text:
Dehydration of methylcyclobutanol
In one of the exams I saw it was written that dehydration of 1-methyl-cyclobutanol yields cyclobutylmethylene (double bond outside the ring) which is the anti Zaitzev product, instead of methylcyclobutene.
Why?
Thanks!

Please draw the products or model them (Avogadro, etc.). Imagine the ring strain.

Please tell us the reaction conditions.

tl;dr
The thermodynamic product of the dehydration reaction of 1-methylcyclobutan-1-ol is 1-methylcyclobut-1-ene, not methylenecyclobutane. This is supported by [1],[3] and my own calculations.

Some details
I was unable to find any reference to why the dehydration of 1-methylcyclobutan-1-ol should yield methylenecyclobutane (3) as the main product. From what I could gather up until now, the major product is 1-methylcyclobut-1-ene (1). I also tried finding suitable mechanisms and/or protonation (hydration) energies for these two molecules according to the following scheme, to compare their relative thermodynamic stability to each other.

Wiberg and Hao report the enthalpy difference between 3 and 1 to be only $\Delta H = -0.54~\mathrm{kcal/mol}$. They obtain this value from equilibrating 1 and 3 over sodium on alumina at various temperatures.[1]  This would translate to a ratio of 1:3 of about 1.0 : 0.4.[2]
Lajunen and Himottu quite similarly report, that 1 is more stable than 3 by about $4.3~\mathrm{kJ/mol} \approx 1~\mathrm{kcal/mol}$. They obtained this value via equilibrium constants measurements in perchloric acid.[3]  This would in return translate to a ratio of 1 : 3 of about 1.0 : 0.2. It is also reported, that the transition states of protonation are quite similar, but lower for 1.
I also conducted some quantum chemical calculations, using DF-BP86/def2-SVP and the more sophisticated MP2/def2-TZVPP. These are in reasonable agreement with the previous findings. The enthalpy differences, approximated at $T=298.15~\mathrm{K}$ and $p=1~\mathrm{atm}$, are $1.4~\mathrm{kcal/mol}$ and $1.3~\mathrm{kcal/mol}$ respectively. This gives a 1:3 ratio of about 1.0 : 0.1. (More values, as well as the proton affinities (PA),[4] yielding 2, are given in the table at the end.)
It is not surprising, that these structures are quite similar in energy. They both contain a four membered ring. In 1 two $\ce{sp^2}$ centres are present in the ring, forcing planarity. As a result of the shorter double bond, two angles are slightly larger than $90^\circ$, while two are forced smaller.

In 3, only one $\ce{sp^2}$ centre is present. However, the dihedral angle is a mere $18.4^\circ$, the out-of-plane angle is $26.8^\circ$, respectively. This is apparently not enough to significantly reduce ring strain. Because of symmetry constraints, i.e. $C_\mathrm{s}$, this arrangement forces three inner ring angles to be larger than $90^\circ$.

The concept of ring-strain, or strain in general, is not well defined and it cannot be measured directly. There are relative methods, to obtain this virtual quantity, sometimes involving hypothetical molecules. There are different reference systems and as far as I am aware none of them are considered the ultimative choice. In this sense, any arguments can be made. It is therefore also almost impossible to derive ring strain from just a pen-and-paper exercise, especially when the molecules are that similar.

References

K. B. Wiberg and S. Hao, J. Org. Chem. 1991, 56, 5108-5110.
How the ratio is determined
M. Lajunen and M. Himottu, Acta Chem. Scand. 1987, 41A, 252-258.
How proton affinity is defined

Appendix
\begin{array}{llrr}\hline
  & & \mathsf{\pmb1} & \mathsf{\pmb3} \\\hline
\text{DF-BP86/def2-SVP} & \Delta E_\mathrm{el}    & 0.0   & 1.3   \\
\text{DF-BP86/def2-SVP} & \Delta E_\mathrm{o}     & 0.0   & 1.6   \\
\text{DF-BP86/def2-SVP} & \Delta H                & 0.0   & 1.4   \\
\text{DF-BP86/def2-SVP} & \Delta G                & 0.0   & 1.6   \\
\text{DF-BP86/def2-SVP} & \text{Ratio} (\Delta H) & 1.0   & 0.1   \\
\text{DF-BP86/def2-SVP} & \text{PA}               & 200.2 & 198.8 \\\hline
\text{MP2/def2-TZVPP}   & \Delta E_\mathrm{el}    & 0.0   & 1.1   \\
\text{MP2/def2-TZVPP}   & \Delta E_\mathrm{o}     & 0.0   & 1.5   \\
\text{MP2/def2-TZVPP}   & \Delta H                & 0.0   & 1.3   \\
\text{MP2/def2-TZVPP}   & \Delta G                & 0.0   & 1.7   \\
\text{MP2/def2-TZVPP}   & \text{Ratio} (\Delta H) & 1.0   & 0.1   \\
\text{MP2/def2-TZVPP}   & \text{PA}               & 399.3 & 398.0 \\\hline
\end{array}

Assumption: The dehydration is probably occurring under acidic conditions.  The hydroxyl group is protonated and then eliminates water to generate the methylcyclobutyl carbocation pictured below, an E1 reaction mechanism is likely involved.  As you note, the carbocation can eliminate a proton from two different types of adjacent carbons; a methylene group (there are two) in the ring or the methyl group outside the ring.

Terms like "Zaitzev" and "Hofmann" elimination generally are used in discusing E2 eliminations. In the E1 case at hand, elimination to form the cyclobutene is statistically favored (there are 4 alpha-methylene hydrogens and only 3 methyl hydrogens), but disfavored energetically.  Cyclobutene is ca. 4 kcal/m less stable than cyclobutane due to increased ring strain from placing the double bond inside a 4-membered ring.  Instead, the lower energy Hofmann product, with the double bond outside the ring, is formed.
As a rule of thumb, a 1.4 kcal/m difference in product stabilities results in a factor of 10 shift in an equilibrium between the two products at room temperature.  In other words, if product A was 1.4 kcal/m less stable than product B, and A and B were in equilibrium, then we would expect a 10:90 A:B product ratio at equilibrium.  In E1 acid catalyzed dehydration reactions the final products are usually in equilibrium.  Since the exo-methylenecyclobutane is roughly 4 kcal/m more stable than the methylcyclobutene isomer, we might expect them to be present in something like a (4/1.6 = 2.5) 99.5 : 0.5 ratio, respectively.  An introductory textbook might well say that the cycobutene isomer is "not present."
EDIT
The information contained in @Martin 's post strongly suggests that methylenecyclobutane is less stable than 1-methylcyclobut-1-ene.  The model compounds (cyclobutane and cyclobutene) I used in the above analysis were apparently inadequate.
Still, this leaves unanswered the OP's question about why elimination from 1-methyl-cyclobutan-1-ol occurs towards the methyl group to produce methylenecyclobutane.  Since the OP didn't specify reaction conditions I assumed (above) that the reaction was run under E1 conditions.  From the relative stabilities Martin has provided, we now see that an E1 mechanism would be expected to produce the "wrong" product, 1-methylcyclobut-1-ene.
Instead of an E1 process, let's examine the possible outcomes under E2 (simultaneous elimination of the leaving group and hydrogen, no carbocation intermediate) conditions.  E2 eliminations occur with the leaving group and the hydrogen atom in an anti-periplanar configuration.  If we use Newman projections and examine the possible conformations that would favor elimination into the ring, we see that an anti-periplanar arrangement of the hydroxyl group and the adjacent ring methylene hydrogens (with the constraint of a cyclobutane ring) is not possible (figure A).  However, the Newman projection for elimination towards the methyl group to create methylenecyclobutane does have an anti-periplanar conformation accessible.  Hence, if the reaction proceeds through an E2 mechanism, we would only expect the Hofmann product, methylenecyclobutane to form.
Perhaps the OP will add the reaction conditions shown on the exam.

Under acid conditions, cyclobutylmethylene (methylenecyclobutane) isn't even the second most abundant product experimentally.  Instead:
45% 1-methyl-1-cyclobutene
35% isoprene
20% methylenecyclobutane
According to The Rate of Hydration of Methylenecyclobutane and the Effect of Structure on
Thermodynamic Properties for the Hydration of Small Ring Olefins J. Am. Chem. Soc., 1957, 79 (14), pp 3724–3729

A solution 0.9 M in 1-methyl-1-cyclobutanol and
  0.3 M in $\ce{HClO4}$ was heated at reflux temperature
  for several hours, during which time a stream of
  nitrogen was swept through the system and the exit
  vapors were passed into a liquid nitrogen trap. The
  olefin sample collected was analyzed by infrared
  spectra in a 10-cm. gas cell at a pressure of 4.9 cm.
  The analysis indicated the decomposition product
  to be a mixture of approximately 45% 1-methyl-1-cyclobutene,
  35%, isoprene and 20% methylenecyclobutane.
  The formation of isoprene readily is
  pictured in the dehydration of 1-methyl-1-cyclobutanol
  in terms of a Whitmore [Beta]-cleavage of a C-C
  bond of the cyclobutane ring.
  An estimate of the relative thermodynamic stability
  of the endo and exo isomers may be made
  from the present results. Assuming the composition
  of the dehydration product given above to provide
  a rough measure of the relative rates of formation
  of the two olefins from aqueous carbinol solution
  at 25 [degrees C] one may calculate from this and the
  relative hydration rates, kp, that the free energy of
  formation of gaseous 1-methyl-1-cyclobutene is on
  the order of 1,5000 cal./mole more negative than
  that for methylenecyclobutane.

Might the isoprene be coming from a secondary reaction (orbital symmetry allowed ring opening) of the primary cyclobutene?  The reaction is run fairly hot.  Your reference is from 1957, they weren't aware of the orbital symmetry pathway back then.

Ron, I don't know much about that, but what you are saying seems reasonable to me. Isomerazation of 1-methylbutene to isoprene has been studied int the 148-200 degree C range: http://pubs.rsc.org/en/content/articlepdf/1962/tf/tf9625800957, but the vapors were quickly passed to the liquid nitrogen trap in the 1957 experiment

@ron the 1996 review article "Chemistry of methyl cyclobutane" Russ. Chem. Rev. 65 131 still cites the 1957 paper and says "Dehydration of 1-methylcyclobutanol gave a mixture of hydrocarbons consisting of 1-methylcyclobutene (~45%), isoprene(~ 35%), and MCB (~ 20%) in a low yield"


text:
Why are ionic compounds bad conductors of electricity in solid state?
I understand the fact that ionic compounds are good conductors of electricity in molten state. But why aren't they good conductors in solid state. Cannot ions vibrate about their mean position and transfer electricity in the same way as they transfer heat?

Electric charge is transferred by physically moving charged particles around. In the case of an electric current moving through a wire (for example), the electrons are moving.
In an ionic compound, the ions are locked in place. They can move around a little bit, but there is not much translational motion - the ions stay in their places on the crystal lattice. In addition, the ions are "happy" with the number of electrons that they have. The ions formed in the first place by giving up or accepting electrons in order to minimize the overall potential energy of the system. If an anion were to transfer an electron back to a cation (for example) the energy of the system would increase, and so in general, transfer of electrons after the compound has formed is not favorable.
In solution or in a molten state, the ions themselves can move around - they become the charge carriers. In a solid, the ions can't move, and so electricity cannot be easily transferred.
You mentioned heat transfer - heat is the transfer of the kinetic energy of atoms and molecules. Heat can still be transferred (in some cases quite easily) in an ionic solid because, as you said, ions can vibrate about a mean position. When this happens they bump into their neighbors, which spreads the kinetic energy around.
In summary, ionic compounds don't conduct electricity very well because the charge carriers can't move through the crystal. They can conduct heat because the kinetic energy itself is the "heat carrier" - it can be transferred without moving ions too far from their mean positions.

While generally true, this isn't always true for ionic compounds. Some do have mobile ions that can carry significant quantities of electricity. These are used in some fuel cell and battery technologies, though the compounds are not that common. See the Wikipedia article on fast ion conductors.

I didn't know about that, @matt_black - thanks for the info!

There are also ionic compounds where you have a mixture of charges in the cations or in the anions, and electrons can "hop" between the mixed-charge ions.  This occuus for instance in FeO, which has a fairly large percentage of Fe(III) and cation vacancies; electrons move between the Fe(II)  and Fe(III) ions.

Ionic compounds do not conduct electricity in solid state because electron motion results in electricity being conducted but in solid state the constitutive particles are strongly bonded or in fixed positions; thus, they can't move from one part to another. In molten state, the ionic compound is split into ions, and ions are charged particles that are free to move so ionic compounds in water or in molten state conduct electricity.

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text:
Titration of bases against HCl
Suppose I take three bases in a solution, $\ce{NaOH}$, $\ce{Na2CO3}$ and $\ce{NaHCO3}$, and I titrate this solution with $\ce{HCl}$. Since the strengths of the bases are different, does it mean that $\ce{HCl}$ would first react with the strongest base completely and then start reacting with the next strongest base?

I would think that it would react with all three of them at once (since all of those bases give off $\ce{OH-}$ which $\ce{HCl}$ can react with), but you would find the $\ce{NaOH}$ go away first since it dissociates more completely.

This is perhaps not the best example, since hydroxide ion will react with bicarbonate ion to form carbonate ion, and also as you titrate carbonate ion note that it will form bicarbonate again. But, generally, yes, the strongest base will be consumed first as you titrate. The titration curve will tell you the amounts of the bases present and the pKas of their conjugate acids.

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text:
What are the effects of adding water to a system at equilibrium?
Given the reaction $\ce{NaCl~(s) <=> Na+~(aq) + Cl^{-}~(aq)}$. $\ce{NaCl}$ is added to water and equilibrium is reached.
Does the reaction shift to the left/right/not shift at all if more water is then added?
It seems to me that adding water would decrease the concentrations of $\ce{Na+}$ and $\ce{Cl-}$, and therefore the reaction would shift to the right. Is this correct?

The reaction $$\ce{NaCl~(s) <=>[\ce{H2O}] Na+~(aq) + Cl^{-}~(aq)}$$ is at equilibrium when approximately $359~\mathrm{g/L}$ sodium chloride are dissolved in water, i.e. the solution is saturated, or $c(\ce{Na+}) = c(\ce{Cl-}) = c_\mathrm{max}$.
If less $\ce{NaCl}$ is dissolved, it is only one phase, which is in equilibrium with itself and all solid salt is dissolved, therefore the equilibrium would be entirely on the right side. At that point adding water would not change the equilibrium at all, it would only dilute the whole system.
When the solution is saturated, there is still solid salt present, the equilibrium is not entirely on the right side. Adding water will disturb the equilibrium, i.e. $c_\mathrm{present}<c_\mathrm{max}$. Therefore more of the excess $\ce{NaCl}$ will dissolve, until the maximum concentration is reached again. If the solution is still saturated after that, nothing will have changed in your solution except for the volume.
If your solution is not saturated anymore, the equilibrium is entirely on the right side and the concentrations will decrease.

I think I may understand your question.
Adding water to $\ce{NaCl~(s)}$ causes it to ionize into $\ce{Na+}$ and $\ce{Cl-}$, now if you think about Le Chatelier's Principle, considering the equation: $\ce{NaCl~(s) + H2O -> Na+ + Cl-}$, the equilibrium will shift to the right, and the concentrations of $\ce{Na+}$ and $\ce{Cl-}$ will increase.
Now if you think about it realistically, since the salt is ionized with water, the more water you add, the more $\ce{Na+}$ and $\ce{Cl-}$ you will get, now if this was considered, it is essentially increasing the concentration of the ions.
However, usually when dealing with water, the system is considered to be in water. So you may see water on both sides of the reaction, in which case adding water would have no effect on equilibrium.


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How to calculate the concentration of all relevant species in a buffer of a given pH?

You prepare $1.0~\mathrm{L}$ of a $0.25~\mathrm{M}$ acetic acid solution with a final $\ce{pH}$ of $6.0$. What are the molar concentrations of all relevant acetic acid species ($[\ce{HA}]$ and $[\ce{A-}]$) given that the $K_a$ for acetic acid is $1.74 \cdot 10^{-5}~\mathrm{M}$?

I am getting confused with this problem.
Since the $\ce{pH}$ is given, I know what the $[\ce{H^+}]$ is.
So now when I try to do the ICE table
$$
\begin{array}{l|ccc}
               & \ce{HA} &  \ce{H+}  &  \ce{A-} \\\hline
\text{Initial} & 0.25    &  0        &  0       \\
\text{Change}  &  -x     &  +x       &  +x      \\
\text{End}     & 0.25 -x &  +x       & +x       \\\hline
\end{array}$$
And from here I begin to assume $[\ce{H^+}] = [\ce{A^-}]$, which I am not sure about.
Then I set $x = 10^{-6.0} = 1\cdot 10^{-6}$ so I get $[\ce{A^-}] = 1\cdot 10^{-6}~\mathrm{M}$ and $[\ce{HA}] = 0.24999~\mathrm{M}$ which I think is incorrect, and to even further ensure my that it's incorrect, when I attempt to check the $K_a$ value with this, it does not match.
My second approach:
$[\ce{HA}] = 0.25~\mathrm{M}$
In this I determine the $\mathrm{p}K_a$ from the $K_a$ which turns out to be $4.759$, which indicates that there should be more $[\ce{A-}]$ than $[\ce{HA}]$.
I now use the Henderson–Hasselbalch equation:
\begin{align}
    6.0   &= 4.759 + \log\left(\frac{[\ce{A-}]}{[\ce{HA}]}\right)\\
   17.40  &= \frac{[\ce{A-}]}{[\ce{HA}]}\\
   17.40  &= \frac{[\ce{A^-}]}{0.25~\mathrm{M}}\\
[\ce{A-}] &= 4.35~\mathrm{M}\\
\end{align}
I feel more confident about my second answer.
Can someone please help me out with this particular problem and perhaps tell me procedure I should use as well as what the correct answer should come out to be and why?

@FredSenese There is a command for chemical formula (using mhchem): \ce{...}. Find more information here and here. It would also be very nice if you could be on the lookout for buzz-words/-statements especially in titles, see also meta.

Okay so using Henderson Hasselbalch:One will get   17.40 = [A-]/[HA]. Following this idea, one can assume that [HA] + [A-] = 0.25M, using algebra, [HA] = 0.25 - X. Now 17.40 = x/0.25-x, again with algebra one can solve that x = 0.236.  So [A-] = 0.236M and [HA]=0.0136M and if you use these to find the Ka, you match the Ka if you use [H+]=10^-6.0

@user109992 Could you write that up as an answer, please. This way you can accept it and it can be upvoted.

The value given for the pH of the solution seems to be incorrect*.
As the reaction is $\ce{HA <=> H+ + A-}$, your initial assumption that $[\ce{H+}]=[\ce{A-}]$ is correct.
From your chart,
\begin{array}{l|ccc}
               & \ce{HA} &  \ce{H+}  &  \ce{A-} \\\hline
\text{Initial} & 0.25    &  0        &  0       \\
\text{Change}  &  -x     &  +x       &  +x      \\
\text{End}     & 0.25 -x &  +x       & +x       \\\hline
\end{array}
it follows that
$$K_a=\frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]} = \frac{x^2}{0.25 -x} = 1.74\cdot10^{-5}.$$
We can solve this for $x$:
\begin{align}
x^2 &= 1.74\cdot 10^{-5}(0.25-x)\\
x^2 + 1.74\cdot10^{-5} x - 4.35\cdot10^{-6} &= 0\\
x &= 0.00207698~\mathrm{M}\\
\end{align}
Thus, $[\ce{HA}] = 0.25~\mathrm{M} - 0.00207698~\mathrm{M} = 0.24792302~\mathrm{M}$ and $[\ce{H+}]=[\ce{A-}] = 0.00207698~\mathrm{M}$.
Also note that in your second attempt, the value $[\ce{A-}] = 4.35~\mathrm{M}$ makes no sense, as the amount of $\ce{A-}$ present should be in a 1:1 ratio to the amount of $\ce{HA}$ consumed in the reaction, and $[\ce{HA}] < 4.35~\mathrm{M}$.
The given $\ce{pH}$ is likely incorrect because the concentration of $[\ce{H+}]$ ions for the given molarity and concentration does not match with the given $\ce{pH}$, and because the given $K_a$ is close to values of $K_a$ provided online.

I don't think there's an error in the question, I think that my first attempt might be wrong in the assumption that H+ = A- in this case.

Okay so I actually figured it out, see my comment to my question

I think your answer would be correct for a pure solution of acetic acid. I would read the question different. After the pure solution is obtained, the pH is adjusted to 6, thus making it a buffer. This part is however horribly implied.

@user13006 I think your answer is great.  Martin could be right about what the person had in mind writing the question too.

Ratio of protonated to deprotonated species
We know the pH and the acid dissociation constant. Using the equilibrium constant expression, we can calculate the concentration ratio of acetate to acetic acid:
$$\frac{[\ce{A-}]}{[\ce{AH}]} = \frac{K}{[\ce{H+}]} = \frac{\pu{1.74e−5}}{10^{-6}} = 17.4 $$
Ambiguity in the question
To complete the exercise, we have to understand the question. They might be saying that  $\pu{0.25 M}$ is the concentration of acetic acid, or this value refers to the total concentration of acetic acid plus acetate. The latter is more likely, as they talk about "the molar concentrations of all relevant acetic acid species". This implies that acetate is viewed as a acetic acid species, and probably was not added to the acetic acid but was made in situ when setting the pH of the solution. In the lab, we would maybe label this solution "acetic acid buffer, 0.25 M, pH = 6" or more clearly "0.25 M acetic acid, adjusted to pH = 6 with NaOH".
Concentration of species
If we take the total concentration of acetic acid plus acetate to be $\pu{0.25 M}$, $\frac{1}{18.4}$ of it will be acetic acid, and $\frac{17.4}{18.4}$ of it will be acetate.
If we take the final concentration of acetic acid to be $\pu{0.25 M}$, the acetate concentration would be $17.4$ as big, i.e. more than four molar. There might not be any acetate salt that soluble in water (assuming water is the intended solvent).
Definitely not a watertight case.

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text:
What happens to the NADH produced by lactate oxidation?
I first want to note that I'm a layman attempting to gain an understanding of metabolism by reading resources on the internet with only a little understanding of chemistry. I've hit a snag:
Oxidation of lactate into pyruvate reduces NAD+ to NADH. Subsequent glyconeogenesis can then turn two pyruvate into a new glucose molecule at the cost of 6 ATP. However, the NADH is presumably used to generate ATP via oxidative phosphorylation producing ~2.5ATP, making an overall reaction of 2 lactate -> 1 glucose cost only ~1ATP. This clashes with other things I have read - that glyconeogenesis costs 3x the energy glycolysis produces - as this would seem to discount the reduction of NAD+ to NADH in glycolysis. I feel I am clearly misunderstanding something, but I'm not sure where.

Welcome to chemistry.se! If you have questions about how to beautify your posts, have a look at the [help]. Do you want to know more about this site, please take the [tour].

What you have read about the direct energy costs/benefits of glycolysis and gluconeogenesis  is not wrong. Glycolysis produces 2 ATP and gluconeogenesis consumes 6 ATP.
The net reaction for coversion of glucose $\ce{C6H12O6}$ into pyruvate $\ce{C3H3O3-}$ by glycolysis is (note that I am keeping track of the charges and the magnesium counterion/cofactor on ADP/ATP):
$$\ce{C6H12O6 + 2NAD+ + 2[MgADP]- + 2HPO4^{2-} ->}$$
$$\ce{ 2C3H3O3- + 2NADH + 2[MgATP]^{2-} + 2H+ + 2H2O}$$
For conversion of pyruvate back to glucose by gluconeogenesis (note that gluconeogenesis usually uses 4ATP and 2GTP - for simiplicity let's consider that a GTP has about the energy equivalence to an ATP):
$$\ce{2C3H3O- + 2NADH + 4[MgATP]^{2-} + 2[MgGTP]^{2-} + 6 H2O ->}$$
$$\ce{C6H12O6 + 2NAD+ + 4[MgADP]- + 2[MgGDP]- + 6HPO4^{2-} +2H+} $$
If you compare the two processes directly, in absence of the source/fate of any of the components, you see that gluconeogenesis consumes 6 ATP/GTP and 2 NADH while glycolysis produces 2ATP and 2NADH. The NADH consumption and production is balanced out.

Why would we compare these two directly and not worry about coupled processes?

Not all organisms undergo oxidative phosphorylation. Most do, but not all.
More importantly, in the context of your question, NADH can do other things. Not all NADH goes to oxidative phosphorylation.
Most importantly, the direct comparison is carbon balanced with the same organic species: glucose and pyruvate. If we try to include all possible variations and coupled processes we find ourselves eventually trying to do a general chemistry level stoichiometric and thermochemical analysis of an entire life-form. I'd rather not. There are lots of feed stocks into gluconeogenesis besides lactate and pyruvate. An amino acid or lipid that can be catabolized into acetyl-CoA can be fed into gluconeogenesis.

As for lactate...

True, pyruvate can be generated from lactate $\ce{C3H5O3-}$, which also produces NADH:
$$\ce{C3H5O3- + NAD+ -> C3H3O3- + NADH + H+}$$
If we start with lactate, we have no net generation of NADH in gluconeogenesis. However, we are also no longer comparing the interconversion of the same carbon species. Lactate is produced from pyruvate (consuming NADH), so we are back to square one. Including lactate in the comparison instead of pyruvate leads to no net NADH production/consumption.


text:
Are matrices a good way to do stoichiometry?
I was never taught in school to use matrices to balance my equation in stoichiometry. But I recently started to review my chemistry and, as I have taken linear algebra, I was suddenly struck by the idea this would be a really efficient way to balance equations. It might save me from making a lot of errors.
Are matrices a common tool for chemists when balancing equations? Does anyone have a good reference for learning about this?

Sure, most of the equation-balancing Javascript calculators on the web use that method. It's overkill to balance a single equation that way, but if you're automating the process or if you have a lot of equations to balance and check it can be convenient.
If you search on "balancing equations matrix" you'll find many simple matrix inversion  approaches for balancing equations that fail if you don't have a single unique solution.  Here's one that's more general:
L. R. Thorne, An Innovative Approach to Balancing Chemical-Reaction Equations: A
Simplified Matrix-Inversion Technique for Determining The Matrix Null
Space, Chem. Educator 2010, 15, 304–308
The basic idea is that coefficients in balanced equations are found in the nullspace of the formula composition matrix (that is, the coefficients are solutions $\bf x$ of the matrix equation $\bf C x = 0$, if $\bf C$ is the formula composition matrix). For example, if I want to find a balanced equation that relates $\rm HCl$, $\rm K_2Cr_2O_7$, $\rm FeCl_2$, $\rm CrCl_3$, $\rm H_2O$, $\rm FeCl_3$, and $\rm KCl$, I can build a formula composition table
$$\begin{matrix} & \ce{HCl} & \ce{K2Cr2O7} & \ce{FeCl2} & \ce{CrCl3} & \ce{H2O} & \ce{FeCl3} & \ce{KCl}\cr H & 1 & 0 & 0 & 0 & 2 & 0 & 0\cr Cl & 1 & 0 & 2 & 3 & 0 & 3 & 1\cr K & 0 & 2 & 0 & 0 & 0 & 0 & 1\cr Cr & 0 & 2 & 0 & 1 & 0 & 0 & 0\cr O & 0 & 7 & 0 & 0 & 1 & 0 & 0\cr Fe & 0 & 0 & 1 & 0 & 0 & 1 & 0\end{matrix}$$
I can find the nullspace vectors using the multistep procedure outlined in that paper, or I can do it very simply in Maxima:
C : matrix([1,0,0,0,2,0,0],
           [1,0,2,3,0,3,1],
           [0,2,0,0,0,0,1],
           [0,2,0,1,0,0,0],
           [0,7,0,0,1,0,0],
           [0,0,1,0,0,1,0])$
nullspace(C);

$$\mathrm{span}\left( \begin{bmatrix}14\cr 1\cr 6\cr -2\cr -7\cr -6\cr -2\end{bmatrix}\right)$$
which corresponds to the equation
$$\ce{ 2CrCl3 + 7H2O + 6FeCl3 + 2KCl\ =\ 14HCl + K2Cr2O7 + 6FeCl2}$$
As chemists, we'd write that equation reversed,
$$\ce{ 14HCl + K2Cr2O7 + 6FeCl2\ =\ 2CrCl3 + 7H2O + 6FeCl3 + 2KCl }$$
Here are a few more nails you can pound with your linear algebra hammer: you can determine the minimum number of chemical equations required in a reaction mechanism, or discover how many equations in a reaction mechanism are stoichiometrically independent. For example, consider the following halogen/hydrogen reactions:
$$\begin{array}{rcl}
\rm H_2 + X_2&\xrightarrow{k_1}&\rm 2HX\\
\rm H_2 + X_2&\xrightarrow{k_2}&\rm 2X + H_2\\
\rm H_2 + X_2&\xrightarrow{k_3}&\rm 2H + X_2\\
\rm H + X_2&\xrightarrow{k_4}&\rm X + HX\\
\rm H_2 + X&\xrightarrow{k_5}&\rm H + HX\\
\rm HX + H&\xrightarrow{k_6}&\rm X + H_2\\
\rm H_2 + HX&\xrightarrow{k_7}&\rm X_2 + H_2\\
\rm 2HX &\xrightarrow{k_8}&\rm X_2 + H_2\\
\end{array}$$
Let's make a formula matrix for the five species in the reactions above:
$$\begin{pmatrix}{H}_{2}\cr H\cr HX\cr X\cr {X}_{2}\end{pmatrix}=\begin{pmatrix}2 & 0\cr 1 & 0\cr 1 & 1\cr 0 & 1\cr 0 & 2\end{pmatrix}\begin{pmatrix}H\cr X\end{pmatrix}$$
Any set of $S-C$ stoichiometrically independent equations is a "complete set" of reactions, where $S$ is the number of species and $C$ is the number of independent equations for the conservation of elements (components). There are five species ($S=5$) and the rank of the formula matrix is 2, so the system has two components ($C=2$). There must be $S - C = 5 - 2 = 3$ stoichiometrically independent reactions for the mechanism to be complete.
Do the reactions above constitute a complete set? We can write a stoichiometric matrix like so:
$$\begin{pmatrix}-1 & 0 & 2 & 0 & -1\cr 0 & 0 & 0 & 2 & -1\cr -1 & 2 & 0 & 0 & 0\cr 0 & -1 & 1 & 1 & -1\cr -1 & 1 & 1 & -1 & 0\cr 1 & -1 & -1 & 1 & 0\cr 0 & 0 & 0 & -2 & 1\cr 1 & 0 & -2 & 0 & 1\end{pmatrix} \begin{pmatrix}{H}_{2}\cr H\cr HX\cr X\cr {X}_{2}\end{pmatrix} = \begin{pmatrix}0\cr 0\cr 0\cr 0\cr 0\cr 0 \cr 0 \cr 0\end{pmatrix}$$
The rank of the stoichiometric matrix is 3, so the eight equations above are equivalent to three independent equations. The mechanism is complete. Whether it's correct is of course a different matter. Stoichiometric constraints don't determine a mechanism; they limit the number of reactions in a particular reaction channel, but you can have more than one channel (more than one mechanism running side-by-side).
Why is this useful---why would you want to know what the number of independent stoichiometric equations ($R$) are? Well, according to Gibb's rule of stoichiometry, $R \le S-C$. If you have $R < S-C$, that tells you that you have a restricted equilibrium: the components must be constrained by conditions other than stoichiometric equations. The restricted equilibrium composition will depend on which species are present initially in that case. There are also experimental design implications, since in a single experiment you can get only $S-C$ reaction rates. You may have to figure out how to favor one channel over another in separate experiments to get all of the rates.
TL;DR: If you're looking for applications of linear algebra in chemistry, chemical kinetics is one playground; there are many others.

Thanks for the article. It was just along the lines of what I was hoping for.

Very nice answer Fred. I believe Leons book on linear algebra includes examples where such systems of equations are worked out. If OP wants, I can upload them for him.

@Jori that would be fantastic


text:
How to find the molar mass of an unknown metal carbonate through a gas evolution reaction?
As part of a pre-lab exercise in Chemistry, we went through the necessary calculations to identify an unknown metal in a metal carbonate. The method was gas evolution by means of hydrochloric acid (see equation below).
$$\ce{HCl + M2CO3 -> MCl + CO2 + H2O}$$
The unknown metal was Alkali, and we had three possibilities: Lithium, Sodium and Potassium as $\ce{M}$.
Using data given to us, we first converted grams of released $\ce{CO2}$ to moles. Then here was the part I was confused on-
To find the molar mass of the metal carbonate, we took the mass of the sample used (the metal carbonate) and divided it by moles of $\ce{CO2}$ released, which was just calculated. The rest here on out was a little algebra. What is the reasoning behind dividing the mass of the sample by released $\ce{CO2}$? What makes it "legal"? Yes, in the balanced equation, the ratio of $\ce{M2CO3}$ to $\ce{CO2}$ was $1:1$.

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$$\ce{HCl + M2CO3 -> MCl + CO2 + H2O}$$

Since you already figured out, that there is a one-to-one ratio of carbonate to carbon dioxide, you can simply write
$$n(\ce{CO2}) = n(\ce{M2CO3}).$$
With the formula for the Molar mass,
$$M(\ce{M2CO3}) = \frac{m(\ce{M2CO3})}{n(\ce{M2CO3})},$$
you can simply substitute one for the other, hence
$$M(\ce{M2CO3}) = \frac{m(\ce{M2CO3})}{n(\ce{CO2})}.$$

Your equation isn't balanced though.
Correct equation: $\ce{M2CO3 + 2HCl -> 2MCl + CO2 + H2O}$

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


text:
Producing carbon nanotubes from graphite
I am searching for the carbon nanotubes for very low cost. I have one idea, which is to use graphite (such as that found in pencils.) Can it be converted to carbon nanotubes? If so, what would be the procedure for this?

Graphite does not contain CNTs, a simple extraction therefore is not possible.
Graphite seemingly can be converted to CNTs by a "simple" oxidative method published by DW Lee and JW Seo on arxiv.org in 2010.
The authors claim to have used commercially available graphite powder (45 µm, Aldrich) of high purity (>99.995%), but did not provide any information on the yields.
Pencils typically contain a mixture of graphite powder and various amounts of clay as a binder.


text:
Is probability of finding electron changing here?
We all know what is a wave function. There is a technique to convert non normalized wave function to normalized wave function.
In case of non normalized wave function integration of square of wave function with respect to 'dt' in a length 'dx' will be any constant other than 1. In case of normalized wave equation the probability of finding will be maximum i.e. 1.
The process of converting a non-normalized wave function to a normalized wave function is known as Normalization technique. In this technique we multiply the wave function with some constant and makes it equal to 1 (condition for a normalized wave function). That means after conversion the probability of finding electron is becoming maximum.
If we multiply a wave function with any constant it's nature doesn't change. So, my question is after normalization of the wave function, the nature of the wave doesn't changes and how come the probability of finding electron becomes maximum i.e. 1???

I swear I read the question many times, but I still not quite sure what it is about.  So I will try to clarify some things hoping that they are relevant.

First off, any complex multiple of a wave function indeed describes exactly the same state of a system. This directly follows from the postulate of quantum mechanics which states that the expectation value of an observable $A$ represented by a self-adjoint operator $\hat{A}$ is given as follows,
$$
\langle A \rangle
=
\frac{\langle \Psi | \hat{A} | \Psi \rangle}{\langle \Psi | \Psi \rangle} \, ,
$$
where $\Psi$ is the wave function. It is indeed easy to show that given $\Psi$ we can multiply it by any complex number $c$ and this would not change the expectation value of any observable,
$$
\langle A \rangle
=
\frac{\langle c \Psi | \hat{A} | c \Psi \rangle}{\langle c \Psi | c \Psi \rangle}
=
\frac{c^{*} c \langle \Psi | \hat{A} | \Psi \rangle}{c^{*} c \langle \Psi | \Psi \rangle}
=
\frac{\langle \Psi | \hat{A} | \Psi \rangle}{\langle \Psi | \Psi \rangle} \, ,
$$
since $c^{*} c$ in numerator and denominator perfectly cancels each other. Also note that the following definition of the expectation value
$$
\langle A \rangle = \langle \Psi | \hat{A} | \Psi \rangle \, ,
$$
holds true only for a normalized wave function $\Psi$ and trivially follows from the above mentioned more general definition.

Secondly, for the simplest case of one particle in one spatial dimension, if one would like to interpret the wave function $\Psi(x,t)$ as a probability amplitude that the particle is at $x$, and consequently, it square modulus, $ \left|\Psi(x, t)\right|^2 = {\Psi(x, t)}^{*}\Psi(x, t)$ as the probability density that the particle is at $x$, he/she have to normalize it as follows,
$$
\int\limits_{-\infty}^\infty d x \, |\Psi(x,t)|^2 = 1 \, .
$$
This is required because one of the axioms of the probability theory requires that

the probability that some elementary event in the entire sample space
  will occur is 1.

The sample space is just a set of all possible outcomes in an experiment, which in our case is the set of events of finding a particle at any possible position, i.e anywhere between $-\infty$ and $+\infty$. Thus, in our case the probability of finding the particle anywhere between $-\infty$ and $+\infty$ have to be 1, and consequently, the wave function has to be normalized. Otherwise its interpretation as a probability amplitude would be inconsistent with the probability theory.

Hope it helps.

Normalization means that the integral of the wave function times its complex conjugate over all space is 1.  It has nothing to do with maximizing anything.  It is only a way of saying "there is one electron (or other particle)".
In other words, if I say "there is an 80% chance Vamsi is at school, 40% chance Vamsi is at home, 50% chance Vamsi it at work, and a 30% chance Vamsi is shopping" it doesn't make any sense.
But if I say "there is an 40% chance Vamsi is at school, 20% chance Vamsi is at home, 25% chance Vamsi it at work, and a 15% chance Vamsi is shopping" it does make sense.

Then What is the difference between a normalized wave function and non normalized wave function??

@vamsi, in short, only normalized wave function can be used to calculate probabilities. See my answer for details.


text:
Calculating boiling point of a solution

What is the boiling point of an aqueous solution that has a vapor pressure of $23.0~\text{Torr}$ at $25~^\circ\text{C}$?
$(P^\circ_{\ce{H2O}}=23.78~\text{Torr};~K_\mathrm b= 0.512~^\circ\mathrm{C\ kg/mol})$

To solve this I used the equation $\Delta T_\mathrm b =m\times K_\mathrm b$. I calculated a molality of $1.88247\ \mathrm{mol/kg}$ which I then used to calculate a boiling point of $-0.9638\ \mathrm{^\circ C}$. This answer is apparently wrong and I don't understand why.

If molality is 1.88, then the change in boiling point is -0.96 degrees C, not the boiling point itself.

But if i subtract the change in temperature from the normal boiling point of water (0 C) then we get a boiling point of -0.96 C, which is wrong.

Water boils at 100 degrees C


text:
Does mercury fulminate really look like crystal methamphetamine?
In the AMC series Breaking Bad , Walter (aka Heisenberg) smuggles in a bag of mercury fulminate under the guise of crystal methamphetamine and threatens to blow the room up if he doesn't get his money.
Does fulminated mercury really have the translucent white/blue crystal appearance as was shown in this scene?
Is the compound that sensitive that throwing it against the wall would cause detonation?

Mercury fulminate $\ce{[Hg(CNO)2]}$ can have a white, gray or greenish-brown color depending upon purity and dryness (it is often kept moistened to prevent accidental decomposition).

Does fulminated mercury really have the translucent white/blue crystal
  appearance as was shown in this scene?

I've never seen a picture where it appears translucent or where it has a bluish color.  The pictures I've seen always show it as an opaque powder.  I suspect that it could be crystallized, but that's probably never done as it would further increase its sensitivity to detonation (e.g. if a crystal cracked, or rubbed against another crystal, the friction, triboelectrification, etc. could cause the material to explode).  Here are some pictures.

Is the compound that sensitive that throwing it against the wall would
  cause detonation?

Here is a link to an examination of the Breaking Bad "detonation by throwing" experiment (with video).  The conclusion is that it can detonate when thrown, but 1) a larger amount of material than what Walt had, and 2) a much harder throw (greater force, greater shock) would have been required to cause detonation.


text:
What's the difference between perfect and ideal gas?
There are all kinds of different dinstinctions in the internet, and I'd like to see what you guys thought.
Actually, I've been asked if the heat capacity of an ideal gas es independent of temperature. I've said no, even though it's practically invariant under small ranges.
Namely the result $C_V/n = \frac{3}{2}R$ is derived from a perfect gas and not an ideal gas and is only an approximation to the latter.
Is this true?
What really is an ideal gas? What's its difference with a perfect gas? Is my answer to the question I was posed correct?
Thanks.

According to Wikipedia, a perfect gas is not the same thing as an ideal gas. Specifically, it seems that all perfect gasses are ideal, but not all ideal gasses are perfect. Both types obey the ideal gas law, but they differ with respect to the temperature and pressure dependence of their heat capacities. None of the current answers seem to have this information.

An interesting discussion on PhysicsForum suggests that what we'd call an 'ideal gas' is actually a 'perfect gas' to engineers. What strains my imagination is how you'd get $C_p$ or $C_V$ dependence without intermolecular interactions.

@chipbuster You can define the molar internal energy of the interactionless gas (monoatomic) as $\frac{3}{2}RT + U(T=\pu{0 K})$ you can get $C_V$ from there and $C_p$ as $C_V+R.$

@David Thanks, shows you how weak my thermo is :P

I'll vote that the difference is ambiguous. The IUPAC gold book defines ideal gas http://goldbook.iupac.org/I02935.html However there is no IUPAC definition for perfect gas so that phrase means whatever the author wants it to mean.

@chipbuster the heat capacities can depend on temperature without intermolecular forces, because they depend on the degrees of freedom of the molecule (even for ideal gases). At high temperatures, more degrees of freedom are active, whereas at low temperatures, some are negligible. So for a polyatomic ideal gas, the heat capacity depends on temperature because the effective number of degrees of freedom depends on temperature. A perfect gas can be defined as only accounting for translational degrees of freedom no matter the temperature, and this would make it identical to a monatomic ideal gas.

An ideal gas is the same as a perfect gas. Just different naming. The usual name for such gases (for which is assumed that the particles that make up the gas have no interaction with each other) is ideal gas, perfect gas is what such a gas is named in Atkins physical chemistry book. Personally I like the perfect gas naming better as it illustrates the perfect nature of the assumptions made about it.
For the simple systems (like monoatomic gases) when we can assume a perfect/ideal gas, $C_{V,m}$ is independent of temperature. For real gases this is certainly not the case.
Note that $C_{V,m}=\frac{3}{2}R$ only holds for monoatomic gases.

@NicolauSakerNeto This answer is misleading. Ideal gas and perfect gas are defined differently and not always the same (perfect gases are special cases of ideal gases, where the heat capacity is temperature-independent even at high temperatures)...

... Your answer is at odds with the perfect gas Wikipedia article and the sources it is based on, which include Physical Chemistry for the Chemical Sciences by Chang & Thoman and the NIST Standard Reference Database Number 69.

Namely the result $C_V/n=\frac{3}{2}R$ is derived from a perfect gas and not an ideal gas and is only an approximation to the latter.
  Is this true?

So, let's look first at where $C_p-C_V=R$ comes from and then look at $C_V=\frac{3}{2}R$ to see what we find.
We start with the definition of heat capacity as being a change in energy per change in unit temperature,$$\Delta H=\int_{T_1}^{T_2}n\cdot C_p\,\mathrm dT$$
Now, I assume heat capacity to be independent of temperature
Then,$$\Delta H=n\cdot C_p(T_2-T_1)$$
Since $H=U+PV$ and pressure is held constant here, we rewrite the expression as $$\Delta U + P\Delta V=n\cdot C_p(T_2-T_1)$$
By the same integration performed above (but with $C_V$) we find that $\Delta U=n\cdot C_v(T_2-T_1)$
Combining those expressions and simplifying,
$$C_p-C_V=P\frac{\Delta V}{n\cdot\Delta T}$$
Using the ideal gas law, with constant pressure, we find,$$\frac{\Delta V}{\Delta T}=\frac{nR}{P}$$
Plugging that in, $$C_p-C_V=R$$

Now, for a monoatomic ideal gas, energy can only be stored in translation.
Invoking the equipartition theorem to avoid having to do math and a little physics, we see that the energy of a monoatomic gas will be,$$U=\frac{3}{2}Nk_\mathrm bT$$
For $N=N_\mathrm A$ particles, we have, $$U=\frac{3}{2}RT$$
So, because $$C_V\equiv\left(\frac{\partial U}{\partial T}\right)_{P,n}$$
We see that, $$C_V=\frac{3}{2}R$$
Conclusions:
We see that in our derivation of the relationship $$C_p-C_V=R$$ we both used the ideal gas law and assumed heat capacity to be independent of temperature.
So, to answer the question quoted at the top of this answer, $\frac{C_V}{n}=\frac{3}{2}R$ is derived from the ideal gas, not the perfect gas.
And, in answer to the other question, our derivation required that we assume heat capacity to be constant with a change in temperature, so it was incorrect to say that heat is dependent on temperature for an ideal gas. It is true, however, that heat capacity varies with temperature for a real gas.
As to whether or not there is a difference between an ideal and perfect gas, I would look at that Wikipedia page posted in a comment above, but it seems superfluous to define something as a perfect gas when an ideal gas is already well understood and the perfect gas essentially behaves the same.
Hope that helps explain some of the math behind this.

You say that Cp-Cv=R was derived from the ideal gas law and assumption of temperature independent heat capacity, which is true. That is the definition of a "perfect gas". However, Cp-Cv=R does apparently hold for an ideal gas as well (more generally than just when the heat capacities are temperature-independent). Nuances, but you weren't incorrect there. However, when you say cv= (3/2)RT is derived from the ideal gas model and not the perfect gas model, I think this is mistaken. It is true for a monatomic ideal gas. However, for polyatomic ideal gases, at moderate temperatures rotational...

... degrees of freedom are accessible and cv will be either (5/2)RT or 3RT. At high temperatures, vibrational modes will be accessible, and cv=(3N/2)RT will be correct (where N is the number of atoms in the molecule). All of this is ignored in the perfect gas model, where cv=(3/2)RT regardless of temperature. However, simple as the ideal gas model is, cv=(3/2)RT is only true for monatomic ideal gases or ideal gases at low temperatures. A polyatomic gas at a moderate high temperature will have cv > (3/2)RT, and this is the difference between an ideal gas and the further "idealized" perfect gas.

As I recall P. W. Atkins etal point out that the interaction between particles in both ideal and perfect gases are constant (i.e. they don't vary with T or P).  The difference is that for a percent gas the interactions are not only constant but they are equal to zero.  For an ideal gas they just are constant.
Also, in regard to the comment made just above (The second point) should be that "The size of the particles is negligible when compared to the volume they occupy".  In fact the distances are large in comparison and that is why the forces of attraction between particles are negligible.

Simply put, an ideal gas follows the ideal gas law. In thermodynamics and statistical mechanics, many physical relations are derived using the ideal gas law, sometimes accounting for Van der Waals forces or other non-negligible effects, depending on the conditions. The way I understand it, a perfect gas is an ideal gas, but is never treated as having non-interacting particles. I don't think the distinction is terribly important unless you plan on doing some thesis or dissertation in the fields of physical chemistry or condensed matter physics...

Well I was asked if Cv has not a function of temperature for the ideal gases. Perfect gases' Cv is not a function of T, but Ideal ones do seem to depend.. Hmm..

Yes, I agree with this.  Physicists define a perfect gas as one that obeys the ideal gas law plus exhibits a heat capacity that is independent of temperature.  An ideal gas, as we engineers call it, is one that obeys the ideal gas law plus exhibits a heat capacity that does depend on temperature.  Real gases actually approach ideal gas behavior at low pressures, and do exhibit temperature-dependent heat capacities.

P.  W.  Atkins says that it is preferable to use the term perfect gas because the particles show no interactions. On the opposite, ideal solutions are constituted by different particles that show the same kind of interactions among them (the toluene/benzene mixture is a good approximation).

In terms of macroscopic thermodynamics (no molecular-based arguments are necessary), the underlying reason for distinguishing between an "ideal gas" and a "perfect gas" is that a mixture (solution) of gases may be an "ideal solution", but each of its components need not be an ideal gas.
In the case of a single component gas phase, it is more precise to use the term "perfect gas", which refers to a gas obeying the equation of state (commonly referred to as the "ideal gas equation of state"):
$$PV = nRT \tag{1}$$
where P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant, and T is the absolute temperature.
From the thermodynamic relation
$$(\mathrm d \mu/\mathrm dP)_T = V/n \tag{2}$$
we obtain the chemical potential of a single perfect gas as
$$\mu_i(T,P) = \mu_i^0(T;P^0) + RT\ln(P/P^0) \tag{3}$$
where $P^0$ is set by convention as $\pu{1 bar}$ (but pre-1990 tables used $P^0 = \pu{1 atm}$.)
So far, we could have used the term "ideal gas".  However, a problem arises in the case of solutions.  The generalization of Eq. (3) to a mixture (solution) of perfect gases is
$$\mu_i(T,P) = \mu_i^0(T;P^0) + RT\ln(x_iP/P^0) \tag{4}$$
where $x_i$ is the mole fraction of species $i$. (Note that thermodynamic manipulations on Eq. (4) readily give the ideal-gas EOS for the mixture
$$PV = n_tRT \tag{5}$$
where $n_t$ is the total number of moles.)
Eq. (4) may be written as
$$\mu_i(T,P) = \mu_i^*(T,P) + RT\ln(x_i) \tag{6}$$
where
$$\mu_i^*(T,P) = \mu_i(T;P^0) + RT\ln(P/P^0) \tag{7}$$
It is then useful to define the general concept of an "ideal solution" (for any phase) as one for which the chemical potential of each species is given by Eq. (6), in which case $\mu_i^*(T,P)$ is the chemical potential of pure species i at the T and P of the mixture. Note that $\mu_i^*(T,P)$ need not be given by Eq. (3).  A solution of perfect gases (whose individual chemical potentials are given by Eq. (4)) is thus a special case of an ideal solution.

What I have learned at high school is that ideal gas and perfect gas are synonyms. That is, the names basically refer to the same type of gas which:

Has negligibly small forces of attraction between its molecules;
Has negligibly small distances between the particles;
Has particles randomly roaming about at high speeds;
Has molecules whose collision with walls of its container are
  perfectly elastic.

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text:
Does ethanol react with potassium to form potassium ethanolate and hydrogen gas?
I'm finding information in Google with the keyword (C2H5OH + K) to find the reaction between $\ce{C2H5OH}$ and $\ce{K}$ and found very little information about it. Just 2-3 related results with contradictory pieces of information.
The first one gave
$$
\ce{K + C2H5OH -> KOH + C2H5},
$$
the second one gave
$$
\ce{2C2H5OH + 2K -> 2C2H5OK + H2},
$$
and the third one gave
$$
\ce{C2H5OH + KOH -> C2H5OK + H2O}
$$
There are no trusted public sources so I'm very baffled. I think that $\ce{C2H5OH}$ will react with $\ce{K}$ to form $\ce{C2H5OK}$ because it looks like the reaction with $\ce{Na}$. But I'm not sure and just want to find a trusted source. Can anyone confirm the true result for this reaction?

You can view water as the simplest alcohol, and indeed, water reacts with all of the Group 1 metals to form hydrogen gas and the corresponding metal hydroxide according to the following equation
$\ce{2M + 2HOH -> 2MOH + H2}$
where $\ce{M}$ is the metal.
The same reaction occurs with simple alcohols to produce the corresponding metal alkoxide and hydrogen gas
$\ce{2M + 2ROH -> 2MOR + H2}$
As you increase the hydrocarbon portion of the alcohol, or reduce accessibility to the $\ce{OH}$ portion of the alcohol (e.g. make the alcohol less like water), you slow down the rate of the reaction.
Here is a nice link that compares and discusses the relative reactivities of the Group 1 metals with water.

Yes, and that's a standard way to dispose of extra Na, K and other alakali metals, since it's a bit less vigorous then dropping them in H20. ;-) See http://www.nap.edu/openbook.php?record_id=4911&page=165.

@DrMoishePippik True, but let's not forget this: https://www.youtube.com/watch?v=HY7mTCMvpEM :D


text:
What is Nano Zinc Oxide?
First of all, I have zero chemical background, but I seek for an easy explanation about Nano Zinc Oxide.
I was told that the name of "Nano Zinc Oxide" couldn't possibly exist. Instead, it should be coined as "Nano Zinc Chelate", since an Oxide particle is quite big and Nano is so small, relatively. It sounds very contradictory.
I know there is, of course, Nano Zinc Oxide, but I don't understand what he really means and  I feel enthusiastic to understand it.

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Nano Zinc Oxide ($\ce{ZnO}$) is a nanomaterial and a nanomaterial is described as

materials of which a single unit
  is sized (in at least one dimension) between 1 and 1000 nanometers
  ($10^{-9} m$) but is usually between 1—100 nm

Nano zinc oxide  can be synthesized using different approaches that included's both chemical (precipitation, reduction etc) and physical (depositon) methods . You can read more about nano zinc oxide and its application in this paper Journal of Physics: Condensed Matter Volume 16 Number 25

This is an image of Nano Zinc oxide grown as a flower (nano-flower) for sensor applications

Nano zinc oxide is a real material used in sunscreens and looks like this.  It is just little crystals of zinc oxide which are about 30 nm across.
Chelate refers to something which is produced by a chemical process of chelation.  Chelation means that a metal atom has more than one chemical bond to surrounding organics.  (That is kind of a bad definition, but I'm trying to keep the lingo simple.)  Chelation can be used to produce zinc oxide, but I don't think this is the process used for manufacturing most industrial zinc oxide.

He is still questioning a lot more which is beyond my ability to understand. At least, I understand its general concept. Thank you again.


text:
Why don't sponges foam up an hour later?
When I add dish soap to a sponge it is able to foam up. But if I leave the sponge alone for an hour, it doesn't foam up, not even if it is wetted.
Why? The sponge should still contain dish soap an hour later.

Evaporation, perhaps?

The sponge is dry up, but the dish soap should remain in.

When the sponge dries, there's not enough water in it to create the air bubbles that cause the foaming.

@HotLicks I see it, but when it be wet again (for example an hour later), the sponge doesn't foaming.

I would push for the migration of this question to chemistry.se since it is concerning soaps and detergents.

Dish soap traditionally contained foaming agents such as sodium laureth sulfate which is chemically stable over long periods of time. But more recently environmentally conscious soap companies are turning to foaming agents that have a shorter half-life. These chemicals tend to break down sooner - perhaps just by being exposed to air. This helps reduce foaming in the settling basins of wasterwater treatment plants, or storm drains that enter the ocean and it takes less water to rinse away the soap.
Read the ingredients of the soap you are using and research to see which are the foaming agents, and what the stable half-life is.


text:
Is there anything akin to molecular chirality for crystalline?
For example, in hexagonal closed-packed (HCP) structure, the atoms may be arranged in ABAB stacks or ACAC stacks. Can these two be considered enantiomers? Do they possess different solid state properties?

There are some crystal structures where the nature of the crystal structure creates chirality and therefore optical activity.
A trivial example (important because of its history: it was involved in the discovery of optical activity and molecular chirality) is tartaric acid. But the crystal is chiral because the molecule is chiral so that might not be what yyou were looking for.
But there are examples of crystals where the structure as a whole is chiral but the building blocks are not. Quartz, for example, is formed from a network of linked SiO4 tetrahedra (not molecules but a giant interlinked network where each oxygen is shared between two silicons). The overall crystal structure, though, is chiral due to small shifts in the rotation of the linked tetrahedra (see wikipedia for a detailed explanation).
I think this is a crystal analogy to molecular chirality.
Only a few crystal structures can do this and the ones you use as examples can't.

Typically, different ways of arranging atoms in a solid are referred to as phases. For a single composition, a material may have different phases depending on temperature or pressure. Different crystal structures are not referred to as enantiomers. For an HCP metal, there is no difference an ABAB stacking and ACAC stacking, other than the naming. If the stacking is ABCABC, then you have a face-centered cubic structure (FCC), not an HCP. The best way to alter solid-state properties such as resistivity, susceptibility, etc., is to change the temperature or introduce defects into the system.


text:
Heat energy of phosphoric acid evaporation
Looking at a system in which phosphoric acid is concentrated in an evaporator which uses a constant recycle stream with a heat exchanger in this stream to heat the acid so that water evaporates out of the acid when it re-enters the main stream, How can I find the Q needed to remove a certain amount of water from the stream in order to concentrate the acid to a specific level?
I am given the amount of water evaporated, both concentrations, and the flow rate of both the entering acid stream and the exiting acid stream. I have also been told that you cannot simply use $Q=mC_p\Delta T$, because this is inacurate for phosphoric acid.
Any help or sources you could link me to would be greatly appreciated.

You need to know the latent heat of vaporization of the solution, which would be the same as that of water when phosphoric acid concentration is low.
See Fig. 10 of this bulletin for latent heat of vaporization as a function of phosphoric acid concentration, but keep in mind this is only strictly valid at the stated temperature.
If you are varying phosphoric acid concentration (concentrating phosphoric acid) you would need to integrate heat of vaporization as a function of phosphoric acid concentration over the appropriate concentration range.
Also, as phosphoric acid concentration gets very high, it is no longer valid to consider the evaporating solution to be only water, it will increasingly contain phosphoric acid as well.

Hello, I am given latent heat and it is assumed that the concentration within the evaporator as well as the concentration in the recycle stream are the same concentration as the exit stream. Can you help with what I should do in this case?

If this is a homework problem, add the exact problem to your question and explain what you need help with

I'm sorry, this is not a homework problem. It's independent study from a professor who posed the question verbally.

As it turns out, I was just confused on the question being asked. You actually can use Q=mC p ΔT to answer the question. However, instead of using it for Phosphoric acid, you use it based on the water, as you know the mass of water being evaporated out, the specific heat capacity of water at this temperature, and the change in temperature of the water. This Q required by the water will be equal to that added to the phosphoric acid by the heat exchanger.
As a side note, I was incorrect in stating that Q=mC p ΔT is not applicable to phosphoric acid. It is applicable, however not reccomended for this problem as you do not know the mass flow rate of the phosphoric acid through the recycle stream, nor is it easy to find the heat capacity of dilute phosphoric acid at the very high temperatures at which this evaporation would take place.


text:
Are there colorful metals besides gold and copper?
Are there yellow metals besides gold? Are there metals with color different than yellow and grey/silver or reddish?

this is not an answer but it is related and might be of interest http://chemistry.stackexchange.com/questions/16633/why-is-gold-golden

Osmium has a bluish-gray tint.
Well; slightly.
Cesium is silvery-golden!,
But don't wear it.

There is no element other than gold and copper with similar colors; see http://periodictable.com/Properties/A/Color.html.
There are many metal alloys that are red and golden, such as phosphor bronze, which can have a golden hue.
In addition, some metallic compounds may be shiny yellow. Freshly sliced potassium is silvery, but soon acquires an oxide/nitride coat that may look yellow (perhaps due to interference of light?). Of course, "fool's gold", FeS2, was often confused with the element. See http://en.wikipedia.org/wiki/Pyrite and Image of pyrite crystal.

Titanium nitride (TiN) is a beautiful gold color.  It is also nice and hard, so is used quite often for coatings, including on things like plumbing fixtures.  So, that lovely gold faucet in the store is likely TiN, not Au.

@Jon Custer it is ceramic, not even an alloy.

@Anixx - yet TiN is metallic, having excellent conductivity mediated by electrons (as do many transition metal nitrides). So, I'll still claim it to be a metallic compound.

Phosphor bronze basically consists of copper. What about yellow alloys that do not include copper or gold?

Whoa! What about cesium?!

Cs is "silver" colored; see http://periodictable.com/Properties/A/Color.html. In air, Cs and other alkali metals may form a surface nitride layer that is yellowish, but look at a freshly cut surface and it's not. Similarly,  Ti is silvery, but the nitride deposited on tools is golden.

@DrMoishePippik Nope. Liquid cesium under an inert atmosphere is clearly fairly golden coloured.

Did you mean metal? There are certainly elements that are not grey (e.g. bromine).

The color of nanoparticles in solution depends on size, so one can create the whole rainbow of colors using a single elemental composition.
See e.g. http://nanocomposix.com/pages/plasmonics for details.
https://en.m.wikipedia.org/wiki/Colloidal_gold shows gold being red and purple.

Strontium seems to be golden too, if of high purity. Look under "Eigenschaften". Beside that, also caesium seems to be of golden color.

Why not link to the English Wikipedia Page? It has the picture right at the top.


text:
What is the geometric formula of CaI2 and AlCl3
Well, I think $\ce{CaI2}$ is the same as $\ce{BeH2}$ and $\ce{AlCl3}$ the same as $\ce{NH3}$, but I'm not sure since $\ce{I}$ and $\ce{Cl}$ have their own pair electrons and I don't know if this affects the structure...

Central atoms/cations are affecting geometry stronger than atoms they are connected with.

$\ce{CaI2}$ is mostly ionic in the solid and aqueous phases.  In the gas phases it is a linear molecule.  See Ionic or covalent? Can some first hints be derived from the solid state
structures of alkaline earth metal halide adducts?
$\ce{AlCl3}$ has an octaheral structure in the solid phase.
In the liquid phase:

Trigonal planar monomers also exist at high temperature.
Source: http://en.wikipedia.org/wiki/Aluminium_chloride


text:
Preparing CdSe electrodeposition bath
What is the best way of preparing an aqueous solution containing 5 mM
$\ce{CdSO4}$, 1 mM $\ce{SeO2}$, 0.5 M $\ce{Na2SO4}$ and adjusting pH to 3 by adding 0.5 M $\ce{H2SO4}$?
What is the conventional way of doing it (in industrial chemistry or simply just the labs)?


text:
Magnetic stirrer with hot plate
I did not know there are products such as a magnetic stirrer with a hot plate, enabling one to apply heat and stirring to a solution with relative ease. Did I understand correctly, that device needs a magnetic bar to be added to the solution the device to work? Does that pose any limits when it can be used (if the magnet would react with something..) How much do these devices increase productivity?

The magnetic stir bars come in all sizes and shapes.  They are coated with PTFE (or similar) so that they do not interfere with your reaction.  Unless your reaction is very large or very viscous, a magnetic stir bar requires much less time to set up and clean compared to a traditional mechanical stirrer.

I have repaired several cheap stirring hotplates. The problem was that they were made with bad glue that eventually wore out, probably wasn't heat tolerant. I used automotive exhaust manifold silicone glue to stick the magnets back on the motor and they worked fine. The silicone had much higher heat tolerance.

@user137 What would be cheap for stirring hot plates? For example, would you have any opinion on this http://www.amazon.co.uk/MAPLE-SCIENTIFIC-HOTPLATE-TEMPERATURE-ADJUSTABLE/dp/B008YC0VPA

Stirring hotplates are very expensive for how simple they are. Expect to spend a couple hundred dollars for a decent one, but if you can find a used one on ebay you might do better. I can't offer an opinion on that particular model, since I've never used it.

Most of the things to be told about magnetic stirring are already in Rons comment. A rotating magnetic field is induced by a rotating permanent magnet inside the hot plate magnetic stirrer. This field drives the magnet inside your reaction flask. To ensure chemical inertness in most situations the magnet is coated with PTFE (Teflon) or protected by a thin layer of glass. Magnetic stirring can replace traditional mechanical stirring is most situations and has many benefits over it: it does not require an additional inlet to the reaction flask, which is very handy for reaction under reduced pressure and is also more easily adjustable to different scales due to the possibility to replace the magnet in the reaction flask. Two exceptions are 1) very large/very viscous reactions or reactions in which excessive amounts of solid are present 2) reactions that involve ferromagnetic compounds that interfere with the rotational movement of the magnet inside the flask (as in Urushibara nickle catalyst preparation).
Vogel has a section on reaction heating and stirring in which this and more is discussed extensively. See pages 71-80.

authors: Ânnô Âl Rïfâï


text:
Will isopropyl alcohol dissolve the coating/paint of a notebook?
The notebook is a Dell Inspiron 6400 and the coating/paint is the silver on the palmrest and back cover.
It says 99,4% on the container of the isopropyl alcohol.

No way to know without trying it. If you're willing to risk it, put a little of the alcohol on a q-tip and rub it on a less visible spot of the laptop to test.

Unfortunately you won't really know until you try it because there doesn't appear to be any info about what they used as a coating. Having said that, isopropanol isn't a superbly strong organic solvent so while it may remove some of the coating, it's not likely to immediately eat through your laptop like how acetone might. Still, follow user137's advice and test it on a very small area before you try anything else, and give it time before you go using it everywhere else. If you don't see any damage and you decide to go through with whatever it is you want to do (I assume cleaning it), just make sure you wipe it off with a damp cloth (water) followed by a dry cloth as soon as you get the results you want to minimize any possible damage.

Thank you. I have contacted Dell and they advise against using alcohol.


text:
Does increasing concentration of copper sulfate decrease or increase its conductivity?
I am a high school student and for my chemistry lab, I did an electrolyis experiment on how the changing concentration of electrolyte (copper sulfate) will change the rate at which the copper is deposited on the cathode.
My hypothesis was that as the concentration is increased, so will the rate of electrolysis. However, my results are completely opposite. When I use 0.1 M solution, loads of copper is deposited on the cathode but as I increase the concentration, the amount of copper deposited becomes progressively less and less. With 1 M solution, there was almost no copper being deposited.
This makes no sense to me. Could somebody explain what is going on? Am I doing something wrong?
I used both copper electrodes. I did not use distilled water, just normal tap water. I used 12 V power supply.
EDIT:
I tried adding a little sulfuric acid (making the pH around 3) to the copper sulfate but it did not make any difference on electrolysis.
I tried to observe the electrolysis very closely today and noticed that there was some sort of oily liquid falling off from the anode. Also, I noticed that initially some flaky brown/black substance fell off from the anode (which may or may not be copper)and soon there was a silvery layer formed on the anode.
Also, I checked using Hanna checker that the overall pH of the solution was around 3 but the pH near the anode read between 9 and 14.
Also, I noticed that when I used a higher concentration, the copper deposition on the cathode was brown and clearly copper. But when I used lower concentration, even though the amount of deposit was a lot more, the substance was probably not pure copper as I had assumed earlier. It was black and flaky. It was certainly different to the deposit I observed when I used higher concentration of copper sulfate.
Does anybody know now what is happening? Is there anything that I can do to fix the problem?

I'm thinking maybe at the higher concentrations your tap water is making a film of $\ce{Cu(OH)2}$ on the surface of your electrode, and that is what's shutting down your electroplating. Your tap water might be mildly alkaline (tap water pH's can range from about $6.5$ to as high as $9.5$).
Try adding a little sulfuric acid to your solutions. If alkaline films are the problem, that will fix it.
It will also increase the conductivity of your solutions tremendously, because hydrogen ions are extremely mobile.
The solubility of copper sulfate decreases with increasing sulfuric acid concentration, so don't add too much acid.

Thanks. Today I have chemistry class and will definitely give sulphuric acid a go. Do you think around 5 ml of diluted acid to 200ml CuS04 will be enough?

I don't know, it depends on the concentration of the acid. If you bring the pH down to around 3 to 5, it should work.

Can you please look at the updates in the question and help me.

Two things are very important in electrochemistry.
1- Cleanliness: Your tap water contains all sorts of impurities that are probably electroactive. My recommendation, take some tap water, don't add any $\ce{CuSO4}$ and repeat the experiment. If you see any deposit, that is not copper. I won't even attempt to make any guesses as to what that deposit may be.
2- Potential control: This is very tricky in the type of setup you are using. In two electrode experiments like you are doing, the solution resistance, the distance between the electrodes and the current you are drawing all effect the actual potential your electrode experiences. With the constant 12V you are applying, the current that is drawn is also a function of the above parameters, so it is very hard to determine what the actual electrochemical potential will be.
Gamry has an excellent application note about 2,3 and 4 electrode experiments here: http://www.gamry.com/application-notes/two-three-and-four-electrode-experiments/ if you would like to read up on the details.

I think the cleanliness of the tap water was probably the reason for the wierd results. In the end, I just decided to do another experiment seeing this one did not work. If I get a chance, I will definitely try electrolysis on only tap water and see what happens. Thanks very much.

Whatever contamination in your tap water should be a great deal less than the concentration of CuSO4, but might be a factor in your experiment.
Conductivity is measured using a non-depositing AC current. The DC resistance or its inverse will depend on many factors, but all will require a measurement of voltage and current simultaneously.
Measuring the rate of electrolysis by visual means is inadequate, since it is obvious to you that the character of the deposit of copper changes. A better measure is determination of mass of copper deposited - but the accuracy required for a short experiment requires a very sensitive balance - and, unfortunately, since copper is sometimes flaking off, measuring the cathode is going to be rough. Measuring the loss of copper at the anode could be a better measurement.
But even better (easily more accurate) is measuring the current. When current flows, it will be carried by Cu++ and SO4-- ions (unless you add H2SO4, in which case, H+ will also be carrying current). Adding H2SO4 complicates the experiment, because you are electroplating copper and electrolyzing water at the same time.
So here is my suggestion: measure the voltage between the electrodes AND the current flowing, simultaneously, at one concentration of CuSO4 (no H2SO4) in water (how about using distilled water instead of tap water to make your results more universal?). Then dump in some more CuSO4.5H2O and follow the change in current and voltage. Both will change. You may adjust the voltage OR current to equal the original voltage or current. In either case, the apparent DC resistance (R= E/I https://en.wikipedia.org/wiki/Ohm%27s_law) of the solution will decrease.

Kohlrausch law states that $\Lambda_m=\Lambda_m^{\circ}-\kappa \sqrt{c}$ with $\Lambda_m = \frac{k}{c}$ (in which $k$ is the electrical conductivity) for strong electrolytes and low concentrations, which is due to the strong interaction of ions. In that light your findings are not inconsistent. I am not sure though about the magnitude of $\kappa$ and whether for the conditions (copper sulfate as electrolyte, 0.1M to 1.0M concentration) you listed the law holds. At least the point I want to make is that it is not obvious or even correct to assume that conductivity depends linearly on the concentration of the electrolyte or that there is even a positive relation at high concentrations.
This is definitely not intended as a definite answer (nice tautology).

Is copper sulphate a strong electrolyte? Also, I read on this website: http://antoine.frostburg.edu/chem/senese/101/redox/faq/cu-deposition.shtml

that increasing copper sulphate concentration upto 1.0 M should increase deposition of copper as per Nernst equation.

@Joe Slater, that's correct, it should (but of course I'd say that, that's me in the link).

@FredSenese wow. What a coincidence. Thanks for your website. It is quite helpful. So kohlrausch law does hold true for 1 M copper sulphate? Are Nernst equation and kohlrausc law derived from one another?

No, they're different things. Kohlrausch's law only works for strong electrolytes that are fairly dilute, as Jori says. Behavior in concentrated solutions is complicated; the concentration is different from the activity (an effective thermodynamic concentration) in concentrated solutions, and you'll also get ion pair formation (direct hookups of $\rm Cu^{+2}$ and $\rm SO_4^{-2}$, in this case).

The solute must ionize, and as the concentration of ions in increased, further ionization is inhibited. The non-ionized molecules are inert, and interfere with reaction rate.


text:
What are the products of the reaction of 1-methoxycyclohexene with dilute hydrochloric acid?

When 1-methoxycyclohexene is treated with dilute aqueous hydrochloric acid, methanol is produced along with another organic product. Identify this product. Explain why methoxycyclohexane is inert under these reaction conditions.

I don't understand why it isn't just a simple addition to the double bond. I get that there's conjugation but is this what makes things different? I don't get why methanol is the leaving group and therefore I cannot predict the other product. Also, it says that methoxycyclohexane (the same molecule without the double bond) is inert under these conditions. Why does it not undergo substitution? The oxygen could surely be protonated and then methanol would be a good leaving group which would give rise to substitution (probably SN1). Why is this not the case?

Vinyl ethers are "masked" carbonyls, therefore in acidic aqueous environment the following pathway takes place: alkene protonation assisted by the methoxy group followed by nucleophilic attack of water with hemiacetal formation. Subsequent intramolecular proton transfer to the methoxy group makes it a good leaving group and cyclohexanone (the product) is formed while methanol is released.

Why wont methoxycyclohexane undergo substitution?

Methoxy cyclohexane could undergo protonation on the methoxy group but the reaction conditions (aqueous HCl) are not harsh enough to promote cleavage of such ethereal bonds. However, use of HBr or HI in organic solvents (since bromide and iodide are much more nucleophilic than chloride anions), one could achieve ether cleavage thus leading to the formation of cyclohexanol from methoxycyclohexane. The mechanism consists of initial protonation of methoxy group and nucleophilic attack of bromide on the methyl group with subsequent release of methyl bromide and cyclohexanol.

You're close.  You are right in thinking that the first step involves protonation of the double bond.  But which end of the double bond will you protonate?  If you protonate the carbon bearing the methoxy group you wind up with a secondary carbocation.  However if you protonate the other end of the double bond you wind up with a carbocation that has the methoxy group attached to it.  Draw a resonance structure and convince yourself that the methoxy group can stabilize this carbocation.
So now we have a carbocation bearing a methoxy group.  The next step is 1) attack by water and ejection of a proton from the water to yield a neutral hemiacetal.  Under acidic conditions the hemiacetal will hydrolyze to produce cyclohexanone and methanol.
Methoxycyclohexane won't undergoe this reaction since there is no double bond to protonate.

Why wont methoxycyclohexane undergo substitution?

The reaction is being run in dilute acid.  SN1 (protonation of the methoxy group followed by elimination of methanol to generate a secondary cyclohexyl carbocation) is unlikely under these weak acid conditions because the cyclohexyl carbocation is not particularly stable (e.g. it is a high energy intermediate).  SN2 is also unlikely since water is a weak nucleophile and methoxide is not a great leaving group.


text:
Why and how does lycopene absorbance increase with heating?
Lycopene, which is a carotenoid and a phytochemical is the main reason for why tomatoes are radio-protective.

Source 1, Source 2
Let's cut to the chase: I was wondering if this claim of this source is of scientific value:
"Heating will result in an increase in amount of lycopene absorbed by the body."
It may be possible that the researcher made a more "every-day" statement for the readers to understand.
I have two questions:

Is the statement scientifically correct?
If so, why and how would such a chemical be absorbed more with the heating process?

Note that if you want to make the argument that this is a biology question rather than a chemistry one, I would bring up these reasons:

The temperature in which the heated lycopene and raw tomato are going to be absorbed in the body are the same. Thus, it might be possible that heating lycopene results in a chemical reaction.
I'm not looking for the pattern of absorption by the body. I'm looking for possible changes or specifications of its structure.

"Heating will result in an increase in amount of lycopene absorbed by
  the body."
  Is the statement scientifically correct?

It is a poorly phrased statement, but it is probably generally true.  It would have been better to say something like, "heating tomatoes will increase the amount of all-trans-lycopene available for ingestion."
Here's a link to a full paper on the subject (click the "View" button near the top-right of the page).  Basically two things are going on at the same time when lycopene containg food is heated.  It is the naturally occurring, all-trans isomer of lycopene that is thought to be beneficial to the human body. Heating lycopene causes it to isomerize to a variety of cis-isomers.  Heating in the presence of oxygen can also cause oxidation.  Both of these processes (isomerization and oxidation) reduce the potency of the lycopene in the broth.
However, in parallel with these deleterious effects, heating also makes more lycopene available in the heated broth or extract.  This is because heating disrupts cell walls and releases lycopene from the cells.  The above link shows that, at least at moderate temperatures, the latter effect predominates and so more active all-trans-lycopene is made available upon heating tomatoes, despite the loss of some lycopene to isomerization and oxidation.


text:
Building an incomplete redox half reaction/reduction
I can't seem to figure out this redox reaction.
$$
\ce{CH3COO- + CO2 -> CH3CH2COO-}
$$
Is the charge on the carbon in the $\ce{CO2}$ supposed to match the charge on the carbon in the $\ce{CH3COO-}$? I am supposed to build a half reaction using $\ce{H2O}$ and $\ce{H+}$ as needed, but I can't seem to figure out how many electrons are actually transferred. What are the charges on the carbon?

It helps to use "fully compressed" formulae so you can easily count atoms and figure out how to balance carbons, oxygens, hydrogens, and charges.  That is, instead of

$\ce{CH3COO- +CO2 -> CH3CH2COO-}$

write it this way:

$\ce{C2H3O2- +CO2 -> C3H5O2-}$

Then it's easy to see that this is not a balanced equation.  Carbon atoms are the same on each side, but not oxygen or hydrogen atoms.  You need two more $\ce{O}$ atoms on the products side.  Those have to come from water since that is the only oxygen-containing species you are allowed to use.

$\ce{C2H3O2- +CO2 -> C3H5O2- + 2H2O}$

Now oxygen is balanced, but hydrogen is still imbalanced.  That can be rectified by adding $\ce{H+}$ into the equation as appropriate.

$\ce{C2H3O2- +CO2 +6H+ -> C3H5O2- + 2H2O}$

Now only charge is imbalanced, which can be fixed by adding electrons to complete the balanced half-reaction:

$\ce{C2H3O2- +CO2 +6H+ +6e- -> C3H5O2- + 2H2O}$


text:
How to choose a colorant for a given substrate?
I know there are volumes of International Colour Index (homepage, wiki, paper), but I can't really extract from there for example what red colorants are compatible with both silk and polyester, etc.
This paper mentions that such compatibility can be estimated using numerical methods:

"Metropolis Monte Carlo methods have been used to determine the surface
  interaction or binding energy for dyes to cellulose." [ref]

Is there a comprehensive database for this?
Is there an analytical way to determine/predict/estimate such compatibility?


text:
How do I make a dysprosium chloride solution from dysprosium oxide?
I know that dysprosium oxide is insoluble in water.
I was told that I need to add concentrated $\ce{HCl}$ to the compound and then evaporate the $\ce{HCl}$ off. From looking at the reaction equation, I can see that I will get dysprosium chloride and water as products.
Once I'm able to dissolve the $\ce{Dy2O3}$, how is it that I'm supposed to evaporate off the excess $\ce{HCl}$? Do I just let it sit, or do I need to gently heat the solution? How can I tell when all the excess $\ce{HCl}$ is gone and I can start adding water to obtain the concentration of dysprosium chloride that I need?

Welcome to chemistry.se! If you have questions about how to beautify your posts, have a look at the [help]. Do you want to know more about this site, please take the [tour].
I have updated your post with chemistry markup. If you want to know more, please have a look here and here. Please do not use markup in the title field, see here for details.

DyCl3 is often prepared by the "ammonium chloride route," starting from either Dy2O3 or hydrated chloride or oxychloride.[2][3] or DyCl3·6H2O.[4] These methods produce (NH4)2[DyCl5]:
    10 NH4Cl + Dy2O3 → 2 (NH4)2[DyCl5] + 6 NH3 + 3 H2O

    DyCl3·6H2O + 2 NH4Cl → (NH4)2[DyCl5] + 6 H2O

The pentachloride decomposes thermally according to the following equation:
    (NH4)2[DyCl5] → 2 NH4Cl + DyCl3

The thermolysis reaction proceeds via the intermediacy of (NH4)[Dy2Cl7].
Treating Dy2O3 with aqueous HCl produces hydrated chloride (DyCl3·6H2O). This salt cannot be rendered anhydrous by heating. Instead one obtains an oxychloride.
source: wikipedia


text:
How pigments actually function as colorants?
Pigments are insoluble colorants that have no affinity to the substrate. So how can they function as colorants if they "can't stick"? Are pigments always applied together with a "binder"?

""that have no affinity to the substrate"" the affinity may be weak, but it is never zero. Aside from the "pigments" precipitated within the fabric as explained by Fred Senese it is simple: pigments are not used as dyes, pigments are made to colorize laquers and plastics.

Consider natural indigo, formerly used to color blue jeans. It is insoluble; reduction converted it into water-soluble "indigo white". Fabric was soaked in indigo white solution; on air oxidation, it became insoluble blue indigo again, with indigo molecules now presumably mechanically entangled in the fiber and held by van der Waals forces.
More often "binders" called mordants are used. They form an insoluble "lake" when they coordinate to the pigment; the lake can bind to fabric fibers. You can read more about the chemistry of the process here.


text:
What tools can be used to estimate binding energies?
I want to estimate the binding energies of various molecules. What are the analytical and numerical tools I can use for this purpose? In particular I'm interested in bonding energies of dyes to substrates (example).

Are you only referring to pen and paper methods, or are you also interested in experimental procedures? It would probably be helpful if you could give some examples of the molecules (and bonds) you are looking into to estimate the size and what methods are suitable.

This will depend heavily on the accuracy you need and the size of the molecules. Binding one protein to another is very different than dealing with the benzene dimer.

@Martin, I've updated the question.

@Martin Isn't Gaussian a common tool to calculate such things? I think OP is referring to computational chemistry not pen and paper methods.

@Jori I consider computational chemistry to be a pen and paper method, at least in principle. The computer just does the job. While Gaussian might be suitable for a lot of things, i believe for the OP's intention it might not be the best fit. I might try to add an answer tomorrow, but I welcome anyone with more insight to be faster.

@Martin How were the results of Gaussian (the very first version) ever verified?  Prof. John Pople simply got himself an army of grad students who worked overtime crunching numbers with paper, pencil and those old cumbersome calculators that had those huge levers you had to 'sha-shunk!' every time you wanted to perform an operation.  I'm so glad I wasn't practicing comp. chem. during those times. ;)

The steps for predicting a binding energy is relatively straightforward in computational chemistry but the process itself can become extremely difficult depending on

The types of molecules involved (size, complexity, etc.)
The desired accuracy of your final answer
The environment that the molecules reside in (gas phase, aqueous phase, etc.)

For very small molecules in the gas phase, the process is easy and routine assuming that electron correlation is trivial.  Aqueous phase environments, complex/large molecules and high accuracy quickly makes the problem more difficult if not impossible.
I have written up a previous answer which addresses how to predict an interaction energy ($E_{\mathrm{int}}$) which is completely different from a binding energy ($E_{\mathrm{bind}}$).  However, much of the concepts will remain the same.  I will try to outline here what in fact is the difference between an interaction energy and a binding energy.  Unfortunately, 'binding energy' is tossed around a lot in the literature and consequently grossly misused.  I hope to clarify the subtle differences between these two terms in my answer to your question.
I will focus my discussion around an elementary system of two water molecules to make this conceptually easier.  In the gas phase, an isolated water molecule will adopt a particular geometry (at 0K and in a vacuum).  We can determine what this minimum energy configuration will be at a particular level of theory by approximating the Schrodinger equation.  The geometry that gives rise to the lowest possible energy for an isolated water molecule will be the geometry that you will observe under those conditions.  Remember that the internal coordinates of water include things like the O-H bonds, and the H-O-H angle.  We don't need to guess at what these parameters will be in the end, solving the Schrodinger equation will determine that.
Now, we know exactly what an isolated water molecule looks like and what the corresponding energy of that water molecule will be.  Both of these components are equally important when it comes to determining a binding energy because (as the name suggests) we are trying to determine an energy.  So the energies that we start off with are crucially important.
We can now consider what happens when two water molecules interact (form a hydrogen bond).  What will be the binding energy of these two water molecules (i.e. what will be the strength of this bond that forms between then?).  It is common to approximate the strength of this bond by simply computing an interaction energy (its computationally easier to determine $E_{\mathrm{int}}$ rather than $E_{\mathrm{bind}}$).  How would we compute an interaction energy?  In this case, you wouldn't need to figure out what geometry or energy an isolated water molecule in its lowest energy configuration.  You would simply need to determine what the geometry of two interacting water molecules (i.e. dimer) would be (and corresponding energy) as well as the energy of each water molecule isolated from each other in the geometry that it adopts in the dimer.
I'll explain.  We can determine the lowest energy configuration of a single water molecule in the gas phase.  This geometry is inevitably different than a water that is not isolated... a water that is interacting with something else such as another water.  If you want to get an interaction energy, you only need...

Energy of the lowest energy configuration of the water dimer
Energy of each water (in the water dimer) isolated from each other but only at the geometry that the water takes on within the dimer.

In this case, you are NOT allowing each isolated water molecule to 'relax' into its lowest energy configuration.  You are simply going to neglect the contributions of this energy relaxation effect to the overall bond strength.  Therefore,
$E_{\text{int}} =  E_{\text{opt-dimer}} - (E_{\text{mon1}} + E_{\text{mon2}})$
where mon1 is the first water molecule in your dimer and mon2 is the second water molecule in your dimer. (opt-dimer means optimized dimer)
However, in the case of a binding energy, you MUST account for the relaxation effects that occur when you break the dimer apart and allow the waters to become isolated from each other (and everything else).  To do this, you simply carry out what I've stated above... you simply optimize each water and then take its corresponding energy.  (Determining an interaction energy means you don't have to do this extra optimization(s)... you would simply do a single point energy on each monomer).  Clearly for a water dimer, you would only do one optimization as each water will become identical when isolated.  For a heterodimer, you would have to optimize each piece separately.
To determine the binding energy, you would use the following equation:
$E_{\text{bind}} =  E_{\text{opt-dimer}} - (E_{\text{opt-mon1}} + E_{\text{opt-mon2}})$
Of course the problem becomes more complex when taking into account basis-set superposition error (you will need to perform counterpoise corrections though there exists a heated debate in the literature about the usefulness of this type of correction).  Also, zero-point vibrational energy corrected binding energies will require that the zero-point energy of the dimer and each monomer will need to be determined (which comes from taking the 2nd derivative of the wavefunction which is computationally demanding).
As far as tools go, you can choose from a number of available quantum software packages.  Gaussian 09 is a popular choice but very expensive.  Q-Chem (paid) has the benefit of having the most beautiful, gorgeous, well-laid out documentation of anything I've ever seen (makes for a good supplemental textbook to boot).  NWChem is opensource.  Google can help you find plenty of options.
A quick note: Basis set superposition error (BSSE) should NOT be referred to as an error as 'error' suggests deviation from the RIGHT answer. BSSE has been shown to actually make  non-corrected answers even WORSE.
One last important note:  When determining differences of energies (such as you do to get an interaction or binding energy), ALL energies being compared must be evaluated at the SAME LEVEL OF THEORY (method/basis set, etc.).

I'm looking at NWChem's sample input files. If I want to compare "binding energies" of various combinations, can I use the SCF method on two molecules separately and substract from the SCF value for the dimer?

Yes, as long as you are comparing energies that were all determined at the same level of theory (method/basis etc.).  Be sure to convert Hartrees to kJ/mol (or kcal/mol if you're still stuck in the 1970s).  Hartrees is unnecessary and inefficient when taking differences of energies (Hartrees are appropriate for absolute energies but absolute energies are completely inaccurate anyway).

For what it's worth, you may consider using SAPT to compute interaction/binding energies directly. It also allows for a physical decomposition of the nature of the interaction. PSI4 contains it, is free, and is apparently pretty fast. http://sirius.chem.vt.edu/psi4manual/4.0b5/sapt.html

@LordStryker ...I like kcal/mol :/

@jjgoings Cautionary note.  The SAPT routines for anything greater than a dimer interaction is broken and shouldn't be used.  3-body and above methods are utterly wrong (last I checked).

@LordStryker Oh, I thought the OP was talking about dimers.

@jjgoings He is, I'm just pointing that out for any future visitors.

No, not just dimers. Eventually I'm interested in binding energies of reactive dyes to cellulose, for example.

Atomdroid is using UFF. Would that be as reliable as SCF for this purpose?

@Sparkler - By dimer I mean a two-body system.  Each monomer in the dimer do not have to be identical.