text:
Does cathodic protection work for air oxidation?
I'm not expert in cathodic protection, but does it work with metals under air oxidation? Is it possible to protect a metal from air oxidation by applying an electrode to it? Or it has to be inside an electrolytic solution like sea water?

Not according to Wikipedia (current version of that page as of writing this answer) https://en.m.wikipedia.org/wiki/Cathodic_protection

Galvanizing, while using the electrochemical principle of cathodic protection, is not actually cathodic protection. Cathodic protection requires the anode to be separate from the metal surface to be protected, with an ionic connection through the electrolyte and an electron connection through a connecting cable, bolt or similar.

So, the portions of your material exposed to air won't actually be receiving any "cathodic protection" formally per se, though galvanization may still help out quite a bit.


text:
Is oxidation always thermodynamically favorable?
In biological systems, energy is often extracted from molecules with reduced carbon backbones — the oxidation of those molecules has $\Delta G < 0$ and that energy can be captured, often in the reduction of other molecules. I'm wondering if it is always the case that oxidation is thermodynamically favorable and reduction thermodynamically unfavorable.
I suppose I'm not asking if the oxidation of any molecule is favorable (I'm pretty sure that's not the case), but for redox reactions that do occur, is it ever the case the oxidation half reaction is thermodynamically unfavorable and the reduction half reaction thermodynamically favorable?

You can't measure the thermodynamics of a half-reaction.

You need two half reactions and the value of $\Delta G$ will depend where they are situated relative to one another in redox potential. Wikipedia has a large table of redox potentials, you can choose some reactions for yourself.

@IvanNeretin - Is it therefore meaningless to talk about the ΔG for the oxidation of a particular molecule? The free energy of succinate in cells is higher than the free energy of fumarate, which is a result of the oxidation of succinate. Should I not say that the oxidation of succinate has a ΔG<0?

Why, that's a full reaction, which is another story. It certainly has some $\Delta G$ and everything.

That is, if we write it as a full reaction: succinate plus oxygen $\to$ fumarate plus water.

As the term redox implies you have an oxidation and a reduction in every redox reaction.
It makes no sense to just talk about a half reaction, but you have to look at the cumulated energy difference of educts and products.
Now you have to remember two things:

You can easily convert between the energy and potential difference of a redox reaction by multiplying with the exchanged charges:
$$\Delta G= Uzq$$
$U$ is the potential (usually measured in Volt), $z$ the number of exchanged elementary charges and $q$ the elementary charge (usually measured in Coulomb).
Energy differences are transitive. That means:
$$\Delta G (A\rightarrow C) = \Delta G (A \rightarrow B) + \Delta G (B \rightarrow C)$$
There is no absolute zero point for the energy in chemical reactions. You can only talk about differences.

What chemists did, was to define a Standard electrode potential.
With this it is possible to measure half reactions compared to this standard. Because of sloppy naming, people call this the potential of the half reaction. But strictly speaking it is just the difference to the standard electrode potential. Since differences are transitive that is not a problem when working with potentials.
This means: The oxidation can be unfavoured compared to the standard electrode potential if the reduction is even more favoured.

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text:
Can green vehicle antifreeze (Ethylene Glycol) be diluted in a glass container?
I picked up a container of undiluted green antifreeze, and oops, I don't have anything to mix it in! I do have a glass container though at the top of my recycling bin that is about a half gallon. I've heard some plastic containers (soda bottles, plastic milk jugs) can't be used for anti-freeze, but what about glass? Two questions arise from this

As Long as I rinse the previous contents (milk) out, can I safely use this glass container for diluting antifreeze and putting it into my vehicle?
Once I'm done can this go back into the recycle bin, or will trace amounts of antifreeze spoil it for recycling?

Out of curiosity, what makes you think there'd be a problem in using a glass vessel to handle ethylene glycol? In laboratories we often keep ethylene glycol in glass beakers and retorts without any issues. Even if trace quantities remain after a thorough washing, I doubt that would adversely affect the recycling process ._.

@paracresol As a laymen (High school chemistry was a long time ago) I tend to assume everything in my garage is 1.) Poisonous 2.) Corrosive 3.) Literal Bottled Evil. At least until I am told otherwise. And concerning the recycle process - That comes from another automotive fluid (oil) which I have been told by my recycling company not to throw in the recycle bin.

I'd just rinse the glass container after using with antifreeze and put it back in recycling. I assume that we're really taking about recycling (so that the glass will get remelted) and not returning the bottle to the milk company. // The general notion is that "chemicals" of any sort should not be put into containers that would be used later for foodstuffs. This is an simple dogmatic rule trying to avoid any sort of contamination problems.

@Sidney Nah it's fine, don't get worked up over it ;P Since Max didn't chew me out for my comment, I guess you can safely go about your garage alchemy. By the way, I liked the "Bottled Evil" bit X'D, Also, you know you can always come over to The Periodic Table (Chat) if you want to discuss anything; guys there are only too glad to help 3:) http://chat.stackexchange.com/rooms/3229/the-periodic-table

As Long as I rinse the previous contents (milk) out, can I safely use this glass container for diluting antifreeze and putting it into my vehicle?

Not sure if you are talking a bout antifreeze for the radiator system or windshield deicers, but both systems are rugged enough that the do not require very pure water to be used. A washed milk container would suffice. The antifreeze wont disrupt the bottle and the bottle wont adulter the antifreeze

Once I'm done can this go back into the recycle bin, or will trace amounts of antifreeze spoil it for recycling?

This somewhat depends on how it will be recycled. If you are putting it in general glass recycling then the bottle will become cullet that will heated to temperatures high enough to burn off any residue and it wont matter.
If you are returning the milk bottle to the producer for reuse then they have to clean them well enough to remove milk reside which is harder to remove than either your antifreeze or the bitterants in antifreeze. Any reputable bottler would expect some amount of non-milk substances in the bottles and clean them sufficiently well to make sure every bottle is clean and safe. Though I would recommended that you do rinse it before furnishing it to them out of courtesy.

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Mechanism of manganese dioxide induced decomposition of hydrogen peroxide
What is the mechanism involved in the decomposition of hydrogen peroxide with manganese dioxide as the catalyst?

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Skeletal formula of methyl chloride
Recently in The Periodic Table, there was a short discussion regarding this structure:

At first sight, I couldn't figure out what is the structure. After a quick google search, I came to know it is trichloromethane or chloroform (chemspider).
Now, I thought of the structure of dichloromethane and found out to be:

From the structures, It is observed that these are the skeletal formula of tri- and dichloromethane and generally in these structures, the letter C is not denoted, the C-H bond in not denoted and C-Me is denoted by a dash.
Now, following these rules, I was now curious of the skeletal formula of methyl chloride. Is it just a chlorine protruding a bond??

I couldn't find out this structure on the internet and this means that this formula is not valid although, this formula follow the basic rules. Am I wrong in my argument or are the rules vague?
Further, hydrocarbons are the basic skeleton of organic compounds. To draw really complex organic compound, we first have to figure what are the hydrocarbon chains. But still, methane and ethane cannot be drawn using skeletal formula. From propane onwards, you can represent them in skeletal formula.
So, there are certain ambiguities regarding skeletal formula (some simple and basic compounds cannot be drawn) but they are widely preferred. Why is this so?

I searched hard and eventually found out the structural formula of ethane. It is a dash as expected but still, it is a dash. Who would mind that it is ethane?

It is a valid generalisation of the skeletal formula principle to draw chloromethane as you did; you may also see it drawn in the following way:
$$\ce{-Cl}$$
In a similar vein, ethane can be drawn as a simple horizontal (or $60^\circ$ ascending or descending) line, ethene as an equality sign and ethyne as a triple line.
However, the concept of skeletal formulae, which was primarily invented to simplify drawing large, complex molecules, does not extend well to these very small ones. A chlorine atom with a line can be mistaken as $\ce{Cl-}$, a horizontal line as a dash, a double line as an equality sign and so on. Thus, while formally valid, you will not see this representation used often for very small molecules. Typically, it would be used from three explicit atoms upwards — as is the case in your second example, dichloromethane.

First of all, you didn't look in the right place to find your drawn structure. The first website I used to look for that compound showed it like this: (middle of the top table). So also basic structures are depicted like this, although writing it out is also not too much extra work, and wouldn't make it too complex.
Why they are preferred? Because you can draw complex structures without overwhelming the reader. It's a "chemical language" that's easy for trained chemists to understand. Of course, writing out all the carbons and hydrogens would be more accurate/complete, but using lines for carbon and not drawing obvious hyhdrogens cleans up structures very nicely.
For trichloromethane  it's perhaps a bit silly to use this notation, but you can imagine how testosterone would look if every carbon and hydrogen atom is written as C or H.

The  formula for trichloromethane (chloroform) is $\ce{CHCl_3}$. The missing H atom in your first figure points towards you. Each angle H-C-Cl is approx $109$ degrees. The figure below is from molecule-viewer.com


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Why can we not compare the coupling constants to chemically equivalent (but not identical) protons to determine magnetic equivalence?
I have been learning about magnetic inequivalence today, and I find it quite hard to undersand. See, for example:

In this case, according to the book I am using, $\ce{H_\mathrm{a}}$ and $\ce{H_\mathrm{b}}$ may be chemical equivalents, but they are not magnetic equivalents, due to the fact that $J_\mathrm{{\ce{a-c}}}$ and $J_\mathrm{{\ce{b-c}}}$ are different. Which I understand, but should we not consider that there is a $J_\mathrm{{\ce{b-d}}}$ which is equal to $J_\mathrm{{\ce{a-c}}}$?
I do not understand why they are magnetically different, but both have to chemical equivalent protons, one in ortho and one in para position.

If, as you say, you understand that Ja-c is different to Jb-c, then you should be able to rationalise that A is different to B. Therefore A will see B as being different to itself and there will be observable coupling between the two. That is the crux of magnetic non-equivalence; not so much as having different coupling to other nuclei around them, but being different to each other and therefore manifesting coupling between themselves.
As always, I'd recommend reading Reich's online NMR resource (symmetry section) to learn more.

The problem is that from the perspective of $\ce{H_{d}}$ and $\ce{H_{c}}$, $\ce{H_{a}}$ and $\ce{H_{b}}$ are different (you can tell which is which by their coupling constants). This difference means that there is a magnetic inequivalence between the two hydrogens $\ce{H_{a}}$ and $\ce{H_{b}}$, so there will be coupling between them.


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Seeking an explanation of synthetic marcasite
At least one widely available commercial product is called 'synthetic marcasite,' and perhaps two or more.  The most prominent is dark gray 'synthetic marcasite' for cheap jewelry.  Possibly different is the synthetic marcasite used in battery research as at Electrochemical Society Proceedings.
I had suppposed the latter has the chemical composition and crystal strucure of natural marcasite. But I find that people also talk about marcasite iron diselenide and ditelluride, FeSe2 and FeTe2, apparently just based on crystal structure.   Does anyone here know more about this?
The jewelry product is dark gray, and very cheap at prices like a few cents per small round stone in lots of 1,000, which you can easily find on line. So I wonder if it is actually one of the things called hematine, which are various ersatz hematites, not matching the composition of hematite but mimicking the appearance.  See for example https://en.wikipedia.org/wiki/Hematine.  Does anyone here know?

From wikipedia In marcasite jewellery, pyrite used as a gemstone is termed "marcasite" – that is, marcasite jewellery is made from pyrite, not from the mineral marcasite. https://en.wikipedia.org/wiki/Marcasite

@MaxW That was true of expensive 'marcasite jewelry' in the 19th century.  But essentially all 'marcasite jewelry' after 1900 has been synthetic stones or just cut steel.  See the talk page https://en.wikipedia.org/wiki/Talk:Marcasite_jewellery .  And the synthetic marcasite does not look at all like pyrite.  It looks like hematite.


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"Adjust pH of the solution to 5.0±0.1 with acetic acid (by potentiometry)"
In a procedure description I'm translating, there's this sentence (I here quote it literally, word for word, as it is in Russian):

Adjust the pH of the solution to 5.0±0.1 with acetic acid (by potentiometry).

The meaning of "(by potentiometry)" is that the personnel should use a potentiometric pH meter diring the adjustment process.
However, I'm not sure what the custom is of indicating this in English.
Maybe "(control using potentiometry")? Or maybe one should add the word "potentiometrically":

Adjust pH of the solution potentiometrically to 5.0±0.1 with acetic acid.

There must be some commonly used turn of phrase for this.

I suspect this is very likely a sentence adapted from of one of the methods from the Russian State Pharmacopoeia (RSP), which has numerous entries for the adverb «потенциометрически» (Eng. “potentiometrically”).
For instance, there is a nearly identical match in the normative section for the preparation of acetate buffer solution, ОФС.1.3.0003.15-1.27 Ацетатный буферный раствор рН 5,0:

К 120,0 мл 6,0 г/л раствора уксусной кислоты ледяной прибавляют 100,0 мл 0,1 М раствора калия гидроксида и 250,0 мл воды, перемешивают. Доводят рН до 5,0 потенциометрически с помощью 6 г/л раствора уксусной кислоты ледяной или 0,1 М раствора калия гидроксида и доводят объём раствора водой до 1000,0 мл.

RSP underwent a vigorous clean-up since the Soviet era when it was primarily original, and currently it is mostly translated from the European Pharmacopoeia (EP), so I decided to take a look at the corresponding entry in  EP 8.0 (currently invalidated, but one can at least find bits of it in the internet), Section 4.1.3. Buffer solutions:

Acetate buffer solution pH 5.0. 4009100.
To 120 mL of a 6 g/L solution of glacial acetic acid R add 100 mL of 0.1 M potassium hydroxide and about 250 mL of water R. Mix. Adjust the pH to 5.0 with a 6 g/L solution of acetic acid R or with 0.1 M potassium hydroxide and dilute to 1000.0 mL with water R.

EP omits the details as to how the pH is controlled, but there are several phrases in the prior text

Adjust to pH … with …, monitoring the pH potentiometrically

referring to the section 2.2.3. Potentiometric Determination of pH, so I think the phrase that you have proposed

Adjust pH of the solution potentiometrically to 5.0±0.1 with acetic acid.

is perfectly fine and will be correctly understood as a more literate alternative to something like "the pH was determined with a pH meter."
A couple of similar textbook-usecases:

Concise Encyclopedia Chemistry [2, p. 803]:

The pH of a solution is determined experimentally with a suitable pH color indicator (see Indicators) or potentiometrically, with a pH meter (see Glass electrode).

Fundamentals of Electrochemistry [3, p. 590]:

The concentrations of hydrogen ions (solution pH) and of a number of other inorganic ions … are determined potentiometrically.

References

Council of Europe; European Directorate for the Quality of Medicines & Healthcare. European Pharmacopoeia 8.0; European Directorate for the Quality of Medicines & Healthcare, Council of Europe: Strasbourg, 2013.
Concise Encyclopedia Chemistry, English language ed., 2nd ed.; Scott, T., Eagleson, M., Eds.; de Gruyter: Berlin; New York, 1994.
Bagot︠s︡kiĭ, V. S. Fundamentals of Electrochemistry, 2nd ed.; The Electrochemical Society series; Wiley-Interscience: Hoboken, N.J, 2006.

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Treatment of ketones with excess Tebbe reagent
The reaction in black below was presented in a set of lecture notes as an example of the use of the Tebbe reagent, however it's not immediately obvious (to me, at least) that the product should be the gem-dimethyl compound shown.

Mechanistically, it appears that the Tebbe olefination has taken place once (standard formation of Schrock carbene, [2+2], retro [2+2] to give the alkene and a Ti=O compound as a thermodynamic driving force). The alkene must then however react with more Tebbe reagent (or more Schrock carbene) to form a metallacyclobutane. It's at this stage that I feel it makes less sense: firstly, what is the involvement of the deuterium oxide, but secondly and more importantly, why doesn't the metallacyclobutane just collapse back down to form the thermodynamically favoured alkene product.
I've used Tebbe in the past to turn ketones into terminal methylene groups, but never encountered what appears to me to be an over-reaction (I'd actually always considered Tebbe to be fairly mild and selective).
After some digging around in Comprehensive Organic Synthesis I, there is some mention of this 'over reaction' (1985JOC50). The authors report that treating the ketone 22 with 1 equivalent of Tebbe reagent affords a mixture of the expected methynlenation product along with some of the gem-dimethyl product (i.e. implying that an excess isn't required, but the reaction is just not overly chemoselective for the cyclic ketone over the cyclic alkene).

What are the typical reaction conditions?

Tebbe reagent and some pyridine to help form the Schrock carbene. Usually use more than 1 eq anyway (i.e. not being super careful to control stoichiometry) but in the JOC paper cited they use a 1:1 ratio and still get some overaddition

I was thinking more along the lines of temperature. Also, I think you already noted this, but you might get an intermediate mixture under thermodynamic control before the quench with water.

@Zhe Perhaps. It seems like one obtains a mixture anyway (similar to Grignard + ester kind of problem?). In any case, the mechanism of the second step is still puzzling me, esp the breakdown using D2O

Isn't it just acid base chemistry at the end?

@Zhe, would you be able to elaborate on how the 'acid-base' chemistry might work.

Metal hydride? I'm talking about the collapse of the matallacyclobutane. There is no hydride on it to my knowledge.

Ugh... I should post comments so late at night. I just meant the organometallic species. So, that C-M is fully analogous to say $\ce{RMgBr}$. If you add water to the Grignard, it will protonate the "carbanion equivalent" species.

the Deuterium oxide step seems to be irreversibly, yielding another titanium oxide Tebbe and the side product. Therefore each molecule that reacts is lost, may the reaction be as unfavoured as it is. also the five membered ring constrain may shift equilibrium of the intermediate step towards metallacycle. maybe less side product with cyclohexanone?
if you activate your tebbe reagent with base you may be better on not doing so, since the overreaction demands another coordination to the metal. this may be less favoured when your tebbe reagent is still in its trimethylaluminium stabilized form.
it is unknown to me if the Tebbe reagent is then reactive enough for the demanded first reaction.


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Hydrogen bonding in solid bicarbonates
Why is there two different kinds of hydrogen bonding in solid bicarbonates $\ce{NaHCO3}$ and $\ce{KHCO3}$? In both cases we consider anions of each for hydrogen bonding .

The anions involved in hydrogen bonding are the same for $\ce{NaHCO3}$ and $\ce{KHCO3}$. But, why is there a difference in hydrogen bonding?

I think, this can only be answered by looking at the crystal structures, and then by saying ‘because this is what the crystal structure looks like.’

I found something interesting in Acta Crystallogr. 1952, 5 (2), 292–292, which is attached below.

As someone rightly pointed out in the comments, we'll know this only if we know the crystal structure of the compound in question.
What's interesting to note is that hydrogen bonding, and the arrangement of atoms in space under influence of the same varies from compound to compound, and is dependent on the crystal structure of the same.


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Hydroxide formation: reaction of group 13 elements with water
I was studying about 13th group elements and I found that in this group Al and Tl react water to form a hydroxide and oxide respectively, but the rest of the elements don't react to form hydroxides.
Can anyone tell me why this is the case?

Can anyone tell me why this is the case?

Though the elements reside in same group, they react with water and form different products. This is mainly due to inert pair effect.

Boron

$$\ce{2B + 3H2O ->[700-800 C] B2O3 + 3H2}$$

Boron react with water to produce boron trioxide(III) and hydrogen.
  This reaction takes place at a temperature of 700-800°C.(source)

Aluminium

$$\ce{2Al + 6H2O → 2Al(OH)3 + 3H2}$$

Gallium

$$\ce{2Ga + 6H2O → 2Ga(OH)3 + 3H2 ~~~~~~(1)}$$
$$\ce{2Ga + 4H2O ->[350 C] 2GaO(OH) + 3H2O~~~~~(2)}$$

Gallium react with water to produce gallium oxohydroxide and water.
  This reaction takes place at a temperature near 350°C. (source)

Thallium

$$\ce{2Tl + H2O ->[atm. O2 ~~~50-70 C] 2TlOH}$$

Thallium react with oxygen and water to produce thallium(I) hydroxide.
  This reaction takes place at a temperature of 50-70°C. (source)

Would you please become more specific?


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Using bleach and water to shine pebbles
I soaked a large jar full of pebbles (from a river) in bleach water for a couple of years and the rocks came out super smooth and shiny when dried. I have tried to repeat this process again but am not having any luck with it this time. Have you ever heard of this and do you know if there is a certain ratio of water to bleach that works? Perhaps I put something else in the water the first time and caused a chemical reaction?
Here is a picture of the rocks that became smooth and shiny after a couple of years in water with bleach added for keeping the water clean and clear. And a picture of the rocks I have done the same to for a year. An interesting note is that fossils such as Petoskey Stones do not change at all.

Are you sure you used the same kind of "rock" (Just because you found them in the same place doesn't mean they're the same) 2) Just why did you go and chuck a few rocks in bleach (and leave it there for a few years)?

I suspect the smoothing may have more to do with the caustic soda (sodium hydroxide) in the bleach rather than the actual bleaching agent (hypochlorite).

Thank you for your answer. You are probably right and I just need to soak the second batch of rocks for another year.

I know I am super late on this but I just found it looking up ways to do this. Concentrated bleach (like the toilet bowl cleaner stuff) will eat away at the rocks in a few minutes depending on the composition of said rocks. The softer stuff usually takes less time. The chemical reaction will cause the stone to "melt away" little by little. There is a video on youtube about it. Something like "using a dremel and acid to polish stones" or some other cheeky name lol. It's worked for me, but be careful and patient. It doesn't work with every kind of stone.

I've collected rocks and tumble polished rocks. It is not possible that just sitting in bleach water tumble polished the stones. You can leave the stones in bleach water till you die and they still won't get polished. The only explanation is that you bought polished stones and put them into the bleach water. Over a period of time you just forgot what type of stones that you started with.
The other point is that "river rocks" (rocks smoothed in a river...) will look much smoother when wet because the water wets the surface and "hides" some of the surface roughness. So dry Petoskey Stones just look sort of mottled gray stones, but when wet the coral pattern shows nicely.

The rocks become polished because of tumbling with grit, not because of chemicals. In the best case scenario the chemicals (either soda or bleach) will not do anything to the rocks and in the worst case scenario they will damage the rocks, forming microscopic pits and destroying any polished surface.
Leaving rocks for prolonged times (years???) just seems absurd to me.

I soaked a large jar full of pebbles (from a river) in bleach water for a couple of years and the rocks came out super smooth and shiny when dried.

Maybe they were polished to begin with and you just don't remember? The rocks in the first picture look tumbled. See that irregular shaped white-orange rock in the top middle? Bleach will never do that. This is tumbled.


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How does selenium protect against mercury poisoning?
I have read that it is not essential to avoid fish with mercury in them if they also have a correspondingly high level of selenium because mercury poisoning is primarily primarily the result of mercury disabling selenium based antioxidants.
How does selenium intake protect against mercury poisoning?  Can enough selenium counteract the effects of high mercury intake and does it matter what kind of exposure took place (breathing vapors etc)?

http://link.springer.com/article/10.1007%2Fs12011-007-8004-8  This looks like it may have your answer. I don't have institutional access at home right now to read it fully, however.

BTW, selenium is a poison itself. So we essentially have two poisons neutralizing each other.

Selenium has a very high affinity for mercury. It strongly binds mercury to itself, thus neutralizing the reactivity of mercury. The complex formed Hg - Se is not absorbed by body and excreted out from the body.
In this way mercury is unable to lodge in fatty tissues, resulting in no damage to the body.


text:
Rutherford's Alpha Ray Scattering Experiment
I understood the result of this experiment that the nucleus is nearly empty and things like that.
 But what I have on mind is that when an alpha particle goes nearer to the thin gold foil why couldn't it capture electrons from gold and convert to very stable helium gas? It could also happen right?
All the textbooks only mention the deviation of alpha particles. Why couldn't the above happen or is there anything wrong about my question?

Your assumption is correct. For alpha particles, the main contribution to the total stopping power can be attributed to the electronic stopping power, i.e. inelastic collisions with electrons. Only a small contribution comes from the nuclear stopping power, i.e. elastic Coulomb collisions in which recoil energy is imparted to atoms.
The stopping power of a material is defined as the average energy loss per path length that the alpha particle suffers when travelling through the material.
According to the International Commission on Radiation Units and Measurements (ICRU) Report 49 Stopping Powers and Ranges for Protons and Alpha Particles (1993), the contributions to the total stopping power for alpha particles in gold are as follows.
Typical low-energy alpha particles with $E=1\ \mathrm{MeV}$:

Electronic stopping power: $3.887\times10^2\ \mathrm{MeV\ cm^2\ g^{-1}}$
Nuclear stopping power: $8.394\times10^{-1}\ \mathrm{MeV\ cm^2\ g^{-1}}$

Typical high-energy alpha particles with $E=10\ \mathrm{MeV}$:

Electronic stopping power: $1.650\times10^2\ \mathrm{MeV\ cm^2\ g^{-1}}$
Nuclear stopping power: $1.315\times10^{-1}\ \mathrm{MeV\ cm^2\ g^{-1}}$

Since the energy that is required for excitation or ionization is only a few $\mathrm{eV}$, an alpha particle with an initial energy of a few $\mathrm{MeV}$ can liberate many electrons on its path. When the alpha particle has sufficiently slowed down due to the stopping interactions, it can finally catch two electrons to form a neutral helium atom. Hence, you can release and detect helium when you dissolve old minerals of uranium or thorium.
In the Rutherford experiment, however, the used gold foil was very thin ($<1\ \mathrm{\mu m}$; note that gold is very malleable and can be beaten thin enough to become semi-transparent).
Even for low-energy alpha particles of only $E=1\ \mathrm{MeV}$, the range in gold is still $r/\rho=3.974\times10^{-3}\ \mathrm{g\ cm^{-2}}$, which can be calculated using the continuous-slowing-down approximation (CSDA), i.e. by integrating the reciprocal of the total stopping power with respect to energy. Considering a density for gold of $\rho=19.3\ \mathrm{g\ cm^{-3}}$, the range $r$ can be calculated as
$$\begin{align}
r&=\frac{3.974\times10^{-3}\ \mathrm{g\ cm^{-2}}}{19.3\ \mathrm{g\ cm^{-3}}}\\[6pt]
&=2.06\times10^{-4}\ \mathrm{cm}\\[6pt]
&=2.06\ \mathrm{\mu m}
\end{align}$$
I.e., the experiment was deliberately designed so that the alpha particles were not completely stopped and converted to neutral helium atoms within the gold foil.

The atom, not the nucleus, is mostly empty. That's a pretty big conceptual difference.
Now, there's no reason to believe that the alpha particles haven't picked up an electron on their way to the gold foil before they even strike the foil. The detector on the other side just detects high energy particles, so we can't actually tell if how they're charged. The deviation of the alpha particles (very slight at best) would still be the roughly the same.


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How to determine which reagent to use H2; lindlar catalyst vs Na;NH3(l)?

I do understand that H2; Lindlar will give a cis double bond and Na;NH3(l) will give trans double bond. However,  I am having difficulty understand how to determine which  reagent will work in this question between H2; lindlar catalyst and Na;NH3(l). The answer is given as H2;Lindlar. Second step is to add Br2 to the alkene. Then a strong base such as NaNH2 will deprotonate OH and the compound will undergo a Sn2 reaction to form the ring. The part I am having difficulty is  choosing between H2; lindlar catalyst vs Na;NH3(l).
Thanks for any help.

This is a really interesting question. I'm not sure I agree with the end product, but I will get to that in a second. There does appear to be a piece missing from this equation, which is a second step (something has to be responsible for the new bromine in the product).
First thing is to identify how to get to the final product from an intermediate. You've correctly ruled out $\ce{NaNH2}$ which is just a strong base. It's frequently used to deprotonate a terminal alkyne (which you do not have), and in this example, all it would do would be to remove the hydrogen on the alcohol.
You've also ruled out $\ce{H2, Pt}$. This is a standard hydrogenation and will reduce a triple bond fully to the alkane. If that happens, you won't have the functional groups to do anything like a ring closure.
That gives two possibilities, hydrogenation with Lindlar's catalyst or a sodium ammonia reduction. Both of these reactions reduce the alkyne to an alkene (and then stop). The difference is that concerted hydrogenation will add hydrogen in a syn fashion to afford the cis-alkene, whereas the sodium/ammonia reduction is not concerted and will give the more stable trans-alkene.
The cyclization happens via addition of bromine. This is a standard reaction: halohydrin formation. Except that the intramolecular opening via the attached alcohol will be fast, so the alcohol attacks instead of water to provide an ether. Assuming this is the mechanism, you have to think about how the relative stereochemistry of the alkene is carried forward.
The key intermediate is the bromonium ion (the three member ring with a bromine). The intermediate retains the stereochemistry of the alkene. You might need a model to identify this, but the correct bromonium ion that gives the indicated product has the hydrogens on the same side. That means that they were originally syn across the double bond, so the double bond was cis. Then Lindlar hydrogenation was the appropriate reaction for the first step.
Here's my issue with this question: I don't actually think you'll get this product. The technical point is that the opening can proceed via a 5-exo-tet attack or a 6-endo-tet attack. Generally, in these types of reactions, the exo attack is favored for stereoelectronic reasons. So my money is on obtaining the 5-member ring instead of the 6-member ring.


text:
How can both kinetic and potential energy increase simultaneously in a open substance?
I have an open test tube containing a solid substance. In a school environment, we  heat it, which increases thermal (kinetic) energy, and this heat breaks its bonds, increasing its potential energy, turning it into a liquid. I understand in thermodynamics, there is a conversation of energy, yet since this test tube is open to other factors, such as air, can this occur?
Also, if this experiment were insulated what would occur?

First off, you are ADDING energy.  What If I grab a ball and throw it up in the air?

I edited my question, do you think you can help with that? thanks

Need a more complete description.  How would you heat it if it's insulated?

Sorry, it is heated due to dropping another substance inside of it. So we have a liquid, if we drop a substance inside (which heats and solidifies it) and quickly covered it/ blocked it from air, what would occur?

The energy is being redistributed. It is not being created.The heat that you are using to increase the energy of your test tube has to come from somewhere (e.g. burning fuel). Conservation of energy holds true in all scenarios. If conservation of energy is not satisfied, you are probably leaving out something.

You are adding energy by heating, this generally does not break chemical bonds (unless the temperature reaches hundreds of degrees) but instead the weak intermolecular potential energy between molecules (sometimes call van-der Waals interaction) and so the solid will melt, then the liquid will boil, obviously :)
The kinetic and potential energy of the sample has increased. Kinetic energy in translational motion of the molecules and internal potential and kinetic energy of the molecules has also increased as they have absorbed some of the heat as vibrational and rotational energy.  (A non-linear molecule with N atoms has $3N-6$ different ways of vibrating, e.g ethane has 18 ways).
The reason that the temperature increases rapidly is due to the fact that if you heat a pan on the stove, or a boiling tube with a Bunsen burner  is that convection of heat away into the air is very poor. So your boiling tube is effectively insulated. If it were completely insulated then less added heat would be needed to reach the same temperature. As an admittedly odd example, you will know that a plumber can join Cu pipes with solder using a blow torch, but only if they are dry. If they contain a even small amount of water more of the heat is carried away (and heats the water) and soldering becomes impossible.
If you heat a sealed tube, it is likely to explode violently due to increase in pressure; so don't do it! (To prevent explosion is why pressure cookers have a pressure release value).


text:
Why does alpha cleavage occur?
I am studying fragmentation patterns in mass spectrometry and I am confused about why alpha cleavage occurs. Why does the carbon-carbon bond adjacent to a functional group break instead of the bond between the alpha carbon and the fucntionalized moiety?

Because then you'd have two unstable fragments. There's simply not enough energy in the system to do that.

I edited this question in an attempt to clarify what you are asking.  Please let me know if I distorted your meaning, and please feel free to revert the edit.

You need to look at the principle modes of decomposition of radicals.

It's worth noting that alpha cleavage occurs primarily for radicals, such as the radical cations formed in electron ionization, and sometimes other ionization techniques (e.g. MALDI).  The fragmentation patterns that result from "even-electron" ions that predominate from e.g. electrospray are quite different.

After ionization occurs, the carbonyl donates its free radical into its own double bond.  This causes the bond adjacent to the carbonyl to split into two electron s, one with spin antiparallel to that of the carbonyl radical and one of parallel spin.  The one of antiparallel spin combines with the free radical to form another pi bond.  This creates a carbon oxygen triple bond and carbon atom with a free radical (alkyl radical).  The carbon atom with the radical is the atom alpha to the carbonyl, hence alpha cleavage.  The oxygen bears a positive formal charge.  Alpha cleavage basically makes substituted carbon monoxides, and side chain radicals.  You will see the "carbon monoxide derivative" on the spectrum.
To do this quickly, just take the original molecule, draw lines showing fragmentation on both sides of the carbonyl, and then draw the carbon oxygen triple bond structure.


text:
Protecting elemental Li from the atmosphere
I would like to make a small decorative object out of lithium, and keep it in an ordinary household (indoors) atmosphere.
The making itself aside, what can I do to protect it long-term from reacting with the atmospheric water and gases? Is there something that could react with it to form a protective outer layer, like oxides do on some other metals? Is it perhaps possible to electroplate it with Mg or Al? Are there other solutions?
To be clear, I don't want to just keep it submerged in oil; I want it to stand on a shelf and hopefully even get handled occasionally by curious hands.
I have no access to advanced equipment or supplies - I'm only armed with a high-school level of chemical understanding, access to ordinary hardware/paint shops, and patience.

I do understand that the question is unusual and has little broad practical utility, but it's still a proper chemistry problem to which I would like to see a proper chemistry solution (or an exhaustive explanation why a search for a solution is futile). Please don't answer along the lines of "your idea is stupid and you should feel stupid".

I’m downvoting because it ‘lacks research effort’.

Rust-oleum (tm).  Spray lacquer.

@Jan, at my level of education, it was already a research effort to find out how readily it reacts with atmospheric N₂ and CO₂ (after hydroxidation). Nobody makes physical things out of Li, so there isn't really any practical information out there, which is why I'm turning to a better-educated site for help. This now sounds like, in order to ask the question "right", I pretty much need to go and keep learning more about chemistry until I know the answer anyway.

Keep it in a jar under mineral oil.

You did give the Wikipedia article a glance, didn’t you?

@Jan, that was the first place I looked.

‘Like all alkali metals, lithium is highly reactive and flammable. For this reason, it is typically stored in mineral oil.’ The sentence is practically screaming ‘your idea is a bad idea.’ To quote you: ‘Nobody makes physical things out of Li’ — yes, for a reason.

"I want it to stand on a shelf and hopefully even get handled occasionally by curious hands." Why? You could say the same about about a rattlesnake, but I would advise against it.

@Zhe, aesthetic reasons. "This rattlesnake figurine is full metal with no cavities, except where venom glands would be - pick it up, see how light it is."

Make it out of zinc instead.

I suppose you could make a hollow glass mould in the shape of your desired object, and then melt a small hole in the mould through which you could you could fill molten lithium.  You'd need to an easy way to work with molten glass under an inert atmosphere and a slew of detailed fire safety protocols (molten lithium is dangerous!!!).  You'd also need to do detailed thermomechanical calculations to make sure that your mould could handle all required thermal stresses.  If this all sounds extremely difficult, that's because it is.

Interesting question!(Though personally I've no idea why you want to make a decorative article out of lithium, but I guess it's best not ask)
You did consider using the process we know as electroplating; yes, it's one of the common, commercial methods of 'passivating' (rendering inert) a lot of metals that are prone to corrosion, like iron for example (Galvanization). However electroplating is something that could never work out with lithium, here's why...

(The above example, courtesy Google Images, suggests the use of copper as the electroplating metal, but since I can't seem to find a more generalized diagram so we'll just go with this)
A quick crash course on the set-up for electroplating:
Your cathode is the metal to be electroplated, while your anode is the metal (copper here) that is used to electroplate whatever you've hooked up as the cathode. Both the anode and cathode are immersed in an aqueous solution of a salt of the metal used as the anode (here, it's copper sulfate).
Clear with that? Good, now here's the catch...
You need to put lithium in an aqueous solution, that is, a solution that uses water as the solvent. In your question you voice concerns over lithium's reactivity towards stuff you typically come across in your surroundings (oxygen and carbon dioxide in the air, moisture/humidity, etc), so I take it that you are well aware of what happens when you dump a large, fresh piece of lithium in water...

So yeah, don't even think about electroplating this.

The making itself aside, what can I do to protect it long-term from reacting with the atmospheric water and gases?

Well, there are a couple of other methods you could use:
1) Coating it in wax.
Coating the lithium in a thin, uniform layer of wax (candle wax, beeswax, furniture wax or any sort of wax that you can easily get your hands on, and remains a solid at room temperatures) would do a pretty decent job of protecting lithium from your ordinary household environment. It shields the metal from both atmospheric gases and moisture. Heck you could even dump the whole thing in water and it wouldn't go off in your face. But evenly coating a "decorative object" made out of lithium with wax while still being able to see the object clearly under the wax will be a bit tricky...even more so, if this "decoration" of yours has any really intricate patterns.
This is where we move on to the next method,
2) Immersing it in a jar full of mineral/paraffin oil
This is far simpler than coating the thing in wax. Get an airtight, transparent jar, fill it all the way to the brim with mineral oil, slowly immerse the object in the jar so you don't trap any air bubbles, screw on the lid tightly and there you have it! True, a major 'disadvantage' here, is that you can't take the object in your hands and examine it closer as and when you wish (which wasn't the case in the waxing method), but maybe a shiny lithium decoration floating around in a jar might make up for this aesthetically.
Which of the two methods you use, is entirely at your discretion, however, a word of caution:
Lithium metal is a fire risk! Do handle it with care. Have necessary safety arrangements in place.

Industrially, this has been done by coating with Parylene.
The Rubber and Plastics Age, Volume 46 says:

Parylene coated lithium metal granules on the left exhibit no reaction with water on which they float

and  New linear polymers says:

Encapsulation of reactive chemicals in thin parylene C coatings provides protection. Parylene-coated lithium metal granules at left exhibit no reaction with water on which they float Unprotected granules at right react instantly and violently with water.

However, I don't know on what timescale this protection lasts.  As others are saying, lithium metal is an extreme fire hazard.  What if the coating is scratched or peels? Be careful!

I would like to make a small decorative object out of lithium, and keep it in an ordinary household (indoors) atmosphere.

You’re trying to open up a really big can of worms. First, let’s start with the making part. The good thing about lithium is that it is soft enough that you can cut it with your average kitchen knife. But that’s about the end of the advantages.
You’ve already read that it reacts readily with atmospheric carbon dioxide, nitrogen and moisture in the air. Hence why you’re asking about some protective coating. But all the difficulties you have with lithium’s reactivity are present and worse when you’re trying to make it. Repeatedly, you will be cutting new edges, adding new blank surfaces that will come into contact with air and moisture for the first time. There will be constant reactions going on.
Let’s assume you somehow managed to handle that. Another sentence of your’s:

I want it to stand on a shelf and hopefully even get handled occasionally by curious hands.

Please don’t let it get handled by curious hands! For one, you should never handle alkaline metals with bare hands (see above their reactivity). For two, any protective coating you may apply to the lithium would get rubbed off given time and curious hands. That would access the underlying lithium metal and all the reactivity problems start again. And when would that happen? Yes, exactly when the curious hand is handling it. Worst case would be spontaneous combustion in somebody’s hand (but admittedly that’s not as likely with lithium as with other alkaline metals).
So if it must stand on the shelf … well … it’s your shelf … But keep a strict ‘do not touch!’ policy for everybody’s safety. (Note that I am not recommending the shelf either.)

I have no access to advanced equipment or supplies

Considering everything, this should be the final straw of your idea. I mean, you also mention:

Nobody makes physical things out of Li

Yes, because it is dangerous, difficult to handle and there are much better solutions out there. Sometimes when nobody does something, it is because nobody came up with the (potentially great) idea yet. Most of the times when nobody does something, it is because there is some very great risk associated with it that is clearly visible to most people. Creating a sculpture out of lithium metal falls into the second category.

I'm flagging this NAA :P

@Martin-マーチン Wow, you got me. I was sure that this was going to be ortho’s comment xD

You could put it inside of a glass vial / melt glass around it under inert atmosphere. Then you can a) see it and it is b) protected from oxidation. Though I don't know how to achieve that with household techniques...

OP might want to watch this video before exposing lithium to molten glass https://www.youtube.com/watch?v=cFGejaYqM-c !

Yeah well its the question if this will also happen under inert atmosphere. @DavePhD That putting very hot oxides on lithium could turn to be dangerous is of course clear, I don't know if the oxidation of the lithium by the glass as shown in the video will also happen under inert atmosphere and less intense heating. In any way, you can melt a vial around the lithium and in fact that is commonly done with the all of the alkali metals (including e.g. Cs which is much more reactive) to provide specimen for inorganic chemistry classes.

The video specifically says the lithium is taking oxygen from the glass itself

@DavePhD Yes, but they a) use a mixture of crushed glass and lithium shavings not normal glass around lithium which greatly enhances the surface area for the reaction b) heat with a bunsen burner c) don't exclude oxygen. I would expect that nothing happens when just melt glass gently around the lithium. And creating a vial will always be possible as then you don't apply the heat onto the lithium metal. Just look it up, evacuated glass vials may be prepared for all kinds of reactive materials including all alkali metals.


text:
Oxidation state of hydrogen in the LaNi5H intermetallic
This question is from an introductory chemistry course:

What is the oxidation state of $\ce{H}$ in $\ce{NaH}$, $\ce{HCl}$, and $\ce{LaNi5H}$?

I've looked in the textbook to no avail.
Google doesn't help me much either. It does tell me that $\ce{LaNi5H}$ is an "intermetallic" compound.
The best I've found is this:

Allen electronegativities also yield zero for the oxidation state of all such $\eta^2$ hydrogen atoms.

What is the justification? Preferably, is there an "intro to chem" level explanation for this?

At an intro chem level, the oxidation state of all elements in an alloy or intermetallic is zero.
At a deeper level of analysis, fractional, integer, and/or decimal oxidation states may be more appropriate, depending on the system in question.

Per Wikipedia:

An intermetallic, also called an intermetallic compound, intermetallic alloy, ordered intermetallic alloy, and a long-range-ordered alloy, is a solid-state compound exhibiting metallic bonding, defined stoichiometry and ordered crystal structure. Many intermetallic compounds are often simply called alloys.

However (emphasis added):

Schulze in 1967, defined intermetallic compounds as solid phases containing two or more metallic elements, with optionally one or more non-metallic elements, whose crystal structure differs from that of the other constituents.

So, in a high-level description, an intermetallic containing hydrogen such as the $\ce{LaNi5H}$ of the question can be considered essentially to be an alloy with hydrogen as a constituent.  Based on this, assigning a $0$ oxidation state to all elements is warranted as a first approximation:

For intermetallic compounds, the ultimate choice of the oxidation state zero at all atoms is best if needed in redox chemistry.  (source | doi)

A more detailed investigation might turn up evidence supporting assignment of nonzero fractional, integer, and/or decimal oxidation states to various atoms:

Subtler estimates and round offs are required for compounds with electrons delocalized over non-equivalent atoms, as expressed by several resonance formulas with weights in arbitrarily long decimal numbers. Without round offs of bond orders in Lewis formulas, decimal values of oxidation states would be obtained for certain bonding connectivities. [...] [U]nambiguous and reasonable fractions of small integers are obtained for oxidation states in compounds such as dithiolate and catecholate (see Section 11) or in (car)boranes such as $\ce{B6H10}$ (see Section 6) and $\ce{B10C2H12}$ (p1041), or when vicinal oxidation states are indistinguishably mixed, such as in $\ce{YBaFe2O5}$ (p1058). Reasonable fractional oxidation states appear also in ions where the charge is distributed over several equivalent atoms such as $\ce{C7H7+}$, $\ce{B6H6^2−}$ (p1040), $\ce{I3−}$, and $\ce{N3−}$ (p1037). (source | doi)

Both nickel and lanthanum have electronegativities lower than hydrogen’s $2.2$. Thus, even if the compound is intermetallic, we can treat it as a hydride and give hydrogen an oxidation state of $\mathrm{-I}$. Oxidation states lower than $\mathrm{-I}$ are highly improbable.
Determining the oxidation states of lanthanum and nickel would be a lot harder, but summed up they must be $\mathrm{+I}$. It would need a structure to determine the rest though.


text:
Do minerals and/or metals found in water tend to settle to the bottom?
I was reading an article that suggested that if one lets water sit for a day in a bottle or pot, then some of the metals in the water, such as lead, will settle to the bottom.
Does anyone know if this theory has been tested?
The reasoning seems somewhat logical, since particles in water tend to sink to the bottom. However, at the molecular level I know that ionic interactions may cause unintuitive behaviors.
I do have a TDS and pH meter. Perhaps I could let water sit for a day, then check the TDS of the water on top. Then compare that with the TDS of water that has been thoroughly mixed. Does anyone see any flaws with this idea?

Excessive toxic metals.
Many city water supplies, especially wells,
  contain some excessive amounts toxic metals or toxic amounts or forms
  of other minerals.  Common toxic metals include arsenic and lead,
  which are found in many pesticides and herbicides and find their way
  into the wells and other water supplies.  At times, these are
  difficult to remove from large amounts of water, so they are just left
  there even if the water violates environmental standards, especially
  in poorer nations.
Other minerals that often contaminate drinking
  water include too much iron, manganese, boron and copper depending on
  the location. One way to remove some of these easily is to let a
  pitcher or pot full of tap water sit for a day and some of the metals
  will settle to the bottom of the pot.
You can then pour off the
  water, leaving most of the minerals in the bottom of the pot.  You may
  lose some beneficial minerals this way, but it is one way to remove
  some toxic metals, as well.

-Dr. Lawrence Wilson
http://www.drlwilson.com/ARTICLES/WATER.htm#INTRO

There is no harm in trying. Go on, do it. Chemistry is an experimental science, they say. Though I wouldn't expect much of anything. Suspensions usually settle; true solutions do not.

Your source has a somewhat dim and unscientific view of standard domestic water that isn't justified in reality (unless you live in Flint, anyway). In western countries municipal tap water has to meet a range of standards about what it contains and usually does (this, BTW, is one of the single biggest innovations in human health in the last 200 years). This means that further filtration or treatment is usually useless. OTOH, supplies which are hard may show changes over time as minerals precipitate (they don't settle, they are in solution). But these changes won't affect health.

@matt_black, I do not agree that water filtration is useless. For example, read about chromium-6 in drinking water. Here is a nice and concise article written by a PhD in chemistry with a lot of background in water filtration: here

@N4v The first rule of Skeptics club is: don't believe anything you read in marketing material. Your source is trying to sell you something. that doesn't mean what he says is wrong but it does mean you can't trust it unless you find independent verification from someone who isn't trying to sell you something.

@matt_black, I understand your logic, but think of these two facts: (1) there are many substances which have been verified through independent research to be harmful. These include things like lead, copper, and disinfection byproducts. (2) The EPA set contaminant limits not based on ideal levels, but based on the balance between health and governmental practicality. Do you disagree with either of the things I just said?

@N4v "Harmful" isn't binary. The extra health gain from reducing some "harmful" contaminants (like chlorinated byproducts that come with chlorinated water) isn't worth anyone's money or effort (especially when the alternative is thousands of times worse (as bugs in your water clearly is). The same applies to most other contaminants.

@matt_black, I see your point. Thank God I'm open-minded so even though I disagreed at first the truth is you may be right. But let's take the example of chlorinated byproducts (a.k.a. disinfection byproducts): why are you confident that any health gain attained from not consuming chlorinated byproducts is essentially infinitesimal?

I have experimented with this phenomenon and found that allowing water to sit for a day may have a non-trivial effect on the TDS gradient in the water. This effect could be due to various reasons, however.
Materials: Glass pitcher, plastic & porcelain & glass containers; tap water from near Chicago, IL, USA; TDS meter
Methods: I poured tap water into various containers and allowed the water to stand for about 30 hours. I then measured the TDS at the top of the water. Then I stirred the water vigorously and let the water sit for about a minute. I took a TDS reading again at the top of the water. I repeated the experiment a total of 4 times.
Results (Updated 12/23/2016):
$$\begin{array}{ccc}
 & \text{TDS Before Stirring} & \text{TDS After Stirring} \\ \hline
\text{Glass} & 138 & 141 \\
\text{Plastic} & 150 & 146 \\
\text{Porcelain} & 153 & 152 \\
\text{Glass} & 147 & 149 \\
\end{array}$$
Conclusion: The results are interesting. I will discuss a few possible reasons behind the results, but I'm not certain.

TDS (after) was higher both times for glass. Maybe this is because when the water sits, a small number of ions adsorb to the glass surface.
For plastic and porcelain, TDA (after) was lower for both plastic and porcelain. This may be due to dust that settled and dissolved into the water, as I did not cover the containers (looking back, covering the containers would probably have made the experiment more controlled). In addition, maybe ion adsorption happens less with porcelain and plastic.

How many times did you measure it? Also, what is the reading for the fresh tap water?

Indeed. Per @IvanNeretin's comment, you should repeat the experiment at least three or four times and calculate mean/st.dev. on the before/after difference.

Have you excluded other explanations, eg.  ions absorbing on the glass wall?

Give me some time and I will repeat the experiment. Easy with the downvotes, though. No need to hate on this. This is a real experiment. @Greg, that is a good question. I will think about that

Are those differences even significant? How do you tell noise form signal here?

The experiment was just for fun. The reasons I mentioned were possible, not thoroughly confirmed.

My well water in California has lots of minerals, especially in the summer and fall, when the water table is low.   I've been letting water sit in tall gallon containers for many years.   Over time, the bottoms of the containers get cloudy with mineral build up.   I pour the water off the top, until it gets down to the last three or four inches in the container.  Then I pour the water that's been on the bottom out, and rinse and refill the container.
When the containers get so cloudy, I use a vinegar solution to get rid of built up minerals.           Seeing how quickly the containers show mineral accretion, I continue to do this.   I keep three glass gallon containers going at the same time, so that two are sitting while I use the last one.
Perhaps it doesn't really do much, but the amount of accretion I see in the jars inspires me to keep doing it.
Thanks for conducting and posting your research!


text:
Is iron(III) carbonate stable in water?
I searched for this onsite already, and also found the following answer on Quora:

In aqueous environment, iron(III) is bound by water molecules and exists as $\ce{[Fe(H2O)6]^3+}$. Due to strong interaction between Fe(III) and some of the aqua ligands, the complex can dissociate into $\ce{[Fe(OH)_n(H2O)6^{n-}]3^{n-}}$ and $\ce{H+}$. Carbonate is a base, so it will capture proton and further facilitate the formation of hydroxyl complexes so much that it will produce $\ce{H2CO3}$ and $\ce{Fe(OH)3}$, the former eventually decomposes into water and $\ce{CO2}$. Therefore, iron(III) carbonate cannot exist in water.
In solid state, most of iron(III) compounds are hygroscopic, which means that they have strong affinity for moisture. And of course, when water gets involved, the above processes start happening, which eventually decomposes iron(III) carbonate, if it could ever be formed as solid.
Therefore, the compound itself is not stable in solutions or in solid states.

Is this correct?  What happens in the following equation then:
$$\ce{2 FeCl3(aq) + 3 Na2CO3(aq) -> Fe2(CO3)3 + 6 NaCl}$$
Will we get different end products?  What will be solid?  Hydroxides or oxides?  Will there be carbon dioxide evolved?  Are any of the products going to be hydrates?

Strongly related but not a dupe: https://chemistry.stackexchange.com/q/38916

No, your equation is not correct. The correct equation is
2FeCl₃ + 3Na₂CO₃ + 3H₂O = 2Fe(OH)₃ + 6NaCl + 3CO₂

It is not clear what do you mean stable. Iron(III) salts generally form an oxo-hydroxo precipitate unless some complexation or other tricks going on. From that point of view, it doesn't matter if carbonate or other non- or weakly complexing ion you have as a counter ion, they are not stable.

Apparently, ferric carbonate might be produced in solution by mixing ferric ions and carbonate ions. This is not the case, because ferric solutions are rather acidic, due to the hydrolysis reaction $$\ce{Fe^{3+} + 2 H2O -> [Fe(OH)]^{2+} + H3O+}$$ If a $\ce{Na2CO3}$ solution is added to a solution containing ferric ions like $\ce{Fe^{3+}}$ or $\ce{[Fe(OH)]^{2+}}$, $\ce{Fe2(CO3)3}$ cannot be produced. Instead, an insoluble basic carbonate is produced with the formula $\ce{FeCO3(OH)}$. The equation is : $$\ce{Fe^{3+} + 2 CO3^{2-} + H2O -> FeCO3(OH) + HCO3^-}$$ or $$\ce{[Fe(OH)]^{2+} + CO3^{2-} -> FeCO3(OH)}$$The basic carbonate $\ce{FeCO3(OH)}$ is a brown precipitate, which is decomposed by heating to boiling temperature according to : $$\ce{FeCO3(OH) + H2O ->  Fe(OH)3  +  CO2}$$
Ref.: R. Delaby, Analyse qualitative minérale, Masson, Paris 1950, p.76.

The reaction
Mixing iron (III) chloride and sodium carbonate solutions results in forming of a brownish precipitate with the liberation of a colourless gas: $$\ce{2FeCl3(aq) + 3Na2CO3(aq) + 3H2O(l) -> 2Fe(OH)3(s) + 3CO2(g) + 6NaCl(aq)}$$
The resulting precipitate is sometimes written as $\ce{Fe2O3\cdot aq}$. Adding an excess of sodium carbonate solution may lead to the formation of $\ce{HCO3-}$ ions in the resulting solution.
Explanation
Iron (III) carbonate is a weak electrolyte since it is formed by a weak acid and a weak base so it gets readily hydrolysed especially in the presence of a mineral acid:
$$\ce{Fe2(CO3)3(s) + 3H2O(l) -> 2Fe(OH)3(s) + 3CO2(g)\ \ \ \ \ \ \ (pH\approx7)}$$
$$\ce{Fe2(CO3)3(s) + 6H+(aq) -> 2Fe^3+(aq) + 3CO2(g) + 3H2O(l)\ \ \ \ \ \ \ (pH<7)}$$
Useful links and references

Ronald L. Rich. Inorganic Reactions in Water (1st ed.): "Carbon oxide species", p. 178.
Wikipedia article on hydrolysis: https://en.wikipedia.org/wiki/Hydrolysis.
Demonstration #1 of the given reaction: https://www.youtube.com/watch?v=XaIPZhOB7CI.
Demonstration #2 of the given reaction: https://www.youtube.com/watch?v=qQp8Odsl5TE.

Please do not give references in Vietnamese language, like 3rd and 4th links (Youtube demonstrations #1 and #2)

Basically, you just add $\ce{NaCl}$ to the solution, since $\ce{NaCl}$ does not change the pH of the solution, the reaction for $\ce{Fe_2(CO3)_3}$ described in the quote does not change.


text:
Do we lose significant figures with digital instruments?
With digital scales, we get an answer to the hundredths or thousandths place, but we do not have the ability to estimate "between the lines".  Is the real significance of a measurement different when it comes from a digital scale, or thermometer or other device than when it is read with the possibility of visual interpolation?
As an example, I might have a digital scale that gives a reading of 25.36 grams.  It could be anywhere between 25.355 and 25.365 grams.  If I read a sliding scale to 25.36 grams, the 6 is an estimate and the true value could range by a few hundredths of a gram while the digital scale at most would have a range of +/- 5 thousandths of a gram.
I realize that you won't have a sliding scale that can match a digital scale anyway, but the same could occur with digital thermometers reading to 0.1 degrees and indicating a range of true value of +/-0.05 while a reported value to the nearest 0.1 degree on a thermometer could be several tenths of a degree off.   So is there any acknowledgement that the reported values have different meanings with regard to significance?

As far as I know, we aren't supposed to interpolate visually, but take the complete reading. No +/- estimation. The provision to compensate for this is to do an error analysis - regarding the least count of the instruments you use and all.

Don't forget that digital instruments record the number in binary anyway. How would you represent $0.1$ in binary? $0.0001100110011\ldots$ And if you were to truncate because there's not enough memory in the balance to  represent the infinite number of repeating figures, then it's not $0.1$ is it?

Related answer concerning interpolation: Using sigfigs when measuring with an instrument with marks other than powers of ten?

@Loong, I thought there was something like that-a quantum model of graduated cyllinders.  We are really just recognizing that the application of gradation divisions are an estimate.  I remember having seen the terms "readability" on a few graduated cylinders and the readability was in values like those in the third column on the table.  I have had some 10 mls with 0.1 ml gradations though.  I assume no estimation allowed between the lines though they are class A.  Also some 100s with 1 ml gradations.

One thing I will mention is that the ability to estimate to 0.5 ml with 1 ml gradations only causes the retention of even more sig figs with such a tool versus a digital tool.  We retain an additional sig fig even though our precision is only to the nearest .0 or .5.  So hypothetically, 25.35 on a sliding scale would be anywhere between 25.325 and 25.375  (a range of 0.05) while 25.35 on a digital scale would be between 25.345 and 25.355 (a range of 0.01).

Relying on the "figures" an instrument gives you beyond 2-3 seems daft unless you've calibrated it, and it also autocalibrates to the environmental conditions (temperature of the strain gauges, I think this is pretty common) and you're also adjusting measurements for atmospheric buoyancy (relevant at 2.9 digits) and the density of your sample...

I think there's a bit of a false equivalency going on here. You're comparing a digital scale with 0.01 g granularity to an analog scale with 0.01 g markings. I don't have definite proof of this, but my guess is that typically a comparable digital scale has finer gradations than the analog version.  That is, if a given scale came in both an analog and digital version, if the digital version had 0.01 g readout, the analog scale would have at most 0.1 g gradations. e.g. most digital themometers have ~0.1 degree readouts, but your standard alcohol thermometer is only marked every 1 degree.

I am actually comparing a digital scale with 0.01 g granularity to an analog scale with 0.1 g markings that are readable to 0.05.  The question is not so much whether the digital and analog are different, but whether the concept of precision or readability is different given that you do not "see" where the digital scale lies between its two nearest output values.  As I said though, with an analog, the markings were already placed with some imprecision so the justification in making a visual interpolation is questionable.

As far as I know the visual interpolation of analogous measurements is bad scientific practice anyway.
We assume that rounding errors work in both directions and compensate each other in the long run. On the other hand visual interpolation by a human can introduce systematic errors.
But putting this discussion away one has to note that even if you visually interpolate, you still digitalise your data (just with a finer grid). By digitalising I do not mean that you write them into a computer, but the abstract mapping of continous data points of $\mathbb{R}$ into a finite subset.
In the end it just is a question of how fine the grid is you are measuring with. And yes, if you don't introduce systematic errors a visual interpolation makes your grid a little bit finer.
On the other hand switching to electronic devices usually gives you orders of magnitude better results.
Some examples:
Let's assume that someone wants to measure lengths of $\mu$m with a normal ruler that has 1 mm spacing.
It is intuitively clear that this won't work.
(Even if you can interpolate.)
Now let's look at another example that was brought up in the comments by @Zhe. The problem that every number in your computer is an element of a finite set. This introduces rounding errors, when saving real numbers (real as in $\mathbb{R}$).
If you execute the following python code or something similar in your language of choice, you will see this rounding error.
import sys
print(sys.float_info.epsilon)
# for python2:
# print sys.float_info.epsilon

On my 64bit machine this is in the order of $10^{-16}$. The standard deviation of the meter is in the order of $10^{-11}$ (wikipedia).
Floating point noise will rarely be a problem, when dealing with experimental lengths.
This kind of analysation has to be done for every experimental measurement device you are using. Regardless of electronics vs. non electronic measurements. The advantage of "old analogous" measurements is that you can intuitively see from the scale how fine the grid is, that you are working with.

Just wondering, are digital scales designed to "round" to the nearest final decimal place on the display or do they trigger the next value when a threshold is passed.  For example, a stopwatch will read 0.01 s only after 0.01 s, and before 0.02 s.  How is a scale designed so that it reads 0.01 grams between 0.005 and 0.015 grams rather than when the 0.01 gram threshold is passed?

Also, while the description of the problem of interpolation makes sense when viewed from the point of view that the tools (like a graduated cylinder) are imprecise, not just the reading of the tool.  Still, I am surprise because MANY college level chem texts describe the method of estimating between the lines-taking the size of smallest increment on the device,  dividing it by 10, and then multiplying that by the fractional distance of the reading between the two increments.

Your first comment: That is implementation dependent, but rounding would make more sense. Your second comment: I can just say what I learned in my lab practicals about visual interpolation. The interpolation scheme you mentioned makes definetely sense to me, but still it may open up the possibility of beautifying data by a human.

It has some base 10 bias (people don't visually divide a span into 10ths very naturally-maybe quarters) and also I think there is a tendancy NOT to chose a value that coincides with an increment mark if the visual does not put the value indiscernibly close to the line.  For example, if one increment is 0.1 and the next is 0.2, observers will tend to avoid reading a value as being 0.10 (or 0.20) because they can detect that it is not precisely on the line, but they can not detect that it is still closer to the increment line than the next 0.01.  "Off the line" readings may dominate.


text:
How many liters of hydrogen, H2, are needed to react with 10 liters of nitrogen gas in the reaction forming ammonia?
How many liters of hydrogen, H2, are needed to react with 10 liters of nitrogen gas in the reaction forming ammonia?
$$\ce{3 H2(g) + N2(g) -> 2 NH3(g)}$$
My try:: Because we have 10 L of nitrogen gas, we have 10/22.4 moles = 0.446 moles of nitrogen gas, and thus need 0.446 * 3 = 1.338 moles of hydrogen gas. As such, we need 2.676 grams of hydrogen gas, or 2.676/0.09 = 29.8 L of hydrogen gas. However this is not correct. Can someone find my problem?
Note: This reaction occurs at STP.

You did it the hard way, but correctly except let's keep units in the density.  You value is off slightly due to rounding of a few terms (with the simplistic assumption that they are ideal gasses).

Well Laplacian, there is a flaw with this question because you need to specify the temperature. PV=nRT and we already know what R is and the volume but we dont know the temperature nor do we know the pressure. So therefore there's too much assumptions that we make and the liter to mole ratio is skewed, you needed to specify whether it was STP or something else

@JohnRawls Sorry, the question was supposed to say that it was at STP (that's where the density came from).

@QuantumAMERICCINO While the question may look like homework to you, it is actually not homework. Please do not randomly tag questions you think are homework as homework.

Doesn't have to be STP as long as both gasses are at same temp miles are proportional to volume.

@LaplacianFourier It does not really matter whether it is actually homework for you or not. It looks like a textbook problem. Tagging it helps pruning duplicates easier. It also helps make searches easier for people that have similar type of questions.

You are on the right track. Dimensional Analysis is your friend:
$$\frac{10\,\text{L  $\ce{N2}$}}{1} \cdot \frac{1\,\text{mol  $\ce{N2}$}}{22.4 \,\text{L  $\ce{N2}$}}\cdot \frac{3\,\text{mol  $\ce{H2}$}}{1\,\text {mol  $\ce{N2}$}}\cdot \frac{22.4\,\text{L  $\ce{H2}$}}{1\,\text {mol  $\ce{H2}$}}=\,?\,\, \text{L $\ce{H2}$}$$
If you do this, you will realize that the quantities cancel exactly and give you a nice clean number.

that won't work if we dont have STP otherwise everything changes

Moles are proportional to volume.  All you need is for both gasses to be at the same temp.  @John Rawls.


text:
Why is the radius of the caesium cation smaller than that of the chloride anion?
I am currently studying the structures of ionic compounds and how we can use radius ratios to predict the possible arrangement of ions in the lattice. My book mentions that the cation to anion radius ratio in caesium chloride is about 0.93, and since this number lies between 0.732 and 1, it suggests a cubic arrangement with a coordination number of 8.
But caesium is in the 6th period of the periodic table whereas chlorine is in the 3rd. A caesium cation would be similar to xenon, which is in the 5th period. And a chloride anion would be similar to argon, a period 3 element.This means that there is a difference of two shells between the two atoms. How is the ionic radius of chloride still larger? In general, how many "shells of difference"  are required for a cation to be larger than an anion?

I think it may be simply nuclear charge. The effective nuclear charge for an electron in a 5p Cs atom is 13.651. On the other hand, the effective nuclear charge for an electron in a 3s Cl atom is 6.1161.
http://www.knowledgedoor.com/2/elements_handbook/clementi-raimondi_effective_nuclear_charge.html#chlorine
I do not think there is an effect similar to lanthanide contraction for cesium.

This is something students often forget when comparing radii: shells are not simply additive.
If you move along eight elements to go from lithium to sodium, you are not only filling the second shell and putting one electron into the third, you are also adding a proton to the nucleus with every additional electron. The larger the nuclear charge is, the greater the electrostatic potential the electrons experience. And the greater the potential the stronger the force that draws them towards the nucleus, thus the more compact an orbital will become. Thus, while each new period will enlargen the atomic radius each added proton will reduce it. Therefore, the atoms do not become infinitely larger in size — although a sodium atom will always be larger than an electronically comparable lithium atom.
A similar discussion can be made for charges. Adding an electron to the mix increases the electron-electron repulsion (more negative charge) while removing one decreases it. Thus, with each positive charge a cation will get smaller while an anion will get larger with each negative charge. All of this results in rather complex calculations; one should always remember that an atomic radius cannot be easily guessed unless it is a straightforward up/down or left/right movement in the periodic table.
As for your second question: an anion may be smaller than a cation if the anion has very few shells and the cation very many. Unfortunately, the Wikipedia article on ionic radii does not quote a size of the hydride anion which would otherwise be a strong contestant. Instead, the smallest anion in the tables is $\ce{F-}$ with a crystal ionic radius of $119~\mathrm{pm}$ in hexacoordinate state. Searching for a cation that is larger quickly gives potassium with $152~\mathrm{pm}$ crystal ionic radius. In effective ionic radius terms (the second table), fluoride is mentioned as having $133~\mathrm{pm}$ while $\ce{K+}$ has $138~\mathrm{pm}$. Thus, by both sets of values the cation in $\ce{KF}$ is larger than the anion.
In crystal ionic radii, chloride is smaller than caesium and fracium while the latter two’s effective ionic radii are larger than chloride’s.
As you can see, there is no hard and fast rule. Fluoride to potassium is a period and two elements difference while chloride to eka-francium is four periods and two elements.

What's the difference between effective ionic radius and crystal ionic radius?

@Kalyan I must refer you to the Wikipedia article. Something about the effective being closer to values Pauling used, I think.


text:
If Substance A is more soluble in water than Substance B, will Substance A be more easily absorbed by the body?
I am comparing the solubilities of artificial vs natural sweeteners for my Chemistry IB Internal Assessment. I must demonstrate personal engagement, so I am relating my topic to weight management in which I am saying that if a sweetener is more soluble in water, then it will be more easily absorbed by the body. Ergo, since the body's need for carbohydrates has been satiated, one will feel less need to eat sugar, thus aiding in weight loss.
I'm not quite sure if the above is true, so I would be very open to any suggestions or corrections. Please indicate any reference material in your answers to my question.
Thank you very much!

Slightly related: How do water insoluble medications get absorbed in the blood stream?

If by 'easily absorbed' you mean quickly then I would guess that it would, since it takes time to dissolve. How big that difference is will clearly depends on the substance.

Of course, if water soluble, it also goes out easily...

Absorption of nutrients on the body does not relate directly to solubility, there are many factors to consider (e.g. fats in milk) and the other ingredients greatly affect this (e.g. iron in meat vs. iron in vegetables). Also, carbohydrates have different effects on appetite depending on the particular molecule; common sugar gives you calories but doesn't really kill your hunger, while fiber isn't absorbed by your body but it certainly satiates it.


text:
Arrow pushing: why does the O become positive when its lone pair becomes a double bond?
In the following image:

The oxygen's top pair of electrons forms a double bond.
But in a covalent bonds, aren't the electrons shared? So won't the oxygen still have these $2$ electrons (that it turned into a double bond) in its outer shell? Leaving its charge as $0$?
So why would its charge change?

I think what is confusing me is what the negative and positive signs mean and how this relates to it's formal charge and its octet.
I know the left structure's oxygen has a formal charge of $0$, and the right hand structure's oxygen has a formal charge of $+1$.
So are plus and minus signs showing formal charge, and not the number of valence electrons?
If so, does the plus sign have nothing to do with the number of valence electrons?

It is formal charge. Formal charges are usually expressed with the charge enclosed by a circle. However, this notation is also used for denoting carbocations. This explains why there's a positive charge enclosed by a circle in the left side of your drawing.

Thank you for the answer @QuantumAMERICCINO . In the carbocation case, does the circled plus sign now refer  to valence electrons and not formal charge?

Yes for carbocations. Electrons are "missing" in carbocations. The carbon in a carbocation does not have a full octect. This is not true for carbanions, however. A carbanion has a full octet. The minus sign in carbanions denote formal charge.

I don't really get what you are trying to say, @Quantum. Those charges are precisely formal charges. In $\ce{CH3+}$ if you cleave all three C-H bonds homolytically (as required for the determination of formal charge) you end up with three electrons on carbon and hence a +1 formal charge.

@orthocresol Interesting. I thought that formal charge entered the picture only after the octets could be completed. What I was trying to say was that the carbocation does not have a complete octet. A carbocation does. Where is the homolytic definition from?

@Quantum https://goldbook.iupac.org/L03513.html "on the formal basis that bonding electrons are shared equally between atoms they join"

Ah, the confusion that layman’s terms can cause. You are assuming sharing to mean ‘both have the same electrons as before, they just share a little bit of them’ — but that’s not really the nature of a chemical bond let alone the idea behind formal charges. Rather, once a molecule forms the electrons are delocalised across the entire molecule and the Lewis structures we draw are a mere approximation that makes things easier for our brains.
When determining things such as formal charges, you should stick exactly to the book and perform it strictly as directed. To determine a formal charge, all bonds must be formally cleaved homolytically, i.e. oxygen only gets two electrons back from the double bond (and another from the other single bond). Then count the resulting electrons and compare to the initial state. Since you will arrive at five electrons for oxygen (which is one less than six), it gets a single, formal positive charge. And indeed physically, there is less negative charge density on oxygen than there was before. (But note that not all formal charges correspond well with actual charge density.)
This is a distinct process from determining oxidation states, whereby you must cleave the bonds heterolytically with the electronegative atom taking it all.
In none of these formal processes you can say ‘but it was originally oxygen’s electrons …’ You cannot make electrons wear pink and yellow hats (not even during Winterbash) and thus you cannot distinguish them.


text:
How does dilution affect result of titration of jam?
We currently do some pH adjustments of jams (target pH of 3.2) using citric acid, however, when there is a large difference between the starting pH and the target pH this can be quite a time consuming process (add to vessel, mix, take sample, repeat).
It's been a long time since I've done chemistry, but I believe that because the pKa of the citric acid is close to the target pH and the jams are a complex mix of buffering agents that it isn't easy to come up with a simple formula for the amount to add (in fact, I observe, even at a similar starting pH, the amount of citric acid required varies a  lot).
My idea was to titrate citric acid into a sample of jam that has been diluted with distilled water for easier mixing:

Jam @ 50% dilution
Citric Acid dissolved at a ratio of 1:9 in distilled water

My thought is that if I keep taking samples and titrating the citric acid with the diluted Jam that I will get a nice correlation between the titration amount and the amount of citric acid required to add to the batch.
My question is:

Will this produce a nice correlation?
Is it possible to do this theoretically?

Thanks,

You can do this at any dilution of both titrate and titrant, as long as you "keep the books" up to date when calculating the results you should be fine. I would probably only dilute the titrate though, unless you have good reasons. (If the dilution of jam is to make it less viscous then that is a good reason for doing it)
Whenever I needed to perform titration something of unknown pH and contents (and I had a large supply of the solutions) I would do a low resolution "sweep" with high concentrates and low tech (old school burettes are usually the fastest, valve fully open and stirring at... vigorous). I'd do two or three of those, just to pinpoint the area of interest. (Sometimes there are several, but then you just do the below for all of them). Assuming one specific point of interest was at 100% addition of titrate, I would add 80% of the titrant to a sample of titrate, then do another titration at 10x dilution of titrant. Three fast samples. Repeat the above, and now at 100x dilution i'd put it in the autotitrator and get some graphs.


text:
What is the noble gas configuration of copper?
Copper has an electron configuration of $\ce{[Ar] 3d^10 4s^1}$. Now sometimes the noble state is written as $\ce{[Ar] 3d^10 4s^1}$ or as $\ce{[Ar] 4s^2 3d^9}$.
Now the first noble state seems to be the same as his normal configuration and the latter seems to have equal electrons but divided in another way.
Now copper has two oxidation numbers $+1$ and $+2$ and as copper is d-element of the $4$ period it follows the $18$-electrons rule. But probably the $18$ electrons aren't created to get the krypton noble state. So how does the noble state configuration of copper work?

"nobel state" is a very unusual notion. Cu looses electrons to be +1 or +2, the electronic state of Krypton would be -7, that makes no sense. Cu +1 has a full 3d and an empty 4a orbital. Check wikipedia von "Hund's rules' and "Aufbau principle".

See "the order of filling of $3\mathrm{d}$ and $4\mathrm{s}$ orbitals" by chemguide for some perspectives.

$\ce{[Ar] 4s^1 3d^10}$  is the correct configuration for copper. Why?
In simplest terms, some elements do not follow the Aufbau principle, so there are some alternate ways that electrons can arrange themselves that give these elements better stability.
Copper is a transition element and transition elements like either half filled orbitals or fully-filled orbitals. This makes them more stable.
For example:
Using the Aufbau principle, you would write the following electron configurations:

$\ce{Cr = [Ar] 4s^2 3d^4}$
$\ce{Cu = [Ar] 4s^2 3d^9}$

However, the actual electron configurations are:

$\ce{Cr = [Ar] 4s^1 3d^5}$
$\ce{Cu = [Ar] 4s^1 3d^10}$

Copper still follows the Aufbau principle (electrons go to orbitals with lowest energies), it is just that the energies change when you are filling the electrons. The assumption that the $4\mathrm{s}$ orbital has a lower energy than the $3\mathrm{d}$ orbital is the Madelung rule instead of the Aufbau principle.


text:
Do unsaturated fatty acids with an odd number of carbons exist?
Our teacher asked us about fatty acid C21:3.
Does this fatty acid even exist?

Biosynthesis adds two carbons at a time, so this less likely than something that's even.

You speak of saturated fatty acids where even number of carbons is more abundant then odd number. But in the case of unsaturated fatty acids, does one exist in nature with an odd number of carbons in its chain?

Interesting question, but the biosynthesis happens via the unsaturated acid anyway. If nature wants the saturated version, it simply performs the appropriate reduction of (what happens to be, at the time) an $\alpha,\beta$-unsaturated carbonyl.

C21:3 is a stable fatty acid that could indicate a variety of double-bond isomers.  It is "unusual" in biology because of the odd number of carbons, but that doesn't preclude its stability and accessibility to organic synthesis.  If someone really wanted to make this acid, they could do so easily.
Such a compound, 5Z,14Z,17Z-heneicosatrienoic acid,  has been reported in the natural environment.  It was found in a deep-sea clam, Calyptogena phaseoliformis, that lives in or near hydrothermal vents and lives off of the organic matter produced by chemosynthetic bacteria that live in close proximity to it.  So in addition to being synthetically accessible, it is also made in living organisms.

Thus, for all reasonable interpretations of "exist" in your question, does this fatty acid even exist?, the answer is yes.


text:
What is the actual voltage to the standard hydrogen electrode and why is it used?
Why is the standard hydrogen electrode chosen to be of 0 voltage, what is its actual voltage and how could that be calculated experimentally?
Also, why is it that we cannot measure the voltage or potentials of the range of electrodes and we must use the standard hydrogen electrode?

The voltage given for any reduction is totally meaningless in the absence of either a comparison to another reduction potential or an oxidation potential. Absolute energies are totally meaningless. Only differences matter.
As usual then, we need references. For example, heat of formation of standard Gibbs energy if formation are set to 0 for elements in their standard forms at standard state.
Likewise, for reduction potentials, we need a reference, and that reference if the standard hydrogen electrode, whose standard reduction potential is 0 by fiat. All other potentials can be determined by comparison.

I see, but why is it that the "the voltage given for any reduction is totally meaningless in the absence of either a comparison to another reduction potential or oxidation potential" and why is it that "only differences matter".

Because absolute potential energies are the only thing that matter. That's how energy works. You can shift the scale however you like, but the differences are still the same and determine everything you want to know.

@Zhe Well, then why don't we connect SHE to ground (zero potential) and measure the "actual" potential of SHE? We can use potentiometer or Wheatstone bridge to measure potential without any current.

@Zhe I'm saying is that - What is the standard potential of the reduction of $H+$?

That's not how reduction potentials work. They have those relative values because the corresponding oxidation potentials are also referenced to the same standard... You cannot just arbitrarily define a new standard like "ground" for these purposes.

Absolute voltages are not meaningless We just are unable to calculate them. They must be real and precise else the definition a of a state function, such as electrode potential, fails because every time we subtract two the precise absolute values subtract out. The hydrogen electrode is hard to use, and I wonder whether it was chosen to be as central as it is or that was serendipitous.


text:
Are molecules on surface of liquid stationary due to surface tension
Are the molecules which are on the surface of a liquid ie the molecules on the top film stationary ie not move about each other under steady conditions ie unlike molecules in the bulk because of surface tension and because surface acts like a stretched film and leaving or coming to the surface for any molecule requires gain or loss of energy ?

While it is true that additional forces (surface tension) are at play at the surface, in general that will not require that the molecules be stationary there. They just see a slightly different potential, and will move slightly differently. For example, consider how a 'stationary' molecule would evaporate?


text:
Reaction rate unit conversion: how to convert from molarity to ppm
How can I convert a reaction rate $k$ with units of $\mathrm{1\over M~~s}$ to $\mathrm{1\over ppm~~s}$?
For context: I want to make a numerical simulation of a system of reactions that happens in air (and in water droplets suspended in the air), I have already built a part of the model for which I used $\mathrm{ppm}$ for my concentrations and thus, e.g., $\mathrm{1\over ppm~~s}$ as the units for the $k$ values of second-order reactions in the system. (Reactions of different orders have different units, obviously, but those details are irrelevant to the question.) The article I have on hand lists a $k$ value for the reaction I need in $\mathrm{1\over M~~s}$ and the reaction takes place in the liquid phase.I assume that the PPM is weight-based although the article in question does not actually specify this.
I am asking the above question because I am not sure if this conversion is necessarily as straightforward as a simple factor of $10^{-6}$.

ppm to moles conversion factor is 1E-6. Similarly, for ppb to moles you need a conversion factor of 1E-9. Since this is just a conversion factor related question and doesn't involve any critical thinking, I think it's OK to help the OP here even if it's a HW problem.

There isn't a direct conversion. "ppm" is typically calculated on a weight basis. The only way to convert from moles would be if the matrix was relatively constant, such as a small amount of product in aqueous solution.

It doesn't really make sense to do this. That unit will for example be terrible for comparison of different rates. Since dependend on the molecular weight different ppm-1s-1 could equal the same effective conversion rate in mole.

I do agree with your concerns about comparing rate in this fashion. The thing is however that, currently I have the model build to work with PPM, as this was the unit I was able to find the kinetics in. However for 1 reaction I intend to add to that, I could only find M-1 s-1. I do intend to convert anyway (making assumptions were needed) and just see if I can get anywhere near a realistic output, at least until better data is available.

The moral of the story is: if you don't want people to think it's homework, don't make it look like homework. When we see a new question, we have no idea who you are, why you are asking, and how many years you haven't had homework for - so the more context you provide, the better the judgments the community can make.

We can do it in two ways:
1) For dilute solutions, ppm and mg/L is equivalent. So, for dilute solutions, the straightforward conversion scheme should look like::
$$ppm=\frac{mol}{L}\times MW\times 1000$$
2) with no assumption::
first, remember that $ppm=y_i \times 10^{-6}$
$$ppm=\frac{mol}{L} \times \frac{x gm}{mol} \times \frac{1 L}{10^{-3} m^3} \times \frac{m^3}{y~gm} \times 10^{-6}$$
here, x is the MW of solute and y is the density of the solvent (for dilute solutions) or solution in $\frac{gm}{m^3}$