Why can radioactive contamination be spread by people? So I am reading a book called "Voices from Chernobyl" where witnesses, nuclear plant workers, firefighters and other persons involved in the 1986 accident give testimony of their experiences.
The very first chapter goes along the lines of the wife of a firefighter that was dispatched for duty on the very day of the incident. He suffered acute radiation and was hospitalized. His wife remain with him throughout the last days before he died. One thing it was mentioned in the book very clearly is how nurses constantly tried to warn his wife not to touch him, hug him or even share objects with him.
In this article, it is mentioned that a person who was exposed to radiation should not be a danger for others once his clothes have been disposed.
So, my question is, I am getting confused between the two readings. Is a person who has been exposed to a big dose of radiation a danger for other people? Why?

Summary: there is not necessarily a contradiction between the two.

Radiation is not contagious, and
a person who has been exposed to ionizing radiation is not dangerous to other people once they are not contaminated with radioactive material any more, but
while they are still contaminated with radioactive material, they may pose a danger (highest danger is to themselves, though).
In any case, other people are dangerous to someone who suffers radiation illness (infection risk).

Moreover,

Thinking radiation contagious is (still!) a fairly widespread mistake, and
I'd think misunderstanding the purpose of the instructions (she being dangerous for her husband rather than the other way round) quite likely in a highly stressful situations such as the husband dying. I.e., I wouldn't expect her to spend time then on double-checking whether she correctly understood the reason behind the instructions and ask, "Sorry, sister, is this because he's dangerous to me or because I'm dangerous to him?".

Thus, it should not be surprising to meet any such misconceptions in such a book. Nor would I be surprised if readers misunderstand the book this way even in case the firefighter's wife is and was perfectly aware that her husband did not pose a radiation danger to her. I'm writing the remainder of this answer mostly assuming that the mistake is in the book rather than on OP's side.
There are basically only two possibilities clear a misunderstanding that happened back then in a book of witness reports: the witness saying "nowadays I know I was dangerous to him, not he to me. But back then I didn't know that."
or the editor putting a footnote explaining the misunderstanding and that it was widespread at the time.

Ionizing radiation and contamination with radioactive substances
Radiation is not contagious in the sense that the firefighter's exposure to ionizing radiation did make him radiate himself. That being said,

Induced radioactivity exists. But that needs very particular conditions to happen to any practically relevant extent. In the sense of the question, you won't be able to produce any measureable effect in a living organism, not even a living organism that is about to die of the radiation.
The firefighter of course radiates like any other living organism. Humans radiate with about 4 kBq due to containing about 15 mg of $\ce{^{40}K}$.

But: chemical contamination, including radioactive substances, can be transferred from one body to another, and they can be incorporated and accumulated where they cause much damage.
(I also wouldn't call that contagious, since the total amount of radioactive material does not increase, but if you consider, say, crystal violet or methylene blue contagious "since if you touch someone, the ones touched by them will be violet/blue", then you can also call the radioactive substances contagious)
However, any radioactive substance contamination that is transferred to a practically relevant amount by touching can also be washed off - and that is the first thing to do for decontamination besides taking off potentially contaminated clothes. Any injury that means that the area cannot be thoroughly washed can also not be touched (for reason of the injury).
Thirdly, during the decay of such radioactive material, other substances form, which may be far more difficult to get rid of (see radon example below). Here, one may say that one can "catch" a contamination that one doesn't easily get rid off again.  However, any such conamination would not be transferred from the firefighter to his wife.

X-rays/γ-rays: these are ionizing electromagnetic rays, i.e. high energy photons. They cause damage when being absorbed, either by directly damaging some biomolecule or by forming OH⋅ radicals/ROS which in turn cause further damage.
The radicals in themselves are nothing very special - they occur all the time as side products of our energy metabolism and we have powerful mechanisms to cope with them. Part of radiation illness is that these mechanisms are overwhelmed.
So, after exposure to X- or γ-rays, we have radicals inside the body, but no "foreign" nuclei, and no body surface contamination. I.e. nothing that became radioactive because of the exposure to high energy photons.

A somewhat different example related to Chernobyl and Fukushima would be incorporation of radioactive  $\ce{^{131}I}$ in the thyroid gland. In particular, if someone with iodine deficiency incorporates iodine, pretty much all of it will end up in the thyroid gland. If that available iodine is $\ce{^{131}I}$, their thyroid gland will subsequently be exposed to large radiation doses. This incorporated radioactivity does include γ radiation of which a part leaves their body.
$\ce{^{131}I}$ is administered in radiotherapy in doses where the patients are e.g. kept in hospital for approximately 2 days (at least here in Germany) so as to not contaminate the wastewater with the radioactive $\ce{^{131}I}$ they excrete in their urine. Such patients are also advised to avoid close contact for e.g. a week after treatment in order to not cause accidental exposure to others, in particular children and pregnant women.. Such radiotherapy treatements use dosages in the 100 - 400 Gy range to the thyroid. And the guidelines are of course with a safety margin.  A quick search for $\ce{^{131}I}$ radiation doeses to the thyroid in  Ukrainian children after Chernobyl got me to Brenner et al: I‐131 Dose Response for Incident Thyroid Cancers in Ukraine Related to the Chornobyl Accident. The largst dose category is > 3.0 Gy, and a diagram has a point a bit below 5 Gy, so 1 - 2 orders of magnitude below the radiotherapy doses.
My conclusion from this is that even in case the firefighter got a $\ce{^{131}I}$ dose to kill off his thyroid, the wife giving the dying husband several close goodbye hugs 10 - 14 days after the exposure would be unlikely to pose a significant threat to her health due to radiation from his thyroid (and under the particular circumstances, the $\ce{^{131}I}$ she ingested after the accident would be a far more important concern for her health).
Again, I would not describe this as "contagious" - but your word use may vary.

In this guest post to cancer letter, R. P. Gale discusses some of his experiences as an MD at the famous hospital 6 (the Soviet Russian radiation clinic) treating the radiation illness patients from Chernobyl with at particular view to the HBO series.

Another error was to portray the victims as being dangerously radioactive. Most radiation contamination was superficial and relatively easily managed by routine procedures. This is entirely different than the Goiania accident, where the victims ate 137-cesium and we had to isolate them from most medical personnel.
Lastly, there is the dangerous representation that, because one of the victims was radioactive, his pregnant wife endangered her unborn child by entering his hospital room. First, as discussed, none of the victims were radioactive—their exposures were almost exclusively external, not internal. More importantly, risk to a fetus from an exposure like this is infinitesimally small.

Valid Reasons for not allowing the wife close to the husband that have nothing to do with radiation being "contagious".

Radiation illness: the bone marrow is rather radiation sensitive, and leukopenia (too low leukocyte counts, a type of immune suppression) are a typical part of radiation illness.
A radiation illness patient is thus at a very high risk from infections.

Radiation illness often comes with burns (the skin is most exposed, and for α and β radiation, almost all damage happens in the skin). Already "normal" severe burns are doubly difficult in terms of infections: the skin damage means that the normal protective barrier against microorganisms is broken down in those areas, and in addition there is a severe immune suppression (after initial inflammatory response). Infections cause half of the deaths after severe burns

Both are very valid safety reasons, just for the firefighter rather than for his wife's safety. Saying that the wife can go close to the firefighter would amount to saying "He's anyways going to die within the next days - it doesn't matter whether he catches an additional sepsis."

"Contagious" radiation as wrong but possibly valid concern after the Chernobyl accident
So from a scientific point of view radiation is not "contagious". Nevertheless, there is a still widespread, though mistaken fear of this. Personally, here and now I count this pretty much in the tin foil hat corner. But OTOH, in the situation in the Ukraine(ian SSR) immediately after the accident I think it a more understandable concern since the possibilities to check whether this concern is valid or not were severely limited. Not only for the general population, but even for the medical staff.  In such a situation, it is a valid decision to err on the side of caution.

How much did the hospital staff know about radiation medicine? Had they any experience with radiation injuries of this kind? There was a major accident, no internet, and a very restrictive information politic already without disaster.
In the video interview (thanks to @TAR86), Alla Shapiro describes that the "hush up" included that the medical staff was deliberately hindered/prevented from accessing medical information about radiation. She also explains that their medical training did not include radiation..
Note however, that she was at a clinic in Kiev. P.R. Gale describes much better expertise at hospital 6 in Moscow (where most of the radiation illnesses were treated).

The post you link says that thorough washing is sufficient for the Fukushima evacuees, and the experts linked above agree for the Chernobyl fire fighters.

However, Shapiro also says that there was a whole lot of this fear in the general population that radiation/radioactivity could be "contagious"/someone who was exposed to radioactive radiation would radiate themselves and thus pose a danger*.
In a situation where probably almost everyone realized that the government tried to hush up a big problem (correct assessment), and in addition any relevant information being locked in the "poison cupboard" of the libraries and not accessible (no way to obtain factual information): would you trust "information" that these patients are not "contagious" wrt. radiation, i.e. information is very much what the government wishes or would you tend to err on the safe side?

In addition: How much did the hospital staff know about what actually had happened and what the firefighter had actually been exposed to?
With those activities to hush up the Chernobyl incident, the hospital staff may have been unsure about what else that firefighter had been exposed to besides high doses of radiation.
Remaining contamination with radioactive material (including incorporated material) can be measured comparatively easily. However, I have no idea whether such instruments were available at the hospitals, say, in Kiev, to measure remaining contamination of their patients: a) the available instruements may have been needed more urgently at the site of the power plant and b) also political coniderations/hushing up may have been standing against that. I'd expect the Moskow hospital to have all kinds of instruments (but that may be my predjudice)

* I'm pretty sure that a non-negligible fraction of the population here in Germany would express that fear if you ask them. Including medical staff, and even after Chernobyl and Fukushima.
If you want an example: have a look at this post (in German) about radiation treatment for food on a web site by the offical consumer protection organizations

 Werden Lebensmittel mit ionisierenden Strahlen behandelt, wird die Strahlenmenge genau dosiert. Die Energiemenge ist so gering, dass die Lebensmittel nicht radioaktiv werden und sich nur leicht erwärmen.

My translation and my emphasis:

When food is treated with ionizing radiation, the amount of radiation is accurately dosed. The energs is so small that the food does not become radioactive and only heats up slightly.

While it is of course true that the food does not become radioactive, that sentence IMHO does insinuate that this could [easily] be the case with higher doses. And one comment (out of a total of 14) clearly indictates that the writer thinks radiated potatoes will radiate themselves.


Reactivity of boron towards acids Boron is a non-metal, i.e. it accepts electrons and acids donate electrons, so boron should react with acid but it doesn't. Why?

To illustrate Oscar's answer further, it should be noted that boron, despite existing in more than ten allotropic modifications, is in general quite chemically inert (especially in crystalline form).

For example, boron doesn't react with hydrogen directly and all boranes are synthesized by other means.
Although boron may be oxidized by fluorine at room temperature, other halogens (even chlorine) require elevated temperatures above $\pu{400 °C}$:

$$\ce{2 B(s) + 3 E2(g) -> 2 BE3}\qquad(\pu{30 °C}: \ce{E} = \ce{F}; \quad\text{above}~\pu{400 °C}: \ce{E} = \ce{Cl}, \ce{Br}, \ce{I})$$

There is no preference in reactivity towards acids or bases either.
Boron indeed doesn't appreciably react either with non-oxidizing acids or with alkali solutions under ambient conditions.

However, concentrated oxidizing acids such as nitric and sulfuric acids as well as aqua regia react all right — especially at elevated temperatures — yielding boric acid:

$$\ce{B(s) + 3 HNO3(aq) ->[Δ] B(OH)3(aq) + 3 NO2(g)}$$

Amorphous boron reacts with concentrated alkali solutions:

$$\ce{2 B (am) + 2 NaOH (conc) + 6 H2O(l) -> 2 Na[B(OH)4](aq) + 3 H2(g)}$$

When fused in air with alkali, $\ce{Na2O2}$ or $\ce{KNO3}$ + $\ce{Na2CO3}$ mix, borates are formed:

$$\ce{4 B(s) + 4 NaOH(s) + 3 O2(g) ->[\pu{350-400 °C}] 4 NaBO2(s) + 2 H2O(g)}$$

All chemical reactions have been adapted from [1, p. 74].

References

R. A. Lidin, V. A. Molochko, and L. L. Andreeva, Reactivity of Inorganic Substances, 3rd ed.; Khimia: Moscow, 2000. (in Russian)


How to determine the volume of water necessary to produce a solution with 4.9 g/l sulfuric acid? The molar mass of sulfuric acid is $M(\ce{H2SO4}) = \pu{98.1 g/mol}$.
  We have $\pu{100 mL}$ of a $0.1$ molar solution of sulfuric acid. What volume $V$ in  $\mathrm{mL}$ of water, $\ce{H2O}$, do we have to add in order to obtain a solution that has $\pu{4.9 g/L}$ sulfuric acid?

I've done the following:

\begin{align}
M(\ce{H2SO4}) &= \pu{98.1 g/mol}\\
M(\ce{H2O})   &= \pu{18.0 g/mol}\\
\text{mass concentration of }  H_2SO_4 &= \pu{98.1 g/L}\\
c(\ce{H2SO4}) &= \pu{1 mol/L}\\
c_{\mathrm{contr}} \times V_{\mathrm{contr}} &=
c_{\mathrm{dil}}\times V_{\mathrm{dil}}\\
\Longrightarrow V_{\mathrm{dil}} &=
\frac{c_{\mathrm{conctr}}\times V_{\mathrm{conctr}}}{c_{\mathrm{dil}}}
\end{align}

The molarity of the diluted water I can't calculate with elementary formulas. So it leaves me with 2 unknowns.

I've tried using so many formulas to no avail.

What formula am I not thinking of? Or what variable am I missing?

We have talked about it in chat and maybe the question is a bit ambiguous. I would interpret the last line as a mass concentration. So what you are actually looking to create is a $\gamma(\ce{H2SO4}) = \pu{4.9 g/l}$ solution. Then the calculation becomes straight forward.
Keep in mind that water is your solvent, so you can't actually dilute it; also writing down the concentration is a bit pointless here.

You can easily convert the mass concentration to an amount concentration. The relative molar mass $M$ is given by the mass $m$ of the substance divided by the amount of substance $n$:
$$M = \frac{m}{n}\tag1\label{molarmass}$$

The mass concentration is defined as
$$\gamma = \frac{m}{V}.\tag2\label{massconc}$$
The amount concentration is defined as
$$c = \frac{n}{V}.\tag3\label{amountconc}$$
You can use \eqref{molarmass} in \eqref{amountconc} and obtain
$$c = \frac{m}{M} \times \frac{1}{V}
=\frac{m}{V} \times \frac{1}{M}.\tag4$$
Substitute \eqref{massconc} back in and you have
$$c = \frac{\gamma}{M}.\tag5\label{amounttomass}$$

Now in the question you have given the following:
\begin{align}
M(\ce{H2SO4}) &= \pu{98.1 g/mol}\\
c_\mathrm{i}(\ce{H2SO4}) &= \pu{0.1 mol/l}\\
V_\mathrm{i}(\ce{H2SO4}) &= \pu{100 ml}\\
\gamma_\mathrm{f}(\ce{H2SO4}) &= \pu{4.9 g/l}
\end{align}
I have given the initial values the subscript $\mathrm{i}$ and the final values the subscript $\mathrm{f}$ (instead of concentrated and diluted).

You are now looking for the volume of solvent you have to add, which is
$$\Delta V = V_\mathrm{f} - V_\mathrm{i}.\tag5\label{volume}$$

Then you are right, you need the following relation
$$c_\mathrm{i} \times V_\mathrm{i} = c_\mathrm{f} \times V_\mathrm{f},
\tag6$$
rearranged and using \eqref{amounttomass}
\begin{align}
c_\mathrm{i} \times V_\mathrm{i} &=
\frac{\gamma_\mathrm{f}}{M} \times V_\mathrm{f},\\
V_\mathrm{f} &=
\frac{M}{\gamma_\mathrm{f}} \times c_\mathrm{i} \times V_\mathrm{i},
\end{align}
and using \eqref{volume}
\begin{align}
V_\mathrm{i} + \Delta V &=
\frac{M}{\gamma_\mathrm{f}} \times c_\mathrm{i} \times V_\mathrm{i},\\
\Delta V &=
\frac{M}{\gamma_\mathrm{f}} \times c_\mathrm{i} \times V_\mathrm{i} - V_\mathrm{i}\\
&= \left(\frac{M \times c_\mathrm{i}}{\gamma_\mathrm{f}} - 1\right)V_\mathrm{i}.
\tag7\label{dilute}
\end{align}

Now simply plug in your values and get the solution (hover over box to reveal):

   $\displaystyle \Delta V = \left(\frac{\pu{98.1 g//mol} \times \pu{0.1 mol//l}}{\pu{4.9 g//l}} - 1\right)\times \pu{0.100 l} = \pu{0.100 l}.$


Would tin (II) sulfide be considered a covalent network solid? Considering that tin has a Pauling electronegativity of 1.96 and sulphur 2.58, and that a bond is considered to be ionic with a Pauling EN difference of approx. 1.7 at the least, would tin (II) sulfide be considered covalent?

This would be astonishing, since you would expect a metal—non-metal bond to be ionic. Indeed, many (non-academic) websites (such as this one) state that SnS is ionic. Evidence for the ionic character of tin (II) sulfide is the fact that it precipitates from solutions containing Sn(II) ions when adding hydrogen sulphide (which forms S(-II) ions) – thus, the two kinds of ions form a ionic compound (right?). Furthermore, SnS dissolves in hydrochloric acid, while typical covalently bonded networks such as silicon dioxide do not.

Finally, this paper states on the first page (which is freely accessible) that evidence for tin(II) salts with typical symmetrical ionic symmetries is completely lacking.

Thanks in advance, and happy holidays!

There is no sharp border between ionic and covalent compounds, much like there is no sharp border between a boy and a man. Some compounds between a metal and a non-metal, like $\ce{LiF}$ or $\ce{NaCl}$, are predominantly ionic, some (think of $\ce{Mn2O7}$ or $\ce{TiCl4}$) have markedly covalent character, and the rest are somewhere in between. As you see yourself, $\ce{SnS}$ is ionic enough to have certain kinds of reactions typical of ionic compounds, but not so ionic in terms of electronegativity or crystal structure.

Now, people in some areas of chemistry would put forth some definitions to tell covalent from ionic. Your "difference of electronegativities of at least 1.7" is one of those. Another is based on $\ce{MgO}$ and would use it as a border. There are a lot of definitions, and it's not like they are all wrong or useless; in fact, they are necessary to keep our stuff in order. One has to remember, though, that they are all slightly different, and all inherently arbitrary (much like the legal age limitations for driving, voting, drinking, and other activities, if we'd talk once more in terms of the "boy or man" analogy).

This must have been said here like a dozen times before, but whatever. Repetition is the mother of learning, they say.


How would you know if ... is H-bond or permanent dipole or induced dipole? For example like HCl, why isn't it a hydrogen bond?

Hydrogen bonding is defined as an electrostatic attraction between a hydrogen atom bearing a partial positive charge in a molecule and an electronegative element of another molecule. The partial positive charge is the result of a dipole, caused by an electronegative element, such as F, O, or N. The other electronegative element to which the partially positive hydrogen is attracted to is also usually F, O, or N.

Here's the IUPAC definition; note that it does not exclude the possibility of hydrogen bonding in molecules such as HCl:

  A form of association between an electronegative atom and a hydrogen
  atom attached to a second, relatively electronegative atom. It is best
  considered as an electrostatic interaction, heightened by the small
  size of hydrogen, which permits proximity of the interacting dipoles
  or charges. Both electronegative atoms are usually (but not
  necessarily) from the first row of the Periodic Table, i.e. N, O or F.
  Hydrogen bonds may be inter-molecular or intramolecular.

http://goldbook.iupac.org/H02899.html


Is it possible to light a match by scratching it on hard surfaces, like they do in movies? In movies cool guys don't need a matchbox − they just light a match by scratching it on whatever hard surface − concrete walls, wooden bar tables, leather belts, etc. Is it just a movie trick, or is there actually a type of matches you can do that with?

So far I've tried scratching usual matches on a ton of different surfaces and the heads just smear without even a smoke.

Most matches these days are safety matches: they're designed to need something more than ordinary levels of friction to ignite, by splitting the combustion materials between the match-head and the striking strip, i.e.  the brownish paint thing down the side of the matchbox or across the front of the matchbook.

Most "regular" matches now are safety matches. The brownish strip has red phosphorous in it. The match head has potassium chlorate $\ce{KClO3}$. Combine those two with friction, and you get fire.

There are still striking matches, with heads of phosphorus sesquisulfide $\ce{P4S3}$, still around. Though largely in countries where consumer protection is not yet well-developed.

With sufficient energy, you can still ignite a safety match without the striking strip. But you'll need more energy than the friction of scratching it on a wall or bit of stubble.

(PS Don't play with fire. Get a friendly neighbourhood chemist to play with fire on your behalf.)


Are all higher cycloalkenes chiral? Are all higher trans-cycloalkenes chiral as trans-cyclooctene is?

This question supplements my previous question: Why is trans-cyclooctene chiral? See also the follow-on question: Do more double bonds cause a bigger number of stereoisomers in cycloalkenes?

Background

One can draw conformations of n-butane with the carbon-carbon bonds oriented in certain directions and the methyl hydrogens pointing in certain directions that are chiral.  However, since rotation about single bonds is typically fast at room temperature, these chiral conformers of n-butane would only be resolvable at extremely low temperatures.  Therefore we do not typically consider n-butane to be a chiral molecule.

Take-away lesson: If a chiral conformation can readily interconvert with an achiral conformation, then the molecule cannot be resolved into enantiomers and will be optically inactive.

However, there are molecules where rotation about single bonds is restricted (for example due to steric interference) and the rotational barrier is high enough that the molecules can be resolved into their respective enantiomers at room temperature.

Molecules where the rotational barrier about a single bond is high enough to allow different conformations to be isolated at room temperature are called atropisomers.

The biphenyl pictured below (in a chiral conformation) is an atropisomer due to restricted rotation about the central single bond connecting the two benzene rings.  The steric interference that would develop between the 4 ortho groups if the molecule were to approach a planar (achiral) conformation prevents the chiral conformer from equilibrating with the achiral (planar) conformer.  Hence this biphenyl has been resolved into its two enantiomers at ambient conditions.

Trans-cyclooctene

Trans-cyclooctene is also capable of being resolved into its two enantiomers due to restricted rotation about single bonds - it also belongs to the class of atropisomers.

In the figure below trans-cyclooctene conformation (a) is planar, therefore this conformation is achiral. The two conformers pictured in (b) are chiral - these conformations are enantiomers of each other. Trans-cyclooctene will be achiral if the two enantiomers pictured in (b) can readily interconvert through achiral (planar) conformer (a).

Build a model or draw trans-cyclooctene and notice the two hydrogens on the trans double bond; one of them is "inside" the ring in the planar conformation and would cause severe steric problems as the methylene chain tries to flip through the planar conformation to the other face of the double bond as pictured below. This steric interference makes it difficult for the molecule to achieve the planar conformation.

(Image source: John D. Roberts, Marjorie C. Caserio, Basic Principles of Organic Chemistry. Excerpt available at http://www.organicchemistry.com/cycloalkenes-and-cycloalkynes/.)

Because of this destabilizing transannular interaction, trans-cycloalkenes with up to around 11 carbons do not exist as planar molecules. They exist in a twisted conformation as pictured above for trans-cyclooctene.  In your model of trans-cyclooctene, notice how you have to pull and stretch the methylene chain in order to convert trans-cyclooctene into its mirror image.  This rotational movement is highly restricted due to the internal hydrogen.  It is this barrier to rotation about singles bonds that allows trans-cyclooctene to be resolved into its two mirror image forms at room temperature.

The barrier to racemize trans-cyclooctene (e.g. pass through the planar conformation (a)) is 36 kcal/mole.  As you might imagine, as we make the chain longer it should get easier and easier to loop the methylene chain around the internal (protruding) hydrogen on the double bond.  Indeed, the barrier to interconvert the two enantiomers of trans-cyclononene is only 20 kcal/mole. Optically resolved trans-cyclononene must be kept below 0°C to prevent racemization.  The barrier to racemization in trans-cyclodecene is even lower at 10 kcal/mole.

  Are all higher trans-cycloalkenes chiral as trans-cyclooctene is?

First off, these rings can have chiral conformations, just like butane.  I think you are really asking, "Are all higher trans-cycloalkenes optically active as trans-cyclooctene is?".  That is, do these larger rings also have difficulty reaching the planar, achiral geometry?

So the answer to your original question is - it depends.  It depends on the temperature you specify.  At room temperature, trans-cyclooctene is the only trans-cycloalkene where the two enantiomeric conformations are stable enough (e.g. the barrier to reach the planar conformation is high enough) to be isolated.  If we lower the temperature to 0°C, then we can isolate the enatiomers of trans-cyclononene.  If we go low enough in temperature, we could isolate enantiomers of n-butane.


Can the Galvanic cell use a different ionic solution? It makes sense that
water-soluble salt when dissolved in water, they conduct electricity. Then when using the old method of the Volta Battery(using copper and zinc) with a different water-soluble salt dissolved in water, can it make electricity too?
Plus, If I use a ammeter to check its current, then does all of the water-soluble salts do not go to zero?

Can the Volta Battery use a different ionic solution?

Yes you can use any soluble salt as long as it does not react with electrodes, and certainly such a battery would be able to produce current. My chemistry high school teacher had a clock which ran on salt water. See for example, https://science.howstuffworks.com/environmental/green-tech/sustainable/water-powered-clock2.htm. This technology is at least 400 years old. volta used common salt as well for zinc and copper.

Now the problem is that you can generate voltage by using copper/ zinc and salt water however one cannot predict the voltage by using the Nernst equation. It requires that the electrode dip in its own ions. Hence such batteries are theoretically difficult to deal with.


Making o-bromoaniline from aniline How can I make o-bromoaniline from aniline (as a major product)?

I know how to make p-bromoaniline from aniline, by acetylation followed by treatment with $\ce{Br2}$/$\ce{CH3COOH}$ and hydrolysis. But in this process the ortho isomer (if formed) will be the minor product.

Didn't you consider the way through sulfanilic acid, the para-position will be protected. In this case both substituents will direct the electrophilic substitution in ortho- position with respect to amino group. Then remove sulfonic acid group by desulfonation and you will obtain your ortho-bromoaniline.


Why Ca 2+ and not Ca +2 I have been learning chemistry for a year or two but I still can't understand this. According to maths we should place the plus or minus sign in front of a number like -2. However, In chemistry we always put the sign behind a number when writing ions like $\ce{Ba^2+}$ ,$\ce{Cu^2+}$, $\ce{O^2-}$ etc. So, I want to know why do we write it that way?

Chemistry has an organization called the International Union of Pure and Applied Chemistry (IUPAC), which sets standards for nomenclature (among other things). Chemists usually follow its recommendations.

IUPAC says

  Ionic charge is indicated by means of a right upper index, as in $\ce{A^n+}$ or $\ce{A^n-}$ (not $\ce{A^$+n$}$ or $\ce{A^$-n$}$).

and

  The iconic charge number is denoted by a right superscript, by the sign alone when the charge number is equal to plus one or minus one.
  ...
  $\ce{Al^3+}$ is commonly used in chemistry and recommended by [74 = see quote above]. The forms $\ce{Al^$+3$}$ and $\ce{S^$-2$}$, although widely used, are obsolete [74], as well as the old notations $\ce{Al^{+++}}$, $\ce{S^=}$, and $\ce{S^{--}}$.

No reason is given, but you'll find a historical account of sign conventions for charge and a defense of IUPAC's choice in this article. He says,

  [IUPAC's convention] avoids confusion with the conventional symbolism for inherently positive and negative numbers and maintains consistency in how we count physical entities. Thus, in counting apples, we say two apples, three apples, etc., not apples two, apples three – that is, the number always precedes the name of the entity being counted. Likewise, when counting charges, we should say two positive charges or three negative charges, not positive charges two or negative charges three. The IUPAC ruling was intended to make the charge number symbolism consistent with this verbal convention.

Personally, I think the charge should be written like any other signed number. Formulas typed from the keyboard without superscripting can be ambiguous; is X2+ to be interpreted as $\rm X^{2+}$ or $\rm X_2^+$? (You should type X(2+) or X2(+), but few people actually do that.) But it's pretty clear what X+2 would mean. Putting the sign first also makes it trivial to parse the charge from a formula typed this way without the need for parentheses.


Are CaOCl2 and Ca(OCl)2 same or different? Are the two compounds same? Also, which one is considered bleaching powder? Please explain their differences on the basis of their structures also.

$\ce{CaOCl2}$ is a calcium chloride hypochlorite. There are two different types of anions: chloride $\ce{Cl-}$ and hypochlorite $\ce{ClO-}$. So its structure is:

$\ce{Ca(ClO)2}$ is a calcium hypochlorite (also called bleaching powder or bleach). There is only one type of anions: hypochlorites $\ce{ClO-}$. So its structure is:


Why is petrol also called octane? Why do people commonly use octane as a synonym for petrol (gasoline)? Isn't petrol a mixture of hydrocarbon distillates--not a pure substance?

Octane is a specific molecule with 8 carbons normally found among the many different molecules in gasoline. Pure iso-octane (2,2,4-trimethyl pentane) was chosen as the reference fuel standard for resistance to preignition for gasoline fuels. So the octane number of a fuel is a rating of the preignition potential of that fuel compared to a fuel containing 100 %  iso-octane molecules. Since iso-octane is a mouthfull, the "iso" part was dropped and people just call it octane.  This is why some vendor calls his product petrol or gasoline or gas or octane or high test, etc...


If given a half reaction, how do you determine if an element can exist in a acidic conditions? I was given the formulas,

$$\begin{alignat}{2}
\ce{MnO4^2- \;&<=> MnO4- + e-}\quad &&E^\circ = -0.56\ \mathrm{V}\\
\ce{MnO2 + 2H2O \;&<=> MnO4^2- + 4H+ + 2e-}\quad &&E^\circ = -2.27\ \mathrm{V}
\end{alignat}$$

I gathered you have to see if it’s non-spontaneous or spontaneous and compare it with the original reaction, but, when I asked my teacher on how to start, he said “multiply the first reaction by 2 and the second by −1.”
I understand the ×2 to balance electrons, but why do you multiple it by −1? I don’t understand this question, any help would be appreciated. Thanks.

Doing so will result in the two half equations

$\ce{2MnO_4^2- -> 2MnO_4^- + 2e^-}$
and

$\ce{MnO_4^2- + 4H^+ +2e^- -> MnO_2 + 2H2O}$

which you then combine into the total equation

$\ce{3MnO_4^2- + 4H+ <=> 2MnO4- + MnO2 + 2 H2O}$.

The question is which way this goes. You can see that you have loads of $\ce{H+}$ on the left-hand side so you can expect that in acidic conditions the equilibrium will be shifted to the right hand side. You can quantify this by calculating the electrochemical potential from the Nernst equation for the two half cells.


Can compounds (or only elements) be referred to as oxidising and reducing agents? In the reaction

$$\ce{3CuO (s) + 2NH3 (aq) <=> N2(g) + 3H2O(l) + 3Cu(s)}$$

the oxidation number of $\ce{Cu}$ goes from $+2$ in the reactant side to $0$ in the products, and so it is reduced.
The oxidation number of $\ce{O}$ is $-2$ and stays the same on both sides.

My textbook refers to $\ce{CuO}$ as the oxidising agent. Is this because the $\ce{Cu}$ in $\ce{CuO}$ is oxidised while $\ce{O}$ is neither oxidised or reduced, so overall it is an oxidising agent?

What happens for a compound $\ce{AB}$ where $\ce{A}$ gets oxidised and $\ce{B}$ gets reduced? Do I call the compound $\ce{AB}$ a reducing agent if the net change in oxidation number (between the two elements $\ce{A}$ and $\ce{B}$) is negative?

My textbook refers to $\ce{CuO}$ as the oxidising agent. Is this because the $\ce{Cu}$ in $\ce{CuO}$ is oxidised while $\ce{O}$ is neither oxidised or reduced, so overall it is an oxidising agent?

No. $\ce{CuO}$ is the oxidizing agent (the thing that causes something else to be oxidized) because the $\ce{Cu}$ is reduced (gains electrons, going from oxidation state $+2$ to $0$) in the course of oxidizing the nitrogen of $\ce{NH3}$, which loses electrons and goes from oxidation state $-3$ to $0$.

Conversely, $\ce{NH3}$ is the reducing agent here, because it causes something else to be reduced (here, the $\ce{Cu}$) and loses electrons in the process.

  What happens for a compound $\ce{AB}$ where $\ce A$ gets oxidised and $\ce B$ gets reduced?

If only the species $\ce{AB}$ is involved, or in the case where two of the same species $\ce A$ react to form both oxidized and reduced products:

$$
\ce{2A -> Ox + Red}
$$

the reaction is called disproportionation.

  Do I call the compound $\ce{AB}$ a reducing agent if the net change in oxidation number (between the two elements $\ce A$ and $\ce B$) is negative?

Strictly in the case of a $1\!:\!1$ compound of $\ce{AB}$ on a mole basis, if the net change in oxidation number is negative, then you would call $\ce{AB}$ an oxidizing agent, because it pulled electrons away from something else (it oxidized that other thing) in order to achieve the net negative change in oxidation number.

For $\ce{AB}$ compounds not in a $1\!:\!1$ molar ratio, a universal principle can't be stated $-$ it depends on the particulars of the mole ratio and the individual changes in oxidation number.


Why must the forward reaction and the backward reaction follow the same pathway? It is often claimed that a (normal) catalyst catalyzes the forward and the backward reaction to the same extent, so the equilibrium position is not shifted.

I would like to ask why must the forward reaction and the backward reaction follow the same pathway?

I could imagine a mechanism for this reversible reaction from $A$ to $D$:

$$\ce{A <=> D}$$
$$\ce{A -> B<=> D}$$
$$\ce{A <=> C <- D}$$

So that the forward reaction and the backward reaction must follow different pathways, as one step of both pathways are irreversible.

Is this scenario impossible in reality? If so, why? If not, is there any example in real life?

A reaction could proceed by many different pathways, for example an extreme way would be splitting the reactants into atoms and combining them to form products, but this is essentially impossible, the energy required would be huge and how would the atoms reassemble just into the product required?

In any reaction the molecules follow the lowest energy pathway, which is to say that the lowest activation barrier between reactants and products will be crossed in preference to any other pathway. The same argument applies to a catalysed as to a 'normal' reaction even though a catalysed reaction proceeds by a different mechanism. The lowest energy pathway also ensures in some general way that close to minimum number of bonds are broken in the reaction.

The reason for this is that the activation energy (barrier height) is usually far higher than average thermal energy and many collisions happen between reactants before by chance there is enough energy to cross the barrier. The probability $p$ of having sufficient energy $E$ is given by the Boltzmann exponential distribution, $p \approx exp(-E/(RT))$. This is a rapidly falling distribution as $E$ increases thus the minimum energy pathway is generally hugely favoured over others.

In biochemical reactions it is possible for the same molecules to be synthesised by different pathways, involving several steps, but the same arguments apply to each stage in the reaction within the enzymes used.Synthetic chemists will have many examples of the same products being formed from the same starting molecules and progressing by different sequences of pathways in a series of $separate$ reactions.  However, the yields will be generally be different; one of the aims of synthetic chemistry is often to optimise yields, but I take the question to ask that the reaction is not that of a separate sequence of steps but in one process.


How does the catalyst in a polyester resin or epoxy work? I would like to understand how the polyester resin or epoxy catalyst works and to what degree it must be mixed with the base material in order to be effective.

Typical product instructions advise to thoroughly mix the catalyst for a specified period of time.

How thorough does this really have to be in order to have the material cure properly? Is the mixing done just to initiate a thermosetting behavior and does that just continue to propagate heat to the rest of the material to complete the curing of the entire mass? If this is the case, wouldn't the base material cure if it were heated sufficiently without the presence of a catalyst?

Or is something else happening when the catalyst begins the cure? And how does mixing effect the quality of the cure?

The degree of mixing is very important in "two part" epoxy binder systems for coatings and adhesives.

In most cases the two liquids are quite viscous and the liquids themselves are better described as suspensions or colloidal dispersions rather than solutions. Unlike say two simple aqueous solutions when combined our two prepolymer liquids will not readily homogenize in a reasonable amount of time unless they are mechanically mixed in some way.  Even if the liquids do combine very slowly they will often gel or vitrify by the molecules at the interfaces reacting before the liquids homogenize.

In many epoxy systems we have, as Beerhunter stated above, some oligomer or polymer with epoxy functionality in one part and some liquid with amine (or sometimes hydroxyl) functionality in the other. Every amine needs to "find" an oxirane ring to attack or the curing process simply can't occur. The curing process works by nucleiphillic attack from amines to open epoxied rings.

The presence of leftover unreacted amine in cured epoxy materials often results in a phenomenon known as amine blushing or amine blooming. leftover amines or polyamides migrate to the surface and may in some cases react with air to form carbamates. This can be disastrous for desired materials properties like tack, adhesion (for the epoxy as a substrate), or cohesive strength of the bulk material.

In summary you can't build a robust three dimensional polymer network without thoroughly mixing because all the reactants can't react. Always mix as thoroughly as possible to get complete conversion.

My experience with polyesters is limited so I will not comment on that.


Applying salt to roads to melt ice--is it special? Most everyone is familiar with applying salt to the roads during winter to melt ice.
I'm wondering if there's anything special about placing salt on the roads or would any other substance, sand for example, work equally well in melting the ice.

Yes, applying salt to melt ice it is special:

Salt is a great ice melter because it causes “freezing point depression.” This means that salt helps in lowering the freezing point and, consequently, the melting point of water (the main component of snow and ice). In its pure state, water freezes at 0°C or 32°F. By using salt, that temperature can be lowered, which forces the ice to melt and prevents the water from freezing or re-freezing. This does not happen with  sand.
After melting, no additional effort is needed to clean the roads, because salt mixes with water. If you use sand, additional work will be needed to remove it from the streets afterwards.
Salt is much cheaper than other chemicals, although arguably more expensive than sand, depending on location.
Salt is recyclable, so no wastage or decompositon of salt is there in such process. Since salt becomes soluted in the melted ice, it gets washed away and absorbed by the ground. This makes salt hard to retrieve after use, and harms the environment by damaging the plants in the area.


Does vibrating an NMR sample produce the same benefits and artifacts as spinning an NMR sample? Does vibrating an NMR sample produce the same benefits and artifacts as spinning an NMR sample?  Are there "vibration sidebands" above the C13 satellites and is there a line thinning effect on the sample spectrum?  Please list any references that you may have read or seen.

Thank You

When you spin a sample you make sure that orientation effects are averaged faster than the relaxation time. This holds true if you spin a sample at 8 s^-1 and H1 relaxes at 1 s^-1. Now, wobbling (aka shaking) will keep orientation of the molecules for the most part (unless it creates spinning motion in the solution). It will change several %, but the general direction will be the same. So, no, shaking will not do the molecule averaging like spinning does.

Btw, why do we spin samples. I mean, if the tube is half decent and you spend 5 minutes on shimming you get about as good of a result. All spinning does is hiding poor XY shims by averaging them.


Carboxylic acid bond lengths in 3-carboxycyclobutane-1-carboxylate Select correct order of bond length of the above bond P, Q, R & S:
(A) P > R > S > Q
(B) Q > R = S > P
(C) S > Q > R > P
(D) None of these

I think the answer should be (D) because in this molecule, both carboxylic acid groups are equally acidic and the $\ce{H+}$ may migrate from one acid moiety to the other. This would be somewhat similar to the idea of resonance, although it isn't actually the same idea.
However, the given answer is (B) which we will get if this structure remains static. Because of the proton transfer, I don't think this can happen. Am I right or not?
Moreover, if we were to use some spectroscopic method to study the molecule, would we get an image in which the molecule has four equivalent bond lengths (due to proton transfer), or not?

It's most certainly true that proton transfer between the two acid groups can occur and is facile (particularly in an intermolecular manner). However, that only means that over a sufficiently long period of time, the four C–O equilibrium bond lengths will all average out to be the same. It does not mean that at any one instant in time, they are all the same.*

As you said, this is not the same thing as resonance. Resonance in a carboxylate group is a phenomenon whereby at all points in time, the two bonds are equal. There is no flip-flopping between two different structures.

Before we get to the answer for the actual question, I'll talk a little bit about your last question.

You bring up the idea of using spectroscopy to measure bond lengths, which is a very good point. The general answer is beyond the scope of this question, but it basically comes down to how "fast" your spectroscopic method is. To put it very simply, if it takes $x$ seconds (or microseconds, nanoseconds, whatever) to obtain the information that you are trying to get, then what you will observe is a time average over those $x$ seconds. Bear in mind that every method of spectroscopy takes a finite time, including the interaction of photons with the system. There is no such thing as an instant method; there are merely faster and slower ones. So it comes down to how fast the system is moving versus how fast your method is.

It's exactly analogous to taking a picture of a moving object, like a fast car.† If you try that with an phone camera with a long exposure time you will probably just get a blur. But a high-speed camera is good enough to take clear snapshots of Formula 1 cars during the race.

Of course, for photons you need to phrase that in a more quantum mechanical manner, using e.g. the energy–time uncertainty principle. But that's the general idea, at least. If you use a method that is "slower" than proton transfer between the acids (perhaps NMR), then what you may find is that all four oxygens (and hence C–O bonds) appear to be equivalent. If you use a method that is "fast" enough (maybe IR?), then you may well find that they are inequivalent.

Back to your question.

Generally, it is the instantaneous (static) picture which chemists deal with. That means that, at least for the purposes of this question, the dynamic motion of the system is not relevant, even though you are very correct to bring up the proton transfer. Equivalently, we could say that these questions are usually meant in the context of a (hypothetical) infinitely fast measurement method.

An infinitely fast method would be able to tell the difference between the $\ce{CO2H}$ and the $\ce{CO2-}$, because at any one point in time, one of them is protonated and one isn't. They can't be in a state where they're both half-protonated.

But an instantaneous method can't differentiate between the two oxygens in the carboxylate group. Those are entirely equivalent by resonance. And therefore you get three different bond lengths: Q > (R = S) > P.

* Of course, we are neglecting vibrational motion of the bonds, which technically mean that at any point in time it's far more likely that all the bonds are different. For example, even in the carboxylate group in which both oxygen atoms are equivalent (due to resonance), if an asymmetric stretch occurs then the instantaneous bond lengths will be different. Here, when we talk about "bond lengths", we are only referring to equilibrium bond lengths. We are also neglecting single bond rotations...

† I am not an expert on photography, so you may want to check out Photography Stack Exchange.


How are unstable transition states determined? The Wikipedia article about the Grignard Reaction says that

  The addition of the Grignard reagent to the carbonyl typically proceeds through a six-membered ring transition state.

This got me thinking: How was it determined that the Grignard reaction has the ring as a transition state? In fact, how are unstable transition states in general determined?

Experimentally, and surprisingly,  there is only circumstantial evidence for most short lived species in any type of reaction, i.e. no one has spectral or structural evidence for a transition state, such as often portrayed in textbooks, other than for example NaI dissociation or atom-diatomic molecule reactions e.g. H+OH  etc. done in ultra-low vacuum.

For longer lived intermediates (picoseconds to longer) short pulse laser pump-probe (flash photolysis) methods have to be used to obtain spectral signatures but give no structure, and this method only works for light triggered reactions which are rare. An alternative is stopped flow, but the time scale is slow, many microseconds so usually too slow for most intermediates, and again does not give structure. NMR is immensely slow also. Time resolved x-ray only works on crystals. So far light driven reactions in proteins have been studied down to a few tens of femtoseconds.  There is a huge and exciting field here to be explored.


Why is sodium sulfate a neutral salt When I have a solution of $\ce{Na2SO4}$, the sodium sulfate should ionize:
$$\ce{Na2SO4(aq) -> 2Na+(aq) + SO4^{2-}(aq)}$$

In my mind, the sodium ions should just float around, but because $\ce{HSO4-}$ is a weak acid:

$$\ce{HSO4- (aq) + H2O <-> H3O+ (aq) + SO4^2- (aq)}$$

The sulfate ion should be a weak conjugate base, and hence the sodium sulfate salt should be a weak base. This was my line of reasoning, until today my friends told me it is actually a neutral salt, which was confirmed by searching online. So why is it a neutral salt? (i.e. what's wrong with my reasoning?)

For a solution of $\ce{Na2SO4}$, the sodium sulfate will ionize as you indicated:

    $\ce{Na2SO4(s) ->[aq] 2Na+(aq) + SO4^{2-}(aq)}$

This reaction is essentially irreversible in that the ions in an unsaturated solution won't suddenly form solid $\ce{Na2SO4}$ again.

$\ce{HSO4-}$ is a not a weak acid, but has a $\text{pK}_\text{a2} = 1.99$ which makes a moderately strong acid. So for  pH > 3 essentially all the $\ce{HSO4-}$ will ionize:

    $\ce{HSO4-(aq) + H2O ->[pH > 3] H3O+ (aq) + SO4^2- (aq)}$

Thus $\ce{SO4^{2-}}$ is such a weak base that reverse reaction doesn't happen when you dissolve $\ce{Na2SO4}$, and a solution of $\ce{Na2SO4}$ in distilled water stays neutral rather than becoming basic.

    $\ce{\require{cancel}H2O + SO4^2- \cancel{->} HSO4- + OH-}$

To think about a different example acetic acid is a weaker acid, $\text{pK}_\text{a} = 4.76$, so the acetate anion is a stronger base than the sulfate anion. A solution of sodium acetate would become weakly basic.

One last point - the statement should be "sodium sulfate a neutral salt in aqueous solution." In other solvents sodium sulfate could act as a base.


N-heterocyclic carbene metal complexes. Question about the delocalization of the nitrogen lone pairs I am wondering why nitrogen lone pairs are delocalized in NHC metal complexes. What I can think of is, when complex is formed, plus charge is formed on the carbon atom and therefore delocalization from the nitrogen atoms. But there should be then "plus" sign like in the picture below.

But these kind of carbene complexes are always depicted without the plus sign. Can anyone please tell me what I am missing out?

Some parts of this answer probably border a bit on personal opinion, but I will at least try to justify having such an opinion.

The first thing to do is to actually examine the heterocycle. What kind of ligand is it?

This is a carbene. It donates two electrons via the lone pair on the carbon in plane with the ring. Notice that there is a smear of positive charge from the delocalization of the $\pi$ system between the N-C-N system. That is balanced by the negative chage on the carbon, however. Overall, this ligand is neutral, so it is an L-type ligand.

What's the overall charge on this metal complex? This is a ruthenium(II) ion (2 chlorides, everything else neutral), so this is a $d^{8}$ metal ion. You have 2 electrons from the NHC, 2 from each chloride (4 total), 2 from the phosphine ligand, and 2 from the carbene (the double bond is back-bonding). That gives 10 electrons from ligands, so this is a fine-looking neutral 18 electron complex.

For a complex that's overall charge zero, why is there a plus charge hanging out by itself? Well, it turns out that we're extremely sloppy with charges. For example, the chlorides are X-type ligands but the phosphine is not. Do we stipulate this difference in the structure? No we do not, though we use it during electron counting. The authors here felt that the smear of positive charge of the carbene, which is how you would draw the structure in isolation, should be shown but chose not to show the negative charge of the carbene, which is the case by convention. That seems somehow inconsistent to me because the overall charge of this ligand is neutral.

Look at these other representations:

Mendeleev Communications, 2019, 29 (1), pp 38-40. (10.1021/om3009298)

Organometallics, 2012, 31 (23), pp 8355–8359. (doi: 10.1016/j.mencom.2019.01.011)

Both representations of the NHC are used: without explicit $\pi$ system and with.

Note that neither of them say anything about the positive charge. This seems more consistent to me, and these authors, at least, agree with me.


Does Victor Meyer test distinguish alcohols and alkyl halides? I read from a source that alkyl halides and alcohols cannot be differentiated by the Victor Meyer test. It says that both react similarly with the reagents and form the same products. Is this correct, and if so what are the reactions for alkyl halides.

Aaron,

You could be correct if your alkyl halides to be ditinguished are 1,2,3 degree alkyl IODIDES.
Because we ultimately are converting 1,2,3 degree alcohols into IODIDES in step-I; followed by reactions with NaNO2/HCl and then NaOH.

@Pritt Balagopal

He had a valid point, when he said "alkyl halides and alcohols are said to react in the same way with the reagents used for Victor Meyer test"


Protonation of Carbonyl oxygen in Fischer Esterification In the Fischer Esterification, why is it the acid's carbonyl oxygen getting protonated and not the hydroxyl oxygen? This protonation occurs in Step 1 of the mechanism below.

I read in some textbooks that this is due to the carbonyl oxygen being more neucleophilic and the hydroxyl oxygen. Why is this so?

There are a few reason why the carbonyl oxygen gets protonated -

1) The carbonyl oxygen is sp2 hybridized which is more electronegative than the sp3 oxygen. So, it has more tendency to give electrons since it is more negative.

2) The lone pairs of carbonyl carbon are more available since they are not in conjugation whereas the lone pairs of the other oxygen are not much available since they are in conjugation with carbonyl carbon.

3) When carbonyl oxygen donates its electron the resulting structure is more stabilized (because of conjugation) than the structure formed when the other oxygen atom donates.

So, that's why carbonyl oxygen has far more tendency of donating electrons than the -OH grp.


Do the products in exothermic reactions have higher kinetic energy than the products? Do they also have lower chemical energy? I understand that in an exothermic reaction energy is transferred to the surroundings. More energy is produced from making bonds than absorbed to break bonds.

I would imagine that the products have lower chemical energy (bond energy perhaps?) but have higher kinetic energy as measuring the temperature of the surroundings is effectively measuring the kinetic energy of the specific molecules? I only doubt myself because on enthalpy change graphs the products are always lower than the reactants in an exothermic reaction, so increasing KE doesn't seem likely. If someone could clear up my confusion that'd be great!

In line with the comment of orthocresol, these potential energy diagrams only display the change in the chemical potential as a function of the reaction coordinate.

To clarify: Let's consider the theoretical situation of a reaction mixture in complete vacuum that undergoes a reaction towards a product state. If we assume that the reaction is exothermic and there is no loss of energy due to irradiation, then this means that the products are heated up by an amount of energy that is gained from the loss in chemical potential (i.e. the exothermic reaction). In other words, if you would plot the chemical potential + the kinetic energy (the two terms combined) as a function of the reaction coordinate, you would obtain a straight line due to the law of conservation of energy, because chemical energy is converted to kinetic energy (heat).


Is there any way to refine Feldspar? Feldspars are mostly composed of three compounds:

KAlSi$_3$O$_8$
NaAlSi$_3$O$_8$
CaAl$_2$Si$_2$O$_8$

I notice they each have Aluminum. Feldspar is far more common than Bauxite (the only Aluminum ore commercially mined). As it stands now, Feldspar is used in glass-making, but I'm wondering if there's some chemical refining process that can extract the Aluminum somehow?---especially that third one, since it has two Al's.

Note: I'm not looking for an economical method. That's proly asking too much. I'm not worried about aluminum reserves, because bauxite is still very plentiful. I'm simply interested in a chemical method to extract Aluminum, or at least Alumina (Al$_2$O$_3$), from Feldspar, purely for reasons of chemical curiosity.

The only thing I've researched is the Bayer Process, but I have no idea if it can be applied or modified for Feldspar.

Aluminium oxide can be isolated from feldspar by first acid digestion with a mixture of hydrofluoric acid and sulfuric acid (get rid of silicon by formation of silicon tetrafluoride), then digesting the residue with sodium-potassium carbonate, dissolving the obtained mass in dilute hydrochloric acid and then precipitating $\ce{Al(OH)3}$ by adding ammonia. The $\ce{Al(OH)3}$ can be calcined to aluminium oxide by heating to over 1100°C.

Note that the described procedure requires a suitably equipped laboratory .


Which of the following pair of molecules have identical bond dissociation energies (approx)? Which of the following pair of molecules have identical bond dissociation energies (approx)?

  $\ce{H2}$, $\ce{F2}$
  $\ce{F2}$, $\ce{I2}$
  $\ce{N2}$, $\ce{CO}$
  $\ce{HF}$, $\ce{O2}$

I think it's either (2) or (3). (Single correct type).

Bot $\ce{F2}$ and $\ce{I2}$ bonds are weak because of inter electronic repulsions and due to large sizes respectively.

In case (3), both molecules have bond order 3.

The answer here has nothing to do with how the molecules relate to each other, e.g. similar bond order or length. This is actually a question that highlights an unexpected break in a trend.

The bond enthalpies for the halogens are:

$$
\begin{array}{|c|c|}\hline
\text{Halogen}&\text{Bond energy (kJ/mol)}\\\hline
\ce{F-F}&\pu{156}\\\hline
\ce{Cl-Cl}&\pu{243}\\\hline
\ce{Br-Br}&\pu{193}\\\hline
\ce{I-I}&\pu{151}\\\hline
\end{array}
$$
Plotting these data we would expect the bond enthalpy of $\ce{F2}$ to be the strongest of them all, something around $\pu{300 kJ/mol}$, but in fact it much weaker than for chlorine.  Indeed it is so weak, it as almost as weak as for iodine.

The reason for this is that fluorine is so small that it breaks the trend.  The atomic radius is so small that the electrostatic repulsion between the nuclei is significant, and the non-bonding electrons surrounding each atom also repel each other due to the small amount of space available.  These forces counteract the strong attraction between the nuclei and the bonding pair of electrons and weakens the bond compared to what might be expected.


What analytical methods are used to quantify C, O, N, and H in a sample? A very common analytical method for detection of more than 70 elements is ICP-OES (Wikipedia page). But apparently, it's not suited for some substances (cited from this paper, section 6.2, page 12):

  The elements that are
  not usually determined by ICP/OES fall into three basic
  categories. The ﬁrst category includes those elements
  that occur either as trace contaminants in the argon gas
  used in the ICP/OES ($C$ from $CO_2$), constituents of the
  sample solvent ($C$, $O$, $H$), or as contaminants from the
  environment or atmosphere ($N$ for example). The second
  category encompasses those elements that require high
  excitation energy, such as the halogens. These elements
  could be determined with poor LOD, however. The third
  category is the family of short-lived radioactive elements
  that are commonly determined by $\gamma$-ray spectrometry.

So, apparently, when I'm working with, say carbon-based fuels, and I want to determine $C$ contents, ICP-OES would not be the method of choice, at least not the standard variety. So, what else would be the typical choice? Or are there modifications to ICP-OES, such as using very pure $Ar$, or $He$, as the plasma gas and sealing the plasma chamber from the atmosphere?

For N:

Total N = N from NO2 + N from NO3 + Total Kjeldahl N

The methods from here are:

Liquid Chromatography: NO2, NO3

Specific electrodes: NO3, Kjeldahl

Spectrophotometry: NO2

This for water including wastewater. I don't know the aplicability for other samples.

Also ICP-MS is worth researching.


Which alcohol will most easily react with HCl to form an alkyl halide? Which alcohol will most easily react with HCl to form an alkyl halide?

A. Primary alcohol

B. Secondary alcohol

C. Tertiary alcohol

D. All alcohols react equally to form alkyl halides

Tertiary alcohols have greater reactivity with hydrogen halides than secondary alcohols, which in turn have comparatively more reactivity than primary alcohols. For tertiary alcohols, the reaction proceeds by an SN1 mechanism which prefers a more substituted alcohol.
Therefore answer is C


Dissociation behaviour of polyelectrolytes Assuming a solution of 100 polycationic molecules, each of them carrying 100 amino groups and an apparent pKa of 6.5. Would it mean that, at a pH of 6.5,

50 molecules will be fully protonated
50 amino groups of each molecule will be protonated  or
50% of the total amount of amino groups present in solution will be
protonated?

The third option would mean that each molecule can have any charge number, from 0 to 100, with 50% of the 10.000 amino groups being protonated.

In other words, is it safe to state that, at a pH below 6.5, all molecules will be mostly protonated?

Edit due to ambiguity:

In other words, is it safe to state that, at a pH below 6.5, all molecules will have more protonated amino groups than unprotonated?

Many thanks in advance!

Ben

Do a thought experiment: Into your polycationic molecule, insert very large spacers in between the amino groups so that they are far away from each other in terms of covalent connection and in terms of spatial distance. Now, break a bond in the middle of each spacer. Do you think this will influence the protonation state of the amino groups?

If the $\mathrm{pH} = \mathrm{p}K_\mathrm{A}$ for your $100^2$ independent amino groups, half of them will be protonated and half deprotonated. If you randomly pick 100 amino groups and turn them into a molecule by repairing the spacers in our thought experiment, you get a distribution of charges (a Gaussian bell curve centered on 50).

If you lower your pH by one unit, about 91% of amino groups will be protonated and 9% deprotonated, and again there will be a distribution of charges for the polycationic molecule/ion.

If, instead of taking a snapshot at a given time, you ask about average protonation state over a couple of milliseconds, you will have a molecular (or ionic, I guess) charge of 50+ for pH 6.5, and about 91+ for pH 5.5.

If the amino groups are not separated by spacers, they will "talk to each other", like the carboxylate groups in citric acid, for example. For citric acid, three pKa values are given, 3.13, 4.76 and 6.40, even though each group has similar chemistry and two are identical. So in that case, things are more complicated. It also gets more complicated if you bundle species with identical charge when figuring out pKa values theoretically (for citrate, there is one neutral species and one 3- species, but three 1- and three 2- species).


Find ratio of number of atoms of $A$ and $B$ However I am thinking of this question to be damn easy but still I'll not say as I am not getting it,

  There are two radio nuclei $A$ and $B$. $A$ is $\alpha$ emitter and
  $B$ is $\beta$ emitter and their decay constant are in ratio $1:2$.
  What should be the number of atoms of $A$ and $B$ at time $t=0$, so
  that probability of getting $\alpha$ and $\beta$ particles are same at
  $t=0$.

So here how can the decay constant be dependent on number of atoms of $A$ and $B$ so that $\alpha$ and $\beta$ particle are in same ratio? I really need a right approach for this one.

There are two radio nuclei A and B. A is α emitter and B is β
  emitter and their decay constant are in ratio 1:2

So B decays twice as fast as A.  If we want the actual number of decay events from A to equal the actual number of decay events from B at t=0, then we must have twice as much A present as B at t=0.

We are given that $\ce{k_{B} = 2k_{A}}$. For the decay events at T=0 to be equal for A and B, then $$\ce{[B]_{o} * k_{B}~=~[A]_{o} * k_{A}}$$
$$\ce{[B]_{o} * 2k_{A}~=~[A]_{o} * k_{A}}$$
or $$\ce{\frac{2}{1}~=~\frac{[A]_{o}}{[B]_{o}}}$$
At t=0, the concentration of A must be twice the concentration of B in order for equal decay events to occur from A and B


Can vapour pressure be used to 'generate' work for free? Given a bottle of water, a closed system, some of the water molecules in the liquid phase will have enough energy to escape that phase, forming water vapour, contributing to vapour pressure.
Now consider the simple setup below:

Water is used as an example, does not have to be water. The work for pumping water is converted to gravitational potential energy of water. But the system seems to be capable of doing extra work on the mass while retaining that gravitational potential energy gained. Where does this capability come from? Doesn't the existence of this capability already violate First Law?
P/S: I know the pressure increment due to formation of vapour will be small, but just because it is small and so negligible does not mean that the 'minor violation' of First Law can be overlooked, so I believe something must have been wrong with the argument above.

I believe the vapour produced in the step 3 of you experiment is due to the particles with high order speed in the Maxwell distribution of speeds curve that escape because of their higher kinetic energies. When escaping, they will reduce the average kinetic energy of particles in the water sample, which will then reduce the net temperature of sample.
This will then prompt the system (water sample) to take energy, in the form of heat, from the surroundings to reestablish the thermal equilibrium. Hence taking back the work done by system on surrounding via expansion of the bottle through heat to maintain its temperature. Although, as you can see in the Maxwell distribution of speed, the particles with enough kinetic energy to escape from the water surface are very less at low temperatures.
So, if you take a system with just enough temperature and manage to isolate it from the surroundings, I think you should see a visible drop in the temperature of water proportional to the work done in expanding the water bottle.
Mathematical description of the above Explanation:
Let's say the total energy change taking place after the particles escape water surface is $\Delta U$ out of which $W$ is used to do work on the surroundings. The remaining energy ($\Delta U - W$) is used to establish an equilibrium temperature inside the bottle and hence is given back to the air-water system inside the bottle.
This step may be happening simultaneously with the next steps I am writing but I am assuming that the air- water system inside the bottle will reach an equilibrium first before any work is done. This is purely to make the math easier and because thermodynamic is the study of equilibriums rather than processes so it won't affect the results as long as this equilibrium is established even when we are studying the final change taken place.
So the net change in temperature inside the bottle is $$\frac{-W}{C_{system}}$$ where $C_{system}$ is the heat capacity of the system.
Net temperature difference between system and surrounding $$\Delta T=\frac{W}{C_{system}}$$
So the system will absorb energy in the form of heat to bring the temperature back to equilibrium $$Q=C_{system}\Delta T=>Q=C_{system}\biggr[\frac{W}{C_{system}}\biggr]=>Q=W$$
So we can see mathematically that the system will take back all the energy it gave to the surroundings by doing work in the form of heat.


Can orbitals in atoms be compressed? Consider an atom (carbon) with a ground state valence electron configuration of:

$$ \underset{3s}{[\uparrow \downarrow]} \underset{3p}{[\uparrow \vert \uparrow \vert \; \; ]} $$

In some molecules such as carbon tetrafluoride, carbon can "expand" to a more energetic electron configuration similar to:

$$ \underset{3s}{[\uparrow]} \underset{3p}{[\uparrow \vert \uparrow \vert \uparrow ]} $$

Although these orbitals would usually be mixed together and also paired together with the opposite atoms they are attached to.

Are there any cases where the reverse can happen?

For example could phosphorous "contract" from

$$ \underset{3s}{[\uparrow \downarrow]} \underset{3p}{[\uparrow \vert \uparrow \vert \uparrow ] }$$

to

$$ \underset{3s}{[\uparrow \downarrow]} \underset{3p}{[\uparrow  \downarrow \vert \uparrow \vert \; \; ] }$$

Looking at the geometry of p orbitals and taking into account molecular orbital theory I think this is always of non benefit unless possibly the materials are put under external strains such as high pressure or heat.

Also, in the cases of atoms like chromium which are "expanded" by default I'd imagine contraction to the typical state could happen.

@bon is right in that you are talking about molecular orbital theory. Speaking broadly, the redistribution of electrons the way you mention happens if the orbitals actually become energetically non-equivalent. This can happen if an atom is surrounded by ligands in certain way. Because orbitals have certain orientations, certain complexes (f.e, tetrahedral) will energetically be more favorable than others (f.e, octahedral).

A simple example is the Jahn-Teller effect: have a look here and here. Essentially, for low-spin d$^7$ configurations you will have something resembling your example, although those are d orbitals; I actually do not know if this happens with p orbitals that you use in your example.

This Wikipedia article on crystal field theory also provides a nice overview of the different ways the orbital energy levels can distort in. However, for better understanding, I suggest a good textbook (I can give a few suggestions if you're interested).


Test for Fe2+ ion in salt using potassium hexacyanidoferrate(III) It's given in my book that $\ce{Fe^3+}$ ion reacts with potassium hexacyanidoferrate(III) to give brown colour.
$$\ce{Fe^3+ + [Fe(CN)6]^3- -> Fe[Fe(CN)6]}$$
Both the reactants have the same oxidation states.

My question is, can I use the same test for $\ce{Fe^2+}$?

First, the hexacyanidoferrate(III) will oxidize $\ce{Fe^2+}$ to $\ce{Fe^3+}$ like this:
$$\ce{Fe^2+ + [Fe(CN)6]^3- -> Fe^3+ + [Fe(CN)6]^4-}$$
Then the reaction will continue to give brown colour.

$\ce{Fe^2+}$ is oxidised to $\ce{Fe^3+}$ while $\ce{[Fe(CN)_6]^{3-}}$ is reduced to $\ce{[Fe(CN)_6]^{4-}}$. These then form the prussian blue precipitate.


Are there reactions where the stirring direction would make a difference? I'm not a chemist. I don't really do anything with chemistry, so I know little of the subject.
I know a bit about chiral molecules. For example, I know that homochirality can affect light polarity.
I'm just curious: Are there chemical reactions where the direction of the stirring (clockwise vs. anti-clockwise) would make a difference? A simple explanation would be appreciated.

No, for three reasons:
Firstly, you need a relatively large molecule and/or a high viscosity and/or a strong shear/extensional flow, before the molecule even notices that it´s in a nonuniform environment. The molecule, if it has an elongated form, will then align perpendicular to the normal vector of that flow field. That only occurs for sufficiently concentrated solutions of polymers, or liquid crystals.
Secondly, you need three vectors to be able to differentiate a left and right hand. You only have the normal vector of the flow field, and the preferred direction of your molecule. Or in other words, your reaction will be indifferent to you watching it from the top of the beaker or through the bottom. ;)
Thirdly, to get a third vector into your system, you need the normal of the flow field to change significantly, in a specific direction (using, say, a propellor instead of a stirring bar), on the lenght scale of your molecules. That is impossible with a macroscopic stirrer.


Is there any difference between medical and cleaning ozone? What are the main differences between medical ozone (auto-hemotherapy, …) and cleaning ozone (for example, the kitchen cleaning devices for vegetables and food). I noticed that they can produce a lower amount of ozone per hour while the medical ones have bigger amount of production.

“Medical ozone” is a mixture of 0.05 to 5 vol% of ozone produced from medically pure oxygen (usually via silent electrical discharge) and 99.95 to 95 vol% of oxygen.
Using air for medical-grade ozone production is strictly forbidden due to formation of hazardous nitrogen oxides $\ce{NO_x}.$
Note that ozone therapy is still considered an alternative medicine.

Ozone used for cleaning doesn't have such constrains and its concentration and production method vastly depends on application area and regulatory standards therein.
For example, for water treatment Russian federal standards prescribe the ozone concentration in the ozone-air mixture from 15 to 22 mg/l based on the required efficiency and level of pollution of the water source, and average contact time of 5 to 15 min.
Prolonged exposure or increased ozone concentration during water treatment may lead to unwanted effects such as problems with coagulation and extensive phenols formation.


Home foundry — melting aluminum, mitigating undesirables I am going to be melting aluminum cylinder head and some soda cans in the home foundry. I am aware the sulphur and the hydrogen present in the melt can make aluminum prone to cracks and brittle.
Are there recommended ingredients that can be put inside the molten aluminum besides flux that precipitates or remove undesirables?

I never encountered S or C as problems in Al. The problem with hydrogen is that it can cause porosity as it is dissolved in the molten Al and comes out as a gas during freezing. Engine blocks and heads are usually Si alloys ( 5 to 12 %) like the wheels. They may have some Cu to make it hardenable. I expect contamination problems would be more likely Fe and Cu ( getting too high ). The cans usually have very little alloy.


What is number of 1H NMR signals for toluene? I am searching the web for the number of 1H NMR signals of toluene and I find two different results. Some say there are two signals while others say there are four (According to me, it should be 4). For the two signals, it suggests that all five aromatic hydrogens are taken as equivalent. A similar problem arises for compounds like ethylbenzene, benzyl alcohol, and anisole. How can I reconcile these different results?

In general, the aromatic region of any $\ce{^1H-NMR}$ spectrum would display a complex coupling pattern even if for mono substituted nucleus such as toluene. However, your $\ce{^1H-NMR}$ spectrum is taken by using a low field NMR instruments such as 60 MHz, you'd not able to get one with good resolution to show these patterns (other factors involve in having broad multiplets instead of resolved peaks are low quality calibration, RF instability, etc.). The following spectrum is taken using a high resolution NMR instrument. The $\delta~7.70-7.70$ region was expanded to show the coupling pattern of 5 aromatic hydrogens. In a low-field instrument, those peaks would overlap to give you a broad multiplet. Read this article to learn about basic principles of coupling patterns.


Potentiometric titration using a pH meter Why is the pH of a buffer solution measured before measuring the pH of an unknown solution?
I do not have any idea about it. I had read a chapter about ionic equilibrium and electrochemistry, but have not been able to find out an answer.

In principle, you should know the pH of the buffer solution. Measuring it before an unknown sample is a straightforward way of calibrating the detector (pH detectors can get a bit funny after a while).

If you put the detector into a pH 5.5 solution and it reads 7, you're going to have to tweak the detector (if possible) or find another.

Measuring unknown samples with a detector that incorrectly measures a known buffer solution is a bad idea but an even worse idea would be to measure unknown samples without knowing your detector was incorrect in the first place.


Validating Phase Diagram of a mixture In lab I did several different runs to create a phase diagram for the mixture between biphenyl and naphthalene. My phase diagram is attached, just for visual purposes. My question is I'm supposed to validate my phase diagram by using the following formula to calculate the enthalpies of fusion for both compounds.

$$\left(\frac{\mathrm dT}{\mathrm dX}\right)_{(T=T_\mathrm m)} = \frac{RT^2}{\Delta H_\text{fusion}}$$
I understand how to do the math, however I'm trying to figure out where that formula comes from to better understand the relationship between the enthalpy of fusion and the phase diagram. Thanks!

A side question is where could I find an accepted value of the eutectic temperature for this mixture to compare my results with?

From the Gibbs-Helmholtz equation we usually write
 $$\left (\frac{\partial \ln(X)}{\partial T}\right )_p = \frac {\Delta H^0_{fus}}{RT^2}$$
where X is the molec fraction of solute. Integrating this equation gives
$$ \ln(X)=\frac{\Delta H^0_{fus}}{R}\left(\frac{1}{T_{fus}} - \frac{1}{T}  \right)    $$
thus a plot of log mole fraction vs reciprocal temperature should produce the enthalpy. The eutectic point is the point where the melting curves meet with the solid phase, its about 0.4 to 0.5 on your figure. On the left should be the melting curve of liquid with solid biphenyl as you plot T vs naphthalene mole fraction, and on the right, liquid plus solid naphthalene. Solid biphenyl + naphthalene should be observed below $\approx 30 $ C at all mole fraction and liquid above the blue dots.


Should I include ions from the solvent in solubility product constant calculations? I'm trying to determine the Solubility Constant $K_\mathrm{sp}$ of $\ce{Ba(NO3)2}$ dissolved in nitric acid solution $\ce{HNO3}$ from experimental data.

$\ce{HNO3}$ will be mostly dissociated so there are many $\ce{NO3-}$ ions already in the solvent before solid $\ce{Ba(NO3)2}$ is added.

When I calculate $K_\mathrm{sp} = [\ce{Ba^2+}][\ce{NO3-}]^2$, do I include in $[\ce{NO3-}]$ the ions already present in the solution before solid $\ce{Ba(NO3)2}$ is added?

For example, say I have dissolved ${0.1 mol}$ of barium nitrate in ${1 L}$ of ${1 M}$ nitric acid solution. There are now ${0.2 mol} + {1 mol} = {1.2 mol}$ of nitrate ion in the solution. Should my $K_\mathrm{sp}$ calculation be

$$K_\mathrm{sp} = 0.1\cdot 0.2^2,$$

or

$$K_\mathrm{sp} = 0.1\cdot 1.2^2?$$

I understand that the common ion effect in this case inhibits solubility somewhat, but it I think my professor told me to include ions from $\ce{HNO3}$ in the calculation, which seems wrong.

Yes, absolutely, all nitrate counts in solubility product regardless of source. The system cannot distinguish what a nitrate ion (or any dissociated species) was bonded to before it dissociated.


Why does a neutral atom of beryllium not have any electrons in a p orbital? Here's what I understand about quantum number and orbitals, please correct me if anything is wrong:

Electrons enter into these different types of orbitals because they have a higher/lower amount of energy, and we know that the number of electrons varies from element to element along with the energy levels of those electrons in a pretty linear way, conveniently allowing us to organize them into blocks on the periodic table where their highest energy level electron is in one of the many types of orbitals (spdf). I know the energy shell occupied by an electron (the principle quantum number) corresponds with the angular momentum quantum number which determines the magnetic quantum number, which determines the type of orbital an electron is in. I know that all of that is because the electrons in higher energy shells/levels have more energy are thus more likely to be in an orbital such as d or f as opposed to an s or p orbital, for example.

Note that I don't know why the orbitals exist in the exact shapes they do (although I imagine it has something to do with the electrons speeding up or slowing down at certain points making it more likely for us to discover it in a certain spot), or what the threshold is for the amount of energy required to make an electron move into the next type of orbital (d to f, for example).

But, I have one question: why does a neutral beryllium atom with 4 electrons in 2 energy shells/levels not have any electrons in a p orbital?

My reasoning is as follows:

If $n = 2$ where $n$ is the energy shell occupied by beryllium's fourth electron, then $l = 0,1$ and $m_l = \text{either }0 \text{ or }-1,0,1$. If $m_l = -1,0,1$. Then, why isn't this fourth electron in a p orbital?

We need to first consider n, the quantum number for the shell, and then l (~the classical angular momentum of the electron), which tells us the shape of the electron density. The first shell (n=1) contains two s electrons with paired spins. That shell is closed, and you begin over again with n=2: the possibilities for n=2 are l=0 (s electron shape, spherical) and l=1 (p electron shape, lobed). Then, since a lobed shape can point in different directions, when l=1, ml can equal +1, 0 or -1 , defining px, py, pz electron shapes, each pointing in one of three orthogonal directions.

So the filling process puts the first two (lowest energy) electrons in the 1s level (when full, it is 1s2); then the next level is 2s (filling to 2s2), then 2p. Since we have run out of electrons, the 2p levels do not get filled, so we don't have to consider 2px, 2py or 2pz designations.

Now, for boron, there will be a p electron. Whether you call it px, py or pz will depend on how you define the directions in a molecule. In an individual free boron atom, 2px, 2py and 2pz all have the same energy and are degenerate.

This is all theoretical. If you ever heated up beryllium so that there were free atoms, they would be so energetic that quantum levels up to 10 or 100 would be occupied, and there would be beryllium ions floating around too!


Do different vibrational modes correspond to vibrational energy levels? Do different vibrational modes correspond to vibrational energy levels? Or all modes can be found in ground state of the molecule? Or each vibration for each bond has its own vibrational energy levels? I don't understand, please help.

yes and no: all vibrational normal modes have a different energy (with the proviso that some may be accidentally degenerate) and hence different absorption frequency. (Fundamental frequencies have vibrational transitions that occur with selection rule $\Delta n=\pm 1$.) All normal modes exist in their ground state with zero point energy (quantum number $n=0$) as well as in other levels based on the same mode ($n>0$) which are populated according the Boltzmann distribution at temperature T. Thus each basic vibrational mode i has its own stack of energy levels, with quantum numbers $n= 0, 1, 2 \cdot\cdot$ and (harmonic oscillator) energy $E_i=\nu_i(n_i+1/2)$ with quantum numbers $n_i$ and frequency $\nu_i$ in  $\pu {cm^{-1}}$. The total zero point energy of the molecule is the sum of all the individual zero point energies above the zero of energy.

The number of vibrational levels populated at a given temperature depends on the vibrational level spacing, thus low frequency modes, say, $500 \pu{cm^{-1}}$ have far more levels populated than a high frequency, say $1500 \pu{cm^{-1}}$ mode.

As the energy increases the number of vibrational energy levels / unit energy can become very large indeed in polyatomic molecules. There are overtones where only one vibrational mode is excited, but with more than one quantum of energy, and combinations where two or more different vibrations of different frequency are excited and so each have non zero quantum numbers.

In water molecules there are 3 vibrations (of frequency approx $3650,1595, 3756 \pu{cm^{-1}}$) if we label them in order with quantum numbers then [000] is the zero point energy $(3650+1595+3756)/2=4500 \pu{cm^{-1}}$,  [010] a fundamental in the second frequency and [020] an overtone in this mode and [110] a combination of modes 1 and 2 with energy above zero of $4500+3650+1595 \pu{cm^{-1}}$


Which of these reduction reagents would me most appropriate to yield this product? I am very confused, I think using reduction would be a good start, but LiAl would go straight to an OH.

1) As you say*, the ketone must be protected.  Protect it. Any propagating naughtiness is irrelevant to the next step.
2) Reduce the ester. Beat on it.
3) Deprotect the ketone.

*Note that you begin with an alpha-ketoester.  In the old Morrison and Boyd that was a French joke.  LAH would have the same problem.  How the mighty have fallen.


Reaction an ROH-resonance-stabilized Group with Hydroxide/Acid-Base Reactions The book says there is no reaction.

However, I do not understand why.

The ROH group is resonance-stabilized with the benzene-like group. Its conjugate base is stable. There is an inductive effect of the cloud of electrons that make the oxygen more electronegative. However, resonance effects are always more important than inductive effects.

So, why is there no reaction?

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

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


Why is the Standard Entropy of a Proton Zero? It was my assumption that the standard entropy was the absolute entropy of a substance at standard state; however, my textbook states various standard entropies ranging from $-144.77\ \mathrm{J/(mol\ K)}$ for $\ce{Al^3+}$ to $0\ \mathrm{J/(mol\ K)}$ for $\ce{H+}$ to $213\ \mathrm{J/(mol\ K)}$ for $\ce{CO2}$. It is clear that these values are not absolute entropies, as the only substance that can have an absolute entropy of $0$ is a perfect crystal at $0\ \mathrm K$ according to the third law of thermodynamics. Why are standard entropies notated like this instead of just stating their absolute form? My assumption is that this is just a way to make calculations involving entropy easier, since they merely involve changes in standard entropies; however, why is the standard entropy still notated $S^\circ$ and not instead $\Delta S^\circ$ (since it is not absolute)?

Indeed, it is clear that these values are not... Wait! What "these values"?

You are putting two entirely different things in the same basket. (If your textbook does so, then it does a poor job.) It is much like treating dogs and cupboards similarly, on the basis of both having four legs. Then, however, many generalizations will fail quite miserably. You feed them meat, but the results are not always great. You call in the carpenter to fix them, but...

Same thing here. See, ions and molecules are very, very different things. You totally can have a mole of $\ce{CO2}$. It is a thing. You can work with it, can measure its heat capacity and stuff, and eventually work out the entropy. Indeed, the textbook value of $\ce{CO2}$ entropy is absolute.

On the other hand, you totally can't have a mole of $\ce{Al^3+}$. It is not a thing. Unless you have some counterions, that is, but in that case you can't tell how much of the entropy comes from the counterion and how much from $\ce{Al^3+}$.

How can we assign entropies to ions, then? Why, it is as simple as that: we assign one ion arbitrarily, then work out the entropies of all its counterions from there, and then turn to their counterions. Just what one ion could that be? Why, let's say it is $\ce{H+}$, because so much of our chemistry is dominated by the acid/base equilibria. It is kind of natural to consider its entropy $0$, much like the absolute elevation of the sea level is $0$. Now there you have them, the textbook entropies of ions, which are not quite absolute.

So it goes.


Will freezing water still expand width-wise in a pipe (possibly breaking it) if it's free to expand lengthwise? Will freezing water still expand width-wise in a pipe (possibly breaking it) if it's free to expand lengthwise?

It might do.

Imagine you have a U-shaped pipe. The two ends are above ground. The pipe runs into the ground for a short length in the middle, then back out again. It's free at both ends but filled with water.

You have a problem there because in cold temperatures, the two above-ground parts are going to freeze first. Then, if it's really cold, the underground part will freeze also. When that happens, the already-frozen above-ground ice won't move and your underground pipe may well go pop. What was initially "free at both ends" became blocked.


How to determine from the formula that there is a phenyl group present in the molecule? Why is the organic compound $\ce{C6H5-CH=CH-CH2Cl}$ named 3-chloro-1-phenylprop-1-ene and not 9-chloronon-5,7-dien-1,3-yne even though there is no specification whatsoever that there is a phenyl group in the organic molecule?
$$\ce{C6H5-CH=CH-CH2Cl}$$
can also be written as
$$\ce{CH#C-C#C-CH=CH-CH=CH-CH2-Cl}$$
and the name for this organic compound would be 9-chloronon-5,7-dien-1,3-yne.
How do we surely know that there exists a phenyl group in the molecule?
Also why is phenyl treated as a substituent and cannot be considered in the longest chain?

It's a matter of convention.

$\ce{C6H5}$ is conventionally used to represent phenyl because it is a common moiety. Your long, highly unsaturated, chain is also $\ce{C6H5}$, but it wouldn't conventionally be written like that without expansion, unless you're setting out to confuse someone.


How to identify if it is a buffer solution Aqueous solutions of $\ce{HNO3}$, $\ce{KOH}$, $\ce{CH3COOH}$, and $\ce{CH3COONa}$ of identical concentrations are provided. The pair (s) of solutions which form a buffer upon mixing is(are)?
A. $\ce{HNO3}$ and $\ce{CH3COOH}$
B. $\ce{KOH}$ and $\ce{CH3COONa}$
C. $\ce{HNO3}$ and $\ce{CH3COONa}$
D. $\ce{CH3COOH}$ and $\ce{CH3COONa}$

I think that option D is correct as it a solution of a weak base and its a salt; that's why it is a buffer solution. But how do I check the other options?

Well, it all boils down to the preparation of buffer solutions in the first place. One of the methods of preparation includes the mixing of a salt solution of an weak acid and a strong base with a solution of a strong acid, with the solution of the strong acid as a limiting reagent, which leads us to conclude positively for the third choice.

Alternatively, you can think which of the mixtures will resist the change in pH on mixing a strong acid or a strong base; for instance in the third choice, nitric acid being a strong acid will help to resist the change in pH when a strong base is added, and $\ce{CH3COONa}$ being a weak acid + strong base type of salt resists the change in pH that could've been caused due to the addition of a strong acid.


Interpretation of 4-benzoyl butyric acid H-NMR Spectrum I am trying to undertake a first order analysis of the spectrum below:

My issue is that every signal appears to have the same coupling constant (I am aware that the value should be reported in Hz, the spectrum just has some of my rough working on it). This does not make sense to me, as I understand that would imply each hydrogen is equally coupled to each other.  I also am at a loss of what type of multiplicity the signal multiplet at 7.5ppm has.

Any help is really appreciated

You should not assume that just because the coupling constants are the same that this means all the nuclei are coupled. It simply means that two sets of coupled nuclei have the same coupling constants. This is not unexpected when you consider that the nuclei are very similar - they are freely rotating alkyl chain methylene groups. And you will find that freely rotating alkyl chain methylene groups have coupling constants of around 7 Hz. This is because the 3J-coupled nuclei (called vicinal coupling) rotate through a range of conformers that have a range of coupling constant values (described by the Karplus equation) to give an average value. Different types of functional group give characteristic coupling constants. Trans-vinylic coupled protons have a coupling of around 15Hz, for instance. So, given the functionality of these alkyl methylenes, you would expect similar coupling constants.

To work out which nuclei are in fact coupled, you would use other information. A set of homonuclear decoupling experiments, or a COSY experiment would give this information clearly, although there is some very useful information in this spectrum. Your signals at 2ppm, 2.5ppm and 3ppm all show significant 'tenting'; they lean towards their coupled partner. That is, you do not get a true Pascal's triangle intensity, with the peaks closer to their coupled partner having greater intensity, as shown below:

The signals at 2.5 and 3 both lean towards the right, indicating they are coupled to something upfield (the signal at 2ppm). Similarly, the peak at 2ppm is leaning to the left, indicating it is coupled to the peaks at 2.5 and 3ppm. You could use your integrals of 2:2:2 to rationalise a -CH2-CH2-CH2- chain. And this would explain the expected splittings you see of a triplet (3ppm), triplet (2.5ppm) and triplet of triplets (2ppm). Of course, if you draw out your splitting diagram for a triplet of triplets with identical coupling constants, it will give you a pentet with intensity 1:4:6:4:1.

To answer your second question, you should try to predict what nuclei these signals represent. I have drawn out some useful props below to assist in the interpretation.

The signals between 7 and 8ppm come from your mono-substituted benzene ring, so you have three distinct environments: ortho, meta and para to the substituted position. They should integrate 2:2:1. The coupling constants for protons around an aromatic ring aren't important for this exercise, other than to say you will only see splitting due to a proton on an adjacent carbon; that is, 3J coupling. So, you have 5 aromatic nuclei in 3 environments, that integrate 2:2:1, and I can see 2 sets of signals that integrate 2:3. Clearly you have two signals overlapping at 7.5ppm. Which signals?

Let's have a think about the splitting we would expect to see from this ring. The ortho protons would be a doublet, as they are coupled to one other 3J proton only.The meta protons would be a doublet of doublets, as they are coupled to the ortho (doublet) and para (doublet) protons. And the para proton would appear as a triplet (coupled to the two meta protons).

What else can we see? The substituent on the benzene ring is a -(C=O)-R. This is a strongly electron withdrawing group, and so if you draw out your resonance structures for this, you will find that the ortho protons are deshielded compared to the meta protons, and so will be downfield of them. The para protonalso  experiences a bit of shielding, and so will also be downfield of the meta protons. So, you can assign the peak at 8ppm as your ortho proton; furthest downfield fits with your electronics adn doublet fits with your expected splitting. The overlapping peaks at 7.5 are the para proton (the left hand side of the forest); this fits with your electronics and (if you were really keen) you could assign some of these peaks to a triplet. The remaining right hand side of the peaks are your meta signals, and again you could assign this asa doublet of doublets. Of course some of the lines for the para and meta signals are overlapping (actually, possibly just one). You could resolve these signals on a higher field instrument quite easily.