text:
How do electrons travel through nodes
I understand this is a basic question, but I'm having such a hard time wrapping my head around it. I'm trying to avoid thinking about it as an actual "particle" but as a wave, but that confuses me too.  Wouldn't a wave be the entire area the electron travels ? Or does the associated wave represent the path and not the electron? If so, where is the electron?
Here's a picture to refer to in your explanation if it helps.

Welcome to Chemistry.SE! Please have a look at this tutorial to acquaint yourself with the way math and chemical formulae can be nicely formatted on this site. Furthermore, when you include formulae or pictures in your questions, please make sure that you define the appearing symbols/variables and declare what is actually shown.

Have a look here and here. Does this answer your question?

From the image above and from what you're asking, I presume you are referring to a particle confined in a certain space region. You can find everything you need on the internet but I wanna clarify a few things about QM:

There is no concept in QM for "path of a particle": position and momentum are incompatible observables, which means that they cannot be measured simultaneously each with zero standard deviation: in fact the product of standard deviations for position and momentum must be greater than $\dfrac{\hbar}{2}$.
$$\sigma_{x}\sigma_{p}\geq\dfrac{\hbar}{2}$$
So it's meaningless in QM to talk about trajectories.
Meaning of the wave function: Let's assume that your system consists of a single particle in a one dimensional space. The wave function $\Psi\left(x,t\right)$
is a complex function that contains all the information about the system. But what is really physical meaningful is its modulus squared $\left|\Psi\left(x,t\right)\right|^{2}$
, which represents the position probability density. What is the position probability density? It's something that multiplied by a certain length (or volume if you're working in three dimensions), let's say dx
, gives you the probability for finding the particle between $x$
and $x+dx$
$$P\left(x,x+dx\right)=\left|\Psi\left(x,t\right)\right|^{2}dx$$
Of course you can integrate the previous eq. to find the probability over a certain region of space $\Omega$
$$P_{\Omega}={\displaystyle \int_{\Omega}}\left|\Psi\left(x,t\right)\right|^{2}dx$$
Stationary state: If the potential is time independent (in your case it's zero), the time dependence of the wave function can be factored out like
$$\Psi\left(x,t\right)=\mathrm{e}^{-i\omega t}\psi\left(x\right)$$
In this case it's easy to prove that the probability density is time independent.
The quantization: For this simple problem, you must solve the eigenvalue equation for the energy
$$-\dfrac{\hbar^{2}}{2m}\dfrac{d\psi\left(x\right)}{dx}=E\psi\left(x\right)$$
By solving it and by imposing that the particle cannot exists outside a certain region (outside the region $\left[0,3\right]$ in your picture), you find a natural quantization condition for the energy: the energy can assume only certain discrete values, which are called eigenvalues, $E_{n}$
. Of course, this means that for every eigenvalue there is an associated wave function, which is called eigenfunction $\psi_{n}\left(x\right)$
. The general solution to the eigenvalue equation is in fact a linear combination of eigenfunctions $$\psi\left(x\right)=\sum_{n}c_{n}\psi_{n}\left(x\right)$$
In the above picture, you can see the first three $\left|\psi_{n}\left(x\right)\right|^{2}$
. They of course represent probability density functions: note that where $\left|\psi_{n}\left(x\right)\right|^{2}$
is zero the probability of finding the particle is of course zero (for example in 0
and 3
, like we requested before).
So where is the electron? Like said before, in QM locating a particle it's a non-sense. All you can do is to compute the average position and momentum, which for a system in the state $\psi\left(x\right)$ are given by
$$\left\langle x\right\rangle ={\displaystyle \int}_{\Omega}\psi^{\star}\left(x\right)\hat{x}\psi\left(x\right)dx={\displaystyle \int}_{\Omega}\psi^{\star}\left(x\right)x\psi\left(x\right)dx$$
$$\left\langle p\right\rangle ={\displaystyle \int}_{\Omega}\psi^{\star}\left(x\right)\hat{p}\psi\left(x\right)dx={\displaystyle \int}_{\Omega}\psi^{\star}\left(x\right)\left[-i\hbar\dfrac{d\psi\left(x\right)}{dx}\right]dx$$


text:
How fast will hydrazine boil out of a tube into a vacuum?
For context I'm talking about ISEE-3 and how long it would take to dump the now useless and hazardous fuel it still carries.
Given a heated tank of hydrazine (somewhere near $0~^\circ\mathrm{C}$) connected by way of a long slender unheated tube to a infinite high vacuum, how fast will the tank vent to ~zero pressure?
My guess is that rather quickly, the boiling action will produce a plug of frozen hydrazine in the tube. Once that happens, the sublimation on the outer face will cause condensation and/or solidification on the other face and the plug will walk back towards the tank until it either reaches it or the solid portion of the plug get to thin in which case it will burst and vent like in the beginning and start over.
The question is, how fast will this happen?

Edit: on further inspection, it looks like the "tube" may be heated after all.

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text:
Net ionic equation with only aqueous ions
When you turn a balanced equation into a net ionic equation, you ignore all the ions that are spectator ions (aqueous on one side as well as the other). You only write the solid and the aqueous ions that form that solid.
So, what happens when you try to write a net ionic equation if ALL the ions are aqueous? Would you write down every ion in the reactants/products, or write nothing?

If all the ions remain aqueous, there is no reaction because nothing was produced. Everything remains in solution before and "after".

If you have a balanced equation like the following:
$$\ce{NaCl_{(aq)} + KBr_{(aq)} -> NaBr_{(aq)} + KCl_{(aq)}}$$
and the solubility products for all the species are big enough that there's essentially no solid, there's really no distinction between the left and right sides of the equation. It's a balanced equation, but nothing has changed. We started with $\ce{Na+_{(aq)} + K+_{(aq)} + Br^{-}_{(aq)} + Cl^{-}_{(aq)}}$ in solution and we still have that on the right side. So there's really not a net ionic equation because there's no net change.

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text:
How to determine number of structural isomers?
I have come across many questions where I'm asked to give the number of possible structural isomers. For example number, structural isomers of hexane is 5, while the number structural isomers of decane is 75.
How can I determine the possible number of structural isomers of a given organic compound?

Is this for a specific class of organic molecule or any organic molecule? Because the numbers you gave suggest that the number of isomers quickly increases with size of molecule, and that's just for simple hydrocarbons. And do stereoisomers count? I tried to come up with an equation for simple hydrocarbons, but couldn't identify 5 isomers for hexane, it's too late at night.

As far as I'm aware, there is generally no simple, formulaic way to do this w/r/t molecules of arbitrary complexity and/or size. Various computational approaches exist; search for, e.g., isomer enumeration.

@user137 This is not for stereo isomer. It is just for simple hydrocarbons. five structural isomers of hexane

See this.

@ashu This only gives you the degree of unsaturation, which is for alkanes always zero.

@Martin Do you consider cyclic alkanes here? they would have a 1.

@user137 I would consider only n-alkanes that obey $\ce{C_{n}H_{2n+2}}$. You are of course correct for cycloalkanes.

If you are given a question like this, I think the purpose is to get you to start drawing different structures and realize which ones are equivalent to ones you have already drawn.

@jerepierre You are right, my teacher told me same thing but sometimes it becomes dead end job like for example decane.

I like that the On-line Encyclopedia of Integer Sequences has a sequence with the number of structural isomers for alkanes with $n$ carbon atoms. The references therein may prove useful in your search for techniques to quantify possible isomers for a given formula.

Good Will Hunting might know.

It isn't easy but it is an interesting research topic
Determining the number of possible structures for a given range of chemical formulae isn't simple even for saturated hydrocarbons. The number of possible structural isomers rises rapidly with the number of carbons and soon exceeds your ability to enumerate or identify the options by hand. Wikipedia, for example, lists the numbers of isomers and stereoisomers for molecules with up to 120 carbons. But the counts are getting silly even at 10 carbons where there are 75 isomers and 136 stereoisomers.
It has been an interesting research topic in computational chemistry and mathematics. This old paper (pdf), for example, list some formulae for simple hydrocarbons among other simple series. Part of the interest arises because of the relationship to the mathematics of graph theory (it seems that chemistry has inspired some new ideas in this field of mathematics partially because enumerating possible isomers of hydrocarbons is strongly related to drawing certain simple trees which is intuitively obvious if you use the standard chemical convention of drawing just the carbon backbone and ignoring hydrogens).
You can look up the answers on the fascinating mathematics site OEIS (the online encyclopaedia of integer sequences). The sequence for simple hydrocarbons is here.
But the mathematical approach oversimplifies things from the point of view of real-world chemistry. Mathematical trees are idealised abstract objects that ignore real-world chemical constraints like the fact that atoms take up space in three dimensions. This means that some structures that can be drawn cannot exist in the real world because the atoms are too crowded and cannot physically exist without enough strain to cause them to fall apart.
Luckily, computation chemists have also studied this. There is, unfortunately, no obvious shortcut other than trying to create models of the possible structures and testing them to see if they are too strained to exist. The first two isomers that are too crowded are for 16 and 17 carbons and have these structures:

If you have any intuition of the space filling view of these, you should be able to see why they are problematic. A research group at Cambridge University has produced an applet to enumerate the physically possible isomers for a given number of carbons which is available here if your Java settings allow it. The results are discussed in a paper available in the Journal of Chemical Information and Modelling.

I've ran regression for these wiki data and for long chains number of isomers rises exponentially and it's about 0.001e^x, where x is no of C atoms.

"But the counts are getting silly even at 10 carbons where there are 75 isomers and 136 stereoisomers." By 75 isomers , you mean constitutional isomers right?

(1) As for the number of alkanes ($\ce{C_nH_{2n+2}}$),  Table 1, which is extracted from the data reported in

S. Fujita, MATCH Commun. Math. Comput. Chem., 57,
299--340 (2007) (access free),
shows the comparison between two enumerations based on
Polya's theorem and on Fujita's proligand method.

The number of alkanes ($\ce{C_nH_{2n+2}}$)
as constitutional isomers (structural isomers) and as steric isomers
is calculated by Polya's theorem (G. Polya and R. C. Read,

Combinatorial Enumeration of Groups, Graphs, and Chemical Compounds, Springer (1987)).
In the process of calculating constitutional isomers,
one 2D structure (graph or constitution) is counted just once.
In the process of calculating steric isomers, one achiral molecule or each chiral molecule
of an enantiomeric pair is counted just once, where achiral molecules and chiral molecules
are not differentiated from each other.
On the other hand,
the number of alkanes ($\ce{C_nH_{2n+2}}$)
as three-dimensional (3D) structural isomers
and as steric isomers is
calculated by Fujita's proligand method
(S. Fujita,

Combinatorial Enumeration of Graphs, Tree-Dimensional Structures, and
Chemical Compounds, Unversity of Kragujevac (2013)).
In the process of calculating 3D structural isomers,
one achiral molecule or one pair of enantiomers is counted just once,
where achiral molecules and chiral molecules (enantiomeric pairs)
are differentiated from each other.
For more information, see an account article entitled
"Numbers of Alkanes and Monosubstituted Alkanes.
A Long-Standing Interdisciplinary Problem over 130 Years"
(
S. Fujita,  Bull. Chem. Soc. Japan, 83, 1--18 (2010),
access free). This account article has discussed the difference between graph enumeration (Polya's theorem) and 3D structural enmeration (Fujita's proligand method) during recursive calculation.
It should be emphasized that graph-theoretical enumerations of chemical compounds
as constitutional isomers (structural isomers) and as steric isomers (based on asymmetry)
should be differentiated from stereochemical enumerations of chemical compounds
as 3D structural isomers and as steric isomers (based on chirality).
Although steric isomers based on asymmetry (graphs governed by permutation groups) and
steric isomers based on chirality (3D structures governed by point groups) give identical
enumeration results, they are conceptually different entities.
This point of view stems from Fujita's stereoisogram approach,
which is described in a recent book
(S. Fujita
Mathematical Stereochemistry, De Gruyter (2015)).
(2) Enumeration of achiral and chiral alkanes of a given carbon content has been
conducted by considering internal branching
(
S. Fujita, Bull. Chem. Soc. Jpn., 81, 1423--1453 (2008)).
Figure 3 of this report is cited below.

The symbol [q, t, s, p] means
the presence of q quaternary carbons, t tertiary carbons, s secondary carbons,
and p primary carbons. Alkanes are categorized into
centroidal and bicentroidal alkanes, which are the 3D extension of
centroidal and bicentroidal trees of graph theory.

This is a start towards a partial answer, but I tried to count the isomers of methane through nonane, and only by moving methyl groups around, each "branch" is only 1 CH3. I noticed a pattern in the number of 1 branch, 2 branch, 3 branch, and 4 branch isomers. Don't know how accurate these numbers are or how well this pattern holds up.
N   Straight    1 Branch    2 Branch    3 Branch    4 Branch
1          1           0           0           0           0
2          1           0           0           0           0
3          1           0           0           0           0
4          1           1           0           0           0
5          1           1           1           0           0
6          1           2           2           0           0
7          1           2           4           0           0
8          1           3           6           1           1
9          1           3           9           4           3

As far as I am aware, there is no straight up formula. Obviously there are very general trends, but not precise enough to tell you the number of isomers.
If you need to decide how many isomers something has, I suggest you either do it manually or use a programme like this: http://www-jmg.ch.cam.ac.uk/tools/isomercount/

We can calculate the number of structural isomers by double bond equivalent i.e by using the formula:
$$1+ \sum_i \frac{n_i (v_i - 2)}{2} $$
where $n$ repesents the number of atoms $i$ of a given valency, and $v_i$ represents these atoms’ valency. Example:
$$\ce{C4H6}: 1+ \sum_i \frac{n_i (v_i - 2)}{2}  = 1+ \frac{4(4-2)}{2}+\frac{6(1-2)}{2}  \\
= \frac{8-6}{2} +1 = 2 $$
Thus the compound may contain two double bonds or a triple bond and there are two possible isomers.

The question pertains towards alkanes and the formula that you have specified will be helpful (to a small extent) only for cyclic alkanes. This might as well be a comment than an answer. And Welocme to Chem.SE!

This answer is blatantly wrong. It calculates the double bond equivalents but the example used ($\ce{C4H6}$; i.e. butadiene, cyclobutene or butyne) already has five possible isomers if only a straight chain is assumed. $5 \ne 2$.

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text:
Would silica have the same effect as activated charcoal?
Silica found in "do-not-eat" packets is a desiccant (I'm not sure about this); and so is activated charcoal. Would consuming it have the same effect if one were to have a stomach ache?

Activated charcoal does not have the effect as Silica- Silica is a desiccant- removes water from the air and anything else. Charcoal is not a desiccant- rather its a catalyst. For stomach gas, some of its effect is conversion of tiny gas to larger bubbles which can easily escape.

oh i always thought charcoal was a desiccant- thanks for clearing it up @user2617804

Activated charocal (Activated carbon) - is highly porous and it IS desiccant. The stomach tablets are created from activated charocal.

I'm voting to close this question as off-topic because it is based on the wrong premises.

The silica gel granules are granular silicon dioxide (SiO2), the same matrial from what is the normal sand.
The pure SiO2 has the same effect to health as when you eat sand.
But because the granules are larger, may cause appendix problems (im not an doc). Anyway, this is not chemical related problem.
It can cause dehydratation or other problems, what is caused by it absorbtion capabality.
The bigger problem is, than the standard silica gel granules aren't pure SiO2.  The common additivas are what causes color change when it is moistured,  like cobalt chloride or others and thise compounds can be toxic!
So, don't eat it!

so...what happens when one eats sand? hmm- I never hope I do have to do that home-experiment

@Nick when you eat sand nothing happens (from the chemical-toxic) point of view. The SiO2 isn't toxic.

so it neither helps or harms you. it's redundant- alright! thank you

@Nick It may harm you mechanically. I mean, it may plug something, and since it is solid, it won't leave easily. In fact, solid sediments of some salts deposited in delicate places may be agonizingly painfull, especially when milled with ultrasound to purge them.

silica gel is not like sand. Only an idiot would eat silica gel.

@MaxW - Silica gel is silicon dioxide. The (common) sand is the same. So their chemical formula is the same: SiO2. The difference is in their physical form - the silica gel is in porous form. So, the danger isn't come from its chemical structure, e.g. it isn't toxic or like (same as sand), the problems arises from its physical form (porous absorbent). Also, as i already said in my answer - the silica gel usually contains other compounds  - and they could be toxic.

@ Max W  ,  when I was a baby I used to eat sand at the beach! I'm ok, it was a bit crunchy tho !

Silica and charcoal are both adsorbents.
However, silica is polar thanks to its plentiful OH groups, it preferentially adsorbs water, and that's why it is encountered in desiccant packages.  In the chemical laboratory one filters stuff through silica pads to separate polar materials out of solutions in non-polar solvents.
Activated charcoal is non-polar; it consists of aromatic rings, and it preferentially adsorbs non-polar molecules.  It is used in deodorizing packages.  In the lab, decolorizing charcoal is used in polar solvents to get rid of coloured impurities.
No, they are not equivalent.


text:
Formation of carbonic acid from breath?
Is it possible that blowing on water (hot/cold) can cause a reaction between the $\ce{CO2}$ from expiration and the water, to form $\ce{H2CO3}$? Is $\ce{H2CO3}$ dangerous?

If you survived drinking cola... ;)

It is possible that carbonic acid will form; however, only in small amounts because the equilibrium of the hydration of $\ce{CO2}$ to $\ce{H2CO3}$ lies mainly on the educt side. As more $\ce{CO2}$ is soluble in cold water, more carbonic acid will be present in the solution at lower temperatures.
$$\ce{CO2 + H2O \rightleftharpoons H2CO3}$$
At $25~\mathrm{^\circ C}$ and in pure water, the quotient of the concentrations is $[\ce{H2CO3}]/[\ce{CO2}] \approx 1.7 \times 10^{-3}$, so most carbon dioxide remains physically dissolved in water. This makes carbonic acid - despite a $\mathrm{p}K_\mathrm{a}$ of $3.6$ for the first dissociation step - a fairly weak acid. It is not a dangerous compound, otherwise it would not be used in carbonated drinks. It is also a natural constituent of blood, where it acts as a pH buffer. In red blood cells, it is produced as an intermediate by the enzyme carbonic anhydrase, which catalyzes the hydration of carbon dioxide.


text:
How can two orbitals constructively and destructively interfere simultaneously?
The molecular orbital theory dictates that when two atomic orbitals form molecular orbitals, then two molecular orbitals must form (i.e number of atomic orbitals = number of molecular orbitals). For this to occur there must be bonding molecular orbitals and anti-bonding molecular orbitals. The bonding MO's are caused when the orbitals constructively interfere and the anti-bonding orbitals form as the orbitals destructively interfere. I understand the concept of interference but I do not understand how two atomic orbitals (one from each bonding atom) can simultaneously interfere constructively and destructively. Moreover, I am under the impression that the anti-bonding orbitals exist even if they're not occupied. As a result, is it correct to think that it's the orbitals themselves that are the waves, not the electrons?
In summary:

How can two orbitals interfere constructively and destructively?
Do anti-bonding orbitals exist when not occupied?
Is it the orbitals themselves that are interfering with each other like waves?

Have a look at http://chemistry.stackexchange.com/a/15117/4945, maybe it helps a little.

Not so long ago I asked exactly the same question. Dropped two bounties on it. Perhaps you shouldn't until you start learning the real stuff (quantum mechanical explanations). Here is the link (it's now on Physics SE): https://physics.stackexchange.com/questions/121671/orbital-plot-to-explain-bonding-and-anti-bonding-molecular-orbitals

Thanks. Do you know the answers to number 2 and 3 in the summary of the question?

They do not simultaneously interfere constructively and destructively.
Let me explain. Since the time-independent Schrödinger equation is not separable for molecules due to inter-electronic interaction terms, we can seek solutions to this problem with the variational method.
Let's examine the case of the $\mathrm{H_2}$ molecule. The most natural/physically meaningful/simple choice is to use a trial function that consists of a linear combination of hydrogen 1s atomic orbitals:
$$E\left(c_{1},c_{2}\right)=\dfrac{\left\langle c_{1}\psi_{1}+c_{2}\psi_{2}\left|\hat{\mathcal{H}}_{\mathrm{e}}\right|c_{1}\psi_{1}+c_{2}\psi_{2}\right\rangle }{\left\langle c_{1}\psi_{1}+c_{2}\psi_{2}|c_{1}\psi_{1}+c_{2}\psi_{2}\right\rangle }=\dfrac{{\displaystyle \sum_{i=1}^{2}}{\displaystyle \sum_{j=1}^{2}}c_{i}^{\star}c_{j}\left\langle \psi_{i}\left|\hat{\mathcal{H}}_{\mathrm{e}}\right|\psi_{j}\right\rangle }{{\displaystyle \sum_{i=1}^{2}}{\displaystyle \sum_{j=1}^{2}}c_{i}^{\star}c_{j}\left\langle \psi_{i}|\psi_{j}\right\rangle }$$
$$H_{ij}\equiv\left\langle \psi_{i}\left|\hat{\mathcal{H}}_{\mathrm{e}}\right|\psi_{j}\right\rangle$$
$$S_{ij}\equiv\left\langle \psi_{i}|\psi_{j}\right\rangle$$
$$E\left(c_{1},c_{2}\right)=\dfrac{{\displaystyle \sum_{i=1}^{2}}{\displaystyle \sum_{j=1}^{2}}c_{i}^{\star}c_{j}H_{ij}}{{\displaystyle \sum_{i=1}^{2}}{\displaystyle \sum_{j=1}^{2}}c_{i}^{\star}c_{j}S_{ij}}$$
After minimizing the energy with respect to $c_1$ and $c_2$, you will end up with two energy levels, each one associated with a wave function:
$$E_{+}=\dfrac{H_{11}-H_{21}}{\left(1-S_{12}\right)}\qquad\psi_{+}=\dfrac{1}{\sqrt{2}}\left(\psi_{1}+\psi_{2}\right)$$
$$E_{-}=\dfrac{H_{11}+H_{21}}{\left(1+S_{12}\right)}\qquad\psi_{-}=\dfrac{1}{\sqrt{2}}\left(\psi_{1}-\psi_{2}\right)$$
In other words, by choosing to model the $\mathrm{H_2}$ molecule with two hydrogen $1s$ atomic orbitals, you find two discrete energy levels, each one associated with its wave function (one anti-bonding and one bonding). So this answers your questions:

The two atomic orbitals do not simultaneously interfere: The MOs are two different wave functions.
Of course the anti-bonding orbital exists when not occupied.

Note that the previous procedure can be generalized for all kind of atomic orbitals.
I want to clarify something about the meaning of the wave function since you're asking: As a result, is it correct to think that it's the orbitals themselves that are the waves, not the electrons?
Let's assume that your system consists of a single particle in a one-dimensional space. The wave function, $\Psi\left(x,t\right)$,
is a complex function that contains all the information about the system. But what is really physical meaningful is its modulus squared, $\left|\Psi\left(x,t\right)\right|^{2}$, which represents the position probability density. What is the position probability density? It's something that multiplied by a certain length (or volume if you're working in three dimensions), let's say dx, gives you the probability for finding the particle between $x$ and $x+dx$
$$P\left(x,x+dx\right)=\left|\Psi\left(x,t\right)\right|^{2}dx$$.
Of course, you can integrate the previous equation to find the probability over a certain region of space, $\Omega$
$$P_{\Omega}={\displaystyle \int_{\Omega}}\left|\Psi\left(x,t\right)\right|^{2}dx$$.


text:
Relation between the virial coefficients and van der Waals constants
I don't know where to begin with this question.

A function $f(x)$ can be expressed as a special power series called the Taylor expansion. The common expansion is $(1-x)^{-1}=1+x+x^2$ Note that the higher terms become neglibible small and usually are not included in expansion. Find the correlation between the virial coefficients and $a$ and $b$ in the corresponding van der Waals equation of state. Start with the equation shown below and compare it with the virial equation of state.
$$ p=\frac{RT\left(1-\frac{b}{V_{\mathrm m}}\right)^{-1}}{V_{\mathrm m}}- \frac{a}{V_{\mathrm m}^2}$$

The virial equation of state is

$$ p=\frac{nRT}{V}\left(1+\frac{nB}{V}+\frac{n^2C}{V^2}+\ldots\right)$$

I get that there are $A$, $B$, $C$… coefficients, and that the $A$ coefficient should be 1.
The best I have is that:
$$\begin{align}
A &= 1 \\
B &= -\left(\frac{b}{V_{\mathrm m}}\right)^{-1} \\
C &= -\frac{a}{V_{\mathrm m}^2}
\end{align}$$
If I were to relate that to the terms in the Taylor expansion, then $p=(1-x)^{-1}$ and then $A$, $B$, $C$… are the terms on the right side of the equation.
Is this the correct approach? Can the virial coefficients have a negative value?

You have to bring both equations into the same form and then compare the coefficients. You seem not to have used the tip with the Taylor expansion and I guess that's why you got an incorrect result.
As you wrote: The virial equation of state is
\begin{align}
p &=\frac{nRT}{V} \left(1+\frac{nB}{V}+\frac{n^2C}{V^2}+ \ldots \right) \\
  &= \frac{RT}{V_{\mathrm{m}}} \left(1+ B\frac{1}{V_{\mathrm{m}}}+ C\frac{1}{V_{\mathrm{m}}^2}+ \ldots \right)
\end{align}
and the van der Waals equation is
\begin{align}
 p &=\frac{RT(1-\frac{b}{V_{\mathrm m}})^{-1}}{V_{\mathrm m}}- \frac{a}{V_{\mathrm m}^2} \ .
\end{align}
Now, you have to bring the van der Waals equation into the same form as the virial equation by factoring out $\frac{RT}{V_{\mathrm{m}}}$
\begin{align}
 p &= \frac{RT}{V_{\mathrm{m}}}\left( \left(1 - \frac{b}{V_{\mathrm{m}}} \right)^{-1} - \frac{a}{R T V_{\mathrm{m}}} \right) \ ,
\end{align}
using the Taylor expansion $\left(1 - \frac{b}{V_{\mathrm{m}}} \right)^{-1} = 1 + \frac{b}{V_{\mathrm{m}}} + \frac{b^2}{V_{\mathrm{m}}^2} + \ldots$ and ordering the terms according to their power in $\frac{1}{V_{\mathrm{m}}}$
\begin{align}
 p &= \frac{RT}{V_{\mathrm{m}}}\left( 1 + \left(b - \frac{a}{R T} \right) \frac{1}{V_{\mathrm{m}}} + b^{2} \frac{1}{V_{\mathrm{m}}^{2}} + \ldots \right) \ ,
\end{align}
so that you can compare the coefficients in both equations to get the result
\begin{align}
 B &= b - \frac{a}{R T} \\
 C &= b^2 \ .
\end{align}

Thank you! This makes a lot more sense. However, Why is there no $a$ in the C term? That is, why is it not something like $b^2-\frac{a^2}{(RT)^2}$?

@JohnSnow That's the result of the calculation. The constants $B$ and $C$ don't have to show any similarity.

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text:
Would equal masses of gold and copper result in the same volume change if added to water?

Would equal masses of gold and copper result in the same volume change if added to water?
  Density of gold: $19.3~\mathrm{g/mL}$
  Density of copper: $8.96~\mathrm{g/mL}$

I was thinking the answer would be no since the density is different between gold and copper but was unsure if this was a trick question since it maybe has to do with the mass of the substance that is placed into the water.

The trick in the question, insofar as there is one, is that neither copper nor gold will float on water. If you just vaguely remember the rule that the mass of fluid displaced by a floating object equals the mass of the object itself, and apply it without thinking, you may get the wrong answer.

Well they could float, depending on the shape, since the question is not specific, but since this is the chemistry SE, I think it's safe to assume that the point of the question is the relation between density, mass, and volume, rather than buoyancy.

No, a trick question would be: What weighs more, a pound of aluminum or a pound of gold? BTW, an ounce of aluminum weighs less than an ounce of gold.

@LDC3 - I'm assuming here that you're referring to a fluid ounce (i.e. measure of volume - one of Imperial, US customary, and/or US food labeling ounce, and probably many others) of gold or aluminum, versus one of the "weight" ounces (e.g. international avoirdupois, troy, metric, apothecaries, Maria Theresa, Spanish, or Tower, and probably many others (for example, the weight of Venezuelan Beaver Cheese is typically measured in Venezuelan Beaver Ounces).  :-)

@BobJarvis An ounce of aluminium would, traditionally, be an ounce avoirdupois (~28.4g); an ounce of gold would be a troy ounce (~31.1g).

"Added to water" is ambiguous.
If the metal objects are shaped and placed such that they float (eg: as a metal boat does, or a light hollow sphere), then they will each displace the same volume of water (the volume of water equivalent to their equal common masses).  So long as they float, they will displace the same amount of water no matter their shape.
If the masses objects are shaped and placed such that they sink (eg: consider a solid sphere or cube), then they will in general displace different volumes.
Bonus.  The second part is slightly more complex than it seems when it comes to calculating how much water they would displace.  If they have no dry internal voids, their respective displaced volumes are simply the mass of each divided by its density.  Others have given the simple calculation.
If they were shaped such that they had dry voids yet were small enough to sink (think of a heavy hollow sphere with overall density greater than water so it sinks, or an overturned metal boat with small air pockets), then the displaced water will be somewhere between the above two calculations (fully floating calc and no-voids calc).
Note that it would be possible to construct the two objects such that they sank and yet displaced the same amount of water, eg: two hollow spheres with the same outer diameter but different internal diameters, both heavier than the same volume of water.  But that would be a very trick question indeed!

You're reasoning and your answer ("no") are both correct; this is not a trick question.
Let's say we had 8.96 gm of gold and the same amount of copper and dropped them both into separate graduated cylinders that were filled to the 10 ml mark with water.  Submerged items displace a volume equal to their own volume. Therefore, this amount of copper should displace
$$\mathrm{\frac{[8.96]}{[density~ of~ Cu]} = \frac{[8.96]}{[8.96]} =1 ml~displaced}$$
whereas 8.96 gm of gold would displace
$$\mathrm{\frac{[8.96]}{[density~ of~ Au]} =\frac{[8.96]}{[19.3]}= 0.46 ml~displaced}$$

Historical anecdote time!
David Richerby made an offhand comment about Archimedes of Syracuse. Archimedes had to figure out whether the king's crown was made of actual gold, but obviously he couldn't damage the crown because that would make the king angry. He had to determine the density, but you need the volume for that, and there wasn't really a way to determine the volume of an irregular object back then.
He was wondering how to do this, and at one point, he had to take a bath. Now, his servant filled the bath to the brim, and when Archimedes got into the water, some of it spilled over the edge. The legend after this goes that Archimedes saw this, jumped out of the water, and still naked, ran out onto the streets yelling "Eureka".
He had the crown weighed and asked for gold equal in mass to the weight of the crown. Then he submerged the crown and the gold into water, he saw that the gold displaced a different amount of water, despite weighing as much. This means that the volume, and thus the density of the object, was not the same, and thus that the crown was not made of solid gold. (note)
So your answer is absolutely correct: the displaced amount of water is dependent on the volume, not the weight.

(note) This is a legend. People who know more about this than me have stated that Archimedes didn't use the volume measurement, since this required measurement accuracy beyond contemporary means. Instead, the more likely method was that he tied them to a balanced scale, then submerged them, subjecting them to a buoyant force equal to the weight of the water displaced. The denser object would encounter less force and thus sink.

authors: Bob Jarvis - Слава Україні


text:
Unknown anion (one carbon and two nitrogen atoms, charge 2−)
What's the name of this ion: $\ce{CN2^2-}$? I think that it has this structure:
$\ce{{}^{-}N=C=N-}$.

The parent molecule is cyanamide.

The ion is usually referred to as either the cyanamide dianion or the carbimide dianion

Since cyanamide also refers to the protonated species, I prefer carbimide (or carbodiimide, which is the derived functional group), although there may not be an official preference


text:
pH of baking soda in water?
If I dissolve 1  g of sodium bicarbonate powder ($\ce{NaHCO3}$) into 400 ml of distilled water, what will the pH of the resulting solution be?

This appears to be a homework type question, please share your thoughts and attempts towards the solution. For formatting help visit the [help] and for more information about this page, take the [tour].

I'm an artist. I'm using baking soda in water to deacidify drawings on newsprint so they can be mounted and framed. (See my other question on the topic.) This is the recipe I'm using, and in case anyone asks me how strong my buffering solution is, I'd like to have more of an answer than "Well, it turns red cabbage juice blue."

Well you're going about it the wrong way. You need to start with so called "acid free paper." https://en.wikipedia.org/wiki/Acid-free_paper The gist is that you can't get the lignin (which makes acid when it decomposes) out of the paper by neutralizing the current state of the paper. Thus making the paper less acidic now will slow the decay,  but it won't stop it. The point is that if you really want your drawings to last, then start off with a better grade of paper.

1 g of sodium bicarbonate in 400 mL of water will give an approximately 30 mM solution with a pH of ~8.4. The interesting thing about the pH of solutions of amphiprotic compounds is that so long as the concentration of the salt is significantly larger than its Ka and Kb, the solution pH only depends on the dissociation constants. Basically any amount of bicarbonate above 10 mM or so will have the same pH in solution, so pH is not a good indicator of how much acid you can neutralize with it.
I'm not really sure what good this information will do you. How much this solution will be able to affect the paper is going to depend on how much paper you use for a given volume of solution and how acidic the paper is. I don't know much about paper—it could be that it doesn't really matter if you go overboard on the bicarbonate. It might be as simple as making a concentrated solution, testing periodically that your bath hasn't become acidic, similar to photographic stop bath, which has a pH indicator that changes colour when the bath is not acidic anymore. If residual bicarbonate is a problem, rinsing with water afterwards is probably advisable.

This answer did me a great deal of good because it led you to my first question, and your answer there is what led me to the really important information I was looking for. (Though I'll probably be following up with a question about the pH of calcium bicarbonate dissolved in seltzer water as soon as I get hold of some of that. Or maybe I'll just see if my local brewing store has litmus paper.)


text:
What is the procedure to create stock solutions?
Until now to execute this recipe, each time i had to count those milligrams/Liter. An enough difficult procedure.
I am posting here this question to help me understand the procedure of creating stock solutions. So the next time i will create that plant substrate, to take a specific volume of that stock solution.
Thank you.

Since it contains agar, you wouldn't be able to make a stock solution with the agar in it; but you can make a stock solution of the other items and add the agar when you need the working solution. Also, since there is a fair amount of sugar in the recipe, it would be best to add that to the working solution as well.
You should be able to make a 100 times concentrate (leaving out the sugar and agar) and then sterilize it in an autoclave. When you need to make a working solution, weigh 5 g of agar and 20 g of sugar, add 10 mL of stock solution, and about 900 mL of water. Boil the solution to dissolve the agar, cool (to about 50°C) and add water (before it sets) to get 1000 mL of solution.

Yes, i knew that i have to leave agar and sugar for the end. Also i thought to create two stock solutions. One for the elements that are in a concentration under 10mg/L and one for the rest. The thing is how should i achieve that 100 times concentration you referred. Should i follow the method Saroj provided?

That is the easiest way to get the correct amounts.

So to close that thread. 100 times concentration means in that i have to create a stock solution of 100M ?

No, because that would be too much. What you want for boric acid is 1000 mg/L instead of 10 mg/L; cobalt chloride is 2.5 mg/L instead of 0.025 mg/L; etc... Multiply the amounts by 100, but keep the volume the same.

For the easy reference,you prepare a concentrated stock solution and you dilute it according to your requirement by the below given procedure.You prepare a solution by dissolving a known mass of solute (often a solid) into a specific amount of a solvent. One of the most common ways to express the concentration of the solution is M or molarity, which is moles of solute per liter of solution.
Example:
Prepare 1 liter of 1.00 M NaCl solution.
First calculate the molar mass of NaCl which is the mass of a mole of Na plus the mass of a mole of Cl or 22.99 + 35.45 = 58.44 g/mol
1) Weigh out 58.44 g NaCl.
   2) Place the NaCl in a 1 liter volumetric flask.
   3) Add a small volume of distilled, deionized water to dissolve the salt.
   4) Fill the flask to the 1 L line.
If a different molarity is required, then multiply that number times the molar mass of NaCl. For example, if you wanted a 0.5 M solution, you would use 0.5 x 58.44 g/mol of NaCl in 1 L of solution or 29.22 g of NaCl.
once your stock solution is prepared you can dilute it to a particular volume to make a new, more dilute solution.  The key equation here is
M1V1 = M2V2
Remember that number of moles you are working with is constant whether in the  stock solution or in the more dilute new solution.  Since molarity times volume equals moles, that it what the equation above is talking about.  M1V1 is the molarity (moles per liter) times the volume of the stock solution and M2V2 is the molarity times the volume of the new solution.  So you are usually given three of the variables in the problem and you simply plug them in and solve for the missing fourth variable.
In terms of actual laboratory practice to make a new solution from a stock solution, first you calculate how many milliliters of the stock solution that you need to make the new solution.  Then measure out the required number of mL's of the stock solution in a graduated cylinder and pour into into a new flask.  Then add fresh water (or whatever solvent you are working with) to the flask until the level reaches the required number of mL's of the new solution.  Then stir the solution to mix evenly.  Now you have a new, more dilute solution of a know concentration.

You didn't look at his recipe so your answer does not help.

I have read this stuff in one link.

The best way of creating a stock solution (even more importante if it is quite diluted) is to start from a highly concentrated solution, and thus dilute it in subsequent steps taking aliquots from the previous one.
It is very important that you check the final dilution to be sure of its exact concentration. That is the use of the "analysis standard" sets that permits you to guarantee the concentration of your stock solution.


text:
What general uses do beta-lactam functional groups have?
I am currently working on a mechanistic study that deals with stereoselective $\beta$-lactam formation. I am well aware that there is a class of antibiotics, that involve this particular functional group. However, the wikipedia article of this compound class does not offer very much to continue, apart from the antibiotics.
Are there any other uses of the $\beta$-lactam moiety in general, apart from drug design?
I would like to be able to view my research in a broader scope. Since my part is more focussed on the reaction mechanism of the synthesis, I have not been too much concerned about the main purpose of this particular functional group, yet. This is of course apart from the obvious drug design process. For me this is just a question of interest.
As I am resurrecting this question, I am in particular looking for some interesting publications about the above mentioned question. However, I am willing to recognise, that the most common use is in drug design. If someone from this area of expertise could provide a source as a starting point for more research in this area.

I would also accept an answer that solely provides literature  to the drug design part of this question.

From a previous life I know that beta-lactams provide access to gamma-aminoacids.  I haven't got literature references handy right now, though.

@AbelFriedman Thank you for that preliminary entry in this amazing raffle ;) I am very much looking forward to hearing from you again - there are still seven days left, if you have the time to spare, at least for this run of the bounty.

@Freddy How360 is okay because they provided attribution to the original post. I have reported Chemistry Ninja to the appropriate parties.

β-lactams, also known as 2-azetidinones, have experienced a resurgence of interest in the last decade or so, with a number of new applications being exploited. Of course, antibiotic activity is well-known, but I hope you don't limit your acceptance of drug design just to this class of chemical.
β-Lactams have found use in a number of fields of medical application, inlcuding gene activation, enzyme inhibition and cholesterol absorption inhibition, and recently there has been growing interest in the use of these molecules as synthetic templates for accessing other target moelcules.
There are lots of sources of references if you go digging, but some key reviews that may interest you include:

TETRAHEDRON Volume 56, Issue 31, Pages 5553-5742 (July 2000). The entire issue is concerned with developments in β-lactam chemistry;
EUROPEAN JOURNAL OF MEDICINAL CHEMISTRY  Volume 45, Issue 12, Pages 5541-5560   (DEC 2010) "2-Azetidinone - A new profile of various pharmacological activities"  gives some very good insights into new medical applications of this class of compound, including anticancer, antiparkinson, antidiabetic, antiinflammatory activities. I think they can also do the washing up and buy your groceries.
TETRAHEDRON  Volume 59, Issue 39, Pages 7631-7649 (Sep 2003) "Recent progress in the synthesis and chemistry of azetidinones" gives a pretty good summary of synthetic approaches to these molecules.
Alcaide and Almendros do a lot of work using 2-azetidones as reactive synthons in catalytic cyclisations. A recent review by them is
ACCOUNTS OF CHEMICAL RESEARCH  Volume 47, Issue 3, Pages 939-952 (Mar 2014) "Gold-Catalyzed Cyclization Reactions of Allenol and Alkynol Derivatives", but they have a heap of other papers also.
Another example, probably somewhat similar, of their use as a reactive synthon is
BULLETIN OF THE KOREAN CHEMICAL SOCIETY  Volume 28, Issue 1, Pages 17-28 (2007) "Indium and gallium-mediated addition reactions" - I have to admit I only read the abstract of this paper, but they appear to use azetidones as synthons in metal-mediated syntheses of a range of compounds.

Finally, the August edition of Chemical Reviews Vol 114, Issue 16 looks at small heterocycles in synthesis and has some interesing papers, inparticular:

CHEMICAL REVIEWS Volume 114, Issue 16, Pages 7930-7953 "Chemical Synthesis of beta-Lactams: Asymmetric Catalysis and Other Recent Advances". This is a very good paper for you, I think.
CHEMICAL REVIEWS Volume 114, Issue 16, Pages 8257-8322 "Four-Membered Ring-Containing Spirocycles: Synthetic Strategies and Opportunities". An interesting application of lactams in spiro molecule formation.
CHEMICAL REVIEWS Volume 114, Issue 16, Pages 7954-8015 "Synthesis and Reactivity of 2-(Carboxymethyl)aziridine Derivatives". This looks at lactams being used as synthons for a range of molecules, here focussing on aziridines.

I hope some of these point you in the direction of what you are looking for.

You have well deserved the extra credits. I have not yet been able to go through all of your linked papers, but they look very promising. Thank you very much for your effort - I did not even know where to start.


text:
"Chemicals" released in headphone with burned cable
My friend works at a headphone company at the support service and today she came across someone who claimed that he threw his broken headphone away because he had heard dangerous chemicals could be released when the cable burned through (because they would be exposed to the outside world possibly) which was according to him affirmed at a local waste disposal service. As I am unable to find any confirmation for this claim online, I decided to go ahead and post the question here. I was a bit unsure whether to post on chemistry.SE, biology.SE or engineering.SE, but went ahead with this one in the end, hope it's not off topic, but if it is then my apologies.

Personally I cannot answer your question, but I think it is on topic here. If you however do not get an answer you can think about flagging and asking for migration.

Burning or Incineration of Plastic cause carcinogenic chemical products if without suitable catalysts... To pyrolyze plastics such as polyethylene(PE) and polypropylene(PP) of olefin series, the catalysts with high cracking property are
needed.

Dioxin- carcinogenic and antibiotic effect to the environment.
Flyash- may produce respiratory illnesses.
Sulfur dioxide
Nitrogen oxide- green plastics.
Toluene- aromatic compound that is related in the burning of cigarettes.
Carbon monoxide- main product of anerobic/ lack of ventilation combustion.
carbonaceous char, oils and gases.
paraffins, olefins, naphthenes and aromatics

Plastic and Diesel
http://www.chemtec.org/proddetail.php?pr...
http://www.p-fuel.com/index_files/faq.ht...
Thermofuel
http://www.cynarplc.com/thermo-fuel.asp
Mael is right that a  headphone set is made of some plastics (which are mainly made of basic carbon polymers) and the metallic part for the electronic conduction.so even i agree it can not be dangerous for human health to a great extent.

A headphone set is made of some plastics (which are mainly made of basic carbon polymers) and the metallic part for the electronic conduction.
The degradation of the plastic could give some hazardous materials for the environment. We cannot be sure of which ones just because each plastic is different. But I do not think it can be dangerous for human health.


text:
Is there any replacement for ethyl mercaptan in LPG?
Ethanethiol, commonly known as ethyl mercaptan, is a colorless gas added to LPG in order to detect its leakage.
Apart from ethyl mercaptan, is there any gas or liquid that may be used as a replacement to this?

Methyl mercaptan is also used as an oderant.

Why would one want to replace it?
The choice of ethanethiol is a reasonable compromise between its toxicity and the extremely low detection limit around 1 ppb (1mg/1000 kg).

I guess they don't like the smell.

it's also worth mentioning that thiols are preferred over amines, to which human noses are also extremely sensitive, because humans desensitize to amines much more quickly than thiols.


text:
OFET vs Traditional Transistors
What is the advance of OFET's (Organic Field effect transistors) over the traditional CMOS Transistors?

The advantages of OFET's over traditional CMOS transistors all derive from how OFET's can be manufactured.  Traditional transistors are manufactured using vacuum deposition techniques.  These are slow, time consuming, batch processes, hence the higher cost for electronic components produced by such methods.
OFET's can be manufactured using standard coating technologies that have been widely used in industry for a long time, and are well understood.  Therefore, the manufacturing equipment needed is readily available. With this coating equipment uniform layers ranging from sub-micron to tens of microns (you don't need thicker layers for an OFET, but the coating technology would allow you to coat much thicker layers if you wanted to) can be coated at high speed on thin, flexible, web supports such as aluminized mylar. Multiple layers can be coated by making multiple passes down the coating line, into a drying oven and then on to the next coater.  Thousands of OFETS can be obtained from a small, coated, mylar spool. Since the process is continuous and high speed the manufacturing cost of the device is much reduced compared to similar CMOS devices.  Each layer contains an active organic compound dissolved in a low-cost plastic (for example, polycarbonate) binder, so material costs are reduced as well.  Another key advantage is that since organic materials, plastic binders and flexible web materials are used, the final product is flexible and can be easily cut to any desired shape.  This allows the manufacturer to use the product in new applications where flexibility (newspapers) and shape (electronic clocks) might preclude the use of rigid CMOS devices.


text:
What is the point of introducing virtual orbitals in Hartree-Fock calculations?
Let us for simplicity discuss RHF formalism. For $2n$-electron system we have $n$ Hartree-Fock equations written for $n$ spatial orbitals $\{ \phi_{k} \}_{k=1}^{n}$
$$
\newcommand{\mat}[1]{\boldsymbol{\mathbf{#1}}}
$$
\begin{equation}
    \hat{F}(1) \phi_{k}(1) = \varepsilon_{k} \phi_{k}(1) \, , \quad k = 1, 2, \dotsc, n \, .
\end{equation}
Once we introduce finite basis $\{ \chi_{q} \}_{q=1}^{m}$ and express spatial orbitals as a linear combination of basis functions $\chi_{q}$
\begin{equation}
    \phi_{k}(1) = \sum\limits_{q=1}^{m} c_{qk} \chi_{q}(1) \, , \quad k = 1, 2, \dotsc, n \, .
\end{equation}
we end up with $n$ Roothaan–Hall equations
\begin{equation}
    \sum\limits_{q=1}^{m} F_{pq} c_{qk} =
    \varepsilon_{k} \sum\limits_{q=1}^{m} S_{pq} c_{qk} \, ,
    \quad k = 1, 2, \dotsc, n \, ,
\end{equation}
which can be rewritten in the following matrix form
\begin{equation}
    \mat{F} \mat{c}_{k} = \varepsilon_{k} \mat{S} \mat{c}_{k} \quad k = 1, 2, \dotsc, n \, .
\end{equation}
The Fock matrix $\mat{F}$ and the overlap matrix $\mat{S}$ are both $m \times m$ square matrices, $\mat{c}_{k}$ is a column $m \times 1$ matrix, $\varepsilon_{k}$ is just a scalar value.

We can then collecl all $n$ $\mat{c}_{k}$ column $m \times 1$ matrices into one $m \times n$ matrix $\mat{C}$ and all $n$ values $\varepsilon_{k}$ into $n \times n$ square matrix $\mat{\varepsilon}$
\begin{equation}
    \mat{F} \mat{C} = \mat{S} \mat{C} \mat{\varepsilon} \, .
\end{equation}

In practice, however, we extend both $\mat{C}$ and $\mat{\varepsilon}$ to $m \times m$ matrices from $m \times n$ and $n \times n$ respectively, which results in having $m-n$ virtual (unoccupied) orbitals.
Taking into account that virtual orbitals are even more unphysical than their occupied counterparts the question is what is the point of such extension of $\mat{C}$ and $\mat{\varepsilon}$? Why do not we just leave them of $m \times n$ and $n \times n$ sizes respectively?

Virtual orbitals are automatically 'given' by the solved eigenvalue equations.  These orbitals are important for some post-HF methods such as CI and perturbation theory.

Indeed, I'd just give that as the answer. If you think about a simple atomic orbital picture (i.e., a minimal basis) there's a conservation of orbitals when you compute the MOs. Since there are typically two electrons per occupied orbital, there will be unoccupied virtual orbitals.

@GeoffHutchison That is absolutely true (even for larger basis sets). See also http://chemistry.stackexchange.com/a/15117/4945

@GeoffHutchison, yes, the question is about why do we expand the matrices. I understand that we can do it, but not have to. At least for HF itself.

@Martin, in answer which you linked you said: "the dimension of the involved matrices have to be the same". But that is wrong, as far as I understand, since they do not have to. You can extend matrices of coofficient and orbital energies to $m \times m$ size where $m$ is the number of basis functions, but you do not have to do so. $m \times n$ matrix of coeeficient and $n \times n$ matrix of orbital energies where $n$ is half of the number of electrons would do the job.

The unitary transformation to form the canonical orbitals guarantees that you have a diagonal $\mathbb{F}$ matrix, and all of them are interdependent on their own solution (pseudo eigenwert), this means they are also dependent on the virtual orbitals. Hence for every function you plug into the formalism you get a molecular orbital.

It's worth pointing out that some solid-state DFT codes do not provide all virtual orbitals by default because there are so many of them and the code therefore uses different methods for the diagonalization. This is more of a technical reason, though. At any rate, one cannot focus solely on the initially occupied orbitals since the inital guess may be bad for the system under consideration. Therefore, one wants to have all orbitals available such that the iterations may switch occupation to the appropriate ones.

Virtual orbitals are automatically 'given' by the solved eigenvalue equations and are necessary for solving said equations. These orbitals are important for some post-HF methods such as CI and perturbation theory and coupled-cluster theory.

On another note, the eigenvalue equations are of course no true eigenvalue equations, since they are dependent on their own solutions.


text:
Formation of 2,4-Dichlorophenol by free chlorine mediated oxidation of Triclosan
I cannot find much guidance in the literature. This is the reaction between free chlorine in water and triclosan.
I understand ether cleavage and electrophilic aromatic substitution (maybe) involved. Any ideas as to what the mechanism of this reaction might be?

Just as ethers can be formed when alcohols are protonated, the reverse reaction, formation of alcohols from ethers can also occur in the presence of acid.  In the case at hand, when chlorine is mixed with water hypochlorous acid is one of the compounds formed, and it can serve as the proton source for the following reaction.  The nucleophile can be $\ce{OCl^-}$, or any other nucleophile present.  In any case this mechanism explains the formation of 2,4-dichlorophenol from TCS, chlorine and water.

Here is a link to the a study entitled, "Formation of chloroform and chlorinated organics by free-chlorine-mediated oxidation of triclosan".  If you click on the "PDF" link in the upper-right corner, you'll get the full text pdf of the article.  The article discusses the reaction pathways to the various products and also comments on the formation of 2,4-dichlorophenol from TCS as well as other reaction products, as you've illustrated.

Thanks! :) The acidic ether cleavage you illustrated above is an SN2 pathway am I right?
And if possible, can you also help to explain the electrophilic substituion of Chlorine from TCS into the Tetraclosans diagramatically as well?
Thanks!!

It's similar, but the aromatic ring blocks the "rear-side approach" normally associated with a traditional $\ce{S_{N}2}$ mechanism.  Instead, a $\ce{S_{N}Ar}$ mechanism is probably involved.  For example, see http://en.wikipedia.org/wiki/Nucleophilic_aromatic_substitution

I'm not sure if this qualifies as an answer since it's more of a guess. Catechols (1,2-dihydroxybenzene derivatives) can be oxidized to ortho-quinones. Applying that transformation here (chlorine in water is certainly oxidizing conditions) gives the oxonium of the ortho-quinone, which can be hydrolyzed to 2,4-DCP.


text:
Represent bonds spanning a unit cell boundary plane
Is there a chemical input format that is-

Supported by Open Babel (and hence supports visualization in avogadro??)
Can represent a bond connecting an atom to its partner in a neighbouring unit cell, for crystals/systems with periodic boundary conditions

Follow up (added later):
I could not understand how cif does it, or cml. But it would be nice if I could get rid of the symmetry folds in the markup given below.
Here is the cif file. The cml output using babel is --
<?xml version="1.0"?>
<molecule id="Calcium titanate" xmlns="http://www.xml-cml.org/schema">
 <crystal>
  <scalar title="a" units="units:angstrom">5.380000</scalar>
  <scalar title="b" units="units:angstrom">5.440000</scalar>
  <scalar title="c" units="units:angstrom">7.639000</scalar>
  <scalar title="alpha" units="units:degree">90.000003</scalar>
  <scalar title="beta" units="units:degree">90.000003</scalar>
  <scalar title="gamma" units="units:degree">90.000003</scalar>
  <symmetry spaceGroup="-P 2c 2ab">
   <transform3>1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1</transform3>
   <transform3>1 0 0 0.5 0 -1 0 0.5 0 0 -1 0 0 0 0 1</transform3>
   <transform3>-1 0 0 0 0 -1 0 0 0 0 1 0.5 0 0 0 1</transform3>
   <transform3>-1 0 0 0.5 0 1 0 0.5 0 0 -1 0.5 0 0 0 1</transform3>
   <transform3>-1 0 0 0 0 -1 0 0 0 0 -1 0 0 0 0 1</transform3>
   <transform3>-1 0 0 0.5 0 1 0 0.5 0 0 1 0 0 0 0 1</transform3>
   <transform3>1 0 0 0 0 1 0 0 0 0 -1 0.5 0 0 0 1</transform3>
   <transform3>1 0 0 0.5 0 -1 0 0.5 0 0 1 0.5 0 0 0 1</transform3>
  </symmetry>
 </crystal>
 <atomArray>
  <atom id="a1" elementType="Ti" formalCharge="4" xFract="0.000000" yFract="0.500000" zFract="0.000000"/>
  <atom id="a2" elementType="Ca" formalCharge="2" xFract="0.006480" yFract="0.035600" zFract="0.250000"/>
  <atom id="a3" elementType="O" formalCharge="-2" xFract="0.571100" yFract="-0.016100" zFract="0.250000"/>
  <atom id="a4" elementType="O" formalCharge="-2" xFract="0.289700" yFract="0.288800" zFract="0.037300"/>
 </atomArray>
 <bondArray>
  <bond atomRefs2="a1 a4" order="1"/>
  <bond atomRefs2="a4 a2" order="1"/>
 </bondArray>
</molecule>

As the developer of both Open Babel and Avogadro I can say the answer is "not yet."
The main thing isn't the file format. It's that Avogadro at the moment doesn't have support at the moment to show bonds across a unit cell.
I believe there was a patch for that, but it was too slow for typical use:
https://github.com/dlonie/avogadro/commits/ENH_intercell_bonds

hmm .. but the CIF and CML files i linked to/quoted above seem to generate bonds across the unit cell boundary when I built a supercell, implying to me that the internal representation does support those bonds!!

I believe openbabel and avogadro should have tags of their own here. But I do not have enough points to make them. Could you help, if you agree?

EDIT: Nope .. I might have been wrong about the CIF and CML files rendering supercells correctly! Thanks for the pointer towards the patch. :)

@DebanjanBasu The question for this was raised on meta and for the time being, we do not want to generate too specific tags. The [tag:software] is already rarely used and more specific ones would be even less used. http://meta.chemistry.stackexchange.com/q/373/4945

Generating a supercell should force bond perception in Avogadro (v1.1.x). That is, if there are bonds across the unit cell boundary, creating the explicit atoms in a supercell should also create new bonds. Rendering bonds across a unit cell boundary is a little bit harder, since you have to generate "ghost atoms" (i.e., determine the supercell, but not actually display it).

I do not immediately see why adding ghost atoms as an extension  is hard, but i should rather look into the internal representation of bonds in avogadro . Thanks!

As pointed out by Geoff Hutchinson, this isn't currently possible by Avogadro and OpenBabel - and he should know best ;)
Since you want to visualize a feature of crystalline systems, you might want to have a look at specialized programmes that predominantly/only operate on CIF files. We have used SCHAKAL, written by Egbert Keller, for this purpose in the past.
Out of my mind, I don't remember how it was done. But it can be done, just like almost anything else you can imagine, such as centroid-centroid bonds between rings, etc.
Be warned: SCHAKAL's GUI violates any design rules (even if you apply Klingon standards). Once you have it compiled and running, switch the GUI off and operate it through chains of unituitive one-key shortcut commands. If you have overcome the horrors, you'll be rewarded with a fantastic postscript output in publication quality.

My suggestion would probably be to use CCDC Mercury, which is pretty easy to use, free (as in money) and available for Linux, Windows, and Mac. Since the CCDC is literally in the business of crystal structures, it's probably the best-tested and accurate at handling crystal issues. I recommend it highly.


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Whats the difference between ionization energy and orbital energy?
If you look at the trend in orbital energies as you go across a period the pattern is clear (orbital energy decreases with increasing effective nuclear charge) and, to my knowledge, it has no anomalies like the trend in ionization energies does when a 2p orbital is filled for the first time with two electrons (when this happens the ionization energy decreases). I expected that the graphs for ionization energy would be mirror images in the x-axis given that orbital energies are negative and measured relative to an ionized electron (which equals 0) albeit with different y-scales because ionization energy is the energy to ionize a mole of a substance in a gaseous state. Is my understanding that orbital energy = -(energy required to remove one electron from that shell) incorrect? Fundamentally, the question is, why doesn't the trend in orbital energies mirror the trend in ionization energies?

Could you include the data (or links to it) for the orbital and ionization energies you are referring to?

First off, the idea that the orbital energy is the negative of the ionization energy is actually Koopmans' theorem. It works qualitatively because of cancelation of errors.
But I digress...
Orbitals are single-electron constructions. That is, in Hartree-Fock theory, an orbital is a single-electron wavefunction assuming the average electrostatic potential of the remaining electrons.
Ionization energy trends are experimental observables and you see effects of multi-electron interactions.
So the short answer is that orbital energies are not physical quantities (i.e., being single-electron mathematical results) and ionization energies are based on multi-electron interactions.
Probably the simplest way to understand the trends in ionization energies vs. orbital energies is to look at Slater's rules and also remember Hund's rules. That is, when we add electrons, we are partially shielding the nuclear charge, but we also have to consider the stabilization due to exchange of electrons and destabilization when we pair electrons in an orbital. These are multi-electron effects, and only some of that can be recovered in a simple one-electron picture.
For the $2p$ orbitals, as you mention, there's a decrease in the ionization potential when you put two electrons in the same orbital. This happens because this new electron does not benefit from the exchange effect (Hund's rules) and has an electrostatic repulsion with the electron in the same orbital. Thus it's easier to remove it than might be expected from a one-electron orbital picture.


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Can every conductor be electroplated using copper and sulphuric acid?
I really like electrolysis. Today I bought 30% sulphuric acid to electroplate something with pieces of copper.

I've poured half water and half acid in a glass. I connected some cuprextit-copper plate to the anode and outdated aluminium coin to the cathode.
After starting the process, I could see that nothing is appearing on the cathode. I scratched the aluminium of the oxides. After repeating the electrolysis, aluminium turned black wherever scratched. Then I found out I swapped the wires.
So, once I had the wires correct, this happened to my cathode:

The dust can be easily washed off.
According to the image from wikipedia, any metal should be "electroplateable". I actually thought that any conductor that doesn't react with the solution can be electroplated.
So what's the reality? Was I just impatient? Is electroplating only limited to some conductors? What about non metal ones? Can I electroplate carbon electrode?

Oh god, I just found out that I had switched cathode and anode. I had half of iron stick dissolved.

The reality is that proper electrlysis used for create nice surfaces is almost black magic. If you want to do it, you should really look it up in literature, because there are many little tricks and additives that no one really know why but works.

While you can use basically any conductive material as an electrode to reduce the metal ions in solution, to actually plate the electrode with the metal, the metal has to be able to attach to the surface. If you have contaminants like surfactants, it can prevent deposition. The trouble with using aluminum as a cathode is that it readily forms an oxide layer that can be difficult to stick things to. I understand that it's possible to electroplate copper onto aluminum, but it's usually done with copper cyanide baths, so you may find it easier to plate copper onto steel or nickel. The other issue is that reduction of the copper ion is not the only possible reaction at the cathode. If your voltage is too negative, you may get significant hydrogen evolution and the bubbles will prevent a solid film from adhering properly. (which looks like it may be the problem in your photo) What you want is to deposit very slowly by keeping your voltage down so that you minimize gas evolution, though you may still have trouble with an aluminum cathode.
As to the sorts of things that can be electroplated, many metals have similar behaviour where it's possible to produce quite high quality films of a second metal over top. Other metals like mercury and bismuth tend to form alloys or intermetallic compounds. It is certainly possible to deposit thin metal films on materials like indium tin oxide, graphite or glassy carbon, thought adherence can be a problem.

I really wanted to electroplate graphite in order to solder on it. The idea with low voltage is a good point. I used 6Volts, which is quite a lot.

I get the feeling that even if you are able to get enough copper onto a piece of graphite to wet solder properly, it won't be very mechanically stable enough. Perhaps something like a silver-filled epoxy would make a good connection.

These glues are quite expensive. I was thinking about very thick layer of copper.


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Why do higher orbitals have more energy?
I have seen in textbooks and videos that an electron must absorb energy (become excited) to enter a farther-away orbital.
The amount of energy that must be gained is equal to the difference in energy between the orbitals. When the electron relaxes and returns to its ground state, it emits that energy.
However, I am not sure I understand why "higher" orbitals are of greater energy, for a couple of reasons.

When I think of the potential energy of the electron, as given by $U = \frac{kQq}{r}$, it seems that $U \propto \frac{1}{r}$, which would indicate less energy as distance increases.
I also know that the electron is attracted to the nucleus due to opposing charge. By this logic, an electron closer to the nucleus would be more stable (have a lower energy state) than an electron farther from the nucleus (which would thus have a higher energy state). This line of thinking is somewhat similar to how a ball that's still high in the air is in a higher energy state than a ball that's closer to the ground.

So #1 is inconsistent with what we are taught about electrons and orbitals, while #2 would support it.
I am attempting to clarify the misunderstanding I must have somewhere (possibly not making a proper analogy) - and better understand why higher orbitals are of greater energy.

When I think of the potential energy of the electron, as given by $U = \frac{kQq}{r}$, it seems that $U \propto \frac{1}{r}$, which would indicate *less* energy as distance increases. You forget that the nucleus and the electron are oppositely charged, so $Qq$ will be negative. Thus, $U \propto -\frac{1}{r}$.

Yes, I did forget that. Realized a little while after posting. Thank you!

@Philipp Could you write this as an answer so we can go ahead and wrap this question up?

potential energy is negative between infinity to nucleus
It varies with .Thus, U∝(−1)1/r so it means the father you go from nucleus the value become less negative that is higher in energy than equilibrium energy.


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What is the best invisible ink for clothing?
I am interested in writing and revealing a message in invisible ink for theatrical purposes. I found a list of potential compounds here but I am having a hard time prioritizing which to experiment/rehearse with.
The message will be written on a lab coat and revealed in the dim lighting of a night-time bonfire so the ink must be dark enough to see easily. Because I will be writing on cloth, I will need an ink that will not run, so that when the message is revealed, it will be legible and not distorted.
Heat-activated inks are preferable because I will have the bonfire on hand for the reveal but there is the danger of revealing the writing too soon.
Chemically-activated inks are less desirable because of the less-theatrical reveal and because many of the revealing-solutions look like they will stain the coat making the message difficult to read but I will keep them as candidates because I have not tested my staining hypothesis and because of the added control they offer.
Blacklight-activated inks are not suitable.
I don't have enough lab coats to test all the compounds listed so my question(s) are:

Which of the inks listed are least likely to run when applied to cloth?
Of the remaining inks, which will appear darkest when revealed?
Of the remaining heat-activated inks, which are easiest/hardest to reveal, in terms of temperature/time required?


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Chirality of heteroatoms
Why is a phosphorus atom with three different substituents chiral, but nitrogen isn't? Nitrogen inverts fast, while phosphorus obviously much slower. How is that explained?

Typical amines ($\ce{NR3}$) are roughly $\ce{sp^3}$ hybridized.  A chiral amine must undergo nitrogen inversion in order to racemize. The transition state for nitrogen inversion is a planar $\ce{sp^2}$ structure.  So in order to invert the ligands attached to nitrogen we must move from an $\ce{sp^3}$ ground state with ca. 109 degree bond angles to an $\ce{sp^2}$ transition state with 120 degree bond angles.
On the other hand typical $\ce{PR3}$ compounds are roughly unhybridized with 90 degree bond angles in the ground state.  In order to racemize they must also invert at phosphorous which again requires an $\ce{sp^2}$ transition state with 120 degree bond angles.  Changing the geometry from 90 to 120 degrees in the case of phosphorous requires more energy that a distortion from 109 to 120 degrees in the case of nitrogen.
P.S. Put a chiral nitrogen in a 3-membered ring and chirality can be preserved because you've now made it so much more difficult to expand the 60 degree bond angle in the ring to 120 degrees.

Also a nitrogen atom that connects two condensed rings has a sufficiently high barrier.

@ron But in general can we prepare the R or S enantiomer of a $\ce{PR_3} compound? Because if it is impossible then the samples we work maybe be R or S and because tthere is no energy difference between the two types of enantiomers we expect a racemic mixture.

It's exactly as you say. Some heteroatoms can be chiral if they have 4 different substituents (or 3 different substituents and a lone pair).
The problem with nitrogen is that it rapidly interconverts at normal temperatures. The inversion barrier is ~24 kJ/mol and quantum effects also play a role in the conversion. The barrier for phosphorous is much, much higher. It's a larger atom so the bonds are usually longer, and the lone pair occupies a larger volume.
There's actually a nice Wikipedia article on nitrogen inversion.
Incidentally, molecules like helicenes are chiral (i.e., with a right-handed or left-handed helicity), but like nitrogen can interconvert between forms through inversion.


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Why is the ionization energy for Hydrogen non-zero?
There are no other electrons to collide, repel and kick Hydrogen's single electron to a distant nucleus. And that a single electron is tightly attracted to the nucleus by the electrostatic energy between them. So it seems to me, that Hydrogen does not require ionization energy at all.
But when I checked on the sources - the Hydrogen's ionization energy is relatively high.
Even Hydrogen makes $NaH$, which just doesn't stay for long and has extreme need for becoming $Na$ metal and Hydrogen gas.
So in this case why can't we say that Hydrogen does not require ionization energy?

I'm unfamiliar with the term ionic energy. By ionic energy, do you mean ionization energy?

I think that [tag:ionization-energy] is a borderline case when it comes to tags, [tag:electrostatic-energy] on the other hand is completely superfluous.

@Martin I have no intention of earning 2reps on tag wiki. I just created them, because the context of my question refers to those.

That's alright, we'll see how they play out...

The ionization energy (IE) of an atom or molecule describes the minimum amount of energy required to remove an electron (to infinity) from the atom or molecule in the gaseous state.

With ionization energy, an electron is not "kicked out" by other electrons, but rather it is "the energy required for the electron to 'climb out' and leave the atom."
Since the electron "is drawn inwards by positive electrostatic potential," it would make sense to infer that the more "tightly drawn" the electron is to the nucleus, the more energy it would require to "climb out."
More of this can be found on the wikipedia page.

Also, Hydrogen does not have any other electrons to 'shield' the one electron from the nuclear charge, therefore the effective nuclear charge will be equivalent to the nuclear charge (not so in multi-electron atoms).

I got the point about "tight, close" attraction between single electron and H nucleus. So how does this single electron acquire energy to fly/shift away from nucleus?

@bonCodigo I'm not sure what you mean. The ionization energy for a hydrogen atom is 13.6 eV (121 nm). If I take a sample of atomic hydrogen and hit it with UV light of 121 nm or higher energy (smaller wavelength), it will ionize.

@GeoffHutchison So I guess that's why Hydrogen makes covalent bonds most of the time, apart from unstable NaH, LiH scenarios?

@bonCodigo It's a bit related. I'd generally revert to electronegativity arguments when discussing whether there are ionic or covalent bonds. The electronegativity of H is almost the same as C, so it does tend to form covalent bonds. NaH and LiH (any hydride) involves $H^-$. In this case, the extra electron in hydride anion is NOT tight or close, making it extra reactive.

The ionization energy of an atome is the minimum energy required to remove an elelectron from the atome in the gaseous state.
In spite of the fact that there are no other electrons to collide, repel and kick Hydrogen's single electron to a distant nucleus. That  single electron is tightly attracted to the nucleus by the electrostatic energy between them. So the electron would require energy to get rid of the electrostatic attractive force of the nucleus and escape from the atome.

Nice- if you don't mind I continue the chat with @LordStryker. Feel free to join and chip in.


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What are the subfields of chemistry that work very closely with biology?
Background information:
I'm not sure if this question belongs here; but i'll risk the down-votes to ask it.
I'm currently doing an undergraduate program, and it's in chemistry. I am torn between doing chemistry and biology and ideally the nicest compromise would be biochemistry. However I have a horrible ineptitude for biochemistry.

My question is, what subfields of chemistry work very closely with biology?

How is chemistry classified in terms of what?  Do you mean subfields in chemistry like, e.g., organic synthesis or polymer sciences or something other than that?  More information is necessary to get you a good answer here.

(as an aside, biochemistry is not easy, so don't let one course derail you from your true interests if that's where they lie)

@jonsca i've edited the question. hmm I don't believe that the area where biology and chemistry meet is only biochemistry; if that made any sense

It's closer, so thank you, but it's still a bit broad, I think.  For instance, what beyond something like http://en.wikipedia.org/wiki/Chemistry#Subdisciplines are you looking for?  There are certainly an almost uncountable number of sub-sub disciplines as well.

@jonsca I've just been on that link you've provided, and I noticed that biochemistry is of a different classification from neurochemistry. Wouldn't biochemistry encompass all of that because it involves our biological molecules and neurochemistry in that sense is a subset of biochemistry if one were to "Classify" them very broadly-

Don't base your choice on somewhat artificial subsets and their intersections in the field of chemistry. Ask yourself what are you good at and what despises you. Do you mind to spend hour after hour in organic synthesis and purification of the intermediates? Do you mind to calibrate a complex experimental setup and lift heavy math? Would you rather prefer to spend the least time in a lab and run calculations instead? Don't neglect the human factor! If you know that you won't get along with a particular supervisor, don't go there - even if the topic might seem interesting.

@klauswarzecha that sounds all good, hence the question to know what a subfield of chemistry roughly consists of so I can ask myself those questions.

@Nick It is curious that they broke that one out, but perhaps that was the choice of one editor.  The others there are fairly sound as subdisciplines.  The biochemistry section does mention neurochemistry.  Perhaps a question on the distinction of those two disciplines would be targeted enough.

@Nick If you go to the upper left corner of the screen it's under the "Stack Exchange" menu.  Sorry about the delay in reopening.

As far as I'm aware, the closest areas are drug design and proteing folding.

I'm in a medicinal chemistry PhD program, and the chemistry and biology are all very thoroughly mixed. And there is a lot of diversity in the research in our department. We have labs working on G protein receptor regulation, computational simulation of proteins to develop inhibitors of bacterial cell wall synthesis, dopamine toxicity and parkinsons, metabolism of PCBs, DNA and RNA delivery, and of course total synthesis of natural products. These labs cover almost all aspects of chemistry, so don't feel you need a specific chemistry to do biology, though biochem and organic are most helpful.

Chemical biology.

In addition to jonsca's link of chemistry subfields, Wikipedia also has a list of biochem subfields. Find what research interests you, perhaps not based on a course you've taken.
You can look up what professors are doing in their respective fields of research by visiting faculty research group sites online. Check out their publications, current projects, probably would be good to join a lab too. (I'm blindly assuming you're interested in research and/or graduate studies)
The list on Wikipedia first lists (for whatever reason these are "main branches" but I'm surprised biotech is categorized under "other branches"...):

Animal Biochemistry
Plant Biochemistry
Molecular Biology
Cell Biology
Metabolism
Immunology
Genetics
Enzymology

Getting into more or less random scopes of research topics which may or may not be within the above fields:

Biotechnology,
Bioluminescence,
Molecular chemistry,
Enzymatic Chemistry,
Genetic engineering,
Pharmaceuticals,
Endocrinology,
Hematology,
Nutrition,
Photosynthesis,
Environmental Chem,
Toxicology
(Probably many, many others...)

Too late to help the OP, but I noticed that biophysics wasn't mentioned.
The OP professed an "ineptitude for biochemistry". Given that they are presumably comfortable with chemistry (wanting a biological field related to chemistry), I'd guess there's a chance the aversion to biochemistry comes from its enormous complexity, and the attendant enormous body of information that needs to be memorized to become proficient.
If that's the case, and what appeals to the OP is the conceptual beauty of chemistry, they might want to consider biophysics, which combines physical/theoretical chemistry (or physics) with biology.


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Can lattice enthalpy only be determined by the Born-Haber cycle?
Since we are able to determine the enthapy of formation experimentally, why it is impossible to determine the lattice enthalpy directly by experiment?

Because you can't take an ionic solid and convert it into a gas of ions. That's not a physically realistic process. You have to infer it using a Born-Haber cycle and other quantities that can be determined experimentally.

what about heating it? I was thinking about this too - perhaps it's because heating it stops it being standard conditions?

What about theoretically calculating it from principles in electrostatics ex: as briefly noted on U Waterloo's webpage here?
For instance, for crystalline ionic compounds: Born-Lande and/or Born-Mayer equations can provide results close to experimentally determined values. The two equations both utilize the Madelung constant, which simplifies the electrostatic calculations for organized crystals. Of course these equations bring up new constants :)
(Kapustinskii took a slightly different approach to determining this.)
EDIT: In retrospect, perhaps this doesn't answer the question since it asks for experimental solutions.


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Evidence of orbitals?
How do we know that there are different types of orbitals? For example, what evidence is there for the existence of $\mathrm{p}$ orbitals instead of there being multiple $\mathrm{s}$ orbitals (for example, why isn't the electronic configuration of sodium $\mathrm{1s^1, 2s^2, 2s^2, 2s^2, 2s^2, 3s^2}$ instead of $\mathrm{1s^2 2s^2 2p^6 3p^1}$)?

http://en.wikipedia.org/wiki/Crystal_field_theory provides evidence for existance of orbitals with lower symmetries.

Let me approach this another way than the others: orbitals are NOT physical objects! They do  not exist in physical sense, they are theoretical constructs, chemical concepts that help understand / visualize / etc. mathematical solutions of Schrodinger / Dirac / Kohn–Sham / etc. equations.
Orbitals are not unique: given linear combinations are equivalent with each other, and there is no "correct orbitals", one can choose whichever they like. Canonical orbitals, natural orbitals etc are all good to go.
What is the evidence they exist? They do not exist, they are just mathematical solutions for given equations, and it is a purely mathematical question if they are good solutions for those equations or not. The theories themselves are consistent with experimental data, e.g. spectroscopic properties, geometries, reactivity.

The answer that I personally was looking for.

@Greg wote, "They do not exist"  Electrons exist and they occupy regions with probabilities defined by equations.  These regions may have different spatial functions.  "Orbital" is a word that can be used to describe that spatial function\distribution of electrons. In that sense of the word, orbitals exist, just as the spatial distribution of the electron exists.  Perhaps this is more a semantics issue.

It is not semantic issue. Which one exist? Canonical ones? NBOs? Or some of their rotations?

"given linear combinations are equivalent with each other" if we are talking about energy

Not all linear combinations are equivalent, only unitary transformations. 2. This also only holds true if we look at the overall energy of the system. If we were to assume that orbital energies themselves are physically meaningful, then this does not hold anymore. 3. The question whether orbital energies are in any way observable harks back to the old question about how to interpret the photoelectron spectrum of methane and its distinct 3+1 intensity pattern. To this day I still haven't found a satisfying answer.

@Antimon 1) OK, clarifying: not all are equivalent, only an infinite subset of them... Still, it doesn’t make them neither unique, nor physical, only the solutions of one-electron given approximations.  2) Orbital energies are only meaningful as approximations. Also, which one? HF? DFT? 3) The question was for proof for orbitals. Just because orbital energies can be useful approximations, it doesn't mean they are proof. Two-body solutions for the Earth’s orbital are useful, but they are not a proof that only two body exist in the universe.

The answer lies with experimental chemistry, specifically successive ionisation energies (i.e how much energy is required to remove the first electron, the second electron, the third electron and so on).

Each point on the graph corresponds to an element. The first one is hydrogen, the second is helium.
The height of each point shows how much energy is required to remove the first electron.
You can note that generally, the energy increases over a period. This is because In each successive element there is one more proton, and this stronger nuclear charge 'hold the outer electrons' more tightly.
Now to explain your question. Observe that within a period (e.g the 3rd dot to the 10th), it is not a constant increase. You can see that between the 4th-5th there is a slight drop, likewise between the 7th-8th.
The explanation for this is the sub orbitals. Some knowledge you need to know is that electron sub shells are only stable when empty, full, or half full (if you need an explanation for this comment on it later).
Let us examine the fourth dot, which represents Beryllium. It has an electronic config of 1s2 2s2. All of its sub orbitals are full, meaning that it is quite stable.
Compare it to the 5th dot, boron. Boron has a config of 1s2 2s2 3p1. Now the P orbital has room for 6 electrons, but this only has 1! It is not happy. It is not full or half full. Because of this it is trying to 'get rid' of the electron to be more stable. That's why it doesn't require so much energy to remove the outer electron.
The decrease between the 6th and 7th is explained by the fact that the p orbital is stable when empty, full or half full. The 7th dot (nitrogen) has 3 electrons in its p orbital (half full). Contrasted with oxygen, which has 4/6. This is not stable, so less energy is required to remove it.
TL-DR:  By analysing the ionisation energy graphs, we can see patterns that can be explained by sub orbitals.
If you need a more basic/complex explanation, comment.

I'd be a bit careful about the whole sub-shells being stable with empty, half-filled, and filled description. Slater's rules suggest these are general tendencies, but not the only story when you get into the transition metals and lanthanides/actinides. But that's a minor quibble.

Oh, I suspect someone could come up with a "s-orbital only" view of chemistry, much like people came up with complicated models for an Earth-centered view of the universe.
Let's start with the simple fact that solving quantum mechanics for a hydrogen atom gives you solutions for s, p, d, f, g.. orbitals, and even the degeneracy (i.e., that there is one s-type orbital, three p-type, five d-type, etc.) based on the angular momentum.
As described above, solutions for the many-electron equations (albeit approximate) match up very nicely with experimental observations for ionization energies, electron affinities, etc.
So an elegant theory and experiment agree to a remarkable degree.
Beyond that, we know that there must be non-spherical orbitals because we see molecular shapes with bond angles. I cannot come up with any way to describe a tetrahedral methane (let alone anything else) without some sort of non-spherical orbital.
Moreover, when we look at reactivities of molecules, we see reactions occurring where we predict lone pairs or radical spin-density, etc.
While orbitals are truly a mathematical construct, we find they are incredibly predictive of a wide range of chemistry. So my response would be "how can we think there are not different kinds of orbitals?"

Given the simplicity of s orbitals, people have tried using 's-orbital only' basis sets and just placing a crap ton of these orbitals all over a molecular space rather than using a smaller number of more 'accurate' atomic orbitals.  Interesting concept.

Minor quibble - the tetrahedron of methane could easily be described as resulting from a spherical C^4- surrounded by four point positive charges. The tetrahedron optimizes the repulsion of the positive charges from each other while keeping them all at the same distance. Not a criticism of your answer in general, just the choice of methane as the example that requires nonspherical orbitals.

Technically you can describe everything with sin/cos functions, too, yet it doesn't mean that the vacuum has tons of orbitals that look like a sic/cos function. Let’s distinguish model from physical reality, especially in places where the model is strictly for tied to an approximation.

While I really like the other answers, the 21th century evidence here is that we have images of these orbitals thanks to atomic force microscopy (AFM).

http://www.themarysue.com/electron-orbital-pictures/

As getafix pointed out, these are actually images of spatial electron density distributions by the examined molecules. The actual evidence here, that the model which uses the orbital term predicts the same distributions as we measured with AFM.

What you evidence of orbitals, can just as easily be called evidence of regions of electron density. ;) So is orbital a real thing or just a description of the spatial distribution of electrons. Electrons are real, regions with a high probability of finding them are real, but are orbitals? food for thought..

@getafix Nice thoughts. :-) This depends on the definition of orbitals. So yes these images are evidence that there is a spatial electron density distribution and that distribution is very similar to the prediction of the limited mathematical model which uses this orbital term. Usually this is enough to prove a hypothesis. E.g. by general relativity such a prove was that by a solar eclipse they were able to check how the gravity of the Sun bends the space and so starlight...

@getafix But yes you are right, I could rephrase the text. So these are not images of orbitals, but images of electron density distributions which are the same as the model predicts.

Well is also somehow similar to fields. While iron filings gives directly the field lines, and in this case more steps are needed, in both case we need a physical handle to picture the underlying thing. Is much of semantic.

Not even electron distribution: only a slice of those close to the  Fermi level.

How do we know that there are different types of orbitals?

In the early days of atomic spectroscopy scientists were able to explain the spectral properties of elements like hydrogen and sodium.  Atoms of these elements had a single electron in the outermost shell and produced spectra that fit with (what we now view as) the relatively simplistic Bohr theory of the atom.  When more complex atoms were examined spectroscopically, fine structure was observed that the Bohr theory could not explain.  These spectral lines were often referred to as sharp, principal, diffuse and fundamental.
As quantum mechanics emerged, the concept of the 4 quantum numbers being needed to describe an electron surfaced.  It was found that the spectral fine structure could now be explained when these various quantum states were taken into account.  To honor the work done by the early spectroscopists (that led to the development of a better theory), scientists used their "s, p, d, f" notation to describe the various values for the angular momentum quantum number $\ell$.
So although the observations made by the early spectroscopists couldn't be explained at the time, it turns out that they were actually observing spectral transitions involving other orbitals (p, d, f) in addition to those involving the s orbital.  The observation of this spectral fine structure requires the presence of orbitals that are different from s orbitals.


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Magnesium powder mixed with CuSO4 solution
$\ce{Mg}$ powder is stirred into $\ce{CuSO4}$, what is the balanced molecular equation for this reaction ? I have $\ce{Mg + CuSO4 -> Mg2SO4 + Cu}$. Also, is the ionic equation the same as the net ionic equation? I believe it is not, because Copper would act as a spectator ion, however, I am not too sure.

Your equation is very close, but you it is not $\ce{Mg2SO4}$. $\ce{Mg^{2+}}$ has a charge of $2+$, and $\ce{SO4^{2-}}$ has a charge of $2-$. They combine in a ratio of 1:1, so the result is
$$\ce{Mg + CuSO4 -> MgSO4 + Cu} ~ .$$
The ionic equation does not include the $\ce{SO4^{2-}}$, and in the case of this equation the net ionic equation is the same as the ionic equation. Ionic equations look at what happens to electrons.
In this case, electrons are being transferred from the magnesium to the copper (since the copper is bonded to $\ce{SO4^{2-}}$ it is in the form $\ce{Cu^2+}$).
$$\ce{Mg + Cu^{2+} -> Mg^{2+} + Cu} ~ .$$
Since the charge on the $\ce{SO4^{2-}}$ doesn't change, it's just a spectator ion.

I also have to note that in some circumstances the reaction can be ... lively.

Ebullient, one might even say.

The answer given is what all the textbooks say, and what the examiner wants. However, magnesium will displace hydrogen from water if copper ions are present which act as a catalyst. So what really happens is bubbles of hydrogen are liberated and white magnesium hydroxide solid is produced.

Did you actually try this? With aluminum metal added to copper sulfate solution, copper is produced: https://chemistry.stackexchange.com/a/128099/79678. A piece of burning magnesium can burn in water, making magnesium fires hard to extinguish. But magnesium powder is not a piece of burning magnesium and it only slowly reacts with very hot water. This is worth a quick experiment.


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Grocery store crystalline citric acid
Which type of citric acid crystalline powder do they sell in grocery stores?
Is the concentration of citric acid in the crystals 100%?Is it anhydrous or a monohydrate?

The box in my kitchen is labelled "reine Zitronensäure" (= pure citric acid) and "lebensmittelsauber".
According to the manufacturer, the latter means that citric acid in this quality can be used to descale (decalcify) kitchen items, but it is not intended to be used as a food supplement and should not be digested!
This may have real technical reasons, i.e. the product may in fact contain impurities resulting from production, processing and packaging. The manufacturer however does not specify the kind or amount of impurities.
On the other hand, this kind of labelling may just reflect different legal treatment (certification, monitoring, etc.) of products intended as food supplement and those only used to clean kitchen equipment.
Concerning waterfree vs monohydrate, neither the package in my kitchen, nor the website of the manufacturer provide any information.
If you know a chemist, this question is easily solved: Let him measure the melting point of a sample. Waterfree citric acid melts at 153 °C, the monohydrate at 100 °C.