text:
What can be used as an electrolyte in water to perform electrolysis other than salt?
I am looking for an electrolyte to use in water to perform electrolysis. It can't be salt, as salt ($\ce{NaCl}$) would cause the chemical reaction of
$$\ce{2H2O + 2NaCl -> Cl2 + H2 + 2NaOH}$$
or
$$\ce{H2O + NaCl -> HCl + NaOH}$$
Which both make products with poisonous chemicals (ex: $\ce{NaOH}$ = Sodium Hydroxide (aka lye) which is poisonous). I am looking for a chemical I can use to create the pure forms of hydrogen and oxygen from the electrolysis process. I understand how electrolysis works, and have tried it before, but am looking for something that could be used as a non-poisonous electrolyte. I read that baking soda could be used, would that produce what I am looking for?

Can't you run electrolysis through pure water?  I know it's not as conductive, but I though you adjust for that with voltage and electrode distance.

I believe that if your concentration of sodium chloride is low, it will release oxygen instead chlorine

You can actually just use salt. It is because that the chemical reaction on the top is for one condition which is that your halide solution ($\ce{NaCl}$) has a very high concentration.
If you halide solution has lower concentration,the hydrogen ion and hydroxide ion in the water will discharged and the two half equation will be
$\ce{4H^+ + 4e^- -> 2H_2}$
$\ce{4OH^- -> 2H_2O + O_2 + 4e^-}$
So, the overall equation is
$\ce{4H^+ + 4OH^- -> 2H_2 + 2H_2O + O_2}$
Thus, is okay to use salt.
Besides Salt
You can also use hydrochloric acid besides salt. Just remember that don't use a high concentration hydrochloric acid as the anode will just discharge the chloride ion instead of hydroxide ion.

I assure you that your worries are without basis. For the hydrolysis of water, the half reactions are:
$$\ce{2H2O_{(l)} ->O2_{(g)} +4H+_{(aq)} +4e-}~~~~\varepsilon^0=-1.23~\mathrm{V}$$
$$\ce{4H+_{(aq)} +4e- ->2H2_{(g)}}~~~~\varepsilon^0=0~\mathrm{V}~\mathrm{(by~definitoin)}$$
The $\varepsilon^0$ of the reaction would be:
$$\ce{2H2O_{(aq)} ->2H2_{(g)} +O2_{(g)}}~~~~\varepsilon^0=-1.23~\mathrm{V}$$
With the proposed half reactions your are suggesting might take place:
$$\ce{2H2O_{(l)} +2e- ->H2_{(g)} +2OH-_{(aq)}}~~~~\varepsilon^0=-0.83~\mathrm{V}$$
$$\ce{2Cl-_{(aq)} ->Cl2_{(g)} + 2e-}~~~~\varepsilon^0=-1.36~\mathrm{V}$$
The $\varepsilon^0$ of the reaction would be:
$$\ce{2H2O_{(l)} + 2Cl-_{(aq)} -> Cl2_{(g)} + H2_{(g)} + 2OH-_{(aq)}}~~~~\varepsilon^0=-2.19~\mathrm{V}$$
(Note that the $\ce{Na+}$ ions are present in the solution, though it is not listed in the reaction)
Because the hydrolysis of water requires less energy than the production of $\ce{Cl2}$, $\ce{H2}$, and $\ce{NaOH}$ from the ions present in the solution, the second reaction will not occur.
The reaction $\ce{H2O_{(l)} + Cl^{-}_{(l)} -> HCl_{(g)} + OH-_{(aq)}}$ would not occur. $\ce{HCl}$ could not evolve from the solution, as there would be no mechanism forcing the $\ce{H+}$ ions onto the electronically stable $\ce{Cl-}$ ions. Rather, it would be much more likely that the $\ce{H+}$ ions would re-react with the $\ce{OH-}$ ions to reform water, meaning no net reaction.
In short, go right ahead using $\ce{NaCl}$ in your hydrolysis reactions. It's actually probably one of the safest salts you could use!

I'd point out that you also have to control the voltage too. With sufficient over voltage you could get chlorine gas with the oxygen. See Wikipedia section https://en.wikipedia.org/wiki/Electrolysis_of_water#Electrolyte_selection for more information.

Actually, you do produce $\ce{Cl2}$. This is an industrial process that was used to produce chlorine gas.

There is an overpotential for the oxygen gas-producing reaction, so at high chloride concentrations, chlorine gas is produced

Mainly anything that contains ions (i.e. electrolytes) is  great to use in the electrolysis of water. Some examples of electrolytes you can use in electrolysis are baking soda (sodium bicarbonate), squirted lemon, hell, I bet you could use Gatorade since it contains electrolytes. Hope this answer helped, I've been researching the electrolysis of water for like 2 weeks now, and I know it can be a little frustrating to understand.


text:
What is dust made of?
I was cleaning my blinds today, and wanted to know what the primary components of dust are. I know that it is made of microorganisms and other particles, but I do not want to guess that it will be $\ce{N}$ or $\ce{Si}$.
I can guess that other materials outside our house might have more than one component for dust. But what about inside the house?

Interestingly, NIST has several dust standards, such as Urban Dust, Trace Elements in Indoor Dust and Organic Contaminants in House Dust. NIST is pretty much the best source for material analyses you can get, but it seems all the certificates are only for trace substances/elements.

Wikipedia says, "Dust in homes, offices, and other human environments contains small amounts of plant pollen, human and animal hairs, textile fibers, paper fibers, minerals from outdoor soil, human skin cells, burnt meteorite particles, and many other materials which may be found in the local environment."

The dust in the house mostly contains sand, dirt and dead skin cells.
First, let's start with the easiest, sand. The chemical formulae of sand is $\ce{SiO2}$, so it contains silicon and oxygen.
Next, dirt has rock, sand, and clay. Sand is made of silicon and oxygen. Rock is mostly made up of graphite (carbon) and some minerals like phosphorus and silicon. Clay is made up of silicon and oxygen and some minor minerals like aluminium.
Lastly, dead skin, like other organic compounds, contains carbon, hydrogen and oxygen, maybe some nitrogen as it may have protein in it.
So basically, dust is made up of carbon, hydrogen, oxygen, nitrogen, silicon, phosphorus and some minor minerals.

Sand? Do you have a source for that? I don't think it is a particularly common component of house dust, which is what Asker seems to be talking about. Of course, the composition of dust varies wildly with where it is collected, since it is composed of any particle sufficiently small to remain temporarily aloft by air currents. Also, rocks are mostly made out of silicate minerals.

@NicolauSakerNeto Well, the content of the dust will obviously be highly dependent on environmental conditions. For people living near beaches or deserts, there will definetly be a considerable amount of sand.

While there are many rocks that have some graphite in them, I'm not sure I would say that rock is mostly made up of it, certainly not over silicates.

There is no standard dust
The short answer is that dust is composed of particles small enough to be easily lofted by air currents.
The chemical composition varies dramatically from place to place; anyone who says "dust consists of x, y, and z" means "I took a sample, and it consisted of x, y, and z."
For example, the main component of office dust is simply paper; it is a very small component elsewhere. Some sources claim skin cells are a large component; most say they failed to detect more than traces.
In natural environments, soil particles and finely crushed, dried vegetable matter are often major constituents; in artificial environments, both are hugely outnumbered by clothing fibres.

Dead skin, like other organic compounds, contains carbon, hydrogen and oxygen, maybe some nitrogen as it may have protein in it. So basically, dust is made up of carbon, hydrogen, oxygen, nitrogen, silicon, phosphorus and some minor minerals (Assan Reza chems. Hons.preparing civil service).

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text:
ammonium tartrate + lead tartrate=?
From Googlebooks First Outlines of a Dictionary of Solubilities of Chemical Substances  By Frank Humphreys Storer:

I find that lead tartrate is dissolved by tartrate of ammonia.

What is the final product?

The addition of ammonium tartrate turns lead tartrate into a water-soluble tartrato complex:
\[\ce{Pb(tart) + tart^{2-} <=>> [Pb(tart)2]^2-}\]

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text:
Is an iron (II) sulphate solution supposed to turn cloudy when silver nitrate is added?
$\ce{AgNO3}$ is not suppose to cause a precipitation when mixed with $\ce{FeSO4}$, but the acidified solution turns slightly cloudy when $\ce{AgNO3}$ is added.
Is this suppose to happen? solutions are all prepared fresh. Any reason as to why? This also happens when silver nitrate is added to magnesium sulphate.
Picture is as attached, the left one has the $\ce{AgNO3}$ added (labelled Ag) while the one on the right has $\ce{Ba(NO3)2}$ added to show the comparison

photo added. any clue as to why this happened? chemicals are also newly bought. expiring in 2016, only been on the shelf for less than 3 months.

According to this resource for solubility rules, $\ce{BaSO4}$ is insoluble and $\ce{Ag2SO4}$ is slightly insoluble, which would explain the the cloudiness of each solution, as well as why the $\ce{BaSO4}$ solution is cloudier than the $\ce{Ag2SO4}$ solution.
According to this table, $\ce{BaSO4}$ has a $K_{\mathrm{sp}}$ of $1.1 \times 10^{-10}$. Comparably, $\ce{Ag2SO4}$ has a $K_\mathrm{sp}$ of $1.2 \times 10^{-5}$, which is considerably higher than $\ce{BaSO4}$.
If you are not aware the $K_{\mathrm{sp}}$ is a mathematical measurement describing the solubility of a salt at a certain temperature (usually $25^{\mathrm{o}} \mathrm{C}$), and is found by multiplying the concentration of the salt's dissolved ions, each raised to the power of their respective stoichiometric coefficients. The higher the $K_{\mathrm{sp}}$, the more soluble a salt is. For example, the $K_{\mathrm{sp}}$ for $\ce{Ag2SO4}$ would be given by the expression: $K_{\mathrm{sp}}=[\ce{Ag+}]^2\cdot[\ce{SO4^{2-}}]$.


text:
What ion remains in the largest concentration at equilibrium: 0.1 M H₃PO₄ and 0.2 M KOH?
Mix equal volumes of $0.1~\mathrm{M}$ $\ce{H3PO4}$ and $0.2~\mathrm{M}$ $\ce{KOH}$. What ion remains in the largest concentration at equilibrium?
My approach: suppose we mix $1~\mathrm{L}$ of each solution. For $\ce{H3PO4}$, $K_{a_1}=7.5\cdot10^{-3}$ and $K_{a_2}=6.2\cdot10^{-8}$, so in terms of the dissociation of $\ce{H3PO4}$, the only ions produced in non-neglible quantities are $\ce{H^+}$ and $\ce{H2PO4-}$. Furthermore the quantities of $\ce{H^+}$ and of $\ce{H2PO4-}$ are both less than one mole, since there is less than 1 mole $\ce{H3PO4}$. We have 2 moles $\ce{OH^-}$ from the $\ce{KOH}$, but since of the $\ce{OH^-}$ neutralizes the $\ce{H^+}$, in particular less than a mole $\ce{OH^-}$ is used in this neutralization reaction. Thus at least one mole $\ce{OH^-}$ remains, at most 1 mole $\ce{H2PO4-}$ remains, and all other ions remain in negligible quantities, so the answer is $\ce{OH^-}$.
Apparently, this is wrong - the answer is $\ce{HPO4^2-}$. I don't see how this is possible because we already established that the amount of $\ce{HPO4^2-}$ formed is negligible from the $K_a$ values, and since we are adding a strong base to a weak acid we know the resulting solution will be basic, so we should expect from the beginning that $\ce{OH^-}$ appears in large concentration.
I don't understand what's wrong with my reasoning?

Actually, the ion with the largest concentration is $\ce {K+}$.
But that is not what they want. Since potassium hydroxide is a strong base, then the concentration of $\ce {OH-}$ starts with 0.2 M (roughly pH of 13.3). To neutralize this, it would take 2 hydrogens from $\ce {H3PO4}$ at the specified concentration to get both concentrations near $10^{-7}$ (pH ~7). So the major anion is $\ce {HPO4^{2-}}$.
You can find the exact concentrations if you need.
Addendum:
Since $K_w = 10^{-14}$, then $\ce {[H^+][OH^{-}]=10^{-14}}$
Adding 1 L of each solution would give 0.2 moles of $\ce {OH^-}$ to react with any liberated hydrogen from the phosphoric acid. Removing one hydrogen from all the phosphoric acid gives 0.1 mole of $\ce {H+}$, which reacts with the hydroxyl, leaving 0.1 mole of $\ce {OH^-}$ (pH of 13.0). Since this is smaller than ${K_{a2}}$, more hydrogen is released from $\ce {H_2PO_4^{-}}$.
Assuming 0.2 moles of hydrogen is released, then there is an equal amount of $\ce {H+}$ and $\ce {OH^-}$ in the solution. To have $K_w = 10^{-14}$, then each of the ions needs to be near $10^{-7}$M.
Calculations:
Find $\ce {[H_2PO_4^{-}]}$:
$$\ce {H+ + HPO4^{2-} <=> H_2PO_4^{-}}$$
$K_{a1}=\frac {[H_2PO_4^{-}]}{[H^+][HPO_4^{2-}]}$ where $\ce {[H+]}=10^{-7}$M, and $\ce {[HPO_4^{2-}]}=0.05$M.
$7.5\times 10^{-3}=\frac {[H_2PO_4^{-}]}{10^{-7}\times 0.05}$
$[H_2PO_4^{-}]={10^{-7}\times 0.05}\times {7.5\times 10^{-3}}$
$[H_2PO_4^{-}]=3.75\times 10^{-10}$M

Find $\ce {[PO_4^{3-}]}$:
$$\ce {HPO4^{2-} <=> H^+ + PO_4^{3-}}$$
$K_{a3}=\frac {[H^+][PO_4^{3-}]}{[HPO_4^{2-}]}$ where $\ce {[H+]}=10^{-7}$M, and $\ce {[HPO_4^{2-}]}=0.05$M.
$4.5\times 10^{-13}=\frac {10^{-7}\times [PO_4^{3-}]}{0.05}$
$[PO_4^{3-}]=\frac {0.05\times 4.5\times 10^{-13}}{10^{-7}}$
$[PO_4^{3-}]=2.25\times 10^{-7}$

I think  the calculations in the second part of the answer,like :
$$K_{a2}=\frac {{[H^+]}{[HPO_4^{-2}]}}{[H_2PO_4^{-}]} where \ce {[H+]}=10^{-7}M, and \ce {[HPO_4^{2-}]}=0.05$M.$$
$$6.2\times 10^{-8}=\frac {{10^{-7}\times 0.05}}{[H_2PO_4^{-}]}$$
$$[H_2PO_4^{-}]=\frac{10^{-7}\times 0.05} {6.2\times 10^{-8}}$$
$$[H_2PO_4^{-}]=0.081$$M

@AdnanAL-Amleh How can you have a concentration of 0.081 M when the phosphate concentration is 0.05 M ?

Sorry, thanks for the interesting and correction; this relation confused me:
$$K_{a1}=\frac {[H_2PO_4^{-}]}{[H^+][HPO_4^{2-}]}$$
Because I failed in reconciling it with this relation:
$$K_\mathrm{a1}  = \frac{[\ce{H3O+}][\ce{H2PO4-}]}{[\ce{H3PO4}]} = 7.5 \times 10^{−3}$$


text:
Why are esters meta directors? Can't the carbonyl donate electrons into the ring?
I'm having trouble understanding why an ester is a meta directing/deactivating group.
I'm not talking about an ester with its oxygen bound to the ring, like an acylated phenol. I mean the R group is bound to the ring and the alkoxy group is bound only to the carbonyl carbon.
It seems to me like the pi electrons in the carbonyl pi bond could be pushed down and delocalized into the ring. Why is this not possible?

As your concept oxygen gets positive charge and that leads to least stability which is as follows:

Yes, nice reasoning.  However, there are no transition states involved here, all of the structures you've drawn in your answer are resonance structures (the 3 different ones for the ester and the 2 for phenol) so they should be connected by resonance (double-headed) arrows.  Simply put, some resonance structures contribute more to the overall description of a molecule than others.

Yes your correct. I thought that directing can happening at any place that ortho para like.... Thats why i mentioned TS. And the double heade arrow can be used only when the charge from, charged particle, is migrating to form other charged particle. So here i have used single headed arrow from non-charged particle to charged particle (first resonance structure). Thanq u for ur valuable review.

Double headed arrows should be used for all resonance structures regardless of whether there is charge separation or not. The single headed arrow indicates a reaction taking place which is not the case.

Yes i agree, but here i mean that the resonance in b/w charged particles. Thank you


text:
Which of these compounds does not liberate heat when water is added to it?
Which of these compounds does not liberate heat when water is added to it: $\ce{KNO_3}$, $\ce{NaOH}$, $\ce{CaO}$, $\ce{H_2SO_4}$, $\ce{Na}$.
I know from experience that dropping sodium tablets into water has explosive results, but I am clueless about the method to solve this problem. I know the answer is $\ce{KNO_3}$, but I don't know why. If someone can provide the method, I think I can do the problem.

It is important to note that if a molecule or formula unit is capable of undergoing hydrogen bonding ($\ce{NaOH}$, $\ce{H2SO4}$), then the formation of this hydrogen bond will be extremely exothermic.
Also, if a substance can react to form a substance that will undergo hydrogen bonding ($\ce{CaO + H2O -> Ca(OH)2}$), this species will subsequently form these hydrogen bonds and liberate heat.
Lastly, as you have stated you know from personal experience, the reaction between a alkali metal ($\ce{Na}$) and water is extremely exothermic— so much so that many of these metals will melt from the heat liberated.
Otherwise, there is no fail-proof method for predicting the enthalpy of solution for any given substance. Generally, the enthalpies of solution are determined experimentally, so unless you are expected to perform all these reactions in a calorimeter yourself, I would recommend something like this: this.

Ok. A couple of things I don't understand. The first two compounds you pointed pointed out - you say they are capable of undergoing hydrogen bonding, but I don't understand what you mean because they are already held together by hydrogen bonding. The OH groups in NaOH and H2SO4 have hydrogen bonding, so when you put them in water the OH bonds break, and wouldn't this be highly endothermic? Second, when you put CaO in water, how do you know that Ca(OH)2 forms? Isn't Ca(OH)2 soluble?

Hydrogen bonding is a special type of electrostatic attraction that occurs between $\ce{H}$ bonded to $\ce{O}$, $\ce{N}$, or $\ce{F}$ and the unbonded electron pairs of a highly electronegative atom. In the case of dissolving $\ce{NaOH}$, the $\ce{OH-}$ that dissociates and the $\ce{H2O}$ participate in hydrogen bonding. In the case of dissolving  $\ce{H2SO4}$, the $\ce{HSO4-}$ and the $\ce{H2O}$ participate in hydrogen bonding. These bonds are the strongest intermolecular bond, and are very stable. The formation of these bonds is therefore extremely exothermic.

Don't you have to consider the energy absorbed in breaking of the hydrogen bonds which hold NaOH and H2SO4 together, compared to the energy released in forming H bonds with water molecules?


text:
Electrodeposition of Ni
How can one electrodeposit Nickel on a substrate using a liquid crystal template?
substrate is Indium Tin Oxide coated plate and the template is cetyl trimethylammonium bromide(CTAB)+WATER+ETHANOL

This seems to be a very specific case. I would say that you should start with a standard Ni electroplating solution and see the result. You could then adjust the parameters of your plating operation and the concentrations of the additives in the solution.
However, I have a bigger concern. How stable is your liquid crystal template under applied potential in an electrolyte? Before trying to deposit Ni, you should try to do simple electrochemistry find the potential window where your template stays intact.


text:
Way to detect lead in gasoline
Is there any way that i can detect lead in gasoline, not by sending of to a lab but to be able to do this by myself?

Why do you suspect that there is lead in your gasoline? In most countries, the use of tetraethyl lead was banned more than a decade ago.

Certainly tetraethyl lead is banned, but in principal, there could still be some contamination. I don't know about levels after refining, but consider that raw petroleum comes out of the ground and carries all sorts of chemical species along.

Personally, I think it's an interesting at-home analytical chemistry question.

@GeoffHutchison You're certainly right on the composition of raw petrol, but I's assume that the main source of lead in refined petrol were tetraalkyl lead compoundds deliberately added as anti-knocking agents.

I think by volumetric  analytic test you can determine by yourself but you need basic lab instrument. BTW I don't know how but according to this book it is possible.

As pointed out in my earlier comment, you won't find tetraalkyl lead compounds in current petrol used for cars. In most countries, it was banned decades ago and replaced by MTBE.
However, the situation seems to be different for fuels used in aviation,   such as AvGas 100 where lead concentrations up to $\mathrm{1\,g\,L^{-1}}$ can be found.
Even the low lead variant, AvGas 100LL contains $\mathrm{0.56\,g\,L^{-1}}$ lead, see product specifications from TOTAL.
In an older technical note by R. E. Mansell and T. A. Hiller, published in Anal. Chem., 1973, 45, 975-975 (DOI), a possible approach for the decomposition of tetraethyl lead and subsequent detection of $\ce{Pb^{2+}}$ is outlined, referring to the ASTM D526-56 method:

[…] recommends refluxing 50 ml of gasoline with 50 ml of $\ce{HCl}$ for 30 minutes to decompose the tetraethyllead.[…]

In the aqueous layer, $\ce{Pb^{2+}}$ can be detected by complexometric titration with EDTA, either directly, or by back titration with a solution of a $\ce{Mg^{2+}}$ salt.

That's a lot of lead O.o

@Martin-マーチン Yes, I was surprised too!

Try to use aqueous sodium chloride.
The lead in the gasoline will with the chloride ion to form lead chloride with is unsoluble white percipitate.
$\ce{Pb^{2+} + 2Cl^- -> PbCl_2}$

Lead ion may react with chloride ion to form lead chloride, but (1) lead chloride is somewhat soluble in water, and (2) tetraethyl lead, the form of lead added to gasoline, will not likely react with chloride ion. But I think this is an interesting question. Kudos for trying to answer it.

Use iodine instead then it is almost completely insoluble


text:
Given the concentration and density of a sample, calculate moles of bleach?
Given the concentration of bleach ($6\%\,\mathrm{m/m}$), density of bleach ($1.07\,\mathrm{g\,mL^{-1}}$), and molar mass of bleach ($74.44\,\mathrm{g\,mol^{-1}}$) how many moles of bleach are in a $4.0\,\mathrm{mL}$ sample? I don't understand where the concentration factors in, but the problem says to use it.

The key issue to address this kind of problems is to determine the mass of the pure compound in the given volume of the solution. So, we'll use first the density to determine the mass of the solution. Then, the mass concentration to determine the mass of the pure compound. After that, we can readily compute the number of moles of the pure compound using the molar mass.
The mass of $1\,\mathrm{mL}$ of bleach solution is $1.07\,\mathrm{g}$.
So, the mass of $4~\mathrm{mL}$ of bleach solution is $4 \cdot 1.07\,\mathrm{g}$
Or, each $100\,\mathrm{g}$ of bleach solution has $6\,\mathrm{g}$ of pure bleach.
So, the mass of  $4 \times 1.07\mathrm{g}$ of bleach solution has $\frac{4\cdot 6 \cdot 1.07}{100}\,\mathrm{g}$ of pure bleach.
The number of moles in  $4\mathrm{mL}$ of bleach solution is: $$\frac{4\cdot 6 \cdot 1.07}{100\cdot 74.44}= 0.0034\,\mathrm{mol}$$

Dimensional analysis is a straightforward, easy, and beautiful way to solve this problem! Simply cancel the units!
$$4\,\mathrm{mL\,(solution)} \cdot \frac{1.07\,\mathrm{g\,(solution)}}{\mathrm{mL\,(solution)}} \cdot \frac{6\,\mathrm{g}\,\ce{NaClO}}{100\,\mathrm{g\,(solution)}} \cdot \frac{1\,\mathrm{mol}\,\ce{NaClO}}{74.44\,\mathrm{g}\,\ce{NaClO}}$$
$$=0.00345\,\mathrm{mol}\,\ce{NaClO}$$


text:
How is it possible, that the hydrolysis of ethers proceed via a SN2 mechanism?
Ethers on hydrolysis with dilute $\ce{H2SO4}$ under pressure produce alcohols.
This is basically a type of acidic cleavage of ethers by the SN2 mechanism.

(source)
There is inversion of configuration.
Now if $\ce{Nu}$ is $\ce{OH-}$ its called hydrolysis of ether.
Is it right, that for SN2 the leaving group must be more stable than the approaching group? In case of ether hydrolysis the leaving alkoxy group is less stable than the entering $\ce{OH-}$ group.
So how does the reaction succeed? Can someone please explain?

The leaving group is not $\ce{RO^{-}}$, but rather $\ce{ROH}$.  The reaction is run in the presence of acid, so the first step is protonation of the ether oxygen.  Neutral alcohol is a much better leaving group than alkoxide.  BTW, depending on the ether substituents the reaction can proceed by SN1 or SN2.

While the mechanism may look simple in the way you described it there, it doesn’t work exactly like that.
First, you are in the presence of (diluted) $\ce{H2SO4}$ so aside from extremely short-lived accidental self-deprotonation of water, $\ce{OH-}$ ions will not be around. Therefore, $\ce{OH-}$ cannot act as an attacking group, whatever the mechanism.
Second, an alcoholate is one of the worst leaving groups in $\mathrm{S_N}$ reactions one can imagine. The only case where you will frequently see leaving alcoholates is for epoxide openings. But that’s not a problem here, we have $\ce{H2SO4}$. We can protonate the ether oxygen so that the leaving group will now be an alcohol.
Third, since $\ce{OH-}$ cannot be the attacking species, we need to choose from the nucleophiles present. Either $\ce{HSO4-/SO4^2-}$ or $\ce{H2O}$ can attack the $\alpha$ carbon (in $\mathrm{S_N2}$ or $\mathrm{S_N1}$ manner). In the former case, a later attack of $\ce{H2O}$ to liberate the sulphate needs to be assumed. Both attacks work, but neither nucleophile is strong, so the driving force must be the liberated alcohol.
Fourth, the attacking $\ce{H2O}$ will create a protonated alcohol. This needs to deprotonate to give the free alcohol. Thereby the proton we used in second (step 1 of the mechanism, if you wish) is reliberated and therefore the reaction is acid-catalysed.
Finally, one needs to consider the possibility of equilibrium. The reaction can proceed in both directions. Usually by the choice of solvent one can direct it into one or the other. Hydrolysis is usually achieved with a large surplus of water, driving the reaction towards the alcohols by Le Chatelier’s principle. The reverse reaction would be triggered by removal of water or high concentrations of either alcohol.


text:
What is the mechanism of trialkylborane oxidation with hydroperoxide?

This is the mechanism of trialkylborane. In this mechanism I can't understand why should the -OH group leave and the -R group enter in its place to form borate ester. Can someone please explain me that step?

The carbon migrates because peroxide bonds are generally weak and dissociate quite easily.

Also, the mechanism is drawn poorly. In the species labeled "Unstable intermediate," the left arrow should not point towards the boron or the bond; the arrow head should terminate at the left oxygen.

@Zhe Yes. In the upper mechanism, curved arrows are drawn in a wrong way. The arrowhead should be placed pointing towards boron in two of the curved arrows.

One aspect is the release of the stable anion $\ce{OH-}$.
Another aspect is that $\ce{B-O}$ bonds are much stronger than $\ce{B-C}$ bonds because the empty p-orbital of boron overlaps with one p-orbital with a lone electron pair of oxygen.
$\ce{B-C}$ $D_{298}^ \circ  = 448 \pm 29\ \mathrm{kJ\ mol^{-1}}$
$\ce{B-O}$ $D_{298}^ \circ  = 809\ \mathrm{kJ\ mol^{-1}}$
values taken from “Bond Dissociation Energies”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.


text:
Acid catalyzed hydrolysis of ester

During acid catalyzed hydrolysis of ester why does the carbonyl oxygen attack the $\ce{H^{+}}$ ? Why doesn't the alkoxy oxygen which is more electron rich and less electronegative not attack the $\ce{H^{+}}$ with its lone pair? Any ideas?

Why do you think that it is more electron rich?

Because one is sp2 hybrid while other is sp3 hybrid.

The difference between the carbonyl oxygen (red in colour) and alkoxy oxygen (blue in colour) is that the highly distorted pi-electron density (green shaded) towards to the carbonyl oxygen, due to more electro-negativity of oxygen that is the feature which is more on this oxygen than alkoxy oxygen . That feature provide the cavity for the H+ ion from acid, and the distortion of sigma- electron density at alkoxy oxygen is comparatively less than the distortion of  pi-electron density at carbonyl oxygen.


text:
Computing pH of a buffer, and concentrations
You pour $50~\mathrm{mL}$ of a $0.0200~\mathrm{M}$ $\ce{HCOOH}$-solution to $150~\mathrm{mL}$ of a $0.0500~\mathrm{M}$ $\ce{HCOONa}$-solution. The $K_\mathrm{a}$ of $\ce{HCOOH}$ is $1.8 \times 10^{-4}$ and $\mathrm{p}K_\mathrm{a}$ is $3.74$.
a) Compute the concentrations of $\ce{HCOOH}$ and its salt after dilution to this $200~\mathrm{mL}$ buffer solution.
b) Calculate the $\mathrm{pH}$ of this buffer solution with the buffer formula.
Attempt at solution: Not sure how to do part a). For part b) I used the following method. We have $50 \times 10^{-3}~\mathrm{L}  \cdot \frac{0.0200~\mathrm{mol}}{1~\mathrm{L}} = 0.001~\mathrm{mol}$ of $\ce{HCOOH}$. Doing this again for $\ce{HCOONa}$ gives $0.0075~\mathrm{mol}$. Hence: \begin{align*} \mathrm{pH} = 3.74 + \log(\frac{0.0075}{0.001}) = 4.61 \end{align*}
But how can I solve a)? I'm having trouble with writing the reaction down. I'm not sure what's happening. Which chemical is neutralizing which? And what are the ions?

We have to compute the new concentrations after dilution:
$$\ce{[HCOOH]}=\frac{0.02\times 50}{200}=0.005\,\, \mathrm{mol/L}$$
$$\ce{[HCOO^- ]}=\frac{0.05\times 150}{200}=0.0375\,\, \mathrm{mol/L}$$
Now, we can compute the $\mathrm{pH}$ of the buffer solution:$$\mathrm{pH}=\mathrm{p}K_a +\log \frac{\ce{[HCOO^- ]}}{\ce{[HCOOH]}}$$
$$\mathrm{pH}=3.74+\log \frac{0.0375}{0.005}=4.62$$

Thanks for the help. How would I write this reaction down? $ HCOOH \rightarrow HCOO^- + H^+$ ? And what does the 'Na' in front of the HCOO mean?

sodium ion is the counter ion of HCOO- to have a neutral compound

In a buffer solution, reactions primarily occur between the externally introduced species and the buffer constituents, HCOONa buffers the incoming H(+) proton from a strong acid and HCOOH buffers OH(-) from a base.
there is, as i believe, no other reactions involved here


text:
Why is the melting of ice a spontaneous reaction?
So here's my question: we know that melting of ice is an spontaneous reaction but is it spontaneous? We have to supply energy in the form heat for the reaction to take place.

First of all,

what exactly is spontaneous reaction?

In very easy language, reaction that occurs in a given set of conditions without intervention is called spontaneous reaction.

Now let us consider melting of ice example,
Take a ice on a plate and leave it for half an hour(conditions). After half an hour we will notice water instead of ice.
This process took place without any intervention from your side. Heat absorbed by the ice from surrounding is due to set of conditions not due to your intervention.
Thus we can consider melting of ice as a spontaneous process(or reaction).

Does this reaction have equilibrium constant? I mean would the ice be completely turned on liquid or some samll quantity will be left back as ice?

I wouldn't call it a reaction, it's a phase change. The term reaction refers to a change in covalent bonding between components of a system.
The phase change occurs in a seemingly spontaneous manner because the environment is capable of donating enough free energy (most likely from enthalpy, could happen otherwise) to overcome the energy needed to activate the process of reorganization of the system (increase in entropy). The other responses have done a nice job in terms of symbolic rigor, so I won't repeat them here.

Does this phase change have equilibrium constant? I mean would the ice be completely turned on liquid or some small quantity will be left back as ice?

When T > 0 °C, ice melts spontaneously; the reverse process, liquid water turning into ice, is not spontaneous at these temperatures. However, when T < 0 °C, the opposite is true. Liquid water converts into ice spontaneously, and the conversion of ice into water is not spontaneous
At T = 0 °C the solid and liquid phases are in equilibrium. At this particular temperature the two phases are interconverting at the same rate, and there is no preferred direction for the process: Both the forward and reverse processes occur with equal preference, and the process is not spontaneously favored in one direction over the other.
When ΔG is negative, a process or chemical reaction proceeds spontaneously in the forward direction.
Also from the wikipedia article on Spontaneous reaction says:

Every reactant in a spontaneous process has a tendency to form the corresponding product. This tendency is related to stability. Stability is gained by a substance if it is in a minimum energy state or is in  maximum randomness. Only one of these can be applied at a time. e.g. Water converting to ice is a spontaneous process because ice is more stable since it is of lower energy. However, the formation of water is also a spontaneous process as water is the more random state.

I'm not sure if you have learnt this, so I just tell the story from the beginning. (I'm not sure, but if this is hard to understand, please tell me.)
There is a universal law usually called as the second law of thermodynamics. It states:

The entropy of the universe tends to a maximum. If a process is reversible, the entropy of the universe will keep unchanged; if it is irreversible, the entropy increases.

Well, it's not the original statement. Historically, from the original statement of the law, a thermodynamic function is considered useful; that's "entropy" I mentioned here.
So what is entropy? It's quite a mystery because at the first set out people do not defined what will it be; while, they just defined how it changes.

An infinitesimal increment in the entropy of a system results from an infinitesimal transfer of heat is $\mathrm{d}S=\delta Q/T$

Ur, I think you will not like this. Hopefully, later, Boltzmann state a hypothesis, he assumed the occupation of any microstate (means the distribution of all the particles we concerned in the space we concerned) is equally probable, then

In most cases, the entropy is proportional to the logarithm of the number of possible microstate.

You propably still think this is hard to understand... so we can say all the above in a easier way (but, I think, a less accurate way):

the second law of thermodynamics teaches us, the randomness of the universe tend to be maximum. If a process is reversible, the randomness of the universe will keep unchanged; if it is irreversible, the randomness increases.

But...we still can't know how to judge whether a process is spontaneous or not yet. However, by the second law of thermodynamics we are easy to know the randomness of universe increases in a spontaneous process.
Thus we get $\Delta S_\text{universe} = \Delta S_\text{system} + \Delta S_\text{surronding} > 0$.
While we also know if the heat transferred to the system from the surroundings is $Q$, then $\Delta S_\text{surronding} = -Q/T$, where $T$ is the temperature of the surrounding. (Since the surrounding is large, we may assume the temperature is unchanged.)
So we get $\Delta S_\text{system} - Q/T > 0$, i.e. $T\Delta S_\text{system} - Q > 0$
if the process is now assumed to be isobaric (means that the barometric pressure of system is unchanged), then $Q = \Delta H$ and we get $T\Delta S_\text{system} - \Delta H > 0$.
Now we can define a new thermodynamic function:

The Gibbs free energy of a system is $G = H - TS$.

Then we can say that in a spontaneous process, $\Delta G < 0$. (If $\Delta G = 0$, it is in equilibrium.)
Maybe you have known it all... uh... so let's solve the question: Why is the melting of ice a spontaneous reaction?
You know in liquid particles will arrange in a messer way then in a solid, so the entropy, or the randonness of the "reaction system" is increased, i.e. $\Delta S_\text{system} > 0$. However, when ice melt, $\Delta H > 0$, so for  $\Delta G < 0$ you need to keep the temperature higher, or supply energy to it. And simliarly, you will find if temperature is low enough, the water will spontaenously freeze. This seems to be the case you are familiar with.
Noticed that sometimes $\Delta G < 0$, however the speed of reaction is too low, so we supply energy for accerlation. For example, the diamond will burn:
$\ce{C_{(s, diamond)} + O2_{(g)} -> CO2_{(g)}},\ \Delta G^\circ= -397\ \mathrm{kJ}$.
But you never see this happen in common situation: because it's too slow.

also note A spontaneous chemical reaction can result in a negative change in entropy.

As long as the net entropy change of system + surroundings is positive, right?

Here is the short answer:
When ice melts, it's molecules (H2O), have to gain energy, that is true. However, you have to look at the Gibbs Free Energy, which is the determining factor in figuring out if a reaction is spontaneous or not. Because disorder increases, the delta S (or entropy) increases. The formula calculating Gibbs Free Energy is deltaG = deltaH - TdeltaS. If the temp is high enough for a given deltaS, the deltaG will become negative.


text:
19F NMR of hexafluorobromine(VII) cation
$\ce{[BrF6]+}$ is an octahedral species, and should have only one fluorine environment due to the symmetry. $\ce{Br}$ has nuclear spin $I=3/2$ and therefore is unlikely to cause splitting due to rapid interconversion between its spin up and spin down states averaging to a zero magnetic effect. Therefore I would expect one singlet in the $\ce{^19F}$ NMR spectrum.
However... the $\ce{^19F}$ NMR is actually two $1:1:1:1$ quartets (at low temperatures). Can someone help me find the flaws in my reasoning?

Related: Symmetry and NMR linewidth with quadrupolar nuclei (in my answer, I just so happened to cover this exact species!)

$\ce{BrF6^+}$ does have an octahedral structure consequently all of the fluorines are equivalent.  As you point out, bromine is a spin = 3/2 nucleus, therefore, it is also a quadrupolar nucleus (all nuclei with spin > 1/2 are also quadrupolar). A spin 3/2 nucleus that is coupled to another nucleus (fluorine in this case) will split the nmr signal for the other nucleus into a 1:1:1:1 quartet.  In the case of bromine we don't normally see this coupling because bromine typically undergoes rapid quadrupolar relaxation (some quadrupolar nuclei do, some don't) which washes the coupling out (e.g. rapid quadrupolar relaxation effectively decouples the nuclei, that's why the protons in $\ce{CH3Br}$ are a singlet, for example).  However, when the quadrupolar nucleus exists in a highly symmetric environment (octahedral, tetrahedral geometry) the electric filed gradient at the nucleus is near zero and quadrupolar relaxation becomes slow (on the nmr timescale) and coupling can often be observed in such cases.
Finally, bromine occurs naturally as a mixture of two isotopes, $\ce{^{79}Br}$ (50.7%) and $\ce{^{81}Br}$ (49.3%).  Both isotopes are quadrupolar with spin = 3/2.  Because of the slight difference in electronegativity of the two bromine isotopes, they will produce a slight difference in chemical shift for the fluorines bonded to them. Hence, this mixture of $\ce{^{79}BrF6^+}$ and $\ce{^{81}BrF6^+}$ will produce two 1:1:1:1 quartets in the $\ce{^{19}F}$ nmr spectrum.


text:
How do popular reducing agents react on compounds that have ester, carbonyl, and unsaturated functionalities?
Suppose we have a chemical compound like ethyl 4-oxo-2-cyclopentene-1-carboxylate

It is $\ce{-COOC2H5}$ bonded to a five-membered carbon ring with one double bond and the ring has another carbonyl group.
Suppose this compound is made to react with these reducing agents:

$\ce{NaBH4}$
$\ce{LiAlH4}$ and $\ce{H+}$
$\ce{BH3}$, $\ce{THF}$, and $\ce{H+}$
$\ce{H2/Pt}$
$\ce{H2/Pd-C}$

Then what will be the product in each cases?
I know some of the reducing agents and I want to know their reactions with a compound which is unsaturated and has an ester functionality $\ce{-COOR}$ and a carbonyl group $\ce{R2C=O}$ bonded to it.
I cannot try to understand it due to my limited knowledge in chemical mechanisms.

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


text:
Catering grade electrolysis of alcohol
Would the electrolysis of alcohol, specifically spirits, result in a potable end product? I'm looking for some novel cocktail ingredients.

This one came up a while ago for beer, but the principles are similar. If the intent is to remove water, it probably won't work and might produce all kinds of weird products that may or may not be safe to consume.


text:
50-90'C color changing chemical to use as warning marker for hot electronics?
I am testing electronics with many potentially unstable contacts, and I wish to monitor if they rise above 50 to 90 degrees Celsius, i.e. using carbon paper that would darken progressively from 50 to 90.
The substance should stay a specific warning color if it goes too high and not change hues again same as thermochromic pigment, so that I can detect high temps.
I hope you know a chemistry trick that could help? Thank you.

wax melts at 60-100 degrees so perhaps i can find a solution using it. Color change would be more safe and visible.

Some "invisible inks" could help you, and most of them are easily available, but I do not know what the optimal temperature for their change is. You might want to experiment with that a bit.

There are commercial sources for thermochromic permanent-change ink, but you might also try thermal printer paper. You might need to check if there is a time-dependence on the color change, as well as temperature dependence. The printer paper is available at Amazon, Staples and elsewhere for less than US$20 / package, but you might just test a cash-register receipt, many of which are on thermal paper. According to this PDF, most start to darken at 79 C.


text:
Are the following expressions for base dissociation constant equivalent?
Given a weak base such as $\ce{C2H5NH2}$, I usually see the expression for $K_\text{b}$ as
$K_\text{b} = \frac{[\ce{C2H5NH3+}][\ce{OH-}]}{[\ce{C2H5NH2}]}$.
Are the expressions
$K_\text{b} = \frac{[\ce{C2H5NH3+}]}{[\ce{C2H5NH2}][\ce{H+}]}$,
$K_\text{b} = \frac{[\ce{C2H5NH3+}][\ce{H2O}]}{[\ce{C2H5NH2}][\ce{H3O+}]}$
equivalent? Do they represent the same process (if so, why do the latter two not work with calculations?) or fundamentally different ways of dissociation? Which one represents what's happening molecularly most accurately? Given the Brønsted-Lowry definition of a base as a proton acceptor I'd think that the second one is most accurate, but I rarely see it used.

$$K_\mathrm b = \frac{[\ce{C2H5NH3+}]}{[\ce{C2H5NH2}][\ce{H+}]}$$
This expression is infact $1/K_\mathrm a$ which is related with $K_\mathrm a K_\mathrm b=K_\mathrm w$
$$K_\mathrm b = \frac{[\ce{C2H5NH3+}][\ce{H2O}]}{[\ce{C2H5NH2}][\ce{H3O+}]}$$
here $K_\mathrm b$ is correct but nowadays any constant concentration (here water) is not written in the expression. As it is assumed that concentration of water is not changing when reaction proceed.

I try to say it clearer and simpler.
The second expression respects to the equilibrium
$\ce {C2H5NH2 + H+ <=> C2H5NH3+}$.
And the third
$\ce {C2H5NH2 + H3O+ <=> C2H5NH3+ + H2O}$.
However, what you want to know is the equilibrium constant of
$\ce {C2H5NH2 + H2O <=> C2H5NH3+ + OH-}$.
The former two both describe how $\ce {C2H5NH3+}$ behaving as a weak acid in aqueous solution, and the third one describes how $\ce {C2H5NH2}$ behaving as a weak base in aqueous solution (hydrolyzation). They are different because in the former two equilibrium $\ce {C2H5NH2}$ are not receiving proton directly from $\ce {H2O}$.
Or you may look it by this view: the last one can be seen as the result of two equilibrium:

$\ce {C2H5NH2 + H+ <=> C2H5NH3+}$
$\ce {H2O <=> H+ + OH-}$

Or you may write as

$\ce {C2H5NH2 + H3O+ <=> C2H5NH3+ + H2O}$
$\ce {2H2O <=> H3O+ + OH-}$

So you can see there is obvious different. The equilibrium constant of $\ce {2H2O <=> H3O+ + OH-}$ is not 1, so the base dissociation constant $K_b$ will not be the same as what you said.
And there is an important convention that in an aqueous solution system, the equilibrium constant should omit [$\ce {H2O}$] because it is an insignificant constant in the most cases.


text:
Burning a diamond produces how much energy?
How much energy is produced by burning a 1 ct. diamond?
I'm trying to figure out how much energy to build the diamonds in a biological process.  See here: https://worldbuilding.stackexchange.com/questions/15314/are-diamond-berries-possible

Still looking for the answer but a question hit my mind: Diamond ignites at 700 to 800 degrees centigrade. Why on Earth would the energy matter if you have to reach those temperatures for the combustion to happen?! Just saying. :P

This is, perhaps, a homework question.

@JonCuster  No, no.  Not homework.  I'm trying to figure out how much energy to build the diamonds in a biological process.  See here: http://worldbuilding.stackexchange.com/questions/15314/are-diamond-berries-possible?noredirect=1#comment37651_15314

@Sam you might add that info to your question. There is some homework policy here. See: Homework policy

OK - the assessed elemental Gibbs free energies are cataloged in 'SGTE Data for Pure Elements', A.T. Dinsdale, CALPHAD 15(4) 317-425. There one finds (at low pressures such as atmosphere) that the free energy of diamond relative to graphite is 1009 + 4.88 T - 0.01 T ln(T) + 135400/T + 33.0E5/T^2 - 9E8/T^3. I'll leave the graphite + oxygen making CO2 part up to you.

Thanks @JonCuster, but i'm not much of a chemist and I don't know what 'T' means here... How do we get to Joules?

T is the temperature. The expression gives the Gibbs free energy of diamond relative to graphite in Joules per mole.

Well it looks for me, the question was addressed by a non-Chemist.  Hence my attempt to answer this accordingly.

You have to know the reaction of the combustion, the starting materials, the product(s), and their relative quantities to balance the reaction equation.  Your case is described by $\ce{C} + \ce{O2} \rightarrow \ce{CO2}$, with "$\ce{C}$" for carbon (here: in form of diamond), $\ce{O2}$ for oxygen and $\ce{CO2}$ to yield the product, carbon dioxide. And the reaction equation is already balanced, i.e. the number of atoms per type on the left hand side (prior the arrow) equals the number of atoms per type written on the right, too.

There is a heat called standard enthalpy of formation, describing how much heat is necessary or liberated to form a compound if one (even theoretically) were to prepare this out of elements.  Normally, you look up these tabulated data for all your products and sum this up, and subtract the sum of all these enthalpies of formation of your starting materials.  In your example, by convention elementary gases (like oxygen, $\ce{O2}$) were attributed a zero value.  By means of thermodynamic, diamond is not the most stable form of carbon, and was attributed a value of ${+1.88}{\,\mbox{kJ/mol}}$ (same reference).  Carbon dioxide was attributed a value of ${-393.5}{\,\mbox{kJ/mol}}$ (reference) and the energy balance of the reaction equation equals to ${-391.62}{\,\mbox{kJ/mol}}$, i.e. if you would burn 1 mole of diamond.  (This is a quantity like dozen, used in chemistry.)

Now if we express the quantity of diamond actually burned (mass $m$ of 1 ct) in this above mentioned unity of mole, we need a proportional factor, the molar mass $M$, that happens to be $12{\,\mbox{g/mol}}$ for carbon (reference).
Now we have the pieces.  Hence we calculate the "chemical quantity" ($n$) of carbon deployed as

$n = \frac{m}{M} = \frac{0.2\,\mbox{g}}{12\,\mbox{g/mol}} = 0.017\,\mbox{mol}$
and than multiply this with the above calculated heat of the reaction, i.e
$\mbox{heat} = 0.017\,\mbox{mol} \times (-391.62\,\mbox{kJ/mol}) = -6.527\,\mbox{kJ}.$
So if you burn 1 ct of diamond in oxygen at 25 degree Celsius (about 77 Fahrenheit) and would collect the heat of the product by cooling the $\ce{CO2}$ to the same 25 degrees Celsius, this offers you 6527 Joules (or about 1559 calories).
Notes: The heat calculated here is the reaction enthalpy. The minus sign indicates energy is "liberated" in the course of this reaction. Actually, if all this heat were completely converted into electrical energy, you were able to switch on a light bulb of 60 W for less than 2 minutes (only 108 seconds).  In the beginning of modern chemistry, diamonds actually were burned by concentrated solar power, for example by Lavoisier (here) although to prove their composition.


text:
Benzylic and allylic carbocation stability order
Most books I have read stop at this order :
Benzylic > Allylic > Tertiary > secondary > primary
However If I want to compare tertiary, secondary and primary benzylic and Allylic carbocations what will be their stability order?

Tertiary benzylic carbocations should be stabilised more due the $+M$ effect of three benzylic groups. The same should apply for allylic carbocations.

Should the answer then be tertiary benzylic>Secondary benzylic>primary benzylic> tertiary Allylic > secondary allylic> primary Allylic > tertiary> primary> secondary

Yes, seems correct to me.

Yeah that "seems" to be correct but I saw one problem where they said that teritary carbonation can be more stable than primary allylic because of a lot of hydrogens that take part in hyperconjugation in the outweighs the mesomeric effect

There is no simple general rule for this. The rules you state are also not "absolute".

Can you clarify what you mean by tertiary benzylic etc. Do you mean three phenyl groups (trityl cation) or one phenyl group and two alkyl groups?

Yes tertiary allyllic carbocation is more stable than secondary because no.  of electrone donating methyl groups are greater than secondary (in tertiary 3 methyl groups  while in secondary  are 2) so they decrease  electrophilicity carbocation but it still it remain electrophillic.


text:
Carbon with two double bonds and oxygen name
I have a compound like this:
$$\ce{CH2=C=O}$$
I wonder if it is a functional group or a normal group. As a functional group I couldn't find out.
Is it a normal compound or a functional group and what is it's
IUPAC name

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) the name of the given compound is the systematically constructed name for ketones.
Therefore, the preferred IUPAC name (PIN) is ethenone.
The name ketene is retained, but only for general nomenclature.

The compound you have drawn is called ketene, since the central carbon is part of both a ketone and an alkene. Ketene is the parent compound of the ketene functional group, where the terminal carbon is substituted with some more complicated structure.
Ketene itself is not stable and dimerizes to diketene by a [2+2]-cycloaddition. Diketene was commercially available and can be converted to ketene by thermal or photochemical conditions.

Ketone - R-CO-R .. I mean that there are two Alkyl group but in the post there is only one. That part is troubling me out

@Darshan Jain It's like you'd insert C between C and O in carbonyl group.

Yes, it is a ketone attached to a highly strained 2-membered ring (a double bond can be viewed as a 2-membered ring).  It is the 2-membered ring analogue of cyclopropanone.

Ketenes do not require photochemical conditions to do cycloaddition. It would be unwise to do so. You would not get that product but 1,2-cyclobutandione instead. Ketenes are reactive enough to do 2+2 with a little stirring. Light is incorrect.


text:
Dissolving Aspartic acid in vinegar?
I am trying to dissolve aspartic acid into vinegar as use as a carbon source in saltwater aquariums.   It does not dissolve.
I was told to increase the pH and it will work?  What is the best way to increase pH with minimal negative effects?
Sodium carbonate seems to work at dissolving; however I may be reducing the acetic acid from 5% to something lower?  Might also be creating an acetate?  Calcium hydroxide could be a better choice for the aquarium but I can't seem to get the mixing ratio right?

Can I point out that in general we are not experts on saltwater aquariums. So if adding aspartic acid to your aquarium is a good idea or not is unknown to us.

I can only see one question in your text: What is the nest way to increase pH with minimal negative effects? Could you elaborate on what you would consider 'negative effects' might be?

The good news is there is no good reason to add vinegar to your aquarium.  You are correct that if you did, the addition of hydroxide to maintain proper pH would then neutralize acetic acid to acetate.
You can dissolve about 4.5 g of aspartic acid into a liter of neutral water. I assume you try to keep your saltwater tank at a pH of around 8.3 however, which should significantly increase it's solubility as aspartic acid is acidic, having a pKa of 3.9. This is the reason you could not dissolve the aspartic acid with 5% acetic acid, which still pretty acidic with a pKa of 4.8.
So the solution is to just use the slightly basic aquarium water itself, plus a dose of patience, to dissolve the aspartic acid. You could place it in a tea-bag next to your aerator or something if you wanted to isolate it and see how quickly it dissolves.


text:
Bromine with 10 electrons
On a test question, my organic chemistry professor gave us a really big molecule and told us to give list the formal charges on each atom. There was a $\ce{Br}$ with 3 lone pairs and a double bond to carbon (which then had a single bond to a $
\ce{H}$ and a $\ce{N}$). On the key, he labeled it with a + charge and I was wondering how is that possible (or if it is a mistake)? Shouldn't a $\ce{Br}$ with that many electrons be highly negative or does it go onto the next orbital allowing it to accommodate for this?

Sounds like a typo to me. Probably should have been two lone pairs, double bond, and positive charge.

It is also highly unlikely, that Br can have 10 electrons.

I suspect that the issue here is with counting the number of lone pairs.
I think the structure you are referring to looks something like the right hand one shown below.
As you can see, this structure can be described by the two resonance forms shown. The double bond arises because the bromine 'donates' one of its lone pairs to form another bond to the carbon, shifting the positive charge onto the more electronegative bromine and hence stabilising the ione. Since this is the 'more stable' resonance contributor it is usually shown as the structure of the ion in drawings but the actual structure is more accurately represented as a resonance hybrid of the two.

Good question! As you state, normally bromine forms negative ions with a charge of -1, as all orbitals are complete except for the 4p5, able to accept one electron. However, there is a positive bromonium ion, first suggested by Irving Roberts and George Kimball in 1937. The generic term for a positive halogen ion is halonium ion.
The first reference has a nice electron models of this odd ion.


text:
Purpose of ammine in Tollens' reagent
In this previous question the mechanism for the reaction of Tollens' reagent was outlined. As I understand it the oxidising agent in Tollens' is the diamminesilver(I) complex $\ce{[Ag(NH3)2]+}$ but the mechanism only showed the silver ion.
What is the purpose of adding ammonia to the silver nitrate to form the diamminesilver(I) complex: why not just use silver nitrate? Is it purely to provide a source of hydroxide ions or is the structure of the complex ion actually important?

Complexation reduces redox potential.

@Mithoron Can you expand upon this a bit. I had a feeling this might be the case but can you give some more data and possibly explain why

Maybe I'll find something :)

The formation of the diamminesilver(I) complex $\ce{[Ag(NH3)2]+}$ due to addition of ammonia has mainly three functions.
1.
The Tollens test takes place under alkaline aqueous conditions. The complexation of $\ce{Ag+}$ with $\ce{NH3}$ prevents precipitation of brown $\ce{Ag2O}$:
$$\ce{2Ag+ + 2OH- <=> 2 AgOH <=> Ag2O + H2O}$$
2.
The Tollens test shall be selective for aldehydes. $\ce{[Ag(NH3)2]+}$ is a milder oxidising agent than $\ce{Ag+}$.
$$\begin{alignat}{2}
\ce{Ag+ + e- \;&<=> Ag}\quad &&E^\circ = +0.799\ \mathrm{V}\\
\ce{[Ag(NH3)2]+ + e- \;&<=> Ag + 2 NH3}\quad &&E^\circ = +0.373\ \mathrm{V}
\end{alignat}$$
The difference in redox potential can be explained using the stability constant of $\ce{[Ag(NH3)2]+}$:
$$K_\text{B} = \frac{\left[\ce{[Ag(NH3)2]+}\right]}{\left[\ce{Ag+}\right]\left[\ce{NH3}\right]^2}$$
$$\begin{aligned}
E&=E_{\ce{Ag+}}^\circ+\frac{RT}{F}\cdot\ln\left[\ce{Ag+}\right]\\
&=E_{\ce{Ag+}}^\circ+\frac{RT}{F}\cdot\ln\frac{\left[\ce{[Ag(NH3)2]+}\right]}{K_\text{B}\cdot\left[\ce{NH3}\right]^2} \\
E_{\ce{[Ag(NH3)2]+}}^\circ&=E_{\ce{Ag+}}^\circ+\frac{RT}{F}\cdot\ln\frac{1}{K_\text{B}} \\
&\approx E_{\ce{Ag+}}^\circ-0.059\ \mathrm{V}\cdot\log K_\text{B}
\end{aligned} $$
3.
The complex formation equilibrium slows down the overall reaction. A slow, controlled reaction is important for creating the desired silver mirror. If $\ce{Ag+}$ is reduced too quickly, colloidal silver metal would appear, which would create a black cloudy liquid.

How have you got from the third to the fourth line of the redox potential explanation with $K_B$

@bon For the standard electrode potential $E_{\ce{[Ag(NH3)2]+}}^\circ$, I assume that the activity $a$ of each solute (except $\ce{Ag+}$) is 1 (i.e. $c\approx 1\ \mathrm{mol/l}$); thus $\left[\ce{[Ag(NH3)2]+}\right]=1$ and $\left[\ce{NH3}\right]=1$.

Ah ok that makes sense


text:
Wavelength extention in AAS
Theoreticly, the width of the spectral line in  AAS (atomic absorbtion spectroscopy) is 10^-5, but in fact there happens an extention of it and it becomes 0.002-0.005. There are some factors which lead to this extention. Most common types of extentions are :

Doppler extention
Pressure extention
Self-absorbance extention.

Does anyone know how to explain in a simple way why do these 3 kinds of extentions take place?
Do these deviations happen all the time in these measurments or are they some deviations that can happen in case the concentration of atoms is to high?

I think you mean "extension", or more commonly called "broadening". Have you tried maybe to consult some textbook or wikipedia? http://en.wikipedia.org/wiki/Spectral_line

From this source broadenings occur due to

Doppler effect
  This effect arises because atoms will have different components of velocity along the line of observation.

Lorentz effect
  This effect occurs as a result of the concentration of foreign atoms present in the environment of the emitting or absorbing atoms. The magnitude of the broadening varies with the pressure of the foreign gases and their physical properties.
Quenching effect
  In a low-pressure spectral source, quenching collision can occur in flames as the result of the presence of foreign gas molecules with vibration levels very close to the excited state of the resonance line.
Self absorption or self-reversal effect
  The atoms of the same kind as that emitting radiation will absorb maximum radiation at the center of the line than at the wings, resulting in the change of shape of the line as well as its intensity. This effect becomes serious if the vapor, which is absorbing radiation is considerably cooler than that which is emitting radiation.

As Greg pointed out in the comment, You can find more detailed explanation on the topic here
Pressure broadening should be dependent on the concentration of atoms. The other type of broadening shouldn't be dependent on the concentration

Thanks a lot, do you have any idea about the last question I made ? Because all these things you mentioned are very likely to happen all the time ?


text:
When is phosphorus a polyatomic molecule?
Phosphorus is listed as a single element in the periodic table; however, I see that it is also a polyatomic molecule as $\ce{P4}$. I am trying to determine the number of atoms of phosphorus in an chemical equation, so I can have a balance equation. So, when would I use the polyatomic version, or the single atomic version if the chemical equation is a word equation, and the subscripts are not listed? For example the word equation is $\ce{P + Cl -> PCl5}$. Is $\ce{P -> P4}$ or just $\ce{P}$? Also, would $\ce{Cl}$ be diatomic too ($\ce{Cl2}$) in this equation?

If there are many allotropes single atom is generally used.

First make sure you're dealing with elemental phosphorus and not something else with P in it. For example, in agriculture it's common to talk about P levels, P concentration, P determination, etc, but that's never elemental phosphorus but actually phosphate ($\ce{PO4^{3-}}$) salts/acid/minerals/etc.

Phosphorus is most common as white phosphorus (but more stable as red phosphorus) and in this form, it exists as a tetrahedron, or $\ce{P4}$. Chlorine is most stable as a diatomic molecule, as you may have learnt it with the other common gases that exists as diatomic molecules

authors: منوعات ملوكي


text:
Determining spin of metal complex
Is there any way to determine the spin of $\ce{[Fe(OH)6]^{4-}}$ without looking at the spectrochemical series?

Are you asking experimentally or theoretically?
Theoretically speaking for general cases you should know what is the energy of the high-spin and low-spin configuration, and in given state, what is the approximate spin and vibrational entropy so you can estimate an equilibrium constant for given temperature. Now this either needs some calculation, or some sort of database from where you can make this estimations.
Experience (i.e. database or whatever series) says that water is a weak ligand, from that you generally guess that water complexes are generally in high spin state.
Experimentally you can determine spin state by measuring magnetic susceptibility, interpreting EPR, measuring 57Fe Mossbauer spectrum etc


text:
Dissolving to Create a Solution with Ethanol
If at 95% Ethanol can dissolve chemical X at 1mg/1mL then theoretically if you have 500mg of chemical X you can dissolve it in 500mL of $\ce{EtOH}$ to get 1mg/mL correct?
My question is, if you have 75.5% $\ce{EtOH}$ for example in a 750mL solution - how many mL will you need to create the 1 mg/mL concentration of quantity of 500mg chemical X?
Is it just 19.5% more of the 500mL volume or do you have to figure out the molarity of each and then go from there?

Knowing that the material (solute)  has a solubility of 1 mg/ml in 95% ethanol does not help you predict its solubility in 75% ethanol. It could be more soluble or less soluble, depending on the properties of the solute.


text:
Why carbon nanotubes can act as a good electrical conductor?
Bulk carbon is an insulator, but at nanoscale (e.g. carbon nanotubes) carbon may become a good electrical conductor. If I am correct this means  carbon nanotube band structure is metal-like. E.g. it has partly filled valence band (like in the rightmost case in the figure below).

At nanoscale materials are influenced by quantum confinement. Because of quantum confinement semiconductor nanoparticle has higher band gap and discrete energy levels compared to bulk semiconductor (see the figure about quantum confinement below).

I am assuming that quantum confinement phenomenon also occurs for carbon at nanoscale. Then, how certain carbon nanotubes can show good electrical conductivity?

I am sorry to say, but your question contains several false assumptions, most importantly:

Nanotubes are good conductors:
No, not all of them. Certain types have metallic/semi-metallic conductance, others are semiconductors. Single wall carbon nanotubes are generally indexed with a so-called chirality index (m,n) that tells you how the carbon sheet is wrapped up into a tube. See wikipage.
Nanotubes behave exactly like 3D nanoparticles in terms of band structure:
No, nanosheets (graphene) and nanotubes behave in different ways and your description of quantum confinements don't work in that way in 1D infinite (nanotubes) and 2D infinite (graphene) cases.
You can derive an approximate band structure of graphene/nanotubes using tight-binding models, which in most cases predicts correctly, whether the structure is metallic or semiconducting. See simulation of nanotube band structure.


text:
If for a chemical reaction the standard gibbs free energy is negative then :
If the standard gibbs free energy for a reaction is negative then the forward reaction will take place and the reaction will not be reversible. Is it true ? I have a problem understanding what is gibbs free energy and its significance in a chemical reaction.
Thanks in advance !

All reactions are in some sense equilibrium reactions. The Gibbs free energy is just a convenient measure of energy that we can use to understand more about our system. In particular you may be familiar with a few equations for the Gibbs energy:
\begin{equation}
\Delta G =\Delta H-T\Delta S,
\end{equation}
\begin{equation}
\Delta G=-RT\ln k
\end{equation}
\begin{equation}
\Delta G =-nFE
\end{equation}
For chemists, these three equations are the most commonly used and depend on what exactly you are investigating, but there are plenty of others too.
If we consider the second of these equations we note that $k$, the equilibrium constant, appears. We can see that as the equilibrium constant increases (and since $x$ follows $\ln x$ in limits) we note that the change in Gibbs energy becomes more negative the larger the equilibrium constant becomes. In other words the more products you have relative to reactants, the more negative the change in the Gibbs energy is, we call this spontaneous.
We could also take the third equation and note that $E$ is the reduction potential for a redox couple ($X^{n+}+e^{-}\rightarrow X^{(n-1)+}$). The more positive the potential for the reduction, the more spontaneous the reaction is.
Hope that helps you, oh and I'm assuming you mean reversible in the reaction sense of equilibria not in the thermodynamic sense of a reversible process?


text:
Calculating pH of an unknown base, and the pKb

The percent dissociation of a base $\ce{X}$ in a $0.002\ \mathrm{M}$ solution is $4.7 \cdot 10^{-3}\ \%$. What is the pH of the solution and what is the $\mathrm{p}K_\text{b}$ of base $\ce{X}$? The autoionization of water may not be neglected.

Attempt at solution: We have the following reactions: $$ \ce{X + H2O(l) <=> XH+ + OH-} \\ \ce{H2O(l) <=> H+ + OH-}$$
Since the percent dissociation is $4.7 \times 10^{-3}\ \%$, this enables us to find the concentration dissociated: $$ \frac{x}{0.002\ \mathrm{M}} \cdot 100 = 4.7 \times 10^{-3},$$ so $x = 9.4 \times 10^{-8}$. Now I did: $$ K_\text{a} = \frac{x^2}{0.002 - x}$$ and I let $(0.002 - x) \approx 0.002$ since $x$ is so small. But this gives me the wrong answer, and it seems I haven't taken the autoionization of water into account yet. How should I solve this problem then?

In such an solution, since the autoionization of water may not be neglected, you should expect $x=9.4 \times 10^{-8}$ do not mean $[\ce{OH-}]=9.4 \times 10^{-8}$. So you can only leave it unknown.
Then, $K_\text{b} = \frac {[\ce{XH+}][\ce{OH-}]}{[\ce{X}]} \approx \frac {9.4 \times 10^{-8} [\ce{OH-}]}{0.002} = 4.7 \times 10^{-5}[\ce{OH-}]$.
However, we still do not know what $[\ce{OH-}]$ is.
If only to consider the autoionization of water balanced, you will get
$[\ce{OH-}][\ce{H+}] = K_\text{w} = 1 \times 10^{-14}$
and this is still not enough.
The answer can only consider by the fact that the solution will be electrically neutral,
$[\ce{OH-}]=[\ce{H+}] + [\ce{XH+}] = \frac {1 \times 10^{-14}}{[\ce{OH-}]} + 9.4 \times 10^{-8}$
$[\ce{OH-}]^2 - 9.4 \times 10^{-8} [\ce{OH-}] - 1 \times 10^{-14}=0$
Since $\Delta = (9.4 \times 10^{-8})^2 + 4 \times 10^{-14} \approx 4 \times 10^{-14}$,
$[\ce{OH-}] \approx \frac {9.4 \times 10^{-8} + 2 \times 10^{-7}}{2} = 2.47 \times 10^{-7}$.
Thus, $\mathrm{pH} = -\log \left(\frac {1 \times 10^{-14}}{2.47 \times 10^{-7}}\right) = 7.39$
$\mathrm{p}K_\text{b} = -\log (4.7 \times 10^{-5} \times 2.47 \times 10^{-7}) = 10.94$.
Sometimes the “electrically neutral” trait will be very useful, this is a good example.

pkb = 10.94 ? At pH = 7.39, this would imply that the concentration of the base is much higher than the concentration of the corresponding acid. However, it is stated in the question that the percent dissociation of the base was only 4.7*10^-3 %. These things do not match.

The method applied in order to solve this problem is in principle ok.
However, the statements given are not correct, leading to wrong conclusions in the calculations.
[XH+]  is not equal to 9.410^-8 (M), but to 0.002 - 9.410^-8 = 0.01999... (M).
[X] is not equal to 0.002 (M), but to 9.4*10^-8 (M).

don't forget that high pKb means low base

Let us call the base for A- and its corresponding acid for HA.
Ctot = [HA] + [A-] = 0.002 (M) => [HA] = 0.002 - [A-].
Equilibrium: A- + H2O = HA + OH-
[A-]/( [A-] + [HA]) = 4.7*10^-3/100 = 4.7*10^-5 =>
[A-] / ([A-] + 0.002 - [A-]) = 4.7*10^-5 =>
[A-] = 9.4*10^-8 (M) => [HA] = 0.002 - 9.4*10^-8 = 0.001999906 (M)
The charge balance of the system is: [H3O+] = [OH-] + [A-]
To compensate for the autoprotolysis of water we multiply all terms of the charge balance with [H3O+] and get:
[H3O+]^2 = [H3O+] * [OH-] + [H3O+] * [A-] =
10^-14.0 + [H3O+] * [A-]
=> [H3O+]^2 - 9.4*10^-8 *[H3O+] - 10^-14.0 = 0
[H3O+] = 1.574 * 10^-7 => pH = 6.802
At equilibrium:
pH = pka + log [A-]/[HA] =>
pka = pH - log [A-]/[HA] = 6.802 - log (9.4*10^-8/0.001999906) => pka = 6.802 + 4.327 = 11.129.
pka + pkb = 14.0 => pkb = 14.0 - 11.129 = 2.87
Answer:
pH ≈ 6.8
pkb ≈ 2.9


text:
Ring expansion from a given cyclic carbocation
How will the cyclobutane ring behave in the case of cyclobutylmethylium (cyclobutylmethyl cation)?

I initially thought there would be ring expansion to a five membered ring so that there may be less angle strain and a secondary carbocation instead of a primary one. But I have also been told that there will be ring contraction for stability purposes. I would be highly obliged if someone could explain the mechanism involved here.

Related: Carbocation Rearrangement Due To Ring Strain And Back Bonding

I think your friend is thinking of the cyclobutyl carbocation which does ring contract to the cycylopropyl carbinyl carbocation (and also equilibrates with the  methallyl carbocation).
However, just as you thought, the cyclobutyl carbinyl carbocation does ring open to the cyclopentyl carbocation (ref_1, ref_2, ref_3).  This rearrangement is driven by carbocation stability (primary to secondary) and relief of the steric strain present in the 4-membered ring, again, just as you suggested.

Note: this question has been asked previously on SE Chem, however I believe the accepted answer is incorrect, as it seems to primarily address the cyclopropyl carbinyl case, and what is said about the cyclobutyl carbinyl carbocation (little to no ring expansion) is incorrect.

What would you think is more stable from the species $\ce{C4H7+}$, the cyclopropylmethylium (that you drew) or the methylcyclopropan-1-ylium? I tend to think the latter, but I am absolutely not certain.

@Martin Good question, the cyclopropylmethylium is primary and resonance stabilized, while the 1-ylium is tertiary with a bit more ring strain.  I'd vote for the former, but I dunno.  Somebody must have done a good calculation.

@ron in general, there's probably some level of delocalized charge between both carbons.

@GeoffHutchison  How would you describe the delocalization in the methylcyclopropan-1-ylium case?

The point would be that under QM, in either case, the charge will not wholly be on one carbon. That said, I ran a high-level CCSD calculation and it's pretty localized on the primary (as you drew): http://i.imgur.com/MhTrSzu.png

Delocalization in the cyclopropylmethylium that I drew is very conformation dependent (see here). In the bisected geometry (looks like that's what you used for your calculation) there is extensive delocalization into the ring.  What did you find in the methylcyclopropan-1-ylium case? Which was more stable?

In concluding Remarks of their work on the structural study of the 2-norbornyl cation, Olah and coworkers (Ref.1) stated that:

More importantly, rather than being rare, two-electron, three (or multi)-center bonding is characteristic of electron-deficient molecules, including nonclassical carbocations.

Later, they found computational evidence for the existence of nonclassical cyclobutylmethyl cation (Ref.2). According to the part of abstract in the Ref.2:

Ab initio calculations at the MP2/cc-pVTZ level show that the cyclobutylmethyl cation is a nonclassical σ-delocalized species, which is distinct from the global minimum $C_2$-symmetric cyclopentyl cation...

However, in reality, they failed to achieve persistent ((long-lived) classical or nonclassical cyclobutylmethyl carbocation, even at lower than $\pu{-80 ^\circ C}$ (Ref.3). Nonetheless, just as you thought, the cyclobutyl carbinyl carbocation readily undergoes ring expansion-rearrangement to give the corresponding cyclopentyl cations (See B (i) in the scheme; NMR evidence are provided in the reference). During the extended study, the authors studied (2-methylcyclobutyl)methanol derivative in superacidic medium and, suggested a mechanism depicted in the following scheme (part A), which clearly shows the reaction goes through nonclassical carbocation:

Moreover, cyclobutylmethanol took a different path under superacidic conditions, undergoing dimerization followed by rearrangement when warm a little bit. Mechanism C has been suggested to explan the outcome. All these evidence suggest that the chemical behavior of cycloalkanes depends on the ring size and the nature of the reagents used.
References:

G. A. Olah, G. K. Surya Prakash, Martin Saunders, “Conclusion of the Classical-Nonclassical Ion Controversy Based on the Structural Study of the 2-Norbornyl Cation,” Acc. Chem. Res. 1983, 16(12), 440–448 (https://doi.org/10.1021/ar00096a003).
V. Prakash Reddy, G. Rasul, G. K. Surya Prakash, G. A. Olah, “Structural Studies of Nonclassical Cyclobutylmethyl Cations by the ab initio Method,” J. Org. Chem. 2007, 72(8), 3076–3080 (https://doi.org/10.1021/jo0701334).
G. K. Surya Prakash, V. Prakash Reddy, G. Rasul, J. Casanova, G. A. Olah, “The Search for Persistent Cyclobutylmethyl Cations in Superacidic Media and Observation of the Cyclobutyldicyclopropylmethyl Cation,” J. Am. Chem. Soc. 1998, 120(51), 13362–13365 (https://doi.org/10.1021/ja9828962).


text:
Do all double displacement reactions lead to precipitates?
I know that most of the double displacement reactions form some kind of precipitates as their products. But do all of the double displacement reactions form precipitates? I have searched on this site as well as on Google, I've found answers, but they are not clarifying my doubts. Maybe I need a simpler explanation.
So, if you answer is yes, they all lead to precipitation, please justify.
If it's no, not all of them give precipitates, please justify with an example.

http://chemistry.stackexchange.com/questions/14885/how-can-you-predict-a-double-displacement-reaction-with-no-precipitate

No, recall the definition of a precipitate "insoluble ionic compound". Take this reaction into consideration $$\ce{ NaOH~(s) + HCl~(l) <=> H2O~(l) + NaCl~(aq)}$$ none of the products are insoluble. You'll learn more when you go over gas evolution reactions.

@MARamezani Please use <=> in mhchem ;) Subscripting the state is also not encouraged any more.

The exception cited above is a neutralization reaction, a special type of double replacement reaction


text:
OD660 readings of yeast
I am working on a yeast fermentation project, and I am getting final OD660 reading of 1.85 for different media: yeast extracts+cane molasses, nitrogen source+cane molasses, and glucose+cane molasses at different concentrations. The final OD reading is the same for all media. Is it normal? Are my readings correct?

I cannot comment on the expected growth rates in different media, but in order to avoid errors in measurement, I would dilute my samples to achieve optical densities in the 0.1 to 1.0 range.

Many inexpensive spectrophotometers cannot measure absorbances over 1.5 or so reliably. It's possible that you're out of the measurement range of your instrument. Try diluting your samples with water first. If you dilute 1:1, then the absorbance value of your sample is double what you measure (assuming the Beer-Lambert law is valid in this situation)
Also, I imagine molasses is very absorbent at that wavelength so it might be obscuring any change from the yeast and it may not be possible to measure the yeast growth by absorbance, with such a matrix.


text:
How does chromium help stainless steel prevent rusting?
Various sources (such as this one) say that stainless steel is stainless because a passivation layer of an oxide is formed, but I haven't found details on why chromium helps create this passivation layer. I can think of a couple possibilities:

Chromium changes the lattice constant of the iron to better match the spacing of the oxide layer.
Chromium migrates to the surface, and with enough Cr a Cr-rich layer forms on the surface which in turn creates the passivation layer.

Are either of these possibilities true, and if not, how does Cr help make the passivation layer?

TL;DR
    Note that the passive layer forms on the surface, there needn't be any change to lattice constant. Chromium needn't migrate , the Cr present on the surface will form the layer to protect it. The key point is how the layer develops from say a single-atom layer of oxide to the usual/maximum width by migration of electron and oxygen in the oxide layer. The process at molecular and atomic level is more complex than you think.
From wikipeddia article on Passivation

There has been much interest in determining the mechanisms which describe how the thickness of the oxide layer on a material increases with time. Some of the important issues include: the relative volume of the oxide compared to the parent metal, the mechanism by which oxygen diffuses through the metal oxide to the metal - oxide interface and the relative chemical potential for the oxide to form. Boundaries between micro grains, if the oxide layer is crystalline, form an important pathway for oxygen to reach the unoxidized metal below. For this reason, vitreous oxide coatings – which lack grain boundaries – can retard oxidation. The conditions necessary (but not sufficient) for passivation are recorded in Pourbaix diagrams.
Some corrosion inhibitors help the formation of a passivation layer on the surface of the metals to which they are applied. Some compounds, dissolving in solutions (chromates , molybdates) form non-reactive and low solubility films on metal surfaces.

The mechanism

All passivating‐type inhibitors are oxidizing agents. However, not all oxidizing agents are inhibitors. A passivating ‐type inhibitor functions by producing local‐action current which anodically polarizes a metal into the passive potential region and thereby provides the means for obtaining a noble mixed potential. This mechanism is independent of whether passivity is caused primarily by oxide or adsorbed films.
The major factors which determine whether a particular system will exhibit passivity are:
  (a) reversible potential of the redox system created by the oxidizing inhibitor;
(b) electrochemical parameters for reduction of the inhibitor on the metal surface (exchange current, Tafel slope, and limiting diffusion current), and
(c) critical anodic current and Flade potential(The potential of a passive metal immediately preceding a final step fall from the passive to the active region.) of the metal. If oxygen is present in the system, some chemicals may function by changing its reduction kinetics.
It is further shown that the amount of inhibitor found associated with the surface is not necessarily related to adsorption. Data are presented for passivation of stainless steel and titanium. The mechanism is discussed in terms of various oxidizing agents including oxygen, chromate, molybdate, and pertechnetate.

Some theories I could find for passivation are Kinetic Theory of Inhibition and Passivation in Electrochemical Reactions and Scaling theory in a model of corrosion and passivation

I found this paper which suggests that the Cr concentration at the surfaces is lower than in the bulk until ~8% Cr at. conc., at which point the concentration increases abruptly to ~20% Cr (and higher experimentally, apparently). So there is evidence that the Cr concentration at the surface may be different from the bulk and that this coincides with the corrosion resistance in Fe-Cr alloys, which means my second guess was closer to being correct.

Reason there is greater concentration of Cr at  surface is surface is (mostly) CrOx, whereas  bulk is roughly overall Cr content of alloy > "enhanced" Cr at  surface.  At higher Cr contents  Cr comp rises in bulk (more Cr in alloy) makes delta between surface/bulk less.


text:
Interstitial sites: edge and face sharing?
In close packed structures there is one octahedral hole per close packed atom. In a HCP unit cell these octahedra form a chain of face sharing polyhedra. Is this this universal in close packing or only common to HCP?
More generally this question is "how do we know when interstitial sites (T± and Oh) share edges or a faces?"


text:
Software for technical calculation of chemical reactors
I'm not a chemical engineer and don't have good background in it.
Just my father asked me to do something for him, as I'm a computer engineering graduate.
This is a software for calculation of technical parameters of reactor. I haven't seen it. I've just printed output of it. Fields are like following:
5  REM**************
10 REM***********
20 REM  Name etc.
~~~~~~~~~~~~
35 REM ********** PRODUCTIVITY********
~~~~~~~~~~~~
155 REM ***** INPUTS*******
~~~~~~~~~~~~~~~~~~~~~~~~
220 DTB
240  DTS
~~~~~~~~~~~~~~~~~~~~
555 REM *********** CALCULATIONS****************
~~~~~~~~~~~~~~~~~~~~~~~~~
~
~
~
********RESULTS*************
DTO=75
G=543.2
Q=XXXX
XXXXXX

Can anyone help me to find what is the software for calculation. I'd be very grateful if you can help.

Years ago a colleague of mine mentioned ASPEN as a typical software (bundle) used to model chemical plants.  Despite this and this lack of examples how the print-out will look like -- and I did not work with it, either.


text:
How to make lead nitrate from pure lead
Is it possible to add copper nitrate solution and lead so that a single displacement reaction occurs to yield lead nitrate. I recently melted lead and added cooper nitrate to the molten lead. I then collected the copper nitrate solution I had poured and repeated the process of melting the lead and pouring the solution over it. I figured that eventually the copper will be displaced by the slightly more reactive lead. Will this method yield lead nitrate?

I am not sure of putting copper nitrate to lead. But the common way to produce lead nitrate is to put lead into nitric acid. It will produce lead nitrate and hydrogen gas.
$\ce{Pb + 2HNO_3 -> Pb(NO_3)2 + H_2}$
If you have lead oxide instead of lead it also can priduce lead nitrate by adding to nitric acid
$\ce{PbO + 2HNO_3 -> Pb(NO_3)2 + H2O}$


text:
How do I write many cyclic voltammetry curves in Nova?
We are making an electrochemical research on supercapacitors that involves checking stability of our material. That requires repeated writing of cyclic voltammetry curves. Our goal is to write at least 2000 of them, but there is a problem with Nova software (version 1.8, to be specific): if we want to write even 50 cycles, Nova complains that "only last 30000 point will be written". We can adjust how many points are plotted, but when it comes to 2000 cycles, it's impossible to extract any information from them (15 points per one curve is just ludicrous).
Has anyone dealt with such problem? Is there a simple solution?

I would say that you should contact Autolab themselves. They must have a solution to this. Can't you take successive data files and overlay them?

Try to use command "repeat n-times", that will make n-data files and each can have 30 000 points.


text:
Equilibrium constant when adding more of a reactant
Suppose I have the reversible reaction:
$$\ce{A +B⇌ C} $$
The reaction is at equilibrium with equilibrium constant $K$
I am told that if I increase the concentration of $\ce{B}$, the rate for the forwards reaction will exceed the backwards one. Fair enough.
I am also told that $K$ will necessarily increase. Why though? Its true that Forward reaction > back ward reaction until we reach a new equilibrium such that more of $\ce{C}$ is produced but I don't see why this implies in any way that the final quotient $\frac{[\ce{C}]}{\ce{[A][B]}}$ will necessarily be any greater.
Certainly if we have a simple reaction:
$$\ce{A⇌ C} $$
and we add more of $\ce{A}$, then the equilibrium constant for the new final final state will remain as it was, ceteris paribus.
What am I getting wrong, because my textbook suggests that $K$ will always increase, no matter what type of reaction I am dealing with (of course, as long as all reactants are in a suspended form, e.g. dissolved or gaseous and the reaction is subject to Le Châtelier's principle).
Please, if possible, keep the answer as a level intelligible to a high-school student. I don't know much about advanced university chemistry.

K will increase initially as equilibrium is disturbed but it will then return to its original value once a new equilibrium is established (provided the temperature doesn't change).

@bon The equilibrium constant $\rm{K_{eq}=\frac{[C]{eq}}{[A]{eq}[B]{eq}}}$ does not change upon addition or removal of species. The reaction quotient $\rm{Q=\frac{[C]}{[A][B]}}$, however, does change immediately after the equilibrium is disturbed, and with time converges to the same value as $\rm{K{eq}}$ once more. The distinction is subtle but important, and causes some confusion between students, so it should be made clear.

@NicolauSakerNeto Yes this is true my comment was poorly worded.

You're correct.  The equilibrium constant will not change.  The equilibrium constant is only defined at equilibrium.  Your analysis of the situation was flawless.

In your book, it was probably mentioned that the Reaction Quotient (Q) will change, K changes only with temperature and there is no other way to change it. Catalysts also effect only the rate (speed) of reaction.


text:
Why is a ketone not capable of hydrogen bonding among themselves?

The ketone has one oxygen atom in it and there are two lone pairs hanging out, both of which make it seemingly capable of hydrogen bonding. But in reality it is not.
Did some research and found this on chemguide:

In methoxymethane, the lone pairs on the oxygen are still there, but the hydrogens aren't sufficiently δ+ for hydrogen bonds to form. Except in some rather unusual cases, the hydrogen atom has to be attached directly to the very electronegative element for hydrogen bonding to occur.

Based on the info, I am assuming that the carbon atom in carbonyl group is not electronegative enough to give a partial charge large enough for hydrogen bonding to the hydrogen atoms, say in pentanone.

Technically they can, but it's very weak.

Short answer: the ketone oxygen can participate in hydrogen bonding
The compound you mention, methoxymethane, is an ether not a ketone.

Ethers don't hydrogen bond very well because the oxygen isn't very polarized.
On the other hand, the carbonyl in a ketone is polarized.  We can draw resonance structures that show this polarization.

(image source)
The negative polarization on the carbonyl oxygen allows it to play the role of hydrogen bond acceptor as illustrated in the following drawing.

(image source)
Edit: If you are asking if the hydrogens on the carbon adjacent to the carbonyl carbon (or any other carbon in the backbone of a simple ketone) can participate as the donor in a hydrogen bond, then the answer to this question is "no". These $\ce{C-H}$ bonds are typically less  acidic and less polar than the $\ce{O-H}$ bond in water.
Therefore, while the oxygen in pentanone can participate in hydrogen bonding with donor hydrogens in other molecules, the hydrogens in pentanone cannot participate in hydrogen bonding.  Pentanone cannot hydrogen bond with itself.

Umm, he's asking about C-H-O hydrogen bonds - ketone as donor

@Mithoron  Thanks for pointing this out.  I've edited my answer accordingly.

"Isn't very polarized"? Can you see just by looking?

And one more question, so what you are saying is in order for a H atom to be a hydrogen Bon donor, donating itself, it has to be DIRECTLY attached to either a sulfur, oxygen, or fluorine atom?

first comment -  for example, the electronegativity of H is 2.1, C is 2.5, so a C-H bond is not very polar. O is 3.5 so an O-H bond is polar. second comment - Yes, N, O, S, halogen are the most common.

Also, one last thing, HF has three lone pairs on F. That means with other compounds containing a H and one of the the three elements, except for itself, it can make four hydrogen bonds. But with itself, it can only two?

No, the number of lone pairs doesn't matter, an atom (N, C, O,F, etc.) can only make one H bond.  Once it makes an H bond, its polarity and electronegativity are reduced and it cannot make any additional H bonds.

https://en.wikipedia.org/wiki/C%E2%80%93H%C2%B7%C2%B7%C2%B7O_interaction https://en.wikipedia.org/wiki/Hydrogen_bond#Bifurcated_and_over-coordinated_hydrogen_bonds_in_water - your answer and last comment are like one side of a coin, but there is another


text:
Can one dichlorotriazine molecule form two bonds with cellulose?
In all the reactions I saw online, dichlorotriazine is shown to form only one bond with cellulose (example 1, example 2; paper).
But cellulose has many potential sites for bonding. Can dichlorotriazine actually give up both Cl atoms for forming two bonds with cellulose?

Yes, see p. 374 here

@ron, that's exactly what I was looking for!


text:
Quantum mechanical explanation for Le Chatelier's principle?
We know from basic chemistry that a reaction reaches equilibrium as described by Le Châtelier's principle. What's the quantum mechanical explanation to this principle? How a change in temperature/concentration causes an equilibrium to shift, from the point of view of wave functions?

Le Chatelier's principle comes from thermodynamics, so I don't see a clear derivation of it from quantum mechanical assumptions. If there is a connection, it would likely be immensely laborious to work out.

@NicolauSakerNeto, I did not ask for a derivation from QM, but rather a separate explanation of the same phenomena.

The main thing that classical and quantum statistical mechanics have in common is probability.  In classical stat mech the equilibrium value of an observable is given by
$$\langle A \rangle = \sum_i^N p_iA_i$$
where $A_i$ is the value of $A$ in microstate $i$ and $p_i$ is the probability of microstate $i$.  The expression for $p_i$ at equilibrium depends on the ensemble (microcanonical, canonical, etc).  One way to define equilibrium is that the probabilities are constant
$$\frac{\partial p_i(t)}{\partial t}=0$$
So if one perturbs the system so that this equation doesn't hold the $p_i$'s will change until it does and Le Chatelier's principle is a heuristic way of predicting whether the individual $p_i$'s increase or decrease.
In quantum stat mech the equilibrium value of an observable is given by
$$ \langle A \rangle = \tfrac{1}{N} \sum_\alpha^N \langle \Psi^{\alpha}| \hat{A} |\Psi^{\alpha}\rangle$$
where $\Psi^{\alpha}$ is the wavefunction for microstate $\alpha$
The wavefunctions are typically expanded in an orthonormal basis ($\phi$'s)
$$ \langle A \rangle = \tfrac{1}{N} \sum_{k,l} \rho_{l,k}\langle \phi_l| \hat{A} |\phi_k\rangle $$
where the density matrix is can be defined by a density operator
$$ \rho_{l,k} = \langle \phi_k| \hat{\rho} |\phi_l\rangle $$
The equilibrium is then defined as
$$\frac{\partial \hat{\rho}(t)}{\partial t}=0$$
So if one perturbs the system so that this equation doesn't hold the $\hat{\rho}$ will change until it does and Le Chatelier's principle is a heuristic way of predicting whether the individual $\Psi^{\alpha}$ contributions to  $\hat{\rho}$ increase of decrease.
The time dependence of $\hat{\rho}$ can be calculated by the time-dependent Schrödinger equation.  So Le Chatelier's principle is a heuristic way of predicting roughly how the time-dependent wavefunctions will evolve at the simulation approaches equilibrium.
To get a more physical picture one can introduce the Born-Oppenheimer approximation in which case the numerical simulation of the time-dependent Schrödinger equation becomes a classical MD simulation (with a QM energy function) from which the time dependence of the probabilities $p_i$ can be computed and Le Chatelier's principle is a heuristic way of predicting whether the individual $p_i$'s increase or decrease.

what's MD simulation?

Molecular dynamics.  If you introduce the Born-Oppenheimer approximation then the time-dependent Schrödinger equation reduces to Newton's equation of motion $F=ma$, where $F=-\frac{\partial E}{\partial x}$ and $E$ is the energy obtained from the time-independent Schrödinger equation $\hat{H}\Psi = E\Psi$.

Typically speaking, QM involves single systems at zero Kelvin. So as @NicolauSakerNeto indicates, temperature and concentration would usually be reflected in statistical mechanics and thermodynamics.
That said, for large systems (e.g., nanoparticles, metals, materials), finite temperatures result in thermal broadening. That is, rather than occupying the absolute lowest energy state, there is some fractional occupation of higher states given by the amount of thermal energy available and the density of electronic (or rotational or vibrational) states.
The individual orbitals themselves are purely mathematical and do not change with temperature - only the occupation of them.

I'm not sure how the answer addresses the question.

The point is that quantum mechanics doesn't have much to say about temperature or concentration. QM is used to describe individual molecules or nanoparticles or proteins or whatever. It's hard to describe "concentration" of one system. Concentration is a bulk response usually treated with statistical mechanics. Stat mech does take energy levels from QM, but temperature won't change the QM properties, only the way the electrons fill in the orbitals.

Your question is a bit strange. Le Chatelier's principle makes sense in a thermodynamics or stat mech context. It's like you're asking me to explain how gravity affects QM.

Quantum gravity takes care of that :)

Quantum chemistry uses an 'isolated system' approximation for convenience.  Le Chatelier's principle is meaningless in this approximation -- it only is sensible when the system is large enough that direct quantum computations would be overwhelmingly impractical. (Aaaand, just now read down to the first paragraph of Ian's answer...)

Quantum mechanics tells us the possible stationary states of any system with a fixed energy. It would be extremely computationally difficult to use quantum mechanics to find the stationary state of a system as large as a solution but in theory this is possible.
Statistical mechanics tells us the probability that the system is in any one of the possible stationary states for some finite temperature. In particular, stat. mech. tells us which of the quantum mechanical stationary states are most likely to be occupied at equilibrium.
Le Chatelier's principle tells us how to find the new equilibrium when a system is displaced from equilibrium. In other words, Le Chatelier's principle tells us which distribution of quantum mechanical stationary states the system returns to after the system is disturbed in some way. Quantum mechanics tells us the possible states of the system, but it can tell us nothing about how those states are occupied. Only statistical mechanics can tell us how those states are occupied. Therefore, quantum mechanics alone cannot explain Le Chatelier's principle; in fact, Le Chatelier's principle holds regardless of the specific interactions between molecules and does not depend on any quantum effects at all. Le Chatelier's principle would hold even if we modeled the interactions between molecules completely classically with a Lennard-Jones or Morse potential.

There is no one because the Le Châtelier Braun principle is outside the scope of quantum mechanics.
The 'principle' formulated by Le Châtelier and Braun in year 1888 summarizes the response of an initially stable system to a perturbation that moves the system away from the equilibrium state. This is irreversibility, because the response of the system is always accompanied by a production of entropy $\Delta_i S >0$. In fact, the thermodynamic stability condition is
$$\Delta_i S = -\frac{C_V (\delta T)^2}{2T^2} -\frac{1}{T \kappa_T}\frac{(\delta V)^2}{2V} - \sum_{ij} \left(\frac{\partial}{\partial N_j} \frac{\mu_i}{T}\right)\frac{\delta N_i\delta N_j}{2} < 0$$
Effectively if perturbing the system reduces entropy, this implies that the original state was one of maximum entropy and the system will react to the perturbation by spontaneously recovering the equilibrium.
Now, it is a well-known theorem that quantum mechanics conserves entropy, and cannot explain irreversibility. To provide a solid foundation to the Le Châtelier Braun principle and similar statements about irreversible evolutions one needs to extend quantum mechanics to irreversible phenomena. This is a difficult task. There are several attempts. You can find the Brussels-Austin school work in [1]. A list of references to the MIT school work can be found in [2].
To provide a quantum explanation to the Le Châtelier Braun principle, one starts with an irreversible quantum equation of motion (for instance that from Quantum Thermodynamics). This equation has two terms: one linear and reversible (that reduces to Schrödinger form in the pure state limit) and other nonlinear and irreversible
$$\frac{d\hat{\rho}}{dt} = \left(\frac{d\hat{\rho}}{dt}\right)_{rev} +\left(\frac{d\hat{\rho}}{dt}\right)_{irr}$$
The irreversible component produces entropy $\hat{s}=\hat{s}(\hat{\rho})$ when there is dissipation (like it happens in the response of a stable system to perturbations)
$$\frac{d\hat{s}}{dt} = \frac{d\hat{s}}{d\hat{\rho}} \left(\frac{d\hat{\rho}}{dt}\right)_{irr} >0$$
and the inverse of this inequality can be used to formulate a microscopic analog of the classic principle.
REFERENCES
[1] http://onlinelibrary.wiley.com/book/10.1002/9780470141588
https://journals.aps.org/pra/abstract/10.1103/PhysRevA.60.861
http://onlinelibrary.wiley.com/book/10.1002/0471619574
[2] http://quantum-thermodynamics.unibs.it/