text:
What is the simplest way to create a precipitation reaction at home?
What is the easiest way to create a precipitation reaction at home, preferably with easily retrievable chemicals?
I need this because I'm making a video for my film education.

Here's one easy way :
Make a solution of salt ($\ce{NaCl}$) in water, and another solution of Silver Nitrate ($\ce{AgNO_3}$), which you can buy easily) in water.
Upon mixing the two, $\ce{AgCl}$ (Silver chloride, a white compound) will instantly precipitate.

Silver chloride is white. The grey colour is due to elemental silver (which AgCl is reduced to)

Silver nitrate is expensive, and not easy to buy without legal authorization. Furthermore, its solutions makes dark spots on skin and clothes, that appears only on the following day. But this spot is impossible to clean.

For a low-hazard experiment with standard household materials, some variation of the following should work:
Chemicals:

$\ce{CaCl2}$ : Available as a type of ice melt (e.g., this)
Acetic acid : White distilled vinegar
$\ce{NaHCO3}$ : Baking soda

Procedure:

Dilute the vinegar in water, at about $1$ part vinegar to $10$ parts water.
Add a small handful of the $\ce{CaCl2}$ and stir to dissolve.
In a separate container, mix $\sim 1~\mathrm{Tbsp}$ baking soda in $1~ \mathrm{cup}$ lukewarm water
If all of the baking soda dissolves, keep adding more until it won't dissolve further.
Let the baking soda solids settle to the bottom of the container, then carefully pour the liquid into a clear container, suitable for the demonstration.

For the demonstration, add the $\ce{CaCl2}$/vinegar solution drop-by-drop to the bicarbonate solution. If I've thought this through properly, a white $\ce{CaCO3}\!\left(\mathrm s\right)$ precipitate should hopefully form as each drop is added.
To note, I specified the dilute vinegar solution to try to avoid premature precipitation of $\ce{Ca(OH)2}$.  If the demonstration solution fizzes too much when adding the drops of $\ce{CaCl2}$/vinegar solution, or if no precipitate forms at all, you could try eliminating the vinegar (skip step $1$) and just dissolve the $\ce{CaCl2}$ in tap water.

Use washing soda instead of baking soda.  You are more guaranteed to get the calcium carbonate by adding a normal carbonate instead of a bicarbonate.

Indeed, @OscarLanzi, I tried this myself with bicarb, and it didn't really work. I'll have to try it again with the washing soda.

Mix solutions of washing soda (Na2CO3) and Epsom Salt (MgSO4) to get a precipitate of magnesium carbonate (MgCO3).

As per magnesium carbonate, this actually produces a hydrated complex of magnesium hydroxide and magnesium carbonate. From the same wiki article, mixing solutions of sodium bicarbonate (“baking soda”) and magnesium sulfate (the typical hydrated form is “Epsom salt”) produces magnesium carbonate.

For an experiment of some practical value, try Epsom salt solution plus soapy water.  Explain that the Epsom salt contains magnesium which is a source of water hardness and show what happens when you add the "hard water" to the soapy solution.  Ewww!  Now students will know why water softening is important.

But hard water is good!

Prepare a solution of table salt in boiling water until no more salt will dissolve. You can then transfer the solution to a different container and allow for the solution to cool. The salt should precipitate out. You should get about 10% of the total salt to precipitate out on cooling.

Because to precipitate means to come out of solution, this word is commonly used in weather, chemistry and even in biology. For a science experiment that is safe for the kids and amazing for adults I recommend DNA extraction. Strawberry DNA extraction to be more precise, follow the online instructions to make a strawberry smoothie. Mix with the extraction buffer (soap and table salt), and save the last and final step for the demonstration. Describe the experiment to the kids or do it with them, once you add the chilled rubbing alcohol to the strawberry wash the strawberry DNA will precipitate out!
Enjoy, Mr. R

The easiest way to make precipitation is using washing powder and water and then mixing it with baking powder and water. When they are mixed together the mixture goes from clear to cloudy and becomes hard water.


text:
How to estimate the concentration of OH that comes from just the autoprolysis of water in a basic solution?
In my analytical chemistry course, a question reads:
What is the pH of a 0.10 M solution of THAM?  Is it valid to neglect the autoprotolysis of water?  Justify and explain your answer by estimating the concentration of $\ce{OH-}$ that comes from just water in this solution.
I'm guessing it has something to do with ionization of water ($\mathrm{K}_\mathrm{w}=1 \times 10^{-14}$), any insight would be greatly appreciated.  The calculated pH neglecting autoprotolysis of water is 10.5.

For general information  - Tris(hydroxymethyl)aminomethane aka TRIS or THAM has a pKa of 8.07 according to Wikipedia. https://en.wikipedia.org/wiki/Tris

What auto-ionization means is best explained this way: If you put in  0.10000000 M of NaOH in pure water, how much hydroxide is there at equilibrium? You can imagine that "initially", adding that amount to a neutral solution gives you $\ce{OH^-}$ = 0.10000010, and that $\ce{[H^+]}$ is 0.00000010. However, the ion-product at this point is much greater than $K_w$, so some of the hydronium and hydroxide must react to bring the product down. If you solved the equation $(0.10000010-x)(0.00000010-x) = K_w$, and used $0.10000010-x$ to get $[OH^-]$, you would arrive at the concentration taking into account auto-ionization. You can see that $[OH^-]$ would just be a very minute amount greater than 0.100000000  because x is very small. The x is the result of taking into account the auto-ionization. As you can see, 0.1 dwarfes x, so we could have ignored auto-ionization and called it 0.1. Similarly, the hydronnium concentration would be different than if you had just divided $K_w$ by 0.1 rather than set up an equilibrium equation--but by so very little.
Now for your problem,
Let me first try to find the $K_b$ for THAM. If the pH of a 0.10 M solution is 10.5, then $\ce{[H^+]}$ $3.16*10^{-11}$. Using the mass balance for THAM, (I'll denote THAM as $\ce{B}$ and the conjugate acid as $\ce{BH^+}$),
$\ce{[B]}$ + $\ce{[BH^{+}]}$ = 0.10 M
$\ce{OH^-}$ = $3.16*10^{-4}$ (assuming 298 K) because  $\ce{H^+ }$ = $3.16*10^{-11}$
If you neglige the autodissociation of water, in other words, that all of the hydroxide in the solution is due to the THAM, your charge balance equation will be that $\ce{[BH^+]}$ is the same as $[OH^-]$ (by stoichiometry). Then, $\ce{[B]}$ = 0.09968, and $K_b$ is $1.00 * 10^{-6}$.
This should tie everything together:
Before, we had said that $[BH^+] = [OH^-]$. Instead, if you took into account the auto-ionization, you get the equation that $[BH^+] + [H^+ ] = [OH^-]$. Why is it justified to ignore $[H^+]$? Because the solution is so basic, $[H^+]$ is so small that if you tried to take auto-ionization into account, the pH would be roughly the same within quite a few decimal places. The change in $[OH^-]$ that comes about due to auto-ionization is dwarfed by the basicity of the THAM addition, just like the 0.100000 M NaOH example that I used. In other words, if you make the solution intensely basic, you can afford to ignore the effects of auto-ionizaton of water, and similarly for acids. In general, it is OK to ignore the auto-ionization process of water if the solution is more than just slightly acidic/basic.
As an interesting note, try to calculate the pH of a $10^{-10}$ M solution of hydrochloric acid without taking into account auto-ionization. You get a pH of 10, which is definitely not right.

I upvoted since this is essentially correct. However there are two checks: $$ [\mathrm{B}] >> [\mathrm{B}^+] $$ so that you don't have to solve quadratic equation and $$ [\mathrm{OH}^{-}] \approx [\mathrm{B}^+] $$ so that you can ignore autoionization of water.

The original question was about the autoionization/protolysis of water.

source: Stack Exchange

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

sort: votes


text:
Arterial blood CO2 content, division and partial pressure
My lecturer cites Boron and Boulpaep's Medical Physiology (2nd edition) for the following claims:

the total $\ce{CO2}$ content of arterial blood is $26.4$ mmol/L or $48\%$ v/v (not sure if these numbers convert)
including $90\%$ as bicarbonate, $5\%$
as dissolved gas and 5% as carbamino compounds (specifically
carbaminohaemoglobin)
the arterial partial pressure of $\ce{CO2}$ is $40$mmHg

How does one get from the arterial partial pressure value to determining these proportions and figures?
What I've tried: Henry solubility constant for $\ce{CO2}$ in water, equilibrium constants for the hydration of $\ce{CO2}$, ionisation of carbonic acid, assumed carbaminohaemoglobin formation went to completion and only consumed $\ce{CO2(aq)}$. Still don't seem to completely match my lecturer's figures.

Even when Douglas equation has been criticized is the best approach I know to calculate CO2 content. You can calculate plasma CO2 content and total blood CO2 content
Douglas 1988 (DOI: 10.1152/jappl.1988.65.1.473)

It's great to have the DOI to a reference, but can you summarize its contents? Otherwise, this doesn't answer the question.

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


text:
Shape of the blood CO2 dissociation curve

What equation determines the shape of the blood $\ce{CO2}$ dissociation curve? I know that since it factors all possible forms of $\ce{CO2}$ carriage, there are a bunch of factors (e.g. Henry's law, formation of bicarbonates and carbaminohaemoglobin, how these are affected by protonation and oxygenation of haemoglobin, etc.) but these don't really seem to explain the non-linear shape. Whereas for oxygen it's fairly straightforward (saturation can be derived using equilibrium kinetics).


text:
Band alignment of heterojunctions
I have seen several band bending diagrams in journals and the web. However, I am not sure as to how they align them. Some diagram appear to align the vacuum level and some align them by Fermi Level. I have found that aligning vacuum levels is known as Anderson's Rules and is applied for semiconductor-semiconductor junctions. However, I do not know what the aligning of Fermi level is referred as. What is this convention called?
Also, what are the rules of thumb on deciding on how to align the bands?

Aligning the Fermi levels is equilibrium. Sadly though, there is no easy rule of thumb - alignments or offsets of the valence, conduction, or both bands are all known.


text:
How to calculate the pH of soft drinks based on titration data?

Your enamel is composed of a calcium hydroxyphosphate ($\ce{Ca5(PO4)3OH}$), also known as mineral apatite. This “apatite” is slowly broken down into calcium phosphate and water as illustrated in the chemical equation below:
  $$\ce{Ca5(PO4)3OH (s, tooth~enamel) + H+ (aq, acid) -> Ca5(PO4)3+ (aq) + H2O (l)}$$
  In this experiment, you are trying to determine the acidity (or the hydrogen ion concentration) of some drinks, therefore, will be using a known concentration of a base to help you in this process. Recall that when acids and bases react, the neutral substance, water, is formed. An example of acetic acid and the base sodium hydroxide is given below:
  $$\ce{CH3COOH  (aq)  + NaOH (aq) -> CH3COONa  (aq)  + H2O (l)}$$
In this experiment you will use a sodium hydroxide solution to complete the titration process. If $\pu{100.0 mL}$ of a $\pu{0.20 M}$ $\ce{NaOH}$ solution is required, determine quantity of solute and solvent as well as a procedure to accomplish this goal of making a solution. Include details such as equipment and specific measurements that must occur. You are testing (lemonade, Gatorade, or ginger ale).

I got that I need $\pu{0.8 g}$ to make the solution.
Then I did the lab and got this data:
\begin{array}{lrrrr}
\text{Drink}
  & \text{Vol. of Drink}
  & \text{Init. Vol. NaOH}
  & \text{Final Vol. NaOH}
  & \text{Vol.-Difference}
\\\hline
\text{Lemonade}   & \pu{25 mL} & \pu{6.0 mL}  & \pu{12.0 mL} & \pu{6.0 mL}\\
\text{Gatorade}   & \pu{25 mL} & \pu{16.5 mL} & \pu{19.5 mL} & \pu{3.0 mL}\\
\text{Ginger Ale} & \pu{25 mL} & \pu{12.0 mL} & \pu{16.5 mL} & \pu{4.5 mL}
\end{array}
With this information I need to answer these questions:

Calculate and record the amount of $\ce{NaOH}$ used for each trial. Write the amount of $\ce{NaOH}$ used for each acid (lemonade, gatorade, and ginger ale).
For each acid calculate the concentration $[\ce{M}]$ for each acid. Show all your work.
Using the molarity of each acid calculated in #2, determine the pH of each solution.

With the volume of both solutions and the concentration of NaOH,  you can calculate the unknown concentration with the following equation:
$$M_{1}V_{1}=M_{2}V_{2}$$
Consider the lemonade trial as an example.
The concentration of your $\ce{NaOH}$ was $0.20~\mathrm{M}$ and in your log, you used a total of $6~\mathrm{mL}$ to reach the endpoint of your titration.
$$\frac{0.20~\mathrm{M}_\ce{NaOH}\times 0.006~\mathrm{L}_\ce{NaOH}}{0.025~\mathrm{L}_\text{Lemomade}} =  {0.048~\mathrm{M}_\text{Lemonade}}$$
By dividing $V_{2}$, which is the volume of your analyte, you determine the unknown concentration. Be sure to convert $\mathrm{mL}$ to $\mathrm{L}$ in any calculations pertaining to molarity. Alternatively, you can convert $\mathrm{mol}$ to $\mathrm{mmol}$ for the same effect.
To calculate the pH of a strong acid, this is easy:
$$-\log [\ce{H+}] = \mathrm{pH}$$
However, we are working with weak acids. Thus, we will need the $K_\mathrm{a}$ in order to determine the actual pH.
$$K_\mathrm{a} =\frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}$$
Continuing with the lemonade experiment, the accepted $K_\mathrm{a}$ of citric acid is $7.4\times{10}^{-4}$. Therefore:
\begin{align}
[\ce{H+}] &= \sqrt\frac{7.4\times10^{-4}}{0.048~\mathrm{M}_\text{Lem.}}\\
[\ce{H+}] &= 0.015~\mathrm{M}
\end{align}
And finally, we take the negative log of the dissociated hydronium to get the pH.
$$-\log{[0.015~\mathrm{M}]}=1.8$$
The pH of the lemonade was 1.8. The accepted equilibrium constants for each of the acids can be found online. The main acid in lemonade is citric acid, so search for the $K_\mathrm{a}$ value of citric acid.

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


text:
Do we use activities or concentrations in the rate expression?
The equilibrium constant of a reaction is a dimensionless quantity due to the fact that we use activities instead of concentrations(in moles/liter).
But, in the rate expression for a reaction, do we still use activities?
If this is so, then isn't the unit of the rate constant always (moles/litre-sec), since these are the dimensions of the rate of a reaction ?

For the rate expression we use concentrations instead of activities. But, the of the rate constant unit is  (moles/litre-sec) only when you have a first order reaction.

The rate expression is always described in terms of molar concentrations. Units of the rate constant will change based upon the overall order of the reaction. This happens such that the rate is always given in M/s.


text:
Why is hydrogen peroxide non-planar?
I had my exam this week and one question asked whether $\ce{H2O2}$ was planar or not. I though that since the oxygen atoms are both sp³ hybridized and have two lone pairs each, the molecule should have a bent shape. I compared it $\ce{H2O}$ which also has the same shape and I knew was planar. But it turns out $\ce{H2O2}$ is not planar. Does anyone know the reason behind this?

There is repulsion of the lone pairs on the adjacent oxygens that cause a barrier to rotation around the single bond.

$\ce{H2O}$ must be planar because three points always make a plane, that isn't a good molecule to use for comparison.

related, if not possible duplicate: http://chemistry.stackexchange.com/q/15754/4945

Think: Why should the the hydrogen atoms stand at a dihedral angle of zero or 180° relative to each other?

Because the O atoms are sp3 hybrid and as a result they adopt a tetrahedral geometry, now the lone lairs also repel each other and also the bond pairs so according to VSEPR theory to minimize the repulsion and attain a confirmation of minimum possible energy, the molecular structure is non planar, although it deceptively looks planar in a flash! The actual structure is like that of an open book

Are you sure it's repulsion and not donation of the lone pairs into $\sigma ^*_{\ce{CH}}$? I know it's not a great acceptor, but it can be done. Repulsion, I would expect, would cause a dihedral angle of 180°, so that would be planar. Think of ethane: Hydrogen there oppose each other. If you went H-C-C-H, repulsion would cause a that to be planar for hydrogens further apart.


text:
Is it possible to efficiently convert hydrogen to methane?
According to this wikipedia page, hydrogen can combine with the carbon found in carbon steel to form some methane inside the steel, weakening it. Is is possible to convert hydrogen to methane by putting it through
graphite powder or activated carbon?
Note that I mainly want the reaction to occur at STP. The flow of the
gas through the graphite will be very slow (at the speed it is generated by electrolysis), so
maybe catalysts not needed?
The purpose of this is to form a more energy-dense fuel than pure hydrogen. Also, it will be much safer since methane is less flammable.

Yes of course, at high pressure and > ~800°C, in a chemical plant, can also produce other hydrocarbons. Ancient technology, Germans did it in large scale in WW2 because they couldn't get at petrol otherwise. "Fischer-Tropsch" process. Don't try to do it at home.

Actually the Fischer-Tropsch process is to produce higher alkanes, alkenes, and alkanols.  Making methane is not FT but is simply referred to as methanation.  It's just the reverse of steam methane reforming.

As an alternative, the Sabatier reaction transforms $\ce{H2}$ and $\ce{CO2}$ into $\ce{CH4}$ and $\ce{H2O}$.

Is it possible to efficiently convert hydrogen to methane?
No. You'll end up putting more energy into the process than you'll get out of burning the methane less the energy of burning the hydrogen that you started from. So although you can make methane from hydrogen, it isn't "efficient."

the op asked if it is possible or not, he didn't ask about efficiency. I am looking for a way to manufacture crude oil from primary elements and your "answer" is not useful at all

Yes. Wood and Wise reported reaction and kinetics between solid graphite and gaseous atomic hydrogen (450-1200°K) to give molecular hydrogen and methane as products, with maximum production near 800°K. They also proposed reaction mechanism.

I'm not talking about atomic hydrogen, rather about molecular hydrogen gas ($\ce{H2}$).


text:
Glass electrode pH measurement
How is the potential difference between the outer and inner surface of the glass bubble in a glass electrode measured if an Ag/AgCl wire is used as the indicator electrode and Ag/AgCl/KCl is used as the reference electrode? In the Nernst equation for the reaction, hydronium ions don't participate.

Any time a semipermeable boundary separates two ionic solutions, the Nernst equations says that a potential difference can develop between these solutions.  Here is some useful info: https://www.electrochem.org/dl/interface/sum/sum04/IF6-04-Pages19-20.pdf

Nice article. I understand how the membrane potential is produced but it is still unclear to me how it is measured by the Ag/AgCl indicator electrode? What change occurs?

None, there is not change

Then how does it work?

You measure an electric potential difference. It can be considered sum of three different effects. If you could isolate them, you will have the electromotive force of both reference electrodes. If they can be isolated, each of them will generate a potential difference at open circuit (when the stationary state is reached) equal in magnitude to the emf. You always will have this difference of potential plus the potential difference at both sides of the membrane. Just add potential differences due to the stationary condition reached because of the emf.


text:
Lattice Energy (enthalpy)
I know that when making a bond the reaction is exothermic (positive)
and breaking a bond the reaction is endothermic (negative).
Lattice enthalpy or lattice formation enthalpy is defined as the enthalpy of formation of 1 mole of an ionic compound from gaseous ions under standard conditions, which is endothermic (negative).
My question is ..
Why lattice formation is negative? It should be positive when making the ionic lattice because we are releasing energy.
Is bond making/breaking different than lattice making/breaking?

You have to consider the system. If energy is being added to the system the sign is positive, if energy leaves the system the sign is negative.

In an endothermic process, energy enters into the system; in an exothermic process, energy leaves the system. You wrote that exothermic means a positive $\ce{\Delta H}$. I don't know if that was a mistake on your part, or if it was a result of your textbook having a different sign convention (do you have an older textbook?).
The formation of 1 mole of an ionic solid from the gaseous ions of its components is always an energy releasing process. When something is imagined to be in the ideal gas phase, it is another way of saying that there are no inter-molecular attractions between the particles. When the gaseous ions become a solid, all of the positive and negative charges interact and form a much more stable compound, thereby releasing potential energy. (Like you said, bonds release energy). Thus, this process is an exothermic reaction by the above definition, releasing energy, and $\ce{\Delta H}$ is always negative.
I think the important thing is for you to first visualize processes in terms of energy entering/leaving the system. Then, try to figure out the signs. This way you won't get confused when some textbooks use a different sign convention (if that is the case with yours), or if different sources define lattice energy (The other definition is the energy needed to break an ionic compound into the gaseous ions, which would be endothermic).


text:
Substitution at thionyl chloride sulfur
What is the correct way of writing the mechanism for substitution at the sulfur atom of thionyl chloride:

Should one write the cleavage of the $\ce{S-O}$ π bond? The argument for the first option is that $\ce{S}$ is already tetrahedral and direct displacement could occur. The argument for the second option is that the π bond will be cleaved first as it is the weakest. Another possibility would be the formation of a tetravalent sulfur species (like the $\mathrm{S_N2}$ substitution at silicon). Is there any experimental or computational evidence regarding the matter?

Both are just as wrong or as right as the Lewis structure of thionyl chloride.

Hah. Nice one. +1

I previously stated in the comments:

Both are just as wrong or as right as the Lewis structure of thionyl chloride.

Indeed thionyl chloride is a nasty little molecule when it comes to the bonding situation. This makes it also incredibly difficult to utilise it in quantum chemical calculations. I have tried to answer a similar question, but eventually gave up due to the complexity of all the involved species.
However, I would like to share a few insights, that might make the whole problem at hand a little bit more comprehensible.
Let us start with the bonding situation in thionyl chloride and its structure. The representation you have used cannot be completely right. As it is drawn, it omits the lone pair at sulfur, implies instead a double bond, and hence suggests that sulfur is trigonal planar coordinated. This is not true.
If one would want to chose a single representation of this molecule it should be a charge separated one, preferably one that indicates the pyramidal structure. A more accurate representation would be as a resonance hybrid. I have performed a calculation on the DF-BP86/def2-SVP level of theory, which I analysed with the Natural Resonance Theory (NRT) of the Natural Bond Orbital (NBO) theory.

What we can already assume from the picture is that the sulfur chloride bonds have significant ionic contribution. This manifests when we look further into the analysis. I include a short version of the analysis, skipping over the core orbitals and truncating minor contributions from polarisation functions. The following is for the major contributor of the bonding situation.
  # (Occupancy)   Bond orbital / Coefficients / Hybrids
 ------------------ Lewis ------------------------------------------------------

  17. (1.99836) LP ( 1) S  1            s( 77.03%)p 0.30( 22.93%)d 0.00(  0.03%)
  18. (1.99504) LP ( 1) O  2            s( 75.06%)p 0.33( 24.93%)d 0.00(  0.01%)
  19. (1.67421) LP ( 2) O  2            s(  2.02%)p48.34( 97.83%)d 0.07(  0.15%)
  20. (1.64705) LP ( 3) O  2            s(  0.00%)p 1.00( 99.85%)d 0.00(  0.15%)
  21. (1.99928) LP ( 1)Cl  3            s( 92.72%)p 0.08(  7.28%)d 0.00(  0.00%)
  22. (1.95239) LP ( 2)Cl  3            s(  1.12%)p88.55( 98.84%)d 0.04(  0.05%)
  23. (1.93382) LP ( 3)Cl  3            s(  0.32%)p99.99( 99.63%)d 0.15(  0.05%)
  24. (1.99928) LP ( 1)Cl  4            s( 92.71%)p 0.08(  7.28%)d 0.00(  0.00%)
  25. (1.95238) LP ( 2)Cl  4            s(  1.11%)p88.67( 98.84%)d 0.04(  0.05%)
  26. (1.93381) LP ( 3)Cl  4            s(  0.32%)p99.99( 99.63%)d 0.15(  0.05%)
  27. (1.99082) BD ( 1) S  1- O  2
               ( 34.00%)   0.5831* S  1 s( 16.81%)p 4.88( 82.01%)d 0.07(  1.18%)
               ( 66.00%)   0.8124* O  2 s( 23.03%)p 3.33( 76.66%)d 0.01(  0.31%)
  28. (1.99354) BD ( 1) S  1-Cl  3
               ( 41.14%)   0.6414* S  1 s(  3.74%)p25.50( 95.33%)d 0.25(  0.93%)
               ( 58.86%)   0.7672*Cl  3 s(  5.84%)p16.06( 93.78%)d 0.06(  0.38%)
  29. (1.99355) BD ( 1) S  1-Cl  4
               ( 41.14%)   0.6414* S  1 s(  3.74%)p25.49( 95.33%)d 0.25(  0.93%)
               ( 58.86%)   0.7672*Cl  4 s(  5.84%)p16.06( 93.78%)d 0.06(  0.38%)
 ---------------- non-Lewis ----------------------------------------------------
  30. (0.07913) BD*( 1) S  1- O  2
               ( 66.00%)   0.8124* S  1 s( 16.81%)p 4.88( 82.01%)d 0.07(  1.18%)
               ( 34.00%)  -0.5831* O  2 s( 23.03%)p 3.33( 76.66%)d 0.01(  0.31%)
  31. (0.36862) BD*( 1) S  1-Cl  3
               ( 58.86%)   0.7672* S  1 s(  3.74%)p25.50( 95.33%)d 0.25(  0.93%)
               ( 41.14%)  -0.6414*Cl  3 s(  5.84%)p16.06( 93.78%)d 0.06(  0.38%)
  32. (0.36856) BD*( 1) S  1-Cl  4
               ( 58.86%)   0.7672* S  1 s(  3.74%)p25.49( 95.33%)d 0.25(  0.93%)
               ( 41.14%)  -0.6414*Cl  4 s(  5.84%)p16.06( 93.78%)d 0.06(  0.38%)

What we learn from this is that all bonds are polarised away from the sulfur, giving the compound significant ionic character. The lone pairs of the oxygen atom are not fully occupied. This will be a bit clearer, once we look at the canonical orbitals. Also very interesting is the fact, that the anti-bonding orbitals (labelled BD*) of the sulfur chloride bonds have non-negligible occupations. This is essentially the electron density missing from the lone pairs of oxygen.
The above table can be nicely put into a picture. You might have to click and enlarge it. The contour value is set to 0.05 a.u. and we still see how massive the lone pair at sulfur is (occupied blue/orange). We can imagine that this electron density effectively could shield a nucleophilic attack at one of the virtual orbitals (red/yellow).

We can further verify the predominantly ionic nature of the bonds by looking at an analysis in terms of the quantum theory of atoms in molecules (QTAIM). Here we can look at the electron density at the bond critical point to compaer their relative strengths and at the value of the Laplacian at that point. If it is positive, then we know it is an area of charge depletion, if it is negative we know that it is an area of charge accumulation. The former we associate with a predominantly ionic nature, while the latter would be found in more covalent compounds. For the three bonds we find the following on the same level of theory as above. The analysis was done with MultiWFN. (Critical points 1-4 are the nuclei.)
#CP    Atoms     El. Density   Laplacian
----------------------------------------
  5    S-Cl      0.106         0.086
  6    S-Cl      0.106         0.086
  7    S-O       0.263         1.251

This nicely agrees with our previous assignments.
Now it's time we look at the canonical orbitals. These are the orbitals wie initially obtain from a calculation. The NBO analysis then transforms them into localised orbitals, which are easier to interpret. Of course this approach destroys the symmetry of such orbitals. For that we have to look a the delocalised pictures.
More specifically, we do not see any contribution of π bonds to the bonding picture. The following picture contains the occupied molecular orbitals starting at number 17 (omitting non-valence orbitals) in the top left corner to the HOMO (No 29) in blue/orange. The last line contains the LUMO (30) and the two orbitals above it (red/yellow).

The first orbital (17) we see, nicely represents the sulfur oxygen σ bond. The next two orbitals give partial bonding to the sulfur chlorine bonds. The fifth orbital is an in-plane π orbital of the sulfur oxygen bond. The seventh orbital can be seen as the out-of-plane π orbital. All other orbitals correspond to linear combinations of lone pairs. What we learn here primarily is that those orbitals do not really have the interpretative value we wish they would have.
The LUMO corresponds the an anti-bonding contribution to the out-of-plane π bond of sulfur and oxygen. This is the orbital where the nucleophilic attack would happen. The LUMO+1 corresponds to the anti-bonding in-plane π orbital and to the anti-bonding configuration of the chlorine sulfur bonds. The LUMO+3 is already the anti-bonding σ sulfur oxygen bond.
From these pictures is seems very likely that a concerted mechanism, direct displacement of chlorine, like the one in your left picture, can not occur.
Your second option is basically already ruled out, since we do not have a π bond of appropriate symmetry.
This leaves the option of addition elimination, a two step mechanism. Or does it?
Well as initially stated calculations of the mechanism, respectively the whole system are incredibly difficult to perform, and even if they would lead to results, I would take with a grain of salt. However, there is a bit evidence, that a tetra-coordinated species could be stable. At least at the DF-BP86/def2-SVP it is a minimum on the potential energy surface. I chose bromine as a nucleophile, because it is fairly simple. This is obviously nowhere near conclusive, it is just a tiny piece in the complete puzzle.

And for final thoughts on the matter I performed a parameter scan of first the bromine sulfur bond and then the sulfur chlorine bond. It appears that on this level of theory the association is barrierless. (Please note that the y-axis has not the same scaling in the graphs.)

But there should always be an alternative thought on the matter. Considering the large ionic contribution of the bonds, one could also assume a dissociation association mechanism, hence creating a strong electrophile, which can possibly be attacked by any nucleophile (even weak ones). The whole process is dependent on many features.
$$\ce{OSCl2 <=>C[-Cl-] [OSCl]+ <=>C[+Nu- ] OSClNu}$$
I was unable to find any literature beyond the phenomenological explanation given in most organic chemistry textbooks. The mechanism of these kind of reactions may remain somewhat of a myth - or an educated guess. Any kind of mechanism that only implies transition states, i.e. electron-pushing or arrow mechanisms, are descriptive at best. Whenever we invoke Lewis structures - even though it works in more than 90% of the times in organic chemistry - we have to question the overall validity of that approach. Lewis structures are easy to understand, but have serious limitations.

Thank you very much for this comprehensive and calculation-supported answer.

Your second drawing would imply a $\ce{S=O}$ double bond, due to available orbitals it would be a d-p π bond. There is considerable debate as to whether these bonds even exist with most of the evidence presented here at chemistry.stackexchange.com pointing towards a ‘no’. That would rule out your second mechanism.
That said, your first mechanism suffers from comparing sulphur with carbon too much. You assume a rather short-lived transition state and direct displacement. I don’t know for sure, but I don’t think that is the case.
Rather, I would propose a two-step mechanism with a medium-lived tetravalent sulphur. In this mechanism, the attack of the nucleophile would turn pseudotetrahedral (one corner of the tetrahedron being a lone pair) into a pseudo-trigonal bipyramid or a see-saw with $\ce{Nu}$ and $\ce{Cl}$ forming a four-electron-three-centre bond. This would then collapse back into a pseudotetrahedral product liberating the better stabilised anion. Close to your first mechanism, but not identical:

'Correct' mechanism is a bit of strange concept here.  Mechanisms are descriptive, not predictive.
They are determined by measuring rates of reaction.  In this case, you have a proposed intermediate of a tetrahedral sulfur (not drawn, but implied) which then eliminates a chloride.  If this elimination is fast as compared to the initial addition you may not be able to determine any different experimentally between the addition-elimination and substitution, because the observed kinetic rates laws would be identical.
That being said conventional wisdom is that this is an addition-elimination mechanism rather than a substitution.


text:
Reaction of hypohalous acids with alkenes
I have read that hypohalous acids react with alkene to give halohydrin. For example the below reaction of hypochlorous acid with 2-methyl-2-butene which I found in Wikipedia.

Now my question is: Since hypochlorous acid is an acid it should give $\ce{H+}$ and $\ce{OCl-}$ ions which it does according to Wikipedia (although it is a weak acid), so shouldn't it add as $\ce{H+}$ and $\ce{OCl-}$ like other hydrogen halides add to alkene? Why would it add as $\ce{HO-}$ and $\ce{Cl+}$?

Let us deconstruct our reagent:
Hypochlorous acids, more commonly, 'bleach', is an oxidant.  Additionally, it is not a great acid, pKa 7.40.
And let us deconstruct our product:
Additionally the chloride is on the least substituted carbon (anti-Markovnikoff) without any indication of added radical source.
In this context it is unwise to view this as an analogous to HCl addition.
Your product could be readily explained by an epoxidation-type reaction followed by a nucleophilic substitution with residual chloride ion.  The regioselectivity of Cl on the less substituted carbon would seem to support that, as well as the absence of a O-Cl bond in the product.

Cyclic chlorinium ion would have OH on the secondary carbon, not the teritiarry

What I am saying is correct as I read it in the "Solomons and Fryhle Organic chemistry". Please follow this link to the snapshot of the concept from the book I mentioned.

I agree that since HOCl is a weak acid it is not correct to compare it with addition of HCl but I am not sure whether it will follow epoxide-type reaction but this link which I found in the duplicate of my question seems to answer my question. Look half way down this link to see the explanation. The reaction goes through a cyclic chlronium ion and then attack by OH- ion, similar to the addition of Cl2 in H2O.

or you can also say that Cl+ ion adds as an electrophile and then OH- ion attacks the more substituted carbon which bears more stable carbocation(Markovnikov's rule), both ways you get the same result.

The link you provided is tangentailly related.  2) Markovnikoff's rule says nothing about carbocations, only substitution.  We infer it is caused by cation stability.  3)  You comment about ~H attacking the more substituted is appropriate, but in the absence of any indication of alkene formation as a likely byproduct, we can't be sure those mechanisms are appropriate.  4) You seem to have some good skills at getting more info, nice.


text:
The mathematical condition for a complete complexation reaction
I'm being asked to find the volume of a $\ce{NH3}$ solution ($0.1\ \mathrm{M}$) to be added to $10\ \mathrm{mL}$ of a $\ce{AgNO3}$ solution ($0.1\ \mathrm{M}$) to assure the 'complete formation of $\ce{[Ag(NH3)2]+}$'. Both constants of formation are given:
\begin{align*}
\beta_1 &= \frac{\ce{[Ag(NH3)+]}}{\ce{[Ag+][NH3]}} =10^{3.3} \\
\beta_2 &= \frac{\ce{[Ag(NH3)2+]}}{\ce{[Ag+][NH3]^2}} =10^{7.2}
\end{align*}
As far as I know, a complete complexation would mean that $[\ce{Ag+}]<10^{-5}$. I tried working it out with the common method of solving those kind of complex equilibria, but the equations I get seem pretty much impossible to solve because of the volume that I have to find, which gets incorporated in them.
\begin{align*}
\dfrac{\beta_2}{\beta_1} &= \dfrac{\ce{[Ag(NH3)2+]}}{\ce{[Ag(NH3)+][NH3]}} \\
\ce{[Ag+]_0} &= \ce{[Ag+]} + \ce{[Ag(NH3)2+]} + \ce{[Ag(NH3)+]} \\
\ce{[NH3]_0} &= \ce{[NH3]} +2\ce{[Ag(NH3)2+]} + \ce{[Ag(NH3)+]}
\end{align*}
$\ce{[Ag+]_0}$ and $\ce{[NH3]_0}$ can be expressed with respect to the volume of the $\ce{NH3}$ solution, but the equations I get are really complicated...
What approximations should I go for besides $\ce{[Ag+]} << (\ce{[Ag(NH3)2+]}+\ce{[Ag(NH3)+]})$?

Their solution is really weird, but much simpler; they invoke this mathematical condition, which makes the problem easily solvable:
$$\dfrac{\ce{[Ag(NH3)2+]}}{\ce{[Ag(NH3)+]}}>10^2$$
Starting from this, I can find $[\ce{NH3}]$ present in the solution, therefore the volume.
Where did it come from?

You seem to be overthinking the problem. The goal would be to have $$\ce{[Ag^{+}] +[Ag(NH3)^{+}]} <<  [\ce{Ag(NH3)2^{+}}]$$ The problem is sloppy with significant figures. Think of having 0.100 M $\ce{AgNO3}$ which would mean that  $$\ce{[Ag+] +[Ag(NH3)^{+}] < 0.001}$$ would be insignificant.

Considering that the concentrations of silver nitrate and ammonia, as well as both constants, are only given to 1 significant figure, I think the "correct" answer is 2E1 mL of NH3 without doing any calculating. ;-)

@MaxW,
In my country, students and teachers haven't really got into the habit of using significant figures; what we do is we consider everything to 3 decimal places. Therefore, a 0.1M solution should be treated as a 0.100M one. ( just noting, their answer is 24.3mL; your answer rounded to one sig fig ;) ) As for your approximation, I applied it, but I still can't get a solvable system of equations; all I'm left with is 2 unknowns ( the volume and $\ce{[NH3]}$ ) in a single equation

I'll work it out later tonight if someone doesn't beat me too it. Sorry my suggestion didn't help. I was just shooting from the hip without really working the problem out. // You may have to consider the fraction of silver in each of the three forms. It'll be messy but doable...

Ok, to start let's just check the given answer of 24.3 mL.
The total solution is 34.3 ml (10 mL Ag + 24.3 ml NH3).
We'll assume that all 0.001 moles of silver are as species $\ce{[Ag(NH3)2^+]}$ so:
$\ce{[Ag(NH3)2^+]} \approx (\dfrac{10}{34.3})0.100 = 0.0291 $
$\ce{[NH3]_{initial} =}(\dfrac{24.3}{34.3})0.100 = 0.0708$
We'll assume $\ce{NH4^+}$ and $\ce{[Ag(NH3)^+]}$ are both negligible so:
$\ce{[NH3] =} 0.0708 - 2(0.0291) = 0.0126$
Now from the equation
$\dfrac{\beta_2}{\beta_1}= 10^{3.9} = 7.94 \times 10^3 = \dfrac{\ce{[Ag(NH3)2+]}}{\ce{[Ag(NH3)+][NH3]}}$
$\ce{[Ag(NH3)^+] =} \dfrac{[\ce{Ag(NH3)2^+}]}{(7.94 \times 10^3)\ce{[NH3]}} = \dfrac{0.0291}{(7.94 \times 10^3)(0.0126)} = 2.91 \times 10^{-4}$
$\ce{[Ag^+] =} \dfrac{[\ce{Ag(NH3)^+}]}{(2.00E3)\ce{[NH3]}} = \dfrac{2.91 \times 10^{-4}}{(2.00E3)(0.0126)} = 1.15 \times 10^{-5}$
fraction of total Ag as $\ce{Ag(NH3)^+}$
$\dfrac{2.91 \times 10^{-4}}{0.0291} = 0.01 $
fraction of total Ag as $\ce{Ag^+}$
$\dfrac{1.15 \times 10^{-5}}{0.0291} = 3.97  \times 10^{-4}  $
Limits on ratios
Given the two equations:
$\mathrm{K}_1= 10^{3.3} = \frac{\ce{[Ag(NH3)+]}}{\ce{[Ag+][NH3]}}$
$\mathrm{K}_2 = 10^{7.2} = \frac{\ce{[Ag(NH3)2+]}}{\ce{[Ag+][NH3]^2}}$
then we can derive a third
$\mathrm{K}_3 = 10^{7.2 - 3.3} = 10^{3.9} = \frac{\ce{[Ag(NH3)2+]}}{\ce{[Ag(NH3)+][NH3]}}$
It is possible to solve for some lower limits on the ratios assuming that the $\ce{[NH3]}$ can at most be 0.1 M. So:
$\frac{\ce{[Ag+]}}{\ce{[Ag(NH3)+]}} \geqslant \frac{1}{\mathrm{K}_1\ce{[NH3]}} = 5.01 \times 10^{-3}$
$\frac{\ce{[Ag+]}}{\ce{[Ag(NH3)2^+]}} \geqslant \frac{1}{\mathrm{K}_2\ce{[NH3]^2}} = 6.31 \times 10^{-6}$
$\frac{\ce{[Ag(NH3)+]}}{\ce{[Ag(NH3)2^+]}} \geqslant \frac{1}{\mathrm{K}_3\ce{[NH3]}} = 1.26 \times 10^{-3}$
So the absolute upper limit on the % of the silver as $\ce{[Ag(NH3)2^+]}$ is 99.874%.
Quick and dirty calculations...
To check this I brute forced some calcs in a spreadsheet.
The calculations below ignores the reactions below (alkali solutions do not produce appreciable amounts of silver hydroxide due to the favorable energetics for the reaction to the oxide):
$$ \ce{NH3 + H2O <-> NH4^+ + OH^{-}}$$
$$\ce{Ag^+ + OH^{-} <-> AgOH(s)}$$
$$\ce{AgOH(s) -> Ag2O(s) + H2O} \text{ (pK = 2.875)}$$
 NH3    [Ag(NH3)2+] [NH3]tot    [NH3]       [Ag(NH3)+]  [Ag+]       mol(Ag)     F*
 20.1   3.32E-02    6.68E-02    3.32E-04    1.26E-02    1.90E-02    1.950E-03   4.87E-01
 20.2   3.31E-02    6.69E-02    6.62E-04    6.30E-03    4.75E-03    1.334E-03   2.50E-01
 20.3   3.30E-02    6.70E-02    9.90E-04    4.20E-03    2.12E-03    1.191E-03   1.61E-01
 20.4   3.29E-02    6.71E-02    1.32E-03    3.15E-03    1.20E-03    1.132E-03   1.17E-01
 20.5   3.28E-02    6.72E-02    1.64E-03    2.52E-03    7.68E-04    1.100E-03   9.11E-02
 20.6   3.27E-02    6.73E-02    1.96E-03    2.10E-03    5.35E-04    1.081E-03   7.46E-02
 20.7   3.26E-02    6.74E-02    2.28E-03    1.80E-03    3.95E-04    1.067E-03   6.31E-02
 20.8   3.25E-02    6.75E-02    2.60E-03    1.57E-03    3.03E-04    1.058E-03   5.47E-02
 20.9   3.24E-02    6.76E-02    2.91E-03    1.40E-03    2.40E-04    1.051E-03   4.82E-02
 21.0   3.23E-02    6.77E-02    3.23E-03    1.26E-03    1.95E-04    1.045E-03   4.31E-02
 21.1   3.22E-02    6.78E-02    3.54E-03    1.14E-03    1.62E-04    1.041E-03   3.91E-02
 21.2   3.21E-02    6.79E-02    3.85E-03    1.05E-03    1.36E-04    1.037E-03   3.57E-02
 21.3   3.19E-02    6.81E-02    4.15E-03    9.69E-04    1.17E-04    1.034E-03   3.29E-02
 21.4   3.18E-02    6.82E-02    4.46E-03    9.00E-04    1.01E-04    1.031E-03   3.05E-02
 21.5   3.17E-02    6.83E-02    4.76E-03    8.40E-04    8.82E-05    1.029E-03   2.84E-02
 21.6   3.16E-02    6.84E-02    5.06E-03    7.87E-04    7.77E-05    1.027E-03   2.66E-02
 21.7   3.15E-02    6.85E-02    5.36E-03    7.41E-04    6.91E-05    1.026E-03   2.50E-02
 21.8   3.14E-02    6.86E-02    5.66E-03    7.00E-04    6.18E-05    1.024E-03   2.36E-02
 21.9   3.13E-02    6.87E-02    5.96E-03    6.63E-04    5.56E-05    1.023E-03   2.24E-02
 22.0   3.13E-02    6.88E-02    6.25E-03    6.30E-04    5.04E-05    1.022E-03   2.13E-02
 22.1   3.12E-02    6.88E-02    6.54E-03    6.00E-04    4.58E-05    1.021E-03   2.03E-02
 22.2   3.11E-02    6.89E-02    6.83E-03    5.72E-04    4.19E-05    1.020E-03   1.94E-02
 22.3   3.10E-02    6.90E-02    7.12E-03    5.48E-04    3.85E-05    1.019E-03   1.86E-02
 22.4   3.09E-02    6.91E-02    7.41E-03    5.25E-04    3.54E-05    1.018E-03   1.78E-02
 22.5   3.08E-02    6.92E-02    7.69E-03    5.04E-04    3.27E-05    1.017E-03   1.71E-02
 22.6   3.07E-02    6.93E-02    7.98E-03    4.84E-04    3.04E-05    1.017E-03   1.65E-02
 22.7   3.06E-02    6.94E-02    8.26E-03    4.66E-04    2.82E-05    1.016E-03   1.59E-02
 22.8   3.05E-02    6.95E-02    8.54E-03    4.50E-04    2.63E-05    1.016E-03   1.54E-02
 22.9   3.04E-02    6.96E-02    8.81E-03    4.34E-04    2.46E-05    1.015E-03   1.49E-02
 23.0   3.03E-02    6.97E-02    9.09E-03    4.20E-04    2.31E-05    1.015E-03   1.44E-02
 23.1   3.02E-02    6.98E-02    9.37E-03    4.06E-04    2.17E-05    1.014E-03   1.40E-02
 23.2   3.01E-02    6.99E-02    9.64E-03    3.94E-04    2.04E-05    1.014E-03   1.36E-02
 23.3   3.00E-02    7.00E-02    9.91E-03    3.82E-04    1.93E-05    1.013E-03   1.32E-02
 23.4   2.99E-02    7.01E-02    1.02E-02    3.70E-04    1.82E-05    1.013E-03   1.28E-02
 23.5   2.99E-02    7.01E-02    1.04E-02    3.60E-04    1.72E-05    1.013E-03   1.25E-02
 23.6   2.98E-02    7.02E-02    1.07E-02    3.50E-04    1.63E-05    1.012E-03   1.22E-02
 23.7   2.97E-02    7.03E-02    1.10E-02    3.40E-04    1.55E-05    1.012E-03   1.19E-02
 23.8   2.96E-02    7.04E-02    1.12E-02    3.31E-04    1.47E-05    1.012E-03   1.16E-02
 23.9   2.95E-02    7.05E-02    1.15E-02    3.23E-04    1.40E-05    1.011E-03   1.13E-02
 24.0   2.94E-02    7.06E-02    1.18E-02    3.15E-04    1.34E-05    1.011E-03   1.10E-02
 24.1   2.93E-02    7.07E-02    1.20E-02    3.07E-04    1.28E-05    1.011E-03   1.08E-02
 24.2   2.92E-02    7.08E-02    1.23E-02    3.00E-04    1.22E-05    1.011E-03   1.06E-02
 24.3   2.92E-02    7.08E-02    1.25E-02    2.93E-04    1.17E-05    1.010E-03   1.03E-02
 24.4   2.91E-02    7.09E-02    1.28E-02    2.86E-04    1.12E-05    1.010E-03   1.01E-02
 24.5   2.90E-02    7.10E-02    1.30E-02    2.80E-04    1.07E-05    1.010E-03   9.93E-03
 24.6   2.89E-02    7.11E-02    1.33E-02    2.74E-04    1.03E-05    1.010E-03   9.73E-03
 24.7   2.88E-02    7.12E-02    1.35E-02    2.68E-04    9.89E-06    1.010E-03   9.55E-03
 24.8   2.87E-02    7.13E-02    1.38E-02    2.62E-04    9.51E-06    1.009E-03   9.37E-03
 24.9   2.87E-02    7.13E-02    1.40E-02    2.57E-04    9.15E-06    1.009E-03   9.20E-03
 25.0   2.86E-02    7.14E-02    1.43E-02    2.52E-04    8.82E-06    1.009E-03   9.04E-03

(1) In the table the column [Ag(NH3)2+] assumes that the 1 millimole of silver is all $\ce{Ag(NH3)_2^+}$. The value is expressed as a concentration so it falls due to the dilution of the extra $\ce{NH3}$. In the column mol(Ag) all the silver species are totaled. The value should be 1.00E-03 moles or 1 mM Ag. Values significantly above that indicate that the calculations are in error. (Note that the way the calculations are done the mM Ag will always be too high.)
(2) The column [NH3]tot is the concentration of all NH3 (mMol NH3)/(total volume)
(3) The column [NH3] is the "free" ammonia. It assumes ammonia is either free $\ce{NH3}$ or $\ce{Ag(NH3)_2^+}$ and thus it ignores the single amine complex $\ce{Ag(NH3)^+}$. It calculates free ammonia as [NH3]tot - 2*$[\ce{Ag(NH3)_2^+}]$
(4) The column labeled F* is the fraction of silver as $\ce{Ag(NH3)^+}$ and $\ce{Ag^+}$. Thus the real decision is what fraction is low enough since the fraction continues to fall as excess ammonia is added.
Conclusion
The choice of the "end point" is arbitrary...
mL        end point
----      -----------------------------------
21.4      97% [Ag(NH3)2^+]
21.6      [Ag^+] 10% of [Ag(NH3)^+]
22.1      98% [Ag(NH3)2^+]
24.3      [Ag(NH3)^+] 1% of [Ag(NH3)2^+]
24.4      99% [Ag(NH3)2^+]
24.6      [Ag^+] = 1*10^-5

Now that I fixed the formulas the numbers make more sense. I had thought with the old calculations that the diamine complex was forming too fast and I didn't stop and try to figure out why my gut was telling me that the calculations were wrong.

Why would they choose such an arbitrary condition without specifying it? I think we're missing something...

@MaW, I'm writing this in case it might be of interest to you or someone else: after exploring other problems and asking some professors of Analytical Chemistry, it seems that there's a quite common arbitrary condition hanging around in our country's analytical problems, which states that: 'a reaction is considered complete when 99.9% of the reactants go into products.' What's interesting is that the author of our problem had decided to change this condition without any reason whatsoever. (I doubt there's even a practical reason for these conditions)

Let $V_0 = 10~\mathrm{ml}$ be the initial volume of $\ce{AgNO3}$, $c_{\ce{AgNO3}} = 0.1~\mathrm{mol\, l^{-1}}$ its concentration, and $c_{\ce{NH3}} = 0.1~\mathrm{mol\, l^{-1}}$ the concentration of the $\ce{NH3}$.
If complete complexation means that $\ce{[Ag+]}<10^{-5}~\mathrm{mol\, l^{-1}}$ we could use it as a target value, i.e.
$$\gamma_1=\frac{\ce{[Ag(NH3)+]}}{\ce{[NH3]}}=\beta_1 \ce{[Ag+]} = 10^{3.3}\cdot10^{-5} = 10^{-1.7} \tag{1}$$
and
$$\gamma_2=\frac{\ce{[Ag(NH3)_2+]}}{\ce{[NH3]^2}}=\beta_2 \ce{[Ag+]} = 10^{7.2}\cdot10^{-5} \,\mathrm{mol^{-1}\, l} = 10^{2.2}\,\mathrm{mol^{-1}\, l}\tag{2}$$
Let $x$ be the number of moles of $\ce{Ag(NH3)+}$, $y$ the number of moles of $\ce{Ag(NH3)2+}$, and $z$ the number of moles of free $\ce{NH3}$ in the solution.
With this we get for the number of moles of silver (neglecting free $\ce{Ag+}$)
$$c_{\ce{AgNO3}} \cdot V_0 = 10^{-3}~\mathrm{mol} \approx x + y \tag{3}$$
and for the number of moles of free and bound $\ce{NH3}$
$$x + 2y + z. \tag{4}$$
The complete volume of the solution is given as
$$V = V_0 + \frac{x + 2y + z}{c_{NH_3}} \tag{5}$$
With this we can write equations $(1)$ and $(2)$ as
$$\gamma_1 = \frac{x}{z} = 10^{-1.7} \tag{6}$$
and
$$\gamma_2 = \frac{y}{z^2} \cdot V = \frac{y}{z^2} \cdot (V_0 + \frac{x + 2y + z}{c_{NH_3}}) = 10^{2.2}\,\mathrm{mol^{-1}\, l} \tag{7}$$
$(3)$, $(6)$ and $(7)$ are tree equations with three unknowns which we can solve.

You made some mistakes. Let $\ce{V_{Ag}}$ be the volume of silver solution used which is 10 mL. Let $\ce{V_{NH3}}$ be the volume of ammonia. In equation (3) the millimoles of $\ce{Ag^+}$ is $\ce{(10 + V_{NH3})(1E-5)}$.  // The OP's problem is that you're pulling the assumption that $[\ce{Ag^+}] = 1 \times 10^{-5}$ out of thin air. Why shouldn't it be $2 \times 10^{-5}$? // Doing the calc you should get 24.6 mL from my table...

@MaxW, That's the assumption that I've noticed to be tackled in most problems with precipitation reactions; which says that if the concentration of the cation or anion becomes $10^5$ smaller, the reaction is complete.

Their argument for such value was that most analysis apparatus wouldn't be able to detect a lower concentration, which seems reasonable (Unless we're dealing with concentrations lower than 0.01 M)

@L3ul - I'll just point out that $10^5$ smaller than $0.1$ is $10^{-6}$ which is the point. You're using some arbitrary choice for the simplification of the problem. // I don't think you can solve the problem without some simplification like that. After fixing all my math mistakes I need to look at that again.

@MaxW: $\ce{[Ag+]} < 10^{-5}$ is the given condition for completeness. Equation (3) gives the total amount of silver which is constant summed over all silver species. If you do the calculation the result is indeed 24.6 ml.

@aventurin - You missed my point. You defined $x$ and $y$ as moles above equation(3). Above equation (1)  you defined $10^{-5}$ as the concentration of Ag+ as mole per liter. You can't add moles and moles per liter.

That's true. Corrected it.


text:
Basicity of heterocyclic compounds
Pyrrol, furan or thiophene don't have any bond pair electrons free to release which is why they shouldn't be basic, but the lecturer of organic chemistry is saying that they are basic as they react with hydrochloric acid to form salts. Which one of the two explanations are correct?

Furan and thiophene are going to be even less basic than their saturated counterparts, which can act as Lewis bases but aren't particularly well known for acting as Brønsted bases (very strong acids can significantly protonate ethers and sulfides, but not weak ones). They do, however, have a free $\mathrm{sp^2}$ lone pair with which they can bind a proton, but only one, unlike the saturated analogues, and an $\mathrm{sp^2}$ lone pair is going to be even less basic than a an $\mathrm{sp^3}$ lone pair because the electrons are closer to the nucleus (i.e. $\mathrm{sp^2}$ is "smaller" and closer to the nucleus than $\mathrm{sp^3}$).
Pyrrole, however, might be expected to be basic, but it also is not, as you rightly expected. This is due to delocalization of the nitrogen lone pair to create aromaticity. The $\mathrm pK_\mathrm a$ of the conjugate acid of pyrrole is $-3.8$, which is lower than that of even sulfuric acid, suggesting pyrrole is a very weak base.
All of those heterocycles are less basic than $\ce{Cl^-}$ or $\ce{H2O}$, so either your lecturer is incorrect or has been misunderstood.

pKa of HCl is even more negative. Also source for  -3.8? Another thing is pKa can be used as measure of "protonating strength"  only for diluted solutions.


text:
Why is the effect of competitive antagonist less than the effect of non-competetive antagonists?
Drugs can  be provided as a competive or non-competitive form of agonists. But the effect of non-competitive antagonists is much more than the other, so what is the cause of that? And are there some exceptions?

Personally, I think the statement "the effect of a competitive antagonist is less than the effect of a non-competitive antagonist" is too general to have much meaning. There's a large number of systems, each with a potentially large number of competitive and non-competitive antagonists, each with a potentially large range of activities. Saying a priori that a non-competitive antagonist is more effective than a competitive one is presumptuous.
However, there is a bit of sense in there, for a particular defined sense of circumstances.
By definition, a competitive antagonist is only active against the non-liganded state of the receptor. The native ligand (agonist) and the competitive antagonist compete for the receptor. In contrast, a non-competitive antagonist can generally work against both the liganded (with agonist bound) and ligand-free versions of the receptor. (How active it is against each depends on the particulars about the antagonist and receptor.)
So in a situation where the antagonist isn't at saturating concentrations (say at a concentration near it's IC50), saturating amounts of native ligand (agonist) will completely reverse the effects of a competitive antagonist, as the agonist will compete off the antagonist. However, the non-competitive antagonist will still have some remaining inhibitory function, even at saturating concentrations of native ligand, as it can bind to and inhibit the ligand-bound (agonist-bound) form of the receptor.
However, there's an apples-to-oranges comparison embedded in there. That comparison assumes that the competitive and non-competitive are each near their respective IC50s. This value can change greatly between drugs, so you can have a competitive inhibitor with a really low IC50, and a non-competitive inhibitor with a really high one, such that if you used identical molar concentrations of each antagonist, the competitive antagonist would be successful in almost completely shutting off receptor activity, whereas the non-competitive antagonist would show barely noticeable activity.

You said 'a non-competitive antagonist can generally work against both the liganded and ligand-free versions of the receptor' Do you mean by liganded version, the agonist binding site? Because in this movie at min. 3:43 is said the opposite: https://www.youtube.com/watch?v=7mJut-vVZRs

@Marijn By "liganded" I meant "agonist-bound form". As your video indicates from 3:50-4:30, the non-competitive antagonist can be bound to (and be inhibitory towards) the receptor both in the presence and absence of agonist binding.


text:
Toxic levels of baking soda
What is the amount of baking soda that would be considered toxic for skin? I ask because there is a lot of misinformation online about using baking soda as a "natural" underarm deodorant, and how it is or is not harmful to use as such. Please clarify.
Thank you for your time in advance.

I wouldn't recommend labelling it "misinformation" until you actually know it to be false.

Here is a material safety data sheet (MSDS) for sodium bicarbonate. http://www.sciencelab.com/msds.php?msdsId=9927258 It is about as harmless as a chemical can be. // I'm unsure about how it would be used as a deodorant. The grain size of typical grocery store purchased baking soda would be abrasive to the skin. // I don't know if you're reacting to the fact that baking soda is a "chemical" or not. Water is a chemical, and oxygen in the air is a chemical. It is true that a strawberry isn't "a" chemical, it is made up of thousands of chemicals.

I used soda bicarbonate as an underarm deodorant for about three years. Not every day but fairly consistently. At a certain stage I noticed aching shoulders, and I stopped it's use. I used a brand that had "aluminium free" printed on the label, which suggests some producers mix aluminium into their soda bicarbonate. Poor soda bicarbonate can't fight in the heavy-weight ring without some additives. When we talk about soda bicarbonate, what other additives are we talking about and blaming on sweet little soda. Without a full spectrum microscope what do I know?

Baking soda is sodium hydrogencarbonate, $\ce{NaHCO3}$, also known as sodium bicarbonate.
The only toxicological data in the GESTIS database on hazardous substances
 is a LD50 of 4220 mg/kg for the oral uptake by rats. There are no data for dermal uptake.
In water, $\ce{NaHCO3}$ is weakly alkaline. I do not know whether it might cause skin irritations, particularly in more sensitive regions, such as armpits.
In occupational safety, I'd recommend to avoid skin contact and use personal protection equipment when handling it.

I'll expand Klaus's answer a bit. By "occupational safety" he means like if you're working in a plant where sodium bicarbonate is being made. In that situation there  would likely be a significant amount of sodium bicarbonate dust floating around. // Taking 1/2 teaspoon from a box to make a cake in a kitchen is safe under ordinary use, and doesn't require any more care than handling salt, sugar or cocoa powder.

Baking soda is used in many kitchens for baking purposes.  Do you bake anything?  If so do you always wear PPE?  Occupational safety people often recommend wearing PPE for everything all the time no matter what, perhaps because their primary concern isn't the safety or convenience of the people they oversee, it is protecting the liability of the firms they work for in the case of lawsuits.

@CurtF. Yes, I do cook and bake - without gloves or a mask ;-) Nevertheless, occupational dermatitis is among the top list of work-related health problems. I agree that occupational safety people who "recommend PPE for everthing all the time" do a lousy job! PPE is the last resort, when other measures (substitution, technical measures, etc.) can not be applied.

@CurtF - I agree with you to a point. I worked for IBM in a chemical environment making chips. I remember commercials on TV showing a lady with makeup and a dress cleaning an oven with spray lye cleaners. She at least did have rubber gloves. On the IBM production line you'd been in a full hastmat suit to do such a thing. But lawsuits are a powerful motivator. So here is a case of the golden rule. Him with the gold makes the rules. So if you wanted to work on the IBM production line, you followed IBM's safety rules. The rules protected both IBM and the workers.


text:
Conductivity of Wüstite
"Wüstite is often represented as $\ce{Fe_{1-x}O}$, where $x$ is some small fraction less than unity. In this nonstoichiometric $\ce{Fe_{1-x}O}$ material
conduction is electronic and the material behaves as a p-type semiconductor, i.e. the $\ce{Fe^3+}$ ions act as electron acceptors, generating holes.
Determine the electrical conductivity of a specimen of wüsite that has a hole mobility of $\pu{3.1E-5 m2 V-1 s-1}$ and for which the value of $x$ is 0.030. Assume that the acceptor states are
saturated, i.e. one hole exists for every $\ce{Fe^3+}$ ion. Wüsite has a sodium chloride crystal structure with a unit cell edge length of 0.437 nm."
My Solution:
4 units of $\ce{FeO}$ for rocksalt structure but for every $\ce{Fe^2+}$ vacancy $2$ $\ce{Fe^3+}$ are needed for electroneutrality, this implies $0.003:1\rightarrow0.24:4$
$$\mathrm{mass}_{\ce{Fe(III)}}=((0.24)(55.85))/((N_a)(10^3))$$
$$=\pu{2.225838592E-26 kg}$$
$$\rho_{\ce{Fe(III)}}=2.225838592\times10^{-26}/(4.37\times10^{-10})^3$$
$$=\pu{266.7 kg m-3}$$
Conductance for a p-type semiconductor is given as: $$\sigma=p(q)(\mu_{h})$$
$$\implies\sigma=(266.7)(1.6\times10^{-19})(3.1\times10^{-5})$$
$$\approx1.3\times10^{-21}\:(\Omega.m)^{-1}$$
I have the feeling this is wrong so any help would be appreciated.

after revisiting the problem I realised where I went wrong. The main mistake with not being careful with units and not reading the qurstion and having incorrect values for hole mobility.
$$\text{mass}_{\ce{Fe(III)}}=((0.24)(55.85))/((N_\mathrm{a})(10^3))$$
$$=2.225838592\times10^{-26}\: \mathrm{kg}$$
$$\rho_{\ce{Fe(III)}}=2.225838592\times10^{-26}/(4.37\times10^{-10})^3$$
$$=\mathrm{266.7\: kg\: m^{-3}}$$
$$\text{No. of holes}=\rho_{\ce{Fe(III)}}/1\:\text{AMU}_\ce{{Fe(III)}}$$
$$\Rightarrow \text{No. of holes}=266.7/(9.296\times10^{-26})$$
$$=2.868975904\times\pu{10^{27}\:holes/m^3}$$
Conductance for a p-type semiconductor is given as: $$\sigma=p(q)(\mu_{h})$$
$$\implies\sigma=(2.868975904\times10^{27})(1.6\times10^{-19})(3.1\times10^{-5})$$
$$\approx1.4\times10^{4}\:(\Omega.m)^{-1}$$

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


text:
Reaction of aluminum sulfate in water
I dissolved aluminum sulfate in DI water and expected to see just a clear water solution with aluminum sulfate dissolved as ions. However, a precipitate was visible. I addd only 0.6g of Aluminum sulfate so there is no way that the precipitated is saturated aluminum sulfate. Furthermore, it does not seen to be aluminum hydroxide, so then what is the precipitate? Aluminum sulfate has the property of filtering out impurities in water, however, I am using DI water so then what is the precipitate?
Also, the precipitate is not aluminum hydroxide because aluminum hydroxide is dissolvable in sulfuric acid; however, when I added sulfuric acid to the precipitate solution, the precipitate did not disappear.

I'll say your sulfate is not what you think it is.

what does that mean?

It's something else. I think you proved that quite convincingly.

i bought the aluminum sulfate from sigma aldrich, so I am pretty sure its aluminum sulfate

Alternatively your water is not what you think it is.

Deionized water contains not much ions, but the ionexchange columns or filters used tend nevertheless to slowly get contaminated with bacteria or other germs. So they should be regular exchanged. Now Al$_2$(SO$_4$)$_3$ has the property to denaturate proteins, this is why it is used also for leather tanning occasionally. So my guess is that its some protein precipitate oringinating from biologically impure water.

According to Wikipedia, "When dissolved in a large amount of neutral or slightly alkaline water, aluminium sulfate produces a gelatinous precipitate of aluminium hydroxide, Al(OH)3."

The IUPAC spelling is aluminium (pg 248)

Try dissolving the precipitate properly! Did you separate and dry the ppt? Or like try removing significant amount of supernatant, it should be Al(OH)3 ....why don't you perform a lake test to confirm the hydroxide?


text:
Why did my lemon and penny battery fail?
I followed instructions in Popular Mechanics publication for making a battery using lemons and pennies but it didn't work. I have a few ideas on why it might not have worked and would like to know which of them are more or less reasonable.
Here is what I did:

I used sandpaper to remove copper from one side of pennies to reveal the zinc core. Supposedly, only pennies minted after 1982 have a zinc core so I made sure to use only such pennies. The instructions call for 5-10 pennies - I did 6.
Deviations from instructions:

I did not thoroughly remove all traces of copper from 2 of the 6 pennies. The amount of copper that was still visible covered less than 1% of the surface.
I used a lower grade sandpaper than the instructions called for.

I cut cardboard discs slightly smaller than the pennies in equal number then soaked them in lemon juice
Deviations from instructions:

the cardboard discs I cut separated into three layers, the outer two layers appeared to be very tightly pressed and minimally absorbent

I made a stack of alternating pennies and lemon-soaked cardboard discs with the copper-side of each penny facing down. On the top of the stack I placed an intact, non-sanded penny then wrapped electrical tape around the stack.
Deviations from instructions:

none that I am aware of

I attached an LED to the top and bottom on the penny stack. I tried both combinations of where to put the positive and negative ends but neither caused the LED to light up.

Why I think it Failed:

By not removing every last trace of copper from the sanded-side of the pennies, I created a kind of short-circuit that discouraged the charges from moving through the stack of pennies.
Since the cardboard was not very absorbent, it might have been particularly porous either and this might have prevented sufficiently free flow of particles between each penny in the stack.
By using a relatively low number of pennies, I guess I set it up to have a relatively low voltage. Maybe I could have overcome problems 1 or 2 if I used more pennies.

I want to know if I my assessment of what went wrong is reasonable and welcome any suggestions of things I might be overlooking.

Here is a link to such a construction... http://www.popsci.com/article/diy/make-led-holiday-light-out-pennies

By not removing every last trace of copper from the sanded-side of the pennies, I created a kind of short-circuit that discouraged the charges from moving through the stack of pennies.

This is the main problem. Until all of the copper is gone from the zinc side of each penny, there is little reason for the two halves of the electrochemical reaction between zinc and copper not to just take place locally.  Your battery likely had one or more ionic short-circuits, just as you say, on these incompletely sanded surfaces.

Since the cardboard was not very absorbent, it might have been particularly porous either and this might have prevented sufficiently free flow of particles between each penny in the stack.

This is unlikely.  As long as the cardboard was soaked through with the lemon juice, it should function fine as an ion-transport medium.

By using a relatively low number of pennies, I guess I set it up to have a relatively low voltage. Maybe I could have overcome problems 1 or 2 if I used more pennies.

This is also unlikely. With five junctions (one between each neighboring pair of pennies), at a theoretical $1.1~\mathrm V$ per junction your stack voltage should have been at least about $5~\mathrm V$. This should be plenty to light a simple LED.

A easy problem would be to have the LED backwards.

@MaxW OP indicated they tried the LED with both polarities (#4 in "Here is what I did").

The other thing that occurs to me is that the edge of the stack of pennies should be dry. So the whole stack shouldn't be wet with lemon juice oozing out. // I was trying to think of a better paper than ordinary cardboard. I think the cardboard from a paper towel roll, or a toilet paper roll would work well.

The ideal of the cardboard from paper rolls was to avoid corrugated cardboard which would probably soak up too much lemon juice and ooze over the edges when you squeezed it down.

@MaxW Just want to make sure I understand: the stack of pennies should be dry for the sake reason every last trace of copper must be sanded off - to avoid letting positive charges migrate to the sides of the pennies. Correct me if I'm wrong or if there's more to it than charge particles moving through the lemon juice.

@Slavatron - I think you understand. The lemon juice should just be on the cardboard between the pennies. You don't want the lemon juice oozing out of the cardboard onto the edges of the pennies.


text:
In an SN1 reaction, 1-bromohexane reacts with NaOH, but would the Na or H be the nucleophile that substitutes the Bromine?
The answer says that the Bromine would be replaced with OH and NaBr would be formed, along with CH3(CH2)4CH2OH, but why? Does it have to do with stability? And if so, how does the OH replacing Br make the product more stable? Thanks.

1-bromohexane does not SN1

NaOH doesn't do SN1

SN1 is very less likely to be followed in 1-bromohexane, rather a SN2 pathway will be followed. OH- replaces Br as Br- as Br is a good leaving group. This has nothing to do with product stability. The haloalkane substrate is reactive so it reacts, as simple as that.


text:
What is the formally correct notation of t-Bu, i-Pr, p-NO2, n-Bu, a-HCl, etc.?
In chemical notation, can I safely assume that the notation of

a single lower-case latin character
followed by a hyphen
followed by an Element upper-case latin character (and possibly more characters)

should always be typeset as

italic lower-case character
hyphen  (i.e. never is a bond)
element following characters

This question is regarding the further improvement of mhchem.
I know that not all of them stand for the same concept, but I assume they share the same typographical representation. Beware, I am no chemist. And because I don't know the name of these notations, I could not find them in the colored IUPAC books.
The notation is quite rare. There were only 21 usages in 43k \ce calls extracted from chemistry.SE:
(p-NO2Ph)_2CO
t-BuO-
i-Pr
t-Bu
R=i-Pr
n-BuNH2
(c-Hex)2BCl

a-HCl  // should this be \alpha?

1s-H  // what is this?

I migrated it to the main site, it will have more traction here and it is a solid question. Notation is something many people struggle with, so having these questions here is very good.

Is there a link for the a-HCl one? That looks really weird imo

If the a-HCl is from here then it's out if context. That was also another abuse of mhchem as it is easier to write ce than mathrm ;) . The question was about pKa values for HF to HI. :)

Note that, in the given examples, “Bu” (butyl) and “Pr” (propyl) are not chemical elements.

Regarding the "1s-H" it's the same abuse of mhchem as an easy version of mathrm as with "a-HCl".

Wait, mhchem is not a chemist? My world just shattered …

tl;dr: That's probably a good approach.
So, there are at least three different things here. None of these abbreviations are standardised as far as I'm aware[1], so whatever is done is going to be just as non-standard as any other approach.

Specific shorthands for alkyl functional groups: t-Bu, i-Pr, n-Bu, c-Hex
I've seen the method you've suggested used for these, and also the first three with superscripted initial characters, e.g. $^{\textrm{t}}\textrm{Bu}$ and $^{\textrm{i}}\textrm{Pr}$, and I'd never seen the last one until I read this question. I've also seen them written simply as tBu, iPr, and nBu. They're abbreviations for tert-butyl, isopropyl, n-butyl (straight-chain), and cyclo-hexyl respectively. (Here are links for IUPAC References for non-cyclic and cyclic alkanes.)
Substitution points on a benzene ring: (p-NO2Ph)_2CO
This example denotes a para- substituted phenyl group (confusingly, benzyl refers to a $\ce{C6H5-CH2 -}$ group). The other options here are ortho- and meta-, which I would personally not find it confusing or uncommon to see abbreviated to o- and m- respectively, and $\ce{{\it p}-NO2Ph}$ seems a perfectly reasonable interpretation of what the person has written here.
Electronic configuration labels: 1s-H
This seems like using an atomic orbital label (1s) as a hyphenated specifier for the hydrogen: I don't think it's a standard form of writing that at all, but the method you propose would presumably italicise the s, which would be fine -- the atomic orbital labels like this are often italicised.
Whatever the a-HCl is: I don't know.
I couldn't find anything that looked appropriate for this one.

Most importantly, I can't think of any counter-examples where that wouldn't be a sensible thing to do.

[1] Edit: I defer to @Loong's superior mastery of the ways of IUPAC in their answer.

My answer does not really invalidate your statement. The abbreviations for organic groups may be common, but they are actually not standardized (as you can see, two relevant documents give different recommendations). The author of the IUPAC Recommendations disparagingly calls them “nicknames”.

In the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), single italic lower-case Latin letters are used in some stereodescriptors (such as ‘c’, ‘m’, ‘p’, ‘r’, ‘s’, and ‘t’), in descriptors for fusion sites (e.g. dibenzo[c,g]phenanthrene), in locants (‘o’, ‘m’, and ‘p’ in place of ortho, meta, and para, respectively; however, this usage is strongly discouraged), and in other structural descriptors (e.g. s-indacene).
However, single lower-case Latin letters that are used for numbering are not italicized (e.g. naphthalene-4a,8a-diol).
All these symbols are typically used in names (which are usually not typeset using MathJax/mhchem) and do not necessarily appear in formulas.
Furthermore, common abbreviations for organic structure fragments may be used in formulas, but not in text. According to the Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008), the permissible abbreviations include:

isopropyl: iPr
isobutyl: iBu
sec-butyl: s-Bu
tert-butyl: t-Bu

However, the ACS Style Guide recommends different abbreviations:

isopropyl: i-Pr
isobutyl: i-Bu
sec-butyl: sec-Bu
tert-butyl: t-Bu

Furthermore, authors may create their own abbreviations, but any abbreviations not included in the recommendations shall be defined clearly when they are used.
Also note (since the question mentions letters followed by an element) that several structural abbreviations in current usage are identical to an element symbol:

Ac (acetyl, actinium)
Ar (aryl, argon)
Cm (carboxymethyl, curium)
Nb (p-nitrobenzyl, niobium)
Np (p-nitrophenyl, neptunium)
Pr (propyl, praseodymium)

Do you know of any counter-example to my proposed formatting rule?

"Graphical Representation ..." notations are on (printed) page 406. It cannot be found with a full-text search.

@mhchem 1. see Table II on page 406 (page 130 of the pdf); 2. I haven’t found any counter-examples yet.

I'd like to point out that they are NOT standard notation. t-Bu stands for tert-butyl, i-Pr stands for isopropyl. These are just short hand abbreviations, mostly used in organic chemistry. They are not even abbreviations of standard, systematic names. Isopropyl, for example is a common name.
If you start going down the road of accepting short-hand organic notations, "slangs", you should be prepared for things like TMEDA, py, TsOH, and similar notations too. I suggest you better put a restriction on your input, as I don't see an easy way to standardize the slang.

If everyone uses this nomenclature and everyone understands it. Isn't it kind of standard even if it is not written in any important IUPAC book? Also what would TMEDA, py and TsOH change for mhchem?

It is well-understood, yes. It might be standard, yes. The thing I am trying to say is that they are not official symbols, so they are subjected to nomrul

...upps, pressed enter. So they are not subject to any rules whatsoever, one might write them however one pleases. For example, I have no idea what a-HCl is, even tho I am a sort of experienced chemist. The problem is, you can not constructnamset of formatting rules, as there is no rules at all.


text:
Studying Chemistry
I know this is a silly statement to say, but I really want to excel in Chemistry. The thing is, I forgot numerous things from elementary Chemistry such as introduction to basic stoichiometry.
Do you know any valuable study sites and books that I can study from beginning to advanced Chemistry?
Thanks.

Please open up a new question instead if you have one. Do not edit this question. It will invalidate community decision and the answers.

Crash course science, which is run by Hank Green (renowned author John Green's brother) has proven quite useful to me, and I was using it up to the end of first year university. The only issue is that he talks incredibly fast, but you get used to it after a while.
https://www.youtube.com/watch?v=FSyAehMdpyI&list=PL8dPuuaLjXtPHzzYuWy6fYEaX9mQQ8oGr

In additional to the Crash Course YouTube channel as mentioned above, I would also recommend the chemistry videos submitted by Khan Academy. They're nicely detailed and very pertinent to first-year chemistry in college.


text:
Electrolysis of concentrated NaCl
I conducted an experiment on electrolysis of concentrated solution of NaCl. Hoffman Electrolysis Apparatus was used for this experiment. I observed a formation of gas bubbles on both anode and cathode. It was expected to obtain hydrogen gas on negative electrode (cathode) and chlorine gas on positive electrode (anode). I also hoped to test these gases: (1) as hydrogen should ‘pop’ with a lighted splint, and chlorine should bleach damp litmus paper. I was unable to demonstrate these tests for gases, as I suppose there was not enough gas formed for test to go efficiently. I wonder, if someone knows what apparatus is better for this goal. I used 9V battery for this electrolysis experiment.
Even though I had no direct proof of hydrogen formation, solution near anode smelled as a chlorine gas, and smell was very strong. I am not sure why wet blue litmus paper didn't get bleached by this chlorine. Blue litmus paper got bleached very fast in another experiment (not related to electrolysis) where chlorine was released, within 30 sec of exposure.
The most interesting part of experiment was formation of brownish chemical species on anode. At times, I didn't see any chlorine bubbles at all, but I could see formation of this brown chemical species. At first, I thought it is an iodine. I tested it with starch and found, that it was not iodine. Today I tested it with alkene and alkane, and I found it was a bromine (decoloration of bromine water was observed with alkene, but not alkane). Yes, I know, that brine is main source of bromine, but is there any way I can buy NaCl without bromine, just for demonstration purposes?

What materials are you using for the electrode?

Platinum electrodes, apparatus is this one: http://www.amazon.com/gp/product/B005QDUSMQ/ref=s9_dcbhz_bw_g328_i3_sh No matter, how platinum they are, they are eaten in a course of this electrolysis chemical reaction, and I am not sure, if a real platinum thing would corrode that fast. I'd rather imagine, it would not corrode that fast (I did experiment 5 times, and before that this set was new), but I have no prior experience with that to be sure about it.

Platinum is quite inert; it is unlikely that such electrodes would corrode if they were pure. (You can try carbon or graphite.) I think your estimates for what gases are produced are correct; collect gases inside an inverted test tube filled with electrolyte. The upside-down test tube should be in a bath of electrolyte. Then, you can use the gases. Note that your chlorine might be mixed with oxygen in some cases.

I don't mind to change set up of apparatus, in fact I ordered two more different pieces of equipment for that and I plan to try them next week, upon delivery. One of this pieces of equipment is with carbon electrodes actually: http://www.amazon.com/Electrolysis-Apparatus-Contained-Water-Device/dp/B00ICSHWTE/ref=pd_sim_sbs_328_1?ie=UTF8&dpID=41FyFGEe18L&dpSrc=sims&preST=AC_UL160_SR132%2C160&refRID=0YRHHMEYMBFSXN4TETFP *

Another one would allow test tubes to be immersed in electrolyte: http://www.amazon.com/gp/product/B00FGDR6US/ref=s9_simh_gw_g328_i1_r?ie=UTF8&fpl=fresh&pf_rd_m=ATVPDKIKX0DER&pf_rd_s=desktop-1&pf_rd_r=1MC9ZSAYH650EEGTWVRZ&pf_rd_t=36701&pf_rd_p=2437869742&pf_rd_i=desktop So I will be able to test both your advices soon. Thank you!


text:
Percolation network analysis with 1H MAS solid state NMR
Assume you have a composite rubber that was produced in an extrusion process and it consists of a basic polymer with some additives.
One of the additives does not get
mixed with the basic resin but forms a percolation network in the resulting resin. Is it possible to get some quantitative information
on this percolation network using 1H MAS solid state NMR spectroscopy?
What if the percolating additive is very similar to the basic resin?
If 1H MAS solid state NMR spectroscopy is in no way suitable for such an analysis, what other methods can you think of to get some
information like this?

I don't really understand the type of quantitative information you are interested to have. Is it related to morphology and microstructure of the final composite material? In this case, SEM and HR-TEM are poweful tools to do that. Is it related to the functional characteristic of the composite? In this case a  good way to examine percolation is by plotting the logarithm of the transport variable of interest (electrical conductivity, thermal conductivity) as a function of the logarithm of the  volume fraction of filler...

The percolation point of the additive has been determined by measurement of resistance before. One further question could be, if there is a way of reducing the percolation point, and a tool that is available to find an answer to such a question is the described NMR spectroscopy.

I see, you want at the molecular level understand the interaction between the filler and the host in order to lower the percolation threshold!

Exactly, thanks for helping me to specify my problem!

Solid-state (MAS) NMR can most likely help you understand more about the interaction between the percolating additive and the resin. If you have not yet carried out solid-state NMR analysis, it can be a useful complement to information that you have determined by other methods. However, it is hard to be 100% sure that it would answer your exact question in terms of guiding you on how to modify the percolation threshold.
Should you do $\ce{^1H}$ MAS solid-state NMR?
Unlike in solution-state NMR, $\ce{^1H}$ MAS solid-state NMR is not yet a routine procedure and has much fewer applications (compared to the widespread use of $\ce{^1H}$ in solution-state NMR) until fairly recently (within the last 5-10 years).
$\ce{^1H}$ has a high natural abundance and high sensitivity. However, unlike in solution-state NMR where the free tumbling leads to averaging of such anisotropic interactions, in solid-state, anisotropic interactions are retained. Thus, the high abundance and sensitivity of $\ce{^1H}$ lead to a dense network of anisotropic dipolar interactions in the solid-state, meaning that a "straight-forward" $\ce{^1H}$ spectrum would yield a broad unresolved hump rather than resolved, informative (sharp) signals.
Thus, the approach taken more commonly is to detect heavy atoms with better inherent chemical shift resolution, such as $\ce{^13C}$. Techniques such as cross polarisation (CP) to tackle the problems of lower sensitivity with heavy atoms have been around for decades and can now be considered routine in solid-state NMR. So, if you may want to consider $\ce{^13C}$ CP-MAS solid-state NMR instead of $\ce{^1H}$ MAS solid-state NMR.
What access do you have to $\ce{^1H}$ MAS solid-state NMR facility and expertise?
$\ce{^1H}$ MAS solid-state NMR has undergone some dramatic technological improvements in the past few years. While it is still not routine, it is now possible to acquire (more) resolved $\ce{^1H}$ solid-state NMR spectra. However, highly specialised hardware is required.
The hardware required is a high magnetic field (700 MHz to 1 GHz) and very high MAS rates (at least 40 kHz but ideally 60-100+ kHz). Both of these types of hardware have been available commercially since about 2009. However, there are not many labs that I know of that have both of these pieces of hardware, and nearly all are national or pan-EU facilities (and unfortunately the ones I can think of nearly are nearly all focussed on structural biology). It is also not straightforward to operate these hardware, so you will need to interest these labs in a collaboration.
There are some labs that are trying to do $\ce{^1H}$ MAS solid-state NMR with lower magnetic fields and/or MAS rates by developing specialist pulse sequences to carry out the $\ce{^1H}$ homonuclear decoupling required. It is not an area of ssNMR that I am very familiar with though I believe that there are people doing this in Israel and India.
Generally, the groups doing $\ce{^1H}$ MAS ssNMR are not going to be doing it like a "service". You cannot just give someone a sample and expect them to give you a spectrum. If you can find a group that specialises in $\ce{^1H}$ MAS ssNMR to collaborate with, then it is worth doing $\ce{^1H}$. Otherwise, if you find a ssNMR group with other interests, you may be able to ask them to carry out a (relatively routine) $\ce{^13C}$ CP-MAS spectrum for you.
Sample amounts and types
I know you did not directly ask about this, but I feel it may be helpful to give an idea of how much and what types of sample you will likely need.
For very fast MAS spinning and $\ce{^1H}$ detection, you need only a small amount of sample. Depending on the density of your polymer it is likely that no more than 10 mg is required. (My estimates are based on biopolymers - mostly carbons and protons, so adjust as appropriate for your polymer.)
For heavy atom detection, more sample is usually advised because that will help with the lower sensitivity. Around 50-100 mg would be a good amount.
It will also be helpful to you (and your collaborator) if you can provide the monomer of the resin, and the pure additive (as monomer if necessary). It can be helpful if you have several rubbers that are more simple "models" of what you are trying to investigate (known to have one particular type of polymer links, etc).
...and what information can you expect to obtain?
In the best case scenario, you will have a fairly resolved spectrum of your polymer sample, and by comparing the spectrum of your monomers and your polymer you can assign nearly all the signals. By looking at which signals have shifted/broadened the most, you can determine the atoms which are interacting (or have become bonded). Knowing which atoms are interacting, you can then have some guidance on what functional groups may allow you on to adjust the percolation point as you wish.
But! If you are used to looking at solution-state NMR spectra, solid-state NMR spectra can look incredibly broad. For polymers, the resolution is often poor and signals are fairly broad (even by solid-state NMR standards). In my experience (detecting $\ce{^13C}$) by CP on a 400 MHz spectrometer with a MAS rate of 10 kHz), the linewidths of polymers I have seen are around 150-300 Hz.
The resolution is a very sample-dependent parameter, and is one of the main reasons why I cannot answer definitely whether ssNMR will help you or not. If, as you said, the additive and the polymer are very similar, their signals may well overlap or be insufficiently resolved for you to determine if there has been any changes.
Moreover, polymers are often heterogeneous. This means that one signal in the monomer may broaden and split into multiple signals in the polymer. If this happens at too many atoms, it will make it difficult to assign your polymer spectrum.
There are ways, again, to overcome these problems. It is possible to employ multidimensional ssNMR spectroscopy detecting $\ce{^1H}$ and/or heavy atoms, leading to heteronuclear correlations or homonuclear double-quantum colrreations. The additional dimension often improves resolution and therefore helps with assignment. However, sensitivity will most likely suffer. Alternatively, the molecular dynamics within the polymer can give insight on where the interaction may be occurring in cases where you have sufficient resolution to assign but it is unclear where the interaction has occurred from observing the chemical shift alone. These approaches are again going into non-routine territory where you will need to find a collaborator rather than having it as a "service".
Here is a review if you are interested in some examples of recent multidimensional $\ce{^1H}$ MAS ssNMR work on polymers. Here is a more dated paper which shows how dynamics can help in identifying the nature of interactions at a rubber-filler interface.

That is a very elaborate and thorough answer! Thank you very much!

You are welcome! I have answered this question a number of times (in real life) so I think there is value to have an answer online somewhere. Let you know if you do decide to go ahead with ssNMR analysis and need any more help.

Great answer, but high speed MAS probes are also available for fields well below 700MHz.


text:
Getting the wrong pH value, according to data and online calculator
I'm doing a titration problem with a week acid and strong base, I used experiment data and an online calculator to check my answer, but my pH I calculated is a bit off. Can anyone see where I messed up? I can't figure out where...
AcOH
0.0984M
0.02500L
0.00246mol
Ka 1.8x10^-5
NaOH
0.096M
0.02580L
0.00025mol
After rxn:
0mol AcOH left
0.0025mol - 0.00246mol = 4.0x10^-5 mol OH excess
0.00246mol AcO-
4.0x10^-5mol OH / 0.0508 L total soln = 7.87x10^-4 M OH-
-log([OH-]) = pOH = 3.104
pH = 10.90
The pH should be 10.52 according to my data and the online calculator.

As you can see from the point that $n_\ce{OH} > n_\ce{AcOH}$, there will be an excess of hydroxide ions. What you want to calculate is the pH for this amount of hydroxide ions within the total volume at this point.
Thus the calculation looks like:
\begin{align}
[\ce{OH-}] &= \frac{n_\ce{NaOH}-n_\ce{AcOH}}{V_\text{a} + V_\text{b}}\\
\Leftrightarrow \mathrm{pOH} &= -\lg ([\ce{OH-}])\\
\Leftrightarrow \mathrm{pH} &= 14+\lg ([\ce{OH-}])
\end{align}
Your error is based on a rounding error ... please don't round intermediate result as all rounding errors will sum up. You just need to round the final results to a proper amount of decimal places.
The correct calculation:
$$\mathrm{pH} = 14 + \lg \left(\frac{0.0024768 - 0.00246}{0.025 + 0.0258}\right) = 10.5194 \approx 10.52$$
vs. your calculation:
$$\mathrm{pH} = 14 + \lg \left(\frac{0.0025 - 0.00246}{0.025 + 0.0258}\right) = 10.8962 \approx 10.90$$


text:
Friedel-Crafts alkylation of five-membered heterocycles
The five-membered heterocycles pyrrole, furan, and thiophene undergo Friedel-Crafts acylation, but I have not seen them undergoing a simple Friedel-Crafts alkylation reaction with a Lewis acid such as aluminium chloride. Why is that so?

Alkylation of the five-membered rings is perfectly possible - nobody says it cannot happen. But there are two problems:

Alkylation further activates the ring, making it more reactive - this is exactly the same problem that makes alkylation an unattractive reaction in benzene rings
The Lewis acid can, and will, lead to polymerisation of electron-rich rings such as the five-membered heterocycles.

With acylation, the first point is moot, and if you choose an appropriate catalyst you can avoid polymerisation. So acylation can be used in certain cases (I'm fairly sure that simply chucking in an acyl chloride with $\ce{AlCl3}$ will not be a clean reaction at all though).
From Joule & Mills, Heterocyclic Chemistry 5th ed., p 300 (chapter on pyrrole):

Mono-C-alkylation of pyrroles cannot be achieved by direct reaction with simple alkyl halides, either alone or with a Lewis-acid catalyst, for example pyrrole does not react with methyl iodide below 100 °C; above about 150 °C, a series of reactions occurs leading to a complex mixture made up mostly of polymeric material together with some poly-methylated pyrroles. The more reactive allyl bromide reacts with pyrrole at room temperature, but mixtures of mono- to tetra-allyl-pyrroles together with oligomers and polymers are obtained.

p 327 (chapter on thiophene):

Alkylation occurs readily, but is rarely of preparative use, an exception being the efficient 2,5-bis-t-butylation of thiophene. (J. Org. Chem., 1984, 49, 4161. DOI:10.1021/jo00196a012)

Presumably the steric bulk of the t-butyl group prevents further reactions.
p 350 (chapter on furan):

Traditional Friedel–Crafts alkylation is not generally practicable in the furan series, partly because of catalyst-induced polymerisation and partly because of polyalkylation.

In addition to your reference, mono- and di-alkylation of thiophene has been successfully achieved using different olefins previously (Alkylation of Thiophene with Olefins).


text:
Does SiCl2 exist and its possible structure
It's known that the elements in IVA all have valence +2 and +4. I also know there exists $\ce{GeCl2}$ and $\ce{SnCl2}$. So my question is does $\ce{SiCl2}$, or even $\ce{CCl2}$ exist? If yes, what about their molecular structure?
In addition, what makes +2 valence in $\ce{Ge}, \ce{Sn}$ more stable?

Via Wikipedia, $\ce{CCl2}$, where carbon is in the +2 oxidation state, does exist but it is fleeting due to its high reactivity.
I was able to find a reference on the analogous dichlorosilylene, and I also found some information on difluorosilylene, also known as silicon difluoride, both of which have silicon in the +2 state and react similarly to a carbene.
So yes, both $\ce{CCl2}$ and $\ce{SiCl2}$ exist where the central atom in the +2 oxidation state.
Tin(II) chloride fill its octet by way of either  bridging chloride bonds in the crystal structure or hydration bonds in water (see Wikipedia images). The structure of solid $\ce{GeCl2}$ isn't given, but I would expect something similar. I'll update if I find anything.


text:
Conversion of carboxylic acid to alcohol
Is it feasible to convert carboxylic acids to alcohols? I know the reverse is done during the oxidation of alcohols (e.g. in the production of vinegar), but is it possible to do it the other way round?
For example: the conversion of methanoic/formic acid ($\ce{HCOOH}$) to methanol ($\ce{CH3OH}$)?

use LiAlH4 followed by hydrolysis,or convert acid to ester then use Ni/H2 or Pd/H2 or pt/H2 or B2H6

Directly using $\ce{LiAlH4}$ works, but borane is probably a better choice:

Borane ($\ce{BH3}$) is particularly good for carboxyl groups and permits selective reduction of them in the presence of many other groups (although the reaction with double bonds takes place at about the same rate in ether solvents). For many years, borane was the reagent of choice for this reduction. Borane also reduces carboxylic acid salts. (March 7ed, p 1515)

Clayden 2ed, pp 531-2 also discusses the borane reduction. Here's a small portion of it:


text:
mechanism in Birch reduction
at first I wondered why 2 H are added to opposite side of benzene like reaction below. because I could imagine 2 H added right next to each other or 1 C away from each other.

so, looking for explanation I saw a molecular orbital of benzene.

It's one of LUMO from benzene and from this I just guessed that maybe hydrogen could be added to 2 unpaired p orbitals in the opposite direction.
but when I learned another Birch reduction I got confused again.

since the E level of LUMO above would get high, I assumed another LUMO maybe come into play.

so from MO right above, I can under stand e- from Na would go to ortho- or meta- position of substituted group.
And here's my question. why does hydrogen go to ortho position first?
additionally, why second Hydrogen goes to the opposite side of first-added hydrogen? and I'd really appreciated any correction about my educated guess above.

It has nothing to do with the shapes of the LUMOs. You've put in a lot of thought, which is good, but no actual research... https://en.wikipedia.org/wiki/Birch_reduction Nearly every website on the reaction will explain why that regioselectivity is observed. Moral of the story: Ask the Internet or your library before trying to figure it out yourself from scratch.

Thank you T_T. I wish I could ask everything I don't know about. Anyway i'll try the source you mentioned!

I've looked it over but i can't seem to find what causes regioselectivity;;


text:
Production of acetone from egg shells and vinegar
I learned from these wiki pages that acetone can be produced from calcium acetate, which can be produced by the reaction of eggshells and vinegar. However, I couldn't find the required temperature. Would it be possible to produce acetone vapours by simply heating calcium acetate over a natural gas flame (~1500 degrees C)?

Did you look in footnotes 3 and 4 for Wikipedia's Calcium Acetate article? https://en.wikipedia.org/wiki/Calcium_acetate#Uses

There was a question about eggshells which seems to have been deleted. Evidently there is a problem with disposing of eggshells and much of the waste eggshell ends up in a dump. This reaction would yield acetone and CaO, but both probably aren't worth the cost of the acetic acid and heat.

I know it's impractical, but I still want to try it because I'm rather interested.

Sorry, didn't mean the cost analysis to be a put down. I thought it was an interesting notion given the eggshell problem. Who knew that getting rid of eggshells would be a problem? The bigger problem with eggshells of course is that there are probably numerous plants generating them. It would cost prohibitive to ship all eggshells to one plant in US for conversion. So you'd need conversion plants near eggshell producers. Interesting problem...

Calcium acetate decomposes starting at around 160 degrees Celsius, according to Wikipedia. So yes, it is definitely possible to do so in a gas flame.


text:
Hydrogen Peroxide - Iron Nail Galvanic Cell?
I am planning to create a simple galvanic cell made up of common household materials.
Suppose I have hydrogen peroxide (to serve as cathode), diluted in water, in one beaker, and I have a bunch of iron nails (to serve as anode), also diluted in water, in another beaker. I put a graphite electrode (in the form of a pencil) in the $\ce{H2O2}$ solution to serve as an inert electrode. Then I'll have a U-tube containing $\ce{NaCl}$ to serve as a salt bridge between the two beakers. I will then connect the two electrodes using alligator clips into an LED. Computing from their reduction potentials, I am supposed to get $2.21~\mathrm{V}$ under standard conditions.
Will the set-up work? If so, do you think it is possible to connect many set-ups in series in order to produce $9$ Volts? I will appreciate all sorts of suggestions.

N.B. household hydrogen peroxide is ~3%, with ~97% water. Greater concentrations can be quite dangerous, while 3% solutions may not produce a noticeable difference in EMF from water.