text:
Dipole moment - calculation of percentage ionic character
Question: The dipole moment of $\ce{HBr}$ is $2.60 \times 10^{-30}$ and the interatomic spacing is $1.41$. What is the percentage ionic character of $\ce{HBr}$?
What I know is that the percentage ionic character is observed dipole moment divided by calculated dipole moment $\times\ 100$.
The calculated dipole moment is $\textrm{charge on electron} \times \textrm{radius of molecule}$.
However, how would I convert interatomic spacing into radius?

The calculated dipole moment is charge on electron * radius of
  molecule

That's off a bit, the calculated dipole moment (calculated assuming one electron has been completely transferred from hydrogen to bromine) is dependent on the bond length (interatomic spacing), not the radius.
\begin{aligned}
\mu_{100\%} &= \mathrm{electron ~charge \times interatomic ~spacing}\\
\mu_{100\%} &= (1.6 \times 10^{-19})\ \mathrm C \times (1.41 \times 10^{-10})\ \mathrm m\\
\mu_{100\%} &= 2.26 \times 10^{-29}\ \mathrm{C\ m}\\
\end{aligned}
As you point out,
$$\mathrm{\% ~ionic~ character} = \frac{[\mu_\mathrm{obs}]}{[\mu_{100\ \%}]} = \frac{[2.6 \times 10^{-30}]}{[2.26 \times 10^{-29}]} = 0.12 = 12\ \%$$


text:
Any possible additives to a drain cleaner I use as a source of NaOH?
It is in pellets. And the box says main ingredient $\ce{NaOH}$ at least $30\%$ (I hope it's more). The box also says "aluminium catalyst". My goal was to take these pellets and dissolve them in water so that I would make a $1\textrm{M}$ solution of $\ce{NaOH}$.
I set everything up and most of it refused to dissolve in the solution. And the solubility for $\ce{NaOH}$ is pretty high. So I added some heat.This made almost everything slowly dissolve. Another thing was these small aluminium bits seemingly reacting and going up and down in the solution. I'm not sure if there was any gas released, but the solution definitely was not still.
I filtered it to remove the small aluminium balls. Any ideas If I have a relatively pure solution of $\ce{NaOH}$, or is it full of some other drain cleaner components. It is transparent-whiteish-little-yellowish. Filter kinda yellowish too.
What could these other components be?
Thanks

If it's a drain cleaner, there may be surfactants to act as detergents and possibly blowing agents to create foam (this is typically seen in the two-part cleaners where the blowing agent is in a separate compartment, and reacts to produce a gas, responsible for the foaming effect in combination with the surfactant). If you can give a full list of ingredients and/or the specific name of the cleaner you're using, it would help to give a good answer.

it says: jedyny oryginalny kret granulki there is no list of contents in a language I understand sadly. I guess it's pretty hopeless and I should start looking for some better drain cleaner. There must be some that are 100% pure, right?

Is this the product in question?

YES it is the right one. Is this magic?

Okay, luckily they provide an ingredient data sheet, and the composition looks to be very simple, so I'll write up an answer accordingly.

The product you've described contains the following ingredients (per the ingredient data sheet on the manufacturing company website):

Sodium hydroxide
Sodium chloride
Aluminium
Mineral oil

One possibility is that the aluminium is there to provide a foaming action, as its oxidation in the presence of aqueous sodium hydroxide liberates $\ce{H2}$ gas, per the following reaction (source):
$$\ce{2Al + 2NaOH + 2H2O -> 2NaAlO2 + 3H2}$$
The aluminium and sodium hydroxide would need to be kept separate and relatively anhydrous in the bottle to avoid reaction, so I wonder if the purpose of the mineral oil (which is hydrophobic) is to act as a moisture barrier and prevent mixing. If this is indeed the reaction that occurs, then the hydrated aluminium salts present are going to be at least somewhat soluble in solution, so I think it's improbable that what you have is pure sodium hydroxide. That said, if the quantity of aluminium is indeed small, this might not be a problem for your purposes.

Thank you very much. Everything makes complete sense now.

@Edza, no problem, glad it was helpful.

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text:
When are we justified in making assumptions in questions

The density at $20^\circ\mathrm{C}$ of a $\pu{0.500 M}$ solution of acetic acid in water is $\pu{1.0042 g/mL}$. What is the molality of the solution? The molar mass of acetic acid is $M(\ce{CH3CO2H})=\pu{60.05 g/mol}$.
The density at $20^\circ\mathrm{C}$ of a $\pu{0.258 M}$ solution of glucose in water is
  $\pu{1.0173 g/mL}$, and the molar mass of glucose is $M(\ce{Glc}) = \pu{180.2 g/mol}$. What is the molarity of the solution?

In question one we can assume a 1 liter solution and in question two we can assume a 100 gram sample. I understand these assumptions are needed to find other information, but when are we justified in making these assumptions?

You can make certain assumptions in these and similar problems because concentration of uniform substances (whether it's expressed as molarity, molality, mass density, or something else entirely) is an intensive property, i.e., it has no dependence on the total amount of material in the object or sample being measured. For example, if you have a homogeneous solution, the solute particles are distributed uniformly among the particles of solvent, and the two form a constant ratio. If I split the sample in half, the solute particles are halved, but so are the solvent particles, so the ratio remains the same.
As it happens, there's no need to assume any particular quantities to solve this problem; instead, you can use algebraic substitutions using the known values given and common definitions to solve for either molarity or molality directly. Let $M = \frac{m_{solute}}{n_{solute}}$ be the molar mass of the solute, $\rho = \frac{m_{solution}}{V_{solution}}$ be the mass density of the solution, $c = \frac{n_{solute}}{V_{solution}}$ be the molarity, and $b = \frac{n_{solute}}{m_{solvent}}$ be the molality, and note that $m_{solvent} = m_{solution} - m_{solute}$. Then, we have:
$$
\begin{align}
b\ & = \frac{n_{solute}}{m_{solution} - m_{solute}}\\
 & = \frac{n_{solute}}{\rho \cdot V_{solution} - M \cdot n_{solute}}\\
 & = \frac{n_{solute}}{\rho \cdot \frac{n_{solute}}{c} - M \cdot n_{solute}}\\
 & = \frac{c}{\rho - M \cdot c}\\
\end{align}
$$
You can rearrange this equation to solve for $c$ (molarity). Note, of course, that units need to match, but it's trivial to convert between grams and kilograms, mililiters and liters, etc.

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text:
How to calculate melting/boiling points at different pressures?
Is there an equation to calculate the melting or boiling points of elements or compounds at a given pressure?
I know water boils below room temperature in a vacuum. I'd like to know if gallium, a metal which melts at 29.77 °C under atmospheric pressure, would be in a liquid state at room temperature if stored under vacuum.

A pressure-Temperature nomograph such as this one by Sigma-Aldrich might help this and related questions.

I don't believe there is an equation that you can use for melting points of a general substance as a function of pressure (since the melting phase transition has a lot to do with the geometry of the molecule and the structure of the solid), but there is one for the boiling point of any pure substance when you are not near the critical point.
The liquid-vapor transition follows a very specific curve on the pressure-temperature plane of a PVT diagram. This is given by the Clausius-Clapeyron relation, which at temperatures and pressures that are not close to the critical point, can be approximated as:
$$
\ln\frac{P_2}{P_1} = \frac{-\Delta H_\mathrm{vap}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)
$$
This is the two-point form, which lets you predict the pressure of the vapor phase at a given temperature if you know the temperature and pressure at another point, and the enthalpy of vaporization.
You can use this to predict boiling points as a function of pressure by recognizing that:

The boiling point is the temperature at which the vapor pressure at equilibrium exceeds atmospheric pressure

This means that if you know the boiling point of a substance at 1 atm (the normal boiling point), and you know $\Delta H_\mathrm{vap}$ for that substance, you can predict the boiling point at another pressure by plugging in values for $P_1$ and $T_1$ (the normal boiling point), $P_2$ (the pressure you are interested in), and $\Delta H_\mathrm{vap}$, the enthalpy of vaporization (also sometimes called the latent heat), then solving for $T_2$ (the new boiling point.)
Like I said, this only works when you aren't near the critical point, but for most applications that's not a concern.
For your specific case, especially since you are probably more interested in the liquid-solid transition at this temperature, you are better off looking at a phase diagram.
Here's one for gallium (source) (original source):

The solid-liquid line has a negative slope at low pressure (as ron pointed out) which means the melting temperature increases as pressure decreases. Depending on how high the vacuum is, and on what temperature "room temperature" is, you might be in the solid phase — but the slope is steep, so even at 0 atm the melting point is still right around 300 K.

There is no general equation for melting points and boiling points that applies to all elements. However, the melting point for gallium has been studied in considerable detail because gallium's melting point has been adopted as a calibration standard. Isotech, a calibration company posted the following on p. 3 of the report [1]:

2.3 The temperature of the melting point equilibrium of pure gallium depends upon the ambient pressure of the surrounding environment, and varies with atmospheric pressure expressed upon the surface of the gallium column as follows:
t C = 29.7646 +/- (-0.002 C per std. atmos.) Eq. 2
(The negative sign of the coefficient is a result of the solidus expansion (liquidus contraction).

Perhaps you could contact Isotech to get further clarification on their formula.
Reference

Henry E. Sostmann. A Technical Guide and Standard for the Use of Gallium Melt-Point Cells; Isotech, 2010. (WayBackMachine — PDF)

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Quaternary ammonium anti-microbials

Also good against fungi, amoeba, and enveloped viruses,[7] quats are believed to act by disrupting the cell membrane.

Quaternary ammonium cleaners have long-alkyl substituent chains. Why is this so?
Does this have anything do with increasing the ability of the cleaners to penetrate the phospholipid bilayer, which has its non-polar portion oriented outward and its polar phosphate backbone oriented inside the cell? Does the alkyl portion dissolve into the lipid bilayer and wreck havoc?
What is the exact mechanism?

A phospholipid has a charged, hydrophilic phosphate head and a neutral, hydrocarbon hydrophobic tail.  In a phospholipid bilayer, two phospholipids combine tail-to-tail such that the two outer, or exposed, surfaces are comprised of the polar phosphate heads.  The bilayer can function as a cell membrane; it can keep what is in "in" and what is out "out".  It can also regulate critical cell functions such as pH and ion balance through the incorporation of proteins in the bilayer that can serve as ion pumps.  Clearly, any disruption to the bilayer could allow the cell contents to leak out and bad stuff to leak in, which would lead to either cell inactivation or cell death.
The mechanism of action of quaternary salts on lipid bilayers is still an active area of investigation, but here is a simplified overview of what is thought to be happening. .  The first step is thought to involve the electrostatic interaction of the positively charged ammonium ion with the exposed, negatively charged bilayer surface, followed by incorporation of the hydrocarbon chain of the ammonium ion into the bilayer. The electrostatic interaction affects membrane surface pressure which, in turn, disrupts cell function.  Shorter chain lengths (n=12-14) are more effective against gram-positive bacteria, while longer  chain lengths (n=14-16) are more effective against gram-negative bacteria.
Here's a reference to a recent thesis (2012) that studies the resistance of bacteria to quaternary ammonium compounds.  Section 1.2 (pages 2-5) provides a clear, concise summary of the latest thinking on the subject.


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Oxygen production from thermal decomposition of metal oxides
Does thermal decomposition of metal oxides always produce oxygen?

A lot of metal oxides decompose under liberation of $\ce{O2}$ when subjected to sufficiently high temperatures. For example, $\ce{HgO}$ decomposes at temperatures above $\pu{500^\circ C}$ into elemental mercury and oxygen:
$$\ce{2HgO ->~ 2Hg + O2}$$
However, disproportionation is another possible reaction pathway. An example is the thermal decomposition of iron(II) oxide to elemental iron and iron(II, III) oxide at temperatures below $\pu{575^\circ C}$ ($\ce{FeO}$ is more stable above that temperature; sources: 1, 2):
$$\ce{4FeO \rightleftharpoons Fe + Fe3O4}$$
Another example is the decompostion of $\ce{SnO}$ at temperatures between $700\mathrm -\pu{1000K}$, which proceeds via an intermediate $\ce{Sn3O4}$ mixed oxide. Tin disproportionates from the oxidation state +II in $\ce{SnO}$ to $0$ in $\ce{Sn}$ and +IV in $\ce{SnO2}$ (1, 3).
\begin{aligned}
\ce{4SnO ->~& Sn3O4 + Sn} \\
\ce{Sn3O4 ->~& 2SnO2 + Sn}
\end{aligned}

In addition to the examples given by Jannis Andreska, metal peroxides, such as barium peroxide and lithium peroxide also release oxygen upon heating:
$$
\ce{2 BaO2 ->[\text{T = 700 °C}] 2 BaO + O2} \\
\ce{2 Li2O2 ->[\text{T = 195 °C}] 2 Li2O + O2}
$$
This was once used to produce pure oxygen.
On the other hand, some, if not most, metal oxides are extremely heat resistant. For example, once sintered at temperatures between $1700$ and $\pu{2000°C}$, magnesium oxide $(\ce{MgO})$ can be heated up to its melting point ($\sim\pu{2800°C}$) without decomposition and can thus be used as a lining for heat sensors.


text:
Is there a database which give reaction constant?
For instance, I have a chemical reaction like :
$$\ce{Fe^{2+} + $0.5$O2 + H^{+} ->Fe^{3+} + $0.5$ H2O}$$
And I would like to use Action-Mass law, but I need to have equilibrium constant for this reaction. And this for many other reaction.
That's why I'm wondering if there exist a nice database with this information ?

There are lots of reaction kinetics databases. A cursory search on the NIST databases for reaction kinetics didn't give me anything for this reaction but feel free to take a look.
(http://kinetics.nist.gov/)

The information you are looking for is out there, but sometimes you might have to work a little for it.
The example reaction you post is a redox reaction. If we split the redox reaction into its two half-reactions, we can look up their reduction potentials (easier to find sometimes than equilibrium constants) and use the Nernst equation to calculate the equilibrium constant. I use Wikipedia for my list of standard potentials, but only because the values are identical to those in textbooks that I own.

Oxidation half-reaction

$$\ce{Fe^{2+} -> Fe^{3+} + e-} \ \ \ E^\circ_{ox}= -0.77 \ \text{V}$$

Reduction half-reaction

$$\ce{O2 + 4H+ + 4e- -> 2H2O} \ \ \ E^\circ_{red} = +1.229 \ \text{V}$$

Full balanced redox reaction

$$\ce{4Fe^{2+} + O2 +4H+ -> 4Fe^{3+} + 2H2O}$$

Cell potential

$$E^\circ_{cell}=E^\circ_{ox} + E^\circ_{red} = -0.77 \ \text{V} + 1.229 \ \text{V} = 0.459 \ \text{V}$$

Nernst Equation

$$E^\circ_{cell}=\frac{RT}{zF}\ln{K_{eq}}$$
$$k_{eq}=e^{\dfrac{E^\circ_{cell}zF}{RT}}$$
In this equation, $z$ is the number of electrons transferred, $T$ is the temperature in kelvins, $R$ is the ideal gas constant, and $F$ is the Faraday constant. At 0 Celsius (273.15 K):
$$K_{eq}=e^{\dfrac{0.459 \ \text{V} \cdot 4\text{ mol}\cdot 96.49\times 10^3 \frac{\text{C}}{\text{mol}}}{ 8.314\ \frac{\text{J}}{\text{K}\cdot \text{mol}}\cdot 273.15\ \text{K} }}=7.565\times 10^{33}$$
If you can look up the enthalpy change or free energy change for the reaction (many of which are available at webbook.nist.gov/chemistry, then you can use a variant of the van 't Hoff equation
$$\Delta_r G^\circ = -RT\ln{K_{eq}}$$
$$K_{eq}=e^{\dfrac{-\Delta_R G^\circ}{RT}}$$


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Which is more soluble in water and why: K₂HPO₄ or CaHPO₄?
If I dissolve $\ce{K_2HPO_4}$ and $\ce{CaHPO_4}$ in water (two different reactions), the latter shows better solubility. Is it because $\ce{Ca^{2+}}$ has ha higher valence of $2$ than $\ce{K^+}$, which has only $1$? Or has it something to do with the lattice enthalpy?

The charge and size of an ion influences both lattice stability and stability of the hydrated ion.

The less stable the lattice the more soluble the ion (less energy required to disrupt the lattice).  The more highly charged and smaller the ion is, the larger are the electrostatic repulsions within the lattice which will act to destabilize the lattice.
The more stable the solvated ion, the more soluble the ion.  The higher the charge on the ion and the smaller its radius, the more stable the solvated ion.

The size of the potassium ion is around 152 pm, the size of the calcium ion is around 114 pm.  So the calcium ion is both the smaller and the more highly charged of the two.  Therefor we would expect the calcium ion to have the less stable lattice and be more highly solvated (stabilized) when in solution.  Both of these factors would lead us to expect $\ce{CaHPO4}$ to be more soluble than $\ce{K2HPO4}$


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Schrödinger equation and wave function
What is the Schrödinger wave equation?
What are radial probability functions, nodes-angular and radial and electron cloud?
I am really confused with all these terms and all my book seems to offer me is a set of very complex words and definitions. I am relatively new to atomic structure and related things, so a brief answer in simple words would really be helpful!
I also would like to know more about Hamiltonian and Laplacian operators and related subjects.

Google and Wikipedia would be an easy way to clarify these definition-based issues.

What book are you using? The fundamentals of quantum theory are already very complex and a definition in simple terms is often difficult and almost always incomplete. "For those who are not shocked when they first come across quantum theory cannot possibly have understood it." (Niels Bohr) "If you think you understand quantum mechanics, you don't understand quantum mechanics." (Richard Feynman)

There are a lot of questions in this one post which is probably why there has not yet been an attempt to answer this post.
1.) What is the Schrödinger wave equation?
The wave equation (i.e. wave function) is a function which contains all the information about a chemical system.  With an appropriate operator, one can extract this information and gain insight into the system.  For example, the Hamiltonian operator will extract the energy of a system from the wave function.  The wave function is an eigenfunction while the energy (in this example) is the eigenvalue of the eigenfunction.  For example, in the Schrödinger equation
$$ \mathcal{H}\Psi = E\Psi$$
$\mathcal{H}$ is the Hamiltonian operator, $\Psi$ is the wave function (eigenfunction) and $E$ is the energy (eigenvalue).  Notice that by operating on $\Psi$ with $\mathcal{H}$ returns $E$ AND $\Psi$, which by definition, makes this an eigenfunction-eigenvalue equation.
2.) What are radial probability functions?
These are probability distributions for the probability of an electron residing some distance from the nucleus of an atom such as the hydrogen atom.  Consider this plot of various radial probability distribution functions of the hydrogen atom.

Here we see that the probability of an electron (in the hydrogen atom) in a 1s orbital is zero at the nucleus (at 0,0) and a maximum a hair shy of $2 a_0$.  Interestingly an electron in a 2p orbital has a small probability of residing close to the nucleus ($< 1a_0$) but is more likely to reside further away from the nucleus than an electron in a 1s orbital.  This makes sense given that a 2p orbital is higher in energy because 'it' is further away from the nucleus than a 1s orbital.
3.) What are nodes in radial and angular functions?
A node is where the function is zero at a particular point.  This simply means in a physical sense that at this point (at the node), the probability of an electron existing at that point is ZERO.  Nodes are a result of the separation of variables technique when solving the appropriate equation(s). When any one component (or variable) is zero then the resulting function will be zero at that point.
4.) What is electron cloud?
Electron 'cloud' is just a nicer away of saying electron density.  Electrons are not strictly particles.  Electrons do not just 'orbit' a nucleus.  Electrons also exhibit wavelike behavior.  (see: double-slit experiment - wave-like behavior and J. J. Thompsons cathode ray tube experiments - particle-like behavior).  The 'cloud' is obtained simply by 'squaring' the wave function.  More technically, its taking the product of the wave function and its complex conjugate.  The 'cloud' is just a probability distribution of where an electron is likely to be (usually given in 3-dimensions).
5.) What is the Hamiltonian operator?
The Hamiltonian operator is the energy operator to the Schrodinger equation.  Because it is Hermitian, it will return a 'real' result (as opposed to a complex result).  There are different versions of the Hamiltonian depending upon what approximations you want to make.  The full Hamiltonian operator can be written as (in terms of atomic units)
$$\mathcal{H} = -\sum\limits_{i=1}^{N}\dfrac{1}{2}\nabla_i^2
              -\sum\limits_{A=1}^{M}\dfrac{1}{2M_A}\nabla_A^2
              -\sum\limits_{i=1}^{N}\sum\limits_{A=1}^{M}\dfrac{Z_A}{r_{iA}}
              +\sum\limits_{i=1}^{N}\sum\limits_{j>i}^{N}\dfrac{1}{r_{ij}}
              +\sum\limits_{A=1}^{M}\sum\limits_{B>A}^{M}\dfrac{Z_A Z_B}{R_{AB}}
$$
Each term on the right hand side of the equation (going from left to right) is defined as

the kinetic energy of the electrons
the kinetic energy of the nuclei
the potential energy of the electron-nuclear attractions
the potential energy of the electron-electron repulsions
the potential energy of the nuclear-nuclear repulsions

We can write the Hamiltonian using a bit more compact notation such as
$$
\mathcal{H} = \hat{T}_{e} + \hat{T}_{N} + \hat{V}_{eN} + \hat{V}_{ee} + \hat{V}_{NN}
$$
where negative signs are assumed for each $\hat{T}$ operator.  The terms denoted here correspond to the respective terms in the equation before this one.
If we take into account the Born-Oppenheimer approximation (the B.O. approximation assumes that the electron motion is orders of magnitude larger than the nuclei they belong to... so much faster that the nuclei are, relatively, standing still with respect to the electrons), we can make two changes to the full Hamiltonian giving us the electronic Hamiltonian.
$$
\mathcal{H} = \hat{T}_{e} + \hat{V}_{eN} + \hat{V}_{ee} +\ \mathrm{constant}
$$
The nuclei are treated as frozen in space when the B.O. approximation is applied for reasons explained above.  Therefore the kinetic energy of the nuclei goes to zero.  The nuclear-nuclear repulsion is now a constant because the nuclei are fixed.  The repulsion will never change.
6.) What is a Laplace operator?
The Laplace operater (or Laplacian) is simply a differential operator of the form
$$
\nabla^2 = \dfrac{\partial^2}{\partial x^2} +  \dfrac{\partial^2}{\partial y^2} +  \dfrac{\partial^2}{\partial z^2}
$$
if you are working with Cartesian coordinates. This is simply a set of second-order derivative operators within the kinetic energy operator.

While I may have addressed each question, there is a lot more depth to these answers that lie well beyond the scope of this post.  I encourage you to find more detailed and insightful answers in well-established texts.  I highly recommend Szabo & Ostlund for a solid introduction to quantum mechanics.  You can snag a paperback version of this excellent work for about $10.


text:
Why is a ketone more nucleophilic than an ester?
I guess the ester is a weaker nucleophile because it does have an additional oxygen atom, unlike the ketone, that is pulling electrons from the C-O double bond towards the carbon atom (this happens vice versa too of course with the ester oxygen). This lowers the ionic character of the C-O double bond (smaller $\delta^{+}$ and $\delta^{-}$ charge on the carbon and oxygen atoms) and thus increases the energy of the C-O $\pi^{*}$ bond, making it less nucleophilic.   Is this correct?

The ester carbonyl carbon is a stronger nucleophile and less prone to nucleophilic attack than the carbonyl carbon in a ketone. I think you are trying to understand why the carbonyl in a ketone typically reacts faster with a nucleophile than the carbonyl in an ester.  Look at the resonance structures drawn below.  Both the ketone and ester have a resonance structure that places positive charge on the carbonyl carbon, however you can also draw a third resonance structure for the ester that removes some of the positive charge from the carbonyl carbon and places it on the ester oxygen. This makes the ester carbonyl carbon less positive (more negative, more nucleophilic, less prone to attack by a nucleophile) than the ketone carbonyl carbon (more positive, less nucleophilic, more prone to attack by a nucleophile).  A nucleophile has electrons it would like to share and will react preferentially with centers of low electron density (more positively charged).  Therefor a nucleophile will react faster with a ketone carbonyl than an ester carbonyl.

EDIT
I like Martin's posts, they make me think - and that is always a good thing. Here is where my thinking has taken me:

I believe Martin's approach (basically frontier MO) is fundamentally correct.  I would be more convinced that his actual numbers are meaningful if he were to calculate 1) the carbonyl carbon LUMO coefficient and 2) the HOMO-LUMO separation for the following series of compounds in order to see if his calculated values correlated with this order of reactivity of carbonyl compounds towards nucleophilic attack (we could use water as the nucleophile) - but Martin, I think such calculations might be a poor use of your time and resources.

acyl halide > acid anhydride > aldehyde > ketone > ester ~ carboxylic acid > amide > carboxylate ion

My argument based on how resonance affects the charge density at the carbonyl carbon is one approach used to explain ("teach") the above order of reactivity.  See for example page 4 on these UCLA notes.  I was just trying to keep it simple.
The issue I have with Martin's approach is, what should the reader do next week when he wants to know if an acid chloride reacts faster or slower than an ester.  While the resonance approach may not be rigorous, it does yield the correct order of reactivity and allows the user to make predictions while only using pencil and paper.
I am considering further editing of this post to use the resonance approach to explain differences in ground state stabilization of the carbonyl series.  Then, since they all go (more or less) to the same tetrahedral intermediate, one could explain the reactivity order based on differences in Eact.  This is another textbook method used to explain the carbonyl reactivity pattern.
Finally, what is the "correct" answer?  Is it the real truth as nature knows it, but in a form that is difficult for the common individual to use and extend; or is it a rough approximation of the truth that can be easily used?  I think the answer to that question depends upon who is asking the question, who is the audience.

However, the enolate of a ketone will be more nucleophilic than the enolate of an ester (for almost all of the same reasons).

Thank-you Ron. I completely forgot about this question until I saw it bumped up today. I think the original problem I had with this explanation is that the resonance structure where the positive charge is on the ester oxygen does not block the mechanism of attack. On nucleophilic attack the negative charge would just "go" to the ester oxygen instead of the carbonyl oxygen. Instead of the I guess you can only explain it then with that the LUMO is more delocalized in the ester, increasing its energy.

@ron Will an amide be more nucleophilic than ester? Also, please explain electrophilic order for an ester, 0° amide, 1° - amide, 2* amide. Please reply.

Looking at the partial charges of esters and ketones, unfortunately ron's answer is only half true. For a simple model I have chosen 3-pentanone and ethyl acetate. You can see, that the carbonyl carbon in the ketone has a smaller positive charge ($q=0.6$) than in the ester ($q=0.8$), so the latter should be more prone to nucleophilic attacks.
This is not the case and the explanation lies within the molecular orbitals. The LUMO of the ketone has a slightly lower ($-1.57~\mathrm{eV}$) energy than that of the ester ($-0.87~\mathrm{eV}$). More important, the coefficient in this orbital is larger at the carbonyl carbon in the ketone compared to the ester. And this orbital will be attacked by the nucleophile, which usually has a high lying HOMO. So the LUMO of the ketone will be more accessible compared to the ester (smaller energy difference).
Another important fact is, that the ester is much more rigid than the ketone as it has one low lying MO that spans the $\ce{O=C-O}$ moiety (MO20), which is obviously not present (as it cannot be) in the ketone (MO18). You can also see, that the $\pi$ contribution of carbon in the ketone is lower.

Thank-you Martin. I completely forgot about this question until I saw it bumped up today. Why is the ketone LUMO lower in energy than the ester LUMO? Because of the delocalization I suppose. The energies you list are relative to what precisely?

Martin, I read this thread once before and then came back to it after some time. I believe that you had posted pictures of the MO's from your simulation, now it looks they are no longer there?

@MikeMurphy I don't know, at my end it displays fine. Maybe there was some issue with imgur. In any case, there is nothing I could do about it. Did you experience it anywhere else in the site? You can also check with other users in [chat]. I hope it works again for you now.


text:
Ethanol and water hygroscopic equilibrium concentration
I am working with a company to help improve their reaction process where they use SDA 2B Alcohol (Karl Fischer titration has it at 99.66% EtOH, 0.3375% toluene). Specifically, I am attempting to determine what theoretical %wt or %vol of water is pulled from atmospheric air into pure ethanol given an infinite amount of time? Does it ever reach an equilibrium state and, if so, what is it?
My gut guess is ~95.5%wt EtOH/4.5%wt H2O, the azeotropic concentration.
Assume constant relative humidity and room temperature. The ethanol is stored in a closed tote (~220 gallons). Initially, the tote is full of as-shipped ethanol (negligible air volume). However, half the ethanol is removed as part of a batch run. So, really, what I'm trying to get at is what happens to the leftover ethanol that is now sitting in a closed tote with ambient conditions.
Please let me know if additional details will help answer the question. Thanks!

Following gut feelings it is much more like a "equilibrium concentration as a function of temperature, humidity level", and most probably around room temperature it doesn't do anything with the azeotropic concentration. Pleas,e also note that the "equilibrium situation" also assumes plenty of alcohol in the air, which happens most probably faster than the humidity going into alcohol.

@Greg I believe this would suffice as an answer. Humidity won't be drawn in, but alcohol will certainly vaporize.

You have a closed system with (say) 100 gallons of (say) pure ethanol and 100 gallons of moist air.  There is not enough water in the air to go into the ethanol and reach the azeotropic ratio -- not by 2 or 3 orders of magnitude (see below).
If you know the relative humidity of the air at the time the tote was opened (and the temperature), you can find the absolute humidity (the mass % of water vapor in the air) and so estimate the maximum amount of water that went into the tote.  Since 100 gallons is about 20 times the volume of a mole of gas, assuming an absolute humidity of 10% (it's very likely way less) the answer is about 2 moles, or about 40 grams. In equilibrium (in the closed system) almost all of the water vapor will very soon go into the ethanol, so you have 100 gallons of ethanol (with a bit of toluene), plus 40 g of water.  That's 0.01%.  Whether this much water ruins your alcohol or not, you will know better than anyone.
Every time you open the tote, you're potentially letting in 40g of water.
I say "potentially" because, since the gas content is not instanty exchanged, much less water than that might go in.  But still, if you open it often, the percentage will keep going up -- asymptotically to the azeotropic ratio, but since the alcohol will also evaporate, as has been pointed out, you probably won't ever get near that ratio.

Hmm, as the alcohol evaporates the approach to a higher (azeotropic ratio) will increase I think.  Granted the amount of air (with water) that enters when opened is small but some alcohol is leaving and diffusion will occur.  The amount of air entering when pouring/pumping out alcohol however may be the half or even more if batch sizes are bigger and less than half remains sometimes.  About 50g/1000l @ 40degC is the maximum you can get in with foggy tropical air.


text:
Can two positive ions be next to each other in a formula?
Let's say I want to find the oxidation numbers of the atoms in a chemical formula.
For example: In $\ce{H3AsO3}$,
Oxygen has a $-2$ charge. They're $3$ of them, so its $-6$
Hydrogen has a $+1$ charge. They're $3$ of them, so its $+3$.
So arsenic must have a charge of $+3$ for all the charges to add up to be $0$? Correct?
However, my professor keeps telling me to keep the positive oxidation numbers away from each other. I thought it only applied when drawing Lewis structures or resonance structures.
Is that a rule when writing a formula?

He could have told that two vowels can't be next to each other in a word. That's about as reasonable.

Can two positive ions be next to each other in a formula?

Yes, there are many examples of this, $\ce{HNO3}$ for instance

So Arsenic must have a charge of 3+ for all the charges to add up to
  be 0? Correct?

Yes, that's correct

However, my professor keeps telling me to keep the positive oxidation
  numbers away from each other. Is that a rule when writing a formula?

It's not a rule that I'm familiar with.  Here is a link to a nice, concise listing of the rules for writing formulas for inorganic compounds.

I thought it only applied when drawing Lewis structures or resonance
  structures.

Yes, it certainly applies in those two cases

I think in $\ce{HNO3}\equiv\ce{(HO-)N(=O)2}$ no two positive partial charges are next to each other, so that would not provide an example, but $\ce{HNO}$ would.

So would you say that I can write $\ce{H3AsO3}$ as $\ce{AsH3O3}$ and it wouldn't make a difference because my Lewis diagram would remain the same?

@Martin  I meant it in the nomenclature sense.  In other words, in the molecular formula an $\ce{H}$ (+1) is written next to an $\ce{N}$ (+5)

@Mark  No, see the rules for constructing molecular formulas in the link I provided.

@Mark There are the official rules for writing sum formulas, like in the link that ron provided. There are some other ways of writing them, like the Hill system that is quite common in Organic Chemistry. Also the Lewis formula is only one representation of a molecule, and there are a lot of examples where this approach fails (see hypercoordinate species).// ron - I guess I did not see that it is a question about the notation, I thought this would be a question about actual charges.

@Martin  In the title of Mark's question he writes "in a formula" so that's how I approached it.

You are a far better reader than me ;) After your comment, I revisited the title and it was obvious after that, and then you got my vote...


text:
How can the density of aqueous solutions be maximised?
What is the maximum possible density of aqueous solutions at room temperature?
The density of the salty water in the Dead Sea at 37°C is 1.24. Would it be possible to increase the density using salts we add to our food, or possibly by using different salts than are present in the sea?

I believe in your first sentence is a temperature too much and a density missing. What do you mean with homesalts? And I do not understand what "... and bringing some from out?" means. Could you please take a look and make your question a little clearer. Thank you.

True. What I meant was what salts could I use to increase the density would the salt that we add in food work or wel have to add the diffetent salts present in the sea.

From what I have found, calcium chloride seems to be one of the better accessible options, with a maximum density of almost 1.4kg/m^3. https://www.engineeringtoolbox.com/density-aqueous-solution-inorganic-chlorides-salt-concentration-d_1955.html

The title should really be "how can you maximise the density of water based solutions". It isn't pure water if there are lots of dissolved salts.

Clerici solution is a solution of equal parts of thallium formate and thallium malonate in water, density is 4.25 g/ml,     Thoulets solution is solution of potassium-mercury iodide in water with density 3.2 g/ml. Don't even think of using them.
Sodium polytungstate $\ce{Na6[H2W12O40]}$ is a relatively safe compound and allows densities up to 3.1 g/ml. $\ce{ZnCl2}$ allows densities up to ~ 1.9 g/ml, but has acidic solutions.
Unfortunately, common nontoxic salts usually either are not dense enough or not soluble enough to make really heavy liquids, but densities around 1.1-1.2 g/ml are quite easy to achieve even with common salt. Note, however, that concentrated salt solutions 'siphons' water from human body through damaged skin and mucous membranes, and to some extent even through undamaged skin, which may lead to desiccation, trauma of mucous membranes and very unpleasant sensations.

I wonder maybe sodium iodide would be a good option to go well beyond Dead Sea density with minimal fuss? It's dense, it's very soluble, it's fairly innocuous, it's a nutrient in microscopic quantities. Don't drink it or add it to food yourself, of course, it is not "salts we add to our food" in that sense.

@SteveJessop Iodide solutions oxidize on air. Other than that, it gives solutions with density up to 1.8 g/ml. Pretty decent.

@prrmeakra I think the crystal structure of sodium polytungstate would be awesome.

I think they use epsom salt (sodium sulfate) in the commercial isolation tanks designed to allow a person to float in water (i.e. salt solution) and meditate without having any tactile input (a phenomenon of the 70s, but I think there still are some around).
The more salt added to the water, the higher the density. Also, some salts are denser than others - so a heavy metal salt like cesium chloride would give a heavier water solution than say sodium chloride.

Epsom salt isn't sodium sulfate.


text:
What is molecular system in general terms? What is a supramolecular system?
What is a molecular system in general terms?
Could the solution of sugar in water be called a molecular system, if we were not interested in sugar-water interaction?
Today Jean Marie Lehn told me that bulk water is classical supramolecular system, but I cannot understand why.

Molecular chemistry is based predominantly on interactions through covalent bonds. It looks at molecules as individual entities, and is limited in its scope of accomodating intermolecular interactions. Your solution of sugar in water could easily be studied as a molecular system, if all you are interested in is the sugar molecule itself. Subtle differences in molecular conformation of the sugar in different solvent systems, for instance, would simply be written off as 'solvent effects'.
Supramolecular chemistry is based on intermolecular interactions. Typically, these are based on weak interactions (non-covalent forces), and include:

ion-ion interactions
ion-dipole interactions
dipole-dipole interactions
hydrogen bonding
van der Waals forces
pi stacking
hydrophobic interactions
and probably some others

It is through these types of interactions of that we can explore phenomena such as host-guest association, self-assembly, metal-ligand chelation, micelles, and even the structure of DNA, as all of these systems rely on the weak intermolecular interactions outlined above. You could also study your aqueous sugar system as a supramolecular system, and here you could consider what predominant intermolecular interactions are at work, and how they affect the overall chemistry of the bulk system.
Water is supposed to exist as water clusters, weak oligomeric and polymeric forms of $\ce{(H2O)_{n}}$


text:
Oxidation numbers of phosphorus pentachloride
I read detailed explanations of what oxidation numbers/ states represent and it boils down simply to this: If it's positive, it shows how many electrons the element lost, and in the case of a negative oxidation number, it shows how many electrons the element gained. That I understand.
What I don't get is, in Phosphorus pentachloride $\ce{PCl5}$, phosphorus forms 5 single bonds with chlorine atoms and 'gains' $5$ more electrons to accommodate a total of $10$ electrons. Shouldn't its oxidation number be $-5$ instead of $+5$?

When considering the oxidation state of covalent compounds, we need to pretend they are ionic. When we do so, we assign all of the electrons to the more electronegative element. The electronegativity of phosphorous using the Pauling Scale is 2.19 and the electronegativity of chlorine is 3.16. Since chlorine is the more electronegative element, we assign all electrons in the bonds to the chlorine atoms. Thus $\ce{PCl5}$ is treated as the fictitious ionic compound comprised of $\ce{P^{5+}}$ ions and $\ce{Cl-}$ ions.


text:
Racemic mixture of meso compound
My solution manual implied that we can have a racemic mixture of a meso compound.
Is this possible?
I was able to rotate the molecule to make it symmetrical. Sure, the bromination in the nucleophilic solvent gives you an anti product, but from my understanding, if you can rotate the molecule so that it is symmetrical, the molecule can be meso.
I'm thinking no because meso compounds by definition have no enantniomers and a racemic mixture has both enantiomers.

This question and answer highlight what is sadly too often the case - solutions manuals are not edited very well.

You are correct and you enumerated the reasons why.  A racemic mixture is a mixture of d and l enantiomers in equal proportion.  A meso compound has no d or l isomers, it is achiral.


text:
How does a Microbial Fuel Cell work?
I am doing a science fair experiment about Microbial Fuel Cells and was wondering how exactly do they work?
I know the basics, that they work similarly to a normal Fuel Cell through oxidation and reduction but am still a little unclear on the full details.

A microbial fuel cell (MFC) is an electrochemical cell that converts chemical into electrical energy, and in which one redox half-reaction is catalyzed by enzymes of microorganisms. The fuel cell consists of anode and cathode compartments which are separated by a membrane that is only permeable for specific cations. At the anode, bioorganic fuel is oxidized to $\ce{CO2}$ and $\ce{H3O+}$. The gained electrons flow from the anode through an external electrical circuit to the cathode, where they are consumed together with protons by the reduction of $\ce{O2}$ to $\ce{H2O}$. The $\ce{H3O+}$ ions are transferred from the anode to the cathode compartment through the membrane.
The fuel is a medium rich in nutrients which can be oxidatively degraded by the microbes, like, for example, sugars (in the following formula, $\ce{C12H22O11}$ stands for a variety of disaccharides):
$$\ce{C12H22O11 + 13H2O -> 12CO2 + 48H+ + 48e-}$$
The electrons from this half-reaction are either transferred directly to the anode (mediator-free MFC) or via a mediator (mediator MFC), which integrates into the electron transport chain of the cell and transfers electrons to the anode by passing the cell membrane. The mediator gains and loses electrons in every transport cycle, and switches between its reduced form (laden with electrons from the cell) and oxidized form (has given up electrons at the anode).
The scheme below (image source: MFCGuy2010/Wikimedia Commons) shows a MFC which uses soil as biofuel. Nutrients from the soil, like acetic acid, are oxidized by microbes in the anodic biofilm to $\ce{CO2}$ and $\ce{H3O+}$. Protons are then consumed in the reduction of supplied oxygen at the cathode, and electrons flow through the external circuit from anode to cathode. In this case, the soil acts simultaneously as nutrient medium and proton exchange membrane.


text:
how are the frequencies at a local maximum of PES like?
On the potential energy surface, if you find a local maximum and calculate its frequencies in Gaussian or something like that, will you get all negative frequencies or all positive frequencies?
I know that at a local minimum the frequencies are all positive and that at a saddle point there are both negative and positive frequencies, but what about a maximum point?

The potential energy surface (PES) is a very complex thing and it is hard to be understood (if we can understand it at all). With quantum chemical approaches we can map out this surface. We differentiate between positive, real and imaginary (usually shown as negative) frequencies. There are no negative frequencies. What you may encounter analyzing the PSE, are one of the following extreme points.

Local/Global Minimum
All derivatives have to be positive, the number of imaginary frequencies (NImag) is zero $i=0$. This corresponds to an observable state of the system/ molecule, i.e. it can be trapped and analysed with spectroscopy. Exciting any of these frequencies, the energy of the molecule will increase. But you already knew that.
Transition State
A transition state is a first order saddle point on the PES. The Nimag is exactly one, $i=1$. It connects two local minima with each other. Try to imagine a molecule, like ammonia. This can undergo inversion, i.e. nitrogen inversion. The two pyramidal forms are local minima, which are connected via a planar form (nitrogen is $\ce{sp^2}$ hybridised), which is the transition state. At that point one of the frequencies is imaginary, which means, that it corresponds to lowering the energy while exciting this frequency. All other vibrations would still increase the energy of a molecule.
You cannot observe a state like this via common spectroscopic methods, but it still has a physical meaning. There are various theories in use to describe reactions of molecules. The most popular is the Transition State Theory, which was developed by H. Eyring and M. Polanyi.
Second Order Saddle point
It formally corresponds to a transition between transition states. Its NImag is $i=2$, so there are two vibrations that lower the energy. It usually has no physical meaning, as the molecule would probably decompose before reaching this state.
If one of these vibrations correspond to a rotation of a moiety, e.g. a methyl group, and is therefore close to zero, it corresponds to the so called rotational space and might be within range of the transition state.
$n$-th Order Saddle point
It is like the above a transition between two $(n-1)$-th order saddle points and has even less physical meaning. You sometimes encounter these saddle points as artefacts of the used calculation model. It is strongly advised to not use these states for any explanation.
Maximum
This is one cannot be found. The PES is usually like a well. The borders of this would be total fragmentation of the molecule. Imagine the homolytic dissociation of dihydrogen (a 2D PES). The maximum value of the PES corresponds to superimposed atoms and its energy value is at infinity.

An additional note. A transition state always connects two local minima. And vice versa, two local minima will always be connected via a series of transition states and local minima. It does not make much sense, looking at any other points than these, as they can most likely not interpreted in a physical meaningful way.

Comment by Ath

What I mean is not global maximum, but local maximum, which doesn't necessarily represent bond breaking...The local maximum I'm interested in is actually the rotation of part of a molecule. I wonder whether the frequencies are all positive or all negative at that point.

There are no maxima on the PES, since the maximum energy corresponds to superimposed atoms. A rotation of moieties is a transition state, or a first order saddle point, and has exactly one imaginary mode.

Thank you for answering my question,Martin. What I mean is not global maximum, but local maximum, which doesn't necessarily represent bond breaking...The local maximum I'm interested in is actually the rotation of part of a molecule. I wonder whether the frequencies are all positive or all negative at that point.

See my edit. It is very important that you do not call those frequencies negative, they are imaginary. There are certain limitations to our programs, so that they usually are printed as a negative frequency to an easier distinguishing from the real ones.

No maxima on the PES? I know transition states are in the valleys, and there are definitely peaks around valleys. The peaks are the local maxima, right?

Those peaks lead to second order saddle points from first order saddle points, or to infinitely high potential energy walls, or to complete dissociation of the system (free atoms), which is a plateau. No maxima are possible.


text:
Finding the Oxidation number
I have got the homework to balance the following redox reaction.
$\ce{Cr(OH)3- + HO2- -> Cr2O7^{-2} + OH- }$(Basic medium)
To balance this reaction first thing we need to do is to find the oxidation number of each element.
My efforts:
I have found out oxidation number of each element on product side.. But it is bit confusing in reactant side.
Can any help me in finding the oxidation number of elements on reactant side ?

Hydroxide is a - 1ion (OH)- so you can figure out the first reactant. For HO2- hydrogen rule overrides the oxygen rule so H+ and Oxygen is -1. Certainly makes more sense than Hydrogen being +3 right?

About oxidation states, I find it easier to follow the extended definition, as it is explained in this question: Electronegativity Considerations in Assigning Oxidation States
Look at the first compound $\ce{Cr(OH)3- }$. What is the most electronegative element? Assume it gets all the electrons to fill the octet.

 Oxygen, it shall have the oxidation state $-2$

What about the other atoms?

 Hydrogen can only receive or lose an electron, in this case it is bonded to the much more electronegative element oxygen, it shall have the oxidation state of $+1$
Chromium gets the oxidation state of $+2$, since $\ce{O}(3\cdot -2) + \ce{H}(3\cdot +1) -\ce{e-}(1\cdot-1)=-Ox(\ce{Cr}) = +2$

Look at the hydrogen peroxide ion, apply what you know about Hydrogen to calculate the oxidation state of oxygen.

 Hydrogen gets $+1$ (see above)
 Charge has to remain, i.e. $-1$
 Oxygen gets $-1$, since $\ce{H}(1\cdot +1) -\ce{e-}(1\cdot-1)=-\frac12Ox(\ce{O})=-1$

Now the hydroxyl ion is straightforward.

 Hydrogen gets $+1$ (see above)
 Oxygen gets $-2$

Look at dichromate and apply the above.

 Oxygen gets $-2$, most electronegative
 Chromium gets $+6$, since $\ce{O}(7\cdot -2) -\ce{e-}(2\cdot-1)=-\frac12Ox(\ce{Cr}) = +6$

In total that will give you:

 $\ce{\overset{{+II}}{Cr}~(~\overset{{-II}}{O}~\overset{{+I}}{H}~)3{}^{-} + \overset{{+I}}{H}~\overset{{-I}}{O}_2{}^{-} -> \overset{{+IV}}{Cr}_2~\overset{{-II}}{O}_7{}^{2-} + {}^{-}\overset{{-II}}{O}~\overset{{+I}}{H} }$


text:
What is the origin of the differences between the MO schemes of O₂ and N₂?
Here are the MO schemes of $\ce{N2}$ (left) and $\ce{O2}$ (right).

Why is the $\sigma$-MO formed by the $p$ AOs energetically above the $\pi$-MO for $\ce{N2}$ but not for $\ce{O2}$?
Can it be explained this way:
In second period elements with more than a half filled $\ce{p}$ orbitals, the energy pattern of the MOs is regular.
In second period elements with less than or half filled $\ce{p}$ orbitals, as the $\ce{s}$ and $\ce{p}$ atomic orbitals have similar energies, so the formed $\ce{\sigma_{g}}$ MO would have similar energy as $\ce{\sigma_{u}}$, but because of electron cloud repulsion it has a big increase in energy, hence it is above the $\pi$ orbitals? And this phenomenom is known as sp mixing.
Or am I wrong somewhere? Is my reasoning corret?

The answer will take some time and I won't be able to get to it before I finished work. So, for the time being: In short, your explanation contains some good and valid thoughts but if you want to really understand the energetic orderings you will have to dive into pertubational MO theory.

This phenomenon is explained by s-p mixing. All the elements in the second period before oxygen have the difference in energy between the 2s and 2p orbital small enough, so that s-p mixing (combination) can occur lowering the energy of the σ(2s) and σ*(2s) and increasing the energy of the σ(2p) and σ*(2p) molecular orbitals. By moving towards right in a period, the s orbital gets more stabilized than the p orbital and the difference in their energies increases, making the s-p mixing for oxygen much smaller.

I think it is an exaggeration to say s-p mixing for oxgyen is negligible in comparision to nitrogen.  Quantitatively, for N2 the sigma bond is about 27% 2s and 73% 2p and O2 is about 13% 2s and 87% 2p.  So there is less mixing for O2, but there is still mixing.  http://pubs.acs.org/doi/abs/10.1021/ja00544a007

“All the elements in the second period before oxygen” does it means all molecules have 1 atom before oxygen or a molecule only have atoms before oxygen? Is it only works with diatomic molecules?

All homonuclear diatomic molecules before oxygen.

S-p mixing is the primary cause of the difference in the molecular orbitals of nitrogen and oxygen, which is influenced by the initial atomic orbital energies.
The lighter second period elements (prior to oxygen) have a relatively small difference in energy between the 2s and 2p orbitals. This allows sufficient s-p mixing to lower the energy of the σ(2s) and σ*(2s) molecular orbitals, and is energetically offset by an increase in energy of the σ(2p) and σ*(2p) molecular orbitals. In the dinitrogen molecular orbital scheme, the dashed lines are there to represent s-p mixing influencing the energy of the four molecular orbitals involved.
The effective nuclear charge increases to the right of the period, stabilizing the 2s orbital more drastically than the 2p orbital. This can be seen qualitatively in the first figure here.
The more stabilized 2s orbital does not s-p mix as effectively, due to the greater energy difference between the 2s and 2p orbitals. As nuclear charge increases, s-p mixing becomes less significant. The change of the molecular orbital ordering between nitrogen and oxygen is the manifestation of this decreased s-p mixing. In the dioxygen molecular orbital scheme the s-p mixing effect is no longer significant enough to alter the relative orbital arrangement.
More information about the details of this difference can be found in most inorganic chemistry textbooks. There are good diagrams showing the gradual change in energy differences across the second row range of homonuclear diatomic molecules. This document highlights the progressive lowering of molecular orbital energies due to the the atomic orbital changes, and presents a correlation diagram linking internuclear separation with molecular orbital energies as well.

The first link, http://www.science.uwaterloo.ca/~cchieh/cact/c120/molitof.html, is dead.

The strength of interaction of sigma and pi-kind developing differently with distance between two atoms.
The pi-interaction, which usually is weaker, is favoured at shorter atomic distance, because it leads to better side overlapping of orbitals. The maximum of energy for pi is by shorter atomic distance than for sigma. $\ce{N2}$ molecule has very short $\ce{N-N}$ distance, which leads to much better p-orbitals overlap and as consequence to lower energy of pi-orbital.

Okay, thank you. Why would pi orbital have bigger energy than sigma in oxygen and not in boron molecule? Oxygen has a double bond.


text:
Can a proton be ejected from an atom?
Consider a neutral atom that has $n$ protons and $n$ electrons. Is it possible to remove a proton from the atom via some technique such as bombardment with another particle or applying an appropriate amount of energy to the system?

Is it possible to remove a proton from the atom via some technique
  such as bombardment with another particle or applying an appropriate
  amount of energy to the system?

Yes, this occurs in nature and can also be done in the laboratory.  In nature the process is known as radioactive decay.  Listed below are two decay processes that will remove a proton(s)

alpha decay: an atom ejects a helium nucleus (2 protons and 2 neutrons)
$$\ce{^239_94Pu -> ^4_2He + ^235_92U}$$
beta-plus decay: a positron is emitted and a proton is converted to a neutron.

In the (nuclear) lab, a process known as phototransmutation can be employed.  If a nucleus is irradiated with high-energy gamma rays (high-energy photons) it can absorb the energy and change to another element by ejecting a proton and\or neutron
$$\ce{^25_12Mg ->C[{h\nu}]\   ^1_1H + ^24_11Na}$$
You can make a rough estimate of the energy required as follows:
$$\mathrm{24.98583692\,amu~ \longrightarrow 1.00727647\,amu + 23.99096278\,amu}$$
comparing the starting and final mass, we find that 0.01240233 amu has been gained in the process.  Since 1 amu = 931.5 MeV, the gamma ray photon would have to have an energy of at least 11.6 MeV.

In the most extreme case, some of the earliest stars in the universe may have been destroyed when temperatures in their cores produced photons with enough energy to shatter atomic nuclei completely and drain the energy that was preventing gravitational collapse: http://en.wikipedia.org/wiki/Photodisintegration#Hypernovae

@ron Is this concept explain by you is applicable for elements having atomic number more than 60 .

@ron Suppose       For example can I remove 1 proton from mercury by alpha decay method or beta plus method?

@murtuza vadharia  There are no naturally occurring radioisotopes of mercury.  However, there are short-lived (seconds to hours) isotopes of mercury that are formed by radiodecay of a parent nuclei and that undergo alpha and \or beta-plus decay.  For details see,  http://en.wikipedia.org/wiki/Isotopes_of_mercury   P.S. If the answer above was helpful, please mark it as accepted, thanks.

There are actually a (very small) number of nuclei that exhibit spontaneous proton emission decays.
Much more common are photon, neutron or alpha initiated proton emission reaction such as the the alpha initiated reaction on Sulfur discussed in this pre-print. Moderate energy alphas incident on light nuclei generate many interesting final states a fact that has been instrumental in figuring out how the nucleus works (which is admittedly usually framed as a physics rather than chemistry question).

That's quite an insightful answer @dmckee!!! Do you work in the overlap between physics and chemistry? I wonder if you might be able to help us launch Materials Modeling which is about computational modeling and scientific software for materials design and spectroscopy / quantum chemistry.

@user1271772 I have suspended my activities on Stack Exchange for the time being, and am in any case more of a experimental and particle guy than a computational or materials guy.

That's more a nuclear physics question. But I will write some anyway.
First, there is a good amount of naturally radioactive nuclei, that emits alpha particle, i.e. $\ce{^4He}$ nucleus. Than, there is a less common $\beta^+$ decay, when nucleus emits positron, transforming one proton into neutron in the process (and emitting neutrino). Next, some nuclei captures electron (usually from lowest electron shell), emitting neutrino and again, transforming one proton into neutron. Proton emission is known, but extremely rare.
Next, there is some amount of nuclear reactions, when an incoming hi-energy particle merges with nucleus, than the resulting excited nucleus emits some other particles. This reactions are extremely specific, each pair of nucleus and particle has their own reactions routes, etc, etc. Example of proton-emitting reaction:
$\ce{\ ^3He + D = ^4He +^p}$
Sometimes, heavy nuclei after capturing a neutron, undergoes $\beta$-decay, transforming into $\alpha$-radioactive nuclei. The net result would be losing a proton and maybe a neutron or two. But this an awfully specific for few nuclei.
TL;DR : no, there is no general technique, but there are some techniques (very costly), than work in some specific cases.

I don't have much knowledge of Nuclear Chemistry but:

is it possible that by any method like bombarding other particle or providing high energy can we remove proton from atom ?

The best example that I can think of right now is alpha decay. An alpha particle consists of two protons and two neutrons bound together into aparticle identical to a helium nucleus.
This is also what we do in nuclear fission/fusion [atom bomb etc.]. But I think adding or removing just one proton might be very hard.

One isotope that is made by removing a proton from an atom is carbon-14.  Nitrogen-14 is bombarded with a neutron and a proton is displaced giving carbon-14, which eventually decays back to the nitrogen isotope.  Our ability to date organic material via carbon-14 thus depends on a proton removal process.


text:
Why is carbon tetravalent?
How does the electron from the 2s orbital "jump" to the 2p, thus leaving 4 unpaired electrons to form four covalent bonds?
Also, does the octet rule not apply to carbon? If it does, how can it bond with only two electrons? It would lack two more.

The octet rule does apply to carbon.  In order for carbon to form a complete octet, it must either gain 4 electrons, lose 4 electrons or form 4 covalent bonds.  Placing 4 units of charge on such a small nucleus would be exceedingly difficult, so carbon typically binds in a covalent manner.
Let's imagine an experiment where we place carbon atoms in a container of hydrogen.  This becomes a chemical reaction, just like any other.  $\ce{CH4}$ will be the final product because we need 4 bonds around the central carbon atom in order for the carbon to have an octet.  As the carbon and hydrogen begin to react, a chemist would say that the atoms begin to move along a reaction coordinate that connects starting materials with products.  As the atoms move along this trajectory they explore many possible configurations.  For example, carbon could form 4 bonds to hydrogen using its unhybridized atomic orbitals.  If this pathway produced $\ce{CH4}$ it would have 2 C(2p)-H(1s) bonds and 2 C(2s)-H(1s) bonds.  The two C-H bonds formed from the carbon p orbitals would have an H-C-H angle around 90 degrees.  The 2 bonds formed using the carbon 2s orbitals would be rather floppy as they could travel over the surface of an s orbital sphere.  The most stable configuration would probably involve these 2 C(2s)-H(1s) bonds arranging themselves as far away from the other 2 hydrogens as possible.  So we'd probably wind up with a distorted tetrahedron with one H-C-H angle around 90 degrees and another H-C-H angle much wider than the 109 degree tetrahedral angle.  Note that the 2 hydrogens involved in the H-C-H 90 degree angle would be very close together resulting in a destabilizing steric interaction.
Let's now consider a second possible path along our reaction coordinate.  In this case let's mix our carbon atomic orbitals (the 2s and 2p orbitals) into a new set of atomic orbitals.  We will "raise" the energies of the two 2s electrons and "lower" the energies of the two 2p electrons such that the 4 new orbitals we generate will have the same energy as the average energy of our initial s and p orbitals.  Equal mixing of the s and p orbitals results in the formation of 4 equivalent new atomic orbitals, they are often referred to as $\ce{sp^3}$ hybridized orbitals.  These 4 equivalent orbitals point towards the vertices of a tetrahedron.  Reaction with hydrogen will produce a $\ce{CH4}$ molecule that has 4 equivalent C(sp3)-H(1s) bonds and all of the resulting bond lengths and angles will be equivalent.
In the case of carbon, the reaction coordinate follows this last pathway.  Why? Because it produces the lowest energy (most stable) version of $\ce{CH4}$ possible.  This lowest energy version of $\ce{CH4}$ does not suffer from unfavorable H-H steric repulsions due to small (90 degree) H-C-H angles.  All of the C-H bonds in this stable version of $\ce{CH4}$ also have stronger (more stable) bonds.  The C(sp3)-H bonds are much more directed and have better overlap that the diffuse, non-directed C(2s)-H bonds we discussed earlier.
If we move from carbon to phosphorous, we find that in the molecule $\ce{PH3}$ the bond angles are close to 90 degrees and the central phosphorous uses its original atomic orbitals (pure P in this case) to form bonds to hydrogen - no further mixing of orbitals occurs in this case.
The bottom line is that nature is smart.  Nature knows how to mix (or not mix) atomic orbitals in order to produce the lowest energy compound possible, and it's method of achieving this is to travel along a reaction coordinate sampling all possible pathways and then letting energetics drive the reaction along the lowest energy pathway.

Look up "orbital hybridization".  It says that atomic orbitals can be combined to make new orbitals for bonding purposes.  Carbon can take 1 s and 3 p orbitals to make 4 sp3 orbitals that can make 4 bonds.

Orbital hybridisation does not exist in the way you suggest it here. No atom can make hybrid orbitals. These are a mere mathematical treatment of a certain bonding pattern in a molecule.

Agreed. Orbital hybridizations were a workaround to make early MO theory play nice with observed geometries. The four $\ce{C-H}$ bonds in methane $\ce{CH4}$ are not degenerate, and thus we do not require the electrons involved in the bonding be described by the same types of orbitals.


text:
Why is HSO4- shown as an example of a weak acid instead of HSO4?
I'm watching some (really awesome) chemistry courses, and I think I have a fairly decent handle on acids and bases (or maybe not, and I'm fooling myself). But the teacher just showed a table of weak/strong acids. $\ce{HSO4-}$  was one of them.
I'm wondering, why hydrogen sulfate ion instead of just hydrogen sulfate? Why is $\ce{HSO4-}$ an acid and $\ce{HSO4}$ apparently isn't? This was the only ion on the table. Or does $\ce{HSO4}$ simply not exist?

The bisulfate molecule doesn't exist in solution. I can only find references to the hydrogen sulfate molecule as part of salts, which implies that hydrogen sulfate is usually found as an ion.
The bisulfate (hydrogen sulfate) ion is indeed an acid. It is a relatively strong weak acid too, with a $\ce{K_a}$ value of $1.2 * 10^{-2}$.
Part of the reason for its acidity has to do with its electronegative oxygens isolationg electron density away from the hydrogen atom bonded to the oxygen. This makes the hydrogen more partially positive and thus more reactive (more likely to be taken by a base).

I have to add, that while $\ce{HSO4}$ does not exist, the doubled molecule $\ce{H2S2O8}$ (peroxodisulfuric acid) does exist and is apparently a quite strong acid (its double ammonium salt is stable solid)

I have to add, that $\ce{HSO4}$ might well exist, just not in aqueous solution - in the gas phase, why not ?

I agree Martin!

@Martin - it would have an odd number of valence electrons, and so it probably wouldn't exist for long!

"Part of the reason for its acidity has to do with its electronegative oxygens isolationg electron density away from the hydrogen atom bonded to the oxygen. " is that actually true? How would that work? Isn't it just acidic because the negative charge can be fairly good delocalized into the $\pi$-bonds? If we'd were to replace one of the two  S-O double bond with a SOH group, I think it would be that acidic anymore...

It is true. Do you think a hydroxide ion with its significant negative charge density would be more likely to attack a hydrogen atom with no partial charge or a hydrogen atom with a partial positive charge? Your reason is also true. The sulfate anion has several resonance structures.

@Jori there is no significant pi bonding in the sulfate anion. I think that was discussed with the phosphate ion here on the site. I am on mobile now, so it is hard to find the reference. However, the electron of hydrogen is already very much localised at the oxygen. The hydrogen oxygen bond is quite ionic.

@thomij Agreed, but it would certainly exist longer than an anion in the gasphase.

" I think it wouldn't be that acidic anymore..." is what I meant to say. Do these structures exist? Replacing one double S-O bond with a SOH group. I think you understood me wrong. I meant that the reason sulfuric acid is so acidic, is not the amount of oxygen atoms close to the SOH group, but the delocalisation of the negative charge when hydrogen takes off. Resulting in a more stable structure. Is the force of the other surrounding oxygen atoms (that are negatively charged) of suck influence over that distance, and wouldn't the strongly positively charged sulfur get in the way?

I'm just thinking here a bit. You guys are probably way better educated than me in chemistry haha. We coudl confirm the truth (or at least get an indication of the truth) if we knew the acidity of the hypothetical structures H3SO4 and H4SO4 (do they even exist?)

I notice that there is no mention of polyprotic acids in the other answers!
$\ce{H2SO4}$ (sulfuric acid) is diprotic - meaning it has two "detachable" protons which can come off in aqueous solution. The first one is strongly acidic - the $Ka = 2.4 × 10^6$ - this means the odds of finding a non-dissociated $\ce{H2SO4}$ molecule in solution are something like 1 out of 2.4 million.
On the other hand, the second proton is weakly acidic - the $Ka = 1.0 × 10^{-2}$. This means that there will mostly be $\ce{HSO4-}$ instead of $\ce{SO4^2-}$ in solution (roughly 100 times as much).
The other answers are correct in saying that $\ce{HSO4}$ does not exist in solution - but it is important to realize that we list the bisulfate ion ($\ce{HSO4-}$) as a weak acid because it is the conjugate base of sulfuric acid, and is in turn the conjugate acid of the sulfate ion. This becomes very important when you start to study acid/base equilibria.

$\ce{HSO4}$ does not exist.  There is $\ce{H2SO4}$ and $\ce{HSO4-}$  Try drawing the Lewis structure for sulfate ion.  You'll see that there are two oxygens with formal charges.  They can be left as is or they can have $\ce{H}$ bonded to them.

I think you mean H- bonded to them (H forms single bonds).

The adequate Lewis structure for the sulfate ion has one negative charge at each oxygen and a double positive charge at sulfur. Everything else is just too far from the truth.


text:
Are there practical chemical reactions that do mechanical work without heat or electricity?
Every day we use chemical reactions to perform mechanical work by creating either heat (e.g. in a piston engine) or electricity (batteries). I know muscles perform work by taking stored chemical energy and converting it directly into mechanical force.
Other than in organic chemistry, are there any examples of chemical reactions that we use to perform mechanical work without the intermediate step of harnessing the heat or electrical potential created by the reaction?

Well, it's really the rapid evolution of gas (e.g., from combustion) that will perform the expansion work in a piston, not the heat per se. Would you discount other (non-combustion) reactions that produce gases to perform $PV$ work?

The heat is doing at least some of the work by expanding non-reactant and production gases. I always thought it was the major portion, but I might be wrong. I guess what I'm looking for is an example where the reactions are not significantly exothermic. Reactions that produce gases would not be discounted.

Sorry, I may have been unclear in my phrasing. I just meant to emphasize that it's the gas itself that performs the work while expanding, though of course the amount of expansion is partially temperature-dependent.

Physical reactions- phase transformations can do work from heat- do you count that as a chemical reaction?

No, I suppose not. And it's the exact opposite of what I'm trying to ask about. I've researched Greg E.'s comment and, yes, the majority of work in a piston engine is performed by evolution of gas. However, the very same reaction is highly exothermic, and so the accompanying heat must be dealt with. I'm looking for a an example of a chemical reaction that can perform mechanical work without generating (much) heat. A battery is one example, but its resultant energy form is electrical, not mechanical, so an extra step is needed.

authors: Белякова Анастасия


text:
How many ligands in Heme?
How many ligands does Fe has in heme ?
I am not sure how many are in heme and how many in cytochrome C.
Its about iron complexes.

Heme: The Iron molecule is bound to 4 Nitrogen atoms, but take a look at porphyrin. All the nitrogens are part of one actual structure. You can visualize the iron molecule in the center of the larger compound's structure, replacing the two Hydrogen atoms in N-H:
The large structure surrounding and coordinating with Fe is also known as a polydentate ligand since it bonds to Iron at more than one site. So there is just one ligand on Fe.

Cytochrome C: is a heme protein, so you may expect the same answer: A single polydentate ligand binds to Iron. This page has a PDB Model of it in a plugin.
As one ligand coordinates to four sites in this case, other ligands typically attach (e.g. oxygen binding in hemoglobin) to fill the octahedral group (up to 6 ligands attached to Fe the coordinate center).


text:
Identical bond angles
Is it possible to incorrectly predict that two bond angles are different from van der Waals repulsions alone?
I'm thinking that because bond lengths differ when different atoms are involved that we could, at least theoretically, have two identical bond angles involving different molecules.
For example, say we are considering the bond angles of molecule $\ce{AXY}$ and $\ce{BZA}$.
X and Y are highly electron dense and also big. The first suggests a big bond angle; the second suggests a long bond length, which in turn suggests the possibility of a smaller bond angle.
Now consider $\ce{BZA}$. Z and A are not electron dense but also large. The first suggests a small bond angle. The second suggests a long bond length and thus a smaller bond angle.
Is it still possible for the two to have identical bond angles?

Is "A" meant to be the same in both molecules?  Are "A" and "B" the central atoms in AXY and BZA respectively?  Can they be linear?

I think the answer to this question is, yes, it is POSSIBLE, at least in non-ideal situations (i.e. non-gas phase, non-vacuum, non-0K, etc.).  In ideal situations?  I would err on the side of saying that it was POSSIBLE but I don't know of a concrete example that I could provide.  Of course, in reality, molecules are not stationary states and are constantly in motion (vibrations, rotations, and/or translations) so it is sort of a moot point from this perspective.


text:
4 compounds in equilibrium with one another - determine their equilibrium concentrations
I know that if I have two compounds in equilibrium and I know $\mathrm{\Delta G}$, then I can use $$\mathrm{\Delta G=-RTlnK}$$
to determine the concentrations of the two components at equilibrium.  I have 4 compounds that are all in equilibrium with one another
\begin{aligned}
\mathrm{\Delta G(1,4)} ~&= ~~~3.41~ \mathrm{kcal/mol}\\
\mathrm{\Delta G(3,1)}~&=-2.21~ \mathrm{kcal/mol}\\
\mathrm{\Delta G(2,3)}~&=-0.46~ \mathrm{kcal/mol}\\
\end{aligned}
I applied the equation sequentially and found that
\begin{aligned}%
[1]~&=~0.34\%\\
[2]~&=~29.1\%\\
[3]~&=~13.5\%\\
[4]~&=~57\%\\
\end{aligned}
Are those the correct concentrations? Is there an "easy" way to do it?

I am linking my question to this post, since they are very related.

In case anyone else travels this road...
Using $\mathrm{\Delta G~=~-RTlnK_{eq}}$ and the 3 $\mathrm{\Delta G's}$ supplied above, we find that
\begin{aligned}
\mathrm{K^{1/4}_{eq}}~&=~0.003\\
\mathrm{K^{3/1}_{eq}}~&=~40.749\\
\mathrm{K^{2/3}_{eq}}~&=~2.163\\
\end{aligned}
We also know that the concentrations of the 4 species must total to 1
$${[1] + [2] + [3] + [4] = 1}$$
so we have the following 4 equations and 4 unknowns
\begin{aligned}%
[1]~ &=~~~ 0.003 ~[4]\\
[3]~ &=~ 40.749 ~[1]\\
[2]~ &=~~~ 2.163 ~[3]\\
[1] + [2] + [3] + [4]~ &=~ 1\\
\end{aligned}
solving for the 4 concentrations yields
\begin{aligned}%
[1]~ &=~~~ 0.23\%\\
[2]~ &=~ 20.27\%\\
[3]~ &=~~~ 9.37\%\\
[4]~ &=~ 70.13\%\\
\end{aligned}


text:
Calculating the molar mass of a solute from freezing point

A $\pu{4.305 g}$ sample of a nonelectrolyte is dissolved in $\pu{105 g}$ of water. The solution freezes at $\pu{-1.23^\circ C}$. Calculate the molar mass of the solute. $k_f$ for water is $\pu{1.86 C/m}$.

My steps:
$$1.23 = 1.86 \times \frac{x}{1.05},$$
where $x = y / 4.305$.
$$0.66 = \frac{x}{1.05}$$
$$x = 0.69$$
$$0.69 = y / 4.305$$
$$y = \pu{2.989 g}$$
But I know this is not correct.

The freezing point depression is based on the molal concentration (moles of solute per kg of solvent).
$$\Delta T_f = -k_f \cdot m$$
You know the freezing point depression of the solution and the cryoscopic constant, so you can calculate the molality:
$$m = -\dfrac{\Delta T_f}{k_f}=-\dfrac{\pu{-1.23 ^\circ C}}{1.86 \frac{\textrm{kg}\cdot ^\circ \text{C}}{\textrm{mol}}}=\pu{0.66 molal}$$
Since molality is the number of moles of solute divided by the mass of the solvent in kg, you can calculate the moles of the solute:
$$n_\textrm{solute}= \pu{0.66 m} \cdot \pu{ 0.105 kg}= \ ...$$
Divide your mass of solute by the moles of solute and you will get the molar mass (and not 2.989 g).


text:
Greater entropy: an atom or macromolecule?
A question that appeared on my last chemistry exam was :

Which of the following has greater entropy A) An atom
  B) A
  macromolecule

The question doesn't specify anything else(i.e. type/size of atom or macromolecule). The macromolecules we learned class were proteins, sugars, nucleic acids(DNA/RNA) or lipids. What is the correct answer and why(hypothetical calculation, if possible) ?

There are many equations that describe the entropy of a system, one is
$$\ce{S = k~ ln ~W}$$
where S is the entropy of the system, k is Boltzmann's constant and W is the number of states a system can exist in.  If we take an atom, it can only move in 3 directions x, y and z - that's it.  If we take a small molecule like ethane, it can also move in these same 3 dimensions but it can also rotate about any one of the atoms, the various bonds can stretch and compress and the various bond angles can expand and collapse.  Clearly, there are a lot more states that ethane can exist in as compared to a single atom, clearly W is larger for ethane than a single atom, clearly S is larger for ethane than a single atom.  As we increase the complexity of the molecule, W and S increase exponentially.  A macromolecule will have a much larger entropy than an atom.
Edit (based on @Gdgames Gamers comment below)
Entropy is a measure of the disorder in a system.  It will always be true that the entropy of substance in the gas phase will always be greater than the entropy of the same substance in the liquid phase, which, in turn, will be always greater than the entropy of the same substance in the solid phase.
Perhaps the person who wrote the exam is assuming that the atom is in the gas phase and the macromolecule is in the solid phase and that based on the phase difference alone the atom has a larger entropy than the macromolecule.  I don't think this is always correct.  First off, the temperature is not specified, nor the state of the macromolecule (is it in solution?).  The macromolecule, depending upon the temperature and state, may or may not translate and rotate through space, but it will vibrate (even at 0K).  I'm just not sure how to compare the entropy contribution from the potential 3n-6 vibrational modes in the macromolecule to the entropy from the 3 translational modes of the atom.  Very poor question by the examiner.

While you are correct, I can't help but feel the question was posed incorrectly in the first place. Knowing that a collection of atoms has more entropy than a single atom feels like it yields no information. A much more interesting question would have been to compare the entropy per atom (or per mole of atoms) when free and when bound in a macromolecule. Of course, the macromolecule has much lower entropy than the sum of its separate constituents.

@Nicolau Saker Neto  If we have a non-linear molecule of n atoms it will have 3 translational, 3 rotational and 3n-6 vibrational degrees of freedom, e.g. it will have 3n degrees of freedom.  n independent atoms would also have 3n degrees of freedom.  So wouldn't the entropy per atom of n independent atoms be equal to the entropy per atom of n connected atoms?

I don't think calculating degrees of freedom alone is enough to evaluate the entropy of a system. A macromolecule can store entropy in vibrational degrees of freedom, which a gas of free atoms cannot, but there is very little vibrational excitation at reasonable temperatures, and I think every macromolecule decomposes well before most of its vibrational degrees of freedom can be explored.

@NicolauSakerNeto I thought the question was lacking also...but this is how it appeared on the exam. I picked the answer to be macro molecule but it was marked wrong..


text:
Calculate the pH of acetic acid for a 0.02 M solution

We have a solution $\ce{CH3COOH}$ (acetic acid) with $c=0.02~\mathrm{mol/L}$  and $K_\mathrm a(\ce{CH3COOH}) = 1.8\cdot10^{-5}$. Calculate the $\mathrm{pH}$ of this solution.

All I know is that it is a weak acid and that
$$\mathrm{pH}= \mathrm{p}K_\mathrm a + \log\left(\frac{B}{A}\right).$$
I know $A$ but there is no information about $B$.

This type of problem is best solved using the ICE (Initial Change Equilibrium) method.
First, write down the chemical equation and the law of mass action for this reaction.
$$\ce{CH3COOH <=> H+ + CH3COO-}$$
$$K_a = \dfrac{[\ce{H+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]}$$
Second, determine the initial concentration of each of the three species. Make a little table to keep everything straight:
\begin{array}{lc} \hline
&\ce{CH3COO-} &\ce{<=>}&\ce{H+} &+&\ce{CH3COO-} \\ \hline
\text{I}& \pu{0.02 M} &&\pu{0 M} &&\pu{0 M} \\ \hline
\end{array}
At equilibrium, some of the acetic acid will be ionized. How much? We don't know right now, so let's have that be a variable; $x$. At equilibrium, the concentration of acetic acid will have decreased by $x$. How much will the concentration of $\ce{H+}$ and acetate anion increase? Since the stoichiometric coefficient of all species is 1, then those concentrations will increase by $x$. Let's add that to our table. Let's also write an expression for the equilibrium concentrations in terms of the initial concentrations and $x$.
\begin{array}{lc} \hline
&\ce{CH3COO-} &\ce{<=>}&\ce{H+} &+&\ce{CH3COO-} \\ \hline
\text{I}& \pu{0.02 M} &&\pu{0 M} &&\pu{0 M} \\
\text{C}& -x &&+x  &&+x \\
\text{E}& 0.02-x  &&x &&x \\ \hline
\end{array}
Now, plug these expressions into the law of mass action:
$$K_a = \dfrac{[\ce{H+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]} = \dfrac{(x)(x)}{(0.02-x)}$$
As long as $x\ne 0.02$, which is very unlikely at equilibrium, we have a solvable quadratic equation:
$$K_a=1.8\times 10^{-5}= \dfrac{(x)(x)}{(0.02-x)}$$
$$3.6\times 10^{-7}-1.8\times 10^{-5}x-x^2=0$$
Some texts will encourage you to make the simplification $0.02-x \approx 0.02$ since $x<<0.02$, however, WolframAlpha can solve the original equation (and so can you).
The equation has two roots: $x=5.91\times 10^{-4}$ and $x=-6.09\times 10^{-4}$. The second one is nonsense (we cannot have negative concentrations).
If you make the assumption that $0.02−x \approx 0.02$, then the positive root is $x=6.0 \times 10^{−4}$, not a big difference given the significant figures. If you knew your initial concentration more precisely, this assumption would be invalid.
Now, we know what $x$ is. What do we still need to know to determine pH? How is $x$ related to pH?

@bennorris method works, but I think my method may save some time.
\begin{array}{lc} \hline
&\ce{CH3COO-} &\ce{<=>}&\ce{H+} &+&\ce{CH3COO-} \\ \hline
\text{I}& \pu{0.02 M} &&\pu{0 M} &&\pu{0 M} \\
\text{C}& -x &&+x  &&+x \\
\text{E}& 0.02-x  &&x &&x \\ \hline
\end{array}
I keep the acetic acid concentration to be the same because I assume the x value will be really small( you can prove it later)
$$K_a = 1.8\times 10^{-5}=\dfrac{[\ce{H+}][\ce{C2H3O2-}]}{[\ce{HC2H3O2}]} = \dfrac{(x)(x)}{(0.02)}$$
$$ x=6.0 \times 10^{-4}$$
You can prove the assumption by dividing your x value with your initial concentration of acetic acid
$$ \dfrac{[\pu{6\times 10^{-4} M}]}{[\pu{0.02 M}]} =0.03 =3\% < 5\%$$
If the percentage you get is smaller than five percent, your assumption will be correct.
This method will save you some time as you don't have to deal withe the quadratic equation and give you a approximate answer

I have to say that if you turned in this answer in an exam I was marking you would score zero.

Yes, but the ICE table should initially be setup as Ben did it. Then make the assumption that $0.02-x \approx 0.02$. The 5% criteria seems to have been pulled from thin air. I'd base the needed precision on significant figures (2) giving a need for about 1%. Lastly the acetate species could have been normalized to 100% which would have given a result very close to the "true" answer.


text:
Deviation from ideal solutions - activities and concentrations
What are some examples of solutions deviating significantly from ideal behavior, and, by extension, showing pHs significantly different from those predicted by usage of concentrations instead of activity coefficients in K expressions?
I know Wikipedia lists two examples but what other examples are there? Do all dilute solutions behave ideally or mostly ideally? Or are there major exceptions in which the use of concentrations instead of activities fails?
Also, is there an easy way to incorporate activities into the calculation of pH for complex systems?

To partly answer your last question, we can replace concentrations with activities in just about any expression, especially in equilibrium calculations. For example $K_a=\dfrac{\alpha_{\ce{A-}} \cdot \alpha_{\ce{H+}}}{\alpha_{\ce{HA}}}$ and $\text{pH}=-\log_{10}\alpha_{\ce{H+}}$. Of course, the harder part is usually determining the activities.


text:
Is oxygen a Lewis acid?
As I recall a Lewis acid is a substance that can accept an electron pair from a donor and oxygen seems to be able to do that. For example:
$\ce{H2 + O -> H2O}$
So, is oxygen a Lewis acid?

Usually the terms Lewis acid and Lewis base are defined within the context of a specific chemical reaction. So in a reaction, if there is lewis acid there must be a lewis base.
Lone-pair acceptors are lewis acids where as lone-pair donors are lewis bases.

Some compounds, such as $\ce{H2O}$, are both Lewis acids and Lewis bases, because they can either accept a pair of electrons or donate a pair of electrons, depending upon the reaction.

In reaction $\ce{H2 + O -> H2O}$, you can say that oxygen is lewis acid but that would be contradictory since hydrogen doesn't have a lone pair to donate.
Also, Oxygen acts as a lewis base by donating a lone pair to hydrogen ion when Acids are dissolved in water.
To be more clear on Lewis Acid/Base concept, see this.

The reaction $\ce{2H2 + O2 -> 2H2O}$ is a redox reaction (also if you go to atomic recombination), there is no Lewis acid or base, but electron transfer and this results in the contradiction you observed.

@Martin : I kinda felt like writing it but I doubted it since it's been a while since I left field of chemistry. :)

Oxygen is a Lewis base (that too a weak one), not a Lewis acid.

REASON: It has lone pair of electrons, which can be donated to electron-deficient species (Lewis acids). The reason which makes it a weak lewis base is that the electronegativity of Oxygen is 3.0 on the Pauling Scale, which increases its tendency to hold on to its lone pair of electrons. Due to this difficulty in parting lp of electrons from Oxygen, it is regarded as a weak Lewis base.

Furthermore, due to its high electronegativity, its repulsion to incoming electrons is very high, and hence it doesnt not accept electrons as easily as a Lewis acid would, which leads to the conclusion that Oxygen cannot be regarded as a Lewis acid in day-to-day terms.


text:
If salt crystals make a cube shape, what is this 'X' marking?
I am an amateur, with very little professional chemistry knowledge. I was doing a demonstration / experiment with my 6 year old child.
We took very warm normal tap water, and dissolved into it a large pile of normal table salt. I didn't measure anything particularly, because I was just showing the basic concept of 'dissolving'. That's all. My kid was amazed to see the white pile of salt disappear visibly, but upon tasting the water understood what had happened.
Next we left the bowl up on a high shelf for 2 weeks, to let the water evaporate. This morning we took a look, and were rewarded by a dry bowl full of beautiful perfectly square salt crystals, many of them between 4-6mm in width.
That the crystals were perfect squares did not surprise me, as I have read that sodium and chlorine atoms form a cube shape at an atomic level.
But I could not answer the question about what the 'X' pattern marking was on each square. I have tried to attach photographs to show what I mean. The 'normal' picture is blurry, but if you look you can faintly see an 'X' on each crystal.
The other 2 pictures are each at 20x, using the Natural History Museum Pocket Microscope.
My question is, what explains those 'X' shapes? If possible I need to translate the explanation into 6-year-old talk :-)

With my knowledge of materials though not of salt crystallisation mechanics, I'd say its grain boundaries. Grains are ordered structures- all the same geometric arrangement. Its nucleated (starting seed) at several  locations and they grow together. At the microscopic level, you have multiple crystals (grains). The lines show the boundaries.

Are you saying that the salt crystal grew from the centre of the 'X', in 4 equal directions at right angles? (ie, instead of looking at the crystal as a square, see it as a diamond shape?)

When salt (NaCl) crystals are grown under normal conditions, the edges of the cubes usually grow faster than the faces (because the edges have more contact with the saturated salt solution than the faces). So the crystals grow with a square-pyramidal indentation on each face. The "X" you observe is your view of the edges of that inverted pyramid. On the other hand, if the crystals are grown really slowly under controlled conditions, this does not occur and you get a more or less perfect cube.

Let's look at structure of $\ce{NaCl}$ ideal cubic crystal

As you can see, atoms on the faces have 5 neighbors, atoms on the edges has four neighbors, and atoms on the vertices of the cube have 3 neighbors. This means, that a) edges and vertices pulls new atoms into crystal structure stronger and b) they 'collect' new atoms from larger volume, than faces. On the other hand, addition of atoms to an edge forms 'step' on the crystal surface

A position near the step has three potential neighbors, so faces grows mostly on steps, and at exceptional speed. Given that, the fastest growing faces usually have 'spiraling' step

The effects of unequal growth are expressed most at significantly oversaturated solutions, otherwise 'steps' travels fast enough over crystal surface to make a surface nearly flat. On the other hand, in case of highly oversaturated solution crystals grow mostly near verticies, edges and surface defects like 'spiraling step', leading to formation of so-known 'dendrites'. The most known example is snowflake (though perfect symmetry of snowflakes is a mystery on its own), the most suited for demonstrations example is growth of ammonium chloride crystals during natural evaporation of concentrated solution.  When the crystal in consideration grows on edges, it may form edges in the form of inverted pyramid, just like in your case. Growth of inverted pyramids is also wide known, and, may lead to quite bizarre forms . In your case. the 'X' is, indeed, an edge of inverted pyramid on the crystal surface.

Thank you for your efforts and great visuals to help explain this. Although you both gave me the answer as an "inverted pyramid", I simply understood the wording and concept of the other answer better. The idea that the "edges grow faster than the faces" is simpler, is easier for me to visualise, and I'm able to explain it to a child better. The stuff here about the spirals threw me, to be honest.


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Other than heating, is there any evidence that normal RF radiation (say, microwave range) can affect chemical reactions?
There's been a lot written (and debunked) about alleged biological effects from RF radiation, even when it's too low powered to cause heating effects or ionization. Is there any chemical reason to believe it's possible for an effect to exist?

Related: http://chemistry.stackexchange.com/questions/24507/can-magnetic-fields-affect-a-chemical-reaction


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Is BCl₅ possible?
Is $\ce{BCl5}$ possible? I saw it  in book. As far as we can see there is no way  to get 5 chlorine atoms with boron.

It appears that $BCl_5$ is not a stable compound and you're right that it is difficult to get 5 chlorides around boron. What you should ask is how can I get a neutral molecule when Boron can only have a charge of +3?

Boron pentachloride is likely not stable except perhaps in extreme conditions, such as under very high pressures. Even then it may be possible that a description such as $\ce{[BCl4^{-}]Cl^+}$ containing a tetrahedral boron anion could turn out to be more accurate than any hypercoordinate structure (a boron atom surrounded by more than four ligand atoms).
However, pentacoordinate and hexacoordinate boron are not unheard of. Indeed they are common in borane clusters, though this type of structure has unusual bonding and perhaps is best left in its own category. Even without considering clusters, it is still possible to create hypercoordinate boron atoms by exploring steric demands of certain ligands. Examples can be found in this article, among several others. A simple diagram exemplifying how to force hypercoordination in boron is given here; notice how a four-coordinate structure requires rotation of the C-C bond between the left and middle aromatic rings, decreasing the extent of $\pi$-conjugation between rings and causing that particular molecular geometry to be less stable relative to the five-coordinate structure. Hexacoordinate carbon has also been prepared with even higher steric forcing.
Additionally, it should be mentioned that "hypercoordinate" is the preferred term instead of "hypervalent". Not only is hypercoordination more general (all hypervalent compounds are hypercoordinate, while not all hypercoordinate compounds need be hypervalent), but hypervalency usually implies some form of octet expansion, which is nowadays not generally believed to be a correct description of bonding.

Kudos for mentioning hypercoordinate! I still wonder if it exists in gas phase. Most certainly not in the hypercoordinate form, but maybe as a Lewis acid base pair $\ce{Cl-Cl->BCl3}$.


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Gauche correction in Bensen group increment
I have difficulty understanding the use of gauche correction in Bensen group increment method to calculate the heat of formation. As an example, 4 times gauche correction (0.8 kcal / mol) is considered in the calculation in the hypothetical equilibrium between the following two isomers. How is that figure decided?

Here is how I do it.

Sight down each C-C bond along the backbone of the molecule and draw a Newman projection
Look at all 3 possible conformations in the Newman projection and pick the conformation with the fewest gauche interactions
Count the number of gauche interactions
Repeat with the next C-C bond along the backbone

Here is a drawing of the 3 Newman projections in 3,4-dimethylhexane that contain gauche interactions (there are no gauche interactions along the $\ce{C1-C2}$ or $\ce{C5-C6}$ bonds.  You can see that there are 4 gauche interactions in the molecule.