text:
Why ground-state configuration is consistent ("additive") from one element to the next?
Ground-state configuration ("electron shells") is consistent throughout the periodic table, for example:
(source)
Why is it this way if the Hamiltonian of each atom is different? In other words, why Al is $\mathrm{[Ne]3s^2 3p^1}$ and not something completely different: (1) at the core, and (2) at the valence.

Well, this configurations arise in what is usually called the orbital approximation and the consistent pattern is due to the very nature of this approximation.
In the orbital approximation the overall electronic wave function describing the atom is represented as the product of the wave functions for the individual electrons known as atomic orbitals. This can be sketched as follows.
$$
\newcommand{\op}[1]{\hat{#1}}
\newcommand{\el}[1][]{_{\mathrm{e}}}
\newcommand{\nuc}[1][]{_{\mathrm{n}}}
\newcommand{\elel}[1][]{_{\mathrm{ee}}}
\newcommand{\elnuc}[1][]{_{\mathrm{en}}}
$$
Electronic Schrödinger Equation
The electronic wave function $\psi_{\mathrm{e}}$ is an eigenfunction of the electronic Hamiltonian $\op{H}\el$
$$
\op{H}\el \psi\el = E\el \psi \el \, ,
$$
$$
\op{H}\el
=
\op{T}\el + \op{V}\elnuc + \op{V}\elel
=
- \sum\limits_{i=1}^{n} \frac{1}{2} \nabla_{i}^{2}
- \sum\limits_{i=1}^{n} \frac{Z}{r_{i}}
+ \sum\limits_{i=1}^{n} \sum\limits_{j > i}^{n} \frac{1}{r_{ij}} \, ,
$$
which indeed is different for different atoms. Thus, the electronic wave function can also be expected to be different. However, the exact analytical solutions of the electronic Schrödinger equation are known only for one and two-electron systems, such as $\ce{H}$, $\ce{H-}$, $\ce{He}$, $\ce{Li+}$, etc., while for all other chemical systems only approximate solutions can be found.
Separation of variables
The electronic Schrödinger equation is partial differential equation and one of the most frequently used analytical methods to solve such equations is technique known as the separation of variables aimed to broke one compound partial differential equations of many variables into a set of simpler partial differential equations of fewer variables. The main disadvantage of the separation of variables is that it is by no means a universal approach: for some equations a solution of the form mentioned above make the separation of variables possible, for others it is of no use.
The most common way of separation of variables for partial differential equation in a function $f(x_1, x_2, \dotsc, x_n)$ of $n$ variables $x_{i}$ is writing a solution as a product of $n$ functions $f_{i}(x_{i})$ each of which is a function of one variable only
$$
    f(x_1, x_2, \dotsc, x_n) = f_1(x_1) f_2(x_2) \dotsm f_n(x_n) \, ,
$$
and then substituting solution of this form into the equation.
Using algebraic manipulation one could try to break the resulting equation into a set of $n$ independent ordinary differential equations for each function $f_{i}(x_{i})$ and if succeed then one could solve these independent equations and by plugging all $f_{i}(x_{i})$ back into the product obtain the solution of the starting equation.
Orbital approximation
So, it is tempting to try to separate the electronic coordinates by representing $\psi\el$ as a product of one-electron wave functions, however, it is fairly well known that this attempt will be successful only if the potential energy can be expressed as the sum of the separate potentials for each particle. This clearly is not the case for the electronic Hamiltonian: the potential energy of the Coulomb repulsions between the electrons $\op{V}\elel$ prevents the separation of electronic coordinates
\begin{equation*}
    \psi\el(\vec{r}_{1}, \vec{r}_{2}, \dotsc, \vec{r}_{n}) \neq \psi_{1}(\vec{r}_{1}) \psi_{2}(\vec{r}_{2}) \dotsb \psi_{n}(\vec{r}_{n})\, .
\end{equation*}
So, exact separation in this form is not possible, but what about approximate one? Can we somehow approximate $\op{V}\elel$ with some model potential which will allow the separation of electronic coordinates? To some extent, the answer is, yes, we can.
In the orbital approximation it is indeed assumed that the many-electron wave function can be written as a product of one-electron wave functions called orbitals. As it was already said, it will work only if the exact $\op{V}\elel$ potential is replaced by the model potential $\op{V}_{\mathrm{MF}} = \sum\nolimits_{i=1}^{n} \op{v}_{\mathrm{MF}}(r_{i})$ which allows the separation. The approximate Hamiltonian is
$$
\op{H}_{\mathrm{MF}}
=
\op{T}\el + \op{V}\elnuc + \op{V}_{\mathrm{MF}}
=
- \sum\limits_{i=1}^{n} \frac{1}{2} \nabla_{i}^{2}
- \sum\limits_{i=1}^{n} \frac{Z}{r_{i}}
+ \sum\limits_{i=1}^{n} \op{v}_{\mathrm{MF}}(r_{i}) \, .
$$
Physically it means that electrons do not instantaneously interact with each other, but rather each and every electron interacts with the average, or mean, electric field created by all other electrons. Hence, another name of the approximation, the mean field approximation.
Even the mean-field Hamiltonian still looks different for different atoms, however, it is now a sum of one-electron Hamiltonian operators
$$
\op{H}_{\mathrm{MF}}
=
\sum\limits_{i=1}^{n}\op{h}(r_{i}) \, ,
$$
where
$$
    \op{h}(r_{i})
    =
    - \frac{1}{2} \nabla_{i}^2
    - \frac{Z}{ r_{i} }
    + \op{v}_{\mathrm{MF}}(r_{i}) \, .
$$
And in contrast to an eigenfunction $\psi\el$ of the true electronic Hamiltonian $\op{H}\el$ an eigenfunction $\psi_{\mathrm{MF}}$ of the mean-field Hamiltonian $\op{H}_{\mathrm{MF}}$ can be written as products of one-electron wave functions (orbitals)
$$
    \psi_{\mathrm{MF}} = \psi_{1}(\vec{r}_{1}) \psi_{2}(\vec{r}_{2}) \dotsb \psi_{n}(\vec{r}_{n}) \, ,
$$
where $\psi_{i}(\vec{r}_{i})$ are solutions of the corresponding one-electron Schrödinger equations
$$
    \op{h}_{i} \psi_{i}(\vec{r}_{i}) = \varepsilon_{i} \psi_{i}(\vec{r}_{i}) \, .
$$
Now this resulting equations that defines orbitals are not that different: there will be just a small difference in one-electron Hamiltonians due to slightly  different nuclear charges as you move along a group. Of course, there will be more and more occupied orbitals as you proceed through the group, but equations themselves and consequently orbitals will be quite simillar.

Thanks! If $\hat h (r_i)$ are added to produce $\hat H_{MF}$, why $\psi_i(r_i)$ are multiplied to produce $\psi_{MF}$?

I expanded the answer a little bit. The key point is that from the get go we want to write the wave function this particular way, but it does make sense only if the Hamiltonian has the form of a sum of one-electron Hamiltonians. Otherwise, we won't get the whole many-particle equation separated into a bunch of one-particles ones.

Sorry, this is too long for a comment. This is an answer to the OP's comment on

If $\op{h}(r_{i})$ are added to produce $\op{H}_{\mathrm{MF}}$, why
  $\psi_{i}(\vec{r}_{i})$ are multiplied to produce
  $\psi_{\mathrm{MF}}$?

The combination of these special forms of $\op{H}_{\mathrm{MF}}$ and $\psi_{\mathrm{MF}}$ is consistent with the time-independent Schroedinger Equation $\op{H}_{\mathrm{MF}} \psi_{\mathrm{MF}} = E_{\mathrm{MF}} \psi_{\mathrm{MF}}$:
\begin{align}
\op{H}_{\mathrm{MF}} \psi_{\mathrm{MF}} &= \sum\limits_{i=1}^{n}\op{h}(r_{i}) \psi_{1}(\vec{r}_{1}) \psi_{2}(\vec{r}_{2}) \dotsb \psi_{n}(\vec{r}_{n}) \\
&= \underbrace{\op{h}(r_{1}) \psi_{1}(\vec{r}_{1})}_{= \, e_{1} \psi_{1}(\vec{r}_{1})} \psi_{2}(\vec{r}_{2}) \dotsb \psi_{n}(\vec{r}_{n}) + \psi_{1}(\vec{r}_{1}) \underbrace{\op{h}(r_{2}) \psi_{2}(\vec{r}_{2})}_{= \, e_{2} \psi_{2}(\vec{r}_{2})} \dotsb \psi_{n}(\vec{r}_{n}) + \ldots \\
&\hphantom{=} \, + \psi_{1}(\vec{r}_{1}) \psi_{2}(\vec{r}_{2}) \dotsb \underbrace{\op{h}(r_{n}) \psi_{n}(\vec{r}_{n})}_{= \, e_{n} \psi_{n}(\vec{r}_{n})} \\
&= e_{1} \psi_{1}(\vec{r}_{1}) \psi_{2}(\vec{r}_{2}) \dotsb \psi_{n}(\vec{r}_{n}) + e_{2} \psi_{1}(\vec{r}_{1}) \psi_{2}(\vec{r}_{2}) \dotsb \psi_{n}(\vec{r}_{n}) + \ldots \\
&\hphantom{=} \, + e_{n} \psi_{1}(\vec{r}_{1}) \psi_{2}(\vec{r}_{2}) \dotsb  \psi_{n}(\vec{r}_{n}) \\
&= \underbrace{(e_{1} + e_{2} + \ldots + e_{n})}_{= \, E_{\mathrm{MF}}} \underbrace{\psi_{1}(\vec{r}_{1}) \psi_{2}(\vec{r}_{2}) \dotsb \psi_{n}(\vec{r}_{n})}_{= \, \psi_{\mathrm{MF}}} \\
&= E_{\mathrm{MF}} \psi_{\mathrm{MF}}
\end{align}


text:
Auger emission probability?
Why is it that lighter atoms are more likely to produce Auger emission and the heavier atoms fluorescence in x-ray spectroscopy? It doesn't make any sense. Lighter atoms should try to resist double ionization, plus there are fewer electrons around which means there is a lesser chance of a photon hitting an electron.

Your second reason does not come in to play at all - there is already a hole down in the electronic structure. The only question is what the atom does about it. Now, why would a lighter ion resist double ionization more than a heavier ion? Then, consider the range of where holes can be in the electronic structure, and whether it is easy or hard to kick an electron out from deep in a heavy ion's orbitals vs letting loose a photon.

There isn't any simple explanation or model for the % of Auger emission. Just accept that is the way atoms behave and move on.

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sort: votes


text:
Comparing the dipole moment of water and ethanol
Why is the dipole moment of water (1.85D) larger than ethanol (1.66D)?
I thought the inductive effect from ethyl would develop more delta - charge on oxygen.. hence increasing the charge and the dipole moment

It's probably due to the slight difference in electronegativity between H (2.2) and C (2.55)
Here's some data on gas-phase dipole moments from the CRC Handbook:

Water - 1.85 D
Methanol - 1.7 D
Ethanol - 1.66 D
n-Propanol - 1.55 D
n-Butanol - 1.66 D

Using some incorrect data, I pointed to a possible odd-even effect. Using the correct gas-phase dipole moments, we see that these are all pretty similar (i.e., +/- ~0.2D).
Certainly water has the highest dipole moment, but H is slightly more electropositive than C, suggesting that as a likely cause.

Side note - my initial thought was to the odd-even effect, so when I quickly found online data suggesting a large dipole moment for methanol and n-propanol, it seemed reasonable. Always better to go to quality sources.

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text:
Surface stability in crystal growth
What is the meaning of "more thermodynamically stable surface" when we speak of crystallographic facet formation? For example in wurtzite crystals, the formation of (10-11) surfaces are more favorable than the formation of (20-21). Why? and How? and What is the physical explanation behind?

Surface atomic position have certain excess energy compared to those in the bulk. Different surfaces have different energies. If the atoms were allowed to vote, they would rather have a crystal with no surface at all, but that just isn't possible.

I think it is reasonable to assume that crystal faces form according to the "more thermodynamically stable surface." However the "more thermodynamically stable surface" is dependent on the various conditions in which the crystal forms, for example temperature, pressure and concentrations. Calcite has hundreds of forms. The gist is that there is no model which can now suitably predict the shape of a crystal.


text:
How to calculate the depression in freezing point of a solution in which there are multiple solutes being added?
In just about every problem I've encountered earlier, all such questions describe the process for calculating the depression in freezing point for a solution wherein there is only one solute. However, on a take-home exam, we were asked to solve the following problem:

What is the freezing point of a solution containing $\pu{0.2 mol}$ glucose and $\pu{0.2 mol}$ $\ce{KCl}$ in $\pu{250 g}$ $\ce{H2O}$?

My thinking is to use the general formula
$$\Delta T_\mathrm{f} = imK_\mathrm{f}.$$
What I'm concerned about is how to reconcile the multiple molalities, as both glucose and $\ce{KCl}$ have the same molality, but ought we add them together? Multiply them? Just a little confused.
Also, he gave us a set of “useful equations”, but on the sheet he did not specify a van 't Hoff factor for $\ce{KCl}.$ This is confusing to me because I'm pretty sure that $\ce{KCl}$ is a strong electrolyte and ought to have one. Should we just assume $i = 2$ since there are two species $(\ce{K+}$ and $\ce{Cl-})$ we are sure to get afterwards?

Use your logic. You need a value with certain dimension to plug into that formula. If you multiply two molalities, what will happen to their dimension? (As for the second question, yes, it's 2.)

Freezing point depression is a colligative property, well defined by this Wikipedia article (emphasis mine):

In chemistry, colligative properties are properties of solutions that depend on the ratio of the number of solute particles to the number of solvent molecules in a solution, and not on the nature of the chemical species present.

This means the problem is the same regardless of the number of different solvent species. You simply add them up to get the total solute concentration and calculate the freezing point depression in the same way you would if they were all the same species.  As you mentioned, you do need to keep track of dissociating species like $\ce{KCl},$ i.e. $\pu{1 mol}$ $\ce{KCl}$ dissociates into $\pu{2 mol}$ of total solutes.

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text:
What makes an atom favor another one when bonding, and should there be enough atoms for both parts?
I'm learning chemistry through Khan Academy and I have 2 pretty basic questions.
Oxygen (O) has 6 electrons on it's last shell, so it needs 2 more to become stable. Hydrogen (H) has 1 electron, so it needs 1 more. O might combine with 2 H's, creating H2O, and it might combine with another O, creating O2. It can also create H2O2.
QUESTION 1: So, given a hypothetical solution (or environment, or whatever), with only O and H atoms, what will make O to favor a bond with another O instead of a bond with 2 H's (or vice-versa)? Both situations will make it stable.
QUESTION 2: Also, if we assume that, in that solution, there's a limited number of both O and H atoms, and after all bonds are made, there's only one O atom and one H atom left, will that H atom bond with the one O atom left, since that will make the H stable (albeit not the O atom), creating a HO molecule ? So, I guess what I'm asking is, will bonds only be made when there's enough atoms for both parts to become stable?
Thanks.

The answer to your question boils down to one thing, bond energies. When a bond is formed, energy is released. When a bond is broken, energy must be absorbed. Elements prefer to be in a low energy state, so if you just added H and O into a container, the reaction with the most negative net bond energy will take place. To calculate the overall bond energy of the reaction, the equation is:
$E_{total} $= (Energy of bonds broken) - (Energy of bonds formed).
(As a side note, the answer is in joules of energy per mole)
For naturally occurring hydrogen and oxygen the bond energy is:
$2H_2 + O_2 $ ->$2H_2O$
= $(436*2 + 493*1)-(463*4)= -487 \frac{kj}{mol}$
(2 H-H bonds broken + 1 O=O bond broken) - (4 O-H bonds formed)
This reaction is favored and would occur under typical conditions if oxygen was in the presence of hydrogen and would release -487 joules of energy per mole. The bond energies for each bond can be found by simply looking them up online. The bond energy for the formation of $H_2O_2$ I calculated turned out to be -149kj/mol. Therefore the formation of $H_2O$ would be favored over $H_2O_2$. In your question, you asked about single O and H atoms, not $H_2$ and $O_2$, which would not occur naturally, but is still very possible. This result however would be the same. The atoms would first bond to form $O_2$ and $H_2$, then would proceed to form water as noted above.
Finally, to answer your last question, if there is one hydrogen and one oxygen left, they will indeed bond to make an $OH$ molecule.

What is the "bond energy for the formation of H2O2"?

Thank you for your answer. So, first the reactions with the most negative net bond energy would be formed, then the reactions with the most negative net bond energy, and so on ? If you want to assure reactions with higher energy state, without using any artificial methods, could you do so by some kind of counting of atoms involved in the process ? For instance, is there like a cuttof: "ok, all H2O are formed, now it will proceed to form H2O2." Sorry if it's a dumb question.

It is better to ask a seemingly dumb question than to remain with some ingrained misperception. See, chemistry is all about "artificial" methods and not about counting (not in this particular sense, at least). By the time all H2O is formed, you will run out of H, or O, or maybe both, so you won't be able to form any H2O2.

There is nothing like an OH molecule. An OH ion, however, is present. (OH-) As per my knowledge, an ion does not exist in free state. It has to be part of another molecule. So no, an OH molecule (so-called) will not be formed. And as Ivan Neretin said:                                                                                See, chemistry is all about "artificial" methods and not about counting (not in this particular sense, at least). By the time all H2O is formed, you will run out of H, or O, or maybe both, so you won't be able to form any H2O2 or OH.

There indeed is such a particle as $\ce{OH}$ (without minus). Surely it is unstable in that it would readily react with many things, but other than that, it is stable. If a lone H in the middle of nowhere encounters a lone O, they would not wait for another H. Instead, they would stick together to form that particle, also called radical: $\ce{\cdot OH}$.


text:
Where to store Sodium Azide?
I have about 100g of Sodium azide and would like to store it properly. Any suggestions as to where? I though of storing it in a locked vault but there are chemicals such as Arsenic sulfide, aniline, and Potassium cyanide, but the presence of metals in the cabinet makes me a bit queezy.
Many thanks in advance.

Have a look at the MSDS of the hazardous materials, which you are planning to store together. At least in Germany, these materials are assigned to storage classes (in German: Lagerklasse (LGK)).
The German regulations are to be found in the document TRGS 510, TRGS stands for Technische Regeln für Gefahrstoffe (Technical Regulations for Hazardous Materials). Document 510 of this series specifies Lagerung von Gefahrstoffen in ortsbeweglichen Behältern (Storage of hazardous materials in non-stationary containers). Within TRGS 510, you'll find a "compatibility table" that specifies, which storage classes may (not) be stored together.
Sodium azide belongs to the LGK 6.1 A (flammable , acute toxic hazardous materials of category 1 and 2).
Materials from LGK 6.1 A may be stored together with other materials of the same LGK, but not, for example, with explosives (LGK 1), strong oxidants (*LGK 5.1 A), organic peroxides (*LGK 5.2), pyrophororic materials (*LGK 4.2), etc.


text:
Determine the amount of substance of calcium chloride used

Determine the amount of substance (in moles) of calcium chloride used in this lab. (Hint: $0.20\rm~M$ means $0.20\rm~moles:1~L$)

I made $96$ milliliters of $\ce{CaCl2}$ which would be $0.096\rm~L$.
$$\rm0.096~L~of~CaCl_2\times\frac{0.20~mol}{1~L} = 0.0192~mol~of~CaCl_2$$
I was wondering if I solved this right.

Yes, you did it correctly, though there is a small issue with significant figures if that is a concern.
$$\mathrm{Molarity}=\frac{\mathrm{moles}}{\mathrm{liters}}$$
So for your values:
\begin{align}
0.20 &= \frac{n}{0.096}\\
n &= \pu{0.20 M} \times \pu{0.096 L} \\&= \pu{0.0192 mol}\, \ce{CaCl2}
\end{align}
However, both your molarity and volume have only two significant figures and you're multiplying those values, so the result should only be reported to two significant figures as well: $n = \pu{0.019 mol}\,\ce{CaCl2}$.

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text:
Conversion of grams to milliliters using kg/m3 density

Convert 100 grams of a solution to milliliters with a density of $\pu{681.9 kg/m3}$. It also has a density of $5.69$ pounds per gallon.

Are any or all of these correct?
\begin{align}
  \pu{100 g} \times \frac{\pu{1 mL}}{\pu{681.9 g//cm3}}
    &= \pu{0.146649~mL}
    &&(\approx \pu{0.15 mL})\\
  \pu{1,000 g} \times \frac{\pu{1,000 mL}}{\pu{681.9 g//cm3}}
    &= \pu{1,466.4906 mL}
    &&(\approx \pu{1,500 mL})\\
  \pu{1 kg} \times \frac{\pu{1 L}}{\pu{681.9 kg//m3}}
    &= \pu{1.4664906 L}
    &&(\approx\pu{1.5 L})
\end{align}
Or, can it be as simple as:
$$\pu{100 g} \times 681.9 = \pu{68.190 mL}?$$
The unit $\pu{kg/m3}$ is what's confusing me. I'm used to $\pu{g/cm3}$ or $\pu{g/mL}$ for densities. Does $\pu{681.9 g/cm3}$ = $\pu{681.9 kg/m3}$, while scaled-up a thousand-fold? Can I consider using the same density with the math involved by substituting $\pu{kg/m3}$ for $\pu{g/cm3}$ in the equation, or do I have to utilize a conversion within the metric system?

Concerning the units in your equation none of them is correct (As all your formulations would end up in a square volume unit).
Definition of volumetric mass density
The density $ \rho $ is given as
$$ \rho = \frac{m}{V} $$
resulting in a mass and density dependent Volume as
$$ V = \frac{m}{\rho} $$

Direct way
Although you could calculate this by just substituting the physical quantities by the given values and units, and then take care of the unit conversion afterward:
\begin{align}
  V
  &= \frac{m}{\rho} \\
  &= \frac{100\,\mathrm{g}}{681.9\,\mathrm{kg}\,\mathrm{m^{-3}}} =\\
  &= \frac{100\,\mathrm{g}\,\mathrm{m^{3}}}{681.9\,\mathrm{kg}} = \frac{100\,\mathrm{g} \cdot \left(10^2\,\mathrm{cm}\right)^{3}}{681.9 \cdot (10^3\,\mathrm{g})} =\\
  &= \frac{100 \cdot 10^6\,\mathrm{cm}^{3}}{681.9 \cdot 10^3} = \frac{100 \cdot 10^3}{681.9}\,\mathrm{cm}^{3} =\\
&\approx 146.65\,\mathrm{cm}^{3} \equiv 146.65\,\mathrm{ml}
\end{align}
I would suggest to convert the density from $\mathrm{kg}\,\mathrm{m^{-3}}$ to $\mathrm{g}\,\mathrm{cm^{-3}}$ as a first step.

Easier way
Of course easiness is something subjective, but I believe this is easier to use.
Calculate conversion-factor for density
First the combined conversion-factor for the units-fraction is calculated
$$ \frac{\mathrm{kg}}{\mathrm{m}^3} \stackrel{\left[1\right]}{=} \frac{\left(10^3\,\mathrm{g}\right)}{\left(10^2\,\mathrm{cm}\right)^3} \stackrel{\left[2\right]}{=} \frac{10^3\,\mathrm{g}}{10^6\,\mathrm{cm}^3} \stackrel{\left[3\right]}{=} 10^{-3}\,\frac{\mathrm{g}}{\mathrm{cm}^3} $$

Start with substituting each units SI prefix:
Care has to be taken with the conversion of $ \mathrm{m}^3 $, since the power applies to the whole substituted expression ($ \mathrm{m}^3 = \left(10^2\,\mathrm{cm}\right)^3 $).
In the last step the exponential expressions are summarized ($ \frac{10^3}{10^6} = 10^{3-6} = 10^{-3} $).

Then calculate the new numeric value of the density in the target unit:
$$ \rho = 681.9\,\mathrm{kg}\,\mathrm{m^{-3}} \stackrel{\left[4\right]}{\equiv} 681.9 \cdot \left(10^{-3}\,\mathrm{g}\,\mathrm{cm}^{-3}\right) \stackrel{\left[5\right]}{=} 0.6819\,\mathrm{g}\,\mathrm{cm}^{-3} $$

Substitute the unit with the previously calculated expression
Multiply the numeric terms

Calculate the volume
Now the unit-adjusted density is used:
$$ V = \frac{m}{\rho} = \frac{100\,\mathrm{g}}{0.6819\,\mathrm{g}\,\mathrm{cm^{-3}}} = \frac{100}{0.6819}\,\mathrm{cm^{3}} \approx 146.65\,\mathrm{cm}^{3} \equiv 146.65\,\mathrm{ml}  $$

Conversion-factors
According to SI Prefixes the following applies:
\begin{align}
1\,\mathrm{m} &\equiv 10^2\,\mathrm{cm}\\
1\,\mathrm{kg} &\equiv 10^3\,\mathrm{g}
\end{align}
And of course:
$$ 1\,\mathrm{cm}^3 \equiv 1\,\mathrm{ml} $$

Formulas
Density is defined as mass divided by volume: $ρ = \displaystyle\frac{m}{V}$,
where $ρ$ is density, $m$ is mass, and $V$ is volume.
Your question has:

Given $m = 100\text{ }\text{g}$,
Given $ρ = 681.9\text{ }\displaystyle\frac{\text{kg}}{\text{m}^3}$,
We want to solve for $V$ in units of $\text{mL}$.

By rearranging the density formula, we have: $V = \displaystyle\frac{m}{ρ}$.
Unit conversions

$1\text{ }\text{kg} = 1000\text{ }\text{g}$.
$1\text{ }\text{m} = 100\text{ }\text{cm}$.
$1\text{ }(\text{cm})^3 = 1\text{ }\text{mL}$.

We can rearrange each unit conversion to become a ratio that equals one. That is to say, whenever we have $x = y$, we can say that $\displaystyle\frac{x}{y} = 1 = \displaystyle\frac{y}{x}$. Therefore:
$$ 1
\:=\: \frac{1\text{ }\text{kg}}{1000\text{ }\text{g}}
\:=\: \frac{100\text{ }\text{cm}}{1\text{ }\text{m}}
\:=\: \frac{1\text{ }\text{mL}}{1\text{ }(\text{cm})^3}
$$
Calculation
We want to compute $V = \displaystyle\frac{m}{ρ}$, which simply involves substituting in the known values. But to get the final answer in the units of $\text{mL}$, we will need to use the unit conversions in a strategic way. Because each conversion is equal to $1$, we can multiply the formula by as many ones as we need until we get the result we want. Namely, $V = \displaystyle\frac{m}{ρ} × 1 × 1 × \cdots × 1$. Here is how we do it:
$V = \displaystyle\frac{100\text{ }\text{g}}{681.9\text{ }\frac{\text{kg}}{\text{m}^3}} = \displaystyle\frac{100\text{ }\text{g}⋅\text{m}^3}{681.9\text{ }\text{kg}}$
$= \displaystyle\frac{100\text{ }\text{g}⋅\text{m}^3}{681.9\text{ }\text{kg}} × 1 × 1^3 × 1$
$= \displaystyle\frac{100\text{ }\text{g}⋅\text{m}^3}{681.9\text{ }\text{kg}} × \displaystyle\frac{1\text{ }\text{kg}}{1000\text{ }\text{g}} × \left(\displaystyle\frac{100\text{ }\text{cm}}{1\text{ }\text{m}}\right)^3 × \displaystyle\frac{1\text{ }\text{mL}}{1\text{ }(\text{cm})^3}$
$= \displaystyle\frac{100\text{ }\text{g}⋅\text{m}^3}{681.9\text{ }\text{kg}} × \displaystyle\frac{1\text{ }\text{kg}}{1000\text{ }\text{g}} × \displaystyle\frac{1000000\text{ }(\text{cm})^3}{1\text{ }\text{m}^3} × \displaystyle\frac{1\text{ }\text{mL}}{1\text{ }(\text{cm})^3}$
$\require{cancel} = \displaystyle\frac{100\text{ }\cancel{\text{g}}⋅\cancel{\text{m}^3}}{681.9\text{ }\cancel{\text{kg}}} × \displaystyle\frac{1\text{ }\cancel{\text{kg}}}{1000\text{ }\cancel{\text{g}}} × \displaystyle\frac{1000000\text{ }\cancel{(\text{cm})^3}}{1\text{ }\cancel{\text{m}^3}} × \displaystyle\frac{1\text{ }\text{mL}}{1\text{ }\cancel{(\text{cm})^3}}$
$= \displaystyle\frac{100}{681.9} × \displaystyle\frac{1}{1000} × \displaystyle\frac{1000000}{1} × \displaystyle\frac{1\text{ }\text{mL}}{1}$
$= \displaystyle\frac{100 × 1 × 1000000 × 1\text{ }\text{mL}}{681.9 × 1000 × 1 × 1}$
$= \displaystyle\frac{100000000}{681900}\text{ }\text{mL}$
$= \displaystyle\frac{1000000}{6819}\text{ }\text{mL}$ (exact)
$≈ 146.6\text{ }\text{mL}$. (rounded to appropriate significant figures)
Notice how we chose conversions so that most of the units in the numerators and denominators cancel out.

For future reference: On Chemistry mathematical and chemical expressions can be formatted using MathJax (and LaTeX Syntax).  If you want to know more, please have a look here and here. | It is better to use actual display style, e.g. $$ ... $$ or \begin{align} ... \end{align} than forcing it in inline style $\displaystyle ... $

@Martin-マーチン I am experienced at LaTex, thank you. My choice of formatting was appropriate in the context of the post.

It's just for your convenience. It looks alright in my desktop browser, but that doesn't extend to mobile. Also two \displaystyle in the same MathJax sequence is unnecessary, only the first one takes effect. So you have a lot of overhead and it takes much longer to load. And please keep in mind: while MathJax uses LaTeX syntax, it absolutely is not LaTeX.

I'm not sure what the problem on mobile is... guessing maybe the math formulas are too long and cannot wrap? As for having two \displaystyle in a sequence, it is indeed redundant in the current text, but I noticed that \displaystyle\frac{a}{\frac{b}{c}} shows up differently than \displaystyle\frac{a}{\displaystyle\frac{b}{c}}, which suggests to me that displaystyle is simply not a global setting.

Well, $$...$$, or (probably better) \begin{align}...\end{align} would at least allow you to control the amount of vertical space between successive lines. That aside, one of the more impactful differences between MathJax and LaTeX is that on Chemistry.SE there are easier ways to typeset quantities with units; you can use $\pu{3.14 J/(K mol)}$ --> $\pu{3.14 J/(K mol)}$, or since you prefer the fractions, $\pu{3.14 J//K mol}$ -> $\pu{3.14 J//K mol}$. https://mhchem.github.io/MathJax-mhchem/


text:
Alcoholic potassium permanganate for oxidation
A question in my textbook:

Though alkaline and acidic $\ce{KMnO4}$ are both oxidizing agents, in the manufacture of benzoic acid from toluene, we only use alcoholic $\ce{KMnO4}$ as an oxidant. Why?

I am not able to figure out the answer. But I think maybe the acidic and alkaline $\ce{KMnO4}$ are too strong and may oxidize the benzene ring. Or the benzoic acid may react with the alkaline medium. (Though I wonder why it wouldn't react with alcohol to give ester)
I also tried digging around the web but couldn't come up with anything satisfactory.

Possible duplicate of Oxidation of toluene to benzoic acid in presence of KMnO4

@ringo Not sure if this is a duplicate or not. Title and wording of the question refer to alcoholic $\ce{KMnO4}$, which is not what the linked question is about. Or, perhaps alcoholic is a typo in biogirl's blockquote?

@ringo I guess my question is a duplicate. I just went through the question but the answer given was not satisfactory.

Practical considerations, mainly that toluene is non-polar, inorganic. Aqueous $\ce{KMnO4}$ wouldn't be miscible so the oxidation can only take place in the interface. By using alcoholic $\ce{KMnO4}$, we can get the $\ce{KMnO4}$ into the organic layer much more easily, thus making the reaction much faster.
To form an ester, you generally need a dehydrating environment. Plus, the oxidation of toluene with $\ce{KMnO4}$ in neutral conditions will produce base which neutralizes the benzoic acid, thus deactivating it making it unable to undergo esterification.


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Making 1,3,5-Tribromobenzene from Phenol
The problem was to make 1,3,5-Tribromobenzene from Phenol.
I thought to do it the following way:

React phenol with $\ce{Br2/H2O}$.
Then add $\ce{Zn}$ dust to get the product.

Now I am in doubt whether the second step would even work.
Are there any better method of doing this ?

I may be missing something obvious, but it's going to be tough to get rid of that hydroxyl. Zn won't do it.

@jerepierre that's what I was also thinking.

Why do you think Zn won't do it?The same reaction works for phenol. What makes it different here?

@KaranSingh I haven't looked into it, but I expect bromine will be reduced off before a hydroxyl group.

First react phenol with ammonia in presence of zinc chloride to obtain aniline.

Then react aniline with bromine water to form 1,3,5-tribromoaniline

Then react 1,3,5-tribromoaniline  with sodium nitrite and hydrochloric acid at 273 K to form 1,3,5 tribromobenzenediazonium chloride.

Then react the diazonium salt with ethanol/phosphorous acid and water in presence of copper to form  1,3,5 tribromonbenzene.

Detail explanation about the reaction, conditions  and the mechanism is given here by actually conducting the experiment.


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Colour due to F-centre is same as that produced in a flame test, why?
I have noticed that the colours that the F-centres impart in Alkali metal halides are the same that the corresponding alkali metals impart to their flames.
$\ce{NaCl}$ when heated in an atmosphere of $\ce{Na}$ becomes yellow.
$\ce{LiCl} \rightarrow$ pink
$\ce{KCl} \rightarrow$ violet/lilac
Why ?

However, ZnO gives yellow color due to F-center formation while its flame coloration is blue-green to pale green.

Why are the colours the same as the metal. Basically, the anion doesn't affect the colour, but the cation does. And since the cation is just the metal missing an electron or several, of course, they exhibit similar behaviour!
Colours we see are determined by the transitions made by electrons "falling" from a high energy state to a lower energy state, and in the process releasing one or some photons with a frequency in what we call the visible spectrum.
When an F-centre is formed in the described conditions, we have effectively extended the lattice with additional cations the same as those already present. But rather than additional anions that match as well, free electrons take the place in the lattice that would be occupied by one.
Now the anion already has its orbitals filled - there is nowhere for an associated electron to move into when excited, nor somewhere for other electrons to enter when relaxed, and thus there is no way for them to release energy as photons that we sense as light. So the anion doesn't affect the colour.
However the cation has empty orbitals when neutral, let alone when positively charged (i.e. has an electron(s) removed). This means any excitation or relaxation will involve these orbitals.
The result is that the electron movements in the  F-centre lattice are the same as those that would occur in the ionisation of the metal which  the cation is derived from, and hence the colours are also the same.
Why those specific colours for the respective compounds? That would be a completely different question, and probably found in a quantum mechanics discussion rather than a chemistry area.

There are some exceptions, as always, due to the way certain orbitals can interact - like those in zinc oxide, for example. As mentioned though, again, this is closer to QM than chemistry.

The colours in case of an F-centre are formed due to excitation (followed by deexcitation) of unpaired electrons which occupy the anion vacancy sites in a crystal and not due to the electrons in the cations..

I thought F-center absorptions tracked the lattice parameter fairly closely. This means that the color is a function of the size of the vacancy the electron fills (determined by the specific anion that's missing). As a counter example RbCl and KBr have nearly identical colors due to f-center, but a flame test can clearly distinguish Rb and K. Likewise the color due to f-center defects in KF is much different than in KBr.


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Determining electrode potential
Can the anode always be considered as having a lower standard electrode potential than the cathode?
When I was studying galvanic cells, I found that for a number of reactions, it always appeared that the anode(i.e negative potential) mostly had a lower electrode potential than the cathode. Can we say that this is always true or are there some exceptions?

By definition, for a cell to be galvanic, you would need the difference $\ce{E^0_(cathode) - E^0_(anode)}$ to have a positive value. So yes, the standard potential for the anode to have a more negative potential than the cathode.
Lower and higher are terms that are best avoided when talking about electrochemistry. More positive and more negative are unambigious.

This is only true at standard state. One could pick electrodes which do not satisfy the equation E˚(cathode)–E˚(anode)>0, yet still have a galvanic cell by setting the concentrations of each species to make  it so that E(cathode) - E(anode)>0. This is the general definition for a galvanic cell.
I agree on the nomenclature point though.

True. One could turn an electrochemical system that is electrolytic in standard state into a galvanic cell by modifying concentrations.

Thank you for helping me out but Can you please give me some example to convert electrochemical system that is Electrolytic in chemical state into galvanic cell by modifying concentrations?


text:
Density and solubility

Do all saturated solutions have the same density?  Based on your data, which salt is more soluble?
\begin{align}
&\ce{NaCl} & \pu{11.812 g/10 mL} \quad &\text{or} \quad \pu{1.181 g/mL} \\
&\ce{CaSO4} & \pu{9.783 g/10 mL} \quad &\text{or} \quad \pu{0.978 g/mL}
\end{align}

I know the first part: all saturated solutions do not have the same density. For the second part I believe that solutions with higher density will be more soluble. So in this example $\ce{NaCl}$ will be more soluble that $\ce{CaSO4}$.
Would I be correct and my answer has to be based on density?

Q: Are solutions with higher density more soluble. A: Yes, in general, but it does depend on the definition of "solubility." You could measure solubility as milligrams solute per milliliter of solution, or millimoles solute per milliliter of solution. The two different measures of solubility might give different results.  For instance freezing point depression is more related to molar concentrations rather than gram concentrations.

Let's think in terms of units. Density is mass per unit volume. Not only will different solutes have different masses dissolved into a unit volume, but the volume of the resulting solution will be different for different solutes.
When people say they can dissolve a unit of mass into a unit of volume (a common measure of solubility) they are not keeping track of the final volume of the resulting solution.
Here are the expressions for the density of the two saturated solutions (we should assume that the temperature and pressure are constant so that the density of the solvent itself is constant).
$\frac{m_{A,sat}}{V_{A,sat}} = \rho_{A,sat}$
$\frac{m_{B,sat}}{V_{B,sat}} = \rho_{B,sat}$
where $m$ and $V$ are masses and volumes, and $\rho$ is the density. $A$ and $B$ are the two solutes.
So if:
$\rho_{A,sat}{}\gt{} \rho_{B,sat}$
that could mean, in one limit where
$V_{A,sat} = V_{B,sat}$,
that
$m_{A,sat}{} \gt {}m_{B,sat}$.
Which is how I'm guessing you're thinking. If the volume change of solvation was similar for two solutes, one could trust that the density of the saturated solution with the (mass) solubility.
But there's another limit where there is a large difference in the volume change of solvation, between the two solutes. Then if:
$\rho_{A,sat}{}\gt{} \rho_{B,sat}$,
as before, and
$V_{A,sat}{} \lt{} V_{B,sat}$,
then it's possible that the unit mass dissolved in the two cases was equal,
$m_{A,sat} = m_{B,sat}$,
so that the two solubilities would be equal.
So no, with the above definitions of solubility and density, one cannot assume that the density of a saturated solution corresponds to the solubility.


text:
Difference between thermal reactions and photochemical reactions
In a photochemical reaction, everything starts with absorption of a photon. I.e. a ground state reactant is excited to the first excited state. Sometimes the reaction barrier on the ground state surface is too high; perhaps lower barriers exist in other dimensions that lead to undesirable products or dissociation. Perhaps the relevant barrier is much lower on the excited state surface, which makes the reaction possible in the excited state.
Is the difference, then, that thermal reactions take place exclusively on the ground state surface, while a photochemical reaction takes place, at least in part, on the excited state surface?

I agree with the other answer; thermal reactions take place on the ground state, but I would add that once an excited state is produced thermal reactions can occur there also. A nice example is the photochemical isomerisation of trans stilbene but other molecules such as octatetraene behave similarly. Photo-induced electron transfer is another example of a thermal reaction in the excited state.
A sketch of the potential energy profile along the reaction coordinate is shown below.

Once in the trans excited state there is a barrier (the transition state) that prevents the molecule from isomerising immediately. The barrier height is approx 15 kJ/mol and there is fluorescence from the excited state (lifetime 70 ps) which is in competition with isomerisation. The long fluorescence lifetime means that thermalisation occurs in the excited state. Once over the barrier there is a conical intersection which is crossed ballistically, that is without thermalisation, and either the trans ground state is produced or the cis ground state.
The small well in the excited trans state is at a greater bond extension than the ground state, meaning that before isomerisation the bond is stretched then it is twisted on the pathway to the cis isomer.
If the cis ground state is excited, this moves ballistically to the conical interaction as there appears to be no barrier to isomerisation.

Yes, you are right thermal reactions take place exclusively on the ground state surface, while a photochemical reaction takes place, at least in part  on the excited state surface.
But there is another difference between them as
photochemical processes offer an advantage over thermal methods by forming thermodynamically disfavored products. Therefore, they overcome large activation barriers in a short period of time, and allow reactions that were inaccessible by thermal processes.


text:
Simpler synthesis?
I have to obtain 1-nitro-2,4,5-triaminobenzene from 1,3-dichlorobenzene. I have thought of:

Reaction with $\ce{Mg}$
Rc with $\ce{CO2}$, then $\ce{H2O}$. By this point I have izophthalic acid.
$\ce{PCl5}$, then $\ce{NH3}$. I have formed the diamide.
Reaction with $\ce{NaOBr}$, which turns the $\ce{-CONH2}$ groups into $\ce{-NH2}$. (Hoffmann degradation) now I have 1,3-phenilendiamine.
Rc with $\ce{Ac2O}$, then nitration, reduction with $\ce{Fe + HCl}$, and again, $\ce{Ac2O}$.
Nitration and then, getting rid of $\ce{AcO-}$ groups with $\ce{H2O}$. Now I should have reached the target molecule.

I am dissatisfied with this synthesis, though. It is a bit too long and I feel like I miss something. I have also thought of nitration or aromatic nucleophilic substitutions with $\ce{NaNH2}$ as the first step of another possible synthesis, but I can't reach the end.
Is the synthesis I have thought of viable? Should I try sth different?
Thank you in advance.
LE: my bad, the target molecule is 1-nitro 2,4,5-triaminobenzene. Thank you orthocresol for pointing out a few things.
LLE: Thanks for your answer, Klaus, good work. You just gave me another idea for a synthesis! Even though there is something I don't like about it.
-double nitration, then rc with 2 moles of $\ce{NaNH2}$. I should have 1,3-dinitro-4,6-diaminobenzene by doing so.
-Zinin reduction, which will transform one $\ce{-NO2}$ group into an $\ce{-NH2}$ one (it does not matter which).
I wonder if 1,3-chlorobenzene after the rc with the first mole of $\ce{HNO3}$ will react with another mole of $\ce{HNO3}$, since it will be very deactivated. It looks too simple to be good.

Why will the Zinin reduction only reduce one of the nitro groups?

Isn't zinin reduction used to reduce polynitro compounds in order to get nitroamines? Yes, of course, it can be used to reduce both nitro groups, but choosing the right reagents leads to the reduction of only one -NO2 group.

Thinking in the deactivating efffect of the nitro group in electrophilic aromatic substitution, running a nitration as one of the later steps in your synthesis makes sense.
Since you're not really pleased with the length of your current approach, let's think about something else.
Please note that this is just a mind game and I didn't really bother to check for the best reagents for each step!
Direct replacement of $\ce{Cl}$ in your starting material by $\ce{NH2}$ via nucleophilic aromatic substitution would be nice, but the condition are probably quite harsh.
Your starting material isn't really activated for that.
The situation might be different after nitration. 1,3-dichloro-substitution should furnish the right product.
Now, it is conceivable that at least the chloro substituent para to the nitro group can be replaced using sodium amide. With a bit of luck, the second one reacts too.
Treatment of the resulting 1-nitro-2,4-diaminobenzene with sulfuric acid might yield ring sulfonation (and turn the amines into sulfonamides).
A further reaction with sodium amide, followed by acidic workup, might eventually give the desired product.

Again, this is just a mind game, I didn't bother to search for exact conditions and/or references!


text:
Materials for growing large quality sugar crystals
I am trying to grow large, solid, quality crystals on a course fabric, and I was wondering what other than sugar I would need to put in my solution, as well as how long it may take, and if food coloring/another dye would change the color of the crystals without destroying them. Any tips or advice would be great.

There is a nice book, about $10, on crystal growing. "Crystals and Crystal Growing" by Alan Holden, Phylis Morrison, Bert Singer
ISBN-10: 0262580500 ISBN-13: 9780262580502

The Exploratorium has directions for making rock candy, the common name for sucrose crystals. In short:

Make a saturated solution of sucrose -- 4 cups sugar dissolved in just 2 cups of boiling water.
Immerse the fabric in the solution, and let it cool slowly (to make larger crystals). You might use a covered styrofoam container to slow the cooling. If crystallization doesn't start in a few days, you might need to drop one seed crystal of sugar into the container.
You can experiment with different types of food color in the sugar solution to color the crystals, but in the process of crystallization, foreign molecules of dye may be excluded from the crystals. The larger the crystal, the less colored it may appear.

Enjoy! Be sure to use sanitary equipment and safe colors, if anyone is to eat this. BTW, ordinary cane sugar is one of the purest substances sold in quantity, better than 99.9% sucrose.


text:
EPR spectrum of a singlet-triplet mixture
Through EPR measurements it is possible to gain information about the spin states of molecules. So a singlet state will give no signal/only noise and a molecule with unpaired electrons will result in an EPR spectrum.
What happens if a measured sample would contain a mixture of singlet and triplet state molecules (of the same species)? Will there be a signal for the triplet state or is there any interaction between both that hinders the experimentalist from measuring the triplet states?

That shouldn't be a problem AFAIK.


text:
Reduction of glucose to hexane with hydroiodic acid
My book (NCERT India) states that n-hexane is formed when glucose is heated in the presence of hydrogen iodide. That is, the following reaction takes place:
$$\ce{C6H12O6 + HI ->[{\Delta}] C6H14}$$
This seems a bit strange to me because I only know of three possibilities that can occur when a compound containing hydroxyl groups is treated with a strong acid:

A substitution reaction, where all the $\ce{-OH}$ groups are replaced by $\ce{-I}$
A dehydration reaction, where $\ce{H2O}$ molecules would be eliminated and it would end up with few double bonds.
Formation of an ether.

There are also other sources which mention this reaction (thanks to Karan Singh, Tyberius, and user55119 for pointing these out). A Michigan State University webpage states that:

Hot hydriodic acid (HI) was often used to reductively remove oxygen functional groups from a molecule, and in the case of glucose this treatment gave hexane (in low yield).

Also, Dietary Sugars (edited by Preedy, V. R.) states (p 80) that

Glucose is reduced with concentrated hydriodic acid in the presence of red phosphorus to form a trace amount of n-hexane. This indicates an unbranched chain of six carbon atoms [...]

It goes on to cite the 5th edition of Principles of Biochemistry by Nelson et al. (published 2008), but I could not find this statement in any recent edition of this book.
Lastly, a similar reaction was carried out by Kiliani (Ber. Dtsch. Chem. Ges. 1885, 18 (2), 3066–3072) where the cyanohydrin of fructose was reduced to to 2-methylhexanoic acid using hot HI and phosphorus.

I would like to know:

Is phosphorus required for this reaction, or not?
What is the mechanism by which this reaction proceeds, with or without phosphorus?

It looks wrong, if HI could reduce alcohols to hydrocarbons chemistry would be different ;)

I don't think it is wrong, I think it is right.  But I'm curious about the book anyway.

HI is a strong reducing agent.  It can reduce iodoalkanes to alkanes through the formation of I2.

https://www.beilstein-journals.org/bjoc/articles/8/36 See scheme 1 and text.

@Zhe can you provide the DOI? I am unable to view the article from your link.

The site seems down right now. Also, not the most reputable journal I've ever seen, (considering I've never heard of it) so use with caution when it comes back.

Hydroiodic acid is a reducing agent.  As Wikipedia says:

Although harsh by modern standards, HI was commonly employed as a reducing agent early on in the history of organic chemistry. Chemists in the 19th century attempted to prepare cyclohexane by HI reduction of benzene at high temperatures, but instead isolated the rearranged product, methylcyclopentane (see the article on cyclohexane). As first reported by Kiliani,[10] hydroiodic acid reduction of sugars and other polyols results in the reductive cleavage of several or even all hydroxy groups, although often with poor yield and/or reproducibility.[11] In the case of benzyl alcohols and alcohols with α-carbonyl groups, reduction by HI can provide synthetically useful yields of the corresponding hydrocarbon product (ROH + 2HI → RH + H
  2O + I2).[8] This process can be made catalytic in HI using red phosphorus to reduce the formed I2.[12]

When $\ce{HI}$ acts as a reducing agent, the oxidized product is $\ce{I2}$.  To regenerate $\ce{HI}$ from the $\ce{I2}$, phosphorus can be used.  Thus, phosphorus is not strictly required for the reaction, as long as you are OK with using $\ce{HI}$ in superstoichiometric amounts and/or getting very low yields.
The overall scheme for $\ce{HI}$ reduction of an alcohol to an alkane is first displacement of the (protonated) alcohol by iodide, leading to formation of alkyl iodide.  Then, $\ce{HI}$ catalyzes the reduction of the alkyl iodide to the alkane, leading to $\ce{I2}$ as a byproduct.  The reduction step apparently follows a radical mechanism, at least under some conditions.
I'm not 100% sure how $n$-hexane would form from glucose.  It is unclear to me how $\ce{HI}$ would catalyze the reduction of the glucose aldehydic (or anomeric) carbon to an alcohol.  But, that is only one carbon out of six.  The other carbons have bona fide hydroxyl groups, and as such can be deoxygenated and reduced by $\ce{HI}$ to form alkyl carbons.  For example, Lv et al. report that hydrocarbons can be obtained from the five-carbon sugar alcohol xylitol in yields above 60%.  Unlike glucose, xylitol has no aldehydic/anomeric carbon, and only has alcohol groups.  This process did use phosphorous acid as an in situ iodine recycling agent.  Also, Yang et al. were able to obtain 5-methylfurfural from fructose by using $\ce{HI}$, without using an in situ agent to recycle $\ce{}$ to $\ce{HI}$.

The book uses that reaction to prove that the glucose is a straight chain molecule

"harsh by modern standards" – are there any better ways to fully reduce glucose to hexane by "modern standards"?

@SasQ Maybe this paper could get you started?  I'm no expert in this area, but it does say we report herein complete reduction of silyl-protected glucose to a mixture of hexane isomers.

@SasQ Maybe also this one: using a ruthenium catalyst and 60 atmospheres of hydrogen, and also temperatures of 500 K, the authors obtained a 70+% yield of hexanes from cellulose feedstock.

Yes, phosphorus is required for this reaction. As stated in your literature and in the comment, red phosphorous is the stoichiometric reducing agent. I assume that $\ce{PI3}$ and $\ce{P2I4}$ are formed. Both are known to be deoxygenating agents (#1, #2).
I propose the mechanism to be like this, where the oxophilicity of phosphorous is the driving force of the reaction. The mechanism starts with an electrophilic attack of phosphorous to an oxygen, which is then substituted by iodide. $\ce{C-O}$ bond cleavage leaves an alkyl iodide. These must then be reduced somehow.

Most of the members of this community are award of $\ce{HI, P}$ reduction and this would be obvious to most. The question is about using just $\ce{HI}$ or literature explaining why phosphorous got dropped.

@A.K. oh i'm sorry. Here's the DOI you requested in the comments above: doi:10.3762/bjoc.8.36
As far as i can say, the Beilstein JOC is a reputable journal.

I think you missed my point.  There are many sources that present this reaction so they have either propagated a myth or glucose is special. That is what we want answered.

I just provided the DOI you asked Zhe for. I have not enough reputation to directly comment in comments of the question.


text:
Reaction of formic acid with bridged bicyclic compound

Write the mechanism for:

I've been able to add part of the product, but am having trouble with the ring formation. Here is my attempt:

Have a look at the adjacent methyl groups attached to the bicyclo[2.2.2]octane. Can you spot the pattern in the product? When the free $\ce{OH}$ is esterified, what happens to the bridgehead methoxy group under acidic conditions?

Could come up with uploadable picture?

@klauswarzecha I don't know if you saw my second picture, but I've already added the formic acids to the methyls and etherified the OH, is there any way I can form a ring with that carbocation? Or am I going about it all wrong?

Maybe start by looking up named sigmatropic rearrangements. That should help get you started

I leave the arrows filling and details of last step to you since it is homework and online Marvin JS did not let me do arrows the way I liked to.

This calligraphy might show you the direction.


text:
Why is vitamin C called an antioxidant?
According to Wikipedia, vitamin C (ascorbic acid) "may also act as an antioxidant against oxidative stress," and is "widely used as a food additive, to prevent oxidation."  But acids are, by definition, oxidizing agents, so how is it that ascorbic acid is used to prevent oxidation?

Most acids (except nitric and some others) have nothing to do with oxidation. Well, I mean, they may oxidize metals, but we are not made of metals.

I dispute your claim that acids are oxidising agents. $\ce{HI}$ is a fairly strong acid that finds use as a reducing agent.

Ascorbic acid is considered an antioxidant because it can supply an electron to a species with an unpaired electron (a "radical", labeled "X" in the below diagram) to stabilize the radical. Of course, since ascorbic acid has now lost an electron, it is itself a radical - but it is a very stable radical due to resonance. Below I have included a diagram of this process below, from UC Davis.

Note how the we can "electron-push" to get the unpaired electron in the middle structure to move to the oxygen on the other side of the ring. This resonance delocalization of the unpaired electron in the ascorbyl radical makes it much more favorable to have the unpaired electron in the ascorbyl radical rather than the X.
Because radicals normally poach electrons from surrounding molecules to pair with their lone electron, they normally cause a chain reaction where their "victim" goes on to pull electrons from another neighboring molecule (oxidation), and so on and so forth. However, because the ascorbyl radical is stable, it is less likely to poach an electron from another molecule in its environment, preventing the chain reaction from occurring. Thus, it is an antioxidant.


text:
Selecting a good test solution for Cyclic Voltammetry
So I have a textbook headed my way for this but need to get a head start.
I have a 3-electrode voltammetry setup but I want to test it for some standard values (for educational purposes) before I start working on my actual project. For now, I'd like to use the equipment I have to run Cyclic Voltammetry tests at a basic, neutral, and acidic pH to verify that it's reading correctly and that I know how to read it.
Problem is, I've been out of Chem for a while. So asking about choosing test chemicals is sort of a multi-pronged question.

Given some chemical, say Potassium Ferrocyanide in a water and KCl solution, how do I calculate the pH? I've seen the equations concerning hydron concentrations, but I'm not too sure how to calculate hydron concentrations (had a little difficulty finding info).
Will performing this calculation tell me anything about the reversibility of the redox reaction, or is that something that it typically found empirically?
Is there anything else I need to be aware of in doing this calculation?

Sorry for the spread question, but I'm just getting started with all of this.
Oh, as a note, I have Pt and AgCl for my electrodes (AgCl wire and I should be getting an AgCl electrode soon).
Thanks!


text:
How do I analyze Mg<->C galvanic cell?
I've made a battery for middle school, and I want to figure out the chemistry, especially so I can say whether the little bubbles in there are $\ce{CO_2}$ or hydrogen.
In salt water, I have a strip of magnesium, and a graphite bar. They connect outside of the jar with wires to an LED, and the cell produces a little electricity. The magnesium strip slowly corrodes, and as it runs it produces little gas bubbles on the magnesium.
I think the bubbles might be hydrogen, but my partner thinks they might be $\ce{CO_2}$. We've tried googling for equations for magnesium and carbon, but haven't found someone who seems to be talking about a cell like this. What's going on? Also, what is the stuff forming on the magnesium? Some of it seems to be flaking down to the bottom of the jar. I think it's some kind of magnesium oxide, but that's a guess.  Can someone guide us? Thanks!


text:
Off taste of over-boiled soup, what is the chemistry?
I had a tin of Asian soup where the label explicitly states "don't boil". It happens, it did boil, and afterwards had a taste that reminded me clearly of the smell I remember from stale soap water, such as from washing clothes without a washing machine.
What is it that caused that? Asking for the substances as well as how they are produced.

Telling us what the listed ingredients on the back of the can were might help.

@ClangorousChimera This is the first time that I regret not having empty tin can as a fetish. The can has been discarded long ago ant this soup is not continuously marketed, so I have to relay on my memory: main ingredient were glass  noodles, with some chicken meat, enriched with east Asian spices. But i did have the same taste with other soups too.

Hmmm well I'm not really sure what it could be. Perhaps you should start of a fetish of having empty Asian soup tin cans until you find that kind that you encounter the soapiness with. This does seam like a puzzling situation.


text:
Wine tasting like soot
I'm not sure if this question is for chemistry.
I had a bottle of (cheap) wine. It had a strong taste of soot, to an extend that I suspected that in the filling process some machine oil might have gotten into it.
I discarded the bottle, but some time later, I've found the same taste, to a lesser extend, in some other wine. Now it looks like my taste buds have become more sensitive to that, as I can taste it now (depending on what, I don't know) quite frequently, mostly on a low scale.
Any idea to explain that?

Guess you should buy better wine now ;)

http://cooking.stackexchange.com/?

@Mithoron define "better wine" as opposed to "wine sold for a higher price" ;-) My "problem" is that some wines sometimes taste like soot, but a month later another bottle from the same market, same brand, same year, same purchase, does not. Looks the difference is in my taste buds.

@bon thanks for the link, maybe if there will be no satisfying answers her I will have a look there. But I don't like to get hyperactive about this.

@bon searching there is  a good idea which I can do without creating yet another account !

The odour and aroma of wine consists of about 600–800 components. Many of these compounds have been described with a range of odour/flavour sensations, even in more than one category, and often have differing flavour impressions, dependent upon their concentrations in aqueous solution or in the air-space above.
The odours and flavours in wine tasting are typically divided into groups, for example: Floral, Woody, Rustic/Vegetal, Balsamic, Fruity, Animal, Empyreumatic, Chemical, Spicy, and Etherish. [Bakker, J.; Clarke, R. J. Wine flavour chemistry; 2nd ed.; Wiley-Blackwell 2012]
What you describe as “strong taste of soot” probably belongs to the Chemical group, which is associated with synthetic chemical manufacture.
Typical compounds in this group that might be responsible for the soot aroma are volatile phenols:
3-methylphenol (tarry/leathery odour)
4-methylphenol (tarry/smoky odour)
2-ethylphenol (smoky odour, phenolic flavour)
2-methoxyphenol (smoky/woody odour)
A special substance that is believed to be responsible for a strong, petrolly kerosene-like aroma developed during aging is 1,1,6-trimethyl-1,2-dihydronaphthalene, which is formed from carotene breakdown.

Thank you for your answer, is there a way to get those chemicals to smell and taste, preferably not to have to buy every one in a 50g charge? Something like  some university or school for food-chemistry providing access to public one  a year ... or something alike?

@GyroGearloose Suggest not to taste pure Cresol...


text:
Where can I access raw values for near infra red absorbance of liquid water?
I have encountered several sites where water's near-infra-red absorption spectra has been plotted as a figure (for e.g, here). Is there a site where I can access these raw absorbance/transmittance values? Eyeballing is prone to errors and inaccuracies.
What I have found: NIST has a nice plot of the IR absorption spectra of liquid water, and I can potentially access the raw data by looking at the plotting functions, but as you can see it has no near-IR data.

I found a database called HITRAN that offers what I wanted. They support a python API that can be used to pull the relevant data. The code I used is written below (you'll need the hapi.py file from here).
from hapi import *
from matplotlib import pyplot as plt
fetch('H2O',1,1,7500,11500)
nu,coef = absorptionCoefficient_Lorentz(SourceTables='H2O')
plt.plot(nu,coef)
plt.show()


text:
Which coloring agents are responsible for the color in red tea and yellow tea?
Have this question in a biochemistry course, can't seem to find it anywhere. Anyone happen to know an alkaloid or something else involved? Thanks a lot!

Take a look on flavonoids ;)

I believe that it is the different polyphenols (Flavonoids are a type of polyphenol) in tea that give it its colors. Black tea has thearubigins that give it a red-brown color. The yellow color in tea comes from theaflavins. The color of the tea also depends on the acidity of the water used, because thearubigins are very weak acids, and the anions they produce have strong colors. If the water you use is alkaline, the color will be stronger. If it is acidic, the color will be lighter.


text:
Synthesis of Potassium Trioxalatoaluminate (III) Trihydrate
We mixed $\ce{Al}$ powder and hot water. Then we added $\ce{KOH}$ to the $\ce{Al}$ solution and it bubbled heavily (producing $\ce{H2}$). When that was over we added oxalic acid dihydrate to the boiling aluminate solution. We filtered and added cold ethanol. How do write these reactions?
I was thinking:
$$\ce{Al + KOH + 2H2O -> K[Al(OH)3] + H2}$$
and when oxalic acid comes in:
$$\ce{K[Al(OH)3] + C2H2O4.2H2O ->K3[Al(COO)3].3H2O}$$
The problem is this site says that the reaction is wrong and What is the reaction between oxalic acid and potassium permanganate?  is a similar question but has a completely different looking answer. I'm lost.

Welcome to Chemistry.SE. I like that you described the steps and your observations! You're not lost, only the stoichiometry is a bit off. That can be fixed :)

You have already concluded that the bubbles indicate that hydrogen gas is formed. Let's assume that two protons are reduced. Each receives an electron and they form a molecule of $\ce{H2}$:
$$\ce{2H+ + 2e- -> H2}$$
Where did the electrons come from? Did the aluminium powder dissolve?
$$\ce{Al -> Al^3+ + 3e-}$$
We are now having two equations, one for a reduction, one for an oxidation (of aluminium).
Let's add these two equations in a way that the electrons cancel out. For that, we multiply the first equation by 3, and the second equation by 2.
$$\ce{2Al +6H+-> 2Al^3+ + 3H2 ^}$$
Something's still wrong with the equation: You didn't throw the aluminium powder into a sea of protons, but suspended it in hot water. Let's add some $\ce{OH-}$ on both sides:
$$\ce{2Al +6H2O -> 2Al(OH)3 + 3H2 ^}$$
This is much better, but still not perfect: You did not describe a cloudy precipitate of a hydroxide being formed, and your first step took place in  the presence of potassium hydroxide:
$$\ce{2Al +6H2O +2KOH-> 2K[Al(OH)4] + 3H2 ^}$$
So far, we have explained the reaction of aluminium metal in aqueous alkaline solution: tetrahydroxoaluminate $\ce{[Al(OH)4]-}$ is formed.
Can you figure out the reaction with the oxalic acid on your own? Have a look at the structure of oxalate (Hint: bidentate ligand) and forget about the permanganate ;)

UPDATE
Let's add some oxalic acid, $\ce{HOOC-COOH}$ now. For the reaction, it doesn't make any difference whether it's a dihydrate, $\ce{H2C2O4*2H2O}$, or not. The reaction is performed in aqueous solution anyway. However, you might want to consider the difference in molecular weight between the anhydrous form and the dihydrate when you're planning to add a stoichiometric amount of oxalic acid.
$$\ce{[Al(OH)4]- + 3HOOC-COOH -> [Al(OOC-COO)3]^3- +2 H+ + 4H2O}$$

So I have three oxalates that are bound to Al, each in a ring form, and 3 K's to make the total charge 0? What happens to the 4 OH's and the 6 H+'s that were taken off oxalic acid to make it into oxalate? I feel like putting them together to form water but I'm not sure since it's not a perfect match. How does the fact that oxalic acid is a dihydrate play into all this? I'm sorry but I need more guidance.

This is perfect! Thank you so much. It looks like I was overcomplicating it by not combining my OHs and H, and the part about dihydrate makes a lot of sense. Thanks again for the detailed response!

I think the reaction takes place in such a way that:
$\ce{Al + 3KOH -> K3AlO3 + 3/2 H2}$
And then after addition of Oxalic acid:
$\ce{K3AlO3 + 3H2C2O4 -> K3[Al(C2O4)3]H2O + 2H2O}$


text:
To extend the life of a cartridge razor blade, should the blade be stored in oil or alcohol?
I've read varied recommendations about extending the life of a razor blade. Some say should stick the blade in oil when not in use, others say in alcohol, others say in a bowl of rice or silica gel.
I thought this community would be able to settle this:
What is the best (scientifically proven) manner to prevent corrosion of a cartridge razor blade?
I believe my Gillette Mach3 razor blades are stainless steel.

[The reason for this question is that cartridge blades are not cheap. Some may suggest going to a DE safety razor. I have tried it. It took me 2x as long to shave, so I returned to the cartridges.]

Just wash it thoroughly, then dry and keep in a (reasonably) dry place. Steel does not corrode in dry air.

@permeakra most air will contain at least some water. While this method might be time-saving, it is not the best manner to prevent corrosion, which is what the OP asked.

@NA > most air will contain at least some water || As long as it doesn't condense, it doesn't matter.

I don't believe corrosion plays any role at all in the life cycle of Gillette blades. To extend the life of a blade, don't shave with it.

@permeakra & Ivan Neretin -- I personally witness rust on my Gillette blades. I store my razor in dry air with no treatment. If you get real close, you can see little bits of orange. This, for example.

Place the razor (blades attached) down into a cup of rice.

Most blades are dulled by either the edge folding over itself, or small amounts of rust forming. To stop the first problem, you can strop the blade, but that's not really chemistry. For the second one, anything that stops oxygen and water from reaching the blade will stop it from rusting.
Silica gel does absorb moisture,so that would prevent rust from forming, but this would be temporary. You would have to replace it, and even then it might not stop 100% of moisture from reaching the blade.
Oil might prevent water and oxygen from reaching your blade, but it would make it... oily. If you are okay with that then it would probably be a fine solution.
Alcohol would prevent water from reaching the blade pretty well, as well as disinfect it. This might be the best solution to extend your blade's life. Just make sure you use medical grade alcohol.
Finally, blow drying or towel drying your blade after every use will help a lot as well.

Alcohol would prevent water from reaching the blade pretty well || alcohol may be not as good as water, but pretty close and contains quite a bit of it.

I currently towel dry the blade. Between oil and alcohol, which one displaces water better? (I don't mind an oily blade.)


text:
When water boils at its boiling point how do we call it as an equilibrium?
On boiling, water becomes a vapor, liquid volume decreases and vapour volume increases.  Since volumes are not constant, how do we call it as equilibrium?

Stop applying heat to it, and wait a bit until the boiling ceases. Now that's an equilibrium.

but in equilibrium means that they should be constant know? but is liquid and vapor constant on boiling point? they are changing then how can we call it as equilibrium?

If you keep heating it nonstop then obviously the temperature will continue to increase beyond the boiling point. Once the temperature is higher than that it is no longer at equilibrium.

You are correct.  The system is not quite at equilibrium when you are adding heat to it and the water is boiling.  However, the liquid and bubble temperatures and pressures  below the liquid surface are very close to the equilibrium values.  In my judgment, @aventurin 's concise answer below is right on target.

An isolated system consisting of liquid water and vapour is always in equilibrium.
If you are applying heat to a system consisting of liquid water and vapour the system is not in equilibrium.

I will give you an answer that doesn't rely with microscopic variables or statistical physics.
There is a misconception about equilibrium definition. Equilibrium is different of "a particular definition equilibrium state". We define the thermodynamic state of a simple system with a function of few macroscopic variables. For example S(E,V,N) (entropy, energy, volume and lets say mass). But you also can define it in therms of (more useful in the context of the question) the Gibbs free energy (G), G = G(T,P,N). In such case the state specification does not require the volume. If you have a composite system in the circumstances that you described (or two phases), you have (for system 1 and 2)
$$ G = G_1 + G_2 = G_1(T,P,N_1)+G_2(T,P,N_2) $$
but as $G$ is an homogeneous function it is trivial to prove that the sum remains constant so we have a lot of states of the composite system with the same total free energy, that it is in fact the minimum energy that the composite system can have.
Then by the incorrect usage of the language we call this set of states "equilibrium state".


text:
Isomers with molecular formula C₄H₇Cl
What are the possible isomeric structures of chlorobutene $(\ce{C4H7Cl})?$
I've come up with seven:

cis-1-chlorobut-1-ene
trans-1-chlorobut-1-ene
2-chlorobut-1-ene
3-chlorobut-1-ene
1-chlorobut-2-ene
cis-2-chlorobut-2-ene
trans-2-chlorobut-2-ene

I can think of nine isomers.

Can you please suggest the two isomers I'm missing?

Wolfram Alpha can find the isomers of a given formula, quite nicely as well.

@AliCaglayan Quite nicely except that it misses two of them? (2 and 3 from orthocresol's figure in my answer) I assume that it is simply extracting from a database except that surely 1-Chloro-1-methylcyclopropane is in any reasonable db? Weird.

@gilleain I am having a hard time finding any literature on 1-Chloro-1-methylcyclopropane or 1-Chloro-2-methylcyclopropane which might mean that WA has not included it because there is no data on it?

@AliCaglayan Fair point. I suspect that it what WA is doing. It makes a lot of sense, actually to only return results about experimentally verified entities, and not just theoretical ones. It's just amusing to me that WA (which can do much more complex mathematics) doesn't just list all possible ones. EDIT: Well there's this paper: http://pubs.acs.org/doi/abs/10.1021/jo00894a001 that talks about photchemical rearrangement of 2-methylallylchloride to 1-Chloro-1-methylcyclopropane

There are 12 different constitutional isomers:

There is also the further possibility of stereoisomerism in some of the compounds above.

3: Two diastereomers (cis- and trans-) are possible.
7, 8, 9: Geometrical isomerism is possible in these alkenes.
11: C-3 (the carbon with the chlorine) is a chiral centre, which leads to two possible enantiomers.

That makes a total of 17 possible isomers.

And that's not counting the cis/trans things.

I hope I didn't miss anything out.

@orthocresol much better answer than mine! Should be a way to transfer answers, I guess. Thanks

Well, you did the work finding the 12 constitutional isomers, I just touched it up a bit :)


text:
Isomers of Octahedral Coordination Compound with Two Bidentate Ligands

Draw all possible isomers of $\ce{[CoNO2(en)2NH3]+}$

My answer:

I reckon that there are 4 isomers, as drawn above. Am I correct?

What about trans isomers (I mean, when NO2 and NH3 are opposite to each other)?

You forgot the linkage isomerism of NO2. https://en.wikipedia.org/wiki/Linkage_isomerism

Note: I am answering this as a homework question, i.e. not actually giving a definite answer.
As a first idea, reduce the degrees of freedom in your complex. You could start by assuming what the number of isomers is for $\ce{[CoCl2(en)2]}$, in which the two non-ethylenediamine ligands are identical. And you can further reduce that complex’ complexity by going to something like $\ce{[CoCl2(NH3)4]}$, in which the ethylene bridges have also been removed. (It may have made more sense to stick with $\ce{NH3}$ ligands in the first step, but then I cannot simply substitute for ammine in the second step, so I chose something entirely different there. What you choose is up to you, as long as you don’t create ambiguity.) For the tetraammine complex, it should be clear that there are two different isomers: cis and trans. Upon readding the ethylene bridge, these two isomers still exist. Do your four structures account for this isomerism?
Next up, ask yourself whether any of the isomers of $\ce{[CoCl2(en)2]}$ can be optically active, that could give you another set of isomers. This is important because of the next step.
This next step involves readding the difference between the two monovalent ligands, i.e. going back to $\ce{[Co(NO2)(en)2(NH3)]+}$. Remember to consider the topicity of the two chlorides you had before in case you run into enantiotopic or diastereotopic chlorides that would give you a new set of isomers.
Finally, remember that nitrite has two or three possible means of attaching to a metal centre: κN, κO and κO,O'. Remember to consider whether all theoretical possibilities are actually possible.
The final answer I came up with is given in a spoiler tag.

 The trans-isomer has a high symmetry of $D_\mathrm{2h}$ if I did the transforming correctly, meaning that it is does not have enantiomers. Also, it renders the two chlorides homotopic, meaning that we get one trans isomer.

 There are two cis-isomers that behave as enantiomers of each other: a Δ and a Λ isomer. These two each have homotopic chlorides again, meaning we get a set of two cis-isomers. To be honest, I only saw that when I tried it out with a modelling kit.

 This means that substituting one of the two chlorides with something else will result in identical compounds, giving us three different isomers. (Note that you have drawn two pairs in which one transforms into the other by simple rotation. You only have two different complexes drawn in the question.)

 The chelating nitrite ion is not possible in your complex since you have a rather small metal centre that only allows octahedral coordination. Therefore, each of the three isomers we established earlier can exist in a κN and a κO form. That gives us a grand total of six possible isomers whereof two pairs are enantiomers of each other and three are constitutional isomers of the other three.


text:
Melting point and boiling point in isomers
We know, that cis isomers are less stable than trans isomers. What I don't get is why the boiling point is changing as opposed to the melting point. The change is melting point makes sense to me considering the stability.
Also the difference of mp and bp in iso and n-butene is not clear to me. How can I possibly explain the change in these physical properties?

Melting and boiling points aren't really related to the stability of the isomers. Perhaps you could give your explanation for the melting point trend.

Melting and boiling point in relation to cis and trans isomers
The factor that affects melting point is, for the lack of a better term- packability. Molecules that are symmetrical can fit into each other more snugly and are difficult to melt. Therefore trans isomers, being symmetrical, always have a higher melting melting point.
As for boiling, molecules that are held together by any sort of intermolecular force are harder to boil. Cis isomers being unsymmetrical will almost always have a net dipole moment, as opposed to trans isomers where the symmetric nature cancels out the dipole moment. Therefore they are slightly polar and attract each other. Therefore cis isomers have a higher boiling point
Edit: I don't know why isobutene has a higher melting point.

I couldn't agree with you on the mp and bp of straight and branched hydrocarbons, as I've read, straigh chains can be stacked better in crystalline lattice resulting and higher mp and they have more surface area than the branched so they are bigger, therefore have higher bp.

I'm very sorry, I had misinterpreted what was in my text book. Thank you for pointing it out. I've edited it.


text:
Crystallization Process
In my lab, my teacher asked me to not cover the beaker or crucible during the crystallization process of different solutions (e.g, FeSO4, NaCl etc). But on the internet, on Youtube, I've come across videos where they suggested that the glassware containing the solution should be covered but loosely.
I accidentally covered my solution with a watchglass and later found crystals formed all around the crucible, inner and outer both surfaces.
So, what's the reason behind it? I think we need to let the water to evaporate and leave the crucible, but why?

Usually you leave a reaction uncovered to induce crystallization, because this allows the solvent to evaporate. As solvent evaporates, the solubility of the product decreases, causing it to exit solution and begin to nucleate crystals. Once nucleation has occurred, crystals precipitate more readily.
Whether partial or totally uncovered isn't always important depending on how volatile the solvent is or how saturated the solution is. Partially covered will slow the rate, especially if the solvent isn't volatile, but it can block loose particles from falling in. On the other hand, a saturated solution doesn't need much evaporation to go below the solubility limit of your product and instigate nucleation. Whatever your case may be, a watchglass is not a sturdy seal against evaporation so it seems like your product was still able to crystallize.
Also, depending on the heat of the reaction and melting point of the product, sometimes evaporation isn't necessary at all, but it's still convenient to allow as much solvent as possible to evaporate in order to isolate dry crystals. If your solvent dries over night then you don't have to remove it the next day, so as a rule of thumb a reaction product is left uncovered unless there's something air/moisture sensitive present.


text:
Confusion about atomic/molecular orbital terminology
I have come across several different orbital terms: atomic orbitals, natural orbitals, split-localized orbitals, molecule-intrinsic orbitals, quasi-atomic orbitals.
I don't know what the difference between these are, or whether some of them describe the same thing. Could someone help clarify what these are?

This looks too broad, also what do you know already in this topic?

The short answer is that orbitals are defined to within a unitary transformation.

Canonical Hartree-Fock equations
The application of variational principle to the Slater determinant leads to the system of Hartree-Fock equations
$$
    \newcommand{\op}[1]{\hat{#1}}
    \newcommand{\mat}[1]{\mathbf{#1}}
    \op{F} \psi_{i}(1)
    =
    \sum\limits_{j=1}^{n} \lambda_{ij} \psi_{j}(1)
    \quad \text{for} \quad i = 1, 2, \dotsc, n \, ,
$$
which can be simplified by choosing a unitary transformation $\mat{U}$, defined by the way it transforms spin orbitals,
$$
    \psi_{k}' = \sum\limits_{j=1}^{n} u_{kj} \psi_{j} \quad \text{for} \quad k = 1, 2, \dotsc, n \, ,
$$
that makes the matrix of Lagrange multipliers $\lambda_{ij}$ diagonal and leads to the system of the so-called canonical Hartree-Fock equations
$$
    \op{F} \psi_{k}'(1) = \lambda_{kk}' \psi_{k}'(1) \, , \quad k = 1, 2, \dotsc, n \, ,
$$
or, in a much more familiar form using symbol $\varepsilon_k$ for $λ_{kk}'$ and dropping the primes
$$
    \op{F} \psi_{k}(1) = \varepsilon_{k} \psi_{k}(1) \, , \quad k = 1, 2, \dotsc, n \, .
$$
The canonical Hartree-Fock equations can be thought of the eigenvalue ones, for solving which there exist a number of well-developed algorithmic approaches, thus, they are so convenient to work with.
Slater determinant invariance to a linear transformation
It can be shown that Slater determinants composed of original and transformed spin orbitals are related as follows,
$$
    \Phi' = \det(\mat{U}) \Phi \, ,
$$
which by taking the absolute values and squaring both sides leads to
$$
    |\Phi'|^{2} = |\det(\mat{U})|^{2} |\Phi|^{2} \, .
$$
This means that the transformed Slater determinant $\Phi'$ describe exactly the same system as the original Slater determinant $\Phi$ in every observable way. Or, in other words, the Slater determinant is invariant to any linear transformation of spin orbitals (except for normalization). The only advantage of the unitary transformation from that perspective is that $|\det(\mat{U})|^{2} = 1$, so that there is no even need for renormalization.
Transformed canonical orbitals
Now, as it was mentioned earlier, it is convenient to do calculations with canonical orbitals, however, once the Hartree-Fock equations are solved, the resulting canonical spin orbitals can be transformed by any arbitrary transformation making no changes with respect to observable properties of the many electron system in question. And it advantageous to use unitary transformations since there is no need for renormalization of the Slater determinant.
So, other sets of orbitals can be constructed by transforming the canonical orbitals, i.e. by forming linear combinations of them. The way these linear combinations are formed (i.e. the choice of a unitary transformation) is dictated by what is the ultimate goal of the transformation. For instance, canonical orbitals are usually fairly delocalized over a molecule, rather than concentrated in limited spatial regions of it as anticipated from general chemistry models. Thus, of a particular interest for a chemists is a transformation which localizes the orbitals. The way localized orbitals are obtained depends on a particular choice of the unitary transformation and there are in fact few different approaches.
But just localizing the orbitals might not be the end of the story. For instance, one might want to not just localize them, but do so in a way that the maximum-occupancy character in localized 1-center and 2-center regions of the molecule. This procedure results in the most accurate possible natural Lewis structure corresponding to a given Slater determinant and is the basis of the natural bond orbital model.

You say that canonical orbitals are quite delocalized in a molecule. I recently ran some HF calculations on 1,2-dioxetane, and looked at the orbitals in order to decide on the active space to be used in subsequent CASSCF calculations. Being used to the "localized picture" from textbooks, I found it difficult to analyze and attribute one molecular orbital to, for example, the $\sigma$ C-O bonding orbital, because they were so delocalized in the molecule. Is this the same kind of delocalization you mention?

@AndersMB, yes, this is exactly what I'm talking about.


text:
percent dissociation
Would these answers be correct?
. Of 100% of the NaCL molecules dissociated, the Van't Hoff factor i, would be equal to 2 for NaCl. We assumed it equaled two when we figured out Kb of the NaCl solution.
a. Now we want it find the percent dissociation of the NaCl molecules by finding i using the equation below (use Kb = 0.512℃/m and your calculated molarity of NaCl solution ) show all work
 i = ( ▵Tb) / Kb x m)
i experimental = 1.8 [ {1.8/2} * 100% = 90% dissociation]    i theoretical = 2 (100% dissociation)
my answer - % dissociation = (1.74/2) *100% = 86.9%
b. . Do you feel comfortable in making the assumption that the NaCl in your experiment was completely dissociated? Why or why not?
I do not feel comfortable in making the assumption that that the NaCl in my experiment was completely dissociate because it lead to a 16.6% difference in my experiment
Thank you

This is one of those questions where I think you'll need to know more chemistry than you probably have been taught.
The freezing point depression formula is only usable (i.e. linear) for "dilute" solutions. As the concentrations of solutes become larger, the relationship becomes non-linear.
So when $\ce{NaCl}$ is dissolved in water all the sodium and chloride ions are truly dissociated. There are no "molecules" of $\ce{NaCl}$ floating around in the aqueous solution.
Imagine the solution this way. Instead of the water molecules bouncing around in the solution more or less randomly, the positive charged $\ce{Na^+}$ ion attracts the more negative oxygen atom of the water molecule leaving the positive hydrogen atoms on the outside of this first "shell" of ligands. Now a second "shell" of water molecules will be attracted to the first (Oxygen to hydrogens) and so on. Now each shell has more and more water molecules but the binding energy decreases as the shell number increases. So the gist is that there is an exponential decrease in the ordering of the water molecules. Ultimately the ordering becomes random again as in the bulk of the water.
Now in dilute solutions these spheres of water around each ion are separated by so much water that they never interact. But as the solution becomes more concentrated then these spheres of water do start to interact. The interaction is measured by what is known as an activity coefficient.


text:
How to determine the work function and electron velocities from data on the photoelectric effect?

A metal received light with wavelength $300~\mathrm{nm}$ in experiment A and light with wavelength $500~\mathrm{nm}$ in experiment B. In one experiment the electron speed was twice the speed of that in the other experiment. Calculate the velocities and the work function ($W$) of the metal.

Because it's the same metal in both experiments we have $W_\mathrm{A} = W_\mathrm{B}$, which means that $E_\mathrm{A} - K_\mathrm{A} = E_\mathrm{B} - K_\mathrm{B}$.
We know that $E=hc/\lambda$ and $K=mv^2/2$. Because $\lambda_\mathrm{A} < \lambda_\mathrm{B}$, we have $E_\mathrm{A} > E_\mathrm{B}$ and $K_\mathrm{A} < K_\mathrm{B}$ which means $v_\mathrm{A} < v_\mathrm{B}$ and $v_\mathrm{B} = 2v_\mathrm{A}$.
Then I made the substituions in my initial equality but when I try to calculate $v_\mathrm{A}$ I obtained a negative number inside my root.

Hint: if $E_\mathrm{A} - K_\mathrm{A} = E_\mathrm{B} - K_\mathrm{B}$ and $E_\mathrm{A} > E_\mathrm{B}$ then $K_\mathrm{A} > K_\mathrm{B}$. Given two numbers you have to subtract more from a bigger one to make them equal. Or, from physical perspective, the bigger is the energy of a photon you shoot, the bigger is the kinetic energy of an electron ejected.

Consider the difference between the kinetic energy in case A and B:
$$K_A-K_B=(E_A-W)-(E_B-W)=E_A-E_B$$ where the work function cancels out because it is the same metal in each experiment.
The right hand side of this is easy to evaluate using $hc/\lambda$, which gives $$E_A-E_B=hc\left(\frac{1}{\lambda_A}-\frac{1}{\lambda_B}\right)=(\pu{1.9878\times10^{-25}J\cdot m})\cdot(\pu{1.333\times10^6m^{-1}})=\pu{2.65\times10^{-19}J}$$
To solve for the left hand side, we simplify to a single variable by recognizing that $v_a=2v_b$ (as Wildcat said, the right hand side is positive and the only way the left hand side will be is if A has the larger velocity): $$K_A-K_B=\frac{m}{2}(v_A^2-v_B^2)=\frac{m}{2}(4v_B^2-v_B^2)=\frac{3}{2}mv_B^2$$
With an electron mass of $\pu{9.109\times10^{-31}kg}$, this gives $v_B=\pu{4.404\times10^5 \frac{m}{s}}$ and $v_A=\pu{8.809\times10^5 \frac{m}{s}}$. The work function can then be solved by evaluating either $K_A$ or $K_B$ and solving for the difference $E_x-K_x=W$ where $x$ is either A or B.


text:
While finding out chiral centers should lone pair be considered as a group?

My book says it is a chiral center.
Should I count lone pairs as one of the groups while finding chiral center ?

Btw, entering even a vague query like "sulfur chiral" on Google gives you the answer... I'm not trying to discourage you from asking questions but honestly you should first and foremost be able to try and help yourself.

I am asking in general. Do we consider lone paira while finding chirality centre

No for nitrogen, yes for sulfur and phosphorus. Maybe try googling "lone pair chiral", the first hit tells you the answer.

Chiral compound does not depends on chirality center. There are compounds with no chiral center but are chiral .

What are you telling me that for? In the compound you used, if the central atom is a chiral centre, then the compound is chiral. If you want to argue about the existence of other forms of chirality then you should not have attached an image of a simple sulfoxide like that.

I found this example in book so i gave it .I know it is chiral I mentioned what my books says .

The question you stated as duplicate is about trigonal S not tetrahedral.

I meam reffered mine as duplicate

@Pi5: firstly, sulphoxide is not tetrahedral. It's trigonal pyramidal. Secondly, Yes. The lone pair is analogous to a 4th group. But sometimes inversions will take place making it impossible to resolve the compound. In case of sulphur, it's not easy for inversion to take place, but for nitrogen, the inversions are rapid.

@orthocresol shouldnt you consider the lone pair as a 4th group of inversions are restricted?


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Raman Spectra of Toluene

This is the raman spectra of Toluene I obtained while working at my lab using a raman spectrometer with laser of 532 nm wavelength. Well, I am a physics student, but I want to know how information about vibrational modes and functional groups are inferred from the raman spectra, let's say from this spectra of Toluene. It would be of great help, since I am very curious about this spectroscopy method. I know physics part of raman spectroscopy, but really struggling with its application part.

Chemists have tables and tables and tables that correlate functional groups to numbers.
For example:
http://www2.ups.edu/faculty/hanson/Spectroscopy/IR/IRfrequencies.html
If we see a number on the graph, a certain type of functional group may be present.  The sharp one at 3056 tells me aromatic hydrogen.  The 2919 is the benzylic hyrodgens.  Since I'm a synthesis guy, knowing the exact types of vibrations occurring there was never important to me.
Be warned, chasing every peak is normally quite frustrating.  Vastly more important is normally the overall look for comparison with a known compound, which is sort of a molecular finger print.

You have asked question about vibrational modes and functional groups with respect to Raman spectra.
General information:
Specific vibrational modes are deduced by normal-coordinate analysis. This gives most probable observable frequency for certain vibration (or vice-versa). Normal coordinate analysis is core idea in spectral interpretation and tables (  one such table  )  for quick interpretation were made decades back using such procedures. All those peaks you see in the spectra are coming from different vibrations.
Other Clues:  One clue is presence of combinations and overtones. These are observed when vibrational quantum number changes by two or more.
Some vibrational modes may have closer frequencies which can cause resonance and result in splitting of peaks. This is another clue to deduce more details.
Isotopic substitution gives ultimately large amount of information of the atoms involved, since the peak position changes considerably.
Important is to understand that a peak here refers to a vibration (and there are different peaks and so different kinds of vibrations) Each vibration tells about the atoms involved and symmetry of vibration (depending upon experiment).
For toluene, we can simply correlate to benzene first. Some interpretation is following:
~3056 cm-1 = CH stretch
~1000 cm-1 = ring stretch (symmetric mode)
~1606 and 1580 cm-1 = splitting by Fermi -resonance
~785 cm-1 = CH bend
and so on.

@Roshan Shrestha  Do you need more details ??


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Will a strained three membered ring open up when exposed to heat or acids/base?
I was wondering if a three membered ring or a three membered epoxide would ever open up spontaneously when exposed to heat, or acids/bases?
I would think at some point, but will it routinely happen to a three membered epoxide?

Well, the whole thing is pretty strained, so yes, it would snap open quite easily.

Definitely, three membered rings are extremely strained. The bond angle is extremely small.

There no other oxiranes (epoxides) but three-membered rings! On the opening with bases, see http://chemistry.stackexchange.com/questions/27288/do-epoxide-rings-react-with-bases/27290, on the opening under acidic conditions see http://chemistry.stackexchange.com/questions/2655/sn1-like-reaction-acid-cataylzed-hydrolysis-of-epoxide and http://chemistry.stackexchange.com/questions/4305/regioselectivity-of-acid-catalyzed-ring-opening-of-epoxides

I realized I added acids/bases into the question, but could anyone tell me if there is any info on spontaneous opening as heat increases in solution


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Why doesn't sodium carbonate decompose?
I was asked to find the hypothetical $\Delta H$ and $\Delta S$ of the decomposition of sodium carbonate and then asked to explain why it doesn't decompose.
I calculated the enthalpy and entropy change and from that I calculated the Gibbs free energy change. I was hoping to get a positive answer, but actually I got a negative answer, indicating that the process is actually spontaneous.
So is the reason that sodium carbonate doesn't decompose due to its kinetic properties, i.e. the reaction is too slow?

What is the decomposition pathway? Don't leave us in the dark! But in general if it's thermodynamically favoured but doesn't happen then it's a kinetic factor.

Your $\Delta G$ must be wrong. Sodium carbonate is thermodynamically stable.

I calculate a positive standard free energy change for the decomposition reaction.

Okay, thanks for all the help. I double checked my values I still get a negative value. The question probably gave the wrong values for the enthalpy and entropy change.

OK.  How about telling us what you got for the free energies of formation of each of the products and the reactant?

@ChesterMiller Thanks for checking it out for me. I got 319.21kJ for enthalpy change and 3.5kJ for the entropy change. But don't worry about it, they probably gave the wrong values.

From the values on the Wikipedia page for sodium carbonate, the Gibbs free energy of formation is negative and thus, formation is thermodynamically preferred over decomposition at $\pu{298K}$. Your given values seem to be wrong after all.
By the way, $\Delta H$ and $\Delta S$ should be the negative of those given on that page for your case, making $\Delta G$ a positive number.