text:
How to calculate molecular dipole moment from a known wavefunction?
Say I have a molecular wavefunction as a set of molecular orbitals and want to calculate the molecule's dipole moment, but don't know how! I searched a lot but couldn't find any practical example.
$$\psi _i=\sum ^N_{i=1}C_i\mathrm e^{-\alpha _ir^2}$$

You could use the variational theorem to determine the molecular orbital coefficients of the molecular orbitals (one electron wavefunctions). If you let you molecular wavefunctions be a linear superposition of basis atomic wavefunctions $\Psi =\sum c_i\psi_i$, with the orbital coefficients you can understand key properties of your molecule. You can rationalise trends in bond polarity too, that can't be explained with other theories!

The dipole moment $\mu$ of a molecule is a measure of charge distribution in the molecule and the polarity formed by the nuclei and electron cloud.
We can perturb our system with an external electric field $\vec E$ and gauge the response of the electron cloud and nuclei by the polarisability, i.e how much the dipole moment changes. In practice the nuclei might be so heavy that their motion is not perturbed, while electrons being light are very mobile. If we imagine that the external field is caused by some other species, and that it itself is not changing, we can call this external constant electric field $\vec E$ at least over the volume of the molecule we are considering. Imagine for arbitrary book keeping that we point it down the $z$ axis. We could also investigate how the dipole moment changes with bond vibrations to discuss IR spectroscopy or if the polarisability changes during a vibration to give Raman spectroscopy.
We can use perturbation theory to expand the wavefunction and the molecular energy in terms of small perturbations of the field. We start by Taylor expanding the energy and molecular wavefunction in terms of the electric field which acts as the perturbation parameter.
\begin{equation}
E(\vec E)=E^0+\bigg(\frac{\partial E}{\partial \vec E}\bigg)_0\vec E+\bigg(\frac{\partial ^2E}{\partial \vec E^2}\bigg)_0\frac{\vec E^2}{2!}+\bigg(\frac{\partial ^3E}{\partial \vec E^3}\bigg)_0\frac{\vec E^3}{3!}+\dots
\end{equation}
\begin{equation}
\psi(\vec E)=\psi^0+\bigg(\frac{\partial \psi}{\partial \vec E}\bigg)_0\vec E+\bigg(\frac{\partial ^2\psi}{\partial \vec E^2}\bigg)_0\frac{\vec E^2}{2!}+\bigg(\frac{\partial ^3\psi}{\partial \vec E^3}\bigg)_0\frac{\vec E^3}{3!}+\dots
\end{equation}
If we use the notation that the wavefunction derivatives are given by;
\begin{equation}
\bigg(\frac {1}{i!}\bigg)\frac{\partial ^i\psi}{\partial \vec E^i}=\psi ^{(i)}
\end{equation}
The Hamiltonian for such a system under the influence of an electric field in the $z$ direction is $\hat H(\vec E)$.
\begin{equation}
\hat H(\vec E)=\hat H^0-\vec E\hat \mu _z
\end{equation}
Where $\hat \mu _z$ is the dipole moment operator that is a summation of the charges of the nuclei and electrons in the molecule. This is the result of Hellmann-Feynman theorem of the energy and the electric field, e.g. $\hat H(\vec E)=\hat H^0 +\hat H^1(\vec E)$,
\begin{equation}
\frac{dE}{d\vec E}=\bigg\langle \frac{d\hat H}{d\vec E}\bigg\rangle=\bigg\langle \frac{d(-\mu _z\vec E)}{d\vec E}\bigg\rangle
\end{equation}
The time-independent Schrödinger equation is now,
\begin{equation}
\hat H(\vec E)\psi(\vec E)=E(\vec E)\psi (\vec E)
\end{equation}
With an energy;
\begin{equation}
E(\vec E)=\big \langle \psi (\vec E)\big|\hat H(\vec E)\big|\psi (\vec E)\big \rangle=\big\langle \psi ^{(0)}+\psi ^{(1)}\vec E+\psi ^2 \vec E^2+ \dots \big|\hat H-\vec E\hat \mu _z \big|\psi ^{(0)}+\psi ^{(1)}\vec E+\psi^2\vec E^2+\dots \big \rangle
\end{equation}
With a little algebra and use of $E^{(0)}=\langle \psi ^{(0)}|\hat H^0 |\psi ^{(0)}\rangle$, as well as the Hermitian properties of the Hamiltonian.
\begin{equation}
E(\vec E)=E^{(0)}+2\vec E\big \langle \psi ^{(1)}\big|\hat H^{(0)}\big|\psi ^{(0)}\rangle -\vec E\big\langle \psi ^{(0)}\big|\hat \mu _z\big|\psi ^{(0)}\big \rangle +\mathcal O(\vec E^2)
\end{equation}
Using the Schrödinger equation $H\psi =E\psi$ and pulling the scalar energy out of the integral;
\begin{equation}
\big\langle \psi ^{(1)}\big|\hat H^0\big|\psi ^{(0)}\rangle =E^{(0)}\big\langle \psi ^{(1)}\big|\psi ^{(0)}\big \rangle
\end{equation}
Since $\langle \phi |\psi \rangle=0 $,
\begin{equation}
E(\vec E)=E^{(0)}-\vec E\big \langle \psi ^{(0)}\big|\hat \mu _z\big|\psi ^{(0)}\big\rangle
\end{equation}
Therefore the expectation value of the dipole moment along the $z$ axis for a molecular state $\psi ^{(0)}$ is $\langle \mu _z \rangle$;
\begin{equation}
\langle \mu _z\rangle =\big\langle \psi ^{(0)}\big|\hat \mu _z\big|\psi ^{(0)}\big\rangle
\end{equation}
To understand the strength of the interaction that causes the transition between the states $\psi ^{(0)}$ and $\psi ^{(1)}$ we use the transition dipole moment which is basically exactly the same but there is a wavefunction from both of the states involved, initial and final!
If you were to repeat this process but retain higher orders (Messy!) you would get (2nd order) the polarisability of the molecule which in essence is the susceptibility of the electron cloud to change with respect to an external electric field (so how the dipole moment changes. Third order would give the hyperpolarisability etc., etc...
As I said, you could also approach this from a really different angle by interpreting the molecular orbital diagram and using computational chemistry (so variational principle etc) to find the molecular orbital coefficients! That would give you a good idea of what is going on!

Hrm. This is a quite thorough formal elaboration of the theory involved, but it doesn't actually describe how one would implement practically the dipole moment calculation. (At least, I am no closer to understanding how I would code a dipole moment calculation given a set of MO coefficients and basis functions.)

This whole derivation is unnecessary. Per definition the dipole moment is the expectation value of the dipole moment operator with the given wave-function.

Thank you both for your comments, I will take this into account and update my answer shortly. I agree with @Brian, this answer does not give a non-specialist sufficient information on how to perform calculations to obtain values for molecules etc. However, I do not feel it is obvious that the dipole moment is the expectation value to someone who is new to this material and not as knowledgable as others. It is my experience that undergrads can struggle in this area if they can not see where something comes from. That's why I always like to give people a little background material. :)

@Greg For those like myself (background in chemical engineering, and not applied mathematics / quantum physical chemistry; but with an interest in a general understanding of the inner workings of quantum computation) it is perhaps not quite as sad...? I very much appreciate the efforts of both AngusTheMan and pentavalentcarbon to lay out the details. Also, frankly: isn't this sort of exposition the entire purpose of StackExchange?

@Brian I appreciate the effort, yet if it is obvious backward thinking, I like to point it out. Understanding axioms is easier then derive them from theorems that were derived from those axioms. I think the entire purpose os SE is to give meaningful answers. A little off topic correction: "quantum computation" is something completely different.

@Greg Practical quantum computation was the topic of the original question -- so, no, it's not at all off-topic. Also, if you're looking for mathematically rigorous developments from axiom to theorem or whatever, you're on the wrong SE site. There are Math.SE and MathOverflow for that.

@Greg I do not think it is "sad" that people do not understand something and I am sorry you feel that way.  I think my answer does more than just show how to use the basic axioms of quantum mechanics however, it shows the OP how to apply perturbation theory, how to take fundamental ideas learnt at school (electric fields) and generate more powerful ideas from them, how to use Taylor expansions as well as the effect of retaining higher order terms in the expansion. Not everyone is clever enough to see things straight away and I am always happy to take the time to help others if I can. :)

@Greg I apologize - on re-reading, I see I did use the wrong term. I meant to say 'quantum chemical computation,' not 'quantum computation' $\approx$ 'quantum computing'.  D'oh!

The necessary formal derivation has already been nicely done by AngusTheMan. I'll start from the last equation:
$$
\langle \mu_{z} \rangle = \langle \Psi | \hat{\mu}_{z} | \Psi \rangle
$$
where $\Psi$ is the variational wavefunction; it can be any molecular state. It's important that it's variational, otherwise the expectation value approach is not exact. So, this works for SCF, CI, and MCSCF wavefunctions, but extra derivatives need to be taken for Moller-Plesset and coupled cluster wavefunctions. More work needs to be done for multideterminental wavefunctions like CI and MCSCF, but the complexity is no different for a single state in each wavefunction. There may be some MO space partitioning I'm neglecting that's required for MCSCF, so I'll restrict my work to a single-determinental wavefunction.
Expand the wavefunction as a linear combination of molecular orbitals (MOs)
$$
\Psi = \sum_{i} \psi_{i},
$$
where each molecular orbital is a linear combination of atomic orbitals (AOs)
$$
\psi_{i} = \sum_{\mu} C_{\mu i} \phi_{\mu},
$$
where $C_{\mu i}$ is the MO coefficient matrix, so our expectation value now looks like this:
$$
\langle \mu_{z} \rangle = \sum_{i}^{\textrm{occ MOs}} \sum_{\mu\nu}^{\textrm{AOs}} C_{\mu i} C_{\nu i} \langle \phi_{\mu} | \hat{\mu}_{z} | \phi_{\nu} \rangle.
$$
The indices $\mu,\nu$ run over all AOs, and the index $i$ runs over the occupied MOs. There's only one index because this is a one-electron operator. I'm also neglecting any complex values here, since we almost always work with real-valued AOs and MO coefficients.
We do one last rearrangement. Replace the MO coefficients with the density matrix
$$
P_{\mu\nu} = \sum_{i}^{\textrm{occ MOs}} C_{\mu i} C_{\nu i}
$$
to give the first explicit "working equation":
$$
\langle \mu_{z} \rangle = \sum_{\mu\nu}^{\textrm{AOs}} P_{\mu\nu} \langle \phi_{\mu} | \hat{\mu}_{z} | \phi_{\nu} \rangle
$$
I say first for two reasons: we usually try and avoid explicit loops like this, and the expression can be broken down further, depending on what molecular properties are of interest; I'll be more clear about this later. Inside the sum there are two terms:

The density matrix $P_{\mu\nu}$, which comes from a converged SCF calculation.
The integral of the dipole operator over two basis functions, $\langle \phi_{\mu} | \hat{\mu}_{z} | \phi_{\nu} \rangle$. Atomic orbitals are represented as atom-centered basis functions. These can be calculated once and at any time, since the quantities here don't change over the course of a calculation.

Each of these terms is represented as a matrix. Since the index $\mu$ runs along the rows and $\nu$ runs along the columns for each matrix, "contraction" involves either a matrix product followed by the trace, or an elementwise product followed by an accumulation sum over all matrix elements. There are other details one needs to be careful about, such as what units the result should be in, which changes the prefactor (programs work internally in atomic units), and what the origin for the dipole operator is, but that's really it.
Well, sort of. I'm actually treating some of the program internals as a black box. If you're familiar with Hartree-Fock, it should be clear where $P_{\mu\nu}$ comes from, but what about the integral? For a general expectation value $\langle A \rangle$ with its corresponding operator, where does $\langle \phi_{\mu} | \hat{A} | \phi_{\nu} \rangle$ come from? If it's already available in the code, then you call a wrapper function that then calls the integral engine to do all the nasty work, and you get back a tidy matrix without having to worry about the details. If $\langle \phi_{\mu} | \hat{A} | \phi_{\nu} \rangle$ isn't present, depending on the complexity of $\hat{A}$, there can be a non-trivial amount of derivation required for the working integral equation, followed by the implementation.
Ignoring any possible contraction of primitive basis functions, expand $\langle \phi_{\mu} | \hat{A} | \phi_{\nu} \rangle$ using the definition of $\phi$ in Cartesian coordinates:
$$
\phi(\mathbf{r}; \mathbf{A}, \mathbf{a}, \zeta) = (x-A_x)^{a_x} (y-A_y)^{a_y} (z-A_z)^{a_z} e^{-\zeta |\mathbf{r} - \mathbf{A}|^2}
$$
where $\mathbf{r} = (x, y, z)$ is the electron position, $\mathbf{A} = (A_x, A_y, A_z)$ is the position of the basis function (almost always atom-centered), and $\mathbf{a} = (a_x, a_y, a_z)$ are the angular momenta for each coordinate, with $l_{\textrm{max}} = a_x + a_y + a_z$ total angular momentum of the basis function. $(0,0,0)$ is an s-function, $(1,1,0)$ and $(0,0,2)$ are d-functions, and so on.
Forming the integral more explicitly gives
$$
\langle \phi_{\mu} | \hat{A} | \phi_{\nu} \rangle = \int\int d\mathbf{r}_1 d\mathbf{r}_2 \left[ (x_1-A_x)^{a_x} (y_1-A_y)^{a_y} (z_1-A_z)^{a_z} e^{-\zeta_a |\mathbf{r}_1 - \mathbf{A}|^2} \right] \\ \times\left[ \hat{A} \right]\left[ (x_2-B_x)^{b_x} (y_2-B_y)^{b_y} (z_2-B_z)^{b_z} e^{-\zeta_b |\mathbf{r}_2 - \mathbf{B}|^2} \right]
$$
Before going any further, $\hat{A}$ must be defined. If $\hat{A} = 1$, this becomes an overlap integral. The dipole operator in the z-direction is given by $\hat{A} = \hat{\mu}_{z} = -ez = -e(z_3 - C_z)$, where $z_3$ is the integration coordinate and $C_z$ is the origin of the dipole in the z-direction, usually taken to be zero. Everything is kept in atomic units until after the integral/density contraction, so drop the prefactor $-e$. We can now generalize this to an arbitrary Cartesian multipole moment operator,
$$
\hat{A} = \mathfrak{M}(\mathbf{r}_3) = (x_3 - C_x)^{c_x} (y_3 - C_y)^{c_y} (z_3 - C_z)^{c_z}
$$
where $(c_x, c_y, c_z)$ determine the coordinate of each multipole, and their sum is the total multipole order; for example, $(1,0,0), (0,1,0), (0,0,1)$ are the x-, y-, and z-directions of the dipole operator. The Cartesian moment operator looks just like a Gaussian basis function where $\zeta = 0$.
Once the form of an operator has been derived, it needs to be implemented as part of an integral package, each of which implement one or more algorithms for computing integrals. Each algorithm is named after the authors of the paper in which they were introduced, and are usually abbreviated. For example, the first one I know of is the Taketa, Huzinaga, O-Ohata paper (THO, DOI: 10.1143/JPSJ.21.2313), where explicit working equations are given for 2-center overlap, 2-center kinetic energy, 2-center electron-nuclear attraction, and 4-center electron repulsion integrals. A working implementation can be found in the PyQuante package. I made an IPython notebook translation of the code snippets on the front page of the official documentation.
Other, more complex algorithms are from the Pople-Hehre (PH), McMurchie-Davidson (MD, DOI: 10.1016/0021-9991(78)90092-X), Obara-Saika (OS, DOI: 10.1063/1.450106), Dupuis-Rys-King (DRK), and Head-Gordon-Pople (HGP) papers. I'm sure I'm neglecting some, including the seminal paper by Boys which introduced the use of Gaussian functions as a substitute for Slater-type functions in basis sets. A good review of these algorithms is found in a Peter Gill paper (DOI: 10.1016/S0065-3276(08)60019-2); he is the original author of the integral code in both Gaussian and Q-Chem.
To bring these things full circle, I wrote some code a few months ago to calculate the dipole moment using pyquante2 and a wrapper that calls an implementation of the Obara-Saika recursive integral algorithm. You can find it here with some comparisons to "industrial" quantum programs.

Thanks all. I really recommend reading the Obara-Saika paper if you're at all interested in integral evaluation. The DALTON package (not sure what the primary integral algorithm is) has an impressive list of one-electron integrals that can be calculated, many of which correspond directly (expectation value times some constants) to molecular properties.

An often used approach, especially for semi-empirical methods, is to use a set of atom-centered charges (most often the Mulliken charges) to calculate the dipole moment. In that case, the molecular dipole moment is given by:
$\vec{\mu} = \sum_a \vec{r} \times q_a $
Note that if the system has a finite charge, this equation has a dependence on the position of the molecular system relative to the origin of the coordinate space. Most program use the center of mass or center of nuclear charge as origin in such cases.


text:
Scission of carbon dioxide bonds using electric arc
I've done my research on this so I've seen the questions asking about the same premise without using plants.
Basically, this is a thought experiment I came up with. Whether or not I perform it depends on the feasibility. This is the reaction I'm thinking of:
$$
\ce{CO_2 \rightarrow C + O_2}
$$
From NIST the formation energy of carbon dioxide back into carbon and oxygen is given as -393.52 kJ/mol.
The basic premise I was thinking of was that $\ce{CO_2}$ could be pressurized in a chamber with electrodes striking an arc, most likely from an HF driver source. The energy imparted to the carbon dioxide could be used to provide the dissociation energy to break the carbon-oxygen bonds.
If the electrodes were designed with sufficient voltage in mind to break the dielectric constant of the gas, a few rudimentary numbers suggest that even at 5% efficiency, a sustained power input of 1kW would break the bond of 1 mol of $\ce{CO_2}$ in about 2 hours.
My concerns with the feasibility of such a process are:

Would it work? (i.e. would using an electric arc be a viable way to perform the bond decomposition? I'm not looking for an industrially viable process, just something that could produce a small end result.)
Would I have to worry about the carbon and oxygen re-forming? Obviously, this would render moot the entire experiment. I know that $\ce{CO_2}$ decomposition doesn't appear to be an equilibrium process, but it's still a point of concern.
If indeed it would work, how would the carbon likely present itself? I was thinking that if it sputtered itself against the vessel, the vessel could be weighed to determine the amount of carbon formed (and therefore the reaction efficiency, once the energy input was quantified).

Is this assuming that you have pure $\ce{CO2}$?

@ringo Exactly. Filled from a supply cylinder.

You will have varying amounts of carbon monoxide and ozone also likely being formed.  I think back-reaction to re-form CO2 would severely limit the attainable conversion, unless you can remove the C or the O2 continually to favor the equilibrium in the direction you want.

@CurtF. Could molecular sieves be used to remove oxygen continually from the vessel while maintaining pressure on the reactants?

Actually, it appears that due to the bond length similarities any sieve allowing diatomic oxygen would also most likely admit carbon monoxide.

Don't forget that $\ce{C + CO2 <=> 2 CO}$ is shifted to the right side with increasing temperature.

@KlausWarzecha If I were to perform this, I'd be attempting to maintain it as low a temperature as possible (that is, as close to 298K). Or are you referring to the arc conditions being the primary area where this equilibrium shift would occur?

@ecfedele The latter.

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Is a negative pH value possible?
Is a negative pH value possible in room temperature...teachers don't tend to teach about the pH of acid or base solutions which are higher than 1M.

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Are the hydrogen atoms of 2-methylindane homotopic, enantiotopic, diastereotopic or unrelated?

I'm very confused. According to the answer key, the circled hydrogens of 2-methylindane (2-methyl-2,3-dihydro-1H-indene) are diastereotopic. I thought they were enantiotopic.
My thought process is the following. They aren't homotopic because replacing one of them with another atom would create a chiral center. Due to them creating a chiral center, I thought it was enantiotopic.
I thought that for it to be diastereoptopic, the carbon with the methyl group off of it would have to be chiral as well, but I don't think it is chiral. The molecule is symmetrical off both sides. Does diastereotopic rely on that carbon being chiral after the new atom replaced the hydrogen? If so, then I understand.
I was under the assumption that the state of that carbon must be chiral before the hydrogen is even replaced.

The process to determine whether two protons are enantiotopic or diastereotopic is to replace each with deuterium and look at the relationship of the resulting compounds. For the indicated hydrogens in 2-methylindane, replacement of each gives a pair of diastereomeric compounds (Figure, top). The compound also contains enantiotopic protons (Figure, center). The carbon where the replacement happens doesn't necessarily become a stereocenter in this process (Figure, bottom).

Can you make one more example depicting Homotopic?

@David in this compound, the only homotopic protons would be the three at the methyl group. I'll add a more interesting case when I get a chance.

I'm late to the party, but here's one example I quite like of homotopic, enantiotopic, and diastereotopic protons. (I double-checked. I hope I didn't get them wrong, these $C_2$ symmetric compounds are confusing.)

Additionally to the way jerepierre showed, which is typically the preferred way of identifying the topicity of groups, you can also attempt to distinguish the hydrogens by symmetry relationships. For that, you will first need to find out the point group of your molecule; 2-methylindane is $C_\mathrm{s}$-symmetric and thus the only symmetry operation available is $\sigma$. For comparison, let’s also consider indane (i.e. without the methyl group) which would represent the point group $C_\mathrm{2v}$.
Homotopic hydrogens are now those that can be transformed onto one another either by a rotation of the entire molecule or by a rotation of a bond. In 2-methylindane, only the $\ce{CH3}$ hydrogens can be transformed onto another in that way, namely by rotating the $\ce{C-C}$ bond by $120^\circ$. In indane, there are 5 pairs of homotopic hydrogens: the phenylic ones on the far left, the phenylic ones top and bottom and the two on the right-hand corner of the cyclopentane ring are the easy ones. The other four benzylic hydrogen form two homotopic pairs, whereby the above-plane upper hydogen is homotopic to the below-plane lower hydrogen and vice-versa. Note that these are the pairs which are transformed onto each other under the influence of the $C_2$ rotation. Homotopic protons have the same colour in the image below.

Enantiotopic are those protons that can be transformed onto each other by using planes of symmetry. In 2-methylindane, as jerepierre showed, there are three pairs of enantiotopic hydrogens. (The final, undisplayed one consists of the top-phenylic and bottom-phenylic hydrogens.) In indane, it’s not as simple due to the two $\sigma_\mathrm{v}$ planes. Most notably, the four benzylic hydrogens are practically all enantiotopic to each other except for the pairings that were homotopic above. Note how the far-right hydrogens do not have enantiotopic partners: they are either transformed onto themselves or onto the other one — but they are already homotopic as discussed above, which is ‘stricter’. Inversion symmetry would also generate enantiotopic hydrogens as would improper rotation.
[2
Finally, all hydrogens that share the same chemical environment but cannot be transformed either by rotation or mirroring are diastereotopic. Indane has no diastereotopic hydrogen pair unless you count the unrelated ones. 2-methylindane has two pairs: The two benzylic ones bound to the same carbon as was shown.
Note that homotopic does not have to mean magnetically equivalent. In indane, the benzylic homotopic hydrogens are magnetically non-equivalent as they have different coupling constants towards third hydrogens.


text:
Naming ionic compounds with multiple cations and anions
I have seen complex ionic compounds that have mixed anions and/or mixed cations. For Example I have seen this:
$$\ce{NaKCl2}$$
Also known as Sodium Potassium Chloride. The only information I can find on this is an ion cotransporter for sodium potassium and chloride ions.
Anyway the structure of a single monomer of this would be $\ce{Na+}$ $\ce{Cl-}$ $\ce{K+}$ $\ce{Cl-}$
I have also thought of this as a possibility:
$$\ce{Na3ClO}$$
This looks very similar to sodium hypochlorite in terms of its formula but has a different structure. The structure of it would be $\ce{Na+}$ $\ce{Cl-}$ $\ce{Na+}$ $\ce{O^2-}$ $\ce{Na+}$
I know that naming mixed ionic compounds is not easy unlike naming simple ionic compounds(which itself can be hard if you don't know the name of a particular polyatomic ion). But how would I go about naming these mixed ionic compounds?

I know that and I know the name of that ion. it is chlorate.

I have removed my comments, as it turned out that I was completely wrong. I had interpreted your question as some kind of beginners nomenclature question, which it clearly is not. And apparently misread half of it. Apologies.

I propose you consult the recommendations of the International Union of Applied Chemistry (IUPAC), in particular the Red Book as reference.  Obviously, you deviated for a reason from the Hill formula of $\ce{NaKCl2}$ (that were $\ce{Cl2KNa}$) some data bases adhere to.
As already presented in rules IR 4.4.2.1, and IR 4.4.2.2, the order of the elements within the formula may be arranged either by their relative electronegativity (starting by the element of least electronegativity), or by alphanumeric order, respectively.  So rules in section IR 5.4 subsequently discriminate between electropositive constituents to be named prior to electronegative ones in the "overall name".  In addition (quote):
"The order of citation is alphabetical within each class of constituents (multiplicative prefixes being ignored), except that hydrogen is always cited last among electropositive constituents if actually classified as
an electropositive constituent."
The same reference provides numerous examples of application useful to look at, too.  Page 76 (of the book, not the pdf) is a good place to start.  As in nomenclature of organic compounds, there are instances you want to point the attention not only to the stoichiometric composition of the compound, rather than to different constitutional isomers.

I deviated from the Hill formula so that I would have the electronegative atoms being last which is usually the case in a chemical formula. I mean for sulfuric acid the Hill formula is H2O4S But this order doesn't make any sense because O is the most electronegative so it should go last. The same thing applies to NaKCl2. With the Cl2 last you can clearly tell that it is Sodium Potassium Chloride. With the Cl2 first you might think "Wait, this is Dichlorine Natride Kalide?" thinking that the Cl2 is a cation and the Na and K are anions when it actually is the opposite.

@caters I completely agree with your comment; there was a valid reason for you to use the convention of "electropositive constitutent(s) first, followed by electronegative (ones)" instead of the Hill formula.  Your question aimed for "How to name this compound?"  Pointing to the Hill formula only aimed to rise attention -- because when you access data bases like Gmelin/Reaxys you may miss your compound if enquired (accidentally) according to an other (formula) convention.

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text:
Why was the prefix 'bi' used in compounds, such as for bicarb of soda?
According to the Wikipedia article IUPAC nomenclature of inorganic chemistry,

he prefix bi- is a deprecated way of indicating the presence of a single hydrogen ion

A very common example is the commonplace 'bicarb of soda', or sodium bicarbonate (or using its correct chemical name sodium hydrogen carbonate). On the Wikipedia page for sodium bicarbonate there is a note that states that:

The prefix "bi" in "bicarbonate" comes from an outdated naming system and is based on the observation that there is two times as much carbonate ($\ce{CO3}$) in sodium bicarbonate ($\ce{NaHCO3}$) and other bicarbonates as in sodium carbonate ($\ce{Na2CO3}$) and other carbonates.

However, the statement above is not particularly clear, especially as carbonates are not the only compounds that used the 'bi' prefix, such as:
Bisulphite $\ce{HSO3-}$
Bisulphate $\ce{HSO4-}$
Why was the prefix 'bi' used to indicate the presence of a single hydrogen ion in compounds?

According to the Online Etymology Dictionary, the word bicarbonate is derived as follows:

1814, bi-carbonate of potash, apparently coined by English chemist William Hyde Wollaston (1766-1828), from bi- + carbonate.

The question then becomes, what is the meaning of the prefix bi- here?  We know it means two, but it also can mean double or twice.  If that is the case, then sodium bicarbonate is sodium double carbonate.
Consider a given quantity of acid, let's say $\ce{HCl}$, that we are going to neutralize with sodium carbonate, $\ce{Na2CO3}$, and an equal quantity of acid that we are going to neutralize with sodium bicarbonate, $\ce{NaHCO3}$.  We add the sodium carbonate to the acid, and as the neutralization reaction goes forward, it produces some quantity of carbon dioxide.  The same process is repeated for sodium bicarbonate, and the carbon dioxide is again collected.
$\ce{Na2CO3 + 2HCl -> 2NaCl + CO2 + H2O}$
$\ce{NaHCO3 + HCl -> NaCl + CO2 + H2O}$
In the first reaction, one unit of sodium carbonate is able to neutralize two units of acid.  In the second reaction, one unit of sodium bicarbonate can only neutralize one unit of acid.  Another way of looking at this is that double the quantity of carbon dioxide is produced from the sodium bicarbonate neutralization.  That is, the compound is a double carbonate.
Similar explanations make sense for bisulfate and bisulfite.  The acid-base reactions that a sulfate or sulfite ion might undergo are already partially completed in the bi- versions of the compound, so it takes twice as much sulfate or sulfite to do the same job as the non-hydrogen form.
The hydrogen phosphates don't fit this pattern as nicely.
Additionally, if the cation in these salts is a relatively light weight one, then the required mass of the bicarbonate or bisulfate is substantially more than that of the non-hydrogen form of the compound, though it's not quite double.  53g of $\ce{Na2CO3}$ could neutralize one mole of $\ce{HCl}$, but it takes 84g of $\ce{NaHCO3}$ to do the same job.  This is closer to accurate for the more massive sulfur and phosphorus based oxyanions.

The use of "bi" for 'two" denotes that there are two different cations (e.g. $\ce{Na+, H+}$ in the examples), but it is confusing, because it could imply that there are two of a single cation, so its use is deprecated. Similarly, in English, biweekly can mean either twice a week or every two weeks.
The use of "di" for "two" is more consistent, i.e. it always means two of a single species, as in carbon dioxide, $\ce{CO2}$. [Perhaps "always" is a bit strong... anyone know of contrary examples?]

Diphosphate, as in adenosine diphosphate (ADP) is the most common contrary example. "Diphosphate" is a side group composed of two esterified phosphates, $\mathrm{H_3P_2O_7—}$, whereas a bisphosphate compound such as fructose 1,6-bisphosphate bears two $\mathrm{H_2PO_4—}$ side groups, which may or may not be bound to the same atom. (The protonation state of all of these will of course depend on pH.)

I'm curious, do you have a source for this answer?  Everything I can find indicates that the answer suggested by santiago in the question is correct.


text:
When, how and why does sodium metal react with cycloalkenes, cycloalkadienes, cycloalkatrienes?
I've been searching for the conditions necessary for the reaction of sodium (Na) metal with cycloalkenes, cyclo-di-alkenes, cyclo-tri-alkenes all over the internet but I cannot find a proper answer. Can someone explain to me the necessary conditions?
For example why does sodium metal react with cyclopentadiene but not with cycloheptatriene?

related http://chemistry.stackexchange.com/questions/29428/acidity-of-unsaturated-five-membered-ring-relative-to-seven-membered-analogue

http://chemistry.stackexchange.com/questions/18805/abstraction-of-a-proton-in-cycloheptatriene

The negatively charged particles are conjugate bases of respective compounds, which (reactants) acts due to presence of active methylene group. Depends on the stabitily of conjugate base the reaction goes forward.


text:
Safe way to extract hydrogen from water through electrolysis
I've done some research on google but most guides I've seen recommend adding salt to the mixture for better conductance which results in the release of chlorine.
Is there a way to perform electrolysis on water that releases hydrogen and at the same time does not release any gases that are harmful to humans?
By that I mean can I use another electrolyte that does not produce harmful side effects (so sulfuric acid is also out, and I do realize it's next to impossible to do electrolysis on distilled water.)
I'm looking for a way to show this experiment to kids in enclosed space without good ventilation.

Welcome to chemistry.se! Please take a minute to go through the [tour] and maybe one more for the [help].

@Martin I do realize you cannot perform destilation on destiled water. I'm looking for different electrolyte or other ideas on how to make the process safe (I know for example Sulfuric acid can be used as electrolyte to, but I don't really want to handle that either).

Read the last line/sentence of the linked post's answer.

also related http://chemistry.stackexchange.com/q/2806/4945

@martin thank you. The answer recommends 4 different salts. Could you tell me which one would be safest to handle? I'm reading about all of them on wiki now, but If you can put that in an answer I'd accept it and you'll get karma :)

Also, sodium hydrogen carbonate (baking soda) will work if dissolved in water.

Some sort of advice, since you are concerned with safety and demonstrating the experiment to children. You should generally not perform an experiment in a badly ventilated space. Please be aware, that the generated hydrogen gas is highly flammable and with the addition of the right amount of oxygen, it can make very loud booms: How explosive is hydrogen gas?
Here are two questions, that are very closely related and already contain the answer to your question:

Can I perform electrolysis without electrolyte? (with distilled water)
How can I produce hydrogen gas efficiently and cheaply?

In general any kind of salt as an electrolyte will do.
In low concentrations about 1-2 mol/L they are usually quite harmless. I personally would prefer diluted sulfuric acid. Any addition of group 2 salts may cause passivation of the electrodes by trapping carbon dioxide from the air.
Diluted sodium hydroxide or potassium hydroxide are also safe to handle.
Probably the safest is sodium sulfate, but it might not be the best performer.
Please wear safety goggles at all times during the experiment, best case: The children wear them, too.

Thank you very much! This is exactly what I was looking for and the more I read about Sodium Sulfate the better it looks. Safe to handle (I could show the crystals) and do not seem to produce any side effects. While I'll obviously open all windows etc. I don't have any way to ensure good ventilation outside of the fan.


text:
What is the effect of the temperature on the no of moles in an equilibrium system?
Here's something that makes me confuse. What will happen to the total no of moles in a system if the temp is raised. Will it increase? If so how is that possible.?
Say we have an equlibrium,
$$\ce{2P <=> Q + R}$$
The temp is $300\:\mathrm{K}$ and total no of moles is 6 moles.
If we increase the temp upto $500\:\mathrm{K}$ will the total no of moles change to 10 moles when the volume and pressure kept constant?

Welcome to chemistry.se! Have a look at the [tour] and the [help] to familiarize yourself with our site.

No, because there are the same number of molecules on either side of that equilibrium.

Oh thnx. But will it increase if the equation doesn't have the same no of molecules on either sides?

The quantity ‘amount of substance’ shall not be called ‘number of moles’, just as the quantity ‘mass’ shall not be called ‘number of kilograms’.

Well, your equation contains the same number of moles either side, so temperature does not matter.
But if the equation had different moles on either side, it should be given if the reaction is endothermic or exothermic.
For eg- if a reaction is exothermic and the temperature is increased, then according to Le Chatelier's principle, the reaction should go in the backward direction (To compensate for the increase in temperature).
So, if the reactant side has more moles than product side, the number of moles will be increased. You can apply similar logic if the reaction is endothermic.

E.G. 2 NO2 <=> N2O4, in which there are twice as many molecules on the left hand side. This reaction is easily reversible near room temperature; see http://chemed.chem.purdue.edu/demos/main_pages/21.1.html.

@DrMoishePippik Therefore we can safely conclude that the reaction is exothermic.


text:
How to find the highest and lowest oxidation state of transitional metal
My teacher taught me the formulas 'nS' and 'nS+(n-1)d' to find the highest and lowest oxidation state of transition elements. He gave the example of $\ce{Mn}$ and showed that the numbers were +2 and +7. But I did not understand it. So can please you give me
a good explanation.

Welcome to chemistry.se! Please have a look at the [help] and take the [tour]. Could you explain, what the variables in your formula stand for, please.

@Martin Although OP should confirm, the "n" stands for the principal quantum number as "s" and "d" refer to the s and d orbitals.

Electron configuration [Ar] 3d5 4s2 - could you be more specific about your lack of understanding? ;)

Well … what about $\ce{Mn2(CO)10}$ and $\ce{[Mn(CO)5]-}$?


text:
Alkylation of conjugated nitriles - regioselectivity
Why does the enolate of the nitrile attack the alkyl chloride from the bond shown in the picture and not from the far end pi bond?

This looks like it can be answered well with orbitals (I’m not the one who could do so, though; it might require calculations).

The hint I have is that it's the product of kinetic control because of bigger HOMO coefficient.

According to Fleming's Molecular Orbitals, Huckel's theory  can't be applied to these systems as it suggests a bigger coefficient of HOMO at the gamma C atom.

I don't like this mechanism, It could depend on coordination of lithium which isn't shown.

Marko, do you have any source for this reaction and/or the proposed mechanism?

I found this article that sheds some light on deprotonation mechanism - it differs significantly from presented in the question and although it's for acetonitrile not crotonitrile it still provides some useful information.
Lithium amide exists as $\ce{(LiNH2)2}$ dimers which tend to coordinate to nitrile groups creating Li-N bonds. In acetonitrile deprotonation occurs intramolecularly with opening of $\ce{Li2N2}$ rings, not pericyclicly, but for crotonic nitrile analogous mechanism isn't possible what suggests intermolecular reaction.

On the basis of our MP2(full)/6-31+G*//MP2(full)/6-31+G* calculations (Scheme 2 and Figure 1), the adduct, which is formed from CH3CN (7) and the dimeric base 8, is stabilized by about 12 kcal/mol relative to the separated reactants. It is likely that in solution this coordination proceeds by replacement of a solvent molecule from the dimeric lithium amide. The existence of a complex comparable to 9 and the dimeric character of solvated lithium amides 6 are confirmed by NMR spectroscopy and structural studies. To allow hydrogen transfer from the nitrile to the amide moiety, the central four-membered ring of 9 must open.  The resulting intermediate 10, an “open dimer”, is also supported by various investigations of related reaction pathways. 7 The cleavage of an Li-N bond requires about 11 kcal/mol, while the transition state for proton transfer, 11, lies only just above slightly more than 1 kcal/mol higher in energy than 10. This low barrier is indicated in a lenghtened C-H bond of the open dimer (1.094 Å, compared with 1.082 Å in acetonitrile). After the formation of another “open dimer”, 12, which is stabilized by 14 kcal/mol relative to the transition state, and extrusion of ammonia, the final intermediate, 14, is a mixed (dianionic) aggregate of LiNH2 and LiCH2CN, comparable to Boche’s QUADAC.

Another article provides some info about anions of nitriles:

The  exceptional nucleophilicity stems primarily from the powerful inductive
  stabilization  of  metallated  nitriles  rather  than  resonance
  stabilization as is the case with enolates.

and influence of nitrile group on carboanion:

Essentially, the  powerful  electron  withdrawal  of  the  CN  group stabilizes the adjacent negative charge in a manner akin to an ylide. Consistent with this inductive stabilization is the range of geometries for the metallated carbon which spans a continuum varying from planar in 12 to pyramidal in the cyclopropanecarbonitrile 13.

Mainly inductive stabilisation suggests that in metallated anion of crotonitrile negative charge is concentrated on alpha carbon and thus it's more nucleophilic then gamma carbon and in low temperature, under kinetic control, attack via alpha carbon should be favoured.

That sounds very promising, please don't wait too long as I have to assign the bounty soon. I am very much looking forward to reading about this mechanism. I would be very happy, if you could include the full citation in your final review.

I am still looking forward to learning more. ;)

It would be nice to see a conclusion.

@Marko Conclusion doesn't  seem exactly satisfying imo, but maybe I'll find sth else later.


text:
What causes triboluminescence and what process occurs to produce light?
I have observed this effect by rubbing together two pieces of quartz in a dark room but am unable to explain this effect. Would it be something to do with the way quartz is bonded as a covalent network?

I'm only a first year chemical engineering student, and we haven't really covered this topic in our course yet, but I will try and answer your question to the best of my knowledge.
(I am sure people more qualified me would provide better responses.)
When we apply certain stresses to a material such as a crystal, we separate charges (which is why it usually is asymmetric... the presence of anistropy allows charge separation.) These charges then can cause electric discharge, which ionizes the surrounding air and releases a flash of light.
There exist symmetric crystals which display this phenomenon as well... and it is suggested that (probably) impurities exist within such crystals which break the symmetry.
Anyway, this is my understanding of the phenomenon, and I don't claim is perfect in any way, but hopefully is helpful to you.


text:
Structure of Ca2B6O11
What is its structure? I found an image of $\ce{B6O11}$ unit, but it looks incorrect:
 The top left corner one.

The oxygen can make max 2 bonds but atom $\mathrm{O8}$ has three!
There are six oxygen with only one bond, that is the unit has $-6$ net charge, but $\ce{2Ca^2+} = +4$ positive charge.

1) Oxygen CAN form 3 or 4 bonds. The most common example is $\ce{H3O+}$ ion This form of bonding is critical to understand basic nature of ammonia solutions, so for exact description of it look that part of the book.
2)Boron also can form more than 3 bonds if at least one its partner provides an electron pair for bonding, not one electron.
As a general rule I prefer to stay away from the chemistry of borates and alumina-silicates as it is ruled largely by ionic interactions and non-intuitive hydration/dehydration.
===========
In structural diagrams of polymeric salts 'unit' often refers to some recurring fragment... however, it is not nessesarily has same stochiometry than formal anion. In alumosilicates the anionic frame consists of $\ce{AlO4-}$ and $\ce{SiO4}$ tetrahedral units. However, they are usually connected by shared vertexes, so the exact formula of formal anion is different. Borates are a bit more complex system, since they may include both $\ce{BO3}$ and $\ce{BO4-}$ units, and oxygen is often connected to hydrogen.
==========
I looked into open cristallographic database. Unsurprisingly, it does NOT feature a compound $\ce{Ca2B6O11}$. However, salts are often written in 'dehydrated' form, so I looked for any borate with $\ce{Ca:B} = 1:3$ stochiometry. Several with rather complex crystal structure were found, so I left only two: one with rather simple structure of typical 'hidrated' borate anion and one with rather complex continous 'dehydrated' boron-oxygen layers. Note: the structures are refined from X-Ray , so they does not reveal hydrogen positions. Isolated oxygens of water molecules and cations are omitted.
$\ce{B3O8}$ unit from mineral Inyoite, $\ce{CaB3O3(OH)5·4(H2O)}$

$\ce{[B3O6]_{\infty}}$ layer from mineral Fabianite $\ce{CaB3O5(OH)}$

and its main building block

The crystal structure of $\ce{Ca2B6O11}$  has actually been deposited to Inorganic Crystal Structure Database in 1980 and modified in 1987. The reference number is #23032.

$\ce{Ca2B6O11}$ is an anhydrous form of several main boron minerals with well-established both compositions and structures, such as:

Colemanite $\ce{Ca2B6O11 * 5H2O}$;
Meyerhofferite $\ce{Ca2B6O11 * 7H2O}$;
Inyoite $\ce{Ca2B6O11 * 13H2O}$.

Water depletion results in a rather noticeable structure rearrangement.
Crystal structure of $\ce{Ca2B6O11}$ [1] is actually deposited in ICSD database (#23032).
Unit cell: $8.68(1)$, $8.56(1)$, $10.78(2)$, $90.00$, $90.00$, $110.0(1)$; space group $P1\,1\, 21/b$ monoclinic, $Z = 4$:
$\color{#804000}{\Large\bullet}~\ce{B}$;
$\color{#FF0D0D}{\Large\bullet}~\ce{O}$;
$\color{#3DFF00}{\Large\bullet}~\ce{Ca}$.
For clarity coordination polyhedra are only shown for $\ce{[BO4]}$.

Cross-linked structure features trimeric linear assemblies of $\ce{[BO4]}$ units connected via vertices into the 3D motif with $\ce{Ca}$ and $\ce{[BO3]}$. In order to answer your question regarding atom bonding, here are the environmental listings for all crystallographically non-equivalent atoms. for the quick reference, maximum number of nearest neighbous of oxygen is $4$, e.g. for $\mathrm{O2}$:

Boron
Forms two types of coordination polyhedra with oxygen: planar trigonal (C.N. = 3) and tetrahedral (C.N. = 4).
    B1   SYMM  O4    O3    O2    O8    B6    B5    B3
O4  1.42 I     -     -     -     -     -     -     -
O3  1.46 I     107.9 -     -     -     -     -     -
O2  1.48 I     113.1 110.4 -     -     -     -     -
O8  1.56 I     108.8 107.7 108.8 -     -     -     -
B6  2.46 2_765 33.3  120.2 79.9  124.8 -     -     -
B5  2.57 2_655 95.4  89.5  136.5 28.1  123.8 -     -
B3  2.64 I     86.3  125.6 27.0  116.8 53.4  142.5 -

     B2   SYMM  O5    O9    O10   O11   B3    B4    B5    B4
O5   1.44 I     -     -     -     -     -     -     -     -
O9   1.45 I     110.5 -     -     -     -     -     -     -
O10  1.46 I     107.9 115.2 -     -     -     -     -     -
O11  1.64 I     111.8 109.4 101.9 -     -     -     -     -
B3   2.49 I     98.6  32.7  145.9 87.3  -     -     -     -
B4   2.57 4_556 119.4 86.3  29.0  116.2 118.0 -     -     -
B5   2.60 I     116.1 82.1  122.7 28.6  58.9  123.9 -     -
B4   2.69 I     21.7  131.1 90.3  104.5 119.5 109.1 119.7 -

    B3   SYMM  O7    O6    O2    O9    B6    B2    B5    B1
O7  1.44 I     -     -     -     -     -     -     -     -
O6  1.46 I     113.5 -     -     -     -     -     -     -
O2  1.48 I     111.1 106.5 -     -     -     -     -     -
O9  1.49 I     108.3 109.9 107.5 -     -     -     -     -
B6  2.30 2_765 26.2  116.5 85.9  125.4 -     -     -     -
B2  2.49 I     95.9  90.0  138.2 31.5  121.1 -     -     -
B5  2.51 I     114.2 27.3  126.0 84.7  129.8 62.8  -     -
B1  2.64 I     85.3  111.3 27.1  126.0 59.5  156.2 137.8 -

     B4   SYMM  O5    O10   O3    O1   B2    B6    B2
O5   1.46 I     -     -     -     -    -     -     -
O10  1.47 4_566 108.7 -     -     -    -     -     -
O3   1.52 3_766 113.6 105.1 -     -    -     -     -
O1   1.53 I     109.6 115.5 104.4 -    -     -     -
B2   2.57 4_566 98.7  28.9  132.6 95.6 -     -     -
B6   2.68 I     134.2 91.5  99.2  27.9 79.3  -     -
B2   2.69 I     21.5  126.3 113.8 89.4 108.8 116.0 -

     B5   SYMM  O6    O11   O8    B3    B1    B2
O6   1.38 I     -     -     -     -     -     -
O11  1.40 I     120.8 -     -     -     -     -
O8   1.41 2_654 118.2 121.0 -     -     -     -
B3   2.51 I     29.0  92.0  147.0 -     -     -
B1   2.57 2_654 86.9  152.1 31.6  115.4 -     -
B2   2.60 I     87.3  34.0  154.2 58.3  169.5 -

    B6   SYMM  O7    O4    O1    B3    B1    O2    B4
O7  1.19 2_764 -     -     -     -     -     -     -
O4  1.50 2_764 130.4 -     -     -     -     -     -
O1  1.51 I     117.5 109.4 -     -     -     -     -
B3  2.30 2_764 32.5  98.1  148.3 -     -     -     -
B1  2.46 2_764 99.2  31.4  136.5 67.1  -     -     -
O2  2.64 2_764 65.7  64.9  157.7 33.9  33.5  -     -
B4  2.68 I     132.9 85.2  28.4  148.8 108.7 132.2 -

Oxygen
     O1   SYMM  B6    B4    O7    O3    O5    O4    Ca1   O10   Ca2
B6   1.51 I     -     -     -     -     -     -     -     -     -
B4   1.53 I     123.7 -     -     -     -     -     -     -     -
O7   2.31 2_764 27.1  137.8 -     -     -     -     -     -     -
O3   2.42 3_766 111.8 37.6  138.6 -     -     -     -     -     -
O5   2.44 I     152.3 34.2  145.8 61.7  -     -     -     -     -
O4   2.45 2_764 35.2  93.0  61.6  77.0  126.9 -     -     -     -
Ca1  2.47 2_764 88.3  115.4 61.7  152.3 91.5  119.2 -     -     -
O10  2.54 4_566 96.2  31.5  106.3 57.2  57.0  73.4  103.5 -     -
Ca2  2.59 3_766 117.7 89.5  127.8 62.3  84.4  105.3 125.9 118.1 -

     O2   SYMM  B3    B1    O6    O9    O7    O3    O4   O8   Ca2   B6
B3   1.48 I     -     -     -     -     -     -     -    -    -     -
B1   1.48 I     125.8 -     -     -     -     -     -    -    -     -
O6   2.35 I     36.5  126.4 -     -     -     -     -    -    -     -
O9   2.40 I     36.5  145.7 61.2  -     -     -     -    -    -     -
O7   2.41 I     34.0  93.3  61.3  59.4  -     -     -    -    -     -
O3   2.41 I     140.8 34.4  160.5 130.5 108.7 -     -    -    -     -
O4   2.42 I     93.5  32.7  104.2 116.9 60.7  57.5  -    -    -     -
O8   2.48 I     130.9 36.7  106.6 167.0 111.8 59.8  59.4 -    -     -
Ca2  2.51 2_654 94.7  91.6  58.7  115.0 107.4 114.3 99.6 56.4 -     -
B6   2.64 2_765 60.1  66.6  80.2  84.4  26.7  85.4  34.0 89.1 107.1 -

     O3   SYMM  B1    B4    O4    O10   O2    O1    Ca2   O8    O5    Ca2  O6
B1   1.46 I     -     -     -     -     -     -     -     -     -     -    -
B4   1.52 3_766 132.5 -     -     -     -     -     -     -     -     -    -
O4   2.33 I     35.5  112.8 -     -     -     -     -     -     -     -    -
O10  2.38 2_755 131.8 36.7  138.2 -     -     -     -     -     -     -    -
O2   2.41 I     35.1  108.5 61.4  96.8  -     -     -     -     -     -    -
O1   2.42 3_766 146.5 37.9  111.4 64.0  143.6 -     -     -     -     -    -
Ca2  2.43 4_557 117.1 93.1  151.4 60.3  99.9  96.4  -     -     -     -    -
O8   2.44 I     37.6  169.6 61.2  141.9 61.4  150.5 91.3  -     -     -    -
O5   2.49 3_766 100.7 32.4  82.8  58.4  84.0  59.7  118.6 138.3 -     -    -
Ca2  2.59 I     90.6  89.6  58.0  124.5 119.2 62.1  137.7 93.5  83.3  -    -
O6   2.68 2_655 81.6  133.1 74.8  144.6 114.6 95.3  96.8  55.5  137.2 53.8 -

     O4   SYMM  B1    B6    O3    Ca2   O2    O8    O7   O1
B1   1.42 I     -     -     -     -     -     -     -    -
B6   1.50 2_765 115.3 -     -     -     -     -     -    -
O3   2.33 I     36.6  126.6 -     -     -     -     -    -
Ca2  2.40 I     99.7  129.5 66.6  -     -     -     -    -
O2   2.42 I     34.3  81.1  61.1  127.3 -     -     -    -
O8   2.43 I     37.5  131.2 61.7  98.9  61.4  -     -    -
O7   2.44 I     93.7  21.7  110.7 142.7 59.5  112.5 -    -
O1   2.45 2_765 145.1 35.4  132.0 96.4  113.6 163.1 56.4 -

     O5   SYMM  B2    B4    O10   O9    O10   Ca1   O1   O3    O11
B2   1.44 I     -     -     -     -     -     -     -    -     -
B4   1.46 I     136.8 -     -     -     -     -     -    -     -
O10  2.35 I     36.4  105.2 -     -     -     -     -    -     -
O9   2.37 I     34.8  166.9 62.7  -     -     -     -    -     -
O10  2.38 4_566 158.5 35.9  140.3 156.8 -     -     -    -     -
Ca1  2.44 I     133.1 90.1  151.3 98.8  61.3  -     -    -     -
O1   2.44 I     102.1 36.2  78.0  136.8 63.6  124.3 -    -     -
O3   2.49 3_766 129.8 34.1  93.8  136.7 58.4  85.4  58.6 -     -
O11  2.55 I     36.6  118.2 58.8  61.4  122.5 134.0 83.7 138.4 -

     O6   SYMM  B5    B3    O2    Ca2   O8    O9    O11   O7    O3
B5   1.38 I     -     -     -     -     -     -     -     -     -
B3   1.46 I     123.8 -     -     -     -     -     -     -     -
O2   2.35 I     145.7 37.0  -     -     -     -     -     -     -
Ca2  2.39 2_654 127.4 100.3 63.8  -     -     -     -     -     -
O8   2.39 2_654 31.2  154.7 153.9 100.1 -     -     -     -     -
O9   2.42 I     90.6  35.5  60.3  118.8 119.9 -     -     -     -
O11  2.42 I     29.8  94.1  122.5 149.7 61.0  62.8  -     -     -
O7   2.43 I     121.8 33.1  60.5  110.8 144.6 58.9  95.8  -     -
O3   2.68 2_654 86.9  147.7 121.3 61.2  57.1  176.4 115.2 124.7 -

     O7   SYMM  B6    B3    O1    O9    O2    O6    O4    Ca1
B6   1.19 2_765 -     -     -     -     -     -     -     -
B3   1.44 I     121.3 -     -     -     -     -     -     -
O1   2.31 2_765 35.3  154.4 -     -     -     -     -     -
O9   2.38 I     141.0 36.5  146.5 -     -     -     -     -
O2   2.41 I     87.6  34.9  119.5 60.0  -     -     -     -
O6   2.43 I     122.4 33.5  151.7 60.3  58.2  -     -     -
O4   2.44 I     27.8  93.6  62.0  116.8 59.9  101.4 -     -
Ca1  2.46 I     97.1  134.5 62.4  98.1  144.3 138.3 120.2 -

     O8   SYMM  B5    B1    Ca2   O6    O4    O3    O11   O2
B5   1.41 2_655 -     -     -     -     -     -     -     -
B1   1.56 I     120.3 -     -     -     -     -     -     -
Ca2  2.35 2_654 143.4 95.5  -     -     -     -     -     -
O6   2.39 2_655 30.6  90.0  173.3 -     -     -     -     -
O4   2.43 I     102.5 33.7  103.9 78.6  -     -     -     -
O3   2.44 I     96.4  34.7  119.2 67.4  57.1  -     -     -
O11  2.44 2_655 29.5  149.4 114.0 60.1  124.5 123.9 -     -
O2   2.48 I     154.1 34.6  62.5  123.5 59.2  58.8  176.0 -

     O9   SYMM  B2    B3    Ca1   O5    O7    O2    O6    O10  O11
B2   1.45 I     -     -     -     -     -     -     -     -    -
B3   1.49 I     115.9 -     -     -     -     -     -     -    -
Ca1  2.32 3_766 139.1 102.6 -     -     -     -     -     -    -
O5   2.37 I     34.7  102.3 126.0 -     -     -     -     -    -
O7   2.38 I     100.6 35.2  103.2 73.9  -     -     -     -    -
O2   2.40 I     150.8 36.0  69.9  134.4 60.6  -     -     -    -
O6   2.42 I     93.3  34.6  127.3 98.9  60.8  58.5  -     -    -
O10  2.46 I     32.6  146.5 110.9 58.2  131.3 164.9 116.1 -    -
O11  2.52 I     37.9  89.4  161.9 62.8  94.4  116.8 58.7  58.0 -

     O10  SYMM  B2    B4    O5    O3    O5    O11   Ca2   O9    Ca1   O1
B2   1.46 I     -     -     -     -     -     -     -     -     -     -
B4   1.47 4_556 122.1 -     -     -     -     -     -     -     -     -
O5   2.35 I     35.7  132.1 -     -     -     -     -     -     -     -
O3   2.38 2_754 157.3 38.2  139.7 -     -     -     -     -     -     -
O5   2.38 4_556 107.1 35.4  138.0 63.2  -     -     -     -     -     -
O11  2.41 I     41.7  135.0 64.8  155.3 101.3 -     -     -     -     -
Ca2  2.41 3_766 126.0 95.1  90.4  60.9  124.0 129.1 -     -     -     -
O9   2.46 I     32.2  90.1  59.1  127.1 79.1  62.3  141.2 -     -     -
Ca1  2.46 4_556 105.9 89.0  131.7 87.6  60.6  67.8  113.0 105.5 -     -
O1   2.54 4_556 98.6  33.0  99.2  58.7  59.4  132.1 93.6  70.8  119.4 -

     O11  SYMM  B5    B2    O10   O6    O8    Ca1   O9   O5
B5   1.40 I     -     -     -     -     -     -     -    -
B2   1.64 I     117.5 -     -     -     -     -     -    -
O10  2.41 I     140.8 36.4  -     -     -     -     -    -
O6   2.42 I     29.3  88.5  117.7 -     -     -     -    -
O8   2.44 2_654 29.5  146.5 153.1 58.9  -     -     -    -
Ca1  2.50 2_764 117.3 114.1 101.9 133.7 95.7  -     -    -
O9   2.52 I     86.0  32.8  59.7  58.6  114.0 144.1 -    -
O5   2.55 I     120.9 31.6  56.4  94.0  145.2 88.3  55.7 -

Calcium
Features two heavily distorted oxo-environments with C.N. = 6 and C.N. = 8.
     Ca1  SYMM  O9    O5    O7    O10   O1   O11
O9   2.32 3_766 -     -     -     -     -    -
O5   2.44 I     106.0 -     -     -     -    -
O7   2.46 I     85.8  71.4  -     -     -    -
O10  2.46 4_566 143.9 58.1  114.1 -     -    -
O1   2.47 2_765 74.3  127.2 55.9  141.8 -    -
O11  2.50 2_765 87.2  147.8 140.0 93.9  84.3 -

     Ca2  SYMM  O8    O6    O4    O10   O3    O2    O1   O3
O8   2.35 2_655 -     -     -     -     -     -     -    -
O6   2.39 2_655 109.6 -     -     -     -     -     -    -
O4   2.40 I     170.9 79.3  -     -     -     -     -    -
O10  2.41 3_766 94.9  152.3 76.8  -     -     -     -    -
O3   2.43 4_567 94.9  128.5 77.7  58.8  -     -     -    -
O2   2.51 2_655 61.1  57.5  125.3 150.1 102.9 -     -    -
O1   2.59 3_766 77.3  98.4  103.5 74.1  131.5 113.7 -    -
O3   2.59 I     129.2 64.9  55.4  89.8  129.3 118.9 55.6 -

Bibliography

Zarakina, N. V.; Brovkin, A. A. Kristallografiya 1976, 21, 502–506. (In Russian)

The oxygen can make max 2 bonds but $\ce{O8}$ has three!

You will have better idea if you see 2D structure instead of 3D

You can clearly see no $\ce{O}$ atom have more then 3 bonds!


text:
How do I predict the spontaneity of an ionic metathesis reaction?
How do I predict the spontaneity of a metathesis reaction?
Is it a matter of summing up the differences between their ions' standard reduction potentials?
Take the salt metathesis reaction between two dissolved electrolytes; sodium chloride and silver nitrate (highlighted here). While the HSAB theory exists, it has not been quantified yet (as far that I know of).
Based my current knowledge of Chemistry, I would expect something like this:
Let the symbol of each element represent its standard reduction potential. Then spontaneity would be given by the following:
$\ce{((-Na) - (-Ag)) + (Cl - NO3)}$
Where a positive value would imply that the reaction is spontaneous, and the magnitude the degree of spontaneity.

Your question is confusing - I have no idea why you're mixing simple precipitation with redox - there's no redox in there!

@Mithoron: Read about salt metathesis reactions here.

I'm well aware how metathesis works  and it's driven by precipitation or vaporisation or complexation

No, the spontaneity of the reaction does not have to do with the reaction potentials. All of the constituent parts of the salts exist as ions even in solid form, so there is no exchange of electrons occurring.
Rather, the insolubility of a salt is determined by the strength of the electrostatic forces between the ions. In some salts, the strength of the attractive forces that water applies is enough to break these ions apart, dissolving the salt. In other salts, the forces applied by the water might not be enough. These kinds of salts are those that are considered to be insoluble.
If you wish to predict whether or not a reaction between salts occurs (i.e. something precipitates out of solution), you should memorize the solubility rules.

This does not answer my question. I am specifically asking about metathesis reactions.

@VatsalManot Which part? I responded to the fact this is not a redox reaction, so the use of reduction potentials is meaningless.

The main question: How do I predict the spontaneity of a metathesis reaction? All you have done is nullify my proposed method.

@VatsalManot Does the list of solubility rules not answer that question? If one of the product salts is insoluble, the reaction occurs. If both of the product salts are soluble, there is no reaction.


text:
Open Database of Phase Diagrams?
Is there some kind of database that has information about phase diagrams of different substances? I'm mostly looking for phase diagrams of elements (silicon, iron, hydrogen, etc.) because I'm trying to write a planetary generator, but I cannot find any information about the mathematics of a phase diagram.
Better yet, is there some way to simply approximate the phase diagrams of different compounds? I'm looking for simulation complexity and variation more than I want realism.

Possibly related: http://chemistry.stackexchange.com/questions/1155/phase-diagram-for-first-10-elements

I believe you can compare many of the models with a search engine.
http://www1.asminternational.org/asmenterprise/apd/AdvancedSearchAPD.aspx


text:
Piperic acid oxidation

There are 2 double bonds $\ce{C=C}$ present within each piperic acid molecule,  thus when piperic acid reacts with hot excess $\ce{KMnO4}$, it oxidised into a carboxyllic acid. Why is $\ce{CO2}$ produced?
The carbon atom bonded directly to the benzene ring will get oxidised into a carboxylic acid and also that the carbon bonded to the $\ce{COOH}$ group will get oxidized also to a carboxylic acid to form a dicarboxylic acid. What changes will take place to the other 2 carbon atoms.

this paper stated that $\ce{CO2}$   could be formed as well, how?

Usually when we consider permanganate oxidations we think of

oxidation of alkyl benzenes to benzoic acid
cleavage and oxidation of a double bond to an acid, through the aldehyde, if there is one hydrogen on the double bond
cleavage and oxidation of a double bond to a ketone if there are no hydrogens on the double bond

However, there are a few special cases such as formic acid and oxalic acid, or their corresponding aldehydes.  These compounds are oxidized by permanganate to $\ce{CO2}$ (for an idea as to why these compounds react differently than, say, acetic acid, see Scheme 1 here).
With this information in hand we can re-examine the permanganate oxidation of piperic acid.

We see that oxalic acid is one of the primary products produced in this reaction.  As discussed above, oxalic acid can then undergo further reaction to produce carbon dioxide.

It's my first time to hear about this. so i should learn that in those cases, CO2 is produced, yes?. Is this applied to some aldehydes?

Yes, you should be aware of the permanganate oxidation of those 2 molecules and their aldehydes.  Since this is your first time here, let me add that if this answer was helpful please consider accepting and\or up-voting it, thanks.


text:
galvanic corrosion in the wild: where is the reduction?
Galvanic corrosion occurs of an active metal (i.e. iron or zinc) in contact with a passive metal (i.e. copper) in a conductive solution. A famous example is the corrosion of iron girders in contact with copper in the statue of liberty.
Although the difference in potential can account for the protection of copper, it does not explain the girders in had a greatly accelerated corrosion rate. In a zinc/copper battery there is a solution of copper sulfate that gets reduced at the copper electrode. However, if all the copper is in metallic form in the first place, no galvanic reaction can occur.
It seems galvanic corrosion in real-life, does not increase total corrosion, but rather concentrates it (at lest for pairs of metals with similar non-galvanic corrosion rates). Copper sheets have a large surface area that initially reacted non-galvanically to form the green color we see. Some of these ions are then reduced at the expense of the oxidizing the girders. The statue of liberty and other infamous examples of corrosion involve large amounts of inactive metals connected to small amounts of active (iron/steel) girders/fasteners/etc.
This logic predicts that the attaching a copper bar to an iron bar of equal surface area will only double it's corrosion rate. That being said, predicting the corrosion rate is not easy so there could be some other effect I am overlooking. Is the surface area logic correct or would there a large increase in the total corrosion?

Comparing galvanic corrosion to the zinc/copper battery is a bit unfair. A fast corrosion rate is still much much slower kinetics compared to the expected rates of the electrochemical reaction in the battery.
The counter reaction in the galvanic corrosion case is almost always reduction of water. This is why it is much worse in conditions near the sea.
Galvanic corrosion rate can be measured and there is an ASTM standard(G71) to make the measurement.
To answer your comment below, I am trying to say that it is impossible to say if it will double or triple or just increase by 10%. The factors are complicated and hard to predict a priori. That is why it is a good idea to measure using the above mentioned standard.

Sea spray certainly would accelerate the process

"attaching a copper bar to an iron bar of equal surface area will only double it's corrosion rate"? That is my main question. Sea water should be much faster in either case.

I edited my answer to respond to the comment as well.


text:
How to find the formula of an binary oxide based on the ionisation energies?
I am a little confused about this multiple choice question and what the requirements are to answer it. It is for a year 11 chemistry student that I am helping and I think I am going in a little too much detail to try to solve it to help them. I am sure that I am overthinking this.

The energies needed to remove electrons in succession from a neutral
  atom of a certain element $\ce{A}$ are given ($\mathrm{kJ\cdot  mol^{-1}}$): $1010$, $2032$, $7056$, $8120$, $9300$.
   The formula  of the oxide of $\ce{A}$ is:
   $\ce{AO}$, $\ce{A2O}$, $\ce{A2O3}$, $\ce{A2O4}$ or $\ce{A3O5}$.

My understanding of the question and what I was trying to do to solve the problem by looking at the ratios between the energies and I was trying to apply
1st ionization energy: $$\ce{X -> X+ + e−},$$
2nd ionization energy: $$\ce{X+ → X^2+ + e−},$$
and 3rd ionization energy: $$\ce{X^2+ -> X^3+ + e−}$$
to the energies below and find a ratio. I thought the ratio was going to be c. I understood that oxide is $2-$, however, I couldn't calculate the $\ce{A}$ value as I didn't know what the element was.

This is a homework question. Please share your thoughts towards a solution otherwise it will be closed in accordance with our homework policy

Hi, my understanding of the question and what I was trying to do to solve the problem by looking at the ratios between the energies and I was trying to apply 1st ionization energy X → X+ + e−, 2nd ionization energy X+ → X2+ + e− and 3rd ionization energy X2+ → X3+ + e− to the energies below and find a ratio I thought the ratio was going to be c. I understood that Oxide is 2- however I couldn't calculate the A value as I didn't know what the element was. I was completely wrong and confused, so thank you.

The point of this problem is to test if you can see where there is a jump in ionization energies. The third ionization energy is considerably larger than the others. Although ionization energies will naturally increase in succession, this is the most profound increase, which would suggest that this atom will tend to lose 2 electrons before obtaining a stable octet structure.
This would mean that its ion would be $\ce{A^{2+}}$. Since the oxide ion is $\ce{O^{2-}}$, the charges will be balanced if the ratio of the ions is 1:1, meaning the answer is $\ce{AO}$.

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


text:
"Middle row anomaly" of the periodic table
I was studying about the periodic table recently, and was reading a topic associated with oxides of halogens, and came across the following line

The bromine oxides, $\ce{Br2O}$, $\ce{BrO2}$, $\ce{BrO3}$ are the least stable halogen oxides (Middle row anomaly) and exist only at low temperatures. They are very powerful oxidizing agents.

So, I went to search this on Google, and found these lines

It refers to the instability of oxides of bromine as compared to relative stability of oxides of chlorine and Iodine at room temperature, the former being stable only at low temperatures.

But the line above is simply restating what the book had already told.
So, is their any specific reason why this happens, or is this only due to the experimental data we have gathered?

According to wikipedia diiodine mono-oxide is very unstable and dibromine monoxide is stable below -40 C. So, which is more stable?

This is due to the transition metal contraction.
Bromine has the electron configuration $\ce{[Ar] 4s^{2} 3d^{10} 4p^{5}}. $The 3d orbital has no radial nodes and is therefore quite contracted (close to the nucleus), so there is relatively little repulsion between the 3d electrons and the 4p electrons. This makes it much harder to acheive high oxidation states of bromine (I, IV, VI for the examples you give) because the ionization energies are higher than you might expect from a simplistic approach to periodic trends.

Why doesn't that argument apply to iodine as well?

@ron he is correct, it does not apply for iodine as 4d orbitals have 1 radial node, while 3d does not have them.

It probably applies for astatine too due to the lanthanide contraction but I'm not sure if the redox chemistry has been investigated yet.

So this argument should rather be made stating the high-than-expected effective nuclear charge in the $\rm{4p}$ and $\rm{5p}$ rows due to the presence of the d block in the fourth and fifth periods, which means 10 additional protons without perfect shielding in each case? The anomalous size of the $\rm{3d}$ orbitals might explain why bromine has the least stable oxides, but I'm never very comfortable with electron subshell repulsion arguments.

Your explanation to me seems equivalent. What is shielding other than electron-electron repulsion ?

Fair enough, I actually hadn't made that connection in my mind before. Shielding is just a way to variationally compensate for the addition of the electron-electron repulsion term in the hamiltonian, after all.

can anyone cite any other examples of this middle row anomally?

@NilayGhosh Instability of AsCl5. Reduction potentials of H2SeO4. Next time, I highly recommend you ask a new question instead of asking in comments though.

@orthocresol, If I ask this  as a new question, it would be regarded as duplicate.

@NilayGhosh It would only be marked as a duplicate if there is another question that specifically lists manifestations of the alternation effect, or middle row anomaly, or whatever you want to call it. This question does not deal with examples of the phenomenon, but rather the reason behind it.


text:
Ground state conformation of dimethyl disulfide
What is the ground state conformation of dimethyl disulfide ?
The two which seem reasonable to me are trans to avoid methyl-methyl gauche interactions or gauche due to sulfur donation into the S-Me sigma antibonding orbital.
I found a discussion in this paper but it doesn't seem to really answer the question definitely.

Your reference says that the $\ce{C-S-S-C}$ dihedral angle is close to 90°, what additional information are you looking for?


text:
How can I choose an indicator for determining the concentration of a base using its pKb values?
I have a base $\ce{M(OH)_{3}}$ having 3 $\mathrm{p}K_{\mathrm{b}}$ values as,
$$
\mathrm{p}K_{\mathrm{b}_{1}} = 0.5\\
\mathrm{p}K_{\mathrm{b}_{2}} = 3.7\\
\mathrm{p}K_{\mathrm{b}_{3}} = 7.7\\
$$
If a strong acid is given as the titrant and phenolphthalein and methyl orange are given as possible indicators, how can I choose an indicator using above $\mathrm{p}K_{\mathrm{b}}$ values ?

Since:
\begin{align}
K_\mathrm{w} &=K_\mathrm{a} \cdot K_\mathrm{b}\\
K_\mathrm{a} &=\frac{K_\mathrm{w}}{K_\mathrm{b}}\\
\log{K_\mathrm{a}} &=\log K_\mathrm{w}-\log K_\mathrm{b}\\
K_\mathrm{w} &=1 \times 10^{-14}\\
\log{K_\mathrm{a}} &= -14-\log K_\mathrm{b}\\
-\log{K_\mathrm{a}} &=14+\log K_\mathrm{b}\\
-\log{K_\mathrm{a}} &=\mathrm{p}K_\mathrm{a}\\
-\log{K_\mathrm{b}} &=\mathrm{p}K_\mathrm{b}\\
\mathrm{p}K_\mathrm{a} &=14-\mathrm{p}K_\mathrm{b}
\end{align}
Using the above formula, the three $\mathrm{p}K_\mathrm{a}$ for the conjugate acid of the conjugate base are: $13.5~(\mathrm{p}K_\mathrm{a_1})$, $10.3~(\mathrm{p}K_\mathrm{a_2})$, and $6.3~(\mathrm{p}K_\mathrm{a_3})$, where $\mathrm{p}K_\mathrm{a_1}=14-\mathrm{p}K_\mathrm{b_1}$, etc.
An important equation to note is that the $\mathrm{pH}$ at the equivalence point is given by the equation: $$\mathrm{pH_\mathrm{eq}}=\frac{\mathrm{p}K_\mathrm{a1}+\mathrm{p}K_\mathrm{a2}}{2}$$
This means that two of the equivalence points exist at $$\mathrm{pH_\mathrm{eq_1}}=\frac{13.5+10.3}{2}=11.9$$ and $$\mathrm{pH_\mathrm{eq_2}}=\frac{10.3+6.3}{2}=8.3$$
The $\mathrm{pH}$ range for methyl orange is $3.0~-~4.4$ and the $\mathrm{pH}$ range for phenolphthalein is $8.2~-~10.0$. Phenolphthalein would be useful for detecting the second equivalence point $(\mathrm{pH_\mathrm{eq_2}})$, since this shift occurs within phenolphthalein's range.
The final equivalence point cannot be calculated from the information given, but given that methyl orange's color transition occurs in an acidic medium, and also the relatively basic $\mathrm{p}K_\mathrm{a_1}$, methyl orange can be eliminated as a possible indicator.

@LDC3 the only $\mathrm{p}K_\mathrm{a}$ close to being within the useful range of one of the indicators is $\mathrm{p}K_\mathrm{a_2}$ to phenolphthalein's range, but it is still not actually within the range. Unless I am misunderstanding how an indicators work (which I guess could be possible) then I don't see how either indicator is useful.

@ringo You want the indicator to change color after all the base has been converted (distant from a $\ce {pK_{b}}$), not when it is being converted (near a $\ce {pK_{b}}$). So the indicators you would choose would not be near any of the $\ce {pK_{b}}$ of the base.

I'll add that the reason to convert to pKa values is that tables for acid-base indicators are usually give in terms of pH not pOH. https://en.wikipedia.org/wiki/PH_indicator

The first step is to determine the $\mathrm{p}K_{\mathrm{a}}$ values for the indicators and convert them to $\mathrm{p}K_{\mathrm{b}}$ values. Then determine how you would titrate the solution using the indicator(s).

Do you mean that the $pK_{b}$ values of $M(OH)_{3}$ have no part to play in choosing the indicator ?

Yes it does, but you need to know when the indicators change colors.

OK. Say I have methyl orange with pH range of 3.0 - 4.4. How do I proceed ?


text:
Nomenclature Precedence Order for Unsaturation and Hydrocarbon Chain
When naming branched alkenes/alkynes, the numbers of the carbons where the branches and multiple bonds are located are supposed to be the lowest possible.  Do the branches take precedence over the multiple bonds, as for example, 2-methyl-3-pentyne?  If the counting of the carbons for this molecule is flipped so that the triple bond is lower in number, it would be 3-methyl-2-pentyne.  Is the first way correct?

Unsaturated bonds take precedence over "branches". Consider these bonds as functional groups and name the compound, it will make you number carbons correctly.
Also the compound you mentioned above can't possibly exist as it has 5 bonds on $\ce{C_3}$.

Thanks, the above structural formula of the molecule was to be drawn on an assignment I was given.  Also, is 2, 3-dimethyl 5-heptyne a possible compound, or do you see something wrong there too?

The naming you did is wrong once again @ChemStudent it should be- 5,6-dimethyl 2-heptyne.

Thank you - this worksheet packet has troubles.  These were the names that were on the sheet with instructions to draw the structural formulas.  I appreciate your assistance.

With regard to numbering of locants within the same compound class, multiple bonds have seniority over simple substituent groups (e.g. methyl, ethyl, etc.) that are used for naming of branched alkanes.
On this matter, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(e) saturation/unsaturation:
  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings);
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(…)

Note that Rule e takes precedence over Rule f.
Furthermore,

P-31.1.1.1 The presence of one or more double or triple bonds in an otherwise saturated parent hydride (…) is denoted by changing the ending ‘ane’ of the name of a saturated parent hydride to ‘ene’ or ‘yne’. Locants as low as possible are given to multiple bonds as a set, even though this may at times give ‘yne’ endings lower locants than ‘ene’ endings. If a choice remains, preference for low locants is given to the double bonds. In names, the ending ‘ene’ always precedes ‘yne’, with elision of the final letter ‘e’ in ‘ene’. Only the lower locant for a multiple bond is cited, except when the numerical difference between the two locants is greater than one, in which case the higher locant is enclosed in parentheses.
(…)

For example, a correct systematic name is 3-methylbut-1-ene (not 2-methylbut-3-ene).


text:
How many molecules are there typically in items made of plastics?
In objects made of plastic (and other polymers), how many molecules would one generally expect to find?
Intuitively, it seems like, say, a plastic shopping bag or PET bottle couldn't be a single molecule, but on the other hand, that's one of the strengths of polymers.
So the question is, I suppose, do items made of plastic consist of a single molecule, or many molecules fused together by way of glues or mechanical action?
Edit: Following a comment, I realise that the core of my question could be clearer, so here's an alternate phrasing:
If I chose an atom in a typical plastic object and gave it a colour, and then gave every atom connected to it by way of covalent bonds the same colour by flood fill, and then chose a colourless atom in the object and repeated the process until no more colourless atoms existed in the object, how many colours would a typical plastic object have?

Most plastic items are composed of many molecules. If you stretch a plastic bag, the molecules line up beside each other (like uncooked spaghetti). Since the bag is not one molecule thick or wide, there must be a lot of molecules. OTOH, under the right conditions the rubber in a tire could be one molecule.

@LDC3 "[A] bag is not one molecule thick or wide[.]"? Molecules come in a wide variety of thicknesses and widths, which is why I'm asking. Although I guess my question could do with a bit of clarification.

The short answer is a very large number and far more than you could reasonably count
Polymerization reactions typically produce very large molecules but not so large that a single molecule would make up as much material as a whole plastic bag or item of any significant size. This source on LDPE tells us that the typical molecular weight of the industrial product is 90,000 (some polymers are much bigger than this).
For non-chemists molecular weights are (simplifying slightly) the weight relative to a single hydrogen atom. Carbon is 12 on this scale and ethylene (the monomer from which the plastic is made which consists of 2 carbons and 4 hydrogens) is 32. This mean that a typical molecule of the plastic is (crudely) made from a little under 3,000 ethylene molecules. The other fact you need to know is that there are about 6 x 1023 hydrogen atoms in a gramme of hydrogen (see the wikipedia definition of Avogadro's number). Simple mathematics tells us that a gramme of LDPE will contain roughly 6.7 x 1018 molecules so that is the order of the number in small plastic item which, I expect, weighs just a few grammes.
So, the bottom line is that even in a small plastic item consisting of a long chain polymer (LDPE is relatively short but other polymers might have chains 10-100 times more at most) there will be something like 1016 to 1018 molecules of polymer, which is a lot.

Very heavily cross-linked polymers could conceivably create single molecules of macroscopic size. As LDC3 mentions, vulcanised rubber is one possibility. I think the molar mass range for polymers can go from anywhere between some $\rm{10^3\ g\ mol^{-1}}$ for oligomers to  $\rm{10^{25}\ g\ mol^{-1}}$ and beyond.

There is nothing very useful or well defined between 10000 kg/mol and one network molecule for the whole object.

Some plastics have their molecular weight (m.w.) specified, e.g. polyethylene and ultra-high-molecular-weight polyethylene. Knowing the type of plastic, and therefore its m.w., and the mass of the object, using Avogadro's constant, you should be able to calculate how many molecules there are.
Measuring a low-density polyethylene sandwich bag would be a convenient place to start.


text:
What chemical characteristics give types of steel their properties?
I am a mechanical engineer, and just like everyone else I had classes on the crystal structure of metals, phase diagrams, and the various heat treatments.
However, even after diving back into that recently and researching online, I can't make sense of how everything fits together. I know the high level, the low level, but can't link the two together.
For example:

How can the phase diagram of steel be linked to its physical properties? Say, given a certain carbon weight fraction, is it possible to tell from the phase diagram if the steel is going to be hard, resilient, ductile? E.g. cementite is hard and brittle, austenite is ductile and soft?
How do heat treatments work? The way I see it, the phase diagram is "steady state", and heat treatments are about spending a specific amount of time in a phase region to add "a little bit of that" etc. (or quickly cooling it to stop the reactions), but I can't link it back to the phase diagram (there is a 3rd dimension, time).

Apologies if this is not a very clear question, but this is exactly what I hope to find: clarification on a quite complex topic.

(Source: Roy Beardmore, http://www.roymech.co.uk/ (defunct, via the Internet Archive)
As carbon content increases, the ability to resist a sudden impact decreases, as measured by the Charpy Impact Test.
As carbon content increases, ultimate tensile strength and Brinell Hardness increases.
As far as a phase diagram:

increasing $\ce{}$ (cementite) content makes the steel more brittle and hard.
Near 0.76% carbon, the steel will be pearlite which is ductile (good for wires).  Below 0.76% carbon, there will be increased alpha-iron (ferrite). Above 0.76% there can be Ledeburite.  Austenite only exists at high temperature.  Depending upon the cooling rate of Austenite, Bainite (slow cooling) or Martensite (rapid cooling) can form.

Thanks for this good answer (and just in time for the bounty!) . So is my intuition of 2] correct? Are durations to spend in certain phase regions extracted from experience, or is there some kind of law that dictates how much of the steel transforms in function of time?

Nevermind, it's in the wikipedia links you've given.

@MisterMystère When a material undergoes some phase transformation with time we call it Ageing. This cannot be figured out from a phase diagram. Its observed from Microscopic techniques after some hours.
Phase Diagram just tells  us at some T temperature and some X composition what is(are) the phase(are) present in our sample.
Heat Treatments- we usually want to either soften(anneal) or harden(quenching) the material for our purpose.

Say we want martensite, we heat material above $727 C$ and then rapidly cool it below the $M_{s} $ and  then $M_{f}$.  But if we just want pearlite/combination of ferrite and cementite we cool slowly(furnace cooling) to room temperature. Note that sometimes cooling rates wont be enough to be considered "equilibrium cooling" and some austenite may be retained.

Please don't just link to the pictures you use, make sure to attribute them correctly. Here we have a typical case of link-rot. I'll try to fix it as well as I can.


text:
Is solubility coefficient affected if ion data is given in Ksp?
Recently I've been looking at this website, which says:

The $K_{\mathrm{sp}}$ of $\ce{Cr(OH)3}$ is $6.70 \times 10^{-31}$
Problem 1: What is the minimum $\mathrm{pH}$ at which $\ce{Cr(OH)3}$ will precipitate?
$$\ce{Cr(OH)3\rightleftharpoons {Cr^{3+}}+3OH-}$$
$$K_{\mathrm{\mathrm{sp}}}=[\ce{Cr^{3+}}][\ce{OH-}]^{3}$$
$$[\ce{Cr^{3+}}]=s$$
$$[\ce{OH^{-}}]=3s$$
Problem 2: What is the minimum $\mathrm{pH}$ at which $\ce{Cr(OH)3}$ will precipitate if the solution has $[\ce{Cr^{3+}}] = 0.0670~\mathrm{M}$?
$$\ce{Cr(OH)3\rightleftharpoons {Cr^{3+}}+3OH-}$$
$$K_{\mathrm{sp}}=[\ce{Cr^{3+}}][\ce{OH-}]^{3}$$
$$6.70 \times 10^{-31}=[0.0670~\mathrm{M}][s^{3}]$$

I don't understand why the site wrote this:

Note that $3s$ is not necessary. In example one, $s$ was assigned to the chromium ion and we knew the hydroxide to be three times greater than that value. Here, the hydroxide is simply an unknown value and it is not expressed in terms of some other unknown value (as it was in Problem 1).

Can anyone tell me why?
Also is also used to other equilibrium constant equation? What did I miss?

Let the concentration of $\ce{Cr^{3+}}$ be $t$ then the concentration of $\ce{OH-}$ will be $3t$.
So, the site says that writing the concentration as $3t$ is not necessary.
What it is basically doing is considering $3t$ equal to some variable $s$, and solving the question. There is nothing wrong in this, is it?
Now, for example we had not done that and calculated the value of $t$ by using the equation
$K_{\mathrm{sp}}=[\ce{Cr^{3+}}] \cdot (3t)^3$
You will then multiply the derived value of $t$ with 3 to get $\ce{[OH-]}$.
From which you will find the $\mathrm{pH}$. So the site says:
"Why not calculate the value of $3t$ directly by using a variable $s$?"
Therefore: $s=3t$
Therefore: $\mathrm{pOH}=-\log{s}$
What you are thinking is also correct, and the answer with that will be:
$\mathrm{pOH}=-\log{3t}$
So you both are doing the exact same thing here.

I'm sorry, but I can't get it. Is using 3s just more good than making new variable (3t)? Could you provide full answer?

@Ridho wherever you used s in the above question, I have used t. Just to make you clear on the concept. Here, 3t=s in the question. Here I am using concentration as t and not s, which we usually do. What the site means to say is that you need the vale of 3t, so why do we need to first find the value of t and multiply it be 3. In short it did that directly

@Ridho I have appended the answer now. The site has just replaced your $3t$ to an independent variable $s$.

Just forget all about 3s, All the s variables above have been replaced by t variables by me.

So I plug the t like this: $6.70\cdot10^{-31}=(0.0670:M)\cdot(3\cdot0.0670:M)^{3}$?

Nope, change the last .670 term with variable t. I think you forgot that. @Ridho

Just do it as you expected the solution to be, you were on the right track. You will get the right answer eventually

Finally, I got it!
$$6.70\cdot 10^{-31}=(0.0670:M)\cdot (3t)^{3}$$
$$6.70\cdot 10^{-31}=(0.0670:M)\cdot 27t^{3}$$
$$\frac{100}{27} \cdot 10^{-31} = t^{3}$$
$$\frac{2.15}{3} \cdot 10^{-10} \approx t$$
Let $3t=s$
$$s=3\cdot \frac{2.15}{3}\cdot 10^{-10} $$
$$s=\approx 2.15\cdot 10^{-10} $$
$$pH=4.13$$


text:
Is it possible to create a 4 inch by 4 inch channel through concrete using acid?
I'm trying to figure out a way to eat through concrete without damage pex tubing inside (its a radiant heated slab). Is there any way using some type of acid (hydrofluoric, muriatic ) to eat a 4 inch wide, 4 inch deep, and 71 foot long channel through the slab that could be tapped out with a blunt chisel? Pex tubing is resistant to most acids so it would not be damaged.
Also, this needs to be cost effective (under $1000).

Offhand, this seems dangerous to the structural integrity of the slab, because the acid would follow the pore structure of the concrete and weaken the wrong places.
However, concrete cutting saw rental is available; e.g. United Rentals
and http://www6.homedepot.com/tool-truck-rental/Walk_Behind_Concrete_Saw_14/MK-CX3/index.html. Two 45-degree cuts ~6" or deeper should do the job.

Thanks, I assumed as much since I had never heard of it being done.  Now I know why.

I'd be worried about willy-nilly cutting such a slab as the OP described. My father-in-law's house was on such a slab. The heated slab had dried out the soil under the slab and the soil has shrunk away from the slab. he could see an open space between the slab and the soil under much of the house.


text:
What would a diagram of the lattice structure of CuSO4 look like?
Just as $\ce{NaCl}$ would look like this:

What would a diagram of $\ce{CuSO4}$ look like?

From the WebElements.com page on copper sulphate, the metal lattice structure looks like:

The yellow pyramids are cooper and the other structures are sulphates.
An alternate view comes from the Wikipedia page of copper (II) sulphate:

The pink spheres represent copper, yellow are sulphur and red are oxygen.
Below is a 2D representation of copper sulphate, from the Royal Society of Chemistry

A similar question with some more insight is found in How do copper sulphate and the like change crystal structure on adding water of hydration?


text:
Calculating the pH of a saturated calcium fluoride solution
How can I calculate the pH of a saturated solution of calcium fluoride ($\ce{CaF2}$)? I am given the following values:
$$\begin{align}
K_\mathrm{sp}(\ce{CaF2}) &= 3.9 \cdot 10^{-11} \\
K_\mathrm{a}(\ce{HF}) &= 6.8 \cdot 10^{-4} \\
K_\mathrm{w} &= 10^{-14}
\end{align}$$
(The $K_\mathrm{sp}$ and $K_\mathrm a$ values are taken from the appendices of Skoog et al. Fundamentals of Analytical Chemistry, 9th edition.)

Here is my approach. The dissociation of $\ce{CaF2}$ is described by
$$\ce{CaF2 <=> Ca^2+ +2F-} \tag{1}$$
Now, if $s$ is the molar solubility of $\ce{CaF2}$ (in $\pu{mol dm^-3}$), then
\begin{align}
K_\mathrm{sp} &= [\ce{Ca^2+}] \cdot [\ce{F-}]^2 \\
&= s \cdot (2s)^2 \\
&= 4s^3 = 3.9 \times 10^{-11} \\
\implies s &= 2.1 \times 10^{-4} \\
\end{align}
The fluoride concentration is then equal to $[\ce{F-}] = 4.2 \times 10^{-4} ~\mathrm{M}$. The pH is then governed by the equilibrium
$$\ce{F- + H2O <=> HF + OH-} \tag{2}$$
and since
$$K_\mathrm{w} = [\ce{H3O+}][\ce{OH-}]; \quad K_\mathrm{a} = \frac{[\ce{F-}][\ce{H3O+}]}{[\ce{HF}]}$$
we find that the equilibrium constant for reaction $(2)$ is
$$K = \frac{[\ce{HF}][\ce{OH-}]}{[\ce{F-}]} = \frac{K_\mathrm{w}}{K_\mathrm{a}}$$
and hence:
$$[\ce{OH-}][\ce{HF}] = \frac{K_\mathrm{w}}{K_\mathrm{a}}\cdot [\ce{F-}]$$
If we assume now that $[\ce{OH-}] = [\ce{HF}]$ (from the stoichiometry of reaction $(2)$), and that the decrease in $[\ce{F-}]$ due to reaction $(2)$ is negligible, then
$$\begin{align}
[\ce{OH-}] &= \sqrt{\frac{K_\mathrm{w}}{K_\mathrm{a}}\cdot [\ce{F-}]} \\
&= \sqrt{\left(\frac{1 \times 10^{-14}}{6.8 \times 10^{-4}}\right) (4.2 \times 10^{-4})} \\
&= 7.9 \times 10^{-8}
\end{align}$$
Adding the $1 \times 10^{-7}~\mathrm{M}$ of $\ce{OH-}$ from the autodissociation of water,
$$\begin{align}
[\ce{OH-}] &= 1.8 \times 10^{-7} \\
\mathrm{pOH} &= -\log{(1.8 \cdot 10^{-7})} = 6.7\\
\mathrm{pH} &= \mathbf{7.3} \\
\end{align}$$

Not $10^{-7}$ mol hydroxide from autodissociation.  When you add an acid or base to water the dissociation has to be less to keep the $K_w$ product.

@OscarLanzi actually, that was from me. I edited it from the original; the original used incorrect values and obtained [OH-] = 6.32e-7. However I was not keen on keeping the incorrect values up, hence the editing. I inserted that extra assumption (1e-7 hydroxide from autodissociation) because I wanted to keep the approach simplistic, in the spirit of the original answer. (As you can see this approach also assumes the F-/HF equilibrium does not disturb the CaF2 equilibrium.)

Hmmm... But why didn't you factor the hydroxide ion concentration from the auto-ionisation of water in the intermediate steps but only in the final step? You could have factored it in by letting the final hydroxide ion concentration be (X+10^(-7)) M and the HF concentration to be X M. Then, solve for X quadratically, using the Kb expression to form an equation like you just did. Also, how do you justify your approximation?

In homework land you are right, but not in real life.
The $K_\mathrm b$ of $\ce{Ca^2+}$ ($\mathrm pK_\mathrm b = 2.43$ according to some sources) is within 1 log unit of the $K_\mathrm a$ of $\ce{HF}$.  In other words, near neutral pH, consideration of the $\ce{Ca^2+/CaOH+}$ equilibrium is almost as important as the $\ce{F-/HF}$ equilibrium.

To begin with, a disclaimer: The approach here only takes into account the equilibria that the question itself cares about, namely $\ce{CaF2}$ dissociation and $\ce{F-/HF}$ acid-base. As pointed out in DavePhD's answer as well as Linear Christmas's comment, for a realistic treatment of the system, extra parameters must be taken into account.

A more accurate calculation would involve setting up five equations for five unknowns and solving them. The first three equations come from the data you provided. All concentrations are in $\pu{mol dm-3}$.
$$\begin{align}
[\ce{Ca^2+}][\ce{F-}]^2 &= 3.9 \times 10^{-11} \tag{1} \\
\frac{[\ce{H3O+}][\ce{F-}]}{[\ce{HF}]} &= 6.8 \times 10^{-4} \tag{2} \\
[\ce{H3O+}][\ce{OH-}] &= 1.0 \times 10^{-14} \tag{3}
\end{align}$$
The next two equations are the so-called mass balance and charge balance equations:
$$\begin{align}
2[\ce{Ca^2+}] &= [\ce{F-}] + [\ce{HF}] \tag{4} \\
2[\ce{Ca^2+}] + [\ce{H3O+}] &= [\ce{F-}] + [\ce{OH-}] \tag{5}
\end{align}$$
Equation $(4)$ comes from the stoichiometry of $\ce{CaF2}$ dissociation. The total concentration of calcium-containing species, times two, must be equal to the total concentration of fluorine-containing species.
Equation $(5)$ comes from the fact that the system must be electrically neutral, i.e. the positive charges are the same as the negative charges. The concentration of calcium ions is weighted by 2 because it is doubly charged.
In this case, I just plugged it into Wolfram|Alpha, where the concentrations $[\ce{Ca^2+}]$, $[\ce{F-}]$, $[\ce{H3O+}]$, $[\ce{OH-}]$, and $[\ce{HF}]$ are represented by $p$, $q$, $r$, $s$, and $t$ respectively. The only solution that is physically sensible is the one where all the roots are positive and real. From this we find
$$[\ce{H3O+}] = r = 7.8 \times 10^{-8}$$
and hence $\mathbf{pH = 7.1}$.

How significantly would the answer change if we included the equilibrium $$\ce {F- + HF <=> HF2-}? \tag {p$K_\mathrm {a}$ = 0.58}$$ I have always steered clear of simplified calculations for $\ce {HF} $ due to this reason, and have advocated for others to do the same. Since this is a saturated solution, I would fear the effect is even more noticable. Hydrolysis of calcium ions might also be important. (I'm not really sure because of the saturation.)

@LinearChristmas I agree that it's undoubtedly more complicated than this. Should have added a disclaimer, but was in a rush... :/ I'll add one now. Obviously it is possible to simply set up more equations and solve them, but one would need the data.

If you ever have the time and find resources/data, feel free to include it (difluoride and hydrolysis). I guess it would be worth a 200 bounty from me. 400, if you also include more complicated equations to take into account departure from ideality (generalised Hückel eq-s, or sth similar). An additional 100 if you go and measure the pH and include uncertainty. (Maybe it's better to post a new question?) Anyways, the offer stands. Have fun ;}


text:
Oxidation number of hydrogen in compounds
In a given compound that contains three elements (e.g. a metal, a non-metal, and hydrogen), how can it be determined whether hydrogen is connected to the metal or the non-metal so that its oxidation number can be assigned accordingly?

You can look at the electronegativity of the atom Hydrogen is bonded too.
For instance, in NaH the electronegativity of H is greater than that of Na (~2.1 as opposed ~.9) so H two would have an oxidation state of -1)
In HCl, we observe that Cl is more electronegative so it will have an oxidation state of -1 and H will have an oxidation state of +1
In borane (even though B is a non-metal) each hydrogen displays an oxidation state of -1 (again electronegativity of H is 2.1 as opposed 2.0 of B)
Hope this makes sense.

Your examples are compounds of 2 elements but the OP is asking about compounds of 3 elements.

When hydrogen is with a metal its oxidation number is -1
When it is with a non metal element its oxidation number is +1
In this case most probably it should be -1