text:
Why are 'table salt and sugar are both terrible candidates for recrystallization'?
I received a comment on one of my other chemistry questions stating that

table salt and sugar are both terrible candidates for recrystallization, albeit for different reasons.

Why are salt and sugar terrible candidates for recrystallization?

What do you know about recrystallization, to begin with?

Meanwhile, if somebody else attempts to answer this, please keep in mind that my comment implied the conditions of a small-scale home experiment. Industrial conditions can be pretty different.

I understand the physical principle of separation by recrystallization. I don't know what properties of sucrose or NaCl are by themselves making those compounds unlikely to be purified successfully by recrystallization.

What are the principles involved, then? See, I need to know what you already know, otherwise how am I to know where to start? Why would a substance dissolve and then crystallize, to begin with?

Solubility is a function of temperature. I want to recrystallize A from a mixture. A has a higher solubility than the other components of the solid. Hot solvent at BP and volume at saturation of A is added to mixture. A good solvent will have certain properties, especially dissolving the whole mixture at its BP. Slow cooling to form seed crystals of A. Continued cooling favors more A being added to existing crystals of A.

Good. Now, is that function of temperature the same for all compounds? Obviously not. Look up the solubility of $\ce{NaCl}$ at different temperatures.

I used Python to plot the solubilities of sucrose and NaCl in water at 1 atm, and I see the solubilities appear to be quadratic functions of temperature with different coefficients.  Sucrose is more soluble and changes solubility 'faster' than NaCl.

I plotted solubility with temperature, I should add.

Quadratic or not, it's not that important. Does it change much (for NaCl, I mean)? How much salt would you use per 100g of water, and how much of that would you eventually lose?

Fair enough about the model structure. Extrapolating, if I keep decreasing the temperature toward 0 K I hypothetically expect pretty much all of the salt to come out of solution. More practically speaking, my solvent will freeze well before 0 K. At 0 C I expect the mother liquor will hold onto up to ~35.65 g of NaCl per 100 g of water, which is considered loss if I stop cooling at that point.

The solubility of $\ce{NaCl}$ has very poor temperature dependency. If you dissolve it at $100^\circ\rm C$ and precipitate at $20^\circ\rm C$, you'll essentially lose about 35 g to purify 3 g. Even using a fridge, you'll still lose more than half of the product, which is quite a bad business. Industrial process relies on evaporation, I think.
Sugar is another story. It does have a nice temperature dependence, but it dissolves too well, even when cold. Just imagine dealing with that awfully viscous syrup, like molasses. It would take forever to filter, and quite a while to crystallize. Industry can handle that, but for home chemistry I'd suggest something more pleasant.

Haha, reminds me of an experiment a colleague told me about, where students in Bochum tried to create a saturated solution of sugar in water (in a coke bottle). Every now and again, they would drop another two lumps of sugar in but it would just keep dissolving.


text:
What is the potential to use for non-dispersion interactions?
Although liquid argon’s intermolecular potential can be described using a Lennard-Jones (LJ) potential, which takes into account dispersion forces (induced dipole-induced dipole interactions), what potential do you use when also including the effects of dipole-induced dipole and dipole-dipole interactions?

Related: http://chemistry.stackexchange.com/questions/27803/why-are-c6-dispersion-forces-named-c6

Because all three of these interactions are of the same order in the interelectronic distance ($1/r^{6}$), they are all captured within the Lennard-Jones potential to the extent of the accuracy of the potential. From the "Limits" section:
The term Van der Waals (vdW) forces roughly includes the following intermolecular forces:

the permanent dipole-permanent dipole force (Keesom force), which is an electrostatic term,
the permanent dipole-induced dipole force (Debye force), which is an induction or polarization term,
the induced dipole-induced dipole force (London dispersion force), which is the dispersion term.

the forms of which are
$$
\begin{align}
V_{\text{Keesom}} &= \frac{-m_{1}^{2}m_{2}^{2}}{24\pi^{2}\epsilon_{0}^{2}\epsilon_{r}^{2}k_{b}Tr^{6}} \\
V_{\text{Debye}} &= \frac{-m_{1}^{2}\alpha_{2}}{16\pi^{2}\epsilon_{0}^{2}\epsilon_{r}^{2}r^{6}} \\
V_{\text{LJ}} &= 4\epsilon \left[ \left( \frac{\sigma}{r} \right)^{12} - \left( \frac{\sigma}{r} \right)^{6} \right]
\end{align}
$$
Note that the LJ potential contains both a repulsive ($1/r^{12}$) and attractive ($-1/r^{6}$) component. This is not the form of the dispersion interaction, but is both a suitable mathematical form that encapsulates a few noncovalent (vdW) interactions together and is easy to evaluate on a computer.
You may also be interested in the form of a more complicated potential, such as the AMOEBA polarizable force field, which uses a slightly different form for the pairwise vdW interaction ($i,j$ are atoms)
$$
\begin{align}
U_{\text{vdw}}(ij) &= \epsilon_{ij} \left( \frac{1.07}{\rho_{ij} + 0.07} \right)^{7} \left( \frac{1.12}{\rho_{ij}^{7} + 0.12} - 2 \right) \\
\rho_{ij} &= \frac{R_{ij}}{R_{ij}^{0}} \\
\epsilon_{ij} &= \frac{4\epsilon_{ii}\epsilon_{jj}}{(\epsilon_{ii}^{1/2}+\epsilon_{jj}^{1/2})^{2}} \\
R_{ij}^{0} &= \frac{ (R_{ii}^{0})^{3} + (R_{jj}^{0})^{3} }{ (R_{ii}^{0})^{2} + (R_{jj}^{0})^{2} }
\end{align}
$$
and also considers up to permanent and induced quadrupole moments
$$
\begin{align}
U_{\text{elec}}^{\text{perm}}(r_{ij}) &= M_{i}^{T} T_{ij} M_{j} \\
T_{ij} &= \begin{bmatrix}
1 & \frac{\partial}{\partial x_j} & \frac{\partial}{\partial y_j} & \frac{\partial}{\partial z_j} & \mathrm{L} \\
\frac{\partial}{\partial x_j} & \frac{\partial^2}{\partial x_i \partial x_j} & \frac{\partial^2}{\partial x_i \partial y_j} & \frac{\partial^2}{\partial x_i \partial z_j} & \mathrm{L} \\
\frac{\partial}{\partial y_j} & \frac{\partial^2}{\partial y_i \partial x_j} & \frac{\partial^2}{\partial y_i \partial y_j} & \frac{\partial^2}{\partial y_i \partial z_j} & \mathrm{L} \\
\frac{\partial}{\partial z_j} & \frac{\partial^2}{\partial z_i \partial x_j} & \frac{\partial^2}{\partial z_i \partial y_j} & \frac{\partial^2}{\partial z_i \partial z_j} & \mathrm{L} \\
\mathrm{M} & \mathrm{M} & \mathrm{M} & \mathrm{M} & \mathrm{O}
\end{bmatrix} \left( \frac{1}{r_{ji}} \right)
\end{align}
$$
along with the other typical bonded terms found in modern force fields.
Here is a nice set of notes that describes the interaction between electric multipoles to different orders.

Just keep in mind that the potentials for Keesom force and Debye force here are all averaged for ALL angles. The exact dependence on the inter electronic distance w/o being averaged are here: http://www.chem1.com/acad/webtext/states/interact.html

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Naming alkyl groups of an ester
$\hspace{6.4cm}$
I would like to know how to name the alkyl groups connected to the above molecule. For example, would the far left alkyl group be considered a methyl ester group? Would the alkyl group bound to the oxygen be a methylene ester group?

You are going about naming them incorrectly, try breaking it into two parts: the conjugate base of a carboxylic acid and an alkyl group.
$\hspace{5.9cm}$
Combining the two, we get:
$\hspace{6.9cm}$
Both the carboxyl and alkyl carbon chains can be numbered, with the carboxyl chain starting at the carboxyl carbon, and the alkyl chain starting at the carbon that places its bond to the ester oxygen at the lowest position. See below:
$\hspace{5.5cm}$
To refer to a specific carbon, simply say "at position $x$ on the carboxyl/alkyl chain."

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text:
Should moderately soluble compounds like calcium hydroxide be denoted (aq) or (s)?
I'm learning to write balanced molecular and ionic equations. For "moderately soluble" compounds such as calcium hydroxide, how do I know whether to write the state symbol as (aq) or (s)? This is the equation I'm supposed to work with:
$$\ce{Al2(SO4)3 + 3 Ca(OH)2 -> 2 Al(OH)3 + 3 Ca(SO4)}$$

Not sure of your context, but the (aq) suffix means that the substance is an aqueous solution. The (s) suffix typically means that the substance is a solid. So $\ce{Cu(s)}$ would be solid copper, but $\ce{Cu^{2+}(aq)}$ would be a solvated $\ce{Cu^{2+}}$ ion in an aqueous solution.

In principle, you cannot know without conducting the experiment. It depends on the concentration of a moderately soluble compound such as $\ce{Ca(OH)2}$ whether it is dissolved or whether there is stuff still sitting at the bottom of the vessel undissolved.
However, you are performing a reaction which is in aqueous solution. This reaction will typically only work if all the reactants are already in solution. Moreover, this is a precipitation reaction. Thus, it wouldn’t make sense to have a non-dissolved reactant to react it with a dissolved one. Given this context, you should assume both reactants are in solution.
$$\ce{Al2(SO4)3 (aq) + 3Ca(OH)2 (aq) -> 3CaSO4 v (s) + 2 Al(OH)3}$$
Note that whether or not aluminium hydroxide precipitates depends on the resulting $\mathrm{pH}$ value. If you started off with excess hydroxide, it may well stay in solution.
The exact answer for a given compound can only be known if you know both concentration and solubility product.

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text:
Solubility of salt in heavy water vs. "regular" water
I would like to know if salt can be dissolved in heavy water ($\ce{D2O}$) as it would be in normal water ($\ce{H2O}$).
If so, what will the maximum salinity be in $\ce{D2O}$? I know the maximum in $\ce{H2O}$ is about 26% of salt by mass.

Well, consider a salt as $\ce{AB_s}$ in water you have :
$$\ce{AB}_s\rightleftharpoons \ce{A_{aq}^{q+} + B_{aq}^{q-}}$$
And $\Delta H_{s}=-\ce{U +\Delta H_{aq}}$ where $\ce{U}$ is the lattice energy.
When the salt needs to be solvate, $\ce{O-H}$-bonds of water molecules around this salt need to be rearranged. More, the "$\ce{DO}$-bonds" in $\ce{D_2O}$ are stronger than "$\ce{HO}$-bonds".$^1$ Then it will be harder to rearranged hence it will cost more energy to solvate the salt which will be then less soluble in heavy water than in water.
I found an article in the JACS and as you can see on the image below the maximum solubility of $\ce{KCl}$ in heavy water is a little less than in water (you can make a linear extrapolation) the difference is not quite importante but it looks it exists.

Look at the second reference $^2$.

$^1$ : https://arxiv.org/ftp/arxiv/papers/0706/0706.1355.pdf, p-5

 Nuclear quantum effects are particularly important in the different
  properties of light (H2O) and heavy water (D2O) where the more restricted atomic vibrations in D2O
  reduce the negative effect of its van der Waals repulsive core so increasing its overall hydrogen bond
  energy.

$^2$ : For the image, http://pubs.acs.org/doi/abs/10.1021/je60070a011

$\ce{O-H}$ bonds in water need to be broken?? If anything, hydrogen bonds are readjusted but no actual bonds are broken in the dissolution of salts …

I'd say he really mean bond energy especially he does not say  bonds in water broken.

@Jan I changed I don't know how to say exactly how to explain but there is an intecration during the solvatation of ions.

This may be far too simplistic but is not that the vibrational zero point energy is smaller in $\ce{D2O}$ compared to water since the  vibrational frequency of $\ce{D2O}$ is $\omega =\sqrt {k\mu} $ where ∗k∗ is the force const, same for $\ce{D2O and H2O} $, and $\mu$ the reduced mass, greater for $\ce{D2O}$  thus the bond dissociation energy is greater.  This would indicate that to solvate with $\ce{D2O}$ by making new bonds to ions need slightly more energy than for water hence lower solubility at same temperature.

a typo above,, I meant $\omega=\sqrt {k/\mu}$


text:
How are solvents chosen in organic reactions?
Just reading through organic chemistry book (Modern Projects in Organic Chemistry by Mohrig, et al.), and it seems like the choice of solvent is usually arbitrary.
One of the experiments designed uses hypochlorite ion as the oxidizer and acetone as a solvent to oxidize 4-tert-butylcyclohexanol, but it isn't clear why this solvent is chosen.
I'm curious if there is a list that gives the suggested solvents for different kinds of reactions, as well as some insight for this particular experiment.

There is no list per se, but there is certainly a process to determine appropriate solvents for a given reaction. For the specific reaction you mention, here are some criteria you would consider to select a solvent:

Because this is an oxidation reaction, choose a solvent that cannot be oxidized, known as an aprotic solvent.
The solvent needs to dissolve the reagents in the reaction: hypochlorite and 4-tert-butylcyclohexanol. Hypochlorite is a polar molecule, and 4-tert-butylcyclohexanol is a little polar thanks to the alcohol group. The alcohol group will be oxidized to a ketone by the reaction, so the solvent should be able to dissolve that reasonably well. Conclusion: choose a moderately polar solvent.

So a polar aprotic solvent is needed for this reaction. There are two more considerations: cost and safety
Obviously, the cheapest solvent with the lowest potential for harm should be selected.
A list of some common polar aprotic solvents for consideration might include:

tetrahydrofuran
N,N-dimethylformamide
acetonitrile
dimethyl sulfoxide
acetone

Tetrahydrofuran could be a reasonable choice, but it forms peroxides over time, which are quite explosive.
N,N-dimethylformamide has some health concerns: GHS07 and GHS08
Acetonitrile also falls into the GHS07 category: it causes cyanide to form in the body at high concentrations.
Dimethyl sulfoxide does not have health or safety concerns.
Acetone also falls into the GHS07 category: it can cause nausea and irritation at high exposure levels.
The health concerns of acetonitrile, dimethyl sulfoxide, and acetone are largely not going to pose a hazard for a laboratory experiment. All three of these can likely be found in any organic teaching lab.
Cost comparison, from Sigma Aldrich:
Acetonitrile: 2 L = $172.50
Dimethyl sulfoxide: 2 L = $208.50
Acetone: 2 L = $94.80
Therefore, choose acetone for this reaction!

Another reason: DMSO is incredibly difficult to get rid of.

To be precise, "cannot be oxidized" and "aprotic" are two rather distinct things. (Though I will admit for many common solvents they do line up.) e.g. acetaldehyde (if for some bizzare reason you tried to use it as a solvent) would be aprotic, but is very easy to oxidize.

Also DMSO may be safe, but it can readily transport reagents/reaction products through skin (hence its use in percutaneous drug delivery); the solute may be less safe.

$\text{aprotic} \ne \text{non-oxidiseable}$. Most importantly, acetone is protic!

I wasn't expecting the answer to get into the economics of the chemical catalog, I wonder if this isn't the biggest factor of all?

@Cayetano Cost is a surprisingly important factor in real life. Of course if it's just pen and paper you're free to disregard it :D

@RyanWard, many scientists might overlook the economics, but at the end of the day money rules. Even if your technology is not the best in what ever metric you need (speed, performance, etc), if you can beat your competitor on price by any significant margin, you have a really good shot at getting the sale!

I did organic chemistry in the lab for 9 years, and never once was there such a list. There are general rules for protic vs. aprotic and polar vs. non-polar, but a lot of times, when a reaction is published, the authors did a study of different kind of solvents and selected the "best" based on yield, reaction rate, ease of purification, or selectivity.

If you consider YANR (your average name reaction), the scientists who published said YANR will typically have done a set of screenings, including a ‘solvent screen’. That means, they will have run said reaction in a number of different common laboratory solvents and compared the yields. The table will then look something like this (disclaimer: all values in the following table made up on the spot):
$$\begin{array}{cc}\hline
\text{Solvent} & \text{yield} \\ \hline
\text{hexane} & 55~\%\\
\text{DMF} & 72~\%\\
\ce{CH2Cl2} & 67~\%\\
\ce{EtOAc} & 59~\%\\
\text{toluene} & 85~\%\\
\textbf{acetone} & 89~\%\\ \hline\end{array}$$
And then they will proceed to note that acetone is their solvent of choice for YANR and proceed to further screenings (additives, catalysts, bases …) before finally recording a full substrate scope table. You may have noticed that this is trial and error.
For those reactions that have been around for a hundred years, nobody really did that optimisation. Instead, the solvent is used that you always have used. Occasionally, a paper turns up in a medium-tier journal that the yield of the reaction can be improved by modifying the conditions in the following ways, including the new solvent FancyDissolve™. Rarely, a top-tier paper turns up doubling the yield and improving the selectivity by orders of magnitude because they chose dichloromethane rather than hexane which never occurred to anybody.
That is the secret behind selecting solvents.

There are many factors. You have to consider the solubility of reactants since the reactants have to mix to react better. There is a solubility chart for several organic solvents. A lab that I used to work in had one pasted on the wall.
Another factor is whether the solvent interferes with the reaction. In other words, you want to minimize side products. An example of this are Grignard reagents: you do not want protic solvents when working with them. Similarly, you do not want your solvent to chelate/react with a catalyst (unless it is a reaction that requires a chelate).
Another one is availabity. This can be cost availabilty or information availability. If there are studies that are similar to the experiment that you are conducting, you will probably use the solvents  similar to he ones that were used in previous studies. Cost may play a factor as well. Why pay extra for an exotic solvent when a cheap common solvent yields ok results? (which sounds like a crime in science...but funding is not always easy to get).

Whether or not being satisfied with OK results depends on what your goal is. If making the stuff is your goal, OK probably isn't good enough. But if making the stuff is only a means to an end then OK is probably fine.


text:
Why is lemon mixed with water considered alkalizing water?
The benefits and facts of alkaline water are often skewed due to the marketing of companies and uneducated hype for the innocent idea of unlimited potential. I wanted to know the specific reactions occurring when mixing water and fresh lemon slices that lead to water becoming alkalized (higher in pH).

It doesn't and I have no idea why you'd think so.

I'm voting to close this question as off-topic because we don't do lists on chem.SE.

I removed the reference to lists

Lemon water should be slightly acidic, not basic, due to the citric acid.

@Todd A reaction mechanism is essentially a list of equations; the particular wording of a question should not play such a big role here. (the question even was tagged with reaction mechanism.) This question is not only very much on topic here, but also asking to fact check some the garbage science claims by the marketing industry. As a science community I believe it is our duty to clear up such claims with real facts. As a side note: "We don't do lists" is essentially the "too broad" close reason. I am voting to leave this question open. I will also upvote it.

@Martin-マーチン - Thanks for clearing that up. Retracting VTC and upvoting.

The claim that lemon water is alkaline is simply wrong.  However, that hasn't stopped people from claiming that it's true:

Lemons are an incredibly alkaline food, believe it or not. Yes, they are acidic on their own, but inside our bodies they're alkaline (the citric acid does not create acidity in the body once metabolized). As you wellness warriors know, an alkaline body is really the key to good health.

Every sentence above is wrong.  Lemons do not turn alkaline inside the body.  They are not an alkaline food.  Citric acid creates neither alkalinity nor acidity in the body when it is metabolized.  An alkaline body is not the key to good health.  In fact, if your body were more alkaline by only 0.5 pH units from its usual state, you would probably die from it.

I wanted to know the specific reactions that occur when mixing water and fresh lemon slices [...]

There aren't really many reactions that occur.  Lemon juice is mostly water.  Mixing the lemon juice present in lemon slices with more water just dilutes it.  Lemon juice is acidic, due to the presence of citric acid and other acids.  Diluting will very slightly lower the acidity.  None of that has any relevance to the pH inside a mammalian body, which is exquisitely regulated by other mechanisms.
I don't know why people make up weird claims that lemons are an "alkaline" food, but those claims, sadly, are out there everywhere.

"I don't know why people make up weird claims that lemons are an "alkaline" food" Because they sell. The people marketing these stuff probably know that it's all rubbish.

I’m an undergrad chemistry major, and I know some about the subject. The reason that people (falsely) believe that lemons somehow “turn alkaline” in the body is because drinking a lot of lemon juice slightly raises the pH of urine. The real reason this occurs is because the citric acid forms a buffer with citrate (which is found in the renal system I believe) and prevents urine from changing pH. It cannot change the pH of your blood. Nothing you eat changes your blood pH. If you eat a whole bunch of antacids (which are alkaline) you would neutralize your stomach acid and definitely have digestive issues, but it would not change your blood pH.


text:
Coordination numbers and radius ratio

According to the image above (source), for a radius-ratio between 0.732 and 1, the coordination number is 8. However, for a radius-ratio of 1, the coordination is 12. This is the same for the boundary values between the other coordination numbers, but I’m focusing on the radius-ratio of 1 for now. I don’t understand how the same boundary value could correspond to two different coordination numbers. If you have a monoatomic metal, then all the atoms constituting it have the same radius and the radius-ratio is 1. In order to achieve the closest packing possible, the metal atoms would form a CCP or HCP structure, wouldn’t they? Those two crystal structures have the greatest coordination number. Yet other monoatomic metals, with a radius-ratio of 1, form can BCC structures, with a coordination number of 8. Why would they form such a structure when their radius-ratio can also correspond to a CN of 12? How can one radius-ratio correspond to two CNs?

Coordination number is the number of the nearest neighbours of an atom, and it is related to the Kepler's conjecture. For CN12 (12 is the 'kissing number' of a 3D sphere), there are two possible ways for the spheres to be 'packed' which are HCP and CCP. Space-filling efficiency they two are exactly the same, and it is why one coordination number can correspond to two crystal structures.
The crystal structure of metal may depend on many features, and the space-filling efficiency is just one of them. Even by the thermal expansion, the crystal structure can change. (Tin is the most familiar example.)


text:
When is diffusion steady-state?
Fick’s first law of diffusion can be used when $\mathrm{d}C/\mathrm{d}t=0$. If the concentration gradient is time-variant, then we must use Fick’s second law. However, when is diffusion steady-state and when is it nonsteady-state? How do I know when the concentration gradient will change with time and when it won’t?

There is no set answer to your problem, it depends on the geometry of the diffusional process. I give two examples.
(a) Suppose there is a volatile solute in a solvent placed in a beaker and the solute evaporates from the surface, then there is a concentration gradient of solute away from the surface as the solute is lost into the air. Similarly, if there is an electrode at which ions are being deposited as a metal, or a precipitate is being formed, then a concentration gradient is set up, but as material is removed no steady state can be set up as there is a finite amount of substance present.
Solving Fick’s 2nd law produces an equation describing the concentration vs. time which is $c_{x,t}=c_0erf(x/(2\sqrt{Dt}) $ where erf is the error function, D the diffusion coefficient, t time and $C_0$ the initial concentration. (The initial conditions are $t=0, x\ge 0, c=c_0$;  $ t>0,x=0, c=0$;$ x \rightarrow \infty, t=0, c=c_0$ ).
The depletion distance, that is the distance from the sink that the concentration is just becoming less than than $c_0$ is given by $\delta = \sqrt { \pi Dt}$ at time t.
(b) In a closed system, for example, a cell with half of it containing a solution and the other half solvent, equilibrium will become established once diffusion into all the cell is allowed. Similarly an insulated bar half of which is initially hot will effectively come to an equilibrium at the average temperature.
Consider a tube filled with solvent and then closed and some solute injected at some place, diffusion will ensure that equilibrium will eventually be reached. Because the initial distribution of concentration  must level out to a constant value, the solution to Fick’s equation must have both a time dependent and a constant part.
At long times the temperature or concentration will become uniform and if $f(x)$ is the amount initially added at position x and $0 < x < L$, the long time value is the average value;  $\frac{1}{L}\int _0^Lf(x)dx$ . The initial condition is given by the shape of the concentration profile $f(x)$.
The concentration of added solute is $c(x, t)$  at position x and time t. Because no material (or heat)  leaves the concentration gradients at the ends of the tube are zero at all times, or
$$\frac{dc(0,t)}{dx} = \frac{dC(L,t)}{dx} =0, ~~~t>0$$
Starting with the diffusion equation
$$\frac{\partial c}{dt} = \frac{\partial ^2c}{dx^2} $$
and separating variables, gives
$$\frac{\partial c_t}{\partial t} = -Dk^2c_t$$
 and
$$\frac{\partial ^2 c_x}{\partial x^2} =-c_xk^2 $$
where k is a separation parameter to be determined.
The solution is $c=c_tc_x$.
The time dependent equation integrates to  $c_t=e^{-Dk^2t}$ but the spatial part needs special attention to incorporate the boundary conditions.
The separation parameter k can have values of $0$, or it can be positive or positive and imaginary. Only when k is positive are the solutions non-trivial and are $c_x=a\sin (kx)+b\cos (kx)$.
The gradient boundary condition ensures that the solution is the cosine term because the derivative, $kb\sin (kx)$, is zero both at $x = 0$ and at L provided that $k=n\pi/L$, where n is an integer.
Next we use k as if it were an integer and only substitute its actual value at the end. The solution is
$$c_k=e^{-k^2Dt}cos(kx)  ~~~~~~~~  k=0,1,2…$$
and the final solution is a linear combination of the k terms including $k = 0$ and this will give a term that is independent of time. The initial condition ensures that
$$ f(x)=\Sigma _{n=0}^{\infty}b_k\cos (kx)$$
giving
$$ b_k=\frac{2}{L}\int _0^Lf(x)\cos (kx)dx$$
where $k=n\pi /L$. The complete solution is
$$ c = \frac{b_0}{2}+\Sigma _{n=1}^\infty b_ke^{-(n\pi /L)^2Dt}\cos (n\pi x/L )  $$
The first term $b_0/2$ is divided by 2 to make it equal to the average concentration at long times, $\frac{1}{L}\int _0^Lf(x)dx$ because by its defining equation $b_0$ is otherwise 2 times too large.
The results below show how the initial profile eventually reaches a constant, time-independent value, the total amount of material, or heat being conserved. Uniform concentrations are found when $ t > \approx  5/ (n\pi /L)^2D $, n is found to be in the range $10$ to $20$.

figure: One dimensional diffusion in a closed tube. The summation was taken to $n=10 $ terms (left) and $20$ (right).  (Left) The solute is injected in the centre and has a gaussian profile ($\exp (-5000(x/L-0.5)^2)$ at $t=0$  Right. Case where the solution is half the cell. The initial profile is a sigmoid curve $(1+\exp (-100(x/L-0.5))^{-1}$. The diffusion coefficient $D = 1$ and $L=1$. The profile at different times is shown.

If you are wondering whether the concentration will be changing with time, you must have some initial concentration distribution that you are considering.  If this initial concentration distribution satisfies Laplace equation (including in the vicinity of the boundaries, taking into account the boundary conditions), then the concentration will not be changing with time.  Otherwise, it will.

Think of the steady state as a snapshot. Fick's first law gives you the flux or rate of diffusion for this snapshot. Now, if you just let the system diffuse, it won't look the same anymore. So, either you treat it like a snapshot or you are always increasing the concentration on one side and decreasing it on the other to balance out the changes from the diffusion process. Then the picture of the system doesn't change, and it's steady state.


text:
How to derive the electron density of Hooke's atom?
Hooke's atom has the following wave function:
$$\Psi( \boldsymbol{r_1}, \boldsymbol{r_2}) \propto \left( 1 + \frac{|\boldsymbol{r_1} - \boldsymbol{r_2}|}{2} \right) \exp\left(-\frac{(r_1^2+r_2^2)}{4}\right).$$
Question: How to calculate electronic density $\rho (\boldsymbol{r_1})$ using this wave function?
The answer, which should contain Error function, is on Wikipedia but not method.
I tried $\rho (\boldsymbol{r_1}) = \int |\Psi (( \boldsymbol{r_1}, \boldsymbol{r_2}))|^2 d\boldsymbol{r_2}$ but got wrong answer. In spherical coordinates:
$\rho (\boldsymbol{r_1}) = \int |\Psi (( \boldsymbol{r_1}, \boldsymbol{r_2}))|^2 d\boldsymbol{r_2} = \int_0^{2 \pi} \int_0^{\pi}  \int_0^{\infty} |\Psi (( \boldsymbol{r_1}, \boldsymbol{r_2}))|^2 r_2^2 d{r_2} \sin(\theta_2) d\theta_2 d\phi_2$
$\rho (\boldsymbol{r_1}) \propto \int_0^{2 \pi} \int_0^{\pi}  \int_0^{\infty} \left( 1 + \tfrac{1}{2} \sqrt{r_1^2 + r_2^2 -2 \boldsymbol{r_1} \cdot \boldsymbol{r_2} } \right)^2 \exp\left(-\frac{(r_1^2+r_2^2)}{2}\right) r_2^2 d{r_2} \sin(\theta_2) d\theta_2 d\phi_2$
where $\boldsymbol{r_1} \cdot \boldsymbol{r_2} = r_1 r_2 \left[ \sin(\theta_1) \sin(\theta_2) \cos(\phi_1) \cos(\phi_2) + \sin(\theta_1) \sin(\theta_2) \sin(\phi_1) \sin(\phi_2) + \cos(\theta_1) \cos(\theta_2)  \right]$
Aso according to the article linked on wikipedia answer should contain Error function. However Mathematica does not yield this result upon carrying out above integral. Where is my mistake?
(Note: crosspost from physics.se).

Welcome to chemistry.se! Please note, that cross-posting is discouraged on the network. A better option would have been to migrate it, but I see no reason why it shouldn't get the attention on physics it deserves.

Also Don't abbreviate Wikipedia as wiki!

PDF to article linked in Wikipedia is here http://www.chem.purdue.edu/kais/paper/ref19.pdf


text:
Dirac notation when bra and ket aren't vectors
So, I know that
$$
\langle \psi_j(x)|\psi_k(x) \rangle =\int_{-\infty}^{\infty} \psi_j^*(x) \psi_k(x) \,\mathrm{d}x
$$
and I was wondering that if this is because of the inner product rule as with vectors? Like $\langle a|b \rangle$, but the $\psi$ isn't a vector, so how would this work in this case?

Wave function for a quantum system is a vector in a Hilbert space, which is a special kind of an inner product space. Specifically, for a spin-0 particle constrained to move in 1 spatial dimension the wave function $\psi(x)$ is an element of $L^2(\mathbf{R})$ space, i.e. the space of functions that are square-integrable on the whole real line $\mathbf{R} = (-\infty;+\infty)$. And the inner product between any two elements $\vec{\psi}_{i} = \psi_i(x)$ and $\vec{\psi}_{j} = \psi_j(x)$ of $L^2(\mathbf{R})$ space is indeed defined as follows,
$$
\newcommand{\inner}[2]{\langle{#1},{#2}\rangle}
\newcommand{\ket}[1]{\,\lvert{#1}\rangle}
\newcommand{\braket}[2]{\langle{#1}\vert{#2}\rangle}
\inner{\vec{\psi}_{i}}{\vec{\psi}_{j}} := \int\limits_{-\infty}^{+\infty} \psi_i^*(x) \psi_j(x)\, \mathrm{d} x \, .
$$

Just note here that, first, it is the matter of convention in which argument the inner product is linear: in physical literature it is often defined to be linear in the second argument, while in mathematically oriented literature it is usually defined to be linear in the first argument. Consequently, mathematicians usually define the inner product between any two elements $\psi_i$ and $\psi_j$ of $L^2(\mathbf{R})$ space in the following way,
$$
\inner{\vec{\psi}_{i}}{\vec{\psi}_{j}} := \int\limits_{-\infty}^{+\infty} \psi_i(x) \psi_j^*(x)\, \mathrm{d} x \, ,
$$
as it is in fact done in some Wikipedia articles referenced above.
And secondly, physicists usually prefer to use even a different notation, the so-called Dirac bra-ket notation, in which elements of a Hilbert space are denoted as $\ket{\psi_i}$ and $\ket{\psi_j}$ (rather than $\vec{\psi}_{i}$ and $\vec{\psi}_{j}$ as it is usually done in linear algebra), while their inner product is designated as $\braket{\psi_i}{\psi_j}$ (rather than $\inner{\vec{\psi}_{i}}{\vec{\psi}_{j}}$).

or even the more consise  notation $\langle i|k\rangle$

@porphyrin, yep. Basically, things inside brackets (or underneath arrows) are just labels, so they can be virtually anything.


text:
Change in volume of two beakers on reaching equilibrium

Two beakers are placed side by side in a sealed container. The first beaker contains 100 ml of pure water. The other an equal volume of an aqueous solution of sugar. What should be the volume of the liquid in the two beakers when all physical change stop, that is when the equilibrium is reached?

I know that the vapour pressure of pure water is higher than the vapour pressure of the sugar solution. But is it anyway going to help me in solving this question.

Now that's a nice one. The problem is simple, but the answer may seem so counterintuitive that people would refuse to embrace it. Also, it is not clear how can we help in solving this question, short of telling the answer straight away. OK, let's try: what would be the vapor pressure inside the container? Mind you, I'm not asking about the exact figures (for which there is no data anyway), but for qualitative description, like "slightly higher than something" or "slightly lower than something".

What do you mean by sugar? Cellulose is a sugar like maltose but their structure are really different.

I would assume this is a solution of sucrose. Any soluble sugar would probably give the same result.

Okay, so you know that the vapor pressure of the pure water is some value, $p_w$. And you know from Raoult's Law that a solution with a nonvolatile solute of mole fraction $\chi_s$ will have a vapor pressure $p=p_w \chi_w$, where $\chi_w = 1- \chi_s$, so $p=p_w(1-\chi_s)$.
Since $\chi_s <1$, then $p < p_w$. How do you use this information?
Well, as the system approaches equilibrium, the two beakers will want to exchange water through the vapor phase to equilibrate the vapor pressures. What will that look like when it happens?  Can it happen?
By this logic, you can determine where the water is in your system. However, you cannot determine the volume(s) in the beakers without the following additional information:

Volume of the enclosing vessel - helps determine what mass of water remains in the gas phase
Temperature of the system - determines initial vapor pressure of pure water
Concentration of the sugar solution.
The non-ideal behavior of the sugar-water solution's volume.


text:
Nomenclature of multiple double bonds?
I know that in a compound, when there are two double/triple bonds its like :
1,3-diene ; 1,3-diyne and so on.
For three :
1,3,5-triene ; 1,3,5-triyne and so on.
But what about 4?
While doing a question I encountered this :
2,4,6,8-(What?)ene
Is it quad,quart,quater(like quaternary carbon) or something else?
Also, it would be helpful if you could extend it to like 5 or as much as you can double/triple bonds.

For four use tetra and after, penta, exa, hepta, octa, nona, deca, undeca and so on :-)

@9-BBN typo: exa > hexa

Yes @DHMO hexa ! :)

The multiplying affixes are given by Table 11 of the Blue Book, referenced in R-0.1.4 and R-4.1:
Number  Numerical term | Number  Numerical term
1       mono-          | 90      nonaconta-
2       di-            | 100     hecta-
3       tri-           | 200     dicta-
4       tetra-         | 300     tricta-
5       penta-         | 400     tetracta-
6       hexa-          | 500     pentacta-
7       hepta-         | 600     hexacta-
8       octa-          | 700     heptacta-
9       nona-          | 800     octacta-
10      deca-          | 900     nonacta-
11      undeca-        | 1000    kilia-
12      dodeca-        | 2000    dilia-
20      icosa-         | 3000    trilia-
30      triaconta-     | 4000    tetralia-
40      tetraconta-    | 5000    pentalia-
50      pentaconta-    | 6000    hexalia-
60      hexaconta-     | 7000    heptalia-
70      heptaconta-    | 8000    octalia-
80      octaconta-     | 9000    nonalia-

Therefore, the answer to your question is 2,4,6,8-tetraene.

Extra information
Note that there is no elision of vowels, specified by R-0.1.7:

R-0.1.7.1 Vowels are systematically elided as follows:
(a) the terminal "e" in names of parent hydrides when followed by a suffix beginning with "a", "i", "o", "u", or "y";
(b) in the Hantzsch-Widman system, the final "a" of an element prefix when followed by a vowel;
(c) the terminal "a" in the names of numerical multiplicative affixes when followed by a suffix beginning with "a" or "o", or a Hantzsch-Widman prefix or stem beginning with a vowel;
(d) the terminal "a" of an element prefix in "ababa" repeating unit names (see R-2.3.3.2) and the terminal "o" of a replacement infix when followed by a vowel;

The relevant part is (c), which states that the final "a" in the above affixes are elided when the next letter is "a" or "o", but in this case the next letter is "e" so it is not elided.
The following are given examples of elision:

benzenehexaol > benzenehexol
[1,1'-binaphthalene]-3,3',4,4'-tetraamine > [1,1'-binaphthalene]-3,3',4,4'-tetramine
1,3,5,7-tetraoxaocane > 1,3,5,7-tetraoxocane

Please note that this nomenclature is 1993 standard.


text:
Radial nodes in molecular orbitals
Do radial nodes exist in molecular orbitals?
Radial nodes exist in atomic orbitals and the number of radial nodes for an atomic orbital can be determined by the general formula $n-l-1$ where $n$ is principal quantum number and $l$  is azimuthal quantum number.
But do radial nodes exist in molecular orbitals, and if they exist, is there a general formula for determining number of them present?

hi, and welcome to chemistry StackExchange.  Have a look at the wikipedia page, you should be able to answer your question. https://www.google.co.uk/search?q=molecular+orbitals&oq=molecular+orbitals&aqs=chrome..69i57j69i60.4959j0j7&sourceid=chrome&ie=UTF-8

Well, if the atom had radial nodes, they are certainly not going anywhere, and so they must appear in the MO. As for counting them, that would be the most strange subject to bother yourself with. Nobody cares about the radial nodes. Anyway, nearly all chemistry occurs outside of the outermost such node.

No, radial nodes are zeros of the radial wavefunction. For it to be possible to seperate the wave function into a radial part and a spherical harmonic the system you are trying to describe needs to be spherically symmetric. An atom is but a molecule can never be. Thus the whole concept of radial nodes does not exist for molecules.


text:
Why does oxalic acid have exactly two water molecules of crystallization?
Why does oxalic acid have exactly two water molecules of crystallization?  In other words, in the formula $\ce{C2H2O4\cdot2H2O}$, why is the number before the $\ce{H2O}$ exactly $2$?
I tried to see this from the structure of oxalic acid. Also, I tried to invoke hydrogen bonds. However, I don't see the reason why there are two water molecules of crystallization from this.
Also, with how many precision is it two?

You might want to start by looking at the crystallization structure of oxalic acid bihydrate. The number is exactly $2$.

It is 2 because of the crystal structure. And crystal structure, mind you, is not something you may infer just by looking at the molecule, or by some simple back-of-the-envelope estimations.

@Jan the left is with serine; the right without.

@DHMO True that …

As @Ivan Neretin states, the crystal structure determines how much water can be accommodated. For example, $\ce{CuSO4}$, cupric sulfate or copper(II) sulfate, may be found as a mineral with three, five or seven molecules of bound water:

Bonattite (trihydrate mineral)
Bluestone (pentahydrate)
Boothite (heptahydrate mineral)

Of course, it can also be heated to form the anhydrous salt.
One could say different crystal forms have room for different amounts of water, or the converse: by adding or removing water, the crystal structure is changed.


text:
Why can't primary alcohols react with sodium bromide via an SN2 mechanism?
I was told that a primary alcohol reacts with $\ce{HBr}$ for example, but not with $\ce{NaBr}$ through the $\mathrm{S_N2}$ mechanism in the same reaction environment. Why?
This reaction can occur:
$$\ce{R-OH + HBr -> R-OH2+ + Br- -> R-Br + H2O}$$
But this reaction does not occur:
$$\ce{R-OH + NaBr -> R-OHNa+ + Br- -> R-Br + NaOH}$$

For $\mathrm{S_N2}$ to occur, the $\ce{-OH}$ has to be "helped along" to leave, by converting it to a good leaving group (good leaving groups = weak bases). If you protonate it with a strong acid, $\ce{Br-}$ can attack, and a molecule of water will be kicked out as the leaving group.
So why doesn't it work with $\ce{NaBr}$? A proton is a strong lewis acid, meaning $\ce{HBr}$ can protonate a small fraction of the $\ce{-OH}$ groups. $\ce{Na+}$ is not, on the other hand; it is too electropositive (just as $\ce{Na}$ tends to give up an electron and become $\ce{Na+}$). It simply cannot hold on to the electrons from a formally neutral oxygen-atom at all. Hence, it's like you are still trying to get $\ce{HO-}$ to leave.

Extra - when oxygen has a formal negative charge

If you go on and study carbonyl chemistry you will find places where $\ce{Li+}$ (which has a better size to interact with $\ce{O-}$ than $\ce{Na+}$) can stabilise the so-called tetrahedral intermediate, and help keep it from collapsing.
This is useful to stop the reduction of, for example, esters at the aldehyde, instead of continuing the reduction to the primary alcohol.
To get the lithium-ion to dissociate, you just add water. As DHMO explained, as soon as you have water, the positively charged group I metal ions are simply solvated so well by the water, that they are virtually inert.

This can be reasoned simply by looking at the acid/base strengths of the species present. Equilibrium will favor the side of the reaction with the weaker acid/base character.
$$\mathrm{p}K_\mathrm{a}(\ce{HBr})=-9 \hspace{1cm} \mathrm{p}K_\mathrm{a}(\ce{H2O})=15.7$$
$$K_\mathrm{eq}=\frac{[\ce{H2O}]}{[\ce{HBr}]}=10^{15.7-(-9)}=10^{24.7}$$
You don't need me to tell you that that's a really, really big number. Let's take a look at the other reaction, though. This will be the same type of comparison, except now we will be comparing basicities.
$$\mathrm{p}K_\mathrm{b}(\ce{Br-})=23 \hspace{1cm} \mathrm{p}K_\mathrm{b}(\ce{OH-})=-1.7$$
$$K_\mathrm{eq}=\frac{[\ce{OH-}]}{[\ce{Br-}]}=10^{-1.7-23}=10^{-24.7}$$
This equilibrium constant is as small as the other was large. So while the reaction you proposed may actually happen (though $\ce{Na+}$ will coordinate rather than form a bond as you show), the formation of the reactants is so favored the products won't be seen.

An $\mathrm{S_N2}$ reaction is a bimolecular reaction. As such, it requires both partners to be sufficiently reactive; that is we need both a good-enough nucleophile to attack and a good-enough leaving group. If we have a slightly better leaving group, we can make do with a worse nucleophile and vice-versa, but overall they have to be strong enough in combination.
Leaving groups in the $\mathrm{S_N2}$ reaction are good ones if they have a good charge distribution (charge per volume), which correlates well with being the conjugate base of a strong acid. If your leaving group is cationic (i.e. becomes neutral after leaving), it is a very good leaving group as there is no charge that needs to be distributed. Hence why protonating a hydroxy group to give $\ce{-OH2+}$ turns it into a good leaving group.
This is also why bromide is a rather good leaving group: Not only is it the conjugate base of a very strong acid $\ce{HBr}$, it is also a very large anion with a small charge and thus a rather good charge distribution. On the other hand, a hydroxide, which would be the leaving group in an $\mathrm{S_N2}$ attack onto an alcohol, is much smaller but still has the same charge. Thus, it is an absolutely terrible leaving group and nucleophilic attacks that liberate hydroxide effectively don’t happen.
Thankfully for synthetic organic chemistry, there are many ways of turning hydroxy groups into good leaving groups; most of these involve adding an acyl residue of some sort that creates a stabilised anion upon leaving with a much better charge distribution. The most popular ones (given as the actual leaving group) are probably mesylate $\ce{MsO-}$, tosylate $\ce{TsO-}$ and triflate $\ce{TfO-}$ — all sulphonic acid derivatives.

As a side note: You wrote an addition of $\ce{Na+}$ to the hydroxy group to generate a charged species. But unlike a proton — which is such a strong Lewis acid that it will never walk alone — a sodium cation will never form that strong a bond with a hydroxy group that they would leave as $\ce{NaOH}$. In fact, $\ce{NaOH}$ if it doesn’t precipitate due to low solubility will always dissociate into $\ce{Na+}$ and $\ce{OH-}$ while water molecules will not in typical organic solvents.


text:
Shouldn't MSDS data on products be publicly disclosed?
I am wondering whether companies are required to disclose MSDS data on their products: toxicity, $\mathrm{LD_{50}}$, etc. Not only that, but I am wondering whether they are required to disclose the methodology that they used. I know that this isn't a legal forum, but some companies that I asked are unwilling to give details on how the toxicity was tested. What are some laws that are in place (regardless of country)? I thought that all MSDS data should be public.

Companies are required to have SDS (don't know why they changed from MSDS to SDS recently, but...) sheets for chemical products they sell. What you see conforms to US federal regulations. It is highly unusual for a company to do their own toxicology (if it is even done). That is done to standards established and controlled by national or international industrial hygiene associations. I have seen (and used) many specialty chemicals where such testing had not been done, so no data was listed since there was none. There is no requirement to perform it for a new chemical.

@JonCuster The thing is that some companies have nearly 30 entries missing. I am looking at some cleaning products...They have No data available on environmental hazard part. Data missing on inhalation. There is no acute exposure limit. Irritation data is missing. Is this even legal..? It says it got approved by the nation's msds lab..

Yes, it is legal. Perhaps not ideal, but legal. If you still have concerns, discuss with your nearest IH person (industrial hygiene).

There is a bit of a misconception here.
Of course, the material safety data sheet (MSDS) is publicly available. Just go to your favourite chemicals supplier, type in your substance and download the corresponding MSDS in your language for your jurisdiction. At least in Europe that is mandatory for all chemicals a company sells; I would be very surprised if there is any developped country where making MSDS publicly available be not mandatory.
On the other hand there is the question of how a value was determined. That is not necessarily public information. Of course, there is a set of standardised tests that companies would perform, but they don’t have to exactly mention whether they used method A or method G. They also have a reason to be nondisclosive about this: imagine an animal rights activist calling you pretending to just be an interested scientist and then using the information they get on the phone to start a name-ruining anti-ChemicalCompany campaign. You don’t want that, do you? And then, of course, there is the general secrecy issue. If you just invented a compound you want to put on the market as a drug, you don’t want anybody knowing anything about it until you’ve started selling it.
Of course, government bodies have a certain reason to know how the tests were perfomed and what the raw results were (otherwise the company could just print a number on their MSDS which the responsible person invented on the toilet). But you’re not government. You’ll have to somewhat trust the government to properly monitor the companies in question.
Of course, the chemist who works safely doesn’t need to know how dangerous something is. They will work with it in a way that they are not affected.


text:
How do I draw a Lewis diagram after drawing the orbital diagram
The question says "Oxygen can form compounds with every period 3 element except argon. Determine which would be ionic or covalent compounds, and draw Lewis diagrams to represent each one." I started with oxygen and fluorine and I'm having a hard time. I don't know if there should be lone pairs, double bonds, or even lone electrons.Here's my attempt at it:

Your Lewis structure for $\ce{OF2}$ is almost correct, but as it stands right now, it is missing one lone pair on oxygen. A Google image search for "oxygen difluoride" should show you the correct answer.

First of all, oxygen follows "the octet rule", which states that certain elements are stable when they have 8 electrons around them. Now this rule is by no means absolute, does not work with d orbitals onward, and should only be used in very elementary chemistry, unless you actually know its cause. In the case of that picture, you are missing 2 non-bonding electrons on the oxygen.
Remember that a covalent bond means that electrons are being shared by the 2 atoms. In the case of O and F, the pauling electronegativities are quite similar, so the electrons are actually "shared".
However, when you try to put oxygen together with group 1 or group 2 elements, you will find very different electronegativity values. What this means is that oxygen has much more affinity to electrons than Na or Mg. The rule I was taught at school was that a difference of 2 or more in electronegativity between the elements results in ionic compounds, meaning the oxygen "steals" the electrons, becoming a negative anion, whereas Na for example becomes a positive one.
The last guideline you should take into account, is that an atom tends to lose or gain electrons because it "likes" to have its most outer shell full. So when you look at the principal quantum number ($n$) of the most outer shell, you should create compounds where it is full. In the case of oxygen, that quantum number is $2$, which can have 8 electrons, hence the octet rule. For Na, you have 1 electron in $n=3$, so if the atom looses that electron, it will become more stable, therefore we have $\ce{Na+}$.


text:
Does NaOH have a higher OH- concentration than NH3?
In my book it's a multiple choice question with only one correct answer.
There are two solutions of $\ce{NaOH}$ and $\ce{NH_3}$ that have similar concentrations and volumes.
These two are among the options:

$\ce{NaOH}$ has a higher $\ce{pH}$ than $\ce{NH_3}$
$\ce{NaOH}$ has a higher $\ce{OH^-}$ concentration than $\ce{NH_3}$

Wouldn't these both be correct?

Given that the solutions have equal concentrations, a solution of $\ce{NaOH}$ (a strong base) will be more basic than one of $\ce{NH3}$ (a weak base), since $\ce{NH3}$ only dissociates partially:
$$\ce{NH3 + H2O <<=> NH4+ + OH-}$$
Loosely speaking, the pH of a solution is defined to be
$$\mathrm{pH} = -\log_{10}[\ce{H+}]$$
and from the ionic product of water, we know that $[\ce{H+}][\ce{OH-}]$ is always a constant (equal to $10^{-14}\ \mathrm{M^2}$ at $25\ ^\circ\mathrm{C}$).
Therefore, a higher pH implies a lower concentration of $\ce{H+}$, and therefore a higher concentration of $\ce{OH-}$. This means that the truth of option 1 implies the truth of option 2, and vice versa, i.e. option 1 is logically equivalent to option 2.
My first suspicion would therefore be a typo in your book. In any case, your reasoning is correct.

Great answer, thank you! I didn't know what the units for $K_w$ were, either. :)

$[\ce{H+}][\ce{OH-}]$ is a product of two concentrations, so the units are simply concentration squared.

Correct me if I’m wrong, I’m new to these, but doesn’t it kind of matter based on what you mean by “higher pH”. Bc a higher pH on the Number scale would mean the solution is basic. However isn’t pH the measure of hydrogen ions in a solution? So by saying “higher pH” you could mean “higher measurement of hydrogen ions” which would then mean the solution is acidic. Someone just posted about this tonight on this question but it was deleted before I could reply. I can kinda see what they were saying tho. @orthocresol (Also how do you tag people in here)

@Hellooo no, a higher pH means less hydrogen ions not more. Please just read about this in any book/the Internet.


text:
What are Orbitals?
I recently learned about orbitals in class, and I am really confused.
I am very confused about what the orbitals look like, what the signs (such as $-2s, ---2p$ etc.) mean and how they work.
What they taught us was how there are little spaces that the electrons can fit through, because they want to get to the center of the atom, and how $2s$ is in the shape of the sphere.
But I am confused how everything is laid out. Is the "sphere" shape inside the shell and around the nucleus, or is it containing the shell. but inside another shell.
They also introduced something about the $x,y,z$ axis and I am just as confused on that too.

These are not signs, but merely placeholders for electrons. Also, there are really no such things as "inside" or "fit"; basically, every orbital is everywhere, and they simply go through each other, much like ghosts. All in all, orbitals are much less real than everyday things.

The basic idea is orbitals are region around the nucleus where we can find electrons with largest probability. I usually visualize a orbitals similar(not same) to a physcial address (eg:BBC Broadcasting House, Portland Place, LONDON, W1A 1AA). Just like we can find BBC broadcasting house with high probability(99.999...9%) on that address, orbitals are space where we can find electons

Oh... very different than how I imagined it...

@Frank, you've received some excellent answers below, they give a good picture of the idea of orbitals (especially when combined). I just want to offer a few words that may help tie it all together: the graphic representations of the orbitals can be thought of as inflated balloons. The electrons that fill them can be anywhere within the balloon (perhaps they are constantly appearing and disappearing at different locations), but at any time most of the volume of the balloons is empty of electrons - the majority of an atom's volume is nothing, but it serves as potential location of electrons.

First thing you need to understand is that orbitals are not actual physical things that exist. Simply put, an orbital is a function that describes the probability of finding the electron with certain energy at certain distance from the nucleus. The shapes of the orbitals are just the boundaries of space where you can find the electron 90% of the time.
Electrons behave both as particles and waves. Their wave behavior is described by what is called a wave function. It has two components - radial and angular. The radial component depends on the distance from the nucleus while the angular depends on the direction. It is the angular component that shapes the orbitals. Just as it is, the wave function does not have any physical meaning, but it's square is proportional to the probability of finding the electron in a particular region of space around the nucleus.
The x, y and z axis simply refer to the Cartesian coordinate system and can describe points in 3D space. Imagine three lines where each of them is perpendicular to the other two. It doesn't matter which one you name x, y or z, this is all arbitrary.
The electrons around the nucleus are not really ordered. The terms shell, sub-shell, orbital are just a way of describing different energy levels and probabilities of finding electrons around the nucleus.
There are four quantum numbers that describe electrons in atoms - principal, azimuthal, magnetic and spin.
The principal quantum number (n) can take integer values from 1 to infinity. It describes the different energy levels with increasing distance from the nucleus. As n gets bigger, orbitals within that "shell" also increase in size.
The azimuthal quantum number (l) can take values from 0 to n-1. It describes the type of orbital the electron resides in. It is also call angular momentum quantum number and as I've said above, the angular wave function is responsible for the shape of the orbitals. Makes sense? For each value of l, there is a different type of orbital - 0=s, 1=p, 2=d, 4=f, 5=g (not really seen occupied in the ground state of any element), and these can go on alphabetically.
The magnetic quantum number (ml) describes the direction of the orbital in space with respect to the three Cartesian axis - x, y and z. It can take values from -l to +l. For s orbital, ml can only be 0 and so there is only one s orbital that can exist for a particular principal quantum number. For p orbitals, ml can be -1, 0 and +1 and so there are three p orbitals that can exist - each along different axis - px, py and pz. The axis labeling get more complicated for d and f orbitals.
The spin quantum number (ms) can either be +1/2 or -1/2 and simply describes the direction of the electron's intrinsic angular momentum.
For n=1, l can only be 0 and so there is only an s orbital in that level, referred to as 1s. For n=2, l=0,1 and there can be s and p-types orbitals. For n=3, l=0,1,2 and so you can have s, p and d orbitals in the third shell.
When you're filling up shells with electrons, you go from the lowest energy level to the highest and this can easily be determined by summing up n+l which would give you the relative energies of orbitals. When n+l is the same for two orbitals, the one with the lower n fills up first. One orbital can house 2 electrons which must have opposite spins. There can never be two electrons with the same four quantum numbers in a system (Pauli exclusion principle).

What is an orbital?
According to the Oxford Dictionary of English [1], orbital is defined as

Each of the actual or potential patterns of electron density which may be formed in an atom or molecule by one or more electrons, and can be represented as a wave function.

Wikipedia  [2] says:

In quantum mechanics, an atomic orbital is a mathematical function that describes the wave-like behavior of either one electron or a pair of electrons in an atom.

If you go to Wikibooks [3], you will get that:

Each shell is subdivided into subshells, which are made up of orbitals, each of which has electrons with different angular momentum. Each orbital in a shell has a characteristic shape, and is named by a letter. They are: s, p, d, and f. (…) Within any particular shell, the energy of the orbitals depend on the angular momentum of orbitals s, p, d, and f in order of lowest to highest energy. No two orbitals have the same energy level.
(…)

Wikipedia  [2] shows these graphical representations:

The shapes of the first five atomic orbitals are: $1s$, $2s$, $2p_x$, $2p_y$, and $2p_z$. The two colors show the phase or sign of the wave function in each region. These are graphs of $ψ(x, y, z)$ functions which depend on the coordinates of one electron. (…)

References
[1] Oxford Dictionary of English, 3rd ed.; Stevenson, A, Ed.; Oxford University Press: Oxford,
U.K., 2010.
[2] Atomic orbital. Wikipedia, The Free Encyclopedia. [Online]; Posted October 29, 2016. https://en.wikipedia.org/w/index.php?title=Atomic_orbital&oldid=746850729 (accessed Oct 29, 2016).
[3] General Chemistry/Shells and Orbitals. Wikibooks, The Free Textbook Project. [Online]; 7 November 7, 2016. https://en.wikibooks.org/w/index.php?title=General_Chemistry/Shells_and_Orbitals&oldid=3143833  (accessed Nov 7, 2016).

The hard part of quantum mechanics is that you can't know how everything is laid out. Elections fit into some distribution about the nucleus defined by the mathematical functions that are the orbitals. You can't practically know where they are per se, just where they are likely to be.
The traditional orbits you learn about are a gross oversimplification.

The form of the orbitals in an atom is determined by the charges and the masses of the electrons and the nucleus. The orbitals determine regions of space where an electron is most likely to be found if we make a measurement of the position of the electron. However make no mistake. The electrons in an atomic orbital can have a well-determined energy, a well-determined total angular momentum and a well-determined projection of the angular momentum on an arbitrarily chosen z-axis. It can also have a spin quantum number. But it cannot have a well-determined position. The electron in an atomic orbital has lost its particle property and has become wave-like. Its charge and mass is smeared out over the orbital and replaced by a density function given by the absolute square of the orbital mathematical function. This density function can be used to calculate the exact Coulomb potential energy of the atom. This Coulomb energy also determines the total kinetic energy (and hence the total energy) of the atom by the virial theorem.


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Why will a strong acid neutralize as much base as a weak acid?
This is a simple concept that I can't seem to understand. Why will a strong acid neutralize as much base as a weak acid, if the acids are of the same volume and concentration? A strong acid will dissociate more in solution and thus have a greater number of $\ce{H^+}$ ions, as far as my understanding goes. The opposite if true for the weak base. So wouldn't the strong acid neutralize more base than the weak acid, because it has more $\ce{H^+}$ ions to neutralize the $\ce{OH^-}$ ions of the base with?

Strong acid is dissociated all but completely: you have a certain amount of $\ce H^+$ and start neutralizing them, one drop at a time. As you do so, their numbers steadily decrease. It feels like sawing a huge log with a hand saw: you see the amount of job, and you do it until it is all done.
Weak acid is dissociated only partially. You see a relatively small amount of $\ce H^+$, but as you start neutralizing them, more molecules of acid get dissociated and their $\ce H^+$ ions stand in place of those you've neutralized. It feels like sawing a thin log, but a peculiar one: it seems to regenerate under the saw.
So at the end of the day, it is the sheer amount of acid that matters, rather than immediate concentration of $\ce H^+$.

The hydroxide ions will take hydrogens from any type of substance that can donate a proton, whether it is dissociated or not.  In fact, even amphoteric substances that tend to have a more basic character will neutralize a solution of strong base to some degree.  Therefore, it doesn't really matter whether the acid is weak or strong.
This also means that polyprotic acids will neutralize more of a strong a base in lower concentrations than strong monoprotic acids, since they have more hydrogens donate.


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Why would the hydrogen sulfide ion be a better nucleophile than a bromide ion?
Why would the hydrogen sulfide ion be a better nucleophile than a bromide ion?
My professor says that hydrogen sulfide ion is a better nucleophile.
I thought I could determine the better nucleophile was based on available electron density and polarizability. The bromine ion has more available electron density and is more polarizable than the $\ce{HS-}$. Why would it be less nucleophilic?

Welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site. Visit the [help] for remaining unanswered questions about how it works.

"Hydrogen sulfide" refers to $\ce{H2S}$. I assume you are talking about  its conjugate base, $\ce{HS-}$. The name that I've most commonly heard for that is "hydrosulfide"; "bisulfide" seems to be used as well.

@orthocresol Actually, hydrogen sulfide is also a correct name for the monoanion to the best of my knowledge. Compare hydrogen sulfate.

There are a number of factors you have to take care of while deciding nucleophilicity. I will try to cover all the important ones.
First we need to clear something, polarizability is the most important factor only if the nucleophile is electrically neutral (uncharged)$^{[1]}$. Of course, this is not hard and fast, but in most cases we generally deal with, this is true. For example, triphenylphosphine is a better nucleophile than triphenylamine, because the former's central atom has a more polarizable lone pair.
In our examples, the ions are charged, so we will put aside polarizability and look at the other factors. For charged species, size of the anion should be small in order for it to be a better nucleophile. Why? Two reasons:

One, the same charge packed into a small volume is attracted more strongly by the electron-deficient area, and it can reach quite close to the positive region. This means that potential energy of the system will fall more, thus increasing stability of the product.
Two, the central atom being small, its overlapping orbital can overlap more effectively with the empty orbital of the electrophile, since the orbital will be smaller in size, and hence less diffuse.

The last factor that remains is the solvent. In fact, the order is same for both the nucleophiles in both the kinds of solvents (protic and aprotic), so I will not go into the analysis of solvent effects. Just a comment, here the solvent effect seems unimportant, seeing that bromide ion is quite behind $\ce{SH^-}$ in the order of nucleophilicity in protic solvents (my source is Solomons and Fryhle and a website mentioned later).
Note: Electronegativity is not important here either, even though it would produce the same order. We generally use electronegativity only to compare nucleophiles with central atoms from the same period.
So ultimately we see that $\ce{SH^-}$ is more nucleophilic than $\ce{Br^-}$.

Most of what I said can be found in the textbook Solomons and Fryhle's Organic Chemistry, and the following link:

$^{[1]}$http://chem.libretexts.org/Core/Organic_Chemistry/Reactions/Substitution_Reactions/SN2/Nucleophile

Here are some more useful links:
- http://chem.libretexts.org/Core/Organic_Chemistry/Alkyl_Halides/Properties_of_Alkyl_Halides/Introduction_to_Alkyl_Halides/Leaving_groups
- http://chem.libretexts.org/Core/Organic_Chemistry/Alkyl_Halides/Reactivity_of_Alkyl_Halides/Alkyl_Halide_Reactions/Substitution_and_Elimination_Reactions_of_Alkyl_Halides/SN1_Substitution_Reactions/Nucleophilicity_and_Solvent_Effects
- http://chem.libretexts.org/Core/Organic_Chemistry/Reactions/Substitution_Reactions/SN2/Substrate

Bromide has more stabilized lone pairs because of higher effective nuclear charge. So the sulfide lone pairs more nucleophilic. They're in the same period, so polarizability is not the prime factor.

Thanks for your answer. I am still a bit confused. Aren't they in different periods? So wouldn't bromine's large orbital make it a lot more polarizable? The test I got the question from had the answer on it, saying the reason is polarizability, but I don't understand why. I know I am missing something.

Oops. They are. I forgot about Cl. Br would be more polarizable, but I think here the group is more important. For example, an alkoxide is a better nucleophile that a chloride.

I totally deserve a -1 on my answer for being an idiot...

The reason is the higher polarizability of the bromide rendering it less nucleophilic than the sulfide. That's the reason. Period.

By the way, what is "available electron density" then there must be also "unavailable electron density" and what could that be?

I also don't see how electron density is directly linked with nucleophilicity.


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Predicting the solution with the highest boiling point
I am facing problem with the following question:

Predict which solution will have the highest boiling point in increasing order?

$1~\%$ solution of $\ce{C6H12O6}$

$1~\%$ solution of $\ce{NaCl}$

$1~\%$ solution of $\ce{ZnSO4}$

$1~\%$ solution of  $\ce{(NH2)2CO}$

I know that more is the van ’t Hoff factor more will be the elevation in boiling point and hence the solution will have more boiling point. Hence 1. should have the lowest boiling point followed by 4. But I am not sure about options 2. and 3.

What do colligative properties depend on?

@getafix It depends only on the number of solute particles and not on the nature of solute particles

You were right that you need to look at the Van 't Hoff factors. Think about which of 2. or 3. is more soluble in water. Also, between 1. and 4., think about whether any association of molecules can occur via H-bonding and how would that affect the Van t' Hoff factor.

I don't think the solubility is the most important factor here. The most important question to ask is what the question means by 1%? I assume this is a mass-to-mass percentage... From there you can apply what you know about colligative properties.

@Zhe yes it I'd mass to mass percentage.but how it could affect if it were some other percentage.

Equivalent mass-to-mass percentages don't imply the same number of particles even if the Van't Hoff factors were the same. Now why would that be?

The first aspect to consider is the van't Hoff factor. Thus #1 is first and #4 is last. Since both #2 and #3 have the same factor of 2, we must consider other properties.
A major part of BP is how many molecules of solvent are exposed on the surface layer. Therefore, the ions with the greater diameter should take up more surface area, decreasing the number of solvent molecules available to 'escape'.. decreasing the vapor pressure and increasing the BP elevation.
My guess is that the polyatomic ion is quite large in comparison and the deciding factor.
from lowest BP to highest
 #1,#2, #3, #4

More is the can't Hoff factor more should be the elevation in boiling point so in on way 4 can have the highest boiling point so I think you are wrong.in my view correct increasing order should be 3>2>1>4

Let's not throw around wild accusations, shall we?


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How to draw a photoelectron spectrum with ionization energies?
I understand the basic concepts of the photoelectron spectrum. However, say I am instructed to draw the PES graph for Silicon. I know how to figure out the relative peaks based on its configuration. But when arranging them on the graph, how can I accurately place them where they should be in ionization energy? That is, the x-axis is labeled 300.0, 200.0, 100.0, etc. Without any other information (and the usage of a periodic table only) how do I know where to put the peaks among the increments? Is there a way to figure out the ionization energies for each peak?
If this is NOT possible, please explain why.

It would seem that you're asking about x-ray photoelectron spectroscopy.  A photon of defined energy is used to ionize electrons in the atoms/molecules of the sample. The kinetic energy of the electrons is the energy of the photon minus the binding energy of the electron. https://en.wikipedia.org/wiki/X-ray_photoelectron_spectroscopy

Thanks. But if I don't have information on what the energy of the photon or electrons are? If the instruction is simple "draw the pes of X element" and the only info I get is 1) already labeled increments on the x-axis and 2) a periodic table, if relevant, then how in the world would I place them appropriately?

Have a look at the units on your axis, if in electron volts (eV) then the energy is in the x-ray region. The lowest energy electrons (1s) will have the largest values as it takes more energy to ionise these. Thus you can get the ordering of  electrons in different orbitals, but without looking up some more information I don't see how you can get accurate values.


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Suitable antifreeze solution

If glycerine $\ce{C3H5(OH)3}$and methyl alcohol $\ce{CH3OH}$ are sold at the same price per kilogram, which would be cheaper for producing an antifreeze solution for the radiator of an automobile radiator?

I think that more depression in freezing point will result in a better antifreeze solution. It depends on molal freezing constant. How can I predict this?

I think the key point of this (probably homework) question, is about the units of price per kilogram, while a product is sold per liter. The density of glycerol is 1.261 g/cm3 and of methanol 0.792 g/cm3 (Wikipedia). So with equal price per kilogram, the price per liter of glycerol is 59% higher than of methanol

You should use the molar mass of the compounds instead of their density

@Lucademics the price is per kilogram and not per litre

Since modern automobile engines contain aluminum, and methanol corrodes aluminum, it should not be used as antifreeze at any price! You might also check if even glycerol requires inhibitors to prevent corrosion.

You are on the right track when you mention molality.
Because freezing point depression is a colligative property of the solvent, in this case water, all you need to do is to find out which solute has more moles per kg, aka molality.
The equation for freezing point depression for a non-ionic solute is given as:
$$\Delta T_\mathrm{F} = K_\mathrm{F}\times b$$
Where:
$\Delta T_\mathrm{F} =$ freezing point depression
$K_\mathrm{F} =$ cryoscopic constant (dependent on solvent, not solute)
$b =$ molality, the moles of solute per kg of solvent
The key point here is that freezing point depression is linearly proportional to solute concentration (molality).
Since glycerol has a molecular mass of $\pu{91.1 g/mol}$, 1 kg will contain 11.0 moles of solute. Methanol has a molecular mass of $\pu{32.0 g/mol}$, so 1 kg will contain 31.2 moles of solute.
So, for any given price per kg, methanol will give you $31.2/11.0 = 2.9$ times as many moles of solute, resulting in a 2.9 times greater freezing point depression.  Note that you cannot calculate the actual freezing point depression without knowing $K_\mathrm{F}$.


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Principle of the Biuret Test
What is the principle of the biuret test? While reading online, it is unclear to me how a bond is created between the $\ce {Cu^{2+}}$ and the proteins? What are the chemical reactions that lead to the color change?
Edit

The biuret test is a chemical test combining two products, $\ce{CuSO_{4}}$ and $\ce{OH^{-}}$ in my case. While exploring online the possible effects on protein, some sites suggest that the $\ce{OH^{-}}$ is responsible of removing on hydrogen from the amine and that is why there is a color change while other suggest that the $\ce{OH^{-}}$ would rather act as a redactor for the copper and the release of electrons from the copper would transfer the copper ion from a cupric form to cuprous one. (Copper-based assay chemistries)
I have also understood that the copper geometry is changing during the reaction, but the idea of the reduction is unclear to me.
I will ask my question again, what are the chemical reactions that lead to the color change?

It is a coordination complex

The $\ce{Cu^{2+}}$ ion is capable of coordinating to the lone pairs of the nitrogens of the amide bonds.
$\hspace{1.4cm}$
The formation of this complex is associated with a certain energy change in the orbitals of the $\ce{Cu^{2+}}$ ion, and hence a color change.

In the typical violet complex that is formed under the strong alkaline conditions of the biuret test (not the blue-green complex that may be initially formed at lower pH), the binding amide groups are deprotonated.


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When Z>1 are repulsive forces dominant or are attractive forces dominant?
According to PVreal = ZRT
If z >1 repulsive forces are dominant or are attractive forces?
Different teachers are telling contradictory answers and after referring a few books, I have become more confused.
Points for ATTRACTIVE FORCES are dominant:

At high pressure the molecules do not strike the walls with full impact as they are dragged back by attractive forces.
When molecules are farther away from each other(as Vreal > Videal) the attractive forces are dominant.
Repulsive force are dominant when he molecules are almost in contact .

Points for REPULSIVE FORCES are dominant:

At high pressure the molecules are close each other that attractive forces are negligible but molecular size is significant.Thus repulsive forces are significant
As volume of real gas is more than volume of ideal gas , the real gas is more difficult to compress.Thus repulsive forces are dominating.

There was a question where P = 12atm T = 350K and Z = 1.2(after calculation). It was given that repulsive force are dominant.But my teacher said it was wrong and it should be attractive forces.

Your arguments for attractive forces are not quite correct, particularly the first one, the other two are partly correct. The points for repulsive forces are more accurate but not completely correct.
The intermolecular potential energy is fixed by the type of molecule and varies with separation between molecules (e.g. a Lennard-Jones potential). The potential well formed by the interaction is short-range. It is effectively zero at about four to six molecular diameters decreases to negative values as the molecules approach  one another (negative energy is attractive) but rapidly becomes positive ( repulsive) when the molecules are effectively in contact.The potential is unaffected by temperature and pressure.
The ratio $p\overline{V}/RT$ is called the compressibility ratio Z and is unity for an ideal gas. ($\overline V$ is the molar volume $V/n$). Ideal gases have no intermolecular potential, or size. Real gases differ primarily in their size and magnitude of their intermolecular potential energy and so deviations from $Z=1$ are to be expected. At all temperatures and pressures the molecules have an attractive potential that tries to pair up molecules, and repulsion on collision when the charge on the electrons in each molecule force them to repel one another.
It is the competition between attraction, repulsion and kinetic energy at a give temperature that determines the shape of the $Z ~ vs. p$ plot.
For molecules at low temperature the line of the graph $Z= p\overline{V}/RT$ vs. p dips down further below $Z=1$ than it does for the same molecules at higher temperatures. This is because at low temperatures the  molecules have less kinetic energy ( so move more slowly) and are more influenced by attractive forces than at higher temperatures. This makes the actual volume of gas smaller than the ideal (as a larger fraction of molecules are paired up), which in turn makes Z smaller.
Similarly, at a given temperature, a larger negative Z is seen in gases with a larger attractive interaction, say $\ce{CO2,  CH4 or ethene}$ compared to those with smaller attraction, e.g. $\ce{He or H2}$ .
At high temperature the molecules have more kinetic energy and so attractive interactions are relatively less important, and repulsion more so as the kinetic energy is larger. This makes the actual volume greater than the ideal and $Z>1$.


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How to prepare a solution of accurate concentration?
How does one accurately dose powders? I would like to know what tools are used and how.
For example, how would one apply 1 gram of a certain powder to a solution in a beaker? Specifically, how do you make sure all of your measured powder (or at least with a very small error margin), gets into the solution, without too much of the powder sticking to the spoon or other utensil.
I figured one might dip a spoon loaded with the powder in the solution, but then the solution might stick to the spoon.

During all the laboratory practicals I been to, we've always ignored the bit of 'powder' (we normally deal with salts though) that sticks to the spatula or scoopula.....because it is negligible under most circumstances. Then again if you're conducting a very sensitive test, using a very small quantity of 'powder', then it would advised to (not always though..) finally stir the solution with the scapula you used to transfer the 'powder' to the beaker in the first place. You mentioned that the solution might stick to the spoon (a very astute observation), but even that's negligible......

^ have a look at your spoon after you're done stirring sugar into your lemonade...how much of the solution do you see sticking to it? A very small amount....and remember, most of the time, that little bit of solution, will have an even smaller amount of the 'powder' in it..... As for 'dosage' .....you weigh it in grams and over the course of your experiment, you gradually whittle it down to 'moles'........I'd make this into an answer if you'd like.....

Sure, that sounds like a reasonable answer. :)

Two things matter if you need accurate concentrations of reagents: your ability to weigh things precisely and your glassware. You can ignore powder sticking to spatulas by weighing the vessel before and after addition of the solid. You can dispense accurate volumes of liquid if the vessel is a calibrated volumetric flask (long thin neck with calibration line for specified volume marked on the thin neck).

This is how I did it in my analytical practicals, works for primary standards with a decent accuracy, works for everything else with a larger error. (You'd want to check concentrations with a primary standard then.)

Determine the mass of the empty volumetric flask
Prepare approximate mass of the substance you want the solution of in a beaker or watch glass or similar
Transfer the prepared substance into the volumetric flask
Determine the mass of the filled flask; the difference will give you the mass of the substance
Fill the flask with the solvent.
Given the precision of the instruments used, you can calculate the concentration quite accurately. If necessary check the concentration with another method.


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Which areas of chemistry require calculus?
I didn't have much chemistry in school/college, and am now building up my knowledge again. I am studying maths in parallel, building up to calculus.
I noticed that the introductory chemistry books (even at the college level) don't use any calculus. However, more advanced books do tend to use calculus.
Can anyone tell me which areas of chemistry require (or make use of) calculus?

linear combinations of atomic orbitals.

Bring me the list of areas, and I'll check those which do. Roughly speaking: quantum chemistry and physical chemistry in general - yes, very much; analytical chemistry - not that much, inorganics - only sporadically, organics - almost none. But remember, in fact there are no areas, everything is connected to everything else.

Frankly, without calculus you don't have a solid grounding in doing chemistry, physics, or engineering. Forty years after taking calculus I still use it and the concepts drilled in taking calculus on a daily basis.

to add to other answers,  understand units,  errors, (e.g. standard deviation and how to propagate),  significant figures, how to fit data (understand least squares) , and understand molecular symmetry, all of these are  just as important in phys chem as in organic chemistry.

In short, in every case where a property varies smoothly when other one does (so, in every area of chemistry). Sometimes you can avoid to use calculus (for example, in cases with very simple linear dependencies), but it would impose serious restrictions.

Analytical chemistry : To predict, for example the $\ce{pH}$, for which moieties will complex and also infer statistics thereof.
Electrochemistry : The Nernst-Plank equation is challenging. But just to calculate the concentration of some things, taking care of different kinds of reactants, depending on what you do you'll need to be good in calculus. Nyquist plots are not really difficult to use but the theory to obtain them is quite complex (it's not a pun) if you look at Nyquist stability criterion.
Organometallic Chemistry : To calculate the oxidation degree of a metal for example, or a TOF (turn over frequency), a TON it refers also to catalysis.
Thermodynamics : You need to have good skills in math to do thermodynamics, also in mass transfer or heat transfer, everything which is close to process chemistry, as per the McCabe Thiele model for distillation (for example) this is not hard but you need.
Quantum Chemistry : Here come the hardest thing (as I know it). You need to be very good in linear algebra, able to solve differential equations, have some skills in analysis to calculate integrals, and so on.
Kinetics :  Calculus is not really hard but some times really weird. Also have some skills in algebra can help and be able to solve differential equations.

If you're picky, you may find other parts of chemistry in which calculus is used, but the main ones are in this post, especially quantum chemistry and process chemistry.
Finally, it also depends on your education or experience level, as the most you will study at a high level will include some good skills in calculus. You'll need to be critic and also to know what is the theory behind a hypothesis to be able to verify if what the software tells you can be consider as true or false.

For first year college chemistry the only calculus, if there is any at a particular school, would be reaction rate equations.
$\mathrm{d}[A] = -k[A]\mathrm{d}t$ ;  $\mathrm{d}[A] =-k[A^2]\mathrm{d}t$ for example
Then second year you take organic chemistry; no calculus.
In the third year, thermodynamics has some multivariable calculus (partial derivatives) and quantum mechanics has partial differential equations.

What about physical chemistry?

@Starlight physical chemistry is thermodynamics, statistical mechanics, and quantum chemistry. Yes, that's where you use calculus.

Calculus is the mathematician's or mathematical scientist's divide-and-conquer tool. When a problem is to too complicated to solve, you break it into manageable chunks and then recombine the solved chunks.
In chemistry, there are two main areas for calculus.
Thermodynamics: mainly in the areas of statistical mechanics (where you need to deal with understanding a very large number of particles) and reaction rates (where you need to deal with continuously changing concentrations)
Quantum mechanics: Many of the fundamental concepts are expressed as mathematical relationships, and the relationships are such that they require calculus to compute.
Notice that calculus is used most in the areas of chemistry that overlap with physics.


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Fundamentals of Li-ion Functionality
I would like to have an in-depth understanding of Li-ion batteries functionality, at atomic level. Can anyone help me to understand it? I have included some known background to support my questions.
For a fully charged battery:

The theory says that all (let's assume an ideal case) the Li-ions are in the anode, right? But are they really ions at this point or simply neutral atoms where the numbers of electrons is equal to the number of protons? In other words, are there ions or atoms in the anode? See figure below for a single Li-"ion":

Do we say that the electrical energy is stored in the electrons or is this a false statement, considering that batteries store chemical energy by definition? If we say that it is only chemical energy, how does one make the conversion between electrical energy and chemical energy? Is it stored in bonds?
Is the anode "more energetic" than the cathode in this case, considering that there are more electrons in the anode?

When discharging:

The theory says that the load "steals" the electrons which travel through the load and then "land" back into the cathode. Is the electrical energy from the electrons used by the load and then they go back to the cathode "empty"? (According to the definition of internal energy, the work is done by kinetic energy due to movement of electrons or molecules).

Fully discharged battery

Now we are in the situation where "all" the ions are back to the cathode. At this point, are there less electrons in the cathode or why is the battery discharged? Popularly, the battery is "out of fuel", but how do we actually translate this into energetic phenomena?

I appreciate any type of help, recommendation or mathematical explanation.
And please don't tell me that I need to go and read the basics of thermodynamics and stuff, because I did that, but none of the books/pages that I found describe these things at an atomic level. However, any recommendation of a book/page that does that will be highly appreciated!
Thank you very much in advance!
Paul

Based on the Wikipedia article, I am inclined towards thinking that the lithium on the electrodes is neutral. The lithium ions move through the electrolyte to facilitate the two half-reactions. The electrons in the half reaction participate in closing the external circuit, so the lithium ions and electrons are moving in the same direction, except one does so inside the battery and the other outside the battery).
Onto your questions:

The lithium in the anode is neutral, not an ion.
There's no traditional bond per se, but it's stored in the energy difference between $\mathrm{LiC}_{6} + \mathrm{CoO}_{2}$ versus $\mathrm{C}_{6} + \mathrm{LiCoO}_{2}$. There is an interaction here, but I wouldn't call it a "bond."
Well, the anode or cathode isn't more energetic, that's not quite the right term. The $\mathrm{LiC}_{6}$ is more energetic than other species. It just happens that this happens to be the anode of a charged battery. Once the battery discharged, the anode is still that side, but it doesn't have more energy.
Electrical energy doesn't work that way per se. We generally say we build up a potential difference $V = \frac{\Delta E}{q}$. The extraction of energy from this difference is dependent on the circuit. If you have an incandescent light bulb, the kinetic energy of the traveling electron is converted to heat and light, for example.
When the battery is discharged, the lithium atoms have moved to the metal oxide. That's the thermodynamically more stable of the two substrates for the lithium to be adsorbed onto.

I've never looked this up before, but this is actually reall cool. We'll basically using the thermodynamics of lithium preferring to be adsorbed onto one electrode over the other. In order for the lithium atom to get there, it has to move as an ion through the electrolyte and the missing electron moves through the external circuit. I think your confusion is here: the system wants to move the lithium atom, so we'll extract something useful out of it by forcing it to move the lithium ion and the electron separately.

Hey Zhe, thank you very much for the answers! Very nice explanation. I would like to extend this a bit. So:

Now it's clear. But I guess that there are still a large number of ions present in the system, right? Especially in the electrolyte.
What do you mean by "stored in the energy difference"? By analogy, the difference in enthalpy gives the energy input/o of the system and the latent heat is the stored heat. But how is energy stored in batteries or where? Can you please give me more details here?
This is related to the previous point.
Good point.

Also related to point 2.

Bottom line, what I would like to understand is the energy conversions. It is clear now that the kinetic energy of the travelling electron is the one that converts into heat and light. This also means that it goes back to the cathode with less energy, making the cathode "less energetic" from this point of view, right?. Now the questions is, do you increase the kinetic energy of the anode (and implicitly of the battery) when you charge the battery? And how is that kinetic energy converted to chemical energy?

I know that most of these questions might sound stupid, but I tend to see all the processes in terms of energy. I see each "moving" component as a source of energy conversion (e.g. the movement of electrons is energy) and that's why I want to understand each path in the battery components, which aren't so many in fact. I would like to reduce the entire system to basic components for a better understanding: one carbon unit cell, one lithium atom, the simplest electrolyte and one metal unit cell.

For 2, in chemistry, we generally consider the change in energy of a reaction $\Delta E$. This is the difference in standard energies of the two sides of the balanced equation. The energy difference is largely due to stabilization of charges via bonds and other interactions.
In terms of storing energy, this is purely potential energy. The potential energy comes from having chemical species that are more energetic than what they could become via the reaction that powers the battery.

Hey Zhe, another question related to point 4. Are the electrons really physically moving? I read these days about electrons and electricity and I can't decide if the electrons move physically or the vibration is transmitted from one electron to another. Apparently they are fermions and studies on electrons would lead to quantum physics. So to make my question clear: Are electrons phyisically moving from one anode to another or only energy is transported by vibration? In the second case, we would not speak about ions anymore, right?

I believe generally, we think of it as net drift. Electrons are physically moving, but that doesn't mean electrons on one end will actually have time to get to the other side, but the migration of electrons en masse is what conducts charge.