text:
Why does water dissociate to H3O+ and OH- rather than H+ and OH-?
Why does water dissociate to
$\ce{H3O+ + OH-}$
instead of $\ce{H+ + OH-}$?
This question came to surface when I was learning about acids and bases, and learned this definition:
$\mathrm{pH=}-\log_{10}[\ce{H+}]$
Which after looking at Wikipedia, looks like a simplified definition, but it got me thinking - how can water have a pH of around seven if it dissociates into $\ce{H3O+}$ instead of $\ce{H+}$? Why does it do this instead of the expected dissociation?

Something still needs to be mentioned, though I can't write an answer with the deserving care. Though of course it's important to underline the reactivity of bare $H^+$ and its tendency to be strongly solvated (by anything), the thing is that even $H_3O^{+}$ is incorrect in water. Exactly how one should represent the hydrated proton in the autodissociation of water is a bit of a nitpick; it doesn't really matter for most purposes, and honestly $H^+_{(aq)}$ is best because it does not erroneously suggest the nature of the solvation is entirely described by the equation. It's just a placeholder.

@NicolauSakerNeto You are so right here, and it is very much of nitpicking. I think it is at least better to consider the first order approximation $\ce{H3+O}$ over the zeroth order $\ce{H+}$, since that is virtually impossible in condensed phase. And the again $\ce{H2O}$ is also only an approximation, the hydrogen bond network is much too strong to be neglected, but I have not yet seen $\ce{H2O_{(aq.)}}$.

You have to analyze the reaction mechanism behind the self-ionization of water to understand better what happen. In fact is not strictly self but a matter of "couple-ionization": you know the water atoms have different electronegativity, the more electronegative oxygen attracts the less electronegative hydrogen of another water molecule forming an hydrogen bonding.

From there the electron from Hydrogen forms a lone pair with the oxygen of the same molecule forming $\ce{OH^-}$. Hydron ($\ce{H^+}$) is in fact a proton, a very reactive specie so is not so simple found it uncoupled and moreover the hydrogen is suppose to be bonded with an hydrogen bonding with the other water molecule so is more accurate use hydronium ($\ce{H_3O^+}$) instead of hydron...
However you don't have to worry about it for your pH calculation:
$\mathrm{pH=}-\log_{10}[\ce {H^{+}}]= -\log_{10}[\ce {H_3O^{+}}]$
Because, in fact, you have the same ratio between basic species ($\ce{OH^-}$) and acid species( $\ce{H_3O^+}$ or $\ce{H^+}$ as you prefer) in both cases!

Noteworthy and somewhat related, traveling protons: http://en.wikipedia.org/wiki/Grotthuss_mechanism

Single water molecule dissociates into  $\ce{H^+ + OH^-}$
As noted in @G M's answer, $\ce{H^+}$ is very reactive species and some people do not like it and write it as less provoking $\ce{H_3O^+}$, for discussion see first paragraph of http://en.wikipedia.org/wiki/Hydronium
In short - $\ce{H^+}$ organizes water molecules around itself and it's behavior is complicated and interesting.
Another point important point stated by @G M is
$\mathrm{pH=}-\log_{10}[\ce H^{+}]= -\log_{10}[\ce{H3O+}]$
which is correct, because the concentration of $\ce{H^+}$  is so low, that using $\ce{H_3O^+}$ does not change the concentration of $\ce{H_2O}$.


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How many p-orbitals are there in each period?

How many p-orbitals are there in each period?a. 1b. 3c. 5d. 7e. 14

I take this to mean "how many p-orbitals are there in each period of the periodic table?" Looking at the p-block on the periodic table, it looks like there are 6 elements with p-orbitals per period. This isn't an answer, so I'm a little lost.

The question you were asked is awkwardly stated, but if you recall that each orbital can accept two electrons you should be able to infer the number of p orbitals from your earlier observation. This number is equal to the number of allowable values of the magnetic quantum number $m$, for $l=1$

Alright, so since p-orbitals require that l = 1, then the magnetic quantum number can assume 3 values (according to the equation x = 2l + 1), the values will be -1, 0, or 1. Therefore, the answer is 3. Thanks for the help!

For the p subshell, the azimuthal quantum number, which describes the angular momentum and shape of the orbitals, is $l=1$. For the associated magnetic quantum number, discrete values in the range $-l, -l+1, ... 0 ..., l-1, l$ are allowed, so the number of different $m_l$ values is $2l+1$.
With $l=1$, it follows that there are $2 \cdot 1+1=3$ p orbitals with the $m_l$ values of -1, 0 and +1, respectively. Each orbital can hold 2 electrons, so the p subshell can contain a maximum of 6 electrons. This also explains why there are 6 elements in the p block of each period. With increasing atomic number $Z$, one p electron is added per element in this block.

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How to figure out the state of aggregation of an unknown substance via properties of a chemical equilibrium in a closed system?
This is a question from the College Entrance Test in China.

In a closed, constant temperature container, reaction $\ce{A(s)<=>B +C(g)}$ reached chemical equilibrium. If the volume of the container is reduced, when the equilibrium is reached again, the concentration of $\ce{C}$ equals the concentration of $\ce{C}$ before the reduction.
Is it possible that $\ce{B}$ is in gas state?

The official answer is impossible.
And the official explanation to this is:

If $\ce{B}$ is in gas state, we have $K=C_\ce{B} \cdot C_\ce{C}$
Otherwise, if $\ce{B}$ is in liquid or solid state, we have $K=C_\ce{C}$
Since the concentration of $\ce{C}$ does not change after the reduction, in order to remain the $K$ value unchanged, $C_\ce{B}$ must not be involved in this process. So $\ce{B}$ is impossible to be in gas state.

But I don't think it makes sense.
Consider a container with a volume of $V$, $b$ mol of $\ce{B(g)}$ and $c$ mol of $\ce{C(g)}$ is inside it.
If   $K_{\text{pre}}=K_{\text{post}}$
, we have $C_{\ce{B},\text{pre}} \cdot C_{\ce{C},\text{pre}}=C_{\ce{B},\text{post}} \cdot C_{\ce{C},\text{post}}$.
When the container's volume is reduced by $\frac{1}{2}$, if $\dfrac{b_{\text{post}}}{b_{\text{pre}}}=\dfrac{c_{\text{post}}}{c_{\text{pre}}}=\dfrac{1}{2}$, the equilibrium can be reached, with all the requirements stated in the question.
Conclusion: $\ce{B}$ can be in gas state.
So, is the official answer wrong or I just didn't make the right conclusion?

What are your thoughts on the problem?

I edited your question to make the chemistry easier to read. You can learn about how to do this at our notation page. Also, this look like it might be a homework question. As jonsca suggests, we do like to see you meet us part way on questions like these (even if it was not directly assigned as a homework problem). Check out our homework policy.

Distinction: you are only considering Kp, which is an equilibrium constant that doesn't take concentration into account. You're going wrong here because the question asks for concentration, which has a unit of $L^{-1}$, not mole fraction, which is just in units of moles.
$$Kc=[C]c[D]d / [A]a[B]b$$
$$Kp=(C)c(D)d / (A)a(B)b$$
Think about their answer to the question. It states that $K=Cc_{pre}$ and $K=Cc_{post}$. The only way for this to make sense is if $Cc$ is in units of concentration because then as the total moles of c go up, the fact that the volume is decreasing doesn't change the value of Kp.
When you say,

If   $K_{pre}=K_{post}$
  , we have $C_{\ce{B},pre} \cdot C_{\ce{C},pre}=C_{\ce{B},post} \cdot C_{\ce{C},post}$
When the container's volume is reduced by $\frac{1}{2}$, if $\dfrac{b_{post}}
{b_{pre}}=\dfrac{c_{post}}{c_{pre}}=\dfrac{1}{2}$, the equilibrium can be reached, with all the requirements stated in the question.

you do not take concentration into account, or the fact that $Cc_{pre} = Cc_{post}$
For then you cannot restate your equation in terms of concentration: $\dfrac{Cc_{post}}{Cc_{pre}}=\dfrac{1}{2}$

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Is ionic bond just formed by "electrostatic" interaction between two oppositely charged ions?
According to the definition of ionic bond:

An ionic bond is a type of chemical bond formed through an electrostatic attraction between two oppositely charged ions.

If we consider the above figure showing formation of ionic bond, due to the transfer of electron from sodium atom to fluorine.It demonstrates clearly, how electrostatic interaction takes place.
In general, electrostatic interaction takes place between two stationary charges.Moreover, electrons would not be static as demonstrated by the above figure.Electrons will be moving at high speed, producing magnetic field around them, due to their motion; and even will also have electric field around them.Thus, we can expect electromagnetic interaction between oppositely charged ions along with electrostatic interaction due to stationary protons.So, does it mean that ionic bond is due to electromagnetic interaction and not just by electrostatic interaction?
[All the statements made are up to my view on the concept.Any correction advisory is welcome]
REFERENCES

http://en.wikipedia.org/wiki/Ionic_bond   (definition and diagram has been extracted from this link)
http://en.wikipedia.org/wiki/Electromagnetic_radiation (provides notion of charge motion effects)
http://en.wikipedia.org/wiki/Electrostatics (gives notion of electrostatic force)

You are absolutely correct that the electron has an angular momentum, which can be thought of like a magnetic field and which interacts with magnetic fields.
In "light" elements, however, the magnetic interaction is almost completely negligible.
The Hyperphysics page on the Zeeman Effect shows that in a hydrogen atom, a magnetic field of one tesla (which is fairly strong) splits energy levels by $5.78 \times 10^{-5} eV$, or roughly one-millionth of the ionization energy of a hydrogen atom.
From just having done calcuations on the anomalous Zeeman effect for homework (you can use the formula in the link I provided and check it yourself), I can tell you that in a potassium atom, the energy levels of a p-electron split by 0.006 Hartree/Tesla, or 1/10,000 of the ionization energy of a hydrogen atom.
In other words, while there is a magnetic field (or what can be treated as one) generated by electron motion, it's fairly safe to ignore it in lighter elements.
As far as heavy elements go...well, you undergo what's called j-j coupling, caused by the fact that the electrons are going so fast that their motion becomes relativistic. To describe that motion requires the Dirac equation, which I have not studied in any level of detail. Instinct says that there are other considerations which are much more significant than magnetic interactions, but I have no proof whatsoever to back it up.

@chipbuster.Did you mean that ionic interaction is due to electromagnetic interaction, and not just due to electrostatic interaction?

Sorry, I'm a little tired right now. What I meant to say is that while there is technically magnetic interaction going on, the bulk of the lattice energy comes from electrostatic interactions.
EDIT: This is definitely not an expert opinion. I have not studied the physics of lattice bonding in detail.

First to the magnetic properties - if you would be happy with Schroedinger equation and paired electrons, which is perfectly appropriate for the example show, then just forget about magnetism at all, it is negligible. All the information needed is in http://en.wikipedia.org/wiki/Electronic_molecular_Hamiltonian, which is valid for all bonding situations.

Yes, in fact electrostatic attraction is an electromagnetic interaction. So there is no need for thinking about different approach. If you mean only magnetic interaction, so you are searching for the relation with magnetism, I have some doubts about the possibility that they are related.

@G M.Up to my knowledge electrostatic interaction is not the same as electromagnetic interaction.Electrostatic interaction deals with charges at rest, which have electric field around them with no magnetic field.Magnetic field will be produced around the charge, only when it starts to move.So, while considering electromagnetic interaction, we will be dealing with electric and magnetic field, whereas in case of electrostatic interaction, we will be dealing only with electric field.

@CURIE is not the same because is one of them."Four fundamental interactions are conventionally recognized: gravitational, electromagnetic, strong nuclear, and weak nuclear."  Electrostatic is inside electromagnetic.

@G M.According to the definition of the electrostatics, "Electrostatics is a branch of physics that deals with the phenomena and properties of stationary or slow-moving electric charges with no acceleration." [this definition has been extracted from the wiki encyclopedia-electrostatics] In case of the electrons, they are not slow-moving charges, they nearly move with the speed of light. So, we can't say that electrostatic force comes under electromagnetic force, under this considered situation of electron revolving with high speed.

@CURIE "Four fundamental interactions are conventionally recognized: gravitational, electromagnetic, strong nuclear, and weak nuclear." Where do you put electrostatic?


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How to study (by heart) all the equations in inorganic chemistry?
Physical chemistry is easy to study as I have to just understand the concepts (like in physics) and apply some equations for calculations. In organic chemistry, reaction mechanisms helpsin remembering a lot of reactions and also doing conversion problems help a lot.
But, in the case of inorganic chemistry it is very difficult for me to remember those equations. My high school syllabus includes a lot of preparation methods of different compounds and reactions showing their chemical properties. But, unlike organic chemistry there is no problems based on it for deep understanding. So, what is actually needed is a good memory power. But, I'm not able to remember all those equations. So, how should I study them?

Every thing, how one knows all the equations in inorganic chemistry, comes from the experience.The time they have spent with that particular concept, asking one self, why one particular compound is prepared by a particular method, and why do they show such properties. Even the ssavec answer comes from his own experience with the concept. Anyone can't access one's experience, one should have little experience about what one is saying.I think the best answer for your question will be your's, which you get after experience.

@CURIE is one year enough for getting this experience? I'm a 12-grade high school student and every student in this level is wanted to remember lot of inorganic equations by heart including the temperature at which they are prepared(if provided in the text) for the final exam.

@RajathKrishnaR.The student who reads one page for max hours, questioning and thinking even about simple concepts is the one who is more experienced, and has better memory in the manner that he feels speaking about complex concept, is as easy as explaining how to prepare coffee, initially learning seems to be slower, but swifts later by the earlier proper understanding he had. He can sit for hours, because of interest, he has, and the joy  he gets from it. On the other hand, the student who reads max pages in one hour, is the least experienced one.

@VINAY , a 12th grader doesn't have that much time to spend because there are other subjects to study as well..

What would you suggest for a high schooler who is currently struggling with the same issue? Since this question is 10 years old, I would love to hear your thoughts on this. Do you think Inorganic Chemistry is all about memorization?

In organic chemistry, everything is based around C, H, O and N. Therefore to understand reactivity (somehow), there is just limited number of options.
In contrast, inorganic chemistry deals with the rest of periodic table, and each of the elements has its special properties. But there are general trends as well, but are more difficult to decipher and are often expressed by set of typical reactions, you should learn.
Usually they concern the stability of (oxidation) states, how to go from less stable to more stable one to obtain the desired compound and few processes how tho achieve the least stable state from the one found in nature.
It is difficult to state some simple rules, but there are several most typical anions and cations, and the chemistry of given element is just interactions with them. During the learning it helped me to have "all" the reactions (several hundreds) on a paper and for each one try to understand what happens and if there are others in the set, which are similar, and why. This helps you to spot the similarities. And - it is better to have more reactions then less, because then you have higher probability to spot the trends.

Organic chemistry also deals with halide functional groups and fundctional groups with sulfur, phosphorus, and boron.

Let me tell you how I have managed to, with about 85-90% accuracy, been able to predict a large number of inorganic reactions. There are some prerequisites to do this. They are as follows:
1. The Periodic Table: You need to learn the stable oxidation states and configurations of ions. For instance, pretty much all sulphides, upon oxidation, become sulphates, because +6 is one of the most stable states of sulphur. Or, Ferrous and Ferric ions are pretty much equally stable, so much so that I like to call them toggler ions, as the switch between each other with ease.
2. Electrochemical Series: This is extremely important. It was simply amazing how much i could put into context after learning about this series. I could now understand why Iron metal produces hydrogen gas upon reacting with an acid. It will help you in countless redox reactions.
3. Solubility of compounds (in water) This will help you to predict precipitation reactions. In lower classes, we are taught these as lousy double displacement reactions, but they are actually pretty inaccurate. A ppt reaction will occur if the initial reactants are soluble, and at least one of the products is insoluble. Le-Chatelier's principle, if you will. Also, there is usually no way to predict solubility, as it is a complex interplay of hydration and lattice enthalpies, so you HAVE to memorize.
Now, a lot of reactions have no apparent logic, so there is no way to predict them. You will have to memorize these.
For instance, in the Rasching's process to prepare hydrazine, Ammonia reacts with the hypochlorite ion to produce chloramine. This cannot be predicted with the above steps. It has to be memorized.
All I can say is, inorganic chemistry needs 55% memorization, and 45% common sense, which includes some intrinsic feeling as to what should be the outcome of a reaction, which will develop after a lot of practice.
Good Luck!

Sir, thanks for the nice answer and experience sharing. It would be very great if you could add two things : 1) List of stable oxidation state of elements ( I can find on internet the oxidation states shown by elements but cannot get the stable oxidation states of elements( especially p- & d-block elements). 2:A good example to illustrate the points. Overall, the answer sound very nice.

I had this problem too, and what helped me was remembering some of the preparation charachteristics for each group in periodic table, because the preparation of elements in a group is pretty much similiar.
For example in 5th group, most elements are prepared from their most important alotropic modifications, more exact from their reduction with carbon. Example quartz $\ce{SiO2 + C ~ = ~Si + CO2}$
So does the $\ce{As}$ oxide, the $\ce{Bi}$ and the $\ce{Sb}$ oxide. Also $\ce{Pb}$ (4th) is prepared by its most important modification which is galenit $\ce{PbS}$.Concentrate in those similiarities in each group and you will have no trouble remembering them. If you don't find those similiarities i can help you.

I would suggest you to try and understand the equations and then learn them by heart.
You can try and write them down when you are done reading them.Once written,you can actually go through it once in 2 days.This will further help you memorise the equations properly!
Best of Luck!

Memorising is a problem however tough we may try. In real life, one cannot go over inorganic reactions every two days. Besides, there are 100+ reactions and hectic trends (with exceptions too)! Your method might work for exam but what about after that?


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Absolute limits on solubility of immiscible solutes?
I know that for a particular, immiscible solute, there is a limit on how much of it can be dissolved in a solvent at a given temperature. However, I was wondering if there is only so much "total solute" that a given solvent can dissolve. As a simple example, suppose you saturate water with salt, then start dissolving sugar into it. Does the fact that it is saturated with salt limit how much sugar it can dissolve? My mental image is one of each water molecule being able to suport only so many solvation structures, regardless of the chemicals participating in the structure.

If you speak about solubility, you are out of domain of simple approximations and the only correct answer is the experiment. For the example given, see http://dx.doi.org/10.1111/j.1365-2621.1992.tb01192.x
The problem is the difficulty to quantify the effect of temperature, which is very important here.

Wow, thanks for the uber-specific, relevant link :) So I guess the answer is YES...the solutes in multi-solute solutions do not behave linearly or independently with reagrd to their phase partitioning. Thanks!!


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Why does MgCl2 lower the freezing temperature, and how does it do so?
I am working on a project in my chemistry class, and even though we've already finished the lab, I'm still confused as to why and how $\ce{MgCl2}$ melts ice/ lowers the freezing temperature of water. Can anybody tell me why this happens?

The $\ce{Mg^{2+}_{(aq)}}$ and $\ce{2Cl^{-}_{(aq)}}$ ions disrupt hydrogen bonding, preventing ice from forming a lattice structure. Therefor, the water must be colder so those weak hydrogen bonds will be acting on water molecules with less kinetic energy (i.e., colder), finally allowing them to create a crystal around the dissolved "impurities" $\ce{Mg^{2+}_{(aq)}}$ and $\ce{2Cl^{-}_{(aq)}}$.
This is a basic colligative property. Its more effective for electrolytes because they dissociate, creating more lattice disruptions. As a general resource, see: Ebbing, Darrell D., General Chemistry, 3rd ed, Houghton Mifflin, 1990.


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Is it customary to express heat loss as a negative number?
In my chemistry class, we had a problem akin to the following:

If an object with mass bla and specific heat bla is placed into a bla g calorimeter with an initial temperature of bla and a specific heat of bla, and the temperature of the calorimeter rises to bla at the end of the experiment, how much heat was lost by the object?

I performed the calculations and got an answer of, let's say, $8 J$. When I submitted this answer, I was told that it was wrong, and the correct answer would have been $-8 J$. Wouldn't this response to the above question mean that the object gained heat?

Sign conventions have always been a bit of an issue in chemistry, especially in introductory courses. I agree with your interpretation in this case, but what might have been meant by the question was to state the heat change of the object, in which case -8J follows standard convention.

It is better to say it as a convention, than saying it as a customary. It is convention that, one hand we consider as right, and the other as left. One may assume the vice versa case. It would not be wrong.

I guess what I'm asking is whether I'm right or wrong.

Note that physicists and chemists use different sign conventions for work, and I've seen some variation in the convention for heat, too.

In cases such as these, I think it can be useful to explicitly write out that "$8 J$ of heat is released", or "considering the convention that negative energies correspond to heat release, the energy involved in the process is $-8 J$" to cut through any ambiguity.

Your assumptions are correct, and from the wording of the question, the "correct" answer doesn't make sense, unless like you said, the object gained heat, which is possible; people have negative profits all the time. Sign conventions are not flawed in chemistry, but you just have to be careful when wording a question. To answer the title of your question, yes it is not only customary, it is necessary to express heat loss as a negative number, but the question still has to be phrased correctly--using the term "lost" serves as a negation in this use of language, the same way the unary operator in -1 serves as negation for a number. If heat loss were not expressed negatively, it wouldn't make sense because the system of an exothermic reaction must lose energy for its surroundings to gain it.


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How does one tell if a specific molecule is acidic or basic?
Let's take the $\ce{KOH}$ molecule into account. I know it is a base from literature, but how would one go about determining if a molecule is acidic or basic simply based on the structure of the molecule? How about amphoteric?
Also, I understand that there are two complementary systems - Lewis and Brønsted-Lowry theories. How do they work and how do they fit together?

I think this is a very good question. Many Bronsted acid and bases are easy to recognize because they have an obvious place to donate or accept a proton (e.g. $\ce{HCl}$ and $\ce{KOH}$ are easy) and some amphiprotic compounds are also easy (e.g. $\ce{H_2O}$ and $\ce{HCO_3^{-}}$). My 'feeling' is, but please let someone give you a more thorough answer, is that Bronsted-lowry acids and bases are easily recognized by proton donating/accepting sites, and that lewis acids/bases are more difficult to 'see coming'

What is unclear, with respect to Bronsted and Acid base and related articles?

Wouldn't it be as simple as saying that entities with low energy LUMOs tend to be acidic, while those with high energy HOMOs tend to be basic?

@NicolauSakerNeto it is not as simple as that since electron transfer reactions can be explained with HOMO/LUMO gaps as well.  The relative energies of HOMOs and LUMOs of products and reactants leads to a qualitative (semi-quantitative?) explanation for acid/base behavior.

@bobthechemist Well yes, low/high LUMO/HOMO species can act in ways other than acids/bases, but I believe it's true that there is no such thing as a base without a high HOMO and an acid without a low LUMO (this is all in relative terms, of course)? Therefore, all bases would have (relatively) high HOMOs, but not all entities with high HOMOs act as bases (and the equivalent for acids).

@NicolauSakerNeto I agree, so long as we are including relative in the discussion.

It is sometimes challenging to determine if a molecule is going to be acidic or basic if the system in which it is reacting is not considered.  An important point to consider when dealing with acids and bases is that acid/base strength is inherently tied to the solvent.  For this answer, I'm going to limit the discussion to acids and bases in an aqueous environment.
It is helpful to consider the terms acid and base as a means to classify substances.  This way, Chemists can explain chemical reactivity and structure-function relationships of substances.  Very early classification systems depended on our senses (acids are sour, bases are slippery to the touch) and more recent classification systems utilize structural characterization tools such as NMR or crystallography.  Many classification systems have been proposed over the years, and only a few of them have found sufficiently widespread use to end up in textbooks used in the standard Chemistry curriculum.  Below are a few systems, taken from Miessler & Tarr's Inorganic Chemistry textbook.  The 2nd through 4th entries are those in common use today.

NAME (YEAR) acid definition [example]; base definition [example]
Liebig (~1776) Acid: an oxide of N, P, S [$\ce{SO_3}$]; Base: Reacts with
acid [$\ce{NaOH}$]
Arrhenius (1894) Acid: Forms hydronium ion [$\ce{HNO_3}$]; Base: Forms hydroxide ion [$\ce{NaOH}$]
Brønsted (1923) Acid: Proton donor [$\ce{HCl}$]; Base: Proton acceptor [$\ce{NaOH}$]
Lewis (1923) Acid: Electron-pair acceptor [$\ce{Ag^+}$]; Base: Electron-pair donor [$\ce{NH_3}$]
Ingold-Robinson (1932) Acid: Electrophile [$\ce{BF_3}$]; Base: Nucleophile [$\ce{NH_3}$]
Lux-Flood (1939) Acid: Oxide ion acceptor [$\ce{SiO_2}$]; Base: Oxide ion donor [$\ce{CaO}$]
Usanovich (1939) Acid: Electron acceptor [$\ce{Cl_2}$]; Base: Electron donor [$\ce{Na}$]
Solvent system (1950s) Acid: Solvent cation [$\ce{BrF_2^+}$]; Base: Solvent anion [$\ce{BrF_4^-}$]
Frontier Orbitals (1960s) Acid: LUMO of acceptor [$\ce{BrF_3}$]; Base: HOMO of donor [$\ce{NH_3}$]

I find the various acid/base systems very enlightening.  Note how From Arrhenius through Lewis there was a broadening of the acid/base classification system; Arrhenius can't be used to describe non-aqueous acids and bases and Brønsted can't be used with aprotic substances.  Yet, the Lewis definition incorporates the previous two (a Brønsted acid is also a Lewis acid; an Arrhenius base is also a Lewis base).  One then may ask, what's up with the Lux-Flood definition then?  This definition is counter to the trend of broadening the classification system, and yet it is useful in describing anhydrous solid-state chemistry and is used to describe geochemical reactions as well as the chemistry of high-temperature melts.  My point being: classification of substances into acids and bases is only meaningful if it helps explain chemical phenomena.
Which brings us to the Usanovich definition, which essentially states that every reaction is an acid-base reaction.  Such a broad definition is not overly helpful, and to those of us with an affinity towards electrochemistry (ahem), is somewhat arrogant :-)
I do believe that determining if a substance will behave as an acid or a base requires a bit of chemical intuition (or a Socratic method).  For example, you may have performed an experiment in which $\ce{KOH}$ served as a base; you know that, like potassium, sodium is an alkali metal; therefore you presume that $\ce{NaOH}$ would be a base as well.
Back to the question at hand
So how do I suggest one use this information to predict whether a substance will behave as an acid or a base using its structure alone?  Personally, I find the Lewis theory as the most useful classification system in answering this type of question.  If the structure of a compound is set before me and I am to predict its acid/base chemistry, I will ask two questions:

Are there any lone pair electrons that can be donated?
Are there any electron-deficient atoms that could serve as electron pair acceptors?

If the answer to question 1 is yes, then the molecule can behave as a base.  If the answer to question 2 is yes, then the molecule is an acid.  If both are true, then I have an amphoteric substance.
In proofreading this answer, I realize I said I would restrict myself to aqueous systems, in which case using the Brønsted system may be more helpful.  In this case, the questions become:

Is there a hydrogen that can be donated? (You'll be right more often than you are wrong if you rephrase this question as "Is there a hydrogen that is attached to something other than carbon?").
Is there a lone pair that can accept a proton?

Your date for Liebig's definition has to be wrong. Justus von Liebif was born in 1803.

@BenNorris Thanks for pointing this out.  The dates are from Miessler and Tarr's Inorganic Chem text, and two entries are provided under the Liebig definition - the second entry has a date of 1838.  The "1776" definition of an acid is the Oxide of N, P, S and the 1838 is "H replaceable by a metal".  I will have to investigate why both are attributed to Liebig.

Shouldn't one consider first how the compound would react towards dissolution in water? For example sodium hydroxide is a salt and will produce sodium and hydroxide ions. $$\ce{NaOH <=> Na+ +\ ^{-}!OH}$$ Hence only the hydroxide ion should be considered a base. (+1 for the extensive collection of definitions)

@Martin If we limit our conversation to aqueous solutions (which is not unreasonable) then yes, we must consider the compounds interaction with water, especially since it is a leveling solvent.  In water, hydroxide is the strongest base, but not the only base.

I think it is not necessary to limit it to aqueous solutions. (It is mandatory up to and including Brønsted.) My statement holds for any solutions and probably even for the gas phase. If you are not dissolving sodium hydroxide, i.e. in toluene, then it will be only lying there, doing nothing.  If you are considering Lewis definition (and after) for NaOH, then you already have an acid-base-pair.

@Martin perhaps I misunderstand your point.  I do mention in my answer that acid/base strength is inherently tied to the solvent so I agree that solvent interactions with the acid/base must be considered, but I'm missing where the confusion is.  As an aside, Bronsted acid/base theory can be used for some nonaqueous solvents, see here for example.

How does one tell if a specific molecule is acidic or basic?

An acid (from the Latin acidus/acēre meaning sour) is a chemical substance whose aqueous solutions are characterized by a sour taste, the ability to turn blue litmus red, and the ability to react with bases and certain metals (like calcium) to form salts.
In chemistry, a base is a substance that, in aqueous solution, is slippery to the touch, tastes bitter, changes the colour of indicators (e.g., turns red litmus paper blue), reacts with acids to form salts, and promotes certain chemical reactions (base catalysis).
We do have $3$ different definitions from Arrhenius, Brønsted-Lowry, and Lewis. We would be in trouble, in some cases, where one definition accepts a molecule as acid or base and other not. So, we do have a common definition for acids and bases as mentioned above.
Coming to amphoteric substance, consider $\ce{H2O}$ which has no taste.
LINKS

http://en.wikipedia.org/wiki/Acid (definition of acid is extracted from this link)
http://en.wikipedia.org/wiki/Base_(chemistry) (definition of base is extracted from this link)

In organic chemistry, whether a compound is acidic or basic can be predicted by looking at the molecular structure and can quite easily be explained by applying the resonance theory.  In essence, if the compound can ionize to form a proton and the resulting anion is stable, then the ionization equilibria will be shifted to the right and the compound is acidic.  The more stable the anion, the more acidic the compound is.
Consider the following examples:

Sulfuric acid   $\ce{H2SO4 <=> 2H+ + SO4^2-}$   the sulfate anion is stabilized by 4 resonant structures, therefore sulfuric acid is very acidic.
Nitric acid  $\ce{HNO3 <=> H+ + NO3-}$   the nitrate anion is stabilized by 3 resonant structures, therefore nitric acid is acidic.
Carboxylic acids   $\ce{RCOOH <=> H+ + RCOO-}$   the carboxylate anion is stabilized by 2 resonant structures, therefore carboxylic acids are acidic.
Phenols versus alcohols.  $\ce{C6H5OH <=> H+ + C6H5O-}$   the phenoxide anion has 4 resonant structures while the alkoxide anion has none.  Therefore, phenols are more likely to ionize to produce a proton than alcohols.  That is, phenols are more acidic than alcohols.
Ketones versus alcohols.    $\ce{R1COCHR2R3 <=> H+ + R1COC-R2R3}$   the resulting anion has 2 resonant structures whereas alcohols have none.  Therefore the α-hydrogen to the carbonyl group is acidic because the enoxide ion is more stable than the alkoxide ion (keto-enol tautomerism).
Amides   $\ce{RCONH2 <=> H+  + RCONH-}$  the resulting anion has 2 resonant structures, but they are not equivalent because one has the negative charge on the oxygen, the other one has it on the nitrogen.  Therefore the anion is not particularly stable so amides are only weakly acidic.  The lone pair of electrons on the nitrogen is only slightly available for bonding because they are withdrawn by the more electronegative oxygen.  Therefore amides are only weakly basic. They are practically neutral.
Sulfonamides  $\ce{RS(O)2NH2 <=> H+  + RS(O)2NH-}$   the resulting anion has 3 resonant structures, 2 of which have the negative charge on the more electronegative oxygen atom.  Therefore sulfonamides are a little more acidic than amides and can form salt with metals (e.g. silver sulfadiazine, sodium sulfacetamide).
Urea  $\ce{H2NCONH2}$ the lone pair of electrons on one nitrogen are withdrawn by the more electronegative oxygen (see amide above), but the lone pair on the other nitrogen is available for bonding.  Therefore urea is basic.
Imides  $\ce{R1CONHCOR2 <=> H+ +  R1CON-COR2}$   the resulting anion has 3 resonant structures, 2 of which have the negative charge on the more electronegative oxygen, therefore imides are acidic (for example phthalimide).

-1 for suggesting that carbonyl compounds (pKa ~20) are more acidic than alcohols (pKa ~15); it's simply not true. Also it's misleading to suggest that the two amine groups in urea are any different. Lastly, resonance is great, but hardly the only deciding factor. HI is very acidic, yet it is difficult to argue that the iodide ion is resonance-stabilised in any manner.

I was not saying all ketones are more acidic than alcohols.  I am just saying that the alpha hydrogen to the carbonyl gp is acidic because of keto-enol tautomerism.  I was not saying that the 2 amino groups in urea are different,  I was just saying that because there are still electrons available even oxygen has suppressed them somewhat.

There's a large gap in the ability to handle Lewis acids with this description.

The resonance theory serves well in explaining the acid-base nature of pharmaceuticals.  Lewis acids and bases are too toxic to be used in medicine,  Therefore it is rarely used to explain acid-base properties of drugs.

I haven't come across any FDA approved drugs that cannot be explained by the Bronsted-LowryTheory.  If you know of any, please let me know.

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How to interpret a MALDI mass spectrum?
I have some questions regarding how to read a MALDI mass spectrum in a paper. I am a biologist by training, so I would like some analytical chemists to help me out.
The researchers ran a western blot, and immunoblotted for a protein of interest. They cut out this band from the western blot, subjected the protein to trypsin digest, and used these peptide fragments for MALDI-TOF analysis. I know for a fact there will be other proteins at the same apparent molecular mass which would have also been included in the analysis, which should be taken as contaminants in the resulting spectrum.
They produced a graph with the absorbance intensity x 1,000 on the dependent axis, and mass:charge (m/z) ratio on the independent axis.
The authors point to two peaks, which they say correspond to two peptide fragments from the tryptic digest of their protein of interest. They provide the supposed peptide fragment sequences and m/z ratio data for these two peaks, and I checked to make sure they are the same as reported. The authors expect to see two peaks: protein of interest is digested to two fragments, A + B.
I have two questions:

Peak 'A' and peak 'B' are of different intensities. Peak 'A' is 20-fold greater than peak 'B'. However, the tryptic digest must yield equimolar amounts of peptide fragments A and B. Is it possible that these intensity peaks can be 20-fold different? Wouldn't the absorbance intensity correlate to total analyzed peptide mass?
Is it normal to multiple absorbance intensity by a (seemingly arbitrary) factor, in this case, 1000? Since the absorbance intensity of peak 'B' is < 0.5 (after scaling up), could this represent noise rather than signal in the analysis?

Could you provide a link or a reference to the article in question? I am a bit puzzled: usually identification of a protein based on only two peptides should be considered very carefully in a peptide mass-fingerprint (PMF) analysis, and is probably not valid if many other proteins are present (as peptides with similar masses could originate from other proteins). Did the authors perform any MS/MS experiment to confirm the identification step?

Please see the wiki articles for the general method:

http://en.wikipedia.org/wiki/Mass_spectrum
http://en.wikipedia.org/wiki/Matrix-assisted_laser_desorption/ionization
http://en.wikipedia.org/wiki/Time-of-flight_mass_spectrometry

Then let's start with question 2: yes, you can and you should. The spectrometer measures in arbitrary units, therefore you can speak only about relative abundances.
And question 1: it depends how easily is the given molecule ionized. So if the fragment A is rich in polar aminoacids, whereas fragment B is mostly hydrophobic and difficult to ionize, than you should expect this big difference in signal intensities.

Thank you. So if we assume that A is twice as massive as B, and are equally ionizable, then there should be a discrepancy in their relative signal intensities. Shouldn't there be some form of internal control, or is that only if one wishes to quantify absolute peptide masses from the spectrum?

If the fragments are equally ionziable, then the intensity should be similar. Of course, the m/z will be twice as large for the bigger one. Maybe I do not fully understand your problem. It's m/z which is the important characteristic, whereas intensity tells you little bit about ionizability, ion stability and quite indirectly about abundance.

Actually you explained the answers to my questions well. I do see the m/z are different by a factor of two. I would assume that ionizability and ion stability would be the same for small peptides, and so I should expect similar band intensities. I think that is enough to be critical of the results I have found in paper.

@leonardo: There are no real reasons for the ionization efficiencies to be similar for all small peptides. Physico-chemical properties of small peptides can cover a wide range, and you cannot assume that "small peptides" is enough to allow for similar ionization efficiencies.

In addition, I would like to point out that the intensities in the spectrum are not only related to ionization efficiencies: the entire workflow has to be analyzed, from the initial peptide extraction from the digested band to the mass analysis. For instance, peptide extraction from a gel can lead to strongly different peptide abundances due to pore sizes in the gel if the peptides have very different sizes. (BTW, did they perform the digest directly on the membrane blotted protein or in gel?) A TOF analyzer can also introduce strong variations in intensity when the m/z values are far apart...

I know this is an extremely late response, but perhaps this will help another person who is looking at this.
for your first question, the degree of ionization in Maldi depends mostly on the acid-base equilibrium between the matrix substance and the analyte, that is, it's Ka value. And as the above answer has already mentioned, a good way to predict this would be to see how many polar amino acids residues there are. So it is entirely possible that even if the theoretical ratio of two peptides are 1:1, you might obtain a different ratio in the spectrum.
Someone in the comments section has asked whether the authors of the paper in question has done any MS/MS experiments. I happen to have the same question. So far, I have heard of anyone doing peptide sequencing using just MS:it is either done by Edman degradation or by MS/MS experiments (through de novo peptide sequencing). Since you mentioned that they did do sequencing, I'm starting to wonder whether you actually meant a MALDI-TOF/TOF. In that case, fragmentation of individual peptides chosen for the second TOF experiment would depend on the collision induced dissociation parameters. Here is a website detailing the fragment types: https://en.m.wikipedia.org/wiki/De_novo_peptide_sequencing


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What is the number of unpaired electrons in cobalt(II) tetrathiocyanate?

Find the number of unpaired electrons present in the d orbital (whose lobes are present along the axis) for the complex $\ce{[Co(SCN)4]^{2-}}$.

Since $\ce{{SCN}^{-}}$ is a weak ligand I did not pair the electrons, but still I got the wrong answer. The correct answer is 0. But how?

"whose lobes are present along the axis" seems very ambiguous to me. Could you clarify what it means?

@ssavec i applied crystal field theory ...... i filled the d orbitals according to hunds rule..... since Cobalt is in +2 oxidation state hence $Co=[Ar]4S^2 3d^7$ hence $Co^{2+}=[Ar]3d^7$ Now i filled up the orbitals.

@chipbuster i think this refers to the $e_g$ set of orbitals because $e_g$ set contains $d_{x^2-y^2}$ and $d_{z^2}$ whose lobes are directed along the axis as the lobes of $d_{xy}$ $d_{yz}$ and $d_{xz}$ are present between the axis.

@Hardik which geometry you assumed, having $\ce{M L_4}$ complex?

@ssavec i got my mistake...... i filled up the orbitals according to octahedral splitting...... instead i had to do it with tetrahedral. Thankyou.

According to the tetrahedral splitting in crystal field theory, the filling of the orbitals is just opposite to the filling in case of octahedral splitting. Thus filling up the elecrons according to it and obeying the hunds rule (since it is a weak ligand) the number of unpaired electrons are 0 in $d_{x^2−y^2}$ and $d_{z^2}$ whose lobes are directed along the axis.

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What does HTO decay into?
I'm aware that tritium undergoes beta decay forming Helium-3. But in the case of HTO, what does it decay into? I don't suppose it decays into HeHO.

Very shortly after a radioactive element decays in a molecule, you could consider that for a brief moment, it does indeed form a molecule with the new element in the same place as its originating radioisotope. So yes, after the decay of $HTO$ you could get $HeOH^+$ via the unusual reaction $TOH → HeOH^+ + e^- + \bar\nu_e$. Of course, though, this is a very unstable compound and it would immediately participate in further chemical reactions, very likely releasing the helium atom at some point. Even if the molecule were relatively stable after having one of its atoms transmuted, the molecule might still decompose as the nuclear decay releases a lot of energy, which can tear the comparatively fragile molecule to pieces.
A curious tidbit is that, in fact, the decay of tritiated compounds has been considered as a means to produce legitimate helium compounds, given that directly reacting helium with other substances has been fruitless other than for very trivial species, such as $HeH^+$. The problem is that tritium has a rather long half life of about 11 years, so it would be quite hard to produce reasonable quantities of helium compounds in a reasonable timespan, along with the likelihood of the helium compound being very poorly bound and the emission of vast quantities of nuclear decay energy which could easily decompose the fleeting substance.

Thanks for your reply. What would HeOH+ be likely to decay into?

That's rather hard to say for sure. An easily seen pathway is for the helium atom to leave as a neutral atom, releasing $OH^+$, which itself would very quickly react with water, possibly giving the $H_3O_2^+$ ion, which would lose a proton to form $H_2O_2$, hydrogen peroxide. There are likely several types of reactions that take place simultaneously, since the initial $HeOH^+$ molecule has such a high amount of energy that even strange reaction pathways can have their activation energy met and will have more stable products than the reactant.


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Is it possible to light a match by scratching it on hard surfaces, like they do in movies?
In movies cool guys don't need a matchbox − they just light a match by scratching it on whatever hard surface − concrete walls, wooden bar tables, leather belts, etc. Is it just a movie trick, or is there actually a type of matches you can do that with?
So far I've tried scratching usual matches on a ton of different surfaces and the heads just smear without even a smoke.

I figured this has to do with friction, but if there's a better site for this question, please comment.

Do you mean the regular matches you need to strike on the side of a matchbox's brownish paint thing? Are they "safety matches"?

@user1306322 See http://en.wikipedia.org/wiki/Match for information. In particular, note the sections on strike-anywhere matches and frictional matches.

Have you done a control experiment by seeing how easy it is to strike a match as intended? In my experience, only a small fraction of the population can actually use matches anymore.

Yes, those matches existed, I have had them myself. Lightening them by rubbing on the sole of your shoe was popular then. But later on they have been forbidden, they were thought to be too dangerous.

Wow this question suddenly makes a certain person feel old...

Most matches these days are safety matches: they're designed to need something more than ordinary levels of friction to ignite, by splitting the combustion materials between the match-head and the striking strip, i.e.  the brownish paint thing down the side of the matchbox or across the front of the matchbook.
Most "regular" matches now are safety matches. The brownish strip has red phosphorous in it. The match head has potassium chlorate $\ce{KClO3}$. Combine those two with friction, and you get fire.
There are still striking matches, with heads of phosphorus sesquisulfide $\ce{P4S3}$, still around. Though largely in countries where consumer protection is not yet well-developed.
With sufficient energy, you can still ignite a safety match without the striking strip. But you'll need more energy than the friction of scratching it on a wall or bit of stubble.
(PS Don't play with fire. Get a friendly neighbourhood chemist to play with fire on your behalf.)

These are "strike anywhere" matches http://www.industrialrev.com/strike-anywhere-matches.html    They need to be stored securely in a fireproof container, and you should never shake or drop the box, or store them near flammable goods. They have special haz-mat shipping requirements too. Oh, and matches of both kinds can be carcinogenic.

Yes.
There's a knack to it but I have gotten matches to light off of bricks, floor tiles, rocks, painted poles.
Strike with force long and hard not gently like with the matchbook


text:
Hot and cool milk
Usually I have the habit of keeping the hot milk for a while, until it becomes cool. I have noticed many a times that, drinking it, after keeping for a while, gives the mixture of hot (relatively less hot than before) and cool milk.

Almost $3$ by $4$ (approx) part of the milk at the top, is relatively found to be hot than the last $1$ by $4$ (approx) part of the milk.
What is the reason for upper part to be hot and lower to be cooler?

By convection heat transfer. Hot milk goes up from the bottom of the cup, while the milk at top download, in contact with air, gets cooler and descents to the bottom of the cup.


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Reference to find dielectric constants and dielectric relaxation time
I am working on a Master degree and need to know the following data :

The static (low frequency) dielectric constant of isopropanol at 25 °C
The Optical (high frequency) dielectric constant of isopropanol at 25 °C
Dielectric relaxation time (τ) of isopropanol at 25 °C

Could someone please provide me with these values and a reference or link to a book contains these data?


text:
During electrolysis, why are the products attracted to the cathode?
Take the electrolysis of Lead(II) bromide:
We can write it as two half-reactions:
$$\ce{Pb^{2+}(l) + 2e^{-} \rightarrow Pb(l)}$$
$$\ce{Br^{-} \rightarrow Br + e^{-}}$$
In the electrolysis reaction, lead is formed at the cathode and bromine is liberated at the anode. But why is it so? I mean, before gaining or losing electrons to become neutral, lead is positive and bromide is negative. So, naturally they should be attracted to the cathode and the anode respectively.
But why do they become neutral after they reach the electrodes. Why do they gain electrons at only the electrodes to become neutral?

You have to think about what exactly a cathode and anode are. You are falling victim to this:
A widespread misconception is that anode polarity is always positive (+). This is often incorrectly inferred from the correct fact that in all electrochemical devices, negatively charged anions move towards the anode and positively charged cations move away from it. In fact anode polarity depends on the device type, and sometimes even in which mode it operates, as per the above electric current direction-based universal definition. Consequently, as can be seen from the following examples, the anode is positive in a device that consumes power, and the anode is negative in a device that provides power. - Wikipedia
It's your conditioning from other chemical nomenclature--"cation" and "anion"--which are absolute about their charges.
So beyond misleading nomenclature, what does cause this to happen?
Take this picture for example: http://img.sparknotes.com/figures/0/02480ae8fc1a41b131a3fdb5a698e9a3/compare.gif
If E°cell > 0, then the process is spontaneous (galvanic cell)
If E°cell < 0, then the process is nonspontaneous (electrolytic cell)
where E°cell = E°(V) reduction + E°(V) oxidation
In your case,
Pb → Pb2+ + 2e- = 0.13
2Br- → Br2 + 2e- = -1.06
E°cell = .13 - 1.06
E°cell = -.93
In your case, the process is nonspontaneous (makes sense, think about trying to reduce ionized bromine), so you're looking at an electrolytic cell, which is battery driven (it needs work put in to operate.)
So, again, to directly answer your question:

In the electrolysis reaction, lead is formed at the cathode and bromine is liberated at the anode. But why is it so? I mean, before gaining or losing electrons to become neutral, lead is positive and bromide is negative. So, naturally they should be attracted to the cathode and the anode respectively.

Lead is formed at the anode because work from a battery was put in that pushed the reaction in that direction. Before work was put in, the reaction would have proceeded in the opposite direction (think exothermic vs endothermic, but in this case spontaneous cell potential vs nonspontaneous cell potential.)

But why do they become neutral after they reach the electrodes.

They become neutral at the electrodes because that is where the reaction happens, and that is where the electrons are transferred.

Why do they gain electrons at only the electrodes to become neutral?

Again, where else could they gain electrons? They are sitting in a solution contain themselves and their respective anions, which they have already reacted with, and will no longer be transferring electrons with.


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Why do they use graphite electrodes in the Hall–Héroult process?
In the Hall–Héroult process of extracting aluminium from aluminium oxide, the electrolyte is a liquid mixture of cryolite $(\ce{NaAlF6})$ and aluminium oxide $(\ce{Al2O3}).$ The cathodes and the anodes are made of graphite.
During the electrolysis process, aluminium is deposited at the cathode and oxygen is liberated at the anode. Some of this oxygen reacts with the carbon in the graphite to form carbon-dioxide, thus slowly burning away the anodes. Thus, the anodes have to replaced periodically.
But if this is the case, why do they use graphite anodes at all? Why don't they use anodes of some other, less reactive material? I'm pretty sure there are many out there that would fit the bill.

According to this source, graphite is used as anode material because carbon is easier to oxidize than $\ce{O^2-}$. The anodic oxidation of carbon
$$\ce{C + 2O^2- \rightarrow CO2 + 4e-}$$
proceeds more efficiently than the oxidation of $\ce{O^2-}$ at an inert anode
$$\ce{2O^2- \rightarrow O2 + 4e-}$$
$\ce{C}$ is a much better reducing agent than $\ce{O^2-}$ whose oxidation state is not changed in the reaction with graphite. As a consequence, anodic current density and the deposition rate of aluminium at the cathode will increase.

And also given that under most conditions carbon dioxide is a gas it leaves no trace in the obtained material ($\ce{CO2 ^}$).

The basic reason for using graphite as an anode is in Hall–Héroult process in the electrolytic reduction of alumina to aluminum metal is because graphite being an allotrope of carbon and an inert electrode reacts with oxygen to give out carbon dioxide which thus prevents the liberation of oxygen as a final product at the anode.
Had there been any other metal oxygen would have been liberated which would have readily reacted with aluminum (as aluminum has a high affinity for oxygen in comparison to other metals, for example, iron—even if iron reacted with oxygen the aluminum will easily displace it) to  form aluminum oxide and the whole process will prove to be useless.
Graphite is also a good conductor of electricity.

It is simple. Graphite anode is used because when carbon react with oxygen liberated at anode it forms $\ce{CO2}$ gas which is a exothermic reaction and maintain the temperature of process and not allow molten aluminum to solidify. Because the density of aluminum in solid form is less dense than cryolite and hence it can't be able to extract from the bottom.

No , carbon dosent reduce alluminium in hall heroult process , it is simply electrolytic reduction and no it is not used to prevent contact of oxygen with alluminium as alluminium is already denser than the solution and sinks to bottom as in this diagram
Actually hall heroult process is done by taking molten mixture of cryolite and fluorspar(95% of the mixture) and about 5% alumina , this is done to reduce melting point of alumina from about 2200 K to around 1000 K , at which they electrolyzed it , the carbon anode is used as fuel to provide energy to keep contents in molten state , you may have heard the fact that about 0.5 kg of carbon is burnt for every 1kg of alluminium produced as it acts as fuel .

Carbon is a better oxidizing agent than oxygen which produces more heat to retain steady temperature


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Oxygen and sulfur bonding
Why does oxygen form a double bond in $\ce{O2}$ but sulfur, also from group 16, forms single bonds in $\ce{S8}$?

This is essentially the same question as to why carbon concatenates through up to four bonds with binding energies low enough to be broken again yet stable enough to last given typical terrestrial environmental conditions. Great for entities like us.
On the other hand, the binding energies of Silicon, Germanium, from the same group IV, for instance, are higher. The binding energies are directly related to the bond length/bond distance, which is great for determining the energies through crystallography, along with proper model fitting.
Sulfur is one of few elements amenable to catenation, - forming long molecules with each other.
So whichever configuration releases the greatest enthalpy is typically the most stable one, but depends upon the initial energies thus giving rise to allotropes - elemental structures of different molecular configurations. The bound system typically has lower potential energy.
If a double bond lowers the overall energy of the bound system more it will typically form over a single bonded one.
Update: I found this (see also this table of dissociation energies):
Update 2: I'd like to add that from a quantum mechanical perspective, which is a viewpoint much closer to nature, the bonds arise from 3D space where electron densities are highest - which means most energetically favorable.
Since for whatever weird reason, nature is really quantized, and I cannot stress that "really" part enough, one gets standing waves for the electron equations around a given nucleus - rather than a continuum - as is the case of a sphere, like say a falling waterdrop to add a few exterior forces into the mix as well. The classical and the quantum object have only in common the "energetically favorable" state.
Somehwat simplified then, those standing waves become "orbitals" when doing the calculations for 3D space. The orbitals change in shape with increasing interference with other electrons and higher potential energies due to larger nuclei , meaning higher positive proton counts.
The energies involved result in very different packing i.e. different lattices. A necessity for life is breaking bonds or "bond accessibility". There simply isn't "enough energy" in today's protein-scaffold based catalyists on earth, to use anything else than the constituents of the periodic table with few protons in case of homogenous bonds (i.e. atoms of the same species like say Oxygen O2).
It does not have to be this way. We from earth are biased by our own biochemistry we observe. Every human generation breaks down the biases of the previous ones and encounters new ones.
Bond energies

1 S8: 1808 kJ/mol
1 O8: 1168 kJ/mol
4 S2: 1408 kJ/mol
4 O2: 1976 kJ/mol

The relative difference in energy between the double bond and single bond is much greater in oxygen than in sulfur, and so is the more stable $\ce{O2}$ diatomic molecule with a double bond, than molecules with multiple single bonds like $\ce{S8}$.


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Why is CuSO4 used as electrolyte while purifying copper?
During the electrolytic purification of copper, the electrolyte is taken as copper(II) sulphate ($\ce{CuSO_4}$). Is there any particular reason for this? Could another salt of copper, say $\ce{CuCl_2}$ work?
I mean, I don't see any reason why it shouldn't. The only difference will be, that at the anode chlorine instead of oxygen will be evolved, which doesn't seem to be such a big thing.
Why, then, is $\ce{CuSO_4}$ preferred?

Handling poisonous chlorine, the cost difference between the slats, relative conductivities, the voltage reqired for electrolysis (considering the reduction potentials and the overpotentials of all the species) are a few reasons to choose copper sulphate.

During copper electrorefining, impure $\ce{Cu}$ from the anode is oxidized and dissolved, while $\ce{Cu^2+}$ is reduced at the cathode, forming the deposit of refined copper. Because anodic impurities which are less noble than copper – most notably $\ce{Ni}$, $\ce{As}$ and $\ce{Sb}$ – also dissolve in the electrolyte, the cell voltage is kept in the range of 0.2-0.4 V to avoid co-deposition of these metals at the cathode.
Copper dissolution is the main oxidation half-reaction at the anode, and the oxidation of anions from the electrolyte needs to be minimized, since it is an unwanted side reaction which reduces current efficiency of the process. In a $\ce{CuSO4}$ electrolyte, this would be oxidation of $\ce{OH-}$ (from water dissociation, so concentration will be low) to $\ce{O2}$, while in a $\ce{CuCl2}$ solution, $\ce{Cl-}$ will be oxidized to chlorine gas, which would present a greater hazard because of its toxicity.
The electric conductivity of the electrolyte is another important factor, which varies with different copper salts and their concentrations. $\ce{CuCl2}$ and $\ce{CuSO4}$ are both readily soluble in water; however, the relative molar conductivities of these solutions will be different. In this study, a copper chloride electrolyte proved to be inferior to copper sulfate and copper nitrate in terms of deposit quality and stability of cell potential.


text:
Why do C-F bond has less dipole moment than C-Cl bond?

Larger the difference in electronegativities of bonded atoms, the larger the dipole moment.

(this statement has been extracted from chemwiki-dipole moment)
The electronegativity difference between carbon and fluorine is about $1.43$ (Pauling scale), $1.649$ (Allen scale). The electronegativity difference between carbon and chlorine is about $0.61$ (Pauling scale), $0.325$ (Allen scale). Although, the electronegativity difference of $\ce{C-F}$ bond is greater than electronegativity difference of $\ce{C-Cl}$ bond, dipole moment of $\ce{C-Cl}$ bond ($\pu{1.860 Debye}$) in $\ce{CH_3-Cl}$ is greater than dipole moment of $\ce{C-F}$ bond ($\pu{1.847 Debye}$) in $\ce{CH3-F}$, why?

The bond dipole moment ($\mu$) of a chemical bond is equal to the magnitude of the separation of charge ($\delta$) times the displacement of the charge (or the distance over which the charges are spearated) $d$, or:
$$\vec{\mu}=\delta \vec{d}$$
Thus, there are two factors controlling dipole moment. The charge separation can be estimated from the electronegativity difference, but not perfectly. Disparities in size and polarizability also effect the separation of charge between two atoms. The displacement is the length of the bond. The $\ce{C-Cl}$ bond length (176 pm) is longer than the $\ce{C-F}$ bond length (134 pm) - data from here.
Thus:
For $\ce{C-F}$, the electronegativity difference on the Pauling scale is 1.43, and the bond length is 134 pm: $1.43\times 134 = 191$ (ignoring units at the moment).
For $\ce{C-Cl}$, the electronegativity different on the Pauling scale is 0.61, and the bond length is 174 pm: $0.61\times 174=106$.
So, my quick and dirty method did not work so well. However the disparity between these two bonds by my method $(106/191)=0.55)$ is smaller than by just comparing electronegativities $(0.61/1.43=0.42)$
 and that is because of the other factor mentioned above: size and polarizability. Chlorine is bigger and more polarizable than fluorine (and carbon). The carbon-chlorine bond is polarized toward chlorine not just because of the electronegativity difference. Chlorine's larger size decreases the effectiveness of orbital overlap in bonding, and chlorine's increase polarizability implies that it can support more partial negative charge than suggested by its electronegativity alone.

I like the idea that chlorine may have a higher partial charge than fluorine given the former's larger size. Indeed, the electron affinities of the elements seems to provide evidence that fluorine simply cannot handle much charge. However, I do believe I've seen a graph in Huheey's book where a fluorine atom can stabilize a partial negative charge more than a chlorine atom all the way to around $-0.8e$, and only then do the relative stabilities switch. Do you think the halogen atoms in the methyl halides have that high of a partial charge?

When I was writing this post, I was looking for some reference, but did not find any. I did not look too hard however, and would love to see one. I was also planning to run some simulations on, for example, the methyl halides, but have not had time to do so yet.

Can larger size imply that it can hold more partial charges? Could you support that with a reference?

Not to waaaay necro this, but I notice the same thing in my textbook recently. It says a C-H bond has a greater dipole than a C-N bond (0.3 D vs 0.22 D). Similar thoughts?


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Mass of a gas sample
What method did John Dalton use to measure the mass of one mole of gases, for example, chlorine and hydrogen, to get their relative masses?

The concept of a mole(c. 1805), and Avogadro's number(1811) occurred after Dalton formulated his law (1801). So he would not have set out to measure a "mole of gas" as the modern formulation sets out. This history of Dalton may be useful to you: Google Book Search Most of his work centered around pressure readings, as his early interests were meterological.

thanks for you answer,the reason I am asking this question is that i want to know how the relative atomic mass of the elements were obtained during that time.
I know that at room temperature and pressure sample of gases of any element with same volume contain the same amount of atoms and we can use it to know the relative masses of the elements compared to hydrogen.
But how did he obtain the mass of the gas sample?

To set 'relative' weights, Dalton took the lightest element known, hydrogen, and set an atomic weight of 1. Then he started to weigh other compounds,e.g., water and determined the ratios of the other elements. From this data, Dalton set the relative weight of several elements.


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How can the Bonding in IF7 be explained using LCAO method?
I know that the VSEPR theory explains $\ce{IF_7}$ (iodine heptafluoride)'s structure as a pentagonal bi-pyramidal one. The valence bond theory can be used to say that it has $\mathrm{sp^3d^3}$ hybridisation (I think). How can the bonding in it be explained using the LCAO (Linear Combination of atomic orbitals) method in molecular orbital theory?
I got this doubt because there are only two "axial" d-orbitals ($\mathrm{d}_{z^2}$ and $\mathrm{d}_{x^2-y^2}$) among the 5 present, which can have head-on overlap with p-orbitals (non-axial orbitals can only have sideways overlap). In the excited state, the iodine atom has 7 half filled atomic orbitals (an s-orbital, three p-orbitals, and 3 d-orbitals) and according to LCAO each of them has to combine with one p-orbital of each fluorine atom to form molecular orbitals by σ and π bonds. But, the formation of only 6 σ bonds can happen between an iodine and 6 fluorine atoms, as σ bonds involve head-on overlap of orbitals, and there are only 6 orbitals capable of doing that (the s and p orbitals, and the two 'axial' d-orbitals). Therefore, the seventh fluorine atom must be in pπ-dπ bond with fluorine.
This doesn't explain "five equivalent bonds and two bonds of equal length (all σ bonds)" as predicted by VSEPR theory. How many σ bonds does it really have?

Have you tried some simple LCAO calculation of this compound? In mopac or similar fast and free software?

i don't know how to do such calculations.im in final year of high school

In the first paragraph, it should be pentagonal bipyramidal structure. I do not clearly understand, what does the excited state has to do with structure. And lastly, there are 5 d orbitals.

oh yeah,pentagonal bipyramidal.I meant that there are only 2 "Axial"d-orbitals(in LCAO sense,other d orbitals have lobes present on the xy,yz,and xz planes).

i rephrased my question a bit.read once again

I recently posted an answer to another SE Chem question involving $\ce{IF7}$, I think the answer will address your question as well.

No 6-centre-10-electron bond.
The bonding situation in $\ce{IF7}$ is explained by Christe et. al. (see ssavec's answer) as a 6-centre-10-electron bond. In the article itself I found no real evidence that supports this statement and I believe it to be completely wrong. The following section is taken from Karl O. Christe, E. C. Curtis, David A. Dixon, J. Am. Chem. Soc. 1993, 115 (4), 1520–1526. and might as well be the culprit of this monstrosity.

The bonding in the square-planar $\ce{XeF4}$, can be described by two semi-ionic, 3-center 4-electron (3c-4e) bonds for the four $\ce{Xe-F}$ bonds and two lone valence electron pairs on $\ce{Xe}$ ($\mathrm{s^2p}_z^2$ hybrids). The 3c-4e bonds involve the $\mathrm{p}_x^2$ and $\mathrm{p}_y^2$ orbitals of xenon. Addition of an $\ce{F-}$ ion to the equatorial plane in $\ce{XeF4}$, results in pentagonal-planar $\ce{XeF5-}$ and the formation of a semi-ionic, 6-center 10-electron (6c-10e) bond involving the delocalized $\mathrm{p}_x^2\mathrm{p}_y^2$ hybrid orbitals of $\ce{Xe}$ and six electrons on the five $\ce{F}$ ligands. The two lone valence electron pairs on $\ce{Xe}$ in $\ce{XeF5-}$ are analogous to those in $\ce{XeF}$.
  The planar $\ce{IF5}$ fragment of $\ce{IF7}$, has essentially the same bonding as $\ce{XeF5-}$, as shown by the atomic population calculations given in Table VII. [...]

They continue to compare the populations of the valence orbitals in question, without stating the what methods they used to determine that.

The electronic structure calculations were done at the ab initio molecular orbital (MO) and local density functional (LDF) levels. [...]

A more accurate description of the bonding in in $\ce{IF7}$
Iodine heptafluoride is one of the classic examples when it comes to describing hypervalency and octet-expansion. Early attempts included d-Orbitals into describing that phenomenon (VSEPR), resulting in a description that yields the correct geometry. We now know that this theory is obsolete and more often wrong than correct.
A very principle problem in describing the bonding in this molecule is assuming that it is completely covalent bound. Christe et. al. acknowledge that this is wrong by stating "semi-ionic" bonds. This is - as we will see - already a stretch.
A more accurate terminology for these kind of molecules is hyper-coordinated; i.e. the central atom has a higher coordination number than common.
The following analysis utilises a calculation on the local density approximation level of theory. The used basis set is of triple zeta quality and employs a small effective core potential for iodine. The density fitting approximation was used to somewhat speed up the calculations. This is often denoted as DF-BP86/def2-TZVPP. The program used is Gaussian 09 rev. D.
Geometry
The performance of the method is not overwhelming. The bond lengths are overestimated by quite a bit. As cited by Christe, observed is $r_\mathrm{obs}(\ce{I-F_{ax}})=178.1~\mathrm{pm}$ and $r_\mathrm{obs}(\ce{I-F_{eq}})=185.7~\mathrm{pm}$. I calculated $r_\mathrm{BP86}(\ce{I-F_{ax}})=190.4~\mathrm{pm}$ and $r_\mathrm{BP86}(\ce{I-F_{eq}})=195.6~\mathrm{pm}$. Other methods, that I could still reasonably perform did not do much better. The MP2 result is $r_\mathrm{MP2}(\ce{I-F_{ax}})=186.6~\mathrm{pm}$ and $r_\mathrm{MP2}(\ce{I-F_{eq}})=191.9~\mathrm{pm}$.
For the purpose of the analysis the DFT method should be enough. (I had a look at the other results, too; they don't differ significantly.)
Quantum theory of atoms in molecules (QTAIM)
I would usually start with looking at the orbitals, but in this case it is more appropriate to look at the electron density itself. Better yet, we look at the Laplacian of the electron density. The depicted QTAIM plots are the Laplacian distribution within the molecule. Solid blue lines indicate charge depletion $\nabla^2\rho>0$, dashed red lines indicate charge accumulation $\nabla^2\rho<0$, red spheres are bond critical points, solid black lines are bond paths, dark red lines are zero flux surfaces (these separate the atoms). See an earlier answer of me for some more information on QTAIM.
We have two bond critical points to consider, one for the axial and one for the equatorial bond. We find low values of electron density at both; $\rho(\ce{I-F_{eq}})=0.125~\mathrm{a.\!u.}$; $\rho(\ce{I-F_{ax}})=0.137~\mathrm{a.\!u.}$. The Laplacian at both points is positive; $\nabla^2\rho(\ce{I-F_{eq}})=0.33~\mathrm{a.\!u.}$; $\nabla^2\rho(\ce{I-F_{ax}})=0.42~\mathrm{a.\!u.}$.
These values indicate predominantly ionic bonding. The graphs nicely show that the atoms appear as separate entities.

With that background it seems wrong arguing the bonding situation by forming covalent bonds, like they are assumed in MO or VB theory.
However, as wrong as it is to neglect ionic bonding, it is equally wrong to neglect covalent bonds.
The Orbitals of $\ce{IF7}$
There are a total of four orbitals with contributions from iodine and fluorine. There are not more than eight bonding electrons. I cannot imagine that Christe et. al. found a different solution, as his level of theory was not better (maybe worse). They either did not know how to interpret these results or simply overlooked it.
Here are the relevant lines from the output.
 Alpha occ 17 OE=-1.196 is I1-s=0.23 F7-s=0.18 F8-s=0.18
 Alpha occ 25 OE=-0.587 is I1-p=0.30 F7-p=0.28 F8-p=0.28
 Alpha occ 26 OE=-0.568 is I1-p=0.31 F4-p=0.19 F3-p=0.19
 Alpha occ 27 OE=-0.568 is I1-p=0.31 F2-p=0.21 F6-p=0.14 F5-p=0.14

(OE is the orbital eigenvalue in Hartree, the contributions missing from 1.00 are at the other atoms, but to small to trigger printing.)

In the LCAO bonding picture we have a totally symmetric bonding orbital from all atoms, an orbital referring to the axial bonds, and two degenerate orbitals for the equatorial plane.
Charges and Wiberg Bond indices
The ionic character of the bonds can also be seen in the charge distribution. While the iodine has a high positive charge $q(\ce{I})=3.24$, the fluorines have a negative charge $q(\ce{F})=-0.46$. These are AIM charges calculated with the MultiWFN program package.
The Analysis with natural bond orbitals give a similar result, $q(\ce{I})=3.20$, $q(\ce{F})=-0.46$.
Looking at the Wiberg bond indices matrix formed from natural atomic orbitals, we find a bond index of $0.54$ for the equatorial bonds and a slightly higher value of $0.56$ for the axial bonds.
Conclusions
It is not possible to determine a number for how many sigma bonds are in the molecule, since this is strictly only possible for covalent bonds. The bonds are here predominantly ionic, with small covalent contributions. The number of bonding electrons does not exceed eight and therefore obeys the octet rule.
VSEPR fails, as it was expected. The term hyper-valency should be avoided and replaced by the term hyper-coordination.

Champion analysis, thanks for digging into it! What software did you  use to generate the Laplacian contour maps, and the orbital figures?

@Brian the analysis is done with AIMAll and the orbitals are plotted with ChemCraft.

Might have to bite the bullet and get a license for AIMAll. MultiWFN works pretty well, but its visualizations are ... substandard.

@Brian Since I have been working a bit with Multiwfn, I see what you mean. The biggest advantage of that program is that it is incredibly fast. I cannot confirm that for AIMAll. I'm not sure how important visualisation eventually is, but I believe that for all intents and purposes of QTAIM Multiwfn works well enough. (It is a million steps up from AIMPAC!) So I guess investing in AIMAll should only be an option is you have money to burn ;)

MultiWFN has a variety of export capabilities, too, including CUBE. If I ever needed to work up some of its output, I'd export and workup in, e.g., VMD.

@Brian or Chemcraft, or Molden, or ..., or ... the possibilities are endless. I just wanted to share my most recent experiences with you so that you don't burn money - and support a good program.

, much appreciated. Also, on looking over your results again, I wonder if instead of ionic bonding $\ce{IF7}$ exhibits significant charge-shift bonding (see doi:10.1038/nchem.327). The values of $\rho$ at both the axial and equatorial BCPs seem awfully high for true ionic bonds.

Molecular orbitals theory does not assume covalent bonding, that is wrong. Molecular orbitals are used in Hartree Fock equation and the hamiltonian accounts coulombic interaction. After solving the equations you obtain the density matrix P, in which you would see if the bond is predominantly ionic. I think the conclusion is wrong.

I think this response has misunderstood the Christe paper. @Martin-マーチン doesn't contradict a 6-center-10-electron bond; in fact, his analysis is entirely consistent with a 6c10e system. The bonding of IF7 requires one 6c10e + one 3c4e bond; together these sum to 8e at the iodine. It also falls directly from group theory, where the equatorial F5 split into A1'+2E1' and the iodine s/px/py transform as A1'+E1'. That leaves 4e (one E1') nonbonding and 6e (A1'+E1') bonding (=10e over 6 centers). The 3c4e bond contributes 2 more bonding electrons (Γ(F2)=A1'+A2'', Γ(pz)=A2'') to complete the octet.

The structure of $\ce{IF_7}$ cannot be understood as simple static pentagonal bipyramid. The full explanation is given in Karl O. Christe, E. C. Curtis, David A. Dixon, J. Am. Chem. Soc. 1993, 115 (4), 1520–1526.
The last sentence of abstract states clearly describes the bonding situation: These features can be explained, however, by a bonding scheme involving a planar, delocalized pxy hybrid on the central atom for the formation of five equatorial, semi-ionic, 6-center 10-electron bonds and an sp hybrid for the formation of two mainly covalent axial bonds.


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How does the inductive effect, explain the reactivity of tertiary alcohols with halogen acids?
The reactions of primary and secondary alcohols with halogen acids ($\ce{HX}$, where $\ce{X}$=halogen) require the presence of a catalyst, $\ce{ZnCl_2}$. With tertiary alcohols, the reaction is conducted by simply shaking with concentrated $\ce{HCl}$ at room temperature. Which mean that, tertiary alcohols are more reactive than primary and secondary alcohols. The reason is said to be, due to the inductive effect.
I don't know how the inductive effect would explain that, tertiary alcohol is more reactive than primary and secondary.

Secondary and tertiary alcohols react with strong acid (and bases) such $\ce{HX}$ following SN1 mechanism.
$$\ce{ROH <=> HX + ROH_2^+ + X^-} \tag{1}$$
$$\ce{ROH_2^+ <=> R^+ +H_2O} \tag{2} $$
$$\ce{R+ +X- \rightarrow  RX} \tag{3} $$
In these reactions, the rate determining step is the formation of the carbocation (2). So the successive attack of the nucleophilic chlorides is dependent on that.

Here is how the inductive effect comes into play. The inductive effect of $\ce{R}$ groups of the alcohols stabilize the carbocation (figure show the stability of carbocations from the more stable to the less) and then increase the chances of a nucleophilic attack!

@G M.Thank you for the answer. Is more stable carbocation, more reactive? if so. How?

@CURIE just think: a specie to react with a reagent must be present, a stable carbocation has more possibility to be present rather a unstable carbocation so more possibilities mean more chances that the reaction take place. You're welcome!

@G M.I haven't yet understood, how the presence of more no of carbon atoms in tertiary alcohols, make them more reactive than primary. What I thought was, if tertiary is more reactive than primary alcohol, it means that carbon oxygen bond is weaker in tertiary alcohol than in primary, which is making the -OH bond in tertiary alcohol to break easily than in primary, when a nucleophile is going to attack. My doubt is, how C-O bond is going to be weaker in tertiary than in primary?

@CURIE if tertiary is more reactive than primary alcohol, it means that carbon oxygen bond is weaker in tertiary alcohol than in primary What are the bases of this assumption? Reactivity is a complex topic related to thermodynamic and kinetic, in this case the slow step (what determinate the velocity of the reaction) is the formation of carbocation so a more stable carbocation (due the inductive effect) is thermodinamically favorite rather an unstable carbocation. Try to cite the exact phrases of the book so we can see the context more clearly.

@G M. I agree with all your explanation that, slow determining step is going to determine the speed of reaction. I assumed that C-O bond to be weaker in tertiary alcohol than in primary or secondary, or in your view inductive effect is making the C-O bond to break easier in tertiary than in primary and secondary. Because, if you think C-O bond to be of same strength in all tertiary, secondary, and primary. When nucleophile approaches carbocation in slow step, C-O bond in tertiary would have same strength (inspite of inductive effect) as in primary or secondary, and reactivity would be same.

@CURIE in my opinion is not related to the bond strength this doesn't matter, what matter is the stability of the species you obtain.

@ImmortalPlayer a tertiary carbocation is more stable than a secondary or primary because of the inductive effects. Basically, the $\ce{-CH3}$ groups have a positive inductive effect, they push electrons onto the central carbon. This results in a slight negative charge buildup at the central carbon, which helps in forming a cation there.

Tertiary alcohols are more reactive because the increased number of alkyl groups increases $+I$ effect. So,  the charge density on carbon atom increases and hence around oxygen atom. This negative charge density tryna push the lone pairs on oxygen atom away. So, the oxygen atom wanted to leave the carbon atom as soon as possible. Hence, the cleavage of $\ce{C-O}$ bond becomes easier. So, a tertiary carbocation is formed which is more stable than secondary and primary. Hence, their formation is favoured by immediate breakage of $\ce{C-O}$ bond in tertiary alcohol. This shows that they are more reactive.

No, this explanation is very wrong.

how jan..........???

This is because when number of electron donating alkyl group on OH-bonded carbon atom increases, polarity of carbon oxygen bond also increases, which further facilitates the cleavage of carbon oxygen bond. Therefore reactivity increases.

Primary alcohols react via $S_N2$ mechanism which involves the carbocation formation
but the secondary and tertiary react via $S_N1$ which is elementary and no carbocation is formed.
hence it appears that both primary and tertiary alcohols should react rapidly but it is not so.
the $\ce{C-O}$ bond weakens as the no of electron donating groups on the alpha-carbon increases this is due to corresponding increase in negative charge developed on $\ce{O}$ atom **

thats why tertiary alcohols are more reactive.

**
further $\ce{ZnCl2}$ is a Lewis acid which coordinates with $\ce{O}$ atom and weakens the bond thus it need to be used for primary and secondary alcohols.

Welcome to Chemistry.se! A SN2 mechanism would not involve the formation of a carbocation, because it is concerted, while in the SN1 reaction a carbocation is formed, as it is stepwise.


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A function for getting the nuclide's nuclear binding energy?
Is there a function $f(A)$ or $f(Z, A)$ that gets the energy of nuclear binding of one nuclide in atom? Or other way to get it? My chemistry book says only to get it from the chart which is very impractical.

Nuclear physics is very complicated and full of unsolved problems. For example, we seem to understand rather poorly the effects of three-body forces between nucleons, and describing them quantitatively is probably out of the question. Therefore, I expect that nuclei binding energies for most isotopes are solely determined empirically and have no good theoretical curve to originate from. Maybe there are statistically fitted curves to the data we have, though they may well not have a simple form and likely are only good approximations for relatively small intervals.
Edit: As it turns out, equations indeed exist, though as expected they are quite unwieldy and are semi-empirical.

Look here: link. It looks like for atoms with higher number of nuclides the function is quite accurate.

@Jantomedes I've edited my answer above, so take a look

I hope to be not wrong, but believe recalling this to be one of the holy grails in theoretical nuclear physics and a testing benchmark for most comprehensive [field] theories on that subject.
A given theory's proper prediction of binding energies/ masses is also a pre requirement towards the concept of a Theory of Everything - should it even be feasible.
I remember as an undergrad having been fascinated for months on end by the obscure Heim theory, trying to approach its mathematics. It took me a while to un-glamour myself and realize that it is virtually impossible for a theory so old and unaccepted to be as good as its advocates state. Statistically speaking, there is always a chance that you can derive at magical results with a lucky mixture of the right constants and formulas. But these new constants would then need to reveal meaning in other natural concepts as well, in order to make sense in the first place.
I don't know much of the status quo, but I would surmise deriving the binding energies is still a matter of more or less competing theories, each with successes in their own right, but none covering the entire chemical elemental spectrum solely based on fundamental physical constants.
As I recall there are several semi-empirical formulas. Nicolau Saker Neto, already pointed out one.
See also:

Standard Model


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Is it true that iron (Fe 0+) cannot exist when there's water around?
So somebody told me that iron metal (zero charge) cannot exist at all when $\ce{H2O}$ is present because it will all turn into rust.
How is this possible? Isn't there some sort of equilibrium?
Does it have to do with the pH/pE? And if iron really can't exist, then what happens if I put an excess of iron into the water?

I think you mean organometallic Fe(0) (i.e. coordinated single atom), and not bulk metal. Can clarify this?

Iron metal can exist in presence of water. As you stated, important point is the $\ce{Fe^0 -> Fe^3+}$ reaction. Water is not an oxidizing agent, therefore is not capable of inducing this process.
To create rust, you need also oxygen, or as you noted, electrochemical potential.


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Why is alcohol boiled with HBr for preparing alkyl bromide?
Alkyl halides are best prepared from alcohols ($\ce{R-OH}$, where $\ce{R}$ is alkyl group), which are easily accessible. The hydroxyl group of an alcohol is replaced by halogen on reaction with concentrated halogen acid ($\ce{HX}$, where $\ce{X}$ is halogen), to get alkyl halides.
Usually we don't need boiling the reactants. But, for the preparation of alkyl bromide requires constant boiling of alcohol with $\ce{HBr}$ ($48\%$). Assume that alcohol is tertiary, so that, we can avoid using catalyst.
What I thought was, alcohol can develop inductive effect, and thus, can get detached with $\ce{OH}$, so that, the bromine from $\ce{HBr}$ can add to $\ce{R}$ group to form alkyl bromide. So, I thought there is no requirement of boiling of reactants, as like we do for preparation of alkyl chloride. Then, why do we need boiling in case of $\ce{HBr}$ and not in case of $\ce{HCl}$?

Can you propose mechanisms for your reactions?

Assuming i) a primary alcohol $\ce{R-OH}$ to be converted into the corresponding primary alkyl halogenide $\ce{R-X}$ ii) by $\ce{HX}$ acid only, I would expect a decrease of reactivity to be observed in this line: $\ce{HI} > \ce{HBr} > \ce{HCl} > \ce{HF}$, with $\ce{HF}$ of lowest reactivity.  This is based on the decrease of the nucleophilicity of the anions of these acids.
Of course, halogenations with these four acids are reactions of equilibria, and these may be driven forward by increasing the concentration of the reactants, for example.  And I do recall two-step protocols where the alcohol is activated by sulfuric acid before adding hydrobromic acid.  But in both variants, rather $\ce{H3O+}$ then a neutral $\ce{OH}$ would leave the molecule.
(However secondary and tertiary alcohols that are of higher reactivity than primary ones under these acidic conditions, are more prone to isomerization, too.  Simultaneously, as hinted by ssavec, instead of a mechanism of type $\mathrm{S_{N}1}$, the of  $\mathrm{S_{N}2}$ may become more favorable.)
For the preparation of alkyl chlorides, the addition of (dry) $\ce{ZnCl2}$ in presence of $\ce{HCl}$ is sometimes used.  More easy to handle are routes relying on $\ce{PCl3}$ or $\ce{SOCl2}$, for example.

$\ce{HBr}$ is a stronger acid than $\ce{HCl}$ (pKa -9 vs. -7), meaning that $\ce{Br-}$ feels more comfortable in dissociated form than $\ce{Cl-}$. In other words, $\ce{Br-}$ is a better leaving group. I think this has a direct relationship to reactivity via $S_N1$.

One reason for using HBr/H2SO4 is to absorb the water produced in the reaction with primary alcohol [Price, C. C., et al. J. Org. Chem, 1946, 11, 281]. HBr is often generated in situ from NaBr/H2SO4 . Gaseous HBr is employed without H2SO4 [Organic Syntheses, Coll. Vol. 2, p.246 (1943); Vol. 15, p.24 (1935); DOI:10.15227/orgsyn.015.0024].  Sulfuric acid is a weaker acid (pKa -3/2) than HBr (pKa -9).


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Can we prepare aryl halides by heating with halogen acids?
My book says, we can't prepare aryl halides ($\ce{RX}$, where $\ce{R}$ is aryl group and $\ce{X}$ is any halogen) by treating phenol with halogen acids ($\ce{HX}$, where $\ce{X}$ is halogen). It is said to be, because of partial double bond character of carbon-oxygen bond in phenols ($\ce{C6H5OH}$), which is difficult to break (I don't know how there is partial double bond character). What I thought was, if we assume there is no partial double bond character between $\ce{C}$ and $\ce{O}$, we can heat the reactants (i.e. phenol and halogen acid), so that $\ce{OH}$ bond eliminates from the phenol (which has the least bond dissociation energy than $\ce{CH}$), and $\ce{X}$ can get dissociated from the $\ce{HX}$, then it can add up to $\ce{R}$ (aryl group) to form aryl halide. So, can we prepare aryl halides by heating with halogen acids?

This Electrophilic aromatic substitution- no substitutions are ever simple (always involve catalysts) since the resonance of the aromatic (inelegantly expressed as partial double bonds) makes the side chain far more stable than its equivalent aliphatic e.g. hydroxyl-cyclohexane in your case. A quick wikipedia search will show the complexity- its not much different between substituting a hydrogen atom or hydoxyl group- either way it first needs to separated from it parent chain- the most difficult step with resonant structures.

Can you propose mechanism for the presented reaction?

It sounds like the questioner wants to perform a nucleophilic substitution on the aromatic ring. Conceivably, the acid can protonate the phenol, creating a relatively good leaving group (oxonium) which could be replaced by the nucleophilic halide ion. However, this is probably unlikely to happen without very energetic conditions, as the aromatic ring strongly repels nucleophiles' electron pairs. The only reaction of that class I know uses aryl diazonium salts. The diazonium group is just about the best leaving group there is, far better than oxonium. Only then can it be replaced.

OH-group in some strongly electron-deficient phenols may be replaced with halogen using $\ce{PCl5}$. As an immediate example  I can come with picric acid, but probably other examples exist. The reaction, however, happens as nucleophilic substitution with oxygen firstly bound to phosphorous by replacing one chlorine.  The net reaction is $\ce{Ar-OH + PCl5 = ArCl + HCl + POCl3}$

The partial double bond character of the C-O bond in phenol becomes obvious from the resonance structures. Contributions from the structures with $\ce{C=O}$ double bond explain why the character of the C-O bond is in between single and double bond, and also show that oxygen lone pairs are less available for protonation to form $\ce{Ph-OH2+}$.

Protonated phenols are therefore fairly acidic, with a $pK_a$ around -6. $\ce{HBr}$ and $\ce{HI}$ have significantly lower $pK_a$ values (-9 and -10, source), so they would indeed be able to protonate phenol. However, $\ce{H2O}$ is unlikely to be eliminated from this compound due to the instability of the resulting phenyl cation. Its positive charge cannot be delocalized into the aromatic system as it resides in an empty $sp^2$ hybrid orbital which is perpendicular to the delocalized $\pi$ bonds (source). Formation of the phenyl cation by elimination therefore requires much better leaving groups like $\ce{N2}$, which can be eliminated from aryl diazonium cations.
Substitution of the protonated OH group by halides with an $S_N2$-like mechanism is hindered by Coulomb repulsion between the $\pi$ electron cloud of the benzene ring and the lone pairs of the nucleophile. This reaction only works with aromatic compounds in which the electron density in the $\pi$ system is decreased by multiple electron-withdrawing substituents, like nitro groups.

We cannot. There are two mechanisms possible for transforming alkanols into alkyl halides, commonly known as $\mathrm{S_N1}$ and $\mathrm{S_N2}$. For the former, a leaving group (in our case: something from the hydroxy group) must dissociate away to create a carbocation which can then be captured by the halide. In the second case, the halide attacks from the rear side of the carbon-hydroxyl bond and displaces it.
Both mechanisms are only possible if the hydroxy group was initially protonated, because $\ce{OH-}$ is a terrible leaving group (but $\ce{OH2}$ is okay).
In case of phenol, the $\mathrm{S_N2}$ mechanism is not practical, since the rear side of the $\ce{C-OH}$ bond is blocked by the phenyl ring. An $\mathrm{S_N1}$ reaction is also strongly disfavoured since the intermediate would be the very unstable (with respect to other carbocations) phenyl cation — the charge is sitting in an $\mathrm{sp^2}$ hybrid orbital in the ring plane and cannot be delocalised across the ring. Hence you can protonate phenol as much as you want, water will never displace itself.
Finally, there is also the $\mathrm{S_NAr}$ mechanism which is essentially a two-step process: attack of the nucleophile onto the carbon in question with loss of aromaticity in the ring followed by displacement of the leaving group regaining aromaticity. This mechanism requires very electron-poor aromatic rings — at least one nitrogen in the ring itself, more are better. In the case of phenol, the aromatic system is generally considered electron-rich (nucleophilic) not electron-poor (electrophilic), thus an electrophilic substitution is possible on the ring but not a nucleophilic one.


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Are orbitals always filled in from closest to nucleus to farthest away?

In which orbital, 4f or 6s, would an electron have a greater likelihood of being near the nucleus.

I figured that the 6s would be closest, since it is filled in first when using the Aufbau principle.
However, does that necessarily imply that it is closer to the nucleus? Or would 4f be closer, since its principal quantum number is smaller?

As you have studied for the quiz, you should know the expected answer.
In full-scope reality, the question has nothing to do with the two explanations you offered, but with the shape of atomic orbitals:

Radial distribution of selected orbitals in $\ce{Sm^{3+}}$, http://pubs.rsc.org/en/content/articlehtml/2004/dt/b407173e