Can you approximate the change in entropy of a reaction?

The molar standard entropy of formation ($S_{f}^{\ominus}$) for many common substances is documented in tables. You can approximate the overall change in entropy of the reaction ($\Delta S^{\ominus}_{reaction}$) with with Hess's law:

$\Delta S^{\ominus}_{reaction} = \sum S^{\ominus}_{f,molar,products} - \sum S^{\ominus}_{f,molar,reactants}$

You notice that there is no $\Delta$ in front of $S^{\ominus}_{f,molar}$, since entropy is an absolute physical quantity. Every system has an absolute entropy.

Don't forget to multiply every $S^{\ominus}_{f,molar}$ with the number of moles that you have. This is only an approximation because the molar entropy of a substance depends on Temperature as well, and the reaction likely doesn't take place at standard temperature.

As a sidenote please be aware that the molar entropy of formation for elements is not zero, although the molar enthalpy of formation is!


As with nearly all STEM-related Wikipedia pages, this one reads quite clear and concise... that is, if you already know at least 75% of the stuff involved. Otherwise it may lead you into the deepest muddle, so you did the right thing by asking this question.

Follow the logic. What is electrolysis used for? It reduces a metal (in particular, zinc) from its salts. You don't reduce what's already been reduced. This means zinc was not reduced before the step 3. In particular, it was not reduced during the cementation. Still, cementation implies that one metal is reduced by another. Which of the two is zinc if we've just concluded that zinc does not get reduced on this step?

Since the naturally found compounds are rarely pure, zinc sulfate after the step 1 contains a few percent of other metals (copper, nickel, etc.), also in the form of sulfates. Should we electrolyze it now, these metals will get reduced along with our zinc and remain in it as impurities, which we don't want. Worse yet, they are mostly less active than zinc, so they will reduce before it and thus concentrate in the produced metal. How do we get rid of them?

Now comes the idea: let's reduce them. We vaguely remember this can be done with a more active metal. But it must not be too active, otherwise it will reduce zinc as well. Also, the oxidized form of this new metal must not spoil our $\ce{ZnSO4}$. Just what metal can that be?

Guess what?

   It is zinc.
 Yes, we add metallic zinc (just a few percent, I remind) and it reduces the impurities as follows:
 $$\ce{CuSO4 + Zn -> Cu + ZnSO4}$$
 The only soluble byproduct is $\ce{ZnSO4}$, which does not upset us one bit.

So it goes.


Avogadro - How can bonds be hidden and only show atoms?

Not in the current version, as far as I know.

If you need it, you could either make a feature request on the Sourceforge page, or you could do something with extensions, such as making a new version of the ball and stick renderer removing the code that renders bonds, or making an extension that adds a button to delete all the bonds.


How to determine the products of a redox reaction

I did some research on this topic (water treatment for TCE) and I come out three answer which is by using peroxide, ozone and permanganate. Since you asked for ozone first, so I will give that first.

What are you missing in the reaction is that water is the reactant of this reaction instead as a product. Here's the equation:

$$\ce{C_2HCl_3 + O_3 + H_2O -> 2CO_2 + 3HCl}$$

For the reaction involved ozone, The reaction will be more effective in an acidic medium and it must perform in a enclosed delivery pipes.

Second, The reaction involving peroxide. Hydrogen peroxide react with TCE will also produce carbon dioxide and hydrogen chloride but water is a product here. Here is the equation:

$$\ce{C_2HCl_3 + 3H_2O_2 -> 2CO_2 + 2H_2O + 3HCl}$$

Same as the ozone reaction, this reaction is more effective in an acidic medium.

Last one is the reaction involving permanganate. We will be using potassium permanganate but you can use sodium permanganate, calcium permanganate or magnesium permanganate. It will react with TCE to produce carbon dioxide, manganese (IV) oxide, metal chloride and hydrogen chloride. Here is the reaction:

$$\ce{C_2HCl_3 + 2KMnO_4 -> 2CO_2 + 2MnO_2 + 2KCl + HCl}$$

This reaction is a bit different compared with the other two. It could be effective in acidic or basic medium between the PH range of 3.5 to 12.

Hope this helps you.


Use the states of oxygen and hydrogen to know the oxidation state of sulphur.


How do I find the moles of NaOH required to reach the equivalence point?

Assuming you are titrating a monoprotic acid, the moles of acid and base should be the same at the equivalence point. To find the moles of base used to reach equivalence, you simply need to multiply the volume of base you added to reach equivalence by the molarity of the base (which you determined from your standardization runs). You also will then know the moles of acid, since they have to be the same at equivalence.


The current version of Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book) and the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) include polyhedra and corresponding polyhedral symbols that describe the geometrical configuration.

When an atom is attached to three atoms or groups (e.g. coordination compounds with a coordination number of three), the geometry may be trigonal planar or trigonal pyramidal. For the structure given in the question, the geometry is obviously trigonal pyramidal, not planar. Thus, the coordination polyhedron is a trigonal pyramid; the corresponding polyhedral symbol is TPY-3.

This coordination polyhedron is not to be confused with a tetrahedron (polyhedral symbol: T-4), which applies to a coordination number of four, or an atom that is attached to four atoms or groups.


One aspect to consider is that the boiling points of alkali metals go down as they get heavier.
From The Encyclopedia Britannica:
Lithium:  bp = 1342°C
Sodium:  883°C
Potassium:  759°C
Rubidium:  688°C
Cesium:  671°C
Francium:  677°C (probably this one is merely predicted)
Thus the heavier alkali metals could be volatile enough to effectively escape as vapors while sodium and the chloride salts remain largely liquid.  Thereby the equilibrium that would otherwise limit the displacement of the heavier metal is pulled forward in favor of that metal being lost to the gas phase.  The vapor would later be condensed to get the metal as a solid.


Why can't alcohols form hydrogen-bonded dimers like carboxylic acids?

The strength of a hydrogen bond somewhat depends on the $\ce{X-H\bond{...}X}$ angle that the hydrogen-bonding hydrogen forms with the two electronegative elements $\ce{X}$. In our case, carboxylic acids or alcohols, $\ce{X} = \ce{O}$ so the angle is $\ce{O-H\bond{...}O}$. The ideal angle for this fragment is $180^\circ$.

As you have drawn for carboxylic acids, it is very easy to allow this linear arrangements. If you wish, you could describe the entire $\ce{(R-COOH)2}$ feature as a benzene-like ring streched in a single direction. Most importantly, the acceptors and donors line up nicely, the carboxyl function features an angle of $120^\circ$ and the $\ce{C-O-H}$ angle (and the $\ce{C=O\bond{...}H}$ one!) is also close to $120^\circ$. These arae the theoretically predicted, unstrained angles.

For two alcohol (e.g. ethanol) molecules to attempt a similar arrangement, we would end up with only four atoms that have to form a rectangle with $\ce{O, H, O, H}$ at the four corners. Thus, the $\ce{O-H\bond{...}O}$ hydrogen bond angle would be much closer to $90^\circ$ — a very bad arrangement. Furthermore, this would put the two electronegative oxygens (which thus feature a negative partial charge) closer together with nothing in-between which would cause destabilisation due to like charges approaching. The hydrogen-bonding hydrogens cannot alleviate this unfavourable interaction since they are at the corners of the square.

The situation for alcohols is much better if they create a network of a number of molecules that allow for much more favourable angles. With four molecules, you could already create a cyclic structure of approximately $100^\circ$ $\ce{H\bond{...}O-H}$ angles and $170^\circ$ $\ce{O-H\bond{...}O}$ angles — much more favourable. Of course, in the actual solution this number will fluctuate strongly as hydrogen bonds are broken and reformed constantly and different ring sizes happen all the time.


What is the relationship between the values of n and ml?

Just work it out the proper way, in my opinion you shouldn't have to resort to shortcuts for such a question.

The magnetic quantum number $m_l$ takes values between $-l, -l+1, \cdots, l-1, l$. As such, as long as $l \geq 5$ then $m_l = 5$ will be valid.

The orbital angular momentum quantum number $l$ takes values between $0, 1, 2, \cdots, n-1$. Consequently if $n \geq 6$, then $l \geq 5$ and $m_l = 5$ is permissible. The smallest possible value of $n$ is therefore $6$.


Feasibility of Nucleophilic Substitutions

Short answer - No. See below for a workaround.

Consider your acids and bases:

  Substituting an alcohol with ammonia directly

$$\ce{NH3 + CH3CH2OH <=> CH3CH2NH2+ + OH-}$$

If this were an acid-base reaction (and it is, at least in the Lewis sense), then which side of this reaction would be favored by equilibrium? Which side has the stronger acid-base pair? Hydroxide is a stronger base than ammonia, so this equilibrium would favor the reactants.

  Using an ammonnium salt

The first step of this mechanism would be a proton transfer to form an activated leaving group and the $\ce{NH3}$ nucleophile.

$$\ce{NH4+ +CH3CH2OH <=> NH3 + CH2CH3OH2+}$$

Again, which side of this equilbrium is favored? Ammonia is definitely the stronger base compared with the alcohol. Estimates of the pKa of $\ce{ROH2+}$ are in the vicinity of $-2$, while the pKa of $\ce{NH4+}$ is $9.25$. The equilbrium constant of this reaction is on the order of $10^{-11}$.

  Workaround

If you are needing to do the transformation

$$\ce{ROH -> RNH2}$$

there is a workaround in converting the alcohol into a molecule like the tosylate ester, which can then react with ammonia:

$$\ce{CH3CH2OH ->[\ce{TsCl,base}] CH3CH2OTs }$$
$$\ce{CH3CH2OTs ->[\ce{NH3}] CH3CH2NH2}$$


Actually you were close, below I have tried to discuss how this occurs:

Glycosidic bonds between monosaccharide units are the basis for the formation of oligosaccharides and polysaccharides such as starch.

It is possible for a sugar hydroxyl group (ROH) bonded to the anomeric carbon (the anomeric carbon is the carbonyl carbon of the open chain form of the sugar and is the one that becomes a chiral centre in the cyclic form) to react with another hydroxyl (R’-OH) to form a glycosidic linkage (R’-OR).

  (A glycosidic linkage is not
  an ether)

Glycosidic linkages can take various forms; the anomeric of one sugar can be bonded to any one of the -OH groups on a second sugar to form an alpha- (as in starch) or beta-glycosidic linkage.

An example of a glycosidic bond (showing formation of maltose):

C-1 of one glucose is linked by a glycosidic bond to the C-4 oxygen of the other glucose
when an ⎯OH (alcohol) of one glucose molecule (right) condenses with the intramolecular hemiacetal of the other glucose molecule (left), with elimination of H2O and formation of a glycosidic bond.

The reversal of this reaction is hydrolysis—attack by H2O on the glycosidic bond. In animal digestion (first phase), it is aided by α (1,4)-glucan 4-glucanohydrolase (salivary α-amylase)

Here is another comparison of the α 1,4 glycosidic linkage and α1,6 glycosidic linkage:

Having discussed about the glycosidic bond, we can refer back to starch:
Starch is composed of two components, α-amylose and amylopectin. α-Amylose is composed of linear chains of D-glucose in α (1,4) glycosidic linkages. The linear linkages in amylopectin are α (1,4), whereas the branch linkages are α (1,6) glycosidic bonds:

Edit

It is also possible for polysaccharide units to form hydrogen bond interactions between neighbouring units, although these are minor and serve to stabilise the helical structure of amylose chains (around  up to 6 residues per turn) but the major bonds are glycosidic bonds

I am sure you now have a clue on how they are bonded. Hope this helps.

References:

Biochemistry (Campbell and Farell)

Lehninger Principles of Biochemistry

Biochemistry (Grisham)


Can nitrate and sulfate ions be oxidised during electrolysis of their molten compounds?

Not only they can, but they must get oxidized. There is no other option. This is the only way the electric current can pass between the electrode and the solution.

Whether or not the products will be some peroxyanions is another question. At high temperatures typical of molten salts, I don't think they will survive. Hence I would expect $\ce{O2}$ + disulfate in the first case and $\ce{O2 + NO2}$ in the second.


Alternative Program for Orbital Population Analysis


The idea of splitting water using electrolysis and pumping it into an engine's air intake has been around for a long time. The idea is that it's supposed to increase the efficiency of burning gasoline. However, even assuming that this is true (there are many reasons why it may not be—modern gas engines are in the range of 95–98% efficient at combustion), electrolysis is very inefficient, so I'm quite dubious of any potential gains.

Hunting around on the internet, it's obvious that a lot of people seem to think this works (and that people who say it doesn't are oil company shills or some such), but I wasn't able to find any compelling evidence to support it. How this kind of system interacts with a modern computer controlled engine may perhaps even make efficiency worse. There doesn't seem to be much academic research into it, though a Google Scholar search brings up a lot of kooky patents. This kind of thing is pretty difficult to test well—tests need to be performed under very well controlled conditions to get accurate results and is thus always subject to confirmation bias and the like.

I did find a single report done by Environment Canada on one of these products for diesel engines and they found that adding the hydrogen generator device didn't make any statistically significant change to combustion efficiency, exhaust emission rates, or fuel consumption. They calculated that the device was about 60% efficient at making hydrogen, at best, and used about 8 mL of diesel fuel every 100 km, so it doesn't really make efficiency worse to any appreciable degree, but doesn't help either. (Source: ERMD Report # 2004-032 If you search, it's on scribd, but I couldn't find any permanent sources)

In short, a water-fuelled car on its own or as a hybrid with gasoline or diesel don't seem to be possible. Water is a very stable compound and it takes a great deal of energy to break it into hydrogen and oxygen gas. At most, you can get the energy you put in back, and realistically, a lot will be lost along the way. This makes a solely water fuelled car a perpetual motion machine of the first kind as it violates the law of conservation of energy—one cannot burn hydrogen and oxygen to produce energy to make the same amount of hydrogen and oxygen consumed and still have energy left over to move a vehicle. A hybrid vehicle faces a similar problem. Because the part of the system that involves water consumes energy rather than contributes it to motion, the gas or diesel part has to produce more energy to make up for the difference. Given that modern gas and diesel engines burn almost all the fuel put into them, there doesn't seem to be any mechanism by which adding a small quantity of hydrogen and oxygen could produce significantly more energy.


Would copper sulfate corrode copper metal over time?

$\ce{CuSO4}$ will not corrode pure copper appreciably over a few months (though there is some dissolution and redeposition through random processes, such as density gradients and thermal motion, it might just reduce a mirror finish to slightly roughened).

However, if the electrodes or wires are brass, bronze or other alloy, or if there are other metals in contact, such as a tin-lead solder joint, there is a chance of some corrosion. In fact, copper sulfate is used to test passivation of stainless steel, and visible corrosion of iron can take place in minutes.


Does radioactivity affect chemical reactions?

I'd like to add to the other answers. Although not directly an answer to your question, I do feel it's relevant. Radioactive decay can cause damage to crystalline solids.

Take the example of apatite $\left(\ce{Ca5[PO4]3[F,Cl,OH]}\right)$, which can host a bit of uranium in the crystal structure. When this uranium decays, the resulting particle damages the host crystal. The crystal can then be polished and etched to reveal these so called "fission tracks":

(source: Stanford Fission Track Thermochronology Laboratory)

Heating of a damaged crystal will cause the damage to "heal" and the tracks to disappear. Thus, radioactive decay can indirectly induce chemical reactions. This routinely occurs in the deep Earth. It also has some very useful applications in geochronology, which you can read about in the Wikipedia article about fission tracks.


How to calculate the number of water molecules of a hydrate based on mass loss upon heating?

Determine what the $\pu{2.388 g}$ is.

   It's the amount of water lost from the crystal structure, assuming the sample was heated enough for this purpose.

What does 'mole' mean? Isn't it just an indicator of the number of atoms/ions/molecules? So, what would be the relation between the quantity sodium sulfate and water molecules?

   Yes, 'mole' is an indicator of quantity. Take a look at the formula: $\ce{Na2SO4.nH2O}$. What does that mean? It means in a unit, there are $n$ water molecules trapped inside the crystal framework, which consists of sodium and sulfate ions.

How do you proceed into finding the amount of substance (in mole)?

   You need to somehow 'remove' any unknown variables to calculate the amount of substance, so you need to remove $n$. Try subtracting the amount of water from the weight of the whole sample:
 $$m(\ce{Na2SO4.nH2O}) - m(\ce{H2O}) = m(\ce{Na2SO4})\\ \pu{4.250 g} - \pu{2.388 g} = \pu{1.862 g}$$

How many grams of anhydrous $\ce{Na2SO4}$ are there? What is the formula to convert that into moles? How many moles of $\ce{Na2SO4}$ are there?

   The formula is $\displaystyle \mathfrak{n}=\frac{m}{M}$ where $m$ is mass, $M$ is molar mass and $\mathfrak{n}$ is the amount of substance. Thus $\displaystyle \mathfrak{n}=\frac{\text{mass of anhydrous sample}}{\text{molar mass of sodium sulfate}} =\frac{1.862}{142} = \pu{0.0131 mol}$.

You found how many moles of sodium sulfate are there. Isn't that going to be the same for water? Plug into the formula; and find $n$.

   We know that for in any unit of Gluber salt, there are 2 sodium ions, one sulfate ion, and $n$ molecules of water. Thus, it's safe to assume that $\mathfrak{n}$ for water is $n$ times that of what we found for sodium sulfate.
 $$\begin{align}0.0131\times n &=\frac{2.388}{18} \\ &= 0.132\bar{6} \\ \Rightarrow n&=10\end{align}$$


[...] I am not sure which is reduced at the cathode. Would it be the sulfate ions?

orthocresol has already pointed out in his comment that there's more in $\ce{Fe2(SO4)3}$ than just the sulfate.

Take a look at the cation:

$$\begin{align}
\ce{Fe^3+& + e- &<=>&Fe^2+\quad &+0.77V} \\
\ce{Fe^3+& + 3e- &<=>&Fe(s)\quad &-0.04V}
\end{align}$$

(Standard electrode potentials are cited from wikipedia.)


Short answer: Both chlorine and iodine in specific concentrations are used for water disinfection.

Iodine has been used to disinfect water for a long time. It is used in concentrations between 2.5 ppm to 7 ppm, but there is some controversy regarding the maximum safe dose of iodine because excess of iodine consumption in humans can lead to various diseases(thyroid disorders being very common). I believe this is the reason we have moved towards more safer options like chlorine for water disinfection.


How to correctly write the condensed formula for salicylic acid

Some people say $\ce{CHCHCHCHC(OH)C(COOH)}$ Or $\ce{(CH)4C(OH)C(COOH)}$

Both of these are definitely wrong since it seems to indicate some sort of alkene.

  Others say $\ce{C6H4(OH)COOH}$

This is much better since it "hints" that $\ce{C6H4}$ is a benzene ring. With only 4 hydrogens and what seems to be two substitution groups that fits. Note there is no indication of the relative position of the groups. So this would represent either 3-Hydroxybenzoic acid or 4-Hydroxybenzoic acid as well.

The basic empirical formula would be $\ce{C7H6O3}$ which would represent probably dozens of molecules.

So the best way to denote the particular molecule specifically is to name it, or to draw out the whole 2D structure.


How do sunscreens protect the skin from UV rays?

Apart from titanium dioxide, the active species in sunscreens are organic compounds from different substance classes, such as cinnamates, benzophenones, benzimidazoles, and triazines.

The do absorb in the 280-315 (UVB) and/or 315-380 nm (UVA) range. This "uptake" of energy results in an electronic excitation, the compounds undergo a transition from the $S_0$ (ground) state to an excited $S_1$ state.

The excited singlet state ($S_1)$ may

deactivate radiationless via internal conversion (IC), the energy of the photon is released as heat or
undergo intersystem crossing (ISC) to a triplet state (usually $T_1)$ with different spin multiplicity

The latter state then might again loose its energy via IC or undergo a chemical transformation, through which the suncreen might be turned to a species that not longer is able to absorb UV radiation.


Find whether a gas cools or heats up on Joule-Thomson expansion

The equation for dH is:  $$dH=C_pdT+\left[V-T\left(\frac{\partial V}{\partial T}\right)_P\right]dP$$


Oxygen can be oxidized. However, it is uncommon due to its high electronegativity. In fact, it is highest second only to fluorine. This makes it extremely difficult to remove electrons from oxygen, thus making it hard to oxidize.

One definitive example of oxygen being oxidized is oxygen difluoride. This exception is even listed on many lists of oxidation state rules (see: oxidation states).

And of course, if you go beyond stable molecules, it is possible to ionize oxygen with an electrical charge.


Can I use lye as a cooking leavener instead of baking soda?

I'm not an expert here, but, as a chemist with a strong gastronomic interest (and a penchant for waffles), can suggest some things which may be helpful. Lye is a caustic alkaline solution, traditionally KOH, and more commonly these days NaOH.

First, my understanding of lye as an agent in cooking is limited to its use in pretzels and "laugen" bread and ramen soups. The pretzel making process involves a final dipping of the dough in a lye solution in order to promote the Maillard reaction during cooking; this is not the same as the leavening process. Ramen noodles are a type of "alkaline noodle" and I believe use the lye to break down the gluten to make the flour more pliable and less elastic. Again, not the same as the leavening process.

The use of lye as a leavening agent comes from taking lye (produced by washing fireplace ash) and boiling with lard to produce potash. The purified version, pearl ash, was one of the first chemical leaveners used in the kitchen. Pearl ash acts as a leavening agent by reacting with an acidic ingredient; fruit juice, sour milk or molasses, to produce CO2.

So to answer your questions:

I can't see how it would make your waffles fluffier than conventional modern rising agents, even if you did get your chemistry right. You still need to get your CO2 from somewhere and substituting your bicarb for lye actually removes this element. You would need to rework your entire recipe, I think.
Yes. Without doubt. More than likely it will degrade the taste of what should otherwise be a delicious dish. Alkaline foods are well known to have a bitter after taste (think kale and broccoli). Even scones made with a little too much baking soda can leave your mouth tasting and feeling like you've just had your annual clean and scale at the dentist.
While lye is used as a cooking agent in some instances, even the food grade solutions are extremely caustic, and, undiluted, are actually classed as poisons in some instances. It is corrosive to the body, especially mouth, throat and stomach, and can cause serious damage to the eyes. This may be stating the obvious, but best to assume ignorance and work up from there. Apart form mooncakes, I am not aware of a dough recipe (or anything else for that matter) that incorporates lye within the dough itself. Unless you have experience handling this ingredient in the kitchen, alternatives are recommended. I cannot tell you how you would determine whether your 'lye reaction' is complete within the dough; certainly gauging the completion of 'the rise' would not be sufficient, and the cooking process may need to be adjusted to accommodate this. Mooncake dough, for instance, is allowed to sit overnight, and even for up to 3 days before cooking. Too long to wait for my waffles!

For your interest, there is a very good comparison of various traditional and modern leavening agents here. Interestingly, the author concluded that the best agent overall was baking soda. Just like modern research chemistry in the lab - someone has done the reaction already.


Why don't heavy and light gases separate in the atmosphere?

That's because of two reasons. One is entropy, the ultimate force of chaos and disorder. Sure, gases would like to be arranged according to their density, but even above that, they would like to be mixed, because mixing creates a great deal of entropy. If you prevent the mixing, then they would behave just as you expected. Indeed, a balloon filled with $\ce{CO2}$ would drop right to the floor and remain there.

On the other hand, if you allow mixing, light gases wouldn't necessarily go upward. Just pierce that balloon with helium, and... Well, you won't actually see much, but in fact, the helium will disperse in the atmosphere and remain there. True, its contents in the upper layers would be somewhat higher, but only somewhat. It is not like a layer of pure helium floating atop all those $\ce{O2}$ and $\ce{N2}$.

The importance of entropy is by no means limited to gases. Think of all that salt in the oceans. Salt is much more dense than water; wouldn't it just drop to the ocean floor? Well, no, it rather wouldn't.

The other reason is the constant action of winds and currents mentioned by Zhe. They stir the atmosphere (or sea water, for that matter) and make it even more uniform than it might have been otherwise.


Is it realistic to have a natural oxygen source in a cave?

According to this page, the CO2 buildup in the cave might also be a problem.

But, if there was a river that ran on the surface, then oxygen and nitrogen would dissolve into the water. The stream could then pass into the cave and release some of the dissolved oxygen into the oxygen deficient environment in the cave. It would also remove the CO2. Of course, if the surface of the planet is frozen, then the river would have to avoid freezing somehow. But perhaps there is a hot spring on the surface.

My intuition is that the rate of oxygen dissolving out of the water would be very slow and it could be increased by increasing the surface area of the water. In the human lungs, there are many, many tiny branches (Alveoli) that allow the lungs to have a huge surface area so that the slow diffusion of oxygen from the air into our blood can go fast enough for us to survive. Perhaps the cave-people can accomplish something similar if they let their oxygen-rich stream pass over a tall, broad waterfall that would cause the river to break up into tiny droplets, thereby increasing the surface area. This has the added benefit of looking super cool in case your story gets made into a movie :)

Other options:
1) they could generate electricity and use that to make hydrogen and oxygen from water.
2) http://www.mpg.de/621120/pressRelease201003241?filter_order=L


Why Does A Brønsted–Lowry Acid Accept Proton from Stronger Acid?

Nitric acid is definitely a strong acid, but that does not exclude the possibility of protonating it with a stronger acid. It can be effectively protonated by a stronger acid such as sulfuric acid. Simply because something has "acid" in its name does not mean it has to dissociate into protons and its conjugate base, or protonate something else.

$$\ce{HNO_3 + H2SO4 <=>H2NO3^+ +HSO4-}$$

Note also that even strong acids in their pure forms - i.e. liquid nitric acid or liquid sulfuric acid - can and do auto-ionize to a fair extent. So if nitric acid can react with itself, of course it can react with a stronger acid.


Impact of Pressure on equilibrium of Tin allotropes

First (as you already did) you need to determine the equilibrium temperature at $1\ \mathrm{bar}$. This happens to be when  $\Delta G^\circ = 0$, so that $T^\circ_{trans}= \frac{\Delta H^\circ}{\Delta S^\circ}=286.36\ \mathrm K$.

Next you can apply the Clausius equation in integrated form

$$ T = T^\circ_{trans} + (P-P^o) \frac{\Delta V_m^\circ}{\Delta S^\circ} $$

I obtain the following value for $\Delta V_m^\circ$:

$\Delta V_m^\circ = (1/5.766-1/7.287)\times \pu{ 118.71\times10^-6 m3/mol} = \pu{4.2973\times10^-6 m3/mol}$

Plugging this and the values you provide for $\Delta P$ ($1089\ \mathrm{bar}$) and the standard entropy changes $\Delta S_m^\circ$ gives $T=\pu{219.89 K}$. Not exactly what you say is the right answer, but, closer.


Conversion of units for concentration: mol/l to ppm

Convert mol-dm^2 to grams/volume.  A decimeter is 10 cm.  A dm^3 is then a liter.  Water has density ~1 g/cm^3.  Multiply the g(Mg)/g(water) by a million.


Can you purify a metal oxide with a laser?

this is an interesting question and I will answer it a bit differently.

There is a well known analysis technique in materials chemistry called ICP-MS (Inductively Coupled Plasma Mass Spectrometry) where you have a sample, and using an inductive electric field, the plasma of the sample containing ions of the elements present is created. This is then passed thorugh a mass spectrometer to know the elemental composition of a material synthesised. It is one of the best known techniques for elemental analysis.

However, the idea of using ICP-MS like technique for purification of metal oxides is novel to me. Instead of induction to introduce electric field and rip apart the electrons from the atoms, you can achieve the same using a laser. The rest idea is same.
However, you can try it, and in case you design an instrument and get a publication for the same, kindly let me know here. I will look forward to it.

References:

https://en.wikipedia.org/wiki/Inductively_coupled_plasma_mass_spectrometry

Kindly contact me in comments in case you have any questions.
Stay Safe. With greetings from India. जय हिन्द. Jai hind.


Why do we use LiCl as a source of chloride ions as nucleophile but not other alkali metal chlorides?

One reason is lithium chloride is more soluble in the organic solvents often used for such reactions, and getting the chloride source into the reaction medium is more important than having it be perfectly ionic.

Let's roll the tape.  Wikipedia identifies lithium chloride as soluble in a variety of polar solvents and gives numerical values for several of them.  The solubility reaches hundreds of grams per liter in some of them, including over 400 grams per liter in methanol.

Much less data are listed here for sodium chloride, but the solubility of sodium chloride in methanol is listed as 14.9 grams per liter under ambient conditions, about 30 times less on a mass basis than lithium chloride.  And the chloridecomponent shows an even bigger discrepancy because of the heavier mass per mole of chlorine in sodium chloride versus lithium chloride.

Clearly lithium chloride's solubility carries over better from water to polar organic solvents.  The reason for this is lithium chloride is more adaptable.  The small size and higher polarizing power of the lithium ion imparts greater covalent character in lithium chloride than in its heavier alkali metal congenres.  Alternatively, in a purely ionic model, it promotes the formation of ion pairs which are uncharged overall because of the equal but opposite charges of the constituent ions.  Either way, the lithium chloride adopts a conformation that resembles a polar covalent molecule, which fits better with the polar but weakly ion-solvating solvents organic chemists generally use than the dissociated ions favored by chlorides of heavier alkali metals.


why distillation of mercury is done at low pressure?

There is one obvious reason and one possible confusing factor.

The production of mercury from cinnabar involves heating cinnabar in a stream of air which causes oxidation of the sulphur and produces a stream of mercury vapour which can be condensed to give the metal. This isn't distillation even though it involves condensation of mercury vapour, but a chemical reaction.

But mercury can be purified by distillation because it is very volatile for a metal. But the boiling point at normal pressure is ~360 °C which is inconveniently hot. Using reduced pressure (say around 1/1000th of an atmosphere) can make this practical at closer to 100 °C which is far more convenient.


treatment of alkyl chloride with aqueous KOH and alcoholic KOH

I have added an image to answer your question, always look at the acid base reactions which are possible to take place, that helps in understanding the reactions and their respective mechanisms.


What are these stains on my stainless steel pans?

It could be that metals in your tap water are plating out;  i.e. the more active iron in stainless steel is being replaced by a less active metal such as copper. Manganese also tends to precipitate, leaving stain, and these, in particular, are dark colored, but can be removed.

Have your tap water tested, either at a nearby school or with a commercial kit(s) that include copper, manganese and lead tests. If there is a high concentration of anions, you might want to install an ion-exchange water softener.


Why do the pennies in the pre/post 1982 penny experiment sometimes dissolve completely?

Your conclusion about formation of the Copper Oxide is correct, but not about the reaction with atmospheric oxygen.

The origin of the Oxygen is from the hydrochloric acid solution. The solution does not mean is pure HCl , it means is a dissolve in water.  There is a redox reaction with water

I leave to you to find the redox reaction.


Tetrahedron is minimum energy conformation for atom with 4 electron domains in VSEPR theory

Sketch of a Simpler Solution
There is one remaining way that is simpler than all of the above, and it has to do with circumscribing a tetrahedron with a cube.
I will leave it to the reader to work on this most beautiful version. The properties of the cube ensure that the distances are maximized, and some more geometry will show that the inscribed shape is indeed a regular tetrahedron. For a hint, check out this excellent answer to this question on the math stack exchange site.


How do I compute the mean dissociation time from the dissociation constant, assuming instantaneous binding reaction rate?

Note that convention usually used Capital K for TD equilibrium constants and small k for kinetic rate constants.
Thermodynamic equilibrium constants say nothing about kinetic reaction rate constants, but about their ratio. There can be 2 reactions accidentally with the same equilibrium constant, but one with both forward and backward reaction rate very fast, the other very slow.
If there is reaction $\ce{AB <=> A + B}$ with the equilibrium constant $K_\mathrm{c}=\frac{[A][B]}{[AB]}$
with the forward reaction kinetic $\frac{\mathrm{d}[AB]}{\text{d}t} = - k_\mathrm{f} \cdot [\ce{AB}]$ and the reaction halftime $t_{1/2}=\frac{\ln{2}}{k_\mathrm{f}}$
and with the backward reaction kinetic $\frac{\mathrm{d}[AB]}{\text{d}t} = k_\mathrm{b} \cdot [\ce{A}][\ce{B}]$
then $K_\mathrm{c}=\frac {k_\mathrm{f}}{k_\mathrm{b}}$
All is assuming the reaction has trivial simple reaction mechanism.
Credit to @Porphyrin:

If $\ce{ A + B -> AB}$ is diffusion controlled, the rate constant would be (for equal sized molecules or approximately so ) $k_\mathrm{d} = k_\mathrm{b} = 8000 \frac {RT}{3η}$
where η is the solvent viscosity,
$k_\mathrm{d}$ evaluates to $≈\pu{5⋅1010dm3/mol/s}$ in low viscosity solvents such as ethanol.

Then :
$$kf=k_\mathrm{b} \cdot K_\mathrm{c}=K_\mathrm{c} \cdot \left( 8000 \frac {RT}{3η}\right)$$
and the reaction halftime is:
$$t_{1/2,f} = \frac {\ln 2}{k_\mathrm{f}} =\frac{\ln {2}}{K_\mathrm{c} \cdot \left( 8000 \frac {RT}{3η}\right)} =\frac{\ln {2} \cdot 3η}{  8000 \cdot K_\mathrm{c} \cdot RT} $$


Would it be possible to produce a non-toxic atmosphere using a coal fire and frozen oxygen?

Yes, it is possible to produce a non-toxic atmosphere using a coal fire and frozen oxygen.

I'll answer your questions last to first.

No, low pressure oxygen will not make flammability worse, fires require something to bond with oxygen (often carbon, causing carbon dioxide). Low pressure will make things less flammable as there is less oxygen to bond with.

Yes, using an open coal fire in a semi-enclosed space will gain toxic levels of carbon dioxide, carbon monoxide and sulfur dioxide (among a few others) quickly. However with a minor modification from a open fire to a fire in a old fashioned chimney pipe stove (venting outside the hut) would eliminate this.

Lastly, Definitely Yes, a coal fire will produce more than enough energy to thaw the oxygen it uses. To quote figures 1 pound of coal will produce roughly 12,500 btu's (British Thermal Units) Assuming the temperature of the frozen oxygen we are thawing is -455°F (the "temperature of deep space") and about 3 degrees below the freezing temperature of helium. To thaw 1 gallon of the frozen oxygen will require roughly 900 btu's (if your stove has a 80% efficiency) and produce 860 gallons of pure oxygen at 80°F and atmospheric pressure. Coal will consume 140% of it's weight in pure oxygen, so for 1 pound of coal 135 gallons of your oxygen will be consumed. Leaving you with 11,500 btu's extra heat and 725 gallons of oxygen. The average Human consumes 25 gallons of oxygen per day making this plenty to survive on.


So G is the expected cyanohydrim, but the second step looks very vigorous (concentrated sulfuric acid and heat).  I would expect two things to happen under these conditions, 1) hydrolysis of the nitrile group to the corresponding carboxylic acid (the reaction conditions are too strong to stop at the intermediate amide, rather the amide is further converted to the corresponding carboxylic acid; keep in mind that an equilibrium will initially exist between the amide and the carboxylic acid, but everytime the acid forms from the amide, ammonia is released and escapes from the system making the reaction irreversible, pushing this reaction further and further to the carboxylic acid side, ultimately only the carboxylic acid will be present.), and 2) acid catalyzed elimination of water from the alcohol portion of the molecule.  Taken together I would expect the final product to be the various cis-trans isomers of "A".


What are the products of the reaction between fluorine and ammonia?

Fluorine reacts with ammonia to give dinitrogen and hydrogen fluoride as the major product.

The Formation of Dinitrogen Tetrafluoride in the Reaction of Fluorine and Ammonia - J. Am. Chem. Soc. 1959, 81 (23), 6338-6339 - states that dinitrogentetrafluoride accounts for 6% of the yield while the major reaction is:

  $\ce{3F2 + 2NH3 → N2 + 6HF}$

The same article quotes in its first paragraph that the result remained the same irrespective of the concentration of the reactants. (In the case of chlorine it matters which reactant is taken in excess)


What's the difference between pharmacotherapy and chemotherapy

I would like to believe that the major difference between these terms is based on their backgrounds and time however nowadays there are used interchangibly, without paying much attention to actual definitions.
Background

The term chemotherapy was coined by Ehrlich at the beginning of the
20th century to describe the use of synthetic chemicals to destroy
infective agents.  In recent years, the definition of the term has
been broadened to include antibiotics-substances produced by some
microorganisms (or by pharmaceutical chemists) that kill or inhibit
the growth of other microorganisms. It can also be  broadened to
include agents that kill or inhibit the growth of cancer cells.

Chemotherapy is the term originally used to describe the use of drugs that are 'selectively toxic' to invading microorganisms while having minimal effects on the host.
The term also embraces the use of drugs that target tumours and, in fact, has now come to be associated specifically with that branch of pharmacology (Oncology). Although, in the public mind at least, chemotherapy is usually associated with cytotoxic anticancer drugs that cause unwanted effects such as loss of hair, nausea and vomiting.
Pharmacotherapy is a more recent term which generally means the treatment of diseases through the administation of drugs. However there is more to it, the term also describes a more patient-focussed approach in which pharmacists treat patients through drug management with one or more specific outcomes.
The specific goals of pharmacotherapy (primary therapeutic outcomes include):
• Cure of disease (e.g., bacterial infection).
• Reduction or elimination of symptoms (e.g., pain from cancer).
• Arresting or slowing of the progression of disease (e.g., rheumatoid
arthritis, HIV infection).
• Preventing a disease or symptom (e.g., coronary heart disease).
Other important outcomes of pharmacotherapy include:
• Not complicating or aggravating other existing disease states.
• Avoiding or minimizing adverse effects of treatment.
• Providing cost-effective therapy.
• Maintaining the patient’s quality of life.
A systematic approach is consistently applied to each case. The steps involved in this approach include:

Identifying real or potential drug therapy problems

Determining the desired therapeutic outcome(s)

Evaluating therapeutic alternatives

Hope this helps
References

RANG AND DALE’S Pharmacology 8th edition: Basic principles of antimicrobial chemotherapy

Pharmacotherapy Casebook


Does the probability function for a molecule to have energy in a certain mode depend on the minimum required quantum energy level of that mode?

There's a couple of conceptual problems at the base of your question that need to be addressed.

1) Free translation of particles in space is not discrete - the distribution of quantum states is continuous, and therefore so is the distribution of energies or of velocities - but rotation or vibration are quantised, so they have discrete distributions, so the formula definitely can't be applied to them. You could apply it to molecules as classical objects, ignoring quantum behaviour, but what you are describing then is not a molecule; you won't get realistic results that you could compare to experimental results.

2) The equipartition theorem states that rotation or vibration degrees of freedom will have the same energy as translation degrees of freedom, but the same energy will not translate into the same velocity.

For instance, in a classical treatment, the energy distribution function per degree of freedom in equilibrium that we obtain from Maxwell-Boltzmann is

$$ f_\epsilon = \sqrt{ \cfrac{1}{\pi \epsilon_i k_B T} } \exp \left[ \cfrac{-\epsilon_i}{k_B T} \right] $$

Taking into account that classical rotational energy is

$$ E_{rot} = \frac{1}{2} I \omega^2 $$

we can derive the classical angular velocity distribution:

$$ f_\omega = \sqrt{\cfrac{I}{2 \pi k_B T}} \exp \left[ \cfrac{-I \omega_i^2}{k_B T} \right] $$

with the moment of inertia $I$ being dependent on the geometry of the molecule. For instance, in the simplest case of a diatomic molecule approximated as a rigid rotor, it is

$$ I = \mu d^2 $$

where $\mu$ is the reduced mass and $d$ is the interatomic distance.

However, as we said, this is not a correct description of an actual molecule, in which vibration and rotation movements are quantised. In a quantum treatment, the allowed values for rotational energy in a linear molecule are

$$ E_{rot} = J \left( J + 1 \right) \cfrac{\hbar^2}{2I} ~ \mathrm{with} ~ J=0,1,2... $$

note that rotational levels are degenerate (with unequal degeneracy $g_j = 2J+1$) and discrete, so we need to use a discrete partition function $Z_{rot}$:

$$ Z_{rot} = \sum_{J=0}^{\infty} \left( 2 J + 1 \right) \exp \left[ \cfrac{-J \left( J + 1 \right) \hbar^2 }{ 2 I k_B T} \right] $$

which in principle we can't calculate analytically. For molecules with a very large moment of inertia $I$ and/or at high enough temperatures $T$, the gap in energy between levels becomes very small and we can approximate this sum as a continuous integral over $J$ (the so-called high temperature limit):

$$ Z_{rot} \approx \int\limits_{J=0}^{\infty} \left( 2 J + 1 \right) \exp \left[ \cfrac{-J \left( J + 1 \right) \hbar^2 }{ 2 I k_B T} \right] dJ = \cfrac{ 2 I k_B T}{\hbar^2} $$

Under that assumption, the (discrete) angular momentum distribution of a rotational degree of freedom can be derived as

$$ F_J = \left( 2 J + 1 \right) \cfrac{ \pi \hbar^3}{ I k_B T} \exp \left[ \cfrac{-2 \pi \hbar^3 \left( J + 1 \right) }{ I k_B T} \right] $$

Note that we're using angular momentum and not angular velocity here because, while the former is a well-defined magnitude in quantum mechanics, it's not clear the later makes physical sense in a quantum treatment of molecules. We could naively calculate a semi-classical "angular velocity" distribution from that angular momentum distribution, but, again, we'd not be actually describing a molecule.

And a final conceptual problem:

3) Rotation and vibration (and electronic) motion is coupled; to give a simple example, in a diatomic molecule, as rotation angular momentum increases, interatomic distance (and therefore moment of inertia) changes; a simple approximation is to introduce a rotational term representing centrifugal distortion, but for higher precision both movements have to be considered at the same time, what makes for much more complex equations.

Keep also in mind that we've restricted ourselves to the simplest cases in rigid diatomic molecules; more complex systems have much more complex mathematics involved and often have to be treated empirically.


What you are asking amounts to the collected knowledge, wisdom, and intuition of the organic chemistry community. Your best bet will be to find an organic chemist, preferably one with a graduate degree, to work with you on this project. Said organic chemist can apply his/her knowledge/wisdom/intuition to each functional group in less time than it would take you to look up the answer. Doing so frees you to work on the algorithm you describe in your other question.

The beginning of the answer to this question takes me two semesters to get through with my introductory organic chemistry students, and that is just for the common functional groups. After covering the basics of organic structure and reactivity, most texts have 15-20 chapters organized by functional group containing the reactions of and syntheses of said functional groups.

The common functional groups covered in most introductory sequences:

alkanes
cycloalkanes
alkyl halides
alkenes
conjugated dienes and other unsaturated systems
alkynes
aromatic ring systems
alcohols
glycols and other diols
phenols
ethers
epoxides
aldehydes
ketones
carboxylic acids
esters
anhydrides
acyl halides
amides
amines
nitriles
thiols
sulfides
acetals
imines
enamines
heterocycles
nitro compounds
diazo compounds
carbohydrates
amino acids
nucleic acids

If you want to know about every functional group, you will need an advanced, probably graduate level, synthesis course. However, even I have not heard of some of the designations on your list, like semithiocarbazide.

  Some good ideas on where to start, if you choose not to consult with an organic chemist.

Check an undergraduate level organic chemistry book out of your local library and get some sense of the subject. In your case, it does not look like you need to know how the reactions work, just that they exist, so you can use the text as a reference source.
Sign up for a free MOOC in organic chemistry. Again, you just want to know that the reactions exist, so you can pick and choose the content from the course.
Acquire a copy of one of the seminal books in organic chemistry, March's Advanced Organic Chemistry. This book is not for the fainthearted but it gives good treatment to the functional groups, including the less common ones.
Find (do not buy - these books are not worth the price tag) a copy of either Comprehensive Organic Transformations: A Guide to Functional Group Preparation by Richard Larock (which is now 14 years old) or the set of Comprehensive Organic Functional Group Transformations, which appears to be many volumes (now 9-10 years old).


The Gibbs energy has not been derived this way. It came from a thought process. About 100 years ago, Gibbs was puzzled by the very existence of spontaneous endothermic reactions. In mechanics, objects always fall down. They never "fall up" spontaneously. They never gets more energy spontaneously. In exothermic reaction, the matter looses spontaneously energy, like a falling object. In endothermic reactions, the atoms and molecules receive energy from the surroundings. They increase their energy. It is surprising. So Internal energy U and Enthalpy H are not equivalent to the gravitational energy. A spontaneous reaction can happen in chemistry with $\Delta$H <$0$ ou >$0$. As a consequence, H or U are not the good criteria to foresee whether a chemical reaction will spontaneously happen.

So Gibbs was thinking : What is the energy that always gets out of a system of molecules and atoms in a spontaneous reaction, even when the reaction is endothermic ? The answer is of course : the electric energy. A galvanic cell can only produce energy, whatever its heat effet. A working galvanic cell may become hotter or cooler, but it can only deliver electric energy. The amount of electric energy delivered by a cell is defined by the difference of a new potential energy, called Gibbs energy $G$, and defined so that $\Delta$$G$ is always negative in a cell. $\Delta$$G$ is obtained form the expression : $\Delta$$G$ = - $zEF$, where $E$ is the voltage of the cell, $F$ is the Faraday, and $z$ is the number of electrons exchanged in the reaction inside the cell. Furthermore, $\Delta$$G$ is defined and calculated in the same way as $\Delta$$H$. Gibbs energies of formation exist like enthalpies of formation.

For example, one may write : $\Delta$$G_{react} = \Sigma$$G_{fin}$-$\Sigma$$G_{in}$ and so on. You probably know the rest to the theory. And for example the relationship between Gibbs energy and the equilibrium constant.

Now if you plot $\Delta$$H$ and $\Delta$$G$ of a given chemical reaction versus $T$. you will notice that $\Delta$$H$ practically does not change much with $T$. But the voltage $E$ and of course $\Delta$$G$ do change much with $T$. The line $H$ vs. $T$ is nearly horizontal. But the line $G$ vs $T$ is inclined. The strange result of this drawing is the fact that both lines are joining one another at $T$ = $0$ $K$. So the difference of these energies is of course : $H-G$ = $T\Delta S$, where $\Delta S$ is the slope of the $\Delta G$ curve.


Finding pH after dilution of acetic acid

$$\ce{CH_3COO- + H_2O<=>CH3COOH + OH-}$$
Approximating that all the $\ce{OH-}$ is from this equilibrium, $[\ce{CH3COOH}] = [\ce{OH-}]$.
\begin{align}
  [\ce{OH-}]_\text{initial} &= [\ce{CH3COOH}]_\text{initial} \\
    &= \pu{10^{-4}M}\\
  \frac{[\ce{OH-}]_\text{initial}[\ce{CH3COOH}]_\text{initial}}{[\ce{CH3COO-}]_\text{initial}[\ce{H2O}]_\text{initial}}
  &= \frac{[\ce{OH-}]_\text{final}[\ce{CH3COOH}]_\text{final}}{[\ce{CH3COO-}]_\text{final}[\ce{H2O}]_\text{final}}
\end{align}
Then you can choose to approximate $[\ce{H2O}]$ as a constant value and omit it from both sides (or not if you need great accuracy).
\begin{align}
  [\ce{CH3COO-}]_\text{initial} &= \pu{1 M}\\
  [\ce{CH3COO-}]_\text{final} &= \pu{\frac{10}{100} M}\\
  x &= [\ce{OH-}]_\text{final} = [\ce{CH3COOH}]_\text{final}
\end{align}
Solve for $x$.


How to experimentally measure number of protons or electrons in an element?

Moseley measured the atomic number in 1913 with relatively modest equipment (by modern standards) but I doubt that you could reproduce his experiments in a typical kitchen.

Atomic number at Wikipedia

Expansion as suggested in comments.

Here is a quote from that article of what he did.

"Moseley measured the wavelengths of the innermost photon transitions (K and L lines) produced by the elements from aluminum (Z = 13) to gold (Z = 79) used as a series of movable anodic targets inside an x-ray tube."

It might be just possible at home if you really knew what you were doing but I doubt that it would be cheap or safe.  What is regarded as safe has probably changed a lot in the last 100 years.


EPDM tube resistant to ozone on the long run?

The problem with EPDM is that it has tertiary carbon sites (and some double bonds for cross-linking). Forget the latter as let's assume the material is already cross-linking.
Normally carbon backbone polymers with substitution need anti-oxidants to prevent the C-H, where there is a side group, from oxidising. These delay the oxidation process and define the lifespan of the material. So after time, the anti-oxidants will be consumed and the polymer will be prone to oxidation, which leads to fragmentation.
Ozone is a great oxidant as you know. When we used high pressure ozone,  we always used fluorinated everything. By using fluorine saturated polymers, there's no sites for oxidation to occur.


Methylammonium lead(II) iodide perovskites in bulk form

Procedure from the paper by Xie et al. [1] (Powder XRD didn't reveal any other impurity phases in the bulk samples):

Synthesis of $\ce{CH3NH3PbI3}$. The as-prepared $\ce{CH3NH3I}$ and $\ce{PbI2}$ (Sigma-Aldrich, 5N) were mixed in stoichiometric ratio and dissolved in γ-butyrolactone (Sigma-Aldrich), and then a 40wt% solution was formed. The solvent of as-prepared solution was evaporated in an oven at 60 °C for 2 days and $\ce{CH3NH3PbI3}$ precipitate was collected.

Similar approach has been utilized by Yamada et al. [2] for the synthesis of monocrystalline samples. From the SI:

Sample Preparation
$\ce{CH3NH3PbI3}$ $(\ce{MAPbI3})$ single crystals were prepared from γ-butyrolactone solution at 100 °C. Purified lead iodide (2.73 g, 5.92 mmol) and methylammonium iodide (0.93 g, 5.84 mmol) were dissolved in dry
γ-butyrolactone (4.8 mL) at 60 °C by stirring. The resulting solution was poured into a screw-cap vial which was placed on a hot plate at 100 °C for three days to produce a single crystal with dimensions of around 5 mm x 5 mm x 5 mm. We obtained a clean surface by cleaving the single crystal with thin knife under an Ar atmosphere before being used for the measurements. The experiments were conducted in an Ar atmosphere. We confirmed that the same results could be obtained with multiple samples that were cleaved and kept in Ar atmosphere.

Alternative method by Hoke et al. [3]; from the SI:

Crystals of $\ce{(MA)PbBr3}$ (orange, rhombic dodecahedra) were first
grown by combining stoichiometric ratios of $\ce{PbBr2}$ and $\ce{(MA)Br}$ in concentrated (9 M) hydrobromic acid. Diffusion of acetone into this solution afforded crystals within 2 days, which were filtered and dried under reduced pressure.

It appears that methods described in [1] and [2] are based on the preparation of the solution of $\ce{MAPbI3}$ for the thin film deposition on $\ce{TiO2}$ layer, which was published earlier by Noh et al. [4] (SI).
References

Xie, J.; Liu, Y.; Liu, J.; Lei, L.; Gao, Q.; Li, J.; Yang, S. Study on the Correlations between the Structure and Photoelectric Properties of $\ce{CH3NH3PbI3}$ Perovskite Light-Harvesting Material. Journal of Power Sources 2015, 285, 349–353. https://doi.org/10/f7c87m.
Yamada, Y.; Yamada, T.; Phuong, L. Q.; Maruyama, N.; Nishimura, H.; Wakamiya, A.; Murata, Y.; Kanemitsu, Y. Dynamic Optical Properties of $\ce{CH3NH3PbI3}$ Single Crystals As Revealed by One- and Two-Photon Excited Photoluminescence Measurements. J. Am. Chem. Soc. 2015, 137 (33), 10456–10459. https://doi.org/10/f7p5s9.
Hoke, E. T.; Slotcavage, D. J.; Dohner, E. R.; Bowring, A. R.; Karunadasa, H. I.; McGehee, M. D. Reversible Photo-Induced Trap Formation in Mixed-Halide Hybrid Perovskites for Photovoltaics. Chem. Sci. 2015, 6 (1), 613–617. https://doi.org/10/gc3g49.
Noh, J. H.; Im, S. H.; Heo, J. H.; Mandal, T. N.; Seok, S. I. Chemical Management for Colorful, Efficient, and Stable Inorganic–Organic Hybrid Nanostructured Solar Cells. Nano Lett. 2013, 13 (4), 1764–1769. https://doi.org/10/gc3g8w.


How to Calculate Lennard-Jones Parameters from TDDFT?

I'm basing my answer on T. Gould, T. Bucko, J. Chem, Theory Comput. 2016, 12, 3603-3613.

In your TDDFT calculation, what you want to compute is the frequency-dependent dipole polarizability, $\alpha_{X/Y}(i\omega)$, that determines the dispersion coefficient, $C_6$.
$$ C_{6,XY} = \frac{3}{\pi}\int d\omega \alpha_X(i\omega)\alpha_Y(i\omega)$$

They also propose some approximate relationships that depend only on the static dipole polarizabilities, $\alpha(0)$.
$$ C_{6,XY} \approx \Xi [\alpha_X(0)\alpha_Y(0)]^{0.73\pm 0.01} $$
where $\Xi = (1.5 \pm 0.1)[\text{hartree} \cdot a_0^{1.62} ]$.

I don't know any tutorials for this, but you could start trying to reproduce some of the numbers on these papers.


How to determine steady state of multi-step reaction with both reversible and irreversible steps?

Can anyone point me in the right direction for how to determine the relative concentrations at equilibrium?

This system will not reach equilibrium as long as there are irreversible reactions removing U2. If someone claims to have found an equilibrium state, the rates associated with k1 and k2 would be equal, but the [U2] would decrease because of those irreversible reactions.

  I know that kt=3kd must be true for the whole system to be in balance, but I have been struggling to figure out how to relate the other rate constants.

Usually, lowercase k is used for rate constants, with the rate of reaction given by a rate law that involves k, but also concentrations of the reactants and perhaps other species. The way you are talking about it, all reactions would be zero order. For zero order reactions, there is no equilibrium (a melting process has no equilibrium - if the temperature is above the melting point, everything melts, and if it is below the melting point, everything freezes).

  I have read that the Keq for multi-step reactions is the product of each single equilibrium constant, but I'm not sure how to apply this with the irreversible reactions in the system.

You can't. There is the concept of steady state, though, where concentrations of intermediates remain constant while the forward and reverse rate of a given reaction are different. In this case, the rates increasing the concentration of an intermediate are equal to the rates decreasing the concentration of that same intermediate, but there is a net reaction. This is typical for living organisms (they can, however, reach equilibrium after they die).


The bonds in cyclopropane 1,1-diol are less strained than the cyclopropanone.

The preferred/lowest energy for tetrahedral bond angles is 109.5 degrees so the closer they can get to that the lower the energy. In cyclopropanes the angles are 60.
The lowest energy angle for the R2C=O system (SP2) is 120 degrees, in cyclopropanone it is 60. This is a very strained and unstable system, anything that reduces that strain is a positive change energywise.

This problem is discussed here see problem 2


That is a good rule of thumb.  It is based on the Arrhenius equation $$k = A {\operatorname{e}^{\frac{-{E_{\text{act}}}}{RT}}}$$ You'll find that the "temperature doubles every 10 degrees" rule  only applies if the following assumptions are valid

The reaction must obey the Arrhenius equation

it must have an entropic pre-term based on collision probability
the reaction must be an activated process (e.g. it must have an activation energy)

The Arrhenius equation works well for simple unimolecular reactions like the ring opening of cyclobutene to buta-1,3-diene.

Now you'll need to ask a weather physicist or a meteorologist if weather can be treated as an activated process.  My guess is that the answer is "no", but that's just a guess.  It seems to me that there is no minimum energy required (activation energy) for the rain to fall or for the wind to blow.


How are the molar mass and molecular mass of any compound numerically the same?

So for example, the relative molecular mass of water is 18.02 u, and it's almost the same as the molar mass of water, which is 18.015 g/mol.

  So my question is, how is 18.02 u × (6.022 × 10^23) = 18.015 g/mol?

That is not quite correct.

The relative molecular mass $M_\mathrm r$ of water is a dimensionless quantity, i.e. the unit for relative molecular mass is the unit one and not $\mathrm u$:

$$M_\mathrm r=18.015\,28$$

It's the average mass per molecule $m_\mathrm f$ that has the unit $\mathrm u$:

$$m_\mathrm f=18.015\,28\ \mathrm u$$

Note that

$$1\ \mathrm{u}=1.660\,538\,921(73)\times10^{-27}\ \mathrm{kg}$$

Also note that you are missing the unit of the Avogadro constant:

$$N_\mathrm A=6.022\,140\,857(74)\times10^{23}\ \mathrm{mol^{-1}}$$

Thus

$$\begin{align}
M&=m_\mathrm f\cdot N_\mathrm A\\[6pt]
&=18.015\,28\ \mathrm u\times N_\mathrm A\\[6pt]
&=18.015\,28\times1.660\,538\,921\times10^{-27}\ \mathrm{kg}\times N_\mathrm A\\[6pt]
&=18.015\,28\times1.660\,538\,921\times10^{-27}\ \mathrm{kg}\times6.022\,140\,857\times10^{23}\ \mathrm{mol^{-1}}\\[6pt]
&=0.018\,015\,28\ \mathrm{kg\ mol^{-1}}\\[6pt]
&=18.015\,28\ \mathrm{g\ mol^{-1}}
\end{align}$$


Can someone explain 3‐(propan‐2‐yl)cyclohex‐1‐ene forming a carbocation?

You are correct that the $\alpha$-carbon would be protonated... and then deprotonated, and back and forth, until eventually, the $\beta$-carbon is protonated, the hydride shift proceeds, and the molecule is trapped in the product state. Since protonating the $\alpha$-carbon doesn't result in some other reaction, it doesn't really matter which secondary cation is more stable. I think the important thing here is recognizing where and when a hydride shift can occur.


What dangerous gases could electrolysis with water and sodium carbonate produce?

My friend pointed out that it also contains chlorine and that's what he's afraid of.

The amount is laughable at. However, using distilled (deionized) water is recommended for other reasons.

  I failed to find out whether these compounds are hazardous. They're in solid state anyway.

They are reasonably harmless. Of course, eating them is not recommended, but copper compounds have low toxity. Move to graphite or platinum electrodes is still recommended.


Does ice increase the humidity of its surroundings?

It depends on the current humidity.

Water vapor certainly does evaporate from ice, as you can see in a vapor pressure chart. If the humidity is 100%, and the air temperature is the same as that of the ice, then the evaporation and condensation are in equilibrium; i.e. water evaporates as fast as it condenses, so there is no net change in humidity.

However, if the ice is warmer than the air, or the air has less than 100% moisture content at that temperature, then ice will evaporate more than it condenses, raising the humidity.

Conversely, if the ice is colder than the atmosphere, it can lower the humidity.


Optical rotation in inversion of sucrose

Not quite sure what your asking, but I think I understand. If I misinterpreted the question please let me know.

The hydrolysis of sucrose to a mixture of the component simple sugars, fructose, and glucose, causes the direction of rotation to "invert" from right to left. This is because sucrose rotates to the right, but the combination of the two simpler sugars rotate to the left. So the rotation at time zero is for the concentration of sucrose, and the rotation expected when the reaction is complete at time infinity can be calculated from the concentration of the sucrose.


how do i find how much energy needs to be removed?

Firstly let us understand that heat removed would be directly proportional to the decrease in temperature and also the mass. $$\delta Q\,\,\alpha \,\, m\delta T$$  Here the specific heat capacity appears as the constant of proportionality.
Therefore we have, $$\delta Q\,\,=\,\,mc\delta T$$
Now all you have to do is to use this formula to convert 8.49g of water at 24.5°C to water at 0°C.
But after this you need to use the concept of latent heat i.e. heat you need to remove to convert water at 0°C to ice at 0°C.As this value is given as 6.01 kJ/mol you will have to remove that much amount of heat for every mole of water(so you need to convert grams into moles).
Now what you have is 8.49 grams of ice at 0°C. Now just use the formula again to convert ice at 0°C to ice at -12°C (using the value of c of ice this time).
At the end you need to add all those heat energies and you are done !!
The sum of heat energies is the amount of energy you need remove from 8.49g water at 24.5°C to convert it to ice at -12°C.
Hope this helps.


What is the role of NaCl in the functionalization of gold nanoparticles?

$\ce{NaCl}$ acts as a charge suppressor so that the negatively charged gold nanoparticle and a negatively charged ligand (with thiol) are not longer charge repelled and can experience a chemical bond. $\ce{NaCl}$ is easily broken into $\ce{Na+}$ and $\ce{Cl-}$ so that the $\ce{Na+}$ can pacify the negative charge on the surface of the gold nanoparticle to become less negative and approach zero charge. Other higher charged salts such as $\ce{Mg^2+}$ impart even greater pacification.
Here are some good references:

J Liu, Y Lu. Preparation of aptamer-linked gold nanoparticle purple aggregates for colorimetric sensing of analytes. Nature protocols, 2006. nature.com

B Du, Z Li, Y Cheng. Homogeneous immunoassay based on aggregation of antibody-functionalized gold nanoparticles coupled with light scattering detection. Talanta, 2008. Elsevier. From this last ref:

in $\pu{10 mM}$ phosphate-buffered solution (pH 7.0) containing $\pu{1.0 mmol L−1} ~\ce{NaCl}$ at room …


Law of Multiple Proportions: What is the significance of small whole numbers?

The law of multiple proportions is largely obsolete these days because we all believe that atoms exist.  Prior to about 1900 this and similar laws were used to show that, for example, the ratio of oxygen in $\ce{CO}$ to oxygen in $\ce{CO2}$ is 2:1.

Of course, that's trivially obvious when we write the formulas, but before we believed in atoms the "small whole number" ratios were used as an example of an experimental situation (the 2:1 ratio) could be explained by atoms (i.e. quantizing atomic masses.)

Today, the law causes more confusion than it cures.  Atoms are here to stay.


"Limit test" vs. "Titration" in a pharmaceutical guide

From what I know, rules for making a monograph are not about giving an exhaustive list of all the analyses possible. Guides describe international agreements about what is recommended and sufficient to identify a substance and its impurities.

A titration is considerably more tedious to perform and validate (see ICH Guideline). If you use titration in a monograph, people referring to it will also have to use this method. Unless you also included a limit test but, if you did so, why bothering carrying out titration? Overall, you would waste your time and, most importantly, others' time.

So the sentence "limit test should be used..." should be read "we agreed that a limit test gives enough information in this context and we prefer it over titration".

HTH


Why would the hydrogen sulfide ion be a better nucleophile than a bromide ion?

There are a number of factors you have to take care of while deciding nucleophilicity. I will try to cover all the important ones.

First we need to clear something, polarizability is the most important factor only if the nucleophile is electrically neutral (uncharged)$^{[1]}$. Of course, this is not hard and fast, but in most cases we generally deal with, this is true. For example, triphenylphosphine is a better nucleophile than triphenylamine, because the former's central atom has a more polarizable lone pair.

In our examples, the ions are charged, so we will put aside polarizability and look at the other factors. For charged species, size of the anion should be small in order for it to be a better nucleophile. Why? Two reasons:

One, the same charge packed into a small volume is attracted more strongly by the electron-deficient area, and it can reach quite close to the positive region. This means that potential energy of the system will fall more, thus increasing stability of the product.
Two, the central atom being small, its overlapping orbital can overlap more effectively with the empty orbital of the electrophile, since the orbital will be smaller in size, and hence less diffuse.

The last factor that remains is the solvent. In fact, the order is same for both the nucleophiles in both the kinds of solvents (protic and aprotic), so I will not go into the analysis of solvent effects. Just a comment, here the solvent effect seems unimportant, seeing that bromide ion is quite behind $\ce{SH^-}$ in the order of nucleophilicity in protic solvents (my source is Solomons and Fryhle and a website mentioned later).

Note: Electronegativity is not important here either, even though it would produce the same order. We generally use electronegativity only to compare nucleophiles with central atoms from the same period.

So ultimately we see that $\ce{SH^-}$ is more nucleophilic than $\ce{Br^-}$.

Most of what I said can be found in the textbook Solomons and Fryhle's Organic Chemistry, and the following link:

$^{[1]}$http://chem.libretexts.org/Core/Organic_Chemistry/Reactions/Substitution_Reactions/SN2/Nucleophile

Here are some more useful links:

- http://chem.libretexts.org/Core/Organic_Chemistry/Alkyl_Halides/Properties_of_Alkyl_Halides/Introduction_to_Alkyl_Halides/Leaving_groups
- http://chem.libretexts.org/Core/Organic_Chemistry/Alkyl_Halides/Reactivity_of_Alkyl_Halides/Alkyl_Halide_Reactions/Substitution_and_Elimination_Reactions_of_Alkyl_Halides/SN1_Substitution_Reactions/Nucleophilicity_and_Solvent_Effects
- http://chem.libretexts.org/Core/Organic_Chemistry/Reactions/Substitution_Reactions/SN2/Substrate


The weight of a 2cc glass pharmaceutical flame sealed ampoule?

I grabbed 10 empty, unsealed ones and threw them on the analytical balance and got mean: 2.4071 g, s.d.: 0.01264 g for Wheaton amber glass pre-scored vials (part number 176796). The variation between vials isn't too bad, but I don't know if they can be filled and sealed consistently enough to determine gas density like that to any useful precision and other brands may be better or worse.


How to find the molecular formula for a hydrocarbon compound from known combustion products?

$$\ce{C_{$a$}H_{$b$} + $\left(a+\frac b4\right)$O2 -> $a$CO2 + $\frac{b}{2}$ H2O}$$

Suppose you had $n$ moles of hydrocarbon, then we have $a\cdot n$ moles of $\ce{CO2}$ and $\frac{b}{2}\cdot n$ moles of $\ce{H2O}$ dividing their moles we'll get $2\frac{a}{b}$:
$$2\frac ab=\frac{330/44}{135/18}=\frac{7.5}{7.5}=1\implies \frac ab=\frac12$$
So the empirical formula is $\ce{CH2}$
Now for actual formula $\frac{70}{14}=5$
Yes the formula is $\ce{C5H10}$.


Why does IR spectroscopy specifically require a change in dipole moment, and why doesn't this apply to other types of spectroscopy?

All the types of spectroscopy you mention have a similar condition for a spectrum to be observed.  This is that the molecule has to be able to interact with the radiation in the first place and then the radiation has to be of the correct energy to change the energy of the molecule from one level to another.

A plain (classical) explanation is as follows. (We implicitly assume that the radiation has the correct energy). As radiation is electric dipole in nature (it consists of oscillating electric and magnetic fields) if the molecule also generates an oscillating electric field it can interact with the radiation; energy is taken from the radiation field and enters the molecule. Thus a molecule may gain a quantum of vibrational or rotational energy or an excited state may be produced.

In the case of a vibrating diatomic molecule only if it is hetero-nuclear e.g. HCl, will it have a dipole and thus an oscillating dipole and oscillating electric field as it vibrates. Thus hetero-nuclear diatomics show an IR spectrum, homo-nuclear diatomics do not. In a polyatomic molecule vibrations can distort the molecule, e.g. benzene is distorted so that it is no longer 6 fold symmetric, so that a  dipole is produced and an IR spectrum may result. What form this takes depends on the molecule's symmetry and group theory point groups are used to determine this.

In a rotating diatomic molecule we usually use a rigid-rotor model. If the molecule has a dipole then this clearly changes as the molecule rotates, produced an oscillating electric field, and so a microwave (rotational) spectrum is produced. A homo-nuclear diatomic e.g. $\ce{O2}$ has no microwave spectrum.

In a UV/visible transition an excited state may be produced but only if the electron distribution in the excited state is different from that of the ground state, i.e. a dipole is produced that can interact with the radiation. A molecule such a benzene has a forbidden first electronically excited state due to its symmetry (which is however observed weakly due to other effects) and so absorbs very little compared to aniline (amino-benzene) or many dye molecules where symmetry does not limit any electronic dipole formation.

Thus although the details vary, the underlying ideas are the same. You should understand now that the statements about spectra in your question are not all correct. The rules used to determine what transitions are allowed or not allowed (called forbidden) are called 'selection rules'.

You can read about this in more detail, as a quantum version, in most undergraduate phys. chem. textbooks.

You may realise that the vibrational frequencies of homo-nuclear molecules are known even though they have no ir spectrum and this is because they can be measured by Raman spectroscopy. This depends on the change in shape/size of the 'electron cloud' (polarisability) a molecule possesses. Electro-magnetic radiation also possesses an oscillating magnetic field but this interacts far more weakly with molecules than its electric counterpart and can often but not always be ignored, the exception being NMR spectroscopy.


Sign of change in enthalpy and change in entropy

TL;DR
In general the entropy of reaction can be written as
$$  T\Delta _r S=  \Delta_r H  + RT\log \left(\frac{Q_e}{Q}\right)  $$
At equilibrium $Q_e=Q$ and
$$  T\Delta _r S_e=  \Delta_r H   $$

Consider a simple reaction that behaves ideally (occurs under ideal solution conditions).
If it is carried out at constant T and p we can write
$$\Delta_r G = \Delta_r G^\circ + RT \log Q \tag{1}$$
where Q is the reaction quotient.
But we can also write that
$$\Delta_r G = \Delta_r H - T\Delta_r S\tag{2a}$$
and
$$\Delta_r G^\circ = \Delta_r H^\circ - T\Delta_r S^\circ\tag{2b}$$
Equation (1) can then be written as
$$\Delta_r G = \Delta_r H^\circ - T(\Delta_r S^\circ-R \log Q) \tag{3}$$
Matching terms in equations (2a) and (3) we have that
$$\Delta_r H = \Delta_r H^\circ \tag{4a}$$
and
$$\Delta_r S = \Delta_r S^\circ - R\log Q \tag{4b}$$
When the reaction is at equilibrium $Q=Q_e$ (the reaction quotient is then equal to the equilibrium constant, here written $Q_e$) and $\Delta_r G = 0$ which means, combining equations (2a) and (4a) that
$$ T\Delta _r S_e =  \Delta _r H^ \circ \tag{5} $$
and
$$ T\Delta _r S^ \circ = T\Delta_r S_e + RT\log Q_e  \tag{6}$$
so that
$$  T\Delta _r S=  T\Delta_r S_e + RT\log Q_e  - RT\log Q \tag{7a} $$
or
$$  T\Delta _r S=  \Delta_r H ^\circ + RT\log\left(\frac{Q_e}{Q}\right)  \tag{7b} $$
Now compare equations (5) and (7b). Equation (5) holds at equilibrium and says, sure enough, that the reaction entropy and enthalpy are equal in sign at this point in the reaction coordinate. However, equation (7b) - which is the more general expression - says that $\Delta_r S$ can in fact differ in sign from $\Delta_r H^\circ$, depending on the magnitude of the reaction quotient Q. It turns out that while the enthalpy of a reaction in an ideal solution is a constant, the entropy of reaction can be tuned by modifying Q.


How to get force constants in internal coordinates in Gaussian?

You have to use freq=intmodes and use the command

formchk -3 xxxx.chk xxxx.fchk

to convert the binary checkpoint file to a formatted checkpoint file, which consists of text. The -3 requests the version 3 of formatted checkpoint files, which support all features of Gaussian 09 and 16.

Source: Ruixing Wang's answer on Researchgate, where I asked the same question.


Ignoring spin, consider an electron in a hydrogen 2p orbital, what is its orbital magnetic moment?

The first equation you have is for spin. Let's look at it in a operator form

To properly treat the upcoming discussion, quantum electrodynamics is introduced to describe the eigenstates of spin. The eigenstates of spin has a direct correlation with spin. Formally we depict it as the following

$$ μ_B =-g \frac {eℏ}{2m}  \frac { \boldsymbol {S}}{ℏ} $$

where μ is the magnetic moment operator, g is the g factor, ℏ is plank’s constant over 2π, m is the mass of the particle and S is the spin operator.

For orbital magnetic moment (your question) you use:

$$\mu =\hbar\sqrt{l(l+1)} \boldsymbol {\mu_B} $$

Recall:

s: l=0
p: l=1
d: l=2
f: l=3

So plugging in:

$$\mu =\hbar\sqrt{1(1+1)} \boldsymbol {\mu_B} $$
$$\mu =\hbar\sqrt{2} \boldsymbol {\mu_B}$$

(Units are $ \boldsymbol {\mu_B}$)


What is the priority for numbering alkyl side chains?

The process for naming this molecule correctly follows the steps listed here:

Identify the parent hydrocarbon, which will be the one with the longest chain. It is correctly identified as the 7-carbon chain, and as such is a derivative of heptane.
Identify the side-chain functional groups. We have 2 methyl groups and an ethyl group.  When listing them, they will be written alphabetically.  So, we have:

X-ethyl-Y,Y-dimethylheptane and not Y,Y-dimethyl-X-ethylheptane
Numbering (which is what you are asking about) is determined by numbering the carbon atoms consecutively from the end of the chain nearer the functional group of the highest priority. In terms of molecule numbering, alkane groups have the same functional group priority. We would only need to consider relative priorities of the ethyl and methyl groups if there was only one methyl substituent, in which case you would have 3-ethyl-5-methylheptane.

In our case, the process is to count in from the ends of the primary chain. Here, the first substituents occur at an equal distance from each end of the chain, both the ethyl and the methyl occur at position 3. We now need to invoke one of the IUPAC sub-rules for nomenclature. If the first substituents occur at an equal distance from each end of the chain, we need to number from the end nearer a second substituent. The second substituent is the extra methyl group, and so this is the end that gains numbering priority.

There are many nice examples of nomenclature on the internet. One very good example, which highlights your question is this page. Have a look at Rule 2 and sub-rule 2


Free Open Source Gas Chromatography Software

One software that I have come across and used is OpenChrom https://www.openchrom.net/

At least for basic tasks it performs well as far as I can tell.

For those seeking pretty chromatograms, OpenChrom can be used in conjunction with R. Open a plot in OpenChrom, export it to a csv file and then plot the data using R.

A script for plotting two overlapped chromatograms in R is below:

filename = "Dodecane-Neat"
filename2 = "Dodecane-Stressed"
PlotTitle = "Dodecane Neat Chromatogram - Pre and Post Test"

infilename = paste0("./",paste(filename,"csv",sep="."))
infilename2 = paste0("./",paste(filename2,"csv",sep="."))
outfilename = paste(filename,"pdf",sep="-rect.")

## Read Data -> File is called 14mielczarek11.csv
undiluted <- read.table(infilename, sep=";", header=TRUE, nrows=10000)
undiluted2 <- read.table(infilename2, sep=";", header=TRUE, nrows=10000)
## Work out Total Peak (want the Chromatogram, not the mass spec)
undiluted$peaks <- rowSums( undiluted[,3:404] )
undiluted2$peaks <- rowSums( undiluted2[,3:404] )

GCxx <- c(undiluted$RT.milliseconds./1000/60, rev(undiluted$RT.milliseconds./1000/60))
GCyy <- c(rep(0, nrow(undiluted)), rev(undiluted$peaks)/max(undiluted$peaks))
GCxx2 <- c(undiluted2$RT.milliseconds./1000/60, rev(undiluted2$RT.milliseconds./1000/60))
GCyy2 <- c(rep(0, nrow(undiluted2)), rev(undiluted2$peaks)/max(undiluted2$peaks))

library(survival)
library(Hmisc)

# A4 is 8.27 × 11.7
pdf(outfilename, height=5, width=8)
par(mar=c(5,5,3,2))

plot("",
tck=-0.02,lab=c(10,10,4),las=1,
xaxs="i",xlab="Time (min)",
xlim=c(0,15),
yaxs="i",ylab="Detector Count Fraction",
ylim=c(0,0.15)
)

polygon(GCxx2+0.017, GCyy2, col= rgb(0,0,1,0.5), border=NA) #Blue - Shift the blue ever so slightly to properly overlap
polygon(GCxx, GCyy, col= rgb(1,1,1,1), border=NA) # White (so we don't get violet)
polygon(GCxx, GCyy, col= rgb(1,0,0,0.6), border=NA) # Red

lines(undiluted2$RT.milliseconds./1000/60+0.017,undiluted2$peaks/max(undiluted2$peaks),
type="l",col=4,lwd="0.05",lty=1) #Blue
lines(undiluted$RT.milliseconds./1000/60,undiluted$peaks/max(undiluted$peaks),
type="l",col=2,lwd="0.05",lty=1) # Red

title(PlotTitle)
minor.tick(nx=5, ny=5, tick.ratio=0.3)

dev.off()

And the results:


Synthesize coumarin from 4-allyl-2-methoxyphenol (eugenol)

Since no-one has offered an answer I'm going to try but as I have no database access some of this is a bit flaky.

Heat (maybe microwave) to give multiple Claisen rearrangement products then separate to isolate 2-allyl, 6-methoxyphenol. This is the flakiest step.
Hydroborate with oxidative work-up to take the allyl to propionic acid.
Ring close (PTSA, Toluene)
Reflux over activated MnO2 to introduce double bond.
Treat with BBr3 to remove Me from MeO group
Zn reduction of OH to H


What's the best way to interconvert between molarity and percentage?

Thanks to LDC3's comments, I understand that the best way to convert from % to Molarity is as follows:

1) multiply density by volume at hand to get weight

2) solve ratio for x, where x is grams of solute (in volume at hand)

(given% / 100%) = (x / weight from step 1)

3) multiply x by molar mass to get moles (in volume at hand)

4) if needed, multiply resulting moles/volume in hand to get Molarity

Example: Let's say I have 100 mL of 38% HCl.

1) 1.19 g/ml * 100 mL = 119 g

2) (38%/100%) = (x/119) where x = 45.22 g

3) 45.22 g * 1mol/36.46g = 1.240 moles

4) 1.240 mol/100 ml = 12.40 mol / 1000 mL = 12.4 M

Now, the result matches the equivalency chart!


Using TBAI (tetra-butylammonium iodide) as a catalyst. Ideas on the mechanism or the driving force?

I was wondering, what is the driving force for the Iodide to bond to the carbonchain instead of the bromide?

It's not so much a driving force as much as it is a dynamic equilibrium of sN2/sN1 reactions. Halides can exchange each other in equilibrium if one is not removed thus processes that remove the unwanted product such as the Finkelstein reaction are use to achieve complete substitution (think Le'Chatlier's Principle).

$$\ce{R-Br +I- <=> R-I + Br-}\qquad\quad\ \ \  \text{(Equillibrium)}$$
$$\ce{R-Br +I- +Na+ <=> R-I + NaBr (s)}\qquad \text{(Finkelstein Reaction)}$$

  Is there a mechanism to explain that?

Given the information you provided, Your reaction is proceeding as follows:

$$\ce{R-Br + I- -> R-I + Br-}\tag{1a}$$

$$\ce{R-I + R'-O- -> R-O-R' +I-}\tag{1b}$$
Where as the reaction:
$$\ce{R-Br + R'-O- -> R-O-R' +Br-}\tag{2}$$

cannot proceed or does not proceed at a fast enough rate and thus the catalyst can improve the reaction rate by first substituting the less reactive halide. The Wikipedia article on Williamson ether synthesis agrees with this:

  However, if an unreactive alkylating agent is used (e.g. an alkyl chloride) then the rate of reaction can be greatly improved by the addition of a catalytic quantity of a soluble iodide salt (which undergoes halide exchange with the chloride to yield a much more reactive iodide, a variant of the Finkelstein reaction).


Role of solvents in ozonolysis and oz0nolysis of alkynes with water

Ozonolysis:

Ozone, an allotrope of oxygen, is highly reactive molecule that can undergo a 1,3-dipolar addition across alkenes, forming a 1,2,3-trioxolanes. This compound is unstable and undergoes a cycloreversion to form a carbonyl compound and a carbonyl oxide. These resultant compounds (the carbonyl compound and the carbonyl oxide) then proceed through a second 1,3-dipolar addition to form a 1,2,4-trioxolane, commonly known as an ozonide. To complete the cleavage of the alkene, the ozonolysis (implies that ozone causes to break (-lysis)) is followed by a work-up designed to break apart the ozonide.  The work-up, which can be oxidative or reductive, determines the oxidation state of the products. For example, cis- or trans-alkenes gives two corresponding aldehydes with reductive work-up while same alkene gives two corresponding carboxylic with oxidative work-up.

The most common reductive work up procedure uses dimethyl sulphide (DMS) in the
presence of methanol. Methanol reacts with the carbonyl oxide to form a hydroperoxy acetal, and this is in turn reduced by the dimethyl sulphide to give relevant carbonyl compound and dimethyl sulfoxide. It is also possible for dimethyl sulphide to reduce the ozonide without the presence of methanol as well, although reation would proceed slowly (A new and convenient method for converting olefins to aldehydes: Tetrahedron Letters, 1966, 7(36), 4273–4278). Other reductive methods: $\ce {Zn/AcOH}$, triphenylphosphine ($\ce {PPh3}$), or thiourea ($\ce {NH2C(=S)NH2}$).

Note: Recently, there is a  publications claiming an efficient and convenient method for reductive workup of the ozonolysis reaction using sodium hydrosulfite. However, the final product could be either aldehyde or acid, thus the method is not selective.

Oxidation: The direct conversion of terminal and 1,2-disubstituted alkenes to carboxylic acids is described as oxidative ozonolysis. This can be easily achieved through treatment of initial ozonolysis products with hydrogen peroxide ($\ce {H2O2}$) or other oxidants.

You may also find answers to your Q3 and Q4 from the reference given above.


Yes, tahini (and cornstarch-water mixtures) are dilatant, non-Newtonian fluids, with viscosity increasing with shear strain. See http://en.wikipedia.org/wiki/Dilatant for definition and https://www.youtube.com/watch?v=D-wxnID2q4A, https://www.youtube.com/watch?v=BleCJJAKkgw for demo.

Tahini is much better in hummus than in a swimming pool, though. ;-)


Why $\ce{HgS}$ and not $\ce{HgO}$?

Because of HSAB. F'x has already answered a previous question of why mercury has propensity towards thiols(or in general sulfur). Quoting the relevant information:

In Pearson's HSAB theory (hard and soft acids and bases), the reason
the $\ce{S-Hg}$ bond is be stronger than the $\ce{O-Hg}$ can be
explained because $\ce{S^2-}$ is a soft basis and $\ce{Hg(II)}$ is a
soft acid, making a good fit while $\ce{O^2-}$ is harder.

If this answer helps, please make sure to upvote F'xs answer.


Open Source tools to draw a set of 2d molecular graphs from sdf file

An SDF file is just a bunch of MOL entries in a single file; $$$$ serves as the separator for the individual entries.

Splitting into individual files AND simultaneous conversion to graphic file formats (PNG, SVG) is possible with Open Babel, either directly on the command line or through scripts/programmes using the bindings available for various programming languages, such as Pybel for Python.

To my knowledge, the direct generation of PostScript files - probably the nicest option when using $\LaTeX$ - is not possible, so you might want to used SVG as an alternate vector-based format.

obabel -isdf lotsofstructures.sdf -m -osvg -xC -xu -xd -O basename.svg

will do the job.

-isdf specifies the input file format, -m does the splitting, -osvg specifies the output file format, and -O basename.svg specifies the prefix and extension for the individual numbered output files that are generated. -xC prevents that terminal carbon atoms are printed explicitly with attached hydrogen atoms. Surely a matter of taste, but I wouldn't want to have lots of $\ce{-CH3}$ in the drawings. -xu prevents the use of colour for heteroatoms and generates a black and white drawing. -xd suppressed the name of the compound to be written and displayed in the SVG file.

This should work for SDF files with 2D and 3D coordinates, but in the latter case I'd suggest to check it carefully; I've seen 3D -> 2D transformations on some molecules that went wrong.


Acidic tendency of copper(II) ion vs. calcium(II) ion

Cu2+ has an incomplete 3d orbital (4s0 3d9) and can therefore be included in the chemistry of transition metals. As it can be coordinated (form dative covalent bonds) with ligands to create complex ions, for example H20 in hexaaquacopper(ii), it acts as a Lewis acid.

But in terms of Brønsted–Lowry theory, perhaps the textbook is referring to the fact that when transition metals are dissolved in water they can donate an H+ ion to another H2O molecule to form H3O+. This can be observed with Fe3+:

[Fe(H20)6]3+ + H2O -> [Fe(H20)5(OH)]2+ + H3O+

The complex ion is acting as an acid by donating a hydrogen ion to water molecules in the solution. The water is, of course, acting as a base by accepting the hydrogen ion.


Is it possible to convert methane and carbon dioxide to acetic acid?

Already methanotrophic bacteria can oxidize methane to formate and the Wood-Ljungdahl pathway can take this formate and a carbon dioxide molecule and coenzyme A and produce acetyl coenzyme A so the catalysts are already there.


How do I create a spreadsheet containing the coordinates and predicted bond angles, molecular properties, torsion properties, etc of a molecule?


How can I predict whether a reaction will occur?

For the first example, I definitely wouldn't want to react $\ce{CuCl2}$ with $\ce{HCl}$, because then I would end up getting the same products again. I'm primarily driven by the motive to create something new, albeit the reaction must be a feasible one. Many factors influence feasibility, and you've mentioned the formation of a precipitate is one of them.
Reactions involving neutralization:
$\ce{HCl}$ would react with $\ce{(NH4)2CO3}$, $\ce{K2SO4}$ and $\ce{Ba(NO3)2}$, as the reactions give out $\ce{H2CO3}$, $\ce{H2SO4}$ and $\ce{HNO3}$ respectively which are weaker acids.
Reactions involving precipitation:
I see a very tempting reaction here:
$$\ce{K2SO4 + Ba(NO3)2 -> 2KNO3 + BaSO4\downarrow}$$
$\ce{BaSO4}$ is a very good precipitate, and I highly emphasize this fact. Just as a fun fact: It can be partially soluble if you try to dissolve it in boiling concentrated $\ce{HCl}$ (or boiling concentrated $\ce{H2SO4}$).
On similar grounds, $\ce{Ba(NO3)2}$ would react with $\ce{(NH4)2CO3}$ to give out the precipitate of $\ce{BaCO3}$. How do I know this? Partly because of the fact that while testing for the presence of $\ce{CO3^2-}$ ions in a laboratory we use the lime water test. Instead of using $\ce{Ca(OH)2}$, baryta water or $\ce{Ba(OH)2}$ is also used at times yielding strikingly similar results.
Also, I know this partly because $\ce{CO3^2-}$ ion is notorious for forming precipitates. It forms a precipitate with a large number of ions, some exceptions (they're the only exceptions I know) being $\ce{Na+, K+, Rb+, Cs+}$ and $\ce{NH4+}$.

However, what happens to the reactions like these?

$$\ce{(NH4)2CO3 + K2SO4 <=> K2CO3 + (NH4)2SO4}$$

Neither of the product is a precipitate. Hey that's easy? $\ce{K2SO4}$ is made up of $\ce{KOH}$ and $\ce{H2SO4}$. And look the base is the same for $\ce{K2CO3}$, that means I've made carbonic acid from sulfuric acid and thus the reaction takes place. Right? Wrong.
In your haste you forgot to take note that I could've used the same logic if I were to deal with $\ce{(NH4)2CO3}$ and $\ce{(NH4)2SO4}$. Wait! This reverses my previous reasoning! Yes it does. Or maybe we get a little confused.
Perhaps these cations are spectators, shall I strip them off? Yeah, let's give it a try too. We're now (sadly) left with:
$$\ce{CO3^2- + SO4^2- <=> CO3^2- + SO4^2-}$$
That's bizarre. What's happening? Nothing. And that's what it is in its true form, the ions basically are just combining with each other, there are four ions, two cations and two anions, four possible combinations. Does that make it a reaction? Perhaps not. It's just an equilibrium, there's no going forward or backward. When we say there's a reaction taking place, we are implying that the equilibrium sways to the the product side, which apparently isn't the case here.


Use of fractional distillation in organic chemistry

The boiling point of an (R)-isomer is the same as the one of the enantiomer, the (S)-isomer.   Since the condensor does not offer a chiral environment to the chiral molecules, there is no discern of them and consequently, neither "single" nor "fractional distillation" separates them.   (You may expose the molecules to a chiral surface to adsorb, though, and separate the enantiomers by chiral gas chromatography; a specialized approach in analytical chemistry and -- because of the scale of amount typically accessible -- much less frequently deployed on a preparative one (a review).)
A successful separation of (cis)-(trans)-isomers by distillation of course depends on how different the boiling points of the pure compounds are.  Looking up the data for (trans)-stilbene (bp 305 C reference) and (cis)-stilbene (307 C at 1 atm reference) suggests other separation techniques may be much more suitable since less demanding.  This especially becomes relevant if you need a technique easy to scale, to manage and to be run continously (e.g., distillation).


How to know if computational calculations are correct

The only guaranteed way to achieve absolute correctness is to use asymptotically better methods and see the asymptotic behavior yourself. This, however, ofter requires prohibitively complex approaches.

Outside of that, one needs an experiment to check your data against. A good way (with limits) is to perform vibration frequency calculations and see if that fit the experiment. (making correction of unharmonic behavior). Another way is to run through similar well-studied cases and fit computational results. Most probably, a systematic error would be found, that can be easily eliminated. Luckily, most probably you are not the first to work with your particular kind of systems, so using google scholar it is usually possible to find a review or an establishing work that would list what work and what does not. Most practical methods have known 'glitches', and the ways to deal with them were considered. In particular, DFT cannot catch dispersion interactions, requiring empirical correction schemes.


Which of these can undergo decarboxylation reaction readily

You have said that  β−ketoacids undergo decarboxylation easily and readily due to formation of a stable intermediate. That intermediate is the carbanion on the alpha carbon. So, any species that can stabilize that carbanion can undergoe decarboxylation easily. In (B), the carbanion formed is stabilized by the double bond beside it so it undergoes decarboxylation easily.
(D) is a special reaction as referenced to in this answer.


Why is change in gibbs energy not zero through the gibbs free energy equation?

The change in Gibbs free energy for a closed system is given by:
$$\Delta G = \Delta H - T \Delta S$$
What you are describing is:
$$\Delta H = T \Delta S$$
which is only true for systems at equilibrium, or when $\Delta G = 0$
Why this is true is a little bit difficult to explain. It is essentially a form of the Clausius Inequality at constant temperature and pressure. The Clausius Inequality is really just a statement of the second law of thermodynamics, and it says that at best, for any possible process you can get a zero change in the entropy of the universe, and usually it will be a positive change. From the perspective of the system you can write that as:
$$\Delta S \ge \oint \frac{dQ}{T}$$
which is basically just saying the entropy change of the system is equal to or greater than the sum of the negative entropy changes of the rest of the universe, for each step in a closed cycle.
If you integrate that at constant temperature you get
$$\Delta S \ge \frac{\Delta Q}{T}$$
At constant pressure (and in the absence of non-expansion work) $\Delta Q = \Delta H$, so we can write:
$$\Delta S \ge \frac{\Delta H}{T}$$
...and a little rearranging gives:
$$\Delta H - T \Delta S \leq 0$$
So this says that any process which can occur on its own (without adding work or heat to it) will obey this inequality. That's a pretty useful thing to know, so we let
$$\Delta G = \Delta H - T \Delta S$$
to give us an easier way to talk about it. Processes will happen (on their own) if $\Delta G$ is negative, they won't happen if it's positive, and when they are at equilibrium, $\Delta G$ will be zero.