text:
Why is gold golden?
Bulk gold has a very characteristic warm yellow shine to it, whereas almost all other metals have a grey or silvery color. Where does this come from?
I have heard that this property arises from relativistic effects, and I assume that it has to do with some distinct electronic transition energies in the gold atoms. But what changes with the "introduction" of relativistic effects, that then changes the energy of the frontier orbitals in such a drastic manner?

See also this related Physics.SE question (and answers): http://physics.stackexchange.com/questions/72368/why-are-most-metals-gray-silver

Do you know, by the way, that not all orbitals are contracted due to relativistic effects? It is known that relativity contracts $s$ and most of $p$ orbitals and either contracts or expands orbitals of higher $l$ quantum number.

I heard somewhere (Science Channel, to be specific) that gold on an atomic level is actually pink. How does increasing the number of gold atoms make it golden?

Wildcat, there are different levels of understanding of any topic.  The ability to study and follow, for the most part, an answer like the one accepted is one level.  The ability to then give a proper and thorough answer to your question in 600 characters or fewer, is quite a different level of understanding.  This may be long-gone water under an ancient bridge at this point but I felt I had throw out my thoughts anyway.

You could more information from https://www.fourmilab.ch/documents/golden_glow/.

Yes, this is a beautiful question.
As you said, in lower rows of the periodic table, there are relativistic effects for the electrons. That is, for core electrons in gold, the electrons are traveling at a significant fraction of the speed of light (e.g., ~58% for $\ce{Au}$ $\mathrm{1s}$ electrons). This contracts the Bohr radius of the $\mathrm{1s}$ electrons by ~22%. Source: Wikipedia
This also contracts the size of other orbitals, including the $\mathrm{6s}$.
The absorption you see is a $\mathrm{5d \rightarrow 6s}$ transition. For the silver $\mathrm{4d \rightarrow 5s}$ transition, the absorption is in the UV region, but the contraction gives gold a blue absorption (i.e. less blue is reflected). Our eyes thus see a yellow color reflected.
There's a very readable article by Pekka Pyykkö and Jean Paul Desclaux that goes into more detail (if you subscribe to ACS Acc. Chem.Res.)

@GeoffHutchison I wish you could answer this in a way I can understand. I have a more general question of "why are some wavelengths reflects while others absorbed?" yet with all your voodoo terms and stuff I'm not even sure what's being said.

I fear it too closely resembles this question.

Is this the same for copper?

@fredsbend For copper, there's no significant relativistic effects in play. It's a similar type of electronic transition ($\ce{3d->4s}$), which is also in the visible - based on the poor shielding of the 3d electrons. So the reason is a bit different even if the result is similar.

Adding to Geoff's source, I very much like this recent paper from Pyykkö with some updated data.

It would help if the answer emphasized why is gold different.

In quantum chemistry, electrons do not have a well defined momentum. Instead, they have a distribution of momenta around the atom. Your answer doesn't really make sense.

Even if there's a distribution of possible momenta (or velocities), that distribution still has an expectation value, which can be evaluated using quantum chemical methods. So yes, we can evaluate expected velocities (see @ron's answer) and we can compare this observable to various measurements.

Moreover, while Heisenberg uncertainty limits some accuracy of knowing momentum, position and both, we can still make fairly good measurements on velocities. It is well established that core electrons do move at significant fractions of the speed of light. (Again, @ron's answer gives the rule-of-thumb formula.)

@GeoffHutchison Just fyi I've just asked Can the color of gold metal be reproduced by modeling only the conduction electron band structure, or do atomic absorption lines contribute? in Matter Modeling SE and referenced your answer there.

See the footnotes I've included if you would like to see more of the detail behind a specific statement.
The figure below compares the reflectance spectrum for silver and gold (let's forget about aluminum, it's not relevant to this discussion; also keep in mind that where reflectance is low, absorbance is high and vice-versa).

The absorption (reduced reflectance) seen at lower wavelengths by silver and gold is caused by an electron transition (caused by the absorption of light) between their $\mathrm{4d \rightarrow 5s}$ and $\mathrm{5d \rightarrow 6s}$ orbitals respectively. Silver absorbs (reflects less) in the ultraviolet, no visible light is appreciably absorbed. Therefore, if we shine white light on silver, none of the visible wavelengths are absorbed, they are all reflected and silver appears as a shiny, white metallic substance. In fact, jewelers say that silver is the whitest of the metals. On the other hand, gold absorbs (reflects less) blue light in the visible region.  When white light shines on gold, the blue light is absorbed and yellow light is reflected (see this article on complementary colors for more detail on how removing certain colors from white light produces different colors).  So our question becomes, why is gold's $\mathrm{5d \rightarrow 6s}$ electronic transition shifted into the visible region?
$\mathrm{s}$-Electrons are strongly attracted to the nucleus because their probability density is higher near the nucleus than other types of orbitals ($\mathrm{p, d, f,}$ etc.). Simple Bohr theory tells us that these strongly attracted $\mathrm{s}$ electrons in atoms with a large number of protons (heavy nuclei) would have to travel at a considerable fraction of the speed of light to avoid "falling into" the nucleus (1). Special relativity takes things a step further and explains that, for these heavy nuclei with their $\mathrm{s}$ electrons traveling over half the speed of light, the electron mass will increase and as mass increases (2), orbital radius decreases (3).
Further, as the orbital radius decreases and the electron is closer to the nucleus it becomes more stabilized and the energy of the orbital decreases. As the various $\mathrm{s}$ electrons (and to a lesser degree $\mathrm{p}$ electrons, but $\mathrm{d}$ and $\mathrm{f}$ electrons don't have enough electron density close to the nucleus so they are not directly affected by these relativistic effects and their orbits are not contracted) move closer to the nucleus, the other non-$\mathrm{s}$ (or $\mathrm{p}$) electrons are better screened from the nucleus. So although the various $\mathrm{s}$ orbitals are contracted and placed at lower energy, the various $\mathrm{d}$ and $\mathrm{f}$ orbital radii shift further out and occur at higher energy due to this increased screening effect.
Hence, in the gold atom, we would expect the $\mathrm{5d}$ orbital to be raised in energy (screening effect) and the $\mathrm{6s}$ orbital to be lowered in energy (relativistic orbital contraction effect). The net effect is a smaller energy separation between these two orbitals and a red-shift in the $\mathrm{5d \rightarrow 6s}$ absorption to the lower energy (compared to the absorption of silver which is not subject to the same magnitude of relativistic effects) blue region of the spectrum, resulting in a yellowish color for light reflected from gold.

(1) The ratio of the speed of an electron traveling in the first Bohr orbit to the speed of light is given by the handy equation:
$$v_\mathrm{rel} = \frac{Z}{137}$$
where $Z$ is the atomic number of the element under consideration and $137$ is the speed of light in atomic units, also known as the fine structure constant.  Consequently, a $\mathrm{1s}$ electron in silver $(Z=47)$ will travel around $\frac{47}{137} = 34\%$ the speed of light, while a similar electron in gold $(Z=79)$ will travel at $58\%$ the speed of light.
(2) The relativistic mass of an electron is given by:
$$m_\mathrm{rel} = \frac{m_\mathrm{e}}{\sqrt{1-\left(\frac{v_\mathrm{rel}}{c}\right)^2}}$$
where $m_\mathrm{e}$, $v_\mathrm{rel}$, and $c$ are the electron rest mass, the velocity of the electron, and the vacuum speed of light, respectively. The following figure provides a graphical representation of the result.

(3) The following equation relates the ratio of the relativistic radius of the first Bohr orbit $R_\mathrm{rel}$ to the normal radius $R_\circ$, to the relativistic velocity of the electron:
$$\frac{R_\mathrm{rel}}{R_\circ} = \sqrt{1-\left(\frac{v_\mathrm{rel}}{c}\right)^2}$$
As the relativistic velocity of the electron increases, the orbital radius contracts (the above ratio becomes smaller).  For silver, the first Bohr radius contracts ~6%, while for gold the contraction is ~18%.

If the effect only relies on the atomic number of the nucleus, how come it is not observed for lead, thallium or mercury?

I know that the effect does manifest itself for mercury, but in other ways.  See, "Why does Mercury have low melting and boiling points?".  I suspect that other heavy atoms show the effect in some way as well.  Since the electron configuration of mercury, lead and thallium is different from that of gold, the 5d-6s transition may no longer be the "operative" absorption.

In particular, mercury, lead, thallium, etc. don't have partially filled 6s orbitals. So a 5d-6s transition is impossible. As @ron said, the relativistic effects show up in different ways (e.g., liquid mercury, low melting lead, inert pair, etc.)

Is this difference in Ag and Au due to lanthanoid contraction?

This effect comes from the band structure of the metal, not atoms.  In simplest terms (see, e.g., Ashcroft & Mermin, Solid State Physics, chapter 1) the electrons in the conduction band have a plasma frequency $\omega_{p}^{2} = 4\pi n e^{2}/m$ - this is a collective excitation of the metal. For $\ce{Cu}$ (reddish), you start exciting the plasmons around $\pu{2eV}$ (red), while for $\ce{Ag}$ you don't hit it until $\pu{4eV}$ (near UV).  $\ce{Au}$ is a bit higher than $\ce{Cu}$ but still in the visible.
If you want an excellent mirror across the visible, you need to use $\ce{Ag},$ not $\ce{Au-Au}$ will absorb the blue.
EDIT - additional information.
Coming from a solid state background, I'll add some more on band theory. Again, following Ashcroft and Mermin's discussion of noble metals (Chapter 15):

In the case of copper (and the other noble metals) at least six bands are required (and six turn out to be enough) to accommodate the eleven additional electrons.  ...
  For almost all waver vectors k the six bands can be seen to separate into five lying in a relatively narrow range of energies from about $2$ to $\pu{5eV}$ below E_F, and a sixth, with an energy anywhere from about $\pu{7eV}$ to $\pu{9eV}$ below E_F. ... It is conventional to refer to the set of five narrow bands as the d-bands, and the remaining set of levels as the s-band.  However, these designations must be used cautiously, since at some values of k all size levels are close together and the distinction between d-band and s-band levels is not meaningful. The nomenclature reflects the fact that at wave vectors where the levels do clearly group into sets of five and one, the five are derived from the five orbital atomic d-levels, in the sens of tight binding...

So it is tempting to jump on the d-band desciption and say, hey, since there are relativistic effects in $\ce{Au}$, than that must carry over into the band structure.  I'm really not convinced.  Yes, while the levels may well derive from d-levels of the atom, that does not mean that the resulting band structure is or is not different from a non-relativistically impacted band structure. As an example, look at the Fermi surfaces of $\ce{Cu},$ $\ce{Ag},$ and $\ce{Au}$ as plotted in Ashcroft and Mermin (Fig. 15-5) - they are all free-electron like, except that contact is made along <111> directions.  It is very difficult to differentiate the Fermi surfaces of $\ce{Cu}$ and $\ce{Au}$, while $\ce{Ag}$ is different (actually more free-electron like). This shows up in the table of Belly-to-Neck areas of the Fermi surface (from De Haas-van Alphen oscillations), where $\ce{Cu}=27,$ $\ce{Au}=29,$ and $\ce{Ag}=51$.
When you overlap the orbitals to make bands, all that matters is their relative placement with respect to the Fermi energy. In the big picture, $\ce{Cu},$ $\ce{Ag},$ and $\ce{Au}$ are all very similar - it is just that our eyes are sensitive to just the small energy region where their plasma frequencies differ, so we see a big difference.
If you are going to ascribe $\ce{Au}$'s golden color to relativistic effects on the orbitals, than I'm going to have to ask how $\ce{Cu}$ has such a similar band structure without the relativistic effects.

While you accurately describe the how, the why part is kind of lacking at the moment. I do not want to deter you, and I like the way the answers are diverse. But I feel you have not addressed the underlying reason.

Copper, silver and gold all have $\ce{d->s}$ transitions (yes, in the band structure). Copper has a $\ce{3d->4s}$ transition, which is also in the visible - based on the poor shielding of the $\ce{3d}$ electrons. So yes, if we're comparing silver to gold, the appearance is definitely relativistic, and you can look through the linked paper above for more science.

I'm happy to hear the physics perspective, but yes, the relativistic effects in Au do carry over into the band structure. For example, if you do DFT band structure calculations with Au, you must use relativistic core potentials.

Here's a quote from Pykko's paper "While the relativistic 5d-to-(6s+6p) interband absorption is calculated to occur at h$\nu$ = 2.38 eV, in the middle of the visible range, in good agreement with experiment, the corresponding non- relativistic band gaps are much larger."

@GeoffHutchison - I think we come at it from different perspectives.  While, no doubt, the wave functions of Au are influenced by relativistic effects, and in DFT or tight binding you need to use an appropriate relativistic-corrected potential, the fact remains that the end result of the band structure calculation looks remarkably like that of the other noble metals. A band structure calculation using improper orbitals will, of course, not result in the actual bands. But, the actual band structure may not owe a lot in general form and function to whether there are relativistic effects at play.

Can you explain why the fact that Ag doesn't absorb blue (what about absorbing other colors) means its excellent for mirrors?

I liked this answer as for the OP could have asked the same using Cu instead of Au

@JawadAlShaikh - the plasma frequency is above the visible (for humans), so Ag does a good job of reflecting at high efficiency across the entire visible spectrum so what is reflected is pretty much what hits the mirror in terms of color balance. Consider a mirror made of silver vs one made of gold (or copper).

You might want to see this: https://mattermodeling.stackexchange.com/q/6189/5


text:
How do you melt metals with super high melting points?
At the Renaissance fair a few years back I was watching a smith forge metal into shapes. During this time a very odd question came to me. I was wondering what the furnace was made of. My logic stated that whatever the furnace was made of must have a higher melting point than the materials he was melting. This quickly turned into an elemental arms race resulting in an odd question of how do we melt stuff like refractory metals (more specifically the one with the highest melting point) so we can melt other things inside of it.
Now I know that (for some odd reason I don't understand) rapid cooling can manipulate the strength of an item. Is there a similar property to manipulate the melting point?
Note: My current best guess (like can be done to make weapons harder) is that we take two elements, melt them, and the resulting compound has a higher melting point.

The level of positive reaction to this not only makes me not feel stupid for my question but gives me a fuzzy feeling inside. Thank you.

The vessel holding the metal doesn't have to be as hot as the metal itself if you have another way of getting energy into the metal. Some refractory metals can be vaporised from cooled containers by directing electron beams or other energy sources into the metal (not for large scale melting but often used in synthesis or vapour deposition processes).

Tungsten's melting point of 3422 °C is the highest of all metals and second only to carbon's, for which melting occurs only at high pressure (there's no standard melting point). This is why tungsten is used in rocket nozzles and reactor linings.  There are refractory ceramics and alloys that have higher melting points, notably $\ce{Ta4HfC5}$ with a melting point of 4215 °C, hafnium carbide at 3900 °C and tantalum carbide at 3800 °C.
Carbon cannot be used to hold molten tungsten because they will react to form tungsten carbide.  Sometimes ladles and crucibles used to prepare or transport high melting point materials like tungsten are lined with the various higher melting ceramics or alloys.  More typically tungsten and other refractory materials are fabricated in a non-molten state.  A process known as powder metallurgy is used.  This process uses 4 basic steps:

powder manufacture - a variety of techniques are available to generate small particles of the material being worked
powder blending - routine procedures are used to blend the constituent particles into a uniform mixture
compacting - the blended powder is placed in a mold and subjected to high pressure
sintering - the compacted material is subjected to high temperature and some level of bonding occurs between particles.

Materials like tungsten may also be shaped through electrochemical machining (ECM) or electrical discharge machining (EDM).

Ta4HfC5 isn't technically an alloy. It is a ceramic (a mixture of tantalum and hafnium carbides). Tungsten has the highest melting point of any metal or alloy. Alloys typically have lower melting points than the metals used to form them.

"transport high melting point materials like tungsten". I would LOVE to see a container pouring out pure liquid tungsten the way we do with iron or bronze casting.

Sorry, can't comment here, but I wanted to more directly answer your question.
Blacksmiths avoid melting their forges because the "heat" that can melt or oxidize iron and steel is actually contained in a ball in the center of the coal. In fact, maintaining coal "structure" is an important skill in blacksmithing.
To clarify better, imagine a hollow in the center of a pile of coal. This is where the temperatures rise past 2000 °F, since the heat reflects back in on itself due to the coal molding into a refractory ball of sorts.
And yes, sometimes your ball does fall apart, or you've structured it poorly - and then you notice that the cast iron drain cover that protects your airflow intake has melted through.

There's an interesting analogue to this in modern fusion reactors. For example, in JET, a thin ring of plasma hotter than the core of our sun (and indeed anything known in our galaxy) is held in place by magnetic levitation. While the ring itself is hot enough to destroy any material it touches, the thermal radiation falloff (iirc due to Planck's law and the Stefan–Boltzmann law) allows them to use supercooled graphite panels to protect the reactor, staying well within the 3550°C melting point of carbon.

We use an levitating furnace to heat samples of refractory ceramics up to approx $3000~^\circ\mathrm{C}$. Its for research purposes, so the samples are small (2 mm) beads. These are balanced on a jet of argon and heated with $\ce{CO2}$ lasers.
Here is a paper which talks about the technique:
D. Langstaff, M. Gunn, G. N. Greaves, A. Marsing, and F. Kargl, Rev. Sci. Instrum.; 2013, 84, 124901. (Mirror)

One could melt them floating on a pool of high-boiling point denser metal, or in space where they can readily be contained. Or one could create a thick actively cooled shell and melt them inside it, melting part of the shell as well. Finally, it's probably not very practical, but one could use an air jet to keep then suspended away from other matter and then melt them with lasers or superheated air.

There are two alternatives to the other answers here, though whether they can be used on a large scale is open to question.
The first is to use an actively cooled vessel to hold the metal and a method of getting energy into the metal not based on the heat of the crucible. Many metal-vapour reactions (used for small scale chemistry research) do this and provide sufficient energy to vaporise even refractory metals using electron guns. See Malcolm Green's site (and this entry "The synthesis of the first zerovalent compounds of the early, refractory transition metal via the development of the electron-gun metal vapour synthesis experiment").
The other method is to use inductive heating of the metal. This can sometimes work even without a vessel at all as a suitable inductive coil will levitate the lump of metal and the induced eddy currents will dump enough energy into it to melt it. There are plenty of youtube videos of this with non-refractory metals such as aluminium but the principle should still work for high melting metals.

source: Stack Exchange

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

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

sort: votes


text:
How does salt and water promote corrosion?
I was always wondering why iron only rusts when it gets wet. I know that for example aluminium oxidises on its own on regular basis.
I also heard many complains that salt used to remove snow from roads increases corrosion rate of car undercarriage.
Recently, I was researching how to produce iron oxide to improve rocket candy fuel - and among with oxidising substances, especially hydrogen peroxide, salt water was suggested.
What do these things have to do with aerial oxygen reacting with iron? I have also noticed that copper roofs turn green. Copper stuff kept away from rain doesn't do that so fast.

The green copper is actually copper carbonate.

I think a tantalising clue can be found in the principle of cathodic protection used in industry.
Cathodic protection is used in the shipping industry to prevent iron/steel hulls from rusting. Essentially you make your hull into a cathode by connecting a (typically) zinc anode to the hull. How do we know the ship's hull is the cathode and zinc the anode?
And zinc in fact?
Lets go back to the standard reduction potential tables. The following reaction has a reduction potential of $\ce{-0.7618V}$:
$$ \ce {Zn^2+(aq)} + 2{e}  \ce{->} \ce{Zn(s)}  $$
Whilst the following has a reduction potential of $\ce{-0.44V}$:
$$ \ce {Fe^2+(aq)} + 2{e}  \ce{->} \ce{Fe(s)}  $$
Clearly the rusting process works in the opposite direction (and thus so do the above equations) which just means we reverse the voltage to get the oxidation value.
A higher reversed voltage means that that process has a greater tendency to occur, which means that zinc here is preferentially oxidised over iron. Hence if we connect an iron hull to a zinc rod, the zinc acts as the anode whilst the hull is the cathode i.e any $\ce{Fe^2+}$ directly formed by reaction with sea-water is pushed back to iron metal by the production of $\ce{Zn^2+}$, a process which liberates two electrons.
This should suggest that some electrolytic process contributes to salt water affecting the oxidation of iron. The most basic explanation is that the sea water acts as an electrolyte, which facilitates this process much more easily than, say, distilled water.
What does this mean however when we don't have an electrochemical cell as above, and only have a single electrode i.e. the ship's hull?
We might like to look up the standard reduction potentials for both chloride and iron, however we quickly see that this isn't the case; both chloride and iron metal have to lose electrons in our case, which isn't going to form an electrochemical reaction.
It's an interesting case that bleach (alkali salts of chlorite, $\ce{ClO-}$)undergoes the following reaction with iron:
$$ \ce {Fe(s)} + \ce{ClO^-(aq)} +\ce{H2O}  \ce{->} \ce{Fe^2+(aq)} + \ce{HClO2(g)} +\ce{2H+(aq)}  $$ which has a favourable overall voltage of around $\ce{2.11V}$.
So we've seen that chloride doesn't directly react with iron, but chlorite does. This is a possible route for corrosion, although it's not obvious by which natural process chlorite comes about.
Another point to note is that seawater just doesn't contain sodium chloride, but myriad other ions which can contribute to this corrosion process,especially sulphate.
$$ \ce {SO4^2-(aq)} + \ce{4H^+(aq)} + \ce{Fe(s)}  \ce{->} \ce{Fe^2+(aq)} + \ce{SO_2(g)} + \ce{2H2O(g)} $$ which occurs with a total voltage of $\ce{0.61V}$, which tells us that the chlorite process is much more favourable thermodynamically.
So in conclusion, there are many different reactions that can utilise seawater to oxidise iron—it's not incredibly clear which ones do contribute, and which ones don't, but the above provides the means for some educated guesswork, so to speak.
As a final point, iron can oxidise in air without water, it just takes a long time. The presence of an electrolyte will speed up the process however. Corrosion also needn't be a reaction with oxygen either, but any process which makes iron ions from iron metal.

Uh-oh. I have a lot to learn. How can reduction by 2 electrons have different voltage every time? Thank you anyway - I'll try to learn so that I can understand the answer completelly.

If we understand voltage as a driving force 'the tendency for electrons to flow' then voltage differences can be due to one electron being removed by a more electronegative element. If we see the differences in half-cell potential between electropositive Li and electronegative F, this gives us a huge cell potential, whereas less electronegative/positive couplings give us smaller voltages. This is at its most basic level however.

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


text:
Introduction to Chemistry Textbook - Theory based

TL:DR conditions below

Highschool to first year undergraduate level
Theory based(Experimentation not wanted)
General - Giving a good introduction to all or most areas of Chemistry
Rigorous - Professional formatting - Not a casual textbook

I am looking for a textbook for someone(me) who has done no chemistry, but isn't too basic as it progresses. What are some of the better textbooks for starting to learn chemistry? I don't know any specific fields, so I would like to get some exposure to a number of different topics. I am a third year Mathematics student, so I am not worried about it being too concise, I am willing to reread areas one hundred times if need be.
One note: I am not interested in doing experimentation at all, and would just like to learn theory, as I have (sometimes) embarrassingly bad knowledge of the field as I didn't take chemistry in highschool(which I greatly regret).
Highschool through to the end of first year undergraduate level preferably, or alternatively multiple textbooks by the same author of successive difficulty.

This is too broad.  We accept book requests, but they must be highly specific (listing out concepts or particular topics that are of interest).

This is a decent book request, just so you can compare.

@jonsca I think this isn't a fair comparison. I don't have any experience, I can't be specific, and I don't know what topics a good introductory Chemistry textbook would cover. That is exactly why I am asking on a website with plenty of more experienced people. Wouldn't it be reasonable to have a single page dedicated to such a question for users to find when they search google(which I had previously).

I would even be happy if you looked on Amazon and were able to narrow it down to a few that you were interested in.  You should be able to tell from the reviews which ones are at an introductory level.  We don't do generic book recommendation questions because they are impossible to maintain and go stale after a while.  Other sites may do them, but we are not other sites.

I'm going to reopen this for now since you've at least attempted to narrow it down.  If the community (some of whom are chemistry professors) decides it's still too broad, I will not stand in the way of them closing it.

If I were still a student, the first thing I would do is go to the library and thumb through some of what they have, noting how well each meshes with your learning style and needs, honestly.

@jonsca That is an interesting idea, I will do that later this week. Thank you.

@Thursday maybe Silberberg? Not the best, but at least it's okay. Others will say Oxtoby, which is very theory based.

Chemistry by Zumdahl is very good. I teach General Chemistry I and this is the book we use. I still reference it frequently. You will want the instructor solutions to accompany it. I let students check this out for classroom use, but you should have it for yourself at least. (I saw for the first time this semester a version that is loose leaf; this is great since it is sort of big. We have a custom version that takes out parts not covered. There is also a nice ebook.) This book is lovely, except for electron affinity.
For introductory chemistry, Zumdahl and DeCoste wrote Introductory Chemistry: A Foundation. We use this for Chem 101. Also very good, but a little too elementary for my taste and there is not as many topics discussed. But this may be what you want. You are probably somewhere between these two.

The next two are probably more rigorous than you would like, but worth mentioning.

Principles of Instrumental Analysis by Douglas A. Skoog is outstanding, but probabbly more advanced than you would like.
Physical Chemistry - Atkins. This book is very dense and is definitly rigorous. This will take an undergrad 2-3 semesters to get through. This may be a little overkill. You will actually see many general chemistry textbooks take exercises from this book. Really need the solutions bad for this book, it is that hard. A great place to find exam questions.

Just want to add my two cents about loose leaf: it's cheap and portable but does not take any abuse. If you want a book that you can continuously reference, I recommend hardcover.

This may be the difference between university and high school. I have trouble getting students to bring the book to class, so it is nice for that. But definitely I agree, hardcovers are much better quality.

"People either like Atkins' Physical Chemistry or loathe it." (c) I am personally on the loathe side and not only with respect to his Physical Chemistry book.

I guess the issue with Atkins is that because it tries to touch base on all topics in Physical Chemistry so it is rigorous to follow. As a physical chemist, I find Atkins way better than any other book including McQuarrie. I think McQuarrie has a great discussion of XRD, but that is about it.

@PhysicalChemist, I must admit that there is no one good book on PChem. One can have 4 good books (one on classical thermodynamics, one on statistical thermodynamics, one on kinetics, and one on quantum chemistry), but not just one book.

Silberberg or Zumdahl Chemistry are great introductory reference books in my opinion. They cover the topics in just the right amount of depth and explain concepts extremely well, I highly recommend it.
Inorganic Chemistry and Physical Chemistry by Peter Atkins also seem to be decent (though very dense) (I haven't used it much though).
I would also suggest Clayden et al.'s Organic Chemistry - it's a great textbook for organic chemistry though a bit intimidating at first- you should probably go through the organic chemistry section of Silberberg or Zumdahl first (nomenclature, functional groups and naming) before they use this.

I'm surprised that no one has mentioned Huheey yet.  This text has less emphasis on descriptive chemistry but explains the underlying concepts extremely well.  As an undergraduate reading it (it wasn't an assigned text) was quite an enlightenment.

It all depends on how much Math you want to do.
The standard "from zero" intro Chem book where I teach would be Hein/Arena's Foundations of College Chemistry. It covers the material just broadly enough, going over the basic beats for everything from gas laws to ochem. The math level is arithmetic/basic algebra.
The most thorough Algebra-based text I found has been McQuarrie's General Chemistry. It has loads of exercises and they purposely don't change editions very often so as to keep the book cheap. The concepts are presented and built on in a way that makes sense, and since it's written by a pchem guy it doesn't skimp on the rigor.
There are two standard Calculus-based Chem texts: Oxtoby's "Principles of Modern Chemistry" and Atkins' "Chemical Principles: The Quest for Insight." Both of these are written by physical chemists and will serve you well as future reference books (they are rather encyclopedic) if you're comfortable with a physical chemistry focus. I've also seen Siska's University Chemistry get good reviews. It's slimmer so there's a lot more required of you in terms of what you need to figure out vs what is explicitly explained.


text:
How to determine the packing geometry of compounds with contradicting coordination number and radius-ratio values?

Determine the packing geometry of the following compounds by using the radius ratio rules. For your convenience assume a coordination number of six for both cation and anion.
Be specific in the your answer, "body-centered cubic with the anion occupying the corners of the cell [1 atom] and the cation occupying the central hole [1 atom].

Below is the table we are given, and we're are choosing between geometries of cubic, body-centered, or face-centered.

The compound I'm on is $\ce{LiBr}$
I did $\frac{r^+}{r^-}=\frac{79}{182}=0.43$ for the Radius-Ratio value.
But from here would I choose square planar because it's closer to 0.414 or do I choose octahedral because we're assuming coordination number 6? And how would I determine the locations of the cation/anion?
My best guess is cubic geometry because the cation is too large to fit tetrahedral holes and octahedral holes, with $\ce{Li}$ and $\ce{Br}$ taking occupying complementary corners.

I'm making the assumption that the coordination number 6 is used as a reference to the atomic radii.
I've currently adjusted my answer based off of other examples I could find to

FCC with OH holes occupied by $\ce{Li^+}$ [4 atoms] and the $\ce{Br^-}$ forms the FCC Lattice [4 atoms (by stoichiometry}] for a 100% occupancy rate.

But I still have an issue as to how to conceptually approach the problem.

You've calculated your radius ratio value and you know you're between 0.414 and 0.732, so cubic is off the table, based on your chart.
You're looking at either square planar or octahedral. Your chart also says that ionic compounds do not use square planar packing and the coordination number would have to be 4. LiBr is an ionic compound (like NaCl) with a coordination number of 6, therefore you're looking at octahedral.
From there you can determine your "specific" answer.

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


text:
Dry adiabatic lifting
If an unsaturated air parcel ascends adiabatically in the atmosphere with an initial temperature, would the final temperature of the air parcel, when it reaches it's destination, be the same at the initial temperature since it does not exchange heat with it's surroundings?

No. When the gas - let's assume it is an ideal gas - rises up it will expand because the pressure $p$ sinks the higher it comes. If it is allowed to expand quasi-statically under adiabatic conditions ($\delta Q = 0$) then it does work $W$ (pressure-volume work $W = - p \mathrm{d}V$ to be more precise) on its environment, and, hence, its internal energy $U$ is reduced, and its temperature $T$ changes because according to the first law of thermodynamics
\begin{align}
  \mathrm{d} U &= \delta Q + \delta W \\
  &\overset{\delta Q = 0}{=} \mathrm{d} W = - p \mathrm{d}V
\end{align}
and also for an ideal gas you have
\begin{align}
  \mathrm{d} U &= C_{V} \mathrm{d}T
\end{align}
where $C_{V}$ is the heat capacity. Hence the temperature changes according to
\begin{align}
  \mathrm{d}T = - \frac{p}{C_{V}} \mathrm{d}V \ .
\end{align}

What does the variable $\mathrm{V}$ stand for?

@JohnSnow I knew I would forget one variable definition :) I hardly thought it necessary to mention any of them because they are all standard thermodynamical notation. $V$ is the volume of the gas.


text:
Standard electrode potential for electrolytic vs galvanic cells
Whys is the standard electrode potential (-) for electrolytic cells but positive for (+)  galvanic cells.

If E°cell > 0, then the process is spontaneous (galvanic cell)
If E°cell < 0, then the process is non-spontaneous (electrolytic cell)

Thus in order to have a spontaneous reaction, E°cell must be positive,
  where:
E°cell = E°cathode − E°anode

^ above taken form wiki.
Also can someone please explain the follow:
In both cells is the electron flow from
anode -> cathode ??
For electrolytic cells isn't the electron flow from the battery (-) anode to cathode (-) of the cell.
Do we then show this by electron flow as cathode (-) ------> anode (+) [in the actual cell itself]??

Potential refers to the amount of energy produced or consumed by the reaction. If the sign for potential is (+) than the reaction is spontaneous (moves on its own). If the sign is (-) then the reaction requires energy to proceed.
In the case of electrolytic cells, the cell is used to decompose a compound. this can require energy depending on the reaction. Think of water and how much energy it takes to produce hydrogen gas and oxygen, that's an example of electrolytic cell doing work.
The definition of an anode is an electrode in which a positive charge flows into an electrical device, this is the direction of the electrons.
For both types of cells direction of electron flow would remain the same, but with electrolytic cells, there would be an additional object, the electrolyte, in the diagram.
Here is a decent picture i found that also does some explanation.

so basically for both galvanic/electrolytic cells
anode -> cathode is the direction of e- flow

Yes, anode gives off the electron and cathode receives.


text:
Redox reactions - alkene to alcohol
How is the "correct" answer not a redox reaction?

Isn't the "correct" choice a textbook example of an organic redox reaction? After all, we are breaking a C-C bond and replacing that with a C-O bond. If that's not redox I don't know what is redox.
Or am I missing something fundamental here?

The reaction in question involves the addition of water across a double bond. The proton from the water is added to one carbon in the double bond - this carbon is reduced.  The hydroxyl group is added to the other end of the double bond - this carbon is oxidized. Even though the oxidation number of one olefinic carbon (the one that winds up with 2 hydrogens attached) changes from-1 to -2, and the other from -1 to 0, overall, for the entire molecule, there is no net oxidation or reduction.

seems like it was a semantics question then!

Perhaps, I don't know how he defined redox reactions in the lecture.  I think it could be argued either way, so it comes back to the definition he put forward in class.  Was this an exam question?  Maybe discussing this point with him would be worthwhile.

redox was defined as the loss/gain of electrons. No nuance regarding "net" oxidation or reduction was made. This is a practice question. I might raise it with the prof but I don't think it'll be a worthwhile endeavor considering that the also stated that bond breaking was exergonic. (I knew that what meant was that subsequent hydration combined with the bond breaking step was overall exergonic but still ...) @ron

The correct term I was looking for above is exothermic. Not exergonic.


text:
What is the lower bound to the temperature at which a fire can burn?
To have a fire, you need 1) oxygen, 2) fuel and 3) heat. You can have different fuels that burn at different temperatures, but you always need some amount of heat to have a fire.
If you can use any fuel, what is the lowest temperature at which you can have a fire with an open flame? Is it possible to have a fire that's so cold that you can hold your hand in it without burning yourself?

Even if the fire is initially set at a low temperature, as soon as you are putting your hand in it, you are technically changing its fuel and the temperature will rise.

Isn't there a type of candle they use in submarines that creates oxygen as opposed to using it to burn?

Fires which burn below about 400°C are known as cool flames, and this is a well known and widely researched phenomenon. Most hydrocarbons and many alcohols can produce cool flames, and the conditions under which they burn depends on the oxygen content available. The most common encounter that most people have with the cool flame effect is through incomplete combustion of a petrol engine, leading to engine knocking.
The lowest recorded cool flame temperatures are between 200 and 300°C; the Wikipedia page references n-butyl acetate as 225°C. You can read a lot more about cool flames on that page. One thing to note is that cool flames are very hard to visibly detect at lower temperatures - both heat and light being two of the by products of the combustion process.
NASA have conducted a number of experiments in space on low temperature combustion of nano droplets of heptane, and this is an interesting area of science still being investigated.
An interesting point to make relates to your question of whether you could hold your hand in the flame and not burn yourself. Clearly the answer is no, but for some cool flames it may take noticeably longer to feel the pain of heat. Paper starts to discolour at about 150°C, but may not burn as ignition temperature is between 220-250°C. For reference, candle flames are about 600-1400°C.

Neither of the previous 2 answers mention a phosphorus flame, which behaves visibly like other flames we are familiar with, but burns at a temperature so cool that you can hold your finger in the flame without getting burned. This only works with white (not red or black) phosphorus. It is (with proper precautions!!) placed into the chamber of an alcohol burner or similar device with the wick removed from the vent tube and warmed gently on a hot plate. The phosphorus (P4) vapor will, if ignited (maybe spontaneously), burn with a very cool flame. The same holds true for phosphine (PH3)and related low molecular weight phosphorus hydrides.

White Phosphorus? Isn't that the stuff the US army uses to light battlefields at night? The stuff called Wiley Pete that is considered a war crime when used as an incendiary weapon? at least, that's what I read in an article on video game protagonists performing war crimes ingame. they used it in Spec Ops: The Line.

Yes white phosphorus ignites at about 30 °C but flame is very hot.

In munitions, white phosphorus burns with flames of 800 °C. https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3188083/

Very late to the party, but $\ce{PSF_3}$ (thiophosphoryl fluoride) burns with a cool flame. As reported in wikipedia (https://en.wikipedia.org/wiki/Thiophosphoryl_fluoride), $\ce{PSF_3}$ was discovered in 1888 by J.W. Rodger and T.E. Thorpe. As stated on the wiki page: "The discoverers were able to have flames around their hands without discomfort, and called it 'probably one of the coldest flames known'." It spontaneously ignites in air, burning with a "greyish-green" flame and producing "solid white fumes". The wiki page also notes, rather dryly, that $\ce{PSF_3}$ "is useless for chemical warfare as it burns immediately and is not toxic enough." Always some pesky drawback ruining what could have been a very sweet lecture demo.

I don't have access to the 1889 Rodger and Thorpe paper. I don't suppose they measured the flame's temperature? Was it "one of the coldest flames known" only in 1889, or is it still?

I don’t know! I think I can get the paper, but I would have a hard time believing any actual temperature (assuming they exist) measurements in it. I know from experience, doing the house gas soapy bubbles demo, that it does not take long (less than 2 seconds) to experience discomfort when my hands were on fire! I know that stunt people in movies have concoctions they use to prevent burns, but I have no idea what they are.

It depends on your definition of fire.  You can have hand warmers based on iron filings, the iron oxidises slowly and gives off heat as it does.  So you have a redox reaction and heat is given off, but its not enough to burn a human hand.
http://www.hometrainingtools.com/a/homemade-hand-warmers
If a "fire" has to involve actual flames or smoke, then of course that's different.

I should have been clearer. I was talking about a fire as in something with flames and smoke, something that John Doe would look at and immediately recognize as a fire.

I don't really think there is a measurable temperature where you can stack a block of $\ce{ClF3}(s)$ on top of a block of $\ce{LiAlH4}(s)$ and not worry about yourself and the room disintegrating due to an immediate and violent conflagration. Certainly nothing happens at ${0 K}$, by definition in fact, but at what miniscule temperature you have to find your running shoes, is hard to tell.


text:
How many atoms of carbon are needed to produce 0.45 mol aluminium?

How many atoms of carbon are needed to produce $\pu{0.45 mol} ~\ce{Al}$?
  $$\ce{3C + 2Al2O3 -> 4Al + 3CO2}$$

I am not sure where to start, whether or not I convert moles straight to atoms or convert moles to grams of $\ce{Al}$, then to moles of $\ce{C}$ and then to atoms? I tried the latter of my suggestions and ended up with $\pu{9.8E24}$, but  I don't know if this is correct.

Stoichometry is done by comparing numbers of things to one another.  We could do stoichiometry with number of atoms or with moles (remember that moles are really just grouping numbers like dozens.)  We cannot convert among species using grams though, since all of the different reactants and products have different masses per particle.

The key to this is by examining the mole ratio. Your balanced equation shows that 4 mols aluminium: 3 mols carbon
If you are familiar with ratios from maths, you can therefore find the specific number of moles of carbon in this reaction.
$4\ce{Al}: 3\ce{C}\implies0.45\ce{Al} :~?$
I think you should be able to get the answer as 0.3375 moles of carbon (again, it's just ratios)! From here, you use moles = mass / mr to get your answer.

You can't reduce aluminium oxide to aluminium using carbon; aluminium, like many metals, forms a stable carbide.  The equation looks like this: $\ce{2 Al2O3 + 9 C \rightarrow Al4C3 + 6 CO}$
That's why in real life aluminium is prepared through the Hall–Héroult process.  Before the electrochemical process was discovered the reducing agent used was sodium.

I downvoted this answer because, while this may indeed be true, I am not sure how this answers a high school stoichiometry question, in which we have to base our answer on another given reaction. This answer of yours is better fits as a comment on the original post. Thank you for your cooperation!


text:
Something I've never understood in chemistry
I've never understood how to systematically write out IUPAC names. For example, how do I write out 6-aminohexanoic acid ? I tried searching for IUPAC related terms, but I didn't get the gist of it.
Can someone explain to me please?

Try http://en.wikipedia.org/wiki/Systematic_name. Check the IUPAC links.

I'd recommend using @LordStryker's approach and then you can come back and edit in a specific question, after which this can be reopened.  Thanks!  Also, please try to write a more useful title, this one doesn't tell me anything about your question at first glance.


text:
How many molecules does it take to have a phase?
A single molecule can't be solid, liquid or gas. It's just a molecule. A mole of something can be any of the three. So, how many molecules does it take for phases to be meaningful?
I realize that this may depend on the molecules in question. The minimum number that can form a solid might depend on the configuration that molecules adopt when solid. The smallest crystal of a lattice for example. How about a gas though? Can two molecules be called a gas if they exist under conditions where a larger quantity of the same moiety would be in a gaseous state?
In short, can a minimum number of molecules necessary to adopt a particular phase be defined?

Could you (theoretically) say that when the average kinetic energy of one particle is at a certain point, it is at a certain phase?

@Noel No, because it is possible to have two phases (e.g. ice in water) or three phases (triple point) at equilibrium, and there is no difference in average kinetic energy (phases at same temperature). Also, phase changes can occur with pressure changes while the average kinetic energy  remains constant (e.g. isothermal expansion).

This is actually an active area of research for water clusters. In principle, for $\ce{(H2O)_{n}}$ there should be a "melting" phase transition, much like for ice to liquid water.
So, in principle, if you had an accurate enough theoretical method, you could do molecular dynamics and see when the melting point of the cluster matches bulk water.
Defining a line between liquid and gas seems tricky to me, but I don't do statistical mechanics, and perhaps there are molecular dynamics techniques for defining a boiling point.
Defining the solid-liquid phase transition is much easier, because the radial distribution function of the system can be monitored to see the transition (i.e., the density changes).
At the moment, I don't think the best dynamics methods get the melting transitions accurately yet, but I think n~20 is about right. I'll revise this when I get a better answer.
EDIT: It turns out that 21 seems to be the number for water drops.
"How many water molecules are needed to solvate one?"
Chem. Sci. 2021

The minimal structure eventually responsible for proper solvation is made of a total of 21 water molecules and includes two complete solvation shells, of which the whole first one is tetrahedrally coordinated to the central molecule.

無
How many grains of sand does it take to make a pile?
The question is nonsensical, because a phase of matter is an emergent property of a collection of atoms or molecules, just like being a pile is an emergent property of a bunch of sand.
Add to that the notion that "molecule" is kind of vague in any case, and you've got a world-class mess on your hands.
Sure, it's quite clear what the unit of C2H5OH is, and that it is a molecule, but for NaCl, what constitutes the unit? What constitutes the "molecule"? Is there one?
What about polymers, or diamonds? Is a diamond with its multitudes of covalent bonds a single molecule? Is it many? Is (C3H6)n n molecules, or one?
I'm afraid the question leads you down a philosophical rabbit hole that goes far beyond the science of chemistry.

I know, and it was kind of tongue-in cheek. I was indeed thinking of the sorites paradox. Still, I was hoping that there may be a more formal definition used in the field or that, at least for gases, it may even be definable to a given number of molecules.

@terdon Typically, phases and whatnot is more about physics than it is about chemistry, meaning that the exact number of bits of stuff and microscopic properties tends to be meaningless, unless you're studying the microscopic properties themselves, in which case you concern yourself with a tiny number of species, so phase becomes irrelevant.

Sorry, this is a very silly objection the the question. Some chemicals don't have clear definitions of a "molecule" (e.g. ionic solids) but this is irrelevant to the very large class of chemicals where the molecule is well defined. For this (large) class, the question is perfectly well sensible.

@matt_black: Phase is still a question of physics, not of chemistry, and when it comes to how many molecules you need to have a phase, that's not even science, that's philosophy. The question has no answer, because it's asking for the bulk properties of individual items, which, again, makes no sense.

@WillihamTotland I disagree, as seen by my answer below. By comparing different sized clusters to the bulk, you can exactly answer this, much like comparing nanoparticle $\ce{Au}_n$ to bulk gold.

@WillihamTotland Disagree. It's chemistry (perhaps physical chemistry but still chemistry). And it is not just not philosophy but can be studied and produce real answers (like how many gold atoms do you need before a cluster behaves just like bulk gold metal?)

360... I'll explain.
As Geoff Hutchison pointed out, one place to look for this type of thing is in the molecular dynamics literature. His answer is certainly correct, but I'll expand on it based on some papers I've looked at in the past.
That number 360 really only works as an upper bound for water when trying to simulate the properties of liquid water or ice.
In molecular dynamics, one treats all the molecules using only classical, Newtonian interactions, but one must still decide what degrees of freedom to give your molecules. For instance, one common model of water is called the TIP4P/2005 model. The details here aren't important besides that they have to tell us how well their model simulates liquid water when they introduce this model.
In that paper which is free online, they simulate liquid water using 360 of their TIP4P/2005 water monomers (with a slight twist I'll explain below) and achieve quite good success in matching the experimental properties of bulk liquid water. For instance, they match the enthalpy of vaporization over a wide range of temperatures, and find a density maximum above the freezing point of water which has often been a very difficult property to simulate.
Now, the twist I was talking about. In molecular dynamics, one must place molecule(s) in a box and run the simulation. This presents a problem, however, in that molecules are never just confined to some arbitrary box. Thus, a common technique is to mirror this box when doing the simulation to make the surface area where there are edges much smaller than the total volume where there are molecules. This means that in this paper I linked to, there only 360 unique monomers, but the properties come from a larger effective number of waters.
Geoff Hutchison mentioned the number 20, which for water I would think is a bit low (though he certainly knows more chemistry than I do), as there are gas phase water clusters composed of 21 (and more) waters. See Cui, J., Liu, H., & Jordan, K. D. (2006). Theoretical Characterization of the (H2O) 21 Cluster: Application of an n-body Decomposition Procedure. The Journal of Physical Chemistry B, 110(38), 18872-18878. for a study of one of these larger clusters (this was actually done by other researchers at Pitt).
The idea, however, is that as these clusters get larger and larger, their properties approach the properties of bulk water. For instance, one could potentially make the argument that locally some of the water molecules in a larger cluster like $\ce{(H_2O)21}$ behave as if they were in bulk water.
Another way of answering the question is through spectroscopic means. Then the question becomes, what is the smallest number of molecules one can associate together and reproduce the absorption spectrum for the bulk phase? With something like water, if one only wants to reproduce the big features in an infrared absorption spectrum, I would put this number at around 60 because this means you're getting 180 unique vibrational frequencies contributing to the spectrum, which, if you blur your eyes, might look about right.
Finally, because I like this question, I must say I disagree with the other rather cynical answer. I think anytime you fall in these rabbit holes of questions which seem unanswerable, it is because the question isn't yet defined well enough.

Also my apologies for answering an old question. I just like this one a lot :)

There's never anything wrong about answering a question, no matter how old it is. That's one of the great things about SE. However, I'm not sure about your 360 rule. For one thing, that 360 were enough doesn't imply that 359 wouldn't be. For another, water is rather special, I'm not sure that the rules found for water would apply to other molecules. Very good point about the spectroscopic properties though. That may well be the right way of approaching this.

Ya that's why I was saying it's an upper bound. I think I remembered to write that... I wasn't sure if I expected that it would take more or less molecules to accurately simulate something with fewer intermolecular forces like propane or something similar. And I only threw in the whole 360 thing just cause the paper I cited used that many molecules so I could at least be certain that should work... It's a very difficult question.

As far as computational chemistry is concerned, you could use periodic boundary conditions to simulate solids and liquids. For crystalline solids, all you need is the contents of the unit cell, which might be as little as one particle. While these system would capture some aspects of a bulk phase, they will lack e.g. the broad distribution of kinetic energies with a defined average, so there will be limitations.
As far as experimental chemistry is concerned, many quantities and concepts that we understand clearly in a bulk phase change. For example, if I split a bulk solution in half, each half contains the same solute concentrations. That is not the case with a phase containing only a handful of solute molecules. If a gas phase has only two molecules, it is difficult to define pressure and temperature. For solids and liquids, surface effects become much more important as a large fraction of molecules will be at the surface.
Even in something as large as a bacterial cell (volume around $10^{-15}$ liter with about $3\times 10^{10}$ water molecules), you are no longer dealing with a familiar bulk situation for molecules at low concentrations (e.g. only one or two molecules of DNA, only a handful of the rarer enzymes etc.).


text:
How does the reaction of HCl and KMnO4 proceed?
I don't get how does this reaction of hydrochloric acid and potassium permanganate proceeds:
$$\ce{HCl + KMnO4 -> KCl + MnCl2 + Cl2 + H2O}$$
What I know:

$\ce{HCl}$ dissociates into $\ce{H+}$ and $\ce{Cl-}$.
$\ce{KMnO4}$ dissociates into $\ce{K+}$ and $\ce{MnO4^{2-}}$

What I don't get: How $\ce{MnCl2}$ and $\ce{Cl2}$ can appear there?
Could someone, please, guide me through this reaction, step by step?

Take a look at the oxidation states. You will arrive at the conclusion that the reaction is of redox nature. I hope you can take it from there.

Correction: "$\ce{KMnO4}$ dissociates into $\ce{K+}$ and $\ce{MnO4^-}$"

As others have pointed out, you don't have a "normal" reaction, you have a oxidation-reduction reaction (if you want to sound especially geeky, calls them RedOx (pronounced "ree-dox") reactions.
The electrons (which we don't normally balance in a equation) feature prominently because there are elements changing oxidation state.
First, split your reaction into its half reactions. Bring the "spectator ions" along for the ride and make sure they balance:
Oxidation: 2 $\ce{Cl-}$ gets oxidized to $\ce{Cl2}$ and gives up 2 electrons. $\ce{H+}$ is your spectator ion. $$\ce{2 H+ + 2 Cl- -> Cl2 + 2e- + 2H+}$$
Reduction: $\ce{Mn^7+}$ needs 5 electrons to reduce it to $\ce{Mn^2+}$. $\ce{K+}$ and $\ce{O^2-}$ are your spectator ions. $$\ce{K+ + MnO4- + 5e- -> Mn^2+ + K+ + 4O^2-}$$
Now, just add water. Oxidation reactions get $\ce{OH-}$; reduction reactions get $\ce{H+}$:
$$\begin{align}
\ce{2OH- + 2 H+ + 2 Cl- &-> Cl2 + 2e- + 2H2O}\\
\ce{8H+ + K+ + MnO4- + 5e- &-> Mn^2+ + 4 H2O + K+}
\end{align}$$
Now we balance the electrons (remember least-common-multiple from math class? You thought you'd never use that, didn't you?)
LCM for 2 and 5 is 10, so multiply the Cl equation by 5 and the Mn reaction by 2.
$$\begin{align}
\ce{10OH- + 10 H+ + 10 Cl- &-> 5Cl2 + 10e- + 10H2O}\\
\ce{16H+ + 2K+ + 2MnO4- + 10e- &-> 2Mn^2+ + 8 H2O + 2K+}
\end{align}$$
Add everybody back together, make more water, and cancel out the electrons:
$$\ce{10H2O + 10 HCl + 6H+ + 2KMnO4 -> 5Cl2 + 10H2O + 2Mn^2+ + 8 H2O + 2K+}$$
Take out the excess water:
$$\ce{10 HCl + 6H+ + 2KMnO4 -> 5Cl2 + 2Mn^2+ + 8 H2O + 2K+}$$
Looks like we're short something to balance out the charges. We can add more chloride spectator ions to both sides of the equation. They're not part of the redox since they are $\ce{Cl-}$ on both sides of the equation:
$$\ce{10 HCl + 6HCl + 2KMnO4 -> 5Cl2 + 2MnCl2 + 8 H2O + 2KCl}$$
Collect terms one last time and we're done:
$$\ce{16HCl + 2KMnO4 -> 5Cl2 + 2MnCl2 + 8 H2O + 2KCl}$$

@JulieThomasWhich equation can we use to calculate  the amount of $\ce{HCl}$ reacted with $\pu{x mol}$ of $\ce{KMnO4}$?
$$\ce{10HCl +6H+ +2KMnO4⟶5Cl2 +2Mn^{2+} +8H2O +2K+}\tag{1}$$
$$\ce{16HCl + 2KMnO4 -> 5Cl2 + 2MnCl2 + 8 H2O + 2KCl}\tag{2}$$

Redox is also pronounced like it's written: "Red-Ox".

I'm assuming that you were expecting a double replacement.  Clearly that doesn't happen here, and it's because of the permanganate ion.  $\ce{KMnO4}$ is a powerful oxidizing agent.  That means that it likes to steal electrons from other species, ultimately producing $\ce{Mn^2+}$ ions and water.
If we look at the species in solution with it, $\ce{K^{+}}$, $\ce{H^{+}}$, and $\ce{Cl^{-}}$, the only one that has electrons to get rid of is the chloride.  If chloride loses its surplus electrons it will turn into chlorine, leaving behind manganese(II) ions and some oxygen ions that need a home.  Those stick to hydrogen ions from the acid, making water. A portion of the chloride ions are not oxidized and remain in solution to balance the charges of the potassium ions.
I've simplified the mechanism, of course.  For example, the permanganate ion doesn't just fly apart into $\ce{Mn^7+}$ and $\ce{O^2-}$ ions once the reaction starts, but the final effect is as if it did.
Oxidation-reduction reactions often have products that are difficult to predict without a significant amount of chemistry experience.  Balancing these reactions often requires some new techniques (beyond inspection) such as balancing by oxidation number and balancing by half reactions.

If you are interested in the mechanism it should look something like this ($\ce{HBr}$ was used instead of $\ce{HCl}$):
$$\begin{align}
\ce{2H+ + Br- + MnO4- &<-> H2MnO4Br} \tag{$K$ fast}\\
\ce{H2MnO4Br + H+ + Br- &-> H3MnO4 + Br2} \tag{$k$ limiting}\\
\ce{H3MnO4 &-> \text{products}} \tag{fast}
\end{align}$$
I suppose that the first two steps can be divided into more steps.


text:
"Carbon snakes" - what physico-chemical process causes the exothermic response?
Making a "carbon snake" requires the mixing of strong sulphuric acid with sugar, as explained in this about.chemistry webpage - the reaction causes rapid dehydration of the sugar and produces steam in the process, which expands and forms the 'carbon snake', following the incomplete chemical reaction:
$\ce{C12H22O11 + H2SO4 → 12C + 11H2O}$  + mixture of excess unreacted water, sulfur dioxide and acid (and presumably a bundle of other chemicals)
My question is, what physical/chemical processes cause this rapid dehydration and exothermic reaction?

Too busy to leave an answer but check this document out: http://ncsu.edu/project/chemistrydemos/Thermochem/DehydrationSugar.pdf

Your equation is not balanced; you did not account for where the sulfur in sulfuric acid went. Some of the sulfur is oxidized into sulfur dioxide if I remember correctly.

@Dissenter yes, I saw that too (that's what I get for copying that formula from the website - clarified it as an incomplete generalised formula)

does any one knows the ingridients of the commercial snakes

$\ce{C_mH_{2n}O_n -> m C + n H2O}$
The hydration reaction of sulfuric acid acts as the driving force because it is highly exothermic (95.33 kJ/mol H2SO4).


text:
Mathematics in inorganic and quantum chemistry
Is mathematics important to learn concepts in inorganic chemistry, such as point groups, point symmetry, because doing these topics requires one to determine the symmetry of the elements and to visualize 3D (such as determining the symmetry of a water molecule?)
As these concepts are derived from mathematics, I wonder whether taking maths courses would be helpful for me to understand these concepts better and if so which maths course would be most useful? Such as linear algebra, abstract algebra?

Linear algebra is, in retrospect, the most stupid thing to NOT learn the basics in. Of course one should have gotten this in highschool already.

For better understanding of molecular symmetry you indeed need some mathematics, namely, some basics of group theory, but aside from that I doubt that you need some mathematics apart form arithmetic and calculus for general, inorganic, organic, and even physical chemistry.
For quantum mechanics you absolutely need to familiarize yourself with the concept of complex numbers and you have to learn some basic approaches for solving differential equations. For an advanced insight into quantum chemistry you need some linear algebra, by which I mean the theory of vector spaces both finite- and infinite-dimensional, not just finite-dimensional part of it, which, in a sense, reduces to matrix algebra.
P.S. Abstract algebra is nothing but a common name for different areas of mathematics which studies different so-called algebraic structures. Say, group theory studies groups, linear algebra studies vector spaces, etc.

Then how about complex analysis, will it be useful for quantum chemistry?

@weijit, sure! But it is very unlikely that you will learn linear algebra without knowing complex numbers, so I forgot to mention that.

What i meant here are those complex analysis that are taught to mathematics majors, which can take one whole semester to learn. Not those simple kinds of complex numbers that are taught in high school

Complex analysis isn't needed for chemistry at all (neither is any analysis), but it helps in understanding calculus formally. For the benefit of others this is the mathematical topic of analysis; it doesn't just mean 'analysing things'.

@Wildcat - I'd add in differential equations for quantum, but otherwise, I agree 100%. For inorganic, a bit of group theory and linear algebra are more than sufficient.

@GeoffHutchison, aren't differential equations studied in calculus?

@Wildcat no, in the US not usually. These are usually two separate courses.

My personal advice would be to do as many maths courses as you possibly can, they never hurt and even if they are not directly relevant to your chemistry career then they sharpen the mind. There's no such thing as too much maths, although my personal opinion aside there are two school of thought on this—that of the chemist, and that of the mathematician.
In perhaps crude terms, chemists often use very simplified versions of mathematics which would probably make a professional mathematician cry. Linear algebra is very important for quantum mechanics. This topic doesn't really differ too much (what I mean by that is a mathematician inverts a matrix just the same as a chemist does). Linear algebra is therefore useful regardless of where you learn it from.
A professional mathematician would tell you to learn group theory from a mathematician's point of view, but this can often be highly inaccessible to beginners (new notation to learn, much stricter logic which unintuitively seems less obvious to the untrained). The second point is that it's also not necessary for your chemistry career to learn all of group theory if all you need to do is follow the literature.
That being said, the more maths you learn, the more open your options are. Taking abstract algebra for example means that you are exposed to ideas which may be useful to your chemistry career, but such ideas haven't been realised in the chemistry community yet. Abstract algebra lays the foundations for topological chemistry and other higher study, of which the benefits to chemistry are only starting to be realised.

In ideal world (where humans do not die) there is indeed no such thing as too much math. In reality life is limited, so one can not learn everything and have to prioritize things.

Although i know mathematics is useful for chemistry, im currently focusing on those aspects of mathematics that have a more immediate applications to chemistry, especially those that would help me to understand advanced chemistry concepts better. Im currently considering complex analysis, will these be helpful for quantum chemistry?

Clearly my approach isn't shared, so @Wei jit, if you're interested in QM, take a look at Freidman and Atkin's book on Molecular Quantum Mechanics. You won't understand all of it, but take maths course on the bits you don't until you do!

Which are  the mathematics course most helpful. (Btw Im currently at Multivariable calculus and i still have 1 year to take mathematics courses before i start taking QM. So i just wonder which maths courses could i still take after i finsish multvariable in this 1 year.)

You'll be pleasantly surprised at how far linear algebra alone will take you if I'm honest, at an beginner's level. Multivariable calculus is good. Try and become ultra comfortable with it though so it becomes very natural. Stick with the calculus too. Your course structure should bring it all together, so don't panic. I think you're a way off from doing Calculus of Variations yet but it is something to think about.

If you are taking quantum chemistry in a year, then why do you ask about inorganic chemistry?

Im have to take inorganic chem soon though.

for inorganic chemistry you need arithmetic and stereometry. Group theory may be fancy, but more or less ignorable as long as you don't draw molecular orbital diagrams in complicated hi-symmetry cases (understanding ones that are already drawn for you does not require this)
for physical chemistry (including understanding of equilibrium) you also need differential and integral calculus and linear differential equations are required. In case of statistical physics approach familiarity with series is of use. complex numbers may be of use in differential equations.
for quantum chemistry you also need partial differential equations, linear algebra and function approximation by series (generalized Fourier series). It is highly recommended to be familiar with complex numbers, but technically speaking it is possible to get basic understanding of quantum chemistry with very basic understanding of complex numbers. For advanced applications calculus of variations and basic understanding of numerical methods, including ones for function approximations, integration, and iterative solution of eigenvalue problems are of use. Group theory is optional, it is touched in simplification of cases with symmetry, but other than that is of no use.
I also recommend to consider memory excercises. You'll have to remember a huge amount of assorted, badly systematized facts and names.

Undoubtedly +1 for memory training! Although, I would say that in chemistry we have to remember a relatively huge amount of assorted, badly systematized data. When one compares this ammount to, say, what they have in biology, it seems quite small in fact. :D That is actually why many years ago I eventually decided to enter chemical university rather then biological one: I hate to memorize a lot of things.

Outside of (retro-)synthetic organic chemistry with its terrible amount of named reactions, chemistry is basically void of badly systemised facts and names. I never even thought about going into biology. :D

For inorganic chemistry and group theory, learn linear algebra and matrix mathematics.
For quantum chemistry, learn how to solve partial differential equations. Depending here on how far and what direction you want to go, you may need either a science approach or a numerical methods approach to the higher-level math methods. At an undergraduate level, a solid course on differential equations should be a good start.

Oh how about probability and statistics?

@wei jit: Not really. Those are good prerequisites for courses that involve data analysis.

You can follow the key aspects of group theory and symmetry for chemistry without ever taking formal abstract algebra courses.
The key aspects are presented in survey and specialized chemistry texts.  Probably all the exercises on different molecules (and IR modes and the like) will actually give you MORE of a feeling for the complexity of physical symmetry than a math first attitude would.
This is not to say it hurts, but you totally don't need math based group theory.
Consider, for instance, Arfken has a remark (in horror) in his advanced math book for physics at the 200+ space groups for crystals.  But a crystallographer or solid state chemist finds them interesting.  [Note that this is not even an aspect of chemistry being very detailed...millions of compounds.  It's straight up from the math itself...could be lattices of twigs for all we care.  But the mathematician just hates this kind of complexity.  A chemist is used to it since he has to organize families of organic compounds and the like.]
Personally I think if you want to dive deeper, would start with specialized books on group theory for chemistry (will have a little more than the touch in your survey inorganic text).  Here there are sort of two schools:
A.  Molecular, with emphasis on point groups and vibrational modes (IR).  Think basic coordination compounds of metals.
B.  Solid state, with emphasis on space groups and crystallography.  Think metal oxides or similar.
Of course you can and will have crystalized groups of molecules as well, so both are relevant.
Only after digging deep into some of the chemical based stuff would I maybe consider the need to go back and get into an abstract algebra based development of group theory (probably starting at set theory and involving things like rings and fields that don't even apply to your need).
And by the way, if you get seriously into crystallography, you can spend a lot of time on that.  And a group theory mathematician won't have a feeling for how to spot flaws in crystal structures (either those that don't make sense chemically OR those that are little logical flaws of the math, e.g. equivalent structures).
If anything learning more vector calc and tensors (and from an applied, simple perspective...not total math theory killer) would help you more.  Even that is basically just needed if you become a crystallographer.  An experimental inorganic chemist working on molecules really doesn't need it--all his xtal structure determinations are outsourced.  Solid state guys tend to do more of this themselves but even here, a lot of it is playing with programs on the computer and the theory of the x-ray geometry is really not needed.  More a careful attitude to look for reasonable chemistry and where solved structures have more uncertainty ("thermal parameters").


text:
Establish the formula of an oxide of chromium

An oxide of chromium used in chrome plating has a formula mass of
  100.0 u and contains four atoms per formula unit. Establish the formula of this compound, with a minimum of calculation.
Express your answer as a chemical formula.

So the question tells me that the mass of the Chromium Oxide is 100.0 amu and has 4 atoms for very unit. Thus, the question wants to know a formula that has 4 atoms? Right?
Hence, if $\ce{Cr}$'s (Chromium) molecular mass is 51.9961 and since the question is stating that we are dealing with an oxide, which would mean that it is a ionic compound (Chromium gives two electrons to Oxygen [15.999 amu]), wouldn't the answer be $\ce{Cr2O2}$; but that would equal 135.9902 grams.
On the other hand, if I am supposed to view a formula unit as an empirical formula, wouldn't that require me to write the simplest formula for a compound?
Yet, $\ce{CrO}$ would give me (51.9961 + 15.999) 67.9951 g/mol and that is not correct, or is it?

Welcome to Chemistry.SE! Please have a look at this tutorial to acquaint yourself with the way math and chemical formulae can be nicely formatted on this site.

Rounding judiciously, the mass of chromium is 52 amu and oxygen is 16 amu.
Our compound is CrxOy and has 4 atoms.  The sum of those 4 atoms' mass is 100 amu.
x + y = 4
52x + 16y = 100
Solve the system of equations to find the answer.  Alternatively, you can just guess and check.
Any species with 2 chromium atoms must have a mass greater than 100 amu, since Cr2 = 104 amu and we haven't included any oxygens.  Thus we know for certain that there is only 1 chromium atom in the compound.  The remaining three must be oxygen.  I leave it to you to check that the answer is correct.
I think your confusion stems from assuming that something called "chromium oxide" must have 1 chromium atom for every 1 oxygen atom, and that isn't the case.

Yes, I thought an oxide was always an ionic compound, that is, nonmetal with a metal.  Ok, thanks.


text:
HCl/diethyl ether + Nitrogen?
Just wondering if anyone can tell me why bubbling nitrogen through a solution of $\ce{HCl}$/diethyl ether catalyst would remove said mix from my solution.
My current thought is that the $\ce{N2}$ will form ammonia and deprotonate the $\ce{HCl}$. But what of the Diethyl ether? I can't think of how it would be removed.
My only final explanation is that it is the $\ce{HCl}$ which is the important species to remove and the diethyl ether is just solvent ... but I would really like a second opinion!

N2 is not going to convert into ammonia from contact with HCl.  Nitrogen is very unreactive stuff.  The removal is as ron described, a purely physical process.

HCl is a gas at room temperature and diethyl ether is a low boiler.  Both will have significant vapor pressures and, therefore, exist in the vapor phase above your  solution.  The HCl and diethyl ether in solution will continuously maintain an equilibrium with the vapor phase. As you bubble nitrogen through the system it will continuously sweep the vapor out, removing HCl and diethyl ether.  But since the HCl and diethyl ether in solution continuously maintain an equilibrium in the vapor phase, they will be continuously swept out until all of the HCl and diethyl ether in solution have been removed.


text:
Thermodynamically Reversible Processes: How does an infinitesimal step make a process reversible and how is maximum work derived from the process?
My book says 'Maximum work can be obtained only from thermodynamically reversible processes,' but why is it so? What is the cause?
Actually to me the definition of reversible process is confusing. It says that at each step during the process, equilibrium is maintained. But, let's say,we have to expand a gas in a cylinder headed by piston from pressure $P_i$ to $P_f$. Suppose,the process is done with innumerable infinitesimal steps. Then, say at each step, $dp$ is changed in the pressure.
Before the step, gas and the piston were exerting same pressure $P_i$ . Now, when the piston is raised, gas does expand to equalize the pressure,which is $P_i - dp$ . So, at the end of the step, the gas & the piston will exert same pressure $P_i - dp$ and hence will be in mechanical equilibrium.
During the step,the gas and the piston was not in equilibrium;it was at the end of the step  the equilibrium is achieved. So, during each step of the process the gas would be at disequilibrium . So,will it really be a reversible process? It is contradictory with the definition. But it is true that during each step, the system is at disequilibrium though infinitesimally small by $dp$ . So, why does the definition tell that at each step,there is equilibrium?
So, I have three questions:

Why is maximum work only obtained from reversible processes?
According to the definition, at each step of the reversible process,there must be equilibrium. But during each step, there is infinitesimal disequilibrium. So, why does the definition say so in-spite of the disequilibrium during each step?
How do infinitesimal steps make a process reversible? I will be very grateful if anyone answer these three questions.

Please help.

Check Carnot cycle.

I haven't seen too many texts deal with the mathematical nuances of reversibility in the Carnot Cycle, in fact none!I know it's a really popular pedagogical tool in classical thermodynamics, but for the life of me I never found it at all helpful!

@user5764 Basically all irreversible paths by definition have to "generate" some new (previously non-existent) amount of entropy (i.e, always make $\Delta S_{univ} > 0$) hence will always have some more inefficiency relative to the reversible processes in terms of doing work (e.g, such as needing more input heat, and for the same amount of work, etc). The rest of the discussion on "infinitesimals/quasi-static/etc" just boils down to (listing) conditions for real processes under which said entropy "generation" can be minimized, well, ideally down to zero for the fully "reversible" case.....

Reversibility is a statement that means, the system and the surroundings are perfectly balanced (in equilibrium) at ANY AND ALL instants in time. This allows us to define $T_{\mathrm{sys}} = T_{\mathrm{surr}}$ and $p_{\mathrm{sys}} = p_{\mathrm{ext}}$ ALWAYS. The latter statement on pressure allows then that we can calculate work in terms of system pressure rather than external pressure. The proof from this point is the mathematics to show, irreversible systems always do less work than reversible systems.
Otherwise, the approach you take with your analysis of pressure in steps is the same one that allowed the tortoise to beat Achilles in a race. It is called Zeno's paradox.
In specific answers to your numbered questions ...
1) Accepting the above definitions allows the mathematics to prove that reversible work is always greater than irreversible work. As a physical answer, we can define reversible work as $W_{\mathrm{rev}}$. Any irreversible process must have a point in time where the system and surroundings are not at equilibrium with each other. Let's use this to get work $W_{\mathrm{extra}}$, and call $W_{\mathrm{irr}} = W_{\mathrm{rev}} + W_{\mathrm{extra}}$. The final proof that $|W_{\mathrm{irr}}| < |W_{\mathrm{rev}}|$ is that $W_{\mathrm{extra}}$ is always work done ON the system over a full cycle (where the system returns to its starting point.
2) You mistake that a process taking (very small but) finite steps as a valid replacement for a process that takes infinitesimal steps. We would never go anywhere in life (literally!) when your approach would be valid. See Zeno's paradox.
3) Infinitesimal steps are not part of the definition of reversibility. They are part of the way to solve a reversible process properly. Reversibility is a definition about the state of system and surroundings during any point in time.

I'll give a mathematical answer, and an analogy also for ease of thought. These are my own intuitions, so they may work for you.
My mathematical reasoning:
A lot of thermodynamic processes are phrased very carefully in terms of reversibility. Now my intuition of this is that reversibility is a consequence of the maths we must use.  Thermodynamics likes to look at continuous processes such that we can use integrals instead of summations in the calculations: this stands to reason since classical thermodynamics treats a collection of particles as moving, say, a piston continuously from one place to another, not in discrete steps.
Thus, one way to think about it is to think of reversibility as being necessary for an integral formulation. If we allow ourselves to try and think in infinitesimal steps, moving an infinitesimal step forward is just an infinitesimal away from going backwards and hence this is described as being reversible. In short, if a process really happens continuously then we must impose the condition of reversibility on it, otherwise the mathematics fails.
My analogy:
Why this gives maximum work can be seen by analogy with a real life classical system: the tug-of-war. If you're in a tug of war with some of your friends who are of equal size to you and of equal strength, you're going to put in maximum work in pulling the rope. If one side is of unequal size and strength it becomes much easier for the stronger side.
To be more serious, if your particles move your piston irreversibly (i.e. they never move backwards along the way) then no work has been expended in going backwards—only forwards. Reversible systems belie the work they've had to do since the physical outcome we actually see—the final outcome—shows the piston in the same place as the irreversible case, however what we haven't seen is the system battling backwards and forwards.
To conclude: if we imagine a set amount of work needs to be done to move a piston irreversibly from a point A to a point B, then the reversible case must provide maximum work since some of the time the system has battled going backwards and has still ended up in the same place as the irreversible case.
EDIT: How to Achieve Equilibrium and still have Movement
I think the source of your confusion lies in the statement 'it was at the end of the step of the equilibrium'. I know what you mean by this, but I don't think it's precise. Equilibrium doesn't mean to describe position as such, but the oscillation of position. Even if you've physically pushed your piston a distance ds, it can still be in equilibrium by the very fact that it can still move forwards and backwards in its next spot.
So in fact the periodicity of position is maintained at each step, which is what we mean by the equilibrium is maintained.
If we imagine a spring oscillating backwards and forwards, the movement of oscillation is what we mean here by a dynamic equilibrium. As long as this oscillation is maintained this equilibrium exists, however if we imagine our particles as springs, then the spring must be extended to reach its next spot. Hence in this sense we have equilibrium and movement can still occur.

Sir, what will you say about the disequilibrium during each step? How should I deal with it? Doesn't it contradict the definition?

I've edited my answer to your response, however the main point to note is: "If we imagine a spring oscillating backwards and forwards, the movement of oscillation is what we mean here by a dynamic equilibrium. As long as this oscillation is maintained this equilibrium exists, however if we imagine our particles as springs, then the spring must be extended to reach its next spot. Hence in this sense we have equilibrium and movement can still occur."

If we consider a chamber with a membrane in the middle separating air and vaccum on either sides,what is my system?If i take the whole chamber as my system,there is no work transfer or heat transfer(chamber is insulated) if i remove the membrane.Therefore this justifies that no work is done by gas in free expansion.But the air and vaccum vary in composition,that is they are not in equilibrium,then taking the whole chamber as my system incorrect.Then what should be my system?Just the air half chamber or just the vaccum half chamber?


text:
Zeolite process
I'm a higher secondary school studet. See this following image, I can't understand how water can rises through the bed of permutit, my textbook says - Hard water introduced from the top of tank and then reaches at the bottom and then rises through the bed of permutit.!!, please tell me how it is possible.

Water pressure from the inlet, and 2. surface tension in the small pores of the zeolite both contribute in varying amounts. Most household softeners that I know of rely on the water pressure for most of the work.

One thing that is not shown in the image is the water control mechanism. The tank can either be making soft water from hard water OR it can be being regenerated by pumping in salt water in the top and the waste salt water out the bottom.  It can't do both at the same time.

Can someone give a reasonable answer ?

Hard water has minerals (mostly calcium and magnesium salts). The hard water is pumped into the coarse sand layer at the bottom. As the coarse sand settles, the water comes into contact with the Permutit zeolite. As the water rises through zeolite, it binds the calcium and magnesium and replaces them with sodium.


text:
What is the constant of integration in a series reaction?
Consider the series reaction
$$\ce{A ->[k_1] B ->[k_2] C}$$
Then
$$\frac{\mathrm d}{\mathrm d t} [\ce{B}] = k_1 [\ce{A}] - k_2 [\ce{B}]$$
We already know that
$$[\ce{A}]=[\ce{A}]_{0} \exp(-k_1 t)$$
Thus far, I have used the method integrating factors, with $u=\exp(k_2t)$, to arrive at the equation:
\begin{equation}
[\ce{B}]= \frac{k_1}{k_2-k_1} \left( \exp(-k_1 t) + C \exp(-k_2 t) \right) [\ce{A}]_{0}
\end{equation}
What would the constant of integration be in this situation?

Did you take in mind that you use a definite integral? You should get a dependence of [B] on [B]o.

As @Marko already stated, the problem seems to be similar to this one. It might be resolved by using definite integrals which should eliminate the unknown constant from you equation.

There might or there might not be a problem with your final equation. Either you have made a mistake while using the integrating factor to solve your differential equation or you made some further substitutions so that your constant $C$ contains something more than just the "standard stuff" I would have expected from the method. Thus it's best I go through the derivation from the beginning so that it'll become clear what I mean.
The method of solving differential equation using an integrating factor is explained well here. You have used the correct integrating factor $u(t)$. For the differential equation at hand, i.e.
\begin{align}
  \frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} &= k_1[\ce{A}] - k_2[\ce{B}] \\
  \frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} + k_2[\ce{B}] &= k_1 [\ce{A}]_{0} \mathrm{e}^{-k_1 t} \ ,
\end{align}
it is indeed
\begin{align}
  u(t) = \exp\biggl( \int_{0}^{t} k_2 \, \mathrm{d} t \biggr) = \mathrm{e}^{k_2 t} \ .
\end{align}
With the integrating factor you can get $[\ce{B}]$ according to
\begin{align}
  u(t) [\ce{B}] &= \int_{0}^{t} u(t) k_1 [\ce{A}]_{0} \mathrm{e}^{-k_1 t} \, \mathrm{d} t + \tilde{C} \\
  [\ce{B}] \mathrm{e}^{k_2 t} &= \int_{0}^{t} k_1 [\ce{A}]_{0} \mathrm{e}^{(k_2 - k_1) t} \, \mathrm{d} t + \tilde{C} \\
  &= k_1 [\ce{A}]_{0} \biggl[\frac{1}{k_2 - k_1} \mathrm{e}^{(k_2 - k_1) t} \biggr]_{0}^{t} + \tilde{C} \\
  &= \frac{k_1}{k_2 - k_1} [\ce{A}]_{0} \bigl( \mathrm{e}^{(k_2 - k_1) t} - 1 \bigr) + \tilde{C}
\end{align}
Rearranging for $[\ce{B}]$ gives
\begin{align}
  [\ce{B}] &= \frac{k_1}{k_2 - k_1} [\ce{A}]_{0} \underbrace{\frac{ \mathrm{e}^{(k_2 - k_1) t} - 1 }{\mathrm{e}^{k_2 t}}}_{= \, \mathrm{e}^{-k_1 t} - \mathrm{e}^{-k_2 t}}  + \frac{\tilde{C}}{\mathrm{e}^{k_2 t}} \\
 &= \frac{k_1}{k_2 - k_1} [\ce{A}]_{0} \bigl( \mathrm{e}^{-k_1 t} - \mathrm{e}^{-k_2 t} \bigr) + \tilde{C} \mathrm{e}^{-k_2 t} \ .
\end{align}
This equation differs from yours a little if your integration constant $C$ would be the same as my $\tilde{C}$. But if you used some substitutions so that your integration constant is $C = \frac{k_2 - k_1}{k_1 [\ce{A}]_{0}}  \tilde{C} - 1$ then your solution and mine would be equivalent. If you indeed made a mistake then I hope my derivation cleared things up a bit and if you had intended the equation to look that way by introducing the substitution then at least you know what integration constant I mean when I am going to explain the method for determining its value in the following.
The integration constant can now be determined by enforcing boundary conditions. In this particular case this means that at the start of your reaction ($t=0$) you require that $[\ce{B}]$ is equal to the initial concentration of your reactant $\ce{B}$, i.e. $[\ce{B}](t\! = \! 0) = [\ce{B}]_0$. So, plugging $t=0$ and $[\ce{B}](t\! = \! 0) = [\ce{B}]_0$ into the equation for $[\ce{B}]$ you get
\begin{align}
  [\ce{B}]_0 &= \frac{k_1}{k_2 - k_1} [\ce{A}]_{0} \bigl( \mathrm{e}^{0} - \mathrm{e}^{0} \bigr) + \tilde{C} \mathrm{e}^{0} \\
  [\ce{B}]_0 &= \tilde{C} \ .
\end{align}
In the case of your reaction $\ce{A -> B -> C}$ the initial concentration of $[\ce{B}]$ will usually be $0$ as $\ce{B}$ is only an intermediate product, so $\tilde{C} = [\ce{B}]_0 = 0$ and the final equation is
\begin{align}
  [\ce{B}] &= \frac{k_1}{k_2 - k_1} [\ce{A}]_{0} \bigl( \mathrm{e}^{-k_1 t} - \mathrm{e}^{-k_2 t} \bigr) \ .
\end{align}

If I may be a little bit pedantic, your answer is not quite complete, as it does not cover the case where $k_1 = k_2$, which leads to "resonance", i.e., $t \exp(-k_1 t)$ instead of $\exp(-k_2 t)$. I lack the training in chemistry to tell whether this kind of "resonance" is interesting or not.

Assuming that the cascading chemical reaction $\ce{A} \xrightarrow{\kappa_1} \ce{B} \xrightarrow{\kappa_2} \ce{C}$ has mass action kinetics, we have the following pair of linear ODEs
$$\begin{array}{rl} \dot a &= - \kappa_1 \, a\\ \dot b &= \kappa_1 \, a - \kappa_2 \, b\end{array}$$
where $\kappa_1, \kappa_2 > 0$ are the rate constants, $a := [\ce{A}]$ and $b := [\ce{B}]$. For the time being, we neglect $[\ce{C}]$.
Integrating the first ODE, we obtain
$$a (t) = a_0 \exp( -\kappa_1 t )$$
Hence, the second ODE can be written as follows
$$\dot b + \kappa_2 \, b = \kappa_1 \, a_0 \exp( -\kappa_1 t )$$
Eventually, we arrive at the solutions
$$b (t) = \begin{cases} a_0 \left(\dfrac{\kappa_1}{\kappa_2 - \kappa_1}\right) \exp( -\kappa_1 t ) + \left(b_0 - a_0 \left(\dfrac{\kappa_1}{\kappa_2 - \kappa_1}\right)\right) \exp( -\kappa_2 t ) & \text{if } \kappa_1 \neq \kappa_2\\\\ (b_0 + \kappa_0 \, a_0 \, t) \exp( -\kappa_0 t ) & \text{if } \kappa_1 = \kappa_2 =: \kappa_0\end{cases}$$


text:
Percent of silver by mass in an unknown sample

Question
  If all of the silver in a $7.563~\mathrm{g}$ sample of a mineral, which was dissolved by using nitric acid, is precipitated as silver thiocyanate ($\ce{AgSCN}$) by using $82.46~\mathrm{mL}$ of a $0.2500~\mathrm{M}$ potassium thiocyanate solution, what is the percent silver (by mass) in the sample?

My Attempt
I'm pretty sure the relevant chemical equation here is $\ce{Ag^+ + SCN- -> AgSCN}$.
Molarity is mols solute over L solution, so the number of mols of $\ce{KSCN}$ used is $0.25~\mathrm{M} \times 0.08246~\mathrm{L} = 0.020615~\mathrm{mol\ KSCN}$. According to the chemical equation above, the molar ratio is 1 to 1 so there is $0.020615~\mathrm{mol\ Ag}$  involved. That accounts for all of the silver in the original mineral sample, I believe. Therefore, there are $2.22370~\mathrm{g\ Ag}$, so the percent silver by mass is $2.22370~\mathrm{g\ Ag} \div 7.563 ~\mathrm{g} = 0.294$. Put into percent, that's $29.4\%~\mathrm{Ag}$ by mass.
I'm unsure about this because  I totally ignored the nitric acid.

The mineral was dissolved in enough nitric acid to do the job, it really doesn't matter how much that was.  The nitric acid just serves to get the silver into ionic form.  The only issue I see is that you're writing CNS instead of SCN, as is the norm (even in compounds.)

Noah--
You're good to go. I was taught to use unit analysis to help me make sure I was working through these problems correctly:
82.46 mL KSCN solution * (0.25 M KSCN/1 L) * (1 M SCN- / 1 M KSCN) * (1 L / 1000 mL) = 0.020615 M SCN- (mL cancels, M KSCN cancels, L cancels)
0.020615 M SCN- * (1 M AgSCN / 1 M SCN-) * (1 M Ag+ / 1 M AgSCN) * (107.8682 g Ag / 1 M Ag) = 2.2237 g Ag ( M SCN- cancels, M AgSCN cancels, M Ag+ cancels)
2.2237 g Ag / 7.563 g sample = 0.294 g Ag per g sample * 100% = 29.4% Ag by mass
I found that it helps if you read your equations out loud you can work through the logic more easily. If you set the equation up as fractions, you can see the units that cancel each other. For the most past, if you can get the right units for your answer, you're doing it right.
And you get to ignore the nitric acid (as well as the potassium) because they don't factor into your analysis.

It's correct. You need to assume that the mineral will be completely dissolved in nitric acid, as in the opposite case, you wouldn't have enough data to calculate the mass of silver.

Nitric acid is used because it can oxidize elemental silver, but it's not important in this case. There's no assumption made because the question specifies that "all of the silver [in the sample]… is precipitated as silver thiocyantate".