text:
What is the relationship between the molecular weight of a polymer and that of polystyrene?
So, I read in a similar chemistry thread that you can get Mn and Mw of the polymer of the interest by comparing the result of the sample obtained from GPC/SEC with the standard known molecule weight polymer like polystyrene.
I don't understand how this works. You compare what of the polymer with what of the polystyrene to determine the molecular weight?? There has to be a constant factor when determining relative values between two things, and what is it in the case of polystyrene and the polymer of the interest?
What do you mean by polystyrene equivalent value? What is the point of determining the relative molecular weight of a polymer based on the polystyrene equivalent value??
Thank you!

You have to have something to calibrate the column with. Polystyrene is typically used since it can be made cheaply (well, relatively cheaply) with a pretty small molecular weight distribution.
Yes, there will be differences between the true molecular weight of a polymer and what the comparison to polystyrene gives you. But does that really matter? Why do you need to know the absolute molecular weight, when a relative comparison works just fine?
It's similar to how we determine the mass of something. At present, there is that infamous block of Pt/Ir in Paris that is the standard kilogram. All mass measurements are made through a series of comparison to that mass.

The problem is that you have no idea what the difference is. We have a rather good idea how many atoms of Pt and Ir are in that block in Paris.

In size exclusion chromatography (SEC), the molecules elute in order of decreasing molecular weight (actually solvodynamic radius - we'll get back to that). The larger molecules (or polymer chains) do not fit into the pores in the stationary phase and thus spend more time in the mobile phase and elute faster. The smaller molecules can get stuck in the pores and thus elute slower.
Now, one of the most common detectors is a differential refractometer, which measures the concentration of analyte in the eluent since the refractive index $\eta$ of a dilute solution depends only on the molal concentration of the solution ($m$), the refractive index of the pure solvent $\eta_0$, and a constant $k_\eta$
$$\eta=k_\eta m +\eta_0$$
This relationship helps us figure out the amount of polymer that is eluting at any given time. However, the system needs to be calibrated so that we can related elution time to specific molecular weights. That's where polystyrene standards come in. Polystyrene is chosen because it can be prepared in low-polydispersity samples of known molecular weight by carefully controlling the polymerization.
We use "polystyrene-equivalent" value or "relative to polystyrene standards" because we know the molecular weights of the polystyrene standards.


text:
What material is an Inkjet transparent printing film (slide) made of?
I don't mean photographic film/film base but rather the A4/Letter sheets of transparent film that were used in slide projectors during lectures. For example see this.

It depends. There are several variants, depending on printer and cleaning solvent to use. However, http://en.wikipedia.org/wiki/Transparency_(projection) states, A transparency, also known in industrial settings as a "viewfoil" or "foil", is a thin sheet of transparent flexible material, typically cellulose acetate, onto which figures can be drawn.

I would have thought polyester was more common as it is for laserjets.

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text:
What's wool made of?
For example, silk is made of fibroin. But what's wool made of?

Unlike silk or cotton, wool does not have a homogeneous structure. Its core is considered to be made of keratin, a generic type of protein found in human hair and epidermis. A detailed PDF on Merino wool can be downloaded from here.
Thanks for encouraging me, through your question, to read that PDF, which explains both the chemical and physical properties in a straight-forward fashion.

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text:
Hydrogen can be stored?
Please tell about the hydrogen gas storage and the pressure created in tank. what is the calorific value of hydrogen gas when it is burnt?

I'll say wikipedia is pretty clear when it gets to explaining Hydrogen storage.

what is the calorific value of hydrogen gas when it is burnt?

286 kilojoules of energy is released per mole of hydrogen burned.

Please tell about the hydrogen gas storage and the pressure created in tank.

In a tank at room temperature, hydrogen is a supercritical fluid, because the critical point is 33K.
In a tank, a long as the tank is strong enough, the pressure could be any value up to at least 10,000 psi, but at high enough pressure it would solidify.
For more information see PROPERTIES OF HYDROGEN (no paywall), especially page III-E for P-V-T information.


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What are the biochemical ramifications of ingesting of distilled water?
Are there any distinct biochemical reactions that are affected by the ingestion of distilled water?
The water is "distilled" in its literal sense.

This was already asked/answered on the biology stackexchange: http://biology.stackexchange.com/questions/2250

http://www.who.int/water_sanitation_health/dwq/nutrientschap12.pdf

@J...  Didn't see this when I posted my answer.  If you'd like to develop an answer around it, I'll delete mine.

@ron no, I think you've covered it - fair wages for fair work.

@BlueRaja-DannyPflughoeft I vote to keep this open.  The answer on the Biology Stack exchange is deeply flawed at best.  It discusses drinking only distilled water and consuming no other substances (you'd starve before you died from being hypotonic...)  There's a massive difference between a water source being "not ideal" and "fatal."

As @DavePhD has noted, drinking small amounts of distilled water does not pose any significant adverse health consequences.  However, the World Health Organization (WHO) has studied the effects of long-term drinking of distilled water and issued a report.  They investigated the following five areas of potential adverse human health impact:

Direct effects on the intestinal mucous membrane, metabolism and mineral homeostasis or other body functions.
Little or no intake of calcium and magnesium from low-mineral water.
Low intake of other essential elements and microelements.
Loss of calcium, magnesium and other essential elements in prepared food.
Possible increased dietary intake of toxic metals.

In their conclusion, they state,

"Drinking water should contain minimum levels of certain essential
  minerals (and other components such as carbonates). Unfortunately,
  over the two past decades, little research attention has been given to
  the beneficial or protective effects of drinking water substances."

and

"Demineralised water that has not been remineralized, or low-mineral
  content water – in the light of the absence or substantial lack of
  essential minerals in it – is not considered ideal drinking water."

(emphasis mine)


text:
What is the Ka of OH- and Kb of H3O+?
What is the $K_\mathrm{a}$ of $\ce{OH^-}$ and $K_\mathrm{b}$ of $\ce{H_3O^+}$? Have these constants been determined?

That's a good question!

Is what that this page says incorrect or incomplete?

Oh, wait. It's incomplete.

He's asking about protonation of hydronium and deprotonation of hydroxide - you won't find these in basic lists.

Sadly. :'(. Let's start a probe.

For $\ce{O^2-}$, the $\mathrm{p}K_\mathrm{b}$ is approximately $-22$, so the $\mathrm{p}K_\mathrm{a}$ of hydroxide is about $36$.
sources: http://www.newworldencyclopedia.org/entry/Oxide and https://www.chem.tamu.edu/rgroup/hughbanks/courses/462/lecturenotes/class6-2.pdf
For hydronium, the reference Fred Senese cited
Protonated hydronium dication...
suggests that at acidities of Hammet acidity function $H_o = -25$ to $-28$, protonated hydronium can be present.

Just stepped in to say "thanks", though this comment is "too chatty"!

I must admit that I expected some more extreme values.

some sources give even less extreme value for hydroxide pKa http://m.docente.unife.it/alberto.cavazzini/analitica-1-e-lab-2014/equilibri-acido-base/tabella-costanti-acido-base

In water both equilibrium constants will be extremely small; the concentrations of $\ce{O^2-}$ or $\ce{H4O^2+}$ ions in water are essentially zero.
According to Cotton and Wilkinson's Advanced Inorganic Chemistry, for the hydrolysis of solid oxide ions you have
$$\ce{O^2- (s) + H2O (l) -> 2 OH- (aq)} \qquad \qquad K_b > 10^{22}$$
so you can never have any significant concentration of oxide ions in aqueous solution. As DavePhD points out, $K_\mathrm{a}$ for the hydroxide ion is less than $K_\mathrm{w}/K_\mathrm{b} = 10^{-36}$.
Now, $\ce{H3O+}$ may have some capacity to act as a base in media like $\ce{HF:SbF5}$ (see this paper on the role of $\ce{H4O^2+}$ in isotopic exchange reactions between hydronium ions). $\ce{H4O^2+}$ can exist in sulfolane solution, too (see this paper).
If $\ce{H4O^2+}$ exists in water at all, it's probably going to actually be two protons bridged and hydrated by a lot of waters, rather than an actual $\ce{H4O^2+}$ ion. It will have a much shorter lifetime than even a hydronium ion. The molecule can exist in theory, though, and its electronic structure has been studied for at least half a century (see Rosenfeld's 1964 paper including SCF calculations on $\ce{H4O^2+}$).

I think the ~picosecond lifetime of hydronium in pure water refers to lifetime in (OH-, H3O+) ion pairs, and the ion pairs don't count toward the 10^-7 M hydronium.  On the occassion that the ion pair separates, the lifetime of hydronium is then 70 microseconds, and only these hydronium that are not part of the ion pairs are the 10^-7 M hydronium.  See page 16 here: http://www.waterjournal.org/uploads/vol1/chaplin/WATER-Vol1-Chaplin.pdf

Ahhh, thanks Dave! I'll come back and fix that when I have time.

Just be careful that all the water doesn't convert to polywater. ;-) All of you scalawags are probably too young to remember the media hooblah  about that one. There were stories that if polywater were dropped into the oceans that it could catalyze the solidification of all the oceans.


text:
Concentration uncertainty due to volume measurement with cylinder
(All volumes were measured using a measuring cylinder which has a $\pm\pu{0.5 cm3}$ uncertainty)

You have a $\pu{50 cm3}$ of $\pu{0.5 M}$ solution to begin with
You then make a  $\pu{0.4 M}$  solution by mixing $\text{40 cm}^3$ of the above solution with $\pu{10 cm3}$ water.

If I'm not mistaken, the uncertainty for the volume should be $\pm\pu{1.0 cm3}$ because the second solution is the result of two measurements (and also uncertainties).
My question is: what is the uncertainty for the concentration?

It depends whether the 0.5 uncertainty represent a property of the cylinder (systematic error) or if the cylinder is perfect and 0.5 represent random error in reading the cylinder. In really life it would be some of each.

Thank you for the reply. In this case, I'm talking about the uncertainty as a property of the cylinder. How would you work out the random error in reading?

1. Let's assume that you're talking about random error. If you specify the error in the volume with $\pm$, this is probably the case---systematic error in this case would have a single sign; for example, if the scale marks on the side of cylinder were shifted by $\rm 0.5\ cm^3$ down from what their true values should have been, you'd have a systematic error of $-0.5\ \rm cm^3$ (the negative sign tells you that your readings are $0.5\ \rm cm^3$ lower than they should have been).
You can say that the error in a sum will be less than or equal to the sum of errors. It's better to say that the uncertainty in a sum is the square root of the sum of squared uncertainties:
$$\delta(V_1+V_2) = \sqrt{\delta(V_1)^2 + \delta(V_2)^2} = \sqrt{\rm(0.5\ cm^3)^2 + (0.5\ cm^3)^2} = 0.70_7\ \rm cm^3$$
For a product or quotient, you can say that the relative error will be less than or equal to the sum of the relative errors in the measurements. It's better to say that the relative error in a product or quotient is the square root of the sum of squared relative errors:
$$\begin{array}{rcl}
\delta(C_{dilute}) &=& \delta\left(\frac{C_{stock}V_1}{V_1+V_2}\right)\\
\frac{\delta(C_{dilute})}{C_{dilute}} &=& \sqrt{\left(\frac{\delta(C_{stock})}{C_{stock}}\right)^2 + \left(\frac{\delta(V_{1})}{V_{1}}\right)^2 + \left(\frac{\delta(V_{1}+V_2)}{V_{1}+V_2}\right)^2
}
\end{array}$$
...can you take it from there?
2. Let's assume you have systematic error, that is, you actually meant something like "The cylinder always reads $\rm 0.5\ cm^3$ too high". You will need a definite sign on a systematic error, not a $\pm$.
There are several ways to proceed here. Let's take the easiest route, which is to estimate the systematic error in the concentration as the difference between the concentration and its perturbed value (the value with all of the volumes having the systematic error added in). If your systematic error in volume was $\delta V = +0.5\ \rm cm^3$, you can estimate the systematic error in the concentration as
$$\begin{array}{rcl}
 \delta{C_{dilute}} &\approx& C_{dilute}^{perturbed} - C_{dilute} \\
&\approx& \frac{C_{stock}(V_1+\delta(V)}{(V_1+\delta(V))(V_2+\delta(V))} - C_{dilute} \end{array}$$
The sign on this error will be important; it tells you whether the systematic error in volume makes the concentration higher or lower than its "true" value.
Note: The above calculation is just an estimate of the error. You can do a better job using derivatives here. I'll show you that if you want to see it.

Thank you very much for the reply. So for the volume uncertainty, we cannot simply add up two ±0.5's? I added them up because I thought that ±1.0 would be the extreme minimum and maximum boundaries. As for the concentration, I have no idea at all before asking. Is the equation Cstock(V1) = Cdilute(V1+V2) derived from nstock = ndilute?

Some people do just add them up to get a MAXIMUM error; you can say $\delta(V_1+V_2) \le \delta(V_1) + \delta(V_2)$ so $\delta(V_1+V_2) \le 1.0\ \rm cm^3$ in this case. Notice the inequality; the way I do it in the answer above gives a better error estimate. And yes, that's where that equation came from.

My value of ±0.5 cm^3 came from the fact that 1 cm^3 was the smallest division of the measuring cylinder. Thus I believe the error is half of the smallest division for analog instruments. I am not entirely sure which method to use but am more leaning towards method 2 of systematic errors. In this case, should the final error be written with signs or ±?

Actually, no, it's random error then because you're estimating the error from the measuring scale; it would be systematic error if the scale on the side of the cylinder was stamped incorrectly by $0.5\ \rm cm^3$. Usually people say you can read a scale to 1/10th of the smallest scale division, so $\pm 0.5\ \rm cm^3$ is pretty conservative.

I see, it seems my understanding on errors was quite off. Thank you for explaining in detail this far, I'll definitely upvote your answer when I can (currently I only have 14 reputation). Sorry for the hassle but I have two more questions. 1) Can you show me the maths behind the random error? 2) What are the values for V1 and V2 in this example?

Ah never mind about the first question. I think I found the method here: link

Yes. That's exactly what I was showing you above. V1 and V2 are 40 mL and 10 mL.

I would use 0.5 $cm^3$ as error on the total volume if you fill the cylinder to 50 mL level. If you add two samples as the problem suggest to do, you should not use the simple formula but compute the derivatives w.r.t. $C_{stock}$, V1 and V2.

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text:
Why does lithium move on the surface of water during the reaction?
If someone asks me to number the most notable "textbook experiments" I would definitely name "the comparison of the reactions of alkali metals with water" among the most favorites.
We just happened to do such a reaction with small pieces (approximately $\text{7.5 g}$) of Li, Na and K. As we were supposed to write down anything that we observe, I came across an interesting phenomenon.
It seems that the answer is so simple, I didn't get it through many resources.

Li piece is moving rapidly (better call it spinning) on the surface of water as it is reacting with it and a subtle explosion sound is heard.

Why does the motion of Li on the surface of the water in the beaker take place? Could the heat released by the reaction be causing current in the water (I don't believe so, as the possibly hot water isn't involved in any convection currents)? Info you might need:
Our chemistry teacher cut a non-oxidized piece of well-preserved lithium. The piece of the reactant wasn't of any distinguishable geometrical shape. After repeating the experiment for several times, I sensed a headache that as I believe is because of release of the impurities of the sample.

Hydrogen evolves, but why does it evolve in such an uneven fashion as to cause motion, and not just jiggling back and forth..

Could there also be some steam being generated from the heat of the reaction (at the lithium surface)? Same effect as the $\ce{H2}$ gas.

I believe surface tension of water might play a role, too.

7.5 grams seems a lot!

What happens when you put lithium (or sodium or potassium) in water? What is the chemical reaction that is occurring?
The alkali metals react with water to produce alkali hydroxides and hydrogen gas:
$$\ce{2Li(s) + 2H2O(l) -> 2LiOH(aq) + H2(g)}$$
The two products of the reaction of water and lithium are lithium hydroxide (which dissolves) and hydrogen. Hydrogen is not very soluble in water, and it is much less dense than air. The motion of the lithium on the surface of the water is created by the escape of the hydrogen gas, which provides upward force.

This is a very speculative answer to your comment on spinning. The spinning presumably occurs because the under surface of the lithium is not flat and also the metal itself is not a perfect cuboid so will not be levitated evenly by the gas produced but will be tilted w.r.t. the surface. There will then be slight force to the side causing it to tilt, if it then momentarily touches the surface then it will start to pivot about this point and so spin as it is levitated again. The angular momentum once started may cause it to continue to do so in the same direction even if it momentarily touches the surface again. As there is hardly any sideways force, as the gas is underneath the lithium pushing it up and flowing round the sides of it as it escapes, sideways motion seems less likely than spinning.
An additional factor is that if it translated across the solution the gas underneath would be less effective to levitate it as the lithium is always moving over water where no reaction has occurred. It would then touch the water and so stop moving until gas built up again. Thus rotation seems to be the lowest energy option.


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What is the oxidation state of nitrogen in nitrate ion and what is its structure?
What's the oxidation state of nitrogen in $\ce{NO3-}$ ion. Simple calculation depicts that it's $+5$. But second period elements never show variable oxidation states. And it's structure is so confusing to me. Some people show dative/co-ordinate bonds, whereas others show a positive charge on nitrogen. Which should be considered as correct one?

Also related http://chemistry.stackexchange.com/q/21856/4945

You are correct that the oxidation state is +5. The second period elements, particularly carbon, nitrogen and oxygen, do show variable oxidation states regularly.
The $\ce{NO3-}$ ion cannot be depicted using a single Lewis structure. However, its resonance structures are shown below:

What this means is that the actual structure of the nitrate ion is somewhere in between the three structures shown below. It is sometimes also depicted like this to show the symmetrical nature of the ion and the distribution of charges across it.

http://en.wikipedia.org/wiki/List_of_oxidation_states_of_the_elements
http://en.wikipedia.org/wiki/Nitrate#Structure

but if Nitrogen can show +5 oxidation state here, then why it cann't form NCl5 (whereas PCl5 is possible). As per my knowledge is concerned, 2nd period elements cann't go for octet expansion due to the abscence of vaccant d-orbital... please enlighten me..

in nitrate the nitrogen uses its p orbitals to bond with the oxygens. however, in a hypthetical $\ce{NCl5}$ molecule the orbital overlap between the chlorine and nitrogen orbitals would be poor due to energy mismatching and steric hindrance between the chlorines and so this molecule is not stable

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text:
Nomenclature of inorganic compounds
As depicted in books, $\ce{NO3-}$ and $\ce{NO2-}$ are called as nitrate and nitrite respectively. Where (-ate) suffix is used for higher oxidation state and (-ite) suffix is used for lower oxidation state (just like:- sulfate and sulfite).
Now since Nitrogen is 2nd period element, hence it can't expand it's octet (no vacant d-orbital). So I didn't get how (-ite or -ate) suffix can be used if nitrogen shows only 1 stable oxidation state i.e; (+3). Can anyone suggest?

Nitrogen can have a variety of oxidation states ranging from -3 to +5. For example:

-3

The nitrogen atom in ammonia $\ce{NH3}$ has an oxidation state of -3 (each hydrogen is +1).

-2

The nitrogen atoms in hydrazine $\ce{N2H4}$ has an oxidation state of -2.

0

The oxidation state of the nitrogen atoms in $\ce{N2}$ is 0.

Nitrate and nitrite

The nitrogen atom in the nitrate ion $\ce{NO3-}$ has an oxidation state of +5. Each oxygen atom has an oxidation state of -2, and the overall charge on the ion is -1.
The nitrogen atom in nitrite ion $\ce{NO2-}$ has an oxidation state of +3.
Nitrogen can have such a range of oxidation states because of its electron configuration, which is $\ce{[He]}2s^2 2p^3$ (5 valence electrons). Nitrogen can "gain" up to three electrons or "lose" up to five electrons.


text:
Why do internal combustion engines only burn to carbon monoxide, not dioxide?
Why don't diesel and regular car engines burn all the way to $\ce{CO2}$?

Most of the carbon-oxygen stuff coming out is CO2.  Some isn't, because of detailed balance of the chemistry, inhomogeneous conditions in the piston, etc.

Seems like another question easily answered with very minor googling.

I think this is sort of covered by What really happens atomically in an explosion? because the burning of the fuel air mixture is basically an explosion. The point is that there isn't a nice simple answer because the reaction is fearsomely complex.
Actually most of the fuel does burn to carbon dioxide. The proportion of carbon monoxide is quite small. In a normally tuned engine there is enough oxygen to burn all the fuel to carbon dioxide but the fuel/air mixture isn't in the cylinder long enough for the reaction to go to completion. You probably know that all modern cars are fitted with catalytic convertors, and what they do is to complete the reaction as the exhaust gases pass through them.

But why not run with excess air to make sure it is all burned? This is especially true with Diesel engines which pump out pure carbon

@DirkBruere: Getting the maximum engine efficiency, i.e. the combination of good mpg with high power, is something of a black art as lots and lots of parameters are involved. An obvious parameter is the fuel/air mixture, and it turns out that the optimal fuel/air ratio results in some carbon monoxide being left in the exhaust gases.

Well, yes. That was my question, sort of. If we get more energy by burning to CO2 why do we deliberately stop short of that? What are we optimizing?

@DirkBruere: Generally, if you run lean (high air/fuel) you decrease power and the engine runs hotter so it generates more oxides of nitrogen. I think the mpg are slightly better with a leaner mixture, but the power falls. Exactly why the currently used ratio is considered optimal I don't know. I'm sure a Google would find articles on the subject.

This is why I posted it in physics - it's not really a chemistry question. Those who voted to move it did not seem to understand.


text:
Why does the inverse relationship between the strengths of an acid and its conjugate base appear to be violated in the case of chloric(I) acid?
Hypochlorous acid is a weak acid with $\mathrm{p}K_\mathrm{a}(\ce{HClO}) = 7.53$. Then why is its conjugate base, $\ce{ClO-}$, a weak base in water? Shouldn't the strength of the base be inversely proportional and shouldn't hypochlorite be a strong base?
If $\ce{HClO}$ is a weak acid, that means it does not readily give up a proton and has a strong pull on them. So when it becomes a conjugate base, $\ce{ClO-}$, shouldn't it readily take protons and therefore be a strong base in water?
Also, when an acid is dissolved in water, and some of that acid dissociates into $\ce{H3O+}$, is the $K_\mathrm{w}$ of water negligible and you only consider the $K_\mathrm{a}$ when finding $\mathrm{pH}?$ But that confuses me because I thought you use 14, the exponent of $K_\mathrm{w},$ when finding $\mathrm{pH}.$

Hypochlorous acid is stronger acid than water so chlorate is weaker base than hydroxide.

Weak acid compared to HCl and many inorganic acids, but not as weak as e.g. acetic acid.

@Greg pKa of HClO is 7.5.  pKa of acetic acid is 4.75.  http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf

@DavePhD is H3O+ a strong conjugate acid?

$K_\mathrm a\times K_\mathrm b = K_\mathrm w$.  Some teachers tell their students that the conjugate base of a weak acid is strong but it's not true.  What is true is that the stronger the acid, the weaker the conjugate base and vice versa.
Yes to your second question.  When calculating the pH of a solution, the hydronium ion concentration is usually controlled by the strongest acid in the solution.  This is the case in the type of problems found in lower level chem classes.  In a higher level class problems with acids close enough in strength so that both acids matter may be encountered.

It's true that as the strength of an acid goes up, the strength of its conjugate base goes down. For example, hydrochloric acid ($\mathrm{p}K_\mathrm{a} \approx -7$) is really strong and chloride anion is a really weak base. On the other side of the spectrum, an alkane ($\mathrm{p}K_\mathrm{a} \approx 50$) is a very weak acid, but its conjugate base will deprotonate nearly anything. So at the extremes it's clear: strong acid gives weak conjugate base and weak acid gives strong conjugate base.
For compounds with $\mathrm{p}K_\mathrm{a}$'s between 0 and 14, such as hypochlorous acid with an intermediate $\mathrm{p}K_\mathrm{a}$, both the acid and the conjugate base are weak. We call them "weak" acids because they don't completely dissociate in water. However, their conjugate bases are also "weak" because they're weaker than hydroxide.
It comes down to just how weak do you mean by "weak".

Ammonia also is weak base and gives weak conjugate acid.So will this follow the same trend as you said for acids

A weak acid $\ce{AH}$ is in equilibrium with its conjugate base $\ce{A-}$ when dissolved in water:
$$\ce{AH(aq) <=> A-(aq) + H+(aq)}$$
In contrast, a strong acid such as $\ce{HCl}$ dissociates fully, i.e. the following reaction goes to completion:
$$\ce{HCl(aq) -> Cl-(aq) + H+(aq)}$$
Likewise, a strong base such as $\ce{NaOCH3}$ accepts a proton from water, i.e. the following reaction goes to completion:
$$\ce{NaOCH3(aq) + H+(aq) -> CH3OH(aq) + Na+(aq)}$$
Neither $\ce{Cl-(aq)}$ nor $\ce{CH3OH(aq)}$ are typically thought of as acid or bases in aqueous solution. They are spectators as far as aqueous acid/base chemistry is concerned.
So the concept of conjugate acid and conjugate base applies only to weak acids and bases. Saying that a conjugate base is strong would mean that its conjugate acid is not an acid at all but rather a spectator species.
Said in a positive way, the conjugate base of a weak acid is also weak because the reaction goes to equilibrium, not to completion (no matter on which side you start). What you might incorrectly call "the conjugate base of a strong acid" is really a spectator ion (or molecule).

There is a widespread inconsistency in how the terms "weak" and "strong" are used for acids and bases, something I posted about here:
Is there a terminology contradiction about whether the conjugate of a strong acid is a "weak base"?
All sources agree that the strongER the acid, the weakER the conjugate base.  However, some sources say that the conjugate of a strong acid is a (very) weak base.  Other sources say that the conjugate of a strong acid is too weak to be considered a base, therefore when they say "the conjugate of a weak acid is a weak base", they mean that the conjugate of a strong acid is not a base at all.
Google turns up dozens of sources which say "the conjugate of a strong base is a weak acid" (or something similar), and also dozens of other sources which say "the conjugate of a weak acid is a weak base" (or something similar).  Taken at face value, these statements contradict each other -- if there is at least one strong base whose conjugate is a weak acid, then it cannot be true that the conjugate of a weak acid is always a weak base.
Nonetheless, despite this inconsistency, most sources agree that a weak acid and a weak base can be conjugates of each other, e.g. the weak base $\ce{NH3}$ and the weak acid $\ce{NH4+}$.


text:
proton NMR deshielding
The protons alpha to the oxygen in methyl propionate are more shielded than their corresponding protons in ethyl propionate.

.

We can see from the NMR data that this is the case as well.
My TA says this is due to the inductively withdrawing effect of the extra carbon in ethyl propionate.
However, this strikes me as blatantly wrong. The carbon should be inductively donating ... I don't see how it pulls electron density away from the specified hydrogens.

In your example we are comparing the electron releasing/withdrawing properties of a hydrogen substituent and a methyl substituent, and the effect this has on proton ($\ce{H_{a}}$) chemical shifts.

Carbon (2.55) is more electronegative than hydrogen (2.2).  Therefore, the methyl substituent will be more electron withdrawing compared to the hydrogen substituent.  This means that the methyl group will remove electron density from the carbon to which it is attached, compared to the hydrogen substituent.  The fact that this carbon has lower electron density in the methyl case, is further transmitted to the protons ($\ce{H_{a}}$) attached to this carbon. Consequently the $\ce{H_{a}}$ protons will also experience lower electron density in the R=methyl case.
Lower electron density at an nmr-active nucleus results in deshielding (downfield shift). Consequently, the $\ce{H_{a}}$ protons are deshielded in the ethyl ester (R=methyl) relative to the methyl ester (R=hydrogen), in agreement with the spectral data you've provided.

Are you saying that the electron withdrawing effects of carbon act through another carbon atom (as is apparently the case in the ethyl ester)?

Yes, we're comparing the effect of hydrogen vs. methyl through a carbon atom and on thehydrogens attached to that carbon atom.

After all, the standard shifts for methyl, methanediyl and methanetriyl groups are 0.9, 1.3, and 1.7 respectively.


text:
What is the Rate of Reaction of Acetaldehyde Dehydrogenase from Saccharomyces Cerevisiae?
I am working on a biochemistry project, and I need to know some information about reaction rates. What is the Rate of Reaction of Acetaldehyde Dehydrogenase from Saccharomyces Cerevisiae?

See Kinetics and Reaction Mechanism of Potassium-Activated Aldehyde Dehydrogenase from Saccharomyces cerevisiae Biochem. J. (1978) 173, 787-798
The kinetics depend upon the particular substrates and cofactors such as the particular aldehyde, NAD vs NADP, the cation activator (K+ or something else). A thiol is also required.


text:
Would solar panels produce energy if uniform temperature in the universe?
If the temperature on the surface of the sun and on the surface on the earth were the same, and the temperature of the universe were uniform, would the solar panels produce energy?

If the temperature of the universe was uniform, solar panels would be the least of your concerns. There would be no life, no thermodynamic processes, nothing. Complete equilibrium and completely boring!

Who would've created those panels is beyond me. Anyhow, welcome to chemistry.SE! If you had any questions about the policies of our community, you can visit the help center or take a tour of the website.

It seems like a question for Physics SE

I'm voting to close this question as off-topic because it is not about chemistry.

Solar panels work on the principle of photovoltaic effect. The only criteria for a voltage difference to be affected is that the light be incident on the solar panel. The energy of the incident light needs to be around that of the band-gap.
According to the question if both sun and the solar panel were of the same temperature, a solar panel on which light rays are incident would indeed generate a voltage. The band gap on the other hand depends on the temperature of the panel. Higher the temperature, smaller the band gap. Therefore, if the light incident on the panel is energetically close enough to the band gap (which depends on the temperature) the panel will generate a voltage difference.


text:
How to cheaply extract iodine from an alcoholic solution?
I have an alcoholic solution of iodine. My goal is to extract the pure iodine without using rare or expensive compounds. How could I do this?

Welcome to chemistry.SE! If you had any questions about the policies of our community, you can visit the help center or take a tour of the website.

Is the solvent alcohol, literally? If so, is it 96% ethanol? Does HCl count as expensive for you? It would be nice of you to provide more info.

Yes it is 96% alcohol, and I do have HCl

Alcoholic solution of iodine

What firstly comes to mind is that you meant "iodine tincture".
This question must have been frequently asked, since answers are available out there with different degrees of reliability. You must first follow the safety instructions:

Do your experiment somewhere that the noxious fumes won't be able to do harm to anyone. It's best to work under a fume hood.
You're gonna be dealing with strong acids. They're very corrosive when in contact with skin, metals or anything you can imagine (except glass).
Iodine tends to sublime. If not contained wisely, you'll get irritated skin, along with damages to your eyes (See about iodine hazards here).

In short, what you should do is:

Pour the tincture in a beaker, or any container. The important thing to note is to have "control" over what's happening in there.
Add water to the solution. The volume of the water you add should preferably be equal to that of tincture.
Add $\ce{HCl}$ to the solution, $\approx \pu{5-7 ml}$.
Add bleach to the solution, preferably equal or more than that amount of $\ce{HCl}$ you added.
Wait...wait...wait. For a relative amount of time, you should leave the solution for the reactions to happen.
You've past the hard part! Congratulations! Now, filter the dilute solution with a coffee filter.

Additional tips:

Iodine crystals sublime, so you shouldn't leave them in open air for long.
You should at least be wearing appropriate eye and skin protection. If you don't want them, do not do the experiment!

Wikihow has additional tips in case of any failures.
Sources:
http://www.wikihow.com/Extract-Iodine
http://www.sci-spot.com/Chemistry/iodine.htm

...Or do exactly the same as above, but replace the 5-7ml of 'bleach' (sodium hypochlorite in an aqueous solution with stabilizers and possibly phosphates 'whitening' effect)  with 3-5ml 3% H2O2. Both methods produce free chlorine gas, which removes the sodium or potassium from the iodine.

Your best bet is to evaporate the alcohol from the solution to get Iodine. Iodine boils at $184.3\,^{\circ}{\rm C}$. But the vapor pressure of Iodine causes it to melt anyway. Note: Iodine does actually melt. To resolve this, take a watch glass and put it onto whatever you are using to heat the tincture. This will make the alcohol drip down the center of the watch glass as the iodine deposits/solidifies on the sides. To vent pressure, I recommend you to use a filtering flask and let some of the liquid condense in another flask. You can then re-evaporate. the liquid to get more Iodine. It should be impossible to to get 100% of the Iodine in the solution out, but is is pretty close.

Since iodine melts, the appropriate description of that temperature is a boiling point, not a sublimation point. There is also a bit of an issue with this, namely the fact that solid and liquid iodine have a significantly large vapour pressure. Simply heating a mixture will cause all the iodine to evaporate along with the solvents. http://chemistry.stackexchange.com/a/37416/16683

He was recommending multiple distillations.  Using a fractionating system might work or even vacuum distillation possibly (not sure which way the vapour pressures will swing).


text:
NO coordination linear vs bent - valence electron
I'd like to count the total valence electron of following neutral complex in the ionic counting (=donor-pair):

Electron count:
$$\begin{align} \ce{Cp-} : \ce{6e-} \\ \ce{CH3-}: \ce{2e-} \\ \ce{PPh3} : \ce{2e-} \end{align}$$
which gives $\ce{10e-}$ from these three ligands.

1. Question: How to decide the charge of NO
a) if I set $\ce{NO+}$ the charge of Re must be +, therefore Re(I): d6
b) if I set $\ce{NO-}$ the charge of Re must be 2+, therefore Re(III): d4
Probably only a) gives tetragonal geometry whereas b) gives square-planar geometry, correct?
2. Question: How many electrons for NO (linear/bent)
a) I expect linear NO is in ionic counting way a $\ce{2e-}$ ligand (in covalent counting (=neutral ligand method) a $\ce{3e-}$ donor).
b) I expect bent NO is in ionic counting way a $\ce{2e-}$ ligand (in covalent counting (=neutral ligand method) a 1e donor).
3. Question: Total valence electron number
I remember: When we go from linear to bent the VE number decreases with -2, therefore I expect:
a) linear goes along with $\ce{NO+}$, therefore: 10 + 6 + 2 = 18 VE
b) bent goes along with $\ce{NO-}$, therefore: 10 + 4 + 2 = 16 VE

Especially with 1. Question  I feel pretty uncomfortable.

square planar is for d8 only (this is mostly Co-group +1 and Ni-group +2, with virtually no examples outside these two columns) or ligands like porphyrins  2)this is somewhat simplified, but linear NO is NO+ (with extra electron given to central atom) and 2 electrons, but stable only with strong electron-donating systems, while bent NO is NO- and also 2 electrons. 3) without NO we have 7(Re(0)) + 5 (Cp) + 1 (CH3) + 2 (PPh3) = 15, 13 from NO. This is consistent with linear NO.

what do you mean with "13 from NO." I expect the difference in VE number between bent and linear is only 1 (17 or 18 VE).

it's a typo, I meant 3 (three) And no, the difference is 2 (if you assume NO to be neutral) - either 3 (linear) or 1 (bent). Another common view is that NO is either NO+ (isoelectronic to linear CO or CN-) or bent NO- (and behaving like amine nitrogen).

square planar is in normal cases only for d8. There are examples for Cu(+II) d9 in [Cu(NH3)4]2+ (reason: weak field) is also square planare. a) Would the case case for a very weak field be true for d10? I expect not. b) So d7 never gives square planar?

Cu2+ is usually octahedral in water solutions. The two remaining ligands are water molecules and pretty easily exchange, but still. $d7$ usually are, to my knowledge, mostly unstable. Actually, I recalled 1 $d8$ outside Co/Ni group: Au3+ in $ce{[AuCl4]^{-}}$ - like complexes, it is square planar.

@permeakra Would you be willing to format your comments into an answer?


text:
Is it possible to form a divalent transition metal ion that has neither s- nor p-shell valence electrons?
Can a transition metal ion with a charge of 2+ form without valence shell s and p electrons?

Do you mean the ion has 0 s and 0 p electrons? Or do you mean the initial atom has no s or p electrons?

Your question is unclear. Please be more specific.

A transition element in its ground state has a configuration of $ (n-1)d^{1-10} ns^{1-2} $
For the atom to have noble gas configuration (which is what I assume you mean) all electrons of the outer orbitals have to be stripped. An oxidation state of +2 implies that only two electrons have been stripped. The lowest number of valence electrons are possessed by the third group elements as far as transition metals are concerned. Their ground state configuration is $ (n-1)d^{1} ns^{2} $ ; removing two electrons gives $ (n-1)d^{1} ns^{0} $ .
Therefore the answer to your question is no, as there has to be at least one electron in the d orbital of a transition metal in +2 state.


text:
What's the name of the problem of finding equilibrium concentrations?
Is there any specific name of the problem of finding equilibrium concentrations with known equilibrium constants and initial concentrations?
Would it be appropriate to call it "direct equilibrium problem", by analogy with the similar problem in chemical kinetics--finding concentrations as function of time for known rate constants--that is called "direct kinetic problem"?

I don't think the phrase "direct equilibrium problem" is used in this context.
"Multiple-equilibrium problem" would be appropriate.
As in the phrase "It involves solution of the multiple equilibrium problem" J. Molecular Biology vol. 247 page 774.
Or "chemical equilibrium problem" if you don't want to imply that there is more than one equilibrium.


text:
Is the term "carboreduction" interchangeable with "carbothermic reduction"?
I am currently studying reactions involving the reduction of metal oxides by carbon at high temperature to produce refractory carbides (such as $\ce{ZrC}$) following equations such as:
$$
\ce{ZrO2 + 3C -> ZrC + 2CO (g)}
$$
I have come across both the following terms for this type of a reaction carbothermic reduction and carboreduction; they appear to be used completely interchangeably. I would like to make sure (as I have not yet found a source to say otherwise) that this is the case in chemistry terminology and not that one may be applied to an equation as above and the other to a similar reaction where the product is actually the pure reduced metal, such as that for iron ore smelting:
$$
\ce{2Fe2O3 + 3C -> 4Fe + 3CO2}
$$

The term "carboreduction" has an IUPAC definition:

Process in which a metal oxide is reduced in the presence of carbon or a carbon-containing compound.

So technically the term does not require heating and isn't limited to elemental carbon.
Carbothermic definitely implies heating.

Thank you for your contribution, it does make sense, have a good day


text:
Why is pyridine used when making tosyl esters from alcohols?
Tosyl chloride is used to make a hydroxyl group into a better leaving group. However, when the reaction of tosyl chloride and an alcohol occurs, a weak base such as pyridine should be used. Why?

The function of pyridine is actually not so simple and not so easy to notice at first glance. There is a fundamental reason why pyridine is used to promote the acylation reaction, which is that it can act as a catalyst.
Despite its basicity and the subsequent formation of its chlorhydrate salt after the tosylation reaction, pyridine is also an excellent nucleophile. Pyridine is, in fact, more nucleophilic than the alcohol, and it attacks the acyl chloride rapidly, forming a highly electrophilic (because of the positive charge) intermediate: N-tosylpyridinium chloride. It is indeed this intermediate the actual tosylating agent which reacts with the alcohol to give the ester. Because pyridine is acting as a nucleophile to speed up the reaction, yet is unchanged by the reaction, it is a "nucleophilic" catalyst.

In reactions of alcohols with SO2Cl2 (thionyl chloride), when the intermediate is formed, HCl is formed along. But pyridine attacks on H+ of HCl and frees Cl-, which attacks on the intermediate and leads to inversion at that carbon. Why doesn't THAT happen here?

The byproduct of the reaction is hydrochloric acid, and with pyridine forms the water soluble pyridinium chloride, which is readily removed in work-up.  Other preparations might use triethylamine instead of pyridine.

In reactions of alcohols with SO2Cl2 (thionyl chloride), when the intermediate is formed, HCl is formed along. But pyridine attacks on H+ of HCl and frees Cl-, which attacks on the intermediate and leads to inversion at that carbon. Why doesn't THAT happen here?

A base is needed to remove $\ce{HCl}$ produced in the reaction. A weak base is needed because strong bases are usually strong nucleophiles as well and would react with tosil chloride themselves. A tertiary amine and not primary/secondary is used, because amines are in general stronger nucleophiles than alcohols, but tertiary amines would give quite electrophilic products themselves. Pyridine is used, because nitrogen in it is in $\mathrm{sp^2}$ state, resulting in significantly lowered nucleophilicity.

This answer is difficult to accept in the context of the pyridinium tosylate intermediate.

@Lighthart Elaborate, please.

Pyridine reacts as a nucleophile with Tosyl Chloride and it is a stronger nucleophile than the alcohols in this reaction.  You, yourself, pointed that out with tertiary amines.  A base is needed for charge balance/acid neutralization, and said base must not irreversibly form an adduct with the tosyl chloride.  Pyridine satisfies this requirement, so any mention of pyridine's lowered nucleophilicity is potentially misleading

@Lighthart Primary and secondary amines reacting with $\ce{TsCl}$ produce cation, that can throw away a proton, so the overall reaction would be $\ce{2R2NH2 + ClTs = {R2NH2}^{+} Cl^- + R2NTs} $, while in case of tertiary amines and pyridine the product would be $R_3NTs^+$ cation that cannot stabilise by throwing away a proton and is an electrophile itself, donating tosil fragment just as tosil chloride. So, unlike secondary and primary amines, tertiary amines can be used in this reaction.

I'm not disputing your statements.  What I'm saying is that comments about the nucleophilicity of pyridine make the issue somewhat more complex, and are not, strictly speaking, necessary for explaining the role of pyridine in the reaction (from the material balance perspective).  The implied formation of a pyridindium tosylate may be interesting from the mechanistic perspective, but it is not clear that such a perspective is relevant to the OP.  Also mixing the comments that tertiary amines are stringer nucleophiles than alcohols with 'pyridine  [has lower nucleophilicity' may be confusing.

@Lighthart No pyridinium tosilate $\ce{{PyH}^{+} TsO^{-}}$ is formed. Not sure, what's the name of the intermediate. Formation of the intermediate is relevant because otherwise it is unclear why secondary and primary amines are not used.

The intermediate you speak (but did nto draw) of is actually called a pyridinium tosylate.  The structure you have drawn is also called a pyridinium tosylate, but when the context is confusing, it is more properly called a pyridinium hydrotosylate.   I don't agree the intermediate (whatever it is called)  is 'why' secondary amines are not used.  The products may form without that intermediate, so clearly, they cannot be part of 'why'.

In my experience with acylations and esterificatio, (both O-acyl and O-propionylations) , a very small catalytic amnt of DMAP (100 mg per mole of product) along with reduced qty of pyridine has allowed the room temp acylation of some normally sterically hindered organics in less than 30 mins. In fact, while working with temp sensitive piperidine based compounds, the acylation was performed at - 10 deg C and had 98% conversion in 35 mins. The undesirable temp sensitive side products were at less than 1% and with some optimization managed to get undesired side products down to less than 0.05%. Of course DMAP is considered more toxic than regular pyridine, but it just with pyridine it can be separated from products by similar methods. We were dealing with esp lipophilic compounds acylated in DCM, and removal of pyridines simply involved a wash with acidic water. Even though the products formed HCL salts. Their products are lipophilic nature kept them in the DCM layer. We never experienced more than 5% product loss during acid wash workup and optimization reduced the workup loss to 2%. Overall, DMAP is impressive and stuff. Even with non temp sens acylations, you can still achieve rapid nearly full acylations in 20 mins at room temp and also the use of a small cat qty of DMAP seems to reduce the pyridine requirements, in our case we reduced pyridine qty by a factor of 5 without detriment to our yield or efficiency In my experience with acylations, using acyl and propionylations, a very small catalytic amnt of DMAP (100 mg per mole of product) along with pyridine has allowed the room temp acylation of some normally sterically hindered organics in less than 30 mins. In fact, while working with temp sensitive piperidine based compounds, the acylation was performed at - 10 deg C and had 98% conversion in 35 mins. The undesirable temp sensitive side products were at less than 1% and with some optimization managed to get undesired side products down to less than 0.05%. Of course DMAP is considered more toxic than regular pyridine, but it just with pyridine it can be separated from products by similar methods. We were dealing with esp lipophilic compounds acylated in DCM, and removal of pyridines simply involved a wash with acidic water. Even though the products formed HCL salts. Their products are lipophilic nature kept them in the DCM layer. We never experienced more than 5% product loss during acid wash workup and optimization reduced the workup loss to 2%. Overall, DMAP is impressive and stuff. Even with non temp sens acylations, you can still achieve rapid nearly full acylations in 20 mins at room temp and also the use of a small cat qty of DMAP seems to reduce the pyridine requirements, in our case we reduced pyridine qty by a factor of 5 without detriment to our yield or efficiency.
The temp sensitive reaction was primary alcohol phenylpipedidine to ester acylation. It is temp and acidic sensitivity was needed to avoid problematic olefination of alcohol forming a undesirable phenylpyridine which at the time was suspected to have possible neurotoxicity mediated thru MAO mediated biotransformation that have been known to have highly specific dopaminergic neuron toxicity and have henceforth been utilized to induce Parkinsonism in animals for the testing of anti Parkinson agents, selegline and others have resulted from this line of research. A related study showing phenylpyridinium toxicity to be highly dependent on N-substitution on the piperidine, any group larger than N-propyl showed no reactivity to MAO B or MAO A enzymes, thereby not undergoing the enzymatic transformation to neurotoxicm. Our piperidine was substituted both on the carbon ring and N-phenylpropyl on piperidine nitrogen so there was no surprise when later testing found the olefin byproducts to have no MAO reactivity. Even taking into acct the lack of toxicity of our undesirable byproducts, the DMAP was an impressive catalyst for acylation and ESTERIFICATION that I still would have chosen to work with it out of the convenience of performing rapid room temp acylations with minimal pyridine and easy workup. DMAP is also highly effective in N-acylations with similar benefits in efficiency and yield. I've read of some studies that also show rapid conversion of N-acylations at room temp with tiny catalytic qty of DMAP.

Welcome to chemistry.se! Please use proper spelling and punctuation. I don't see how this text answers the question posed.


text:
How could click reaction take place at low concentration?
I have DBCO-functionalized resin, and the DBCO group is validated with FAM-azide.
But I cannot use this resin to react with FAM-Azide-labled primer, which has very low concentration (20 μM, it is already very concentrate at DNA-scale).
I think that might be caused by concentration, because the reaction between DBCO-resin and FAM-azide is also not significant at such low concentration.
I have already tried long reaction time and higher temperature.
How could I get this work?

A couple of thoughts: (1) Can you get your FAM-azide conjugate to react with anything? I realize this might be a difficult question to answer, but seeing as you've confirmed the reactivity of the DMCO-resin, it's the next logical question. (2) Use more resin.

@jerepierre Hi, thank you for your reply. I have negative control that shows FAM-azide cannot react with COOH-functionalized beads. (2) Do you mean I need more resin to react with primer? Base on the concentration, the DBCO on the resin already exceeds a lot.

Have you tried reacting the FAM-azide with some other alkyne, whether on solid support or not? You need some indication that the azide is reactive at all. You might also consider a Staudinger type conjugation to solid support. I asse that there's a resin available for that.


text:
White of an egg
What happens when you dissolve white of an egg in water?
Firstly, what's white of an egg? Is it the white part? Or implies whole of an egg? I couldn't relate much of it with colloids, either.

The white clear liquid in the egg is nothing but albumin. It helps in supplying nutrition to the yolk and gives strength to the yolk during the formation of chicken. White part doesn't imply whole of an egg.
       The composition of that white liquid is 90% water and 10% protein.
       If you dissolve it in water, then it is nothing but a colloidal solution. We can call it as emulsion as it is 'protein in water '.
    Hope helpful..

Thanks a lot! That was quite simple. I got one more question for you: what happens when you pass sulfur vapors through cold water.

@user208322 If you have a different question, please post it as a new question, not a comment.

The white of the egg, albumin, should be dissolved in cold water as in hot water it clogs.


text:
Resonance structure of cyclobutadiene?
I studied cyclobutadiene as an anti-aromatic compound. But I am unable to draw the resonance structures for cyclobutadiene. Can anyone help me?
What is the driving force for the resonance to start?
Is it possible that a compound is having conjugate double bond system but not satisfying resonance structures?
Is a compound called as anti-aromatic if it is not undergoing resonance?

possible duplicate of Is cyclobutadiene anti-aromatic?

Not quite a duplicate

@KlausWarzecha, it's not a duplicate. The problem was the poorly-stated title of this question.

@KlausWarzecha I think some of the question are answered in that question but still is can't be duplicate.

Antiaromaticity is a concept describing an explicitly fictional situation, no molecule is antiaromatic, because all molecules will react to avoid the situation.
So the answer to the implied question is that cyclobutadiene has very little resonance because the molecule is prevented from being antiaromatic (by the nature of the universe and its physics).
Elaboration:
To avoid this situation, butadiene is very reactive; it will undergo an intermolecular Diels-Alder reaction avoid this antiaromatic state.  When reaction is prevented something much more interesting happens.  When frozen in an argon matrix, x-ray crystallography of cyclobutadiene indicates the molecule is highly rectangular, not square.  For dienyl resonance, there is some changing of bond lengths.  In butadiene, the 2,3 bond shortens.  If this were to occur in cyclobutadiene, then there would be degeneracy in the orbitals leading to a Frost diagram that would describe an antiaromatic state.  So if reaction is prevented, the molecule distorts its own geometry.
The fictional resonance structures you would draw are double bonds on the top and bottom, and double bonds on the sides.  However, generally, one might assume there is no resonance for this molecule.

I disagree with the first statement of this answer. Antiaromaticity must be a real concept if molecules go out of their way to avoid it. Perhaps there aren't any molecules that exhibit antiaromaticity, but that's not what this statement implies.

Molecules " will go to great lengths to avoid ..."? Avoid an explicitely fictional concept?

People used to go to great lengths to avoid dragons as well.  Although I must concede, as jerepierre indicates, there may be a better way to say it.

What you mean is that there are no compounds that exhibit antiaromaticity. Molecules will avoid antiaromaticity in a number of ways, such as skewing out of plane (eg cyclooctatetraene) or acting as two separate pi systems instead of a conjugated system (eg cyclobutadiene). The only time that there's trouble is when an exam, let's say, has a question to pick out antiaromatic compounds. In that case, the correct response would be that none exist for the reasons that you describe. But we use the concept of antiaromaticity to explain why cyclobutadiene doesn't behave the way we expect (conjugated)

You
mix up resonance and (anti-)aromaticity the latter being
interpreted today by molecular orbital symmetry.
This is unhistoric.
Resonance is a concept predating Woodward-Hoffman and Fukui.
Resonance meant that delocalizing the electrons of bond(s)
lowers the energy irrespective of the kind of orbitals.
This works for conjugated double bonds, but for aromatic
compounds it is not wrong, but insufficient.
The amount of stabilisation of eg benzene is more than
conjugation can explain.
For antiaromatic compounds oldfashioned resonance is simply wrong,
and drawing "resonance stuctures" is misleading,
because the resonance destabilisizes the molecule.
If we had a friendly Laplacian demon clamping the 4 carbon
atoms of cyclobutadiene on the corners of a square with a
bond length similar to benzene, we would see that
cyclobutadiene is a triplett diradical.
Maybe somebody will synthesize a molecule where a
cyclobutadiene core is clamped in in such a manner by some
planar ring structures "around" some day?


text:
Structures present in liquid water
I know this may be a too generic question, but here it goes anyway.
It is already know that solid water (ice) organizes its molecules in a lattice (in a variety of 16 different crystals and one amorphous phase). See, for instance, http://en.wikipedia.org/wiki/Ice .
Now, what about liquid water? Is it the case that there are a (limited) number of local structures, like stable clusters or lattices configurations, combining and dissociating over relatively short periods of time, that dominate the liquid phase, loosely resembling the organization of a solid phase (the principal difference being that the hydrogen bonds would be weaker and would break more easily because of thermal agitation)? If so, what are those frequent structures?

Do You want to know whether textbooks say the truth? What else is the aim of your question?

Yes, there are various water clusters leading to local order on the scale of a few molecules holding stabilized orientation with one another for measurable timescales.  This article(1) is a pretty good introduction to the major varieties in liquid water.  This article(2) talks more about various structures available in any phase, including the ices, which gives some pretty good context.
(1) Keutsch, Frank N., Saykally, Richard J., Water clusters: Untangling the mysteries of the liquid, one molecule at a time, 2001, PNAS, 98, 19, 10533–10540, doi:10.1073/pnas.191266498
(2) Ludwig, Ralf, Water: From Clusters to the Bulk, 2001, Angewandte Chemie International Edition, 40, 10, 1808-1827, doi:10.1002/1521-3773(20010518)40:10<1808::AID-ANIE1808>3.0.CO;2-1

There is no lattice or psuedo-lattice structure associated with water while it is the liquid state. Look at it as a medley of H2O molecules combining, breaking apart, and recombining with Hydrogen bonds as the main mediator of attraction.

Polar molecules (like the above water molecules) have a weak, partial negative charge at one region of the molecule (the O atm) and a partial positive charge elsewhere (The H atm) These bonds that form between water molecules account for some of the unique characteristics of water.

You are correct that there is no lattice present in liquid water - but this does not mean that there can not be localised structures -and indeed there are...


text:
Friedel-Crafts Alkylation and hybridization
Why must the alkyl halide be sp3 hybridized with respect to the carbon connected to the halogen for the alkylation to occur?
Why, for example, can't we use acetylide with a bromine substituted for hydrogen on one side in conjunction with aluminum tribromide to alkylate benzene?
If anything this seems to be more favorable as an sp carbocation is likely less stable than an sp3 carbocation.

You can do this.  As @jerepierre has noted the intermediates involved are higher in energy and consequently more reactive once generated.  But under controlled conditions the reaction does work.  You just need to use a stronger lewis acid than an aluminum trihalide.  Google "Friedel Crafts alkenylation".

If anything this seems to be more favorable as an sp carbocation is likely less stable than an sp3 carbocation.

That's the problem, sp2 and sp hybridized carbocations are less stable than sp3. Those carbocations are difficult to form and will likely undergo rearrangement faster than attack of the aromatic ring.
Modern organic synthesis has much better ways of making sp-sp2 and sp2-sp2 carbon-carbon bonds by palladium catalyzed cross-coupling.

Here's another way to look at it. The Friedel-Crafts alkylation reaction may be an electrophylic substitution with respect to the aromatic compounds, but it is a nucleophilic substitution with respect to the alkyl halide ($\text{S}{\text{N}}2$ for primary and $\text{S}{\text{N}}1$ for secondary and tertiary halides), although the leaving group needs to be activated so the reaction can occur in nonpolar solution. Alkenyl and alkynyl halides do not work in FC reactions because they cannot be $\text{S}_{\text{N}}1$ substrates.


text:
Does the boiling point of an aqueous solution depend on what the solute is?
Consider a situation where we have two aqueous solutions which have the same concentration (e.g. 1 M), but the solute is different. For example, one is a solution of glucose, and the other is a solution of ethanol.
I am aware that dissolving something in water can lead to changes in boiling point (e.g. elevation). But would the boiling points of these two solutions be affected in the same manner, given that the concentration of the solute is the same? Or will the exact boiling point depend on the chemical species that is dissolved?

No, in water the two solutions will not have the same boiling point.
Ethanol is a volatile liquid with boiling point lower than water, and lowers the boiling point.
Water forms an azeotrope with ethanol.  There are positive and negative azeotropes, so when two or more volatile liquids are dissolved, you cannot easily predict whether boiling point will decrease or increase.
There can also be cases, like adding HCl to water, where as solute is added, first the boiling point increases to a maximum (108 °C around 20% HCl) above the boiling point of water, then decrease to below the boiling point of water (48 °C for 38% HCl).
Glucose is non-volatile and increases the boiling point of water.  The boiling-point elevation formula can be applied to non-volatile solutes in volatile solvent.


text:
How to make sense of the 'Relaxation Method' for measuring reaction kinetics?
I get the concept that one can measure the rate of reaction by measuring the rate at which an equilibrium state reaches another equilibrium state after the system is perturbed (usually by heating). However, I do have an issue with it. Consider the reaction $$\ce{A <=> B}$$
My lecture handout suggests that the rate of change of $[\ce{A}]$ is given by:
$$\frac{\mathrm d[\ce{A}]}{\mathrm dt}=-k_\text{forward}[\ce{A}]+k_\text{reverse}[B]$$
However, surely it's just $\frac{\mathrm d[\ce{A}]}{\mathrm dt}=-k_\text{forward}[\ce{A}]$. The equation in the lecture handout would make sense at equilibrium because that is when $\frac{\mathrm d[\ce{A}]}{\mathrm dt}=0$ but I think it seems to insinuate that this is always valid (without explicitly saying so). Am I right or is that equation always valid? If it is: why?

You have two simultaneous reactions,
$$\begin{array}{rcl}
\ce{A} &\xrightarrow{k_1}& \ce{B}\\[6pt]
\ce{B} &\xrightarrow{k_{-1}}& \ce{A}
\end{array}$$
and $\dfrac{\mathrm d[\ce{A}]}{\mathrm dt}$ equals $-k_1 [\ce{A}]$ for the first reaction alone and $k_{-1} [\ce{B}]$ for the second alone. To get the total $\dfrac{\mathrm d[\ce{A}]}{\mathrm dt}$ for the reactions running simultaneously, you have to add the contributions from each reaction:
$$\dfrac{\mathrm d[\ce{A}]}{\mathrm dt} = -k_1 [\ce{A}] + k_{-1} [\ce{B}]$$
This will always be true; at equilibrium you'll have $\dfrac{\mathrm d[\ce{A}]}{\mathrm dt} = 0$ and
$$K = \frac{k_1}{k_{-1}} = \frac{[\ce{B}]}{[\ce{A}]}$$
If you did it your way, this wouldn't be the case.

Perfect. Thank you very much. Seems obvious now.


text:
Replacement For RF sputtering
In my experimental work, I'm following an article. It said that:

A 30 nm ZnO seed layer was deposited on the glass substrates by RF sputtering of a ZnO target at room temperature.

But I haven't any tools for "RF sputtering". Is there any replacement for this technique?

If you need a high quality thin film of $\ce{ZnO}$, especially on insulating substrates, you're mostly limited to sputtering techniques, PVD (including things like pulsed laser deposition), CVD (plasma-enhanced CVD, spray pyrolysis, and atomic layer deposition are common). Unfortunately, all of these require fairly sophisticated instruments. More accessible methods include sol-gel synthesis and electrodeposition.
The sol-gel synthesis of $\ce{ZnO}$ only requires a spin coater and an oven that can do 600 °C. It's suitable for coating large substrates, but if you really need a layer that's only 30 nm thick, it may not be the best technique.
$\ce{ZnO}$ can be electrodeposited onto conductive substrates from aqueous solutions. This requires a potentiostat and controlling thickness and orientation can be tricky. One possible method: $$\ce{Zn^2+ + NO3- + 2e- -> ZnO v + NO2-}$$
I should add that if you don't have access to any of these tools, there are certainly commercial suppliers of $\ce{ZnO}$ coated glass substrates.


text:
Solubility of N-[(4-hydroxy-3-methoxyphenyl)methyl]acetamide at different pH

The molecule below is a simplified analog of capsaicin. Explain in which of the following options, it should be most soluble:

aqueous HCl (pH = 1)
aqueous NaOH (pH = 14)
pure water (pH = 7)

  N-[(4-hydroxy-3-methoxyphenyl)methyl]acetamide, National Center for Biotechnology Information. PubChem Compound Database; CID=606880

I chose aqueous NaOH, because then the molecule will be completely deprotonated, which will make it more soluble. Please tell me if this is correct.

I would agree that it is most soluble in NaOH.
However, the molecule will only deprotonate the OH proton not the NH proton.  I wouldn't give your explanation much, if any, credit because you aren't saying which proton would deprotonate, and you are implying the molecule will lose all protons.
Also, at low enough pH the N will protonate, i.e. it would gain a second proton.


text:
Why do some salt ions in solution conduct electricity better than others?
My experiment involves a potentiometric titration, and I have used $\ce{AgNO3}$ as a titrant and KCl/KI/deionised water only as an analyte.
In the image below of my data, the potential difference as recorded by a multimeter drops from 0.6V to 0.15V.

What causes the drop in potential difference when the silver nitrate and whatever analyte (all exhibited the same pattern of a sudden drop in PD) reaches its endpoint in the titration?
Why is there a higher potential difference when using KI as analyte than KCl?

What is Your problem? Conductivity as stated in the headline or potential?

Both essentially, how do they correlate to each other?

As you add $\rm AgNO_3$ to $\rm KCl$, solid $\rm AgCl$ forms. The silver ion you add will at first be completely consumed by the chloride. Once all of the chloride is locked up in the precipitate, the silver ion concentration will rapidly increase. The Nernst equation shows how the potential is expected to change when this happens. If you're using a silver/silver chloride electrode with a reference electrode, the indicator potential is
$$\rm E = E^\circ_{\rm AgCl,Ag} - \frac{R T}{\cal F}\ln\frac{K_{sp}}{a_{\rm Ag^+}}$$
To see whether you'll get a drop or a rise in $E$, ask yourself what happens when silver concentration rises. If you've got the leads hooked up to your multimeter properly (negative lead to the silver cathode, positive lead to your anode), shouldn't you be seeing an increase in potential? What exactly are your $x$ and $y$ axes on that graph?
You'll get a bigger change when you form $\rm AgI$, because $\rm AgCl$ is about a thousand times more soluble than $\rm AgI$. The $\rm AgI$ will lock up relatively more ions when it forms (the $K_{sp}$'s of $\rm AgCl$ and $\rm AgI$ are $1.8\times 10^{-10}$ and $1.5\times 10^{-16}$, respectively).
Addendum 1: You say you're just using a silver electrode; the indicator potential is
$$\rm E = E^\circ_{\rm Ag^+,Ag} - \frac{R T}{\cal F}\ln\frac{1}{a_{\rm Ag^+}}$$
(but if you dip a silver wire into a chloride solution you'll get a very thin layer of AgCl and it'll begin to be a silver/silver chloride electrode (Source: G. D. Christian, Analytical Chemistry, 4th ed., Wiley 1986, p. 282)).
You also say you don't have a true reference electrode, just a graphite electrode... the fact that its potential can change as well complicates things. Your cell potential will depend on what the oxidation process is at the anode.
Addendum 2: You asked for a source for the $K_{sp}$'s. CRC Handbook 58th edition gives $K_{sp}$'s at $25^\circ$C for AgCl and AgI as $1.56\times 10^{-10}$ and $1.5\times 10^{-16}$; the 84th edition gives $1.77\times 10^{-10}$ and $8.52\times 10^{-17}$; Keller's Basic Tables in Chemistry gives them as $1.0\times 10^{-10}$ and $1.0\times 10^{-16}$ for "temperatures near 298 K". They're very temperature dependent and they depend on conditions like ionic strength as well, so it's not surprising that the values you're finding on the web are all over the place.  I took the values above from a genchem text that happened to be lying on my desk.

Thanks for the response Fred. So what does the chloride ion have that the nitrate ion doesn't? Otherwise, there'd be no change surely?

Do you mean, why does AgCl precipitate when AgNO$_3$ doesn't?

No, why can the nitrate ions not conduct the electrical current like the chloride ions can?

OK, the issue is not conductivity so much as what ions your electrodes can sense directly to produce potential changes. If you're using a silver indicator electrode it mostly only "sees" changes in the concentration of the $\rm Ag^+$ ions.

Oh that makes more sense. Just to clarify if you will then, what dictates when that potential difference drop occurs? Is it when all the Ag+ has bonded to all Cl-?
Also, with regards to attempting to verify that value derived from the Nernst equation, unfortunately the reference electrode was just graphite (hence no reference electrode SEP to use with Nernst)

I edited my response into the answer...

The cathode was a silver bar, and the anode was graphite (it's what I had to work with sadly). My x and y axis labels are Volume of Silver Nitrate added/cm^3 and Potential Difference/V respectively. I got a PD drop in every titration I did, which was consistent with some I've seen from other sources.

Sorry to continue this, but could you provide a source for those Ksp values? I'm seeing a wide array of values from various sites.

...added to answer.


text:
Acetylation reaction
In the Friedel - Crafts acetylation of bromobenzene, how is it possible to obtain a black-coloured product?
When the acetic anhydride was slowly added to the reactants - $\ce{AlCl3}$, DCM, bromobenzene - a small, hard, yellow lump formed.
Would this be $\ce{AlCl3}$?
When the acetic anhydride was added, the reflux was initiated. The lump appeared to disappear, leaving behind a dark red solution.
However, when the reflux was turned off, the solution appeared black. I'm trying to think of potential products, such as ortho and para bromobenzene or acetic acid or $\ce{Al(OH)3}$ but why so black?
Does anyone have any ideas?

I assume "DCM" represents dichloromethane, if so...
There is an old paper that describes the formation of a low molecular weight polymer from the double Friedel-Crafts alkylation of benzene by dichloromethane. That's probably what's happening in your reaction as a minor pathway.  The dissolved polymer and tar is what's producing your black solution.
I actually ran that reaction once, I wanted to prepare that polymer.  It didn't work for me, but at the end, after the aluminum chloride had been removed, I did have a dark solution.

Thanks ron . Can you link that paper to me?  When you say ''tar'', what do you mean?

@Edward  I looked for a link before I posted and couldn't find one, the paper may be to old for an on-line citation to exist.  I'm guessing it was Journal of the Amer. Chem. Soc., maybe the 1940s or 50's.  If someone searches Chemical Abstracts with the terms Friedel-Crafts, dichoromethane, benzene, polymer, I think you'd find it.  When I rotovap'd the black liquid down, up wound up with a tar-like solid.  I wasn't able to isolate a clean, low-molecular weight polymer (oligomer) like the paper described.  I figured the tar was just a mess of higher MW polymers.

Okay. I was confused by a further two things. The addition of half-saturated NaCl - does that reduce the surface tension thereby allowing the organic droplet to ''fall'' as such?        And washing with water can remove AlCl3 by way of Al(OH)3 , H+ and Cl- ions...but why is HCl gas not produced?

And the bridge between the two benzenes? Would that be the culmination of SN1?

"half-saturated NaCl" - this must be in the product isolation step?.  Organics are less soluble in saltwater than they are in pure water, so less of the product remains in the aqueous layer and more goes into the organic layer during extraction. 2) HCl is generated during the reaction.  When the AlCl3 reacts with the anhydride and aromatic, 3 equivalents of HCl are generated along with aluminum salts containing the unreacted part of the acetic anhydride molecule - the acetate.  If there is any excess AlCl3 remaining after the reaction is complete, it will generate HCl when mixed with water

The bridge is the result of CH2Cl2 undergoing 2 Friedel-Crafts alkylations with 2 aromatic rings.  While it is not a true SN1, the Friedel-Crafts alkylation does involve a carbocation-AlCl3 complex.

yes, I understand that 3HCl are formed during the reaction, but when water is used to wash the organic product, why isn't HCl gas produced?

Typically, only a small excess of AlCl3 is used.  In this case, there will be a small amount of AlCl3 remaining after the reaction.  On contact with water during workup, HCl will be generated, but only a small amount.  It will probably just stay dissolved in the aqueous layer (test it with a pH strip).

Hi ron. I was wondering, in the mechanism for the acetylation reaction, how is HCl produced? I know it's produced during the hydrolysis step, but before that? Is there even a mechanism for this?

I see that you've asked this as a separate question; that was probably the best way to go.  BTW, if the above answer and comments were helpful, please consider marking the question as accepted. Thanks!


text:
What's the type of bonding in La@C60?
Endohedral fullerene is the name given to a brand new chemical species with the following attributes:

A metal, usually a transition metal, is "trapped" inside the fullerene, like $\ce{C60}$ and $\ce{C82}$. These transition metals are often La, Sc, Ce, Y, or, Ba and Sr etc.

Without any "binding" forces, the atom couldn't be possibly trapped inside the fullerene.
Here it's been discussed that the bonding is the type of "donor-acceptor". Later within, it was explained that

Though there isn't empirical proof, It's believed for these species to have metallic structure.

I have two questions:

Is there another simpler name for this "donor-acceptor" bonding? My thoughts lead to wierd names!
How "close" is this type of bonding to metallic bonding?

Edit: Citation: The second article:
Publisher: [AIP]; German journal of chemical physics. Volume 115, number 15, page 7215. (15 Oct 2001) (Juelich GMBH or something; Couldn't figure the name out, maybe German fellows would help.)
DOI: 10.1063/1.1406500
Authors: R. Klingeler, G. Kann, I. Wirth, S. Eisebitt, P. S. Bechthold, M. Neeb, and W. Eberhardt (Dr. Rudiger Klingeler is the main author)
Title: $\ce{La$@$C60}$: A metallic endohedral fullerene.
Unfortunately the other mirror links are behind paywalls. (But this article is available in Researchgate, in case you've got institutions to be able to create an account)
$%edit$

Is this a clathrate, similar to methane hydrate? A skim of the PDF did not find use of the word "clathrate", but that is another name for "caged" molecules. See http://en.wikipedia.org/wiki/Methane_clathrate

"Cage", "trapping", "inside the lattice". These terms are applied for both, I can see. But the article mentioned a "donor-acceptor" bond with some metallic characters. I'm even more confused now, not to mention more eager to hear answers.

:) Not that "type"! I mean, there should be a chemical name for the force that allows La to stay there. Unless, it's like a ball in a box that's in rapid motion "hitting" C atoms. (And that isn't mentioned in several articles I skimmed)

""Unless, it's like a ball in a box that's in rapid motion "hitting" C atoms"" What else? Whether the motion is "rapid", do You have figures?

That's what's bothering me. You can ignore my upper comment. It's a guess. I haven't seen anywhere anything that can even be close to proving it. If La is somehow remaining there it's either because of a force or because "it simply can't get out". I was making up chemistry for what it would have been if the second was true. :)

I edited in the doi link to the publisher, I hope you don't mind. It was behind a paywall for me so thank you a lot for the research gate link.

That the question says: "Without any 'binding' forces, the atom couldn't be possibly trapped inside the fullerene", shows you are misunderstanding fullerenes.
Atoms such as Helium and $H_2$ can be confined within the interior of the fullerene.
There are even reports of extraterrestrial helium trapped in fullerenes during astroid or comet impact.
The helium can move around within the fullerene.  It is sterically confined rather than bonded.
In the case of La, the metal ion does not remain neutral, but loses about 3 electron to the fullerene.  So, in addition to being sterically confined in the cage, there is the interaction between postively and negatively charged species.
As calculated in a later article Computed Structure and Energetics of
La@C60 International Journal of Quantum Chemistry, Vol 104, 272–277 (2005), the La has a charge of somewhere betwen 2+ and 3+.  In the lowest energy state, the La is off-center with respect to the fullerene, located between the center and the center of a hexagon face.  Also, the fullerene cage is distorted from icosahedral by Jahn-Teller distortion.
The La ion is located 2.62 Angstroms from the nearest carbon atom (4.6 Angstroms from the furthest), when at the potential miniumun, but is expected to move within the fullerene unless the temperature is very low.  The reference also says "there is obviously a relatively large negative charge on some carbon atoms of the cage", without specifying how this charge is distributed.
Overall I would say, whereas in the case of helium and dihydrogen being confined in the fullerene is completely because of van der Waal repulsive force of avoiding the walls, in the case of La, there is attractive electrostatic force (ionic interaction) between the La3+ and negatively charged carbon atoms.
See also Theoretical investigation of structural and thermodynamic properties
of lanthanum carbides LaCn (n = 2–6) which finds strong ionic interactions between La and Cn due to charge transfer.

It is unlikely to find an uncharged metal atom hovering in the centre of a fullerene cage.
$\ce{C60}$ is a good electron acceptor. This has been widely demonstrated in a lot of studies on photoinduced electron transfer in donor-acceptor dyads, where the donor was a metalloporphyrine attached to the fullerene. Here, excitation of the donor led to an intramolecular redox reaction in which the fullerene radical anion $\ce{C60^{-\cdot}}$ was formed.

If you're interested in this particular topic, look for some earlier works by Devens Gust, Thomas A. Moore and Ana L. Moore.

Taking that into account, full electron transfer, resulting in caged ion pair, or at least some charge transfer is conceivable.
The term "donor-acceptor" thus seems appropriate.


text:
Determining the value and units of the rate constant
Consider this initial-rate data at a certain temperature for the reaction described by
$$\ce{2NOBr(g) -> 2NO(g) + Br2(g)}$$
\begin{array}{cc}\hline
\ce{[NOBr]_0 (M)} & \mathrm{Initial \,rate \, of \,Br_2}\, (\pu{M/s}) \\ \hline
0.600 &1.08 \times 10^2 \\
0.750 & 1.69 \times 10^2\\
0.900 & 2.53 \times 10^2\\ \hline
\end{array}
I'm not understanding how to find initial rates from the data given and then use that to find the rate constants. I also don't really understand how this information can give me the orders of reactions, for example when does the number double, stay the same, or quadruple, etc.

Welcome to chemistry.SE! This is a homework question. Thus, we should make sure that we aren't doing homework for you. You should provide some info so that we make sure you're "aware of the underlying concepts".

I guess I'm not quite understanding to find initial rates and rate constants. I also don't really understand the orders of reactions and which number and when does the number double, stay the same, or quadruple.

Hmm. I think I'll leave this for the community to decide. Maybe a good answerer will be able to do this with hints. I'll also edit your question to include a "reference-request", so that the answer will include a page link for more studying.

I've moved key ideas from your comment, @JeannaT, into your question, which gives folks a place to start on this question without it needing closed.

Alright, so in the reaction
$$\ce{2NOBr(g) -> 2NO(g) + Br2}$$
the rate law is
$$\text{rate} = k[\ce{NOBr}]^x$$
where $k$ is some constant and $x$ is the order of the reaction in respect to $\ce{NOBr}$.
By seeing how the initial rate changes when we change the concentration of $\ce{NOBr}$, we can determine the value of $x$. We can use any two of the three. I'm going to use the first and third trials. If we divide them we get
$$\frac{\text{rate 3}}{\text{rate 1}} = \frac{k[\ce{NOBr}]_3^x}{k[\ce{NOBr}]_1^x}$$
$$\frac{2.43\times10^2}{1.08\times10^2} = \frac{0.900^x}{0.600^x}$$
The $k$'s cancel out.
$$2.25 = 1.5^x$$
$$x = 2$$
The rate is second order in respect to $\ce{NOBr}$, and the rate law is written $\text{rate} = k[\ce{NOBr}]^2$. If you double the concentration, the rate will quadruple.
$$\text{rate before doubling concentration} = k[\ce{NOBr}]^2$$
$$\begin{align}\text{rate after concentration} &= \left(2[\ce{NOBr}]\right)^2\\
&= 2^2[\ce{NOBr}]^2\\
&= 4[\ce{NOBr}]^2\\
&= 4 \times \text{rate before doubling concentration}\end{align}$$
A tripling of the concentration will increase the rate by a factor of nine, a quadrupling of the concentration increases the rate by a factor of 16, and so on.

The ratio of 0.9 to 0.6 is 1.5.
The ratio of 2.43 to 1.08 is 2.25.
$1.5^2=2.25$
The initial rate is proportional to [NOBr]$^2$


text:
Why are potentials of cells with magnesium electrode always lower than expected?
I have prepared the 1M water solution of magnesium acetate. Then I placed $\ce{Mg}$ and $\ce{Cu}$ electrodes in it. I predict there such reactions take place: $$\ce{Mg-2e^- \rightarrow Mg^{2+}}$$ $$\ce{H_3^+O + e^- \rightarrow\frac12 H_2 + H_2O}$$
The gas bubbles emitting on both electrodes probably confirm my prediction (I did not check is it hydrogen or not).
So I expect to obtain the potential difference about 2.3 V. But really I observe something like 1.62 V.
Then I changed the acetate solution to a water (simple water which we drink) and observed almost same value of 1.60 V. Adding there many crystals of $\ce{Mg}$ acetate (I added them till they could not dissolve more) almost did not change the voltage!
When I added much acid to the solution the potential difference became larger - up to 1.9 V.
So, I have two questions: 1) why the real voltage is so different from the expected value? 2) Why changing the $\ce{Mg^{2+}}$ concentration almost does not change the voltage?
You may say I have dirty magnesium electrode etc. But if you search e.g. experiments with potatoes (when one puts $\ce{Mg}$ and $\ce{Cu}$ strips in potato) you can observe that everywhere it is mentioned that such $\ce{Mg-Cu}$ cell gives 1.7 V.
I have several ideas why can it happen: a)some story with overpotentials, b)oxide film on magnesium surface.
What is the real reason?

At least one significant reason is that the voltage you "expect" is probably for a standard state of one mole per liter of $\ce{H+}$, i.e. at pH 0.  Neither the magnesium acetate nor the plain (tap?) water you used for the experiments come close to this pH. Also, the standard state for $\ce{H2}$ would be one bar of atmospheric pressure.  Probably this condition was not achieved either.
You can calculate the expected potential at any pH using the Nernst Equation.
The half-cell reactions in your system are

$\ce{Mg^0 -> +2e- +Mg^{+2}}$ (this has a potential of 2.372 V not 2.7 V as a mistakenly wrote in a prev. version of this answer)
$\ce{2H^+ +2e- -> H2}$

The total reaction is thus $\ce{Mg^0 +2H+ -> Mg^{+2} + H2}$
So applying the Nernst equation to a one molar magnesium solution at pH 7 (i.e. $\ce{H+}$ concentrations of $10^{-7}$ mol/L) would give:

$E_\text{cell} = E^{\ominus}_\text{cell} - \frac{RT}{zF} \ln Q$
$E_\text{cell} = 2.4 \text{ V} - \frac{RT}{zF} \ln \frac{1}{(10^{-7})^2 }$
$E_\text{cell} = 2.4 \text{ V} - 0.41 \text{ V} = 1.9 \text{ V}$

Obviously, this correction gives you a predicted voltage that is still higher than what you measure closer what you measured.  But you reported using acid and getting a higher potential, so it's clear that the pH effect matters for your system.  But something else must be going on too.  Some other possibilities for other minor effects that tweak the voltage a bit more:

Hydrogen partial pressure is unaccounted for
overpotential
your cell is not open but is conducting current
dirty electrodes etc.
other unaccounted for electrochemical reactions (copper oxidation, etc.)

""At least one significant reason is that the voltage you "expect" is probably for a standard state of one mole per liter of H+, i.e. at pH 0. "" Normal potentials are always for a solution of the same metal ion 1 molar and neutral, if not otherwise stated.

Georg, please have a look at the wikipedia page for the standard hydrogen electrode.  The SHE is the standard that is usually assumed in electrochemistry, unless otherwise stated.  It is approximated in the laboratory by using 1 mole per liter of strong acid -- i.e., pH 0.

Also, if you see this table of standard reduction potentials, you will see that Mg is listed at the potential stated in the OP, 2.7 V, and that preface to the table says "standard electrode potentials are given in the table below in volts relative to the standard hydrogen electrode".

You are not correct about 2.7V: 2.7V is potential for $Mg-1e^-\rightarrow Mg^+$ but here we expect $Mg-2e^-\rightarrow Mg^{2+}$ with standard potential 2.3V.
So the expected voltage seems to be ~2.3-0.4=1.9 what looks closer to the truth :)
Thank you for your answer!

Thanks for the comment Martino.  I mis-read the wikipedia table and then assumed (without checking) that the number matched the OP!  I'll edit my answer to eliminate the mistake.

Can we say that the bubbles of H2 are at 1 atm pressure if they have not mixed with air above the electrolyte surface?

@JamesGaidis even if they do mix, they aren't at one atmosphere of partial pressure because the total atmospheric pressure is 1 atm, so mixing with the atmosphere will dilute the H2.  It could be that the bubbles at the electrode are at 1 atm or even higher; depends on the current, how fast the bubble rise to the surface, how many bubble there are, the solubility of H2, and many many empirical factors.

I run a lab that produces the same confounding results. We connect the standard magnesium half cell to 3 other half cells (silver, copper, zinc) and in each case the result is consistent. The data booklet we provide students has the magnesium half cell listed as -2.37 V, our measured results puts it at -1.60 V instead. It's not a concentration issue - we are using 1.0 M solutions and the temperature is near enough to 25 C in the room. Our lab tech thought the magnesium was oxidized and so the supposed standard magnesium 1/2 cell is not what's actually reacting.


text:
What causes hydrogen abstraction in the radical chain mechanism of alkane halogenation?
In the first propagation step of the radical chain mechanism of alkane halogenation, why does the halogen actually abstract hydrogen from the alkane? My organic chemistry textbook so far has went into details on MOs, hybridized orbitals, and general thermodynamics of reactions but has been lacking in the by and large why of reactions. Is the $\ce{C-H}$ bond broken by sufficient collisions from halogen free atoms and the hydrogen goes to the halogen as a result of $\ce{F}$/$\ce{Cl}$/$\ce{Br}$ being more electronegative than carbon, or what should cause this to happen?

why does the halogen actually abstract hydrogen from the alkane?

Look at the two tables of bond energies on this webpage.  The second or bottom table shows the $\ce{C-H}$ bond energies for various carbon containing compounds.  In the top table we see that the H-Cl bond strength is 103 kcal/m.
If we start with a chlorine radical, then if we break a C-H bond and make a Cl-H bond, the energetics of the reaction will be
$$\ce{\Delta H=bonds~ broken~-~bonds~made}$$
$$\ce{\Delta H=C-H ~bond~ broken~-~H-Cl ~bond~made}$$
$$\ce{\Delta H=C-H ~bond~ broken~-~103~ Kcal/m}$$
Look at the top row in the bottom table again.  As long as our chlorine radical breaks any C-H bond that has a dissociation energy less than 103 kcal/m, our reaction will be exothermic, and the reaction will continue.
Edit

As I understand, the first propagation step in the free-radical
  monochlorination of methane is slightly endothermic but is driven to
  completion by the exothermic nature of the second propagation step

The first propagation step involves breaking a $\ce{C-H}$ bond (103 kcal/m) and making an $\ce{H-Cl}$ bond (103 kcal/m), so it is roughly thermoneutral according to the above link.
$$\ce{CH3-H + Cl{.} -> CH3{.} + H-Cl}$$
$$\ce{CH3{.} + Cl-Cl -> CH3-Cl + Cl{.}}$$
The second propagation step involves breaking a $\ce{Cl-Cl}$ bond (58 kcal/m) and making a $\ce{CH3-Cl}$ bond (81 kcal/m), so it is quite exothermic.
These steps are consistent with my earlier argument.  You are correct that using free energies would be better (more accurate) than enthalpies, but the Tables I could find were based on enthalpies - I suspect they provide us with a reasonably accurate "back of the envelope" calculation and view of what is going on.
As to the actual mechanism of the reaction, once we've created a chlorine radical, we've created a reactive species.  It is reactive because it doesn't have an octet of electrons.  By undergoing the above-reactions it can regain its octet and become a stable entity.
Here is a diagram of a reaction coordinate (note we are using free energy as the Y-axis in this diagram).  Generally the more exothermic (exergonic for free energy) a reaction, the lower the activation energy required to reach the transition state and pass over from reactants to products.  So our reactive chlorine radical bounces around probing different parts of the reaction coordinate (e.g. different $\ce{C-H}$ bonds, different angles and approaches to these bonds) until it encounters a bond from the right direction, right angle, etc, such that the collision has enough energy to pass over the top of our reaction coordinate and become a product.

As I understand, the first propagation step in the free-radical monochlorination of methane is slightly endothermic but is driven to completion by the exothermic nature of the second propagation step--by this, does your explanation hold? I'm inclined to think that a better determination--from a thermodynamics standpoint--is Gibbs free energy. I was really looking for more of an answer on, for example, mechanisms of the bond being broken beyond thermodynamics (e.g. if what I said on collisions & EN is correct). Please explain if I missed something about what you said, as I have much to learn.

First
mistake you do is speaking of "the radical chain reaction".
There are many reactions with different mechanisms which
are radical chains.
Re the cause: All reactions take place because there is
some energy (more specific: delta G) gain. Any step
in that radical chain has to fulfil this requirement.

Oh yes, that's right; I did mean free-radical halogenation. "Radical chain mechanism" is a bit ambiguous, sorry about that! I truly do appreciate your time, but your answer doesn't really satisfy what I'm looking for based on what I outlined in my comment on Ron's answer. Could you possibly expand beyond free energy? I may be overthinking this... but it's the first mechanism in ochem and I'd like to have a tight grasp on it.

The initiation step of all these radical halogenations is the homolytic cleavage of a halogen molecule: $\ce{Cl2 ->[{h \cdot \nu}] 2 Cl^{.}}$ These radicals rapidly move away from each other so immediate recombination is not an option. Also, the radical concentration is generally rather low, so it is unlikely for the radicals to find other possible recombination partners. That is what gets the chain started.
So the radical has the choice of capturing one of the other species present in the mixture which are the alkane $\ce{C_$n$H_{2$n$ +2}}$ or unreacted halogens $\ce{Cl2}$ (or previously reacted haloalkanes). In all cases, one bond would be cleaved and one formed, giving a new radical species. If the radical happens to meet $\ce{Cl2}$ then nothing apparantly happens, only if you colour one chlorine atom yellow, the second blue and the third red would you observe a change.
If the halogen radical meets an alkane and the Gibbs free energy change is favourable (refer to ron’s answer for numbers) it will react. It cannot directly bind to carbon, because that would induce a $\ce{H^{.}}$ radical which is much more unstable than anything else — $\Delta G$ is too high. Instead, it cleaves a $\ce{C-H}$ bond by abstracting hydrogen: $\ce{Cl^{.} + H-C_{$n$}H_{2$n$+1} -> Cl-H + ^{.}C_{$n$}H_{2$n$ + 1}}$. This new radical then goes looking for reaction partners and usually finds a $\ce{Cl2}$ molecule first according to what we write — but it could also find another alkane and transfer the radical from one alkane molecule to another.


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Is there an example of an opaque solution which can be made transparent by adding another solute, such that no chemical reaction occurs?
I am further interested in knowing exactly what the requirements are for a liquid to be transparent.

Welcome to chemistry.SE! If you had any questions about the policies of our community, you can visit the help center or take a tour of the website. I think this question may be flagged as too broad or primarily opinion based.

really? I thought it was quite specific. I am just looking for one example of a solution which is impermeable to a wavelength of light, such that when you add another solute (which does not react with any of the other solutes) becomes transparent to that wavelength.

I'm leaving it to see how it plays out. The example part may be and only may be, primarily opinion based. And the "too broad" part could be when you ask for why something is transparent. You can add a "reference-request" to avoid that part. Ah well, I think I'm being too nitpicky in here. It must be a symptom of the cold I caught. :)

When I "dissolved" phsophatidylcholine in water it was cloudy, because it made micelles. When I added triton X-100, a surfactant, it became transparent, which I assume because of smaller micelles. Not a chemical reaction, just a change of state I think.