text:
How does electron localization/delocalization work?
I can't find a simplified explanation on this topic. My textbook just adds to the confusion and all I'm really left with are more questions rather than answers.
And adding to that question, does polarity have an effect on electron localization/delocalization?

In general terms, looking through the meaning of the terms, it seems that, delocalization of electrons refers to something which is not restricted to a particular atom (we can expect this in metallic bonds). Whereas, localization is the opposite of that, where we find electrons restricted to a particular atom (we can expect this in covalent bonds). I am not strictly speaking, but seeing through the terms, I have this opinion.

Draw the LCAO orbitals.  Delocalization is the summation of discrete electron occupations of orbitals, such as a 1,3 shift (a pi-bond flip).  Remember that LCAO is a terrifically wrong black box. Its only virtue is that it simply, beautifully works (until the Woodward-Hoffmann rules).
Given propene, abstract a methyl proton.  The LCAO anion is half at both ends by a 1,3-shift of the pi bond.  MO shows degenerate orbital occupation still with (nearly) no additional density at the central carbon.  Abstract an allylic proton from 1-butene.  The allylic ends are no longer 50:50 occupied by the anion for the different electronic properties of two attached protons versus a methyl group and a proton.

There is no simplified explanation, and especially not of "how it works". The behavior of electrons is governed by quantum mechanics. So electrons do what they do, because it is energetically favorable. It should become quite clear upon obtaining master degree in theoretical chemistry ;)
As a teaser, I'd at least show the pictures:

Localized sigma bond, localized means it is as compact in space, as possible, now connecting two carbon atoms in ethane:

Delocalized pi electrons in benzene, shared by 6 atoms:


text:
How do I calculate the cell voltage using the following half-reactions and solubility products?
The concentrations of $\ce{NaCl}$ and $\ce{KI}$ are at $\pu{0.120 M}$ and $\pu{0.090 M}$, respectively, in the following electrochemical cell:
$$\ce{Cu_{(s)}|CuI_{(s)}|I^{–}_{(aq)}||Cl^{–}_{(aq)}| AgCl_{(s)}|Ag_{(s)}}$$
How do I calculate the cell voltage using the following half-reactions:
\begin{align}
\ce{Cu+ + e- &<=> Cu(s) & E^\circ &= \pu{0.518 V}} \\
\ce{Ag+ + e- &<=> Ag(s) & E^\circ &= \pu{0.7993 V}}
\end{align}
The solubility products $(K_{sp})$ for $\ce{AgCl}$ and $\ce{CuI}$ are $\pu{1.8e–10}$ and $\pu{1.0e-12}$, respectively.

I think you need the Nernst equation. This is something we were just introduced to today in my class, so I’m not so sure about the notation you’re using in your question at the top, but you should be able to correct my answer to fit your needs if you understand the equation I’m about to implement.
$$E_\text{cell} = E^\circ_\text{cell} - \frac{RT}{zF}\ln Q$$
Where $E^\circ_\text{cell}$ = what the $E_\text{cell}$ normally is, under STP with the solutes at a concentration of $1\ \mathrm{mol\ dm^{−3}}$.
$$E^\circ_\text{cell} = E^\circ_\text{cathode} − E^\circ_\text{anode}$$
Since, the reduction potential of copper is lower, it will be the reducing agent in this reaction.
$$E^\circ_\text{cell} = -0.7993\ \mathrm{V} + 0.518\ \mathrm{V}$$
$$E^\circ_\text{cell} = -0.281\ \mathrm{V}$$
Where $R$ is the familiar gas law constant:
$$R = 8.314\ \mathrm{J\ K^{-1}\ mol^{-1}}$$
$T$ is the absolute temperature (it’s always assumed to be $298.15\ \mathrm{K}$):
$$T = 298.15\ \mathrm{K}$$
$z$ is the number of moles of electrons transferred in the half-reaction:
$$z = 1$$
And $F$ is Faraday’s constant, which is measured in coulombs per mole:
$$F = 9.649 \times 10^4\ \mathrm{C\ mol^{-1}}$$
And $Q$ is the also familiar reaction quotient, but as I discovered, it takes some work to calculate, and you have to use the quadratic equation.
$$Q=\frac{[\ce{red}]}{[\ce{ox}]}$$
$$Q=\frac{[\ce{Cu+}]}{[\ce{Ag+}]}$$
$$K_{\text{sp},\ce{CuI}}=10^{-12}=[\ce{Cu+}][\ce{I-}]$$
$$1.0 × 10 ^{-12} = [X][0.090 + X]$$
Do the quadratic formula
$$X = 1.11 \times 10^{-11}\ \mathrm{mol\ l^{-1}}$$
$$[\ce{Cu^+}] = 1.11 \times 10^{-11}\ \mathrm{mol\ l^{-1}}$$
$$K_{\text{sp},\ce{AgCl}} = 1.8 \times 10^{-10}\ \mathrm{mol\ l^{-1}} = [\ce{Ag+}][\ce{Cl-}]$$
$$1.8 \times 10^{-10} = [X][0.120 + X]$$
Do the quadratic formula
$$X = 1.50 \times 10 ^{-9}\ \mathrm{mol\ l^{-1}}$$
$$[\ce{Ag+}]=1.50 \times 10^{-9}\ \mathrm{mol\ l^{-1}}$$
$$Q=\frac{1.11 \times 10^{-11}\ \mathrm{mol\ l^{-1}}}{1.50 \times 10^{-9}\ \mathrm{mol\ l^{-1}}}$$
$$Q=0.00740$$
So, plugging this all into the Nernst equation, the answer should be:
$$E_\text{cell}=-0.281\ \mathrm{V}-\frac{8.314\ \mathrm{J\ K^{-1}\ mol^{-1}} \times 298.15\ \mathrm{K}}{1 \times 9.649 \times 10^4\ \mathrm{C\ mol^{-1}}}\ln(0.00740)$$
$$E_\text{cell}=-0.155\ \mathrm{V}$$
I should add that since $R$ is a constant, $F$ is a constant, and $T$ is usually something like $298.15\ \mathrm{K}$ or just not measured in elementary chemistry, the equation can just be
$$E_\text{cell}=E^\circ_\text{cell}-\frac{0.0257}{z}\ln Q$$
Which will give the same answer:
$$E_\text{cell}=-0.281\ \mathrm{V}-0.0257\ \mathrm{V}\ln(0.00740)$$
$$E_\text{cell}=-0.155\ \mathrm{V}$$

Generally you can neglect X and avoid the quadratic formula. .1 plus a very small number is still .1.

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text:
How would substitution reaction results differ between reacting halides with a tertiary bromoalkane vs reacting it with a primary bromoalkane?
Say, you conducted two separate reactions; in one, you combined $\mathrm{50~mmol}$ saturated $\ce{KCl}$ solution with $\mathrm{10~mmol}$ 1-bromooctane (assume heating), and in the other you combined $\mathrm{10~mmol}$ saturated $\ce{KI}$ solution with $\mathrm{10~mmol}$ 1-bromoctane. In both cases $\mathrm{0.5~mmol}$ hexadecyltributylphosphonium bromide was used as a catalyst.
So, due to relative charge densities and the fact that the electrophillic carbon is primary, you would have a low yield of 1-chloroocatane vs starting material in reaction 1 and you would have a high yield of 1-iodooctane vs starting material in reaction 2, both cases obtained via SN2 reaction.
Now, suppose you would have repeated the same two reactions, only this time you would have reacted the $\ce{KCl}$ and $\ce{KI}$ with 2-bromo, 2-methyloctane, how exactly would the results differ from the results of the previous two reactions, or would they have been the same?
I know that the reaction would have no "choice" but to react via the SN1 (or perhaps E1) route, as the electrophillic carbon is tertiary and thus very sterically hindered. Also, I know that $\ce{I-}$ ions are good nucleophiles, but that doesn't matter in this case, does it?
Perhaps this reaction would go slower than the previous, owing to the fact that it requires the $\ce{Br-}$ to dissociate, and therefore there would be lower yields of both haloalkane products. But then again, wouldn't it go faster owing to the fact that it is in a polar protic solvent, and that the formed carbocation would be tertiary?

There are (at least) four major factors to consider in predicting the results of these reactions: (a) the nature of the substrate, (b) the strength of the nucleophile, (c) the quality of the leaving group, and (d) solvent effects. Halides are able to function, to varying degrees, as both good leaving groups and strong nucleophiles.
In protic solvents, negative charges are stabilized by the $\ce{\delta^{+}}$ of the solvent hydrogens bonded to some electronegative atom (conventionally limited to $\ce{N, O, F}$). Since nucleophiles carry either a full or partial negative charge, protic solvents tend to reduce the reactivity of nucleophiles. Because smaller nucleophiles are more easily and thoroughly solvated than large and polarizable nucleophiles, the effect on them is more drastic. Consequently, in protic solvents, the relative nucleophilicity of the halides is generally given as $\ce{I- > Br- > Cl- > F-}$. In aprotic solvents, the reactivity is precisely the opposite. Hence, the proper selection of solvent is crucial. In your first example, there is no particular reason I can think of to suppose that the yield of 1-chlorooctane would inevitably be low. Indeed, in an aprotic solvent, the usual choice for SN2 reactions, $\ce{Cl-}$ can be expected to be a stronger nucleophile and poorer leaving group than $\ce{Br-}$. I would expect the yield to be good, especially with an excess of $\ce{KCl}$.
On the matter of the substrate, with a tertiary carbon, SN2 is all but completely impossible. In that situation, elimination reactions are typically favored if some base is present. The halides, however, are extremely weak bases (being the conjugate bases of strong acids), so if a non-basic solvent is selected, substitution by an SN1 mechanism seems most probable. Selection of a relatively non-basic, non-nucleophilic protic solvent would be ideal. (Off the top of my head, DMSO comes to mind, which is only weakly nucleophilic, especially by comparison to the halides.) You're right that the strength of the nucleophile is not the deciding factor in an SN1, since the rate-determining step is the loss of the leaving group and the resultant carbocation formation. In a protic solvent, $\ce{I-}$ should theoretically be more effective as a nucleophile than $\ce{Cl-}$. However, I would still expect the latter reaction to work, particularly if the concentration of $\ce{Cl-}$ is high. I'm fairly confident that the difference in reactivity between, e.g., $\ce{Cl-}$ and $\ce{Br-}$ would be largely negated if the former is available in vastly larger concentration in the solution.
Edit: On a second reading, I just noticed that you labeled the halide solutions as aqueous. I don't know if that's deliberate, but I don't think that would make sense. For one thing, the reagents (namely, the haloalkanes) would be largely insoluble in water. Second, because water is an amphoteric species, it could be potentially reactive.

We conducted this experiment in lab, and the professor said that for iodide there should be mostly product and for chloride there should be mostly starting material. Also, in the instructions it said add saturated KCl or KI "solution," wouldn't that mean that it was aqueous? And in the final steps we had to separated out the "aqueous" layer using petroleum ether in a sep. funnel. Furthermore, we used a quartenary phosphonium bromide as a catalyst. What are the implications of this? How exactly should the results differ between tertiary and primary electrophiles for reactions with both halides?

@user32134, yes, those solutions are aqueous. Hypothetically, I think it would be ideal to carry out the reaction in an organic aprotic solvent and sequester the K+ ions with, e.g., a crown ether. I don't have hands-on experience with that, though, just speculating. With mixed aqueous/organic phases and a large amount of water present, I think iodide should indeed be favored. My comment about the chloride was predicated on the assumption that you'd be using just an aprotic solvent. Apologies if my response was too academic, I didn't realize this was in reference to a particular lab protocol.

@user32134, as for the use of the quaternary phosphonium salt, it's presumably the phase transfer catalyst, there to enable the migration of the halide anions into the organic phase. I haven't worked with phosphonium salts for that purpose, though I've conducted reactions in which ammonium salts were used analogously. I don't know whether the salt you used has varying affinities for one halide anion vs. the other, or how it performs by comparison to different phase-transfer catalysts. I'd be curious to know how phase-transfer catalysts compare to chelation by ligands.

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text:
Teeth friendly alcohol drinks
What is the best alcohol for your teeth according to the pH?
Is it true that if the acidity of the drink is higher (and ph is lower) than it's worse for your teeth (enamel erosion + bacteria friendly acidic environment) than if the alcohol drink is less acidic?
Does vodka have the highest pH of all the alcohol drinks out there and therefore is the best for your teeth?

Why the downvote? Should I move it to biology SE ? But I personally think that biochemistry is much closer to chemistry than biology.

Maybe the title could be better.. more related to chemistry than medical health... (But this is a my opinion...)

Ethanolic KOH, clearly. Great for your teeth, terrible for your rest of you.

Worth mentioning that alcohol has a tendency to reduce salive excretion which in turn increases growth of bacteria.

Is quite reductive think that pH is the onlyfactor that determinate the health of your teeth, the content of carbohydrate have a strong effect too.
However in fact to my knowledge clear vodaka have one of the highest pH and very few carbohydrates so indeed is one of the most "teeth friendly alcohol drink".
These is take from Victor R. Preedy, Ronald Ross Watson Academic Press, 24/dic/2004 Comprehensive Handbook of Alcohol Related Pathology with the adding of Vodka and Coca Cola.
\begin{array}{|c|c|c|c|}\hline
Alcolic \space beverage & Carboydrate \space content(\%) & pH \\ \hline
Beer               & 3.0-5.0   &  4.1-4.5    \\ \hline
Wine & 0.0-12.9 &  2.8-3.8     \\ \hline
Sweet \space liquor           & 30.0-31.0 & 3.3-3.9 \\ \hline
Strong \space alcohol & 0.0-1.2 &   6.5-6.9\\ \hline
Vodka                & very \space few       & 6.0-7.0        \\ \hline
Coca Cola & 10.6(?) & 2.8\\ \hline
\end{array}
However keep out  from Vodka lemon!

Sugar content will be the determining factor in tooth decay, my dentist informed me when I asked her this very question that it's not the inherent acidity of the drink / food stuff that does the damage.
What does cause damage is the bacteria in your mouth digesting simple sugars over a long period of time. From Wikipedia;

Tooth decay disease is caused by specific types of bacteria that
  produce acid in the presence of fermentable carbohydrates such as
  sucrose, fructose, and glucose.

I took this advice as meaning I should take all my drinks neat of course.

I think he told you just the A and forget to tell you the B. And that is that the enamel is destroyed by acid stuff in general (as all human cells) but most of them can be repaired or replaced by other cells. Everytime you take a drink of a soda you get 20 seconds of acid attack on your teeth. Which in time will eat away at your enamel and then cause a cavity. The increased acid strips the natural protective layer on your teeth and the enamel begins to slowly lose its natural minerals and therefore breakdown, causing decay.

@Derfder You are right that pH matters, but mostly wrong that it matters much. If you drink lots of coke or some fruit juices, the acid will erode enamel, but you have to drink a lot to do more damage than the sugar content will cause. Most alcoholic drinks will not be either so sweet or so acid as coke (unless they contain coke, of course) so will cause far less damage. Sweet alcoholic drinks will most damage you because of the sugar not the alcohol.


text:
Reaction of Benzene with 2-methylpropene?
I'm studying organic chemistry at the moment and it's really confusing. What could ever happen between benzene and an alkene? and why is $\ce{H2SO4}$ there?

Sulfuric acid protonates the alkene to give the highest substituted carbocation, tert-butyl.  That electrophilically adds to a benzene carbon atom, to give the obvious next cation.  A proton is eliminated to obtain tert-butylbenzene.  Akyl groups activate benzene rings toward ortho, para attack (mild electron donation via  hyperconjugation), thus 1,6-di-tert-butylbenzene, certainly if the addend is in excess.  Especially stable carbocations can dissociate and wander around the benzene ring under Friedel-Crafts conditions.  If you push it, you get 1,3,5-tri-tert-butylbenzene.
Steric hindrance strongly discourages further reaction.  You can go to the literature to find how many adjacent tert-butyls can be jammed onto a benzene ring - and how it is done.
With due respect, you are in deep trouble for passing the course if the Friedel-Crafts reaction throws you.   Either look at things with the LCAO model, look in a few more textbooks, or talk with folks who have grasped it.  You must understand it.  You cannot survive substituent effects if you cannot flip bonds.  This is easy stuff compared to what will follow.  Chess can be well-played by computer.  Organic synthesis is done in wetware, though computers help help.
http://www.chemtube3d.com/model/electrophilic_aromatic_substitution/friedel-crafts/alkylation/benzene%2Btbu/mechanism.png
Close enough

First, I think the name of your molecule must be malformed. That said, the reaction of an alkene with strong acid can generate a carbocation. That carbocation is electrophilic, and can react with benzene (as the nucleophile) in an electrophilic aromatic substitution reaction. The final product is the (potentially poly-)alkylated benzene.

THANKS FOR ANSWERING! which molecule is wrong? I typed what it said, or showed an an ethylene molecule with two methyls hanging off of it, and H2S04 is also in the equation maybe as a catalyst I dont know, so just want to clarify the benzene dosent open up any of its double bonds right? its just the ethylene and it opens up creating the carbo cation right?

@Unistudent9, if I understood you correctly, the alkene you're referring to is one of the isomers of 2-butene. The benzene does not react directly with the acid. Rather, the proton transfer to the 2-butene produces a carbocation, which is attacked by the benzene, resulting in alkylation of the benzene ring. Look up electrophilic aromatic substitution for a detailed mechanism.

I took the liberty and corrected the title of your question, assuming that the "ethylene with two methyls" is 2-methylpropene (isobutene). Over here, this classic reaction is known as Friedel-Crafts alkylation.


text:
Derivation of relationship between Gibbs free energy and electrochemical cell potential
Why is $\Delta G = -nFE$?
I don't understand what the motivation is behind this definition. Was it derived or just given? The textbook provides no justification for this equation. In fact, much of the book associated with the Gibbs free energy provides no justification and just says, 'This is how it is. Now go and solve some problems.'

There are two ways to understand this equation.
One is to realise that (reversible ideal case) $\Delta G = W_\text{non-exp}$ (non-expansion). Therefore, in an ideal chemical cell, if the potential difference between the electrodes is $E$, to move one mole electrons across the external circuit will be $FE$, which must be equal to the decrease in gibbs free energy of the system. Hence for $n$ mole electrons transferred at the same potential, $W_\text{non-exp} = \Delta G = -nFE$.
The fact that $\Delta G = W_\text{non-exp}$ can be derived as under:
\begin{align}
\mathrm{d}S
  &= \frac{\delta q}{T}  && \text{(reversible case)}\\
\mathrm{d}U
  &= \delta q + \delta W_\text{non-exp} + \delta W_\text{exp}\\
\delta W_\text{non-exp}
  &= \mathrm{d}U - \delta W_\text{exp} - \delta q \\
  &= \mathrm{d}U + p\,\mathrm{d}V - T\,\mathrm{d}S
  && \text{(const. $p$ and $T$)} \\
  &= \mathrm{d}H - T\,\mathrm{d}S \\
  &= \mathrm{d}G
\end{align}
Another approach is to use
$$\mu_\mathrm{electrochemical} = \mu_\mathrm{chemical} + zF\phi,$$
where $\phi$ is the electric potential at the point, $z$ is the charge on the species. Using this and
$$\mathrm{d}G_{T,P} = \sum_{i=1}^n \mu_i\,\mathrm{d}n_i,$$
we get the required equation for $z = 1$ for an electron, and considering the electrochemical potential of both the electrodes. I am unsure of a rigorous derivation for the electrochemical potential but Wikipedia has some basics on it.

I'm surprised your textbook did not derive this equation from the reaction isotherm relationship between $\Delta G$ and the reaction quotient $Q$ and the Nernst equation. The derivation is not hard.
Reaction isotherm equation:
$$\Delta_\mathrm{r} G =\Delta_\mathrm{r} G^\circ + RT\ln Q$$
Nernst equation:
$$E_\mathrm{cell} = E^\circ_\mathrm{cell} - \frac{RT}{nF}\ln Q$$
If we solve both equations for $RT\ln Q$, we get your equation (almost).
\begin{align}
  RT\ln Q
    &= \Delta_\mathrm{r} G -\Delta_\mathrm{r} G^\circ\\
  RT\ln Q
    &= nFE^\circ_\mathrm{cell} - nFE_\mathrm{cell}\\
  \Delta_\mathrm{r} G -\Delta_\mathrm{r} G^\circ
    &= nFE^\circ_\mathrm{cell} - nFE_\mathrm{cell}
\end{align}
Why is my equation not as simple as the one you started with? Your equation is at equilibrium, and I assumed that we might not be at equilibrium. At equilibrium, the following are true, which simplify the relationship.
\begin{align}
Q &= K\\
\Delta_\mathrm{r} G &= 0\\
E_\mathrm{cell} &= 0
\end{align}
At equilibrium, $\Delta_\mathrm{r} G = 0$ because the reaction has achieved a minimum energy state — the chemical potential $\mu_i = \left(\frac{\partial G}{\partial N_i}\right)_{T,P}$ is also $0$ because there is no net change of state at equilibrium. Similarly $E_\mathrm{cell} = 0$ at equilibrium. There is no change of state, and thus the redox reaction has ground to a halt.
At equilibrium, the final relationship is
$$\Delta_RG^\circ = nFE^\circ_\mathrm{cell}$$

I guess his textbook is going the other way, and derives the Nernst from the one above (since the first is a more or less trivial relationship between the electric work and enthalpy).

The chemical potential is not zero. I would remove that statement since it is incorrect and doesn't add value to your explanation.

Thermodynamically if you calculate the equilibrium constant the equilibrium quotient depends on number of moles of both reactants as well the number of moles of products that is reaction concentration at the equilibrium.more the free energy of reactants the reaction will proceed in a spontaneous manner the more feasible the reaction . Likewise the more the product concentration the backward reaction wii be feasible. The equilibrium  remaining zero the forward and backward reaction rate will be same

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text:
Derivation of relationship between equilibrium constant and Gibbs free energy change

Why is $\Delta G=\Delta G^o+RT\ln Q?$

It feels like all online sources were written for introductory Chemistry students! Where do I find a rigorous proof of this identity? Greatly appreciate it!

http://en.wikipedia.org/wiki/Chemical_equilibrium this contains a pretty rigorous treatment of gibbs-helmholtz equation.

In the following section:
The chemical potential of a reagent A is a function of the activity, {A} of that reagent. How do we get $\mu_A=\mu_A^o+RT\ln [A]$?

see the thermodynamic derivation part here.http://en.wikipedia.org/wiki/Nernst_equation#Using_thermodynamics_.28chemical_potential.29

I edited the question to remove MathJax from the title. Copious Mathjax in titles makes questions hard to locate using searches (both internal and external). Let me know if the new title is inappropriate.

Using the fundamental equations for the state function (and its natural variables):
\begin{align}
  \mathrm{d}G &= -S\mathrm{d}T + V\mathrm{d}P\\
  V &= \left(\frac{\partial G}{\partial P}\right)_T\\
  \bar{G}(T,P_2) &= \bar{G}(T,P_1) + \int_{P_1}^{P_2}\bar{V} \mathrm{d}p
\end{align}
Here $\bar{x}$ represents molar $x$, i.e. $x$ per mole
\begin{align}
  \bar{V} &= \frac{RT}{P}\\
  \bar{G}(T,P_2) &= \bar{G}(T,P_1) + RT \ln\frac{P_2}{P_1}
\end{align}
Defining standard state as $P = \pu{1 bar}$ and $\bar{G}=\mu$
$$\mu(T,P)=\mu^\circ (T) + RT\ln \frac{P}{P_o}$$
consider the general gaseous reaction $\ce{a A + b B -> c C + d D}$
$$\Delta G=(c\mu_\ce{C} + d\mu_\ce{D} - a\mu_\ce{A} - b\mu_\ce{B})$$
for "unit progress" in reaction. Using $\mu_i = \mu^\circ_i + RT\ln \frac{P_i}{\pu{1 bar}}$
\begin{align}
\Delta G &= (c\mu^\circ_\ce{C} + d\mu^\circ_\ce{D} - a\mu^\circ_\ce{A} - b\mu^\circ_\ce{B}) + RT \ln\frac{P_\ce{C}^c P_\ce{D}^d}{P_\ce{A}^a P_\ce{B}^b}\\
\Delta G &= \Delta G^\circ + RT\ln Q
\end{align}

Gibbs free energy is the free energy at constant pressure and temperature, so why did you use dT and dP in your equation? It would seem that this violates the definition of the Gibb's free energy. Also, why did you take the path where entropy and volume remains constant, it can take any path.

This discussion is further continued in a new post [https://chemistry.stackexchange.com/questions/104950/how-to-derive-the-relation-between-gibbs-energy-and-equilibrium-constant/104964#104964]

In this equation $\mu(T,P)=\mu^\circ (T) + RT\ln \frac{P}{P_o}$ is $P$ the total pressure or the partial pressure of the gas in the system?

@ado sat it's partial pressure

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text:
Why do liquid crystal displays (LCDs) visually distort under pressure?
A venerable question, as old as LCDs themselves. I assume it has something to do with pressure-induced changes in the path length across the display, but that's just a guess. Of particular interest to me as I press my fingernail into my (TFT LCD) phone screen1 are the observations of:

A symmetrical bilobate distortion in the direction of movement of my fingernail.
Ripples of colour distortion that radiate outwards at about a centimeter per second.
A trail of slight colour distortion that with a persistence of a few seconds.

Any ideas as to the cause of this rich behaviour?
[1] Don't worry, it's a nokia n95 and as such will likely outlive the human race if not the solar system.

Liquid crystal displays work by using liquid crystals where the direction of alignment of the molecules in a plane rotates as you go from layer to layer. Usually this is done with twisted nematic crystals, however cholesteric LCs can also be used to achieve similar effects (I don't think smectic LCs are useful here as they don't align).
To create a liquid crystal cell, twisted nematic liquid crystal is filled between two crossed polarizers1, and a backlight is sent through. The important thing about liquid crystals is that the specific rotation is easy to control by varying the potential difference (more potential difference leads to more anisotropic disorder).
Now, if you make a grid of these and put color filters on them, we get an RGB LCD display.
What happens when you press on them? You change the concentration/length of the liquid crystal, as well as making it bulge out slightly2.. This changes the optical rotation it induces, causing the amount of light that comes out to change (as the amount of light is dependant on how aligned the rotated light is with the second polarizer)
I'm not too sure of the exact reasons behind the phenomena you list, but I can guess:
Firstly, the color changes are caused because each type of crystal cell is affected differently by the pressure. I think there's a difference in concentration between the different color cells to better suit human eyes. When I do it to LCD displays accessible to me, I only get a darkening, not a color change, so it seems to not be a universal feature of LCD displays. If the rotation of each color cell is affected differently by pressure, then pressure can easily cause color distortions as each color will not darken by the same amount.
I think the ripples are caused by high pressure where you change the optical rotation of the inner cells (the ones closer to your finger) so much that they go through a at least one full 360o rotation, leading to regions of both bright and dark light.
The persistence is because liquid crystals are viscous — they have to be, as they consist of long molecules. They seem to align quickly to potential differences (potential differences align the molecules without moving the liquid), but when we add pressure (where the liquid itself distorts) I think it takes some time to bounce back. The ripples move due to this same persistence.
1. If you're ever at a 3D movie, have a look at your phone screen with the 3D polarised glasses on. It looks pretty cool!
2. If this did not occur, then the rotation would stay the same as $\theta\propto lc$ and $lc$ does not change if there is no bulge.


text:
Why do puffed rice stick to wet hands?

My mom gave puffed rice to eat, I had quite wet hands. I saw that, each puffed rice pieces started hanging to the hand, when I brought my hand in contact with them. I thought about electrostatics initially to be the cause. Actually, the electrostatics usually deals with rough surfaces. Here due to hand being wet, it has got smooth surface. Then, what might be reason? I started thinking, suddenly I thought it to be because of hydrophillic nature of kadale puri (puffed rice). Is it really because of hydrophillic nature of puffed rice?

you should better use the term 'Puffed Rice' rather than 'kadale puri'..

I think it's a combination of hydrophilic surfaces trying to spread water out and water's high surface tension trying to pull itself back together, creating a sort of elastic tension. Likely more of a physical process than a chemical one.

In many cases of stickiness, where obviousness is not a given (thus mostly excludes dipole-dipole int.), it is a slew of usually weak intermolecular forces such as the London dispersion force which are increased by special material properties such as surface area and resonance assistance.

This thesis, (Texture and Hydration of Expanded Rice, Clive Norton, September 1998), has a vast amount of information on rice, including puffed rice.
Rice is mostly starch, which when heated is gelatinised. This causes "the breaking of hydrogen bonds between poly-α-1,4-glucan chains, leading to loss of crystallinity" allowing those hydrogen bonding sites to absorb more water. The loss of crstallinity also leads to air pockets making puffed rice sponge like. The starches become "interspersed with gas pockets" which are filled by water through capillary action.
So, puffing makes the starch more hydrophillic, exposing hydrogen bonding sites, and the air pockets act as capillaries. The puffing also increases the ratio of surface area to weight. These factors cause the rice to be liftable with a "water chain".

Hydrophilic surface and capillary action is sertainly part of the answer together with the surface tension of water trying to minimize its surface area. But another part is probably the very low weight of puffed rice! That makes the effect of the beforementioned forces more noticeable. If the rice grains would be heavier (denser) you would probably not notice the same stickyness.


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Is it possible to melt diamonds into a liquid?
Is it possible to melt diamonds into a liquid? I mean if you heat diamond at open air it will start to burn around 700 degrees Celsius, reacting with oxygen to produce carbon dioxide gas. In the absence of oxygen it will transform into graphite, a more stable form of carbon, long before turning into graphite. My question is, is it possible to melt diamonds into liquid? If so, then, how? If not, then why?

"...I mean if you heat diamond at open air it will start to burn around 700 degree celsius, reacting with oxygen to produce carbon dioxide gas....." If it burns reacting with oxygen, producing carbon dioxide, then it mean there is no diamond or diamond liquid, it has all got converted into  carbon dioxide. I don't know whether what I am saying makes sense or not, but that is what I thought when I read this question. There might exist liquid diamond during heating, but it gets converted to carbon dioxide, in a specific time interval.

@CURIE To avoid oxidizing the diamond, it would have to be heated in a vacuum or inert atmosphere, yes. Exposed to oxygen, it would burn completely far before it ever got close to melting. But this is not particularly special, for many substances would react with our atmosphere before melting/boiling.

See this related question about liquid carbon… the answers include a phase diagram of carbon, which answers your question

Liquid carbon does indeed exist, but perhaps surprisingly, relatively little is known about it. It only exists above around $4000\ \mathrm{K}$ and $100\ \mathrm{atm}$, which are not trivial conditions to sustain and probe. There certainly are many theoretical studies into the properties of liquid carbon, though. You can find a phase diagram for carbon here, though it is likely just a rough guide. The specific bonding properties of carbon atoms in the liquid phase still seems uncertain to some extent because carbon is very versatile and can create very dissimilar structures (easily seen comparing its two most common phases, graphite and diamond). This source suggests that bonding in the liquid varies continuously between linear acetylenic chains, graphite-like and diamond-like structures, depending on the exact conditions.
Edit: Now I realize that obtaining liquid carbon and melting diamonds are two slightly different things. Liquid carbon is trivial to obtain compared to melting a diamond. As you mention, heating diamond at relatively low pressures will cause it to convert into graphite before it could melt. The only way to inhibit that would be to exploit the slight difference in density between graphite and diamond; as diamond is denser, the increased pressure will stabilize it slightly, decreasing the tendency for diamond to revert back to graphite.
You'll need a lot of pressure though! Rigorously speaking, according to the phase diagram above, to obtain liquid carbon directly from melting a diamond would require a pressure of at least $10\ \mathrm{GPa}$, which is about $10^5\ \mathrm{atm}$. To put that into perspective, it's about $1/30$ of the pressure at the centre of the Earth. Amazingly, we actually can reach such absurd pressures in the lab by compressing two diamonds against each other. However, if the conditions required conditions for melting diamond (and sustaining it in molten form) are reached, it would probably compromise the integrity of the instrument and cause catastrophic (possibly explosive) failure, so I don't think anyone will be purposefully melting diamonds any time soon! Other techniques could be used (laser inertial confinement, light-gas guns, etc), but I believe most would only transiently produce liquid diamond.

Carbon, which makes up diamond, can certainly exist as a liquid and phase transition between the diamond phase and the liquid phase seems to be possible based on this phase diagram. Whether diamonds melting can actually be observed depends on the kinetics though: The phase transition may be too slow to be observed on a human time scale.

While transformation of diamond into graphite is very slow even in moderately high temp., melting isn't such a kinetically hindered thing. Over 4000 K accelerates things very much and melting happens fast or not at all.


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What chemicals cause the acrid smell of burning food?
When roasting vegetables in a oven, after a while there is a nice caramel smell. After that, if they start burning, there is a smokey smell, and if they continue to burn, the smokey smell gets more and more acrid. The smell really sticks in the back of your throat/sinuses, and induces gagging if it's strong enough. What chemicals make up the gasses and particles that cause that acrid smell?

The chemistry of cooking is quite complex. Sugars present in food can suffer multiple dehydration reactions resulting in a complex mix of degradation products, including unsaturated ketones/aldehydes, some of which are noted to have an acrid smell/taste, such as acrolein (though acrolein itself seems to be more related to burning fat than sugars).

If only the Maillard reaction products were also classified by acrid-smell :)

Take a look at this link about Maillard reactions:
http://www.landfood.ubc.ca/courses/fnh/301/brown/brown_prin.htm
And what occurs if continue burning: pyrolisis
http://www.healthy-food-site.com/burnt-food.html


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Do alkanes evaporate faster than alcohols?
If so, is it because alchohols have hydrogen bonds?
Which type of compound evaporates faster, and why?

Yes, alkanes do evaporate faster than their corresponding alcohols (i.e. an alkane boils at a lower temperature than an alcohol with the same number of carbon atoms). This effect is due to hydrogen bonding between the alcohol hydroxyl groups.
Wikipedia says:

Because of hydrogen bonding, alcohols tend to have higher boiling points than comparable hydrocarbons and ethers. The boiling point of the alcohol ethanol is 78.29 °C, compared to 69 °C for the hydrocarbon hexane (a common constituent of gasoline), and 34.6 °C for diethyl ether.


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Biopolymers as alternative for PVC and PU?
I'm working on a project making a product line more sustainable, environmentally friendly and ready for recycling.
Now I'm searching for good and suitable biopolymers for a part of the product that is made from PVC and one for PU (Polyurethane). For PES I already have chosen PLA. Is there someone who can help me select the right biopolymers with similar characteristics of PVC and PU?
Please let me know, thank you!

Ethylene vinyl acetate (also known as EVA) is the copolymer of ethylene and vinyl acetate and is a replacement for PVC.
Biomass > (fermentation) > Bioethanol > Ethene
Ethylene + acetic acid + 1/2 O2 → Vinyl acetate + H2O

Tung oil can be used to create a plastic like water proofing film as a replacement for PU

Thank you for your reply! :)
Maybe I did not ask my question the right way. By biopolymers I mean biobased or biodegradable polymers, like PLA, PHA.. Materials derived from renewable sources such as corn starch, grasses and sugarcane.

Your question is hard to answer because biopolymer is vague, as well as "derived from renewable sources". Almost everything can be derived from a renewable source, it just is more expensive than using petroleum.

For instance, Clostridium acetobutylicum makes acetone and butanol during fermentation. Those substitutes I listed can be derived from renewable sources. Ethene can be made via the dehydration of ethanol, which of course can be made from fermentation. Acetic acid is also a byproduct of fermentation of certian bacteria. Tung oil is oil pressed from the seed of the Tung tree. Try clicking the links in the answer to see more information.

If production of PEVA from renewable sources seems too daunting, wood is probably a good replacement for PVC. Wood is actually a cellulose polymer. You could coat it with Tung oil if you wanted to waterproof it.

Bioethanol is an ecologic menace of the worst kind. Much worse imo than  using regular ethylene from petrol chemistry to make PE.

@Karl
It's not viable as a economic or energy efficient alternative, but it does  have value in self-reliance and building rugged supply chains capable of surviving war and economic collapse.

@Motes It destroys brazilian rainforests and therefore increases CO2 emissions, instead of reducing them. But who cares with "green" governements that enforce 10% ethanol in car fuel.

The common properties of PVC vary wildly with the concentrations of plasticizers and filling reinforcement.
The common properties of PU vary wildly with the choice of isocyanate and cross-link density. PU comes in foams, fibers and hard plastics.
Modifications, copolymers and alloy blends are common with both.
When selecting a plastic, it is useful to gather requirements rather than emulate another plastic.
Some physical properties to look into are: Glass transition, melting/vicat softening point, degradation onset temperature, thermoset/thermoplastic, solvent resistance, transparency, UV-resistance, food/medical-safety, tensile strength and Young's modulus.
With regards to manufacturing some properties to look into are: pot life, working time, tack time, demold time, cure time, cost at scale, shipping and sourcing.
With regards to your question you're also considering if they are biobased(ASTM D6866), their recyclability and their compostability.
The only bioplastic that is highly recycled is HDPE. The only other plastic that is highly recycled is PET. There are probably more recycling operations for other plastics, but they are much smaller in scope.
Compostibility is dominated by starch and cellulose plastics.
As a start, you should visit Matweb for the basic properties of various plastics.


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How to Understand Concentrations Measurement
I was helping out in my beginner's chemistry class today, as I'm a "lab assistant" of sorts, and was reading on a bottle of HCl: 1M. I'm assuming that is the concentration of how much hydrocloric acid is in the bottle, in moles, but how would you determine the concentration of the solution (thanks Ove)? What is the actual percentage of HCl to water(and whatever else is in the bottle)?

What do mean by "how do you compute that"? How to prepare a 1 M (1 mol/L) solution? Or how to determine the concentration of an existing solution?

This is a quite basic chemistry question in my opinion you should read wiki molarity page than ask what you don't understand.

The unit M is an abbreviated form of mol/L: 1 M = 1 mol/L, i.e. there is 1 mol of HCl per liter of solution. The M is pronounced "molar". For example, a solution with concentration 1 mol/L is said to be a "1 molar" solution.
In order to calculate the fraction of HCl in a 1 M solution, some information is needed. The density of the solution varies with concentration, so I looked up 1 M HCl on Wolfram Alpha and found $\rho = 1.016~g/cm^3=1.016~kg/L$. Also, the molar mass of HCl is $M_{HCl}=36.46~g/mol$. Thus, 1 L of solution has a mass of 1.016 kg and 1 mol of HCl weighs $$m_{HCl}=1~mol\cdot36.46~g/mol=36.46~g$$ The mass of water in the solution must therefore be $$m_{H_2O}=1.016\cdot10^3-36.46=979.5~g$$
From this, the mass fraction of HCl in the solution can be calculated:
$$w_{HCl}=\frac{m_{HCl}}{m_{tot}}=\frac{m_{HCl}}{m_{HCl}+m_{H_2O}}$$
Similarily, the molar fraction can be determined:
$$x_{HCl}=\frac{n_{HCl}}{n_{tot}}=\frac{n_{HCl}}{n_{HCl}+n_{H_2O}}=\frac{n_{HCl}}{n_{HCl}+\frac{m_{H_2O}}{M_{H_2O}}}$$


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Functional group naming order?
I'm currently studying some organic chemistry at highschool level. I understand that when naming an organic compound, the names for substituents should be done in alphabetical order. For instance 3-bromo-3,4-dimethylhexane (B in bromo before M in methyl), or 3-chloro-3-ethyl-4-methylhexane (C in chloro before E in ethyl before M in methyl).
However, it doesn't seem to be the same for functional group suffixes. For instance, for one compound I wrote but-1-amin-3-ene, while the correct answer was apparently but-3-en-1-amine. What's more, these are referred to as suffixes, while what I thought was prop-3-amin-1-ol was in fact 3-aminopropan-1-ol.
So is there some kind of pattern or rule as to the order of writing functional group "suffixes" in organic compounds? At first I thought perhaps it follows the priority the groups have over each other in numbering the parent chain carbons (i.e. alcohol > amine > alkene, etc), but this doesn't seem to be it either. It's possible that it'll be explained further on in the course, but I want to make sure I haven't missed anything.

Generating the systematic names of organic compounds follows the rules for IUPAC nomenclature.
In this system, a name has several parts:

Prefix (meth-, eth-, prop-, but-, pent-, etc.) - which indicates the number of carbon atoms in the parent chain.
Infix (-an-, -en-, -yn-) - which indicates the presence or absence of carbon-carbon multiple bonds in the parent chain.
Suffix (-e, -ol, -amine, -oic acid, -al, etc.) - which indicates the highest priority functional group in the parent chain.
Substituents (bromo-, methyl-, hydroxy-) - indicate the presence of other things on the parent chain.
Locants (the numbers) - indicate where all of these things happen.

You are confusing the roles of infixes and suffixes.

Infixes

The base infix is -an- indicating no carbon-carbon multiple bonds. The presence of alkenes or alkynes in the parent chain changes the infix to -en- or -yn-, respectively. No other functional group can change the infix, as there are no carbon-carbon bonds of higher order than 3 or lower order than 1. It is also possible to have a complex -enyn- infix if you have both, but let us not worry about that now. In your example compounds:

3-bromo-3,4-dimethylhexane - the infix is -an- because there are no carbon-carbon double or triple bonds.
but-3-en-1-amine - the infix is -en- because there is a carbon-carbon double bond.
3-aminopropan-1-ol - the infix is -an- again because there are no carbon-carbon multiple bonds. The amine and alcohol functional groups do not consist of carbon-carbon multiple bonds so they cannot change the infix.

Suffixes

The base suffix is -e, signifying a hydrocarbon with some substituents. Many functional groups change the suffix of a molecule's name, but the name can have only one suffix. If two or more suffix-changing functional groups are found on the parent chain, then the one with the highest priority is chosen to be the suffix and all others become substituents. The common functional groups by decreasing priority are:
functional group     suffix         substituent name
carboxylate anion    -oate
carboxylic acid      -oic acid      carboxy
ester                alkyl -oate    alkoxycarbonyl or alkylcarboxy
acid halide          -oyl halide    halocarbonyl
amide                -amide         amido
nitrile              -enitrile      cyano
aldehyde             -al            oxo
ketone               -one           oxo
alcohol              -ol            hydroxy
thiol                -ethiol        mercapto
amine                -amine         amino
hydrocarbon          -e             varies by group

The name 3-aminopropan-1-ol is correct. Both the alcohol and the amine cannot change the suffix. Only the higher priority group (alcohol) changes the suffix. The other (amine) becomes a substituent. As mentioned before, neither group changes the infix.

Substituents

In addition to the substituents listed above, several functional groups are always substituents (which are listed alphabetically in front of the parent name):
functional group     substituent name
alkyl                alkyl
alkyl halides        halo
ethers               alkoxy
nitro                nitro
sulfides             alkylthio

There are some other common substituents that deal with exception cases, like phenyl, aryl, vinyl, methylene, formyl, allyl, and acetyl.

Thanks a lot, exactly what I needed. My school's resources listed double-bonds as a functional group with the suffix -ene, and only briefly touched on the concept of infixes (although it didn't even use that term). Neither did it specify at all that functional groups get treated as substituents when there are more than one type. Once again thanks a lot, this has made it much clearer to me!

@Ben           Wouldn't be better to also include that if more than one functional group of same type is present then base suffix e should be retained as in ethane-1,2-diol?


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How to prevent petrol from being flammable?
Is there any way to prevent petrol from being flammable without losing it's solvent properties? Preferably some affordable way that is applicable for DIY.
In my case, petrol is used for bearing cleaning, and there is a risk of inflammation due to accidental spark.
If no affordable way exists, what octane number is safer (less flammable)?

For cleaning, diesel oil or kerosene instead. Not so volatile. Also cheaper.

Thank you for proposition, not sure about diesel but kerosene seems to be hard to buy at my area.

You aren't going to like it, but working at $-45^\cdot C$, below the flash point of gasoline is one way.

By the by, breathing in petrol fumes is also super bad for you.

Makes sense. I was thinking about using something automatic like http://www.bont.com/quad/products/bearingCleaner/ with gasoline and custom remote control, but it seems to be too dangerous/harmful. Probably will switch to citrus cleaner.

Keeping open flames away might help.

Nearly every solvent is flammable, some exceptions are halogenated hydrocarbons. But plain hydrocarbons are pretty much all flammable, though some are more dangerous than others like e.g. diethyl ether.
Chloroform is non-flammable, but due to the toxicity I would not recommend to use it. There are some other non-flammable solvents for cleaning mentioned in this presentation from NASA on the search for non-flammable solvents. But it seems that many candidates are being phased out because they are a hazard to the ozone layer.

I think you are right. Now I consider using citrus cleaner to clean bearings. Hopely it will be able to remove oil and old lubrificant nearly as effective as petrol.

@Dmitriusan citrus would cause corrosion.  This may be a case of "don't do it without proper precautions".

Chloroform is both toxic and dangerous for a routine solvent. But dichloromethane has most of the good properties and few of the bad ones.


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Is hybridization used in ab initio valence bond calculation?
Many general chemistry textbooks introduced the concept "hybridization" to construct a symmetry-adapted VB-type wavefunction. In the textbooks, usually the minimal basis is used and without optimizing all resonance forms.
My question is, in the ab initio valence bond calculation with extended basis set and optimizing many resonance forms,
(i) do people still use hybridization to construct the wavefuntion?
(ii) if the answer to question (i) is yes, does the result differ from the textbook description? For instance, the lone pair electron in the $\ce{H2O}$ molecule? (is there any unitary equivalent result like orbital localization in MO-based method?)
I asked the question (ii), since NBO analysis tends to give non-equivalent composition for lone pairs in $\ce{H2O}$, one sp, the other p. I am wondering what is the answer in ab initio valence bond method.

Do I understand correctly that you are looking for a Valence Bond Description? If so, you might have already read Ab initio valence-bond calculations of $\ce{H2O}$? I personally don't quite understand the benefits of VB theory compared with other approaches as CASSCF, MBPT, CC, CI or DFT - but that might well be because I do not understand VB theory completely.

@Martin Proponents of VB feel that it is intuitively superior to MO type calculations because the calculations are performed with configurations that resemble "Lewis structures".  However, VB configurations are generally non-orthogonal, and so anywhere there is an overlap of configurations computation, it has to be done explicitly. Thus MO based methods beat the pants off of VB in terms of computation speed.

@Martin (cont.) In the full limit of VB-CI and MO-CI, the theories are equivalent. Since both methods yield the same observables, MO theory wins out from a practical level.  (Unless one desires a VB-like description of the wavefunction, which I think the poster desires, hence the NBO analysis)

@Eric You are very right with your first statement. However the second statement is only half true. VB theory is naturally capable of describing strongly correlated electrons. In MO theory you only generate one configuration, so this case is impossible to consider. Therefore you need more advanced methods like MCSCF. So for any easy molecule, complete description, your statement holds, for more complex systems you should rather state: VB = MCSCF.

@Martin I am assuming CI is full, multi reference CI, "full limit"

@Eric You are right of course, I somehow missed that you added these two letters.

@Martin no problem, my comment needed clarification that it is multi reference CI, rather than e.g. CISD from a single reference.

(i) do people still use hybridization to construct the wavefuntion?
Yes and no. No: Only atomic orbitals are included in the base set. Yes: if the wavefunction is optimized, linear combinations of atomic orbitals are allowed, and we all know that one S-orbital + some P-orbitals give hybridized orbitals.
For $\ce{H2O}$, the lowest 3 molecular orbitals are generally these (see below):
At first, this result seems unlike the SP$^3$ bonds between hydrogen and carbon. There is a trick, however. It is called "localization of the orbitals", meaning that we generate linear combinations of molecular orbitals to generate new ones. Below, I'll give you an example in which we linear combinations of molecular orbitals $\varphi_2$ and $\varphi_3$ generate orbitals that look more like the SP$^3$ orbitals from text books.

(ii) if the answer to question (i) is yes, does the result differ from the textbook description? For instance, the lone pair electron in the $\ce{H2O}$ molecule? (is there any unitary equivalent result like orbital localization in MO-based method?)
As you may have noticed above, the results differ slightly, but I can elaborate.
Lone pairs are given by $\varphi_4$ and $\varphi_5$ below.

For $\varphi_4$, note that p$_z$ is along the z-axis, which is in the plane of the oxygen and hydrogen atoms: it is the 2-fold rotation axis.
For $\varphi_5$, note that p$_y$ is along the y-axis, which is perpendicular to the plane of the oxygen and hydrogen atoms: it is out of the plane.
For $\varphi_6$ and $\varphi_7$, note that they are unoccupied orbitals (i.e. lumo and lumo+1)
Now, orbital localization can be applied, linear combinations of $\varphi_4$ and $\varphi_5$ will give 2 lone pair orbitals, that stick slightly out of plane. I haven't tried drawing those yet, as it is not so easy to do so in this representation in 2D, so I'l try to switch to another representation...

As you see, those localized molecular orbitals closely resemble the SP$^3$ lone pair orbitals on oxygen, as predicted by valence bond theory. The only difference is the shape & probability of the small counter-part of the lobe; I included phase, not probability density here.
Does this answer your question?

I think this completely misses the point of the question. You are describing a molecular orbital approach and the interpretation with localised orbitals. So far so good, unfortunately the question is about ab initio valence bond theory and the use of hybridised orbitals within the construction of the wave function. (Also I think the localisation schemes you apply in your images do not add up. And especially in $\varphi_2\pm\varphi_3$ the angle does not make any sense. Also in none of the localisation algorithms will the oxygen lone pairs be equivalent.)

I can answer your first question definitively from my experience in quantum chemistry. Yes, hybrid orbital basis sets are still used, albeit minimally. Typically, these basis sets are used in QM/MM in hopes to scale down the QM computation. There are severe limitations in this approach. You tend to fix the geometry to hybrid orbitals, as you can only overlap certain orbitals. It becomes "ball and stick" chemistry. For this reason, it is not popular. Better to use a traditional QM technique on a smaller region where you will have reactivity.
I can attempt to answer your second question from a comprehensive standpoint as I have never done ab-initio calculations myself using hybrid orbitals. The results should be similar for geometries of simple molecules and break down for increasingly complex molecules. This is because non-VB electrons can undergo orbital overlap.

It is news to me that there are uses of valence bond calculations in QM/MM approaches. They certainly would not scale down the calculation cost. All in all, I have troubles how this would answer any part of the question.


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Why do alkyl bromides and iodides develop colour when exposed to light?
My book says

Alkyl halides are colourless when pure. However, bromides and iodides develop colour when exposed to light.

Why do they develop colour when exposed to light? It is important to note that alkyl fluorides, and alkyl chlorides are not said to develop colour.

This is because of spontaneous breakdown in case of iodides and bromides which have a lower bond-dissociation enthalpy as compared to fluorides and chlorides. This low enthalpy means that even at normal temperature or on exposure to sunlight, the halides (iodides and bromides) decompose to a certain extent forming free iodine and bromine, both of which are coloured and hence impart colour to the original halides.

@Satwikpasani.Hi Pasani, why do free iodine and bromine (radicals) impart color, whereas bonded don't?

@CURIE Colour is a property of the electronic distribution of the chemical species. Ion and its atom in a bonded molecule state are completely different chemical species, and therefore there is no correlation in their chemical porperties. $\ce{Cl-}$ is essential for life while $\ce{Cl2}$ is poisonous. The same is true for $\ce{Na}$. $\ce{Cu^++}$ is blue while metallic copper is pink. The same holds true for $\ce{Fe^++,Fe^3+}$.

This is because of the fact that in presence of light iodine and bromine absorbs light and excitation of the electron from the ground state to the excited state takes place which on returning back to the ground state radiates energy. This radiating energy comes under the visible region. Therefore they are co lured.

which on returning back to the ground state radiates energy. So this is fluorescence?! Are you sure? I don't believe that is right...

The color one sees with a white light source is the complement of the absorbed light [that is also the reradiated light] The color is the light that is NOT absorbed.


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How to determine whether a compound is gas or liquid or solid?

My book says, methyl chloride, methyl bromide, ethyl chloride and some cholorofluromethanes are gases at room temperature, whereas higher members are liquids or solids. Why?
With this question, I got another question for which the former would be subset. How to determine whether a compound is gas or liquid or solid (assume room temperature)? Thinking about this question, I thought molecular force between the molecules is going to be the key for answer of this query door. But how to determine, whether a compound has stronger intermolecular force between the molecules or not?
We have another way to check if a compound is gas or solid or liquid, i.e. entropy (randomness), which is given by heat gained or lost divided by temperature. In general, compound which has greater randomness will be gas. We can expect most of the compounds to be gases, which have standard molar entropy greater than $200\ \mathrm{J K^{-1} mol^{-1}}$ at $298\ \mathrm{K}$.
Entropy concept brings many questions under light, for example, $\ce{H2}$ gas has standard molar entropy of $130.7\ \mathrm{JK^{-1}mol^{-1}}$, which is lesser than $\ce{Br2}$ liquid, which has standard molar entropy of $152.3\ \mathrm{JK^{-1}mol^{-1}}$. We would had expected $\ce{H2}$ gas to have greater randomness than $\ce{Br2}$ liquid, because gases should have greater randomness than liquids. Anyway it becomes difficult, to determine a compound as solid or liquid or gas using entropy.

LINKS

http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch4/gases1.html#roomt (This link helps to determine a compound as gas, but leads to still more questions like: Why most non metals are gases? Why all the gases are covalent bonded that contain two or more non metals? Why gases are defined to have low molecular weights, in spite of knowing that molecular interactions matter? Can’t ionic bonded compounds be gases?)
I will try to find list of solid and liquid compounds, but now I have just the above link. If any one have links related to this particular topic, they are welcome.

Whoa whoa whoa. One question at a time please.

As the molecular weight increases (more halogens or carbon atoms), so does the electron density and consequently the van-der-waals forces(dispersion forces) between the molecules rises. These forces increase intermolecular attraction and therefore increase the boiling point. T^his is valid in case of most other organic compounds' homologous series, not just alkyl halides.


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Why is the reactivity of primary alkyl halides with nucleophiles (SN2 mechanism) greater than secondary and tertiary alkyl halides?
For the reaction between alkyl halide and a nucleophile, following the SN2 mechanism, the reactivity of alkyl halides is in the order: primary halide > secondary halide > tertiary halide.
If you consider the reactivity of alcohols with halogen acids to form alkyl halides, the reactivity of alcohols is in the order: tertiary alcohol > secondary alcohol > primary alcohol.
Why is there a difference in order of reactivity in both the cases?

The order of reactivity by substitution in these two reactions is difference because they have different mechanisms. The substitution of an alkyl halide by a strong nucleophile in a polar aprotic solvent is an SN2 mechanism. A $\ce{C-Nu}$ bond forms and a $\ce{C-X}$ bond breaks at the same time:
$$\ce{CH3CH2Br + NaOH ->[\text{acetone}] Na}\bigg[\ce{CH3CH2(Br)(OH)}\bigg]^{^\ddagger_{-}}\ce{->CH3CH2OH +NaBr}$$
All else equal, the rates of these reactions are controlled by sterics on the alkyl halide, which control how well the nucleophile can react the $\ce{C-X}\ \sigma^*$ orbital. Thus:
$$\text{Methyl}>1^\circ>2^\circ>>3^\circ$$
The reactions of alcohols with hydrohalic acids follow a different mechanism (mostly). These reactions happen in a protic medium ($\ce{HX}$ is protic) with a poor leaving group ($\ce{OH}$), at least initially. These are conditions that favor the step-wise SN1 mechanism, where $\ce{C-X}$ bond breaking occurs before $\ce{C-Nu}$ bond forming. A carbocation is formed as a key intermediate.
$$\ce{(CH3)3COH + HBr -> [(CH3)3COH2+]Br- -> [(CH3)3C+]Br- + H2O -> (CH3)3CBr + H2O}$$
According to the Hammond postulate, the stability of this carbocation influences the rate of reaction. More stable carbocation intermediates suggest lower energy transition states for the formation of those carbocations, which implies lower activation energies and faster rates. Thus:
$$3^\circ>2^\circ>>1^\circ,\text{Methyl}$$

@BenNorris.Thank you for the answer. By my other question it follows that tertiary alcohols require no catalyst (more reactive), whereas primary and secondary alcohol (less reactive) require catalyst for the reaction with alkyl halides. Does the use of catalyst really imply that primary and secondary alcohols are less reactive than tertiary and they follow $S_N1$ mechanism?

The need for a catalyst to make the reaction productive does imply that primary and secondary alcohols are less reactive, as expected based on the $S_N1$ mechanism. The catalysts work by changing the mechanism to be more $S_N2$-like.

@BenNorris Good answer, but I think you could include steric hindrance as a reason too. Tertiary halides are quite hindered compared to the Primary ones. This also explains why neopentyl halide doesn't undergo Sn2.


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Ions and Polyatomic Ions
Before a sodium atom and a chlorine atom combine together to form the ionic compound $NaCl$, they are just neutral atoms. However, when an electron is transferred from the outer energy level of the sodium atom to the chlorine atom, both of these gain a charge on them; the sodium atom or ion has a charge of $1+$ and the chlorine ion has a charge of $1-$. Both of these ions attract each other and an ionic bond is formed between them.

But what happens when an atom of an element say $Ca$ combines with a polyatomic ion say $OH^-$. Is this calcium atom an ion, is it already a charged atom when it interacts with the hydroxide ion or does it gain the charge by any transfer of electrons to the $OH^-$ ion and becomes $Ca^{2+}$ to bond with the $OH^-$ ion?

I think the polyatomic ions say $OH^-$ is already an ion and not an uncharged compound of elements that gains charge as in the case of $NaCl$, and the bond between an ion and a polyatomic ion is due to the attractive forces between them if already they have opposite charges.

Does $2Na+Cl_2$ give really $2NaCl$? For the second question I'm quite sceptic if you can us $Ca$ without a reaction with the water of the solution in which $OH^-$ is present... However theoretically is a good question!

@GM I guess for the second you could imagine a reaction with the molten metal hydroxide.

What about the answer to my question?

Your question comes from an odd angle so it's a bit hard to answer. You can get $NaCl$ by reacting $Na$ and $Cl_2$, so neither "start out" with charge. Also, you can get polyatomic ions such as $OH^-$ from a neutral polyatomic compound, like in the reaction $\ce{Ca + 2H_2 O → Ca(OH)_2 + H_2}$. There's no relation between "starting out with charge" or not and a species being polyatomic or not.

As I commented above, you could consider $Na$ vapour reacting with molten $CsOH$, where the sodium is directly exposed to hydroxide ions. However, the reaction actually happens between $Cs^+$ and $Na^0$, yielding the exchange reaction $\ce{Na^0 + CsOH → Cs^0 + NaOH}$. The hydroxide is merely a spectator ion, and you could well write the equation simply as $\ce{Na^0 + Cs^+ → Cs^0 + Na^+}$. You could also attempt to react $Na$ with $Cl^-$ directly in molten $CsCl$, but again all that happens is the exchange $\ce{Na^0 + CsCl → Cs^0 + NaCl}$ with no direct participation of $Cl^-$

@NicolauSakerNeto nice comments, I think this could be an answer.

I risk not directly answering the question as I have not perfectly understood it.
In your first example, you consider the reaction between two neutral atoms, $\ce{Na}$ and $\ce{Cl}$. An electron is transferred generating $\ce{NaCl}$ where the species are now charged ($\ce{Na^+ \  Cl^-}$). Then you follow up with a hypothetical reaction between a neutral atom ($\ce{Ca}$) and an already charged species ($\ce{OH^-}$). The two reactions are fundamentally different, but not because only the latter has polyatomic ions; the difference is in the very nature of the reactions you constructed (two neutral species reacting in the former, compared to one neutral species and an ion reacting in the latter).
The most direct way to obtain $\ce{NaCl}$ is the reaction between sodium metal (composed of neutral atoms, $\ce{Na^0}$) and chlorine gas (composed of neutral diatomic molecules, $\ce{Cl2}$). One then has:
$$\ce{2Na^0_{(s)} + Cl2_{(g)} → 2NaCl_{(s)}}$$
To obtain $\ce{Ca(OH)2}$, a similarly simple reaction would be that between calcium metal (composed of neutral atoms, $\ce{Ca^0}$) and water (composed of neutral molecules, $\ce{H_2O}$), where the water serves as a source of hydroxyl ions according to the equation:
$$\ce{Ca^0_{(s)} + 2H2O_{(l)} → Ca(OH)2_{(s)} + H2_{(g)}}$$
Now imagine you want to force a reaction between a neutral atom and an ion. There are many cases of this, but we can stick to something similar to what you wanted. In order to expose a substance directly to ions, a conceptually easy way is to add it to a molten salt. For now, let us consider pumping sodium vapour into molten caesium hydroxide ($\ce{CsOH}$). You seem to expect some interaction between the neutral $\ce{Na^0}$ and charged $\ce{OH^-}$, but in fact it is $\ce{Cs^+}$ that reacts with the sodium, according to the following equation:
$$\ce{Na^0_{(g)} + CsOH_{(l)} → NaOH_{(l)} + Cs^0_{(g)}}$$
Here we have simple exchange reaction. The hydroxide ion is actually only a spectator, and can be omitted without loss of understanding in an ionic equation:
$$\ce{Na^0_{(g)} + Cs^+_{(l)} → Na^+_{(l)} + Cs^0_{(g)}}$$
If we consider the reaction between sodium and molten caesium chloride instead, where chloride ions would be present, we have again a simple exchange reaction:
$$\ce{Na^0_{(g)} + CsCl_{(l)} → NaCl_{(l)} + Cs^0_{(g)}}$$
Indeed, it can be described by the same ionic equation above.
(Disclaimer: I actually am not sure the reactions between sodium and caesium compounds works as above. I don't have any understanding of the relative electrochemical potentials in molten salts, though I believe I've read that the reaction with molten chloride is a legitimate way to obtain metallic caesium. But that is mostly beside the point, either way. I avoided exemplifying with calcium and calcium hydroxide because I think the reaction wouldn't work, and the hydroxide is thermally unstable)


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Calculating percentages of microstructures in equilibrium phase diagrams (engineering)
So I have an equilibrium phase diagram of steel and I am asked to 'Calculate the proportion of pearlite in the microstructure of 0.4 wt% C steel just below the eutectoid temperature (727 °C).'
I have been using the forumlas
$Wa=\frac{Cb-Co}{Cb-Ca}$%
$Wb=100-Wa$%
$Ca=weight$% of (Carbon in this example) at the first intersection point on the eutectoid line.
$Cb=weight$% of (Carbon in this example) at the second intersection point on the eutectoid line or the bottom of the 'V'. I don't know how to explain any better sorry.
So looking at my diagram, I see that $Ca=0.022wt$% Carbon and $Cb=0.76wt$% Carbon, and $Co$ is just 0.4%.
So $Wa=\frac{0.76-0.4}{0.76-0.022}$%
$=0.487805%$%
This is where I am confused. What exactly is Wa? Is that how much ferrite/austenite/cementite that is in the pearlite then? Or how much pearlite is at that point compared to ferrite? And in any case, what does that mean Wb is?
Any help would be much appreciated. Thanks

Assuming you mean this phase diagram, it looks like your calculations are correct.
Based on my understanding of the diagram, at $727\ ^\circ \text{C}$, you have 100% pearlite at $C_b = 0.76\ \%C$ and 0% pearlite at $C_a=0.022\ \%C$. $W_a$ then appears to be how far $C_o$ is between $C_a$ and $C_b$, so $W_a$ should be the amount of pearlite.
The units of $C_a$, $C_b$, and $C_o$ are $\% C\  by\  mass$. That means when you calculate $W_a$ initially, you are calculating the fraction of pearlite and not the percent of pearlite. All of the $\% C\  by\  mass$ units cancel:
$$W_a=\frac{0.76\ \%C-0.4\ \%C}{0.76\ \%C - 0.022\ \%C}=\frac{{\%C}(0.76-0.022)}{{\%C}(0.76-0.022)}=0.487805\implies48.7805\%$$

So Wa is my pearlite amount? And what happens to the amount of pearlite after point Cb? Does it gradually go down until the Cementite area?

@user88720 That is exactly my question..Tell me something why did you choose $0.76 % C$ on the other end(I mean why did we not take the complete end that is $6.67%C$).

@user586228 - We use 0.76%, because that is the 100% pearlite line. To the left is a mix of pearlite and ferrite $\alpha$. To the right is a mix of pearlite and ledeburite. At 6.67% you have a crossover from a mix of $\ce{F3C}$ (cementite) and ledeburite to 100% cementite. There is no pearlite at this line. Since the goal was to determine the proportion of pearlite at 0.4%, we need a limit of 100% pearlite on one side (happens at 0.76%) and 0% pearlite on the other (0% C).

@BenNorris How did you predict the crossover at 6.67%?We are generally taught to draw the phase diagrams particularly upto 6.67%.Is there any intuition that helps us predict this?Or is it purely by knowledge?


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Interpretation of the following mass spectrum

I'm given the above spectrum. $M^+ = 102$, probably $\ce{C_6H_14O}$ or $\ce{C_5H_10O_2}$.
I'm thinking of ethyl propionate. (that picture also shows the fragments I expect). I can account for the following peaks:
m/z = 73: Loss of ethyl from acyl side (expected lowish intensity)
m/z = 57: Loss of alkoxy to form $\ce{CH3CH2CO+}$ (moderate intensity)
m/z = 45: Alkoxy ion (weak)
m/z = 29: ethyl carbocation, moderate intensity.
I can't account for 74, since according to my book the alkoxy side would not proceed through normal Mclafferty, but the Mclafferty + 1 rearrangement... That should yield m/z=75.
I proposed methyl butyrate, and of course that's not right. My instructor wrote to pay attention to the m/z = 57 (M-45) and m/z = 85 (M-17). This fits well with carboxylic acids, but again cannot account for the m/z = 74. McLafferty of carboxylic acids would result in a fragment of m/z=60...
Can someone help me out, I've come to a dead end for hours now.

I was on to the wrong structure. My main issue with the carboxylic acid was the fragment m/z=60 that it would produce by McLafferty. I didn't have that, so I thought that left me with t-Bu-COOH, which was the only structure with no $\gamma$-hydrogens. Besides I couldn't explain the m/z=74 peak with an acid. On the other hand I couldn't explain the M-17 peak by ethylpropanoate.
But of course, if the carboxylic acid has a methyl substituent at the $\alpha$-carbon, then that methyl group would also be part of the McLafferty fragment, therefore resulting in m/z=74!
The spectre is of: 2-methyl-butanoic acid
M-17 and M-45:
Alpha cleavage of carbonyl => loss of hydroxyl radical or carboxyl radical

m/z = 74:
Loss of ethene by McLafferty rearrangement

m/z = 45:
$\ce{COOH^+}$ fragment

M-15 and M-29:
Loss of methyl radical or ethyl radical

m/z = 29:
Ethyl carbocation

For m/z 74 it could be methyl ester:  in the loss of a $\ce{CH2=CH2}$ by McLafferty rearrangement.
$\ce{CH2=C(OH)OCH3+}$

This doesn't seem likely to me, especially considering the intensity. Do you suggest a mechanism :-)?

May be I'm wrong, but I've thought in the lost of a CH2=CH2 by a McLafferty rearrangement.

Okay I just read it again. It is the ethyl ester that fits with many of the peaks, but I was troubled because I thought the book said it would only undergo McLafferty+1 rearrangement (with double proton transfer, M/Z=75), but apparently that's just an additional mode of fragmentation to the normal McLafferty arrangement. This gives [CH3CH2C(OH)O]+ m/z=74. I would like to explain the m/z=85 though, now that my instructor pointed my attention to it. hmm

Possibilty: Loss of a hydroxyl group (m = 17).


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Why do hairs (rubbed with plastic comb) fallen on water seems to be sticking?
I was observing from many days, how hairs seems to be sticking to water, when made to fall on it. I made an attempt to know what really was going on. I kept a jug of stationary water under the light, and kept my dry hairs combing, to make hairs fell on it. I observed once again that hairs appeared to be sticking to water.

I started thinking about the cause for this observation. At first, what went to my mind was about hairs being rubbed with comb. Usually hairs get positively charged (I don't know exactly. I thought it because plastic becomes negatively charged when rubbed with most of  other suitable surface) when rubbed with plastic comb. We also know that water has more negative charge because of oxygen atom (I came to know about this from the diagram of wiki encyclopedia-electronegativity). So, I thought positively charged hairs might be attracted by negative charge nature of water molecules, which might be making hairs to stick with water.

Is it really because of positively charged hairs and negatively charged water molecules, which is making hairs to be sticking to water?

While indeed quite polar, water is exactly electrically neutral (as is any substance in most situations without an external potential being applied), so your explanation is not appropriate. Surely this question will be best answered in Physics.SE.

@curie can you explain what you ment by "hairs seems to be sticking to water". Whether you mean hairs are sticking together in water!

It doesn't have to do as much with the combing as it does with surface tension. Water on the hair clumps them together, because smaller clumps have more surface area (net surface area) and thus more surface energy. Breaking these large clumps, therefore, requires energy input, which gives us the impression that they "stick" together.
As for sticking to the water individually, that's surface tension again. The surface is able to withstand a bit of weight before collapsing and letting the object drop through. Hairs are oily, and thus hydrophobic. The repulsion potential energy that would be formed / needed when it sinks outweighs the energy required to bend the surface (and get enough tension force to balance gravity). This is how water skeeters work.

Maybe it's surface tension. See if hairs without electrical charge react the same way.


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How can I figure out the electron configuration of Br-?
It's easy to do atomic electron configuration, but ionic is confusing.

Welcome to Chemistry.SE. What about ionic configuration is confusing? The more we know about what trouble you are having, the better an answer we can give. For example, if you tell use what you think it is or where you are getting stuck, we can help you with the specific problem you are having.

The electronic configuration for $\ce{Br-}$ is:
$$\mathrm{1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6}$$
Because it have one more electron than bromine, which ends its electronic configuration with $\mathrm{4p^5}$.

I suppose that you know how to find the atomic electron configuration for Br which is $\mathrm{1s^22s^22p^63s^23p^64s^23d^{10}4p^5}$. Now for ions, it is really simple. All you have to do is look at if the ion is a cation or an anion. If it is an anion then add the same amount of electrons as the number of the charge. If it is a cation subtract the number electrons as the number of the charge.
In the case of $\ce{Br-}$ it is an anion of charge minus 1 meaning that it will have one more electron than a normal bromine atom. So really its electron configuration will be the same as krypton which is the atom after bromine. So all you have to do is add an extra electron in the last orbital, the 4p orbital. Therefore its electron configuration will be: $$\mathrm{1s^22s^22p^63s^23p^64s^23d^{10}4p^6}$$

But these configurations are for the ground state right?


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Why are tertiary alkyl halides obtained in higher yield than secondary or primary during free radical halogenation of alkanes?
From my textbook:

Halogens ($\ce{Cl_2}$ and $\ce{Br_2}$ ) react with alkanes in the presence of ultraviolet light to form haloalkanes.This reaction is free radical substitution and gives a mixture of mono, di, or polysubstituted haloalkanes which are difficult to separate into pure components. Moreover, the yield of any one compound is low because of the formation of other compounds. For example, in case of butane, a mixture of 1-chlorobutane (28%) and 2-chlorobutane (72%) is obtained, where the secondary alkyl halide is obtained as the major product rather than the primary one.
       $$\ce{CH_3CH_2CH_2CH_3->[\ce{Cl_2,~UV~light}]CH_3CH_2CH_2CH_2Cl + CH3CH_2CHClCH_3}$$
    In general, the ease of substitution of various hydrogen follows the sequence:
      $${\rm tertiary > secondary >primary}$$

After reading through all the above statements in my book, I now have the following questions:

Why are tertiary alkyl halides obtained as major products instead of secondary or primary ones? It seems that a hydrogen from a tertiary carbon can be more easily replaced by a halogen than that of a secondary or primary carbon. If it is so, why is hydrogen from a tertiary carbon more easily replaced than a hydrogen from a secondary or primary carbon?
Why are fluorine and iodine not considered in case of free radical halogenation (the first two lines of the above passage extracted from the book mentions only chlorine and bromine)?
Why is a tertiary alkyl halide not produced in the above example? Even if we think it might be produced in a lesser amount, there should have been more of it than the primary or secondary alkyl halides. It seems the reason for tertiary alkyl halide not being in the equation is because it might not have formed or it might have formed in a lesser amount than primary or secondary alkyl halides. If it is either any case, it would violate the order mentioned in the book.

Do you know the reaction mechanism for free-radical halogenation of alkanes, the intermediates formed therein?

http://web.chem.ucsb.edu/~neuman/orgchembyneuman.book/11%20RadicalSubstitutionAddition/11FullChapt.pdf

The reason for the ordering is that tertiary radicals have a lower energy (and are thus easier to form) than secondary radicals, which are in turn easier to form than primary radicals. This is because of a phenomenon called sigma conjugation - the weak overlap between the half filled orbital and the adjacent side group orbitals gives a slight lowering of energy.
This means that it is easier to remove a hydrogen to give a tertiary radical intermediate than to give a secondary radical intermediate and so on.
This is similar to Markovnikov's rule.
The reason a tertiary product is not formed is that for butane there is no tertiary centre to begin with - nowhere that a Cl radical can react to give a tertiary butyl radical. Forming a tertiary product would require a rearrangement of the carbon chain.

Why are tertiary alkyl halides obtained as major than secondary or primary?
  It seems that hydrogen from tertiary carbon can be easily replaced by halogen than from secondary or primary carbon. If it is so, why is hydrogen from tertiary carbon can be easily replaced than from secondary or primary?

The mechanism involve:
$$\ce{X_2->[\Delta][\text{or UV-light}]2X.}\\\ce{X.+RH->HX +R.}\\\ce{R.+X2->RX +X.}$$
where last two steps keep on repeating cyclically.
The difficult step is step is as follows which decide the rate of the mechanism:

The relative rates of primary, secondary, tertiary alkyl halides depends on two factors: probability and Activation Energy.Probability factor for a typical case of propane is $3^\circ:2^\circ:1^\circ=0:6:2$(directly equal to number of hyrogens of that type.)
The activation energy of $\ce{R-H + X.->R. +H-X}$ in KCAL/MOL are:
$$\begin{array}{c|c|c}\ce{R&X=Cl&X=Br\\CH3&4&18\\$1^\circ$&1&13\\$2^\circ$&0.5&10\\$3^\circ$&0.1&7.5}\end{array}$$
The more stable the radical, more easily it is formed.Factors that stabilize the free radical tend to stabilize the incipent free radical in the transition state:
$$\ce{-C-H+X. -> [C^{\delta.}\bond{...}H\bond{...}X^{\delta.}] -> -C.+H-X}$$
A certain factor (delocalization of the odd electron) causes the energy difference between radicals.
Stability of free radicals:$3^\circ>2^\circ>1^\circ>\ce{CH3}$
Ease of formation of free radicals:$3^\circ>2^\circ>1^\circ>\ce{CH3}$

Why is fluorine and iodide not considered in case of free radical halogenation (as you can see, the first two lines of the above passage extracted from the book, mentions only chlorine and bromine)?

Iodination is reversible but it may be carried out in the presence of an oxidising agent such as $\ce{HIO3,HNO3,HgO,}$etc., which destroys the hydrogen iodide as it is formed and so drives the reaction to the right:
$$\ce{CH4 +I2<=>CH3I +HI\\5HI +HIO3->3I2 +3H2O}$$
Iodines are more conviniently prepared by treating the chloro- or bromo- derivatives with $\ce{NaI}$ in methanol or acetone solution:
$$\ce{RCl+NaI->[\text{acetone}]RI+NaCl}$$ This reaction is possible because $\ce{NaI}$ is soluble in methanol or acetone, whereas sodium chloride and bromide are not. This reaction is known as Finkelstein or Conant-Finkelstein reaction.
Direct Fluorination is usually explosive; special conditions are necessary for the preparation of the fluorine derivatives of the alkanes.One such example is Swarts reaction.

Why is tertiary alkyl halide not produced in the above example? Even if we think it might be produced in less amount, it should have been greater than primary alkyl or secondary alkyl halide, is n't (because of the order)? It seems the reason for tertiary alkyl halide not being in the equation is because of either, it might not have formed or it might have formed in lesser amount than primary or secondary. If it is either any case, it would violate the order mentioned in the book.

See carefully there's no tertiary alkylhalide possible, because no tertiary hydrogen is available. As radicals do not rearrange, the structure never changes!

To address your second question, fluorine and iodine are not considered for this reaction because the first is too reactive while the latter is too unreactive.
Fluorine will react harshly with alkanes, producing an unselective mixture of fluorocarbons. The chain propagation steps of the reaction are so exothermic that it proceeds explosively, and special measures are necessary to control the reaction (e.g. low reactant concentrations).
In contrast, iodine will react very slowly, if at all. Although the activation energy for the reaction (+151 kJ/mol) is the lowest of all halogens, the first chain propagation step (+142 kJ/mol) and the overall reaction enthalpy (+54 kJ/mol) are very endothermic, which disfavors radical iodination of alkanes (reference).

You asked " Why are tertiary alkyl halides obtained as major than secondary or primary?" As pointed out in the above comments, the order of reactivity is tert.>sec.>prim. But it is not necessarily true that the tertiary chloride is the major product. Tertiary hydrogens are ~5.5 times more reactive than primary hydrogens in radical chlorinations. In isobutane there are 9 primary hydrogens and only 1 tertiary hydrogen. The the product ratio is tertiary/primary = 5.5/9. Primary product 1-chloro-2-methylpropane (62%) is the major product and 2-chloro-2-methylpropane (38%) is the minor product.


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Most economical method to convert potassium oxide to potassium nitride.
I have a compound potassium oxide($\ce{K2O}$) and I am trying to convert it into potassium nitride ($\ce{K3N}$).
Here are a few possible reaction methods :

Split potassium oxide into the constituent elements potassium and oxygen  (Not very good  as a lot of energy is required)
$\ce{K2O + N2 -> K3N + O2}$ (however this won't happen(?) because nitrogen has a lower electronegativity value than oxygen does)

and so on.
Does anyone know how can I convert the oxide to the nitride, releasing oxygen in the process, by not having to use a lot of energy-requiring processes?

The second reaction can happen- Thermodynamics are altered by concentration differential. Lots of Nitrogen and little K2O. It will need catalysing to break the nitrogen to atoms which is a kinetics problem. Potassium Nitride is not all that stable. Just look at Sodium Nitride in wikipedia.

@user2617804 How would I be best to begin on doing more research for the process that you mentioned above?

@user2617804 - Potassium nitride $\ce{K3N} \ne$ potassium azide $\ce{KN3}$

Do you really want to go there?
Based on this entry from Wikipedia the compound is synthesized and stable only below room temperature, and on top of this requires elemental potassium:

This compound was produced by the reaction of potassium metal and liquid nitrogen at 77 K (−196.2 °C; −321.1 °F) under vacuum:[1]
$\ce{6K + N2 → 2K3N}$
This compound decomposes back into potassium and nitrogen at room temperature.

You may want to check that the target is actually $\ce{K3N}$ versus the more plausible $\ce{KN3}$.
Cited Reference

Fischer, D.; Cancarevic, Z.; Schön, J. C.; Jansen, M. (January 2004). "Zur Synthese und Struktur von K3N". Zeitschrift für anorganische und allgemeine Chemie (in German). 630 (1): 156–160. https://doi.org/10.1002/zaac.200300280


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Finding the percent of calcium carbonate in commercial antacid by using back titration
I'm an IB Chemistry student and I had to do a back titration lab where 1g of a powdered antacid tablet were mixed with 50mL 1M HCl, and the mixture was titrated using 0.2 M NaOH; the NaOH was to neutralize the leftover HCl that hadn't reacted with the antacid. The active ingredient in the antacid is CaCO3.
It took me about 40mL of NaOH to titrate the leftover HCl to endpoint; 40 mL at 0.2 mol/L translate to 0.008 moles of NaOH. In the equation for reaction between NaOH and HCl the ratio of the two reactants is 1:1; therefore there were 0.008 mol of unreacted HCl in the mixture.
There were 50mL of HCl (1 mol/L) in the mixture originally; 50mL * 1 mol/1000mL = 0.05 mol HCl total. Subtracting from this the molar quantity of unreacted (leftover) HCl yields: 0.05 - 0.008 = 0.042 mol HCl that did react with the CaCO3 in the antacid.
The reaction between CaCO3 and HCl is as follows:CaCO3 + 2HCl --> CaCl2 + CO2 +H2O
Therefore, molar ratio between CaCO3 and HCl is 1:2. 0.042 moles HCl reacted, therefore 0.021 mol CaCO3 were present in the sample. Multiplying this by the molar mass yields 0.021mol * (40+12+48)g/mol = 2.1 grams CaCO3.
That's a problem - i can't be having 210% composition by mass (the sample was 1g; I calculated 2.1 g CaCO3 present). So somewhere along the line i made a mistake, but I've checked my math so I don't know if it was a calculation error or if my method is dead wrong? Can anyone help me with this?

The calculations seem correct and the methodology should work. Could it have been some sort of experimental mishap? Also, are you sure the compound in the antiacid tablet is calcium carbonate, and that there is no other active ingredient? There could be another compound with higher capacity to absorb protons per gram than $\ce{CaCO3}$ in the mixture, such as $\ce{Mg(OH)2}$. Other than that, I'm stumped.

That's what's annoying me, the lab is titled "to find the percent of caco3 in commercial antacid by using back titration", and our teacher told us to focus only on the CaCo3 . And everyone in my class got volumes of NaOH (to endpoint) around 40mL, so I don't think it was something wrong with my experiment

That's highly suspicious then. It is entirely possible for the antiacid bought to not have any $\ce{CaCO3}$, as it may have been purchased without due attention to that detail. Is there any chance you can see the package for the antiacid, or do you know the brand?

In chemical analysis usually do three times the same experimet, so you can minimize experimental errors.
After that, in order to find where is the error you can check:
Has everyone in you class use the same HCl and NaOH solutions (taken from the same bottle i.e.?
The antiacid tablets are all the same (given by your teacher)? If don't, could you post the composition?
Have you measured correctly the volumes used? To obtain that 210%, it seems tha you have used around 25 mL of HCl solution instead 50 mL.

Yeah we did several trials, still got around 40 mL. And yes, everyone used the same solutions, and the antacid tablets were supplied by the teacher, but he didn't give us the label on them. It's possible that the point of the lab is to conclude that CaCO3 isn't the only active ingredient - in IB we have to evaluate the experiment and how applicable it is, so perhaps there was a second reagent, such as Mg(OH)2 as @Nicolau suggested, that couldn't be determined from such a simple experiment.

Yes, but with a molar mass of near 57, does not fully explain the volume of HCl used. If all the tablet was magnesium hydroxide, maybe could obtain such titration values.

Another thought, if your teacher told to focus on CaCO3, maybe we can discard other HCl reactive compounds. But 1 g tablet of CaCO3 contains 0,01 mol of CaCO3 wich reacts with 0,02 mol HCl. 50 mL of HCl 1 M contains 0,05 mol, so 0,03 mol didn't react. 0,03 mol HCl needs 0,03 mol NaOH (150 mL).

@mdan Indeed I should have been a little more careful suggesting another possible compound. Your unknown antiacid needs absorb at least $0.042\ mol$ of protons per gram, or in other words it needs to have a molar mass of at most $23.8\ g$ per mole of protons absorbed.

Looking at a list of possible compounds in common antiacids, the lowest ratio I can find is for $\ce{Al(OH)3}$, for which the ratio is $26\ g/mol(H^+)$. Therefore it seems none of those candidates can work. However, $\ce{MgO}$ is just below the required value ($20\ g/mol(H^+)$). So unless the tablet was made of almost pure $\ce{MgO}$, I can't think of anything else except some experimental problem, possibly erroneously determined acid/base concentrations.

Well the antacid itself needs to absorb 0.042 mol (H+) per gram; if the tablet is a combination of CaCO3 and some other compound, could that work out? I don't think I need to be getting very specific with what possible compounds could be present in the tablet, I just need to talk about errors in my experiment and possible causes but not that specific. Thank you!

@mdan Yes a mixture will work, but notice that even pure $\ce{MgO}$ will almost reach the $1\ g$ limit ($0.84\ g$ actually), so while there could be other compounds, they can't be present in a very large proportion relative to $\ce{MgO}$ or else their total mass would go over the limit. This is all of course assuming every other step of the experiment was performed perfectly, only then does it become mathematically impossible for the antiacid to be mostly $\ce{CaCO3}$, or indeed practically anything other than $\ce{MgO}$.

Well, this is embarrassing. The 0.008 moles of HCl that didn't react were for a 10 mL sample; I had forgotten that for each titration with the sodium hydroxide, a 10 mL sample of the completed reaction mixture (between HCl and antacid) was used. Therefore, total unreacted moles HCl in the 50 mL that were used overall = 5 * 0.008 = 0.04 mol, subtract that from the 0.05 mol present in 50 mL and get 0.01 mol HCl that reacted. Therefore 0.005 mol CaCO3 reacted, which in turn corresponds to 0.5 grams per 1 gram, so 50% of the tablet is calcium carbonate. Sorry for causing so much trouble!

Don't worry. but always remember to take note of all the data of your experiments.

Also, always keep in mind the steps you did in your experiment (it's useful to draw an scheme).


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Producing gold nano flakes
In terms of producing gold nano flakes, Guo et al. [1] added aniline to the solution of chloroauric acid in ethylene glycol at 95 °C. However, I couldn't figure out the step by step chemical reaction in it. Can anyone please explain me the reaction procedure or guide me to the relevant source?
References

Guo, Z.; Zhang, Y.; DuanMu, Y.; Xu, L.; Xie, S.; Gu, N. Facile Synthesis of Micrometer-Sized Gold Nanoplates through an Aniline-Assisted Route in Ethylene Glycol Solution. Colloids and Surfaces A: Physicochemical and Engineering Aspects 2006, 278 (1), 33–38. https://doi.org/10.1016/j.colsurfa.2005.11.075.

I believe that ethylene glycol in this reaction is used as a solvent and Aniline is used as a reducing agent. Ethylene glycol solution heated with chloroauric acid produces $Au^{3+}$ ions (first 20 min) which is then reduced to $Au^{0}$  (3hr reaction) in the presence of aniline. Because there is no stirring during those 3 hrs (so less agitation of the reaction mixture)there is less formation of gold nanoparticle and more gold nano flakes are formed.
May be this paper will help you, it has a similar mechanism


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Which is the best rocket fuel?
I searched on internet about rocket fuel and the most common fuel is $\ce{KNO3}$ + sugar (rocket candy), but I think that I could use another fuel methods, like using can with gas or using $\ce{HCl + Al}$.
My question is: Which one ($\ce{KNO3}$ + sugar, can with gas or $\ce{HCl + Al}$) gets more performance? Can I increase the performance by increasing the pressure and the durability?

Please take the necessary precautions when working with these substances and, when in doubt, please consult with someone that is an expert in this area.

When i refer to best is to the performance.

Do you mean rocket fuel for home experiment?

What do you mean by performance?

I used Zn + S as a teenager. It is a lot safer than Al, and you can mix it cold.
Stick to solid propelants because it's simpler, safer and smaller. You don't want an explosion but something very quick burning.
Mix the two fine powders well.
You can mix them in a thick plastic bag or wide mouth bottle and shake until you see an even color in all the mix with no lumps in sight.
(You can make a paste with alcohol 100% but I never tried this.)
Then tap it in with wood or plastic. You don't want a spark! Nozzle reduction is 1/3 and not smaller. The mix is 3:1. Light with a filament, it is safer than a spark.
In those days all I had was a book but you can find a lot of info out there.
We switched to this after a (not too serious) accident with rocket candy.
Safety first allways, the rocket can go anywhere!

Why do you say that? Could the Hydrogen released on the reaction HCL + AL be inflammable?

I searched about that and Hydrogen is Highly inflammable.

Stick to solid propelants because it's simpler, safer and smaller. You don't want an explosion but something very quick burning.

Yeah, to make a solid propelants with Zn + S you mix it cold and stick with a hammer?

Mix the two fine powders well and tap it in with wood or plastic. You don't want a spark! Nozzle reduction is 1/3 and not smaller. The mix is 3:1. Light with a filament, it is safer than a spark. I'm remembering a lot...

There are references like Model rockets. BE VERY CAREFULL!

Yeah safety first! I searched about ZnS and it is extremely sensible to hot. When you sai that i can mix it cold, how can i do that?


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Speed of ionic reactions?

Why are ionic reactions extremely fast,  as compared to other gaseous
  reactions?

The ionic reactions (precipitations, neutralizations etc.) are extremely fast (somewhere about $10^{12}$ times faster) than gas-phase reactions and other ordinary reactions in solutions (inversion, substitution etc). Why is this so?
My thoughts
One possible reason might be the relative freedom of steric orientation in case of simple ionic reactions as compared to organic reactions and complicated gas reactions, but this isn't the case with simple gas reactions or formation and dissociation of ionic complexes, which do require specific steric orientations. The activation energy could be very low for ionic reactions, if this is so, then what is the reason behind it?

Reaction rate is inversely proportional to activation energy for a given process. The examples given have very low barriers and often the diffusion in solution is the rate limiting process.
If you think of reaction of anion with cation, than there should be no barrier at all, because the interaction between oppositely charged species is strongly attractive.

In ionic reactions there is no electronic transfer between ions and the number of collisions between ions with opposite charge in high.
When reaction occurs between neutral compounds, there is electronic transfer between atoms or molecules, and also there is bonding reorientation, son the speed in such reactions are lower than in ionic reactions.

Reactions of ionic compounds are very fast because ionic compounds exist as ion in aqueous solutions.therefore no bond is to be broken.only few bonds are formed.EXAMPLE# by adding AgNO3 solution to NaCl solution,white ppt of AgCl are formed immediately.in this reaction,both AgNO3&NaCl are ionized.therefore no bond is broken,few are formed.thus reaction occurs rapidly

This doesn't really add any new insight beyond what has already been said.

It's not true - all those ions are solvated, particularly Ag+