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How to find the number of hydrogen atoms in 0.5 mol of hydrogen gas?
How do I calculate the number of hydrogen atoms in $\pu{0.5 mol}$ of hydrogen gas?
I presume the answer would probably somehow use the equation
$$n = \frac{m}{M},$$
where $n$ is the amount of substance, $m$ is the mass, and $M$ is the molar mass. However,  I'm not sure how to calculate it.

There is no need to use
$$n = \frac{m}{M},$$
since you already have been given the amount of substance. You would need to use the equation, if you were given the mass $m$, e.g. as $\pu{0.5 mg}$.
If you want to calculate the number of hydrogen atoms in $\pu{0.5 mol}$ hydrogen gas, then you should consider that they are diatomic molecules, $\ce{H2}$.
The unit mole is represented by the Avogadro constant and is
$$N_\mathrm{A} = \pu{6.02214076×10^23 mol−1}
\approx \pu{6.022×10^23 mol−1}$$
Each hydrogen molecule has two hydrogen atoms, therefore
\begin{align}
n(\ce{H}) &= 2 \cdot n(\ce{H2}),\\
N(\ce{H}) &= n(\ce{H}) \cdot N_\mathrm{A},\\
N(\ce{H}) &= 2 \cdot n(\ce{H2}) \cdot N_\mathrm{A},\\
N(\ce{H}) &= 2 \times \pu{0.5 mol} \times \pu{6.022×10^23 mol−1},\\
N(\ce{H}) &= \pu{6.022×10^23}.
\end{align}


text:
Why does the chlorination of indene occur with syn selectivity in heptane?
In the chlorination of indene to form 1,2-dichloro-2,3-dihydro-1H-indene in heptane as solvent, the syn dichloride is formed as the major product, as reported in J. Org. Chem. 1980, 45 (25), 5150–5155):

The explanation I was given for this was that "in a non-polar solvent, the ion pair collapses quickly". What does this mean, and why is the stereoselectivity reversed, as compared to typical halogenations which proceed via a halonium ion intermediate to give anti dihalides?

Key
The reaction is run in heptane, a non-polar solvent.  A non-polar solvent will not stabilize or support ions as well as a polar solvent.
Background
When the $\ce{Cl2}$ begins to interact with the indene an initial ion pair is formed.  In the case of chlorine, it is not always clear whether

a cyclic halonium ion (the 3-membered structure on the bottom left)
an open ion (the structure on the bottom right)
or an equilibrium between the two ions occurs

[Side Note: In electrophilic bromination reactions involving alkenes, bromonium ions are very common intermediates usually leading to trans addition.  However, chlorine isn't as polarizable as bromine and in this particular case with indene, the open ion is extremely stable due to its benzylic nature. Hence the involvement of only a chloronium ion is questionable in this case.]

Analysis
As mentioned above, ions are not very well stabilized in the non-polar solvent. As a result

the two ions (one positive, one negative) will tend to stay close together  (tight ion pair) and stabilize each other
and further due to this lack of stabilization they will have a short lifetime leading to quick collapse to products.

Both of these factors (chlorines located close to one another, short lifetime doesn't allow much movement of the second chlorine) favor syn collapse - addition of the second chlorine from the same side that the first chlorine is on.
If we were to repeat the reaction in a polar solvent that would allow the ions to separate and have a longer lifetime, more trans product from anti addition would be expected.

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text:
How do I calculate the color of a compound?
I know this is complicated, but I'll ask. How do you calculate the color of a compound with pen and paper? Everyone knows that iron(II) sulfate is yellowish. How do I know without looking at it?

This is an awesome question, I wish I knew how to answer this well but I can only point you in the right direction. It comes down to looking at the orbitals and seeing which transitions an electron can make, since this is exactly what corresponds to the frequency of light as you may be aware of $E=h \nu$. Transition metals tend to get their colors from the d-orbitals and organic compounds get their colors from conjugated $\pi$-systems in general. I wish I could discuss and explore this more with you but the site doesn't allow this kind of thing apparently.

Since this is quite a broad question, I will also only be able to provide an answer in very broad terms.
The short answer is: You cannot. At least not with (a real) pen and (real) paper.
The slightly longer answer is:
With a simple pen and paper approach this is virtually impossible. Even the use of more elaborate methods like MO theory can lead you to completely wrong results. Electronic transitions involve excited states, which can be incredibly hard to calculate. In most cases you can get an approximate spectrum by using time-dependent density functional theory. This can be right for the wrong reasons - it is still somewhat a crude theory. The best results you will get with multi-reference methods or big automated computations, which may take a long time on a supercomputer.
In other words: There is not easy, general, rigorous way to determine the colour of a compound.

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text:
Le Chatelier's Principle: Temperature dependence of endothermic and exothermic reactions (chemical equilbrium)
What effect does temperature have on endothermic and exothermic reversible reactions? Please also explain the reason why it responds like this.
And what do people mean when they say something like "increasing pressure or volume shifts the reaction to left/right"? What do people mean by "left/right reaction shift"?

In a reaction, there is an exothermic and a endothermic side. Increasing the temperature will favour the endothermic side.
So if your reaction is exothermic, that means it must release heat if it shifts to the left. Then, since the left is exothermic, it shifts to the right (endothermic) to minimise this disturbance.
If your reaction is endothermic, that means it must absorb heat if it shits to the left. Then, since the left is endothermic, if you increase the temperature it shifts to the left (endothermic) to minimise this disturbance.

The above is much easier to understand if you think of 'heat' as a product or reactant. In an exothermic reaction, heat is on the product side of the equation. So if you increase this component, to minimise this disturbance you must shift the equilibrium to the LEFT to decrease the 'heat' amount - and vice versa.

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text:
Plot Binary Eutectic from Temperature of Melting and Enthalpy
I'm a little confused about plotting a binary eutectic diagram (temperature vs mole fraction), and finding the eutectic point.  I understand the eutectic point is the lowest temperature that a mixture of the two substances will melt, and that this is somehow related to minimizing the Gibbs free energy of the system. However, I can't figure out how to actually solve for the eutectic point or how to graph it based on the melting temperature and enthalpy of fusion of two compounds.
I found a formula $$\frac{X_A}{X_B} = \frac{T_B-T_0}{T_A-T_0}$$ where X is the mole fraction of each component, but I was getting unrealistic answers for certain mole fractions, e.g. for $$X_A=.9$$ I was getting a value greater than the melting point of A.  And for $$X_A=X_B$$ this equation can't be solved.  Also, this formula doesn't take into account the enthalpy of fusion, so doesn't seem to relate to the free energy.
The volume & entropy of the system is unknown, so I wasn't sure how to set up an equation to solve for the Gibbs free energy of either component, but I was thinking that if I had an equation, I would be able to  solve a system of equations for the lowest free energy and plot it (somehow?).
Can anyone explain the relationship between the making a eutectic diagram and the free energy of components in the system?

Edit: Thank you for the answer! That was really helpful conceptually in relating the free energy of a mixture to that of the pure substances, but I actually have a problem from the class that says:

Calculate and graph the binary eutectic melting diagram for the system diopside - anorthite using the data given below, and assuming for both diopside and anorthite that the activity in the liquid is equal to the mole fraction.
\begin{array}{lll}
\text{Diopside}\, (\ce{CaMgSi2O6}): & \Delta H_\mathrm{fusion} = \pu{20,000 cal/mol} & T_\mathrm{fusion} = \pu{1390^\circ C} \\
\text{Anorthite}\, (\ce{CaAl2Si2O6}): & \Delta H_\mathrm{fusion} = \pu{29,000 cal/mol} & T_\mathrm{fusion} = \pu{1553^\circ C}
\end{array}

So it must be possible to find based on these parameters right?  Is the fact that the activity is equal to the mole fraction make the difference in solving it?

The reason that you can't figure out how to solve for the eutectic based on the melting temperature and enthalpy of fusion of the two constituents is because, well, you can't. Those bits of information, while useful in constructing the phase diagram (well, at least the melting temperatures), are not what determines the eutectic. You will have a eutectic if and only if there are interactions between the constituents in the liquid phase that reduce the Gibbs free energy below the ideal value.
Lets make this more concrete. Take elements A and B, each with a solid and liquid phase, no intermediate phases. Taking just one phase (solid or liquid) for the moment, what happens to the Gibbs free energy when we mix them together? The first bit you get is the weighted combination of the pure phase free energies, that is:
$G_{mixture} = x_{A}G_{A} + x_{B}G_{B}$.  But is that enough? Well, no, we forgot the entropy of mixing term. So, for an ideal mixture,
$G_{ideal} = x_{A}G_{A} + x_{B}G_{B} + RT(x_{A}\ln x_{A} + x_{B} \ln x_{B})$
What does this look like? Well, taking 'fake' elements, A that melts at 950K and B that melts at 1000K, just using ideal mixtures for solid and liquid, one gets a pretty boring diagram that looks like:

with the slight separation in solidus and liquidus from the slight asymmetry in the melting points. Hmmmm... - no eutectic (since the ideal mixing impacts both liquid and solid in the same way, that is to be expected).
OK, now lets move to a regular solution model. The difference is the addition of an enthalpy term, that describes whether an A atom would rather see A or B around it (and vice versa). In the simplest way, this can be represented by a heat of mixing term using one parameter, $\Omega$. $\Omega$ is positive if the atoms don't want to mix (so, it can cause a miscibility gap), or negative if they actually prefer to mix. So, for a regular solution one gets:
$G_{regular} = x_{A}G_{A} + x_{B}G_{B} + \Omega x_A x_B + RT(x_{A}\ln x_{A} + x_{B} \ln x_{B})$
Setting $\Omega$ to -1kJ/mol, the previous diagram turns in to:

and you see that a eutectic has now formed at about $x_B = 0.25$.
Now, back to the parameters you thought were important:
The melting temperature is kind of important, in so far as an asymmetry between them shifts the eutectic of the regular solution towards the lower melting element.
The enthalpy of fusion is not a factor at all in determining where phase boundaries are.
The interaction term(s) are highly relevant, since they determine where and how deep the eutectic is.

You are almost certainly being asked to make several assumptions:

the solids that form are "pure" solids with only one or the other component in them.
the liquid is an ideal liquid, with an ideal entropy / free energy of mixing
the specific heats of the fluids and solids can be approximated as being equal.

This said, the free energy of all phases is known.  Calculate the chemical potentials in each phase.  Set the chemical potentials for both compounds in two or more phases equal to each other.  When there is a solution for two different phases, that is the liquidus / solidus.  When there is a solution for three phases at once, that is the eutectic.
Oh, except that if you have many compounds all together, you need to have all phases in equilibrium for a eutectic.


text:
When you fill a bath, Is it different that you fill it directly or non-directly?
Yesterday I was taking a shower that this thought occurred to me that when you are filling a bath, flask, etc with water in two different ways in specified time, Will the amount of water in bath be different?
I did a experiment myself in 'Crocodile Chemistry' and got these results as seen below:
First level was pouring water from a faucet directly to the bath from a specified height for 10 seconds. The result was that approximately 250 cubic centimeter water was poured into it.

Second level was pouring water from a faucet to a 'already filled beaker' and filling the bath with overflowed water of the beaker at the same height and the same time. The result was similar to the first level.

I saw all these but I'm not sure that if they're completely accurate, So I wanted to ask somebody that has experience in this manner: Do these experiments show the right results or it's different in the real world?

This seems more physics related really but aside from the water droplets that will stick to the side of the beaker in the second option, if the same amount of water passes through the tap, then the same amount of water must end up in the bath (assuming the beaker is perfectly full).

thanks for reminding me of droplets that stick to the beaker, I didn't think of that but as for your second opinion I must say that there is a time limit in this question.

If the beaker is at or below the outlet of the faucet, the two values should be nearly identical.  There are very small effects, like water sticking to the outside of the beaker as it overflows or the surface of the water changing shape constantly (splashing out more at some times and less at others as a result.)  Those are insignificant for practical purposes.
If the top of the beaker is above the outlet of the faucet, that is, the faucet is submerged in the beaker, then there will be a small addition to the height to which the water must be pumped.  It's only centimeters, but the pressure increase and flow decrease is entirely measurable and would have a small but significant and predictable effect over the course of running a bath.


text:
Is buckminsterfullerene aromatic?
According to Wikipedia,

The $\ce{C60}$ molecule is extremely stable,[26] withstanding high temperatures and high pressures. The exposed surface of the structure can selectively react with other species while maintaining the spherical geometry.[27] Atoms and small molecules can be trapped within the molecule without reacting.

Smaller fullerenes than $\ce{C60}$ have been distorted so heavily they're not stable, even though $\ce{M@C28}$ is stable where $\ce{M\,=\,Ti, Zr, U}$.

Some of us have heard and learned about the "rules" of aromaticity: The molecule needs to be cyclic, conjugated, planar and obey Huckel's rule (i.e. the number of the electrons in $\pi$-system must be $4n+2$ where $n$ is an integer).
However, I'm now very skeptical to these so-called rules:

The cyclic rule is violated due to a proposed expansion of aromaticity. (See what is Y-aromaticity?)
The must-obey-Huckel rule is known to fail in polycyclic compounds. Coronenefigure 1 and pyrene figure 2 are good examples with 24 and 16 $\pi$ electrons, respectively.
Again, Huckel fails in sydnone. The rule tells you that it's aromatic, while it's not.

The planar rule is not a rule at all. We're talking about "2D" aromaticity when we're trying to figure out the $n$ in $4n+2$. The "3D" rule is as following:

In 2011, Jordi Poater and Miquel Solà, expended the rule to determine when a fullerene species would be aromatic. They found that if there were $2n^2+2n+1$ π-electrons, then the fullerene would display aromatic properties. - Wikipedia

This would mean $\ce{C60}$ is not aromatic, since there is no integer $n$ for which $2n^2+2n+1 = 60$.
On the other hand, $\ce{C60-}$ is ($n = 5$). But then this rule strikes me as peculiar because then no neutral or evenly-charged fullerene would be aromatic. Furthermore, outside the page for the rule, Wikipedia never explicitly states that fullerene is not aromatic, just that fullerene is not superaromatic. And any info on superaromaticity is unavailable or unhelpful to me; including the Wikipedia "article" on that topic.
So, is $\ce{C60}$ aromatic? Why, or why not?

"The molecule needs to be cyclic, conjugated, planar and obey the Huckel's rule." This is redundant, because Hückel's Rule is "Monocyclic planar (or almost planar) systems of trigonally (or sometimes digonally) hybridized atoms that contain (4n + 2) π-electrons (where n is a non-negative integer) will exhibit aromatic character. [...]" see the goldbook. The definition acc. to IUPAC is much more fuzzy.

Aromaticity is not binary, but rather there are degrees of aromaticity.  The degree of aromaticity in benzene is large, whereas the spiro-aromaticity in [4.4]nonatetraene is relatively small. The aromaticity in naphthalene is not twice that of benzene.
Aromaticity has come to mean a stabilization resulting from p-orbital (although other orbitals can also be involved) overlap in a pi-type system.  As the examples above indicate, the stabilization can be large or small.
Let's consider $\ce{C_{60}}$:

Bond alternation is often taken as a sign of non-aromatic systems.  In $\ce{C_{60}}$ there are different bond lengths, ~1.4 and 1.45 angstroms. However, this variation is on the same order as that found in polycyclic aromatic hydrocarbons, and less than that observed in linear polyenes.

Conclusion: aromatic, but less so than benzene.

Magnetic properties are related to electron delocalization and are often used to assess aromaticity.  Both experiment and calculations suggest the existence of ring currents (diamagnetic and paramagnetic) in $\ce{C_{60}}$.

Conclusion: Although analysis is complex, analysis is consistent with at least some degree of aromaticity.

Reactivity - Substitution reactions are not possible as no hydrogens are present in $\ce{C_{60}}$.  When an anion or radical is added to $\ce{C_{60}}$ the electron(s) are not delocalized over the entire fullerene structure. However, most addition reactions are reversible suggesting that there is some extra stability or aromaticity associated with $\ce{C_{60}}$.

Conclusion: Not as aromatic as benzene

Resonance energy calculations have been performed and give conflicting results, although most suggest a small stabilization. Theoretical analysis of the following isodesmic reaction

$$\ce{C_{60} + 120 CH4 -> 30 C2H4 + 60 C2H6}$$
suggested that it only took half as much energy to break all of the bonds in $\ce{C60}$ compared to the same bond-breaking reaction with the appropriate number of benzenes.
Conclusion: Some aromatic stabilization, but significantly less than benzene.
This brief overview suggests that $\ce{C_{60}}$ does display properties that are consistent with some degree of aromatic stabilization, albeit less than that found with benzene.

Yes The book MARCH'S ADVANCED ORGANIC CHEMISTRY talks sbout 3D aromaticity. And also a question was asked in the stack exchange previously  https://chemistry.stackexchange.com/questions/88637/what-is-three-dimensional-aromaticity

But although less stabilising than normal aromaticity

There are quite a few conjugated $\pi$-systems out there; some of them are pretty stable, some are less so. The ultimate way to find out is to go and solve the eigenvalues problem for the corresponding matrix; in some cases this can be done manually, with pen and paper.
There are simple cases with analytical solutions. One of them is the case of cyclic polyenes. The result can be summarized in the Huckel's 4n+2 rule. Indeed, it applies exclusively to the cyclic (that is, not polycyclic) conjugated planar systems. Historically, these were the only systems called aromatic. Later, the concept began to stretch more and more.
With fullerenes, there is no simple parametric solution that would fit a more-or-less wide range of systems. The whole idea of $2n^2+2n+1$ electrons is downright wrong, exactly for the reasons you specify (odd number of electrons, to begin with), if not for any other. There is no simpler way than to solve the eigenvalues problem for a particular molecule and see. It turns out that $\ce{C60}$ is stable, and so is $\ce{C70}$ and many other fullerenes, including some isomers of $\ce{C80}$, but not the $I_h$-symmetric $\ce{C80}$, which may however be stabilized by adding six electrons. You see that the question of fullerene stability is highly complicated, and can't be resolved by considering the number of electrons alone.
Whether or not to call the stable fullerenes aromatic is but a question of convenience. I'd rather not, for their reactions are somewhat more like those of alkenes.

C$_{60}$ is antiaromatic as it does not follow Hirsch's rule for aromaticitity of spherically symmetric molecules, $2(n+1)^2$. If we add up all the $\pi$ electrons, we can see that there are 60, which is not twice a perfect square. Here is a great paper explaining spherical aromaticity.
Is C60 buckminsterfullerene aromatic?
In summary, if we look at the MOs of C60, we can see that for C$_{60}^{10+}$, the g-subshell is completely filled, and thus the species is aromatic. If we move to the neutral molecule, it has a partially filled subshell, and thus is not aromatic, which is in line with Hirsch's rule.

While assigning the label 'aromatic' is itself rather meaningless mathematically, saying $\ce{C60}$ is antiaromatic is a bit too much. Yes, it's estimated that 80 percent of the enthalpy of formation is strain energy, but still, fullerene is a very stable molecule.

You need to define what you mean by stable. Fullerenes are not thermodynamically stable, they just don't have any decomposition routes with small activation barriers. From the conclusion of the C60 paper
"C60 is ‘the most strained molecule ever isolated’ with the
astronomically unstable experimental heat of formation of
610±30 kcal mol 1 (over 10 kcal/mol strain per carbon)!"

@Jonathon I'm not sure I agree. Cook a bunch of carbon and take a mass spec, and you see a peak at 720 amu. There are plenty of examples before Smalley et al realized it was C60. If it was so unstable, other products would be likely. Empirically, there must be some stability.

non aromatic or antiaromatic?


text:
Bonding between rings and iron atom in ferrocene
How are the 10 vertical bonds in Ferrocene (sandwich structure) connected, by what type of bonds?

What is the real structure of ferrocene? Concerning the 10 vertical extra long bonds?

What do you mean by 'real structure'? If I'm not mistaken, we're looking right at it.

The "bond lines" in the picture you show are only there to hint that there is some bonding between the rings and iron. But they are not to be interpreted in the sense of bonds in Lewis structures, so don't take them literally. For molecules containing transition metals Lewis structures are not very accurate in describing the actual bonding (this might be interesting in that context).

Yes, but the question I am asking is how are the rings in 3-D SPACE ATTACHED TO THE Iron atom. The discovers & others for Ferrocene openly admitted they could not discover the actual COVALENT bonding type and the bond angles were not discoverable from their x-Ray diffraction patterns. I am looking for replays that can postulate the bonding TYPE, bonding angles and bond lengths, what fraction of the bond is from Fe & from Carbon. Might it be possible that the bonding is of the Pauling "bent-bonding" type. Also, what is the 3-D geometrical shape of the Iron Atom within the so-called sandwich?

How does Iron form 10 or more bonds of any type? Can Iron form 20 bonds to Iron? What is the maximum # of bonds Iron can form from one iron atom? What is the geometrical 3D structure of the Iron atom within Ferrocene?

@ChuckBoldwyn We know the structure of ferrocene. The C5 units are flat and symmetric and the iron forms the filling in a sandwich where the rings are parallel to each other on either side of the iron atom. You seem to be trying to impose a bonding structure made up of single bonds that doesn't (and probably can't) explain this structure. Only a more complex view of the bonding makes sense involving delocalised pi-orbitals from the rings and d-orbitals from the iron. No simple picture with single bonds captures this.

Matt, so you are unable to give a positive, knowing explanation to my question(s)  concerning the 3D bonding structure of 10 bonds from 10 Carbon atoms to one large Iron atom. Also you are only talking "p" orbital bonding when I am talking all sp3 bonding from the 10 Carbon atoms. When the Carbon atoms bond to the Iron, all Carbon orbitals are sp3 orbitals. Think Pauling, sp3 only, bent-bond, banana bonds, not molecular orbitals, not molecular orbital theory.

Also, The orbitals do not form a straight on, direct bond, as shown in the above graphic, but they meet at an angle, indirectly so as to form 10 Pauling type "bent bonds" or
" banana bonds" .
Whatever the hybrid orbitals coming from the Iron atom should not be a problem in the bonding scheme just mentioned. Maybe you can review the Linus Pauling " Bent-Bond Postulates", like in the double bonds of ethylene and all alkenes and all alkynes, all bent bonding, all sp3 orbital bonding,

@ChuckBoldwyn So does the Pauling picture provide any intuitive explanation at all of the aromaticity of the rings? Molecular orbital theory does with no weird assumptions.

@ChuckBoldwyn Drawing these lines this way boils down to convention.  See the later part of my answer here: https://chemistry.stackexchange.com/questions/116519/how-to-you-show-a-3-center-2-electron-bond-in-a-lewis-structure/116526#116526

Related; https://chemistry.stackexchange.com/questions/132633/benzene-as-a-pi-acceptor/132721?r=SearchResults#132721

The "real" structure of ferrocene is an iron atom sandwiched between two flat, parallel, pentagonal C5H5 rings. The diagram above shows the atomic positions from a crystal structure but the bonds are merely a convenience and don't accurately summarize the way the bonding works though the picture does accurately summarize the atomic positions in the crystal structure.
The real bonding requires some understanding of molecular orbital theory and how the d-orbitals of iron interact with the delocalized pi-orbitals in the unsaturated ring. Some pictures of the orbitals contributing to the bonding are shown here.
Ferrocene creates some flurry of new thinking about bonding theory. Observations, for example, show the rings rotate easily in the plane of the rings and this very clearly shows that any picture with 10 sigma bonds is not a good explanation for the overall structure. Plus the chemical reactivity is far more similar to benzene than to an unsaturated system which, again, implies it isn't all sigma bonds.

You are assuming all "p" orbitals & their bonding are only being used in bonding to Iron.....How about all sp3 orbitals & bonding coming from the carbon atoms attaching to whatever orbitals are extended from the Iron atom all as bent bonds. Could a structure, just proposed be possible or likely. Could this be a result of all free-radical bonding resulting in the Pauling proclaimed " banana bonds or bent bonding type?

I didn't say anything about p-orbitals. The only way to really explain the structure is to think of molecular not atomic orbitals for the whole ring (which has delocalised pi-orbitals which interact with iron's d-orbitals). And we are not proposing a structure; we know the structure from x-ray and NMR evidence. All the carbons are equivalent and the structure looks like a sandwich with an iron filling.

Are not pi orbitals made up of "p" orbitals overlapping vertically? I think when the cyclopentadieyl ring converts from sp2 carbon atoms to all sp3 carbon atoms, one would get simple cyclo pentane forming 5 sp3 bonds with the multi- valent Iron atom

@ChuckBoldwyn The ring bonding doesn't "convert" to sp3 and the ring is aromatic with delocalised electrons. Pictures of the bonding that ignore this make no sense of the stability of structure of the molecule.

The real problem is : Why is this structure more stable than similar sandwiches with cycles containing 4 or 6 carbon atoms ?

@Maurice Why not posting this as a new question; preferentially setting sandwich complexes with 4, 5, and 6 carbon atoms on each side next to each other (same transition metal and oxidation state), and a check if these are are Hueckel aromatic.

The pictures we use can inspire questions that are more complicated than they have to be. I like the hamburger picture of ferrocene:

OK, there are some people who like a picture of electron lobes:

Without going into the mathematics of all the orbitals, you can see that the filled orbitals on cyclopentadienide ion will have two lobes to donate into d-orbitals on ferrous ion. And there will be an electrostatic attraction between the cyclopentadienide ions and the ferrous ion.
The use of points (atoms) and lines (bonds) connecting them distorts the filled-space reality of the actual molecule. Counting out 10 Fe-C bonds doesn't suggest the best explanation of the bonding. Of course, when you look at the filled-space models, you can't see exact bond lengths and angles, but it may give you a better idea of atoms fitting together, with molecular orbitals providing some sort of glue.

The second figure you show is not electron density, just one random MO

@Greg: Thanks. You are quite correct. I will change it to say "lobes", unless you suggest a better word - and change the picture, too!

Since the OP is talking about a VB-based approach to describing the bonding in ferrocene, I'm going to ignore any purely MO-based tactics in this answer.
Ferrocene is made by combining two cyclopentadienide anions and one ferrous cation. Each cyclopentadienide anion has one lone pair(which actually corresponds to a Hückel bonding orbital; still writing "lone pair" since I'm ignoring any purely MO-based tactics such as Hückel theory), and these two pairs can each be shared to the ferrous cation.
By now, the iron has 10 (6 from the itself, 2 each from the two bonds) valence electrons near itself. But wait- iron wants 18 electrons so it can be like krypton. Luckily, there are exactly 8 more electrons total nearby- the total of 4 pi-bonds of the two cyclopentadienide rings. Each pi-bond turns into a 3-center-2-elctron bond.
By now, there are 4 bonds near each of the 8 carbons (one C-H, two C-C sigma and one C-C-Fe 3c-2e) with no lone pairs, and 6 bonds near the iron (two C-Fe 2c-2e each, and four C-C-Fe 3c-2e each) with three lone pairs. Since (4 times 2) means octet and (6 times 2) + (3 times 2) means decaoctet, everybody is happy. Note that, since nobody has a formal charge here(dative bonds do not affect formal charge), this resonance canonical satisfies the Pauling rule of electroneutrality. In short, there are two 2c-2e carbon-iron bonds and four 3c-2e carbon-carbon-iron bonds.
Now, there initially are two cyclopentadienide anions, and the lone pair on each ring can reside on any of the five constituent carbon atoms of the ring. Therefore, there are 25 such canonicals, the combination of which constitutes the majority of the ferrocene molecule.
Explaining the pi-backbonding in the ferrocene molecule using VB theory is more complicated, but it can be intuitively understood by the following scheme: starting from a resonance canonical described above, a lone pair on the iron, and a 3c-2e bond involving both the iron and two or the carbons(total number of electrons involved : 4), magically rearrange to two iron-carbon bonds(total number of electrons involved here : 4).
Before: 4 bonds near each of the 2 carbons (one C-H, two C-C sigma and one C-C-Fe 3c-2e) with no lone pairs, and 6 bonds near the iron (two C-Fe 2c-2e each, and four C-C-Fe 3c-2e each) with three lone pairs
After: 4 bonds near each of the 2 carbons (one C-H, two C-C sigma and one C-Fe) with no lone pairs, and 8 bonds near the iron (four C-Fe 2c-2e each, and four C-C-Fe 3c-2e each) with two lone pairs
Again, the two carbons and the iron each satisfy the octet/decaoctet, and again, the dative Fe->C bond does not change the zeroness of the formal charges here, and hence the results are also equally valid resonance canonicals(have to add an MO note here: since the Fe->C bond is dative, the oxidation numbers do not change even after the formal donation of the iron's electrons to the more electronegative carbon).
All of this is an explanation purely in basic VB theory. In modern VB theory, this slightly fails. For example, NBO calculations appear to indicate that there is one half-filled "lone pair" on each cyclopentadienyl ring and no bonds between the ring carbons and the iron, and with the two remaining electrons being used up as parts of partially-filled "lone pair"s of antibonding character on the central iron atom. However, since ferrocene must contain at least one N-center-2-electron bond, it is likely that the intuitive approach above matches the purely VB-model of the bonding in ferrocene more than the NBO calculations do.


text:
Using significant figures when measuring with an instrument with marks other than powers of ten
If an instrument does have markings that are powers of ten, then finding the precision of the instrument is easy. For example, if a graduated cylinder had markings for every $\pu{0.1 mL}$, or every $\pu{mL}$, or every $\pu{L}$, then it is easy to tell that those measurements will be to the hundredth mL, to the tenth of a $\pu{mL}$, and to the tenth of a $\pu{L}$, respectively. These would be valid measurements with the correct number of significant figures.
See this image, for example:

These markings are to the tenths and ones. This is simple.
However, what if that graduated cylinder went up by something like $\pu{5 mL}$? Would its precision be to the tens or to the ones (in other words, will the measurements be to the $\pu{0.1 mL}$ or to the $\pu{1 mL}$)? What would be the valid measurement?

I think it's a personal judgement call more often than not.  In the case of 5mL increments, I would not feel comfortable estimating the value to 1/50th of a division, so I'd go to either whole milliliters or perhaps half milliliter estimates.

The problem is that significant figures are too simple for the real world. Clearly, if you compare a graduation of 1 mL to 5 mL (a thin vs a thick graduated cylinders so the space between labels is the same), you can't express that sufficiently with significant figures, but you can by saying (+/-) 0.1 mL and (+/-) 0.5 mL. If this is of concern, then it is time to move on from significant figures to something more sophisticated.

Usually, you estimate (interpolate) the measurement between the graduation divisions. You should be able to estimate the measurement to one tenth of the smallest division. However, you cannot indefinitely improve the accuracy (the significant figures) of the measurement by interpolation. Volumetric instruments, such as graduated cylinders, volumetric flasks, bulb pipettes, graduated pipettes, and burettes, – like any other measuring instrument – have a limited accuracy. The dimensions, graduating divisions, and accuracy limits of volumetric instruments are standardized. For example, the requirements of measuring cylinders are specified in the international standard ISO 4788. Two classes of accuracy are specified: Class A for the higher grade and Class B for the lower grade. The nominal capacity, graduation divisions, and accuracy limits are shown in the following tables.
$$
\textbf{Graduated measuring cylinders with spouted neck, tall form, Class A} \\
\begin{array}{lll}
\hline
\text{Nominal capacity} & \text{Graduation divisions} & \text{Accuracy limits} \\
\mathrm{ml} & \mathrm{ml} & \pm\ \mathrm{ml} \\
\hline
5 & 0.1 & 0.05 \\
10 & 0.2 & 0.1 \\
25 & 0.5 & 0.25 \\
50 & 1 & 0.5 \\
100 & 1 & 0.5 \\
250 & 2 & 1 \\
500 & 5 & 2.5 \\
1000 & 10 & 5 \\
2000 & 20 & 10 \\
\hline
\end{array}$$
$$
\textbf{Graduated measuring cylinders with spouted neck, tall form, Class B}\\
\begin{array}{lll}
\hline
\text{Nominal capacity} & \text{Graduation divisions} & \text{Accuracy limits} \\
\mathrm{ml} & \mathrm{ml} & \pm\ \mathrm{ml} \\
\hline
5 & 0.1 & 0.1 \\
10 & 0.2 & 0.2 \\
25 & 0.5 & 0.5 \\
50 & 1 & 1 \\
100 & 1 & 1 \\
250 & 2 & 2 \\
500 & 5 & 5 \\
1000 & 10 & 10 \\
2000 & 20 & 20 \\
\hline
\end{array}$$
For example, a measuring cylinder of Class A with a nominal capacity of 500 ml has graduation divisions of 5 ml. By interpolation, you might be able to estimate the measurement to one tenth of the division, i.e. to 0.5 ml. However, the accuracy limits of the measuring cylinder are ±2.5 ml (at any point on the scale).
Various sources of error are inherent in calibration and use. When the greatest possible accuracy is desired, the measuring cylinder should be used as closely as possible to the manner in which it has been calibrated. Most importantly, the capacity of the measuring cylinder varies with change of temperature. The particular temperature at which a measuring cylinder is intended to contain its nominal capacity is the reference temperature of the measuring cylinder. Usually, the reference temperature is 20 °C.
Note that measuring cylinders are calibrated ‘to contain’ (In); i.e. the contained (not the delivered) quantity of liquid corresponds to the capacity printed on the measuring cylinders. To set the meniscus precisely, fill the cylinder with the relevant liquid by means of a plastic tube to a distance of a few millimetres above the selected graduation line, so that the walls of the measuring cylinder considerably above the graduation line are not wetted. Wait 2 min to allow liquid in the cylinder to drain. Then withdraw the surplus of liquid by means of a tube. Careful swaying may be necessary to refresh the meniscus shape.

I think the rule of thumb for the "reading error" being $\pm\,0.1$ of the smallest division applies here, so for your case, it'd be $5\;\mathrm{mL}\pm\, 0.5\,\mathrm{mL}$. See this page, entitled "Math Skills - Scientific Notation", from Texas A&M. Specifically, it is noted that

A rule of thumb: read the volume to 1/10 or 0.1 of the smallest division. (This rule applies to any measurement.) This means that the error in reading (called the reading error) is 1/10 or 0.1 of the smallest division on the glassware. If you are less sure of yourself, you can read to 1/5 or 0.2 of the smallest division.

Please note that the linked page contains a misleading example in which a beaker is used for measuring the volume with an uncertainty of ±1 ml. This is inadmissible! Beakers, Erlenmeyer flasks, and the like are not volumetric instruments. They are not precisely calibrated, and the scale serves only as an approximate guide.


text:
How to predict a good resonance structure?

I have to draw curved arrows to convert the above molecule to its resonance structure but I'm not entirely sure how. I believe the sulfur atom has one lone pair and the oxygen atom has 2 lone pairs. But how would I be able to predict the location of the curved arrows?
I took a look at their formal charge and both seem to be 0.

Well, major resonance structures rarely involve tetrahedral carbons, so we can probably rule out any electron movement in the S-C single bonds. What do you think about the electron distribution in the S=O double bond?

As another hint: Add all the electron pairs into the structure, it might give you an idea. (Also, the resonance structure you have drawn is no longer considered major ;))

oxygen is more electronegative so would a pi bond go to the oxygen atom and make that O-?

Yes, but those would be only formal charges reflecting it's single dative bond


text:
Are there general rules for Base + Metal Compound reactions like there are for Acids?
Are there general rules for Base + Metal Compound reactions like there are for Acids?
For example, acids have some general rules.
**
Acid + Metal => Salt + Hydrogen Gas
Acid + Metal Carbonate => Salt + Water + Carbon Dioxide
Acid + Metal Oxide => Salt + Water
Acid + Metal Hydroxide => Salt + Water
**
Are there equivalent rules for bases?
Base + Metal => ?

On the same level of generalization, the rule is "no reaction", but this rule has quite a few exceptions (probably more than the rules you quoted above).
Water-soluble bases would react with amphoteric metals to produce hydrogen and the corresponding salt:
$$2\ce{Al + 2KOH + 6H2O \to 2KAl(OH)4 + 3H2}$$

How can a metal be 'amphoteric'? Metals undergo redox reaction instead of acid-base reactions, don't they?

Just like that.

So if I write in an examination that certain metal is amphoteric, will it be accepted by a general examiner? Or calling a metal amphoteric is something done in informal talk and should be avoided in the exams?

@Aumkaar I think it will be accepted.


text:
Burns from boiling water and steam

Why is a burn to the skin caused by steam more serious than a burn caused by the same amount of boiling water at the same temperature?

The temperature is the same, which implies that the kinetic energy of the particles in both the steam and the boiling water are the same; and yet the steam is more dangerous. Why so?

Let’s consider the following cases:

getting $1\,\mathrm{mol}$ of $100\,\mathrm{°C}$ water on one’s skin
getting $1\,\mathrm{mol}$ of $100\,\mathrm{°C}$ air on one’s skin
getting $1\,\mathrm{mol}$ of $100\,\mathrm{°C}$ water vapour on one’s skin

With the slightly irrealistic assumption that all of these liberate all their thermal energy to the skin while cooling down to $40\,\mathrm{°C}$.
1st case
Assuming isobaric conditions (constant pressure), the heat energy liberated from $100\,\mathrm{°C}$ water that is cooled to $40\,\mathrm{°C}$ can be calculated as follows:
$$\Delta Q = \Delta T \cdot C_p(\ce{H2O}) \cdot n(\ce{H2O})$$
$$\Delta Q = 60\,\mathrm{K} \cdot 75.327 \frac{\mathrm{J}}{\mathrm{K\,mol}} \cdot 1\,\mathrm{mol}$$
$$\Delta Q \approx 4.5\,\mathrm{kJ}$$
2nd case
A similar formula holds true. However, now we are considering a gas all the way, not a liquid. (And ‘air’ is technically not a single substance, but we can work with it.)
$$\Delta Q = \Delta T \cdot C_p(\mathrm{air}) \cdot n(\mathrm{air})$$
$$\Delta Q = 60\,\mathrm{K} \cdot 29.19 \frac{\mathrm{J}}{\mathrm{K\,mol}} \cdot 1\,\mathrm{mol}$$
$$\Delta Q \approx 1.7\,\mathrm{kJ}$$
3rd case
Unlike the previous cases, we start off with steam, which will condense to water. Therefore, we need to add the heat of vapourisation to the equation.
$$\Delta Q = \Delta T \cdot C_p(\ce{H2O}) \cdot n(\ce{H2O}) + \Delta H_{\mathrm{vap}} \cdot n(\ce{H2O})$$
$$\Delta Q = 60\,\mathrm{K} \cdot 75.327 \frac{\mathrm{J}}{\mathrm{K\,mol}} \cdot 1\,\mathrm{mol} + 40.66 \frac{\mathrm{kJ}}{\mathrm{mol}} \cdot 1\,\mathrm{mol}$$
$$\Delta Q \approx 45.2\,\mathrm{kJ}$$
You will note that the third case delivers a lot more heat energy than the first two cases.
(Part of the simplifications I used means that I grossly overestimated the heat transferred by hot air. Which is also why one can put one’s hands under a hair dryer and be fine with it.)

After a bit of research, I discovered the following logic:
The boiling point of water is 100 degrees Celsius. At this temperature the water will begin to change from the liquid state to the gas state. In order for this change to take place, additional energy is required. In fact, every gram of liquid water requires 540 calories of heat energy just to convert it to steam. This is called the Heat of Vaporization. During the conversion process, temperature does not increase. It is, therefore, possible to have both liquid water and steam that exist at 100 degrees Celsius. While they both would exist at the same temperature, the steam would have a lot more heat energy due to the additional 540 calories per gram of heat energy that has been absorbed.
This energy release, when made contact with skin, causes a much worse burn than if the same amount of boiling water were to hit your skin where it would decrease in temperature (to your skins temperature) but would not have to go through a phase change. The loss of energy that is released from steam hitting your skin occurs quickly and in a small localized area, therefore causing damage to your cells.


text:
Biodegradation time for biodegradable shopping bags?
How long does it take for the plastic/polythene shopping bags marked biodegradable to biodegrade?

It depends on the environment and on the definition of degrade.
In an anoxic landfill, a very long time before any change is noticed. Even in a wet environment, open to air and sunlight, only the cornstarch additive decomposes.
Initially, it was thought that adding cornstarch would help speed decomposition (and provide an outlet for surplus farm products), but studies such as this show that though some speed in decomposition is gained (shown by water absorption), tensile strength of the plastic is decreased by adding starch, so yet more plastic is needed to make the material sufficiently strong (and the bags still rip).
Also, though the corn-starch additive can decompose rapidly, this study shows that the polyethylene, though now fragmented, does not further decompose. This leads to run-off of microparticles into lakes and oceans.
So if your definition of biodegrade means how long does it take for the bag to shred, according to yet another study performed in a heated environment with added oxygen, about a year. At the end of that time, though, perhaps 80% of the initial mass remains, mostly polyethylene fragments. Much of this eventually is washed into oceans, is eaten, and winds up in us, as well.


text:
Do carbon monoxide and graphene oxide react?
I want to know what happens if  $\ce{CO}$ and graphene oxide (GO) react.
Through the internet the reaction of GO with $\ce{H2O}$ is available but I did not find anything about carbon monoxide.

People: Please add better tags if appropriate.


text:
History of octanol-water partition coefficient
We are aware that one of the basic information about a material is the logP value it has. The logP is worked out from the partitioning between the octanol and the water phases. Does anyone know who was the first to introduce this octanol-water partition coefficient? I had a look through the literature but was not able to find the author of that rule.

Solid-state-chemistry? Really??

well, no relevant tags available (the ones I thought of at least). I tried to add some myself, but my reputation is still very low to do so. Thanks for the new tags :)

I didn't create any new tags, I only added existing ones which you could have done yourself.

The most oft-cited publication is Partition Coefficients and Their Uses. A. Leo, C. Hansch, and D. Elkins. Chemical Reviews 71(6), 525-616 (1971). This manuscript was noted as a "Citation Classic" (see text here) in which A. Leo writes (in part):

In recognition of the effort of designing
  a method of calculating log P (o:w) from
  structure, the early version of which was reported
  in this article, the Chesapeake section
  of Sigma Xi presented me with its Excellence
  in Science Award for 1980.”

Note bene: In the last article I've linked, mention is made that the phenomena was first investigated by Bertholet in 1872 and "put on a firm basis by Nernst in 1891." That said, the 1971 publication of Leo $\it{et\; al}$ is the first modern/comprehensive listing of octanol-water partition coefficients and the place where the symbology (and definition) were written.


text:
Difference between ionization enthalpy and work function?
The work function is stated as the energy required to eject an electron out.. And ionisation enthalpy is stated energy required to move an electron out of the gaseous atom. I don't understand if there is any difference.

So are there any differences between a gas and a solid? If so, what does that mean for the electrons?

Is the difference only about the state of the substances? So these two things are pretty much the same?

Conceptually they are very similar. But, one is a property of the atom (electron orbitals), and the other is a property of a solid (band structure formed from interactions of all the electrons). So, the work function of graphite and diamond are different, even though they are both carbon allotropes.

To broaden the comment, the first ionization energy for a carbon atom is about 11.2eV, the work function for diamond is around 7.2eV (fluorine terminated), and the work function for graphite is around 4.8eV. Graphene seems to be in the 4.9-5.1eV range. Surface states matter.

You mean, as long as its a gas, it will have a constant ionisation enthalpy value, but if it is a solid, then energy required to eject the outer electron , now being the work function, varies depending on the surface?

And the bulk band structure leads to different energy levels for the solid overall than for an isolated atom.


text:
Why is ATP, but not ADP or AMP, a sole source of energy?
Why are three phosphate groups (such as in ATP) required for release of energy, but not two (ADP) or one (AMP) phosphates? How does breaking the phosphate bond release energy?

Your premise is flawed - ADP is used as a power source in some reactions.

Let me add to iad22agp's excellent answer.
Often times in introductory biochemistry classes, the difference between phosphoanhydride and phosphate ester bonds is not adequately explained.
ATP (and ADP) contain phosphoanhydride bonds.  In these bonds, two phosphoric acid groups are condensed into one:
$\ce{R~-H2PO3 + R'~-H2PO3 -> R-HO2P-O-PO2H-R' + H2O}$
In contrast, AMP and many other phosphate-containing biomolecules such as glucose-6-phosphate glyceraldehyde-3-phosphate, contain phosphate ester bonds, where a phosphate group is condensed with an alcohol:
$\ce{R~-H2PO3 + R'OH -> R-HPO2-O-R' + H2O}$
Just as "regular" (i.e. carboxylate) esters like ethyl acetate are more stable to hydrolysis than carboxylic anhydrides (e.g. acetic anhydride), phosphate esters are much more stable and "lower energy" than phosphoanhydride bonds.  They don't have enough energy to power many of the key reactions of biosynthesis.  So that explains why AMP isn't a good energy source.
(By the way, mixed phosphate-carboxylate anhydrides are as high or higher in energy as phosphoanhydride bonds.  Examples from biochemistry include acetyl phosphate or phosphoenolpyruvate.)
So ATP has two phosphoanhydride bonds.  And as iad22agp mentioned, the electrophilic nature of the central P atom makes the outermost phosphoanhydride bond especially prone to hydrolysis.  However, ADP still contains a high-energy phosphoanhydride bond.  Consider this reaction:
$\ce{ATP + AMP <=> 2 ADP}$
The equilibrium constant for this reaction is about $2.82$, meaning the rightward "forward" direction is favored, but not by a whole lot.  So the "extra" energy of ATP's particularly high-energy phosphoanhydride bond is not really that much higher than that of ADP's regular run-of-the-mill phosphoanhydride bond.
There are somewhat rare examples where ADP powers key metabolic reactions but ATP does not.  That shows it is nearly as good an energy source as ATP.
The reason that ATP is so prevalent is probably because of evolutionary reasons.  Also, another trick that enzymes used with ATP is sometimes to hydrolyze the interior phosphoanhydride bond, leading to AMP and pyrophosphate.  Because pyrophosphatase enzymes quickly degrade pyrophosphate to two free ortho-phosphate  groups, this trick effectively makes what would have been a reversible reaction effectively irreversible.  That trick doesn't work with ADP because it has only one phosphoanhydride bond.

Think of ADP and ATP as substituted phosphoric anhydrides. ATP has a strong tendency to hydrolyze due to the electrophilic nature of the central phosphorus atom which bears two electron-deficient phosphate leaving groups. (note that ADP has only one, and is a poorer source of energy, whereas AMP has only a normal phosphate ester linkage.) ATP is kinetically relatively stable to hydrolysis in the absence of an enzyme (which would simply waste the energy as heat), but a wide variety of enzymes are able to couple the ATP hydrolysis reaction so as to drive various reactions or other processes.


text:
Energy a system at room temperature can supply?
"This reaction requires XYZ activation energy, therefore it cannot happen at room temperature." is a sentence I find often in organic chemistry textbooks. Activation energies are steadily given in textbooks too. Even though it is a nice piece of info, which gives a feeling of how (not) sluggish a reaction can be, it would be nice to know how much energy can such a system actually supply? Like, I don't know - 30 kJ/mol or 35 kJ/mol.
How much energy do particles in a system at normal conditions (25°c, 1 atm) have? I am note sure if there even is an answer, taking into consideration Maxwell-Boltzmann distribution, and an endless array of reactions... but maybe there is?

R = 8.314 J/K/mol is a good place to start...

Echoing Jon Custer's comment:

R = 8.314 J/K/mol

Multiply by 298 K, and you have a rough-and-ready rule-of-thumb number:
The "thermal energy" at room temperature is about 2.5 kJ / mol.

Or about 23 meV if you prefer those units, particularly for semiconductor physics...


text:
How to write the balanced equation of the reaction of potassium nitrate, carbon, and sulfur?
Given the following reaction equation:
$$\ce{KNO3 + C + S -> SO2 + CO2 + N2 + K2O}$$
Here is my solution:
\begin{align}
& 2\times & \ce{2 N^{+V} + 2 * 5e- &-> N2^0}\\
& 5\times & \ce{C^0 &-> C^{+IV} + 4e-}\\
\end{align}
That is why $$\ce{KNO3 + 5 C + S -> SO2 + CO2 + N2 + K2O}$$
Then I think it should be $$\ce{12 KNO3 + 5 C + 10 S -> 10 SO2 + 5 CO_2 + 6 N2 + 6 K2O}$$
Am I right?

I strongly object against having free $\ce{K2O}$ among the products. Surely it will react with those acidic oxides.

Ok, but this is how it is written in our book.

Well, then leave it this way. Also note that the equation is not unique. Your ratio of C to S is 1:2. It could just as well be 1:1. It could be anything.

Your equation balances so it could be right. However, there is no unique solution to your given equation because the $\ce{S}$ and $\ce{C}$ in the reactants and the $\ce{SO2}$ and $\ce{CO2}$ in the products allow you to dispose of any excess oxygen from the nitrate. Another balanced equation with different stoichiometry is:
$$\ce{4KNO3 + 3C + 2S -> 2SO2 + 3CO2 + 2N2 + 2K2O}$$
Now the interesting part. The equation you are given is for the burning of some form of gunpowder, although one without any charcoal, which is usually described as $\ce{C7H4O}$. The products of burning gunpowder are complicated to predict and strongly dependent on the exact composition of the powder sample being used.
As suggested in the comments by @IvanNeretin, the $\ce{K2O}$ is likely to react with the acidic oxides $\ce{CO2}$ and $\ce{SO2}$ to form a mixture of carbonates and sulfates. There are many possible equations showing this (one is shown below) but the exact stoichiometry will depend of the ratio in which the reactants are mixed in the powder:
$$\ce{10KNO3 + 3S + 8C -> 2K2CO3 + 3K2SO4 + 6CO2 + 5 N2}$$

Well I see a good reason, electronically for this reaction being balanced in more than one possible way.
I would like to draw your attention on the electrons transferred in these redox reactions.
As the redox reactions balance they need to have equal no. of electrons lost [by reductants] and gained [by the oxidants] at the same time.

To make the electrons transferred equal, a lowest common multiple of electron is required. Which fixes up to 20, i.e.
4 KNO3 gain 20 electrons;
and C & S taken together lose 20 electron, wherein they can have a number of choices - the simplest of which could be:
1C + 4S
2C + 3S
3C + 2S
4C + 1S
2.5 C + 2.5 S
2.75 C + 2.25 S etc. etc.
Each of which will balance the equation.
This would be true theoretically and well as experimentally is we collect a sample of detonated gun powder to analyse the amounts of CO2 & SO2 - subject to the different (delta)G values, which could have preference for a given reaction under given conditions over the other.
As @bon said, K2O formed in each of such cases would be combining with the acidic gas mixture [if NOT allowed to escape] to produce K2CO3 & K2SO3 [NOT the SULFATE, bon]

Can you please stop fully capitalising words? It is generally considered to be the online equivalent of shouting. If you have not already, please do read the policy on Being Nice.

Sorry! No offence intended.

You can [edit] your post at any time. While you're at it, please have a look at the possibilities with MathJax and mhchem here and here.

@CheMistry If you want emphasis use bold or italic or both.


text:
Why are the C-C bonds in benzene closer in length to double bonds than single bonds?
The $\ce{C-C}$ bonds in benzene are $\mathrm{140~pm}$ long which puts them in between single bonds and double bonds (given at $\mathrm{147-154~pm}$ and $\ce{134-135~pm}$ respectively, sourced from here and here) as expected. However, since the two resonance structures for benzene contribute equally, leading to six identical bonds, one might expect that the bond lengths would be exactly intermediate between a single and a double bond, but in fact the bond lengths seem to be closer to a double bond than a single bond. Why is this the case? My teacher asked me this a while ago when I was preparing for university interviews and I have never thought of a satisfactory answer.

The "single" bonds in benzene involve the overlap of $\ce{sp^2-sp^2}$ orbitals, whereas alkane single bonds involve the overlap of $\ce{sp^3-sp^3}$ orbitals. This leads to the former bonds being shorter.

Extending my earlier comment, the single bond length in 1,3-butadiene is around 145 pm ( presumably the bond would be a bit longer if we eliminated the resonance overlap between the 2 double bonds by placing butadiene in a conformation where the 2 double bonds are oriented 90° to one another). So the benzene bond length is more or less in between an $\ce{sp^2-sp^2}$ single bond (butadiene, 145 pm) and an $\ce{sp^2-sp^2}$ double bond (135 pm).

Butadiene is a questionable system, as it includes some amount of conjugation. Cyclooctatetraene is a bit more relevant. It gives 1.47 for single bond and 1.33 for double bond, while definitely no having any significant conjugation.

https://chemistry.stackexchange.com/questions/37688/why-are-the-c-c-bonds-in-benzene-closer-in-length-to-double-bonds-than-single-bonds

one might expect that the bond lengths would be exactly intermediate between a single and a double bond

They actually are. The relevant reference structure, that can provide us lengths of single and double bond for $sp^2$ carbons with not conjugation involved is cyclooctatetraene. The article on the X-ray structure of cycloocta-1,3,5,7-tetrene gives lengths 1.47 and 1.33 angstrom, with their average (1.40) being the exactly value of C-C bond in benzene you provided.

Why should they be bang in the middle? Or to better phrase the question: What reasoning are you applying to assume an averaged bond length?
Bonds don’t work as we laymen like to write them, with either a single line or a double line; and a double line being equivalent to two single lines. Rather, bonds — and most importantly, their lengths — are the result of equilibrated forces: The repulsion of two atoms pushing them apart and the bond-forming energy drawing them together.
Most notably, a double bond is not shorter than a single bond because more or less or whatever energy is involved, but because the molecular orbitals are best (i.e. lowest energy) at that particular length.
When comparing the aromatic benzene bonds to single or double bonds, we have to notice:

benzene’s carbons are sp² hybridised while single bonds’ carbons are sp³
there is a significant $\pi$-bonding across the benzene ring still happening, even though that amounts to only $0.5$ additional bond order.

Therefore, benzene’s bonds are much closer to double bonds in nature by the orbitals forming them. It is only reasonable that the resulting bond length is closer to that of double bonds than that of single bonds.
Remember finally, that ‘bond order’ is only a concept counting bonding and anti-bonding electrons. So even formal single bonds could be much shorter than expected (even though I do not have an example on me at the moment).

Relevant to this discussion is the prediction coming from the Hückel model.
As many readers know, for a single homoatomic cycle the Hückel model resolves into the Frost circle, which has a radius equivalent to two isolated pi bonds. Putting in the orbital positions for a six-membered ring and filling the lower half, modeling benzene, we find that the total pi- bonding energy on this model is equivalent to
$2+2×(2\cos60°)=4$
isolated pi bonds, not three. The delocalization of the optimal number of conjugated pi electrons around the ring has effectively created an extra bond beyond those formally drawn into the structure.
This effect not only accounts for the stability of the benzene ring (and aromatic rings in general), it also means that in effect there is more than half a pi bond to go with the sigma bond at each carbon-carbon linkage. So the observed properties of the linkages will be closer to double bonds than to single bonds.


text:
Is Potassium Silicate better than Sodium Silicate (water glass) in a refractory?
I would like to know whether Potassium Silicate is a better binder than Sodium Silicate in making refractory bricks.
A search on the internet on a 'vs' question relating to the two says that Potassium Silicate is more soluble, has lower efflorescence, lower tack, something about atmospheric dissolvers, none of which makes any sense to me. The only thing of interest was apparently Potassium Silicate can be used at marginally higher temperatures ($\rm50\, ^{\circ} C$).
Can I substitute Potassium Silicate for Sodium Silicate?

Your points are good, but potassium silicate also has a lower viscosity than sodium silicate for most/all modulus ratios; so your handling and workability parameters should also improve.


text:
Can you call a table frozen?
Someone has argued that the table sitting in front of us is 'frozen'. I do agree that the table has a freezing point. The table has a melting point, but that is beyond the ignition point and the boiling point is even further beyond that. If an object only exists in one state can you call it frozen? Secondly, I think that there is some disconnect because I was arguing more from a sociological stand point. I think that the definition is more subjective. If I call a wooden bowl frozen, it would hinder the meaning of the sentence from an outsiders stand point. What do you think? Can you say that the table in front of you is frozen?

A wooden table does not have a melting point per se. While there is a product called 'liquid wood', it is not molten wood.

Honestly, this is not a chemistry question. The answer is based on what you interpret "frozen" to mean.

I'm voting to close this question as off-topic because it is a philosophical, not a chemical question.

The technical definition of 'freezing' implies a phase transition from liquid to solid, and therefore one assumes your interpretation requires that a table can at some point be a liquid.
Under normal circumstances, being a solid does not necessarily imply something is frozen, because not all materials can exist as liquid. Heating some materials will simply lead to (i) phase transition through sublimation or (ii) combustion or decomposition (a chemical reaction) leading to a both gaseous and solid products.
Presumably, the table you are pondering is made of wood. Wood is a mixture, composed of mostly cellulose, hemicellulose and lignin. The ratios depend on the wood type, and there are a few other bits in there also. Importantly, wood, even tables, still contain water. Under standard conditions one might expect to find a table (normal pressures), simply heating will not cause your table to melt. Above about 100oC, decomposition products are slowly released from the bulk wood; noncombustible carbon dioxide, water vapour and trace organics. Cellulose decomposes above about 250oC, and hemicellulose somewhat below that. Lignin begins to decompose above about 150oC, but has a large decomposition range.
Now, if you could heat a table under immense pressure, you may be able to influence the decomposition rates to a point where you could possibly get these materials to melt rather than simply decompose. Heating wood under moist high pressure conditions is known to impart an element of plasticity to some woods, so that it can be bent. I'm not confident that you could melt it though.
So I would call your table simply solid, not frozen.


text:
What happens when 3-(2-chlorophenyl)propanenitrile is treated with sodium amide?

I'm having some trouble figuring out what happens in the above reaction. Based on the strongly basic conditions I think that a benzyne mechanism should be operative, but I'm not sure what else should happen, or whether the nitrile is involved in the reaction. What is the product and how is it obtained?

You are correct that a benzyne is formed as an intermediate in this reaction. However, the nitrile is not an innocent bystander. Recall that protons α to a nitrile group are acidic: this is because the negative charge can be delocalised onto nitrogen. The pKa of a typical nitrile is around 30, lower than that of ammonia, which is roughly 40 (see: Evans pKa table). So, sodium amide is capable of deprotonating α to the nitrile.

This anion can now act as a nucleophile in the benzyne mechanism which you are no doubt familiar with. Note that the intramolecular attack of this nucleophile is likely to outcompete intermolecular attack of the amide ion. Also, since the nucleophile must attack intramolecularly, there is no question about which triply bonded carbon it attacks; it can only reach the position ortho to itself:

This specific reaction to form a benzocyclobutene is described in, for example, J. Med. Chem. 1966, 9 (5), 656–659.


text:
Saturated and unsaturated elements?
I was reading about Group 15 and came upon the following:

The penultimate shell in N contains 2 electrons (saturated), in P contains 8 (saturated), in Arsenic 18 (saturated) while antimony and bismuth contain 18 electrons (unsaturated) each.

What does the author mean by saturated and unsaturated?

Well, are the shells full seems only option IMO

Although I've never heard this terminology before, I'm pretty sure I can see what it means.
As you can see from this diagram, each increase in the principal quantum number (n), adds a new type of orbital to that shell.

The electron configurations of the group 15 elements are:
\begin{array}{|c|c|} \hline
Element & Electron~configuration \\ \hline
\ce{N} & 1s^2~2s^2~2p^3 \\ \hline
\ce{P} & 1s^2~2s^2~2p^6~3s^2~3p^3 \\ \hline
\ce{As} & 1s^2~2s^2~2p^6~3s^2~3p^6~3d^{10}~4s^2~4p^3 \\ \hline
\ce{Sb} & 1s^2~2s^2~2p^6~3s^2~3p^6~3d^{10}~4s^2~4p^6~4d^{10}~5s^2~5p^3 \\ \hline
\ce{Bi} & 1s^2~2s^2~2p^6~3s^2~3p^6~3d^{10}~4s^2~4p^6~4d^{10}~4f^{14}~5s^2~5p^6~5d^{10}~6s^2~6p^3 \\ \hline
\end{array}
$\ce{N, P}$ and $\ce{As}$ all have full penultimate shells, whereas $\ce{Sb}$ and $\ce{Bi}$ do not -  the $4 f$ and $5 f$ orbitals are unnocupied respectively. Therefore 'saturated' seems to refer to elements where the penultimate shell is fully occupied, and 'unsaturated' to elements where it is not.


text:
Gas expansion through adiabatic turbine
Gas expands through an adiabatic turbine.  The inlet stream flows at 100 bar and 600 K, while the outlet stream is at 20 bar and 445 K. The constant-pressure molar heat capacity of the gas is $30.0 + 0.02T$, where the molar heat capacity is in J/(mol K) and the temperature is in K.  The process is at steady state.
Assume the gas is non-ideal and follows the following equation:
$$p=\frac{RT}{V_\mathrm m-b}+\frac{ap^2}{T\left(V_\mathrm m-b\right)}\tag{1}$$
with $\displaystyle a=0.001\ \frac{\mathrm{m^3\ K}}{\mathrm{bar\ mol}}$
and $\displaystyle b=0.00008\ \frac{\mathrm{m^3}}{\mathrm{mol}}$
For this gas the differential molar enthalpy change is:
$$\mathrm dH_\mathrm m=C_{\mathrm m,p}\mathrm dT+\left[V_\mathrm m-T\left(\frac{\partial V_\mathrm m}{\partial T}\right)_p\right]\mathrm dp\tag{2}$$
Assume that the ideal expression for the molar heat capacity given is still valid, but that it is only true as $p$ goes to 0.
And the question is; what is the shaft work of the turbine?
This question is from a thermodynamics course and to solve it students were sort of guided through a solution.
First substitute equation (1) into equation (2) and find a general expression for the differential molar enthalpy change of the non-ideal gas.
Then draw a $pT$ diagram of an appropriate path to evaluate the enthalpy change of the process, and just integrate along the path to find the answer.
I still can't figure out how this was supposed to be done.
Thanks to anyone who bothered to read/think about this problem as it is kind of long.

You are aware that, from the open system version of the first law of thermodynamics, the shaft work per unit mass must be equal to the change in enthalpy per unit mass, correct?  So you need to determine the change in enthalpy per unit mass between the gas entering the turbine and the gas exiting the turbine.  Do we agree so far?
We know that enthalpy is a function of state.  So we need to figure out a convenient path between state 1 at 100 bar, 600 K, and state 2 at 20 bar ,445 K such that the equation provided can easily be applied to determine the change in enthalpy.  Consider the following 3 part path:

Hold the temperature at 600 K and lower the pressure to say 1 bar, where the gas behaves as an ideal gas.  Use the equation provided with dT set to 0 to evaluate $\Delta h$ for this step
Hold the pressure constant at 1 atm, and use the ideal gas heat capacity to calculate $\Delta h$ for cooling the gas down from 600K to 445K.
Hold the temperature constant at 445K and raise the pressure from 1 bar to 20 bars.  Use the equation provided with dT set to 0 to evaluate $\Delta h$  for this step.

The sum of the enthalpy changes for these three steps adds up to the enthalpy change for the overall change.


text:
How to connect proteins via disulfide bonds computationally?
I want to connect proteins together to form a dimer. As seen in the picture, the monomers come close to each other along the edge of the higher-order structure (forget about the sulfate there). How can I find out, where I need to introduce an Xaa->Cys mutation so that the most stable disulfide bond is formed between the proteins? It is easy to get a visual clue where about a mutation might be good, but is there a good computational method to calculate the energies and scan through some possibilities?
(I would also go with hydrophobic or van de Waals interactions. Basically, I just want to glue those proteins together.)

If you have a structural model for the interaction, i.e. atomic coordinates, then you could use our software called Disulfide by Design.  See the following links.
http://cptweb.cpt.wayne.edu/DbD2/
http://www.biomedcentral.com/1471-2105/14/346

Perhaps you could give an overview of how the program works and what it is useful for. At the moment this seems rather like spam.

I don't think it looks like spam. Actually, it's rather interesting. But so far it seems that I can only calculate energy minima of residues that are in the same protein, not in multiple ones ...?

Far from spam.  I am the author of the software and publications. The original algorithm has been used in more than 90 published research papers, including a number in Cell, Science, and Nature. The above link to the paper in biomedcentral describes the software. You can find additional information on the underlying algorithm in my original paper here: http://bioinformatics.oxfordjournals.org/content/19/14/1852.full.pdf

Also . . . the software has been used in several studies that linked separate proteins or subunits.  Here are the PubMed IDs for a few examples: 18583346, 19425108, 24315638.    The energy calculated in DbD is associated only with the actual bond.  It is a relative value used for selecting bonds most likely to form.

You might find useful our recent review article on disulfide engineering: http://www.sciencedirect.com/science/article/pii/S0014579313008685


text:
Can all compounds that plants and animals produce be synthesized in a lab?
I guess a better way to ask the question would be, Are there compounds that exist in biology that humans don't have the capacity to produce in a lab?

Long story short, we can synthesize anything bioorganic, but in some cases that would be prohibitively expensive.

This is dependent on the limitation of the technology. Theoretically you can synthesize all the enzymes responsible for the biosynthesis of a molecule, immobilize them and create a reactor that will synthesize the desired product. But why waste money when you can do it easily using micro-organisms.

I suspect, in its current form, this question is largely opinion based. Currently, as it stands in September 2015, there are definitely limitations. But never say never.
On the small molecule scale, Maitotoxin is a famous example of a molecule that has yet to been synthesised in the lab. But perhaps only for a matter of time (or money)
3D printing and automated synthesis is expanding the scope for small molecule synthesis, giving more flexibility to synthesising recalcitrant targets.
On the larger scale, human DNA contains up to about 300 million base pairs. Currently, the largest DNA synthesised in the laboratory is about 600,000 base pairs, so still a way off. However, with increasing capacity in areas like artificial gene synthesis, complete replication of entire DNA is not unforeseeable. One day.

I work in a natural products synthesis lab, so your question is exactly about what I do every day.
There is the term ‘paper chemistry’, which describes reactions that one has drawn on a paper (a rather easy task in a chemist’s lab life) but which one hasn’t actually performed yet. The vast majority of a Ph.D. student’s work is paper chemistry: Thinking of how to synthesise a molecule, attempting the reactions, finding out that something unexpected happens, rationalising it, facepalming for the silly idea one had and moving on to the next paper chemical reaction. With paper chemistry, everything is possible.
In reality, it usually takes a Ph.D. thesis for a molecule of average complexity. Some colleagues are lucky and get it done in a year because everything worked, some are unlucky and don’t get through with their project in four and a half years because nothing worked. Or it worked, but only in $10\,\%$ yield. Or it worked but the protective group fell off and we need to reistall it; all those little nastinesses of everyday lab work.
Nature manufactures its biomolecules, both primary and secondary metabolites, extremely efficiently; usually with yields close to $100\,\%$ and in a very atom-economic way of synthesis. Lab chemistry usually deals with yields of less than $95\,\%$, separation of undesired side products, and uses bulky protective groups. Many natural product university classes show two pictures of how nature does it (assembly-line style) and how chemists do it (a smith with a hammer) just to emphasise these differences.
The reason for this efficiency difference is 4.5 billion years of optimising that nature has gone through, while mankind has had only about 150 years time to optimise its chemistry. Maybe our efficiency will catch up one day. Maybe not.
On the long run, I don’t believe that there is a molecule out there that cannot be synthesised in vitro. On the other hand, there are examples like maitotoxin (19 years as of now) that show just how long it can take for just a single molecule. I believe I read about a molecule whose total synthesis took 24 years in Classics in Total Synthesis III — that’s the linear equivalent of 8 Ph.D. theses! So it is probably up to mankind’s endurance whether the extremely complex molecules will be synthesised or not.

"some are unlucky and don’t get through with their project in four and a half years because nothing worked" And then they say: "Negative results are also results."

Hard question, but I suspect the answer is yes. The original definition of organic molecule was one that could only be manufactured in an organism, which turned out to be false. It may be true that there are molecules we do not yet know how to synthesize, but that does not preclude our figuring it out in the future! We managed to synthesize vitamin B12, for example, which was a major feat. So there may well be compounds that we cannot synthesize today, but that is not good indicator of what we could do in the future.


text:
Naming functional groups in a molecule
Did I name the functional groups that are present in the following molecules correctly?  All I was given is the picture below. I have drawn my attempt in the picture.

Edit1: response to k--

Edit2: response to Mithoron

second version is worse than first...

@Mithoron.        Can you please tell me where I messed up?

Basically what k told in comment. Functional groups are heteroatoms with hydrogens or  carbons double/triple bonded. Carbons with only single bonds are part of alkil (methyl itp.) groups, not other groups. Substituted aromatic rings  belong to aryl (phenyl, phenylene itp.) groups.

Btw you can make better pictures with IsisDraw, ChemSketch, Inkscape or other free programs.

@Mithoron        I included an updated drawing, if you could check it I would appreciate it. Sorry I didnt use those chemistry drawing programs you suggested, but I didnt think I had the time to fuss with them. I am bad at learning new things when I have alot on my plate. Thanks for the advice.

Better, but you should cut in half single bonds neighboring to heteroatoms and double bonds. Alkyl and aryl groups aren't functional so you can skip them and or put ethylene (-CH2-CH2-), isopropyl  (CH3)2CH- and phenylene -C6H4 if you want to be hardcore ;)

First molecule:
The ester group is correct - you have a carbonyl (C=O) carbon bonded to an oxygen and both sides are bonded to an R group (any carbon/hydrogen group).
Phenol is incorrect - for a phenol group, the hydroxy/OH must be directly bonded to the ring, like so:

So instead, what you've labelled comprises of two functional groups: the ring is an arene group, and the OH is a hydroxy/alchohol group.
You haven't labelled the two branch alkyl group on the left of the molecule - this is called an isopropyl group (it has three carbons, so it's basically a propyl group with something attached to the middle carbon). This group might not always be considered a functional group though (maybe ask your lecturer?). Or it can just be called an alkane group.
Second molecule:
The amide group is pretty much correct, except it should only include at most the atoms bonded directly to the nitrogen and carbonyl carbon.
The alkene should only include the double bond (while the whole chain could be considered an alkene, the alkene functional group only contains the double bond).
One that you haven't included is the ether group, R-O-R.
To learn these I would suggest just using a table of common functional groups until you start to remember them. I think the Wikipedia page is good because you can look at your molecule and see what unusual atoms you have (e.g. nitrogen, oxygen) and see if any of the groups under that heading look like they're in your molecule, except it might be missing a few e.g. phenol.

Could you see if my edited response was accurate? Thank  you. And my teacher did give me a table of common organic functional groups, I guess I just have problems telling where one ends and another begins. He suggested memorizing them for our exam though, which I plan to do.

I would say that functional groups end where the R group begins, so e.g. if you have the functional group R-OH, the actual functional group is just the OH bit plus maybe the first atom the OH is connected to, rather than the rest of the alkyl chain like you've circled.
Sorry if I wasn't clear but you had the amide group right the first time - with the nitrogen and the C=O part. The ether group I was talking about is actually the oxygen in the ring: R-O-R. The rest seems good though.


text:
Why using smaller portions are better for extractions
Why is using one smaller portion more efficient than using one large portion for extraction. No math answer please thank you

When you say 'portion', are you referring to the amount of just the sample, or of both the sample and the extraction solvent? For example, are you asking about the difference between $10\ \mathrm{mL}$ sample extracted with $100\ \mathrm{mL}$ solvent, versus $100\ \mathrm{mL}$ sample extracted with $100\ \mathrm{mL}$ solvent?  Or, the difference between $10\ \mathrm{mL}$ sample extracted with $10\ \mathrm{mL}$ solvent, versus $100\ \mathrm{mL}$ sample and $100\ \mathrm{mL}$ solvent?

It is not. Using ONE large portion (of solvent I assume) will give you more product than using ONE smaller portion. Unless you have a different definition of efficiency (which is fine but you will need to use maths to explain this I'm afraid)

Suppose you have 1 liter of water solution of something you want to extract, 1 liter of solvent, and a partition coefficient of 1 (not very practical, but this is just to make the math simpler). You mix them all together, the compound distributes between water and solvent at 1:1 ratio, then you separate the solvent and end up with 50% of your compound. Can we do better than that?
OK, now let's divide the solvent in two equal portions. The first portion of 500 ml takes $1\over3$ of our compound, then the second portion takes $1\over3$ of the rest, or $2\over9$ of the original amount, thus making the total of ${5\over9}\approx55\%$, which is somewhat better than $50\%$.
Believe it or check it yourself, splitting the solvent into more portions will get us even more of our compound. You may call 5% a marginal gain, but with more favorable partition coefficients, the figures may grow quite a while.
And finally, if we split the original solution as well, and apply the portions in the right order, we may get pretty close to the industrial process.

without showing math how will it be more efficient. What is the process

Showing efficiency without math is pretty much like walking without legs, or talking without tongue. As for the process, imagine two liquids flowing in opposite directions, all the while being in contact with each other and exchanging the substance; now approximate that with finite partitions the best you can.

Because extraction is about maximizing surface area contact between the two solutions, and you simply get more surface area contact with smaller volumes. You can mix two smaller portions faster and more thoroughly than with large portions.


text:
Products of silicate synthesis
I'm attempting to make a hard refractory cement to apply over kiln brick made of sodium silicate and Perlite. This creates a very good kiln brick, however they are susceptible to abrasion.  I've been making my own sodium silicate, and thought I might be able to make it harder with the addition of some oxides.
I added silica gel and water; then I added magnesium oxide and sodium hydroxide.   There was a extremely strong exothermic reaction.
What would this be called? I imagine it's called something like $\ce{Na2SiO4Mg}$.  Tomorrow I will try some combinations of sodium silicate with Aluminum Oxide, and magnesium oxide. What would these be called?
If you have a recipe for a hard refractory cement to protect the Perlite mix from abrasion I would appreciate the input.

im curious, did you find a solution? I could used an abrasion resistant refractory coating as well

I imagine it's called something like $\ce{Na2SiO4Mg}$.  Tomorrow I will try some combinations of sodium silicate with Aluminum Oxide, and magnesium oxide. What would these be called?

Much simplier than that. Sodium hydroxide dissolves incredibly exothermally:
$$\ce{NaOH (s) -> NaOH (aq)}\qquad{\mathrm{\Delta} H = \pu{-44.5 kJ mol-1}}$$
This is what heated your mixture.

Expanded perlite is a bubbly glass, so if a kiln brick has a rough surface, there is insufficient fine material in the mix. Impacts of anything on the surface will break the thin glass bubbles. Of course, adding lots of fine material to the mix will strengthen it, but also increase its thermal conductivity, so that may not be a good way to improve abrasion resistance.
The surface can be improved by painting on a mix of finely ground perlite (previously expanded, so as to remove bound water) with sodium silicate to fill in surface imperfections. If you adjust the sodium silicate to perlite ratio, and paint 5 of the 6 sides, you can set the painted brick on its unpainted side and heat it in your kiln to a bit above your service temperature, whereupon the surface paint will melt and set, as opposed to dry and react. Adding $MgO$ or portland cement to the paint will give a higher melting final product, although you may be able to get an initial melting that then stiffens up during the heating. Now whether this smooth, but brittle coating is more abrasion resistant will depend on the impacts it must withstand - the brick is still not really tough, just stronger.


text:
Enthalpy and Internal Energy for Isothermal Expansion
Considering a system with constant atmospheric pressure , i.e a massless piston sitting in a cylinder containing water. At constant temperature say $\rm 100\,^{\circ} C$, the water turns into vapour and pushes the piston up.
Now shouldnt the $\Delta U=0$ as $Q = -p(V_2 - V_1)$, considering the Ideal Gas Behaviour
And then as $\Delta H = \Delta U + \Delta n\cdot R\cdot T$, then shouldn't $\Delta H = \Delta n\cdot R\cdot T$?
Then according to the question in image below, why isn't $\Delta U = 0$?

Refer to my answer here. http://chemistry.stackexchange.com/a/39719/16683

You are right that an in in ideal gas, internal energy is a function of temperature only, and that in this problem, temperature is not changing.  However, I think you are confused about how broadly the ideal gas law applies to this problem.
The question states that the ideal gas law applies to the water vapor.  But the question is about a phase change of water.  Let's break down some of the components of the problem.  In the question we have:

Liquid water.  The ideal gas law does not apply to liquid water.
Water vapor.  The ideal gas law does apply.
A phase change of liquid water to water vapor.  $\ce{H2O(l) <=> H2O(g)}$  The ideal gas law does not apply to the process of the phase change, simply because processes are not gases and cannot be modeled by the ideal gas law.

Thus only one of three "components" of the problem is an ideal gas.
As a look at any reasonable steam table will tell you, the internal energy of water vapor is higher than the internal energy of liquid water.  This difference is the heat of vaporization (at constant volume).

Thanks a lot sir, i was infact wondering that phase change and higher energy of steam could be responsible for that internal energy.
But does it mean that ΔU for water alone and steam alone is zero and the ΔU for the system is the difference between internal energies of water and steam?

you cannot define "$\Delta U$ of water" or "$\Delta U$ of steam" here, your problem is that you don't understand exactly what the term $\Delta U$ applies to. You apply it to a process that a system, or part of a system, is undergoing. There is only one process here, and that is water turning into steam. You cannot break up the process and say the water on the LHS is one process and that the steam on the RHS is another process and that the overall $\Delta U$ is the sum of both processes.

Now that water turned to steam and that it is expanding, what is going on inside the system. How is internal energy changing, I find this topic really hard to understand.

The internal energy per unit mass of the liquid water and the internal energy per unit mass of the vapor are not changing.  What is changing is the number of moles of liquid and the number of moles of vapor.  The total internal energy of the system is $$U=n_Lu_L+n_Vu_V$$where the n's are the numbers of moles and the u's are the internal energies per mole (Note that there is a discontinuous change in the internal energy per mole in going from liquid to vapor).  Similarly for the total volume of the system, we have:
$$V=n_Lv_L+n_Vv_V$$
  If $\Delta n$ moles of liquid evaporate, the change in total internal energy of the system and total volume of the system will be:$$\Delta U=(u_V-u_L)\Delta n$$
$$\Delta V=(v_V-v_L)\Delta n$$
So, from the first law,$$(u_V-u_L)\Delta n=Q-P(v_V-v_L)\Delta n$$
Rearranging this equation gives:
$$Q=[(u_V+Pv_V)-(u_L+Pv_L)]\Delta n=(h_V-h_L)\Delta n$$
where the h's are the enthalpies per mole of saturated liquid and saturated vapor.  We define the heat of vaporization as the enthalpy per mole of water vapor minus the enthalpy per mole of liquid water $(h_V-h_L)$.  So the amount of heat added is equal to the heat of vaporization times the number of moles evaporated.


text:
How to reduce Fe(III) impurity in Iron(II) sulfate?
I have bought 50 kg $\ce{FeSO4}$, but it was stored on air, and it has turned to brown. Is there a cheap method to recover some $\ce{FeSO4}$?

You speak of Iron(III) sulfate in your title, yet your question's body only contains Iron(II) sulfate. Care to clarify?

Quite possibly, most of your $\ce{FeSO4}$ is still good. Whether or not some admixture of $\ce{Fe}^{3+}$ will do any harm, depends on your application.

@inɒzɘmɒЯ.A.M I was confused by the title at first but the question itself is clear enough, some of the iron(II) was oxidised by air.

FeSO4×7 H2O is green/blue, but when I opened the bag I saw a brown/yellow stuff, with some green parts. So I think it has oxidated. It was not good idea to store outside. I've read that when FeSO4 oxidates Fe2(SO4)3 + Fe2O3 is the product. I need ironsulphate for gardening (lowering the pH in soil). Do you think it is still usable?
Thanks for reply.

@user21150 Please add the information (that you want to use it for gardening) to your question, because people don’t want to read the comments to be able to answer. There is an edit link just below the tags.

Please let us know how much you want to use at once, and how big an area that you want to spread the compound on. We need to separate beaker experiments from "gardening."

I recommend you dissolve your iron sulfate in hot water, dissolving as much sulfate as possible in as little hot water as possible. While keeping it hot, dip some coiled iron wire in it as well as some steel wool, and stir gently. The metallic iron should reduce your iron III back to iron II.
Once the solution is clear green, remove the iron wool, but keep the iron wire in it, and let it cool down as slowly as possible, while protecting it from air the best you can. Finish the cooling by putting it in a fridge
You should be left with beautiful large, green crystals of highly pure iron(II) sulfate heptahydrate, and a clear solution of relatively dilute iron(II) sulfate.
Since you have 50 kg of it, you will likely have to proceed in several batches, so I recommend you keep that solution for the next batch to minimize product losses.
Once you are done recrystallizing your iron sulfate, you can heat up that solution to a boil, until some crystals start forming on its surface. Then, immediately stop the heating and cool down that solution again. You'll have some lower grade iron sulfate, but that should still be perfectly usable for gardening.
And if you want to keep some of the prettiest crystals you made in the process, I recommend you keep them in a sealed container with a small piece of moist cloth. Otherwise, they may tend to slowly dehydrate and turn into powder.

For gardening application, do you think it need any processing at all? From the description it seems most of the material is iron(II), and I don't think gardening would need high purity material. Also, in this scale it sounds like an awful slow reduction reaction.

Hi! Thanks for the advice. I'll try it spring. I've spreaded 25 kg yet on 1000 sqm. I made solution of pH 4.5. It was 1 month ago, but it seems that the soil pH is still the same (~7.8). I'll try the recrystallizing, and write the results.

I have purified $\pu{1 kg}$ of old iron(ii) sulfate, which was yellow to brown color, using this method:
At room temperature add $\pu{10 ml}$ of diluted $\ce{H2SO4}$ (about $10 \%$) to $\pu{40 ml}$ of distilled water, scale up to prepare enough quantity of such mixture (you can use $37 \%$ battery acid in first step, in second step I have used chemically pure $96 \%$ sulfuric acid, of course recalculate the amounts to obtain the same concentration).
Dissolve as much as possible impure iron(ii) sulfate in such acidified water (prepare saturated mixture), stir very well, let impurities settle down and after decantation filter using very fine filter paper (slow filtration).
Then make first crystallization. If it does not start, you have probably not saturated, but diluted mixture, you can promote crystallization with evaporating (do not boil!), ventilation or cooling (do not freeze!). Finally you should obtain small light-green crystals.
Then wash crystals with ethanol ($96 \%$).
If you are not doing second crystallization immediately, then store intermediate product in vacuum bag to avoid reaction with air (without any desiccant!).
For second crystallization (i.e. re-crystallization), prepare the same acidified water, but with high purity $\ce{H2SO4}$ (p. a. or chemically pure), heat to max. $\pu{55-60 °C}$, and dissolve intermediate product, prepare saturated mixture.
Filter until hot using fine filter paper to crystallization dish and put dish to thermally isolated box, leave it overnight. You should obtain very nice and pure crystals. If not, the solution was not saturated, you can help using cooling (do not freeze), or ventilation, or if you are not satisfied with the result, you can do another recrystallizations.
Gently dry the crystals and put them in vacuum bag, without desiccant.


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How to calculate atomic number of an atom when wavelength is known?

In the emission spectrum of an atom, an emission line in UV region is observed for a transition from third orbit to first orbit with the wavelength of 64.1 Angstrom. What will be the atomic number of the atom?

I tried using the formula to calculate change in energy
$$\Delta E = R_h \left(\frac{1}{n_i}+\frac{1}{n_f}\right)$$
where $\Delta E$ is change in energy, $R_h$ is the Rydberg Constant $\left(2.18 \cdot 10^{-18}\right)$, $n_i$ is the initial orbit (i.e 3rd orbit) and $n_f$ is the final orbit (i.e 1st orbit).
On finding the value of this I equated it with another formula:
$$E = -\frac{R_h Z^2}{n^2}$$
where $E$ is the energy, $R_h$ is the same as above, $Z$ is the atomic number and $n$ is the orbit.
I used this formula to find out the energy at 3rd orbit and then 1st orbit and took $Z^2$ common. Then I subtracted both the obtained values but failed to find out the value of $Z$ ...

I tried calculating change the energy using Rydberg Constant and also subtracting the energy at third orbit by the energy at first orbit, but I don't where to use the value of wavelength.....

@GanguDas Note that the Rydberg constant actually is $R_\infty = 10,973,731.568,508\ \mathrm m^{-1}$. You are using the Rydberg energy $R_\infty hc = 2.179,872,325 \times 10^{-18}\ \mathrm J$.

@Loong yeah I need to use the second value

Wikipedia (Bohr Model > Rydberg formula) tells us:

The energy of a photon emitted by a hydrogen atom is given by the difference of two hydrogen energy levels:
  $$E=E_i-E_f=R_\mathrm{E} \left( \frac{1}{n_{f}^2} - \frac{1}{n_{i}^2} \right)$$
  where $n_f$ is the final energy level, and $n_i$ is the initial energy level.
Since the energy of a photon is
  $$E=\frac{hc}{\lambda}$$
  the wavelength of the photon given off is given by
  $$\frac{1}{\lambda}=R_\infty \left( \frac{1}{n_{f}^2} - \frac{1}{n_{i}^2} \right).$$
This is known as the Rydberg formula, and the Rydberg constant $R$ is $R_\mathrm{E}/hc$, or $R_\mathrm{E}/2\pi$ in natural units.

Wikipedia also tells that it can be extended with $Z^2$ for hydrogen-like elements:

The formula above can be extended for use with any hydrogen-like chemical elements with
  $$\frac{1}{\lambda_{\mathrm{vac}}} = R_\infty Z^2 \left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$$
  ... where $\lambda_{\mathrm{vac}}$ is the wavelength of the light emitted in vacuum, $R_\infty$ is the Rydberg constant for this element, $Z$ is the atomic number, i.e. the number of protons in the atomic nucleus of this element and, $n_1$ and $n_2$ are integers such that $n_1 < n_2$, corresponding to the principal quantum numbers of the orbitals occupied before and after.

Now you can see, that your first equation can be derived from your second equation. You also can see, that you a) forgot the squares in your first equation and b) made a sign error.
$\lambda = \mathrm{64.1 \cdot 10^{-10}\ m}$
$R_\infty = \mathrm{10973731.568508\ m^{−1}} \left(= \frac{R_E}{h\ c} = \frac{\mathrm{2.18\ 10^{-18}\ J}}{h\ c} \right)$
With all this in mind, you should be able to solve the equation to get $Z$.

 $$Z^2 = \frac{n_f^2\ n_i^2}{\lambda\ R_\infty\ (n_f^2 - n_i^2)} \xrightarrow{Z>0} Z \approx 4$$