text:
Reaction an ROH-resonance-stabilized Group with Hydroxide/Acid-Base Reactions

The book says there is no reaction.
However, I do not understand why.
The ROH group is resonance-stabilized with the benzene-like group. Its conjugate base is stable. There is an inductive effect of the cloud of electrons that make the oxygen more electronegative. However, resonance effects are always more important than inductive effects.
So, why is there no reaction?

If the OH group gets deprotonated it is not resonance stabilized by the naphthalin ring because it is not directly attached to it. What reaction are you expecting, exactly? Just deprotonation or something else?

The conjugated base of the depicted alcohol is not resonance-stabilized. The negative charge cannot be delocalized into the aromatic ring because the alcohol group is not a phenolic OH group, whose oxygen lone pairs are conjugated to the aromatic $\pi$ electrons, like shown below. Instead, it is attached to an $sp^3$-hybridized benzylic carbon which is not part of the conjugated system.

You might have mistaken the depicted compound with 2-naphtol, whose $\ce{pK_{a}}$ (9.51) is much lower than that of aliphatic alcohols and which can therefore be deprotonated by $\ce{NaOH}$. Aliphatic alcohols typically have $\ce{pK_{a}}$ values of about 16-18 (source), so there will be no deprotonation by sodium hydroxide.

These would be the resonant structures of your deportonated compound:

As you can see, there will be no delocalization at al... If you try to move the electrons just a bit, you obtain a very unstable structure, which will not contribute to the resonance hybrid.
However, it is possible to deportonate it with a stronger base, like NaH, or LDA.


text:
A chemical compound with these conditions?
I am searching for a chemical compound which meets these five conditions:

slightly soluble in water
layered strutcture
ionic conductor
thermochromism
lead must be a part of it.

I'm not sure is it lead (II) oxide because it has the last three condition. The first two, I am not sure about it

I will satisfy 4 of the five conditions, but you will have to research the fifth. Lead(II) iodide has a layered structure, is slightly soluble in water, is ionic (but only weakly). I hope that this helps.

Actually, lead(ii) iodide is a bright yellow solid at room temperature, that reversibly becomes brick red by heating. (see: http://en.wikipedia.org/wiki/Lead(II)_iodide) so it would satisfy all 5 conditions!

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
Where is the center of symmetry in the PbO crystal?
I'm very confused as to where the center of symmetry is for the PbO  (litharge form). I know that it is in the P4/nmm space group, and that it's centrosymmetric, but beyond that I'm clueless. Thank you.

The best thing to do is load up the CIF at the Crystallography Open Database (COD), e.g.
http://www.crystallography.net/1010977.html?cif=1010977
JSmol used for the visualization lets you visualize the symmetry and adjust the packing. In this case, look for the inversion center, e.g.:

The inversion center itself is the yellow ball. But it's probably easier to see it in the inverted Pb centers I've circled in yellow. You can see that there's an exact inversion center there.

One of the tricks with inversion centers in the solid state is that they don't need to be in the center of the unit cell. There may be a screw axis or glide plane that creates the inversion.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
How to find the balanced reaction equation for the decomposition of potassium dichromate in non aqueous conditions?
I can't find a method to balance a reaction: the heating of potassium dichromate, which forms potassium dichromate (III), dichromate oxide (III) and oxygen.
I can't add water products as $\ce{H+}$, $\ce{{}^{-}OH}$ because we are not in solution.
So, how can I balance it?
Who oxidises in this redox? The oxygen inside the salt? The oxygen of $\ce{H2O}$ which is in the air?

I think that part of your difficulty lies in incorrect naming of the products which are simply potassium chromate(VI) and chromium(III) oxide together with oxygen. The equation then becomes:
$$\ce {4K2Cr2O7 -> 4K2CrO4 + 2Cr2O3 + 3O2}$$
As for what is oxidized and reduced; the chromium is reduced from chromium(VI) to chromium(III) in going from $\ce{K2Cr2O7}$ to $\ce{Cr2O3}$ and some of the oxygen is oxidized from −2 in the dichromate to 0 in the oxygen gas.

Thank you, you're right there was an error in the products. But is there a method to balance it or you did it by trial and error?

Small nitpick, the oxidation state in potassium chromate is not three, but six: $$\ce{4\overset{+I}{K}_2\overset{+VI}{Cr}_2\overset{-II}{O}_7 -> 4\overset{+I}{K}_2\overset{+VI}{Cr}\overset{-II}{O}_4 + 2\overset{+III}{Cr}_2\overset{-II}{O}_3 + 3\overset{\pm0}{O}_2}$$

The "nitpick" is absolutely correct and I have ammended my answer accordingly. Apologies for not proof-reading my answer carefully enough. The comment helpfully shows how I ballanced the equation. Provided you know the reactants and products (which you did) look for the oxidation numbers which change; in this case some of the chromium: down 3 and some of the oxygen: up 2. To balance the equation these must match, so for every 3 oxygen atoms in O2 there must be two chromium atoms in Cr2O3. Hence 2Cr2O3  + 3O2. You then adjust the other two to make this so. I have run out of characters. Sorry.


text:
What is the balanced ionic redox reaction of nitrite ions with manganate(VII) ions in acidic solution?
What is the balanced ionic redox reaction of nitrite ions with manganate(VII) ions in an acidic solution?
My suggestion is:
$\ce{5NO2-   +   2MnO4-   +   6H+   ->   5NO3-   +  2Mn^{+2}  +  3H2O}$
Is that right?

Thanks for adding in your efforts at solving the problem.  What were your steps to arrive at that answer?  If those answering can see your reasoning, they can steer you in the right direction if you are incorrect.

related: http://chemistry.stackexchange.com/a/15720/4945

Your suggestion is certainly part of the picture, but as is usually the case with nitric acids, a number of side reactions can and often do take place depending on concentration and temperature. In acid solutions, for example, nitrite spontaneously decompose giving nitric oxide together with dinitrogen tetroxide. My feeling is that under optimum conditions your suggested reaction would predominate


text:
Thermodynamics of Solid-Liquid
If you have an ice cube in water and the ice cube does not melt after reaching equilibrium with the water, does the water have temperature 0 degrees Celsius?

Yes, when ice is present at equilibrium with liquid water (at 1 atm), the system is at 0 °C.
If you continue to remove heat from the water (such as in your freezer), the temperature of the system will remain at 0 °C until all the liquid (water) becomes solid (ice) after which, continuing to remove heat will result in a drop of temperature below 0 °C. Conversely, if heat is added to the system, the temperature of the system will remain at 0 °C until all of the solid (ice) has melted into liquid, after which, the temperature will increase above 0 °C. The amount of (latent) heat required to melt ice is called its enthalpy of fusion (334 J/g). This energy is converted into intermolecular bonds (a change in internal or potential energy of the system) rather than a change in kinetic energy (that is, a temperature change).
Note that at a much lower pressure (0.006 atm) and slightly higher temperature (0.01 °C), pure water will be in three states or phases (gas, liquid and ice). This is known as the triple point of water, as illustrated in the phase diagram of water, shown below:


text:
Why are there different chemical structures for epinephrine (adrenaline)?
The Wikipedia structure is:

Note the empty bond at far-right.
"Journals on the Web" has:

CH3 added at far-right.
And rxlist.com shows:

There's an extra H going into the page! Is one of these incorrect, or are they all consistent but w/ some implied bonds or something?

There are often shortcuts taken in drawing complicated molecules, particularly if there's not any interesting chemistry going on at those functional groups.
The $\ce{-CH3}$ is implied at the end of the "stick," just like how the carbon atoms aren't explicitly labeled in the benzene ring.
Hydrogens are rarely explicitly drawn for the same reasons, given that you can infer the hybridization by inspection.
See Skeletal formula for more details.  Briefly:

Carbon atoms are usually depicted as line ends or vertices with the assumption that all carbons have a valence of 4 and carbon-hydrogen bonds, usually not shown explicitly, are assumed to complete each C valence. A skeletal formula shows the skeletal structure or skeleton of a molecule, which is composed of the skeletal atoms that make up the molecule.

They are all consistent. With line diagrams, all bonds to "nothing" are actually implicit bonds to carbon (this is why the benzene ring and carbon tail can be drawn without $\ce{C}$ in each of your diagrams). In addition, hydrogens are added implicitly until each carbon is saturated. So they are all the same structure.

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


text:
How is the crystal direction important for etching silicon surfaces?
I am a student of masters study with a background only in electrical engineering, but I have studied Miller indices and crystallographic planes and directions, personally, to help myself understand the process of silicon etching which is required in my current course of study. But, I am unable to understand the following written in my text book:

"When Silicon is etched anisotropically, i.e. depending on crystal direction, and a {1 0 0} surface direction is used, cavities will result which have the same {1 0 0} orientation at the bottom and are determined laterally by {1 1 1} surfaces. The reason is that {1 1 1} surfaces are etched off using a minimal etch rate and this way becomes the most resistant"

Could someone please help me understand this concept.

Hello and Welcome to Chemistry.se! While your question is straightforward, I think it would be much appreciated if you [edit] in the source for your quote.

Certain types of wet etching agents have differential reactivity on different crystal surfaces. Consider that the Si-Si distance and coordination environment will be quite different.
Let's start with an isotropic etch - it just eats silicon. Well, if you started with a pinprick (i.e., just a point and not a gap in the mask) you'd etch out a hemisphere. Normally, we etch through gaps in the mask, so you'll get rounded sides.
With an anisotropic wet etch you can get flat, angular side-walls defined by crystal planes.

For instance, potassium hydroxide (KOH) displays an etch rate selectivity 400 times higher in <100> crystal directions than in <111> directions.

OK, so let's consider that etch through a gap in a mask into a <100> surface. As we go down 400 atomic layers of Si, we'll clearly have some etching along the side-walls (e.g., an "undercut"). But also consider that depending on the size of the feature for the etch, you'll go down into the <100> but not completely uniformly. Different wet anisotropic etches will have different selectivity rates between <100> and <111> and thus you get different aspect ratios.
Incidentally, dry etching (e.g., RIE, plasma) can give more vertical side-walls.


text:
Are there carbon-silicon alloys?
Both carbon and silicon expose a lot of metallic properties like metallic lustre, electrical and heat conductivity, etc, so often considered metalloids. The both form alloys with other metals such as iron.
On the other hand, the usual silicon carbide is not an alloy at all. It is tranparent, electrical insulator etc.
I wonder whether there is a different carbon-silicon compound that has more metallic properties?
Carbon (graphite):

Silicon:

Silicon carbide:

Here you can find a phase diagram for $\ce{Si/C}$ system. It does not have zones with homogeneous non-stoichiometric solids. So, there is not thermodynamically stable Si/C isomorphic alloys. However, since the liquid likely to be homogeneous, it is likely for fast cooled liquid to form amorphous alloys. Indeed, google search provides plenty of links for amorphous $\ce{Si/C}$ alloys. However, it is also possible for cooling liquid to form polycrystalline alloy, say, with $\ce{SiC}$ and $\ce{C}$ grains, similar to pig iron.

I already saw the diagram, thanks. So do they mix with each other in and only in liquid state?

Note that the diagram covers only one pressure.

@Anixx condensed phase is rarely significantly affected by pressure change (water is a pathological case). And yes, judging from the diagram it is that way.

@Anixx At least as far as pressure does not skyrocket. When it is skyrocketing, weirdest things may happen.

One might consider reaction-bonded silicon carbide(see also Ref. 1), which is essentially a composite of silicon carbide and elemental silicon, sometimes also containing graphite for sealing applications.  To make reaction-bonded silicon carbide, a porous part containing silicon carbide and carbon is permeated with molten silicon.  The reaction between the silicon and the carbon binds the material, as does leftover molten silicon which solidifies as it cools.  The resulting material is lightweight and has a favorable properties for a variety of high-temperature and mechanical applications, including some cases (such as burner tubes in furnaces) where it competes with metallic alloys.
Reference

Sebastian Meyers, Lien De Leersnijder, Jef Vleugels, Jean-Pierre Kruth (2018).
"Direct laser sintering of reaction bonded silicon carbide with low residual silicon content:.
Journal of the European Ceramic Society,
38(11),
2018,
Pages 3709-3717,
ISSN 0955-2219,
https://doi.org/10.1016/j.jeurceramsoc.2018.04.055.

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


text:
Resonance and reductions with sodium borohydride

I know that sodium borohydride doesn't work particularly well on resonance-stabilized carbonyl groups (i.e. esters) but what about this carbonyl in the center of the above picture? This appears to be somewhat resonance-stabilized ... why would sodium borohydride still work  here? Sodium borohydride is a nucleophilic reagent. On the other hand the second resonance contributor I've drawn is pretty crappy; yes - it's part of an aromatic system - but it's also a carbocation. And the R' group is inductively donating (although I suspect this is a fairly minor point).
Or does it just work in the above synthesis but not with a yield comparable to something such as sodium borohydride and formaldehyde?

Where did you find this synthesis? Could you give the source? Are they giving any more details? Maybe using the softer kind of reducing agent is due to selectivity reasons...

@Martin This is the second to final step in the synthesis of Allegra (fexofenadine).

@Martin http://www.chemdrug.com/databases/SYNTHESIS/SYN/23/23156601a.gif

Sodium borohydride is definitely strong enough to reduce an acetophenone derivative. It is likely that the resonance stabilization does reduce the reactivity of the ketone. It's not often that a sodium borohydride reduction requires 35 hours. In the cases that I have run, usually less than one hour is sufficient.
I did some quick searching of whether an isolated ketone could be reduced in the presence of an acetophenone, and I did not find an adequate example of a case where only the isolated ketone was reduced. That says to me that the reactivity difference is not that great. Perhaps someone else has come across this though.


text:
How to efficiently reduce of ilmenite to pure titanium?
I am looking for a process in which reduces ilmenite to pure titanium with the highest yield possible. We are to use a process of our choice or one that we have created in conjunction with FactSage to calculate the yield of titanium.
the materials we have available are:

Pure Metals

$\ce{Al, Ca, Mg, Na, Si}$

Pure Non-metals

$\ce{Ar, C, Cl2, H2, N2, O2, S }$

Molecules

$\ce{CH4, CO, CO2, HCl, H2O, H2S, NH4}$

There are a few process that come to mind such as reduction using Magnesium, $\ce{H2SO4}$, high temperature $\ce{CO2}$, but any help on which would be the most efficient combined with the highest yield would be great.

@Micheal it would be better, if you would use the mhchem package to format chemistry, rather than HTML.

Thermite Reduction is easiest.

Ilmenite mainly consists of $\ce{TiO2}$ and iron oxides. The idealized formula of ilmenite is approximately $\ce{FeTiO3}$.
Note that the aluminothermic process would yield ferrotitanium alloys, and the reduction with carbon would yield titanium carbide.
The titanium ore can be converted to titanium tetrachloride according to
$$\ce{TiO2 + 2Cl2 + 2C -> TiCl4 + 2CO}$$
or
$$\ce{2FeTiO3 + 7Cl2 + 6C -> 2TiCl4 + 2FeCl3 + 6CO}$$
(If a large content of iron oxides causes problem, it can be partially removed first by partial reduction with carbon.)
Titanium tetrachloride (boiling point: 136.5 °C) can be purified by fractional distillation.
Finally, titanium tetrachloride can be reduced to titanium sponge with magnesium (Kroll process):
$$\ce{TiCl4 + 2Mg -> Ti + 2MgCl2}$$
The remaining $\ce{Mg}$ and $\ce{MgCl2}$ can be removed with hydrochloric acid.


text:
Equivalencies and catalytic hydrogenation
Can we actually control catalytic hydrogenation by introducing specific molar quantities of hydrogen gas? My prof insists on no. No we cannot use a certain equivalency of hydrogen gas to stop at a desired product.
Yet in the book I keep seeing examples of specific molar quantities of hydrogen gas being used ...

Using the equipment that I have and working on the scale that I tend to work on (<1 mol typically), I can't deliver only one equivalent of hydrogen gas. We can measure the amount of hydrogen consumed as the reaction proceeds. The problem you have shown could be interpreted that way, i.e. 1 mol is consumed but more may be present.
On industrial scale it may be possible to deliver a more accurate amount of hydrogen. I'll leave it to others that might have more experience on large scale to confirm or deny.
There are alternative sources of hydrogen that can more easily be delivered accurately on small scale. Either ammonium formate or cyclohexadiene can be used to generate hydrogen in the presence of palladium. The former liberates ammonia and carbon dioxide as byproducts, while the latter forms benzene as byproduct.

Why can't you deliver only one equivalent of H2 gas? Can't you just put in 1 mole of H2 gas into the reaction chamber?

@Dissenter Delivering a gas to a reaction is not as simple as a solid or liquid. You can just weigh out a solid. You can measure the volume of a liquid or weigh it out. Usually solids and liquids are dissolved in the solvent, and we know that the measured quantities are available to react.

@Dissenter The gas must be delivered out of a gas cylinder. For simple hydrogenation reactions, we can just fill a balloon and attach it to the reaction vessel. Some of the hydrogen will dissolve and be available to react, but the rest will remain in the atmosphere. As hydrogen is consumed, more will dissolve, but eventually we'll get to a point where the pressure is low enough there won't be an appreciable amount in solution. I suppose we could account for that, but I've never seen anyone do that in practice, again with the caveat that I don't work on production scale.


text:
Enones and hydrogenation
Are enones resistant to hydrogenation? My solutions manual thinks so.

I don't recall discussion regarding enones in my text. Nonetheless, could the resonance stabilization of the enone be preventing its hydrogenation? In that case, why is the alkene hydrogenated; both the carbonyl and the alkene are resonance-stabilized to some extent!

I read your question as asking why the C-C pi bond is reduced while the C-O pi bond is not. It appears to be possible to change the selectivity of hydrogenation based on the catalyst used (reference), but for simple palladium on carbon, reduction of alkenes (even conjugated to carbonyls) is much faster than an aldehyde or ketone. Coupled with complementary hydride reagents, which are selective for carbonyls, synthetic chemists can achieve the desired reduction without worrying about reaction of other reducible groups.


text:
Calculating energy level from wavelength

The transition of an electron from a higher level to a lower level results in the emission of a photon of wavelength $350.0\ \mathrm{nm}$. If the energy of the higher level is $-3.24\times10^{-19}\ \mathrm J$, calculate the energy of the lower level.

By using $E = h\nu$
I find $E = 5.679\times10^{-28}\ \mathrm J$
How do I use this value to solve the qus?

All you need are two equations and two constants (speed of light and Planck constant):
\begin{eqnarray}
\mathrm{E} & = & h\nu\\
h & = & \mathrm{6.62606957 \cdot 10^{−34}   J\cdot s}\\
\mathrm{c} & = & \lambda\nu\\
\mathrm{c} & = & 299792458\ \mathrm{m\cdot s}^{-1}\\
\end{eqnarray}
Using these data, the energy of a photon with $\lambda = 350\ \mathrm{nm}$ calculates to:
$$\mathrm{E} = \frac{6.62606957 \cdot 10^{-34} \cdot 2.99792458 \cdot 10^8}{350 \cdot 10^{-9}} \cdot \frac{\mathrm{J \cdot s \cdot m}}{\mathrm{s \cdot m}} = 0.05676 \cdot 10^{-17} \mathrm{J} = 5.676 \cdot 10^{-19} \mathrm{J}$$
This is the energy difference to the higher level.

Thanks for your reply @Klaus Warzecha. Am I right to say that after substituting wavelength = 350nm into the equations, I'm already finding the energy of the lower level?

@Kenny Almost, you're calculating the energy of a photon with $\lambda$ = 350 nm. This is the difference to the other level for which the energy was given.

Thank you for your reply. I got totally messed up with my units.

@Kenny My pleasure :) Units are a pretty helpful thing - unless one confuses imperial and metric measures and therefore looses a Mars Climate orbiter. Fortunately, NASA openly stands to this historical error and even cites it educational materials. Good on them!


text:
Do all salts taste salty?
Recently, I am learning the production of soluble and insoluble salts. My friend and I have done this experiment at the school lab.
We wanted to taste them to see whether they are salty are not. The teacher luckily stopped us from doing that.
So without tasting them, I would really like to know whether all salts are salty.

A very interesting case is lead(II) acetate, as well as some other lead salts.

Old mantra from the folks trying to make Salt Substitutes: "There seems to be only two chemicals that taste salty to human beings, and one of them is poisonous."  I believe that the poisonous one is Potassium Chloride.  ... Hmm, according to Wikipedia, KCl is bitter and mildly salty, and only poisonous at high doses...

Another salt which can be bought in some specialist bakery stores is Hartshorn, also known as ammonium bicarbonate. I've tasted ammonium carbonate before, and it has a sort of bitter, watery/salty taste (I know that doesn't make much sense, but I'm not sure how else to describe it).

@RBarryYoung It is not KCl, it is LiCl. KCl is has no salt taste and you can find it in table salts substitutes. Please don't call it poisonous.

@Greg Ah yes, thanks.  Lithium Chloride does make more sense.  KCl's status was already clarified in the same comment above.

In the olden days, chemicals were taste-tested when they were newly discovered (or when they were resynthesised to make sure they matched what was reported before). So if you look at papers from before 1900 (and probably later ones, too), you will often find descriptions of how a sample tastes. For example, while not technically a salt boron phosphate $\ce{BPO4}$ is odour- and tasteless.

I can't help but link this YouTube video.

No. There are sweet, bitter, and various other salts. (Likely, there are tasteless salts too). Pure salty taste is as far as I know exclusive for table salt, though I wouldn't bet on it.
Lead and Beryllium salts are said to be sweet, though toxic. Epsom salt, $\ce{MgSO4}$, is bitter. $\ce{CuSO4}$ has an incomprehensible, persistent metallic taste. (Based on personal experience. Copper salts are slightly toxic, but not extremely, so I survived with no consequences.)
Salts with hydrolysing cation (various alums) are acidic in addition to other notes.

It seems you are dared to try them although they are toxic.

@Simon-Nail-It This was a good lesson why it is necessary to wash you hands after work in a lab and before you eat.

Actually, the chloride salts do lean to a salty taste, especially when a "lightweight" cation (Periods 3-4, including Na) is used. Calcium chloride is an additive to pickle brine, used to give a salty taste without increasing sodium content. Potassium chloride is a similar "salt substitute" in small doses, as is magnesium chloride (which can be a danger to wildlife when using it as a road de-icing agent, specifically because it tastes salty but becomes toxic more quickly than table salt). The "flyweight" alkalis, lithium and beryllium, are quite toxic in any quantity you could taste.

Soap is a salt.

Salty liquorice (salmiak) is pretty salty, but it contains  ammonium chloride, not table salt.

Saltiness is perceived when alkali metal enter taste buds.
From wikipedia:

Saltiness is a taste produced primarily by the presence of sodium
  ions. Other ions of the alkali metals group also taste salty, but the
  further from sodium the less salty the sensation is. The size of
  lithium and potassium ions most closely resemble those of sodium and
  thus the saltiness is most similar. In contrast rubidium and cesium
  ions are far larger so their salty taste differs accordingly.[citation
  needed] The saltiness of substances is rated relative to sodium
  chloride (NaCl), which has an index of 1. Potassium, as potassium
  chloride - KCl, is the principal ingredient in salt substitutes, and
  has a saltiness index of 0.6. Other monovalent cations, e.g. ammonium,
  NH4+, and divalent cations of the alkali earth metal group of the
  periodic table, e.g. calcium, Ca2+, ions generally elicit a bitter
  rather than a salty taste even though they, too, can pass directly
  through ion channels in the tongue, generating an action potential.

Ammonium chloride is used as a food flavoring in salty liquorice, which is quite popular here in northern Europe. The pure stuff has a very intense and distinctive, vaguely bitter flavor, that's hard to describe to someone who hasn't tasted it; it's a bit like trying to describe the taste of table salt as something other than "salty". Products flavored with it do, of course, usually also have sugar, salt and other flavorings, often including liquorice root, that blend with the taste.

When you can't be bothered to write your own answer and instead just copy/paste the relevant section of wikipedia, at least put in the minimal effort needed to remove the [citation needed], or include the links, or at the very least hit enter once to preserve the paragraph breaks and make it a little easier to read. That will make it easier for you to argue that it "takes time to research an article".

Well, believe it you must check this article @terdon I have edited it little bit according to what I think was needed. Obviously the [citiation needed] I once thought to remove but I didn't want to convey any wrong knowledge to my fellows who are reading it and letting them know that you don't have a good source for it. A wikipedia article would definitely have some links like : [1]  linking to their reference which were indeed deleted.

permeakra is quite right with his counterexamples of salts that don't taste purely salty, but I'd like to expand on why. We haven't fully identified and elucidated the receptors involved in taste, but they can be broadly classed into tastes people are familiar with: salty, sweet, sour, bitter, and umami.
Salty and sour receptors are both known to be mostly dominated by ion channels, where salty receptors allow small cations to pass and sour receptors are sensitive to pH in a similar fashion. The other receptors are g-protein coupled receptors. The trick with taste (and smell, for that matter) is that there is not a single receptor type for each possible chemical one can taste, i.e. there's not a receptor entirely selective towards $\ce{Na+}$. The way it works is that there are many types of taste receptors that are somewhat selective to specific classes of molecules and a given molecule may induce a response in many different receptors. The brain takes all the data from all these receptors; combines them; integrates information from the nose, eyes, ears, tactile nerves in the mouth, etc.; and then produces the sensation of taste.
Thus, though both $\ce{Na+}$ and $\ce{K+}$ will stimulate some of the same saltiness receptors, they also stimulate other receptors and in different amounts, making them taste different. In general, small cations can stimulate the ion channel receptors for saltiness, so many salts do taste salty to a certain extent, but many also stimulate other receptors. Because what we perceive as taste comes from the interaction of all these different receptors (as well as other senses), one salt may taste very salty and another quite bitter, despite both activating saltiness receptors; the responses are not independent (which is why it's possible to mask overly salty soup or whatever by adding sugar).

The short answer is no, as suggested in the various comments. Not being a biologist I cannot give any underlying theory to explain this, but from personal and reported tasting I can give some examples: Copper sulphate, iron(II) sulphate, aluminium sulphate among others are very bitter. One you could try yourself would be sodium carbonate (washing soda) or sodium bicarbonate (baking soda). Interestingly Lead acetate as mentioned in one of the comments has a sweet taste which makes it particularly dangerous due to its toxicity. Even potassium chloride (used as a substitute for salt to reduce sodium intake) tastes quite different.

Sodium bicarbonate does taste salty, due to the presence of sodium ions. Please don't try tasting sodium carbonate, at least not as anything but a very dilute solution -- at high concentrations, it's strongly caustic.

In response to the above comment, I would first like to point out that tasting is very different from eating or drinking. I have personally tasted sodium carbonate and while it is certainly alkaline I suffered no ill effects apart from the very unpleasant soapy taste. As to sodium bicarbonate this is quite harmless  and I would suggest that its taste bears little similarity to salt.

I actually went and tasted a pinch of baking soda before posting that comment, and it sure does taste salty to me. It takes a second or two to come out, but that sodium flavor is definitely there. (As for sodium carbonate, you may be right. All I know is that I for sure wouldn't want to put anything that alkaline in my mouth, not even briefly.)

As mentioned, while not all salts taste salty, many do, and not just salts with sodium ions. If you want to try another edible salt, try to get ahold of ammonium chloride, which is used (both as a powder and otherwise) in sweets in the Nordic countries, the Netherlands and Germany. Its taste can definitely be described as salty, but not the same kind of salty as sodium chloride.

In my opinion, the taste of ammonium chloride is due to a combination of saltiness plus an element of umami.

In this context, maybe "sal ammoniac" would be a more proper name.

Ionic forms of some organic compounds that you will perfectly know, which can also be considered salts, have sweet taste, like sodium saccharin or sodium cyclamate, which are used as "artificial" sweeteners. The latter is considered to be about 50 times sweeter than sugar itself.

I wrote an answer to this question eight years ago concerning the taste of ammonium chloride, but as I've tasted quite a few different salts since then I'd like to mention some prominent examples, demonstrating that salts can have quite a wide variety of taste profiles. All the salts were purchased in bulk from reliable suppliers to the culinary industry, and I've tried all both as powders and dissolved in water. In the following, "salty taste" is short for "tastes like sodium chloride".

Ammonium chloride (NH4Cl), which I discussed in my other comment, tastes quite salty indeed. Of the salts I've tasted, it's the only one that tastes roughly as intense as sodium chloride at roughly the same mass – most edible salts seem to taste less intense. Erik Kowal suggested that its taste is a combination of saltiness (like sodium chloride) and an element of umami, and it's possible that there's an element of umami there (although its taste doesn't have much in common with pure MSG). Personally, I'd rather describe its taste as a mixture of salty and sour and an indescribable airway-clearing pungency that's hard to describe. It's certainly very salty, though: if you tasted the typical Scandinavian mix consisting of ammonium chloride, sugar and liquorice powder you'd swear there was sodium chloride in there, but there (typically) isn't.

Potassium chloride (KCl) is used as a replacement for (sodium chloride) table salt, and is the most common potassium salt used as a dietary supplement for people with low potassium levels and to prevent potassium deficiency in people taking certain kinds of medications that reduce potassium levels. My main impression of potassium chloride is that it has a much less intense taste than sodium chloride (both as a powder and dissolved in water). It has a mild salty taste, and even milder bitter undertones. When equal parts (by mass) of NaCl and KCl are dissolved in water, the taste of the NaCl almost conceals the presence of KCl, but the combination is less salty tasting overall.

Magnesium sulphate (MgSO4; also known as Epsom salt, used most prominently as bath salts but also sometimes in food and brewing and occasionally as a dietary magnesium supplement) combines a medium intensity salty taste with an intense bitter flavour. The bitter taste lingers in the mouth for several seconds. Not very pleasant, although not nauseatingly vile (depending on your tolerance to vile flavours) and tolerable at low concentrations.

Magnesium citrate (C6H6MgO7; medically used as a laxative; commonly used as a dietary supplement) only has a mild, "round", slightly soapy taste at first, but develops an aftertaste that I'd describe as that of stale or even rotten milk. Quite vile indeed, and the aftertaste lingers for a long time and is hard to conceal. No salty notes to its taste at all. Trimagnesium dicitrate is also relatively common, but I haven't tried it.

Trisodium citrate (Na3C6H5O7; usually just sodium citrate) is commonly used in food as a flavorant or preservative. A decade ago, trisodium citrate was popularized by a recipe (that came out of the "molecular gastronomy" movement) for a cheese sauce consisting only of water, cheese and trisodium citrate, where the trisodium citrate acts as a stabilizer to prevent the fats in the cheese from separating. Trisodium citrate has an inoffensive taste (at equal mass) consisting of a mild salty flavour, a slightly sour flavour and maybe a tiny bit of soapiness. The taste is easily concealed, and when trisodium citrate is present in food it's hard to notice even if added in greater amounts than table salt usually is.

Sodium bicarbonate (NaHCO3; baking soda; also commonly used as an antacid) has a taste that I would describe as sour, soapy and slightly bitter – not a very strong tasting salt but absolutely noticeable in sufficient amounts, as anyone who has added too much baking soda while cooking or baking can attest to. I described it as "sour", which I stand by, but sodium bicarbonate is obviously quite basic and not acidic! This is an example of the fact that there is no strict relationship between general chemical properties like acidity and how something tastes (in this case, "sour"). Certain online comments would have it that acidic compounds always taste sour and basic compounds always taste bitter, but many of the most bitter compounds are insignificantly basic and sodium bicarbonate certainly doesn't just taste bitter.


text:
Why is mercury much more toxic when inhaled than injected?
Following wikipedia:

Cases of systemic toxicity from accidental swallowing are rare, and attempted suicide via intravenous injection does not appear to result in systemic toxicity,[1] though it is still causes damage by physicaly blocking blood vessels both at the site of injection and the lungs.

Shouldn't both inhalation and IV injection of elemental mercury produce the same result, namely, mercury in the bloodstream?

I guess the answer is that mercury atoms stick with each other very well (e.g. high surface tension in comparison to most other liquids $\gamma_{293K}^{\ce{Hg}} = 0.47 \text{ N/m}$, where  $\gamma_{293K}^{\ce{H2O}} =  0.073 \text{ N/m}$ [1]) and that this continues to happen in the blood stream. Such clumps of mercury are not able to pass the blood brain barrier (BBB) effectively, especially in comparison to finely divided dissolved vapor, which is transported trough the BBB by the formation of a complex with cysteine.

The simple answer is that elemental, metallic mercury isn't very soluble in either the blood or in the contents of the digestive system. If ingested, metallic mercury will simple pass through the gut with little effect on the body and with very little being absorbed into the places where it can do harm. The same is true in the blood.
Mercury causes harm when it interacts with essential enzymes in the body and makes them malfunction. For it to do this, mercury atoms or ions have to be able to reach a significant concentration in bodily fluids. There are two ways for this to happen easily. Mercury vapour is absorbed into the bodily system (but this requires some effort such as combining bad ventilation with finely divided mercury droplets and heat or bad ventilation and a long time period as early mercury chemists discovered). Many mercuric compounds, especially organic ones, are also readily absorbed (either because they are soluble, as many salts are, or because they are volatile, as many organic mercury compounds like methyl mercury are.)
In fact the worst case of human mercury poisoning (the Minamata disaster) resulted from the release of methyl mercury into the environment where it accumulated in fish ultimately eaten by people.


text:
Is there a simple way to separate deuterium oxide from tap water?
I understand that tap water contains minute amounts of deuterium oxide (heavy water). Is there a simple way to separate, concentrate the $\ce{D2O}$?

No, there is no simple way.  Wikipedia or a Google search (heavy water separation process) will lead you to some references.  The main point is that the natural abundance of D is only ~110 parts per million, so you need very large scale industrial processes to separate the D from H.  Also, since the abundance of D is so low, there is almost no $D_{2}O$ in nature - it will almost all be $DHO$.

Agree with Custer.  You need a super-elaborate distillation process, since HOD apparently boils at 101.4.

The biggest production plant used the Girdler Sulfide process, which exploited equilibria between deuterated hydrogen sulfide and deuterated water.  That's probably the way to go (but not in your house.  Hydrogen sulfide is nasty AND toxic).

It depends how pure you want the $\ce{D2O}$ to be and what you consider simple :)
Electrolysis of water strongly favors "H" being converted to hydrogen gas rather than "D", by a factor of about 8 to 1 (depending upon the electrodes).
Inexpensive Equipment for the Preparation and Concentration of Pure $\ce{D2O}$ Ohio Journal of Science volume 41, number 5, pages 357-365 (1941) explains in great detail how to carry out the electrolysis method.  The article says the process is suitable for "those who have occasion to prepare or concentrate heavy water for research, demonstration or
advanced student laboratory use".
In a first stage, $\ce{NaOH}$ solution (initially $\pu{0.5 M}$) is subject to electrolysis, until only 1/32 of the original volume remains.  This increased the original concentration of deuterium by a factor of 12.
By then the $\ce{NaOH}$ is too concentrated, and $\ce{CO2}$ is added (as dry ice) to neutralize the solution, and the first stage solution is distilled to get rid of the sodium carbonate. About $5~\%$ of the $\ce{NaOH(D)}$ concentrate would be saved and added to the distillate to provide electrolyte for the next stage.
Then a second stage of electrolysis is performed. By the end of the second stage they reached about $10~\%$ deuterium. The process can be repeated to achieve a desired deuterium level.  They used three stages.  Yield can be improved by capturing the deutrium/hydrogen gas of the late stages, burning it to make water, and feeding this back to an earlier stage.

Basically you freeze water partially. DDW has a higher freezing temperature. Stick a bowl of water in freezer. When the top freezes over wait another 15 minutes. Take out of freezer and poke a hole in ice. The water that comes out is around 50 ppm. If you do this a few times you can get it down even further BUT: this method does not remove all impurities from ddw. If you start with filtered water first (like a brita fliter on your faucet) you'd be much better off.

What’s DDW?$%..$

Haha, I'm sure this would work in principle, but you need to optimise the process quite a bit before you get, say, 50% D2O that way in a reasonable number of steps. And btw. you remove NO impurities this way. They always accumulate in the liquid phase.

Because, Deuterium is twice as Heavy as Protium (the more abundant form of Hydrogen) it gives the molecule of water a greater moment of inertia and slows down its average rate of rotation.   Consequently it does not bond to an growing ice crystal quite as rapidly as the lighter molecules of water.  So if you bring water to a boil (to break up polymerization and increase the number of single water molecules) and then cool it rapidly to the point of fusion so there is less time for them to participate in polymer groups the ice will have a greater percentage of light molecules than normal water and the water that is not frozen will have a slightly higher percentage of deuterium.  Instead of 110 ppm you might get it to 112 parts per million.  So the water that didn't freeze is brought to a boil again and then rapidly cooled to the point of fusion removing even more of the lighter molecules.  A very slow process with the only difficulty being the rapid cooling needed to avoid excessive polymerization. It requires far simpler apparatus than distillation or electrolysis as given in the first answer, but is much less efficient and the boiling phase would eventually use more energy.  However, the question gave simmplicity as a criteria and this boil and freeze process is simple.

Obscure description with rather misused technical terms.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the [help].

I think you mean freezing point not point of fusion. Point of fusion is something quite different.

@bon how is it different? https://www.merriam-webster.com/dictionary/fusion%20point


text:
Creation of ester from acetylene derivative

I took a much different approach from the solutions manual. My synthesis:
1) H2 and Lindlar's catalyst to stop at the alkene.
2) Oxidative cleavage to form benzoic acid. Works here because we have a benzylic hydrogen.
3) Addition of excess allyl alcohol with catalytic concentration of acid and heat. A Fischer esterification step.
This is also one fewer step than what my solution's manual has. The solutions manual went up to the acid chloride and down to the ester, which is a nice step (avoids the heat). Does mine also work on paper?

No, your approach doesn't give the right product. There's a carbon missing between the benzene ring and the carbonyl carbon. See the difference in the products below.
I don't know exactly what your solutions manual suggested, but I've included a conservative approach through phenylacetaldehyde. As you suggest, it's not absolutely necessary to go through an acid chloride to make an ester from a carboxylic acid. The Fischer esterification will work. Also, for many carboxylic acid to ester conversions, you can simply react the carboxylic acid with thionyl chloride and the desired alcohol all in the reaction together, rendering it a one-step transformation.

bake starting material in Wilgerodt-Kindler reaction (it works on phenylacetylene too)
hydrolyze to phenylacetic acid
esterify

Could you possibly add some more detail here? At the moment this is just a statement with no real explanation of the reasons behind it and as such it will probably be deleted.


text:
What conditions would make SN1 products the major products over E1 products, and vice versa?
It seems that if a carbocation intermediate is formed through solvolysis, then the product(s) are going to always be a pretty even mix of SN1 and E1 products.
Are there any conditions that greatly favor one mechanism? I thought that a very bulky solvent - like t-$\ce{BuOH}$ - would favor E1. However, the carbocation is $\ce{sp^2}$ hybridized, so there still wouldn't be a lot of steric hindrance to prevent SN1, right?

In a practical way, you would have to consider each particular case, there are tons of reasons that could make a reaction to yield mostly E1 or SN1 product (it will be always a mixture, that can vary from 50:50 to 0.01:99,9, I hope I'm making myself clear, they are always competing reactions).
But there are several factors that contribute to make one of the two more favorable than the other, for example:

The kind of the reactant employed:
If you use hydrogen halides, ($\ce{HCl, HBr, HI}$) you will mostly obtain SN1 products.

If you use acids like $\ce{H2SO4, H3PO4~or~TsOH}$ you will mostly obtain E1.
Why is this? The reagents/acids of the first case, are very good nucleophiles. The reagents lisetd on the second case, are awful nucleophiles.
A good nucleophile is key for a substitution. But for elimination you only need something that can abstract a product in the alpha position to the leaving group.
On the other hand, very strong and small bases, can favour an elimination through E2 mechanism, with almost no substitution taking place. E1 mechanisms is usually observed only under mild basic conditions.
Solvents, like you say, also can favour one type or another:

In some cases, solvent acts a base for E1 reactions. (i.e. $\ce{EtOH -> EtOH2+}$). Bulkiness of the solvent can cause both E1 and SN1 (in case that the solvent itself is the nucleophile, which usually happens) to be slower: it has to approach the compound that undergoes the reaction anyway. Actually is easier to approach a hydrogen than a carbon, so probably will favour SN1 over E1, BUT, do NOT take this as a determining factor at all. The alpha carbon from which you need to abstract the proton might be more or less hindered.

Solvents with low nucleophility also decreases the % of substitution reaction.
Both E1 and SN1 are favoured in protic solvents, in general.
To sum up, you will always have both products, but yes, there are factors that can be modified in order to maximize one of the two reactions. There are some general considerations, like the ones I posted, but in practice, each reaction is a world.

Overall, nice answer. But your statements If you use hydrogen halides, (HCl, HBr, HI) you will mostly obtain Sn1 products. and A good nucleophile is key for a substitution. need some comment: Hydrogen halides, at least HBr and HI (HCl not so much) are good nucleophiles and good nucleophiles abet only $\mathrm{S}{\mathrm{N}}2$ reactions. The rate of an $\mathrm{S}{\mathrm{N}}1$ reaction remains mostly unaffected by choosing a good over a bad nucleophile because the intermediate carbocation is so reactive that it reacts even with very weak nucleophiles.


text:
faraday laws of electrolysis
I came across this question:
A quantity of electrical charge that brings about the deposition of 4.5 g $\ce{Al}$ from $\ce{Al^{+3}}$ at cathode will also produce the following volume (STP) of $\ce{H2}$ (g) from $\ce{H+}$ at cathode:
A)22.4
B)11.2
C)5.6
D)2.8
If we use faradays first law:
For $\ce{Al^{+3}}$
4.5-=(27*c)/ (96500*3)
c=96500/2
For $\ce{H2}$:
m=c/(96500*2)
m=0.25
Hence V=2.8
If we use 2nd law, we get mass of $\ce{H2}$ produced as 0.5
Hence, V=5.6
So then what was the mistake
(Please reply if I have made calculation mistake, I generally do that)

Your data are rather disagreeable to read, simply because you never mention the units. Is the volume in liters ? in cubic meters ?

I am not sure what your answer was (2.8 or 5.6dm3) and therfore am not sure what the mistake is. I suggest an alternative application of Faraday's law which strikes me as simpler. The law states that the same electrical charge will liberate the same mumber of moles of different atoms divided by the charge on their ions.
In this case the No. of moles of aluminium = 4.5/27 = 1/6
therefore No. of moles of H atoms = 3*1/6 (because Al has three times the charge of H+ ions)
therfore No. of moles of H2 molecules = 3/2*1/6
= 1/4. therefore volume of H2 at STP =22.4/4 = 5.6 dm3. which is one of the answers that you quoted.

authors: Месауд Кермише


text:
Can you eliminate cigarette smoke odor between walls?
If holes are made in dry walls is there a substance that can be sprayed in the walls to eliminate cigarette  odor?

Getting rid of odor is a difficult task in general, particularly since molecules stick to surfaces and often stick in surfaces.
There's a good reason that "Febreze" works well. It contains a cyclodextrin ring that binds many odor-causing molecules. So the molecules are still there, but bound in a form you can't smell anymore.
(I remember an anecdote that they had to develop special fragrances to work with Febreze, but I can't find that anywhere.)
I suspect spraying paint primer inside the walls could work. This is often used to fix smoke damage from fires.


text:
Are there attempts to investigate phase states at negative pressure?
Phase diagrams of various compounds and systems usually include only positive pressure. But I wonder whether it is possible to somehow investigate them under negative pressure?
Are there materials that change phase if mechanically stretched? For instance given that metallic form is sometimes achieved at high pressure, can a negative pressure applied to a metal produce a non-metalic form?
Or maybe the effect of negative pressure may be achieved by electrically uniformly charging the material?

How do you define negative pressure?

I think there is a little misconception here.
There are no negative absolute pressures. When someone talks about negative pressures, they are referring to values under atmospherical pressure, this is,between 0 and 1 atm.
There are many ways of creating this vacuum, and of course, that matter has been studied under these conditions, for example, this kind of negative pressures are applied to get the boiling point of substances reduced, so thermolabile compounds can be vaporized (or the solvent which contains them) under its decomposition temperature.


text:
Soap versus Shampoo
Soap and shampoo are both cleaning agents in our daily life.
Why is soap in solid form and shampoo is in a liquid form, though? Is it because of the chemical components inside them or another factor? Which cleaning agent has a higher efficiency in cleaning grease from the body?

Well, whether these products are liquids or solids typically has more to do with the additional ingredients than the actual cleansing agent.
Soap, chemically speaking, is a surfactant composed of the salt of a fatty acid, typically lauric, palmitic or stearic acids. It's typically produced by adding a strong hydroxide base to liquefied plant or animal fat, which liberates the fatty acids from the triglyceride via hydrolyzation, where they then bind with the free cations from the base forming an amalgam of salts.
Shampoo, or more specifically the detergents present in it, are similar, but produced by a different process, to create a surfactant that is less likely to bond to the mineral compounds in hard water, and so is more effective than soap when using it with tap water from most regions. Sodium lauryl sulfate and sodium laureth sulfate are the two frontrunners here; the former is very powerful, and doesn't foam as much, so it's typically used for laundry detergent. The latter is gentler on human skin (less irritating) and so is typically used for various hand and body cleansers including shampoos. The two split about 50-50 in dish soaps, with a third surfactant, lauramine oxide, also making an appearance as a heavy-hitting degreaser.
Both of these substances act in a similar way; they have a hydrophilic (water-attracting) side, and a hydrophobic (water-repelling) side. When dissolved in water (which happens slowly unless the surfactants are already dissolved), the hydrophobic side is attracted to anything that isn't water, and so it will loosely bind via various van der Waals forces to mineral compounds and organics such as those found in dirt, grease, grass stains, oil, lipstick etc. These molecules end up surrounding small particles of these substances on all sides in a bid to "shelter" their hydrophobic end, forming a "micelle". The hydrophilic side exposed to the water, then, allows the micelle to float away in the water and resists the substance in the core of the micelle reattaching to the surface it had been on (dishes, clothes, skin, hair etc), so it can be easily rinsed away.
Now, the form in which a detergent is most useful typically determines whether its sold or packaged as a liquid or solid. Overall, its state is most directly determined by the amount of water in the product. If you've ever had a housemate that never wipes off or closes the nozzle on a bottle of dish detergent, you will have learned that these "liquid" soaps don't have much trouble being solid, either. The liquid products sometimes have an alcohol that helps keep the water from evaporating (and some, like cetyl alcohol, are what make certain products opaque as well as thickening them to give the appearance of a cream), but if marketing wants the product to be a solid, all the chem-engineers have to do is run the mix through a warm (160-250*F) oven to evaporate excess water, leaving behind a pliable solid that can be pressed into bars or minced into soap flakes such as for laundry powders. As far as efficacy in any given form, as I said, liquids tend to disperse in wash water more easily than solids (which is good in the case of a bar soap, not do good with a dishwasher or laundry detergent), though the powder forms are designed to dissolve well, usually through use of some sort of "filler" that spaces out the active ingredients and dissolves easily (and also allows for "everyday" units of measure, so it feels like you're adding enough soap to the water to get your clothes or dishes clean, when in reality a tablespoon of the concentrated active ingredients is more than enough as evidenced by the little gel packs).


text:
Is There a Difference in the Definition of Complete and Stoichiometric Combustion?
My question has to do with some combustion related definitions. When the term complete combustion is used, is it assumed that the combustion is stoichiometric (reactants combine perfectly), but dissociation does not occur, such that only carbon dioxide and water is the product? If so, does the term stoichiometric combustion mean the same thing except dissociation effects do occur?
Edit: Minor species of combustion as a result of dissociation

Stoichiometric, theoretical or complete combustion, is the idealized process where a hydrocarbon is completely reduced to carbon dioxide and water, consuming oxygen (or, if other elements are present, they get all also oxidated, like sulphur compounds yielding sulphur dioxide, but as you can see, not to its maximum oxidation state, sulphur trioxide is not produced in real combustions)
It is called stoichiometric for obvious reasons, stoichiometric amounts of the hydrocarbon and oxygen / water carbon dioxide are consumed / generated. It is supposed to be a complete reaction.
But is also called theoretical, because it does not take place like that in real life. In practice, long hydrocarbon chains do not break down completely as efficiently as it's expected, subproducts are obtained, not only carbon dioxide is formed, but also carbon monoxide, unburned elemental carbon and like in every chemical reaction, it does not proceed to completion.
Also, if there is not enough oxygen (not at least an stoichiometric amount), incomplete combustion takes place, yielding almost exclusively carbon monoxide (toxic stuff).
In some cases, such as the fuels combustion processes, another process called pyrolysis, occurs before combustion. In this incomplete combustion, products of pyrolysis remain unburnt and important contaminants are produced, like the product of partial oxidation of ethanol can produce acetaldehyde, which is really toxic (Considerably small LD50) and very very flammable (flash point under 0 Celsius degrees).

From this it looks like stoichiometric and complete combustion are synonymous? Is there a special name for stoichiometric combustion where dissociation of combustion products CO2 and H2O are considered (neglecting the other real effect mentioned, assume that fuel and oxygen are all consumed)?

Sincerely, I'm not sure what you mean when you say "dissociation of combustion products are considered". Yu mean CO2 and H2O being dissociated into its elements? That'd never happen...

By dissociation effects, I mean minor species that form like CO, NO, OH, H, H_2, etc. as indicated on the graph I added in my question, from the thermal decomposition of CO2 and H2O. Complete combustion to my understanding assumes no decomposition, and that only CO2 and H2O are formed.

These components are not products of a complete combustion, they can be generated in an incomplete one. CO, NO, H2... End up converted into CO2, H20, NO2, the products of complete combustion

In the context of the textbook I'm reading, the effects of reactant decomposition is neglected and it only considers combustion product dissociation. My book considers complete combustion to be stoichiometric and free of dissociation (products are H2O and CO2 only). However, these products do dissociate to form the minor species as noted in the graph at stoichiometric (equivalence ratio = 1). I'm looking for a term that is defined as: stoichiometric combustion with product dissociation (if such a term exists) or does the term stoichiometric combustion already assume this.

I dont think a term like that does exist, since in a combustion, you will never get the formation of H2 from H2O. That would be a reduction process, and conbustion is by definition an oxidation. If H2 (or any other of your examples) is produced during the process, it can either remain like that (in an incomplete oxidstion), or get burned/oxidized to give H2O (complete oxidation), but it does not spontaneously happen in the opposite direction.

Per my book, the graph above was generated from calculations based on atom/energy balance and dissociation of CO2 and H2O to equilibrium state that includes CO2 CO H2O H2 H OH O2 O NO N2 and N, so I can only assume that H2 from H2O is one of the equilibrium reactions. Since the definition of stoichiometric reaction says nothing about what happens to the products post combustion, only mole ratio of reactants, would this term cover “complete combustion with product dissociation” as it, in a sense, is inclusive of complete combustion (stoichiometric, no dissociation)?


text:
How to find the mass spectrum of Cl3+?
The setting of this problem is on the moon of Jupiter: Io.

Given that chlorine exists in two isotopes: $\ce{{}^{35}_{17}Cl}~(33.3\%)$ and $\ce{{}^{37}_{17}Cl}~(66.7\%)$, what is the mass spectrum of $\ce{Cl3+}$?

What I know now is that there are 3 peaks in the mass spectrum of $\ce{Cl2+}$ because there are 3 possible combinations ($\ce{{}^{35}_{17}Cl}$ + $\ce{{}^{35}_{17}Cl}$ / $\ce{{}^{35}_{17}Cl}$ + $\ce{{}^{37}_{17}Cl}$ / $\ce{{}^{37}_{17}Cl}$ + $\ce{{}^{37}_{17}Cl}$). Therefore, there are 4 possible combinations and 4 peaks for $\ce{Cl3+}$.
However, I do not know how to make a mass spectrum graph with only this much information.

Consider Pascal's Triangle.

Okay, let's tackle this. First of all you worked out correctly that there are four different combinations, resulting in four different peaks. You have also given the probability for finding a $\ce{{}^{35}_{17}Cl}$ is $a=33.3\%\approx\frac13$ and the probability for finding a $\ce{{}^{37}_{17}Cl}$ is $b=66.7\%\approx\frac23$. Since you have three positions to fill, this will result in a binomial formula of third order, $(a+b)^3$. You can solve this easily with Pascal's triangle, $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$. You spectrum should therefore look like this table:
\begin{array}{lrr}\hline
   & \text{m/z} &\text{abundance}/\%\\\hline
\ce{[{}^{35}_{17}Cl_3]^{+}}                & 105   &  3.7  \\
\ce{[{}^{35}_{17}Cl_2{}^{37}_{17}Cl]^{+}}  & 107   & 22.2  \\
\ce{[{}^{35}_{17}Cl{}^{37}_{17}Cl_2]^{+}}  & 109   & 44.4  \\
\ce{[{}^{37}_{17}Cl_3]^{+}}                & 111   & 29.7  \\\hline
\end{array}
The following snowball diagram should further demonstrate this:

I don't know mass spectro stuff but I guess in this line - "since you have three positions to fill, this will result in a binomial formula of third order" you probably meant "since you have four positions to fill..." (as there are four rows in your table)

@GaurangTandon No. Three atoms, three positions, three rows in the diagram, four different isotope combinations.

Aren't there eight possible combinations of isotopes? (as I see from the bottom row of your graph, and as is also easily observed from the fact there are three atoms in $\ce{Cl3+}$ each having two choices)

@GaurangTandon there are four different masses you can measure, because positions are irrelevant for that. There are six possible combinations, as there are symmetry equivalent ones.

"Six possible combinations" of what...?

@GaurangTandon Last row count all and substract the summery equivalent ones, you have 6 different ones left; four different sum formulas.


text:
Regioselectivity of electrophilic aromatic substitution in imidazo[1,2-a]pyrazine

I am supposed to rationalise the regioselectivity of bromination of 8-chloroimidazo[1,2-a]pyrazine, as depicted in the scheme.
The left ring kind of looks like a pyridazine. If it's anything like a pyridine it'll be bad at electrophilic substitution without activation, so bromination at C–5 or C–6 is out. I guess it has to be one of the carbons in the five-membered ring. But why is it C–3 and not C–2? I am trying to draw out resonance structures and reason that way.

You are on the right track. Pyridine and similar nitrogen analogs of benzene are deactivated with respect to the electrophilic aromatic substitution reaction. You are also right that a resonance structure can help explain the regioselectivity of the reaction on the five-membered ring.
Electrophilic attack at C–2 (bottom line) directly gives an intermediate with non-aromatic resonance structure 2a. Moving electrons around, we are able to draw an aromatic resonance structure 2b, but this structure also contains a non-octet atom (carbocation), repulsive neighboring positively charged atoms, and charge separation.

On the other hand, electophilic attack at the observed site gives an intermediate 3 that maintains aromaticity in the six-membered ring and complete octets in all atoms.

We can predict that the intermediate arising from attack of the electrophile at C–3 would be more stable than the corresponding C–2 substitution intermediate. Since this step should be endothermic, the rate of this step will be influenced by the stability of the resulting intermediate (Hammond's postulate). From there, we can proceed through the usual addition/elimination mechanistic sequence to get to the product.


text:
Molecular Orbital Theory and No. of bonds
The order of filling of molecular orbitals $\ce{O2}$ by MOT is this :

But from it, how can I deduce that there is one sigma and one pi bond in an oxygen molecule ?

If I am not mistaken, the $2\sigma_g$ should be higher in energy than $\pi_u^x,\pi_u^y$.

@martin That's for molecules having electrons less than or equal to 14 , like nitrogen

Overall bond order equals number of electrons that occupy bonding orbitals minus numer of electrons that occupy antibonding orbitals divided by two.
Now you can apply the same concept indivudually:
For s molecular orbitals, both electron pairs "cancel out".
For p molecular orbitals, you have 2 electrons in sigma bonding orbitals and none in sigma antibinding orbitals, which means (2-0)/2 = 1 sigma bond.
Also, you have 4 electrons within pi bonding orbitals and only 2 electrons within pi antibonding orbitals, which means (4-2)/2 = 1 pi bond.
This yields the result that you expected.

Actually, there is not a sigma and a pi bond. The MO scheme clearly depicts that - the bonding situation is a little bit more complicated.

Yeah, right, but that just makes the OP question nonsense. In MO theory there are no individual bonds, just that, molecular orbitals filled with electrons.

@martin I am studying MOT as a part of high school chemistry and this was a question in one of the tests. But actually, are MOT and VBT mutually exclusive theories ? Should I not try to find anything in common ?

@biogirl They are not mutually exclusive, quite the opposite is the case. They are complementary. So they are just views from different angles. If you want to know more about this matter I encourage you to ask a new question, it is too extensive to be dealt with in the comments.


text:
Finding pH after dilution of acetic acid

We have $\pu{10 L}$ of solution of $\ce{CH3COONa}$ with concentration $C = \pu{1 M}$ and $\mathrm{pH} = 10$. Then we add $\pu{90 L}$ $\ce{H2O}$. What will be the pH of the new solution?

My work:
\begin{align}
  \ce{CH3COONa &<=> CH3COO- + Na+}\\
  \ce{CH3COO- + H2O &-> CH3COOH + OH-}
\end{align}
We have $\pu{1 M } \ce{CH3COO-}$ and let $x$ denote the concentration of $\ce{OH-}$. If the $\mathrm{pH}$ of the solution is $10$, it's implied that
$$\mathrm{pOH}=4\iff-\log_{10} x=4 \iff x=\pu{10^{-4} M}.$$
I know I have to use that the amounts of substances didn't change, so $$C_1V_1 = C_2V_2,$$
but I don't know what values I have to put in there. Is the $x$ I found my $C_1$?

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


text:
What possibilities are there for refining aluminium on the moon?
Engines burning aluminium and liquid oxygen have been investigated for use on the moon because there is plenty of both elements in the soil - about 13% and 40% of highland soil, respectively. Propellant thus has the potential to be sourced from the moon directly, but it isn't practical unless a refining process can be found that uses the moon's environment to extract pure aluminum efficiently using local resources. A viable way to extract the oxygen is known already, using a pyrolysis process.
Is there a way of taking advantage of the moon's environment to do this effectively? You have at your disposal:

A hard vacuum
A month-long day that is strong sunlight for two weeks then a two week night - it is estimated that a furnace using concentrated sunlight could reach 3000 K.
1/6 Earth gravity
No moisture - chemicals are reduced
Powdery soil that is loose to a depth of 10 or 15 cm
Some soil samples from Apollo 16 were a quarter alumina by weight

A source of fuel in space is a prerequisite for a larger ongoing human presence there. A rocket launched from Earth can't manage to be more than about 5% payload. Shipping fuel from Earth to any off-world base by rocket would be prohibitively expensive (even if the rocket is reusable). This is why there has been so much interest in sources of water ice in space - it can be electrolysed into liquid hydrogen and liquid oxygen, which makes excellent propellant. But you have to go to the asteroid belt to get it (there are indications there might be significant reserves in craters at the moon's poles in permanent shade, but this is inconclusive). Once there is a way to refuel in space, all sorts of things become possible that are beyond our reach right now. An option that can compete with the price of mining asteroids has business potential. All such undertakings are going to require vast initial outlays that won't pay off for a decade or more. The process being considered here only has to meet that standard.
This question is a partner to a question at SE's Space Exploration site - How feasible is it to use aluminum and liquid oxygen as future propellant sourced from the moon?. I realized it was largely a chemistry question and so decided to ask about it here. There are two references in the other question: one from Wickman Spacecraft & Propellant, and one from the Artemis Data Book. I'm afraid neither discusses the chemistry and I haven't found anything else. The AIAA has several papers on it in their catalogue.

Given the insurmountable transportation costs of the newly-mined raw material, I don't see how anyone will be willing to foot the bill.  Therefore, I think the possibility for this refining to occur is essentially zero.

The purpose of this is to create fuel to be used in space. I should probably edit the question to reflect that.

Okay, now i've added a paragraph on the value of propellant sourced in space.

While my comment was meant to be tongue-in-cheek, I'm glad to see the addition of content to your post.  It makes it much more interesting.

That's how this process works, right? Glad to get comments like this. And i'm a little thick about word-play sometimes.

Aluminium oxide is solid, so aluminium+oxygen will be a very bad propellant.

@permeakra the aluminum+oxygen provides the heat. The oxygen is in liquid form. By making a mix of the two with extra oxygen, the heat of the Al + O2 reaction causes the remaining oxygen to rapidly expand. In this way you get the pressure for high exhaust velocity. The alumina is swept out with the oxygen and thus also provides some forward momentum.

My theory was to use electromagnetic separation, aka the Calutrons used in the Manhattan Project.  Fine powder feedstock (check), into an arc discharge to vaporize and ionize (lots of good designs for high volume ion sources out there), electrostatic acceleration, then electromagnets to mass separate.  Power with a solar panel array.  No moving parts (except for the powder feed).  No vacuum systems needed.  And you separate all of the elements at once. Now, it only works during sunlight, but during the lunar night you could do maintenance and collect the separated elements.

Alright, in a few hours i'll be hunting for material online about all this - Calutrons, electrostatic acceleration, electromagnetic mass separation, vaporization by arc discharges. This gives me lots to go on, i appreciate it. Can you suggest good sources of information? It needs to be available online, but i'll buy something if it has what i want to know in it.

Wikipedia has a write-up on the Calutrons. Separation of uranium isotopes is really hard compared with refining ores (the difference in U-235 vs 238 and others is a small mass fraction).  Other useful info might also be found for AMS (atomic mass spectroscopy), used for, e.g. C-14 dating.  Again, they want a very precise, single pass measurement. Cascades of Calutrons might be more efficient. Van der Graaf or Cockroft-Walton accelerators are also on Wikipedia.  Ion source are probably best found in the technical literature which you may not have easy access to.

...and wouldn't understand anyhow. I've gotten used to just letting what i'm reading wash over me and absorbing what i can. My theory is that eventually that can add up to enough working knowledge if i round it out by learning what i really need to know. Once i have looked at that stuff i may return with more questions. :)

@kimholder   Meteoric impacts on The Moon should have vaporized lots of anorthite, but did it separate  into corundum and other minerals ? And ionization of anorthite will not separate it into minerals either , will it ? So why the mass separation ?  I think only high enough temperature (and pressure) will do the job.  www.minsocam.org/ammin/AM65/AM65_272.pdf

The ion source, designed properly, would rip it all apart.

Ionic liquid refining of aluminum is a promising solution which uses 3 kWh of electricity per kg instead of around 15 kWh for existing methods.

The Gibbs free energy for the decomposition of 1 mol of Al2O3 is about 1.5MJ (for 54g of Al.) That makes 27.8MJ/kg = 7.7kWh (though a consumable carbon anode can reduce this by about a third due to the reaction Al2O3 + C --> 2Al + 1.5CO2). I fail to see how a new method can do anything to change basic thermodynamics and electrochemistry. Your linked paper is not about production of bulk aluminium, but rather a special high purity grade, in which the Aluminium seems to be produced from AlCl3 rather than Al2O3.

I don't think this is going to be a good idea.
Notice that the Al and O existing in the Moon's soil are not there are free elements, but are tightly bound in minerals.
On Earth, Al is extracted from bauxite - this is naturally occurring Al2O3. This extraction process is extremely energy intensive. In fact, they transport the bauxite mined elsewhere to Iceland because the electricity in there is so cheap, instead of doing it on the spot of the mine, or nearby.
The Al and O that occur on the Moon are mostly in the form of the mineral anorthite plagioclase: CaAl2Si2O8. Extracting the Al from there would be even harder! You first have to extract the Al2O3 out of the CaAl2Si2O8, and only then you can get the Al out. This mineral is not unique to the Moon. In fact, anorthite, albite (NaAlSi3O8) and all minerals in between are the most common minerals in Earth's crust, and we still don't use them as a source for Al.
That said, burning Al as a fuel would not be a good idea because the end result is a solid, not a gas.

The point about solid reaction product was made before, but i never answered because i wasn't able to get a more professional opinion than mine. Basically, there has to be a lot of extra oxygen in the mix. The heat of the Al + O2 reaction rapidly converts it from liquid to gas and provides thrust. As for the difficulty of refining ore, there is no easy solution to fuel on the moon. With great luck there will be lots of easily accessible water ice in polar craters, barring that, this could well be the best solution.

You could also burn aluminum with ammonium perchlorate, making a solid fuel mixture whose combustion does give off gases (nitrogen, water vapor, maybe volatilized aluminum chloride).  Now all you need is to get said ammonium perchlorate on or to the Moon.  Oops.

The standard process for electrolysing $\ce{Al2O3}$ to $\ce{Al}$ should work fine on the moon. Note that on Earth, a consumable carbon anode is used to reduce the voltage required, giving the reaction (Hall–Héroult process)
$$\ce{Al2O3 + 1.5C → 2 Al + 1.5 CO2}$$
An alternative, non-consumable anode would be used on the moon, to avoid the need for anode material and to give oxygen instead of $\ce{CO2}$.
The process for purifying $\ce{Al2O3}$ would be rather more difficult. On Earth, this is done by dissolving bauxite in sodium hydroxide solution, filtering out insoluble iron salts, cooling, precipitating out $\ce{Al2O3}$, and discarding the liquor containing silicates.
This process is highly intensive in water and sodium hydroxide. Major advances would be needed to make it less intensive in these materials, and to make the electrolysis step more tolerant of impurities.
I think we are better off looking for water to use as feedstock for our propellant factory. It's far easier to purify (all you have to do is distill it).


text:
How do I write the deterministic rate equation for bimolecular reaction with similar particles?
I was wondering how to write down the deterministic rate equation for a bimolecular reaction with similar particles.
e.g.
$$
\ce{A ->[k_+] 2B}
$$
and
$$
\ce{2B ->[k_-] A}
$$
Now the rate equations for the above reactions are:
$$
\mathrm{\frac{\delta A}{\delta t}= - k_{+} A + k_{-} B^2}
$$
$$
\mathrm{\frac{\delta B}{\delta t}= 2 k_{+} A - 2 k_{-} B^2}
$$
Now I have included a factor of 2 in the second equation to conserve $\mathrm{A + \frac{B}{2}}$.
But I don't understand the reasoning for this factor of 2 in the second equation.

There is really only one independent equation here because
$\frac{dB}{dt} = -2\frac{dA}{dt}$
The "second equation" in the OP is just -2 times the first equation.
Every molecule of A destroyed creates 2 molecules of B.
Every 2 molecules of B destroyed creates 1 molecule of A.

I understand that. but http://chemwiki.ucdavis.edu/Physical_Chemistry/Kinetics/Rate_Laws/Reaction_Mechanisms/Elementary_Reactions They mention the rate of the bimolecular reaction with similar species without a factor of 2.

I don't see anything at that site that contradicts what you did.  Certainly you could replace "2k" with a different constant "c" and still have a true equation.


text:
Why is ammonia so much more basic than water?
The only reason I can think of is that the lone pairs in the oxygen in water are in lower energy orbital relative to the nitrogen in ammonia due to increased effective nuclear charge in oxygen. However, as far as I am aware, ammonia is SIGNIFICANTLY more basic than water. Surely the orbital energy cannot account for this.

Basicity and acidity can be accounted for by the stability of the conjugate base and the conjugate acid. Try taking a look at $\ce{H3O+}$ and $\ce{NH4+}$'s $\mathrm{pK_a}$'s.

How is that explained though?

Hybridization changes from sp sp2 in water, sp2 to sp3 in ammonia. So in the first case there might be less relative repulsion with nucleus. However, the comparison might not be appropriate.

When it comes to comparing ammonia and water, the acidity or basicity can be explained because of the central atoms electronegativity. When comparing atoms of the same periodic row, the more electronegative an atom is, the less it will want to donate an electron pair.
The electronegativity values from the Pauling Scale for $\ce{N}$ and $\ce{O}$ are
$$\ce{N}=3.04$$
$$\ce{O}=3.44$$
So based off of these values we can see that nitrogen is less electronegative, it less strongly it holds onto its electrons, and thus it is a better electron donor. This allows the addition of a $\ce{H}$ to be more stable.
The atomic reasoning behind this is that, as a general periodic trend, as you move across the periodic table, from left to right, the nucleus is increasing, which increases it's charge, producing a stronger attraction to its valence electrons.

Be careful with this approximation; if you continue to the left on that row, you find carbon, with less electro negativity, but it does not have a lone pair to share, so it won't behave like a base in methane (nor an acid), and if you move even more to the left, you find the weird boron, which will have an empty p orbital which, in principle, would make borane behave like a Lewis acid (I say in principle, because actually it shows an amphoteric behavior).

Edited my answer to reflect that my statement is based off of a trend.

Also, my statement was about electron pairs and your comment seems to not take into account molecules where carbon or boron would have lone pairs. Carbon would definitely have lone pairs if it was in a molecule like $\ce{CH3-}$, and you can sure bet that $\ce{CH3-}$ is a strong base. The same would be applied to a compound like $\ce{BH2-}$.


text:
How do I list these elements according to their radius?
$\mathrm{Z=33,\ Z=37,\ Z=9,\ Z=31}$
Since we shouldn't look at the periodic table, I found their groups and periods, what should I do next?

Of course you should look at the periodic table. Groups and periods are the just components of the periodic table.

Step 1: google atomic radius chart
Step 2: use chart.

We have these elements:(just for ease, I'm naming them, you can just avoid it)

$$
\begin{array}\\
Z=33&\text{Arsenic}&\ce{As}\\
Z=37&\text{Rubidium}&\ce{Rb}\\
Z=9&\text{Fluorine}&\ce{F}\\
Z=31&\text{Galium}&\ce{Ga}
\end{array}
$$

Now, let's see the periodic table (the useful region, thanks to Martin) :
$$
\begin{array}{ll|cccccccccccccccccc}
&\text{Gr.} &1 &2 &13  &15  &17 \\
\text{Per.}&\\\hline
2&&. & .& .& .& \color{red}{\ce{\tiny^9F}}\\
4&& .& .&  \color{red}{\large\ce{^{31}Ga}}& \color{red}{\small\ce{^{33}As}}& .\\
\hline
5&& \color{red}{\huge\ce{^{37}Rb}}\\
\end{array}\\
$$
Now consider this:

While going down or towards left, size increases. Equivalent, is to say the size is larger towards bottom left corner

It wouldn't be now difficult to conclude that size is as follows:
$$37>31>33>9$$

I try this question before. First you find out the electron configuration and the group and period for each substances
(I will name each substances for clearing confusion A=33, B=37, C=9, D=31)
Electron arrangement
A= 2.8.8.15 (Group 4, Period 15)
B= 2.8.8.18.1 (Group 5, Period 1)
C= 2.7 (Group 2, Period 17)
D= 2.8.8.13 (Group 4 ,Period 13)
Now we have two ways, either you can sketch the table out since the numbers are small or you can do it my way.
We know that the radius of a substance increase from right to left and from up to down. Meaning the smaller the group number and the period number the smaller the radius.
If you want to find the substances in an ascending sequences according to their radius. First, we will look at their group number. C has the lowest group number so it has the smallest radius, then we can see that A and D have the same number of group, thus we now check on the period number. We can see that D is smaller than A. Meaning, D has a bigger radius than A . Lastly check back the group number and we have B with the highest group number so its radius is the biggest.
Now, we can get the sequences which is C→A→D→B.
If you want the desending sequences , do it backwards. Hope that it will help


text:
Which group does germanium belong to?
$\mathrm{Z= 32}$
$\mathrm{1s^2\ 2s^2p^6\ 3s^2p^6d^{10}\ 4s^2p^2}$
According to me it belongs the $\mathrm{IV\ B}$ group since it has the $\mathrm{d}$ completed, but it belongs to $\mathrm{IV\ A}$. But we know that in the $\mathrm{A}$ group only the $\mathrm{s}$ and $\mathrm{p}$ elements belong. Why does this happen?

My periodic table has Ge as a IV B element, with IV A being Ti, Zr, and Hf. So I'm not sure which periodic table you are perusing.

According to the IUPAC, the is not such thing as groups IV A and B anymore.
All the elements you are analyzing belong to the 14th group of the periodic table, each one in one period, from the 2nd to the 6th.
In regards to your other question, you need to look for a trend on the elements of the group.
What happens as you go down the group?
I think you've already figured out that the number of electrons around the nucleus significantly increases, and this has an important effect on the radius of the atom.

Here is a good periodic table, and the one you should be using: IUPAC periodic table

Thanks for the link.  Now I can come into the 21st century...


text:
Femtosecond spectroscopy
Hi I am new to laser pump-probe spectroscopy and want to understand the concepts better. At the moment I am struggling to understand the following: When one takes absorption measurements with lasers in a pump-probe manner, the spectra that are collected need to be normalized. What does the Normalization mean and how does one do it? Also, why are relative changes important as opposed to absolute changes (in absorption) because I think they should give the same information?

I'm not sure exactly what you mean by normalising as it can mean more than one thing. At each probe wavelength ideally you measure two signals simultaneously from the same probe pulse which is split into two parts before entering the sample. One part probes the sample at the same place that it is excited and the other probes unexcited sample. You then record the transmission of these two signals and calculate optical density change or work with transmission depending upon choice.  It is clearly necessary to normalise to the pump laser intensity since a more intense pulse will produce a bigger effect. Second the probe pulses have to be monitored and adjusted as necessary (without exciting sample) so that these signals are kept equal or at least in a constant ratio so they can be normalised later.
Pump-probe is just like any other transient absorption measurement (used to be called Flash-Photolysis) and a difference signal is measured. This produces a transient spectrum which might be positive or negative. If you have the exact optical density the you can construct the spectrum of a transient species knowing the ground state spectrum. If there is no ground state absorption where you have a transient spectrum then this must be positive and is the transient spectrum. If it is negative it might be due to transient gain, i.e. the probe causing stimulated emission from the sample. The schematic shows one way the experiment can be set up. The absorption change at t = 0 is exaggerated for clarity; a 1% change is a huge signal.

If possible, please give citations for the source of images.

@ankit7540 it looks he must did them by himself.

yes my own images. Feel free to use them.


text:
What is the difference between molarity and concentration?
I learned these two words in school but still I don't understand what the difference is between them. From my understanding, molarity and concentration means the number of moles in a certain solution. So, what is the difference between the two?

Concentration is any measurement of the quantity of a solute that is present per unit of solution (in general, there are exceptions like moLality, which measures the number of moles of solute per kilogram of solvent).
Molarity is just one way to express concentrations, which, more specifically, is the number of moles of solute per liter of solution.
But you can give values for concentrations in many different ways, like for example, grams of solute per liter of solution, percentages (grams of solute divided by the grams of solution multiplied by 100), etc.
You can find a wide range of ways of measuring concentrations, including molarity here

The correct and standardized quantity name is amount-of-substance concentration.
In chemistry, however, the name is generally abbreviated to the single word concentration, it being assumed that the adjective ‘amount-of-substance’ is intended.
The quantity symbol is a lower case $c$.
The quantity ‘amount-of-substance concentration of $\ce{B}$’ is defined as
$$c_{\ce{B}} = n_{\ce{B}}/V$$
where $n_{\ce{B}}$ is the amount of substance of $\ce{B}$ and $V$ is the volume of the solution.
The dimension of the amount-of substance concentration is
$$\dim c = \mathsf{L}^{-3} \mathsf{N}$$
The coherent SI unit for amount-of substance concentration is ‘mole per cubic metre’ (unit symbol: $\mathrm{mol/m^3}$).
However, the usually used unit is mole per litre ($\mathrm{mol/l}$) or any decimal multiple or submultiple rather than mole per cubic metre ($\mathrm{mol/m^3}$).
In the older literature, this quantity was often called molarity. According to current IUPAC recommendations, the traditional name ‘molarity’ should be avoided due to the risk of confusion with the quantity molality. Current international standards do not list ‘molarity’ as a valid special name for the quantity amount-of-substance concentration.

Yay for amount of substance! Nay for number of moles! Thank you very much for this. However, I am not able to understand what $\dim c = \mathsf{L}^{-3} \mathsf{N}$ actually means. I $\mathsf{L}$ a length and $\mathsf{N}$ a number? And is the $\dim$ operator not usually reserved for vector spaces? Is concentration a vector space?

ah, I guess I found it... http://physics.stackexchange.com/q/77690/104362

Concentration is the ratio of the amount of solute per amount of solution.
Molarity is a unit of concentration that specifically relates the number of moles of a solute per liter of solution.

Concentration is a general term representing the quantity of solute in a solution of known volume. Now the solute quantity in solution can be described in different ways. for example molarity, normality, formality, ppm .
different terms used where they are to be applied


text:
Role of Alcohols(EtOH,MeOH) in nitro reduction used for Metal(Sn,Fe)

I'm trying to reduce an aryl nitro group.
I understand that these reactions are radical reactions by the metal, but I don't understand
why alcohols are used as the solvent. Are they used as a source of protons, or just because they are good solvents?

LHM...Look at this mechanism that I posted as an answer to an earlier SE Chem question/answer relating to the reduction of nitrobenzene by metals.

In most reactions that occur on metal surfaces, many of the mechanistic details are not fully understood (see @Martin's comments in the earlier post).  Nonetheless, this mechanistic scheme shows several steps where a proton ($\ce{H+}$) is required.  These proton transfers are necessary to complete the reduction process.  This is why these reactions are typically run in solvents, like ethanol, that have available protons that can be donated in order to complete the reaction...ron

Thanks for your answering my question.
If you know mechanism of ethanol give a proton to nitro?
In condition, it's react in RT so i think O-H bond is hard to break. or nitrogen's lonepair electons catch the O-H proton?

Notice that in some of the steps requiring proton transfer there is a negative charge on oxygen or nitrogen.  This makes the oxygen or nitrogen more basic, and makes it easier for the oxygen or nitrogen to remove a proton from the solvent.  Notice too that acid is usually involved in these reductions and can also serve as a source of protons.

I do not believe that ethanol is mandatory. You surely need a proton source, but that's why those reactions are performed in protic acidic media, like your first example, with HCl, or what I think were the original conditions of Bechamp reduction, using acetic acid.
Probably ethanol makes a good solvent for phenol rings.
Edit: this is the Bechamp original paper. I couldn't manage to find it in English. I'm not French speaker either, but just in case you are curious :-)
As far as I'm concerned, nowadays catalytic hydrogenation is used to perform this reaction (at least at industrial scale)
About the mechanism, I'm sorry but I can't give you a solid statement on this.


text:
Will an iron nail sunk in a highly concentrated salt solution rust?
Assume I have a salt solution with the same concentration as the dead sea and I put a iron nail into the solution (assume it sinks). Will it rust? If yes, why? If no, why not?

IMO, very vague question

Rust is iron oxide, $(\ce{Fe2O3.nH2O})$
It forms from iron(III)hydroxide made when iron reacts with oxygen in the presence of water. The chemical reaction can be summarized as follows:
$\ce{4Fe(s) + 6H2O +3O2 ->4Fe(OH)3(s)}$
Now, water 'naturally' contains some dissolved oxygen (8-14%), depending upon the source of the oxygen (photosynthesis from underwater plants, diffusion from the atmosphere) balanced against oxygen demands (such as fish breathing and decomposition of organic matter). The temperature of the water also affects the solubility of oxygen (gases are less soluble at higher temperatures).
The ocean (which is typically 3.5% salt) typically has 20% less dissolved oxygen in it than freshwater. This is because the water molecules are attracted to the dissolved ions from the salt (solvation), which has the tendency to decrease the weak affinity of non-polar oxygen molecules to water, thereby driving dissolved oxygen out.
Now, the salinity of the Dead Sea is about 290 (that's 29% salt!), which is over 8 times the salinity of the ocean!  At such high levels of salinity, most of the oxygen is 'salted out' of the water, which means that the levels of dissolved oxygen are very low, especially below the surface, so an iron nail totally immersed in water with 29% sodium chloride (salt) will not rust.


text:
How to recrystallize boric acid into flakes?
I am putting 100 g of boric acid, 400 g of water and trying to crystallize at low temperature and letting cool slowly, however it recrystallizes as needle like texture. I need soft large crystals like mica. Can someone help out?

The ratio seems about right, though the solution could be a bit more saturated as the solubility of boric acid in $\pu{100 mL}$ of boiling water is $\pu{27.5 g}$, so for $\pu{400 mL}$ of water you can theoretically use $\pu{110 g}$ of boric acid.
From my experience it is crucial that initial nucleation occurs at the surface, then one gets flake-alike crystals. That's said, you need to start to cool down the solution very slowly, preferably using a warm water bath and a closed vessel with large cross-sectional area (wide beaker with a watch glass on top should do). Don't rush the cooling process, especially at the beginning – crystallization should begin on surface, not in the volume, otherwise you get a bunch of crystals of different morphologies.
P. S. When drying the crystals at the end after rinsing them with cold water on a filter make sure not to exceed $\pu{130 °C}$, otherwise boric acid will start to decompose and form metaboric acid $\ce{HBO2}$, then subsequently upon further heating, $\ce{H2B4O7}$, and, finally, $\ce{B2O3}$ [1, pp. 119–120].
Reference

Patnaik, P. Handbook of inorganic chemicals; McGraw-Hill handbooks; New York, 2003, ISBN 978-0-07-049439-8.


text:
What happens when adding swimming pool salt to a freshwater pool?
Why does adding swimming pool salt to a freshwater swimming pool cause a white sediment to build up and stick to the floor of the pool?

Because the salt did not dissolve?

First of, what do you understand by "swimming pool salt"? There are millions of different "salts".

Salt used in swimming pools is exactly like table salt.  The only difference is, the salt is ground more finely to motivate faster, more complete dissolution.  Clearly some salt did not dissolve fast enough.  Pools that use chlorinators (a device built into the pump's pipeline) generally require salt to produce chlorine.


text:
What is the correct chemical formula for monosodium glutamate?
Recently, I am interested with home materials and I discover that the chemical symbol for MSG or monosodium glutamate is $\ce{C5H8NO4Na}$.
My problem is shouldn't it be $\ce{NaC5H8NO4}$ as we always write the metal first before the non-metal?

It should, but in organic molecules salts, often the counterion is written latter, to indicate its position on the molecule acting like so, for example, in sodium acetate, $\ce{CH3CO2Na}$, this would be something like a "semicondensated" formula.

Agreed. Usually the organic (carbon containing) part of the molecule is written in the Hill notation and the counter ion added to the end. Usually this is done, because it makes ordering in a Database easier. Note that the correct Hill formula for MSG would be $\ce{C5H8NNaO4}$. Compare to Sigma-Aldrich Data Sheet


text:
Why does AlCl3 dimerise to Al2Cl6 at low temperatures?
I know that it has something to do with HOMOs and LUMOs (I can't what the HOMO/LUMO for AlCl3 is, though).

At low temprature, $\ce{AlCl3}$ exists as a close packed lattice of $\ce{Cl-}$ with $\ce{Al^3+}$ occupying octahedral holes. On heating $\ce{Al2Cl6}$ species are formed.

I agree with @KishoreSShenoy, OP has it the other way around, but why doesn't it dimerize to form a  macromolecular structure?

Perhaps, we can employ the Gibbs free energy equation for spontaneous reactions...
$$\Delta G=\Delta H-T\cdot\Delta S$$
Since the dimerisation process involves the formation of bonds, $\Delta H$ should be negative, while $\Delta S$ will be negative too, since two $\ce{AlCl3}$ molecules become one $\ce{Al2Cl6}$. As such, when temperature is low, the magnitude of $T\cdot\Delta S$ will be smaller, and $\Delta G$ will be negative or more negative, thus favouring the dimerisation process.

There are certain stable compounds where the octet rule is not obeyed, that is, their valence
shells contain more or less than 8 electrons. The compounds containing less than 8 electrons in the valence shells are called electron deficient compounds such as $\ce{AlCl3}$, where the aluminum atom has 6 valence electrons. It has been suggested that two aluminum atoms complete their octet by accepting a pair of electrons from chlorine atoms:

So, that is why $\ce{AlCl3}$ dimerises to $\ce{Al2Cl6}$ at low temperatures.

This does not answer why this happens only at low temperatures.

But why at low temperatures? Also; does it have anything to do with HOMOs or LUMOs at all? Thank you, though.

I think that the highest occupied molecular orbital (HOMO) of one AlCl3 (consisting of non-bonding electron pair on  chloride)  interacts with  the lowest unoccupied molecular orbital (LUMO) of the other molecule AlCl3 (consisting of the unoccupied case on Al). This interaction leads to the formation of a sigma bond between Al and Cl.

The  increase of temperature may decrease the gap between the HOMO and LUMO, and this may lowers the interaction and disfavor the bond formation.

Why isn't this dimerization macromolecular? It seems like it should be