm{b,eq}}{V_\mathrm{a} + V_\mathrm{b, eq}},
$$
where $C_\mathrm{b}$ is the concentration of the base, $V_\mathrm{b, eq}$ is the volume of the base at the end point and $V_\mathrm{a}$ is the volume of lactic acid.
$$
C' =
\frac{\pu{0.1010 mol L-1} \times \pu{13.1 mL}}{\pu{25 mL} + \pu{13.1 mL}}
 = \pu{0.03473 mol L-1}
$$
The $\ce{pH}$ of the solution at the end point is given by the equation:
\begin{align}
  \mathrm{pH}
    &= \frac{1}{2}(\mathrm{p}K_\mathrm{w} + \mathrm{p}K_\mathrm{a}
      - \mathrm{p}C')\\
    &= \frac{1}{2}(14 + 3.86 + \log 0.03473) = 8.20
\end{align}

To calculate the purity percentage of lactic acid, we calculate the amount of substance of pure lactic acid: At the end point
$$n_\mathrm{a} = n_\mathrm{b}.$$
This means:
$$
n_\mathrm{a}
  = C_\mathrm{b} V_\mathrm{b}
  = \pu{0.1010 mol L-1}\times \pu{13.1E-3 L}
  = \pu{0.001323 mol}
$$
The mass of pure lactic acid is:
$$
m
  = \pu{0.001323 mol} \times \pu{90.1 g mol-1}
  = \pu{0.1192 g}
$$
Or the mass of $\pu{0.5 \%(w/v)}$ lactic acid is:
$$
m'
  = \frac{\pu{0.5 \%(w/v)} \times \pu{25 mL}}{100\%} = \pu{0.125 g}
$$
The purity percentage of lactic acid $P\%$:
$$P\% = \frac{m\times 100\%}{m'}=95.36\%$$

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


text:
Estimate the concentration of HCl and volume of 6 M HCl required to complete precipitate each of the given salts
In our chemistry lab we performed the following experiment below.

Given that $K_\mathrm{sp}(\ce{AgCl}) = 8.2 \times 10^{-11}$ and $K_{\mathrm{sp}}(\ce{PbCl_{2}}) = 1.7 \times 10^{-5}$
Procedure:
  $0.2\,\mathrm{mL}$ each of $\mathrm{0.05\: M \: \ce{Ag+}}$ ions and $\mathrm{ 0.05\: M \: \ce{Pb^{2+}}}$ ions were added into a centrifuge tube. The solution was diluted to $1\,\mathrm{mL}$ $6\,\mathrm{M}$ $\ce{HCl}$ was added to the solution till it completely precipitates. The quantity of $\ce{HCl}$ required was recorded. The precipitate and supernatant were separated and kept for subsequent experiments.
Observations: Precipitate formation was observed almost instantly. It took 6 drops of $\ce{HCl}$ to completely precipitate the contents.
Estimate the concentration of $\ce{HCl}$ and volume of $6\,\mathrm{M}$ $\ce{HCl}$ required to complete precipitate each of the given salts.

Here is my approach:
\begin{align} c_{1}V_{1} &= c_{2}V_{2} \\
\text{Let } c_{1}=0.05\:\mathrm{M}, V_{1}&=0.2\:\mathrm{mL}, \text{and} \ V_{2}=1\:\mathrm{mL}\\
[\ce{Pb^2+}] &= 0.2 \cdot 0.05\ \mathrm{M} \\
[\ce{Ag+}] &= 0.2 \cdot 0.05\ \mathrm{M}\\
K_{\mathrm{sp}}\:(\ce{AgCl}) &= [\ce{Ag+}] [\ce{Cl-}] \\
8.2 \times 10^{-11} &= 0.2 \cdot 0.05\ \mathrm{M} \cdot s_{1} \\
s_{1} &= 8.2 \times 10^{-9}\ \mathrm{M}\\
K_{\mathrm{sp}}\:(\ce{PbCl_{2}}) &= [\ce{Pb^{2+}}] [\ce{Cl-}]^{2}\\
1.7 \times 10^{-5} &= 0.2 \cdot 0.05\ \mathrm{M} \cdot (s_{2})^2\\
s_{2} &= 0.041\ \mathrm{M} \end{align}
Thus to precipitate $\ce{AgCl}$ and $\ce{PbCl_{2}}$ completely, the concentration of chloride ions in solution must exceed $8.2 \times 10^{-9}$ M and $0.041$ M respectively).
I am not sure of my computations up till this point, and I need some help on how to proceed further.
EDIT: My calculations  tell me that volume of $6\,\mathrm{M}\,\ce{HCl}$ necessary is $0.00683\,\mathrm{mL}$. In the lab we have estimated the volume of one drop of our pipette as $0.037\,\mathrm{mL}$, based on this I calculated that the number of drops of hydrochloric acid necessary to complete precipitation is $0.18$ drops. But in the actual experiment I performed 6 drops of $\ce{HCl}$ were dispensed and based on the computations we did, everything should have precipitated. But after centrifugation and separation the supernatant still had lead ions which were precipitated as lead chromate. Why is this so? (This is the main reason why I suspected my calculations and turned to this website for help).
EDIT2: I now see the flaw in my computations. what I have calculated is in fact that amount of HCl necessary to start precipitation. Thanks to everyone for their help.

Does this count as a point?

Hi getafix, if you'd like more guidance on formatting be sure to check out our page here and here.

Your calculations are looking good. I didn’t check the numerical values, I’ll just assume you’re able to do your algebra and use a calculator correctly.
What you got are two different concentrations required to start to precipitate all the cations in your solution. However, you are asked to completely precipitate the cations. Well — that is a nigh impossible task unless you define a concentration threshold which you want to reach. Your threshold could be $1\,\mathrm{mM}$, $1\,\mathrm{µM}$, $1\,\mathrm{nM}$ or whichever you see fit. Use this desired concentration to recalculate the concentration of chloride ions required for both cations. Then continue on.
I recalculated the value in the spoiler tag for $1\,\mathrm{µM}$ residual cation concentration.

 $4.12\,\mathrm{M}$ chloride, equivalent to $0.687\,\mathrm{mL}$ or 18 drops.
Backwards calculation tells me, that you had a residual lead concentration of ca. $9.58\,\mathrm{µM}$, explaining your second result.


text:
Periodically oscillating chemical reactions?
I've seen before chemistry demonstrations where solutions are mixed with one another and subsequently where the resulting product goes through periodic color changes: for example blue to orange and back to blue again. The reaction seems to go on for awhile at a regular period.
My question - can this reaction be considered resonance in the sense that energy is flowing back and forth between two chemical states? In such reactions are there actually two chemical species between the color changes or just energy level/temperature differences?

Are you referring to a clock reaction? http://en.wikipedia.org/wiki/Iodine_clock_reaction

@LordStryker shame that the Wikipedia article does not cover multiple-cycle iodine clocks =C

@Jan You should edit it into the article!

@LordStryker If I knew what I needed to mix together I might. But that would involve me researching stuff first so it'll be low-priority on my to-do list ;)

A couple of points:
The demo is probably an oscillating chemical reaction, in which the concentration of a given species increases and decreases repeatedly over time. However, the energy of the overall system is dissipating as that is happening - not oscillating.
Resonance, on the other hand, is a (widely misunderstood) concept that applies to a single chemical species and not a bulk reaction mixture. Resonance does not consist of energy flowing back and forth between two chemical states, but rather is meant to represent a single energy state that is a hybrid between two (or more) classical Lewis structures. A double headed arrow between the two classical structures indicates resonance stabilization — mostly resulting from delocalization of electrons relative to the classical structures. The double headed arrow does not indicate oscillation between those two structures.

authors: ﺷﺮﻳﻒ حازم خلف


text:
Is solubility in Qsp affected by coefficient?
Related to my previous question: Is solubility coefficient affected if ion data is given in Ksp?

$200\,\mathrm{mL}$ solution of $0.02\,\mathrm{M}\,\ce{AgNO3}$  is added to $200\,\mathrm{mL}$ $\ce{CrO4^{2-}}$ and $\ce{PO4^3-}$ ions. Find out both of $Q_{\mathrm{sp}}$.

Actually the question is asking "Will it precipitate", but I skip it.
Here is my half approach:
$$200\,\mathrm{mL}=2\times 10^{-1}\,\mathrm{L}$$
$$V_{\mathrm{total}}=400\,\mathrm{mL}=4\times  10^{-1}\,\mathrm{L}$$
$$[\ce{Ag+}]=[\ce{CrO4^2-}]=[\ce{PO4^3-}]=\frac{0.02\,\mathrm{M}\cdot 2\times 10^{-1}\,\mathrm{L}}{4\times 10^{-1}\,\mathrm{L}}=0.01\,\mathrm{M}$$
$$\ce{Ag2CrO4 -> 2Ag^{+} +CrO4^{2-}}$$
$$Q_{\mathrm{sp}}(\ce{Ag2CrO4})=[\ce{Ag+}]^{2} \cdot [\ce{CrO4^2-}]$$
$$Q_{\mathrm{sp}}(\ce{Ag2CrO4})=[2s]^{2}[0.01\,\mathrm{M}]$$
I know that to find $K_{\mathrm{sp}}$, the $s$ needs to be multiplied (and to the power) with the coefficient from the reaction, but how about $Q_{\mathrm{sp}}$? Does the $s$ also need to be multiplied?
I asked this because my teacher said that only to the power is affecting the $Q_{\mathrm{sp}}$.

You don't need to multiply s, since $Q_\text{sp}$ is the result of multiplication of ions concentration (and you don't need to multiply with the coefficient in any $Q$). So the right way to answer $Q_\text{sp}$ is
$$Q_{\text{sp}}\:(\ce{Ag_{2}CrO_{4}})=[\ce{Ag^{+}}]^{2}[\ce{CrO_{4}^{2-}}]$$
$$Q_{\text{sp}}\:(\ce{Ag_{2}CrO_{4}})=[0.01]^{2}[0.01] = 10^{-6}$$

But Ksp is also the result of multiplication of ions concentration. What's the different between them?

K_sp and Q_sp differ in that K_sp is the solubility product at equilibrium (and Q_sp = K_sp at equilibrium). Q_sp is the solubility product when the system is not at equilibrium.


text:
Combustion of halo-organic compound
My question is based on the Preparatory Problem Set for the 47th IChO Problem 26 - Holy War against Four Horsemen of the Apocalypse (it is on page 36). It is written that cooling of Y combustion products from 150 to 0 degrees Celsius does not result in any condensed phase, and addition of silver chloride to the supernatant of this combustion product provide some precipitate, so I concluded that this organic compound have either (or both) chloride and bromide, since iodine is too large for this substance. After calculating some possible answers, I've concluded that compound Y is $\ce{C3H2BrClO2}$.
My question is, if I combust compound Y ($\ce{C3H2BrClO2}$), will I get some products that aren't condensed when I cool these combustion products?

I'm surprised that you don't get any water formed. Most organics give some water on combustion.


text:
Why does the gauche conformation of ethane-1,2-diol show hydrogen bonding?
Why does the gauche conformation of ethane-1,2-diol show hydrogen bonding even though there is no 6-membered ring system? If we don't consider one of the $\ce{H}$ (which is what is usually done), then there is only a 5-membered ring formed. How then does hydrogen bonding take place?

(Also, I feel that there must be some reason apart from 6-member ring for hydrogen bonding which I'm not aware of)

What's the deeper reason for you to go looking for a six-membered and only a six-membered ring system?

Ok thats what i've learnt from my teacher till now.If you know of any other reason please let me know!

How is it that your reputation never increases?

Cyclopentane has a relatively low strain energy, something on the order of 6.5 kcal/mol more than cyclohexane (reference).  In the case of ethane-1,2-diol, the strain energy in the 5-membered ring formed by hydrogen bonding will be even less because some of the $\ce{H-C-C-H}$ and $\ce{C-C-C-C}$ interactions that produce the strain in cyclopentane are no longer present; the oxygens don't have the hydrogens attached like a methylene group does and the hydrogen used in the hydrogen bond is also much smaller than a methylene group.
A typical $\ce{O-H..H}$ hydrogen bond can stabilize a system by ~7 kcal/mol (reference).  So to form a 5-membered ring in ethane-1,2-diol and create a hydrogen bond it will cost something less than 6.5 kcal/mol to overcome the strain energy in the ring, but we get 7 kcal/mol back due to hydrogen bond formation - a net stabilization.
$%edit$

Updated your first reference.


text:
Using Distribution Coefficient to Determine Mass and Percent of Acid Extracted
The $K_\mathrm{D}$ (distribution coefficient) of coffee water at $25\,°\mathrm{C}$ is $7.8$ in a methylene chloride/water system. If there is $2.0\,\mathrm{g}$ of caffeine in $200\,\mathrm{mL}$ of water, calculate the grams and the % of caffeine that can be extracted with a single $100\,\mathrm{mL}$ portion of methylene chloride.
My attempt:
$$\frac{x}{\frac{2\,\mathrm{g}}{200\,\mathrm{mL}}}=7.8$$
$$\frac{x}{0.01\,\mathrm{g\,mL^{-1}}}=7.8$$
$$x=0.078\,\mathrm{g\,mL^{-1}}$$
$$x=0.078\,\mathrm{g\,mL^{-1}}$$
$$0.078\,\mathrm{g\,mL^{-1}}\cdot 100\,\mathrm{mL}=7.8\,\mathrm{g}$$
But I think I'm wrong. Am I supposed to maybe use the molar mass of caffeine ($194\,\mathrm{g\,mol^{-1}}$)?

You got it almost right ;-)
The distribution coefficient gives the ratio of equilibrium concentrations in a biphasic mixture.
It is crucial to realize that a caffeine concentration of $2.0\,\mathrm{g}$ in $200\,\mathrm{mL}$ of water is the initial concentration, but not the concentration after a single extraction step! The latter will be lower, since a part of the caffeine went into the organic layer.
It doens't matter whether you express the concentrations in $\mathrm{g\cdot L^{-1}}$ or $\mathrm{mol\cdot L^{-1}}$ since $K_D$ is a dimensionless constant.
$$K_D = \frac{{[\mathrm{caffeine}]_{\ce{CH2Cl2}}}}{{[\mathrm{caffeine}]_{\ce{H2O}}}} = \frac{x}{10-x}\ \mathrm{g\cdot L^{-1}} = 7.8$$
$$x = 7.8\cdot (10-x)\,\mathrm{g\cdot L^{-1}} = \mathbf{8.86} \,\mathrm{g\cdot L^{-1}}$$
Note that $x$ is the concentration in $\mathrm{g\cdot L^{-1}}$. In order to calculate the absolute amount of caffeine in the organic layer or the remaining amount of caffeine in the water phase, you have to consider the volumes.
By the way, there's an article by G.S. Klebanov, L.N. Mednikova, and A.D. Ovcharova on this topic: Extraction of caffeine from aqueous solutions, published in
Pharmaceutical Chemistry Journal, 1967, 1, 221-223 (DOI).


text:
Determining phase from intermolecular interactions and kinetic energy
If the energy of intermolecular attractions are much less than the average kinetic energy of the molecules, what is the physical state of the substance?

Interesting that plasma is an option. Plasmas are actually strongly self-interacting, in a way that is hellish to describe mathematically; the interactions are long-ranged (electric force goes as $1/r^2$), there are strong higher order effects (magnetism) and there are both attractive and repulsive interactions between particles. No wonder plasma physics is a tough nut to crack. It's been keeping nuclear fusion power generation out of reach for decades!

Most likely gas.
In fact, the definition of an ideal gas is a gas in which its molecules have no intermolecular interactions (except elastic collisions).
As the strength of intermolecular attraction increases, the molecules will be brought closer together. Molecules in liquids are much more densely packed than molecules in gases.


text:
What is torsional strain?
I am having difficulty in understanding the concept of torsional strain. Can someone explain it to me in a more elaborate manner?
I mean why should bond restrict movement in eclipsed conformation? Doesn't that make it unstable! I tried this Wikipedia article but I'm still having difficulty in understanding the concept.

For the case of a single bond, consider a simple case like ethane, which can take an eclipsed conformation, a staggered conformation, or anything in between depending on the torsion angle of the C-C bond.  Each carbon has a region of especially high electron density where it's bonded to a hydrogen.  When the torsion angle is changed from staggered to eclipsed, these regions approach eachother and repel, and thus the eclipsed conformation is higher in energy than the staggered.
In the case of a double bond, as in ethylene, the energy associated with torsional strain is much higher because the P orbitals involved in the pi bond have to be oriented the same way.  Thus, applying torsional strain involves breaking this pi bond, and has a high energy.

Well, you don't have to break the $\pi$ bond to apply torsional strain. Consider biphenyl, where the lowest-energy dihedral angle is definitely not flat. I'd consider some torsional strain. There are even "torsional balance" molecules designed to measure this.

How energy is increased if nothing adds energy to the molecule? Also when in chemistry we speak about energy what is this energy? Potential or kinetic or both?


text:
Does 'Stoichiometry' refer to the mole ratio of atoms in compounds and the mole ratio of compounds involved in chemical reactions?
For my intro to chemistry class, I was asked to answer whether 'stoichiometry' refers to the mole ratio of atoms in compounds and the mole ratio of compounds involved in chemical reactions?
To the best of my knowledge those ratios are an essential part of stoichiometry, but stoichiometry is a discipline, and not a ratio, so the answer would be false.
Is that correct?

That's right -- stoichiometry is a field which uses these numbers and the relationships between reactants and products, but not the numbers itself. The mole ratio is just the mole ratio, whereas stoichiometry is the entire discipline of using mole ratios to calculate relative quantities of reactants and products in reactions.


text:
Configurations of Reactants in Second Order Nucleophilic Substitutions
The question asks me to draw the product for the reactions. However, I am stuck and do not understand how to draw the reactants with configurations correctly.
(a) (S)-2-Chloropentane and NaSH
I thought the reactant I drew was in S configuration. Also, I thought since it was a chirality center, inversion of configuration will occur. Can someone please explain this in detail?

I tried to pinpoint you in the right direction through comments, but apparantly I failed. The bottom line is that the molecule you drew and the one in the textbook answer are identical. Here’s why:

In the top row, you can see the (S)-2-chloropentane, which I have drawn as you have on the left, slowly rotating around an axis equivalent to the reaction arrows to give you the same molecule drawn in a different orientation.
Now, if you do an $\mathrm{S}_{\mathrm{N}}2$ reaction, all you need to do is replace the chlorine atom and invert the stereochemistry which I have done for both cases. You can now see that both sides are identical.
It does not (and must not) matter whether you draw the 2-chloropentane with the substituted carbon to the right or to the left. Both ways are identical. Usually, you will chose an orientation that is consistent with further or earlier steps of your synthesis, but if you only have one one reaction that point is moot. (It matters a lot more in total synthesis, where it helps to draw reactands in a way that show their conformation in the final molecule.) Furthermore, there is a IUPAC style guide out there somewhere, which probably says one of the two ways is to be preferred if no other reasons speak for the other … but that’s more relevant for journal publications, not for practising or for exam answers.
So just for total consistency’s sake: Everything you did was correct; just remember that molecules can be drawn with different orientations, and learn to turn them around before your inner eye.

I think I'm rather shaky in my understanding and so am totally relying on the textbook solutions to verify my understanding of the material, but now I feel a little more confident about being able to present the information in a different manner. Thank you for your thorough explanation, and clear diagrams!


text:
Method of deriving some properties or possible method of synthesis of a specific monomer
This is kind of like two questions but since they're related I have put them together.
Consider a monomer under a hopefully anatomically correct name 'Phenyl Carbolithia Divinylene Oxide [$\ce{C11H9LiO4}$] (For ease of understanding.)

Would it be possible for me to find or estimate some basic properties (physical and chemical)? Or the 3D structure of the molecule (As opposed to the one I plotted, which is based on my estimation of dipole with minimal reasoning.)
By separating the molecules into parts:

Lithonate (Lithium Carbonate)
Benzene (Phenyl group)
Ethylene (Vinylene group)
Divinyl Ether

Perhaps now it is possible to make some predictions based on some functional groups? (Such as ether.)
Though theoretical chemistry has shown that such achievement in predicting a physical structure is unlikely, I am wondering if anyone would know specific properties of this molecule and the way they have deduced it in which I haven't consider.
The second part of the question may lead to the solution for the first part, by synthesizing such molecule.
My humble attempt at theorizing the chemical synthesis with my limited knowledge on the subject is based on condensation polymer. Perhaps having the separate parts of molecules mention previously dehydrated to form a chain, the molecule may be created if a similar polymer is formed except to have the carbon-oxygen backbone chain to form a ring instead.
I also have a slight inquiry on whether it is possible for the double bond to exist in this monomer; and whether my nomenclature for this substance is anatomically correct.

Sorry, but organic naming doesn't work like that - this compound should be named as substituted heterocyclic compound

I doubt that your "monomer" is very stable. Orthoesters, $\ce{R'-C(OR)3}$, are hydrolyzed under acidic conditions: $\ce{R'-C(OR)3 + 2 H+ -> R'-COOR + 2 ROH}$. The fate of your monomer is:

Could further explain what is it that makes the electron the oxygen group bonding to carbon instead of the sole oxygen atom as an ion?

@čaritisio When $\ce{RO-}$ leaves the molecule, a benzoate (ester of benzoic ester) is formed. Here, the $\ce{C=O}$ double bond is in conjugation with the $\pi$-system of the benzene. This is definitely lower in energy and I wouldn't be astonished if the formation of the ester would happen spontaneously.

I agree with Klaus that your compound wouldn't be stable, although there could be some hope, if instead of lithium, you'd have alkyl group. Your compound is actually two molecules of ketene bound to lithium benzoate and even if you'd somehow create it, it would easily break down via mechanism starting with step presented in Klaus' answer. $\ce{RO-}$ is here enolate - good leaving group.


text:
Relevance and Determination of the reflection symmetry +/- in Spectroscopy?
If there are two electrons located in a shell (which contains two subshells) the molecule can either have a singlet state (if both electron are in the same subshell; because of the Pauli principle) or a triplett (if the two electrons are in the two different subshells).
The sign + or - means symmetric or antisymmetric with respect to reflection at a mirror plane that contains the molecular axis (for $\Sigma$ states only) of a diatomics.
Let's look how to determine if a state is of + or - symmetry:

$\mathbf{For~oxygen~in~the~ground~state:}$
$\Psi_{\mathrm{a}}^{\mathrm{R}}(q_1,1_2) = \sqrt{2}i\chi (z_1,\rho_1)\chi(z_2,\rho_2)\sin(\varphi_1-\varphi_2) \qquad \mathrm{for~the}~^3\sum~\mathrm{term}$
and
$\Psi_{\mathrm{s}}^{\mathrm{R}}(q_1,1_2) = \sqrt{2}\chi (z_1,\rho_1)\chi(z_2,\rho_2)\cos(\varphi_1-\varphi_2) \qquad \mathrm{for~the}~^1\sum~\mathrm{term}$

(image)
It's easy to determine if a state is of +/- symmetry if we know the mathematical description of the radial wave function $\Psi^R (q_1 ,q_2)$ (with the two coordinates $q_i$ of the two electrons), because exchanging $\varphi_2$ with $\varphi_1$ in $cos(\varphi_1-\varphi_2)$ part stays the same and therefore + and for the $sin(\varphi_1-\varphi_2)$ (function the sign changes and therefore we get - symmetry.
What is my question now?
I've some troubles to determine the +/- symmetry of a state (of a diatomics for light atoms and therefore only for s and p orbitals) if I have no mathematical description or tabular data given.
I guess it should be possible to directly derive the +/- symmetry by knowing the electronic configuration.

A specific example of oxygen.

I read that in the two electrons in a subshell (that means the same molecular orbital shell and the same spatial orbital) of either $\lambda$ = +1 or -1 and therefore a singlet gives a state with + symmetry.
I'm wondering now: Has a singlet state always + symmetry and a triplet state always - symmetry? I'mn interested to understand why this is the case. I guess this has something to do with spatial and spin part.
I don't know if I'm on the right track, but I think for the ground state two electrons in the same spatial orbital (=same subshell) must have opposite spins (because of Pauli principle) and I guess this means having antisymmetric spin-part. Because the product of spin and spatial part must be antisymmetric I expect the spatial part to be symmetric, what is labeled with g(from german gerade). But I guess this has nothing to do with +/-.
For the first excited state of oxygen [electronic configuration ...$(1\Pi_u)^3(1\Pi^*_g)^3$] there exists six states: $^1{\Sigma} ^{-}_u$,$^1{\Sigma} ^{+}_u$,$^3{\Sigma} ^{-}_u$,$^3{\Sigma} ^{+}_u$,$^1{\Delta} _u$,$^3{\Delta} _u$.
This is also one case where my rule singlet gives + symmetry not work. I'm now thinking in which case we have $^1{\Sigma} ^{-}_u$ and in which case we have $^1{\Sigma} ^{+}_u$. The two electrons have opposite spin. One electron has $\lambda_1=+1$ and the other has $\lambda_2=-1$. I guess that the sign +/- depends if the electron with $\lambda_1=+1$ is in a $(1\Pi_u)$ or in a $(1\Pi^*_g)$ orbital.

Symmetry properties from a table.

Instead of determining the + or - symmetry we can use tables which shows the possible terms for a specific electronic configuration:

$\pi_g \otimes \pi_g = \sigma_g^- \oplus \sigma_g^+ \oplus \delta_g$

(image)
This relation shows that two electrons both in $\pi$ orbitals results in these three different states. Sometimes this terms are written in upper case letters, sometimes they are written in lower case letters. This seems to be arbitrarily choosen.

Practical relevance of the +/- symmetry in Spectroscopy.

In Spectroscopy there are two important selection rules: The Parity selection rule and the Angular momentum selection rule.
I  read that a transition between two states must undergo a change of the parity from + to - and vise versa. This is the Parity selection rule. The +/- is the basis for the parity selection rule.
Therefore I would expect that the g/u symmetry (symmetric/antisymmetric with respect to an inversion on the centre-of-mass) is the basis of the Angular Momentum selection rule.
Is this true?


text:
What solvents can get rid of oil stains?
What kind of solvent would you use to remove an oil based stain?
I need help for a chemistry lab report and I tried to look it up but nothing seems to be helping me

My suggestion would be to look for dry-cleaning solvents...

I use lipase to do the job

Depending on where you live different dry-cleaning solvents are indeed the best choice. Perchloroethylene $\ce{Cl2C=CCl2}$ is probably still the easiest one to get.


text:
Which of the following releases the smallest amount of energy upon gaining an electron?
$\ce{C}$, $\ce{Rb}$, $\ce{Ga}$, $\ce{Ge}$
I am confused in the sense that I don't know what it means by "released the smallest about of energy upon gaining?" would it be $\ce{Rb}$ because the shell is so big that when it loses an electron the overall energy doesn't really matter?

Alright, this is similar to a question that I have answered before but I will just give you the reasoning for this phenomenon. I will not give you the answer since this appears to be a homework question.
Why an atom gains energy when you take off an electron?
Well for this, think of the attractions between the nucleus and the electrons, so when you must strip away an electron from this stability, you must add energy to do so.
Why an atom releases energy when you add an electron?
This one might be a little hard to grasp, but at the end it is not too bad. Essentially the molecule is becoming more stable. It is getting near to having a full octet. (Ignoring noble gasses). In other words the kinetic energy added by gaining the electron is released when the atom is achieving a stable state.
This is the general reasoning behind this, not too complicated, but I hope you will be able to understand it. Just, if you can't, remember the first one and think of the opposite for the second one.

Now, take a look at this question, to help you with your question even more. You will then be able to deduce the correct answer.

Just to add clarity: It’s not only noble gases that gain energy if an electron is added. (You knew that and it’s not relevant to the asker’s specific question but before any confusion of any kind arises anywhere ;).)

Ok, alright. :)


text:
What are some general tests that would determine an unknown compound?
I know a little bit about unknown compounds but I am still new to the topic here. I know that you could use filtration, chromatography and distillation. You can also combine the unknown with others to form a compound. And you can measure properties and such and such to figure out what the compound is.
If I were to do this, what should I watch out for, and what would be the best way to proceed? I would appreciate it if you don't go "PhD" on me.
This question will be ambiguous for most people, but I am looking for a general idea and what you would normally do, instead of "what I would do in this situation".

Normally you use some combination of NMR, UV-VIS and mass spectroscopy.

If its an inorganic compound, one can use the now outdated method of Inorganic Qualitative Analysis. This is especially useful if you're not familiar with spectroscopy (which I am not). However keep in mind that this method has many limitations.

As usual, it depends on what is available.
Mass spectrometry, $^{1}\ce{H}$ and $^{13}\ce{C}$ NMR, IR, UV/VIS spectroscopy, would be obvious choices in organic chemistry nowadays. XRD of a suitable crystal would be nice to have!
Back in the days, (undergrad) laboratories didn't have sophisticated analytical instruments and you were happy to find a melting point apparatus, a refractometer and an IR spectrometer.
You would have used IR spectroscopy to identify funcional groups, possibly supported your hypothesis by using staining reagents for TLC, like Dragendorff's reagent and then started to synthesize derivatives from which you knew that they show sharp, well-defined melting points.
You would have converted

amines to sulfonamides
carbohydrates to osazones
ketones and aldehydes to 2,4-dinitrophenylhydrazones or semicarbazones

and compared the melting points of the derivatives with tabulated values. Those were the days…

Your last comment reads almost like you would be sorry about that we have advanced  O.o

@Martin-マーチン No, not at all. I don't miss the 60 MHz Varian NMR (CW, ink plotter). And remembering that one had to measure the refractive index of rather volatile acid chlorides almost brings back the tears - but those were not the tears of joy ;-) But it is still remarkable how much was achieved in natural products chemistry and structure elucidation in those early days.

Is anything about the material known?  And how much of it do you have?
If you think the material might be inorganic, an adsorption, emission, or mass spectroscopy might be the first step, as that will tell you what elements are in it.  If the sample is large enough, and can be formed into crystals, then x-ray diffraction will likely give a thorough idea of the structure.
If it's organic, NMR is probably the most powerful technique, and only requires a minuscule amount of the sample.  Mass-spec and IR-spec are also commonly used.


text:
Liquid-Liquid phase separation
Consider a container that contains a mixture of many liquids. Can two of these liquids phase separate without having any significant effect on other liquids inside the container? Is there a feasible scenario where the above can actually happen? I am interested in a scenario where the two liquids only separates themselves from each other and not the rest of the mixture.
More precisely:
Assume the container contains a mixture of liquids A and B with many other liquids (state 1). Can these two liquids separate (triggered by change in the pressure, temperature, ...) so that at the end (state 2) the liquids in the container are divided into two regions with the following properties:

The first region does not contain liquid B. It contains liquid A with the exact same mixture of the other liquid (same as state 1).
The second region does not contain liquid A. It contains liquid B with the same mixture of other liquids (same as state 1).

Indeed we want other liquids in the container to be neutral to the phase separation of the liquids A and B.

I don't think what you are asking for is thermodynamically impossible, but I think it is very unlikely.  Let's say you start off with a single liquid phase containing molecules A, B, X, Y, and Z.
Let's also say that when temperature changes, two phases form.  One phase is mostly A and the other phase is mostly B.  Let's call those two phases $a$ and $b$.  Your question is, is there some condition where the concentrations of molecules X, Y, and Z in phase $a$ is the same as it is in phase $b$.
In general the distribution of a solute between two phases is governed by a parameter called a partition coefficient or a distribution constant.  The distribution of molecule X in phase $a$ vs phase $b$, for example, could be noted as $K_{ab, X}$.  For your condition to be satisified, the distribution coefficient of X would have to be 1.0, i.e. it would have an equal preference for both phases.  The same would be true of molecules Y and Z, i.e. $K_{ab, X}=K_{ab, Y}=K_{ab, Z}=1$ is required for your condition.
A common type of liquid phase separation is oil and water: they don't mix.  I can't think of three molecules that each have an equal preference for water vs. oil.  These molecules would be somewhat hydrophobic and somewhat hydrophilic.  Feel free to look through a giant compendium of octanol-water partition coefficients published by NIST to look for three molecules X, Y, and Z that all have octanol-water partition coefficients of 1.0 by yourself, though.

Separation can be done in two ways:

Natural:
Natural separation term is not a good one but i'm gonna use it here. If at equilibrium (When the free energy of all the chemical constituents are same), those two specific liquid like to separate themselves out from the rest of the mixture, than you just need to give it enough time to reach equilibrium to separate them out. And this is the only method i can think of which will have no significant effect on the liquid in that container.
Forced:
Forced separation: Now if the above mentioned equilibrium condition needs different temp. and pressure than room temp. and pressure. Then by simply changing temp. and pressure you can achieve the above separation. Or you can utilize the different in various properties to separate them, e.g. Viscosity, boiling point, melting point etc.
Distillation: If they have separate boiling point than you can use distillation column and collecting the liquid at various height will give you separated solvent.
Solvent extraction, Filtration, mixer settlers, centrifugal force, ion exchange, membrane filtration, using bio separation like algae etc you can also achieve liquid liquid extraction.

I am interested in a scenario where the two liquids only separates themselves from each other and not the rest of the mixture. Is it also possible?

I tried to answer the above comment but couldn't due to character limitation. So, i'm posting it here as another answer:
To answer your question we need more data. For example if we want to separate mixture of sand and sugar, we can utilize the solubility difference of these two substances. To separate water and salt, we use heat to vaporize water but that doesn't mean salt is non-volatile. Salt can be vaporized but at very high temp. (As far i can remember mp of NaCl is around 800 deg C and bp is 1500 deg C. So, if you use temp. greater than 1500 deg C to separate water and NaCl, then you will end up with a mixture of vapor of water and salt. So, in room temp. your desired separation will happen or not or do we need to exploit any physical properties, we need to know their properties. If you know the mixture components, than use CRC or Perry's handbook to find out their behavior. Even you can also find multiple component phase diagram, which will tell you everything about the mixture. Or if their is no data available, you can use software like HYSYS to exploit the QSPR.

I revised the question. Please note that I am interested in liquid-liquid phase separation.


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Does the carbocation rearrangement in a SN1 reaction specify a change in degree?
I am asked to determine whether the following process is likely to involve a carbocation rearrangement or not:

I think it is possible through the rearrangement given above (though the rearrangement cannot become a secondary or primary carbocation). Can someone please clarify if the carbocation rearrangement specifies a change in degree?

I feel your wording is a bit unclear. Do you mean to ask that whether this rearrangement is possible and whether it will be called a carbocation rearrangement even though there is no change in the degree of the carbocation?

How do you say the rearrangement cannot become a secondary or primary carbocation? And as Binary Geek there is no change inthe degree of the carbocation provide sime clarity. By what reference your thinking it is possible ?

What I get of your question, the process you have made is not possible. It will help if you c;arify.

Try to draw the mechanism for the second transformation and you'll have your answer.

Yes @BinaryGeek I mean to ask whether it will be called a carbocation rearrangement even though it doesn't become a primary or secondary carbocation, and instead remains a tertiary carbocation.

@imaginov I suggest you to add your last comment into your question to clarify it.

@jerepierre That's why I added ~ ? to the arrow when I replaced the drawing ;-)

The most common carbocation rearrangements involve [1,2] hydride or alkyl shifts. As a typical result a secondary carbocation is converted to a more stabilized tertiary carbocation.
In the course of these reactions, the migrating group and the positive charge change place, they move to opposite directions.
This is not the case in your example.
I fail to see how your transformation can be described as a cationic rearrangement!

Looks like the OP added some info  in the comments after you answered the question. It would be good if you do so :)


text:
Differentiation between zinc, aluminium, and magnesium ions in solution
If I have three aqueous ionic solutions in which I know that the cation is $\ce{Al^3+}$, $\ce{Mg^2+}$, or $\ce{Zn^2+}$, how do I find out which is which?
I was thinking to add $\ce{OH-}$ in the form of $\ce{NaOH}$, or something else to produce a precipitate, and testing these precipitates in other liquids, but I'm not sure what other liquids to try.

First, you add sodium hydroxide to the three different solution. (Remember to safe some of the solution, because we need it for the next test.)
$\ce{Zn^2+}\text{ ion}$
The zinc ion will react with the hydroxide ion in the sodium hydroxide solution to form zinc hydroxide which is white. So, you see white precipitate.
$$\ce{Zn^2+ +2 OH^- -> Zn(OH)2}$$
Adding more sodium hydroxide, the zinc hydroxide will form the zincate ion which is colourless. So, you will see a clear solution.
$$\ce{Zn(OH)2 + 2 OH^- -> Zn(OH)4^2-}$$
$\ce{Al^3+}\text{ ion}$
The aluminium ion will react with the hydroxide ion in the sodium hydroxide solution to form aluminium hydroxide which is white. So, you see white precipitate.
$$\ce{Al^3+ +3 OH^- -> Al(OH)3}$$
Adding more sodium hydroxide, the aluminium hydroxide will form aluminate ion which is colourless. So, you will see a clear solution.
$$\ce{Al(OH)3 + OH^- -> Al(OH)4^-}$$
$\ce{Mg^2+}\text{ ion}$
The magnesium ion will react with the hydroxide ion in the sodium hydroxide solution to form magnesium hydroxide which is white. So, you see white precipitate.
$$\ce{Mg^2+ +2 OH^- -> Mg(OH)2}$$
However, in this case, it won't form any else by adding more sodium hydroxide. So, you will still see the white precipitate even though you have continued to add sodium hydroxide.
We can differentiate $\ce{Mg^2+}$ solution form the three solutions and now we are left with $\ce{Zn^2+}$ and $\ce{Al^3+}$. Take another sample from the remaining two solutions and add aqueous ammonia solution ($\ce{NH3}$) in the samples.
$\ce{Zn^2+}\text{ ion}$
The zinc ion will react with the ammonia solution to form zinc hydroxide which is white. So, you see white precipitate.
$$\ce{Zn^2+ + 2NH3 + 2H2O <=> Zn(OH)2 + 2NH4^+}$$
Adding more ammonia solution, the zinc hydroxide will form  tetraamminezinc(II) ion which is colourless. So, you will see a clear solution.
$$\ce{Zn(OH)2 + 4NH3 -> [Zn(NH3)4]^2+ + 2OH^-}$$
$\ce{Al^3+}\text{ ion}$
The aluminium ion will react with the ammonia solution to form aluminium hydroxide which is white. So, you see white precipitate.
$$\ce{Al^3+ + 3NH3 + 3H2O <=> Al(OH)3 + 3NH4^+}$$
However, the aluminium hydroxide will not further react with ammonia. So, you will still see the white precipitate even though you have continued to add ammonia.
In conclusion, first put sodium hydroxide in the three solutions (sample in test tube) and keep adding it until only one of the solutions still has white precipitate. That solution is  $\ce{Mg^2+}$. Next, take the remaining two solutions (sample in test tube) and  keep adding it until one of the solution still has white precipitate. That solution is $\ce{Al^3+}$. And finally the last solution is $\ce{Zn^2+}$.

Why not use potassium ferrocyanide for zinc... bluish white ppt (in acidic medium)?

You can use another way if you want to

Apart from the methods, Ringo already described, you can do a few other tests.
Aluminium
This is loosely translated from the German chemgapedia.de. Look at the pretty pictures.
Probably the easiest test you can do is reacting it with Morin in ethanoic acidic medium. It forms a yellow-green chelate complex, which has strong fluorescence under UV light. The colour and UV activity should be persistent when adding glacial acidic acid (water free), but vanish when adding concentrated hydrochloric acid.

Unfortunately sodium is also fluorescent, so keep your equipment clean and if you use a base, use potassium hydroxide instead. Also perform blind tests with all the chemicals involved, to see if they are clean enough.
Other elements that can show fluorescence are $\ce{In^3+, Ga^3+, Sc^3+, Zr^4+, Be^2+}$, usually very pH dependent.
From my own experience in first year of university, this is a very efficient test. Another test, that can be performed from the salt, is reacting it to Cobalt blue. If you have a solution, you have to precipitate aluminium first: $$\ce{Al^3+ + 3 {}^{-}OH ->C[NH3/NH4Cl] Al(OH)3 v}$$
Put the precipitate on a magnesia groove and add some cobalt nitrate. Glowing the mixture will result in the characteristic cobalt blue:
\begin{align}
\ce{2 Al(OH)3 &->[\Delta] Al2O3 + 3 H2O ^}\\
\ce{Co(NO3)2 &->[\Delta] CoO + 2NO2 ^ + 1/2 O2 ^}\\
\ce{CoO + Al2O3 &->[\Delta] \color{\navy}{CoAl2O4}}\\
\end{align}
This is an almost fool proof method for aluminium and is not influenced by other elements. And it looks quite nice.
Zinc
Zinc is probably one of the only elements that form a white precipitate with sulfide ions.
$$\ce{Zn^2+ + H2S -> ZnS v + 2 H+}$$
This precipitate can be used to form Cobalt green. The procedure is the same as for the above Cobalt blue.
$$\ce{ZnS + Co(NO3)2 + O2 ->[\Delta] \color{\green}{ZnCo2O4} + 4 NO2 ^ + SO2 ^ }$$
The only interference with this test is, that using too much cobalt nitrate will produce the black cobalt oxide, which can make it hard to see the green colour.
$$\ce{3Co(NO3)2 ->[\Delta] \mathbf{Co3O4} + 6 NO2 ^ + O2 ^}$$
Again you can have a look at the pretty pictures at chemgapedia.de.
Magnesium
A common problem with magnesium is, that all tests only work well and specific if you have a pure solution. Most heavier elements react more readily with the chemicals, than magnesium itself does. Another problem is, that salts from magnesium have no flame colouration at all.
I am again loosely translating from chemgapedia.de.*
You can precipitate magnesium with 8-hydroxyquinoline:

It forms a yellow-green complex. This is only a specific test, if you know you have no heavier elements in your test tube. For example, aluminium forms the complex Tris(8-hydroxyquinolinato)aluminium: $$\ce{Al^3+ + 3 C9H6NOH -> Al(C9H6NO)3 + 3 H+}$$
Another possibility is precipitation as magnesium ammonium phosphate (struvite). And yes, you guessed it, this happens with all other alkali and earth alkali elements (except lithium). But if you are certain you only have got magnesium, you can grow nice crystals. The pH for this reaction should be buffered at around 8-9 with ammonia/ ammonium chloride.
$$\ce{Mg^2+ + NH4+ + {}^{-}OH + HPO4^2- ->C[H2O][NH3/NH4Cl] Mg(NH4)PO4⋅6H2O v}$$
There are two more tests, that I would consider best suitable, because of the pretty colour.
The reaction of magnesium hydroxide with titan yellow in presence of sodium hydroxide will produce a red precipitate. It is important to do a control/ blind experiment, as older solutions of titan yellow will form a precipitate with sodium hydroxide alone.
The reaction of magnesium hydroxide with azo violet (Magneson) or Magneson II (Magnezon II) in presence of sodium hydroxide will produce a blue precipitate. Again test with sodium hydroxide to observe the violet solution. This reaction also occurs with $\ce{Ca^2+, Al^3+}$.
Testing for magnesium is as far as I remember is some kind of a lottery when in combination with other elements (qualitative analysis). Good luck.

* Please note that the first reaction equation has depicted hydroxyquinoline with two circles for the aromatic system, which is wrong. In benzene one circle stands for 6 $\pi$ electrons. So that would lead to hydroxyquinoline having 12 $\pi$ electrons, two too many.

$\ce{Al^3+}$

The chemical Aluminion can be added to a solution to test for the presence of $\ce{Al^3+}$ ions. This is done by first adding $\ce{NaOH}$ to the solution to precipitate out $\ce{Al(OH)3}$. In the presence of Aluminon, the $\ce{Al(OH)3}$ forms a lake dye through chemisorption, causing a pinkish red precipitate to form. Only $\ce{Be^{2+}}$ cations and $\ce{Fe^{3+}}$ cations show a similar reaction, so there is no need to worry about misidentifying in your case.

$\ce{Mg^2+}$ and $\ce{Zn^2+}$

The flame test is easy enough to perform! Dip a splint into the solution of interest, place it under a bunsen burner flame, and observe the color. Different metals produce different color light when excited under flame, allowing our mystery metal to be identified. $\ce{Mg}$ produces a bright white light, while $\ce{Zn}$ produces a blue-green to pale-green color.


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Why do Grignard reagents or Organolithium compounds not displace halides as easily as acetylide ions do?
This is a snapshot from LG Wade Organic Chemistry. I have a doubt with a particular statement in the book. It says that Grignard reagents do not displace halides as easily as acetylide ions do.
But shouldn’t it be the other way round? Acetylide ions, having an sp-hybridised carbon, are much more electronegative compared to the sp³-hybridised carbon in Grignard reagents. So acetylide ions should have less tendency to displace halide ions by attacking with their lone pair compared to the alkyl group in Grignard reagent. Shouldn’t it?
Please help me sort this out!

According to you, the sp carbanion is more stable and less reactive ?

Then your statement is contradicting the book's one.I request you to elaborate. @user97554

An acetylide ion proceeds in a nucleophilic substitution pathway (SN2 in the picture due to primary halide). The substitution involves the attack of a highly nucleophilic sp-carbon on a highly electrophilic primary carbon. The enthalpy of the C-C bond formation is relatively low due the pKa of the nucleophile being ~ 25. The ion is treated as a "soft" nucleophile that prefers to substitute alkyl halides instead of acting as a base and conducting an elimination reaction.
A grignard reagent is essentially a highly polarized bond between Carbon and the Bromide(Halide)/Magnesium complex. The electronegative difference between Carbon and Magnesium is 1.3 (http://www.thecatalyst.org/electabl.html <-- Source of values). The Carbon atom in a Gringnard is treated as a carbonanion. This carbonanion characterization is highly unstable in nature and treated as a "hard" nucleophile with pKa of ~ 50. The highly basic character of a Grignard reagent often results in an elimination reaction or no reaction at all. The transition state to substitute the alkyl halide is less stable than the Magnesium/Bromide(Halide) complex. This is due to a ligation formation between the solvent and the Magnesium atom. The solvent (ex. diethyl ether) shares non-bonding electrons with the cation and creates a stable electronic atmosphere with a lower energy than a possible C-C bond with the alkyl halide.
Substitution isn't preferred with a hard nucleophile and therefore, the acetylide ion is more likely to preform a nucleophilic substitution reaction with an alkyl halide. Note that the electronegative difference between Carbon and Magnesium supersedes the hybridization of the sp or "triple" bond complex. The bond is so high in energy that the nucleophilic character of the grignard far surpasses the electronegative character of the sp-hybridization of the acetylide ion.


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Determining (approximately) the amount of a strong base needed to change a buffer's pH by a certain number of units
I'm measuring buffer capacity by titrating $\ce{NaOH}$ into a buffer of $\ce {H3PO4 + NaH2PO4}$. I'm measuring buffer capacity by recording the amount of $\ce{NaOH}$ required to increase the $\mathrm{pH}$ of the buffer by 2 units, and then taking the ratio.
My problem is that I have no idea how much $\ce{NaOH}$ I'll need to change the $\mathrm{pH}$ by that much. If I've underestimated the strength of my buffer, I might have to reduce the drop in $\mathrm{pH}$ to just one unit. I need to have a rough idea so I can organize how much of each I'll need to prepare.
Basically, is there any way I can predict (roughly) how much $\ce{NaOH}$ I'll need?
Details:

$1\:\mathrm{M}$ $\ce{H_3PO_4}$ and $1\:\mathrm{M}$ $\ce{NaH_2PO_4}$
Buffer volume: $50\:\mathrm{mL}$
$1\:\mathrm{M}$ $\ce{NaOH}$

If you can calculate the theoretical buffer capacity, you might be able to come to an approximation. But theoretical capacities usually differ from the experimental ones.

How would I go about that though? I can get the pH with the Henderson-Hasselbalch equation, but where do I go from there?

You can calculate the quantity of sodium hydroxide, required to change the $\mathrm{pH}$ of your system by two unites,  by calculating the concentration of sodium ion in the charge balance equation:
$$\ce{[Na+] +[H+]=[OH- ] +[H2PO4^{-}]} \,\,\,\,(1)$$
In fact, you know the initial concentration of sodium ion in the buffer system before adding sodium hydroxide (from the known concentration of $\ce{NaH2PO4}$), and you calculate the final concentration of sodium ion  after adding sodium hydroxide by the above equation. So you can deduce the required quantity  of sodium hydroxide.
As for the  other ions in the charge balance equation, they are calculated as follows:
At first the  $\mathrm{pH}$ of the buffer system $\mathrm{pH=pK}_a=2.15$.

If you want to change the $\mathrm{pH}$ of your system by two unites, then the final $\mathrm{pH}$: $$\mathrm{pH}= 4.15\,\,\,\,(2)$$
Now you can estimate the concentrations of ion hydronium and ion hydroxide.

The use of Henderson-Hasselbalch equation gives: $$\ce{[H2PO4^- ]= 100[H3PO4 ]}\,\,\,\,(3)$$

The equation of matter conservation gives: $$\ce{[H2PO4^- ] + [H3PO4 ]= C_0=1 \mathrm{M}}$$ As the buffer is $25\,\mathrm{mL}$ of $1\,\mathrm{M}$ of each of the components, so there's a total of $1\,\mathrm{M}$ in the $50\,\mathrm{mL}$ solution due to dilution (not as you mentioned in your comment).
Now, you can calculate the concentration of ion $\ce{H2PO4^{-}}$
from equation (3) and (4):$$\ce{[H2PO4^- ] = \frac{100}{101}\mathrm{M}}$$

If we substitute the numerical values in equation (1):
$$\ce{[Na+] +10^{-4.15}=\frac{10^{-14}}{10^{-4.15}} +\frac{100}{101}} $$
$$\ce{[Na+]_0 + [Na+]_{\mathrm{added}} \approx 10^{-4.15} + \frac{10^{-14}}{10^{-4.15}} +\frac{100}{101}} $$
$$0.50 +\ce{ [Na+]_{\mathrm{added}} \approx 10^{-4.15} + \frac{10^{-14}}{10^{-4.15}} +\frac{100}{101}} $$
$$\ce{ [Na+]_{\mathrm{added}}} \approx 0.4902 \mathrm{M} $$
$$n=\ce{  [Na+]_{\mathrm{added}}} \times V_{\mathrm{sol}}= 0.4902\times 50\times 10^{-3} =24.51 \times 10^{-3} \mathrm{mol} $$
The volume of sodium hydroxide is $$V_{\mathrm{added}} = 24.51 \times 10^{-3}\times 1= 24.51 \times 10^{-3} \mathrm{L} $$
I suggest to use a  solution of sodium hydroxide more concentrated, so as to neglect the effect of the added volume on the initial concentrations of the buffer system.

I think I understand what you're getting at, but I'm still really confused. I understand all of the theory individually, but how do I actually use it together solve for the number of moles of NaOH I'll need? By the way, the buffer is 25mL of 1M of each of the components, so there's a total of 2M in the 50mL solution.

Hey, just want to say thanks. I did the experiment today and your numbers were spot on.

You have a phosphate system so compute
$$r_1 = 10^{(pH - pK_1)}$$
$$r_2 = 10^{(pH - pK_2)}$$
$$r_3 = 10^{(pH - pK_3)}$$
$$f_0 = 1/(1 + r_1 + r_1*r_2 + r_1*r_2*r_3)$$
$$f_1 = f_0 r_1$$
$$f_2 = f_1 r_2$$
$$f_3 = f_2 r_3$$
$$Q = (f_1 + 2 f_2 + 3 f_3)$$
Do this for closely spaced values of pH near your buffer design pH. Obviously, use a spreadsheet, MatLab, Igor or some other program.
If the total phosphate in your buffer is 1 mole then $f_0$ is the number of moles of phosphoric acid, $f_1$  the moles of $\ce{H2(PO4)-}$, $f_2$ the moles of $\ce{H2PO4^{-2}}$ and $f_3$ the moles of  $\ce{PO4^{-3}}$. $Q$ represents the moles of negative charge on the phosphate species. If you add acid (protons) the balance between the 4 phosphate species shift and so does $Q$. The buffering capacity at any pH is the change in $Q$ corresponding to a small change in pH. If you differentiate $Q$ with respect to pH you will obtain a curve of the buffering capacity in units of $Eq·pH^-1·mol^{-1}$. It will exhibit three peaks at the three pK's where the buffering is maximum. Multiply this curve by the total moles of $\ce(PO4)$ you put into the buffer and you have Eq/pH. The Eq of acid to be added to the buffer for a shift of ${\Delta pH}$, when that is small is ${\Delta pHEq/pH}$.The same amount of base will shift the pH by the same amount but upwards.
For a larger pH shift the amount of acid or base required to go from $pH_1$ to $pH_2 is $Q(pH_2) - Q(pH_1)$$.
If $pH_2 > pH_1$ then $Q(pH_2) > Q(pH_1)$ and $Q(pH_2) - Q(pH_1) > 0$. If,for example, this is the case and you find  $Q(pH_2) - Q(pH_1) = .010$ Eq (10 mEq) that means that you would need to add 10 mL of 1 N  $\ce{NaOH}$ or 1 mL of 10 N.

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Available orbitals for promotion
I am under the impression that elements from the third period onwards can promote electrons into d orbitals to form hybridised orbitals. Why would an element in the third period promote an electron to the 3d orbital when we learn that 4s orbitals are filled first because they are lower in energy (is it to do with orbital penetration?)
Why also can a 2nd period p block element not promote to 3s?

There is a considerable amount of debate as to whether d-orbitals of main group elements (I assume we’re talking about them here; the discussion is rather pointless for d-block elements) are used in hybridisation or not.
Originally, the octet expansion, a.k.a. using them, was postulated by Linus Pauling because he was unhappy to use charge separation to describe molecules such as sulphur trioxide. However, it is entirely possible to draw Lewis structures that do not violate the octet rule and hence do not require d-orbital hybridisation for practically every case. There may be more mesomery required in some cases and less in others.
I’m not too good at repeating the arguments of supporters of main group elements’ d-orbital hybridisation theory, because I belong to the other group. The main argument against d-orbital participation is, as you noticed, that they are far removed on an energy scale; even behind the next s-orbital (3s, 3p, 4s, 3d) — hence the order of elements in the periodic table. I also have troubles understanding why phosphorous would only allow one d-orbital to hybridise, sulphur two and chlorine and the heavier halogens three. This makes no sense to me. (It seems to make sense for others, though.)
In my opinion you are absolutely correct in asking that if a 3p-element can use 3d orbitals, why should it not use 4s for hybridisation, and consequently why 2p cannot use 3s. Assuming 3d orbitals to be used, I could not answer your question satisfactorily.