can sodium persulfate etch or damage a titanium heater?

If the surface of the immersed element is all titanium, the answer appears to be that it won't get corroded unless the temperature is much higher than 50$^\circ$C; see http://www.azom.com/article.aspx?ArticleID=1336

The case is changed if there are joints or parts made of something else, as in this illustration: http://www.saltwatereddies.com/content-popup_image/pID-2199/popup_image.html

I would think that would not be so safe.

Of course, if you've been breaking glass heating elements right and left, you have little to lose by trying the titanium one...


What is the most "important" resonance structure of SCN⁻?

I performed a quick calculation on the DF-BP86/def2-SVP level of theory and analysed it with Natural Resonance Theory (from the Natural Bond Orbital Theory).
This results in the following major contributors to this wave function:
$$\left[
\underset{(1)}{\overset{67.49\%}{\ce{^-S-C#N}}} \ce{<->}
\underset{(2)}{\overset{21.25\%}{\ce{S=C=N^-}}}
\right]$$

The third contributor is a weird structure with a 'long-distance-bond' between the sulfur and the nitrogen with $7.22\%$. All other contribution are neglected/discarded.

While your structure 3 is indeed a valid contributor, its actual contribution will be very small. Forcing the program to use it as a structure, it resulted in an error, as it was unable to match the orbitals to that structure. The reason for that is likely that the overlap between the sulfur and the carbon is too poor to actually be considered a good contributor. This would also explain the smaller contribution of the second structure.

Here are the localised (according to NBO) molecular orbitals:

(Colour code: blue/orange - occupied molecular orbital [Lewis];
red/yellow - virtual molecular orbital [unorccupied, non-Lewis])

The total contributions in terms of atomic orbitals to the above:

     (Occupancy)   Bond orbital / Coefficients / Hybrids
 ------------------ Lewis ------------------------------------------------------
   8. (1.98209) LP ( 1) S  1            s( 79.86%)p 0.25( 20.13%)d 0.00(  0.01%)
   9. (1.77474) LP ( 2) S  1            s(  0.00%)p 1.00( 99.91%)d 0.00(  0.09%)
  10. (1.77474) LP ( 3) S  1            s(  0.00%)p 1.00( 99.91%)d 0.00(  0.09%)
  11. (1.96342) LP ( 1) N  3            s( 52.42%)p 0.91( 47.54%)d 0.00(  0.04%)
  12. (1.99743) BD ( 1) S  1- C  2
               ( 44.86%)   0.6698* S  1 s( 20.64%)p 3.81( 78.61%)d 0.04(  0.75%)
               ( 55.14%)   0.7425* C  2 s( 51.41%)p 0.94( 48.47%)d 0.00(  0.12%)
  13. (1.99846) BD ( 1) C  2- N  3
               ( 41.02%)   0.6404* C  2 s( 48.44%)p 1.06( 51.48%)d 0.00(  0.08%)
               ( 58.98%)   0.7680* N  3 s( 48.02%)p 1.08( 51.75%)d 0.00(  0.23%)
  14. (1.99735) BD ( 2) C  2- N  3
               ( 43.98%)   0.6632* C  2 s(  0.00%)p 1.00( 99.91%)d 0.00(  0.09%)
               ( 56.02%)   0.7485* N  3 s(  0.00%)p 1.00( 99.80%)d 0.00(  0.20%)
  15. (1.99735) BD ( 3) C  2- N  3
               ( 43.98%)   0.6632* C  2 s(  0.00%)p 1.00( 99.91%)d 0.00(  0.09%)
               ( 56.02%)   0.7485* N  3 s(  0.00%)p 1.00( 99.80%)d 0.00(  0.20%)
 ---------------- non-Lewis ----------------------------------------------------
  16. (0.01904) BD*( 1) S  1- C  2
               ( 55.14%)   0.7425* S  1 s( 20.64%)p 3.81( 78.61%)d 0.04(  0.75%)
               ( 44.86%)  -0.6698* C  2 s( 51.41%)p 0.94( 48.47%)d 0.00(  0.12%)
  17. (0.01384) BD*( 1) C  2- N  3
               ( 58.98%)   0.7680* C  2 s( 48.44%)p 1.06( 51.48%)d 0.00(  0.08%)
               ( 41.02%)  -0.6404* N  3 s( 48.02%)p 1.08( 51.75%)d 0.00(  0.23%)
  18. (0.22011) BD*( 2) C  2- N  3
               ( 56.02%)   0.7485* C  2 s(  0.00%)p 1.00( 99.91%)d 0.00(  0.09%)
               ( 43.98%)  -0.6632* N  3 s(  0.00%)p 1.00( 99.80%)d 0.00(  0.20%)
  19. (0.22011) BD*( 3) C  2- N  3
               ( 56.02%)   0.7485* C  2 s(  0.00%)p 1.00( 99.91%)d 0.00(  0.09%)

               ( 43.98%)  -0.6632* N  3 s(  0.00%)p 1.00( 99.80%)d 0.00(  0.20%)

On the terminology.
Alchimista already explained most of this, however, I cannot stress enough: There is no such thing as a most stable resonance structure. Therefore when you say common, you probably mean large contribution to the wave function, and when you say rare, you probably mean little contribution. None of the resonance structures can be independent from each other, as they are all hypothetical.

Please read more about that here: What is resonance, and are resonance structures real?


How can I find the emf of a cell given the individual reduction and oxidation reactions and their respective potentials?

This is my first answer in stack exchange if some mistake comment on.
$$E^\circ_\ce{Fe^3+/Fe} \neq E^\circ_\ce{Fe^3+/Fe^2+} + E^\circ_\ce{Fe^2+/Fe}$$

but

$$\Delta_r G^\circ_1 = \Delta_r G^\circ_2 + \Delta_r G^\circ_3$$

for the three reactions:

$$\ce{Fe^3+ + 3/2 H2 -> Fe + 3H+}\tag{1}$$

$$\ce{Fe^3+ + 1/2 H2 -> Fe^2+ + H+}\tag{2}$$

$$\ce{Fe^2+ +  H2 -> Fe + 2H+}\tag{3}$$

With $$\Delta_r G^\circ = - n F E^\circ , $$ we can establish a relationship between the reduction potentials:

$$n_1 F E^\circ_1 = n_2 F E^\circ_2 + n_3 F E^\circ_3$$

Cancelling F and solving for $E^\circ_1$ allows us to calculate the reduction potential of the iron/iron(III) half reaction:

$$n_1 E^\circ_1 = n_2 E^\circ_2 + n_3 E^\circ_3$$

$$E^\circ_1 = \frac{n_2 E^\circ_2 + n_3 E^\circ_3}{n_1}$$

$$ = \frac{1 \cdot \pu{0.77 V} + 2 \cdot (\pu{-0.44 V})}{3}$$

$$ = \pu{-0.367 V}$$


Ivan Neretin is correct in stating that no particle in a (nondegenerate) state can have a permanent electric dipole moment in the laboratory frame. In fact, a permanent dipole moment would require a violation of time (T) and parity (P) symmetry. This is because any such dipole should be directed along the angular momentum vector of the molecule and time-reversal and parity operators have a different effect on the rotational angular momentum and direction of the dipole.

What chemists call a permanent dipole moment refers to the dipole moment in the molecular point group to which the molecule belongs. Only when a polar molecule is subjected to an electric field (e.g. an EM wave or DC field), this field will mix rotational states of opposite parity in such a way that the molecule "orients" itself in the field. You can only speak of the inversion of ammonia in the presence of an electric field (the eigenstates of the Hamiltonian are time independent in the absence of a field or other perturbing force).

Although degenerate states could in principle have a permanent dipole moment, it was shown by Klemperer et al. (J. Phys. Chem. 1993,97, 2413) that the degeneracy of these states is removed by higher-order terms in the Hamiltonian.


For Hess' law type questions, when do you take the inverse of the constant and when do you switch the sign?

Equilibrium constant

When a reaction is at equilibrium, the species will have concentrations called the equilibrium concentrations. They do not depend on whether the reaction is written in one direction or in the reverse direction, and whether all the stoichiometric coefficients are multiplied by the same value. However, your equilibrium expression will be different (switching concentrations from numerator to denominator if reversing the chemical equation, or changing exponents when changing stoichiometric coefficients). As a result, the value of the equilibrium constant will change while the equilibrium concentrations themselves will remain the same.

Enthalpy of reaction

For a given reaction, the enthalpy change associated with making - say - 1 mol of species A is a given. If you use 1 mol of species A in the reaction instead, the enthalpy change will be of the same magnitude but opposite sign. If I make a different amount of species A, the enthalpy change will change in a manner proportional to that amount. In a similar manner, if I change the chemical equation for $\ce{A -> B}$ to $\ce{2A -> 2B}$, the molar enthalpy of reaction will double as well. The same goes for Gibbs energy or for entropy of reaction.

Cell potential

For a given reaction with certain concentrations of species in the two half cells of a voltaic cell, I will measure a certain cell potential. If I switch the role of reactants and products (in the equation), I will still measure the same potential (the direction of the actual reaction does not change). If I multiply all coefficients in the chemical equation by the same number, it still correctly describes the reaction, but this will not change the potential.

Mathematical relationships

$$\Delta G^\circ = - R T \ln(K)$$

The way logarithms work, if you multiply $\Delta G^\circ$ by a number, you have to raise K to the power of that number. If you add Gibbs energies, you have to multiply equilibrium constants.

$$\Delta G^\circ = - z F E^\circ $$

If you multiply $\Delta G^\circ$ by a number, z (the number of electrons transferred) will also increase by that number. Therefore, $E^\circ $ does not change.

Multiplying all coefficient in a chemical equation by (-1) changes the role of reactants and products. Thus, once you know the rules for changing coefficients by a common factor, you don't need to memorize another rule for reversing the equation.


Formation of (Gamma) Methylparaconic acid

The mechanism looks OK. Though catalytic NaOEt would be a better base than NaOH as the EtO- is regenerated as the reaction proceeds.

The opening of the lactone under basic conditions is reversible but the reformation of the lactone is favoured. A mild basic hydrolysis - 1eq LiOH in 19:1 THF/H2O - should give you your target.

Self condensation is always a possibility with acetaldehyde - this is why few reactions go with 100% yield in real life


Calculating pH of diprotic and amphoteric solutions

I'll address your questions in 3 parts:

(A) Why amphoteric

(B) How to solve the problem without making unexplained assumptions

(C) Why the solution in the book uses the approximate equality
    $\ce{[H3AsO4]=[HAsO4^{2-}]}$.

PART (A)

An amphoteric species (molecule or ion) is one that can either gain a proton with relative ease, or lose a proton with relative ease. "Relative ease" means, for example, that it can happen in aqueous solution. If your species will gain a proton in the presence of enough $\ce{H3O+}$, and if it will lose a proton in the present of enough $\ce{OH-}$, it can certainly be called amphoteric.

Translated into equilibrium constants, if your species has a pKa between 0 and 14, and its conjugate acid (what you get when you protonate it) also has a (necessarily lower) $pK_a$ between 0 and 14, your species will act in aqueous solution as either as an acid (losing a proton in the presence of strong base) or as a base (gaining a proton in the presence of strong acid).

The $pK_a$ of (neutral) arsenic acid is between 2 and 3 (because the equilibrium constant for removing the first proton, $K_1$, is between $10^{-2}$ and $10^{-3}$). The $pK_a$ of the version that lost one proton, $\ce{H2AsO4-}$, is around 7 (from the value of $K_2$). So clearly $\ce{H2AsO4-}$ will gain a proton if the pH is below 1, say, and it will lose a proton if the pH is above 10. This justifies why it's amphoteric.

PART (B)

In a monoprotic acid like HF, all the F appears either as F$^-$ or as HF (two species). You can write down 1 equilibrium equation, a mass balance for F and a charge balance, and solve all three equations simultaneously; when you combine the three you get a quadratic equation that can be solved exactly.

But with a multiprotic acid, things get complicated fast. In this problem you have three equilibrium equations, plus the mass-balance and charge-balance equations.  That's 5 equations in 5 variables.

Fortunately it's possible to combine 4 of them easily, using only high-school algebra. They are the 3 equilibrium equations and the charge balance. This reduces the problem to two variables (the pH and the concentration of one of the arsenic ions) in two equations.

Let's write and combine the 4 equations. There are four arsenic species with varying charges; I'll call their concentrations $A_0,A_1,A_2,A_3$ -- the index denotes the charge (which is 3 minus the number of hydrogens). So $A_0=[\ce{H3AsO4}]$, $A_1=[\ce{H2AsO4-}]$, and so on.

By definition, $A_1[\ce{H^+}]/A_0=K_1$, or equivalently $A_0=A_1[\ce{H^+}]/K_1$. This relates two concentrations, $A_0$ and $A_1$. Similarly $A_1=A_2[\ce{H^+}]/K_2$ and $A_2=A_3[\ce{H^+}]/K_3$.

The important thing is that all these four concentrations are proportional and given one you can find all others if you know the pH. (We don't know the pH, but I'll address that in a moment.)

The charge balance (expressed in moles present in 1 L of solution) is simply
\begin{equation} 0\,A_0+ 1\,A_1+ 2\,A_2+ 3\,A_3+K_w/[\ce{H^+}]= [\ce{H^+}]+0.1.
\end{equation}
On the left we have the negative charges, on the right the positive ones. Notes:

The concentration of each arsenic species is multiplied by its charge, obviously.
On the left side, $K_w=10^{-14}$ is the dissociation product of water, and $K_w/[\ce{H^+}]$ represents the concentration of OH$^-$; it will turn out to be negligible, but we don't know that a priori, and including it doesn't make the problem any harder.
The 0.1 on the right comes from Na$^+$ ions; being an alkali metal, sodium does not exist in water solution except in ionic form, so we know [Na$^+$] exactly. This explains something that will be important later: The 0.10 M solution of $\ce{NaH2AsO4}$ is exactly what you obtain by dissolving 0.10 moles of NaOH in one beaker, 0.10 moles of $\ce{H3AsO4}$ in another, and mixing the two, assuming the combined volume is 1 L.

Now we express the left side of (1) in terms of just two
concentrations, $A_3$ and $\ce{[H^+]}$: $$
\begin{aligned}A_1+2\,A_2+3\,A_3&=(\ce{[H^+]}/K_1)\,A_2+2\,A_2+3\,A_3\\
&=(\ce{[H^+]}/K_1+2)\,A_2+3\,A_3\\
&=(\ce{[H^+]}/K_1+2)(\ce{[H^+]}/K_2)\,A_3+3\,A_3\\
&= \bigl((\ce{[H^+]}/K_1+2)(\ce{[H^+]}/K_2)+3\bigr)\,A_3
. \end{aligned} $$

The last line is the key; it allows us to relate the concentration of one arsenic
species to the pH.  With this (and moving the $K_w/\ce{[H+]}$ to the
right)
the charge balance becomes
$$ \bigl((\ce{[H^+]}/K_1+2)(\ce{[H^+]}/K_2)+3\bigr)\,A_3
=\ce{[H^+]}+0.1 -K_w/\ce{[H+]}$$
or again
$$ A_3=
=\frac{\ce{[H^+]}+0.1 -K_w/\ce{[H+]}}{\bigl((\ce{[H^+]}/K_1+2)(\ce{[H^+]}/K_2)+3\bigr).}$$

This look complicated, but it's easy to put into a spreadsheet. The last step is this: The right [H$^+$] is the one that, with $A_3$ calculated as above, obeys the mass balance, $$ A_3 + A_2 + A_1 + A_0 =0.1.$$

At this point we could declare the multiple-choice problem solved, by cheating a bit: take the various proposed values of  [H$^+$] and see which one works.  For answer (a) the value of the sum $ A_3 + A_2 + A_1 + A_0$ is 0.62; for (b) it is  0.128; for (c) it is 0.10165, and for (d) it is 0.100002.  So the best answer is (d).

What if you didn't have multiple choice? Then you'd want to draw a graph or a table, taking (say) every integer value of pH and connecting them.  Here is such a graph:

The horizontal axis is pH and the vertical one is total arsenic concentration.  The correct pH is where the vertical coordinate is exactly 0.1, and you can get a good idea of its value by eyeballing, or solve by successive approximations.

Note that this method allows you to solve with no more effort the following problem: what is the pH when you add 0.1 mol/L of $\ce{NaH2AsO4}$ (as in the original problem), plus 0.05 mol/L of $\ce{H3AsO4}$?  The only difference is that now you'll be looking for the point on the graph where the vertical coordinate is 0.1+0.05 = 0.15, instead of 0.1.

PART (C)

Why does the book say that $\ce{[H3AsO4]=[HAsO4^{2-}]}$, approximately? And is this a good approximation?

The equality comes from the claim that the "dominant reactivity" of $\ce{[H2AsO4-]}$ is $$\ce{2\,H2AsO4- \leftrightharpoons H3AsO4{}+HAsO4^{2-}}.$$ If $\ce{H2AsO4-}$ exchanges protons only with itself, and not with, say, water, then every proton lost by one $\ce{H2AsO4-}$ ion means a proton gained by another $\ce{H2AsO4-}$, and the result is that $\ce{[H3AsO4]}$ and $\ce{[HAsO4^{2-}]}$ are generated in equal amounts.

So given the claim about "dominant reactivity", the equality is true, and by the way, the equality implies that the pH is the average of $pK_1$ and $pK_2$. (This follows, by taking logs, from the analysis that I'm guessing your book gives right after the part of the solution you quoted.)

The problem with the claim is that it does not follow from the amphoteric nature of $\ce{H2AsO4^-}$ alone. It requires two additional conditions: You're starting from only $\ce{H2AsO4-}$, and in relatively high concentrations. I'll illustrate why these conditions are necessary:

(1) You have to start with only $\ce{H2AsO4-}$ (plus a counterion that doesn't participate further, like Na$^+$). If you started with some $\ce{H3AsO4}$ in addition, you wouldn't have the equality $\ce{[H3AsO4]=[HAsO4^{2-}]}$, even approximately. Or, to use the example you gave, if you start with H$_2$S the amphoteric character of HS$^-$ wont help you calculate the pH of the resulting mix. On the other hand if you started by dissolving NaHS alone, it would.

(2) You have to start with a strong solution. If the original question had specified a 0.1 mM solution instead of a 0.1 M solution, the resulting H$^+$ concentration would not be $2.5\times 10^{-5}$ anymore but $3.3\times 10^{-6}$, and the equality $\ce{[H3AsO4]=[HAsO4^{2-}]}$ would be off by a factor of 60!

I hope this helps answer your very interesting and perceptive questions.  Don't hesitate to ask for further clarification if necessary!


Any home made tricks to fill up a baloon that floats?

There are some options, not all easily available or safe: Helium is the most used and it most certainly makes them float, being safe at the safe time. However, it's not an home-made trick. Methane is more easily available (in many cities, methane is piped into homes for domestic heating and cooking purposes) but it's very unsafe for such an use, as it could explode. Hot air is also used (think of hot-air balloons) but I wouldn't know how to make it last for a long time as the air would cool down to room temperature very quickly. Lastly, there's hydrogen which could also be produced with some reagents available at home, but it's as dangerous as methane.


I think what you may find most helpful is to know a bit of the history of element discovery and atomic theory.

The first pure substance containing only the element oxygen to be isolated was dioxygen ($\ce{O2}$), in 1774, though it was called "dephlogisticated air" until 1777 when Lavoisier used the term "oxygen" for the first time. This was some 30 years before John Dalton even proposed the first empirical atomic theory. Back then, we only barely had an understanding of stoichiometry, such that Dalton famously claimed the molecular formula for water was $\ce{HO}$. The fact that dioxygen is a substance made of molecules containing two atoms of oxygen probably wasn't widespread knowledge until at least 1811, with the gas stoichiometry experiments of Amadeo Avogadro.

Basically, for a point in time, we knew that there was a substance composed of a single type of atom, which could not be broken down into anything simpler. This fit the then-prevalent definition of an element; "a pure substance that could not be decomposed into any simpler substance". We knew that Lavoisier's "oxygen" had to be $\ce{O_n}$, for some n, but we had no reason to assume $n \neq 1$ for decades. By the time we figured out $n = 2$, the name "oxygen" was already widely used to refer to dioxygen. The fact that $n = 3$ also forms a stable compound in ambient conditions (ozone) would also not be known until 1867. A similar story happened with (di)nitrogen (octa)sulfur, (tetra)phosphorus, and so on. The only elements which form stable monoatomic substances in reasonable conditions are the noble gasses.

There is an interesting aspect to consider behind all this. There are some (such as Eric Scerri) who claim we are doing Chemistry a disservice in muddling together the properties of elements and the pure substances which they make. Nowadays our definition of an element is solely dependent on the number of protons inside an atomic nucleus, with no reference to reactivity or in what form the pure substance can be found. In this sense, the elements do not have "reactivities", "melting points", etc.; these are all properties of the pure substances. The only true properties of the elements are things such as electronic distribution, ionisation energies, and so on. However, it is common to see periodic tables stating the melting and boiling points the pure substances of each chemical element, and even Wikipedia bundles the physical properties of dioxygen with the atomic properties of elemental oxygen. For better or for worse, we're stuck with this subtle ambiguity in nomenclature.


What methods could be used for determining the concentration of Zn2+ in a solution?

Mass spectroscopy(MS)
HPLC-MS, if separation is required
Atomic emission spectroscopy(AES)
(Probably)
Polarography/voltammetry

The list is not exhaustive.


Peroxides contain two oxygens connected by a single bond. X-ray or neutron diffraction will show that the oxygens in $\ce{MnO2}$ are too far apart to be bonded, and therefore it is not a peroxide.

Peroxide is well known as a ligand to transition metals (though whether it is better described as peroxide or superoxide is sometimes not always clear, e.g. in some cobalt species). An example which you might come across in the lab is chromium(VI) oxide peroxide, $\ce{CrO5}$.


Conversion of grams to milliliters using kg/m3 density

Concerning the units in your equation none of them is correct (As all your formulations would end up in a square volume unit).

Definition of volumetric mass density

The density $ \rho $ is given as

$$ \rho = \frac{m}{V} $$

resulting in a mass and density dependent Volume as

$$ V = \frac{m}{\rho} $$

Direct way

Although you could calculate this by just substituting the physical quantities by the given values and units, and then take care of the unit conversion afterward:

$$ V = \frac{m}{\rho} = \frac{100\,\mathrm{g}}{681.9\,\mathrm{kg}\,\mathrm{m^{-3}}} =\\= \frac{100\,\mathrm{g}\,\mathrm{m^{3}}}{681.9\,\mathrm{kg}} = \frac{100\,\mathrm{g} \cdot \left(10^2\,\mathrm{cm}\right)^{3}}{681.9 \cdot (10^3\,\mathrm{g})} =\\= \frac{100 \cdot 10^6\,\mathrm{cm}^{3}}{681.9 \cdot 10^3} = \frac{100 \cdot 10^3}{681.9}\,\mathrm{cm}^{3} =\\\approx 146.65\,\mathrm{cm}^{3} \equiv 146.65\,\mathrm{ml} $$

I would suggest to convert the density from $ \mathrm{kg}\,\mathrm{m^{-3}}  $ to $ \mathrm{g}\,\mathrm{cm^{-3}} $ as a first step:

Easier way

Of course easiness is something subjective, but I believe this is easier to use.

Calculate conversion-factor for density

First the combined conversion-factor for the units-fraction is calculated

$$ \frac{\mathrm{kg}}{\mathrm{m}^3} \stackrel{\left[1\right]}{=} \frac{\left(10^3\,\mathrm{g}\right)}{\left(10^2\,\mathrm{cm}\right)^3} \stackrel{\left[2\right]}{=} \frac{10^3\,\mathrm{g}}{10^6\,\mathrm{cm}^3} \stackrel{\left[3\right]}{=} 10^{-3}\,\frac{\mathrm{g}}{\mathrm{cm}^3} $$

Start with substituting each units SI prefix:
Care has to be taken with the conversion of $ \mathrm{m}^3 $, since the power applies to the whole substituted expression ($ \mathrm{m}^3 = \left(10^2\,\mathrm{cm}\right)^3 $).
In the last step the exponential expressions are summarized ($ \frac{10^3}{10^6} = 10^{3-6} = 10^{-3} $).

Then calculate the new numeric value of the density in the target unit:

$$ \rho = 681.9\,\mathrm{kg}\,\mathrm{m^{-3}} \stackrel{\left[4\right]}{\equiv} 681.9 \cdot \left(10^{-3}\,\mathrm{g}\,\mathrm{cm}^{-3}\right) \stackrel{\left[5\right]}{=} 0.6819\,\mathrm{g}\,\mathrm{cm}^{-3} $$

Substitute the unit with the previously calculated expression
Multiply the numeric terms

Calculate the volume

Now the unit-adjusted density is used:

$$ V = \frac{m}{\rho} = \frac{100\,\mathrm{g}}{0.6819\,\mathrm{g}\,\mathrm{cm^{-3}}} = \frac{100}{0.6819}\,\mathrm{cm^{3}} \approx 146.65\,\mathrm{cm}^{3} \equiv 146.65\,\mathrm{ml}  $$

Conversion-factors

According to SI Prefixes the following applies:

$ 1\,\mathrm{m} \equiv 10^2\,\mathrm{cm} \\
1\,\mathrm{kg} \equiv 10^3\,\mathrm{g} $

And of course:

$ 1\,\mathrm{cm}^3 \equiv 1\,\mathrm{ml} $


How can an electron shield another electron of the same subshell?

Shielding is simply the cancellation of some of the effective charge of the nucleus.

It's a lot easier to envision an electron to shield another electron that, by time average, is likely to be farther from nucleus, e.g., $1s$ shielding $3p$.

But for 2 electrons in the same subshell, on average, you can expect one to be closer to the nucleus than the other roughly 50% of the time. While it's true that a good deal of the time that other electron may be on the other side of the nucleus, for d- and f- orbitals, we're looking at more than 1 electron, so there's some partial shielding. Partial, because the shielding electrons don't actually shield a good portion of the time due to relative distances from the nucleus.


How does calcium chloride act as a sequestrant (food science)?

Food Scientist here. Calcium chloride is typically used in high concentrations in pickles to keep them firm (give that desired bite) and also add a slight salty flavor without increasing sodium content. It is also used in potato chips to reduce formation of acrylamide. Calcium chloride is highly soluble in water and will dissociate into Ca2+ and 2Cl-. It is used to sequester metal cations. The metal ion complexes with the free Cl-. It has also been found that calcium ions may reduce iron absorption, and is a divalent cation that can form complexes of its own. Calcium chloride can stabilize colors, flavors, antioxidants, and also improve oxidative stability. Fats and oils readily oxidize in the presence of metal ions, so a sequestrant such as calcium chloride is used to form a complex with the metal. Additionally CaCl2 is used to stabilize gels such as those formed with alginate. Another popular sequestrant, or chelating agent is EDTA.


Solution, hardener and metal required to cast key duplicates

From the video, the material looks like some kind of silicone. There aren't many other materials that have that translucent elastomeric quality. If that's the case, the larger of the two mould solutions will be mostly some kind of silane precursor such as SiOH. The second part will contain the catalyst for the polymerization. Depending on the exact chemistry used it could be perhaps tin-based. My guess for the "hardener" is that it's simply more catalyst. Two part silicones with longer working lives are more common than very rapid cure ones so adding more/stronger catalyst as the third part may be just to make it cure quickly. Many silicones can withstand temperatures of 200+ °C.

As for the low melting point alloy, there are a number of things it could be. Given that the application is keymaking and people touch keys with their hands, I hope it's safe to assume that the alloy doesn't contain lead or cadmium. The video says its melting point is around 90 °C. The two closest common alloys without toxic metals are 58% Bi, 42% Sn @ 132 °C and 66.3% In and 33.7% Bi @ 72 °C, from the wikipedia page on fusible alloys, so it's likely some alloy with similar composition to one of these. Bismuth and tin are far cheaper than indium, so it wouldn't surprise me if the alloy was some mixture of those two elements.


It is a very good questions from a student who seems to be thinking deeply. I have been thinking to write on this topic for the last 7-8 years. In short the poles of the batteries are electrostatically charged and one can actually "sense" this charge with the help of sensitive electroscopes which were known in the time of Volta.

The label anode or cathode is not defined with respect to the electrostatic sign as explained in the post by Maurice.

Think of a large parallel capacitor. Connect one terminal of the plate to one positive terminal and the other one the negative terminal. The capacitor plates would show a electrostatic charge.

By all means, using one single cell you would not be able to detect the electrostatic charge, like a charged comb. Only specialized electroscopes can sense that.


Is it feasible to make dendritic crystals at home?

The single displacement reaction between silver nitrate and metallic copper will yield dendritic crystal growth of metallic silver. Normally, a copper wire is placed into silver nitrate solution. The reaction that occurs is:
$$2AgNO_3(aq) + Cu(s) \rightarrow Cu(NO_3)_2(aq) + Ag(s)$$


How is the electron "trapped in the crystal" in irradiated NaCl?

What you're describing is thermoluminescence, which is used as a paleochronology method (aka Trapped Electron (TE) dating, afaik). The wikipedia article has a good synopsis. Simply, electrons get bound to potential wells generated in the radiation-defected structure. When you heat the crystal up you allow it to anneal and return to a more regular structure. The defected crystal is stuck in an energetic local minimum.

You can also do this trick with X-rays. Calcite (or was it aragonite?) exposed to X-rays will turn light blue for a while.


Let's take a close look at what the reaction entails. You want to convert $\ce{CO_2}$ to $\ce{CH_4}$ using $\ce{H^+}$. It is possible that the hydrogen ion will co-ordinate with $\ce{CO_2}$ to produce $\ce{[CO_2H]^+}$.

Unfortunately it is very unlikely to have a second hydrogen ion attaching to the same oxygen to produce $\ce{[CO_2H_2]^{2+}}$. Now for the impossible step of water leaving and producing $\ce{[CO]^{2+}}$. This step would require a large amount of energy and $\ce{[CO]^{2+}}$ would be extremely reactive. Since it is highly charged, you would not have any hydrogen ions approaching the moiety.

At this point, any further progress towards $\ce{CH_4}$ is fantasy. To stabilize the reaction, you need to add electrons to the carbon (which you don't have a source).


As a high school student, here’s how I considered it:

The wavefunction of an orbital describes the region of space that an electron residing in that orbital would be in about 90% of the time.

The wavefunctions would have a radial component (depending on period), which is why the 3p orbitals Arena larger than the 2p etc.

The wavefunction would also have an angular component, which would describe its shape.

In the case of the p orbitals, the x, y and z labels simply means that their angular components are x/r, y/r and z/r respectively (hence the subscript). Do note that we usually use the radial forms of the wavefunctions though.

Hope this helps.


How to estimate the concentration of OH that comes from just the autoprolysis of water in a basic solution?

What auto-ionization means is best explained this way: If you put in  0.10000000 M of NaOH in pure water, how much hydroxide is there at equilibrium? You can imagine that "initially", adding that amount to a neutral solution gives you $\ce{OH^-}$ = 0.10000010, and that $\ce{[H^+]}$ is 0.00000010. However, the ion-product at this point is much greater than $K_w$, so some of the hydronium and hydroxide must react to bring the product down. If you solved the equation $(0.10000010-x)(0.00000010-x) = K_w$, and used $0.10000010-x$ to get $[OH^-]$, you would arrive at the concentration taking into account auto-ionization. You can see that $[OH^-]$ would just be a very minute amount greater than 0.100000000  because x is very small. The x is the result of taking into account the auto-ionization. As you can see, 0.1 dwarfes x, so we could have ignored auto-ionization and called it 0.1. Similarly, the hydronnium concentration would be different than if you had just divided $K_w$ by 0.1 rather than set up an equilibrium equation--but by so very little.

Now for your problem,

Let me first try to find the $K_b$ for THAM. If the pH of a 0.10 M solution is 10.5, then $\ce{[H^+]}$ $3.16*10^{-11}$. Using the mass balance for THAM, (I'll denote THAM as $\ce{B}$ and the conjugate acid as $\ce{BH^+}$),

$\ce{[B]}$ + $\ce{[BH^{+}]}$ = 0.10 M

$\ce{OH^-}$ = $3.16*10^{-4}$ (assuming 298 K) because  $\ce{H^+ }$ = $3.16*10^{-11}$

If you neglige the autodissociation of water, in other words, that all of the hydroxide in the solution is due to the THAM, your charge balance equation will be that $\ce{[BH^+]}$ is the same as $[OH^-]$ (by stoichiometry). Then, $\ce{[B]}$ = 0.09968, and $K_b$ is $1.00 * 10^{-6}$.

This should tie everything together:

Before, we had said that $[BH^+] = [OH^-]$. Instead, if you took into account the auto-ionization, you get the equation that $[BH^+] + [H^+ ] = [OH^-]$. Why is it justified to ignore $[H^+]$? Because the solution is so basic, $[H^+]$ is so small that if you tried to take auto-ionization into account, the pH would be roughly the same within quite a few decimal places. The change in $[OH^-]$ that comes about due to auto-ionization is dwarfed by the basicity of the THAM addition, just like the 0.100000 M NaOH example that I used. In other words, if you make the solution intensely basic, you can afford to ignore the effects of auto-ionizaton of water, and similarly for acids. In general, it is OK to ignore the auto-ionization process of water if the solution is more than just slightly acidic/basic.
As an interesting note, try to calculate the pH of a $10^{-10}$ M solution of hydrochloric acid without taking into account auto-ionization. You get a pH of 10, which is definitely not right.


Why is the carbon monoxide triple bond stronger than the nitrogen-nitrogen triple bond?

I agree in part with Mithoron and MaxW's comments: these are different molecules, not even sharing a mutual atom (e.g. C-O vs C-C), so direct comparison is restricted.

However, Pauling's concept of electronegativity does explain why, in general, the heteronuclear A-B bond is stronger than the average of the homonuclear A-A and B-B bonds. the difference in the electronegativity of the two atoms in the bond lend a somewhat ionic character to the bond, strengthening it.


Misleading features of Lewis Dot Representation

Lewis dot structures predate the advent of modern quantum theory and quantum chemistry. There is no Schroedinger equation associated with them -- in fact placing a dot on an atom (seems to me that it) violates the concept that electrons must be indistinguishable.

There are no slots, there are no "octet rules".  There are only particles, and kinetic and potential energy that depend on their masses and charges. They take a configuration that minimizes the total energy with respect to their location in space.

So my advice is to learn Lewis dot structures as a heuristic device that can be figured out for the purposes of passing General Chemistry and speaking with the vast majority of chemists who have continued to suffer with learning about octet rules, resonance structures, etc. and the "astonishment" when these rules fail. They are hard to learn, because one can not draw on experiences learned about other physical systems that depend on e.g. QM, electrodynamics, etc.

There is a subject known as Valence Bond theory, that uses (spin-adapted) Lewis structures as a basis. But if you knew what went into constructing a proper quantum mechanical valence bond wave function, you might as well say "sure, anything so long as it forms a complete basis."  i.e. we might as well talk about molecules being made up of millions of plane waves.

You did mention orbitals: "Orbitals" only arise due to the fact that they are complete bases of Spherical Harmonics and Laguerre Polynomials that were used by Schroedinger to solve the case of the hydrogen-like atom -- a spherically symmetrical system.  They do not really apply to molecules, and it is common to use other bases to describe molecules such as Gaussian-Type orbitals, plane waves, etc.  Atomic orbitals, molecular orbitals (constructed from linear combinations of AO's) and plane waves have never been observed.  Mathematically, orbitals may be complex, and leaves me wondering why we place so much stock in them except as useful complete bases with convenient boundary conditions, e.g. periodic with a unit cell or decaying in the limit of infinite distance.  (In contrast, the electron density is real and observable.)


Finding moles of different compounds using balancing

Never having heard of the Principle of Atom Conservation (POAC) before I'm curious.

Looking up the concept, I understand it simply to mean that atoms can't be created or destroyed in a chemical reaction. (So we're ignoring radioactive decay for example.) I start blindly with a variable for each reactant and product, and will then look for linear relationships in the variables.

$$\ce{$a$\,K2Cr2O7 + $b$\,H2SO4 + $c$\,SO2 -> $d$\,Cr2(SO4)3 + $e$\,K2SO4 + $f$\,H2O}$$

So with $a$ $\ce{K2Cr2O7}$ molecules, I must get $a$ $\ce{Cr2(SO4)3}$ molecules to balance $\ce{Cr}$, and $a$ $\ce{K2SO4}$ molecules to balance $\ce{K}$. So:

$$\ce{$a$\,K2Cr2O7 + $b$\,H2SO4 + $c$\,SO2 -> $a$\,Cr2(SO4)3 + $a$\,K2SO4 + $f$\,H2O}$$

Now $\ce{H}$ is only in $\ce{H2SO4}$ on the left and $\ce{H2O}$ on the right so $f=b$.

$$\ce{$a$\,K2Cr2O7 + $b$\,H2SO4 + $c$\,SO2 -> $a$\,Cr2(SO4)3 + $a$\,K2SO4 + $b$\,H2O}$$

To balance $\ce{S}$:

  $$b + c = 4a$$

To balance $\ce{O}$:

  $$7a + 4b + 2c = 16a + b$$ which reduces to $$3b +2c = 9a$$

At this point just balancing atoms leaves you with 2 linear equations and 3 unknowns. So there are not enough equations to explicitly solve the problem using linear equations. However relationships can be established.

\begin{array}{cccc}\\
3b& +2c & = & 9a \\
-2(b& +c & = & 4a)\\ \hline
 & b & = &a \\
\end{array}

$\therefore c = 3a = 3b$

The final step in this problem is to realize that an integer solution (a Diophantine problem in mathematics) is desired. So $a=b=1$ and thus $c=3$. So the final chemical equation is:

$$\ce{K2Cr2O7 + H2SO4 + 3SO2 ->Cr2(SO4)3 + K2SO4 + H2O}$$

PS - Many thanks to user Mathew Mahindaratne for suggesting a cleaner solution in the comments.


Is it possible to blow out a candle flame using pure oxygen?

It depends on the wick.   A flammable wick, exposed to oxygen, will burn, but
a non-flammable wick, saturated with liquid wax, could be chilled by the gas flow,
and when the evaporated wax is blown clear of the wick, the candle flame
could extinguish for lack of fuel.   Basically, the oxygen flow might
(at sufficiently high velocity) remove the vapor fuel so that the flame
becomes a detached fireball, and the wick and wax of the candle cool.
The oxygen flow just has to be faster than the flame propogation in the
fuel-oxygen mixture.

A solid flammable wick, though, won't be displaced, just burn faster,
and on discontinuing the oxygen, the candle will still be lit.


Difference between Excess property and property changes of mixing

The difference is subtle, however they are different as you suspect.

You can look at the property of mixing as the change in property resulting from mixing the pure components (with everything at the same temperature and pressure) together to form the mixture.

The excess property is deviation between the actual property and the property in an ideal mixture at the same temperature and pressure.

They are the same for volume and enthalpy because the volume of mixing and the enthalpy of mixing in an ideal solution are both 0. This is not the case for G or S.

In other words the difference arises because $M^{id}=\Sigma x_i M_i$ is not true for all properties. It is only true for enthalpy and volume. So the second term in your expression for the excess property above, cannot in general be written as $M^{id}=\Sigma x_i M_i$ (if it could, the excess property and mixture property would be the same). It should instead be $M^{id}=\Sigma x_i M_i + M_{mix}^{id}$.

The extra term for mixing the components would be be given by $$
\frac{G^{id}_{mix}}{RT} = \Sigma x_iln(x_i)
$$
and
$$
\frac{S^{id}_{mix}}{R} = -\Sigma x_iln(x_i)
$$
for Gibbs free energy and entropy, respectively.


sp, sp2 and sp3 hybrid orbitals are actual solutions of Schrodinger's equation or only a heuristic?

As soon as you move to systems with more than one electron them all orbitals are approximations (i.e. not real). Hybridized orbitals are no more "heuristic" than atomic orbitals.

The transformation between hybrid and atomic orbitals can be mathematically, though not uniquely, defined (e.g. as the minimization of inter-orbital electrostatic repulsion) and the transformation does not affect the total electron density.

See for example http://molecularmodelingbasics.blogspot.dk/2010/03/canonical-and-localized-molecular.html and http://molecularmodelingbasics.blogspot.dk/2010/04/double-bonds-banana-and-otherwise.html


Should one pronounce “periodic” the same in “periodic acid” and “periodic table”?

I am an Australian English speaker and yes, this distinction is made in practice.

Possibly the only reasonable opportunity to use the pɪə pronunciation in the name of a compound is in the case of the entertaining molecule periodane, which is actually named after the periodic table. This molecule (and later a number of different plausible isomers) was identified computationally by a methodology called 'mindless chemistry' which optimises randomly generated molecular graphs. Periodane is a stable configuration of each atom on the second row of the periodic table, with the exception of neon (although some people are working on that).


There is a manual available that states (page 111).

  TranSiesta
  defines the Left Electrode to be the first atoms specified in the SR
  .fdf
  file,
  and the Right Electrode to be the last ones.  The transport direction has to be considered
  to be the the third cartesian axis, the
  z
  axis.  The Left Electrode atoms must have smaller
  z
  components than the Right Electrode atoms.  It is also crucial that the atomic positions
  specified at the left (right) EL calculation must be equivalent to the left (right) electrode
  part of the SR setup.  Here, equivalent means that they can be made equal by a simple
  translation  in  space.   It  is  also  possible  to  use  buffer  atoms.

There is more information that I think may address further doubts in the manual as well. I am not very familiar with the program. So, this is as much as I can help you. You could try asking the mailing list, though SIESTA's mailing list does not get many responses to be honest.


The decarboxylation of phenylacetic acid via addition of acid

The decarboxylation of carboxylic acids having allylic or benzylic carbons can also be explained by what may be called the vinylog concept. One of the most common decarboxylations is that of β-keto acids as illustrated by the decarboxylation of β-keto acid 1 to afford cyclohexanone 2. Hagemann's ester (3) is a γ-keto ester, which upon aqueous acid hydrolysis gives acid 4. This acid suffers decarboxylation to afford 3-methylcyclohex-2-en-1-one (5). The red double bond in 4 simply extends the properties of a β-keto ester to the γ-position in a vinylogous (allylic?) fashion.
Nitrobenzene 6, which may be the compound in question (if not, ignore the methyl groups), is doubly vinylic (benzylic?) as indicated by the two red double bonds in structure 6. While this compound does not have a keto group in conjugation with the aromatic ring, the nitro group functions as a viable substitute for a keto group. Protonation of the N=O bond on oxygen leads to decarboxylation and rearomatization to nitrobenzene 7.
The green σ-bonds in structures 1, 4 and 6 must have overlap with the π-framework of the carbonyl (in 1), the γ-position of enone 4 and the aromatic ring of nitrobenzene 6 for decarboxylation to occur.
Finally, nitroacetic acid, as its magnesium salt, decarboxylates at 80 oC to provide nitromethane and CO2.


Can covalently bonded hydrocarbons be influenced by a magnetic field?

Just because there are no unpaired electrons doesn't mean magnetic fields can't influence something. Unpaired electrons tend to lead to strong interactions with magnetic fields (the dioxygen in air has two and is paramagnetic, an effect best seen in the liquid phase where a magnet will attract the cold liquid).

But compounds with paired electrons also interact with magnetic fields, though the effects are usually weak. But the repulsive effect is sometimes strong enough to levitate a diamagnetic substance (see this youtube video showing how strong magnets can levitate plates of graphite which is strongly paramagnetic).

It is unlikely that such effects would be easy to see in gases. But you are talking about combustion and the most likely effect will not be on the neutral gas but on the flame where the gas is not neutral but a plasma of charged particles. Charged particles or molecules will have a much stronger interaction with a magnetic field though it might still require sensitive apparatus to detect how the field makes a difference.


Work done in reversible isothermal expansion

I agree with getafix, if you would like an answer that is more tailored to you, you should show us exactly what you've done.

However, I am going to make a (hopefully educated) guess that what you did was to pull $p_\mathrm{ext}$ out of the integral. That is incorrect, because $p_\mathrm{ext}$ is not a constant here.

This process is known as an isothermal expansion - isothermal because the temperature remains constant throughout - and expansion because volume is increased. In thermodynamics it is very important to note which variables are held constant, because then that lets you decide which formula is appropriate to use, or how to derive such formulae).

Since the process is reversible, the external pressure must always be equal to the pressure exerted by the gas, which can be calculated via the ideal gas law $pV = nRT$. Therefore, you have (where 1 and 2 denote the initial and final state respectively)

$$\begin{align}
w &= -\int_1^2 p\,\mathrm{d}V \\
&= -\int_1^2 \frac{nRT}{V}\,\mathrm{d}V
\end{align}$$

and now since $T$ is a constant, you can take it out of the integral (along with $n$ and $R$ which are also constants)

$$\begin{align}
w = &= -nRT\int_1^2 \frac{1}{V}\,\mathrm{d}V \\
&= -nRT\ln\left(\frac{V_2}{V_1}\right)
\end{align}$$

Since you weren't given numerical values for $V_2$ and $V_1$, you can't evaluate a numerical value for $w$. The closest you can get is therefore

$$w = (-24.9~\mathrm{kJ})\ln\left(\frac{V_2}{V_1}\right)$$


Count carbon and oxygen atoms in the book's answer versus the original molecule, and infer that the book got it wrong.  One of the carbon atoms in the book's answer should be oxygen.  Question for you:  Which one?


How to convert P2O5 concentration to H3PO4 concentration?

Your question is a valid analytical chemistry question which has deep roots in history. No, the reason is not that $\ce{P2O5}$ reacts with water to from phosphoric acid and that is why $\ce{P2O5}$ values are quoted as suggested in the comments below the question. This is a very old historical tradition. In older analytical chemistry, chemists had only two main tools to analyse something namely, titration and gravimetry. Oxygen was also used to determine atomic weights, and people were interested in the ratios of element:oxygen ratio. One person even got a Nobel Prize as well. Generally, when the precipitates were heated, what was left was oxides. In really old books (at least 100 year old) you will notice that many analyses were quoted as oxides. For example, this is true in fertilizer industry to quote potassium as $\ce{K2O}$ and so on. Now if you strongly heat a precipitate phosphate, one will not get a $\ce{P2O5}$ but the convention to report elements as oxides continues.

Second part: I will let you solve the rest of the problem: Start from here

1 mol $\ce{P2O5}$ = 2 mol $\ce{H3PO4}$

and do wt percentage conversions to moles. A 70% $\ce{P2O5}$ solution in water means 70 g $\ce{P2O5}$ is present in 100 grams of solution. Do mol conversions. You will also need the density of the solution.


Antioxidants are molecules that help pair up electrons in the cell.

The way I see it, antioxidation in the body is all about getting electrons to pair up. dioxygen, $\ce{O2}$ is a molecule with an overall even number of electrons (12), but the two highest energy electrons are in different molecular orbitals, so it is actually a diradical. Species with unpaired electrons, like ground state oxygen, cannot react with the body's even electron count organic molecules. This is why the body must employ metals, which have unpaired electrons in order to reduce oxygen to water to form cellular energy in the mitochondria. On its way to being reduced to water, oxygen can form reactive species which can react with organic biomolecules, like $\ce{O2-}$, $\ce{HO-OH}$, and $\ce{HO.}$ These are known as Reactive Oxygen Species (ROS).

ROS cause oxidative stress, which is a process that damages important biomolecules like lipids and nucleic acids through oxidation. Unstable ROS like peroxides can initiate radical chain reactions, leading to oxidation of important molecules in the cell. Two good examples of ROS damage are lipid peroxidation, which damages the cellular membrane and nucleobase oxidation, which leads to DNA damage. I will show mechanistic examples for both:

Lipid peroxidation:

$\hskip1.8in$ (1)

formation of 8-oxoguanine:

$\hskip0.9in$ (2)

This creates a hydrogen bonding situation where guanine can match with adenine instead of cysteine:

$\hskip1.4in$ (3)

This is an example of base pair mismatch, a common form of DNA damage, potentially leading to genetic mutations.

Because of these possibilities, it is very important for the body to have safety mechanisms to protect itself against these (and other) forms of oxidative damage. There will always be some concentration of ROS in the body's cells, but the body's anti-oxidation systems should help minimize this.

"Antioxidant" is a broad class of chemicals, including both small molecules and peptides. The general idea is that they are molecules that prevent or reverse oxidation of important biomolecules, or that terminate radical chain reactions. In the lipid peroxidation reaction shown above, the final step is propagation of the radical chain reaction. Since radicals always form new radicals when reacting with non-radicals, the chain reaction will only terminate in one of two ways: either the ROS concentration gets soo high that two radicals bind (very bad), or an antioxidant like vitamin E or Vitamin C will "quench" the chain reaction. Vitamin E will "scavenge" ROS in a process coupled with reduction by ascorbate (vitamin C anion):

$\hskip1.6in$ (4)

Since vitamin E is fat soluble, it can help to protect the lipid bilayer. The semidehydroascorbyl radical that's produced is fairly stable, with a half-life of $10^-5$ s. Now, we have a less reactive radical. Processes like this are known as radical "quenching". two molecules of semidehydroascorbyl radical will quickly disproportionate to dehydroascorbate and ascorbate. This is a process where two odd numbers of electrons turn into two even numbers. 1 + 1 -> 0 + 2

Another sort of antioxidant is the sort that dimerizes. Because one electron needs to be donated to each radical to form a non-radical, one way of removing them is to convert two radicals and two non-radicals into three non-radicals. This is a process where two odd and two even numbers of electrons turn into three even numbers. 1 + 1 + 2 + 2 -> 2 + 2 + 2

Glutathione is probably the most important antioxidant in the body's anti-ROS arsenal. Glutathione can convert a molecule of of $\ce{HO-OH}$ to water in a processes catalyzed by the enzyme glutathione peroxidase. A peroxide is not a radical, but it's a high energy ROS which can easily form two $\ce{HO.}$. The overall equation for the reaction is as follows:

$$\ce{2GSH + H2O2 → GS–SG + 2H2O}$$

Glutathione can also turn the dehydroascorbate we mentioned previously back into ascorbate (vitamin C) through catalysis by the enzyme dehydroascorbate reductase. All oxidized (dimerized to GS-SG) glutathione is finally converted back to reduced glutathione by glutathione reductase.

Depending on your level of knowledge of chemistry and biochemistry, some of the specifics here may be difficult to understand, but the key is simple: The commonality between all antioxidants is that they are able to convert odd numbers of electrons into even numbers of electrons in processes like these:
1 + 1 -> 0 + 2 or 1 + 1 + 2 + 2 -> 2 + 2 + 2 and probably many others.

Antioxidants are just molecules that help pair up electrons.

ROS comes into play mostly as exhaust from the mitochondria, the location where oxygen is turned into water. Antioxidants take care of it, and many of them are regulated through the action of glutathione, which is dimerized on oxidation, then split apart again during enzymatic reduction (along with the cofactor NADPH). Through a system of enzymes and coupled reactions, the body tries to maintain a low ROS concentration.

(1) Wikipedia: https://en.wikipedia.org/wiki/Antioxidant#/media/File:Lipid_peroxidation.svg

(2) Jena, N. R.; Mishra, P. C. "Mechanisms of Formation of 8-Oxoguanine Due to Reactions of One and Two OH. Radicals and the H2O2 molecule with Guanine: A Quantum Computational Study" J. Phys. Chem. B, 2005, 109, 14205-14218

(3) Wikipedia: https://en.wikipedia.org/wiki/8-Oxoguanine#/media/File:8-oxoG_forming_Hoogsten_base_pair_with_dA.svg

(4)Valko, M.; Izakovic, M.; Mazure, M.; Rhodes, C. J.; Telser, J. "Role of oxygen radicals in DNA damage and cancer incidence" Faculty of food technology, Slovak Technical University, 2004


Name of round piece of glassware with thin neck?

We used to have glassware like that in our lab.  It is basically an "unfinished" piece of glassware. In a nutshell, it was used to handle (collect, transport) gases.

It was easy and inexpensive for the glassblower to make several of those bulbs at a time; he'd make a variety of sizes.  We'd keep them in a drawer and then before we used them we'd do some glassblowing ourselves and add the expensive part - a stopcock with a ground glass joint.  Then, having already measured the volume, we could attach one to a vacuum line (no vacuum) with the stopcock open and feed a gas into the vacuum line, close the stopcock and remove the bulb.  Knowing the pressure, temperature and volume, we knew how many moles of gas were in the bulb.  We could then attach the bulb to a reaction flask and open the stopcock, so that a known amount of gas was used in a reaction.

Here is a simple picture, imagine the tube on the bottom is the vac line.

These bulbs were also used to trap gaseous reaction products by attaching the bulb with its stopcock and joint to a vac line along with a reaction flask.  The bulb was cooled to the desired temperature and when the reaction produced gases, they could be collected in the cooled bulb.


As to your question a to whether a hydrophobic compound can also be
  hygroscopic, the simple answer is "yes to a point." Organic compounds
  can elicit both behaviors simultaneously when they have varying
  functional groups with a wide enough separation. BioDiesel is an ester
  with a very short side and a long side, the short side can exhibit
  polar qualities due to the presence of oxygen while the long chain
  exhibits nonpolar qualities. The polar side is what attracts the
  water, while the long chain repels the water. Cells of living
  organisms utilize a similar method to control the amount of materials
  crossing the membrane. I hope this explanation helps. Das Nerd 20:08,
  28 October 2006 (UTC)

https://en.wikipedia.org/wiki/Talk%3AHygroscopy

The composition of petrol diesel is very different, that's why it is not as hygroscopic as biodiesel.

As far as I know hygroscopy depends mostly on polarity. The water molecule is a polar, so it likes to mix with other polar compounds, like KOH, glycerol, salt, sugar, H2SO4, etc. Apolar compounds repel water. Air consists mostly of apolar components, so when the water in the air meets with polar things like sugar, it starts to move to that phase, because it has lower energy in there.

E.g. silicagel which is a common desiccant has this structure:

where you have polar $-OH$ chemical groups on the surface, that's why it is very hygroscopic.

While silicone oil contains molecules with this structure:

where you have apolar $-CH_3$ chemical groups on the surface, which makes it not hygroscopic.


Difference between the terms “analyte” and “analyate” and their applicability

The unabridged Oxford English Dictionary, available by subscription, is the ultimate source to confirm the usages of English words. I checked the word "analyate" and it does not exist in the entire dictionary. The word analyte is also very recent (mid 1950s). I am not surprised that wrong spellings are published in high-end journals. If you study the statistics of wrong references, that is even more shocking. How many times it has happened that you check a citation and it turns out that the citation never mentioned anything like that.

Here is an excerpt from the OED

  Origin: Apparently formed within English, by compounding. Etymons: analyse v., -lyte comb. form.
  Etymology: Apparently < ana- (in analyse v.) + -lyte comb. form, perhaps after electrolyte n.
  Chemistry.
  Categories »

  A substance that is analysed or determined, esp. quantitatively.

  1955   C. M. Crawford in Chem. & Engin. News 5 Dec. 5262/2   There is
  not, as far as I am aware, any short word denoting ‘the substance
  being analysed for’ in use... My proposal is that the word ‘analyte’
  be understood to convey the above meaning.


What would the name of this organic compound be?


What is the use of chloroform in the procedure of determining peroxide value?

Acetic acid is required for this method for exactly the reason you gave.  It also has the fortunate property of being miscible with many polar and non-polar solvents.  Because this is a method for the analysis of oil samples, it is necessary to chose a non-polar solvent capable of dissolving the sample that is also miscible with the acetic acid.  Chloroform is one solvent that fits that description nicely.

As a quick side-note, the use of chloroform in this test procedure is actually now discouraged by the Standard Methods for the Analysis of Oils, Fats and Derivatives.  Isooctane seems to be the preferred solvent, apparently for health and safety reasons.


There is a difference in preparing a solution and determining its concentration. As you said, NaOH solutions absorb carbon dioxide over time. It is possible to remove the absorbed carbon dioxide by boiling, but it is just as easy to use the solution as is and re-determine its concentration when you use it.

The concentration of sodium hydroxide solutions can be determined by titrating a known amount of dried potassium hydrogen phthalate.

If you really need an exact concentration of the solution, make it slightly more concentrated, standardize it and then dilute it to get the concentration desired. It would only be accurate for a few hours.

  I am told that NaOH is hygroscopic.

The powder and pellets of sodium hydroxide are hygroscopic, but the solutions of sodium hydroxide already have water, so it is unlikely more will be absorbed.


How can it be proved that (S)-bis-[2-(t-butylsulfinyl)phenyl] diselenide is a chiral non-racemic product?

Compounds with the general structure $\ce{R-SO-R'}$, i.e. containing a sulfinyl $\ce{-SO-{}}$ functional group attached to two carbon atoms, are called generically ‘sulfoxides’. Sulfoxides may be considered as a tetrahedral system composed of a central atom, ligands, and a lone pair of electrons (or phantom atom).

When the two organic residues $\ce{R}$ and $\ce{R'}$ are dissimilar, the sulfur atom is a stereogenic center.

The compound bis[2-(tert-butylsulfinyl)phenyl]diselenide contains two chiral sulfinyl groups and thus two stereogenic centers.

Therefore, three stereoisomers exist:

bis[2-((R)-tert-butylsulfinyl)phenyl]diselenide,

bis[2-((S)-tert-butylsulfinyl)phenyl]diselenide,

and the meso compound [2-((R)-tert-butylsulfinyl)phenyl][2-((S)-tert-butylsulfinyl)phenyl]diselenide


What sort of molecules or polymers are used in OLED displays?

A typical organic light emitting diode (OLED) structure consisting of a glass substrate, a indium titanium oxide (ITO) anode, and an organic bilayer diffused into a metallic cathode, is given below:

The organic bilayer consists of:

a hole transport layer (HTL), typically of N, N'-diphenyl-N, N'-bis(3-methylphenyl) 1, 1'-biphenyl-4, 4' diamine (TPD) or N,N′-di(naphthalene-1-yl)-N,N′- diphenylbenzidine (NPB)
an emitting/electron transport layer (EML/ETL), typically of tris(8- hydroxyquinoline) aluminum (Alq3)

Source: http://espace.library.uq.edu.au/eserv.php?pid=UQ:8598&dsID=SMS.pdf


Does this reaction require catalyst/solvent?

Just three points why omitting a solvent might not be a good idea without experimental evidence that support the contrary:

There is a reason many organic syntheses are run with reagents in a concentration in range of 0.1 to 1 mol/L; to moderate both rate (kinetics) and heat (thermodynamics) of a reaction.  The omission of a solvent for an already exothermic reaction may lead to a "runaway", where the heat suddenly generated transforms your products into a tar.  The issue regarding the heat balance often becomes important when scaling the reaction to larger batches in process chemistry.
Secondly, your product target equally is an aldehyde.  At least potentially there already is a problem (chemoselectivity) discerning one of the starting materials (salicylic aldehyde) from your product (2-formylphenyl benzoate) in respect to benzoyl chloride.  In case of a highly concentrated (neat) reaction lacking any solvent, if the reaction of the acyl chloride with the aldehyde is exothermic, you may reach a temperature of your reaction mixture so elevated that both reaction pathways became accessible instead of only one (and other not envisioned here, too).
According to the information about who sells this pricy commercial product at PubChem and how -- per gramm, instead of per millilitre -- it appears safe to assume that the product of your reaction is a solid.  Having a solvent dissolving both your starting materials as well as your product equally may offer you a reaction more complete, because the progress of the reaction is not hampered by an interface of a liquid phase (your starting materials) and a solid (your product).  It equally may lead you to a product more clean, since these grains won't contain (this much) of starting materials (occlusion).


Regioselectivity in radical chlorination of a carboxylic acid

I am going to say: no, this does not work in practice. In fact it does not even work in theory.

In simple theory you would say that that the α C-H bond is the weakest because the resulting radical is stabilised by conjugation into the C=O. But under radical conditions, we don't observe α-chlorination in the lab. Is there a better way to explain this?

The chlorine radical

The problem is twofold. Firstly, the chlorine radical, $\ce{Cl.}$, is much less stable and therefore extremely unselective in radical chlorinations. In physical organic chemistry, we quote something called the Hammond postulate, which says that in this case, the transition state for hydrogen abstraction resembles the starting material more than the product. Therefore, the stability of the product does not significantly affect the activation energy, and the rate, of the hydrogen abstraction.

On top of that, it is what we would consider an "electrophilic radical"; it has a very low-energy SOMO, and therefore has a tendency to abstract a more electron-rich hydrogen atom. Essentially, it means that the radical behaves more like an electrophile than a nucleophile. In the transition state for hydrogen abstraction, the chlorine wants to have a significant build-up of negative charge on itself (remember, the product of hydrogen abstraction is going to be $\ce{HCl}$). It therefore goes for the hydrogen that is most willing to give it this negative charge, and that is the most electron-rich hydrogen atom.

An example

Consider the radical chlorination of propionic acid, i.e. $\ce{R} = \ce{CH3}$ in your diagram above.

The α-hydrogen, compared to the β-hydrogen, is comparatively electron-poor because of the electron withdrawal by the $\ce{-COOH}$ group. So, it turns out that the rate of abstraction of the α-hydrogen is much slower: $k_\alpha/k_\beta = 0.03$ (Moody & Whitham, Reactive Intermediates, p 12). Here's an article which discusses the selectivity observed in the chlorination of propionic acid under different conditions: Tetrahedron, 1970, 26, 5929.

As written in Carey & Sundberg 5th ed., Part A, p 1022:

  Radical chlorination shows a substantial polar effect. Positions substituted by EWG [electron-withdrawing groups] are relatively unreactive toward chlorination, even though the substituents are capable of stabilizing the radical intermediate. [...] Because the chlorine atom is highly reactive, the reaction is expected to have a very early TS [transition state] and the electrostatic effect predominates over the stabilizing effect of the substituent on the intermediate. The electrostatic effect is the dominant factor in the kinetic selectivity of the reaction and the relative stability of the radical intermediate has relatively little influence.

As an example, they use the chlorination of butyronitrile, which is the same as propionic acid except that the EWG is now $\ce{-CN}$ instead of $\ce{-COOH}$:

As @Loong mentions in the comments, in the specific case of propionic acid, there is a further problem: there are three β-hydrogens for it to abstract, and only two α-hydrogens. Therefore, even in the absence of everything else I have described, we would expect there to be 1.5 times as much chlorination at the β-position. This certainly does not help at all with the selectivity! The importance of statistical factors is most clearly seen in the radical chlorination of simple alkanes, where all the hydrogens are pretty much similar. I shan't go into detail here, but you can look at this answer (by Loong again), and there are examples on Clayden 2nd ed., pp 987-8.

The solution

If you wanted to selectively chlorinate propionic acid at the α-position, you would use a polar mechanism that goes via the enol. The paper above has some examples of conditions that work. All of them involve suppressing the radical reaction by adding radical scavengers.


Why doesn't the boiling point of an aqueous solution depend on intermolecular attraction?

GOOD QUESTION! That is the nastiest p-chem question I've seen.  Atkins, Physical Chemistry 2nd edition (the readable one), pp 228-232pp.  Boiling point elevation is presented as purely an entropy effect, given two assumptions:

1) The solute is non-volatile.
2) The non-volatile solute does not dissolve in the solid solvent.
Both assumptions are explored in Chapter 10.

A solute increases the entropy of the solution but does not change the entropy of the vapor at a given P and T.  A higher T is then necessary to give the same entropy difference.  I am not satisfied that this wholly addresses a strongly interactive solute versus a non-interactive solute versus a strongly negatively interactive solute,

http://www1.lsbu.ac.uk/water/kosmos.html
Kosmotropes and chaotropes

http://www.amazon.com/Physical-Chemistry-Peter-Atkins/dp/1429218126
Atkins 9th edition is current.
"There is almost no chemical concept that Atkins can't make unnecessarily confusing"
http://www.amazon.com/Physical-Chemistry-A-Molecular-Approach/dp/0935702997/ref=zg_bs_16052591_2
Library - it's pricey.
http://www.amazon.com/Thermodynamics-Engineering-Approach-Yunus-Cengel/dp/0073398179/ref=dp_ob_title_bk
Library - it's very pricey.


Why are Tyrosine and Tryptophan considered hydrophobic?

Unlike that in organic chemistry, polar and non-polar compounds determination is done differently in biochemistry, probably because most biochemical reactions are done in aqueous medium. Therefore, the classification of them as polar and non-polar is somewhat different from source to source. In general, amino acids, which can make hydrogen bonding in aqueous solutions are called polar, and ones repel from water are non-polar. According to Ref.1:

All polar amino acids have either an $\ce{OH}$ or $\ce{NH2}$ group (when in aqueous environment), and can therefore make hydrogen bonds with other suitable groups.

Following image shows 20 amino acids categorized to polar (10) and non-polar groups (10):

All 10 non-polar amino acids are hydrophobic and 10 polar amino acids are hydrophilic. Polar amino acid residues have a tendency to be on the outside of a protein, due to the hydrophilic properties of the side chain (Ref.1). Keep in mind that tryptophan has indole function, but its lone pair of nitrogen is involved in the aromatic system. Thus, it make only weal H-bonding, which are not good enough to categorize as "polar."
References:

Bruce Alberts, Alexander Johnson, Julian Lewis, Martin Raff, Keith Roberts, Peter Walter, In Molecular Biology of the Cell; 5th edition, Garland Science: New York, NY, 2008, p. 125-129 (ISBN-13: 978-0815341055).


What is the purpose of subtracting the blank when constructing an analytical calibration curve?

Good to see interesting analytical chemistry questions after a long time.
I can give a qualitative example. Suppose you wish to collect the spectrum of a star. What time would be the best to carry out the measurement? You would agree that night time would be the best, because constant unwanted sunlight background is missing at the time of measurements. Subtraction of the blank serves a similar purpose, you wish to remove the background signal from your analytical signal so that you can assess the magnitude of your analytical signal. The blank also helps to serve to establish the minimum concentration that can be detected reliably. This is why blank subtraction is carried out
One of the commentators on your question has written a whole book on this subject.
You did not mention the analytical instrument you are using, it would have helped to discuss more. Anyway, calibration curve is another story. In this case, you certainly can make two calibration curves, and you are very correct, their slopes are the same. The sample concentration will turn out to be the same from both.
With blank correction, your equation should have a form y= mx with negligible intercept.
Without blank correction, your equation has the form y=mx+c, where c is the intercept, the average blank reading.
I may ask another thought provoking question which is certainly not discussed anywhere.
For example, what if I square all the signal intensity and then plot a calibration curve as a function of concentration. The calibration curve will still remain linear.
a) What will happen to the magnitude of slope?
b) Will the calibration curve sensitivity increase? "Apparently, yes!"
c) If the sensitivity increases by this simple mathematical manipulation, why shouldn't we do it?
Coming to the second part of the question:

If I try to quantify an analyte in a given sample with a different
matrix than the one used in the standard (different solvent, for
instance) maybe the "blank of the sample" so to speak would be
different than the blank of the standards (F0).

In this case, you should not/ perhaps cannot use a simple calibration curve. One of the key assumptions of a calibration curve is that the matrix of the sample is the same as the standard. If you violate this rule, then we cannot apply this technique.
Consequently, you will have to use a technique called Standard Addition Method. It produces very accurate results, when the sample matrix is significantly different from the standards.


Proportion of gases in Ideal Gas Mixture?

Your interpretation of specific gravity is incorrect:

  Specific gravity (SG) refers to the ratio between a given density and that of a reference.  Usually it is defined for liquids with respect to water and gasses with respect to air so in this case

    $0.6=\frac{n_{CH_4}\rho_{CH_4}+n_{C_2H_6}\rho_{C_2H_6}}{\text{air}}$

  where $n_i$ are proportions by mole factor. This is the incorrect assumption

    $0.6=\frac{\rho_{CH_4}}{\rho_{C_2H_6}}$.

  Gasses are always a little dicier since the reference is much more important with changes in temperature and pressure affecting the resulting density.

Example

  The density form of the ideal gas laws goes like:

    $\rho = \text{mass / volume = molecular weight * pressure / constant / temperature}$.

  Therefore for constant temperature and pressure, the specific gravity
  is equal to the ratio of molecular weights.  This follows because
  $\frac{\rho_1}{\rho_2} = \frac{MW_1}{MW_2}$ at constant conditions.

Equations and Solution

    $M_{\text{air}} = 28.97$ (I got $28.84$ using the simple $0.21*32+0.79*28$ and this agrees with  this)

    $\text{SG}=0.6=\frac{n_{CH_4}\rho_{CH_4}+n_{C_2H_6}\rho_{C_2H_6}}{\text{air}}$

    $1=n_{CH_4}+n_{C_2H_6}$

  so by the first equation and the second equation the density of the mixture is $0.6 * 28.97 = 17.38$. The third equation holds only if $0.9*16 + 0.1*30 = 17.4$ so

    methane is 90% and ethane is 10%.


How can Ellingham diagrams be linked to Gibbs energies of formation?

The Ellingham diagram doesn't actually use molar Gibbs energies of formation $\Delta G_\mathrm{f}^\circ$ per se; it is more accurate to say that it uses molar Gibbs energies of reaction $\Delta G_\mathrm{r}^\circ$. The difference is that the formation energy is only relevant to one specific chemical equation, for example:

$$\ce{Ca + 1/2O2 -> CaO} \qquad \qquad \Delta G_\mathrm{r}^\circ = \Delta G_\mathrm{f}^\circ(\ce{CaO})$$

in which the stoichiometric coefficient of CaO is equal to 1.** On the other hand, for any (balanced) equation with any stoichiometric coefficients, it is valid to define a Gibbs energy of reaction:

$$\ce{2Ca + O2 -> 2CaO} \qquad \qquad \Delta G_\mathrm{r}^\circ = 2\times \Delta G_\mathrm{f}^\circ(\ce{CaO})$$

which is related to the energy of formation, but is not the same thing, as evidenced by the factor of 2.

In the Ellingham diagram, every reaction has the same stoichiometric coefficient for $\ce{O2}$, which is typically 1. This is needed to make sure that different reactions are comparable. Let's say, for example, you want to see whether the reaction

$$\ce{C + 2CaO -> CO2 + 2Ca}$$

is feasible. This is done by checking the sign of $\Delta G_\mathrm{r}^\circ$: if it is negative, then the reaction is feasible, and vice versa. The point is that this $\Delta G_\mathrm{r}^\circ$ can be calculated by subtracting two reactions together:

$$\begin{align}
\ce{C + O2 &-> CO2} & \Delta G_\mathrm{r}^\circ &= c_1 = \Delta G_\mathrm{f}^\circ(\ce{CO2}) \\
\ce{2Ca + O2 &-> 2CaO} & \Delta G_\mathrm{r}^\circ &= c_2 = 2 \times \Delta G_\mathrm{f}^\circ(\ce{CaO}) \\ \hline
\ce{C + 2CaO &-> 2Ca + CO2} & \Delta G_\mathrm{r}^\circ &= c_1 - c_2 \\
\end{align}$$

but these two equations add up nicely only if the coefficients of $\ce{O2}$ in both equations are the same. What the Ellingham diagram does is to plot the Gibbs energies of reaction, $c_1$ and $c_2$: if $c_1 < c_2$, then the reaction is feasible. It doesn't plot the Gibbs energies of formation, because comparing those wouldn't tell us anything about the sign of $c_1 - c_2$.

As a final remark, note also that the equation

$$\Delta G_\mathrm{r}^\circ = -RT \ln K$$

holds true for any reaction, whether or not it actually corresponds to a formation reaction.

** Having a stoichiometric coefficient equal to $x$ does not mean the same thing as $x$ moles of the compound are produced in the reaction. The coefficient is purely a mathematical expression which tells us the stoichiometric relationship between different species in the reaction. It does not correspond to a real-life reaction, where a defined quantity of reactant is added to a defined quantity of product. To illustrate this, let's say you go to a lab and mix 0.4 mol of HCl to 0.4 mol of NaOH. You're asked to write a balanced equation for this. You can write

$$\ce{0.4 HCl + 0.4 NaOH -> 0.4 NaCl + 0.4 H2O,}$$

and that would be correct, but it is hardly the only correct possibility: the more conventional

$$\ce{HCl + NaOH -> NaCl + H2O}$$

is equally correct, even though the stoichiometric coefficients (1 in all cases) do not match up with the actual amount of substance used in the reaction (0.4 mol). Note also that the units are different: stoichiometric coefficients are dimensionless, but amount of substance is measured in moles. The difference is subtle, but one well worth pondering about, as mixing these two up can lead to a lot of misconceptions in thermodynamics.


Nucleophilicity in polar aprotic solvent (DMSO)

let me start with nucleophilicity of halides. In protic polar solvent, $\mathrm{F^-}$ is more solvated (hydrogen bonding) than $\mathrm{I^-}$ and hence less reactive.

In polar aprotic solvent $\mathrm{F^-}$ solvation is less and therefore more reactive.

Such behavior may not be applicable here, since phenoxide is resonance stabilized while other compound is not. Therefore phenoxide $\ce{2}$  is less nucleophilic then cyclohexanolate $\ce{1}$.


Why does benzyne dimerise and not trimerise?

As per Wikipedia,

  Biphenylene is quite stable both chemically and thermally, and behaves in many ways like a traditional polycyclic aromatic hydrocarbon. However, both the spectral and chemical properties show the influence of the central [4n] ring, leading to considerable interest in the system in terms of its degree of lessened aromaticity.

This implies that biphenylene is rather stable, despite its anti-aromaticity.

As for triphenylene, I'm not sure, and it doesn't say so anywhere, but perhaps the free energy of formation of triphenylene is much more, as it requires more molecules to be put together, leading to the reaction being non-spontaneous. However, benzyne does trimerize as well, forming triphenylene.

So benzyne can actually undergo both dimerization and trimerization, forming two different, but stable products.

As requested, a more primary source is available here. This site cites other sources, whose links I have provided in the quotation.

These lines say more or less the same thing:

  Several recent publications (2 3 4 5 6) concluded that biphenylene possesses signiﬁcant strain energy of the order of 50–60 kcal/mol. These numbers are at least confusing since molecules that possess such tremendous strain can hardly exist. In the meantime, biphenylene is a stable molecule that has been synthesized a long time ago while its chemical properties show typical aromatic behaviour.


The starting compound is an organic compound known as acrolein. $\ce{CH2=CH-CH=O}$
Now your first reaction would be:
$$\ce{CH2=CH-CH=O ->[CH3SH][1,4 addition]CH3-S-CH2-CH=CH-OH} $$
Now, this compound tautomerizes to form methional
$$\ce{CH3-S-CH2-CH=CH-OH <-> CH3-S-CH2-CH2-CH=O}$$
Now, the second step in your reaction is a named reaction known as Strecker Amino Acid Synthesis. This converts an methional (in this case) into methionine.

Now, onto the question at hand, it asks about certain properties of methionine. Methionine is an essential amino acid. So A is correct. Since methionine doesn't have any extra acid or base groups, it is neutral.
As for the Lassaigne Test, if the compound had only nitrogen, it would have shown a prussian blue color. However since there is sulphur as well, it shows a violet colouration. So, the answer is A.


Intramolecular reactions are always preferred over intermolecular.

Remember that these reactions are reversible so even if the intermolecular attack proceeded, the product would be unstable with respect to intramolecular attack by the free OH that is within the molecule.


Negative charged intermediate in acidic conditions - what's the problem?

Acidic conditions — whether you are using the Brønsted or the Arrhenius definition of acidity — is characterised by readily available protons. Remember the general dissolution ionisation of a Brønsted acid in water  as shown below:

$$\ce{HA + H2O <=> H3O+ + A-}\tag{1}$$

The proton is readily liberated and transferred to species which are more happy to accept it — for strong acids, that means it is transferred to water, a really weak base. A proton typically gladly attaches to anything that has a negative charge, because those species are typically more basic than water in this example. Therefore, postulating an enolate or any other oxygen-centred anion in acidic solution is not considered correct since there will be enough protons to quench it.

However, this does not mean that no anions can exist in aquaeous solution. For example, a typical method to transform alcohols into iodides is refluxing in aquaeous hydrogen iodide. Here, $\ce{HI}$ is a very strong acid that very readily dissociates to give iodide anions, $\ce{I-}$. These, even though negatively charged, do exist in acidic conditions and are able to attack nucleophilicly. That’s all due to the high acidity of $\ce{HI}$.

$$\ce{R-OH + H3O+ + I- <<=> R-OH2+ + H2O + I- -> R-I + 2 H2O}\tag{2}$$


C + O2 is equal to C + O, how is that possible

When thinking about chemical reactions it is very important to know which chemicals may react with each other. Jun-Goo Kwak already pointed out the nature of oxygen.

A quick reminder: The ground state of elementary oxygen is the triplet biradical $\ce{o2}$, which is a gas. This is what we have on the surface of the earth. Carbon however comes in many  different forms in nature. The most popular and often is graphite. Other forms include diamond, fullerenes and graphene.
At one point in your life you haw almost certainly come in contact with graphite: coal. As the principle repeating unit is carbon itself, its formula will be written as $\ce{C}$.

As for the binary combination of oxygen and carbon, there are also many different modifications. The most important of them are carbonmonooxide ($\ce{CO}$) and carbondioxide ($\ce{CO2}$). Like Uncle Al stated there are also suboxides known, which are usually byproducts of incomplete combustion (if not targeted explicitly).

Having said all that, if you burn coal, the following main reaction will happen (1):

$$\ce{C + O2 -> CO2}$$

However, given the right conditions (excess carbon) also carbonmonoxide  may be formed (sum of reaction, 2):

$$\ce{2C + O2 -> 2CO}$$

The reaction itself will detour via the Boudouard reaction, which is very important in blast furnace processes. First forming carbondioxide via 2 and then convertin excess carbon to carbonmonoxide via 3:
$$\ce{C + CO2 <=> 2CO}$$


IUPAC nomenclature: "Smallest sum of locants"?

The guideline that you have mentioned (‘While naming any compound, the numbering should be done such that position of substituted groups gives the smallest sum.’) does not exist in the IUPAC recommendations.
The corresponding section in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) actually reads as follows:

P-14.3.5 Lowest set of locants

The lowest set of locants is defined as the set that, when compared term by term with other locant sets, each cited in order of increasing value, has the lowest term at the first point of difference; for example, the locant set ‘2,3,5,8’ is lower than ‘3,4,6,8’ and ‘2,4,5,7’.
(…)

Therefore, the smallest sum of locants can lead to wrong results.
For example, for 2,7,8-trimethyldecane, the smallest sum of locants would yield the wrong name ‘3,4,9-trimethyldecane’, since $2+7+8=17 > 3+4+9=16$.

Likewise, for 2,2,6,6,7-pentamethyloctane, the smallest sum of locants would yield the wrong name ‘2,3,3,7,7-pentamethyloctane’, since $2 + 2 + 6 + 6 + 7 = 23 > 2 + 3 + 3 + 7 + 7 = 22$.

For 1,6,7-trimethylnaphthalene, the smallest sum of locants would yield the wrong name ‘2,3,5-trimethylnaphthalene’, since $1 + 6 + 7 = 14 > 2 + 3 + 5 = 10$.

For 1,6-dimethylcyclohex-1-ene, the smallest sum of locants would yield the wrong name ‘2,3-dimethylcyclohex-1-ene’, since $1 + 6 = 7 > 2 + 3 = 5$.

Anyway, the IUPAC recommendations have more than 1500 pages. Hence, you should be careful when applying any simplified or abridged version of the nomenclature rules.


In chemical thermodynamics, activity (symbol a) is a measure of the “effective concentration” of a species in a mixture, in the sense that the species' chemical potential depends on the activity of a real solution in the same way that it would depend on concentration for an ideal solution.
(From Wikipedia.)

So the main point is that that activity is just a measure of the deviation from the ideal solutions, because the laws you are using in physical chemistry  assume that solution is ideal. But usually it is not true. Laws can work for real life with a certain precision, but you should always remember about their "ideality" and the non-ideality of our world.


Think of it simply as spheres in a box. The spheres are your molecules. The higher the temperature, the higher the energy the molecules have and the more they move around. So when they have high enough energy for the molecules to move around a lot and stay away from each other they are a gas. (This is a simplified answer). If you decrease the energy they have (cool the system down), they move about less and there's a certain point the molecules aren't moving far away enough from each other to stay as a gas, so they liquify. As Raditz_35 said, look at a phase diagram. So when you put a gas under pressure, the molecules are getting "squished" together and they don't have a lot of room to move around, so they can't move away from each other enough and thus liquify. This is how they store Hydrogen as fuel. It's too dangerous as a gas and they put the Hydrogen under pressure to store it as a liquid. Hope this helps.


Can you artificially create molecules with a machine?

This sounds to me like molecular beam chemistry. Within a vacuum chamber, you can create and collimate a focused "beam" of moving atoms or molecules and direct it to hit a stationary target or another beam. You can use spectroscopy to study the species that are formed. The idea is to look for and characterize reactive intermediates or transition states between starting materials and products. But if you were to create a familiar species such as water in this manner, it is not "artificial" water, but the real thing.

Maybe there is a physical chemist out there that can elaborate further on molecular beams.


First, you can use phenolphthalein indicator solution, $\ce{Ca(OH)2}$ and $\ce{NaHCO3}$ will turn red. Then use 6M $\ce{HCl}$ and the $\ce{NaHCO3}$ bubbles.

$\ce{NaOH}$ to identify $\ce{Pb(NO3)2}$: it generates precipitations. Use $\ce{Pb(NO3)2}$ to identify $\ce{KI}$: generate golden precipitations.

$\ce{H2SO4}$ with $\ce{BaCl2}$ gives white precipitations, $\ce{Na2SO4}$ with $\ce{BaCl2}$ likewise. So the rest is $\ce{NaCl}$.


Free radical substitution - bromination of ethane

If there is an excess of ethane, you would be getting the monosubstituted bromoethane, $\ce{CH3CH2Br}$, as the main product.

When ethane, $\ce{CH3CH3}$, is in excess, for any bromine radical formed, most of the reacting particles around it would be an unreacted ethane molecule (high concentration of $\ce{CH3CH3}$), this means that there is a greater chance of a bromine radical colliding with and reacting with $\ce{CH3CH3}$ to form the monosubstituted bromoethane.

This is in contrast with when bromine is in excess, where the excess of highly reactive bromine radicals mean that most of them react fully with the available ethane molecules to form fully substituted $\ce{CBr3CBr3}$.


Acrylamide is a Michael acceptor. All Michael acceptors are potentially carcinogenic because DNA can act as a Michael donor. This Michael reaction can damage the DNA, which can ultimately develop into cancer. Of course how well a certain Michael acceptor reacts with DNA depends on its chemical structure (i.e. the carbon chain and other function groups as well as its conformation), and hence not every Michael acceptor is equally carcinogenic.

If you worry about eating too much French fries I would first be concerned with my caloric  and sodium intake. For that there is solid evidence that it greatly reduces human health if taken in excess.


How can I prepare two phosphate buffer of different concentrations but give same pH?

For a phosphate buffer with $\mathrm{pH} = 7,$ the two dominant species are $\ce{H2PO4-}$ and $\ce{HPO4^2-}$. The relevant $\mathrm{p}K_\mathrm{a}$ is $7.2$ (this is $\mathrm{p}K_\mathrm{a2}$). From the Henderson-Hasselbalch equation, using $\mathrm{pH} = 7,$ the ratio of $\ce{HPO4^2-}$ to $\ce{H2PO4-}$ is about $0.631.$ You can see this in the alpha diagram below:

In this figure, the green trace is the $\ce{H2PO4-}$ fraction (of total phosphate) present at $\mathrm{pH}$ values from $0$ to $14.$ The blue trace is the $\ce{HPO4^2-}$ fraction (of total phosphate) present at $\mathrm{pH}$ values from $0$ to $14$. So, at $\mathrm{pH} = 7,$ the blue curve is below the green curve, and the ratio of alpha fractions is $10^{-0.2}$, i.e., about $0.631.$ This is directly from the Henderson-Hasselbalch equation. Since all the phosphate species are in the same buffer volume, this is also the ratio of the respective concentrations.

The difference between the $\pu{0.01 M}$ phosphate buffer and the $\pu{0.1 M}$ phosphate buffer is that the $\pu{0.1 M}$ buffer is ten times higher in both $\ce{HPO4^2-}$ concentration and $\ce{H2PO4-}$ concentration. So the ratio is the same as in the $\pu{0.01 M}$ buffer and the $\mathrm{pH}$ is the same for the two buffers. Of course, the more dilute buffer has lower buffer capacity, but that is another issue.


Can water be liquefied or solidified just by adjusting the temperature, regardless of the pressure?

If I actually understood your question correctly, yes there are regions above which there will be only solid state (rather than no solid state). This is limited to the molecular region.

The Roman numerals indicate various ice phases.

From the above phase diagram (which can be found here) the solid-liquid equilibrium line doesn’t touch the y-axis as you suggest. But there will be only solid phase above the $10\ \mathrm{GPa}$ region(approximately).

In the phase diagram you drew, if (a big if) such a PD exists for some element/molecule, then there will be no solid above the point where solid-liquid line meets y-axis.

As pointed out in the comments, when we reach high temperatures and pressures we won’t be seeing the molecular water. It will dissociate into atoms/ions as shown in the figure.


Why only structures corresponding to valleys on PES are taken into account?

The vast majority of thermodynamics we consider is at equilibrium, we is the issue here.

At equilibrium we will only find our S states at the bottom of the valleys (or minima) populated.  This this because they have now force acting on them to drive them to another configuration.

An important relationship to remember with potential energy surfaces is:
$$
F = -\frac{dE}{dx}
$$
Force is the negative derivative of potential energy with respect to some coordinate.
This means that the driving force is given by the slope of the PES and is pushing down the PES.
Then a molecule with a $\phi = 150^\circ$ experiences a rotational force pushing it towards the minimum at $\phi = 180^\circ$.

Answer:

The result is that any molecule not at a minimum will be quickly driven towards the nearest minimum, so only the minima will be occupied and need to be summed over.

Side Note - Maxima

Maxima (the peaks) also experience zero force.  However this is only true at the exact peak itself.  Any point infinitesimally far away from the maxima experiences a force pushing it down the hill into a minima.

Thermal fluctuations and collisions with other molecules will continually slightly nudge the molecules around on the potential energy surface.  At maxima this will push any molecules there off the peaks and down into minima.  In minima, the force will bring the molecules back into the minima.


Metering Device decreases pressure yet produces saturated liquid?

I'm a bit confused by your question -- your diagram shows the path of refrigerant through the cooling system, but you seem to be discussing water's state changes.

Liquid refrigerant under pressure enters the metering device. As it leaves the device at a lower pressure, it boils, producing some vapor, but not evaporating completely. The rest of the liquid evaporates as it passes through the evaporator coil, absorbing heat as it does so.


Friedel craft alkylation halo-carbon bond cleavage

In Friedel-Crafts alkylation, the halogen-carbon bond breaking depends on the strength of bond between halogen atom and the Lewis acid used. The more stronger is the bond, the easier it is to generate a carbocation for the reaction.( In all cases, the carbocation exoplicitly needn't have to generate, a strong $\delta$+ charge on the electrophilic carbon is sufficient enough to have a $\ce{S_E2}$ or $\ce{S_E1}$ type reaction. Here, the Lewis acid used is  $\ce{BCl_3}$ where there is a vacant  $\ce{2p_z}$ orbital on the  $\ce{B}$ atom,(Assuming, the molecule is oriented in a triangle in the  $\ce{xy}$-plane.)and the Lewis base is the given reagent 1-chloro-3-fluoropropane., which has two Basic centre. But donation through  $\ce{F}$ will correspond to a stronger  $\ce{2p$\pi$ -2p$\pi$}$ overlap with Lewis acid,rather than weaker  $\ce{3p$\pi$ -2p$\pi$}$ overlap in the case of donation through  $\ce{Cl }$ atom. Thus bonding with  $\ce{F}$ will correspond to a very huge amount of  $\ce{-I}$ effect of the adjacent carbon atom which will weaken the bond between them by pulling the electron pair towards  $\ce{F}$ and easily favour the attack by the $\pi$-electron cloud of Benzene ring.Also, the repulsion between $\pi$-electron cloud and the  large sized $\ce{Cl}$ atom is very less during the formation of  the above compound. But, if it would have bonded through  $\ce{Cl}$, there won't be a great electrophilic nature of the adjacent carbon due to less effective bonding in the case of  $\ce{Cl}$(corresponding to even lesser  $\ce{-I }$ effect),so attacking will correspond to a highly energetic transition state, and also, there will be a slight electrostatic repulsion with the  $\ce{F}$ atom of the  $\ce{$\pi$ }$-electron cloud.
Due to such above mentioned phenomena and energetically favoured conditions, bonding through cleavage of  $\ce{C-F}$ bond is more preffered over cleavage of  $\ce{C-Cl}$ bond.


What is the oxidation state of nitrogen in ammonium nitrate?

There are 2 different nitrogen atoms in $\ce{NH4NO3}$, one having oxidation number -3 and one with +5. They are different, having separate oxidation numbers independently of each other. It is perfectly fine to have several atoms of same element having different oxidation state in same compounds, for example in thiosulphate  $\ce{SO3S^{2-}}$ one (central) sulfur atom has oxidation number +4 and another has 0, and $\ce{Fe3O4}$ is actually $\ce{\overset{(II)}{Fe}\overset{(III)}{Fe2}O4}$ and many more.


Why is N₂ stable but HCN and C₂H₂ unstable?

I think this question can be answered reasonably well with thermodynamic considerations.  When we say something is stable, it is (or should be) assumed that the compound does not engage in a chemical reaction.  It is important when we discuss the stability of compounds to focus not only on the compound itself but what the products of the reaction are.

The three triple bonds you mention ($\ce{N#N}$, $\ce{C#N}$ and $\ce{C#C}$) all have very high bond dissociation energies (from various sources such as this one they are $\approx950, 940\text{ and }960\:\mathrm{kJ/mol}$) so all of these triple bonds require a large amount of energy to break.

Now, when we think about the stability of acetylene, I suspect most people are thinking about its stability towards ignition (which has been studied, by the way).  That gives us a reaction we can explore, namely the combustion of acetylene in the presence of oxygen to produce carbon dioxide and water.  WolframAlpha can calculate the thermodynamics of this reaction, giving a result of $-2600\:\mathrm{kJ/mol}$; clearly a very exothermic reaction.

Now let's look at the combustion of nitrogen in the presence of oxygen, which could form nitrogen dioxide.  Again, WolframAlpha (Note, if you happen to play around with WA, make sure you put your reaction in quotation marks or it won't be interpreted properly) gives us a result of +66 kJ/mol or an endothermic reaction.

What do we make of these results?  Focusing on the products we see that for the acetylene combustion, there is a large amount of energy released when creating the products.  The same is not true for the products of nitrogen combustion.

In summary, when considering stability, we must think about we mean by the word, since that meaning will lead to a chemical reaction, and the tools we use to describe chemical reactions help us quantify that stability.


The difference in perception of taste is altered both by the lower air humidity as well by the lowered cabin pressure while on flight, the later typically kept to what corresponds to an altitude of about 2'400 m above sea level (source).

Regarding the actual changes in taste, BBC Future once nicely shed light on context, causes and outcome; in probably more accessible words than ongoing research articles like this about it.