text:
How to Find Volume from Moles?

What volume of $\pu{12.0 M }\ce{HCl}$ is required to make $\pu{75.0 mL}$ of $\pu{3.50 M }\ce{HCl}$?

I didn't know how to solve it at first until I multiplied $\pu{75mL}$ by $\pu{3.50 M}$ and got $\pu{262.5 mM}$ and then divided that by $\pu{12.0M}$ to get the answer. I was just wondering if there was an equation that I can use for these kinds of questions? Or is it just simple math? Ex. $\pu{75.0mL} \times \pu{3.50 M }\ce{HCl}$.

The equation you are looking for is the definition of concentration
$$c=\frac nV$$
where
$c$ is concentration,
$n$ is amount of substance, and
$V$ is volume.
You can rearrange this equation to solve for concentration, amount of substance, or volume as required.
Furthermore, you know that the amount of solute does not change when you dilute a solution, i.e.
$$n_1=n_2$$
and thus using the above-mentioned equation
$$c_1\cdot V_1=c_2\cdot V_2$$
You can rearrange this equation to solve the problem that is given in the question:
$$\begin{align}
c_1\cdot V_1&=c_2\cdot V_2\\[6pt]
V_1&=\frac{c_2\cdot V_2}{c_1}\\[6pt]
&=\frac{3.50\ \mathrm{mol\ l^{-1}}\times75.0\ \mathrm{ml}}{12.0\ \mathrm{mol\ l^{-1}}}\\[6pt]
&=21.875\ \mathrm{ml}\\[6pt]
&\approx21.9\ \mathrm{ml}
\end{align}$$

"$\pu{12 M }\ce{ HCl}$" means that there are $\pu{12 mol}$ of $\ce{HCl}$ per liter of solution (or 12 mmol per milliliter).
You are looking for a volume of $\pu{12 M }\ce{ HCl}$ that contains ($\pu{3.5\times 75 = 262.5 mmol}$ ) of $\ce{HCl}$.
Since you need $\pu{262.5 mmol}$, and the $\pu{12 M }$ solution has $\pu{12 mmol}$ per $\pu{mL}$, then you need $\pu{262.5mmol}\div \pu{12M} = \pu{21.875 ml}$ of $\pu{12 M }\ce{ HCl}$.
This is to explain why it worked, but I suggest you to learn what the equations mean, rather than just learning "how to do it": once you understand what you are doing, it's not hard.

I'd add that using dimensional analysis is a good way to keep things straight until you're comfortably just grinding the numbers.


text:
Modeling 3-center-2-electron bonds in Avogadro
Is it possible to model 3-center-2-electron bonds in Avogadro?
It seems to allow users to add however many bonds they would like to an atom, but I can't seem to find documentation on whether this is valid.  When modeling $\ce{B2H6}$, for example, I can draw a $\ce{B-H-B}$ bond, but does Avogadro understand that this is a bridge bond?
For example, below is an $\ce{(AlH3)5}$ cluster I modeled in Avogadro. Two pairs of bridge bonds appear merged after UFF optimization, which doesn't seem right. This is what led me to wonder if modeling the bridge bonds the way I did was incorrect.

somewhat related http://chemistry.stackexchange.com/questions/55646/why-the-optimization-in-gaussian-shows-no-bond-with-some-atoms-in-diborane

What do you mean by "merged?" Maybe this is better handled through the Avogadro mailing list or forum

I'm really sorry for the confusion.  I've added numbers on the bridge bonds to the picture for clarity.  In the lower left corner, only 1 Al-H-Al bond is visible between each pair of Al atoms, while there are 2 everywhere else.  I'll try to word it better on the forum.  Thank you for the link!

I believe that you are going to need some QM  in order to optimise the geometry. Search the literature or ask on the forums to see what is the lowest level of theory you would need to obtain a reasonable geometry (if that is all you need). There is this, for instance: http://pubs.acs.org/doi/abs/10.1021/j100170a027?journalCode=jpchax  A bit old, though. Once you have your optimised geometry you can check if Avogadro can "understand" it. You can also try Gabedit. If it does, you are done. Otherwise you can manually edit it and save it to a format that stores bond information (such as mol2).

First of all, I am not an Avogadro developer, just an occasional user. My view of this matter is the following:

As 3D molecular drawing/editing software, Avogradro, does not need to "understand" B-H-B bonds. If it allows you to draw them in a sensible way, it is already fulfilling its function as a drawing/editing program.
Avogadro is also an interface to a number of codes, either internal or external, that allow you to perform tasks such as geometry optimisations and conformational sampling. Now, there becomes relevant whether or not those methods can "understand" B-H-B bonding. Ab initio and other quantum mechanical methods should afford a reasonable treatment of such molecules, provided that an appropriate level of theory is used, whereas for semi-empirical and molecular mechanics methods it will depend on whether or not a given parametrisation model or a given the force field (version), respectively, have been parametrised to take such situations into account.

In summary, if you are able to draw and edit the structure in Avogadro, the real question is whether or not you are going to do something else with it within Avogadro and whether or not your method of choice is appropriate to treat your molecular systems.

Thank you for your answer!  I was indeed concerned about the validity of auto-optimization on a molecule with lots of bridge bonds.  I'm working with (AlH$_3$)$_n$ clusters, and sometimes they don't look right after optimizing.  I guess I'll continue for now. :)

I'm the developer of Avogadro.
Beyond the other answer, which is excellent, I want to emphasize two things:

There is no rigorous definition of a "bond" in computational chemistry, particularly quantum chemistry. We like to visualize bonds as chemists, and Avogadro has a procedure for automagically interpreting bonding, and for allowing users to edit the bonds in the visualizer.
The UFF force field method used by default in Avogadro does have a "bridging hydrogen" atom type for cases like diborane.

There are also some nice ways of visualizing 3-center 2-electron bonds, using QTAIM if your software generates a .wfn file. (Under Extensions -> QTAIM -> Molecular Graph). Using QTAIM, the dotted path indicates the bent 3-center 2-electron interaction.
e.g.:

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text:
Are there any common and identifiable hazards to being a Chemistry Teacher?
I'm a High School Chemistry teacher. I have only been teaching for a few years and know very few other chemistry teachers. Of the ones I do, One has died of cancer, one is a cancer survivor and one is being tested and currently in cancer therapy while an equal number remain healthy (but have taught fewer years). That's scarry.
Could it be coincidence or some other more localized cause? Yes. Although cases were up to 100 miles apart. I realize the sampling size is small and this is all very anecdotal but I'm looking for answers nonetheless.  It seems to me that a possible culprit might be the haphazard way that stock handling and chemical lab supply making is done in High Schools as compared to higher education institutions or research facilities. Many schools don't vent their stock rooms even though it's required (there are no consequences) and teachers take the stock put it out on the counter and measure out what they need and make solutions. They don't bother with gloves, lab coats or aprons either. In a professional laboratory this would be done in a fume hood to minimize long term exposure to many substances with proper protective clothing.
Are there any readily identifiable health hazards for a High School Chemistry teacher based on the above information?

Unless you are working with exotic chemicals and high energy radiation, there should not be a problem.

Take a look at the list of carcinogens here http://www.cancer.org/cancer/cancercauses/othercarcinogens/generalinformationaboutcarcinogens/known-and-probable-human-carcinogens

Groups of cancer in people working together always causes interest, but clustering is a result of  randomness because regularly spaced events cannot cluster. Cancers do seem to be linked to lifestyle so you would have to remove non chemistry causes first from your study of chemists, and that I would suggest is almost impossible.

see these studies http://www.ncbi.nlm.nih.gov/pubmed/8480770 and http://www.ncbi.nlm.nih.gov/pubmed/7252610 and http://www.cdc.gov/niosh/oerp/pdfs/OERP-INFO-a.pdf

I have been teaching chemistry at the college level (if you include my years as a TA in graduate school) since 1999.  While there are certainly dangers involved in dealing with hazardous chemicals, I would not say that the job is unusually dangerous if you follow safe practices and use decent equipment.  I have known many chemistry teachers and professors who lived well into their 80's and beyond.  You're probably going to find certain workplaces which, for whatever reason, are more dangerous than most.  I have heard of (but not verified) an incident at a local college where five female biology professors were diagnosed with uterine cancer, an event which is extremely improbable to occur by chance.  Obviously, there is a substantial need to investigate such a situation and see if there is a common carcinogen present which is causing the cancer.  Such a situation could conceivably occur in the vicinity of a chemistry lab, although it seems unlikely if all standard precautions are followed.
The greatest source of danger is likely to be the students themselves.  When you have many young people together who are (1) easily distracted, (2) not particularly aware of their surroundings, and (3) lacking in manual dexterity (e.g. clumsy), they are far more likely to be a source of injury than the chemicals on their own.  So long as you are willing to invest the time in enforcing safety in your labs on a consistent basis, I think you're going to be just fine.

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text:
Solubility increases as Temperature increases.. Ks increases with temperature meaning dissolving is endothermic... But what about NaOH?

Question: I was self-studying and came across some confusion across a question. My confusion lies in this: Solubility increases as temperature (T)  increases, Ks increases with T, meaning dissolving is endothermic. But what about $\ce{NaOH}$?
The problem I am putting forward is more to do with the equilibrium Ks, especially using the Le Châtelier's principle. We all know that the solubility of ionic substances increases as increases in temperature. That is a fact, we don't have to argue with that. We also know that dissolving $\ce{NaOH}$ is exothermic, so we don't need to discuss that. However the two statements (observations) contradict the Le Chatelier principle, because if the dissolution of $\ce{NaOH}$ is exothermic, the solubility should decreases with increase in temperature! Why doesn't it?

You should also consider self-dissociation whe considering hydroniums and hydroxides.

Let me address this statement first:

We all know that the solubility of ionic substances increases as increases in temperature. That is a fact, we don't have to argue with that.

While this is frequently the case, there are plenty exceptions, and only one exception is required to prove the statement false. Lets consider the solubility of $\ce{Ca(OH)2}$.
$\begin{array}{rr}
\pu{0^oC}: & \pu{1.89g/L}\\
\pu{20^oC}: & \pu{1.73g/L}\\
\pu{100^oC}: & \pu{0.66g/L}
\end{array}$
$\ce{Li2SO4}$ is another example that follows this trend of decreasing solubility with increasing temperature, and there are plenty others.
Although the heat of a chemical reaction often follows Le Chatelier's principle, there are exceptions.  Based on the facts as laid out above and in the OP, the dissolution of $\ce{NaOH}$ is a case in which Le Chatelier's principle does not apply to the heat of the reaction; the dissolution of $\ce{NaOH}$ is exothermic, yet it's solubility increases with temperature.


text:
Why it is only carbons that have hydrogens attached (and not just hydrogens directly attached) to the charged carbon which result in hyperconjugation?
For the following compound:

Looking at figure 2, my understanding of hyperconjugation is that carbon 2 is more electronegative than the hydrogens attached to it, and so pulls charge towards carbon 2, giving it a partial negative charge. This negative charge then stabilises the positive charge on carbon 1.
Figure 2

My book says, for the compoound $\ce{C+H2CH2CH2CH3}$ there is only one hyperconjugation centre (i.e. the $\ce{CH2}$ attached to the C+).
Why can't the hydrogens attached to the C+ act as hyperconjugation centres? E.g. won't these hydrogens be less electronegative than the C+ and so the C+ will attract the electrons closer to it , somewhat stabilising the positive charge on the cabron?

Hyperconjugation happens because of an overlap of orbitals.
Consider this: in (uncharged) 2-methylpropane, the central carbon has sp3 hybridisation (and a hydrogen bound to it).
When the carbocation of the image is formed, the hydrogen leaves, the hybridization becomes sp2 (hence, planar), and now there is a free pi orbital available.
Here is what causes hyperconjugation: pi orbitals are quite larger than how they are drawn in books, and they can overlap partially with the orbitals of the adjacent hydrogens, and get part of their charge.
In the example of butane, the thing is the same: when the carbocation is formed, the carbon and the two hydrogens attached become planar (sp2), and a perpendicular pi orbital is formed. It cannot interact with the hydrogens attached to the charged carbon because they are on its nodal plane. Instead, it overlaps partially with the orbitals of the hydrogens on the adjacent carbon.


text:
Why is incubation in dark place necessary?
While estimating the  total protein through Lowry method one needs to incubate the test tubes in dark once after adding alkaline solution and once after Folin-ciocalteu solution. My teacher said it is necessary for proper colour development. But I am curious to know what is the chemistry behind it?

Sigma and this document say that the mixture is light sensitive, and it decomposed if exposed to light or stored for long periods. It's not that the reaction needs darkness to occur, just that the reagent will lose strength if the light hits it too hard, hindering the reaction with proteins and giving you less color.

Studies have not yet defined the exact mechanism of Lowry method. However, incubation in dark is suppose to promote the stability of copper tartrate complex with Folin-ciocalteu solution and thus their reaction/interaction with protein which ultimately leads to the oxidation of aromatic residues. You can read chemistry behind Lowry method for more details.

As folin reagent is used in reaction.. Folin reagent is sensitive to pH change.,The pH range is 10 to 10.5.
If it kept in light then due to temp. It's pH get change... So it is kept in dark

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text:
Why are alkyl halides more soluble in alcohol water solution rather than pure water?
Why are alkyl halides more soluble in alcohol water solution rather than pure water?
I could not find any satisfactory explanation using Google.

a brief answer on solubility in general  is given at https://socratic.org/questions/how-is-molecular-polarity-related-to-solubility.

As others have mentioned, it generally comes down to "like dissolves like".  Water is highly polar, low molecular weight alcohols like ethanol a bit less so.  By increasing the ratio of alcohol to water in a solution you are effectively making the solution less polar as a whole.  There are effectively fewer hydrogen-bonding opportunities in an alcohol water solution than in pure water.  In general, alkyl halides tend to fall in the low-to-intermediate polarity range.  Methylene chloride for example is only very slightly soluble in water, which makes it an excellent candidate for performing extractions.  By adding alcohol to water, you are giving the solution as a whole a more non-polar character, and increasing the potential for favorable interactions between the alkyl halide and the solvent.


text:
Naming of a Hydrocarbon with a Benzene Functional Group

What would the name of this hydrocarbon be?
I thought it might have been $\ce{3-phenyl-1-butene}$, however, after searching through a number of chemical websites, I found that 3-phenyl-1-butene isn't the correct name (the structures coming up had the benzene ring attached at the 4th carbon from the $\ce{C-C}$ double bond).

First, by your logic, it would be 3-phenylbut-1-ene. Note that the 1 is right before the ene, indicating that is where the double bond is. It is a perfectly understandable name, one that many chemists would use for this compound.
The other way to name it is to consider it a substituted benzene. Then you get (but-3-en-2-yl)benzene. This is the preferred IUPAC name for the compound, because benzene is the base molecule by IUPAC priority rules (rule P-52.2.8 of the 2013 edition of the Blue Book: “Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name.”).

Thanks to @Loong for pointing out in chat that “(but-3-en-2-yl)benzene” is the correct name, not “3-buten-2-ylbenzene”

Great to have you back!

(1-Methyl-allyl)-benzene is the correct IUPAC name. Benzene is the parent part of this hydrocarbon and 1-methy-allyl is the substituent. So, 3-phenyl-1-butene is wrong.

Please note that, according to current IUPAC nomenclature, the name ‘allyl’ is retained for general nomenclature only. Substitution is allowed, but not by alkyl or any other group that extends the carbon chain. Thus, the name ‘1-methylallyl’ is no longer a correct IUPAC name. (In the 1993 recommendations, the name ‘allyl’ was still retained for use with no limitation on substitution.)


text:
Why did the Rio 2016 pool water turn green?
One of the swimming pools at Rio 2016 Olympics has turned green:

While proliferation of algae is a likely culprit, there have been some alternative explanations offered by local organizers. CNN has the following quote from a Rio 2016 official:

A change in alkalinity
Mario Andrada, the communication director for the Rio 2016 local organizing committee, says a sudden change in alkalinity is the culprit.
"We expect the color to be back to blue soon," Andrada said, adding there is "absolutely no risk to the athletes or anybody."

How could a change in alkalinity occur on such a rapid time scale (a few hours)? And why would it directly change the color of the water? I can imagine that a more alkaline water could lead to precipitation of green metal hydroxide compound, but you would need a massive concentration of e.g. nickel to start with — does not seem very likely.
This article (http://www.independent.co.uk/sport/olympics/rio-2016-green-pool-drained-as-olympic-organisers-reveal-hydrogen-peroxide-is-cause-of-colour-change-a7189616.html) claims, that the treatment of the water via hydrogen-peroxide is what has caused the rapid change in the chlorine levels (which in turn made it possible for the algae to pollulate coloring the pool). Is this chemically possible? If so, how (or if not, why not)?
Also, the organizers claimed, that if was safe to enter the pool even after it had turned green. Was it really safe, or were the organizers lying about the hazards to cover up what they had done?

This article (http://www.independent.co.uk/sport/olympics/rio-2016-green-pool-drained-as-olympic-organisers-reveal-hydrogen-peroxide-is-cause-of-colour-change-a7189616.html) claims, that the coror change was caused by algae, after the chlorine in the water was neutralized by hydogen-peroxide. Could anybody explain how's that possible and was it really safe, to enter the pool after it has turned green?

Although not strictly alkalinity, this article quotes a changed pH level and an expert gives two possible explanations

One: A low or high pH can bring out minerals. "Calcium would show itself as cloudy water, copper would show up as a blue-green," he says.
Two: A too-high pH would render the chlorine ineffective, which could then allow for the proliferation of algae. (Hernandez says it's doubtful, though, that the green color is mainly from algae. It typically doesn't bloom that fast in such large pools.)

I can see an indirect effect (allowing algae to bloom), but the direct effect I'm skeptical… what concentration of metals would it require to begin with?

@F'x Very low. Remember, water is mostly transparent. It doesn't take a lot to change the colour of something that is already mostly colour-less (or white). And different countries (or cities, or even districts in the same cities) have different water supplies, with different kinds of impurities (either straight from the source, or from the water infrastructure itself - say, using copper pipes to transfer water).


text:
Chemical Equilibrium - finding equilibrium constant given degree of dissociation
$\ce{N2O4}$ is dissociated $20\ \%$ at 27 degrees Celsius and 1 bar. Find the equilibrium constant of that reaction. What percentage of $\ce{N2O4}$ dissociates at $13.15\ \mathrm{kPa}$?
An explanation of the solution is below, but there are two things I do not understand.

Why do we have $1 + \alpha$ in the denominator in equations ${\text(6)}$–${\text(9)}$?
How do we get the fraction $\dfrac{(2\alpha)^2}{1-\alpha}$ in equation $\text{(11)}$?

If the degree of dissociation is $\alpha$, then the amounts of substances present $n_0$ dissociated is $n_0 \alpha$, and the remaining amount of undissociated substance $n_0 (1-\alpha)$. Since each molecule $\ce{AB}$ dissociates into molecules of $\ce{A}$ and $\ce{B}$, the amounts of each substance is
$$\begin{align}
n_{\ce{A}} &= n_{\ce{B}} = n_0 \cdot \alpha \tag{1}\\
n_{\ce{AB}} &= n_0 \cdot (1-\alpha) \tag{2}
\end{align}$$
In the case of the dissociation of dinitrogen tetroxide, $\ce{N2O4}$, it will be
$$\begin{align}
n_{\ce{A}} &= 2n_0 \alpha \tag{3}\\
n_{\ce{AB}} &= n_0 \cdot (1-\alpha) \tag{4}\\
\sum{n} &=n_0 (1+\alpha) \tag{5}\\
x_{\ce{A}} &= \dfrac{2\alpha}{(1+\alpha)} \tag{6}\\
x_{\ce{AB}} &= \dfrac{1-\alpha}{1+\alpha} \tag{7}\\
x_{\ce{NO2}} &= \dfrac{2\alpha}{(1+\alpha)} =\dfrac{0.40}{1.20}=0.3333 \tag{8}\\
x_{\ce{N2O4}} &= \dfrac{1-\alpha}{1+\alpha}=\frac{0.80}{1.20}=0.6666 \tag{9}\\
K_p &= \dfrac{(P_{\ce{NO2}})^2}{P_{\ce{N2O4}}}=P\cdot \dfrac{( x_{\ce{NO2}})^2}{ x_{\ce{N2O4}}} = \mathrm{ 1\ bar \cdot \dfrac{(0.3333)^2}{0.6666} = 0.1667\ bar} \tag{10}
\end{align}$$
We obtain the degree of dissociation of $\ce{NO2}$ at a pressure of $13.15\ \mathrm{kPa}$:
$$\dfrac{K_p}{P} = \dfrac{(2\alpha)^2}{(1-\alpha)}=\mathrm{\dfrac{0.1667\ bar}{0.1315\ bar}=1.268} \tag{11}$$
From here, we get:
$$
4\alpha^2 = 1.268 (1-\alpha) = 1.268 -1.268\alpha  \tag{12}$$
Which becomes
$$
4\alpha^2 =1.268\alpha – 1.268 = 0  \tag{13}$$
For the degree of dissociation, we get:
$$\alpha =\dfrac{-b + \sqrt{b^2} -4ac}{2a} = \dfrac{-1.268 +\sqrt{1.268^2}+20.288}{8} \\= \dfrac{-1.268 +4679}{8} = 0.426  \tag{14}$$
Check your results:
$$K_p = \dfrac{(P_{\ce{NO2}})^2}{P_{\ce{N2O4}}}=P\cdot \dfrac{( x_{\ce{NO2}})^2}{ x_{\ce{N2O4}}} = \mathrm{0.1315 \cdot \dfrac{(2\alpha)^2}{1-\alpha} =0.1315 \cdot \dfrac{0.852^2}{0.574}= 0.1663\ bar}  \tag{15}$$

It looks like alpha $\alpha$ is the degree of dissociation. The way it is explained in your image is a little odd, so let me see if I can do a better job. What seems like it might be problematic for you is that the explanation you have been given skips some algebra along the way. Below, I am showing all of the algebra up until your points of confusion.
For your reaction: $\ce{N2O4 <=> 2NO2}$
In the first part of your question, we need to determine the equilibrium constant
$$K_p=\frac{P^2_{\ce{NO2}}}{P_{\ce{N2O4}}}$$
We know the total pressure, but we do not know the partial pressures of each species. Fortunately, thanks to Raoult's Law, we can substitute in mole fractions since $P_i =P \chi_i $.
$$K_p=\frac{P^2 \chi^2_{\ce{NO2}}}{P\chi_{\ce{N2O4}}}=P\frac{\chi^2_{\ce{NO2}}}{\chi_{\ce{N2O4}}}$$
We aren't given any number of moles, but let's say that initially there was some number of moles of $\ce{N2O4}$ before any dissociation occurred. We'll call that number $n_0$. We can write expressions for the number of moles of $\ce{N2O4}$ and $\ce{NO2}$ after dissociation using $n_0$ and $\alpha$. Since $\alpha\%$ of $\ce{N2O4}$ is dissociated, $(1-\alpha\%)$ remains undissociated:
$$n_{\ce{N2O4}}=n_0 (1-\alpha)$$
The moles of $\ce{NO2}$ are $n_0$ times $2\alpha$, since $\alpha\%$ of $\ce{N2O4}$ was converted into $\ce{NO2}$ and there are two $\ce{NO2}$ molecules for every $\ce{N2O4}$ molecule:
$$n_{\ce{NO2}} = 2n_0 \alpha$$
The total moles (need this for mole fraction) is
$$\sum{n}=n_{\ce{N2O4}} + n_{\ce{NO2}} = n_0 (1-\alpha) + 2n_0\alpha = n_0 (1-\alpha +2 \alpha) = n_0 (1+\alpha)$$
When we calculate the mole fractions of each species, the $n_0$ term cancels out, which is good since we do not know what its value is:
$$\chi_{\ce{N2O4}}=\frac{n_{\ce{N2O4}}}{\sum{n}}=\frac{n_0 (1-\alpha)}{n_0 (1+\alpha)}=\frac{1-\alpha}{1+\alpha}$$
$$\chi_{\ce{NO2}}=\frac{n_{\ce{NO2}}}{\sum{n}}=\frac{2n_0\alpha}{n_0 (1+\alpha)}=\frac{2\alpha}{1+\alpha}$$

Now you know where $1+\alpha$ came from. These values are plugged into the law of mass action for $K_p$ to determine the equilibrium constant.

In the second part of the question, you use the value of the equilibrium constant and the new pressure to determine a new value of $\alpha$. It all starts with the law of mass action and we manipulate some algebra from there. First the law of mass action using mole fractions:
$$K_p = P\frac{\chi^2_{\ce{NO2}}}{\chi_{\ce{N2O4}}}$$
Next, substitute in the expressions for mole fraction in terms of $\alpha$.
$$K_p = P\frac{\chi^2_{\ce{NO2}}}{\chi_{\ce{N2O4}}}=P\frac{\frac{(2\alpha)^2}{(1+\alpha)^2}}{\frac{1-\alpha}{1+\alpha}}$$
The term $1+\alpha$ appears in both the numerator and the denominator and can be canceled.
$$\require{cancel}
K_p = P\frac{\frac{(2\alpha)^2}{\cancel{(1+\alpha)}(1+\alpha)}}{\frac{1-\alpha}{\cancel{(1+\alpha)}}}=P\frac{(2\alpha)^2}{(1-\alpha)1+\alpha}=P\frac{(2\alpha)^2}{1-\alpha^2}$$
Now we move $P$ over to the left side of the equation to separate the values we know ($K_p$ and $P$) from those we do not know ($\alpha$).
$$\frac{K_p}{P}=\frac{(2\alpha^2)}{1-\alpha^2}$$

Now you have something like your second equation that you are confused about. Perhaps there is a mistake in the original source?

I still do not get it. Why haven't we squared  1 + alpha when you have written that xNO2 is squared? Shouldn't then the fraction go (2alpha/1+alpha) ^2 ? Why do we square only the numerator?

The equation is wrong, both terms should be squared as you noted. The bottom term should be 1-alpha^2 .

Good catch both of you.


text:
Is there a formula to tell how many conformers of a molecule to generate?
Let's say we want to generate conformers of molecules (e.g., ligands for docking in a protein pocket).
Is there a formula, like a function of the number of rotatable bonds, to decide how many conformers (maximum) to generate for a given molecule ?

If you are docking, shouldn't you be more concerned whether you find the lowest energy docked conformer(s)? Since the PES is very rough w.r.t. conformation, you would probably use a stochastic optimizer. Depending on the optimizer, the optimization path itself gives you an idea of the density of states (if your interested in free energies for instance).

I haven't seen any rigorous benchmarking or guidelines.
Consider that the total number of conformers goes up like $3^n$, where $n$ is the number of rotatable bonds, and ~3 is the approximate number of symmetrically unique conformers for each rotatable bond.
For small molecules with <3-4 rotatable bonds, using 100 conformers seems to be enough (i.e, only a small deviation from the crystal structure geometry). There's a more complete discussion in Ebejer's paper "Freely available conformer generation methods: how good are they?": J Chem Inf Model. 2012 52(5) pp. 1146.
(Slide from Jean-Paul Ebejer)

Beyond 3-4 bonds, it's less clear if sampling more conformers is a clear win. I would guess the answer is "yes," but I have not seen any papers (yet) dealing with such flexible molecules. One may need better tools, and also better force fields or computational methods.

I'll point out that $3^n$ = 81, for 4 rotatable bonds, suggesting greater sampling is needed. One can't sample an exponential number of conformers, but for 10 rotatable bonds, 100 is clearly insufficient. ($3^{10} \approx  60,000$)

Hi Geoff and thanks for the pointers. Why do you choose 3? This is super low. You are considering a rotational discrete step is 120 degrees.

@FrancoisBERENGER - somehow only saw this comment now. Consider that some conformers are symmetric with others. I've always used $3^n$ as a rule of thumb, but we're doing some exhaustive sampling and I might revise my answer here in a few months.

In isolation, enumerating very large conformer databases is meaningless. A better approach would be to sample rigorously but to look carefully at which conformers are "the same" based on an rmsd cutoff. Also I would be wary of including conformers with too high an energy. So define an energy window, say 10 kcal/mol and a similarity cutoff early.

@Francois, I've run into the same issue while being a newcomer to computational chemistry. I, however, have the pleasure of access to Spartan and Gaussian at school- and GAMESS at home. Allow me to share what method I've come to favor lately. Keep in mind, I may be wrong, but I may be right.
I am investigating phenolic compounds with alkyl tails no more than five carbons long. In Spartan, I will change the torsion angles from their default of 180 or 120 deg(i.e., 2 or 3 rotations) to forty degrees. I have also learned that in older versions of Spartan, there existed a keyword "THOROUGH" which would automatically increase the number of default torsions by 3- this made me feel more comfortable choosing 9 across the board. What I've found with these alkyl tails- especially for butyl and pentyl, as you could likely imagine- quite a few conformers are produced.
Essentially, I am after the varying conformations of the -OH groups as I am looking at bond dissociation energies, and have become aware that hydrogen bonds will differentially affect polyhydroxylated compounds in ways which require close attention. The conformation of the alkyl tails don't mean much to what I'm doing, so I simply view the conformers which are lowest in "tail energy" and rid my set of the ones which have roughly equal -OH orientation, but with varying tail orientation. This allows me to locate many conformers of -OH groups, all with the lowest "tail energy". I keep those conformers, of course, and perform semi-empirical restricted Hartree-Fock calculations, so far with 3-21+G* basis set, to provide me with further information on the equilibrium geometry before I continue with higher level theory in Gaussian 09.
Here is the kicker, and mostly the reason for my post- Spartan also allows one to display the weighted Boltzmann probabilities (terminology?), or rather the equation discussed here, and especially here, to find which conformers are most likely to exist. I have yet to decide at what probability I can truly drop one of my preferred conformers, but for the sake of the discussion, I will just dream for a moment that P<0.05 would be more than enough reason to drop a conformer (for reasons to obvious in context).
I don't want to say much more, on account of my lack of experience, but I do hope that I have provided some insight, in lack of a mathematical equation of which OP inquired.

A straight forward formula to know the number of possible conformers (C) is:
$$
C = \prod_{i=1}^{n} \theta_i
$$
Where n is the number of rotable bonds and θ are the number of allowed angles for each rotable bond. For example, if you have a molecule with n=4 rotable bonds, the number of possible torsional angles being 2, 3, 4, 6 the number of conformers is:
$$
C = \prod_{i=1}^{4}\theta_i = \theta_1\theta_2\theta_3\theta_4 = 2x3x4x6 = 144
$$
There are some cautions that must be considered (repeated conformers depending on the values and combinations of θi, suitable choice of θi, etc.) but it might work as a first approximation.


text:
What's the difference between radial probability and probability density?
I understand that radial probability is the probability of finding an electron at some distance from nucleus in a shell. But what does that exactly mean? What does it tell us about the atom and electrons?

Probability density is probability per volume.  Probability density at a given point means probability per volume in the limit that the volume is infinitesimally small.
Radial probability distribution at a given radius is the probability per distance that the event occurs in a infinitesimally thin spherical shell at that radius. (The distance being the thickness of the shell).

Knowing the radial probability of finding an electron as a function of distance from the nucleus tells you a lot about the atom. For instance this electron density can be used to calculate the potential energy and to get an idea about the shell structure. You can estimate many different properties like cohesive energy, band gap, optical properties etc.  Take a look on the Thomas$-$Fermi theory, the Hartree$-$Fock method or Density Functional Theory (DFT) to get an idea.

The 1st question was answered by dave.The 2nd question was ' what the radial probability tells us about the electron/atom'. I tried to give some hints, little downvoter ..


text:
Acids, Bases and Salts?
I'm a 7th grader, and this is what is printed in my Chemistry textbook:
An Acid is a chemical compound that is sour in taste and is soluble in water.
A base is a chemical compound that is bitter in taste.
A salt is a compound that is form alongside water when an Acid and a Base reacts.
Come on. That can't be what Acids and Bases are! What is an Acid, a Base and a Salt?

This understanding is over 200 years out of date.    http://chemistry.stackexchange.com/questions/43989/why-are-acids-usually-depicted-in-fiction-as-green/44046#44046    -  Can you post which chemistry textbook you are using and when it was published.

related http://chemistry.stackexchange.com/questions/35041/acids-bases-salts-etc

http://www.ratnasagar.com/bookdetail/bookdetail.aspx?MainID=TTAx&SubCat=Y2F0MTg= this is the book.

They do not normally expect 7th graders to achieve understanding of Arhenius, Bronsted, Lewis, etc.  Thus that explaination. You can search up these to read up. These are theories defining what acids and bases are.

@RishiNandhaVanchi I am familiar with them now. This post is ~5 years old. I'm in Grade 12 :).

I will try to make the answer as non-technical as possible, to make it understandable for you.
Some substances are composed of atoms, like $\ce{H}, \ce{O}, \ce{C}, \ce{Fe}$, etc. Each of these atoms actually exist out of even smaller particles, called protons, neutrons and electrons. Protons are positively charged ($+1$), neutrons are, as you might have guessed, neutral, and electrons are negatively charged ($-1$).
The protons and the neutrons are together in the core of the atom, the electrons are in a cloud around it. Because atoms are not charged, an atom has equally many protons as neutrons. A picture will help: (Source: User:Fastfission at Wikipedia)

However, in some cases, those atoms take an extra electron or give a electron to another atom. They are then called ions. For example, a normal sodium atom $\ce{Na}$ has 11 protons and 11 electrons, but a sodium ion $\ce{Na^+}$ has 11 protons and 10 electrons. Similiarly, a normal chloride atom $\ce{Cl}$ has 17 protons and 17 electrons, but a chloride ion $\ce{Cl^-}$ has 17 protons and 18 electrons.
Now we know enough to define what a salt is: A salt is a particle that is is composed of one or more positively charged ions and one ore more negatively charged ions. For example $\ce{Na^+Cl^-}$, more commonly just written as $\ce{NaCl}$, is common salt, i.e. the kind of salt people do on some of their food (at least in some countries).
There are two special ions, $\ce{H+}$ and $\ce{OH-}$. If you have liquid water, then these two are inside. Now, we call something an acid if there is more $\ce{H+}$ inside than $\ce{OH-}$, and we call something a base if there $\ce{OH-}$ inside than $\ce{H+}$.
I'd also like to note that the book definition is dangerous: An Acid is a chemical compound that is sour in taste. Acids, or chemical compounds in general, are of course not things you should taste to test what they are! They might be poisonous.

So practically, a base is a hydroxide of an element, right?

@AravindSuresh Usually, but not always. If you add $\ce{OH-}$ to $\ce{H+}$ (note that it is not possible to actually have those seperated, there will be other positive and negative ions around), then you get $\ce{H2O}$, which is not an acid nor a base. A reaction might look like $$\ce{HCl + NaOH -> NaCl + H2O}$$ Note that you have an acid (HCl) and a base (NaOH) on the left side, and on the right side we have a salt and water.

Aravind, As you can see from the answers and comments you have opened Pandora's box [look her up]. What you have astutely inferred at an early age is that textbooks, teachers, the internet, and everything are incomplete in explaining anything. I will try to explain acids and bases in the manner your text tried. Remember this is to describe the acids and bases you probably encounter daily or occasionally not the chemical and physics reasons that there are acids and bases. First! Every substance likely to be encountered is made of atoms [except possibly a lightning bolt or electric shock they are made of electrons, but the electrons came from atoms and are trying to get back to atoms, it is prudent to not get in their way.]
The common properties of acids and bases involve water and aqueous solutions but are not limited to water. We will stick to water and aqueous solutions. Water is both an acid and a base for 2 reasons The first is that it reacts with itself to form very small amounts of H3O+ and OH- ions,  2H2O = H3O+ + OH-, and second because water can donate protons, be an acid, or accept protons, be a base. We have established that an acid in water is either a solution of H3O+ ions [commonly called hydrogen ions or H+ ions] or is a compound that can possibly donate protons to a base. A base is a compound that either forms hydroxide ions in water or can accept protons from an acid.
There are two main classes of acids and bases. The first is Strong; these react completely with water and are essentially solutions of H3O+, acids; or OH-, bases. Common strong acids are hydrochloric-HCl [stomach acid], nitric-HNO3, Sulfuric-H2SO4; common strong bases are Sodium hydroxide-NaOH, potassium hydroxide-KOH, possibly calcium hydroxide-Ca[OH]2. The second are the Weak acids and bases. These only react partially with water to give hydronium or hydroxide ions. Some weak acids are acetic [vinegar], citric [lemon, orange juice] lactic [sour milk, yogurt]. Some weak bases are bicarbonate ion, HCO3-; Ammonia, NH3.
Acids and bases react to give water usually leaving the counter ions in an almost neutral solution. Removing the water results in a compound called a salt. the term comes from common table salt NaCl that can result from the reaction of NaOH and HCl. Salts usually dissolve in water as ions Positive ions from the base and Negative ions from the acid. Some common salts are Sodium chloride NaCl common table salt, Ammonium chloride NH4Cl found in cough drops, sodium bicarbonate NaHCO3 baking soda, sodium carbonate Na2CO3 washing soda, sodium hypochlorite NaOCl bleach, Calcium carbonate CaCO3 limestone, marble, chalk, Tums antacid. Salts are everywhere.
Common acids: Stomach acid-hydrochloric, acetic acid-vinegar, oxalic acid-spinach and other greens, Citric- lemons oranges fruits, lactic-sour milk, carbonic soda water rain water.  Some common bases: sodium hydroxide-lye drain cleaner bleach, Ammonia- some cleaners, bicarbonate ion-baking soda antacids, Carbonate ion-washing soda.
Acids, bases, salts, And Chemistry are everything about you. Observe, ask questions and look for answers. Best of everything.

Taste is not the best way to test for acids and bases, Ok for food, it is dangerous elsewhere. Litmus paper is the old standard acid is red base is blue. There are natural indicators tea is lightened by acids, Red cabbage is red or blue [i forget which check it out]


text:
Why does mesitylene absorb at a longer wavelength than benzene (UV-Vis)?
According to UV-Vis spectroscopy, mesitylene (1,3,5-trimethylbenzene) absorbs at $\lambda_\text{max} = 210\ \mathrm{nm}$ while benzene absorbs at the slightly shorter wavelength of $\lambda_\text{max} = 204\ \mathrm{nm}$.
Why is that? One would think that $\pi$-ring's hyper-conjugation with the $\sigma^*$ orbitals of the methyl hydrogens would help stabilize the system thereby increasing the gap between the HOMO and LUMO (in turn decreasing absorption wavelength).
An opposite effect is observed when nitrobenzene ($\lambda_\text{max} = 252\ \mathrm{nm}$) is compared to 2-nitrotoluene ($\lambda_\text{max} = 250\ \mathrm{nm}$) and 2-isopropylnitrobenzene ($\lambda_\text{max} = 247\ \mathrm{nm}$). In these cases, the addition of larger and larger alkyl groups seem to be decreasing the absorbed wavelength (therefore increasing gap between HOMO and LUMO).
Is that because of sterics? Or because the the strongly electron withdrawing nitro group (conjugation) is being "counteracted" by the alkyl electron donating groups and therefore electrons are more concentrated in the ring rather than being spread though the molecule?
EDIT:
My question in more general terms: I understand HOW more conjugation results in a decrease in absorption frequency (due to molecular orbitals overlapping constructively vs destructively therefore decreasing gap between HOMO and LUMO). What I don't understand is HOW pumping electron density into a conjugated system (or withdrawing) effects the HOMO and LUMO. An explanation from a MO perspective would be greatly appreciated.

The beginning of an answer might be found here. Electron donating/withdrawing groups have different effects whether they are located on an odd/even carbon from the chromophore.

@SteffX Right, but why is that? Is there a review article or textbook section that explores these concepts in more detail?

The general observation is that alkyl substitution intensifies and shifts the benzene spectrum to longer wavelengths. The effect is similar to that of groups with lone pairs such as $\ce{NH2}$ and O-alkyl. So I don't think that opposite effects are observed as you suggest. Some data are given below.
The 0-0 band transition, from  v=0 vibrational level to v' =0 in the first excited state of benzene is at 262 nm (due to symmetry this is a forbidden transition so its position has to be calculated. The maximum in the S0-S1 transition in benzene spectrum is at 245nm, and S0-S2 at 203 nm).  Toluene does not have a forbidden 0-0 transition, and has a 0-0 band at approx 266 nm (with maxima at 261 nm and S0-S2 at 207 nm), 0-zylene 0-0 is at approx 273 nm, so the substituents do stabilise the excited state very slightly. This has been attributed to 'hyperconjugation', i.e. some charge being donated by the methyl group to the ring, or the alkyl groups behaves as it it had a lone pair. (The idea of hyperconjugation is still hotly contested). Toluene shows a slight increase in intensity as the symmetry is broken $\ce{D_{6h} to C_{2v}}$ but otherwise the effect on the electronic spectrum is slight. Aniline has a relatively broad and intense  absorption maximum (not 0-0) at 280 nm, nitrobenzene max very broad spectrum with a maximum at approx 268nm, but with a long shoulder in absorption stretching out to almost 400 nm, which is thought to cover the n-$\pi^*$ transition.
With very broad and structureless spectra (almost all the molecules you mention except benzene and toluene) the maximum intensity does usually always correspond to the 0-0 energy of the transition, due to Franck-Condon factors (overlap of wavefunctions which control intensity of transitions) and the fact that extra groups add considerable vibrational intensity to transitions and so broaden them. Thus comparing absorption peaks is not a reliable way of comparing state energies. (Experimentally the only ways are to measure vapour  phase absorption spectra or measure in a solid matrix  at 4k if possible.) Secondly spectra are stabilised differently in different solvents, even relatively non-polar ones and this can lead to spectral shifts.
The general sort of handwaving explanation, that is nevertheless useful,  is to suppose that the molecule has energy levels that can be described like those by a  particle on a ring. The bigger the ring the lower the energy levels become and so the lower the energy of the electronic transition. Thus incorporating a substituent that involves conjugating the electron over more atoms, increases the ring size very slightly and so  produces a longer wavelength transition.


text:
What's the structure of "inorganic benzene"?
Can anyone give me a possible structure for a compound called "inorganic benzene"?

@Agriculturist How is that a dupe?

@bon I see your point - missed the difference between benzene and inorganic benzene.

"Inorganic benzene" also known as borazine has a molecular formula of $\ce{B3N3H6}$.

It is a planar structure. Both B and N are sp2 hybridised. It is called "inorganic benzene" as it is isoelectronic with benzene.

What are the sources of pics?


text:
Comparison Of Lewis Acidity
Which is a stronger Lewis Acid $\ce{AlCl_3}$ or $\ce{SiCl_4}$?
I am confused as in $\ce{AlCl_3}$ , Al has an incomplete octet so it must be a stronger Lewis acid as compared to $\ce{SiCl_4}$, but the answer is just opposite.

Well, answer is wrong so I guess there was a mistake or you misunderstood sth.

agree, AlCl3 is much certainly a stronger Lewis acid.

AlCl3 + RCl → AlCl4(-) indicating that AlCl3 is a Lewis Acid
but
SiCl4 + RCl → SiCl5(-) ??????  So, this tells me that AlCl3 > SiCl4 in terms of electron accepting behavior.
Now, don't get me wrong: pentacoordinated Si is well known. One of the failures of the Lewis Acid concept is that, as far as I know, there is no unambiguous way to measure it. It is much more qualitative than quantitative, despite years of attempts to do so. You may know that Phosphorous is typically pointed to as the first example of the breakdown of the octet rule. (see Hypervalence) After Period 3, the concept has little use, imho. Since Lewis Acidity is tangled up with Lewis structures, perhaps it is not too surprising that the concept has some serious problems. Usually, we condsider both electronic effects as well as steric effects to 'explain' Lewis Acid/Base strength. One popular alternative is Hard.Soft.Acid.Base (HSAB) "theory". Anyway, comparing a tri-coordinated species with a tetracoordinated species and predicting which will form a bond with an electron donor is really difficult. Did you know that AlCl3 doesn't really 'exist' in the solid or liquid state? And even in the gaseous state, the dimer is common. The structure of the (solid) dimer (Al2Cl6) has two Al-Cl-Al bonds. Anyway, as far as I know AlCl3 is the stronger Lewis Acid in most apple-to-apple comparisons. To be honest, I don't think that if the Lewis Acid concept weren't a historically established meme, that it wouldn't be invented today. Personally, scales should actually be founded on measurement. The Lewis Acid strength scale is just sorta kinda correlated to what we can actually measure.


text:
Why 3-chloroprop-1-en-1-ol is unstable?
In the treatment of HCl with acrolein 3-chloroprop-1-en-1-ol is the first product which is (my textbook says) unstable and turns into 3-chloropropanal.
Why is this 3-chloroprop-1-en-1-ol unstable? is −I effect of chlorine responsible?

This earlier answer might provide some useful background

3-Chloroprop-1-en-1-ol is in equilibrium with the aldehyde, 3-chloropropanal (aka $\beta$-chloropropionaldehyde).
According to Organic Syntheses, this compound "is a very unstable substance which polymerizes rapidly, especially in the presence of traces of hydrochloric acid".
But I do not see it being reduced to 3-chloropropanol.

Reactivity/instability
In 3-chloropropanal, or more precisely in its enol form 3-chloropropenol, the chlorine atom is in allylic position (i.e. on a carbon atom adjacent to a $\ce{C=C}$ double bond) which makes it more reactive towards nucleophilic substitution. Another molecule of the enol can then substitute it and the resulting product is
$\ce{Cl-CH2-CH=CH-O-CH2-CH=CH-OH}$
which still has an allylic chlorine atom, so it can react further, hence the polymerization.

I have edited the question. Can you explain it more clearly. Your use of 'it' is puzzling me.

Enols are virtually always in equilibrium with their keto tautomer.  In this case, the predominant compound you would expect to encounter is 3-chloropropanal.  There are exceptions to this (1,3-diketones for example) but this is generally the case, particularly when acid or base is available to catalyze the reaction.


text:
Reaction between NaOH and CO2
So I wanted to know what the reaction between sodium hydroxide and carbon dioxide can be, and upon research I got 2 answers. The first one is
$$\ce{CO2 + NaOH(aq) -> NaHCO3(aq)}$$
and the second one is a two-step reaction, first with water then sodium hydroxide:
$$\ce{CO2 + H2O(l) -> H2CO3(aq)}, \\
\ce{H2CO3(aq) + NaOH(aq) -> NaHCO3(aq) + H2O(l)}$$
So my question is which one of these answers is right?
To me, I do not understand why the 2nd option takes into account water because really concentrated $\ce{NaOH}$ would/should be written just as $\ce{NaOH}$, which to me makes the 1st option the right one. Am I wrong?

Short answer is that you're right, but quite a bit can be told about carbonic acid and existence of H2CO3 molecule.

could you please give me the longer explanation?

As Mithoron mentioned, H2CO3 is a unstable compound and was only isolated as a pure compound recently.  You cannot obtain H2CO3 directly through chemical reaction.

can you tell how much water will be in solution in the 1st equation?

This question is a great opportunity to talk about state symbols, ionic bonding, and multi-step reactions.
Very often when we write an equation for a chemical reaction we only look at the starting material and the products, like when you wrote the first form:
$$\ce{CO2 + NaOH (aq) -> NaHCO3 (aq)}$$
Multi-Step Reactions
Firstly, this form doesn't necessarily tell us what molecules are interacting with one another.  Is one molecule of carbon dioxide colliding with one sodium hydroxide molecule to make a sodium bicarbonate molecule?  In this case, no.  Two different reactions happen in sequence.

Carbon dioxide and water react to make carbonic acid.
$$\ce{CO2(aq) + H2O(l) -> H2CO3(aq)}$$
Carbonic acid reacts with hydroxide ions to make bicarbonate ions.
$$\ce{H2CO3(aq) + NaOH(aq) -> NaHCO3(aq) + H2O(l)}$$

Reactions in Aqueous Solution
Secondly, the state symbols tell us important information about the reaction conditions.  The (aq) state symbol tell us that the reaction is happening in aqueous conditions.  That means the reactants are dissolved in water which can have a significant effect on how the reactants behave.  Here the water reacts with carbon dioxide to form molecules of carbonic acid, $\ce{H2CO3}$, in solution which is an important intermediate in this reaction.
$$\ce{CO2(aq) + H2O(l) -> H2CO3(aq)}$$
The aqueous conditions also mean that ionic compounds are separated into their constituent ions.  So $\ce{NaOH(aq)}$ actually means a solution of separated positive sodium ions, $\ce{Na+}$, and negative hydroxide ions $\ce{OH-}$, which can each react on their own.
As you might notice, that means our solution contains molecules of carbonic acid and basic hydroxide ions.  A neutralisation reaction occurs.  The basic hydroxide ions remove a proton from the carbonic acid to make water and a bicarbonate ion.
$$\ce{H2CO3(aq) + OH-(aq) -> HCO3-(aq) + H2O(l)}$$
Because the sodium ions don't react with anything, we can leave them out of the equation.  This is similar to how we don't explicitly include water in the one step equation - because we would have the same amount of water on each side of the equation.   If we add the sodium ions we get something equivalent to the equation for the second step.
$$\ce{H2CO3(aq) + Na+(aq) + OH-(aq) -> Na+(aq) + HCO3-(aq) + H2O(l)}$$
In water, $\ce{NaOH(aq)}$ and $\ce{Na+(aq) + OH-(aq)}$ are the same because sodium salts dissociate into their ions in water.
Adding equations
To see the equivalence between the multi-step form and the shorter one step form we can add together the equations for the two steps to get an idea of what is happening overall.
$$\ce{CO2(aq) + H2O(l) -> H2CO3(aq)}\\+\\
\ce{H2CO3(aq) + Na+(aq) + OH-(aq) -> Na+(aq) + HCO3-(aq) + H2O(l)}\\
\implies \ce{CO2(aq) + H2O(l) + H2CO3(aq) + Na+(aq) + OH-(aq) ->\\ Na+(aq) + HCO3-(aq) + H2O(l) + H2CO3(aq)}$$
There's a lot going on in this equation, but we can make it more readable but taking away $\ce{H2O(l)}$ and $\ce{H2CO3(aq)}$ from both sides, just like any mathematical equation.
$$\ce{CO2(aq) + Na+(aq) + OH-(aq) -> Na+(aq) + HCO3-(aq)}$$
In aqueous solution, sodium salts dissociate into their ions so we can write the ion pairs as the aqueous sodium salts.
$$\ce{CO2 + NaOH (aq) -> NaHCO3 (aq)}$$
So we've gone full circle, from a single summary equation, to the two reactions involved, back to the summary equation once again.

No both are right.
If you add the latter equations, then you will get the former.
If sodium hydroxide solution was highly concentrated like you said, then both equations are wrong since sodium hydroxide solution will strip away the second hydronium as well forming carbonates.
All state notations with (aq) implies that it is dissolved in water. If it was pure NaOH, it would be in a solid state under room temperatures. It is also noteworthy to know that NaOH must be contained in a dry environment as it will spontaneously absorb water from the air.

oh okay but would the second one have access water molecules while 1 is just all the NaHCO3 in ion form. In the sense that even the 2nd one will have Na and HCO3 ions, but will it also have access water molecules?

Both are aqueous solutions. As I've stated in the answer, state notations are important. Ionic compounds with sodium usually dissociate all the way in water, but to know how much an aqueous substrate dissociates, you need to know its equilibrium constant.

okay, how would you calculate concentration of sodium hydroxide to use to make sure HCO3- remains in solution and not carbonate ions?

You'd need to know the ratio of NaOH and CO2 dissolved in moles.

Which one dominates would also depend on how you perform the experiment.  Bubbling CO2 into concentrated NaOH would favour the carbonate more than adding a NaOH solution to water continuously saturated with CO2

The mechanism of reaction between the acidic oxide and the alkali depends on the concentration of the alkali solution.
(a) When the alkali ($\ce{NaOH}$) solution is very dilute ($\mathrm{pH} < 8$), carbon dioxide will first react with water to form carbonic acid ($\ce{H2CO3}$) slowly. The acid thus formed then reacts with the alkali to give sodium hydrogencarbonate ($\ce{NaHCO3}$).
$$\begin{align}
\ce{CO2 + H2O &= H2CO3}\\
\ce{NaOH + H2CO3 &= NaHCO3 + H2O}
\end{align}$$
(b) When the alkali solution is a fairly concentrated one ($\mathrm{pH} > 10$), carbon dioxide directly reacts with it to form the bicarbonate, which further reacts with the alkali to form sodium carbonate ($\ce{Na2CO3}$) as the main product by complete neutralization.
$$\begin{align}
\ce{NaOH + CO2 &= NaHCO3}\\
\ce{NaHCO3 + NaOH &= Na2CO3 + H2O}
\end{align}$$
Thus, only when the concentration of the alkali solution is quite low, the reaction proceeds via the formation of carbonic acid. But the acidic oxide is not completely neutralized in this case.


text:
What is the mechanism of saccharification by H2SO4?
We have been estimating total carbohydrate by Umbreit method/ Anthrone method in college. The website I have been reading says that $\ce{H_2SO_4}$ and heat is used to treat a polysaccharide solution yield hexoses and pentoses which further undergo dehydration to form hydroxymethylfurfural and furfural respectively. These compounds when react with Anthrone form coloured compounds.
I have been wondering what the mechanism of hydrolysis of polysaccharide is?

Hmm, you're asking about most boring of those reactions :( Also don't write that you found nothing, what's the point in that?

http://chemistry.stackexchange.com/questions/49703/hydrolysis-of-sucrose-over-sn1ac-or-sn2ac the mechanism given in the question by OP is correct. Don't think this qn deserves a -1. Googling "polysaccharide hydrolysis mechanism" or "glycosidic bond hydrolysis mechanism" don't give many good results outside of papers which might not be accessible to OP. The best search term seems to be "sucrose hydrolysis mechanism" but you would have to know that the mechanism of hydrolysis of sucrose (a disaccharide) is the same as that of polysaccharides.


text:
Ratio of energy required for electrolysis vs creating water
I had an idea a while back for a machine that would be totally green. Powered by a Hydrogen Fuel Cell. A big question came up: Where do I get the Hydrogen? I assumed that I could just electrolysis of water to create Oxygen (which would be released into the air) and Hydrogen (which would be used in the cell). But, then I remembered a key factor of bonds: Creating a bond gives the nuclei a more stable configuration, which releases energy. Breaking a bond requires an input of energy.
Let's say I use electrolysis to split water. This will take energy. Then I use a Hydrogen Fuel Cell to generate electricity. This process gives me the same amount of water that I just split.
Q: Which process uses/gives more energy? Would the above example result in a net gain or loss?

If your electricity generation, usage, transmission, and storage are perfectly efficient, you will break even exactly. Otherwise, you will come out at a loss. It is impossible to gain energy from this process.

Theoretically, you will get the same amount of energy out of putting hydrogen and oxygen through a fuel cell as it takes to produce them by electrolysis, and you can reach this result by adding up all the enthalpies of formation involved using Hess' Law and realizing that the fuel cell is just the reverse process of electrolysis.
It is therefore impossible to have a net energy gain by this process as it would violate the conservation of energy.
It is, however possible to lose energy during this process and in the real world, energy is lost during both steps.
Electrolysis:

Takes a net input of energy to produce hydrogen and oxygen
Much of the electricity added is wasted to produce heat due to electrical resistance and overpotential losses.
Current large scale electrolysis systems have efficiencies of at best ~70% [1]
Storage of the gases is also energetically wasteful as they need to be compressed. (or perhaps the hydrogen stored as a metal hydride)

Fuel cell:

Produces a net output of energy by consuming hydrogen and oxygen.
Suffers from the same types of electrical losses as the electrolysis process, resulting in sub-100% conversion efficiency

Fuel cells are still useful devices as they can produce electricity without combustion, without the limitation of sunlight, etc., but they require a source of hydrogen.
Most hydrogen is produced from natural gas because producing it by electrolysis is so inefficient and requires energy from somewhere else.
In the future electrolysis combined with fuel cells might become useful for temporarily storing electricity for load balancing solar panels or other green power sources, and there are some projects using hydrogen generation for grid storage purposes where excess power is stored as hydrogen in underground reservoirs, but they use hydrogen combustion generators to get electricity back as fuel cell technology has not advanced sufficiently. Example in Newfoundland

Electrolysis vs reproduction of water would probably create a net loss in energy.  However, a significant problem in alternate energy sources is consistent availability.  Thus we struggle with energy storage, i.e. the battery.  Batteries create problems of their own, i.e elements to produce batteries, toxic byproducts, etc., etc.  However, the production of pure hydrogen, inconsistent with high an low volume levels dependent on the electrical source (wind, solar, whatever) over the long term results in a stable source of energy with a byproduct of pure water upon combustion.  Produce hydrogen as often as possible with available electrical sources, compress into tanks and produce an easily stored, stable, clean energy source.  While stored, we enjoy surplus oxygen.  After use, we have the purest water on the planet.  This seems simple.  Tell me the overwhelming problem with this scenario.


text:
Why is the lanthanide element Thulium malleable?
I was doing some research about Thulium and every website says that it is malleable, but doesn't explain why.
What chemical properties does Thulium possess which makes it malleable?

All the lanthanides are soft metals; there is nothing special about thulium in that regard, AFAIK. For that matter, so are the alkali metals, and gold can be scratched with a fingernail.
Most pure metallic elements are malleable. "This is because of the ability of the atoms to roll over each other into new positions without breaking the metallic bond." Brittleness (the opposite of malleability and ductility) often occurs because of impurities such as carbon, or a metal alloy which separates on cooling. These impurities "pin" dislocations in the metal, preventing them from moving smoothly. It makes alloys harder, as well as more brittle. The early technologists discovered that the alloy bronze was harder than pure copper.

Bcc metals generally are not very malleable. Think tungsten (or pure iron for that matter). Fcc and hcp (like thulium) have much nicer slip systems for plastic deformation.

Pure tungsten is ductile and malleable: look at a filament of an incandescent lamp. It did take some time to make W of that purity, so Edison had to use C, but by WWII, it was used to make filaments and nosecones of bazooka shells. See https://en.wikipedia.org/wiki/Tungsten#Physical_properties


text:
Why don't covalent compounds have ions?
When a covalent bond is formed, elements gain shared electrons. Wouldn't this mean covalent compounds contain ions? According to BBC and other sources they don't have any ions at all... How is this so?

Covalent compounds don't have ions because otherwise they would be classified as ionic and not as covalent. It is as simple as that.

Any bond between two elements has a certain covalent character or some ionic character. When the ionic character is much greater, the compound is said to be ionic and vice versa. An example of an ionic compound is common salt or $\ce{NaCl}$ while diamond is a covalent compound formed by carbon atoms which are $\mathrm{sp^3}$ hybridised. When covalent character is high, electrons mostly occupy the bonding orbital between the two nuclei. The electrons are said to be more in a "shared" state. In ionic compounds like $\ce{NaCl}$ , $\ce{Na}$ almost gives its electron due to high metallic character and $\ce{Cl}$ almost accepts due to high electronegativity. In such case, much positive charge is found on $\ce{Na}$ while high negative charge is found on chlorine. In diamond, however no charge is found on carbon (although in a molecule like $\ce{HCl}$, $\delta-$ charge is found on $\ce{Cl}$ and $\delta+$ on $\ce{H}$ due to electronegativity difference).

It is difficult to understand what do you want to ask.


text:
Gay-Lussac law explanation
I was wondering if someone could explain me what exactly is Gay-Lussacs law?
And my textbook says that Berzelius and Dalton concluded based on that law that all gases have same amount of particles. How?

Gay-Lussac's Law, in its most basic form, says that, assuming constant volume (V) and amount/moles (n), the pressure (P) of a gas is directly proportional to the absolute temperature (T) of a gas.  We can write this mathematically in a couple different ways:
$$ P = kT$$
 (assuming constant V and n), where k is some constant.  More often, we use it in the form below, where we know the pressure and temperature under one set of conditions, and then change those conditions (again, keeping V and n constant):
$$
\frac{P_1}{T_1} = \frac{P_2}{T_2}
$$
As for the next statement, "all gases have the same amount of particles", as stated this makes no sense.  Perhaps what you meant to say was that, under identical conditions, equal volumes of an ideal gas contain equal number of moles (and therefore equal numbers of gas particles).  For example, it can easily be shown using the Ideal Gas Law that one mole of an ideal gas occupies 22.4 L at STP.


text:
How to quench unreacted sodium azide from reaction mixture?
The reaction involved synthesis of azide formation from alkyl bromide in DMF using sodium azide. After the reaction has taken place, how do I quench the unreacted sodium azide from the reaction mixture?

Why do you want to "quench" it?

I happen to read about safety precautions regarding handling, using sodium azide. So I came across few web pages, where they warn you not to dispose it into the drain but try to quench it. So I thought about to get an opinion on this. Anyways  I took my reaction mixture, extracted in ethyl acetate/water. And it worked well.

Excess of sodium azide can be quenched using nitrous acid.
$$\ce{2NaNO2 + H2SO4 → 2HNO2 + Na2SO4}$$
$$\ce{2NaN3 + 2HNO2 → 3N2 + 2NO + 2NaOH}$$
For further reading, Click here.


text:
How prevalent is the use of old style names for ions (e.g. ferric, ferrous) outside the classroom
In teaching my preparatory chemistry courses, I'm constantly wondering whether or not I should be having students memorize the "old names" of common multivalent monatomic cations, such as plumbic for lead (IV), plumbous for lead (II), etc.  Every year I teach the course, this information seems increasingly outdated, as I find that I'm coming across these less and less in my general reading.  To the readers I'd like to pose this question: In your practical experience, particularly in industry or other fields outside chemistry, do you still employ this style of referring to ions and compounds?

Im sure the answer will always be opinion based, but certainly I've never been taught these old names. The only time I see them day-t0-day is on labels of old chemical containers, but even then theres usually an accompanying molecule formula. I feel like students hate having to remember things (nomenclature wise) without trying to coax them into learning trivial names that are neither endorsed by IUPAC nor used by anyone under the age of 50.


text:
Defining an acid-base reaction from the perspective of the Arrhenius theory of acids and bases
$$\ce{NH3 + HCl -> NH4Cl}$$
The above equation is the reaction of ammonia with hydrochloric acid (all substances are aqueous). I understand that ammonia counts as a base because it can react with water to give hydroxide ions, and $\ce{HCl}$ is an acid. By logic, this should be an 'acid-base' reaction and this link
says that it is indeed an acid-base reaction but how is this so if no water is produced?
I am learning acid-base theory from the beginning and can understand why this is an acid base reaction from the perspective of the Brønsted-Lowry theory and the Lewis theory, but I am now attempting to understand the Arrhenius theory, so please explain to me, in terms of the Arrhenius theory, why this is an acid-base reaction and a 'neutralistaion' if there is no water produced and no hydroxide ions which take place in the reaction what so ever?

Unless you are considering this equation as a gas phase reaction then substances are by convention always dissolved in water, unless some other solvent is specifically stated. Thus you know immediately that there are $\ce{ H^+}$ (or $\ce{H3O^+), NH4^+ , Cl^- and OH^-} $ ions present.

I understand that ammonia counts as a base because it can react with water to give hydroxide ions, and HCl is an acid.

This is where you went wrong.
The Arrhenius theory is the oldest of the common acid-base theories. According to Arrhenius:

An acid is a compound that will dissociate in aquaeous solution forming $\ce{H+}$ ions.
A base is a compound that will dissociate in aquaeous solution forming $\ce{OH-}$ ions.

As per the Arrhenius theory, ammonia is not a base. Also, contrary to all later theories (Brønsted-Lowry, Lewis and Pearson), Arrhenius did not account for any acid-base reactions; instead, he intended to define only those substances whose solutions reacted acidicly or basicly. It is therefore meaningless to define an acid-base reaction according to the Arrhenius theory.
What Arrhenius did account for was neutralisation reactions. Indeed, if an Arrhenius acid and an Arrhenius base are mixed, the form water and a salt, which can be easily shown in the equation below:
$$\ce{(H+ + A^-) + (B+ + OH^-) -> HOH + A^+B^-}$$
But it was not until Brønsted’s and Lowry’s theory that it was realised how a reaction should always be acid-base and not one or the other.

I'll just add that there are multiple theories about how to characterize acids and bases. By learning the different theories and where they are applicable you learn about the nuances of chemistry.

Arrhenius theory is based on the interaction of the two substances in water. Since there is no water in your equation, it can't be applied directly. However, we can make some partial equations to work around it:
$$\ce{NH3(aq) + H2O(l) -> NH4+(aq) + OH-(aq)}$$
$$\ce{HCl(aq) + H2O(l) -> H3O+(aq) + Cl-(aq)}$$
Now the neutralization equation:
$$\ce{H3O+(aq) + OH-(aq) -> 2H2O(l)}$$
If you add the three equations you end up with your original equation:
$$\ce{NH3 + HCl -> NH4Cl}$$

Thank you very much, so is it just a case of considering more than one reaction taking place, then adding together the equations? The way that you've done it, it seems to me like the 3 equations can be regarded as the mechanism by which the reaction takes place.

It's a neat trick in chemistry: you can break down a reaction equation into as many partial equations as you want in order to explain it better. The cool part is that they don't need to be able to happen in the real world in the way you represent them. The ion-electron method is a popular case where half the equations are mere fantasy, but in the end it all wraps nicely (http://chemistry.bd.psu.edu/jircitano/redox.html). Just remember that this way of doing chemistry doesn't represent the mechanism by which a reaction occurs. It's only a trick to understand better what's going on.

But how can I possibly understand better what's going on, when it isn't even going on? And the ion electron equations, aren't they merely a representative of what happens when you have specific ions in a solution, irrespective of the compound? Also, if the ammonia and hydrochloric acid reaction doesn't happen in the way in which you represented it, then this isn't actually an acid base reaction. Where am I going wrong? I'm so confused, all of chemistry seems wrong

I used this method to show you how it's an acid base reaction according to the Arrhenius theory. If you apply the Lewis theory, an acid is a substance capable of accepting a pair of electrons to form a coordinate covalent bond, while a base donates that pair of electrons to form that bond. That is what happens between the H+ of HCl and NH3, to form NH4+ (+Cl- since it's a salt). Don't worry about seeing on paper things that aren't real. Every field of science has its own little tricks, like when physicists assume that lenses are negligibly thin. It's wrong but it makes math easier.

Another question, if I may. I understand that the Arrhenius theory was rejected because it meant that a substance which was described as an acid in water, wouldn't be described as an acid in gas phase (for example), which was silly. Firstly, if the reaction takes place directly as a collision between NH3 and HCl, then why can't the gas phase reaction just be regarded as an acid-base reaction? Secondly, why can't we just dissolve a substance in water in order to see if it increases the concentration of H+ or OH- ions then permanently define them as acids and bases, irrespective of the phase?

Also, when I said mechanism, I didn't mean curly arrows thing, but I meant to ask whether the steps which you mentioned actually physically exist. All the reactions you mentioned are in equilibrium. Reaction 3 is the natural equilibrium which exists in water. Reaction 2 causes an increase in H3O+ ions and a decrease in water molecules while reaction 1 causes an increase in OH- ions concentration and a decrease in water concentration. This decreases the equilibrium constant of reaction 3, which means that it does indeed take place. In that sense, is that why it's a neutralization reaction?


text:
What is the number of unpaired electrons in Fe 2+?
Specifically, how can I find out the number of unpaired electrons in any ion?

First draw the electronic configuration for Iron.
Now since we are given with Iron(II) we take out two electrons from the outermost orbit so as to give the iron a +2 charge.

As evident from the configuration, in total there are 4 unpaired electrons.
Using the same logic you can count the number of unpaired electrons for any ion.

Did you create those images for the $e^-$ configurations yourself?

no , took a picture from google and then used paint to edit it

Not just any ion, but ground state ions.
Following Aufbau and Pauli rules until you reach the transition metals.They fill up 3d orbitals after 4s orbitals are filled. But when they are ionized, their electrons fill only the 3d orbitals in ground state.
So Iron (II) ion will have 4 unpaired electrons.


text:
Reaction of succinic acid with ammonia
When succinic acid is treated with excess $\ce{NH3}$, what will the mechanism be? I feel both $\ce{COOH}$ groups should be converted to $\ce{CONH2}$, but this is not the product.

Did you actually mean to say $\ce{COONH2}$? Because that is about as wrong as it gets …

I actually think you mean $\ce{CONH2}$, the amide, but feel free to correct me if I'm wrong.

Yup.. Its amide

@Jan It's not as simple as it seems... the ring actually closes. (Picture taken from here)

@DHMO I knew. But OP suggested a peroxide product which was just badly wrong ;) (Actually, it is as simple as it seems because five-membered rings are kinetically favoured and intramolecular attacks are faster than intermolecular ones, etc. etc.)

@Jan there's another problem: we don't need to make the acyl chloride before making the amide?

@DHMO No, amides are the ‘thermodynamic hole’, they will be formed from acid + $\ce{NH3}$ given enough time. All conditions out there are just to accelerate. It works better, if you start at an ester though.

There are no conditions given! In which case, why not ammonium bis-succinate?

From Synthesis of Succinimide on Thomas' Chemistry:

Add 186 g of succinic acid to RBF, place the beaker in an ice bath and start adding 220 ml of 26% ammonia slowly. Assemble short path distillation without a condenser, add some boiling chips. Heat the mixture under a small flame to distil of all the water. When the temperature reaches over 100°C the mixture darkens. Switch the receiver and collect everything that comes off before yellow fumes temp 280-300°C. Cool down to room temperature and place the receiver into a freezer for 24h, recrystallize the crude succinimide in hot ethanol. Then filter the crystals and wash with 40 ml of cold ethanol.

(Note that RBF = round-bottom flask)

The catch is that the amide can attack itself to close the ring, forming succinimide.
There are two rounds of amidification via a nucleophilic addition/elimination mechanism.

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


text:
The natural equilibrium of water
Why can the equilibrium reaction which exists in water be simply represented as
$$\ce{H2O <=> OH- + H+}$$
rather than
$$\ce{2 H2O <=> OH- + H3O+}$$
Doesn't this change both the equilibrium constant AND the enthalpy change?
Why would this still give the correct answer when you use the ionic product of water in solution?

There is no such thing as bare $\ce{H+}$ in water. When anybody talks about it, they actually mean $\ce{H3O+}$. That's why there is no difference.

The former equation is just a simplified version of the latter.
The important factors are the concentrations (activities) of the positive and negative species.
However, since these values are small ($10^{-7}\,\mathrm{mol\cdot l^{-1}}$) compared to the concentration of water ($55,56\,\mathrm{mol\cdot l^{-1}}$) this distinction would not affect the value of the dissociation constant either.

It is said that water is incorporated in the equilibrium constant. [H2O] will DEFINITELY have a different value from [H2O]^2...

I should probably delete my last sentence or stick to activities rather than concentrations. Since the activity of pure condensed phases is 1 per definition, and thus $[\ce{H2O}] = 1$, it would become clearer. I'll change that...
https://en.wikipedia.org/wiki/Thermodynamic_activity


text:
Why alkali metals have bcc structure?
It is well known that metals are typically, well, metallic and usually favor the most dense packing, like hcp or fcc cells. This may be easily rationalized as dense packing of metal ions in a sea of electrons, since metals typically have low electronegativity. The exceptions, however, do exist.
However, the simplest metals, where one would except the metallicity to define everything, have bcc cell, and not the densest packing. I'm speaking about alkali metals. Interestingly, alkali earths favor dense packings (hcp for Be, Mg and fcc for $\ce{Ca},\ce{Sr}$) except for $\ce{Ba}$ and $\ce{Ra}$ (again, favor bcc). So, the question is: why alkali metals have bcc cell and what guards relative stability of bcc/hcp/fcc cell stability in $\ce{Li/Ba}$ square? Bonus points for the fitting description for the p-block metals. I'm pretty sure there is no simple answer for d-block, as valent interactions involving d-electrons become the deciding factor there.

Be, Ca, and Sr are bcc at higher temperatures. Bcc is not quite as dense as fcc or hcp, but isn't so far off. And, it has a different coordination number, so is favored in some cases. Simple predictions of crystal structures are bound to fail in multiple cases for multiple reasons. Seeking simple heuristics in the periodic table is a losing proposition.

@JonCuster That's pretty much obvious and is not helpful at all. I'm not seeking for a general heuristics, I'm seeking for simple explanation for several selected simple cases.

The point being that even those examples aren't 'simple', even though you think they should be.

@JonCuster Are you an oracle to know what I'm thinking? I do have a chemical education and know that the question IS complicated. I'm ready for a complex answer. I do not appreciate wasting my time on someone's attempts to figure my though process. Please, avoid such attempts in the future.

Then perhaps you could spend some time clarifying the question - it is true that I am not an oracle. Perhaps I'm particularly dense (close packed?), but what you are looking for as an answer is not coming across clearly to me. As one example, why would one expect 'densest packing' to be what all metals, particularly simple ones, aspire to?

(Note that, like you, I have seen many naive questions on the periodic table and why it is not rigidly monotonic). Depending on what direction you really want to head in, I have a number of starting points to suggest to dig more deeply in to the band structure of the alkali metals. I'm not quite sure which are the most relevant for your query. It starts stepping deeply in to solid state physics pretty quickly, in any direction.

For the moment I will focus on Na, although I may mention other elements in comparison.
First, it is true that the STP stable phase of Na is bcc. However, the energy difference between the bcc, fcc, and hcp phases is quite small. As one example, the SGTE database (A.T. Dinsdale, CALPHAD 15(4) 317-425 (1991)) gives the Gibbs free energies (relative to BCC) in Joules/mole as FCC = -50 + 1.3T, and HCP = -104 + 2T. These are really small; normally one would expect kJ/mole or 10's of kJ/mole type differences. Some experiments indicate that Na transforms to hcp below 51K. It is also known that at lower temperatures (35K), Na undergoes a martensitic phase transition. So, bcc is stable at STP, but not very from a big-picture thermodynamics view.
For a more detailed view, lets start with Ashcroft & Mermin's Solid State Physics book. They start their description of the alkali metals (Chapter 15) with

The alkali metals have singly charged ions (whose core electrons from the tightly bound rare gas configuration and therefore give rise to very low-lying, very narrow, filled, tight-binding bands) outside of which a single conduction electron moves. If we treated the conduction electrons in the metal as completely free, than the Fermi surface would be a sphere of radius $k_{F}$...

Then we get

De Haas-van Alphen measurements of the Fermi surface confirm this free electron picture to a remarkable degree of precision, especially in Na an K, where deviations in $k_{F}$ from the free electron value are at most a few parts per thousand... Thus, the alkalis furnish a spectacular example of the accuracy of the Sommerfeld free electron model. It would be wrong to conclude from this, however, that the effective crystalline potential is minute in the alkali metals. What it does suggest is that the weak pseudopotential method (Chapter 11) is well suited to describing the conduction electrons in the alkalis. Furthermore, even the pseudopotential need not be minute, for except near the Bragg planes the deviation from free electron behavior occurs only to second order in the perturbing potential...

In other words, they (including Na) really are 'simple' metals, but even simple metals might not be as simple as one might expect.  Note that nowhere is the question of fcc vs hcp vs bcc brought up, but this is a textbook. However, being bcc does not prevent the electrons from being close to free.
Moving on to more recent work on the crystal and electronic structure, I'll turn to On the Constitution of Sodium at Higher Densities (which has Ashcroft as a co-author, likely resulting in the excellent writing as well as the references to Wigner and Seitz' work on sodium from the 1930's). Some quotes relevant to the problem at hand include:

Under atmospheric conditions, the ground state electronic
  structure of sodium therefore conforms to the classic
  nearly free electron system of Wigner and Seitz ...,
  having a single valence electron and an equilibrating density
  where the core volume is certainly an exceedingly
  small fraction of the cell volume. The pseudopotential
  for sodium at this density is quite well approximated as
  local, and under moderate compression a purely local picture
  also continues to suffice: The primary Fourier component
  of the pseudopotential $V_{G}$ diminishes relative to
  the growing bandwidth. But beyond an approximately
  3-fold compression in volume this simple interpretation
  no longer holds. In Fig. 1 we plot the band structure of
  fcc sodium at $r_{s} = 2.3$ and compare it with that of lithium
  at the same density. The bands near points K, W, and L
  of Na are quite different from those of Li, and, interestingly,
  the Fermi surface does remain nearly spherical in
  Na at these elevated densities.

and

We have seen here that at the relevant length scales the electron-ion interaction in sodium is no longer weak under compression ..., and its putative "simple" metallic behavior at low densities appears to be an accident of the relative core and unit cell volumes present at one atmosphere.

Stepping deeper in to the simulations, one might look at First-principles study of the structural properties of alkali metals for pseudopotential calculations of bcc, fcc, and hcp in Li, Na, and K. Another interesting paper might be Structural phase stability in third-period simple metals, where calculations on fcc, hcp, and bcc for Na, Mg, Al, and Si are shown. One point of interest in this paper would be Figure 9, where they show the calculated interatomic pair potentials extracted from their calculations, including showing where the nearest neighbors fall. For Na, while the fcc nearest neighbors are closest to the minimum of the pair potential, the first and second bcc neighbors both are in that first minimum. This gives 14 nearest neighbors for bcc in the big minimum vs only 12 for fcc, yielding the possibility (and reality) that the bcc phase can be slightly more stable than fcc.
So, to wrap up:

Being bcc does not limit the simple nearly-free electron behavior for metals.
For sodium, the energy differences between bcc, fcc, and hcp are really small. The observed stability of bcc at STP is the result of slight differences in the effective pair potential (as well as dynamic effects such as the electron-phonon coupling).
Lower temperature or higher pressure push the balance towards other phases.

I would be happy to expand discussion on specific points if so desired.

Oookay, that's enlightening. So, the energy difference between fcc/hcp and bcc is quite small and actually favors fcc. However, entropy plays for bcc and overwhelms energy difference at relatively low temperature, thanks to very low energy difference. The reason for it, as I understand, is that in bcc atoms have more 'swing', with larger displacements. This also allows for electron-phonon interaction , lowering bcc energy a bit. Did I sum it up right? O, and thanks for the paper.

@permeakra - Yup, that is the basic summary. It all comes down to some slight variations in the pair potential. The 'free electron'-like behavior (spherical Ferrmi surface) holds for both bcc and fcc (and likely hcp), so that really doesn't impact anything.  Sorry if I came across as a pain in the comments above, but there are a few subtleties flying around on this question...

Nice. I  put a bounty, and reward it once the time is up.

Great answer! In addition, recently published research shows, from DFT calculations, that dispersion interactions and phonon contributions play a main role in stabilizing the bcc phase in Group 1 metals even at $0$ K. Indeed, higher temperatures further stabilize the bcc phase for these metals. However, comparing with Group 11 (also 1 valence electron), the cohesive energy differences are like ten times greater than Group 1, favoring fcc. Further, bcc becomes unstable for LJ potentials, so many-body effects should contribute as well.


text:
How do we count charges in this example of a Redox reaction equation?
My brother(younger) had a doubt regarding this chemistry book example so I let him here :
Question :
Balance the equation : $\ce{H+ + MnO4- + Fe^2+ -> Fe^3+ + Mn^2+}$
Solution steps(as given by the textbook):
Step 1 :  $$\ce{MnO4-  ->  Mn^2+}$$
Step 2 : $$\ce{MnO4-  -> Mn^2+ + 4 H2O}$$
Step 3 :  $$\ce{MnO4- + 8 H+  -> Mn^2+ + 4 H2O}$$
Step 4 : $$\ce{MnO4- + 8 H+ + 5 e- -> Mn^2+ + 4 H2O}\tag{1}$$
Step 5 : $$\ce{Fe^2+ -> Fe^3+ + e-}\tag{2}$$
Step 6 : $5\times(2) + (1)$ and thus we have the balanced equation:
$$\ce{MnO4- + 8 H+ + 5 Fe^2+ -> 5 Fe^3+ + Mn^2+ + 4 H2O}$$
Explanation given in the textbook for step 4:  we equalise the charge by adding 5 electrons on left.
My doubt regarding step 4: how do I know that in step 4 that 5 electrons do actually equalize the charge in the given equation? More accurately how do I know that there is any surplus or deficit of charge in any given equation atleast in this case?

@Nij Please check out formatting hints for chemical equations. Using plain MathJax for chemical equations is wrong since the elements should be typeset upright, not in italics.

If you look at the charges in step 4, the left side adds up to 7 and the right side adds up 2. The MnO4 on the left has a -1 charge, and the 8 hydrogens add a +8 charge, adding up to a +7 charge on the left overall. The Mn on the right has a +2 charge so that is +2.
To balance out this charge, they must be equal. How can +7 be made equal to +2? By subtracting 5. In this case, adding 5 electrons creates a -5 charge that means both sides of the equation have a +2 charge.
You just need to add up the charges like this for a given equation to balance it.


text:
Why do Hydrogen and Helium have molar volumes higher than an ideal gas?
In my chemistry textbook, there is a table of the molar volumes for different gases. Most of them are below 22.42 l/mol, the ideal gas molar volume, but I noticed that hydrogen and helium do not. Hydrogen is 22.433 l/mol and helium is 22.434 l/mol. Why is this? What makes these particles, with mass, have higher molar volumes than the basically mass-less particles of an ideal gas? And why does the molar volume increase when going from Hydrogen to Helium?

Mass is hardly relevant here. The deviation from ideal molar volume is an interplay of two factors, both of which are absent in ideal and present in real gases: (1) volume of molecules themselves and (2) attractive interactions between molecules.

And of those two factors, the nonzero volume may outweigh the attractive force in the case of hydrogen and (to a greater extent) helium, giving a positive deviation from the ideal gas volume--whereas attractive forces dominate in the other gases, causing a negative deviation from ideal volume.

Are these tiny differences between hydrogen and helium really significant?

@ChesterMiller The major differences that come about from the tiny differences in IMF's are in the boiling points (H2: -252.9 degC, He: -268.9)  [values from wikipedia]

The ideal gas approximation has nothing to do with mass. Ideal gases just have neither molecular volume nor intermolecular interactions!

The ideal gas law gives a very good approximation at STP, so at this point the question is mostly trifling over decimal places. For a nonideal gas, increasing the pressure and decreasing the temperature will tend to increase negative deviations from ideality: the molecules are interacting and want to stick together. Increasing the temperature while increasing the pressure, however, can introduce positive deviations from ideality, where the molecules repel each other on average, compared to a non-interacting ideal gas. If you give them enough energy, the molecules slam into each other so hard that they never even feel the underlying attractive potential.
A very useful principle here is that of corresponding states. If we take the temperature and pressure and divide them by the critical temperature and pressure of the gas we're using, we get the "reduced" temperature and pressure. The principle of corresponding states says that all gases should behave the same when they're at the same reduced temperature and pressure. If we do this for hydrogen, helium, and nitrogen at STP, we get:
$$\begin{array}{cr}
\text{Gas} & T_\mathrm{crit}/\pu{K} & P_\mathrm{crit}/\pu{bar} & T_\mathrm{r} & P_\mathrm{r}\\ \hline
\text{H}_{\text{2}} & 32.9 & 12.9 & 8.3 & 0.08 \\
\text{He} & 5.2 & 2.3 & 52.5 & 0.44  \\
\text{N}_\text{2} & 126 & 33.9 & 2.2 & 0.03
\end{array}$$
From this you can see that from hydrogen and helium's perspective STP conditions are quite hot indeed! Since we've introduced corresponding states, we can look at a chart of the "generalized compressibility factor" and estimate the deviation from ideality. This link gives such a chart and from it we can see that at such high reduced temperatures, we expect a positive deviation from ideality (Z>1) and thus a larger molar volume than the ideal gas.

As hydrogen and helium's molecular weight are too less, the intermolecular attractions are also too less. So the a/V^2 is negligible.  So the v.d.o equation becomes P (V - nb) =RT . By which, it can be said as the pressure of these elements are too less and from Boyle's law , the volume is higher.


text:
Differences in specific free radical reactions enthalpies

Iodination of alkanes using iodine $(\ce{I2})$ is usually an unfavorable reaction. Tetraiodomethane $(\ce{CI4})$ can be used as the iodine source for iodination, in the presence of a free-radical initiator such as hydrogen peroxide. Propose a mechanism (involving mildly exothermic propagation steps) for the following proposed reaction. Calculate the value of $\Delta H$ for each of the steps in your proposed mechanism ($\ce{cp = cyclopentyl}$).
  $$\ce{cp-H + CI4 ->[H2O2, heat] cp-I + HCI3}$$
The following bond dissociation enthalpies may be helpful:
  \begin{array}{lr}\hline
\text{bond} & \Delta H / \pu{kJ mol-1}\\\hline
\ce{I3C-I}  &  188 \\
\ce{HO-OH}  &  213 \\
\ce{HO-I}   &  234 \\
\ce{cp-H}   &  397 \\
\ce{I3C-H}  &  418 \\
\ce{cp-I}   &  222 \\\hline
\end{array}

From the book: L. G. Wade Organic Chemistry; 8th edition; Pearson: 2013; exercise 4-57.
The correct answer is given as:
\begin{align}
\text{Initiation:}\\\hline
\ce{HOOH &-> 2HO*}& \Delta H &= \pu{+213 kJ/mol}\\
\ce{HO* + I-CI3 &-> IOH + CI3}& \Delta H &= \pu{-46 kJ/mol}\\[2ex]\hline
\text{Propagation:}\\\hline
\ce{cp-H + *CI3 &-> cp* + H-CI3}& \Delta H &= \pu{-21 kJ/mol}\\
\ce{cp* + I-CI3 &-> cp-I + *CI3}& \Delta H &= \pu{-34 kJ/mol}\\\hline
\end{align}
The overall enthalpy is $\Delta H = \pu{-55 kJ/mol}$ for the propagation steps.
In the book, there is also an overall table of bond dissociation enthalpies.
One important one I used here is the one of water, $\pu{498 kJ mol-1}$.
I have a significantly more exothermic process here:
\begin{align}
\text{Initiation:}\\\hline
\ce{HOOH &-> 2HO*}& \Delta H &= \pu{+213 kJ/mol}\\
\ce{cp-H + OH* &-> cp* + H2O} & \Delta H &= \pu{-101 kJ/mol}\\[2ex]\hline
\text{Propagation:}\\\hline
\ce{cp* + I3C-I &-> cp-I + I3C*}& \Delta H &= \pu{-135 kJ/mol}\\
\ce{I3C* + cp-H &-> cp* + I3C-H}& \Delta H &= \pu{-21 kJ/mol}\\\hline
\end{align}
Here the initiation is less endothermic, and the propagation is significantly more exothermic. So why is this not given as the correct answer? Am I missing something here? I'm quite new at organic chemistry, so I think I must be wrong, not the textbook.


text:
Co vs Rh atomic radius
Why is the empirical atomic radius for both Co and Rh 135 pm, the same?
Source

Why not? Why can't two elements have a similar radius, particularly considering how fuzzy a concept it is in the first place?