text:
Why is salt needed when using vinegar to clean pennies?
Let's say you have a solution that is vinegar and it has salt (NaCl) dissolved in it. Then you place old dull pennies in it. Then you look at it 5 minutes later and the pennies are clean and looking like new again.
I understand acid can etch the copper compound but why is the salt necessary for this process and what exactly does it do?

I do not believe that the salt is necessary, but you should definitely try this at home. Copper(II) oxide is not soluble in water by itself, but Copper(I) oxide is soluble in acid. Vinegar is mostly water but a small percent of it is acetic acid. - You can find that under solubility on wikipedia.

The salt might increase the conductivity and facilitate a redox process. Or it could be there for the common ion effect. $\ce{CuCl2}$ is much more soluble than $\ce{CuO}$. @MDB Does the solution turn blue?

I'd say the sodium chloride is just a catalyst to speed up the reaction given how dilute vinegar is. It forms CuCl2 in minute quantities which enables the neutralisation to speed up.

This experiment has been done literally millions of times by high school and middle school science students and the result is very consistent. Salt plus water plus vinegar is the combination which cleans the copper. THAT is an extremely strong experimental argument for why adding salt to vinegar shifts the equilibrium (dramatically). It's also common knowledge that HCl (hydrochloric acid) is stronger per mole than acetic acid. Try it yourself, it's very easy to verify.

The copper surface is oxidized, which slows down the formation of copper acetate. When you add chlorides to the solution, the chloride penetrates the oxide layer and creates a hole in that layer that allows acetate to react with the copper. To find more information on the effects of chlorides on the surface of copper look up copper corrosion and chlorides.

Hi and welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site and check out the [help] for further questions about it.

Hi Dale, I noticed that this reaction caused the release of gas. What is it's chemical composition? I assume it's not dangerous considering I've seen this experiment shown on multiple kid's science websites.

So salt only affects copper oxides?  Or iron rust also?

I think that what is happening, is that the salt adds sodium ions to the solution, thereby forcing the equlibrium in dissosiation of acetic acid to acetate and H+ ions further to the right.
It happens because Na gives it's electrons easier away to water than Cu does, and because acetic acid is a weak acid, normally giving only 1 of 100 H+ ions / acetate ions per 100 molecules of CH3COOH.
An exess of salt will keep the copper clean, even with no acid, and force Cu to first release the green Cu2+ from it's surface, and next for solid copper itself to oxidice, slowly.
So You can clean copper by just adding salt and water, even though the process is slow.I have tried this with copper nails. In a soultion of of water and salt, where salt remains solid at the bottom, the nails are still shiny orange, and the solution has become almost emerald green, after a couple of months.
I am not sure of the reaction mechanism here, but I have become more convinced that what I claim above really is happening.
Try doing this: Add liquid soap to two glasses, and pour in the same amount of water. Stir the solution. Add the same amount of vinegar as water and soap. Watch what happens. Then add salt (sodium chloride), but just to one of the glasses. Stir and wait for 15 minutes. Observe the very different results.
"You are the salt of the Earth ..."

Salt lowers the pKa of acetic acid? Do you have a source for that knowledge?

The salt is added to vinegar to speed up the process.
Addition of NaCl to vinegar shifts the equilibrium of acid dissociation, i.e. $\ce{CH3COOH <=> CH3COO- + H+}$, to right therefore concentration of $\ce{H+}$ & $\ce{CH3COO-}$ increases. Due to this, $\ce{NaCl}$ reacts with $\ce{CH3COOH}$ to give $\ce{HCl}$
$\ce{NaCl + CH3COOH <=> CH3COONa + HCl}$
Cooper Oxide (base) reacts with acids to form water and copper salts.
Acetic acid ($\ce{CH3COOH}$), which is a weak acid, will react slowly with Copper oxide as compared to $\ce{HCl}$ which is strong acid.
So when use vinegar-salt solution to clean penny, Copper Oxide parellely reacts with HCl and CH3COOH both and gives Copper salt and water.
$\ce{CuO + CH3COOH -> 2H2O + Cu(C2H3O2)2}$
$\ce{CuO + 2HCl -> H2O + CuCl2}$

I'm still not entirely convinced that adding NaCl will shift the equilibrium. It's definitely possible, but I don't see a good argument for why it should happen that way.

@chipbuster, I'd be strongly inclined to agree. Judging by pKa, HCl is around twelve orders of magnitude more acidic than HOAc, so even in a totally saturated solution the quantity of HCl generated would have to be miniscule. Unfortunately, I can only hazard a guess myself as to the reason that addition of salt accelerates the dissolution. If the solution is saturated, then the lower concentration of water itself could possibly drive the equilibrium forward by favoring dissolution and formation of water from the oxide in the process. That's entirely speculative, though.

There is a slight effect due to the Debye-Huckel effect.

To a first order approximation though, how strong would such an effect be?

still it will show enough effect because HCl is much more acidic than acetic acid.

I'm wondering how large the effect would be of of the shift. If we get 1ppm HCl, that does us no good.
I'm also curious as to why you don't get $\ce{CuCl2}$ if HCl is doing any significant portion of the work. Copper is most certainly not a hard acid.

It would increase the ionic strength of the solvent, hence favouring the formation of ionic species more, and that is what would shift the equilibrium for acetic acid <-> acetate + H+. Usually it's a small effect.

As @Brian said, there is an effect on the apparent pKa, but it has to do with the ionic strength changing the activity coefficient (cf. Pitzer model or Specific Ion Interaction theory), not the formation of HCl or NaOAc. The magnitude of the effect is only 0.1-0.2 units on pKa. (Plotted here: https://www.researchgate.net/figure/The-dependency-of-the-pKa-of-acetic-acid-on-the-ionic-strength-at-18-C-in-water-Data_fig3_256333708). More detail can be found in this book chapter: https://link.springer.com/chapter/10.1007%2F978-1-4419-8690-0_6.

Copper(II) oxide is basic and reacts with acids to form a copper salt and water.
A strong acid, like HCl, reacts faster than a weak acid - like acetic acid (which is the main acid component of vinegar).
If one adds NaCl to vinegar, you get HCl:
$$\ce{CH3COOH +NaCl<=>CH3COONa +HCl}$$
Copper(II) chloride dissolves in the water.
So, the salt is added to speed up the process.

The equation above isn't correct.  The sodium chloride is totally ionized, the acetic acid is only partially ionized and the amount of hydrogen ion in solution is thus relatively small.

Thank you, @PaulJ.Gans Essentially what is required is more H+ in solution. I think that the addition of NaCl might shift the equilibrium of acetic acid dissociation towards the right is this equation: $$\ce{CH3COOH<=>H+ +CH3COO-}$$ so that you get more H+.

To a first approximation the addition of $\ce{NaCl}$ won't shift the equilibrium at all.  There is a slight effect due to the Debye-Huckel effect, but that's mostly negligible.


text:
Homemade Rocket Fuel and Engine
I want to put an action figure in orbit but I need to know how to make a secure yet not too  expensive rocket fuel engine. The design of the engine must very primitive and not too complicated.
I've read some information about hydrazine and hydrogenperoxide but it is too dangerous to use and make at a noncommercial scale. Although I studied some chemistry in highschool, I don't have so much knowledge about this matter. So far my best bet is blackpowder based fuel engine. What do you suggest?

http://www.wired.com/2011/12/to-build-a-diy-spacecraft-is-a-daunting-task/ This is the perfect place for you to find out the right answer !

The best way to get into space without anything complicated is to use a balloon. Getting into orbit requires sophisticated control systems to manage the rocket thrust: just having powerful thrust doesn't do it; the fuel isn't your limiting factor.

Note: I'm not a modeler myself, but tested several recipes of rocket candy and read rocket modeler forum for a while.
First of all, you have to think if you will be satisfied with solid propellant or want a liquid fuel engine.
Building home-made liquid fuel rocket engine is very hard. One need to find pumps, make combustion chamber from something very heat-proof and reasonably stable chemically. Anyway, some successful reports may be found on the internetz. Note, however, that if you can choose safe liquid fuel, like kerosene or some alcohol, there is no safe oxidant for liquid-fueled rocket engine. The closest thing to safe oxidant I can imagine is nitric acid, that require work on fresh air or in specially designed room and tends to form explosives on contact with many things, like milk protein (yep!), some of which are volatile
Solid-fuel model propellants are much easier to prepare and use, there are numerous articles in the internetz about such fuels. The most common one is rocket candy, prepared from potassium nitrate and sugar. I tried it to see how it works (I didn't build any rockets), but instead of sugar I used sorbitol. Rocket candy based on sorbitol can be melted and then can be casted in any form. wiki article contains both info and links to sites with well-testes recipes by successful modelers.
Another possible option is making missile model based on some classical jet. There is quite simple pulse jet and there are some books about home-made turbine jet designs.
Anyway, I recommend to search nearby rocket modeling club and find supervisor.

I dont know if this old post is viable but here goes!  This post covers an commonly asked question/goal. Most of the responders to the OT are sadly correct, it's nearly impossible for a average builder to find or to design and produce a LF rocket. I have scoured the internet for ten years looking for a obtainable LF rocket engine, or kit or someone that has a (model) liquid fueled (LF) rocket engine with the ISP etc to put even a tiny payload into LEO. There is only one LF kit I know of (the Solitaire ?) and it hasn't nearly the power to achieve the escape veloicty etc for my goals. Even a solid fuel design of the highest efficiency would be huge and no longer in the realm of 'model rocketry', high power or otherwise. My last design was a three stage all solid fuel motors. It had a fuel grain nearly 15' long and 2.5' OD (tapering to 12") for stage three. All three stages used nozzles that were chrome moly outer shell wound with  carbon carbon/graphite thread meant to be semi ablative. It would have had  gimbaled nozzle linked to a gyroscope for stability. Even with all the above my calculations indicated I could expect only an sub orbital ballistic shot that would obtain a apogee of about 130km. The space shuttle type propellant would not produce enough ISP to produce the escape veloicty to pull out of earths gravity well. Close but no B-nanner, lol! I am fairly sure if I added clustering first stage and an exotic fourth stage LEO would be achievable.
Why didn't I build it? First too much red tape. It would be more illegal than building any destructive device. Second MONEY!!!! Even if I did the most of work myself 100K would just get me started. on the three stage version! But hang in there! There are some fun stuff being done with Hi altitude balloons and rockets, or just balloon carrying telemetry. Also Hybrid rockets are interesting but again they fall short in the power needed for  orbital applications.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the [help]. At the moment this reads more like a comment than an actual answer - could you elaborate a little more. With a bit more rep, you will be able to post comments on any question/answer.

Ok so first of all you cant get into orbit with anything "simple"... Remember that as the altitude increases, the pressure decreases, the makeup of the atmosphere changes, temperature changes and so on.
Now on to rocket fuel, if you want to go high, I suggest sinking some time into a hybrid engine. A hybrid engine uses a solid fuel and a liquid or gaseous oxidizer. There are some good example videos on YouTube of engines that use a hollowed acrylic rod for fuel and use either nitrous oxide or straight O2 to produce a very effective, throttle-able engine. However if you just starting out, I suggest rocket candy, its a cheap, homemade solid fuel+oxidizer mixture that is great for experimenting with. You can usually use sucrose and either potassium nitrate or sodium nitrate, I suggest the sodium nitrate as it is slightly lighter, but then again I'm a perfectionist.

I think that if you want a orbit, you should ask NASA, because the biggest ever rocket homemade reached " just" 56 kilometers. For NASA space starts at 100 km. A orbit asks for multiple stages with several tons of fuel, and the lowest orbit is higher than the edge of space, so its almost impossible to send a rocket into orbit.

I don't know how hard it would be, it might not be so hard than it appears, I mean, the only think i want is put in orbit (elliptical probably) a small mass then the only think i need is some V increment in a direction.  With an  arduino autopìlot rocket and an aerostatic balloon (10 km) as first stage maybe there will be  possible. I haven't calculated yet.

"The reason it's hard to get to orbit isn't that space is high up. It's hard to get to orbit because you have to go so fast." - xkcd what-if 58, Orbital Speed https://what-if.xkcd.com/58/ I'd honestly recommend playing some Kerbal Space Program or Orbiter to get an intuition on why going into space is hard.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
Why are there quantised energy levels in the vibrational energy of a molecule?
Why are the energy levels for vibrational energy in a molecule discrete as opposed to continuous?
I don't understand how a vibration can't have continuous amounts of energy that depend on the frequency of any infrared photon.
Can someone combine this question onto my stack exchange account: olly_price, because I can't sign in on this computer.

Really short answer: because energy is quantized and all energy changes are not continuous but discrete. It's in your question - "any infrared photon". Photons are discrete packets of energy, the molecular phenomena they interact with must also be discrete. I don't remember enough quantum mechanics of vibrational states to provide a satisfying explanation, however.

Am I incorrect in thinking that only infrared photons of certain frequencies can change the energy?

"Why" is a good question, and one that science has yet to fully answer. We generally have a good understanding of "how" things work at the subatomic level, based on over a century of observation followed by theory and math backed up by experimentation. However, the really basic, naively simple questions, like "where do particles get the charges, spins and masses that define them" are the questions scientists are asking now.
As for the discrete quantum energy states theorized by quantum theory, those are the result of a lot of math that explains what we see much to well to be coincidence. Basically, an electron can be thought of as orbiting its nucleus, but it's really more like a vibration. Similar to a vibration, there are frequencies above the "fundamental" that produce "harmonics"; standing waves with wavelengths of a perfect integer fraction of the fundamental, and perfect integer multiples of frequency, which are self-stabilizing. Electrons can only exist in these discrete energy states because if they didn't, they'd "break orbit" and go zooming off as free radicals, or have an unstable or unenergetic path that sends them crashing into the nucleus.
What's weird is that that doesn't ever happen as far as we know; not only do electrons exist in stable states, they only exist in these states, and we can't get them to exist in any energy state between two discrete states.
Max Planck first posited these quantum energy states existed while attempting to solve a conundrum of classical physics. Back to the vibration analogy, consider a guitar string. By damping the string at precise points on the string, we can induce clear harmonics. however, the higher the harmonic, the harder you have to pluck to produce a wave of the same amplitude. As the wavelengths get smaller, the energy you have to put into them to obtain a constant amplitude (and thus the energy inherent in the wave; Newton's third law) increases exponentially, diverging to infinity as the wavelengths approach zero. Now, this was obviously wrong; observed visible spectra of a glowing piece of metal as it is heated clearly shows a "peak" in the frequency of light produced, and at frequencies above this peak, amplitude of emitted light drops off sharply.
Planck solved this "ultraviolet catastrophe" by describing what was going on as similar to this string having a number of springs of different length attached to it, with weights on the other end of each spring. As the string vibrated, it in turn induced vibrations in the springs (known technically as "harmonic oscillators"), and as the frequency increases, it approaches, matches, and then falls away from the harmonic frequencies of the springs, increasing their relative amplitudes and then making them fall off as the vibrations begin to cancel out, while simultaneously increasing the energy imparted to a different string whose harmonic was being excited. The relative harmonic frequencies of the springs are what determine the colors of light seen, and the total energy of the string is divided among the springs, keeping the same total energy as frequency increases instead of diverging.
To make the math work, Planck had to describe the springs as each receiving an integer portion of the energy of the entire system, and described these energy levels as proportional to the length of the spring. The proportionality is defined by a slope known today as Planck's Constant. Thus, Planck asserted that all energy existed in discrete amounts, which he called "quanta", hence "quantum theory".
While this worked, it seemed to do so by convenience, until Einstein then further developed photon theory: light was made of of photons, "packets" of energy that could be quantified in proportion to the photon's frequency, based on Planck's Constant. This in turn explained puzzling properties of the "photoelectric effect", namely that the threshold where electrons began to be ejected from metal when exposed to light (forming an electrical charge) was dependent not on intensity of light, but on the frequency of the light. Einstein's photons explained that, because an electron was likely to only be hit by a single photon of light, no matter how many there were, the energy imparted to the electron could only come from that single photon, and as a photon's energy increases with frequency, there is a minimum frequency required to impart enough energy with a single photon to eject an electron.
Planck's constant, and the highly definite nature of how atoms interact with photons, continues to show up in math that accurately describes things we observe about matter, such as how materials emit very specific wavelenghs of light while absorbing all others. That was explained by the Bohr model of the atom, which was the first model to introduce quantum theory to atomic structure. The Bohr model was based partially on a mathematical model first proposed by Johann Balmer and then improved on by Johannes Rydberg, showing that the emission spectra of hydrogen was predictable by an equation with integer solutions. Bohr showed that the energy states of electrons, which can be altered by absorption of photons, must exist in certain integral states, defined by an equation derived from Rydberg's but that also includes, you guessed it, Planck's Constant. The famous "double-slit experiment" showing that light is both wave and particle, and the resulting mathematical models by Louis de Broglie showing the movement of electrons as a series of stable standing waves of vibration in the electromagnetic field of the nucleus (electrons, in fact, are similar to photons in that they are as much a wave as a particle), reproduced the hypothetical energy states of Bohr's model by a different method, and in math, when you get independent confirmation of your results by a completely different mathematical model, you know you're on to something.
Now, that's the theory to date (well, actually I stopped about 100 years' worth of research ago, but them's the basics of the discovery of discrete quanta). What we are still trying to explain is why these quanta actually exist. A spaceship can have an infinite number of stable orbits around the Earth, and there are countless stable orbits of planets and planetoids around the Sun; why then does an electron stabilize so deterministically in these discrete states? Our best answer is that the electron's wave motion is only stable in these discrete states; otherwise the wave would interfere with itself or others and cancel out. However, we have never seen this "turbulence" happen and can't seem to force it to happen; it seems to be a fundamental quality of matter.

There is a general result in quantum mechanics that a bound system has quantized energy states. Atoms in a molecule are clearly bound to each other, so they have quantized energy states.
More quantitatively, one easy model for the vibrational energy of molecules is to model the atoms as a set of masses, and the bonds as a set of springs connecting the masses. To see where this comes from, back up for a second to classical mechanics. The classical harmonic oscillator has the equation $\ddot x + \omega^2x = 0$, with $\omega = \sqrt{k/m}$. $k$ is the stiffness constant for the spring, and $m$ is the mass, but physicists usually keep everything in terms of $\omega$. This represents a mass on a spring, attached to an immovable wall. The energy of that oscillator is $E = KE + PE = \frac{1}{2}kA^2 = \frac{1}{2}m\omega^2A^2$, where $A$ is the amplitude of the oscillation. $A$ is determined by the initial conditions, and can be any continuous value. So the energy spectrum is continuous.
To translate this to the quantum mechanical simple harmonic oscillator, we write the Schroedinger equation as $(-\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + \frac{1}{2}m\omega^2x^2)\psi = E\psi$. The solutions to this equation are the discrete set of wave functions $\psi_n(x)$ with energies $E_n = \hbar\omega(n + \frac{1}{2})$. Again, this is a mass $m$ attached to an immovable wall via a spring. If you replace the wall with another (movable) mass of $M$, the only complication is to replace the $m$ dependence in $\omega$ with the reduced mass $\mu = \frac{mM}{m+M}$. In the case of molecules, the mass terms $m$ and $M$ have the obvious meaning of the mass of the relevant atom. The spring stiffness $k$ corresponds to the bond energy (though I'm not sure of the exact relationship). In practice, the energy is what you actually measure.
Molecules with more than two atoms are more complicated. There is one characteristic frequency $\omega_i$ (the normal modes for each degree of freedom in the molecule. Actually solving for them involves solving a set of coupled equations. As a simple example, consider carbon dioxide, and restrict the motion of the atoms to be only along the bonds; there are two bonds, so there are two normal modes. One normal mode corresponds to the carbon atom remaining stationary while the two hydrogen atoms move oscillate. The two "springs" are in phase with each other: they are compressed at the same time, and extended at the same time. The other normal mode corresponds to the two hydrogen atoms remaining stationary while the carbon atom oscillates. In this normal mode, the two bonds are out of phase: one is compressed when the other is extended, and vice versa. Any motion along the bonds can be written as a sum of these two normal modes, and they are the only vibrational energies the molecule can absorb. More generally, if you allow the atoms free range of motion in space, there are nine degrees of freedom, so there are nine normal modes. Three correspond to motion of the center of mass of the atom in space. Two are rotations of the molecule as a whole around the center oxygen (it "can't" rotate around its axis). None of these contribute to the vibrational spectrum. Two are the springs I described above, and are called the stretching modes. The final two are due to the oxygen atom oscillating out of the line connecting the three atoms, and are called the bending modes.

That's not quite right.  An atom has three degrees of freedom.  That is, three independent directions of movement (in three dimensions, of course).  Put two atoms together into a molecule and there are six degrees of freedom.  With carbon dioxide there are nine, and so on.  In molecules the degrees of freedom come in three types: translational, i.e. movement of the center of mass in each of three independent directions, rotational around three independent axes of rotation (linear molecules like carbon dioxide have only two) and the rest are vibrational.  So $\ce{CO2}$ has FOUR vibrations.

@PaulJ.Gans, you are completely correct. I was trying to simplify things, but ended up simplifying too much. I have added some material.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
How does localized bonding theory and hybridisation work?
So I understand molecular orbitals and how to do VSEPR models, but I seem to be struggling with understanding localized bonding theory and how to do hybridizations.  After drawing the initial lewis structure then molecular orbital, I do not know where to go in terms of whether something has a $\pi$ or $\sigma$ bond.
Also, how do I know when an atom has a carbon attached as an extra sigma bond or not? I understand how to know if it is $\mathrm{sp}$, $\mathrm{sp}^2$, etc. but I do not know what that means.

First of all, 'hybridisation' is a hypothetical concept, i.e. orbitals don't really mix to form new orbitals, but 'hybridisations' are very successful in explaining structures of molecules and their physical properties. So knowing that a compound has $\text{sp}^2$ hybridisation will just help you arrange the constituent atoms around the central atom. Physically it has no other meaning. It's just like vector cross product in Physics (for example the torque on a rotating wheel acts along the axis, i mean what does that even mean physically).
Secondly regarding how to know if an atom has a $\pi$ or a $\sigma$ bond: observing many structures(Lewis structures) and recognizing certain trends is the way out.

Some rules of the thumb that will help you draw structures are as follows:
1. Identify the central atom as the one that can make the most bonds.
2. Hydrogen makes only one bond. If Oxygen is present in the same compound, the Hydrogen is     generally attached to the Oxygen forming a $\ce{O-H}$ group.
3. Second period elements cannot expand their octets(i.e have more than 8 electrons in valence shell)
4. Mostly second period elements can make $\pi$ bonds.(That really solves part of your query). Although in some rare cases third period elements may also make $\pi$  bonds.
5. Never neglect the lone pairs(or non-bonding electrons) of the central atom.
Remember that these are only thumb rules and may be violated sometimes.

I don't see why you regard vector cross products as just useful hypotheticals, the direction of the torque vector is the direction the axis of rotation points in, it gives us the rotation sense (clockwise or anticlockwise)

@Amadeus I realize that myself from 9 years ago might have been overly critical about vector cross product. On the other hand, I think I'm gonna leave the remark as is, till I can think of another parallel.

I'm not very fond of VBT, but I'll try my best:
In the 20th Century, Chemist Linus Pauling decided to try and "hybridise" atomic orbitals to explain molecular geometry. We you probably know, the s orbital is spherical and the 3 p orbitals are dumbbell-shaped, meaning that the "unhybridised" orbitals SHOULD form bonds 90 degrees to each other (clearly not true).
To explain molecular geometry, Pauling decided to use a couple of mathematical concepts to explain molecular geometry, and this is what he postulated:

The greater the overlap between atomic orbitals, the stronger the bond
Due to the shape of the molecular orbitals, a pure s orbital has bond strength 1 and a pure p orbital has bond strength sqrt3.
The maximum bond strength is obtained when an orbital has 25% s character and 75% p character, and this is what we call "sp3" hybridisation.

In the case of electron deficient molecules (i.e. BF3), sp2 hybridisation is observed instead; this is because hybridised orbitals have higher energy than non-hybridised s orbitals, and the greater the p character, the higher the energy. Hence, in this case forming another bond (i.e. forming a Lewis adduct) would cause the hybridisation to change to sp3, increasing bond strength.
There are a couple more things that Pauling postulated regarding lone pair s character etc., but I don't think you should concern yourself with those.
Last thing- VBT fails miserably once you study it further. For example, the sp3 orbitals are not orthogonal to one another, and hence cannot be stationary-state solutions to the Schrodinger Equation. At the end of the day, Molecular Orbital Theory (MOT) is much more accurate.
Hope this helped.

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


text:
Resonance in carbocation derived from allenes
During a reaction if addition on one of the $\pi$ bonds of an allene (containing even number of $\pi$ bonds like $\ce {H2C=C=CH2}$) takes place then a carbocation is formed. To decide the position of the positive charge, we look at the stability of the carbocation. Example:
$$\ce {H2C=C=CH2 ->[\ce{H3O+}]} \quad?$$
In the above reaction the proton ($\small{\ce{H+}}$) attacks the electron rich $\pi$ bond (say the left one) and can form one of the following carbocation:
$$\ce{H2C+\bond{-}CH=CH2}\quad\quad \text{or }\quad\quad \ce{CH3\bond{-}C+=CH2}$$
$$ (\text{I})\quad\qquad\qquad\qquad (\text{II})$$

Now the major product (acetone) is formed as a result of the carbocation on the right (II), which apparently is less stable than the one on the left (I) as the positive charge in I is in resonance/conjugation with the $\pi$ bond. But our teacher said that since in allenes the overlapping p-orbitals (say $\ce{p_z}$) of one $\pi$ bond are perpendicular to the overlapping p-orbitals of the other (say $\ce{p_y}$; which makes $\ce{p_x}$ in the direction of internuclear axis $x$) and hence the positive charge cannot resonate with the $\pi$ bond (formed by overlap of $\ce{p_z}$) as the positive charge is effectively an empty p-orbital (the $\ce{p_y}$) and hence the carbocation (II) is relatively more stable.
But once addition has taken place the $\ce{C_1-C_2}$ $\pi$ bond no longer exists and the $\sigma$ bond is free to rotate so the empty p-orbital/positive charge can freely rotate and become coplanar with the p-orbitals of the $\pi$ bond. But why does this not happen?
The reason cannot be steric hindrance as the two groups on $\ce{C_1}$ are the smallest, $\ce{H}$.

Another thing to consider would be that the allene under acidic medium would first isomerize to propyne, followed by electrophilic addition.

The answer to your question might need a few clarifying points. My answer will mostly focus on the protonation (and deprotonation) of allene based on gas-phase experiments and theoretical calculations. A few references that point to elements that will be used concerning the allyl and 2-propenyl cations are as follow : G. van der Rest et al. Eur. Mass Spectrom. 3 323 (1997) and K. Raghavashari et al. J. Am. Chem. Soc. 103 5649 (1981).
Relative stability of the allyl and 2-propenyl cations
Both articles make it clear that the allyl cation (I) is the most stable structure for the $\ce{C_{3}H_{5}^{+}}$ carbocations. It is around $\pu{21 kJ mol-1}$ more stable than its 2-propenyl (II) isomer. The other possible isomers are all higher in energy.
Protonation of allene
According to the first reference above, the protonation of allene on an extremal carbon to lead to the allyl cation requires to pass a high energy barrier, whereas no high barrier is present to lead to the 2-propenyl cation. Figure 4 gives the structure of the interaction complex between $\ce{H_{3}O^{+}}$ and allene, showing clearly that the incoming proton is initially in interaction with the $\pi$ orbital on one side of the allene. Their is obviously no conjugation contribution between this protonation structure and the other $\pi$ orbital, since the proton arrival plane is perpendicular to the other $\pi$ orbital.
From this position, formation of the 2-propenyl cation (II) requires only a minimal energy barrier, requiring only a bending of the $\ce{C-C-C}$ bond and bending of the $\ce{CH_{2}}$ group. Once this is done, the $\ce{CH_{3}}$ group can rotate freely, but one should notice that it does not involve the $\ce{sp^{2}}$ vacant orbital located on the central carbon. This orbital will always remain orthogonal to the remaining $\pi$ orbital.
From this position, one could also consider direct protonation of the central carbon, followed by a rotation of the $\ce{CH_{2}}$ group since no conjugation is present at this point. But here, one should consider that stabilization by conjugation can only proceed after formation of a (non stabilized) primary carbocation. These species are much less stable than secondary carbocations (on the order of $\pu{75 kJ mol-1}$), thus leading to the presence of a large barrier for the direct formation of an allyl cation.
A last pathway that one should consider is a 1,2 hydride transfer between (II) and (I). Although hydride transfers are generally easy between carbocations, this transfer proceeds through a three-center system which can be considered a proton attached to a $\pi$ orbital. Thus, this leads us back to the initial problem, which is that no conjugation can serve as a driving force leading to the allyl cation. And only once the unstable primary carbocation is formed can a rotation of the $\ce{CH_{2}}$ group lead to formation of the allyl cation.

The link to the article gives page not found. Could you update the link?

@OsheenSachdev Just updated the link. Thanks for the notice.


text:
What are the common names of this two types of plastic?
I need to do some plastic molding design. I want to know the names/types of plastic which are commonly/widely used in industries (please see the pictures below).

I am going to take a stab and say it looks like tupper-ware so it is probably polyethylene

The most common types of plastics in use today are polyethylene (PE) and polystyrene (PS). Each of these comes in several variants.
It is difficult to tell just from a photograph exactly what plastics are used in the products shown.
Also, various additives are often added to plastic to enhance certain properties such as high-impact, chemical resistance, recyclability and flexibility.
The type of plastic is sometimes marked on the item itself to help with recycling, typically in the form of an extruded imprint, or you could try to contact the manufacturer.
Both PE and PS are suitable for moulding and extrusion. You can also do some simple physical and/or chemical testing to determine what the plastics contain, although this will involve destroying at least a part of the item. Low density polyethylene (LDPE) typically has a lower melting point $(105-130^{o}C)$ than polystyrene $(200-240^{o}C)$.
LDPE is also softer and slightly rubbery whereas polystyrene is typically more rigid & crackly when crushed. Without physically touching the items to get a proper feel for what they are made of, as a first guess, I would say that the colourful containers such as plastic bins and laundry baskets in your first photo look like some sort of polyethylene, whereas the clear plastic packaging in the second photo looks more like polystyrene.

One easy way to distinguish polyethylene from polystyrene is it's density; of all the commonly used thermoplastics only polyethylene and polypropylene are less dense (and will therefore float in) fresh water. (However, plastics manufactured as a foam, like Styrofoam made of polystyrene and foam-core plumbing pipe made of ABS will also float). Another easy distinguishing feature as mentioned above is the "Resin Identification Code" used for recycling, it's a triangular symbol with a number between 1 and 7 identifying the plastic; 2 and 4 correspond to polyethylene while 6 corresponds to polystyrene.
Lastly, if you're really unsure and not opposed to destructive testing you could consult a chart like this and evaluate quantitatively the way the plastic burns:
http://www.loyresearchgroup.com/uploads/1/0/2/2/10227236/new_id_chart_.pdf

You can tell the difference between polyethylene (PE) and polystyrene (PS) by rubbing a non-visible part with a tissue and a bit of acetone. PE is inert to acetone while PS is soluble and you will see the surface tarnishing.


text:
Which plastic can be partially broken by pressure easily?
I need to create a container which hold ~150 to 500 ml water. My plan is to create something like a inverse cone. I will apply pressure (~20psi) on the top of the cone and I want to create a hole at and only at the tip of the cone.
I plan to make the tip of the cone thinner, so that it will  be easier to be break. However, I am new to plastic. Would anyone please give me some suggestion of material that I should use?
Thanks

Are you actually trying to make something that will break under pressure, or are you just trying to make something that shape and thinking that pressure would be a good method?

From my experience with plastic (not much but some) it's not good idea to create a hole with applying pressure to it. I think it would be better to heat the tool which you would use or use curving drills such as applied to wood.

I think he wants to create an automatic drain. Plastics respond to stress concentration- change in direction worse than metals- the point of the cone would most likely to fail. Just the forming of a range of plastics would put a lot of stress on the point. Try high density polyethylene- its really all the design. How is this container going to be supported?

If you need good control, put a spring-loaded or cantilever-loaded valve at the tip.  Rupture diaphragms typically have a needle point at the far end. The bubble deforms with pressure, contacting the point that then ruptures the diaphragm that is under tension.  It is much easier to control and reproduce deformation than it is to engineer catastrophic failure at maximum deformation.


text:
Which materials/compounds expand in response to cold temperatures?
Which materials or compounds expand in response to cold temperatures?
This material or compound should expand at a slow (but constant rate), from about 0–20 months when in the following temperature range −25 °C to −10 °C.
Also, the compound should not react with plastic or be toxic to humans or the environment.
Please explain why the compound does this as well.

This seems to me to be a chemistry question. I found one compound http://pubs.acs.org/doi/abs/10.1021/ja106711v ,  Cubic ScF3 from the references in the wiki article http://en.wikipedia.org/wiki/Negative_thermal_expansion#cite_note-greve-7

Isn't the typical example is water and ice?

In old time printing using type, the alloy used for the type was designed to have a zero coefficient of thermal expansion over a wide range so that they type did not change shape when cast at a high temperature and then cooled off.
It is very likely that slight modifications of that formula would yield an alloy with either a positive or negative thermal coefficient.

Perhaps one of the most studied materials to exhibit negative thermal expansion is cubic zirconium tungstate $(\ce{ZrW2O8}).$

It's already been mentioned that certain liquid-solid phase transitions are accompanied by negative thermal expansion (NTE). This also includes elements ($\ce{Si, Ga, Ge, Sb, Bi, Pu}$), alloys (Wood's metal) and numerous materials (zeolites, quartz, tugstates etc.) that obey virial expansion due to the way atoms are packed in a crystal lattice:
$$\frac{p}{RT} = \frac{1}{V_m} + \frac{B_{2V}(T)}{V_m^2} + \frac{B_{3V}(T)}{V_m^3} + \ldots$$
where $p$ - pressure, $R$ - gas constant, $T$ - temperature, $B_{iV}(T)$ - $i$-virial coefficient, $V_m$ - molar volume.

There is also another physicochemical phenomenon called Gough–Joule effect. If mechanical stress is applied to the elastomer or rubber string which is subsequently heated, the polymer contracts as a result of the intensive growth of the role of the entropy factor (the maximum entropy in polymers occurs in the coil-like ordering of the molecules). This effect is also utilized in the rubber-band heat engine (Feynman wheel).

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


text:
Why is the relative atomic mass of carbon not exactly 12?
Relative atomic masses of atoms of all chemical elements are numbers without units, being the value of proportion compared to  $\frac{1}{12}^\text{th}$ the mass of the carbon atom.
But the relative atomic mass of carbon is never 12! Instead, it is 12.01 or, more accurately,  12.011.
Why is this so?

Simply because the atomic mass is defined as 1/12 of the mass of 12C. Others isotopes of carbon (13C mostly, with an abundance of 1.1% approximately) account for an average atomic mass slightly above 12.

Yes, the mass of an element on the period table is the weighted average atomic mass over all naturally occurring isotopes. It is why some elements, like chlorine (35.45) have masses that are even less integer-like.


text:
What are the g-block's predicted properties?
I have been researching for a Physics/Chemistry exam and thought: what will the future periods in the periodic table (periods 8 and above) entail? Each block contains its own properties that allow for accurate predictions of elements in these blocks. Each block has very unique properties and benefits society in major ways. The g-block is the theoretical block thought to exist in period-8 though no elements have been produced in this period for any sustained time. My question is:
What are the predicted properties of the G-block if any?  Who and what methods have they been predicted by?
I'd also like to know if there are any ways of "stabilizing" these elements so that they take more time to decay.

First of all, from the perspective of chemistry, it is the group (column) that matters more than the period (row), and subshells act at the period level.  Second, the elements beyond the actinides have virtually no significant chemical research owing to their extreme radioactivity.  So there may not be many predicted properties for the g-block beyond the simplistic theoretical assumptions about groups, which are much less relevant for a hyper transitional period like the g-block.
As for you question about potential stability, Glenn Seaborg made some predictions here, which you can read that this Wikipedia page: Island of Stability.  In general, though, these elements are not predicted to be stable.

It is the group that matters, however the f-block elements, particularly the lanthanide series, do have a lot of similarities to each other as the whole f-block is technically part of Group 3 (or IIIB if you prefer).

@Ben Norris: Yes, glad you added this. That's what I was getting at (badly) in the last sentence of the first paragraph.


text:
Where can I find a downloadable spreadsheet of element properties?
Sorry for the FAQ, but I really don't know where to find a full spreadsheet for element properties. I mean a table that has one row for each element, and the columns are its properties, such as name, atomic number, density, melting point, etc. I know that there are many sites or references for that, but I need it in tabular format, because I want to analyze data, make charts, etc.

Something like this?

Or if someone knows programming, they can web-scrape source code of this website.(Don't know if it is legal)

Yes, @AdityaSriram, That's exactly what I wanted

just visit "http://images-of-elements.com/element-properties.php" and "select all" and copy to excel

Well, here's a CSV file that I parsed out of the JSON data provided by Paul Nathan's website, which in turn was produced from gPeriodic data in response to this question. gPeriodic is FOSS, so I can only assume that the element data contained within is good to share, though I have no idea of its provenance.
The parser uses the python json module to read the data, which I then wrote into comma separated values, which should be readable by any halfway competent spreadsheet program.
Here it is: Pastebin Link
save it as elementdata.csv and you're good to go.
Some gotchas:

This reproduces the gPeriodic data, warts and all. I make no guarantees regarding its correctness, recency, etc. I just munged the data.
Some of the data is augmented with tildes, notes about temperature/polymorph/state etc. I've left these as they are but you may need to trim them if you want to plot them as numeric values.
Argon had an atomic radius of '2-', whatever that means. I cut it.
Excel is joke software and habitually interprets numbers wrapped in parentheses as negatives, because apparently some accountants decided that surrounding numbers in parens is a more sensible option than using a minus sign. Needless to say, this is wrong. I've addressed the problem by wrapping the affected numbers in angle brackets.
There are no ionic radii in the source data. At all. Not my fault.
The columns inherit the units of the source data. I've left the units out on purpose because I wanted each column to have no spaces in the name for ease of processing in R or whatever. Furthermore, the units in the source file have some inconvenient characters from an encoding perspective.

Tried plotting Z versus covalent radius in R from this data - looks alright:

Some of the data points are missing, some are not read by R because they are wrapped in <> - pre-process to your heart's content.

Thanks, man but @AdityaSriram got me the excel file that I had described. Another time, thank you.

@MostafaFarzán - No offence intended to Aditya, but the data that I link to contains considerable additional information that may be quite interesting to plot. Moreover, as a .csv, it can be opened in excel as well as literally hundreds of other programs on all platforms, making it considerably more portable than an .xls, for which interoperability has always been a low priority for Microsoft.

Sorry, I didn't know that excel can open these files. The file was cool, but like that xls file, needs some works before charting.

"No offence intended to Aditya" -> None taken.

This data hasn't been updated with the 2016 element names.

Fine, here's my data taken from the reference mentioned in my above comment.
I cannot vouch for it's accuracy.
This is in the .csv file format for its wide use as enlightened by Richard Terrett. You can combine  certain ranges from this csv/spreadsheet to the one prepared by Richard, because it contains certain properties not contained in Richard's source.
Also there are certain properties in Richard's file which are absent/improper in mine so this file is best used in conjunction with Richard's file.

Not bad! I assume this was obtained via scraping the ChemReference site?

Yes from the (chem *cough* refer *cough* ence site). First copied the source HTML, used Python-Beautiful Soup to get just the periodic  part then wrote a program to gather data/properties from the ,, and  tags into dictionaries finally output the dictionary contents into a csv file.

This data hasn't been updated with the 2016 element names.

I found a nice json database here (it may have errors, I wouldn't know, but there are links to the original sources).
Best visualised with Chernoff faces IMO,

library(jsonlite)
library(ggplot2)
library(ggChernoff)

# https://github.com/Bowserinator/Periodic-Table-JSON
d <- fromJSON('PeriodicTableJSON.json')[["elements"]]

ggplot(d, aes(xpos,10-ypos)) +
  geom_chernoff(aes(fill=category, size = phase, smile = density, brow = molar_heat)) +
  geom_text(aes(label=symbol), vjust=2.5) +
  guides(fill=guide_legend(ncol=2)) +
  theme_void() +
  scale_brow_continuous(breaks = c(10, 25, 50))+
  theme(legend.position = 'right', panel.background = element_rect(colour="black"))

I've collected a lot data on the elements from several sources and built a small python package around it called mendeleev. The data is contained in a SQLite3 database from which you can easily extract csv files of individual tables.
Most of the data if properly referenced so you can trace back its origin.
Moreover since the interface is written in Python you can easily plot and analyze the data if needed.

Here is a Google Sheets spreadsheet with basic data:
https://docs.google.com/spreadsheets/d/1vEOwbbl-l805_i2oWSVCFxfmFDstwkuRjkZQfBrTh6k/edit?usp=sharing
Hope this is useful!


text:
Why does radium have a higher first ionisation energy than barium?
I'm wondering why radium appears to buck the general trend that first ionisation energies decrease as you move down a group in the periodic table: barium (the group 2 element preceding it) has a first ionisation energy of $\pu{502.9 kJ/mol}$, whereas radium has a slightly higher first I.E. of $\pu{509.3 kJ/mol}$ (from the Wikipedia, although my textbook agrees).
Is there any explanation for this at present? (I imagine that quantum mechanics may be involved in some way, but I'm not entirely sure how)
There are most likely other examples of the trend being broken, but this is the only one I've come across so far and I'm curious to know why this is the case.

The same trend-breaking can be observed with Cs and Fr. Also the trend of decreasing first ionisation energy doesn't seem to hold with the transition from 5th period to 6th period transition metals (as well as some of the main group elements) and the transition from lanthanides to actinides (see: http://chemreference.com/), so I guess it must have something to do with the additional f electrons, but I have no ideas what it is exactly.

I think it's also important to mention relativistic effects here.  They already start becoming quite visible after $Z=70$, and $\ce{Ra}$ lies a good bit after that.
In very heavy atoms, the electrons of the $\ce{1s}$ orbital (actually, all orbitals with some electron density close to the nucleus, but the $\ce{1s}$ orbital happens to be the closest and therefore most affected) are subjected to very high effective nuclear charges, compressing the orbitals into a very small region of space. This in turn forces the innermost electrons' momenta to be very high, via the uncertainty principle (or in a classical picture, the electrons need to orbit the nucleus very quickly in order to avoid falling in). The momenta are so high, in fact, that special relativity corrections become appreciable, so that the actual, relativistically corrected momenta, ($p_{\text{relativistic}}=\gamma p_{\text{classical}}$) are somewhat higher than the approximate classical momenta. Again via the uncertainty principle, this causes a relativistic contraction of the $\ce{1s}$ orbital (and other orbitals with electron density close to the nucleus, especially $\ce{ns}$ and $\ce{np}$ orbitals).
The relativistic contraction of the innermost orbitals creates a cascade of electron shielding changes among the rest of the orbitals. The final result is that all $\ce{ns}$ orbitals are contracted, getting closer to the nucleus and becoming shifted down in energy. This is relevant to the question because the $\ce{7s}$ valence electrons in $\ce{Ra}$ are more attracted to the nucleus than one would expect from a simple trend analysis, since they rarely take into account the increase of relativistic effects as one goes down the periodic table.
Thus, the first (and second) ionization energy of $\ce{Ra}$ becomes higher than expected, to the point that there's actually a upward blip in the downward trend. Eka-radium ($Z=120$) would have far stronger relativistic effects, and can be expected to have a significantly higher ionization energy compared to $\ce{Ra}$. In fact, relativistic effects will conspire to make the group 2 metals slightly more noble! Though the periodic table becomes such a mess near the super heavy elements that it's hard to say whether it'll be a clearly visible trend, or just one effect to be combined with several others.

This is a good theoretical explanation of the effects noted in my answer. Basically, any element after the lathanides will have a higher first ionization energy than the one above it for this reason.

Well, this is actually not the case for heavy elements in groups 15-18 (Bi, Po, At, Rn and those below them). Here, relativistic effects actually destabilize two of the three valence $np$ orbitals (technically, the $np$ subshell has its degeneracy broken due to strong spin-orbit coupling, forming a stabilized $np_{1/2}$ orbital and two destabilized $np_{3/2}$ orbitals), so that even with increased effective nuclear charge, all ionization energies up to $np_{1/2}^2$ require less energy than for the elements above. This effect is most apparent in Bi, because Pb has slight relativistic nobility.

"relativistic contraction" I am not sure what you mean by contraction?. Is it like the probability of finding electron shrinks to a smaller space? Or the orbitals become closer to the nucleus?

@Mockingbird I'm not too sure what the difference is between the two options you've given, but basically the average distance of the electron to the nucleus, in all $s$ orbitals, is smaller when you include relativity.

This behavior can be attributed to the same phenomenon as that which causes the lanthanide contraction. The electrons in $f$ subshells are very poor at nuclear shielding, so the $s$ electrons in the next higher shell are closer (on average) to the nucleus than you might expect. If these electrons are closer to the nucleus, then the atom exhibits a smaller radius than expected and these $s$ electrons are more difficult to remove. According to the Wikipedia article on the lanthanide contraction, there are both quantum mechanical and relativistic causes for the lanthanide contraction.
This behavior also occurs in hafnium (72), which is right after the lanthanides. Hafnium's first ionization energy is $\pu{658.5 kJ/mol}$, while zirconium (40, right above hafnium) is $\pu{640.1 kJ/mol}$. Barium is before the lanthanides and radium is after them. Any pair so split will probably exhibit this new trend in ionization potentials. Go and check it out!

Elements 57+ have electrons in f orbitals.
Radium has a full 4f sublevel these electrons as less effective at shielding than d electrons (f

You can find more detailed information on the poor shielding of f block electrons here.  http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Descriptive_Chemistry/Transition_Metals_and_Coordination_Complexes/The_Lanthanides/Lanthanide_Contraction

Looks like your answer has been cut short. In addition, it would be nice to incorporate some of the link into your answer (writing in your own words would be nice, though quoting is fine)

It is because the 4f orbital of radium shields the 7S orbital less effectively from nuclear attraction, therefore a higher ionization energy is required for their removal.

Well it's all about the shielding effect,f subshell have many holes and thus the nuclear influence on the outer s electrons increases,so valence electrons are much tightly bounded to the nucleas.hope you got the answer


text:
Why do shape and size have the greatest effect on the optical properties of nanoparticles?
The LSPR properties of nanoparticles depend on a number of factors, including shape, size, material, crystal structure, dielectric environment, etc. However, it has been consistently shown that shape and size have the greatest effect on the optical properties of nanoparticles. Exactly WHY, however, is this true? Does anyone have an explanation at the molecular level?

As Surface Plasmon Resonance phenomenon is mainly seen in metal nanoparticles, I don't know whether this answer will help you? The size dependent optical phenomenon of semiconductor nanoparticles (e.g., $\ce{CdS}$) can be explained using the effective mass approximation model.

where r is radius of the particle (the rest details of the above equation can be find in the link).
so the energy gap is inversely proportional to the radius of the particle.

When the radius of the particle  decreases, the band gap increases so it will emit high energy wavelength (i.e., wavelength in blue region) and when the particle size is large the energy gap will be less so it will emit light in red region (lesser energy).

Thanks for your answer! However, isn't the effective mass approximation model only applicable to spherical nanoparticles (i.e. quantum dots)? I'm writing a paper on the synthesis of silver nanoplates and nanodisks, and I was wondering in general why shape and size have the greatest control over LSPR modes. Are you familiar with the Discrete Dipole Approximation method? I believe that it is an alternative (for arbitrary geometries) to Mie Theory, which only applies to spherical shapes. If so, could you explain the reasoning behind how, given any shape/size, DDA can derive the LSPR modes?

@jojoma42 yes the above equation is for spherical nanoparticles and i am sorry i dont know about discrete dipole approximation method

@jojoma42 i have found these links for DDA 1) http://ftp://ftp.astro.princeton.edu/draine/papers/pdf/JOSA_A11_1491.pdf and http://www.physics.ohio-state.edu/~stroud/optics1.ppt DDA is an approximation technique which is used to find approximate value of scattering and absorbtion of irregularly shaped nanoparticles by breaking up nanoparticles into smaller volumes with dipole movement. The relathionship of size and absorbance can be find in this link http://pubs.acs.org/doi/full/10.1021/jp984796o . Hope this is useful to you

Optical properties of nanoparticles come from electronic spectra of one excited electron trapped within. The energy levels in the spectra are derived from lengths of waves, that resonates with energy well within particle. Of course, this lengths depends on size and form of the resonator. Note: this IS oversimplification, but it allows to grab the concept.


text:
How can I dispose of chlorine and sodium hydroxide as table salt and water?
If I used chloralkali electrolysis of brine to produce hydrogen, and can't use the other products (or not all of them), what route(s) can I take to revert them back to brine? I figured, since I have to put in energy to take the brine apart, it should have some desire to reform.
With the $\ce{H2}$ taken away, the remaining products are:
$$\ce{2 NaOH (aq) + Cl2}$$
If I just mix them, I get sodium chlorate or hypochlorite. Can I process those further?
Both $\ce{H2O}$ and $\ce{NaCl}$ have a negative $\Delta H_\mathrm{f}^\circ$ ($\pu{-285.8 kJ/mol}$ and $\pu{-407.0 kJ/mol}$ respectively), and $\ce{Cl2}$ is already in its standard state, so we would need to overcome the $\Delta H_\mathrm{f}^\circ$ of the $\ce{NaOH}$, which is negative as well ($\pu{-470.1 kJ/mol}$). Can I use ion filters to some benefit?
Routes that allow retrieval of some of the energy spent for the electrolysis are favored (i.e. the more exothermic, the better).

Also, it seems, at some point, I'm left with a couple of oxygens, which would make for a nice byproduct.

It is better to use alkali metal sulfate or carbonate.

@permeakra - you mean, instead of NaCl? The brine would be a given. To produce hydrogen, I guess there are better ways than electrolysis altogether, anyway, right?

yes, there are some. At home the simplest way is to dissolve metal (zinc or iron) in dilute acid. In industry, hydrogen is produce using process $\ce{CH4 + 2H2O -> 4H2 + CO2}$

It looks as if the simplest way would be to mix the products, and let the mixture warm up, so we get sodium chlorate in water. We can then have that crystalize and heat it until it disintegrates into $\ce{NaCl}$ and $\ce{O2}$. That would need no additional substances and have a clean end result. The only remaining question would be the energy budget, but the forming of the chlorate is exothermic and additional heat might be available.
Alternatively, we might find something useful to oxidize with the $\ce{NaClO3}$.


text:
How can I produce hydrogen gas efficiently and cheaply?
I have been intrigued by one of the principles of chemistry lately- electrolysis. In my knowledge, electrolysis is a quite dangerous operation as the decomposition of water produces hydrogen and oxygen gas, both which are flammable. However, its potent to be volatile isn't what I am concerned with; rather, it is the production of these gases sufficiently and cheaply that I have been wondering about.
Thus, what is the cheapest way available right now to build an efficient electrolysis apparatus that could be used for a relatively long amounts of time? In addition to this, what are the side effects with electrolysis of water? In addition to this, could solar panels be used in the process of electrolysis? I want this electrolysis apparatus to be as independent from the grid as possible!

Note: oxygen is not flammable, it merely enhances the combustion of other flammable stuff.

Flammable means "easily set on fire", I think oxygen qualifies. Sometimes words go beyond the concepts they represent.

Electrolysis of water involves the splitting (lysis) of water molecules using electricity.
$\ce{2H_{2}O(l) \rightarrow 2H_{2}(g) + O_{2}(g)}$
From the above equation, we see that the production of hydrogen will occur twice as fast as oxygen (on a volume basis). The electrode materials must be inert (non-reactive with water or its constituents). Platinum is good but very expensive, so titanium and/or nickel alloy catalysts are sometimes used. The current flows into the anode where oxygen gas is produced, whereas hydrogen gas is produced at the cathode. Generally, the higher the current, the higher the rate of production of hydrogen gas.
Since this is not a spontaneous reaction, energy is required from an external source in order to drive it. This is provided by a direct current (DC) electrical power supply, such as a battery or solar cell. The amount of energy required can be determined from the Gibbs energy of formation of water, which is -237.2 kJ/mol (at 298.15 K or 25 °C).
The theoretical minimum voltage needed for the reaction to occur is the standard electrode potential of oxygen, which is +1.23V. In practice, a slightly higher voltage (about 1.48V) is needed because of an over-potential of about 0.25V which is used up as heat.
Pure (demineralised) water can be expensive and has very low electrical conductivity, so some impurities in the water are desirable. Salt (sodium chloride) dissolved in the water will increase its conductivity dramatically, thereby reducing the current necessary to produce a produce hydrogen at a given rate. However, the chlorine gas can be produced at the anode (instead of oxygen) which is highly corrosive and hazardous. Minerals such as magnesium or calcium (present in sea water) will form insoluble precipitants on the cathode which could build up a 'passive' layer, reducing the rate of hydrogen production. Some acids or bases such as sulfuric acid or sodium hydroxide are therefore sometimes preferred.

Just a note, the easier way of typing chemical equations is via mhchem. See http://meta.chemistry.stackexchange.com/a/88/22

Speaking of chlorine gas, I have heard that they were used during World War II in gas bombs; therefore it would be impractical for me to do electrolysis with a solution containing sodium chloride . What other ionic compounds could I add to tap water (or any ordinary, dirty water) that will produce conduct electricity but is non-toxic at the same time? Thanks in advance.

Chlorine gas was used on the battlefields of WW-1 by Germany, being heavier than air, would settle in the trenches of their enemies, forming hydrochloric acid and mucous in the lungs and thereby suffocating/choking the soldiers. Use of sodium hydroxide can be used as an electrolyte instead of sodium chloride so as not to produce chlorine gas. Also, using a manganese oxide coated anode is preferred for oxygen production at the anode (instead of chlorine) with high efficiency.

You could also use sodium bicarbonate (baking soda), which is cheaper, but probably quite a bit less efficient.


text:
Chemical software for solving reactions
I'm looking for a software which can solve chemical equations. For example something like this (given with result):
\begin{align}
    \ce{CH4 + 2O2 & → CO2 + 2H2O} \\
    \ce{C3H6 + HCl & → C3H7Cl}
\end{align}
Is there are such software? I'm studying chemistry as a hobby and want to check if I'm doing the reactions correct.

I'm sorry, but this question is of the "make-a-list" type and thus needs to be closed. Please read the [faq] before posting :)

Other questions like this one were not closed and seem to be helpful, see: http://chemistry.stackexchange.com/q/24/1208 Why not re-open this question?

No, there is no such thing as chemical reaction are not that easily predictable. You can get result right in some cases, but it is impossible to count all factors.


text:
Bent molecular geometry of water
Could you please explain how the bent molecular geometry of water is due to hydrogen bonding? I was under the impression that it was because of the lone pairs (VSEPR Theory).

Your impression is correct. The bend in the water molecule doesn't have anything to do with hydrogen bonding, as evinced by the fact that isolated water molecules retain their bent geometry.

Your understanding is correct. The usual explanation given in introductory classes is VESPR theory. The oxygen atom has four bond-like "things" coming out of it. Two of those are the covalent bonds with the hydrogen bonds. The other two are the two lone pairs. VESPR theory calls this a steric number of four.
The slightly more advanced explanation comes from molecular orbital theory. Oxygen's valence electrons are all $2\text{p}$ electrons; hydrogen's electron is $1\text{s}$. The only way to mix those orbitals to get enough bonding orbitals is by using two $\text{sp}^3$ hybrid orbitals to generate the two bonds. The $\text{sp}^3$ orbitals are arranged in a tetragonal geometry.


text:
Is the vinegar and baking soda experiment safe for children?
To be more specific, could the carbon dioxide fumes produced in the reaction be harmful to a small child? I don't have a strong background in chemistry, and I was prompted to ask this question when it was suggested by a friend that a mixture of 2 cups of vinegar and baking soda is commonly used to euthanize rodents. I'm aware rodents are more sensitive than humans, but I was concerned that a small child could inadvertantly breathe enough fumes to cause harm. I think it would be a great experiment to introduce children to chemistry if not unsafe. Thanks for any advice or guidance.

A basic safety rule in chemistry is "don't breathe fumes". Maximum waft a sample of evolved gas to the nose with your hand. Given a controlled environment, adequate size vessels and small quantities and adequate ventilation, it could be a good introduction to chemistry. However - start in the way you should continue - safety glasses, safe operating procedures etc.. BTW, red cabbage makes a good indicator for this.

I did this experiment with my kids (2yo & 5yo) outside for this reason. We also tied a balloon around the outside of the bottle to keep it isolated and to watch the balloon grow. But to reassure you, one would need to be in an isolated chamber with concentrated CO2 for harmful effects to occur. Remember that CO2 is already present in our bodies and in the air we breathe too, so concentration is everything.

Unfortunately this anecdote doesn't add much to the present answer and is more suitable as a comment.

It's pretty safe. The primary source of damage to people from CO2 is asphyxiation. That means you don't get enough oxygen quickly enough. Its danger is similar to helium. You need ridiculously high concentrations of CO2 for it to be harmful and you need prolonged exposure.
You probably drink carbonated drinks. Do you worry about them killing you when your nose gets close to the cup?
I think the much bigger danger is these chemicals getting in people's eyes. Even then, it wouldn't be super dangerous.
Eye glasses are important, though. Lab coat is not really necessary for safety, but it could preserve clothing.
There was already an answer but I thought this might be helpful.
Depending on the quantities of baking soda and vinegar used, it may be perfectly safe to do it indoors, even. In terms of CO2 at least.


text:
Kinetic energy of electrons emitted in Photoelectron Spectroscopy
In multiple sources (example), I found the information that the kinetic energy $E_k$ measured using (Ultraviolet / X-ray) Photoelectron Spectroscopy is given by:
$E_k = h\nu - B.E._F - \phi_{spec}$
where $h\nu$ is the energy carried by the incident photon, $B.E._F$ the energy difference between the energy level of the electron and the Fermi energy and $\phi_{spec}$ the work function of the spectrometer.
As far as I understood the measurement process, the actually measured quantity in Photoelectron Spectroscopy is the kinetic energy $E_k$ of the electron. This energy is measured by the radius of the electron in a homogeneous electric field. Thus, the kinetic energy of the electron after the emission would $E_k = h\nu - B.E._F - \phi_{sample}$, this independent of the work function of the spectrometer. So where does this difference come from?

From Wikipedia (emphasis mine):

Because the energy of an X-ray with particular wavelength is known
  (for aluminum Kα X-rays, Ephoton = 1486.7 eV), and because the emitted
  electrons' kinetic energies are measured, the electron binding energy
  of each of the emitted electrons can be determined by using an
  equation that is based on the work of Ernest Rutherford (1914):
\begin{equation}
     E_\text{binding} = E_\text{photon} - \left(E_\text{kinetic} + \phi\right)   \end{equation}
where $E_\text{binding}$ is the binding energy (BE) of the electron, $E_\text{photon}$ is
  the energy of the X-ray photons being used, $E_\text{kinetic}$ is the kinetic
  energy of the electron as measured by the instrument and $\phi$ is the
  work function dependent on both the spectrometer & the material. This
  equation is essentially a conservation of energy equation. The work
  function term $\phi$ is an adjustable instrumental correction factor
  that accounts for the few eV of kinetic energy given up by the
  photoelectron as it becomes absorbed by the instrument's detector. It
  is a constant that rarely needs to be adjusted in practice.

So, it is the absorbtion of the photoelectron by the detector of the spectrometer that causes the work function term. The work function of the measured sample should be part of $E_\text{binding}$ as this is the ionisation energy of the electron. However, I think the equation you quote might have it slightly wrong because if $B.E._F$ is the energy difference between the energy level of the electron and the Fermi energy then there is indeed some portion of the work function of the sample missing since $B.E._F$ is not equal to the ionisation energy.


text:
A first-order reaction at infinite concentration
I have a first-order reaction, for the binding of two molecules, of the form: $\text{k}_\text{f} = \ce{k_1 (M*s)^{-1}}$.  What happens at infinite concentration?  How is this upperbound typically handled for approximations in chemistry?  Do we need to measure additional kinetics parameters in this regime?
If anyone has any specific knowledge, I'm particularly concerned with DNA hybridization, where we have $\text{k}_1 \approx 10^6$.  Will we simply converge on the elementary step time for the formation of a base $\text{k}_+ \approx 10^6$ to $10^7$ (Craig et al., 1971; Porschke & Eigen, 1971)?  It seems reasonable to me that as $\text{C} \to \infty$ we'd have $\text{k}_\text{f} \to \text{k}_+$, which, I'd guess is the point where the rate limiting step is the time for the molecule to sample different rotamer/etc. configurations.  Perhaps it isn't an accident that these orders of magnitude match up?
Also, should we think about the entropy of complex formation in the same way in the limit of infinite concentration?  My guess for this would be "yes" since the entropic penalty of holding two molecules together, derived from studies on monoatomic gases, should be concentration independent as long as diffusion/mixing is allowed.  In practice, however, we'll probably have a gel at these higher concentrations.

I would love to try to answer this, but I don't know what you mean by "infinite concentration".  Clearly there can be no such thing so you must mean something else.  Can you edit your question to show what you mean?

I'd have to agree with Dr. Gans, infinite concentration is an odd concept to consider.

""I'd have to agree with Dr. Gans, infinite concentration is an odd concept to consider. – Lighthart"" Maybe in a black hole? :=)

When you go to the limit of infinite concentration (obviously you can never get there), a reaction cannot stay first order, because you will always run into some other rate-limiting mechanism thus making the reaction zeroth order in the highly concentrated species.
To give an example, for the binding of 2 molecules if both molecular species are 'infinitely' concentrated, the rate will not be limited by molecules 'finding' each other (as is normally the case for first order reactions), but it will be limited by the finite time binding process itself takes.


text:
Derivation of formula for the degree of unsaturation
Degree of Unsaturation (DU) of an organic compound tells us about the total number of $\pi$ bonds (both double bonds and triple bonds) and number of rings/cyclic chains in the compound.
Degree of unsaturation forms an integral part of the four tools used to identify the structure of an organic compound (some of the others being structure after hydration and ozonolysis (there's one more I don't remember)).
Wikipedia states the formula of degree of unsaturation as:

$$\text{Rings + $\pi$-bonds} = C - \frac{H}2 - \frac{X}2 + \frac{N}2 + 1$$
  where $C$ = number of carbons, $H$ = number of hydrogens, $X$ = number of halogens and $N$ = number of nitrogens

A more general formula is stated as

$$DU = 1 + \frac{1}{2}\sum{n_i(v_i - 2)}$$
  where $n_i$ is the number of atoms with valence $v_i$.

What I'm interested is the underlying concept from which this formula is derived.

Well, that is simple.
A Carbon (C) makes 4 bonds, a Hydrogen (H) or a halogen (X) makes 1 bond and Nitrogen (N) makes three bonds. So in a molecule with $c$ Carbons, $h$ Hydrogens, $x$ halogens and $n$ Nitrogens, we will have $4c+h+x+3n$ "bonding capacity".
Assuming there are no rings, no double/triple bonds, we must have $c+h+x+n-1$ bonds in between these atoms [think like how many dashes you can draw between $m$ $\ce{O}$'s: $\ce{O-O-O}$ two dashes for three $\ce{O}$'s apparently].
Since each bond uses one "bonding capacity" from each of the two atoms its binding, $c+h+x+n-1$ bonds are using $2(c+h+x+n-1)$ bonding capacity. So the remaining capacity is:
$$(4c+h+x+3n)-2(c+h+x+n-1)=2c-h-x+n+2$$
If this is larger than zero, it means our molecule is unsaturated, i.e. still has a capacity to make bonds. And our assumption that "there are no rings, no double/triple bonds" is wrong. This bonding capacity is used in forming new bonds either in rings or in pi-bonds.  Since each bond uses two capacity one from each side we should have $c-h/2-x/2+n/2+1$ "unsaturated" bonds.

Nice explanation +1


text:
Basis sets that can be used for macrocyclic complexes
What could be the possible basis set that can be used for a metal base macrocyclic complex in gaussian09?
I am using transition metals such as $\ce{Ni}$, $\ce{Co}$, etc.  Also, I have tried using DFT and HF theory for optimization but I am not getting appropriate results.
I am basically looking for binding of metal with the ligand in these coordinate complexes on basis of which i can use them for a particular application.
The basis set that I am using right now for DFT is Opt=(Cartesian,maxcycles=500)  b3lyp/gen  nosymm   pseudo=read SCF=(conver=5,vshift=400,maxcycles=1000)
Edit: Earlier I was using a specific basis set for transition metals and now I am using lanl2 as a single basis set for the DFT calculation. I am also using hf cal in which the sto3g basis set is used but iam getting two consistent errors in optimization which are convergence failure and number of step exceed error processed link 9999.

Could you also indicate what you want out of the calculations? For example, if you care about excited states, you may choose a different basis set than if you don't.

I approved an anonymous edit which I assume is you from another computer, however the line at the beginning of your file is not actually the basis set but rather a collection of directives that tells the calculation how to run. The basis set is a set of mathematical functions that are used to describe the AOs of each atom. In your case, the gen keyword tells G09 to read a basis definition included at the end of the file, as Deathbreath has mentioned. If the definition is missing your calculation will fail.

Additionally, Pseudo=read also means that the calculation will expect a pseudopotential definition. The other stuff in the directive line consists of geometry optimisation instructions (Opt=), a functional (B3LYP), manually disabled symmetry (nosymm) and SCF convergence information (SCF=).

replace scf=... with scf=xqc, if that fails try scf=qc and report back

It might be helpful if you would post your current input file.  There are many places where things could go wrong, and all we can do is conjecture what the solution should be.

In order to check on the availability of basis sets you may want to look at the Basis Set Exchange. After determining which basis set suits your problem (this will likely involve some literature search, Nickel, DFT, basis set, in Web of Science should do the trick), you can enter basis sets into Gaussian using the Gen keyword. BSE will export any selected basis sets in Gaussian format, so that shouldn't be a problem.

You didn't say anything about size of the system. Also what computer hardware is available for your computations.
For TM-complexes you need to get better basis than lanl2 and definitely you need much better basis for your ligands. sto3g is a joke. Start with 6-31G(D,P) and LANL2DZ. For good data go for 6-311+G(d) for non-metal and LANL2TZ+f for Ni and CO. Another possibility is def2-svp (it has all 4th raw metals in as well as non-metals).
General remark. I hope that you have pre-optimized your ligand before you started with the TM-complex.

Small remark: you would better start from 6-31+G(d,p) rather then 6-31G(d,p). See, for instance, http://dx.doi.org/10.1021/jp021590l. I quote from there: "Considering the performance vs cost tradeoff, we recommend that diffuse functions always be included on nonhydrogenic atoms in DFT and hybrid DFT calculations unless the sole goal of the calculation is absolute bond energies or absolute ionization
potentials."


text:
Why cannot HCN react with ethene?
I expected ethene ($\ce{C2H4}$) to undergo electrophilic substitution due to $\ce{HCN}$ but it doesn't happen, as I am told by my textbook.
I couldn't figure out the reason, as I found out that $\ce{HCN}$ does react with $\ce{CH3CH2Br}$ by nucleophilic substitution.  Upon ionisation, the proton of $\ce{HCN}$ is an electrophile,while the cyanide group is a nucleophile.  Correct?

Yes H+ is a electrophile while CN- is a nucleophile.

then is it a problem with acidity/basicity of ions?

substitution or addition?

Ethene can't undergo electrophilic substitution; only addition is possible The nucleophilic Substitution in $\ce{CH3CH2Br}$ is a different reaction. If you can be a more specific, we will be able to help in a better way.

The reaction between Ethene and $\ce{HCN}$ can happen catalytically and under high pressure (http://pubs.acs.org/doi/abs/10.1021/ja01650a034). Since the Cyanide ion is not as nucleophilic as the Bromide ion the double bond of Ethene needs to be activated by a transition metal catalyst.

@Philipp thanks,it says that addition of $\ce{HCN}$ to ethene occurs in the presence of dicobalt octacarbonyl.Thanks a lot.Can you please type it as the answer below so that we can close the question?

The "classical" way

Direct addition of $\ce{H-X}$ to alkenes proceeds via the addition of a proton to the $\ce{C=C}$ double bond in the initial step, i.e via a carbocation. This works nicely for strong mineral acids, but not for the weak acid $\ce{HCN}$ with the rather weak nucleophile $\ce{CN-}$.
Synchronous addition?
Q: Can't $\ce{H-CN}$ add synchronously add to $\ce{C=C}$?
A: NO! Such a reaction would involve the interaction of the HOMO (highest occupied molecular orbital) of one reaction partner with the LUMO (lowest unoccupied molecular orbital) of the other. The LUMO of an alkene has a nodal plane orthogonal to the $\ce{C-C}$ bond, while there is - obviously - no nodal plane but a $\sigma$ bond between hydrogen and carbon in $\ce{HCN}$.
The synchronous addition of $\ce{H-CN}$ to $\ce{C=C}$ is symmetry-forbidden!
Catalysis?
Yes! Since the first catalytic hydrocyanation using $\ce{Co2(CO)8}$ (see Philipp's excellent comment above), other catalysts have been developed, namely $\ce{Ni(0)}$ based species using a trypticene-derived chelating phosphine ligand (see image).

Ethene does not normally undergo nucleophile substitution reaction. Halogenation is an exception as the mechanism through which it goes forms a cyclic intermediate of Halonium ion and hence it substitutes electrophilically to $\pi$-bond.
But in $\ce{CN}$, no such case arises. (probably due to it being less nucleophilic than halogens)

Summarizing Phillip and Siddahrtha's comments.
Under normal conditions ethene normally undergoes addition since the double bond is nucleophilic.  In order to have a substitution reaction on ethene H $^{-}$, would need to be the leaving group.
$C_2H_4 + HBr \rightarrow CHCHBr + H^{-} $
This is not energetically favorable under normal conditions.  Under high pressure and with a catalyst the reaction would be possible.

I'm not sure this is relevant, since this electrophilic substitution is an acidic reaction.  Also, please check the balance of your equation.


text:
Why do sulfur and oxygen form sulfur dioxide?
Both sulfur and oxygen have 6 electrons in their outermost shell. Since both only need two electrons to stabilize, why doesn't just one atom of oxygen react with one sulfur atom, forming sulfur monoxide SO instead of sulfur dioxide $\ce{SO2}$?

One atom of oxygen is a fictional species outside of a plasma.  Oxygen is naturally diatomic, and the process by which it oxidizes sulfur is radical-based, leading to both oxygen atoms reacting.  The fact that SO2 forms instead of SO is a result of the [redundant answer] thermodynamics favor SO2 over SO.
Given that simple fact here is a quantum mechanical rationalization: Sulfur can expand its octet, which means in SO2 it has 12 electrons 'around' it.  This allows resonances forms where O- has a significant contribution to resonance, while S still has at least 8 electrons around it.  In SO no such resonance forms exist, meaning the highly electronegative oxygen must share its electrons with with sulfuer to avoid exposing a six-eletron sulfur species.  Hence SO2 should be a better structure than SO overall.

Its just like answering formation of $\ce{CCl4}$ against $\ce{CCl2 }$
for $\ce{SO}$ and $\ce{SO2}$

It is highly unstable
It becomes more stable after forming a double bond with O ( maybe resonance )

Structures:
 ,
It has an electronic configuration of $\ce{[Ne]} \: 3s^2\: 3p^4$
It has 4 bonds due to uncoupling of p electrons (a little energy is spent), but due to the formation of four bonds, a large amount of energy is released.

in most reactions I've come across, SO2 is evolved. I've rarely seen SO anywhere. So, in what kinds of reactions is sulphur monoxide formed?

FROM WIKIPEDIA     Production and reactions
Production of SO as a reagent in organic syntheses has centred on using compounds that "extrude" SO. Examples include the decomposition of the relatively simple molecule thiirane 1-oxide: as well as more complex examples, such as a trisulfide oxide,
C2H4OS → C2H4 + SO
The SO molecule is thermodynamically unstable, converting initially to S2O2. SO inserts into alkenes, alkynes and dienes producing molecules with three membered rings containing sulfur.

Generation under extreme conditions
In the laboratory sulfur monoxide can be produced by treating sulfur dioxide with sulfur vapour in a glow discharge. It has been detected in single bubble sonoluminescence of concentrated sulfuric acid containing some dissolved noble gas

Uses/occurence
-Ligand for transition metals
-Astrochemistry
-Biological chemistry

If SO is unstable like you said, why doesn't SO2 get formed in such cases? sorry I'm quite a novice here

It would be better if you explained why $\ce{SO}$ is unstable compared to $\ce{SO2}$. Currently the answer doesn't have much conceptual content.

I don't think this is a good analogy.  The two phenomena are completely different.  See answer below.


text:
Why does the conversion of SO₂ to SO₃ require high pressure?
The conversion of sulfur dioxide to sulfur trioxide in the contact process for the manufacture of sulfuric acid requires the following condition:

Temperature: 450 degrees
Pressure: 1-2 atm, although High pressure would fasten the reaction, creating those pressures is not economical
Vanadium Pentoxide ($\ce{V2O5}$) as catalyst
and excess of oxygen

$\ce{2SO2 + O2 -> 2SO3}$
My textbook says that high pressure favors the rate of the reaction. The reason it states is that the volume of product is less than that of the reactants.
Can somebody explain how the volume of the product being less causes high pressure to be required?

You might want to read something about Le Chatelier's Principle (http://en.wikipedia.org/wiki/Le_Chatelier%27s_principle)

so according to the same principle , if he remove generating S03 as the rate it generate, it will definitely speedups the process?

As Philipp stated in the comments it is Le Chatelier's principle in play.
The principle in layman's language states that

If you try to bring about any physical change in a system the system will try to but not necessarily succeed in cancelling the change.

The reaction you have given is an equilibrium reaction where all the compounds are in gaseous state.
Now if you count the total number of moles on the left-hand side it will be $\ce{2SO2 + O2}$
i.e. $\ce{2 + 1}$ that is $\text{3}$.
And on the right-hand side there are only $\text{2}$ moles of $\ce{SO3}$.
So ultimately if you consider the reaction to proceed in forward direction 3 moles of gas are converting into 2 moles of gas and hence the volume decreases in forward direction.
If you think according to Gas Law(or logically) an increase in pressure generally corresponds to a decrease in volume.
Now use Le Chatelier's principle to say that if I increase the pressure on the the system ,in order the system
to counteract to the change will decrease it's own volume and it has no other way of achieving that but to proceed in the forward direction.

But if the query arises that high pressure should favour any and every reaction take a look at the example below:
$$\ce{PCl5 <=> PCl3 + Cl2}$$
Here the gaseous moles on the left-hand side is $\ce{1}$ and on the right-hand side $\ce{2}$. So increasing pressure makes the reaction proceed in the backwards direction as there are less number of gaseous moles on the left-hand side.

In fact, in higher level this idea is what is called $\Delta\text{n}_\text{g}$ i.e the difference($\Delta$) of gaseous($_g$) moles($\ce{n}$).
$$\Delta\text{n}_\text{g} \quad = \text{gaseous moles on right-hand side - gaseous moles on left-hand side}$$

If $\Delta\text{n}_\text{g}$ is positive, then on increasing pressure reaction proceeds backwards and vice-cersa

so doesn't that mean high pressure would favour all reactions in terms of amount of yield? So isn't it right if I say, higher the pressure of any reaction- higher will be the yield?

@Ghost I edited the answer to incorporate your query.

The reaction you've provided is reversible, and hence it's able to proceed backwards. Now my question is , will high pressure favor each and every 'irreversible' reaction?(since it cant proceed backwards)

Le Chatelier's principle is not applicable on irreversible reactions. If the reaction is irreversible it will proceed all the way forward and changing the pressure will have minimal to null effect on the reaction.


text:
Why do the names of most chemical elements end with -um or -ium?
Why do the names of most chemical elements end with -um or -ium for both primordial and synthetic elements?

There is no -um only -ium

@NeilRoy I recognise that -ium is far more common, but molybdenum, lanthanum, and tantalum are counterexamples to your point.

Also stannum, plumbum, lots of them.

Zinc and wolfram (tungsten) do not end with um.

To expand on @BelieveInvis's answer -- in the early 19th century, when the Royal Society was really in the swing of things, the dominant language of scholarship was still Latin. Since Latin didn't have words for the new metallic elements, new words were coined from the existing terms for the substances and given Latinate endings.
From the OED's entry on -ium:
The Latin names of metals were in -um, e.g. aurum, argentum, ferrum; the names of sodium, potassium, and magnesium, derived from soda, potassa or potash, and magnesia, were given by Davy in 1807, with the derivative form -ium; and although some of the later metals have received names in -um, the general form is in -ium, as in cadmium, iridium, lithium, osmium, palladium, rhodium, titanium, uranium; in conformity with which aluminum has been altered to aluminium.
So, I think after that, other elements were simply given the suffix to fit the generally useful naming scheme, and then, metal names which were already in common use kept their common language names (e.g. gold as opposed to aurum) simply by force of usage.

The -ium suffix is a Latin suffix which forms abstract nouns, thus it is used to form chemical elements' name from its naming origin, such as minerals (calcium from calx) or person names (gadolinium).

To confirm and add a reference to the other two answers.

The deliberate naming of chemical elements such that they end in -um or -ium originates in $1811$. More specifically,

The Swedish chemist Berzelius, who recognised the value of logical naming, proposed (in $1811$) that the names of the elements should all be Latin in form.

Some elements had optained such a name prior to $1811$. Some were changed (albeit temporarily), and to some were simply added the suffix.

A number of names already given to metallic elements, e.g., uranium, chromium, barium, were of this form. He [Berzelius] used the Latin names ferrum, stannum and stibium for iron, tin and antimony, and gave zinc the "Latin" name zincum.
The names sodium and potassium were changed to natrium and kalium. All metallic elements discovered after this date ($1811$) have been given names ending in -ium (occasionally -um). Examples are cadmium, lanthanum, lithium, thallium, radium.

This is also why helium has the suffix.

The name helium is anomalous. This element, which was first discovered in the sun, was assumed to be an alkali metal, but when it was discovered terrestrially it was found to be an inert gas.

W. E. Flood. The Origins of Chemical Names. 1963, Oldbourne, London. (page xii)

Well, they end in -um because it's a good Latin neuter singular of the second declension: that is, it's a way you can refer to 'stuff' of the sun (helios), or of Magnesia (where magnesium and magnetic stuff was found), of lime (calx, calcis). Some elements have -ium because they're derived from stems that have an -i at the end, like the ones listed above. But others, like lithium, don't actually have an -i at the end of the stem etymologically (cf. Ancient Greek lithos). There, it seems instead that the people in charge of the naming reanalyzed the original pattern of adding a plain -um to the end of the stem, and instead added -ium on analogy to several elements formed from Greek or Latin -i stems.
And to make it official, the IUPAC (chemistry's governing body) now basically requires any new element to end in -ium (since 2002), so that's pretty much all we're going to see from here on out.
(Source: http://media.iupac.org/publications/pac/2002/pdf/7405x0787.pdf)

Though 117 Tennessine and 117 Oganesson do NOT end in -ium. How did that happen?

"the suggestion for the element 117 was tennessine, with a symbol of Ts, after "the region of Tennessee". From the wiki article - it's named after Tennessee and they (probably) tried to keep the name very close, phonetically, to Tennessee. Also Tennessium or Tennesseeium both sound awful and/or like they're named after sports.

@ErikE look above in the periodic table in the same group:  tennessine, astatine, iodine, …; vs  oganesson, radon, xenon, …. The mentioned IUPAC rule is not general: Any new metallic elements should be given names ending in -ium.


text:
Why does the boiling and freezing-point of non-polar molecules rise with increasing molecular mass?
In the section about fuels, my textbook mentions that the boiling and freezing-points of alkanes (which are non-polar) increase with an increase in molecular mass due to London dispersion forces.
Now, I understand what these are and why they become relevant when talking about non-polar molecules, but I can't figure out how they are connected to the boiling and freezing point of the matter. How do these London dispersion forces influence the boiling and freezing point?

Short answer:   Boiling point and freezing point depends wholly on the inter molecular forces/interaction. The greater the inter molecular forces, the greater the boiling point.

In general, the larger the molecule, the larger the accessible surface area of the molecule and the more surface polarisable electron density. There are more atoms that can participate in dispersion interactions and the effect stacks up. To melt or boil a substance you need to overcome the intermolecular forces holding the molecules together and dispersion forces -  whilst weak even with respect to hydrogen bonding - can add up to large interaction energies. One of the more interesting examples of strong dispersion interactions is the gecko's foot, which permits the reptile to easily stick to sheer surfaces due to the exceptional surface area of the footpad, which is covered in many many tiny hairlike protrusions.

But how does the molecule's mass contribute to it's intermolecular forces?

@EdwardStumperd - I'm going to assume that the example given in your textbook is a homologous series of alkanes or similar, in which case more massive molecules correspond to later members in the series, with greater chain lengths and correspondingly larger volumes and areas.


text:
What is the process for purifying KOH?
I'm partnered with a jewelry production company whose plating baths require $\ce{KOH}$. We're located in Indonesia, where import fees are absurdly high, and reliable suppliers are rare. A local shop sells $\ce{KOH}$ at \$5/kg, but whatever impurities are present in their mix cause enough problems to spoil the entire bath. (Expensive mistake when doing gold plating.) The only alternative we can find is a Merck redistributor, who sells $\ce{KOH}$ at close to \$100/kg, which is ridiculous.
Is there a relatively simple process we could learn (of the DIY variety) for removing impurities, turning non-lab-grade $\ce{KOH}$ into lab-grade $\ce{KOH}$?

According to Armarego, et al. (Purification of Laboratory Chemicals, 6th edition), they only provide a procedure to purify a solution of $\ce{KOH}$, likely owing to its very high solubility.  Further, the only impurity of note they cite is carbonate (from standing in contact with air), so if this is not your impurity then this procedure will not help.
If this is your impurity, they recommend adding a saturated solution of a barium salt, either $\ce{BaCl}$ or $\ce{Ba(OH)2}$, shaking, and then setting aside to allow the insoluble $\ce{BaCO3}$ to settle.  It sounds like you are working on large scale, so it is important to recall that the procedure just described should be done in several small batches to work best.
Not certain this is what you are looking for, but I hope it's a start.

I read somewhere that non-lab-grade KOH usually contained heavy metals. We'll try a barium salt process and see if things improve :) Thanks!