text:
What are similarities and differences among shells, orbitals, subshells, and energy levels?
I think I know what these terms mean:

each row on the periodic table has a new shell of electrons,
a subshell is e.g. 1s, 2s, 2p, 3s, 3p, etc.,
and energy levels are the same as shells,
whilst orbitals are the same as subshells.

Is this correct or totally wrong?

I am using the chemistry textbook by Masterton/Hurley (Cengage) as an outline for this answer. It has a side note that reads "Our model for electronic structures is a pragmatic blend of theory and experiment". In the spirit of that quote, there is some experimental data mixed in with the concepts.

Each row on the periodic table has a new shell of electrons

Yes, and you see that in the plot of atomic properties vs. atomic number. For example, atomic radii (image source)increase significantly as you go from noble gas (last element in a period) to group 1 metals (first element with the next shell occupied). The quantum number that defines the shell is the principal quantum number n.

A subshell is e.g. 1s, 2s, 2p, 3s, 3p, etc.

Yes. The periodic table is organized into s-block and p-block (main group elements), d-block (transition metals) and f-block (lanthanides and actinides) according to which subshell is partially filled (or just filled). The quantum number that defines the subshell is the second quantum number, $\ell$.

Are energy levels the same as shells?

This is true only for the hydrogen atom and other one-electron systems ($\ce{He+}, \ce{Li^{2+}}$) in the absence of a magnetic field:
$$ E \propto -\frac{Z^2}{n^2} $$
with $Z$ the charge of the nucleus and $n$ the principal quantum number.
For systems with more than one electron, distinct subshells in the same shell correspond to slightly different energy levels. One method to measure these energy levels is photoelectron spectroscopy, with some data on aluminium 2s and 2p shown below (image source).

In general, the energy levels depend on the atomic number (number of protons in the nucleus) and the electron configuration (how many electrons are present, and whether they are at ground state or in an excited state). Going from a hydrogen atom to a larger multi-electron atom, the energy levels of distinct shells can be roughly described as an electron bound to a nucleus surrounded by inner electrons with a combined effective charge $Z_\mathrm{eff}$ (smaller than the actual charge of the nucleus because the attraction by the nucleus is offset to a degree by the repulsion by the other electrons):
$$ E \propto -\frac{Z_\mathrm{eff}^{2}}{n^2} $$
For the innermost shells (n = 1), $Z_\mathrm{eff}$ is almost the same as the nuclear charge, whereas for the outermost shells, $Z_\mathrm{eff}$ can be much lower.
For example, comparing 1s electrons across the periodic table, the binding is tighter and tighter as the atomic number increases. On the other hand, the outermost shell (valence shell) in elements of the same group (e.g. Be, Mg, Ca, Sr, Ba, Ra) corresponds to similar energy levels, even though they differ in the principal quantum number. This explains why elements in a group have similar chemical behavior.

Are orbitals the same as subshells?

No, there are seven different orbitals in the f-subshell, five different orbitals in the d-subshell and three different orbitals in the p-subshell (compare size of blocks in the periodic table). The three orbitals in the p-subshell, for example, are either designated (+1), (0), and (-1), or alternatively $p_x$, $p_y$, and $p_z$. For the s-subshell, there is a single orbital, so only in this special case is the subshell the same as an orbital. The quantum number defining the orbital in a subshell is the third quantum number $m_\ell$.
While the different orbitals of the p, d, and f subshells have distinct, sometimes intricate shapes, a filled subshell corresponds to a spherical distribution of electrons.

How do these concepts help to understand chemical bonds?

We study electronic structure because it explains properties and reactivity of atoms and molecules. It might seem that atomic orbitals of hydrogen are of little use in understanding the electronic structure of larger atoms and of molecules. It turns out, however, that they can be used as building blocks for the more complicated systems, both qualitatively (talking about sigma and pi bonds and lone pairs) and quantitatively (computational methods use atomic orbitals as basis functions).
Surprisingly, the changes in electronic structure going from atoms to molecules are moderate, with spherical electron distribution around the atoms explaining most of the electron density observed, for example, in crystallographic studies. The small deviations from this are called electron deformation densities, and they give an experimental view of the subtle shifts in electron density upon bond formation. The picture below shows a contour diagram of the total electron density on the left (near the atoms, there would be many more contour lines, but they are not shown), and the deformation density on the right (lines: higher density than atomic density, dashed: lower density than atomic density, dot-dash: same density as atomic density).
(Image source)

If you are interested in deformation densities and the electrostatic interpretation of the chemical bond, this is a good starting point: https://www.chemistry.mcmaster.ca/esam/Chapter_7/section_3.html

Historically, the first of these concepts to emerge was the shell of the Bohr model. This very nicely rationalized the observed spectral lines of the hydrogen atom, by restricting electrons to be in orbits with fixed energy levels. In modern parlance, the shell corresponds to the principal quantum number $n$.
Later experiments (also, it's enough just to look a the spectral lines for helium) suggested that this was not enough and that each shell had multiple energy levels within it. These levels within the shell are called subshells. In modern parlance, this corresponds to a fixed value of $l$, the azimuthal quantum number. We refer to the different subshells by letters ($s$, $p$, $d$, $f$, $g$, ...), and these correspond to respective values of $l$ (0, 1, 2, 3, 4, ...).
At this point, it is helpful to draw a division because the concept of the orbital is only loosely associated with the prior concepts. The previous concepts come from models that are considered out-of-date by modern standards. This is because quantum mechanics provided a new model whose principal features are incompatible with the fundamental assumptions of these prior models (for example, orbits are not compatible with the uncertainty principle) or simply unaddressed at all (e.g., spin).
Quantum mechanics arrives at a solution that is the following (or at least, this is roughly what one of the major formulations says). The behavior of matter is governed by the Schrödinger equation. For a hydrogen-like atom, we can parametrize the solutions of the Schrödinger equation for that lone electron with several quantum numbers ($n, l, m$). The $n,l,m$ fully describe the probabilities of finding the electron in any point in space, and therefore, it roughly defines where the electron is likely to be. This region is not an orbit or shell, since the electron is not really restricted in its distance from the nucleus. It is therefore an orbital.
The orbital has some features that are similar to the older models (it has to since we still need to rationalize older experimental outcomes). It has a fixed energy level corresponding to $n$ (corresponding to a shell). Different orbitals may be associated with the same shell but still have different energy levels dictated by $l$ (akin to the subshell).

There is an arcane lore of quantum mechanics, that says that movement of electron in a hydrogen-like atom (with exactly one electron) is fully specified by four numbers:

Principal quantum number (n) (1, 2, 3 ...) - specifies a shell
Azimuthal quantum number(ℓ) (1 .. n) - specifies a subshell within the shell
Magnetic quantum number (m) (-ℓ .. 0 .. ℓ) - specifies an orbital within the subshell
Spin quantum number (s) (+/- 0.5) specifies one of two possible positions of electron on the orbital

While movement of electrons in non-hydrogen-like atoms is somewhat different, the form of movement, i.e. orbitals, are mostly same, so it is generally said, that electrons in a multielectron atom may be assigned the same numbers (with a side-note that electrons cannot be distinguished) and no two electrons can have all four numbers the same.
'Energy level' is a completely different concept: it is a group of states possible for some quantum chemical system and having the same energy. This term may (randomly) pop-up in discussions where the electron is viewed as a singular particle with other parts of the system being 'framework'. While somewhat working, such considerations are incorrect and should better be avoided.

There is no movement of electrons in orbitals in a strict sense. It is just a probability density of electron around a positively charged  nucleus.

@M.Farooq Orbitals in particular and wave function in general is quantum equivalent of trajectory. In particular, each orbital corresponds to some specific kinetic energy.

Interestingly, just two days ago, I was discussing similar questions with physical chemists. Both explicitly said that orbital is nothing more than a probability density around a nucleus. The kinetic energy is no longer classical and hence the concept of trajectory, moving path or anything else related to orbitals is rather a misconception.

@M.Farooq You shouldn't discuss it with phys. chemists, but with physicists. Quantum mechanics is derived from classical mechanics in Hamilton formulation by direct replacement of trajectory with wave function and fitting replacement of impulse and coordinate variable with respective operators. Since it is directly derived, terminology is inherited and thus, "movement of electrons" is still named this way even if the wavefunction is stationary.

Of course, you understand it. I was writing for the students (perhaps the original poster). They have tons of misconceptions regarding orbitals. You might enjoy the article "The nature of the chemical bond—1990: There are no such things as orbitals!" https://pubs.acs.org/doi/abs/10.1021/ed067p280

But they are a valid conceptual tool in many cases: Jenkins

Shells are the major grouping of the orbitals (1, 2, 3, 4, etc). Subshells are the groupings inside each shell (s, p, d, f, etc). Energy levels correspond to the orbitals and vary with the element and the number of electrons in the atom ($\ce{Fe}$, $\ce{Fe^{2+}}$, $\ce{Fe^{3+}}$).


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Determining central atom in NCl3
Less electronegative atom is the central atom. $\ce{N}$ is more electronegative than $\ce{Cl}$ as $\ce{Cl}$ has larger atomic radius. So why in $\ce{NCl3}$, $\ce{N}$ is the central atom?

Cl is slightly more electronegative according to the tables I've seen: http://en.wikipedia.org/wiki/Electronegativity
But I wouldn't say it's an absolute rule that the less electronegative atom is central.  For example, $\ce{NH3}$ has the N central and $\ce{Cl_2O}$ has the O central.

The relative positions of chlorine and nitrogen on the electronegativity scale can switch depending on the definition of electronegativity being used. Allen scale electronegativities are quite recent but also popular, and there nitrogen is higher than chlorine.  But yeah, the asker is just taking a rule of thumb too strictly.

Acc. to Pauling's scale, both have electronegativity 3.0 but as $\ce N $ has lesser atomic radius,it is more electronegative.But here it seems contradiction as a lesser electronegative atom ought to be the central atom.But it is also accurate that in nitrogen tri chloride , nitrogen is the central atom .Is the 'less electronegative atom is the central atom ' a wrong concept?

How can I determine accurately  the central atom ?

@user36790 I would say that the example of Cl and O basically proves there is no simple set of rules, since Cl can be in the center for chlorate, perchlorate and other compounds, while $\ce{Cl_2O}$ has the O central. Also, the symmetrical choice or lone atom in the center, isn't always correct as there is NNO. The rule "H is never in the center" you can usually count on, although a hydride can bridge two metal or boron atoms. I would say that carbon and nitrogen tend to be central slightly more than their electronegativities suggest and halogens non-central. SeCSe, CAt4, NI3.

Being "central" has nothing to do with electronegativity.  It is a matter of geometric symmetry. The atom at the "center" of a molecule is the atom with the highest geometric symmetry. But if you look at benzene, there is no atom at the "center" of the molecule.

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Theoretical principles of molecular simulations
I am pharmacy student and would like to work in the field of drug design. We learn to work with the programs but thats not enough for me. I want to learn the theoretical principles of computational chemistry and develop software themselves, or at least parts of it.
However, I have a few questions:
Where do I start with the chemistry part?

solving the Schrödinger equation
Hartree–Fock method
Density functional theory

Which mathematical principles do I need?
Are there comprehensive literature for beginners and what would you recommend? A good book would certainly be the best start.
I need a good starting point, especially in terms of mathematics and hope you can help me.

Statistical mechanics,  partial differential equations

I don't want to be an ass, but  "computational chemistry" in drug design is at least as much chemoinformatics as quantum chemistry. Actually, I hardly saw any quantum chemistry in this field: statistical and force-field based methods are the dominant.

Computational chemistry is a synthesis of many different subjects, each of which is very complex on its own. So, if you really want to learn the theoretical basis, you should be prepared to put in a lot of time and effort. I don't want to discourage you, I just want you to have realistic expectations.
The fundamentals of computational chemistry are physical chemistry and computer science. The fundamentals of those are math, physics, and logic. So, depending on what your background is, you might have to start with math and physics.
Math:

Calculus
Vector Calculus
Differential Equations
Statistics

Physics:

Classical Dynamics
Classical Electrodynamics

Once you have taken those courses (if you haven't already) you can start with physical chemistry. You could conceivably teach yourself those subjects, but it would be very difficult - a free online course would be a good option if you can't take a university course. I believe Khan academy covers all of these subjects.
Physical Chemistry

Statistical Mechanics
Quantum Mechanics

I always liked Atkin's physical chemistry book (although I don't know if the newer editions are still good). McQuarrie's quantum chemistry and statistical mechanics are very good as well and are classics in the field.
You asked:

Where do I start with the chemistry part?

solving the Schrödinger equation
Hartree–Fock method
Density functional theory

Quantum mechanics in physical chemistry will cover solutions to Schrodinger's equation for all of the cases where a closed-form solution exists. This will give you an understanding of the basic principles, so you should start with that. To solve "real" systems, you need more advanced approximation techniques, and Hartree-Fock and DFT will get you started. That is a good order to study them in, as well.
For computer science, a lot depends on which software you will be using or what languages you want to use. Most of the core, highly optimized algorithms are written in fortran or C. Some go as far as assembly for the most important parts. Most of the software I have seen (non-core) is written in fortran or C++. My experience is a little out of date (I last worked in the field about 7 years ago) so that may no longer be true. At any rate, you will need at least a basic programming course in the language you want to use (C would be good to start with, since you will really learn the fundamentals). I would also take some object-oriented courses, so courses involving more advanced C++ and java would be good as well.
Key concepts (beyond programming and architecture) that you will need are:

Stochastic methods (Monte Carlo)
Linear Algebra
Numerical integration

With all of the above, you will have a good set of base skills and knowledge on which you can build.
You asked:

Are there comprehensive literature for beginners and what would you recommend? A good book would certainly be the best start.

I am only familiar with one book that might be suitable for beginners in computational chemistry, so perhaps others can give better suggestions. The one I used was Molecular Modelling by Leach. It gives a good overview of common techniques, but doesn't go too in-depth. However, it was last updated in 2001 - so it may be out of date. The state of the art moves pretty quickly in terms of detailed application, although the fundamentals tend to stay pretty consistent over time. Another classic text which covers the basics in some depth is Allen & Tildesley's Computer Simulation of Liquids.
Both of these books focus more on the modeling aspects of molecules as opposed to chemical reactions, so if you are more interested in simulating reactions as opposed to statistical mechanics or protein-ligand interactions, you would need different books.

Disclaimer: I don't have significant experience in the field, but I did touch it a little.

I want to learn the theoretical principles of computational chemistry and develop software themselves, or at least parts of it.

The idea is noble, but extremely naive.

would like to work in the field of drug design.

Lucky for you, you don't actually need any advanced math staff to work in this specific field. Sure, it may be handy, but not necessary. Most drug-related simulations may be performed using molecular mechanics/molecular dynamics approach, that is quasi-classical in nature.
general programming notes
the most common language in scientific community is, unfortunately, fortran. It is a very old language with huge amount of inherent problems. I advice to learn programming in some more sane language first and then move to fortran/c/c++. Freely available SICP (structure and interpretation of computer programs) based on scheme is a good choice. Sure, it uses quite esoteric scheme, but it does discourage bad habits.
minimal list, in order

classical newtonian physics (forces, acceleration, speed - all in vector form, their relations and electrostatic interactions)
theory of programming: big-O notation, algorithmics, space-sorting, indexes in space etc. Specifically, MM/MD needs a way to estimate total forces in many-body system with electrostatic interaction where thousands and millions of bodies may be present. This requires effective algorithm to discard small interactions, which in turn requires effective algorithm producing list of close pairs.
optional: genetic algorithms.
optional: machine learning.

after learning that, it is time to start to read actual articles/reviews. Areas of interests are QSAR (relies on machine learning) and docking (relies on MM/MD approach to 'fit' molecule to receptor.)

in case you still have resolve to code something, it is time to actually code. take some existing package and tune it.
in case you want to work with huge systems or hi-performance computing, learn

theory of parallel programming: threading libraries, synchronisation primitives

common parallelization technologies: mpi, openmp.

GPGPU : OpenCL or Cuda

maximal list, in order. Overkill for drug design. In addition for minimal list

linear algebra.
analytical geometry
advanced calculus (multidimensional integrals, partial derivatives and related stuff)
differential equations
function approximation using linear combination of other functions
quantum mechanics, nonrelativistic case.
any general quantum chemistry book - qualitative theory of bonding, hartree-fock equations, black magic of MP2 and DFT

up to this point the topics are useful for any chemical student. Specialized staff follows.

general programming

algorithmics

big-O notation

numerical errors, estimating them.

IEEE floating-point format, use and limitations

numerical methods, specifically

fast fourier transform

multidimensional integration

iterative diagonalization methods

at this point it becomes possible to code something simple. In case hi-performance computing is of interest (it is a separate area, that is not chemistry related per se)

parallel programming,

see minimal list

distributed storage and processing

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How would you know if ... is H-bond or permanent dipole or induced dipole?
For example like HCl, why isn't it a hydrogen bond?

Hydrogen bonding is defined as an electrostatic attraction between a hydrogen atom bearing a partial positive charge in a molecule and an electronegative element of another molecule. The partial positive charge is the result of a dipole, caused by an electronegative element, such as F, O, or N. The other electronegative element to which the partially positive hydrogen is attracted to is also usually F, O, or N.
Here's the IUPAC definition; note that it does not exclude the possibility of hydrogen bonding in molecules such as HCl:

A form of association between an electronegative atom and a hydrogen
  atom attached to a second, relatively electronegative atom. It is best
  considered as an electrostatic interaction, heightened by the small
  size of hydrogen, which permits proximity of the interacting dipoles
  or charges. Both electronegative atoms are usually (but not
  necessarily) from the first row of the Periodic Table, i.e. N, O or F.
  Hydrogen bonds may be inter-molecular or intramolecular.

http://goldbook.iupac.org/H02899.html

Can you answer me one last question....why is hydrogen bonding stronger than permanent dipole and not permanent dipole stronger than h-bond.....tell me all reasons...need it :)

Is this a homework question?

No Sir my homework question was list 3 intermolecular attractive forces and list them in order of increasing strenght? that ive answered it...but teacher might ask question beside that to see some knowledge..

Chemistry Speech in class 60s to answer the question and the remaining if time left then he would ask some question of what you know of this question so im trying to sum it up.

Try giving some reasons in your OP and I'll see if they're right.

ok umm...H-bond has high boiling point,great polarity of bond, and close approach of dipoles.

highly polar ecuase of 0, N, F

if those are true i dont know how to say it properly...or if anything else is missing ill add thanks

So am i kinda wrong?...cuz im going to bed if you please.


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How can water with a very low hardness have substantial buffer capacity?
My freshwater fish tank has a pH of 5.8 and a KH value in excess of 40 or 720+ppm while still maintaining a water hardness of 20 ppm or GH< 1 now I am a math/phsyics student but I have taken both first year chem class at university so I expect to follow most basic ideas pertaining to chemistry but this doesn't make a lot of sense. Most water with a high alkalinity also has a basic pH around 8-8.5 normally I'd say that a high KH value is simply saying I have an acidic solution that is very resistant to a pH change i.e. a large buffering capacity but my water is very soft to the point of almost not having anything in it ( GH<1) or less than 20ppm of minerals how can water be so acidic have such a large buffering capacity and have nothing in it?
(the reason why I am asking here is all the aquarium forums i have tried can't seem to figure out what's going on in my tank)
EDIT:
This answer poseing an intresting possibility my tank has Turtles in who obviously are advanced enough to excreet urea instead of amonium i Titrated out a reaction and Got KH value of 28 and GH value of 4.5 i also added several air stones ( my buddie has a bio chem degree and figured it could of been too much $CO_{2}$ we managed to get the Ph up to ~6.4 and drop the KH value ot be more closely in line with accpetable parameters ( a value of 8 still oddly high compared to the GH value of 4.5)

Is your tank exposed to the atmosphere? Carbon dioxide can dissolve in water to form a fairly acidic solution.

It will take time for the bacteria in the aquarium to adjust to the new pH level. When it does, you will probably see the KH value decrease.

Distilled water may have pH 5-6 because it dissolves significant amount of carbon dioxide. Carbon dioxide in water is in equilibrium
$$\ce{CO2 + H2O <=> H2CO3 <=> H+ + HCO3-}.$$ Since amount of dissolved carbon dioxide is much larger than formed acid, the solution acts as buffer.
Tap/sweet water usually contains fair amount of K, Na, Ca, and Mg cations, that traps significant amount of carbon dioxide forming $\ce{HCO3- /CO3^{2-}}$ buffer, that is alkaline.

Are you saying that these cations are attracted to the hydrogen carbonate ion and the carbonate ion?

Said cations are dissolved from minerals by exchange with protons produced by mentioned equilibrium. Only limited amount of carbon dioxide can be dissolve in water and be in equilibrium with specified amount hydrocarbonate, and maximum amount of hydrocarbonate in a solution with notable amount of metal cations is determined by amount of cations that does not have other (non-(hydro)-carbonate) anion partners.

Can you give an example of this proton exchange?

@Dissenter for examle, anorthite on prolonged contact with atmosphere slowly gives up its calcium , producing kaolinite  $CaAl_2Si_2O_8 + 2 H_3O^+ = Ca^{2+} + Al_2Si_2O_5(OH)_4 +H_2O$. The process occurs at very slow rate, but fast enough on geological scale, and requires only rainwater to occur. Other volcanic aluminosilicates behaves similarly.

In an aquarium, there could be several non-mineral buffers. The most likely one is a phosphate buffer. There would also be carboxylic acid (amino acids) and probably urea, as well as carbon dioxide.


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Why May K Expressions Be Manipulated As Such?

Why is the above massaging of various K expressions valid? Something in me says "Hess's Law." But then doesn't that deal with enthalpy? What's the true reason?

Your intuition is correct. The reason equilibrium constants from different reactions can be strung together and multiplied comes from a generalization of Hess' law which covers Gibbs' free energy changes (apparently also called Bordwell thermodynamic cycles). Enthalpy is a thermodynamic function with a special property; it is a state function. That is, enthalpy changes in cycles are determined solely by the initial and end states of a system, and not the trajectory in which the initial state travels to the end state (as happens for functions such as work $W$ and heat $q$). Any state function can be manipulated in the way enthalpy is used for Hess diagrams. Thus, by adding reactions, you can add their respective reaction free energy changes $\Delta_rG$.
Let us now take two reactions, with free energy changes equal to $\Delta_rG_1$ and $\Delta_rG_2$. There is a relationship between a reaction's free energy change and its reaction quotient $Q$, namely:
$$\Delta_rG = \Delta_rG^o + RT\ ln\ Q$$
, where $^o$ indicates standard conditions. For reactions 1 and 2, we therefore have:
$$\Delta_rG_1 = \Delta_rG^o_1 + RT\ ln\ Q_1$$
$$\Delta_rG_2 = \Delta_rG^o_2 + RT\ ln\ Q_2$$
Adding the reactions, we obtain a reaction 3 where the same formula applies:
$$\color{\red}{\Delta_rG_3} = \color{\navy}{\Delta_rG^o_3} + \color{\green}{RT\ ln\ Q_3}$$
However, since we can add the free energy equations for reactions 1 and 2, reaction 3 is also represented by:
$$\Delta_rG_1 + \Delta_rG_2 = \Delta_rG^o_1 + \Delta_rG^o_2 + RT\ ln\ Q_1 + RT\ ln\ Q_2$$
$$\color{\red}{\Delta_rG_1 + \Delta_rG_2} = \color{\navy}{\Delta_rG^o_1 + \Delta_rG^o_2} + \color{\green}{RT\ ln\ Q_1 Q_2}$$
Clearly, we have $Q_3 = Q_1 \times Q_2$. Considering the equilibrium conditions of all reactions ($\Delta_rG = 0$), then the reaction quotients are defined as the equilibrium constants $K$, and similarly one finds $K_3 = K_1 \times K_2$.


text:
Why is it that neither silver nor copper react with a strong acid?
$\ce{Fe}$, $\ce{Mg}$, $\ce{Ni}$, $\ce{Pb}$, $\ce{Sn}$, and $\ce{Zn}$ all react when they (in solid form) are submerged in an acid solution with the presence of a strong acid like $\ce{HCl}$, but silver and copper do not.
I have examined the electronegativity values and both $\ce{Ag}$'s and $\ce{Cu}$'s electronegativity value is only different from $\ce{Ni}$ by $\mathrm{0.1}$ on the Pauling scale. I don't see a pattern; the only thing I see which is common between copper and silver is that they both may have an oxidation number of +1.
So, why is it that $\ce{Ag}$ and $\ce{Cu}$ are immune to the H cations in the acidic solution?

Electronegativity is rather poorly defined for the $d$ and $f$ block elements, so invoking it is probably not a good idea.

This is a good question. I'd like to know myself.

@NicolauSakerNeto Can you elaborate? I realize a lot of chemistry that is taught in earlier years is not quite accurate. I don't mind learning more advanced theory.

I suspect this has to do with reduction potentials. Silver is less reactive than hydrogen.

The simple answer: the electrode potential $E$ for the reaction between, e.g., solid copper and $\ce{H+}$ to yield $\ce{Cu^{+2}}$ and hydrogen gas is negative under standard conditions. Since $\Delta G_{cell} = -nFE_{cell}$, negative electrode potential implies a positive change in free energy, which indicates a nonspontaneous reaction (i.e., net work has to be done on the system for the reaction to occur, for example by application of external current). This may not be a satisfactory answer, though, since it leaves the question of why the reaction is spontaneous with other metals.

Both silver and copper react with nitric acid. So, don't say a strong acid like HCl, instead say they don't react with concentrated HCl.

@Greg E: Why not post your comment as an answer?

@LDC3 What's the difference? Besides the odd conjugate acid, all acidic solutions are ones with high concentrations of hydronium--no? I haven't tried nitric acid, but based on the fact that they don't react with a solution of HCL, I would assume they don't react with nitric acid.

@trb456, I don't think it would be a fully satisfying answer as it currently stands. Perhaps when I have a bit more time I'll consider the pertinent thermodynamic and physical parameters (i.e., things like ionization energies, enthalpies, etc.) and write up a complete answer that addresses the question at a more fundamental level.

Nitric acid is an oxidizer as well. The reaction with copper is $Cu + HNO_3 \rightharpoonup Cu^+ + H_2O + N_2O_4$ (unbalanced).

@Klik, the difference is that, with nitric acid, there are alternative redox reactions available that are more favorable, i.e., the formation of $\ce{NO2 + H2O}$. Other pathways also exist, depending on concentration and other conditions.

@LDC3 you might want to clarify that it's the nitrate ion in nitric acid that's pulling the weight; nitrogen is pretty versatile at oxidizing stuff because of the variety of oxidation states it can adopt (NO, NO2, NO3-, N2O4, N(3-), N2, etc).

@Dissenter The same can be said for chlorine ($Cl^-, ClO^-, ClO_2^-, ClO_3^- and ClO_4^-$). Oh wait, $Cl^-$ is already the most reduced state, whereas $NO_3^-$ is in an oxidized state.

True, and the N in NO3(-) has a +5 oxidation state; nitrogen can definitely adopt lower oxidation states (i.e. +2 in NO).

This will involve some degree of hand-waving. Let me first limit the analysis to the transition metals so that more effective comparisons can be made.
The reaction between a metal and an acid requires the metal to be oxidized, i.e. lose electrons to the acid, creating positively-charged metal ions in solution. The general idea is that as you go along the transition metals in a period from left to right, there is an increase in effective nuclear charge felt by the valence $ns$ electrons (due to the additional $(n-1)d$ subshell electrons poorly shielding the $ns$ electrons from the nucleus) making the valence electrons harder to remove. This trend can be seen to some extent in the increase in ionization energies of the elements, for example, and also in the standard reduction potentials for metal cations of same charge (they become more positive, in other words making oxidation and hence attack by acid more difficult). Thus, metals at the left of the transition metals tend to be more easily attacked by acids than the metals at the right.
There is a rather interesting exception though: zinc strongly breaks the trend, being far easier to oxidize than copper, the element before it, even though zinc has a much higher ionization energy, clearly indicating its electrons are held tighter (cadmium also exhibits this anomaly, to a milder extent). What gives? The thing is that the tendency for a metal to oxidize is the result of several combined factors, one of which is also how strongly bound the solid metal is in the first place. If a metal contains atoms which are strongly bonded to each other, then oxidation tends to be more unfavourable as it would require these bonds to break. When comparing copper and zinc, it is clear that the latter has far weaker metallic bonding in the solid (their melting/boiling points are $1360\ \rm{K}$ / $2835\ \rm{K}$ and $695\ \rm{K}$ / $1180\ \rm{K}$, respectively). Thus, if you include the energy necessary to separate the atoms from the solid before ionizing them, it turns out that the process is easier for zinc than copper.
Comparing different rows of the transition metals is a bit less clear, in part because not all elements have ions of the same charge which can be directly compared. The general tendency is that the metals become tougher to oxidize as you go down the rows. For the transition metals in the sixth and seventh period this is probably a consequence of lanthanide/actinide contraction and relativistic effects, which decrease the energy of the valence $ns$ orbitals.
Now for some other elements. Magnesium is very reactive towards acids because it is both a metal containing relatively weak bonds and because its ionization energy is comparatively low, being an alkaline earth element. For tin, the metallic bonding is not weak, but its ionization energy is not too high, so it will oxidize in acids, though not as strongly. Lead is somewhat less reactive to acids, possibly because its oxidation tends to stop at +2 instead of tin's +4, creating a significantly softer cation which isn't as well solvated by water.

So if you were given the periodic table and asked to predict another metal that wouldn't react in a solution of HCL, what element would you choose, and which figures would you look at to base your judgement on? It seems electronegativity numbers are not the be-all-end-all. Would you say melting and boiling points are a better predictor of this characteristic for redox reactions?

Well it's hard for me to "predict" whether a metal will react with acid or not because I already know to some extent which metals are attacked and which ones aren't. But as I said, reactivity towards acids is a combination of several factors, so looking at a single one is a bad idea. The only safe way (but even then not sure-fire) to predict reactivity towards acids would be to look at the experimental reduction potentials, as mentioned by Greg in his comment; the more positive the reduction potential, the harder it should be to dissolve by acids.

Experimental reduction potentials--I will definitely keep that in mind. I always thought that reactions between elements and compounds of a few elements would be easy to predict. Does the difficulty in predicting such reactions come from the geometry of the atoms? I.e. the position of the electrons and shape of the orbitals. Anyways, I appreciate the answer you've given me. Thanks!

Even though theoretically any problem can be solved to arbitrary accuracy, reactions are rather hard to predict from first principles (i.e. taking into account only the most fundamental assumptions) because molecular simulation calculations are computationally very expensive, and successively coarser approximations must be made as the system size increases in order to keep the problem tractable. The geometry of atoms in a system is sometimes one of the factors which contribute to complexity, though usually only in larger molecules (a few tens of atoms upwards, depending on symmetries).

In order to avoid getting bogged down in the dense calculations, we can sometimes isolate simpler factors which can be used on their own, after calibration against experimental data (not first principles). However, these higher-level analyses often have a reduced degree of correlation between prediction and practice, and in order to compensate, one may seek to correlate with several parameters simultaneously, again making predictions harder and problems more subtle, but increasing the range in which an analysis is applicable.

Best to look at a Pourbaix diagram in order to make a guess. Even though the thermodynamics (std. redox potentials) may indicate that a metal reacts with a certain acid solution, the metal or alloy may form a stable oxide film that impedes or reduces the reaction rate (i.e. kinetics of the reaction must be considered as well)


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Calculating Redox Potential; Coupled Reactions
For the reaction of silver and hydrochloric acid to form silver chloride: would the following be the correct way to couple the below two half reactions?
$\ce{2Ag(s) + 2Cl^- ->2AgCl(s) + 2e^-} ~~E^0 =-0.222 ~V$
$\ce{2e^- + 2H_3^+O -> H_2 + H_2O}{ ~~E^0 =0.000~ V}$
Add the above:
$\ce{2H_3^+O +2Ag(s) + 2Cl^- -> 2AgCl(s) + H_2 + H_2O}{ ~~E^0 =-0.222~ V}$
Implying that the reaction is non-spontaneous as we have a "negative potential"?

Just for future reference, standard electrode potentials will not always determine correctly whether a reaction will happen or not, because most reactions are not performed in standard conditions. The proper course of action is to perform some correction by taking into account non-standard conditions, often with the Nernst equation, and then analyze the reaction free energy change.

$2~Ag + 2~H^+ ~(+2~Cl^{-}) \rightarrow 2~Ag^{+} + H_2 ~ ( + 2~Cl^{-})$
As two half-reactions, it is seen that the silver is oxidized:
$2~Ag \rightarrow 2~Ag^{+} +2~e^- ~~ E^{\circ}= 0.80\,V$
And the hydrogen is reduced:
$2~H^+ + 2~e^- \rightarrow H_2 ~~ E^{\circ}= 0.0\,V$
To calculate the reduction potential you have to solve the following equation:
$\Delta E = E_{acceptor} - E_{donor} = E_{2H^{+}/H_2}- E_{Ag/Ag^{+}}$
$\Delta E =  0\,V - (0.80\,V) = -0.80\,V$
The reactions doesn't run voluntarily if $\Delta E < 0$


text:
In the lattice model, what's a nearest neighbor?
I was reviewing my grad Thermodynamics textbook to go over a concept that I never learned well: the Lattice model.  One set of equations1 has:
\begin{align}
zN_1 &= 2N_{11} + N_{12}\\
zN_2 &= 2N_{22} + N_{12}\\
\end{align}
where $z$ is the coordination number, or the number of nearest neighbors (defined twice).
My trouble is a simple thought experiment:

pure solution of $1$
$z = 10$
$N = 100$

So then the first equation becomes $10{\times}100=2{\times}N_{11}+0$, so $N_{11}=500$. Five-hundred nearest pairs, right?  But if I actually had a solution of 100 species 1 each with 10 neighbors, then I'd expect (ignoring boundaries) to have about $100{\times}10=1000$ pairs of $1-1$.
I can't seem to figure out what is meant by $N_{11}$.  Could someone explain it?
1 J. Prausnitz, R. Lichtenthaler, and E. Gomes de Azevdeo. Molecular Thermodynamics of Fluid-Phase Equilibria, 3rd Edition, p.331, Eq.7-68.

A nearest neighbour in general terms is literally that: Find the closest atom of any given element, that is your nearest neighbour distance for that element in the lattice. The number of nearest neighbours for that element is the number of atoms that are at this distance from your starting atom.
$N_{11}$ is the number of nearest neighbour pairs of atoms of element 1, and $N_{12}$ is the number of nearest neighbour pairs between any atom of 1 and any atom of 2 etc. For the $N_{11}$ or $N_{22}$ case, you have to consider that each atom of 1 or 2 in the equation has a contribution to $zN$, hence the term being multiplied by 2 when viewed from the perspective of element 1 or 2.
In the case of your thought experiment, your expected value neglects edges (you can't just neglect edges!) so will be too high by a good margin.
The dissertation of Alexander Zhou1 covers these equations (p.34) and explains them slightly differently (and more clearly I think) than de With's textbook.2

Alexander Zhou. Explicit Non-Random Contribution in Lattice Fluid Equation of State. Explizite nicht zufällige Wahrscheinlichkeitsverteilungen in gitterbasierten Zustandsgleichungen. Doktor Ingenieur, Technische Fakultät der Universität Erlangen-Nürnberg, Nürnberg, 2013. See UB Friedrich-Alexander-Universität Erlangen-Nürnberg (FAU)
Gijsbertus de With. Liquid-State Physical Chemistry: Fundamentals, Modeling, and Applications; John Wiley & Sons: 2013. See Google Books

It appears that you are linking "This link may help" to content inside a book. You may be lucky enough to reside in a country where this resource is available (it is not where I am), but in general it would be preferable to include a citation in readable form, so that other means of obtaining that text could be performed.

Looking back at this question, it looks like I accepted your answer, but forgot to +1 it.  Ohh... right, I probably didn't have the rep to +1 back then.  Hah, silly me.

The discrepancy between 500 and 1000 is due to double counting: the method of multiplying 100 by 10 counts each pair twice: one for each atom it is a part of. So we get the total number of pairs by dividing that number by 2: 500 pairs, still neglecting edges.


text:
Does resonance affect the polarity of the molecule?
Ozone is a resonating structure, is it because of its resonance and bent structure that it is a polar molecule? Or Is it because of the formal charges whereby the formal charges determine the direction of the dipole moment. If yes, how does this work with resonating structures. Initially, I thought that resonating structures automatically are polar, however, the azide ion is a clear counter-example. I am curious, how one deals with such a case, should we take each resonating structure by itself and consider the superposition in the end.

Edit:
Ultimately, my question is how can you explain ozone's polarity without formal charges because on a wikipedia page, they explain it using formal charges. If that's the case, shouldn't resonance be considered. If yes, how?

Concerning your ultimate point, presumably the "polar" you mean is the dipole moment (correct me if I am wrong). N$_3^+$ is linear. Its eigenstates are parity eigenstates. Therefore it has no dipole. O$_3$ is V-shape. The only possibility to let O$_3$ has vanishing dipole is to let its electronic contribution cancels exactly with its nuclear part. It could only happen in very special cases, e.g. the total density is a perfect superposition atomic density.

This is highly unlikely.  Since covalent bond will produce superposition of wavefunction, thus deviates electronic density.

When talking about resonance it is important to know that this is only a concept to visualise certain bonding features. It is very important to understand, that the actual bonding situation is a mixture of all possible resonance structures. The Molecule itself exists at all times as this mixture, hence neither of these resonance representations can be observed. One can regard it also as a superposition of these structures.
Resonance structures had to be introduced into the Lewis formula context, when it was obvious, that one formula can no longer describe a bonding pattern remotely accurate. While the Lewis formalism is great for many things, as it keeps chemistry visualisable and simple, it unfortunately has its boundaries. Only for molecules with very simple bonding patterns it is somehow possible to guess some physical properties (qualitatively).
Formal charges are also only a concept, or a result from the limited depiction of the Lewis representation - hence the name formal. They do not usually represent any physical charge. However, sometimes, the physical charge will coincide with the formal charge. Therefore it is usually not possible to derive any actual dipole moment from them.
Charge separation within molecules depend of the geometric arrangements of the nuclei and the electrons around them, i.e. the electron density. It is a physical property which can be observed and also predicted by means of quantum chemical considerations. Molecular Orbital Theory is a very fundamental concept which allows to derive these properties (at least on a qualitative level).
The bent arrangement of ozone is a result of its electronic structure. And the dipole moment is a result coming from that.

Upon the extension of the question and some comments:
Like Thomij already stated, the formal charges are only for bookkeeping. They always have an integer value. A partial charge, however, is a physical observable, resulting from the difference from the electron density and the field of the nuclei.
In all cases the overall formal charge has to add up to the overall physical charge, which can also be calculated from all partial charges. These partial charges can also be calculated from quantum chemical calculations, since you are approximately solving for the optimal electron density at a certain nuclear arrangement.
There is absolutely no possibility to predict a dipole moment without knowing the electron density (or wavefunction) at a given nuclear geometry. A Lewis formula can be generated from these facts, showing the simplest but still well matching bond description. This representation has to sometimes include formal charges. The linked wikipedia page does not employ this concept to explain the polarity, they rather use partial charges.
The partial charges in ozone result from a superposition of the molecular orbitals (click here to enlarge). These are the valence orbitals obtained from BP86/cc-VPDZ:

Especially the HOMO (very right) and the HOMO-3 (4th from right) and HOMO -6 (4th from left) contribute massively to a slight charge separation in ozone.

I disagree; overall formal charge represents a real physical quantity - the overall charge on the molecule.

@dissenter The formal charges must add up to the overall charge, but the overall charge is not the same thing as a formal charge. Formal charge is a bookkeeping method for assigning charges to individual atoms simply as a way of visualizing electron density distribution.

What's the difference between overall charge and overall formal charge? Numerically, they're equivalent right? I think what Martin and you are getting at is that formal charge =! partial charge (which is a lot more useful).

Thanks for your answer, but I don't think it answers my question.

Ultimately, my question is how can you explain ozone's polarity without formal charges because on a wikipedia page(http://en.wikipedia.org/wiki/Chemical_polarity), they explain it using formal charges. If that's the case, shouldn't resonance be considered. If yes, how?

@user29568 I hope this edit resolves your questions.

So, the wikipedia page should say partial charge instead of formal charge?

Partial charge is also not an observable. The partitioning of electron densities is ill-defined to begin with, which is why many different schemes exist.

@Kexanone Indeed the partial charge is not an observable, but please do not confuse ill-defined with unambiguous. Many different schemes exist, this is also true, but most of them are well-defined. There is just not one unambiguous way to do it. But I kind of get where you are coming from and this answer could probably should be improved.

@Martin Yeah, ambiguous would be the correct word.

Yes, ozone has a dipole moment because it is bent and has resonance structures that involve formal charge separation.  Resonance can affect the polarity and dipole moment of a molecule, but only if at least one of the resonance structures involves formal charge separation, and even then there are cases where the overall molecule will not have a dipole moment.  Some examples will help.  Let's start with benzene. Resonance structures can be drawn for benzene, but none of them involve charge separation, therefore benzene is not polar and does not have a dipole moment. So just being able to draw resonance structures does not mean the molecule must be polar and have a dipole moment.  Next let's examine cyclohexanone.  In this case two resonance structures can be drawn and one of them involves charge separation.  Therefor, we would expect this molecule to be polar and have a dipole moment. For 1,4-cyclohexanedione we can also draw resonance structures with formal charge separation, however in this case the symmetry of the molecule causes the local polarization at each carbonyl to cancel and overall, the molecule has no dipole moment.

By formal charge separation ,do you mean difference? Then, does that mean that the net dipole vector is bigger if the difference is bigger?

In some of the resonance structures I've drawn there is a "plus" and a "minus" charge - that is called "formal charge separation".  The further apart the charges and the more that resonance structure contributes to the overall "true" picture of the molecule, the bigger the net dipole.

So formal charge and formal charge seperation are two different things?

Not really, I could say "the resonance structure shows the formal charges" or I could say "the resonance structure shows the formal charge separation."


text:
In a redox reaction, why does more than one oxidation state of an element form?
Consider the reaction 100ml 10M $\ce{NaOH_{(aq)}} +$ 100ml 10M $\ce{NaHSO3_{(aq)} +}$ 100mL 0.01M $\ce{KMnO4_{(aq)}}$
In the picture posted here, the left columns give information on the Molarity of species present before 100mL 0.01 M $\ce{KMnO4}$ is added. The right column is after $\ce{KMnO4}$ is added.

In the resulting solution, $\ce{Mn}$ is oxidized to various states: $\ce{Mn^2+ Mn^4+ Mn^6+ Mn^7+}$ (where $\ce{Mn^7+}$ is the original form).
I would like to know, why do different oxidation states of $\ce{Mn}$ appear? How can we predict the concentration of oxidation states?

Maybe this answer is of any help.

Actually, $Mn^{7+}$ is reduced to $Mn^{6+}$, $Mn^{4+}$ and $Mn^{2+}$ and $SO_3^{2+}$ is oxidized to $SO_4^{2+}$.

They all appear because they are all finite amount of energy apart, but note that some concentrations are many orders of magnitude greater than others, and the concentration of $\ce{MnO4-}$ is so low that the table shows a zero.
Predicting the concentration requires being able to read tables of potentials. In standard tables like http://bilbo.chm.uri.edu/CHM112/tables/redpottable.htm  you will find equations such as $$\ce{MnO2{}+4H+ +4e-  \rightleftharpoons Mn + 2H2O}\quad (1.23\mathrm{V})$$ and $$\ce{Mn^{2+}{}+2e^-  \rightleftharpoons Mn}\quad (-1.185\mathrm{V})$$ (the numbers in parentheses are the standard reduction potentials, $E^\circ$, always measured in volts).
With this and the Nernst equation you can relate the concentrations of any set of species that form a balanced reaction.  In a complicated mix like this you don't know a priori the concentration of any species, only the certain sums (for instance, for all sulfur-containing ions) but you still can, perhaps iteratively, solve the corresponding set of algebraic equations and find the individual concentrations.  All the equilibrium equations have to be satisfied simultaneously; if they aren't the system is not in equilibrium.
(In this answer I've used the word "concentration" but the equations written with concentrations are only approximately satisfied; to account for second-order effects the notion of "activity" was introduced, and while it is an essentially empirical notion, it's still useful if the dependence of activity on concentration for a particular problem has been tabulated.)


text:
What is the extent of a reaction between an acid and its conjugate base?
What is the extent of say, this reaction?
$\ce{H_2CO_3 + HOCO_2^- <=> HOCO_2^- + H_2CO_3}$
K for the reaction appears to equal 1, and given the stoichiometry of the reaction, this implies that the reaction reaches an extent of 50% at equilibrium.
However, is this actually true?

Yes, K=1, but why do you say "this implies that the reaction reaches an extent of 50% at equilibrium."? K=1 doesn't imply anything like that.

Why not? For this reaction Keq = [product]^2/[reactant]^2 = 1, and if you take the square root of Keq, you get 1 = [product]/[reactant]. Only way for that to be true is for [product]=[reactant]. Now I do understand that for non-symmetric reactions K = 1 does not imply 50%.

I am sorry, but since [R1] = [P1] and [R2]=P2] the actual equqtion is K = [P1][P2] / ([P1][P2]) = 1, and it is true for any P1 and P2. You already rewrote this equation assuming that P1=P2, yet not assuming that the concentration of a material is same with itself.

The reaction you've shown is termed an "identity reaction", the products are the same as the reactants so $\ce{K_{eq}}$ must equal one.  In the particular case you've drawn the reaction involves a simple proton transfer and will likely be diffusion controlled.  Other examples of identity reactions are shown below.  That they are actually occurring can be demonstrated using isotopically labeled substrates.  The last one is one of my favorites, the degenerate rearrangement of bullvalene into itself via a Cope rearrangement.  My drawing of bullvalene doesn't do it justice, just google bullvalene and prepare to be fascinated.  If you number each of the ten carbon atoms in bullvalene you'll find that there are over 1.2 million possible arrangements.  Sequential Cope rearrangements will convert each of these arragments into every other possible rearrangement.  Connectivity in bullvalene is lost, each carbon atom is rapidly bound to every other carbon atom.  At a bit over 100 °C, both the proton and carbon-13 NMR of bullvalene show only 1 sharp peak.

Thermodynamics only applies to statistical behaviour over macroscopic quantities, rather than individual proton shuffles. When considered on this scale and with this level of abstraction, no reaction has taken place and so this isn't a meaningful question.

Excuse me, I don't follow what you mean by no reaction has taken place.

@Dissenter, the individual molecules consumed and formed are identical, the bonds broken and formed are identical (both intra- and intermolecularly), hence there's no change in enthalpy. There's also no net change in concentration of any species, and no change in physical states specified, so there's no change in entropy. $\Delta H$ and $\Delta S$ are both zero, so $\Delta G$ is zero. Since $K_{eq} = e^{\frac{-\Delta G}{RT}} = 0$ the reaction is at equilibrium, forward and backward reaction rates are equal, and there is no discernible macroscopic change.

@Dissenter, sorry, that should read $K_{eq} = 1$, not zero; apologies for the typo.


text:
Why do cotton clothes take a longer time to dry as compared to synthetic clothes?
This may be in relation to the $\ce{-O-H}$ group present in cellulose, but I would like to have a more specific answer.

I think this has something to do with the hydrophobic nature of synthetic clothing materials. Cotton "absorbs" water while synthetic clothing allow water to evaporate quickly, and like you say, it might just have to do with the hydroxyl group and hydrogen bonding.

Cotton fiber is primarily cellulose:

The red atoms are oxygen, and the white atoms are hydrogen. Whenever you have hydrogen attached to something extremely electronegative (such as oxygen), there is a potential for hydrogen bonding to occur:

As you can see, the electronegative oxygen atom draws electron density away from the more electropositive hydrogen atom, which results in a very polar bond. The positive charge on one molecule attracts the negative charge on another, and a very strong physical bond forms.
Now compare cellulose with something like nylon:

Since water can hydrogen bond to cellulose, it "sticks" much better - we say that it is hydrophilic. Since it can't form hydrogen bonds to nylon, we say nylon is more hydrophobic.
The same is true for most synthetic polymer fibers, and as a result, cotton absorbs water much more readily than other fibers, and tends to "hang on" to the molecules for longer - which means cotton fibers take longer to dry.

Cotton is very absorbent so the reason it takes longer to dry is physical not chemical. It has capillaries which is why it is so absorbent.

Do you have a reference for this?


text:
Balancing redox equation
So I had these two reactions that confused me .. I tried the standard way of balancing it (Putting oxygen number everywhere, +1,+2,-2 etc.) and then go on with it but it didn't work
So anyways here are the 2 reactions:
$$\ce{3 Cu + 8HNO3-> 3Cu(NO3)2 + 2NO + 4H2O}$$
$$\ce{3Ag + 4HNO3 -> 3AgNO3 + NO + 2H2O}$$
I've no idea how do you get those coefficients (numbers).. Please, is there an easy way to do it (perhaps it has to do with electrons added/given? ) and if so - where do I start?

This answer might help. Or this answer might help.

There are a few basic principles that are useful to keep in mind when you want to balance a redox equation.
1) OILRIG - Oxidation Is Loss, Reduction Is Gain (of electrons)
Let's look at the basics of the first equation:  $$\ce{Cu + HNO3-> Cu(NO3)2} + ?$$
On the left hand side we have Cu(0) metal and N(V), on the right hand side we have Cu(II) ion and N(V), so from OILRIG we know that the copper is being oxidised... but we know from OILRIG we have to put the electrons from the Cu(0) somewhere...
2) Hydrogen ions are always +1, and oxygen ions are always (there are a few exceptions but this is a good general rule) -2
... So the only thing that can change is the nitrogen. The nitrogen must be being reduced.
Nitrogen actually has a large number of possible oxidation states, taking any value between -3 and +5, so how do we decide which one it goes to in this case?  This depends on the temperature and the concentration of the acid. We can probably assume room temperature and bench concentration of acid (otherwise conditions need to be specified), and that the nitrogen isn't going to gain 5 electrons and become nitrogen gas, which means that we are going to form an oxide. Of the oxides, the most thermally stable is NO (nitrogen is in oxidation state +2 here).
What does that look like now?
$$\ce{Cu + HNO3-> Cu(NO3)2 + NO}$$
To balance the electron transfers, we need 3 Cu(O) going to Cu(II) to donate 6 electrons for 2 N(V) to go to N(II). Balancing for electrons and nitrogens, we now have:
$$\ce{3Cu + 8HNO3-> 3Cu(NO3)2 + 2NO}$$
Ok, now we need to look at balancing oxygens and hydrogens.
3) Use water if you can to balance oxygen and hydrogen. If not, start looking at hydroxide or hydrogen ions.
Here we need to somehow account for 8 hydrogens and 8x3 - 3x3x2 + 2 = 4 oxygens... That's 4 water molecules, so we finally have the answer you provided at the start!
$$\ce{3Cu + 8HNO3-> 3Cu(NO3)2 + 2NO + 4H2O}$$

Good explanation; I would add why one would use water to balance oxygen and hydrogen over hydroxide or hydronium ion (if that is the case, and in what conditions).


text:
Properties of powdered metals
I have seen certain metals such as Aluminium used in a situation where it is in a powdered form and "compacted" into a shape under intense pressure (in a mould), and it ends up looking mostly like solid metal. It is also heated afterwards, I assume to form a complete metallic bond.
I have a few questions on this:

How are oxide layers on the powder affected when it is compacted? Eg. Can the $Al_2O_3$ layer, or at least the O atoms from them, be "inside" the moulded object when it is compacted?
Metals in a powdery form aren't as reflective as in their "solid" form (also please let me know the right word for that). Why is this?
Are most metals in their powdered form pure, eg. Not like a salt, but a single element?
I know gold powder, obtained from chloroauric acid, becomes gold in its metal form when melted down. It looks brownish and takes up far more volume than the gold it contains. Is this pure Au or a compound/salt?
Also, in 4. with the gold powder, it actually forms rather large clumps from the chloroauric acid. How is this possible assuming it's a pure element?

Thought I better get all the questions done in one go. Thanks in advance

Start with a search for powder metallurgy- note its not limited to metals- it also combines metals and ceramics such as AL2O3. "solid" form- the usual term is dense where powder metallurgy has porousity.

Hi baharini, I think you should split your question in this form really require too much information to be answered....

Let me try to answer the part 1. of your question.
The process of compacting powdered Aluminium is also called Sintering. And as you can guess, the thin oxide layer of Alumina inhibits the binding of Aluminium grains between the powder particles.
So, while sintering, we need to either:
1) Break part of the Alumina film or,
2) Reduce part of it to Aluminium,
 in order to facilitate formation of a metallic bond between the powder particles.
For 1), we use a nitrogen atmosphere as it enhances the destruction of Alumina.
For 2), we use because magnesium as a alloying element because magnesium, even in small amounts, concentrates on the surface of powder particles
and supports sintering by local reduction of Alumina.

In conclusion, I would say that compacted Aluminium does have Alumina inside the molded object. But metallic bonds between powders are formed after partly removing the oxide film by the above mentioned methods.


text:
Bonding in the phosphate ion
I'm looking for an explanation of the bonding in the phosphate (PO43−) ion:

(Image courtesy of Wikipedia)
Phosphorus (15P) - being the fifteenth element - has fifteen electrons, five valence electrons and the following electron configuration:
1s2
2s2 2p6
3s2 3p3
This being so, by what means does it form its bonds? Naively, one might expect it to form 3 single covalent bonds and be happy. But it doesn't, it forms five bonds, presumably using each of its five electrons. My first question is - why would it do this? It seems to me that this would result in the Phosphorus atom having ten electrons in its outer shell (four from the double bond with the oxygen and two each from the single bonds with the other three oxygens).
Apart from the oxygen with a double-bond, we have three oxygens forming single bonds. Does this not leave them one electron short?
Finally, why are the oxygens shown as carrying a negative charge?

First of all, let me state the obvious: Phosphorus is awesome.
After we got that out of the way we can focus on why.

There are many different modifications of phosphorus in nature. With increasing thermodynamic stability they are $$\ce{P_{white} -> P_{red} -> P_{violet} -> P_{black}}.$$
Apart from this there are many low molecular different allotropes, like $\ce{P4 (white)}$, $\ce{P6}$, $\ce{P8}$, $\ce{P10}$, $\ce{P12}$. And because that is not enough, there are chain like polymeric allotropes, too.
Apart from this it is possible to formulate many different cations and anions, that are derived from the above molecular structures. Just to name a few, there are $\ce{P3+, P5+, P7+, P9+}$ mainly observed in gaseous phase an $\ce{P^{3-},P2^{4-}, P3^{5-}, P4^{2-}, P7-,...,}$ usually in combination with alkali metals. Most amazingly it may form anionic polymer strains of the general form $\ce{[P7-]_{\infty}, [P15-]_{\infty}}$.
But now for the most important part, phosphorus has stable oxidation states in compounds, ranging from $\mathrm{-III}$ to $\mathrm{+V}$. Here are a few examples:
$$\ce{\overset{-III}{P}H3,\ \overset{-II}{P_2}H4,\ [\overset{-I}{P}H]_{n},\ \overset{\mathrm{\pm0}}{P4},\ H3\overset{\mathrm{+I}}{P}O2,\ H4\overset{\mathrm{+II}}{P2}O4,\ H3\overset{\mathrm{+III}}{P}O3,\ H2\overset{\mathrm{+IV}}{P2}O6,\ H3\overset{\mathrm{+V}}{P}O4}$$
While dealing with these compounds it is usually completely unnecessary to describe bonding with hybrid orbitals.
In case of phosphane $\ce{PH3}$ it would be wrong. Assuming $\ce{{}^{sp^3}P}$ one would expect $\angle(\ce{H-P-H})\approx109^\circ$, while it is found to be $\angle(\ce{H-P-H})=93.5^\circ$, which is almost the same angle as the $\ce{p}$ orbitals are having towards each other.
In general your assumption is correct, that it is possible to form only three covalent bonds to reach a stable configuration. And that will most likely be the case when phosphorus forms compounds with more electropositive elements.
Now dealing with oxygen, means dealing with a much more electronegative element, i.e. $\ce{En(O)}\approx3.4$, $\ce{En(P)}\approx2.2$. That also means that bonds are much more polarised towards the oxygen.
Analysing the phosphate anion $\ce{PO4^{3-}}$ it is crucial to recognise its symmetry, which is tetrahedral $T_\mathrm{d}$. In this arrangement it is perfectly safe (but not at all necessary) to describe phosphorus as $\ce{sp^3}$ hybridised.
A Natural Bond Orbital analysis (BP86/def2-TZVPP) reveals that there are 4 equal $\ce{P-O}~\sigma$ single bonds and each oxygen has three lone pairs. The contribution of the $\ce{d}$ orbitals to bonding is well below $1\%$ and can be interpreted as numerical noise (use as polarisation functions) of the DFT method.
\begin{array}{rlrr}\hline
       &             &\mathrm{\%P (hyb)}  &\mathrm{\%O (hyb)}\\\hline
3\times&\ce{Bd(O-P)} & 24 (\ce{sp^3}) & 76 (\mathrm{sp^{2.3}})\\
       &\ce{Lp(O)}   &                & 100 (\mathrm{sp^{0.4}})\\
2\times&\ce{Lp(O)}   &                & 100 (\ce{p})\\\hline
\end{array}
This is consistent with the partial charges, i.e. $q(\ce{P})=2.2$, $q(\ce{O})=-1.3$. Therefore a more accurate Lewis formula is with charge separation.

The corresponding NBO reflect the bonding picture one would expect given all details from above. It should be noted, that natural bond orbitals are a linear combination of the canonical orbitals and do not have a physical meaningful energy eigenvalue. The two top rows represent the $\ce{p}$ lone pair orbitals of oxygen, the third row represents the $\ce{sp^{0\!.4}}$ lone pair orbitals, the fourth row gives the $\sigma$ bonding orbitals. (The last row is the orientation of the molecule, core orbitals are not displayed.)

The corresponding canonical orbitals which have a physical meaningful eigenvalue are delocalised over the whole molecule, hence they are not providing a simple understandable bonding picture. While NBO fail to respect the symmetry point group, canonical orbitals are constructed to obey this principle. (Here shown from highest energy, top, to lowest energy, bottom, core orbitals not displayed.)

But since each oxygen has a negative formal charge, wouldn't phosphoric acid instead be H4PO4, rather than H3PO4? Since each negatively-charged oxygen can bond with one hydrogen

@TanYongBoon then you would have charge imbalance and it would be $\ce{H4PO4+}$, but in principle I see no reason why it couldn't be protonated there, too.

Ok. But which would be the dominant species in aqueous solution? It would still be H3PO4 as H4PO4+ would be a strong acid and fully dissociate in water to give H3PO4? Not sure of what I'm saying...

@TanYongBoon In aqueous solution of $\ce{H3PO4}$ there will be multiple equilibria with all protonation stages present. The dominant species will probably be $\ce{H2PO4-}$ or $\ce{H3PO4}$. Very unlikely without another acid is $\ce{H4PO4+}$. Without another base, $\ce{HPO4^2-}$ and $\ce{PO4^3-}$ are also quite unlikely. The formal charges are just a book-keeping tool, in the phosphate ion they are delocalised. But when you add protons, then they will be counteracted.

Phosphorus ($\ce{_{15}P}$) - being the fifteenth element - has fifteen
  electrons, five valence electrons and the following electron
  configuration:
$\ce{1s^2,~2s^2,~2p^6,~3s^2,~3p^3}$
This being so, by what means does it form its bonds? Naively, one might expect it to form 3 single covalent bonds and be happy. But it doesn't, it forms five bonds, presumably using each of its five electrons. My first question is - why would it do this? It seems to me that this would result in the phosphorus atom having ten electrons in its outer shell (four from the double bond with the oxygen and two each from the single bonds with the other three oxygens).

Very observant! What happens is that phosphorus is hybridized in the phosphate ion. Hybridization allows the central phosphate atom to form more bonds than you might expect from its ground state electron configuration.
Within the framework of hybridization, one may look at the bonds in the phosphate anion as having a great deal of ionic character.
The electronegativity of phosphorus is 2.19; the electronegativity of oxygen is 3.44. The difference is 1.25 - or rather sizable. As a result, electron density is largely withdrawn from the central phosphorus. Hence the negative charges you see on the three oxygen molecules in your picture.
Other depictions, which reject a large degree of d-orbital utilization in hybridization, might give a negative formal charge to all the oxygens and a positive formal charge to the central phosphorus by eschewing the $\ce{P=O}$ double bond.

Apart from the oxygen with a double-bond, we have three oxygens
  forming single bonds. Does this not leave them one electron short?
Finally, why are the oxygens shown as carrying a negative charge?

You could draw double bonds for all the oxygens but there aren't that many valence electrons to be distributed. The structure shown above is a compromise between number of valence electrons and electronegativity considerations - we wouldn't want a surfeit of electrons around the central phosphorus since that's not as thermodynamically favorable as having the more electronegative oxygen support these electrons.
In addition, the sum of the formal charges on an atom must add up to the overall charge on the atom. Phosphate anion is triply negative charged; note how there are three oxygens bearing a single negative charge.

The oxygen's are not 'short' of electrons if you're referring to the filling of their shells. The sigma bonded oxygens have 1 bonding pair and 3 lone pairs which gives a full valence shell. The negative formal charge occurs because oxygen has 6 valence electrons in the ground state and 2/2+6=7 electrons in this structure. 6-7=-1

Thank you very much, both of you, for some very detailed replies. I'm going to try to synthesise them into a single answer, which I hope will be accurate. Here is how I understand it, armed with the new knowledge contained within the posts above:
In the phosphate ion we have a central Phosphorus atom, with five valence electrons.
This is bonded to four oxygen atoms, which have six valence electrons.
Five P electrons plus 4 times 6 O electrons gives 29 electrons. We also have a charge of 3- indicating that we need to take into account 3 additional electrons, giving us 32 electrons overall.
We could model this situation by proposing that the P atom exists at the centre, with four O atoms bonded to it. One O is double bonded to the P and the other three Os are single bonded, giving 10 electrons to the P (which it is happy with).
In the case of the double-bonded O we would have 2 lone pairs of electrons (makes 4) and a further 4 electrons from the double bond. Oxygen has 6 valence electrons and so the 2 lone pairs are its own electrons and its other 2 valence electrons form the double bond with two of the P electrons. The double-bonded O has eight electrons and is happy.
In the case of each single-bonded O we would have 3 lone pairs (presumably the oxygen's own six valence electrons?) plus an extra electron (presumably from elsewhere? I don't really understand this) which participates in a single bond with a P electron. This gives us eight electrons, which the oxygen is happy with.
This is the model shown in the wikipedia drawing:

Now, if we draw the Lewis structure:

We can see all of this put together and we can also see that, while the double-bonded O is neutral, having six electrons assigned to it, the three single bonded O atoms have seven electrons assigned and therefore carry a -1 charge.
There is no reason, however, why we should select the top O atom as the one to have a double-bond. This is not at all what we find in reality.
We can build a simplified model of reality, where the double-bond (and its electrons) is exchanged between each O atom:

And where there is no double-bond at all:

All of this, however, is really nothing more than an extreme simplification. The reality is a hybrid. It is possible to propose that we have four sigma bonds with four sp3 hybridised orbitals from the phosphorus and a fifth delocalised pi bond with the d orbital of the phosphorus. This gives our P 10 electrons and each O 8 electrons.
But the reality seems to be that, in fact, the d orbital is not involved at all and instead we have four sigma bonds formed by sp3 hybridised orbitals from the phosphorus.
And here's what I just don't understand. When I drew the Lewis structure, I added three extra electrons from somewhere. This would make sense if, as martin says, each oxygen has three lone pairs, we can imagine a sea of the five P valence electrons and these extra three electrons spread evenly with two forming each sigma bond. This gives each oxygen eight electrons and it would give the P eight electrons too. The higher electronegativity of Oxygen means that the O atoms attract the electrons more strongly and each atom receives a negative charge, whereas the P attracts them less strongly and receives a positive charge. I'm really not sure I've understood the arrangement of electrons though, I would like to know how the electrons are apportioned in the ion, where they came from and how the overall charge comes about.

You should ask another question. In short: It does not matter at all where the electrons came from. For each anion you will most likely have an cation and that can be the source for the extra electrons.

I would like to clarify, that in the phosphate anion all oxygens are equal, no double bonds are formed. Phosphorus and oxygen both fulfill the octet rule. The bonding picture I provided in my answer in no simplification, it is the best representation for this molecule. A Lewis structure, however, is a very crude simplification.

Simply put: To make PO4 you must start with a neutral P and three O with a -1 charge and one O with a neutral charge.
The phosphorus can make 4 bonds by hybridizing
The phosphorus hybridizes to sp3 by loosing one electron to the neutral O. This creates four oxygen atoms that have a -1 charge and one Phosphorus atom with a +1 charge total charge -3
Once the bonds are formed they are all equal and the molecule is stable.
It concerns the valance electron availability the neutral O has 6 and gains one while the other three O -1 have 7 each
and the P has 5.
total of 32
When the P hybridizes it then has 4 valence electrons
 while the neutral O gains one valence electron, thus achieving 7.
the total remains the same.
It doesn't mater what position the neutral O attaches to.
The neutral O will have transferred the electron before attaching.
In phosphoric acid H3PO4 phosphorus undergoes hybridization to produce 5 valence electrons by transferring an electron to a higher hybrid level.
Phosphorus gains stability by rearranging its valence electron shells.
Phosphoric acid striped of the hydrogen protons will leave you with phosphate.

I happen to look at this site just now.
The reason is all a matter of hydridization.
Look at the electron configuration of Phosphate and that of the next element that use next orbital i.e. K.
The hybridization would bring in 3s, 3p, and 4s together. To complete fill them, number of electrons or bonds required would therefore be 5.
I guess that there are 5 sp3 hybrid orbitals.
Because of electron negativity of Phosphorus and oxygen, there are negative charge on the oxygen atoms that are singly bonded, while the doubly bonded oxygen, being closer could have insignificantly charged negatively.

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text:
Why does the cyclization of open-chain glucose occur via the C5 hydroxyl group?
In converting glucose from its acyclic (open-chain) form to either cyclic form (alpha or beta) why is the C5 hydroxyl group used instead of the bottom-most hydroxyl group?
Also, are these structures considered isomers? All sources I've seen only say "form" or "representation."

Due to the internal ring strain the best possible conformation is a six-membered heterocycle in this case called pyranose . A five-membered heterocycle may in rare cases also be formed, called furanose. (From left to right: α-D-Glucopyranose, β-D-Glucopyranose, α-D-Glucofuranose, β-D-Glucofuranose; courtesy of wikipedia)

Including the linear chain form they are all isomers (wikipedia), as they have the same chemical formula. More specifically the relationship between $\ce{linear <-> pyranose <-> furanose}$ is called constitutional or structural isomer.
The $\ce{\alpha-D <-> \beta-D}$ forms are diastereomers, or more specific diastereomeric conformers.

The cyclic form of glucose uses the C5 oxygen for bonding since it forms a 6 member ring (more stable then either 5 or 7 member ring).
They are isomers since they have the same chemical formula.


text:
Finding overall order of reaction
Recently I've been going through an exercise of Chemical Kinetics and Nuclear Chemistry and right now I am stuck on one question that was previously asked in JEE(Advanced)-2013. The question says,

In the reaction,
$\ce{P + Q -> R + S}$
The time taken for $75\%$ reaction of $\ce{P}$ is twice that time taken for $50\%$ reaction of $\ce{P}$. The concentration of $\ce{Q}$ varies with reaction time as shown in the figure. What is the overall order of the reaction?

And here was my approach to the question,
In the question different data have been given for different reactants (i.e. $\ce{P}$ and $\ce{Q}$). For $\ce{P}$ it has been given that it take $\ce{P}$ double the time to react $75\%$ than that from $50\%$. So assuming the time taken for $50\%$ reaction to be $t$ and that of $75\%$ to be $2t$, we can easily find that order of reaction w.r.t P is $1$. As,

$$\frac{t_{\frac12}}{t_{\frac34}}=\frac{\log 2}{\log 4}=\frac12$$

Further w.r.t $Q$ there is a graph given which is something like in form $\ce{[Q]=[Q_o]-mt}$. So comparing with the equation of zero order reaction which is like $\ce{[A]=[A_o]-kt}$ (where $\ce{[A]}$ is the concentration of reactant and $k$ is rate constant), we can conclude that order of reaction w.r.t to $Q$ is $0$.
So I concluded that overall order of reaction will be

order of reaction of $P +$ order of reaction of $Q$
$\implies 1+0=1$

So was this the right approach from my side or there is any kind of error in any step?

The question seems self-contradictory to me.  If P and Q are actually reacting stoichiometrically with each other, and [Q] decreases linearly with time, then [P] must decrease linearly with time, which contradicts the "time taken for 75% reaction of P is twice that time taken for 50% reaction of P" aspect of the question.

@DavePhD it's not necessary that the order of reaction w.r.t each reactant will be equal to it's stoichiometry. Check this http://www.chemguide.co.uk/physical/basicrates/orders.html

I agree that the order of an overall reaction with respect to a reactant does not have to be the reactant's stoichiometric coefficient.  What I am saying is if P is actually reacting with Q, then the observed d[Q]/dt will be c(d[P]/dt) where c is a constant.  In this question, after a certain time [Q]=0.  For [P] to decrease exponentially, and therefore continue reacting for an infinite amount of time, P and Q are necessarily not reacting with each other.

@DavePhD: What are your thoughts on this reaction:
$\ce{ArN2+ + X- -> ArX + N2}$ ?
This reaction is first order with respect to $\ce{ArN2+}$ and zero order with respect to $\ce{X-}$
Source:
https://en.wikipedia.org/wiki/Order_of_reaction#First_order

@Abcd If there is no X-, the reaction will not occur.  It will appear to be first order, until [X-] is very low.  Then the second order nature will be observed.  It is pseudo first order. http://cbc.chem.arizona.edu/~salzmanr/480a/480ants/pfo3oarr/pfo3oarr.html

Ah @DavePhD but there are so many reactions like this .. are all of them "pseudo" first order? I think this should be a question on main.

@Abcd yes, ask a new question about that.  But certainly a reaction can not proceed when a required reactant has zero concentration.

@DavePhD We are talking about order wrt a reactant not concentration of any reactant ... Anyway, I'll ask tomorrow surely.

As summarized in the IUPAC definition of reaction rate, the rate of change of one reactant must be directly proportional to the rate of change of each other reactant.
If the concentration of one reactant decreases linearly, the concentration of every reactant must decrease linearly (the reaction rate is a constant, and the overall reaction order is zero).
The text of the JEE question contradicts its Figure (which shows linear decrease of [Q]), by saying that the "time taken for 75% reaction of P is twice that time taken for 50% reaction of".
Instead, since [Q] decreases linearly, [P] must decrease linearly, and if 75% of an initial [P] reacts it must take 1.5 times as long as for 50% to react.
It's likely that the author of the question intends that the answer is "first order", but really there is no correct answer.


text:
Why does mercury have low melting and boiling points?
Many metals have relatively high melting and boiling points, but mercury has relatively low melting and boiling points. What are the possible reasons for this?

http://onlinelibrary.wiley.com/doi/10.1002/anie.201302742/abstract

Mercury is different!  It is not as reactive as its neighbors in the Periodic Table, it doesn't conduct heat and electricity as well as other metals, and it is a liquid unlike other metals.  The electronic structure of mercury is $\ce{[Xe] 4f^{14} 5d^{10} 6s^2}$, so the first thing we notice is that all of its orbitals are full, there are no unpaired electrons in mercury.  This helps explain mercury's reluctance to react with other materials or form bonds with other mercury atoms.  Most metal atoms share their outer electrons with other metal atoms, in fact, all of these "shared" metal electrons exist as a diffuse "sea" of electrons.  It is the extended nature of this sharing that makes metals good conductors of heat and electricity.  The fact that the metal electrons are shared provides a strong bonding interaction between metal atoms that gives metals their solid structure.  None of this applies to mercury.  With its filled shell electron configuration it is very reluctant to form bonds even with other mercury atoms.
One additional, interesting effect makes mercury even more reluctant to share its electrons.  Special relativity suggests that for larger nuclei (and mercury with atomic number 80 is large enough for the effect to apply) electrons will begin to travel closer to the speed of light.  In turn, their mass will increase and as mass increases, orbital radius decreases.  The electron is closer to the nucleus, more attracted to the nucleus and less available for bonding.  Here is a nice reference that explains this relativistic effect and how it applies to mercury in plain English and a bit more detail.
These two effects, filled outer electron shell and contraction of the outer orbital closer to the nucleus, combine to make mercury reluctant to form bonds, even with other mercury atoms.  It is this lack of electron bonding between mercury atoms that makes it melt and boil at such low temperatures.

Is the second effect, actually the inert pair effect?

@Marko  Yes, it is.

This qualitative explanation would make one suppose that copernicium should have even lower melting point. But in practice, it's predicted to be higher: 283 K (mercury melts at 234 K).


text:
What is defined by part by weight in the definition of equivalent weight?
According to my book, equivalent weight is defined as the ‘number of parts by weight’ of an element that interacts with 1.008 parts by weight of hydrogen. It is dimensionless while gram equivalent weight has gram as units. But in https://en.wikipedia.org/wiki/Equivalent_weight, equivalent weight and gram equivalent weight are same. I’ve two questions: Which one is correct, my book or Wikipedia? And what is ‘number of parts by weight’?

The term "parts by weight" is how you would specify in a recipe the relative proportions of the ingredients. So if the recipe called for one part by weight of flour and two parts by weight of sugar, you could use 1 ounce of flour and two ounces of sugar (to make a small batch) or 1/2 pound of flour and 1 pound of sugar to make a large batch.
So the "recipe" for water calls for exactly 1.008 parts by weight of hydrogen and 8.000 parts by weight of oxygen. If you use too much of either ingredient, there will be an excess of it after the reaction is run.
"Parts by volume" would apply if you were combining liquids and measuring them by volume rather than by weight.


text:
Electronegativity of Potassium (Pauling)
Just a little question, I was taught that the electronegativity of potassium[K] is 1, but I recently discovered most internet resources say 0.82. Some even say both. What is the actual electronegativity of potassium, 0.82 or 1?

It depends which electronegativity scale you choose to use.  There are a lot of different scales for comparing electronegativities of the elements.  I've included a list of 11 of them below if you're interested.  In any case here are the electronegativities reported for potassium using 5 different electronegativity scales.  All values are in "Pauling units" so that the numbers can be compared.  The Pauling scale is the most commonly used, so 0.82 would be a reasonable estimate for the electronegativity of potassium.

Pauling 0.82
Sanderson 0.45
Allred Rochow 0.91
Mulliken-Jaffe 0.73
Allen 0.734

On the Pauling scale, electronegativities range from around 0.7 to 3.98 (fluorine), so as expected, potassium is at the electropositive end.  None of these scales report a value of 1.
List of 11 Different Electronegativity Scales

Pauling Scale: (1932)  Obtains values by thermochemical methods
Mulliken Relation: (1934)  Defines a relation that depends upon the
orbital characteristics of an atom in a molecule. Mulliken
electronegativity is the numerical average of the ionisation
potential and electron affinity
Gordy Scale: (1946)  Defines electronegativity in terms of the
effective nuclear charge and the covalent radius. (Zeff)e/r
Walsh Scale: (1951)  Relates electronegativity to stretching force
constants of the bonds of an atom to a hydrogen atom.
Huggins Scale: (1953) Alternative to Pauling's thermochemical
procedure
Sanderson Scale: (1955)  The ratio of the average electron density of
an atom to that of a hypothetical "inert" atom having the same number of electrons. This ratio is a measure of the relative compactness of the atom
Allred-Rochow Scale: (1958)  Defines electronegativity in terms of the effective nuclear charge and covalent radius. Like the Gordy scale but uses (Zeff)e/r2
Jaffe Scale: (1962)  Uses the electronegativity of orbitals rather
than atoms to develop group electronegativities for molecular fragments (e.g. $\ce{CH3}$ vs $\ce{CF3}$) that take into account the charge of a group, the effects of substituents, and the hybridization of the bonding orbital
Phillips Scale: (1968)  Defines electronegativity in terms of the
dielectric properties of atoms in a given valence state
Martynov & Batsanov Scale: (1980) Obtained by averaging the
successive ionisation energies of an element's valence electrons
Allen CE Scale: (1992) Configuration energy (CE), the average
one-electron valence shell energy of the ground-state free atom, is
used to quantify metal-covalent-ionic bonding


text:
How do you perforate a teflon sheet?
I have a sheet of thin unperforated teflon that I need to perforate by tomorrow for a lab I work in. None of my coworkers know how to perforate the sheet. Does anyone know how I can do this using basic tools from my house?

Can you provide some further infos about the number of holes and the diameter and the application?

You should contact a specialized company, however you should also provide more infos about the diameter and the number of the holes you want. This is an excerpt from FedTech:

laser cutting is the method of choice for small perforation, as laser
  cutting has a smaller, tighter cutting tolerance and thus has the
  capability to create tiny, intricate perforations. However, if you
  prefer, abrasive waterjet cutting can be used as well but for larger
  perforation (.030" and up, respectively). If you prefer to use
  non-abrasive waterjet cutting, you can create perforations up to .004"
  in diameter, however this form of waterjet cutting is restricted in
  cutting certain types of material. With non-abrasive waterjet cutting,
  you can cut up to about 12" thick from the following: plastic, rubber,
  foam, composites, silicone, foam, felt, cork, Teflon (PTFE), neoprene,
  and other soft materials


text:
What is relationship between molecular weight distribution of PE and its strength?
I'm trying to figure out the relationship between the molecular weight distribution of a polyethylene plastic and its strength and heat sealability.
(The molecular weight distribution is described as being narrow, which apparently is helpful in terms of strength and heat sealability).

Do you mean low density Polyethylene or high density polyethylene? Strength is related to degree of crystalinity- higher percentage, higher strength. Narrow distribution should prefer higher percentage but processing come into determining that too.

I have found an old paper that might answer your question, which is:
"The Effect of Molecular Weight Distribution on Polyethylene Film Properties"
P. J. Perrono, P. B. Lederman, Polymer Egineering and Science, 1972, 12, 5, 340-345 (DOI: 10.1002/pen.760120505).
The properties of polyethylene evaluated in this paper as a function of the molecular weight include film gloss, haze, tear resistance and film impact strength.
This study shows that broadening the molecular weight distribution increases haze and reduces film gloss. Further, it was shown that a linear relationship exists between film gloss and external haze. Both values are measures of surface irregularities in the film which are affected by the drawing characteristics of the polymer. A broader molecular weight distribution results in increased impact strength as measured by the Dart Drop Impact Test. This is, it is believed, a result of the increase in long chain branching of the higher molecular weight fractions of the polymer which cause a higher degree of molecular weight entanglement at the branch sites. In contrast the tear strength is reduced as the molecular weight distribution broadens because of the low molecular weight fraction in the broad spectrum material which tend to decrease resistance to tear.
Hope this helps.


text:
Double Deionized Water and Superheating
I was told that dd-water will superheat when heated up in a microwave because of a lack of ions.
1) Doesn't water auto-ionize? I asked this; the answer was it auto-ionizes "too slowly" in dd-water to prevent superheating.
2) This begs the question: what are the kinetics of the auto-ionization of water? What's the rate?
3) This also begs the question - what is superheating? What does ions have anything to do with superheating?

This has a lot to do with nucleation.  Just like crystallization and initial rain drop formation, for boiling to occur an initial site within the solution for a bubble of vapor to form on is required. So if you have very clean water that has been deaerated in a highly polished vessel there is an increased chance for superheating to occur due to the reduction in potential nucleation sites.  Tap water can superheat too, it is just less likely to do so because it typically contains more impurities that can serve as nucleation sites. This why boiling chips are commonly used in lab distillations.  They provide a site for vapor nucleation and usually prevent superheating of the material to be distilled.  Introducing other sites for nucleation, such as dissolved bubbles, solid contaminants, rough spots on the wall of the container all serve to lessen the probability of superheating.  I'm not sure what ions would have to do with the superheating process other than they are often representative of microscopic solid impurities that can provide that surface for the initial nucleation event.