text:
Why is the Vitamin B complex, a "complex"?
I often come across the term "Vitamin B Complex" in my biology classes and innumerable times on the back of multivitamin packets, but what does the term "complex" here, even mean?
I'm still in high-school, and the only times we ever come across the term "Complex" in our Chemistry classes is when we're dealing with Coordination Compounds, but I don't think that's what our textbook is getting at when it says "Vitamin B Complex". I compared the structures of a few B-vitamins, but they don't appear to have anything to do with the coordination Chemistry kind of "Complexes".

My teacher tells me, that the term "Complex" here is supposed to indicate the existence of a wide range of B-Vitamins (Like $\ce{B1, B2, B3,}$ etc.). But I didn't find that 'explanation' satisfying, since there are all sorts of other vitamin categories (such as D and K) that have quite a few subtypes ( $\ce{D1, D2, D3, K1, K1, K3,}$ etc) but I've never heard of a Vitamin D Complex or a Vitamin K Complex for that matter.
So my question stands:-
What does the term "Complex" mean, with regard to the Vitamin B Complex? Do  Vitamin D Complexes and Vitamin K Complexes exist as well?

Your teacher is right. You or anyone else can try to sell a bunch of D vitamins under the brand name "Vitamin D Complex", and that would be technically OK for pretty much the same reason. Whether or not you will succeed on the market is a question outside of realm of chemistry.

I was not quite right an hour ago, and neither was your teacher. See, the reason for the "complex" term is that they (though chemically unrelated) often occur in the same food, and so were historically discovered together. Only gradually the individual active compounds were separated and identified. This is also the reason behind their naming (i.e., why is it $B_2, B_3\dots$, rather than $C,D\dots$).

If you google "D Complex" or "K Complex" you will find some instances. I think the reason no-one boldly claims to contain "Vitamin B." is because they're widely known to be multiple ones, which may be related to the fact that they're clearly chemically unrelated to each other.

Why question about semantics got so many upvotes? It's just a "non-chemical" meaning of complex, like a complex of buildings - simply saying there's many vitamins in there.

I have been studying Vitamins for a while (Nutritional Biochemistry), and though I agree with the explanation of your teacher and with the answers already given,
Part of the reason has to do with their discovery:
The general definition of vitamins is described:

The vitamins are a disparate group of compounds; they have little in
  common either chemically or in their metabolic functions.
  Nutritionally, they form a cohesive groupof organic compounds that are
  required in the diet in small amounts (micrograms or milligrams per
  day) for the maintenance of normal health and metabolic integrity.
  They are thus differentiated from the essential minerals and trace
  elements (which are inorganic) and from essential amino and fatty
  acids, which are required in larger amounts.

In addition to systematic chemical nomenclature, the vitamins have an apparently illogical system of accepted trivial names arising from the history of
their discovery.
For several vitamins, a number of chemically related compounds show the same biological activity, because they are either
converted to the same final active metabolite or have sufficient structural similarity
to have the same activity.
I believe initial classification was based on two forms- fat solubility and water solubility. It must have been elucidated through experiments with milk:

When it was realized that milk contained more than one accessory food
  factor, they were named A (which was lipid-soluble and found in the
  cream) and B (which was water-soluble and found in the whey). This
  division into fat- and water-soluble vitamins is still used, although
  there is little chemical or nutritional reason for this, apart from
  some similarities in dietary sources of fat-soluble or water-soluble
  vitamins

Indeed the  "Vitamin B Family" has a lot of compounds (riboflavin, pantonthenic acid, biotin, thiamin, niacin, cobalamin etc)  that are structurally unrelated,
but were grouped possibly due to their discovery as they either had similar chemical property-water solubility abeit some forms of vitamins (e.g ascobic acid) are water soluble.

This division into fat- and water-soluble vitamins is still used,
  although there is little chemical or nutritional reason for this,
  apart from some similarities in dietary sources of fat-soluble or
  water-soluble vitamins. Water-soluble derivatives of vitamins A and K
  and fat-soluble derivatives of several of the B vitamins and vitamin C
  have been developed for therapeutic use and as food additives

As the discovery of the vitamins progressed, it was realized that “Factor B”
consisted of a number of chemically and physiologically distinct compounds.
Before they were identified chemically, they were given a logical series of alphanumeric
names: B1, B2, and so forth
From a pharmaceutical perspective...
The "complexity" of vitamin B compounds simply suggest a combination of different compounds that fall under this family. Because vitamin
compounds are required in relatively large amounts compared to fat soluble counterparts, it makes sense to combine them as supplements rather than packaging
7+ pharmaceutical products as "Vitamin B1", "Vitamin B2", "Vitamin B3"... suplements, it creates more confusion and non-compliance instead their systematic names are used for example pyridoxine injection (vitamin B6), Nicotinamide tablets but where multivitamin are prescibed its just a "complex" of there compounds usually 5-6 of these "B" vitamins.
It is important to note that such compounds are requirement in very small quantities/day, and overdosing (through such  these compounds can cause undesirable side effects).
Other forms of vitamins (fat soluble) are equally important but most of them are either stored in body in reserves e.g Vitamin A, others like Vitamin K are synthesised in the body by bacteria hence are needed in very small amounts
This is the chief reason why there are fewer supplements of such vitamins (although they may be needed in special cases).

For a compound to be considered a vitamin, it must be shown to be a
  dietary essential. Its elimination from the diet must result in a
  more-or-less clearly defined deficiency disease, and restoration must
  cure or prevent that deficiency disease.

Several forms of vitamin D?

Vitamin D is not strictly a vitamin, rather it is the precursor of one
  of the hormones involved in the maintenance of calcium homeostasis and
  the regulation of cell proliferation and differentiation, where it has
  both endocrine and paracrine actions.

The name vitamin D1 was originally given to the crude product of irradiation of ergosterol,
which contained a mixture of ergocalciferol with inactive lumisterol (an isomer of ergosterol) and suprasterols. When ergocalciferol was identified
as the active compound, it was called vitamin D2. Later, when cholecalciferol was identified as the compound formed in the skin and found in foods, it was
called vitaminD3.
Remarks
The "Vitamin B" naming of these compounds must have been through discovery, and no clear experiments had accurately produced identity of these compounds, there were named as they were discovered but since they have been identified they they now have systematic names
abeit vitamin B still being used today and are formulated as "vitamin B complexes" in pharmaceutical products (perhaps to avoid confusion) hence systematic names are used (folic acid, pantonthenic acid, biotin, thiamin, niacin, cobalamin etc) I have never come across complexes of other Vitamins. Remember for a compound to be named a vitamin it must fit the description above, but I am not disputing the fact that other compounds with similar biological activities exist as "K" group.
References

Nutritional Biochemistry of Vitamins (Bender)
Nutritional biochemistry (Brody)
Modern Nutrition in Health and Disease (Rosset al)

This is a fantastic answer! +1'd it!

Thanks, just a small edit, (yes there are supplements for individual vitamins but they are given systematic names e.g nocotinic acid, pyridoxine and not usually vitamin Bx etc I guess the reason is the same as explained. There are actually injections for pyridoxine (B6), Nicotinamide given as 50, 100mg tablets as well. But the main point I guess when given as multivitamin supplement they as refered just as "complexes"

It is essentially exactly what your teacher says: the term vitamin B complex is used because the structures assigned a name ‘vitamin Bx’ don’t really have anything in common although they are all somewhat important.
As to why nobody ever uses vitamin K complex or vitamin D complex: well, at least in my country those vitamins are very under-represented when compared to ABCE. ABCE vitamins are those often seen as the most important, the most valuable or similar.[citation needed] They are often advertised as being included in supermarket products. The advertising industry probably doesn’t want to ‘establish’ yet another thing they could add but would be rather cost-intensive.

All vitamins are equally important, in that we can't live without them. I guess those A,B,C are most often missing or lacking in our food, hence the need for the supplements. Also, this is the reason why they got discovered first and got named by the first letters of the alphabet.

"Vitamin D" is also often advertised as being included in supermarket products, most often whole milk.

@Random832 Your profile doesn’t really tell me in which corner of the world you live, but I can say for sure that I would be really hard-pressed if asked for any ‘vitamin D included’ ad. And drinking a lot of milk, I feel safe to say that it’s not on $95~%$ of the whole milk sold in Germany.

In the US, the label on the package usually says "Vitamin D Milk" as the name of the product - I didn't mean like in an ad they put in the paper, but on the package itself. I think it's a practice like putting Iodine in salt. (I don't know of other fat percentages have it added, but it's only named that way on the package for whole milk) - like on the shelf there's fat free skim milk, 2% reduced fat milk, and "vitamin D milk" which is whole milk.

In J. Biol. Chem., 1913, 15, 167-175,
an article titled The Necessity of Certain Lipins in the Diet During Growth,
E. V. McCollum, one of the pioneers in vitamin research did indeed use the term "complex" and wrote:

Whether the resumption of growth is the result of supplying in the ether extract of egg or of butter, some indispensable organic complex of the chemical nature of the lipins, or is the result of a stimulating action of some substance accompanying the lipins, cannot be decided from the data at present available.

 (Emphasis, mine)

Although he clearly uses "complex" there, that sounds more like a hand-wavy use of the word to refer to a reality that was not yet fully understood. I can't read the rest of the paper because it's behind a paywall, though, so I can't confirm this theory. I know that McCollum eventually determined that there are multiple B vitamins, but I don't think he had made this discovery as early as 1913. I'm betting there is another, later paper that would contain a more significant use of the word "complex".

I didn't know that "complex" here meant a combination of the B vitamins; I had always assumed that it meant vitamin B12, which is a coordination complex itself:

B12 is an amazing molecule! AFAIK though, the remainder are just plain old organic molecules.

@orthocresol Yes, the rest are "boring". I am not one for synthetic chemistry, but the fact that this was made still blows my mind. Woodward is captivating to listen to.

And yes, that is the correct answer, not mine, and I know about the listings, but I felt it would get lost as just a comment, where images can't be displayed.


text:
electroplating process
I'am not familier with chemisty,& trying electroplating on brass, by using C4 H6 CUO4.H2O.But the result is unsatisfied,am I using right chemical or I have to choose another one.I yes,anyone may guide me the right one.

Welcome to Chem.SE! Thanks for posting your question here. Would it be possible to provide more information about your apparatus, and what it is about the result that's unsatisfactory? It's hard to know what might be the problem without more details. Also, if you haven't already, please check out the tour and help pages for more information about the site more generally. Enjoy!

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text:
Nitrogen fertilisation expressed as nitrogen equivalent
In a plant pathology paper I found levels of fertilisation applied in the experiment expressed as nitrogen equivalent.
Reporting the exact phrase:

Nitrogen fertilisation with 8.6 g of nitrogen equivalent …

I found that equivalent weight for an element would be its atomic weight divided by its valency so, if I'm not wrong, equivalent weight for N would be ~4.67 g
What does this mean?
How much fertiliser was applied e.g. using ammonium sulfate?

This is referring only to the mass of elemental nitrogen available, rather than the total mass of fertilizer. In your case, ammonium sulfate is 21% nitrogen (see the table here for other fertilizers and here for more information on fertilizer composition), which gives ~40.95g ammonium sulfate.

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text:
Why does the Gibbs Free Energy change determine spontaneity of a reaction?
Here is a doubt that arose in my mind while studying the chapter Thermodynamics of the 11th standard. (It might be silly, but I would love it if you try to understand what I mean to ask before down voting. Also, I want the doubt cleared.)

We have been told that if the value of $\Delta G$ or Gibbs' free energy change is negative or zero the process is spontaneous or in equilibrium, respectively. If otherwise, it is non-spontaneous.
Why? What does $\Delta G$ exactly signify that if it is positive the process won't happen by itself. I mean, I know what $\Delta G$ is equal to, but why does spontaneity of the reaction depend on its sign?

System may alter energy by exhangeing mass from the surroundings in case the system is open:)

By the second law of thermodynamics, your main goal is to ensure that $$\Delta S_{\mathrm{universe}} \geq 0$$
$$\Delta S_{\mathrm{universe}} = \Delta S_{\mathrm{sys}} + \Delta S_{\mathrm{external}}$$
$\Delta S_{\mathrm{sys}}$ is measurable on your system. $\Delta S_{\mathrm{external}}$ is a bit more challenging, since you're not going to count the microstates of the external system, so you compute indirectly by heat transfer to the surrounding. Thus,
$$\Delta S_{\mathrm{external}} = -\frac{\Delta H}{T}$$
Combine:
$$T\Delta S_{\mathrm{universe}} = T\Delta S_{\mathrm{sys}} - \Delta H$$
or
$$\Delta G = -T\Delta S_{\mathrm{universe}} = \Delta H -T\Delta S$$
And with the sign flip, you're looking for $\Delta G \leq 0$.
Importantly, I should point out that this is not a closed system in the sense that we assume constant-pressure and not constant-volume.
If you're looking at a process that is constant-volume, you need to replace $\Delta H$ (which has pressure-volume work compensation) with $\Delta U$ (which does not) and replace $\Delta G$ (Gibbs free energy, constant pressure) with $\Delta F$ (Helmholtz free energy, constant volume).

Thank-you....so what I find is: Both of you are trying to prove ∆S(universe,final)<0, and I guess because it signifies more stability. Right?

We're not trying to prove that at all. That's the second law of thermodynamics. Both answers are trying to derive a different formulation from that basic statement. It has nothing to do with stability. There's no energy involved. It has everything to do with the number of microstates of the universe.

The condition for spontaneity is outlined by the Second Law of Thermodynamics, i.e.
$$\mathrm{d}S_\text{isol} > 0 \tag{1}$$
where the subscript isol indicates an isolated system.
A closed system and its surroundings necessarily comprise an isolated system (if either matter or energy leave the system, it must be to the surroundings, and vice versa. Therefore, together, the closed system and its surroundings are isolated.) The changes of entropy of the system and of the surroundings must therefore be positive:
$$\mathrm{d}S_\text{syst} + \mathrm{d}S_\text{surr} > 0 \tag{2}$$
Conventionally, the surroundings are taken to be an infinite heat reservoir, held at a constant temperature $T$. Furthermore, the system and surroundings are taken to be in thermal equilibrium, i.e. the system also has temperature $T$. Any heat transfer occurring from system to surroundings is therefore reversible. For example, if $\mathrm{d}n$ moles of a reactant decomposes exothermically, then the temperature of the system will be raised from $T$ to $T + \mathrm{d}T$, and heat is therefore transferred from the system to the surroundings. However, since the temperature gradient is infinitesimally small, the heat transfer is reversible.
From the point of view of the surroundings, which is chemically inert, the only thermodynamic process taking place is heat transfer, which is reversible. The entropy change of the surroundings is therefore given by the equality
$$\mathrm{d}S_\text{surr} = \frac{\mathrm{d}q_\text{surr}}{T} \tag{3}$$
Furthermore, any heat transferred out of the system must be transferred into the surroundings, and vice versa. Therefore,
$$\mathrm{d}q_\text{syst} = -\mathrm{d}q_\text{surr} \tag{4}$$
Combining equations $(2)$ through $(4)$, we can obtain
$$\mathrm{d}S_\text{syst} > \frac{\mathrm{d}q_\text{syst}}{T} \tag{5}$$
which should not really be surprising, since our system is undergoing a spontaneous chemical process.
At constant pressure, $\mathrm{d}q_\text{syst} = \mathrm{d}H_\text{syst}$. Therefore
$$
\begin{align}
\mathrm{d}S_\text{syst} > \frac{\mathrm{d}H_\text{syst}}{T} \tag{6} \\
\mathrm{d}H_\text{syst} - T\mathrm{d}S_\text{syst} < 0 \tag{7} \\
\mathrm{d}G_\text{syst} < 0 \tag{8}
\end{align}
$$
as desired. Note the imposition of the conditions of constant $T$ and $p$.
If $\mathrm{d}G_\text{syst} = 0$, then the system is at equilibrium. Whether any process with $\mathrm{d}G_\text{syst} = 0$ is considered "spontaneous", I am not entirely sure.

I think it is fair to say that $\mathrm{d}S_{isol}=0$ also implies a spontaneous process, and equilibrium (at least in my course) is considered two-way spontaneity.

Fair enough. As I said I was not really sure, so thanks for chipping in.

Perhaps a third person will also have a say, so that we can both be sure ;)

My answer uses non-strict inequality.

"The changes of entropy of the system and of the surroundings must therefore be positive" Why? Shouldn't it be zero?


text:
Finding out pKa of acid from molar conductivity
I'm reading about the electrical properties of solution where there is a problem like that:

The molar conductivity of $0.0250\ \mathrm M$ $\ce{HCOOH(aq)}$ is $4.61\ \mathrm{mS\ m^2\ mol^{-1}}$. Determine the $\mathrm pK_\mathrm a=-\log K_\mathrm a$ of the acid. (limiting ionic conductivity of $\ce{H+}=34.96\ \mathrm{mS\ m^2\ mol^{-1}}$ and limiting ionic conductivity of $\ce{OH-}=19.91\ \mathrm{mS\ m^2\ mol^{-1}}$)

I have to solve the problem using the equation below:
$$\frac1{\Lambda_\mathrm m}=\frac1{\Lambda_\mathrm m^0}+\frac{\Lambda_\mathrm mc}{K_\mathrm a\left(\Lambda_\mathrm m^0\right)^2}$$
Here which limiting ionic conductivity should I use to solve the problem?  $\ce{H+}$ or $\ce{OH-}$?

You need to add the limiting ionic conductivities for $\ce{H+}$ and $\ce{OH-}$ together to get the limiting ionic conductivity for all the ions in solution ($\Lambda_{0}$, which will replace $\Lambda^{0}_{\mathrm m}$ in your equation).
This arises from a simplification for calculating $\Lambda_{0}$ in weak electrolyte solutions (such as yours) according to Kohlrausch's Law in which it is stated:

Each ionic species makes a contribution to the conductivity of the solution that depends only on the nature of that particular ion, and is independent of the other ions present.

from which we can then estimate $\Lambda_{0}$ as:
$$\Lambda_{0} = \sum_{i}\lambda_{i,+}^{0} + \sum_{i}\lambda_{i,-}^{0}$$
For your problem:
$$\Lambda_{0} = \underbrace{(34.96 + 19.91)}_{54.87}\ \mathrm{mS\cdot m^{2}\cdot mol^{-1}}$$
we solve (I am omitting units here for clarity, and have confirmed via dimensional analysis that the answer is correct):
$${1\over 4.61} = {1\over 54.87} + {4.61\times 0.025\over K_{\mathrm a}\times (54.87)^{2}}$$
which simplifies to yield the answer (spoiler alert):

$$K_{\mathrm a} = 1.926\times 10^{-4} \implies \mathrm{p}K_{\mathrm a} = 3.72$$

and can be favorably compared to the value of 3.77 given here.


text:
Hydrogen Peroxide Heat of Dilution
Based on some data here, I am getting the heat of dilution of 100% $\ce{H2O2}$ as approx. 1100 cal/mole.

That's about 4.6 kJ/mol if I did my math right.
Is this a correct value? Seems too low to me.
e.g. For the corresponding heat of dilution for $\ce{HCl}$ I get 74.8 kJ/mol from Wikipedia.

HCl dissociates into ions which both needs solvatation; in short, HCl reacts with water quite fiercely. On the contrary, H2O2 just sits there and does nothing, except for a few relatively insignificant hydrogen bonds. So no, the huge difference is not unexpected.


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Why is H2SO4 a good dehydrating agent while other strong acids are not?
I understand that $\ce{H2SO4}$ is a very good dehydrating agent because it is a strong acid, so that it quickly donates $\ce{H+}$ to hydroxide ions in water to form $\ce{HSO4-}$ and $\ce{H3O+}$ .But we know that there are a lot of other acids stronger than $\ce{H2SO4}$ like $\ce{HClO4}$ which are not used as dehydrating agents. They can also readily give $\ce{H+}$ to hydroxide ions in water and act as dehydrating agents but we don't use them for this purpose. Why is that?

As pointed out by Maurice and myself, the argument provided in the answer by iad22agp is rather incorrect. Sulfuric acid is a dehydrating agent just because it is available in concentrated form whereas rest of the common acids like HCl(aq) are not. No, it is not a concentration effect. First of all, it is not possible to have HCl(aq) more concentrated than 38% wt/wt under normal conditions. This is already a saturated solution of the gas.
Secondly, liquified HCl cannot be a good dehydrating because as everyone knows HCl g fumes heavily in the presence of moisture. It cannot fulfill the criterion of a dehydrating or dessicant. The same arguments apply to conc. nitric acid as well.
The two main reasons which make conc. sulfuric acid a good dehydrating agent are (a) it has a very low vapor pressure and (b) secondly the hydration of sulfuric acid is quite exothermic as compared to other common acids. Maybe a quantum mechanical study reveal "why" this is so. Acid strength alone cannot be used to predict the dehydrating capability. One has to think about the vapor pressure as well.
General chemistry book also preach that sugars are dehydrated by conc. sulfuric acid. It is actually a wrong notion to say that sulfuric acid extracts water out of sugar. There is no water molecule in sugar. This is a chemical reaction between sugar and conc. sulfuric which generates water and carbon. The true demo for sulfuric acid dehydration is the dehydration of hydrated copper sulfate. Adding a few drops on the blue crystals makes them white (anhydrous) copper sulfate almost immediately.

Just a note in favour of textbooks. Once one know what it is spoken about, it is more than acceptable to say that reaction between sulphuric acid and, eg, a sugar, is a dehydratation one. Of course, it has nothing to do with a real extraction or with drying. Indeed water isn't there, but it is a dehydratation reaction. Beside this, plus 1 for fixing the thread.

@m.farooq (b) secondly the hydration of sulfuric acid is quite exothermic as compared to other common acids.  Is available comparison of with concentrated HNO3 and HClO4 ? Is possible this exothermic nature is just more commonly known for H2SO4, as it is more common acid form?

The properties of sulfuric acid in reactions have very little to do with its properties as a dehydrating agent. In a closed dessicator, for example, the acid will be separate from the substance to be dried. All that matters here is that any water vapour in the vessel will be rapidly absorbed by the acid as that is highly favoured by the thermodynamics of water mixing with the acid. And the low vapour pressure of the acid won't contaminate the substance.

@Poutnik, I don't have the enthalpies but if you recall even the hydration of sulfuric trioxide is very exothermic but when HCl gas dissolves in water, it is just warm. The same must be true for nitrogen dioxide. There is indeed something special about sulfuric acid and water combination.

@M.Farooq Hm, I guess hydration comparison of SO3, NO2, HCl is not much fair and relevant. NO2 is not HNO3 anhydrid and HCl is out of game. Add there P4O10, used to prepare N2O5 from HNO3 or in other extreme dehydrations. Acids of P(V) are not particularly strong dehydration agens.

@Poutnik, Right, but I was saying there is something special about the interaction of sulfur trioxide and water and sulfuric acid and water. The common factor is both release a lot of heat. The fundamental question in why?

@M.Farooq I bet dissolution of N2O5 or Cl2O7  must be special as well. All  N2O5, Cl2O7 or SO3 can be prepared applying P4O10 on the respective acid. We are used to about pure H2SO4, but not to pure HNO3 nor HClO4.

Nitric and perchloric acids are sold at 70% aqueous, so most of their dehydrating ability has already been consumed, while sulfuric acid is sold as 98%. At 98% concentration, nitric and perchloric acids are likely to be very good dehydrating agents, but these are also very strong oxidizers, less stable, etc.
Likewise, commercial hydrochloric acid is sold as a 38% solution in water, so it already has had its dehydrating ability consumed. However, I imagine that liquid hydrogen chloride would be a very good dehydrating agent.

H2so4 also has strong oxidising properties. If the oxidising nature of nitric acid is the reason we do not use it as dehydrating agent then h2so4 should also not be used.

Oxidizing properties of H2SO4 may be called strong, but they are nowhere near those of HNO3 and HClO4.

I dried $\ce{NiCl2 . 6 H2O}$ in a dry $\ce{HCl}$ gas stream in an inorganic lab course to give anhydrous $\ce{NiCl2}$. So yes, $\ce{HCl}$ even in gaseous form is a good dehydrating agent.

iad22agp's argument is that the dehydrating power is due to the minimum amount of water in the commercial solution of the acid. I am not sure about the value of this argument. For example acetic acid is available with a concentration equal to 100%. It is not a strong dehydrating agent.

This is a completely wrong argument. I agree with Maurice.

@Maurice Acetic acid strength is not comparable to mineral acids.

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text:
Why only two modes of energy exchange?
In the  thermodynamics chapter of the 11th standard, I read something that I have a hard time believing to be true:
"There are two ways by which a system can interact or exchange energy with its surroundings:
(i) Heat (ii) Work."
This is what I found in various books, exactly quoted from one of them.
Why only two? Why aren't there more ways for a system to interact or exchange energy with its surroundings? If there are, what are they?

Let's look from a different angle: you tell us what other ways there might be. If you won't name any, then at least you'll see why only two.

@IvanNeretin Well, these are forms of energy being used as 'modes of energy exchange'. Right? So, what if we try and use other forms of energy as modes of energy exchange?

Indeed, the system may exchange energy with its surroundings via light, or electric current, or by gravitation waves, or by whatever else. I guess all these subtleties are just omitted for clarity. Anyway, entry-level thermodynamics is mostly about gases in cylinders with pistons.

@IvanNeretin Okay. So other modes of "energy exchange " are possible?

It is a matter of convention, really. Maybe the authors of your book classify all these as merely forms of heat and work.

@IvanNeretin OK. Then, I mean, can you please give an example where light, for example is being used as a mode of energy exchange.

Think of the Sun as a thermodynamic system, and ourselves as a part of its surroundings.

@IvanNeretin Oh. Yes!

@IvanNeretin but radiation is a form of transfer of heat.

Yeah, sure. And gravitation waves are a form of work. A matter of convention, like I said.

@IvanNeretin So, technically, every exchange thing's heat and work!?

This is a matter of convention, how many times do I have to stress that? If your definition of heat includes light, fine. If it does not, that's OK anyway.

Oh.Ok. I got it now. Sorry for the inconvenience. And, thanks a lot! (:


text:
Basic behaviour of bromide ion with sulfuric and phosphonic acid
I  recently came across these two reactions:

$$\begin{align}\ce{2KBr + 3H2SO4 &-> 2KHSO4 + Br2 + SO2 + 2H2O}\tag{1}\\[0.6em]
\ce{KBr + H3PO4 &-> KH2PO4 + HBr}\tag{2}\end{align}$$

I understand that the bromide ion being the conjugate base of a strong acid $\ce{HBr}$ is a very weak base and therefore it cannot take $\ce{H+}$ from a very strong acid $\ce{H2SO4}$ in the first reaction. But the second reaction is an acid-base reaction.
I checked the $\mathrm{p}K_\mathrm{a}$ values of $\ce{H2SO4}$ and $\ce{H3PO4}$ to find that $\ce{H2SO4}$ is a way better acid than $\ce{H3PO4}$, then why is that bromide ion could take $\ce{H+}$ from $\ce{H3PO4}$ but not from $\ce{H2SO4}$?
The first reaction is a redox reaction and I understand that redox reactions take place when acid-base reactions are not possible because acid-base reactions are quick. Then why is the first reaction not an acid base reaction?

https://en.wikipedia.org/wiki/Hydrobromic_acid In first reaction you'd need sufficiently concentrated acid and in second too, I guess, as in very diluted solutions nothing would separate.

A common misconception is that the more acidic component will always protonate the less acidic componend and displace it from its salts. That misconception would be exemplified by equation $(1)$
$$\ce{HBr + KH2PO4 -> KBr + H3PO4}\tag{1}$$
This is incorrect. There is a constant acid-base equilibrium and even bromide ions will be protonated to a non-neglegible extent. The principal difference between $\ce{HBr}$ and $\ce{H3PO4}$ is that the former is a gas at room temperature and standard pressure. Thus, in an open vessel, any $\ce{HBr}$ in solution is again automatically in equilibrium with $\ce{HBr}$ in the gas phase. Thus, the equilibrium should better be written in form $(2)$:
$$\begin{multline}\ce{KBr (aq) + H3PO4 (aq) <<=>\\ <<=> HBr (aq) + KH2PO4 (aq) <=> \\ <=> HBr (g) ^ + KH2PO4}\tag{2}\end{multline}$$
As you can see, the gaseous $\ce{HBr}$ can diffuse away. This reduces the partial pressure of $\ce{HBr}$ above the solution, which affects the right-hand side equilibrium by Le Chatelier’s principle, reducing the concentration of $\ce{HBr}$ in solution, which in turn affects the left-hand side equilibrium, reducing the concentration of $\ce{KBr}$. Thus, if left to equilibrate in a well-vented area, large portions of $\ce{HBr}$ will diffuse away leaving a solution of potassium phosphate or potassium hydrogen phosphate.
The principle alluded to in the first sentence is thus correct, if rather than using acid strength it is quoted as:

The less volatile acid will displace the more volatile acid from its salts.

It is worth noting that an acid-base equilibrium $(3)$ also is observeable in your first reaction. However, the redox reaction is faster than the removal of hydrogen bromide. It is also irreversible, hence those are the final products obtained.
$$\ce{KBr + H2SO4 <=> HBr + KHSO4}\tag{3}$$

You said the redox reaction in the case of H2SO4 occurred faster than acid base reaction. But why is that, sulphur is in its most stable +6 oxidation state in H2SO4 and it has to go to +4 oxidation state in the redox reaction ? There should not be a great driving force for this redox reaction.

@Blue I did not say that the redox reaction be faster than the proton transfer. I said that the redox reaction is faster than diffusion of HBr out of the solution. Big difference. Also, do not fool yourself into thinking a ‘most stable oxidation state’ exists; these are only the most stable within given circumstances. What counts is the overall $\Delta G$ term, which, for redox reactions, is negative if and only if $\Delta E$, the potential difference, is favourable.


text:
Relative basicity of cyclic amines

Can anyone explain this trend (the amines are listed in order of increasing basicity)?
Actually, I thought of understanding this by thinking of inductive effect of the alkyl groups attached to the amine group but since they are cyclic I am confused how to apply the inductive effect here or I want to know whether there are other ways of dealing with this problem.

Well I would love to know the answer aswell! Well the the option with methyl group has to be obviously the most basic as it is a 3° amine. But what about the others?

Inductive effect in this sequence in the reason of the increased basicity only in the case of N-methyl piperidine compared to piperidine. Increased basicity in the sequence of azetidine < pyrrolidine < piperidine is connected with the ring strain. Upon the decrease in the C-N-C angle, sp3-hybridized N atom "rehybridize" to increase p-character of orbitals involved in the C-N bonds formation, leading to the increased s-character of orbital involved in the N-H bond and the lone pair of electrons. Since s-electrons are closer to nucleus, this rehybridization leads to the increased stability of the N-H forming orbital and the lone pair in more strained cyclic amines making the lone pair being 'less exposed' to protonation (less basic).

An alternative explanation is hyperconjugation between the lone electron pair on nitrogen and the sigma antibonding N-C orbital. The electron accepting ability of the orbital increases as the ring size decreases because the p character increases in the same manner (the energy drops).


text:
Is Jet Fuel/Kerosene still toxic after drying out?
A friend of mine is currently living in India. The people she is living with just treated the wooden door with either kerosene or jet fuel (its not clear which of the both, as the language barrier makes communication rather difficult). We do now wonder if touching or even being close to this door will be still dangerous after the kerosene/jet fuel dried out.

There's kerosene in toothpaste. So I would assume not. Nevertheless, you should ask the nearest expert about it.

For all practical purposes, jet fuel is kerosene. Once evaporated, there should be no issues.

Let's say "much less". If it was "kerosene" in the sense of "household fuel" i'd say mostly harmless. After it stops smelling (i.e. stinking up the whole room) it's probably safe. But then there is no telling what crook sold them this "kerosene".

Its India, they use kerosene for their lamps and cooker -.-

Then it's safe. After drying out, it is simply not there. The same applies to most common liquid fuels. There was once a nasty thing called leaded gasoline, but it is now banned pretty much everywhere.

Kerosene is a common timber treatment medium, and is often used as the carrier solvent for wood preservation, with additives such as antifungals, insecticides and wood preservatives present. Unless it is a low grade kerosene (so NOT jet fuel), paraffin residues are not likely to remain after evaporation. But the questions remains...what else was in the kerosene?

Kerosene is not a pure compound. To quote Wikipedia:

Regardless of crude oil source or processing history, kerosene's major components are branched and straight chain alkanes and naphthenes (cycloalkanes), which normally account for at least $70~\%$ by volume. Aromatic hydrocarbons in this boiling range, such as alkylbenzenes (single ring) and alkylnaphthalenes (double ring), do not normally exceed $25~\%$ by volume of kerosene streams. Olefins are usually not present at more than $5~\%$ by volume.[10]

[10]: American Institute of Petroleum (September 2010). "Kerosene/Jet Fuel Assessment Document" (PDF). EPA. p. 8. Retrieved October 28, 2016.

Because kerosene is a complex mixture, it is hard to say something specifically. However, the compounds noted are basically long chain alkanes and substituted cycloalkanes, a smaller fraction being alkylated aromatic rings. Benzene and naphthalene do not seem to be significant contributors.
None of these are classified as toxic. Most importantly, while alkylated benzenes and alkylated naphthalenes may still intercalate into DNA making them potential carcinogens, they are metabolised by a completely different pathway from benzene proper, making their metabolites much less dangerous. (Rather than the very reactive 1,2-epoxybenzene, various benzoic acids are the primary oxidation products.) That doesn’t mean they’re harmless, but toxic is the wrong designation and should be reserved for more dangerous compounds.

A general rule for all compounds is:

As long as you smell them, they can affect you.

The desired effect of painting a door with kerosene is, however, that the kerosene penetrates the door and stays there. The least-volatile constituents will remain on the door itself, protecting it from damage. If all the kerosene diffused away, the method would not be used because it would be worthless. Therefore, after the door has vented properly, letting the volatile components disappear, there is practically nothing left that will be harmful to any noteworthy extent.
In fact, I am include to proclaim that your average coloured paint may well contain more harmful chemicals that the kerosene does.
Yes, you can touch it, no you do not need space suits to be safe.

But is it in toothpaste? :)

@ToddMinehardt I don’t know, I never GC’ed my toothpaste …? ô.o


text:
Mechanism Permitting Proton Sponge Effect
The proton sponge effect is utilized as a drug delivery method to release contents packaged in micelles (usually coated in PEG or another protective polymer) into the cytoplasm of a cell after being endocytosed.
The micelle is able to attract hydrogen ions (which attracts chloride ions) into the endosome, decreasing the internal pH of the endosome. Water also follows the inward flow of ions and causes the endosome to burst, releasing micelle and contents into the cell.
I am wondering how ions are able to cross the endosome lipid membrane in order to attract water molecules and create swelling. I am aware that endosomes may have membrane bound hydrogen ATPases, but this doesn't explain chloride entry.

Feels like this might be more bio.SE than chem.SE...

It's a mix of both. I felt chem.SE was appropriate because I am asking about ion and lipid properties rather than a larger biological function. Lipids, ions and chemical-biology are all tags on this site as well, so I can't see why it should be removed.

It is true that ATPases pump protons across a membrane, and ions such Na and K also need membrane bound protein pumps; Na+ out of the cell and K+ into it. The crystal structures of some of these are known. see Brookhaven data base PDB code 2ZXE or 1K4C for examples. I'm not aware of anion pumps but they may exist.

@porphyrin Thank you for the references to the PDB structures, I will check  them out.

When I first read proton sponge effect, I thought of 1,8-bis(dimethylamino)naphthalene …

@Jan that's a pretty interesting molecule, I will read about it soon when I have time.


text:
Can a free radical be created by chemical reaction of non-radical species?
Here's my thinking:

Atomic oxygen in its ground state is a free radical because there are two unpaired electrons in p-orbitals.
When species react new orbitals are formed, and perhaps a situation can occur where the amount of degenerate orbitals is changed by a chemical reaction. This would implicate the filling of these orbitals, following the aufbau principle and hunds rule of filling.
Therefore it is thinkable that two reactant species (i.e. not homolytic bond fission) which are not radicals, can react in such a way that the new electron orbitals are filled in such a way that all electrons are no longer paired => radical species formed from non-radical species.
Likewise it is conceivable that a radical species can react with a non-radical species (i.e. not an even number of radical species terminating eachother) to form only non-radical species.

Is my argument and results 3 & 4 correct? Can someone give an example of 3 and/or 4?
Edit: My main takeaway - when considering the feasibility of a radical creation/destruction, look at conservation of spin

NOTE TO ALL: As of the time of this comment, the answers from Klaus, myself, and porphyrin were posted before the OP edited his question to include the two constraints excluding homolytic bond fission for #3 and radical termination reactions for #4. Klaus updated his answer in response to a comment from OP. Jan's answer was posted after the above edits to the question were made.

Allow me to re-reiterate that while your’s and Klaus’ were, Porphyrin’s was not. (Albeit by 3 minutes)

A free radical cannot be created in a simple, one-step process from non-radical reactants other than by bond dissociation (which you explicitly excluded).
A fundamental principle of quantum chemistry is the conservation of spin. Over the course of a simple reaction, the total spin quantum number $S$ must remain constant — excluding more complex effects like spin-orbit coupling. Non-radical compounds have a singlet state, i.e. their total spin quantum number $S = 0$ or $2S+1=1$; all electrons are paired. Typical organic radicals such as the tert-butyl radical have one unpaired spin and are therefore in a doublet state $S = 1/2$, $2S+1 = 2$. Diradicals such as $\ce{O2}$ are in a triplet state with $S = 1$ and $2S+1 = 3$. To arrive at a product state with nonzero $S$, you would either need to generate a pair of opposing spin compounds (simple bond dissociation) or start of with a nonzero $S$.
This is why reactions that start with peroxides often generate singlet oxygen rather than triplet oxygen. By complex mechanisms and emitting a faint red glow, this then slowly isomerises to triplet oxygen.

So because of the same conservation law, 4. is also impossible?

to continue with spin ideas, it is spin allowed for two singlet states to combine to produce two triplets, which at a push might be considered as radicals, for example $\ce{^1M + ^1O_2 \rightarrow ^3M + ^3O_2}$

@porphyrin Doesn't that completely contradict the conservation law Jan stated? You start with S=0 (singlet states) and end with S=2 (two triplet states)

@Adroit It doesn’t contradict much like the creation of two dublet radicals from a singlet molecule (i.e. homolytic bond-cleavage) doesn’t contradict.

@Jan I don't fully understand.  Does that mean there must be Equal amounts of electrons With opposite spins, in order to cancel out => S=0 before and after?¨

@Adroit Pretty much, but I’m not good enough at quantum mechanics to fully answer that.

@Jan Also, isn't this an example of raidcals being created by non-radical reactants without it being homolytic bond dissociation? So when you say "A free radical cannot be created in a simple, one-step process from non-radical reactants other than by bond dissociation ", you mean a single radical cannot be formed?

@Adroit, it easier to start with two triplets. Total spin has series $|S_1+S_2|..|S_1-S_2|$ which is $2,1,0$ spin or multiplicity $5, 3, 1$. Assume pair collide then complex formed has spin multiplicity one of $5, 3, 1$. Thus for singlet spin complex  $^3A+^3B <=> ^1(^3A ^3B) <=> ^1A + ^1B $ is spin allowed.  For triplet complex products are $^3A+^1B$ and probably no products from quintet complex. Triplet-triplet annihilation is well known in photochemistry as is singlet -singlet annihilation but a intense laser is needed as singlets have a short lifetime wrt triplets.

As for #3, photoinduced electron transfer (PET) reactions do what you want.
Imagine that 2,4,6-triphenylpyrylium tetrafluoroborate ($\ce{P+BF4-}$) is irradiated at $\lambda = 350~\mathrm{nm}$ in the presence of an alkene and a nucleophile ($\ce{ROH}$).

In the excited singlet state, $\ce{^1P+^*}$ is a strong electron acceptor, that oxidizes the alkene to the corresponding radical cation an is itself turned to a pyranyl radical, $\ce{P^{\cdot}}$. (Here is the first radical.)
Then the alcohol adds to the alkene radical cation, the adduct deprotonates spontaneously. (Here is the second radical.)

I don't fully understand your answer. There are some concepts i have yet to learn, like "singlet state" and "adduct". Furthermore I think I might have been misunderstood (see corrections in bold). Is your example mechanism an example of non-homolytic bond fission leading to non-radical species forming radical-species? If so, please try to make the mechanism more clear for me.

Apologies :) I'll add a sketch of the process that is not a homolytical fission to make myself more clear.

Can a free radical be created by chemical reaction of non-radical species?

Yes.
One straightforward example of this is the unimolecular dissociation of hydrogen peroxide into two hydroxyl radicals:
$$
\ce{H2O2 -> 2 HO^\bullet}
$$
Related, if you consider this homolytic dissociation reaction as occurring in the presence of a free radical scavenger like ABTS, you end up with two relatively stable radicals in the final solution:
$$
\ce{2 ABTS^{2-} + H2O2 -> 2 ABTS^{2-} + 2 HO^\bullet -> 2 ABTS^{\bullet\,-}
             + 2 OH-}
$$

Likewise it is conceivable that a radical species can react with a non-radical species to form only non-radical species.

Agreed, but only if you have an even number of electrons in the total system. If there's an odd number, you're guaranteed to have a radical in there somewhere, at all times.
That said, one example is the dimerization equilibrium of nitrogen dioxide with dinitrogen tetroxide:
$$
\ce{2 NO2^\bullet <=> N2O4}
$$

Good answer with use of understandable examples. I do, however, think my question was not phrased clearly enough, as I have been misunderstood, For point 3. I meant specifically non homolytic bond fissions. For point 4. I meant a odd number of radical species reacting with non-radical species to  create only non-radical species.

@Adroit An odd number of radical species (well, monoradical species, I guess) can never react to form all non-radical species. It's mathematically impossible when you have an odd number of electrons.

Also, @Adroit, it's bad practice on SE sites to make edits to existing questions that change the meaning of the original question, as it leaves any answers posted prior to the edit as being no longer relevant to the question. IMO you're better off leaving the question as it was, then asking a new question with more precise wording, with a link back to the prior question.

A radical species need not necessarily have an odd number of electrons? Think for example of O(g) or O2(g). They are radicals because of degenerate orbitals being filled by hunds 1. rule

@Adroit , quite right, I follow you now; that was why you specifically mentioned $\ce{O2}$ in #1. I still recommend reverting this question, and asking a new, better-clarified one, as I was not the only person confused by what you posted.

How exactly do i "rollback"?

@Adroit Click the "edited x min ago" link underneath your question to go to the 'edit history' page.  There you'll find a 'rollback' link associated with your original version of the question.

I appreciate you pointing out bad SE behaviour on my part. I will however chose not thos revert his post as i feel adding a new, nearly identical question on the page will result in clutter. I think my question was Clear enough, if not as clear as possible (as it hopefully is now).

@Adroit No problem at all, it's your call to make. Sorry to have been nipping at your heels so much!

Yes very,very many reactions:
$\ce{H_2 + O_2}$ produces H, O, OH and $\ce{HO_2}$ radicals;
$\ce{H_2 + X_2}$, where X is a halogen, produces H & X radicals,
$\ce{CH3CHO \rightarrow \cdot CH3  + \cdot CHO}$   and
$\ce{ \cdot CH3 + CH3CHO \rightarrow CH4 + \cdot CH3CO}$
and many others. These and many more are all radical chain reactions, either initiated by high temperature or light, and the rate laws are often misleadingly simple.
In a chain reaction one radical reacts with non-radical to form a new  radical and a non-radical as in the example with $\ce{\cdot CH3}$ and acetaldehyde above.

First example isn't great because $\ce{O2}$ is basically a diradical.

@zhe, yes it is paramagnetic. It produces OH radicals via $\ce{H2 + O2 \rightarrow 2OH\cdot}$. An alternative is $\ce{RH + O2 \rightarrow \cdot R + \cdot HO2}$ where RH is a hydrocarbon or $\ce{Br2 + CH4}$ which produces Br and $\ce{\cdot CH3}$ radicals

All these examples either start from radicals or include homolytic bond cleavages which was specifically excluded by OP.

@Jan This answer was posted before OP edited his question to add the constraint excluding homolytic bond cleavage.

@hBy2Py That is not correct. The edit to the question was made at 20:42:53Z, this answer was posted at 20:45:31Z — almost three minutes later. I noticed that your answer was indeed posted before the edit, which is why I did not downvote it.

@Jan Ah, my apologies, I didn't notice this answer until after the edits had been made -- I was in the midst of the discussion w/OP in the comments of my own answer at the time. For my information, how did you determine the exact time this answer was posted? All I can find is the humanized "17 hours ago" reference.

@hBy2Py Hover above the 17 hours ago with your mouse; a tooltip will appear with the exact posting time. Works here, on the questions page and in the revision histories.


text:
Chemical in (s) mixed with chemical in (aq) vs. chemical in (aq) mixed with chemical in (aq)
Hopefully the title is comprehensive enough.
But I am wondering, does it make a difference if you take a powder and mix it with an aqueous solution or take a solution with the very same chemical formula as the powder and mix it with the same aqueous solution?

Short answer is yes. A lot of times you are trying to control the exothermicity of a reaction by doing a slow addition. This is kind of hard with a powder, and if you dissolve it in a lot of solvent, you can add it slower than you would be able to with a powder.
Without the slow addition, the temperature may increase and decrease your yield or selectivity.

Wouldn't the reaction go quicker with the aqueous solution? Because the chemical is already dissolved and the ions floating around?

(1) You can add the aqueous solution much more slowly. (2) The aqueous solution is significantly diluted over the solid. So, not necessarily. It is possible and would depend on the reaction.


text:
What's the difference between polyurethane finish and polyurethane plastic?
The plastic version is made by mixing two liquid chemicals (MDI & polyol) which react immediately and become a solid. Polyurethane finish is applied as a liquid and dries, no mixing involved.

Or are they not related at all?

Prompted by Veritasium YouTube video.

I'd like to know the answer to this too.


text:
Colour changes in coordination complexes
What causes coordination complexes to change colour as they are heated? Do the ligands detach from the central metal or do the electrons held in the d orbitals leave? If both happen, which happens first?
Specifically, I am trying to explain the colour changes observed when potassium ferrioxalate $(\ce{K3[Fe(C2O4)3]})$ was heated. The colour change is from green-yellow to brown/red.

Many things can happen when you heat a coordination complex: ligand loss, reactions in the ligand structure, loss of hydration water, isomerization, oxidation/reduction of the metal center etc.
In the specific case of potassium iron(III) oxalate trihydrate, as reported in this paper, the thermogravimetric analysis shows that the first structural change happens at 113 ºC with the loss of three water molecules (from the trihydrate salt), yielding potassium iron(III) oxalate, a yellow-orange solid.
When the temperature reaches 296 ºC (in air), iron(III) is reduced to iron(II) to form iron(II) oxalate, an orange solid.
Further heating leads to the decomposition of iron oxalate into FeO at around 470 ºC. FeO is a brown/red solid (probably the compound you mentioned in your question).

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


text:
Why are free radicals so reactive?
Why are free radicals are so reactive? They can break almost any bond, including $\ce{C-H}$ bonds, which are fairly stable. Don’t they have an activation energy or something? Do they not also experience steric effects when colliding with reactants ?

They do experience steric effects.

Related: http://chemistry.stackexchange.com/questions/8586/molecule-with-13-total-electrons/8587#8587

Radicals are so reactive because they require so much energy to form. When we speak about radical reactivity, "more reactive" generally means a more exothermic hydrogen atom abstraction step. The more exothermic this step, the more the transition state resembles the reactants (Hammond's postulate), with a correspondingly lower character of the carbon-centered radical product. This makes the reaction less sensitive to the stability of the carbon-centered radical.
This being said, some radicals do have an activation energy, and some don't. The $E_a$ is essentially zero for all hydrogen atom abstractions by fluorine atom, and the free energy barrier arises solely from the pre-exponential factor of the Arrhenius equation ($k=Ae^{-E_{a}/(RT)}$), which is near 13 for the halogenations listed below.

The activation energies for abstraction by chlorine atoms are also exceedingly small, but there is a slight preference for more highly substituted $\ce{C-H}$ bonds. Lastly, the activation energies for abstraction by bromine atoms are substantial, clearly showing a preference for abstraction of 3°, 2°, and 1° $\ce{C-H}$ bonds in that order. It is because of these differences in activation energies that we observe the varying regioselectivities of the different halogenation reactions$^{[1]}$.

$[1]$ Anslyn, E. V.; Dougherty, D. A. Modern Physical Organic Chemistry; University Science: Sausalito, CA, 2006.


text:
Can a reaction of the form A + B → AB ever be endothermic?
I understand that forming a bond between $\ce{A}$ and $\ce{B}$ is an exothermic process, with negative enthalpy, but won't there be energy required to get $\ce{A}$ and $\ce{B}$ close enough together for the bond to form?  So we would get an energy hill with activation energy on one side and heat of formation on the other side.
Is it ever possible in a chemical reaction to have more energy required to get the reactants $\ce{A}$ and $\ce{B}$ close enough together than the energy released in forming the bond?  In terms of atoms I am thinking that the repulsion between two atoms may reach a peak at some distance greater than the bond length, so we might get an energy hill diagram where:

What about a situation like breaking and forming bonds on elemental crystals like graphite and diamond?  If graphite pieces $\ce{(A)}$ are sloughed off of a large piece of graphite $\ce{(B)}$ this obviously requires energy to break the bonds, but the graphite flakes don't spontaneously re-bond to the large piece upon light contact.
Likewise, a fractured diamond does not re-bond on contact. Also, in formation of a diamond, why is there a need for a continual input of heat?  If the energy of formation is greater than the energy needed to activate the reaction, then wouldn't the released heat of formation cause other diamond bonds to form without more input of energy.  This indicates to me that some energy is required to begin the process of combining $\ce{A + B}$ to form $\ce{AB}$, even if the actual bond formation is exothermic - that there is a two sided energy hill in reactions of $\ce{A + B -> AB}$, though the activation portion may be smaller than the formation portion.
What about an analogous $\ce{A + B -> AB}$ nuclear fusion reaction with heavy species above iron?  Clearly in this case there is a large input of energy needed to fuse elements above iron, and a small release as the nucleons reach a distance that allows the strong nuclear force to dominate.  In this case the energy hill has a greater required activation energy than an energy of formation.  How is this fundamentally different from chemical bonds?
I understand that a critical component is that nuclear attraction becomes stronger than electrostatic repulsion at a critical distance, but might there be some instances of chemical reactions where $\ce{A}$ and $\ce{B}$ have to pass through a less stable conformation to get to the more stable bonded state?

So there are three possible answers:

$\ce{A + B -> AB}$ reactions can be described energetically solely by the heat of formation of the bonds between $\ce{AB}$.
$\ce{A + B -> AB}$ can be energetically described by an energy hill, but the heat of formation will always exceed the activation energy, or the energy required to move $\ce{A}$ and $\ce{B}$ through any intermediate states that are less stable than $\ce{A + B}$ (far apart).
$\ce{A + B -> AB}$ can sometimes be exothermic where the energy required to move $\ce{A}$ and $\ce{B}$ into an arrangement that will allow bonding can exceed the heat of formation of the bond between them.

If anyone can explain how I can take an image that I draw and save on Google docs and turn it into an image that can be directly formatted into a question, I would appreciate an explanation.

Easiest way I know is to take a screenshot, paste into Paint (or whatever image editor), crop, and save; then insert wherever into the post.

Well, I draw them in paint.  How do I insert them into the post?  I tried to copy and paste.

Scroll down to the bottom on this help centre article. To not make your question invalid over time, you should use the built in upload form from stack exchange (who has a special agreement with imgur). Usually you can also enter web-addresses to directly upload a picture from the web; however, this doesn't work with google drive (for reasons I don't know). In these cases you need to download the picture to your computer and upload it from there with the normal dialog.

$\ce{He + He -> He2}$ would be endothermic, I guess. That's why it just doesn't occur.

@Ivan Neretin: I saw on the internet that there is definitely a 3 gram sheet.

@JosephHirsch A 3 gram sheet of what?

$\ce{He2}$ http://chemistry.stackexchange.com/questions/63562/how-do-i-calculate-the-volume-of-a-molecule

Ah, that one. Well, yeah, sometimes those textbook examples tend to create a fantastic universe with their own rules and laws.

A bond does not simply form because two atoms are close enough. You also have to bring them into the right electronic state to do so. Which takes various amounts of energy. So yes, an A+B -> AB reaction can be endothermic.
Also beware of mistaking the abscissa in those energy diagrams as a distance: It's usually just a symbolic "reaction progress", or it is a line on a multidimensional hypersurface describing the energy potential with respect to all angles, distances, electronic parameters etc. in the reaction.

I know it's not a distance, though distances could be part of it given variable strength of static electric field.

Yes it is possible for a $\ce{A + B -> AB}$ reaction to be endothermic but it will generally not occur.
Whether the reaction proceeds spontaneously depends on the change in Gibbs free energy $\Delta G = \Delta H - T\Delta S$.
${\Delta H}$ is the enthalpy (negative for exothermic reactions), ${T}$ the temperature and ${\Delta S}$ the change in entropy (a measure of disorder in the system).
If ${\Delta G \le 0}$ then the reaction will proceed.
When you combine two molecules, such as $\ce{A + B -> AB}$ you are forcing order on the system (molecules must now stay close to each other), so the entropy goes down.
So in an endothermic combination reaction you have ${\Delta H > 0}$ (endothermic) and ${-\Delta S > 0}$ (decreasing disorder). This means that ${\Delta G > 0}$ and the reaction will not happen.
The exception is high pressures (e.g. heavy nuclear fusion) where a decrease in volume caused by the fusion can provide the necessary driving force (It's included in the ${H}$ term)
A decomposition reaction $\ce{AB -> A + B}$ however will increase the disorder (molecules may now move about freely), allowing endothermic reactions to proceed provided that ${\Delta H < T\Delta S}$.

While you are correct, as far as I tried to follow, you have also completely missed the point of the original question. ;-) When you think about formation of single molecules, thermodynamics is hardly an issue.


text:
Why doesn't a fractured salt crystal stick back together?
If you crack a large crystal of a salt like NaCl, why don't the pieces spontaneously bond back together on contact? The bonding is an exothermic process and the charged ions are presumably stuck approximately right where they were before the facture.

Have you considered how accurately you should put together the fragments so the charged ions facing to the right ones again?

No, but I figured that you certainly are going to be able to get positive ions and negative ions in the same vicinity.  And the same would work for metals.  I mean I know id doesn't work, but wondering if there is a precise explanation.  Does it take some energy to get them arranged to bond with each other.  It is connected with my question about whether a pure synthesis reaction can be endothermic.

I have to read this one http://chemistry.stackexchange.com/questions/5915/why-do-salts-dissolve-why-do-nacl-crystals-not-stick-together?rq=1 to see if it mine is a dup.

You would have to get the two surfaces (or rather their crystallographic axes) together at the atomically perfect original angle to each other. Otherwise no overall (ionic) binding energy will result: Even if the surface is perfectly even, half the ions will be at places where they feel repulsion rather than attraction by the other side.
With polished metal surfaces, this works much better, also because metal atoms are (in "average") more mobile on a surface than ions.
Another point is humidity. Many surfaces, and certainly ionic substances, adsorb a layer of water from the air, practically instantly. It'll take quite a while before that has diffused out again when you press the surfaces against each other.

Not only humidity but also oxygen and nitrogen. Every solid surface has a layer of air gases adsorbed on it.

@vapid No they don't. I've done the BET experiment numerous times, it needs significant cooling (LN2).

Well yes, it depends on temperature and pressure. To get a layer of adsorbed gas you typically need very low temperatures, especially for smooth surfaces. But even at normal pressure and temperature gases are adsorbed on isolated spots (cracks, pores, etc.) on a surface. Try heating an MS flight tube soon after venting (when pressure is still a bit high) and you will see a nice surge of nitrogen signals.

Ah. You said "layer" above. ;-)

OK, my mistake;-)


text:
Why wouldn't you see blue or purple when you heat metal and only either red or yellow?
Question
Why wouldn't you see blue or purple when you heat metal?
Background and thought process
When metal gets heated the electrons gets excited and thus it emits the color that we see. But my question is why don't we see the other colors? If we heat metal to an extreme temperature it might turn white but not blue or purple. Why is this?

Please have a look at black-body radiation.

After looking at the link provided by Klaus think about melting and boiling points of elements.

And even if we could get the metals as hot as 8000 K or 10,000 K to get them to turn purple, we would not see it with our eyes.  Before we get potentially blinded, our color-sensing cones would be oversaturated and thus the differences between blue/purple and red/yellow radiation is largely washed out.

https://chemistry.stackexchange.com/questions/64041/why-wouldnt-you-see-blue-or-purple-when-you-heat-metal-and-only-either-red-or-yellow

Because hot bodies emit a continuous range of light not a single wavelength
All warm objects emit a range of wavelengths that depend on how warm they are. The source of the light is a variety of mechanisms not all of which involve electronic transitions (some will involve transitions between molecular or solid state vibrations or rotations.) The important point for the perceived colour of the light they emit, though, is that the emitted radiation is continuous.
The chart below show the mix of light wavelengths from bodies at different temperatures:

The peak of the emission curve changes with the temperature. So the mix of light you see does change colour somewhat with the heat of the object but will never be a pure colour like blue or purple. Tungsten light bulbs, for example, reach temperatures of 2.5-3K °C and have peak emissions in the infra-red; their light has more red than blue in the visible so they are perceived as somewhat orange when compared to sunlight (which follows the curve for a body of 5-6k °C, the sun and has peak emissions in the middle of the visible spectrum).
The differences between different sources of black body radiation are harder to perceive than you might think because the eye and brain are good at adjusting what you see to a common perceived colour mix and it is only when comparing different light sources that the differences in the mix of light become easy to see (something also obvious when comparing other types of light bulb to tungsten bulbs).
So a hot object at ~7k °C will emit more blue light than red light but it will still be emitting a lot of red light so you will tend to see blue-tinged white rather than pure blue. Astronomers can, when their eyes are well adjusted to darkness, tell the subtle colour differences of different stars (which range in temperature from maybe 2k °C to maybe 20k °C). But the effect is small.
Of course, if you heat something enough you will turn it into a vapour or plasma and there will be enough energy to excite the outer electrons. The light resulting from transitions of outer electrons is often very narrow in wavelength and can be used to identify the atoms in the vapour. Low pressure Sodium street lights (the bright orange ones) are a result of this effect (the excited outer electrons sodium atoms emit yellow-orange light very characteristic of sodium). Many other elements have characteristic lines that can be just about any specific colour under similar conditions).
But normally when you heat things up you just see the continuous black-body emissions and that can't be a single colour.


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What kind of molecules do I smell when draining cooked potatoes or rice?
When I drain potatoes after cooking them, they exhibit a nice smelling odor. It is not very strong. Now I know that some perfumes use natural molecules and one of them could come from potatoes (or rice.) But I forgot what the name was for this molecule, or perhaps someone knows some perfumes using these kinds of compounds.
It would be interesting to know what the functions of such molecules are in the organisms. Just an enzyme, or possibly an attracting force for some reason?

What a good question. But unanswerable by me and my prior knowledge and skill in google-fu. I have some general answers. This is a complicated area, since smell is both a physical issue and an individual and unmeasurable issue. Olfactory science is at a surprisingly basic level, given how far we are generally. I tried investigating what compounds in rice provide odour, the answer seems to be; way more than I could possibly have thought. Check out this article - while it is rough to digest a scientific articl unless you have basis knowledge just look at that list at page 505-506. There are many aromas in rice!
As for what functions they perform - looking at the list it is probably impossible to generalize down to what functions such a wide array of compounds do. Certainly, some compounds that are odorous has a primary function of being odorous, attracting pollinators and attracting mates. For rice, which has been around humans since time immemorial (at least 9k BC) they have probably been through a selection process, the most benevolent smelling rice plants being selectively bred, but this is speculation on my part. Other compounds have other functions and just happen to have a strong odour. Generally, to be odorous, the compound must have some volatility. What can also be said generally and very broadly is that alcohols and aldehydes / ketones are more often odorous than other compounds. There are also certain inorganic compounds that are (and often strongly so), Ammonia, Sulfide and the "garlic" series (S-Se-Te) for instance.
Good question. If you learn more, let us know.

I think I've found the particular molecule: 2-acetyl -1 pyrroline. Should this be used in perfumes?

@Marijin -- I'm not sure I follow why you as "should it be used". If you mean can it be used, asking whether it is safe for people, some cursory Googling doesn't show any signs that this molecule has identified hazards.


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Why is 1H-borepine aromatic?
For a compound to be aromatic, each atom in the ring must be a part of conjugation. Boron here has no lone pair, so how can 1H-borepine be aromatic as I was told by someone?

1H-borepine is isoelectronic to the cycloheptatrienyl cation. In both cases, you have a planar ring of seven atoms, whose hybridisation is close enough to $\mathrm{sp^2}$ to allow for a third p-orbital to be perpendicular to the ring plane. And in both, you are then filling these seven π orbitals with 6 π electrons, resulting in a Hückel-aromatic system ($4n+2 = 6$ for $n=1$).
There is no requirement for each atom along the chain of conjugation to have a lone pair; for example, in benzene no carbon atom has a valence lone pair. The only requirement for Hückel aromaticity is through conjugation, i.e. each atom must have an accessable p-type orbital that can overlap with the neighbouring atoms’ p-orbitals in a favourable π manner. However, neutral aromatic systems with empty p-orbitals are rare if boron is not used as most non-metals do not have favourable electronic configurations.

We can make contributing structures in cyclopropenone where a $p$ orbital on carbon is formally empty, the carbon is positively charged (cyclopropenyl cation), and the negative charge is on a substituent (oxygen).  Such an aromatic zwitterion structure is not really accurate, but cyclopropenone shows properties consistent with a zwitterionic contribution (basicity, dipole moment).

@OscarLanzi Interestingly, a similar (but opposite) effect is apparant in fulvene in which, when calculated, the exocyclic double bond seems to be almost more accurately represented by an exocyclic cation and a 6 pi electron ring system.


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Measuring turbidity using a spectrophotometer
I am conducting a lab investigating on the effect of aluminium sulfate concentration on turbidity. In my high-school, we only have a spectrophotometer. Is there a way to measure turbidity with a spectrophotometer? Do I need to dye my solution or alter it in some other way? How do I prepare my blank?

You can use a spectrophotometer to measure turbidity since the suspended particles will scatter light and so a reduced amount of light is detected dependent upon how much scatter occurs. The relationship is $I_{trans}=I_0\exp(-k\cdot C)$ where k is a proportionality constant depending on the particle size and shape, particle refractive index, sample path-length and C is the mass concentration in what ever units you choose. Usually in a spectrophotometer  the log is produced by the machine as optical density ( OD ). Thus $OD = -\log_{10}(I_{trans}/I_0)= kC$.
It is essential when comparing samples that the particle size is the same otherwise k varies between measurements and as C is also varied this confuses any measurements. Gum arabic or gelatin are often used to stabilise a suspension. The more finely divided is the suspension the greater is the ability to scatter light. Often experiments are made by comparison to a standard suspension say of Ca oxalate, $\ce{BaSO4}$, or Zn ferrocyanide.


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Are PCC and Collin's reagent completely identical in their oxidizing action?
Are PCC and Collin's reagent completely identical in their oxidizing action? Is there any advantage (or disadvantages) over the other?

In the Wikipedia article of PCC, it is written that:

PCC offers the advantage of the selective oxidation of alcohols to aldehydes or ketones, whereas many other reagents are less selective.

Further, on the Wikipedia article of Collin's reagent, it is written that:

This complex is both difficult and dangerous to prepare, as it is very hygroscopic and can inflame during preparation. It is typically used in a sixfold excess in order to complete the reaction. Nowadays, PCC or PDC oxidation have largely supplanted Collins oxidation for these reasons.

Okay, so PCC wins in preparation part.


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Hydrogen electrolyzer output pressure flow
This is for pressurised hydrogen storage number crunching. I have a question about the pressure and flow at the hydrogen output of an industrial size alkaline water electrolyzer. These electrolyzers are rated as hydrogen Nm3/hr and output pressure. So lets say the electrolyzer in question is rated at 15 Nm3/hr and output of 8 atm. Meaning it creates $15~\mathrm{m^3}$ of hydrogen at 1 atm at 68 °F (20 °C) every hour and has a pressure of 8 atm at the electrolyzer hydrogen output.
Now my question is if there is a back pressure of 8 atm at the output ($\ce{8atm <=> 8atm}$) is there no flow or would there be a flow of $1.875~\mathrm{m^3}$ ($15~\mathrm{m^3}$ at 8 atm) according to the combined gas law?


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Plant Coloration Stabilizer
Question
What chemical is used to stabilize berry juices used as wood stains?
Aim & tries
I am trying to make a stain for a snowboard design and coloration of the wood. I used pure pokeberry juice, but almost all of the color vanished after sending through the heat compressor. Now I need to find a way to stabilize the berry substance and I am unsure of where to start.

If someone could please comment and help me out I would highly appreciate it. I tried to ask my Chemistry teacher in high school but she would not help me due to her "not having time to delve into my question". So if anyone has anything helpful, please share it with me.

You're looking for a mordant or fixative, such as potassium alum, $\ce{KAl(SO4)2·12H2O}$.
Though I do not know how well this will work on wood, mordants are often used on textiles. Check for cotton and linen mordants in particular, since they are cellulosic material, like wood (though wood also has lignin, which may inhibit uptake of the stain).
You might ask your teacher if she thought you are too young to dye.

I will ask her. And thank you this is very helpful and I plan to run as many tests as possible with as many different solutions as possible. Your comment is appreciated.


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Raman Spectroscopy: How to calculate their constants?
$^1{\ce{H2}}$ is interrogated with a laser of wavelength lambda = 532 nm and vibrational Stokes transitions are observed at S(0) = 14302.206 cm$^{-1}$, Q(0) = 14638.442 and O(2) = 14992.823 cm$^{-1}$.

Estimate ωe (cm$^{-1}$)
Determine B1 (cm$^{-1}$)

I know the band origin = Q(0). I also know from rovibrational spectra that
$ ω_0 = ω_e -2ω_eX_e$.

Could someone please help me answer this question and explain?

If there are any additional resources I could read on these topics that would be great too.

You could improve this question by adding more detail to the question. E.g what does the notation stand for? what is the name of the equation? how did you derive it?

In Raman spectroscopy the difference in frequency between the exciting line and the measured signal frequencies contains the information you want. These values, made positive if necessary, give the ro-vibrational transition frequencies.
You are assuming that the molecule is vibrating with an anharmonic potential, such as a Morse potential. The energy levels are given by $E_n= \omega_e(n+1/2) - x_e\omega_e(n+1/2)^2$ (in wavenumber units) with quantum number $ n = 0, 1, ...~$, and $\omega_e$ the frequency and $x_e$ is the dimensionless anharmonicity.
In addition to the vibrational energy there is rotational energy whose levels are $E_J =B_eJ(J+1)$ for rotational constant $B_e$ (in wavenumbers) and quantum number $J=0, 1, 2,...$ and ignoring centrifugal distortion. You will also have to use different $B_e$ values for the $n=0$ and $n=1$ vibrational levels since the bond length is slightly different between (anharmonic) vibrational levels.
You know that the rotational-vibrational levels appear in groups  O, P, Q, R, S in increasing frequency and that Raman has O, Q & S bands only. In the Q band $\Delta J = 0$ between vibrational levels n to $n+1$ and is $-2$ and $+2$ for O  & S branches respectively. The Raman selection rules are $\Delta n = \pm 1$ and $\Delta J = 0, \pm 2 $.
If you look at a rotation-vibration energy level diagram for IR transitions you should by adapting this be able to work out your values for your Raman transitions.
There is a figure showing energy levels at
 https://en.wikipedia.org/wiki/Rotational%E2%80%93vibrational_spectroscopy
The Q band transition ($\Delta n =0, \Delta J =0$) in your notation is $\omega_0 = \omega_e - 2x_e \omega _e $ and the other branches calculated similarly with $\Delta J = \pm 2$ as appropriate.

Good explanation is given by @porphyrin's answer and I am going to get into details now.
The laser is at 532 nm hence in absolute wavenumbers this comes to $18789.9\hspace{1ex} cm^{-1}$.
$S(0)$ appears at $14302.2\hspace{1ex}cm^{-1} \rightarrow 4487.6 \hspace{1ex}cm^{-1} (Ramanshift)$
$Q(0)$ appears at $14638.4\hspace{1ex}cm^{-1} \rightarrow 4151.4 \hspace{1ex}cm^{-1} (Ramanshift)$
$O(0)$ appears at $14992.8\hspace{1ex}cm^{-1} \rightarrow 3797.1 \hspace{1ex}cm^{-1} (Ramanshift)$
*Just the difference from the incident laser in wavenumbers.
Next, rotational constant is easy to estimate roughly,
$E_{J}=6\times B_{0}$ for $J=0$ to $J=2$.
$354.3=6\times B_{0}$
Rotational constant, $B_{0}=59.05 \hspace{1ex}cm^{-1} $

In this equation, $E_n=ω_e(n+1/2)−x_eω_e(n+1/2)^2$ , we already know $E_n$ and $n$. ($n=1$ and $E_n$=energy of the Q branch in wavenumbers, relative to laser). then $\omega_e$ calculation is straightforward.


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Dissolving copper minerals in acid
I have copper minerals in rock that has been pulverized. These copper minerals are pure (native copper), chalcocite, chrysacolla, malachite, tenorite, cuprite and azurite. Does anyone know which acids will dissolve which minerals? I need this information as a lab has proposed to use $\ce{HF}$, $\ce{HNO3}$ and $\ce{HClO4}$ to perform an assay and I'm not sure it will work 100%.

most should dissolve in HNO3; though chrysocolla  should require HNO3/HF mixture. Be cautious dissolving malachite and lazurite, both should dissolve readily producing a lot of CO2 gas. chalcocite would likely produce significant quantities of  SO2 and NO2. And HF is not sweet water either.

Any idea as to what range Cu will be as wt %? // I also assume you're sampling a large amount of ore (dump truck size load) ground in a ball mill rather than trying to sample the "raw" ore rock by rock.

If I remember correctly, we (an analytical laboratory) would generally get all copper minerals into solution using a nitric/perchloric digestion, depending on the sample matrix.  Unless the sample matrix as a whole were highly silicious in nature an additional HF digestion was not required. As commented above though, $\ce{HNO3/HF}$ may work for you also.

Hi in your opinion what assay would you use for tenorite. I have doubts whether you can use the same acid for native copper and tenorite. How long does the acid take to dissolve the native copper?

Copper metal will be gone in a jiffy.  Tenorite should have no problem at all going into solution refluxing in nitric/perchloric for several hours.  Is that what you meant by assay, the digestion method or an analytical testing method for the copper after the prep?


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Determining Charge of Protonated Alkaloids
During the extraction of many alkaloids, you lower the pH in order to protonate the compound, making it polar and water soluble. With Mescaline, you must precipitate it as a salt.
What I don't understand, is how can you figure out the mole ratio between the Mescaline and the conjugate base of the acid (for example acetate). My understanding is you'd have to figure out the charge of the protonated mescaline. How? Thanks
Edit:
The reason is not so I can ionize all of the mescaline. I want to know this so I can calculate the molar mass of the mescaline salt. That way I can take the weight I have of product, and calculate (using the molar mass) the weight equivalence of freebase mescaline, for dosage purposes. Also, it's just interesting knowledge that I'd like to know.

I don't think anyone is counting. You don't use exact equivalents of the acid. You just add hydrochloric acid (probably 1 or 3 M) until the pH is low enough.

Mescaline has one amine group so what would be there to count?

Yes, but I'm not counting for that reason. I'm counting so I can estimate the molar mass of the salt.

Are you sure you're trying to calculate the molar mass of the mescaline salt? Isn't that known, depending on the acid you use?
It seems to me the key you need is the mole ratio. For a weak acid-base conjugate pair, you can use the Henderson-Hasselbalch equation:
$pH = pK_a + log(\frac{[B]}{[A]})$
where [B]/[A] is the ratio of concentrations of conjugate base to conjugate acid. If calculated for a single solution containing both forms, the total solution volume cancels and this is the mole ratio. In a more convenient form,
$\frac{B}{A} = 10^{pH - pK_a} = \frac{K_a}{[H^+]}$
(The first equality is more practical, the second is useful for theoretical manipulations.)
Note that when $pH = pK_a$, $B = A$ (the analyte will be 50% protonated and 50% deprotonated).
For mescaline (more precisely, its conjugate acid), $pK_a =$ 9.56, typical of an amine base/alkaloid. So at a pH of 9.56, approximately half of your mescaline should be protonated. You could extract this, measure the weight of the salt, use the known molar mass to convert to moles of the protonated mescaline (I recommend using a monoprotic acid such as acetic if possible to simplify the situation). By the Henderson-Hasselbalch, the moles of protonated mescaline you extracted at its $pK_a$ will equal the moles of freebase mescaline remaining in your organic solvent.

Rule of thumb is to go 2 pH units below the pkb of the compound you're going to protonate, in order to get ~100 % protonation (and vice versa; 2 units above pka for complete deprotonation).
So for mescaline: pkb: 4.44, you would be pretty safe to add HCL to pH ~2.

Thanks! I know that much, but that's not why I'm trying to find the charge. I'm trying to calculate the molar mass of the salt. Also it's just useful information I'd like to know.

Ok, and you are using which acid for salt formation?

Debating between acetic acid and citric.


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Multiple pKa's of Vitamin C
Layman here. I was looking at vitamin C serums in skincare and wanted to know more about it. It has been awhile since taking chemistry in college so my understanding is rusty.
I want to better understand this diagram (Source: U of Iowa) that talks about the different forms vitamin C:

All I remember of $\mathrm{p}K_\mathrm{a}$s was that high $\mathrm{p}K_\mathrm{a}$s means the acid is very strong and dissociates all the way while lower $\mathrm{p}K_\mathrm{a}$ means a weaker acid. So I'm a little lost on how to interpret this diagram in terms of the $\mathrm{p}K$ values, ie the $\ce{OH}$ group is deprotonated from $\ce{AsCH2}$ to $\ce{AsCH-}$ but I'm not sure how the $\mathrm{p}K$ value fits in all of this? Layman me wants to say at $\mathrm{pH}=4.1$ is when deprotonation happens, but I have a feeling that isn't right.

That's backwards. Lower $\mathrm{pK}_{a}$ values correspond to stronger acids.

$\mathrm{pH} = 4.1$ would imply that there are equal concentrations of the neutral and singly-protonated species in solution via the Henderson-Hasselbach equation.

@Zhe, oops my mistake, you're right. I got pka and Ka mixed up. Okay so when Ph=pka in this case when pH= 4.1 there are equal amounts of AscH2 and AscH-, correct? And would it be correct to say that at ph below 4.1, the form is mostly AsCH2 and at ph's above 4.1, the form steadily increases to be more and more of AsCH- ?

That's correct. Take a look at the equation in ringo's answer if you're still confused.

There are three more, chemically partially relevant $\mathrm{p}K_\mathrm{a}$ values — although the corresponding anions do not exist in water.

You actually have it backwards. A low $\mathrm{p}K_\mathrm{a}$ is a strongly dissociating acid, and a high $\mathrm{p}K_\mathrm{a}$ is a weakly dissociating acid. It would do you well to remember the origin of $\mathrm{p}K_\mathrm{a}$ \eqref{pKa}, as well as the Henderson–Hasselbalch equation \eqref{HHeq}:
$$K_\mathrm{a}=\ce{\frac{[A-][H3O+]}{[HA]}} \Rightarrow \mathrm{p}K_\mathrm{a} = -\log(K_\mathrm{a})= -\log \left( \ce{\frac{[A-][H3O+]}{[HA]}}\right)\tag1\label{pKa}$$
$${\mathrm  {pH}}={\mathrm  {p}}K_{{\mathrm  {a}}}+\log _{{10}}\left({\frac  {[{\mathrm  {A}}^{-}]}{[{\mathrm  {HA}}]}}\right)\tag2\label{HHeq}$$
From this we can surmise a physical meaning behind what the $\mathrm{p}K_\mathrm{a}$ of an acid is—it is the $\mathrm{pH}$ of the solution when $\ce{[A-]=[HA]}$. This means that for $\mathrm{pH}>\mathrm{p}K_\mathrm{a}$, $\ce{[A-]>[HA]}$. Since ascorbic acid is a diprotic acid, it has two $\mathrm{p}K_\mathrm{a}$ values, one for each deprotonation.

okay this helps a lot. I would like some clarification on this relationship: pH>pKa,  [A−]>[HA]. Not sure if I have this right, so if I dissolve AsCH2 in a solution where the ph is higher than 4.1, let's say 6.0 for example. The solvent solution (let's say slightly acidic water) at ph= 6.0. pH>pKa , [A−]>[HA] says there there would be more of AsCH- than AsCH2. Is it b/c there would be more OH- ions in the water which then deprotonates ASCH2 leaving more ASCH- ?

The $\mathrm{p}K_\mathrm{a}$ of an acid is a constant and does not change. This means that you can think of $\mathrm{pH}$ as a variable that responds to changes in $\ce{[HA]}$. When the $\mathrm{pH}$ is higher than $\mathrm{p}K_\mathrm{a}$, the other term in the Henderson–Hasselbalch equation has to be positive. Logarithm terms are only positive when the argument is >1; therefore, $\ce{[A−]>[HA]}$.

I would say that it is because there aren't enough $\ce{H3O+}$, so there will be a shift in equilibria to balance this. Think of the reaction in terms of $\ce{AscH2 + H2O <=> AscH- + H3O+}$

I get what you are saying from a math standpoint when you were mentioning how pka is a constant and a log value has to be > 1 so A- > HA. But you mentioned how pH is a variable that responds to changes in HA. But I was thinking how the pH of a solution can also be the driver of what form Vit C can take on. I have to break this comment into two, read below:

Here's how I was thinking about it. I take a Vit C powder in the AscH2 form and I dissolve it in alkaline water of pH=13. Water dissociates freely into hydroxide and H+, there is more OH- in this case b/c the water is more basic. I know OH- is a very strong base so it then deprotonates AscH-  so that there is more AscH2-. The HH equation you showed me says when ph=pka is when we have equal [ ] of both. But our alkaline water is greater than the pka2 value, so I can intuitively just guess that there is greater than 50% AscH2- in the solution. I mean this logic is correct also or no?

@Arcuvia Yes, your logic is correct given that the addition of the acid doesn't drop the $\mathrm{pH}$ below the $\mathrm{p}K_\mathrm{a}$. Usually the $\mathrm{pH}$ of a solution is altered once you've already added the acid.


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How to measure and transfer sticky solids
What is the best way to measure and transfer a sticky compound in the lab, in particular for air/moisture sensitive reactions? I have a really sticky solid Wittig salt which I'm going to use in a Wittig reaction.

PTFE (Teflon etc., e.g. http://www.sigmaaldrich.com/catalog/product/aldrich/z115282) may help if the salt is hygroscopic, since both polar (i.e. salt solution) and non-polar substances do not stick well. Silicone tools might also be of use.

Could you wrap it in plastic wrap?

Glovebox - you can fill it with a neutral dry gas. I would also suggest you checking the purity of your substrate - sticky solids usually mean impurities or solvent presence.

Ideally, you want your sticky solid to already be in a Schlenk flask or similar. If it is unfortunately still in a normal bottle, take the weight of a Schlenk flask, fill in enough of the sticky solid and re-weigh. Note the difference, this is what you have in your flask.
Assuming that the sticky solid has a low enough vapour pressure (spoilers: practically all solids do), evacuate the Schlenk flask under high vacuum and backfill with argon (or nitrogen, if that is your inert gas of choice. Add a known quantity of your (dry, stored over molecular sieve and under inert atmosphere below a septum) solvent. Ideally, you either want an amount of solvent that gives you an easy-to-handle concentration (e.g. $10~\mathrm{M}$) or an amount so that what you need is in an even volume of liquid (e.g. $1~\mathrm{ml}$).
Finally, transfer the solution into your reaction flask.
If you need the sticky solid in different solvents and it is air-stable, you can evaporate the dry solvent at the rotavap. If not, keep it in solution under argon (or nitrogen). If the latter is the case, do your colleagues a favour and choose a standard concentration, not a good filling amount for exactly your reaction.

I do wonder why this answer has been deemed ‘not useful’ or ‘unclear’.


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Confusion over percentage volume
We've been told our lab report has to be written with "% v/v" instead of volume. I understand how to calculate it, however I'm unsure if I'm doing the right thing when it comes to using it later on. Example:

2ul 'W', 4ul 'X', 2.2ul 'Y', 1.8ul 'Z' is mixed in an eppendorf. It is incubated for 1 hour. After incubation, 0.95ml of 'B' is addded and then 150ul of the final mixture spread on plates.

I know the first bit would be:
20% v/v 'W', 40% v/v 'X', 22% v/v 'Y', 18% v/v 'Z' is mixed in an eppendorf.
But I dont know what to do when it comes to after the incubation?
I'm generally pretty good with this stuff but I've never had to think of it this way. Would, when 0.95ml is added, I do 0.95/0.95+2ul+4ul+2.2ul+1.8ul?

My assumptions:

'ul' Refers to microlitres, be careful that the addition of B is in millilitres.
You are trying to calculate the % v/v that is being removed from the total solution at the end?

So before incubation the total volume is 2ul + 4ul + 2.2ul + 1.8 ul = 10ul (0.01ml)
You are adding in 0.95ml, totalling 0.96ml.
You are then taking 150ul (0.15ml)
This equates to (0.15/0.96)*100 %
= 15.6%


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Why does ozone cleave alkenes?
I am studying additional reactions in organic chemistry and I am confused as to why ozone cleaves double bonds? I already know the mechanism of what happens but can anyone give a more conceptual understanding of why this occurs?

The mechanism actually tells you why this occurs. It's a step-by-step description. The overall reaction is favored. Each step is favored or provides easy access to a favored step.

Not all reactions are driven by... wait, I think I already said that.

Analysis of the frontier molecular orbitals of the alkene and ozone molecules easily and elegantly predicts the occurrence of the first step of this reaction:

The actual cleavage step forms a carbonyl and carbonyl oxide—a pair of molecules that are electronically identical to the starting reagents, but are thermodynamically favored over their regenration:
$$ \small
\begin{array}{lcc}
\hline
& \text{Bond Dissociation} & & \text{Bond Dissociation}\\
\text{Bond} & \text{Enthalpy}\ \mathrm{(kJ\ mol^{-1})} & \text{Bond} & \text{Enthalpy}\ \mathrm{(kJ\ mol^{-1})} \\
\hline
\ce{C-C} & 347 &\ce{C=C} & 614\\
\ce{O-O} & 146 & \ce{O=O} & 495\\
\ce{C-O} & 358 & \ce{C=O} & 745^{[1]}\\
\hline
\end{array}
$$

Formation of the secondary ozonide is also thermodynamically favored:

$^{[1]}$Bond Dissociation Enthalpy Tables


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Electron flow: How does osmium tetroxide react with alkenes?
This is a question that has bothered me for some time. I have always thought that the part of one molecule will only attack part of another molecule if it has a higher positive charge. So with this in mind, why does the alkene double bond attack the $\ce{O}$ atom of $\ce{OsO4}$. Isn’t the $\ce{O}$ atom negatively charged?

I am not sure whether it is the alkene that initiates bond formation. However, take a look at the mechanism. It is a cycloaddition. For all we know, the oxygen from the osmium tetroxide may be the one initiating the formation of the bond. I will try to look some journals and post what I find.

Not all reactions are driven by simple attraction of oppositely charged atoms. Consider alkene hydrogenation, acetylene trimerization, or the Diels–Alder thing.

I have a suggestion that could explain the attacK of electron pair of alkene pi bond to the oxygen of osmium tetroxide : The electron pair of alkene pi bond repels the electron density of the oxygen when the osmium tetroxide approaches the alkene, creating a temporary δ+ on the oxygen that is closer to the pi bond of alkene. The electron-rich alkene then attacks the electron-poor oxygen.
Dimitrios Kariotoglou

Are these your opinions? Please put in references to literature, or else the answer is just speculation. And this explanation seems quite far-fetched to me. I could be wrong, but the right way to refute my disapproval will be to find literature and add it in. Good-day.

Yes. But check the formal charge on the Osmium. It is so very positive that it actually steals electron density from the Oxygen.


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On a periodic table it shows that Zn and Cd and Hg all have 0 electron affinity. Why is that?
Question
On a periodic table it shows that Zn and Cd and Hg all have 0 electron affinity. Why is that?
What I've noticed
I've noticed that these three elements are the end of the D orbital and they are completely full, i think that it has to do something with that But i am not entirely sure
Part two of question
Why is electron affinity expressed in negative numbers?
Ex: Cs=-45.5kj/mol

They all have 2 electrons in their highest occupied s sublevel and 10 electrons in their highest occupied d sublevel (both sublevels are full).  As a result, they can not complete any orbitals by adding an electron and they can not put an electron in a sublevel that is occupied.  All of the other transition metals will either add an electron to complete the highest s sublevel, or into an already occupied d sublevel, or (in the case of palladium) add one to an s sublevel that is empty.
Cd, Hg and Zn thus are unique in that they will all have to put an unpaired electron alone into a p sublevel.


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Why can hydroxide not be used to synthesise ethanol from ethyl amine?
In the following reaction we have to find A and B.

I thought that A could be $\ce{Br_2}$/$\ce{KOH}$ resulting in a Hoffmann bromide reaction.
For B, I thought I would need a base such as $\ce{KOH,NaOH}$. But the answer is $\ce{HNO2}$.
Why?

Think about what forms when a primary amine reacts with HONO...

@EashaanGodbole I don't know

Well, your choice of reagents is definitely not going to do anything. The amide is a terrible leaving group and will not undergo substitution. You should look at diazonium...

@Zhe does it is good substituting compound in almost every cases

I think that actually highlights your misunderstanding of the substitution reaction. You must have a good leaving group. Having a good nucleophile doesn't hurt, but if the leaving group is bad, no amount of nucleophilicity is going to force that substitution.

@Zhe so in this case amide is bad leaving group . Then why only HNO$_2$ is used for substitution why not any other nucleophile or base ?

You totally ignored my hint. That reagent converts the amino group into a leaving group...

@Zhe after that we should add H$_2$O for alcohol

Add water? It's aqueous...

Hydroxide may be a somewhat good nucleophile but the amide anion is one of the most lousy leaving groups you can imagine. If you want to remove an amino group in an $\mathrm{S_N2}$ reaction, it can only be eliminated as an ammonium cation — but protonating it means we can no longer use hydroxide as a base. Under basic conditions an amino group will never be a leaving group.
Instead, you must turn the amino group into something that is actually a good leaving group. If you include nitrogen and let your thoughts wander, you may arrive at the diazonium group $\ce{-N2+}$. Upon nucleophilic attack, this will be released as nitrogen gas which is great both thermodynamically and kinetically. This leaving group is so good that even relatively bad nucleophiles such as water will react with it.
However, you need to produce it from the amine. This is where nitrous acid comes in. Under acidic conditions, $\ce{HONO}$ will react with primary amines to give diazonium salts according to the following general equation:
$$\ce{R-NH2 + HONO + H+ -> R-N2+ + 2 H2O}\tag{1}$$
Since nitrous acid is produced by acidifying aquaeous solutions of nitrite salts, you have all three components in hand if you ‘just’ add $\ce{HONO}$ or $\ce{HNO2}$.