text:
Titration of H2SO4 w NaOH: Solving for the molarity of H2SO4?
I am given $\ce{H2SO4}$ in a reaction vessel of about $50~\mathrm{mL}$. I need to solve for the molarity of $\ce{H2SO4}$. A $10~\mathrm{mL}$ sample of $\ce{H2SO4}$ is removed and then titrated with $33.26~\mathrm{mL}$ of standard $0.2643\ \mathrm{M}\ \ce{NaOH}$ solution to reach the endpoint. How do I solve for titration of the $50~\mathrm{mL}$ sample?
Obviously I can use the formula:
$$M_i \times V_i = M_f \times V_f$$
Which brings me to
$$M_i \times 10~\mathrm{mL} = 0.2643~\mathrm{M} \times 33.26~\mathrm{mL}$$
Thus:
$$M_i = (0.2643~\mathrm{M} \times 33.26~\mathrm{ml}) / (10~\mathrm{mL})$$
This will find the molarity of the $10~\mathrm{mL}$ sample of $\ce{H2SO4}$.
Now, how do I find the molarity of the $50~\mathrm{mL}$ sample of $\ce{H2SO4}$ from this? Would I just do five times the $10~\mathrm{mL}$ sample's molarity?

Actually, ×=× is for dilution. It seems to be like you're dealing with a stoichiometry problem. This would require dimensional analysis to determine the molarity of the final product... but I could be wrong.

The molarity would be the same whether you have $5~\mathrm{mL}$ of $\ce{H2SO4}$ or a swimming pool full of it.
However, if you wanted to solve for moles of $\ce{H2SO4}$ in $50~\mathrm{mL}$, you would have to multiply the number of moles in the $10~\mathrm{mL}$ sample by $5$.
Note: Make sure you're working with molarity and not moles. I'm in analytical chem right now and often we're multiplying the number of moles in our sample by the total volume of the volumetric flask from which the sample was drawn, so we're doing calculations similar to this.

A couple things to think about:

How many protons can one molecule of sulfuric acid give? Will this affect the amount of NaOH it takes to neutralize a given amount of sulfuric acid?
Molarity is the number of moles in a Litre of solution. If I double the volume, it doubles the number of moles. Does this change the ratio of moles to litres?


text:
Products of Lead(II) nitrate decomposition
It is known, that decomposition of Lead(II) nitrate is one of the ways of generating $\ce{NO2}$ for lab use.
I recently did this in order to acquire $\ce{NO2}$ in liquid form (ambient temperature was ~0 °C), and got quite strange results. First, liquid $\ce{NO2}$ had 2 visible layers, usual brown one at the top and transparent at the bottom:

Also, during condensation liquid initially had green color (I know that Graham condenser is not safe in this configuration - I wouldn't see this color change in conventional condenser setup).

Can anyone clarify why I got such variety of products instead of pure $\ce{NO2}$?

in case you want purer $\ce{NO2}$ without lead compound involved consider reaction $\ce{2HNO3 + NaNO2 \rightarrow 2NO2 + H2O + NaNO3}$

BarsMonster, I updated my answer, hopefully some helpful information

The colorless area may be $\ce{N2O4}$ which is in equilibrium with $\ce{NO2}$. $\ce{N2O4}$ is colorless and has a greater density and therefore is found at the bottom and green compound might be $\ce{N2O3}$ which is in equilibrium with $\ce{NO2}$ and $\ce{NO}$.

Edit :
Boiling point of $\ce{N2O4}$ is 21.69 °C, so the temperature at which you perform experiment is also critical. Moreover, there is high probability that $\ce{N2O4}$ might be present in liquid state too.

Hmm... Is N2O4 stable enough to form a separate layer (i.e. shouldn't NO2 form inside N2O4 layer and vice versa)? I had an impression that It should have been very hard to separate N2O4 and NO2 "gravitationally"...

You cannot seperate them as they are in equilibrium as if you try to remove NO2 from here much of N2O4 will convert to NO2 to establish the equilibrium so you cannot seperate them.on basis of gravitationally as you are maintaining same pressure conditions

ONE ore thing
Boiling point 21.69 °C of N2O4 so the temperature at which you perform experiment is also critical . And there is high preobability that N2O4 might be present in liquid state too.

Check if the green condensate is $\ce{N2O3}$. It has been reported to be blue-green.
In the presence of water vapor and oxygen, the nitrogen dioxide that you are trying to obtain is in equilibrium with nitric acid and nitrogen trioxide, as explained here.
As reported in the Journal of Industrial and Engineering Chemistry, volume 12, number 6, page 531-538 (1920), condensation of gaseous nitrogen dioxide in a spiral condensor at -5 °C in the presence of oxygen and water vapor resulted in liquid with distinct upper and lower layers, the upper layer being a concentrated nitric acid layer (essentially no water) and the lower layer being a diluted nitric acid layer (27 %) water.
So the reason you are getting side products is likely that water vapor and oxygen are also present and causing equilibrium reactions that yield nitric acid, nitrous acid and nitrogen trioxide.
Reference

CCXIV.—The dissociation of gaseous nitrogen trioxide, Bernard Mouat Jones,  J. Chem. Soc., Trans., 1914, 105, 2310-2322, DOI: 10.1039/CT9140502310

That looks very close. Important difference though is that there should have been no water in the system, Lead (II) nitrate decomposition goes at 250C, when no humidity is left. So yes, there was plenty of oxygen but no water.

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text:
Is thin layer chromatography suitable to characterize distillate samples?
Would I be correct if I state: Thin layer chromatography (TLC) is a separation method for non-volatile mixtures. Because distillate samples are pure, this would not be appropriate.
I'm unsure and would like some insight into this.

Your statement is unnecessarily complicated and not easy to follow. You can use TLC as a quick way to assess the purity of a sample. It doesn't matter if your sample is a distillate or not. It only has to be non volatile as you said and soluble in some solvent or a mixture of solvents that can (the solvent/s) make it move on the TLC plate.
I assume you want to use distillation to separate a mixture and TLC in order to assess the purity of your fractions. Obviously TLC becomes an unsuitable method for low b.p. distillates

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text:
Dehydration of secondary alcohol via E1 mechanism

Write the mechanism for this reaction:

I am trying to do a dehydration of an alcohol using the reagents sulfuric acid and acetic acid. I don't know how I am supposed to use the acetic acid, instead I've been using water. This is my proposed mechanism:

But it doesn't include the acetic acid, and it doesn't say in the question whether it is done in solution with water or not but I only know how to do it with $\ce{H2O}$. My main issue is that I need to put the acetic acid into the mechanism.

You can use $\ce{HSO_4^-}$ to pick up the proton in the bottom drawing. Going back to the original question I can't see what the role of AcOH is. I would have thought that $\ce{AcOH/H_2SO_4}$ is a very nice way to make an acetate (esterification) but if you plan to dehydrate then I am not sure what AcOH helps with. Not much help sorry, hopefully someone else could answer this

Your proposed mechanism looks reasonable. I think that acetic acid primarily serves as an acidic solvent for the reaction. However, in the presence of concentrated sulfuric acid, $\ce{CH3CO2H}$ can be protonated by $\ce{H2SO4}$, like in the first step of this esterification. The resulting resonance-stabilized cation is strongly acidic ($\mathrm{p}K_\mathrm{a} = -6.1$ according to this source), and should readily protonate the alcohol. This would be a possibility to include acetic acid into the mechanism.
$$\ce{CH3CO2H + H2SO4 <=> CH3C(OH)2^{+} + HSO4-}$$

(Image source: ChemGuide)
So in the first step, the alcohol can either be protonated by $\ce{H2SO4}$ or $\ce{CH3C(OH)2^{+}}$, and in the end, the eliminated proton can either be picked up by $\ce{HSO4-}$ or $\ce{H2O}$.

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text:
How do I isolate calcium from calcium carbonate?
I have the ability to use many kinds of reactions, and the use of most common acids/bases.

An easy method is to add vinegar to the calcium carbonate to remove the carbonate and get calcium acetate (in solution). What do you want to use the calcium for?

The answer is you cannot do it almost for sure.
To obtain $\ce{Ca}$ metal from $\ce{Ca (+II)}$, you need to perform a reduction process. For this purpose, you would need a very strong reducing agent, because calcium is very close to the bottom in the Ellingham diagram (used in metallurgy for this same purpose: obtain metals from its salts).
So you will need to perform an electrolysis of molten calcium chloride, which is not cheap, and by any reason not doable at home or at an standard lab (just a quick view of calcium salts melting point will give you a hint about why is this...)


text:
Bromination of 1,3,5-cyclooctatriene followed by nucleophilic substitution by excess dimethylamine

Write the mechanism for the following reaction (bromination of 1,3,5-cyclooctatriene followed by nucleophilic substitution by excess dimethylamine):

My answer so far:

I have only shown one of the reactions, I presume it is the same for adding the other $\ce{Me2NH}$ on the other side.
How do I get rid of the second bromine, or does just one of the Br's add? If it is only one, why, because you usually use $\ce{HBr}$ for that?

This is a quote from Solomons and Frylhe Organic Chemistry 10th Ed, Chapter 13, p.614

Reactions of this type are quite general with other conjugated dienes. Conjugated trienes often show 1,6-addition. An example is the 1,6-addition of bromine to cycloocta-1,3,5-triene.

A plausible mechanism for the bromination step would be the initial formation of a bromonium ion which is attacked distally by bromide:

Sequential nucleophilic substitution with dimethylamine then leads to the desired product.
Echoing Martin's comment on Siddharth Porwal's answer, there is a distillation step between the bromination and the substitution with dimethylamine. It is not possible for a putative bromonium ion to survive. Also, the isolation of the 1,6-dibromide also disproves the possibility of 1,2-addition mentioned in RE60K's answer.
The reaction above was originally reported by Cope et al.:[1]

Bromine (3 g) in 20 ml of chloroform was added to a stirred solution of 2 g of 1,3,5-cyclooctatriene in 20 ml of chloroform at −15 to −25° in a nitrogen atmosphere in a period of one hour. The product was distilled rapidly under reduced pressure; b.p. 95° (0.3 mm), n25ᴅ 1.5954. [...] Previous work had shown that dibromocyclooctadiene prepared by the two routes gave the same bis-dimethylaminocyclooctadiene by reaction with dimethylamine.

Reference

Cope, A. C.; Stevens, C. L.; Hochstein, F. A. Cyclic polyolefins. v. preparation of bromocyclooctadienes and 1,3,5-cyclooctatriene from 1,5-cyclooctadiene. J. Am. Chem. Soc. 1950, 72 (6), 2510–2514. DOI: 10.1021/ja01162a047.

Two possible mechanisms exist:
Either the molecule, undergoed 1,6 bromine addition or 1,2 addition, in which the first case easily gives the product and in case of second, the final addition can be 1,6. It was 1,6 addition in the first or second step is hard to say, as I'm concerned.

Don't be confused if the final structure is looking as if it had no rotation, it had michael addition.

@eaxdpioynyeantial  Do you have an explanation why, according to your mechanism, one bromine would react entirely by an SN2' pathway, while the other bromine would react entirely by an SN2 pathway?  It seems unlikely to me.

Why wouldn't the 1,2-dibromide be formed as well?  In fact it should be more likely here than in 1,3-butadiene and 1,3,5-hexatriene because the ring confers the same thermodynamic stability on the 1,2- and 1,6-dibromo products.

I wonder if the equilibrium with bicyclo[4.2.0]octadiene is a factor. Perhaps the bicyclic isomer reacts preferentially and provides the product of 1,6-addition exclusively?


text:
How much water is required to dilute glacial acetic acid to given pH?
I have 100% acetic acid (glacial) with a pH of 2.4 and I want to know how much DM water is required to dilute the acetic acid and obtain a solution with a pH of 4.3.
As per observation 0.0001 M is equal to pH 4.3. So how can I implement this calculation using the Nernst equation?

The question is flawed; the $\mathrm{pH}$ of 100 % glacial acetic acid is not $\mathrm{pH=2.4}$. (Note that 100 % glacial acetic acid contains no water; therefore you cannot apply the usual equations for dilute aqueous solutions to calculate the $\mathrm{pH}$ of 100 % glacial acetic acid.)
Also note, since acid dissociation is not a redox reaction, you cannot apply the Nernst equation here.
The quantity $\mathrm{pH}$ is actually defined in terms of the activity of $\ce{H+}$ ions; however, for simplicity’s sake, we may use the old concept of $\mathrm{pH}$ with the definition in terms of the concentration of $\ce{H+}$ ions
$$\mathrm{pH}=-\log_{10}[\ce{H+}]$$
or
$$[\ce{H+}]=10^{-\mathrm{pH}}$$
The equilibrium of acid dissociation is usually written as:
$$\ce{HA <=> A- + H+}$$
where $\ce{HA}$ is a generic acid that dissociates into the conjugate base $\ce{A-}$.
The equilibrium constant $K_\mathrm a$ for this reaction is defined as
$$K_\mathrm a=\frac{[\ce{H+}]\cdot [\ce{A-}]}{[\ce{HA}]}$$
Usually, values are given for the logarithmic constant $\mathrm pK_\mathrm a$, which is defined as
$$\mathrm pK_\mathrm a=-\log_{10}K_\mathrm a$$
or
$$K_\mathrm a=10^{-\mathrm pK_\mathrm a}$$
The corresponding equilibrium for acetic acid is
$$\ce{CH3COOH <=> CH3COO- + H+}$$
The value of the logarithmic constant for acetic acid at 25 °C is
$$\mathrm pK_\mathrm a=4.756$$
Thus, the value for the acid dissociation constant is
$$K_\mathrm a=\frac{[\ce{H+}]\cdot [\ce{CH3COO-}]}{[\ce{CH3COOH}]}=10^{-\mathrm pK_\mathrm a}=10^{-4.756}\approx1.75\times10^{-5}$$
The chemical equation shows that, one $\ce{CH3COO-}$ ion and one $\ce{H+}$ ion are produced for every $\ce{CH3COOH}$ molecule that dissociates. This can be represented by subtracting a value $x$ from the original acetic acid concentration $[\ce{CH3COO-}]_0$, and adding this value $x$ to the concentrations of $\ce{CH3COO-}$ and $\ce{H+}$, i.e. when neglecting the $\ce{H+}$ ions from the autoprotolysis of water:
$$\begin{align}
[\ce{CH3COOH}]&=[\ce{CH3COOH}]_0-x\\[6pt]
[\ce{CH3COO-}]&=x\\[6pt]
[\ce{H+}]&=x
\end{align}$$
Thus, the equation for the acid dissociation equilibrium can be rewritten and solved for the original acetic acid concentration $[\ce{CH3COOH}]_0$ as follows:
$$\begin{align}
K_\mathrm a&=\frac{[\ce{H+}]\cdot [\ce{CH3COO-}]} {[\ce{CH3COOH}]}\\[6pt]
&=\frac{x\cdot x}{[\ce{CH3COOH}]_0-x}\\[6pt]
&=\frac{x^2}{[\ce{CH3COOH}]_0-x}\\[6pt]
K_\mathrm a\cdot\left([\ce{CH3COOH}]_0-x\right)&=x^2\\[6pt]
[\ce{CH3COOH}]_0-x&=\frac{x^2}{K_\mathrm a}\\[6pt]
[\ce{CH3COOH}]_0&=\frac{x^2}{K_\mathrm a}+x
\end{align}$$
Since $[\ce{H+}]=x$, $[\ce{H+}]=10^{-\mathrm{pH}}$, and $K_\mathrm a=10^{-\mathrm pK_\mathrm a}$, the required acetic acid concentration $[\ce{CH3COOH}]_0$ can be given in terms of the desired $\mathrm{pH}$:
$$\begin{align}
[\ce{CH3COOH}]_0&=\frac{x^2}{K_\mathrm a}+x\\[6pt]
&=\frac{[\ce{H+}]^2}{K_\mathrm a}+[\ce{H+}]\\[6pt]
&=\frac{\left(10^{-\mathrm{pH}}\right)^2}{K_\mathrm a}+10^{-\mathrm{pH}}\\[6pt]
&=\frac{\left(10^{-\mathrm{pH}}\right)^2}{10^{-\mathrm pK_\mathrm a}}+10^{-\mathrm{pH}}\\[6pt]
&=10^{\mathrm pK_\mathrm a-2\mathrm{pH}}+10^{-\mathrm{pH}}
\end{align}$$
For example, for the given acetic acid solution with $\mathrm{pH}=2.4$:
$$\begin{align}
[\ce{CH3COOH}]_0&=10^{\mathrm pK_\mathrm a-2\mathrm{pH}}+10^{-\mathrm{pH}}\\[6pt]
[\ce{CH3COOH}]_0&=10^{4.756-2\times2.4}+10^{-2.4}\\[6pt]
&=0.907630545\\[6pt]
&\approx0.9
\end{align}$$
and for the desired acetic acid solution with $\mathrm{pH}=4.3$:
$$\begin{align}
[\ce{CH3COOH}]_0&=10^{\mathrm pK_\mathrm a-2\mathrm{pH}}+10^{-\mathrm{pH}}\\[6pt]
[\ce{CH3COOH}]_0&=10^{4.756-2\times4.3}+10^{-4.3}\\[6pt]
&=0.000193338\\[6pt]
&\approx0.0002
\end{align}$$
Finally, since
$$c_1\cdot V_1=c_2\cdot V_2$$
you may use the concentrations to calculate how much the given acetic acid solution with $\mathrm{pH}=2.4$ should be diluted in order to obtain the desired acetic acid solution with $\mathrm{pH}=4.3$.

authors: 37 กวิดา ธนิตย์จินดา


text:
Potassium carbonate and water/alcohol/toluene?
Recently I did a distillation. At a certain point an azeotropic mixture of alcohol, water and toluene evaporated and was caught in a beaker with anhydrous potassium carbonate. I had to shake it and then pour it back into the reaction mixture (while filtering out the solid).
Why did I have to catch it in a $\ce{K2CO3}$-mixture?
(The product was diethyladipate)

Was it simply a drying agent? I can't imagine it reacting with one of the solvents...

Welcome to Chemistry.SE! To acquaint yourself with this page, take the tour and visit the help center. Furthermore this tutorial shows you how math and chemical formulae can be nicely formatted on this site.

The synthesis you are referring to is outlined here. Potassium Carbonate acts as a drying agent as you said as the later part of the distillation requires a specific temperature to remove the toluene and alcohol.
The water comes from the concentrated sulphuric acid solution added with apidic acid. The residue is also rinsed with alcohol which is a strong indicator that water needs to be removed.


text:
Colloidal Silver?
There are a lot of tutorials on the net on how to make colloidal silver. Many of them basically go like this - Hook up two 99.99% silver coins or wires to three 9V batteries, and put the silver in saline, an inch apart, for a few minutes. Does this make real colloidal silver or silver ionic silver or silver chloride?
Here's an example:
http://www.instructables.com/id/How-to-make-COLLOIDAL-SILVER-Easy/

Welcome to Chemistry.SE! To acquaint yourself with this page, take the tour and visit the help center. Furthermore this tutorial shows you how math and chemical formulae can be nicely formatted on this site.

I hope you don't plan to ingest colloidal silver, it can turn you blue.


text:
Why HF plus H2O is not a buffer solution?
I think $\ce{HF}$ can be hydrolysis and get $\ce{F-}$ as a conjugate base, so this solution should be a buffer solution. But the correct answer is that this solution is not a buffer. Please explain it to me. I'm very confused.

What are you talking about?
$\ce{HF}$ in water solution is acting like moderately strong acid. It's because:
$$\ce{HF + H2O \rightleftharpoons H3O+ + F-}$$
$$\ce{H3O+ + F- \rightleftharpoons [H3O]F}$$
which is undissociated.Then:
$$\ce{F- + HF \rightleftharpoons HF2-}$$
As you can see the large affinity of $\ce{F^-}$ to $\ce{H^+}$ won't allow it to dissociate.
The $K_a(HF)(6.6x10^{-4})$ is strongly favoring reactants.
So the $\ce{HF}$ in $\ce{H2O}$ isn't buffer because there isn't almost any $\ce{F-}$ to be considered as conjugated acid. So if you add some acid there will not be enough $\ce{F-}$ for it's neutralisation.
So if you add some acid or base it will react with water and equilibrium constant will be changed and that will lead to drastic change of pH.
So the buffer is solution which consist of:

salt of strong acid and weak base and the weak base (for example $\ce{NH_{4}Cl}$ and $\ce{HN3}$ ).
salt of weak acid and strong base and the weak acit (for example  $\ce{CH3COONa}$ and $\ce{CH3COOH}$).

Hope i helped you. If you have any other questions please ask.


text:
How does dimethyltryptamine cross the blood-brain barrier?
Is there a specific chemical reaction that enables dimethyltryptamine to cross the capillary edothelial cells?

Although serotonin and dimethyltryptamine (DMT) have similar chemical structures (see Jori's answer), serotonin is not transported across the blood brain barrier (BBB) ("serotonin does not cross the blood–brain barrier").  On the other hand, not only does DMT cross the BBB, it is one of the few compounds that is actively transported across the BBB.  While most compounds that cross the BBB do so by being lipid soluble (relatively small non-polar compounds) some molecules are assisted across the barrier through a variety of mechanisms.  Some of the known assistive paths include

glucose carrier: transports glucose and several other sugars across the BBB
phenylalanine carrier: transports a number of neutral amino acids across the BBB
arginine carrier: transports several cationic amino acids across the BBB
lactate carrier: transports several monocarboxylic acids across the BBB
adenosine transporter: transports purine nucleosides as well as uridine
organic cation transporter: transports a variety of cationic compounds, including monoamine neurotransmitters

There are more classes and sub-classes within each.  The following two references provide nice, readable overviews

Small Molecular Drug Transfer across the Blood-Brain Barrier via Carrier-Mediated Transport Systems
Drug transport across the blood–brain barrier

I don't know which of the above transporters is active in the case of DMT, but I strongly suspect it is the "organic cation transporter".  Here are some further references that relate specifically to DMT active transport across the BBB

Takahashi T, Takahashi K, Ido T, Yanai K, Iwata R, Ishiwata K,
Nozoe S (1985) 11C-labeling of indolealkylamine alkaloids and
the comparative study of their tissue distributions. Int J Appl
Radiat Isot 36:965–969
Yanai K, Ido T, Ishiwata K, Hatazawa J, Takahashi T, Iwata R,
Matsuzawa T (1986) In vivo kinetics and displacement study of
a carbon-11-labeled hallucinogen, N, N-[11C]dimethyltryptamine.
Eur J Nucl Med 12:141–146

They were found in this article, "Bufotenine: Toward an Understanding
of Possible Psychoactive Mechanisms"
P.S.  We've all read fairy tales where kissing a frog turns the frog into a prince.  When the skin of the Bufo toad (Colorado river toad, sonoran desert toad and other members) is licked, a psychedelic experience results.  It is not illegal in the US to own the toad, although the Bufotenine is a controlled substance!  Anyhow, just wondering if this, or something similar, is what the fairy tales are based on.


text:
Dissociation of water into H+ and OH-: Does the law of mass action hold at nanoscale?
Suppose water under neutral conditions is confined in a virtual spherical nanocontainer with a radius of 25 nm. To calculate the number of hydronium ions, one uses water dissociation constant which is derived from the law of mass action. In this case, considering the temperature at 25°C I come up with a number which is smaller than one.

A logical suggestion here is probability. Is there any limitation to the law of mass action, i.e. does it break down at the nanoscale level and quantum mechanics and thus probability kick in? If yes, is there any boundary to specify things with respect to the size of the system?

There is definitely a breakdown of mass action at very low numbers of molecules.  This can happen either because concentrations are tiny, or because the system of interest is tiny.
In either case, your idea to turn to probability is an excellent one, and is how these systems are modeled.  But the reason doesn't necessarily have to do with quantum mechanics.  It's just a consequence that matter is not really a continuum that can have any concentration, but is in fact granular.  At the scale of atoms, there can either be an atom, or not be an atom.  You can't have 0.5 of an atom.
The right boundary at which mass action breaks down depends a bit on the particular system, in particular, if reactions are occurring that depend on H+.  In the less complicated case of no reactions depending on $\ce{H+}$, then the boundary between mass action being valid and not is actually fairly easy to calculate. If you know the average or expected value of $\ce{H+}$ ions in your water sphere (which you do, since you calculated it) the actually observed number of ions will follow a Poisson distrubtion.  If the mean number of atoms is $N$, then the standard deviation of atoms is $\sqrt{N}$, so the relative uncertainty in the mean number of atoms is proportional to $\frac{\sqrt{N}}{N}=\frac{1}{\sqrt{N}}$.  According to your calculation, $N$ is less than one, so the relative uncertainty is greater than 100%!  But if you enlarged the sphere so the expected number of ions was $N=10$, then the uncertainty would be about 32%, still pretty bad.  If you enlarged the sphere so $N=10000$, then the relative uncertainty is only 1%, meaning that mass action predictions for this system are pretty close to accurate!  Getting $N$ tht high requires a sphere about 10,000-times bigger in volume, which means the radius would have to be about 38 times bigger.  But keep in mind this result is only true for the water dissociation problem.  So each chemical system has its own scale where mass action breaks down.
For systems where reactions are occurring that depend on $\ce{H+}$, this simple logic breaks down.  Complex probabilistic simulations need to be done.  Depending on the details of the reactions, the chemical master equation or the chemical Langevin equation would have to be used.
One time I calculated the concentration of frogs in an aquarium was about 1 zeptomolar.  I don't think we could assume frogs followed the law of mass action!

Isn't Heisenberg's uncertainty principle stating the exact opposite of what you say in paragraph two?

Martin, I'm not sure what you mean.  Can you elaborate?

It may be worth saying that the limit at which deterministic kinetics represent the system reasonably well is called the Thermodynamic Limit. Also, van Kampen's System Size Expansion is a work worth citing.

A detailed analysis of the significance of the mass action law at the limit of very few molecules can be found in the book The Law of Mass Action. Section 3.7 ('Breakdown' of the law of mass action) on page 99 even specifically analyses the case of water dissociation in very small droplets, though the discussion and some of the results derived within are general.
It takes statistical thermodynamics to prove, but as Curt points out well, it's a consequence of molecules/ions coming in discrete (integer) amounts. We're used to doing calculations in systems involving many billions of species, so we can get away with treating the number of any species $A$ as continuous, even though it's an approximation; we unconsciously equate the most probable amount of $A$ with the expected amount of $A$.
One of the consequences of making the distinction clear is that systems with zero molecules of a reagent or product can still be in equilibrium, even though a naive application of the mass action law would result in undefined equilibrium constants. What differentiates then an equilibrium constant of $10^{100}$ from $10^{200}$ (assuming the same amount of species are involved) is how often the equilibrium can be found with zero molecules of the products; the higher the equilibrium constant, the less likely any given measurement will indicate the presence of at least one molecule of reactant.
This can be an unintuitive concept to grasp, and since its derivation is somewhat involved, it can be hard to explain. However, there is a simpler case which shows up in physics, based on the granularity of light (photons) rather than the granularity of matter (atoms), which is useful as a stepping stone. I present it below:

Imagine a point light source in a perfect vacuum with a power of $100\ W$, emitting light at a certain wavelength, say $500\ nm$. Via the formula $E=\frac{hc}{\lambda}$, it is possible to calculate that the light source is therefore emitting about $2.52 \times 10^{20}$ photons per second. As is well known, light intensity falls off with the square of the distance $r$ from the source, since the photons are spread out over a spherical shell with area $4\pi r^2$. Thus, for example, at a distance of $r=100\ m$ from the source, the $2.52 \times 10^{20}$ photons per second are spread out over an area of approximately $31416\ m^2$, which means the photon flux is that distance is $\Phi = \frac{2.52 \times 10^{20}\ photons\ s^{-1}}{4\pi (100\ m)^2} = 2.00\times 10^{15}\  photons\ m^{-2} s^{-1}$.
Now we place a perfect photon detector with a $1\ m^2$ cross section facing the light source. At a distance of $100\ m$, as previously calculated it will detect $8.02\times 10^{15}$ photons arriving in one second. The distance between the light source and detector is then increased greatly to $10^7\ km$. At this new distance, we calculate a photon flux of $\Phi = \frac{2.52 \times 10^{20}\ photons\ s^{-1}}{4\pi (10^{10}\ m)^2} = 0.200\ photons\ m^{-2} s^{-1}$. Thus, the detector, with a $1\ m^2$ cross section at $10^7\ km$ away from the source, will count 0.2 photons per second. There is no such thing as a fifth of a photon, so what does this number mean? Of course, this number is meant as a statistical time average. If the detector is left open for $10^5$ seconds, then we can expect it to detect $2\times 10^4$ photons.
But what happens if we force the detector to stay open for only one second at a time? Once again, there is no such thing as a fraction of a photon, so the detector must display an integer number of counts. How can the expected amount of 0.2 photons per second possibly be met in this case? Easy: sometimes the detector will measure photons, and sometimes it will not. In fact, if the test is repeated many times, we will find that the detector will count exactly 0 photons with 81.9% probability, exactly 1 photon with 16.4% probability, exactly 2 photons with 1.6% probability, and so on, such that the probabilities obey a Poisson distribution with an expected value of 0.2.
In much the same way as photons are discrete, molecules are also discrete. Thus, when a reaction at equilibrium contains less than one molecule of a given species, what is actually meant is that there will be times when the system will be measured to have zero molecules, and other times more than zero. The smaller the fraction of expected molecules, the less probable it is to measure the system while containing more than zero molecules of the species.


text:
Cubic lattice: How corner atom of a cube is shared by 8 atoms?

Each atom is at the corner of a cube and 8 cubes meet at each corner. Therefore each atom is shared by 8 cubes.

But I can't imagine it in space, so please tell how I visualize it.

As it stands, this is a geometry question. I assume this is related to a particular compound, probably ionic. Can you give an example compound?

you can consider any cubic compound.

I've answered your question, based on the assumption you mean simple cubic lattice. Please include "cubic lattice" in the title so others can benefit from it.

I'm voting to close this question as off-topic because it's more of a geometry question.

Surprisingly, this simple question does not seem to have been asked before. http://en.wikipedia.org/wiki/Bravais_lattice shows all the two dimensional Bravais lattices in extended form (but only the unit cell for the three dimensional ones.)
Here is the 2-dimensional image from wikipedia. As you can see in the simple square lattice, each sphere is a the corner of a square and 4 squares meet at each corner. Therefore each sphere is shared by 4 squares.

The simple cubic lattice is the same, except that the atoms are found at the corner of the cubic unit cell. Therefore each atom is shared by 8 unit cells. There are eight eighths (equal to one whole atom) in each unit cell.
I am not aware of any elements that crystallise in the simple cubic lattice. It is not energetically favoured because it is not spatially efficient. The more complex face-centred and body-centred lattices are preferred. Caesium Chloride, Bromide and Iodide form interpenetrating simple cubic lattices with the halide ion at the centre and the caesium ion at the corners of the unit cell (or vice versa depending on your viewpoint.)
Here are the relevant wikipedia links
http://en.wikipedia.org/wiki/Caesium_chloride
http://en.wikipedia.org/wiki/Caesium_bromide
http://en.wikipedia.org/wiki/Caesium_iodide

I am not aware of any elements that crystallise in the simple cubic lattice. It is rare but Polonium actually crystallizes in a simple cubic structure (according to  Wikipedia).

@Philipp that's bizarre, I didn't think that would be possible. It does say it's the only known example of a simple cubic structure on a single atom basis. The image on that article is actaully better than the one I have included. It also links to another page I didn't find before: http://en.wikipedia.org/wiki/Cubic_crystal_system

Simple cubic lattices are also in some compounds.  Cesium chloride looks like it's bcc because of its eightfold coordination, but the nonequivalence of alternating cesium and chlorine atoms renders it actually simple cubic.

Here let me plot that for you in Diamond (Software):

You might want to say "Diamond software" or some such thing, to avoid readers mistaking this for a (wrong) structure for a carbon allotrope.


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can i calculate equilibrium constant of this reaction?
Good day,
i had organic chemistry labs. And we had this reaction:
$$\ce{(CH3)3COH + HCl -> (CH3)3CCl + H2O}$$
My question is:
how do i explain my yield of reaction is only 40%?
My theory is:
reaction is reversible so the K isn't so high and that lead to yield 40%.
If my theory is right please explain me how to calculate K?
I had these informations:
V$\ce{((CH3)3COH)}$ = 10 ml
V$\ce{(HCl)}$ = 20 ml
HCl was concentrated.
Thank for answers.

Welcome to Chemistry.SE! To acquaint yourself with this page, take the tour and visit the help center. Furthermore this tutorial shows you how math and chemical formulae can be nicely formatted on this site. Finally, we have an important policy: your questions (especially homework questions), should show your own work or thinking that you have already done in an initial attempt to answer the question.

Information about the equilibrium constant can be found here.

$\ce{H2O}$ is a very poor nucleophile and we are dealing with a tertiary substituted carbon, so we can minimize the scope of possible reaction mechanisms to the $\ce{E 1}$ and $\ce{S_{n}1}$ type mechanisms. $\ce{E 1}$ will compete here with $\ce{S_{n}1}$, forming tert-Butyl chloride and isobutylene respectively. That is probably where most of your loss is coming from. Other reasons can be indeed that a bit is left in the equilibrium, but I don't think that would be much since $\ce{H3O+}$ is a much better leaving group than $\ce{Cl-}$. Note that is chemistry papers the yield obtained is almost never a 100% of theory.


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Why does HCl both precipitate and dissolve Silver?
Adding $\ce{HCl}$ to silver nitrate solution causes silver chloride to precipitate. Adding more $\ce{HCl}$ causes the silver to re-dissolve. What is happening to the silver chloride when it re-dissolves?

The precipitation reaction is:
$$\ce {Ag+ + Cl- -> AgCl_{(s)}}$$
The next reaction is:
$$\ce {AgCl_{(s)} +Cl- -> AgCl2- }$$

It might be useful to include the equilibrium constants for the two reactions: AgCl's k_sp is 1.77e-10 (so 5.6e9 for the equation as written), and the k_f for AgCl2- is 1.1e5.


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Difference between any arene and an aromatic compound?
I can't seem to find what the difference between these two is. In my text book, it says that 'arenes are aromatic hydrocarbons containing one or more benzene rings' which to me suggests that arenes are a subgroup of aromatic hydrocarbons and aromatic hydrocarbons need not contain one or more benzene rings. However later on in my book it says that 'any compound with the benzene ring is classified as an aromatic compound', which to me is the exact same definition as that given for an arene. Is there a difference between arenes and aromatic compounds?
Thank you in advance :)

There are non-hydrocarbon compounds (heterocycles) that are aromatic, such as pyridine, pyrrole and furan. It would not be correct to refer to these as arenes. Azulene (an aromatic hydrocarbon with a 5-membered and 7-membered ring) is a grey area for me.

Thanks! I think I get it- arenes are a subgroup of aromatic compounds which have a benzene ring. Aromatic compounds need not necessarily have a benzene ring

Arene — Compound which contains a benzene ring
Aromatic Compounds — Compounds having aroma (wait, the modern definition is different, I know)
Every arene is an aromatic compound but every aromatic compound need not be an arene.
Mathematically,
$$\text{arene} \subset \text{aromatic compounds}$$
therefore anything which contains a benzene ring has to be aromatic
Some examples of arenes:

Getting into aromatic compounds, the modern criteria for aromaticity are:

The molecule is cyclic (a ring of atoms)
The molecule is planar (all atoms in the molecule lie in the same plane)
The molecule is fully conjugated (p orbitals at every atom in the ring)
The molecule has $4n+2$ π electrons ($n=0$ or any positive integer)*

Examples (non-benzenoids):

Now, benzenoids (compounds which contain benzene) do match with this. Therefore, benzene derivatives are aromatic compounds.


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Mechanism for reaction between aluminium and iodine
What should be a suitable mechanism for the reaction between aluminium and iodine, catalysed by water?
In my initial approaches, I had first thought of repeated attacks of the lone pairs of oxygen in water on aluminium, then formation of $\ce{Al(OH)3}$ by losing $\ce{H+}$, then $\ce{H+}$ attacking $\ce{I2}$, and formation of $\ce{HI}$, and finally formation of $\ce{AlI3}$.
But actually $\ce{Al2I6}$ is formed, so how can I accommodate its formation in that?
Is anything wrong in the above suggested mechanism? If the actual mechanism is something else, please write it in your answer.
A Lewis dot structure of the mechanism would be appreciated.

There are a lot of problems in your proposed mechanism, which you would notice if you actually wrote the equations out.

The biggest problem is of course, my initially thought mechanism does not give me $\ce{Al2I6}$ , and that is why I asked this question.

$\ce{Al2I6}$ is just the dimer of $\ce{AlI3}$; if you generate the latter in a reaction equation you can stop there (potentially add $\ce{2 AlI3 <=>> Al2I6}$ at the end)

How is this a homework question? Why is this put on hold even though I have literally explained my initial approach?

@Jan I don't see the reason this would just pop up in the end. Mechanisms usually incorporate the step of instability which leads up to dimerazation

@Rohinb97 In a nutshell, $\ce{AlI3}$’s aluminium is not coordinatively saturated (i.e. could take one more ligand). Iodine is happy to supply that missing saturation and the resulting structure is a double tetrahedron connected along a side with two iodine bridging between the two aluminiums.


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How does B2Cl4 have two perpendicular c2' axes?
I have been looking at the symmetry of $\ce{B2Cl4}$ and was cannot understand how it has two perpendicular $C_2'$ axes?
I understand it has a $C_2(z)$, $S_4(z)$ and two dihedral planes bisecting each of the $\ce{Cl-B-Cl}$ bonds. However my book says it also has an additional two $C_2'$ axes in the $x$ and $y$ planes centred at the middle of the $\ce{B-B}$ bond. I just can't see it. Any help? ( I understand it will be difficult online but I would very much appreciate your assistance.

Diboron tetrachloride has the same symmetry as allene, which I will use to demonstrate the other two $\ce{C_2^{'}}$ axis. The back $\ce{CH_2}$ group is turned 90° in comparison to the front $\ce{CH_2}$ group. This conformation is most stable as it minimizes steric interactions between both groups.
In this case, a picture tells more than I thousand words I think.

The two $\ce{C_2^{'}}$ axis are turned by 45° in comparison to the C-H bonds. It therefore also has two dihedral planes, thus belonging to the $\ce{D_{nd}}$ point group.
If you have difficulties picturing this in your head that is not strange. Many people make mistakes in assigning point groups. You get a better visualization by practicing online.


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chlorine formed in hydrochloric acid
Is it plausible that free chlorine could be formed in concentrated hydrochloric acid.
The background to this is that I have bought some 30% HCl to remove calcium carbonate scale from a reverse osmosis water purifier. The latest batch of HCl seems to have a very slight yellowish tint which I had not noticed on previous occasions. Having used some of this acid I find that the performance of the membrane seems to have deteriorated (these membranes are notably sensitive to hypochlorite).
Could some factor (perhaps exposure to direct sunlight) be responsible?
Is there a way that I could detect the presence of dissolved chlorine?
Can anyone suggest an alternative explanation?
Thanks,
Chris.

You could buy the chlorine test strips from a pool supply store to test for hypochlorite. You may need to neutralize the acid before testing, so ask if the test is accurate in a strong acidic solution.

The color could be due to other impurities, such as iron. That might also be harsh on delicate RO membranes, perhaps by precipitating a floc on the membrane.
If it is iron, neutralizing with clear alkali (NaOH or KOH) to pH 6-7 would precipitate a floc. This floc might be more difficult to see than the off-color, so try a Tyndall test. Take a flashlight with a very narrow beam of white light and shine it thru the neutralized solution. If the beam lights up like a beam of light in a smoke-filled room, you have a precipitate. There are inexpensive LED flashlights that are perfect for this. Red lasers might not show the scattered beam as well.
I have to warn you - after you see the Tyndall effect in an otherwise clear solution, you'll want to test everything you come across!


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Do disaccharides undergo anomerization?
I am interested in disaccarides of D-glucose linked 1-4 style. Anomerization is the interconversion of alpha-D-glucose into beta-D-glucose. I realize that this could only be possible on one of the glucose sub-units: The other glucose is obviously unable to inter-convert due to the glycosidic bond preventing ring-opening.

I do not see what would prevent it:

Not only is there no reason to think anomerization does not occur in the disaccharide you mention, it most definitely does occur.
As long as water has access to carbon-1 of a glucose unit, there will be an open-chain form at that site, and then when it re-closes, the possibility of alpha- and beta-anomers.  This will be the case for di-, tri- tetra-, oligosaccharides, polysaccharides, etc... and applies to monosaccharides of any carbon length or type (aldoses or ketoses).
The reason is that hemi-acetals or hemi-ketals are intrinsically unstable in water. By contrast, when that carbon is a full acetal or full ketal (a so-called non-reducing sugar) then it is stable in water, and anomerization does not occur.  Examples include trehalose and sucrose.

There is no reason to think that anomerization on the terminal glucose unit does not occur. Although there may be a slight difference in the equilibrium constant between a monosaccharide and a disaccharide due to the change in environment.


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Hexafluorosilicid acid attacks glass?
When HF meets glass, Hexafluorosilic acid is formed (assuming some water is present). It appears that Hexafluorosilic acid and further attack glass? How is this so, considering that all F atoms are "used up" (already bonded to Si atoms)?
http://www.cdc.gov/niosh/ipcsneng/neng1233.html

$\ce{H2SiF6}$ can decompose into HF in the presence of water fairly easily.  I wonder if HF is acting almost as a catalyst, with the fluoride chipping away at the glass, then breaking off, then chipping again, and so on.  If the hexafluorosilicic acid were put into water and started to decompose, it ought to provide a source HF to attack the glass, in the right environment.

So HF+SiO2 -> H2O*H2SiF6 (or something like that) is reversable?

Glass isn't exactly $\ce{SiO2}$, but that was my idea.  It would effectively be converting glass (big messy molecule that it is) into independent molecules of silicon dioxide.  To be clear, that was just my best guess.  I've never worked with the stuff and didn't find a mechanism for how it attacks glass when I looked.

@Jason: Glass dissolves (your term "independent molecules") up to 100ppm in water but could be more soluble in acid (but HCL would not attack glass due to the slow kinetics) Keep in mind "dissolve" actually indicates a reversable reaction, so it m

may need a catalyst to proceed to this equilibrium at a reasonable rate (they used finely divided particles). The fluoride probably increases the solubility.

Actually, Si-Cl bonds are unfavorible so HCl won't attack glass at all. However, Si-F bonds are not as unfavorable (compared H-F bonds).

yes, Jason Patterson might be right. $\ce{H2SiF6}$ decomposes to HF and  is further chips the glass. This whole process occurs spontaneously or within a short period of time. Because HF is concentrated, it almost readily form $\ce{H2SiF6}$ and whole process favours forward reaction and so the reaction is not reversible. btw $\ce{H2SiF6}$ is hexafluorosilicic acid.

@Jason If you let 70% HF eat as much glass as it can, then evaporated some HF (keeping [H2O] the same), the SiO2 would get super-saturated. Could this be used as a bath to coat things in glass/quartz?

@KevinKostlan It might work, but it seems like a fairly hazardous method, plus you'd only be able to coat things that were impervious to HF.  If you've never played with it, I would suggest buying a small quantity of sodium silicate (water glass.)  I've used it a number of times with it resulting in a glass-like coating on substances after it dries.  It would take some time to get the technique down, but it might be what you want.  It's also very safe if used with even a small amount of common sense.


text:
General methods of Analysis for FTIR Results
What can we say about a molecule from the position and number of bands in FTIR spectrum. I know that the more symmetric a molecule is, the less number of bands it has. What else can we say, for example about the position of the band? Does it show the frequency of vibrations and therefore the strength of the bonds?
For example, the spectrum of MoF4 has one band at 674 cm -1. XeF4 has 3 bands at 586, 291, and 161. How can we compare the structure of these materials from the above information?

I understand that this link has almost everything in detail for XeF4 : http://www.users.csbsju.edu/~frioux/group/XeF4.pdf  ; My question is more on how to differentiate between the two compounds by checking the IR bands, since one may expect the structures to be close. The MoF4 is a bit shaky though, I am not sure if the compound exist in stable form.

IR is not a good spectroscopy for differentiation. What I mean is that, if you have two samples of which you know nothing at all, the experiment you should run is not IR.
IR gives you some additional information once you already know a lot about your powder.
However, in your case, knowing the formulas and the structures, you can expect MoF4 to show higher frequencies in comparison to XeF4 due to the mass effect I discussed shortly in my answer.

Let me start by saying that the XeF$_4$ is a standard example of group theory applied to IR, which is generally treated in Uni courses, for which you can find plenty of notes on the web. For these reasons, I will not explain it, but you can rather put a bit of effort and read it yourself.
In general, the level of information you can extract depends on the nature of the molecules you analyse. For example, for organic molecules, the analysis is qualitative and limited to functional group identification. This is because usually they have a high number of atoms, thus there will be many vibrations. Since the chemical composition is quite homogeneous, all these vibrations have similar frequencies, and they overlap. Thus a quantitative analysis is difficult, and the only information you extract concerns some particular bond whose signal is isolated from the mess (usually carbonylic and carboxylic bonds, triple bonds, X-H bonds). An additional problem lies in conformational flexibility.
Now, for molecules like yours, which are small, have few vibrations, are symmetric etc, you can achieve a very detailed level of analysis. But, as I said, I will not discuss it, since you can read it yourself with a quick google search.
The basic thing to keep in mind is the frequency of oscillation of a classical harmonic oscillator, given by
$$ \omega = \sqrt{\frac{k}{\mu}} \\
\mu = \frac{m_1 m_2}{m_1 + m_2}$$
where $k$ is the force constant of the bond, and $\mu$ is its reduced mass.
So, what this tells you is that stronger bonds vibrate at a higher frequency. It also tells you that bonds involving lighter atoms vibrate at a higher frequency. Another thing that you can deduce is that bending frequencies are lower than stretching frequencies.
So, to give you a hint, what you can say immediately by looking at those frequencies you reported is that the 586 cm$^{-1}$ vibration is a stretching.


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What is the difference between organosilyl, alkoxysilane and silane compounds?
What is difference between organosilyl, alkoxysilane and silanes?
How do these groups make a surface hydrophobic?

Silane is the equivalent of methane for silicon, $\ce{SiH4}$.
Silicon containing organic derivatives of many kinds exist.
Generally speaking, organosilanes are those in which $\ce{Si-C}$ bonds exist, like for example, tetramethylsilane ($\ce{TMS}$, used as a reference on $\ce{{}^{1}H~NMR}$).
Alkoxysilanes are those compounds that contains an organic group bonded to a silicon but not directly, through an oxygen atom instead, like so: $\ce{R-OSi(Me)3}$, common structure on silyl enol ethers.
A surface is hydrophobic if it's repulsions towards water overcome its attractions, and this is the case for many silicon compounds, but not all of them. For example, glass (silicon dioxide, $\ce{SiO2}$) is hydrophilic, due to the possibility of formation of hydrogen bonds between the glass oxygen and water molecules.

Welcome the Chemistry.se! Since you are quite new to the site, I thought you might be interested in this meta post on how to use math/chemistry here. I have updated your post with this markup.


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What does 0.12 N mean on a bottle of sulfuric acid?
I just bought a bottle of sulfuric acid in order to simulate acid rain for an experiment. I was wondering what $0.12\:\mathrm{N}$ on the bottle means. I need a simplified version of the explanation as I'm currently a freshman in high school taking biology, I have never taken a chemistry course before. I've read that the $\mathrm{N}$ means normal, but the explanation included a ton of jargon that I couldn't understand.
The time in question: does the $0.12\:\mathrm{N}$ mean concentration, as in $12\%$ concentration?
How would you guys suggest making simulated acid rain with this? My guess is that I would keep titrating the sulfuric acid into the water until the $\mathrm{pH}$ gets to my preferred $\mathrm{pH}$.

0.12N is indeed 0.12 normal. In short, there are 0.12 moles of hydrogen ions per liter. Since this is sulfuric acid, it is 0.06 moles of sulfuric acid per liter.
0.12 moles of hydrogen ions per liter would give a pH just less than 1. To obtain a pH near 3, you would need to dilute 1 mL of this solution to 100 mL. For a pH of 4, dilute 1 mL to 1000 mL.

Perfect thank you! I also appreciate you giving me an estimate on how much acid I should use to get a certain pH.

density ( g/L) x % by mass / 100 = g/L of H2SO4
g/L / 98 g/mol = molarity
N = Molarity x 2
I have calculated grams of pure H2SO4 in 1L .12N solution:
Molarity = .06
g/L = .06 x 98 = 5.88 g/L
And then calculated g/L in the 37% solution:
1280 x 37 / 100 = 473.6 g/L
Then calculating how many mL of 37% solution would contain 5.88g:
1000 / (473.6 / 5.88) = 12.42 mL


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Why is iron used in the body?
I'm writing a report on the spectrophotometric determination of iron. In my introduction I want to briefly explain the importance of iron in the human body.
Wikipedia states that "Iron is essential to life due to its unusual flexibility to serve as both an electron donor and acceptor"
1.) So is iron the only element that can do that? If not, what is it about iron that makes it the only element found in hemoglobin?
2.) What is it about being able to gain and lose electrons make it essential to life? I think it has something to do with oxygen being able to bind with iron. But there are so many elements that can bind with iron! What is special about iron?

1.) So is iron the only element that can do that? If not, what is it about iron that makes it the only element found in hemoglobin?

No, iron is not the only metal that will fit inside the porphyrin structure. But its ability to be an electron acceptor and donor makes it very useful for oxygen and carbon dioxide to bind to the hemoglobin. Since iron is also very abundant, it was easy to incorporate it and utilize its properties.
Other macrocyclic structures include chlorophyll with magnesium in the center and vitamin B12 (cobalamin) with cobalt in the center.

2.) What is it about being able to gain and lose electrons make it essential to life? I think it has something to do with oxygen being able to bind with iron. ... What is special about iron?

Iron is abundant. The iron in the hemoglobin can weakly bind both oxygen and carbon dioxide.
You can find more information on the Wikipedia page.

Not sure carbon dioxide binds to hemoglobin.  Crummy ligand for that.  Carbon _mon_oxide binds to hemoglobin (big problem) through its exposed carbon atom, but AFAIK carbon _di_oxide just ends up in the carbonic acid/bicarbonate buffer.

@OscarLanzi IMO I believe that the lungs expel more carbon dioxide than what would defuse from a carbonate buffer.

I would like to clarify this Q & A.
Both Fe(II) and Fe(III) are found in metalloproteins that are essential to life. Fe(II) can serve as an electron donor, and Fe(III) an electron acceptor in life.

"So is iron the only element that can do that?" No, for electron transfer, many biological transition metals (Co, Ni, Cu...) can also participate in redox reactions in life (but not Mg(II) in chlorophyll)
However, Fe(II) in hemoglobin is not a good example for "electrons donor/acceptor". Although the oxidative state of Fe(II) in heme changes slightly when binding to O2 or CO, it never oxidizes to Fe(III). (If it does, it loses its ability to bind O2). The ferrous ion in hemoglobin is important in oxygen transport.
"So is iron the only element that can do that?" No, for oxygen transport, copper ion in haemocyanin can also transport oxygen in the blood of spiders, crabs and octopuses. When oxygenated, they are blue.
"What is it about being able to gain and lose electrons make it essential to life?" Because biological redox reactions (e.g. cellular respiration) are essential to life. Each biological metal is important in its own specific biological function. Their roles are the product of evolution. For oxygen transport, neither heme nor iron is indispensable for that function in other life forms.
Iron is abundant in earth crust mainly as iron(III) oxides. However, iron is NOT abundant in bioavailability. Ferric and Ferrous hydroxides are so poorly soluble in the environment that many species have developed sophisticated iron acquisition systems such as siderophore to fight for this "scarce" element as an essential nutrient.

Great answers, but I would like to add one additional use of iron at the cellular level.
In the electron-transport-chain (ETC), within the mitochondria of each cell, are Iron-Sulfur clusters that are responsible for the oxidation of carbohydrates in aerobic respiration.
If you were to use search-terms, "ETC and citric acid cycle respiratory complexes", you should be able to find lots of info on this.


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Reactions that occur at the anode electrode of electrolysis dilute H2SO4
How to know that reaction that occur at the anode electrode is $\ce{2H2O -> 4H+ + O2}$?
Why the reaction is not the oxidation of  $\ce{SO4^{2-}}$?
This problem does not give the E0 of $\ce{SO4^{2-}}$ too.

How to know that reaction that occur at the anode electrode ...

There are two kinds of anions: oxidizable (ones that contain central element in lower oxidation states), like $\ce{ClO3-}$, $\ce{Cl-}$, $\ce{NO2-}$, and unoxidaziable, where central element is in hightest oxidation state ($\ce{ClO4-}$, $\ce{NO3-}$. Fluoride obviously is unoxidizable as well.  As a rule of thumb, it is usually assumed that unoxidizable anions do not react in electrolysis. A rationalization for it is that in such cases an oxygen should be oxidized, and oxygen in water reacts faster. For the sake of basic chemistry course it is enough.
However, it is not entirely true: electrolysis of cold solution of ammonium sulfate in sulfuric acid results in formation of ammonium persulfate $\ce{(NH4)2S2O8}$, where anion has structure $\ce{ (O3SOOSO3)^{2-}}$ Similiarly, electrolysis of very cold solutions of carbonates produces various percarbonates. A synthetic method exists, that is based on electrolysis of carboxylic acid. The anions on oxidation form $\ce{RCOO\cdot}$, that further loose $\ce{CO2}$ and resulting radicals pairs, resulting in net reaction $\ce{2RCOO- \rightarrow R2 + 2CO2 + 2 e-}$ However, in general, formation of peroxocompounds requires low temperatures, usually below zero.
The exact mechanism of the process highly depends on the nature of the electrodes, but likely includes formation and transformation of surface particles on the electrode, making reasoning difficult and electrode-specific. Since commonly anodes used have oxide coating, it can be more-or-less safely assumed, that the process includes dissociative adsorption of water with subsequent oxidation of adsorbed (hydro)oxide atoms.


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Alkane nomenclature - choosing the parent chain - maximize or minimize substituents?
Given structure:

Here's my go at it. The longest carbon chain has 10 Carbon atoms, so the root -- decane. Next: let's take the substituent in position 3, we get 3 ethyl. And finally, we get 5-(2, 2-Dimethylbutyl). Adding them all up:
5-(2, 2-Dimethylbutyl)-3-ethyldecane.
However, the solution in my book is given as follows:

And the answer given is: 5-(2-Ethylbutyl)-3, 3-dimethyldecane
What I don't get is: Why is the main chain taken by me incorrect?

The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-44 SENIORITY ORDER FOR PARENT STRUCTURES
A parent structure is defined (P-15.1) as a parent hydride, for example methane, a functionalized parent hydride, for example, cyclohexanol, or a functional parent compound, for example, acetic acid. For choice of the preferred IUPAC name based on a senior parent structure, see P-45.
(…)
P-44.3 SENIORITY OF ACYCLIC CHAINS (THE PRINCIPAL CHAIN)
In an acyclic compound consisting of individual atoms, alike or different (an acyclic chain), the chain on which the nomenclature and numbering is based is called the 'principal chain'. When there is a choice for the principal chain, the following criteria are applied, in the order listed, until a decision is reached.
The principal chain:
(…)
(b) has the greater number of skeletal atoms;
(…)

P-45 SELECTION OF PREFERRED IUPAC NAMES
P-45.0 INTRODUCTION
Two or more names may result based on the same senior parent structure selected according to P-44, because of different substitution patterns or multiple occurrences of the same senior parent structure. (…) Application of
the criteria in this section will generate the preferred IUPAC name.
(…)
P-45.2.1 The preferred IUPAC name is based on the senior parent structure that has
the maximum number of substituents cited as prefixes (other than 'hydro/dehydro') to
the parent structure.

In accordance with P-44.3 (b), the principal chain is a decane.
It could be either “5-(2-ethylbutyl)-3,3-dimethyldecane” or “5-(2,2-dimethylbutyl)-3-ethyldecane”.
Because of P-45.2.1, the preferred principal chain corresponds to 5-(2-ethylbutyl)-3,3-dimethyldecane, which has a principal chain with three substituents (5-, 3-, and 3-), whereas the principal chain of “5-(2,2-dimethylbutyl)-3-ethyl” has only two substituents (5- and 3-).

The simple rule is that if there are two chains of equal sizes, then the chain with more number of side chains is to be selected as part of the parent chain...
Here, the two chain are $\ce{-CH2CH(CH2CH3)CH2CH3}$ and $\ce{-CH2C(CH3)2CH2CH3}$ .
The second one has two side chains ($\ce{-H3C}$ and $\ce{-CH3}$) while the first one has only one side chain ($\ce{-CH2CH3}$). Therefore it is the second chain that is selected as part of the parent chain. So, the answer is 5-(2-ethylbutyl)-3,3-dimethyldecane [and not 5-(2,2-dimethylbutyl)-3-ethyldecane].
(also note: the chain which is not selected is numbered such that the carbon atom attached to the parent chain has the lowest number, therefore it is 2-ethylbutyl as $\ce{-CH2CH3}$ is present on the second position of the carbon side-chain.)


text:
Why is the structure of 3, 4-Dimethylhexane given as follows?
Question: name the IUPAC name of the following structure:

And here's what I could come up with is this: we start from the first carbon atom from the extreme left and end with the 2nd substitute while moving right (considering the first C atom at the extreme left to be in 1st position). After that, we have the first substituent in the 2nd position and we get 2 ethyl and then we have '3 methyl' and the root will be pentane. So, we get 2-ethyl, 2-methyl pentane.
However, here's the answer given in my book 3, 4-Dimethylhexane. Any idea where I've gone wrong?

Yes, you started at the wrong atom. You should start on the ethyl group since it will give you a longer chain.


text:
Why are acetals stable to bases and nucleophiles?

Why are acetals stable to bases and nucleophiles?

Could it be due to electronic effects? Lone pairs on the two oxygen atoms create an unfavorable environment for the approaching nucleophiles and bases. Or could it also be steric factors?

Keep in mind that in these reversible reactions of carbonyl compounds, the mechanism for the forward and reverse reactions must go through the same intermediates. The result is that if you can't make a particular functional group under particular conditions, that functional group can't be destroyed under the same conditions.
Hydrates (as shown in the second reaction in ron's answer) and hemiacetals can be formed/decomposed under either acidic or basic conditions. In the hemiacetal formation under basic conditions, an alkoxide (methoxide below) attacks the carbonyl. Protonation gives the hemiacetal. In order to proceed to the acetal, the alkoxide would have to displace the hydroxide in an SN2 reaction. Hydroxide is not a good enough leaving group for SN2 reaction, so it is impossible to form the acetal under basic/nucleophilic conditions.

Hemiacetals can be decomposed to a carbonyl and alcohol under basic conditions. Deprotonation of the hydroxyl starts the mechanism, and the resulting anionic oxygen can kick out an alkoxide as a leaving group to give the carbonyl compound. Note that in this carbonyl chemistry hydroxide and alkoxides are competent leaving groups in an anionic elimination mechanism, such as the one here.

In an acetal, without the free hydroxyl group, alkoxide/hydroxide has nothing to deprotonate and there are no suitable leaving groups for SN2 reaction. There are no reasonable pathways for a nucleophile or base to react with the acetal.

I have two questions to ask which came to my mind as I read your answer: first, what makes an alkoxide a competent leaving group in this case? Is it due to the basic medium around it? And second, if an alkoxide ion is capable of acting as a leaving group - why can't a hydroxyl ion simply attack an acetal C and replace the alkoxide to form a hemiacetal in an SN2 reaction - this can be repeated to finally form a ketone.

@Charles very good question, which merits a new question. If you ask it, I'd be happy to answer and you'll probably get other useful responses.

@jerepierre Has the question been asked eventually?

I haven't seen the question asked.

Related: https://chemistry.stackexchange.com/questions/41331/why-does-a-carbonyl-reform-in-ester-reactions

In both acid-catalyzed

and base-catalyzed acetal formation

the first step involves the alcohol (or alkoxide) behaving as a nucleophile and attacking the electrophilic carbonyl carbon.  At the end of the sequence, when the acetal has been formed, the nucleophilic alcohol has been transformed into an ether, but the nucleophilic center, the "alcoholic"  oxygen, remains intact.
Where would a second nucleophile attack?  The second nucleophile would be electrostatically repulsed by the still nucleophilic acetal oxygen.  It could attack an adjacent carbon in an $\ce{S_{N}2}$ process, but under basic conditions the only possible leaving group, alkoxide, is not a very favorable leaving group.

Ron has highlighted the presence of nucleophilic acetal oxygens which "electrostatically repulse" attacking nucleophiles while jerepierre has pointed out the absence of good leaving groups on the acetal. Both points seem reasonable. However, I would like to add an additional point based on stereoelectronic effects. This is similar to the effects responsible for the inertness of dichloromethane and chloroform towards $\ce {S_N2}$ reactions, as discussed here.
There are stereoelectronic effects at play in acetals, as illustrated on p. 804 of Clayden, Warren & Greeves (2012). These effects was mentioned to explain the gauche conformation of the acetal. However, I believe the reasoning can be extended to explain the relatively low reactivity of acetals towards nucleophiles. The nucleophile's HOMO would seek to interact with the acetal's LUMO, which is its $\ce {C-O}$ $\sigma^*$ MO. Due to the presence of the interaction between the lone pair orbitals on the acetal oxygens with this LUMO, the molecule's LUMO is raised in energy, making it more energetically inaccessible for the nucleophile to attack it. Thus, nucleophilic attack at the acetal carbon occurs much less readily.

Reference
Clayden, J.; Greeves; N.; Warren, S. Organic Chemistry (2nd ed.). New York : Oxford University Press Inc., 2012.

authors: Αντώνιος Κελεσίδης


text:
What is the oxidation number of the oxygen in BaO2
According to the book for a binary compound, first we assign the element with greater electronegativity its oxidation number (oxygen always -2 except in peroxides). So, we have
-2 x2 =-2. So, each oxygen would have an oxidation number of -2. The book says the answer is -1.
The only thing I can think of, is since barium is an alkaline metal, it has an oxidation number of +2, so each oxygen would be -1. But, it contradicts itself because in the rules. It says for a binary compound, "the element with greater electronegativity is assigned a negative oxidation number equal to its charge in simple ionic compounds of the element."

Did you look on the web for barium dioxide to find information about the bonding?

In compounds, barium is universally encountered as Ba(II).  I'm unaware of any compounds where barium has any other oxidation state.

"oxygen always -2 except in peroxides"

This exception is in effect here, because $\ce{BaO2}$ is a peroxide consisting of $\ce{Ba^{2+}}$ and $\ce{O2^{2-}}$ ions. The oxidation number of Ba is +II, and the oxidation number of each of the oxygens in the peroxide anion is -I. This fits with the charge of the peroxide anion ($2 \times -1 = -2$), and as $\ce{BaO2}$ is a neutral compound, the sum of all oxidation numbers is 0. Moreover, there are even more exceptions to the rule of thumb cited above, for example, the superoxide radical anion $\ce{O2-}$ with a fractional oxidation numer of $-\frac{1}{2}$, or the dioxygenyl cation $\ce{O2+}$ with a formal oxidation number of $+\frac{1}{2}$ for each oxygen.
In contrast to peroxides, "simple ionic compounds" of oxygen are oxides, which contain the anion $\ce{O^{2-}}$. The charge of the oxide anion is $2-$, so the oxidation number of oxygen in those compounds is -II.

I think we also have to account for the negative half oxidation number in case of super oxides in the general law at the head of your answer.

@SatwikPasani I have edited my answer to take this aspect into account.

Oxidation numbers are an extremely useful form of chemical accounting, but should not be considered as having any actual reality. For instance fractional oxidation numbers are perfectly possible, but you cannot have half an electron. Anoxidation number of +7 is assigned to chlorine in perchlorates, but the idea that you could possibly strip 7 electrons off a chlorine atom is unthinkable (add up the first 7 ionisation potentials). It is the second part of the quote that you give is misleading and in many cases incorrect. The point to be born in mind first is that the oxidation number of a given element in a compound is the charge it would have IF the compound was entirely ionic. The sign of the charge is based on relative electronegativity of the bonding atoms. Group I and II metals (except in very rare cases) are assigned charges of +1 and +2 respectively. Oxygen is most commonly -2, but in the case of peroxides it is assigned an oxidation number of -1 in order that the combined oxidation numbers come to zero in this case or to the total charge on the group of atoms if they happened to form an ion.


text:
ICP-OES data treatment: correct for dilution and inweight, the units
Let's say the ICP-OES software gives me a concentration of 186,49 ppm (mg/L) of Mn. I diluted the acid digested sample to 14.2 mL. I weighed in 10.17 mg of the powder I analyzed. I need to calculate the actual concentration.
It is quite logical in my head what I need to do: multiply by diluted volume and divide by inweight. However, I don't understand my units.
When I multiply 186.49 mg/L with 0.0142 L, I get 22,648 mg. When I then divide this by 10.17 mg, I end up with 0.260. But this is unitless. How do I know what unit I have? I thought I converted the w/V concentration to w/w by this procedure, but the units does not follow.
Is it so, that I end up with a ratio (not sure what kind of ratio)? Then I think I understand. 0.26 would then represent 26 %, meaning my sample contains 26 % Mn. But how do I know that this is in w/w, and not w/V as I started with. I just don't get the units to add up.

I think you have a typo in your calculation (also, using both dots and commas as the decimal point is rather confusing), but you're on the right track. The unfortunate thing about elemental analysis is that %, ppm, ppb, ppt, etc. get misused in a confusing way. These units are supposed to be used for only dimensionless ratios, the units you start with, ppm is actually mg/L which represents mass concentration, not mass ratio. Your calculations are correct:
$$\left(186.49 \frac{\mathrm{mg}}{\mathrm{L}}\right)\left(0.0142\mathrm{\ L}\right) = 2.648 \mathrm{\ mg}$$
gives the mass of the Mn in your sample, so when you divide by the sample mass:
$$\frac{2.648 \mathrm{\ mg}}{10.17\mathrm{\ mg}}=0.260$$
you get the mass fraction of Mn in your sample, and you know it's a mass fraction because you have mass units in both the numerator and the denominator, leaving you dimensionless. You can express it just like that, or you can convert it to 26.0% (and this is a dimensionless ratio, so it actually makes sense to use % here), or else express it as the mass of Mn in an arbitrary sample mass: 26.0 g/100 g.

Thanks a lot! So, if I so wanted, I could write 260 promille, or 260000 ppm Mn?
I will second the confusing use of ppm. The instrument gives me ppm, but it is the actual units I am interested in. ppm, ppb, etc are fine for reporting LOD's and LOQ's. Thanks again, Mr. Dryden.

Yeah, you can use ppm if you want, though it's ambiguous and discouraged because it doesn't say whether you're using mass fraction, volume fraction, or mole fraction. mg/kg or the like is more explicit.


text:
Painting metal, does it matter if all rust is removed before?
In the context of automobile repair work, assume a metal surface has rust on it. Does it matter if all of the rust is removed before the metal surface is painted? If it does matter, then why?
It appears that at least some will claim you can do this, e.g. the company Krylon.
How I've tried to figure this out myself:
I'm assuming that the rust-covered surface area will affect how well the paint will stick to the surface. Maybe this alone can be a problem in terms of durability and sealing the metal from the atmosphere?
If it is governed by a more complex process, what process(es) should be read into further? However, I also assume that (as the aforementioned link suggests) that making the surface as smooth as possible with loose rust and debris removed will yield better results -- i.e. results that approximate that of having no rust at all there? Will it never be as good as simply having all rust removed? I know I'm asking in a way that is difficult to quantify, but "how good" will it get, compared to having all the rust removed?
After studying the Wikipedia article on rust I also imagine that from the dehydration equilibria we have that even a small area of rust can:
$$\ce{Fe(OH)2 <=> FeO + H2O}$$
Even assuming the rust is properly sealed from the atmosphere it appears it can create new $\ce{H2O}$ under this protective layer.
But we need oxygen for the key reaction to occur:
$$\ce{O2 + 4e^- + 2H2O -> 4OH^-}$$
From this how stuff works I have:

As the acid is formed and the iron dissolved, some of the water will begin to break down into its component pieces -- hydrogen and oxygen.

How exactly is the water broken down into hydrogen and oxygen? From a chemistry textbook (Masterton & Hurley) I have:

For example, water does not spontaneously decompose to the elements by the reverse of the reaction referred to above
  $$\ce{2H2O(l) -> 2H2(g) + O2(g)}$$ nonspontaneous.

But electrolysis can certainly create these elements. But can it happen in this environment? Sealed under a surface of paint? What does the chemistry look like?
From chemical formula I have that we can get hydroxide from:
$$\ce{4e^- + 2H2O(l) + O2(g) -> 4OH^-(aq)}$$
However still, from where does new oxygen get introduced into the system, assuming the atmosphere is actually completely isolated from this internal system taking place under the surface of the paint?
Should it matter, assume the environment is approximately 1-atmosphere pressure and that the temperature varies between -10 to 40 degrees Celsius.

I have been removing most of the rust with grinding and wire brush, there are still small areas of dark color tightly adhered corrosion. I wash with soapy water, hot if possible, rinse, dry and paint. I have seen soapy water recommended by several different paint companies. It seems like the soap results in some kind of almost protective coating on the steel so it does not rust as it dries. I have always assumed that if the remaining rust is thin tightly adhered spots the paint will stick to it. I have also assumed that the rust itself is relatively inert and will not spread or further degrade

[con't] unless more moisture gets thru the paint to it. Clearly this is not as good as sandblasting to 100% white metal but it has worked for me. I have always been a little suspicious of claims of converting rust to an inert layer when it seems like it is already pretty inert?

When I worked on an offshore platform, we studied resting and painting in detail as it was a constant process there. Our sources stated that even without pin holes in the paint coating, any remaining hydroxide chemicals in the metal would pull moisture through the paint by osmosis. For that reason, we always use a neutralizing metal prep solution on the metal before painting. The phosphoric acid in the metal prep solution would neutralize any chemicals remaining in the metal. I believe the instructions said to rinse with clear water, then let dry after metal prep...

How exactly is the water broken down into hydrogen and oxygen?

As the acid is formed and the iron dissolved, some of the water will begin to break down into its component pieces -- hydrogen and oxygen.

That's extreme oversimplification. The reaction common for active metals (not including iron) is
$$\ce{Mg + 2 H2O ->  Mg(OH)2 + H2 ^ }$$
Iron does not react this way. However, it does react with acids, say, with hydrochloric acids
$$\ce{Fe + 2 HCl(aq) -> FeCl2(aq) + H2 ^}$$
The reaction is quite slow even for concentrated nonoxidizing acids. However, in presence of oxygen and in diluted acids two subsequent reactions occur, the one noted above is followed by
$$\ce{4FeCl2 + 10H2O + O2 = 4Fe(OH)3 v + 8HCl }$$
The second reaction effectively removes iron from solution, significantly fastening the overall reaction.  However, even without acid present, even in slightly basic conditions iron slowly reacts with water in presence of free oxygen, forming rust. The process is hindered only at pH above 10 or at absence of free oxygen. Since natural water is usually slightly acidic due to presence of dissolved carbon dioxide, iron always rust on contact with water and atmosphere. This, actually, is a reason why concrete layer over steel frame is regulated: concrete always contains water, and surface layers quickly looses basicity due to reaction with atmospheric carbon dioxide, so steel near surface of concrete quickly rusts, increasing in volume and tearing the concrete from inside.

In the context of automobile repair work, assume a metal surface has
rust over it. Does it matter if all of the rust is removed before the
metal surface is painted? If it does matter, then why?

Yes, it does. Rust is weak, so any paint over it will hold weak. As long as the paint film over rusty part is broken, water and atmosphere contacts the metal and start to erode it, resulting in more paint film be destroyed. So it is critical to remove all rust from the surface being painted. Abrasive blasting, if you have the equipment, is probably the best way. While some paints may be applied over rust, it should be a very thin layer, and even then I would prefer traditional paint over cleaned surface anyway.

So the process that allows the rust formation to continue even after paint has been applied over the rust is basically that the paint does not seal properly? In theory, if the paint does not crack, then there is no chemistry that allows the metal to continue rusting?

@AttributedTensorField if it is a proper anti-rust paint (i.e. water- and air-proof) that formed dense film and ignoring the air and water that were sealed under paint, than there is no such chemistry I'm aware of.  Theoretically speaking, oxygen and water have non-zero diffusion speed even through quite dense films, so the process may continue even then, but the time of failure would be very large.

Most paints will prevent oxygen and water to come into contact with the metal. Usually, rust under paint will allow the paint to break off. Some paints, like Rust-Oleum, can be applied over rust, but most cannot.
Also, with rust, you usually have an uneven surface which shows through the paint and gives an impression of a poorly done paint job.

Sometimes the best chemistry, is to simply PREVENT the chemistry of rust in the first place.
So in addition to the already well covered chemistry issues above, I think it's very important, that before coating the underlying surface, that that surface is both smooth and solid.  Old guys refer to it being 'sound'.
So I've found that one important reason to remove all rust, often by sanding, is to help achieve 'smooth and solid'; in other words not to affect the chemistry of the surface, but to adjust the physical characteristics of the surface.

I've found that in actuality almost all coats of paint have tiny holes in them that show up later.  These tiny holes are perhaps from bubbles or debris in the paint as it is applied.  A few years goes by and tiny rust spots start to appear, then they get bigger with time.
By carefully preparing the surface each time before it's coated you can minimize the chance for these pin-holes.  In other words, by again making sure the surface is smooth, so that dirt from the previous coat don't stick up, and so that the surface is clean to not give any bubbles anything to hang onto.
Multiple coats helps, but only if each new coat is again carefully applied.

BTW, another source of holes in paint where rust starts to form are places where impacts occur, like where rocks hit fenders, or where surfaces connect and expansion and contraction can break the paint joint in time.

Something you apparently missed on the Krylon webpage. "If the surface still has extensive rust after all loose rust has been removed, start with Krylon® Rust Protector™ Rust Converting Primer or Krylon® Rust Tough® Rust Fix. This primer chemically changes rust into a waterproof, paintable surface. Spray directly over the remaining rust to protect against further corrosion."
 Those products contain acids that are supposed to etch, seal, and "convert" the rust.
Other claims by Krylon and Rustoleum are exaggerations in any climate but a high desert.

If you can't clean down to "white metal" , you will need to use conversion coating , as has been noted. It is phosphoric acid with manganese and secret stuff - depending on which brand. "Navel Jelly " is the generic conversion , I think. Conversion coatings do very well in salt spray tests but I can't remember and real numbers. However white metat is certainly the best.

Best solution is use of a first rate primer over clean metal, with all rust removed. Rust is really bad for coatings. If you've got rusting or blistering under paint, then you need to remove the paint and scrape, sand, scour, or blast off the rust. I've had both good and bad results using a phosphoric acid based liquid (naval jelly) to dissolve the rust. I tried to figure out why so much variation and I couldn't - and at the time I was working as a corrosion product development chemist! Meaning, it's a crap shoot. 20 or 30 years ago the rust conversion COATINGS were mostly just visual effects. (SO there are rust dissolvers, meant to be rinsed off, and rust conterters (made with paint resins), meant to form a primer coat (or sub-primer coat). Both are confusingly claimed to "convert" rust. How do I prepare a steel surface for painting? Remove all rust. Rinse with water. Rinse with deionized water. Rinse with acetone (gets rid of most of the water). Rinse with toluene (gets rid of most of the acetone). Dry thoroughly. Keep dry. Prime. Coat. Coat. Coat. YMMV. Could a "film forming" rust converter help? Maybe in some circumstances. Same with naval jelly. But if you can't remove all of the thicker rust, you're well and truly screwed. Chances are nothing will prolong the life of the part (or coating) any better than frequent stripping and recoating. Safety note: both solvents are flammable. Use outdoors or with plenty of ventilation (and away from heat, flames and sparks (like cell phones)).


text:
Definition of major, minor, and trace elements in analytical chemistry
What is the definition of major, minor, and trace elements in a sample? I am unable to find this anywhere on the internet. Is it something like:
Major elements are those > 0.1 %
Minor elements are those > 100 ppm < 0.1 %
Trace elements are those < 100 ppm

In geochemistry, major elements are usually considered to be above 1%, minor elements are between 0.1% and 1%, and trace elements are below that and usually measured in ppm. Note that this applies per sample. For instance, silicon  would be the only major element in sand, whereas it would be a trace element in pure limestone.

In, in sum, we don't have a set of definition for each term. I will judge myself, then. Thanks for your help.

The only one I could find a IUPAC publication for is trace element which is about 100 ppma or less. These terms don't seem to have any universally-agreed quantities. There may be some broadly-accepted ranges for different fields (probably everyone agrees that over 10 or 20% is definitely major), but I guess nobody's ever felt the need to standardize them.

I see. Thank you for confirming this lack of definition, at the very least!

There is no definite classification of what are the major, minor and trace elements are. While the numbers you mentioned are indeed the rule of thumb, they are rather loosely applied. It depends on field (geochemistry, solid state physics, analytical chemistry). There are however several guidelines.

Major elements are the elements that define the material in question. Some examples are:

Gold and silver in electrum,
Yttrium, aluminum and oxygen in YAG,
Hydrogen and oxygen in the ocean.

If you change the major elements, you are essentially changing the material. If you replace the hydrogen with potassium then instead of water you will get potash, which is not the same material and has vastly different properties (although this example might be a bit extreme).
Major elements are usually measured in percentage (either mass or molar), and are commonly above 1% of the chemical composition of the material.
Trace elements are elements that occur in such small concentrations that they do not change the essence of what a material is. Some examples:

Copper and platinum in electrum. It's still a gold-silver alloy, just with some other metals. The properties (such as colour) may change slightly.
Neodymium, erbium, etc. in YAG. YAG is used to generate lasers, and doping it with trace amounts of other rare earth elements may change certain properties of the laser (wavelength). However, it's still a YAG.
Almost every other element other than H, C, Cl and Na in ocean water.

Trace elements usually have concentrations of below 0.1%. In this case it's easier to measure them in units of ppm (parts per million, this mostly refers to mass). Remember: 1% = 10000 ppm. In extreme cases ppb (parts per billion) may be the appropriate unit.
Minor elements are everything in between. Technically, this means things between 1% and 0.1%. You might think of sodium and chlorine in ocean water as an example of minor elements (even though they're both slightly above 1%). To be honest, other than a mention in introductory textbooks about analytical chemistry and geochemistry, I rare see this term used in the professional literature.


text:
Why would ethanol or water give poor yields in a peptide deprotection hydrogenolysis?
A homework question asks:

The following reaction was carried out using ethanol or
  water as solvents, but gave poor yields in both cases.
  Suggest why this might be the case and how these
  problems could be overcome.

I've gathered that this is the deprotection step of a dipeptide synthesis reaction (Phe-Ser). The palladium catalyst and hydrogen gas remove the Cbz group from the n-terminal and the benzyl ester from the c-terminal, with a carbamic acid intermediate that spontaneously decarboxylates to the product.
I don't really understand why ethanol or water would give poor yields as a solvent, since I've seen multiple references where they use exactly those as solvents for this reaction (or methanol, or THF). The only hint I could find was a suggestion that using acetic acid may be necessary when deprotecting a Cbz group from an amine, since protonating the nitrogen facilitates the reaction. A couple of references said that an aliphatic amine could poison the catalyst and require the addition of dilute acid to protonate the amine...I'm lost on why or how the poisoning occurs.
A nudge in the right direction would be greatly appreciated.

This is a vague question (not your fault obviously). A poor yield might mean that you recover starting material, or that you get a different main product which could be a result of the desired product not being stable (for example if it cyclises). The poor yield might be a result of poor solubility, of poisoning the catalyst as you mentioned or maybe the yield will be improved if you heat the reaction and/or do it under pressure. I guess the answer is something that was mentioned in lecture but in general when you do research you should not be trying to answer vague questions.


text:
Why is there an exponent 4 after the brackets in sp3?
My professor wrote an electron configuration for carbon as: 1s2 (sp3)4
I thought it was just 1s2 sp3 where did the 4 come from?

The electron configuration of carbon is canonically written as 1s22s22p2. What you now can do is invoke orbital hybridization and combine the electrons in the 2s and 2p orbitals. Then you get sp3 hybrid orbitals with a total of 4 electrons in them. The 1s electrons stay where they are. To mark the new hybrid orbitals as separate, you put brackets around them.
You then end up with an electronic configuration of 1s2(2sp3)4.


text:
How can an excited helium dimer be formed?
Why does a collision between an excited helium molecule ($\ce{He^*}$) and a ground-state helium atom create $\ce{He2^*}$?

The electronic configurations are $\mathrm{(1s)^2}$ for $\ce{He}$ and $\mathrm{(1s)^1(2s)^1}$ for $\ce{He^*}$.
Put this information in an energy diagram and see what the molecular orbitals look like, especially how they are filled. (Further hint: Filled-unfilled orbital interactions are generally favorable.)
I have attached a graphic:

This leads to a bond order of $\frac{1}{2}(2-1+1) = 1$, and therefore $\ce{He2^*}$ is stable with respect to $\ce{He + He^*}$. The excited dimer is known as an excimer.
As an aside: The relative energy levels of the molecular orbitals are very hard to get right out of the blue, so take them with a grain of salt. As was kindly pointed out by Martin, the excitation of helium will initially form a singlet excited configuration, depicted above. This should be able to relax to a triplet excited state.

Just a little nitpick. I think the excited Helium should not be described as a triplet, as excitation usually does not flip the spin, keeping it a singlet.

@Martin I mostly see these diagrams in org chem, where they don't care all that much about spin-correctness. I've amended my answer, hopefully to your satisfaction. Thanks for the nitpick :)

Lovely, I appreciate the appreciation :D. Most of the times it might not even matter and a triplet Helium is of course also an excited state, probably just not the first.

@Martin-マーチン the triplet state would probably be the "first excited state" in that it is the lowest-energy excited state, but yeah, probably not the initially formed excited state ;) Anyway, I will take the liberty to edit the diagram.

In any chemical collision process, one of the most important physical parameters to consider is the angular momentum. Throughout a simple chemical reaction, the total angular momentum of the system is preserved. In the case of a simple diatomic reaction, we need to conserve both the spin and the orbital angular momentum.
Consider the ground state of orthoheium He (1s)$^2$. This atom has two spin paired s-electrons. Therefore the total spin is 0 and the total orbital angular momentum is also 0.
Consider now the ground state of parahelium, He(1s)(2s). This atom should have (in the simple case) have two same spin electrons in each of the 1s and 2s orbitals. Therefore the total spin is 1, and the orbital angular momentum is 0.
So, we need to create a final molecular orbital picture with orbital angular momentum 0 (easy: all $\sigma$ orbitals) and a total spin of 1. This can only be achieved if our final molecule also has two unpaired, aligned spins. The lowest energy configuration possible is therefore (1$\sigma_g$)(1$\sigma_g$)(1$\sigma_u^*$)(2$\sigma_g$), which is indeed the excited state helium dimer He$_2^*$.

Why should the excited Helium be a triplet? Can you back that up?

Ah, what I should have said was that the "excited state" helium atom was actually the ground state of parahelium. The question does, after all, ask how two group state helium atoms can create He* (ie ground state of orthohelium and parahelium). I will update.

Orthohelium is the triplet with S=1 and Parahelium is a singlet with S=0. And I still do not understand why the excited Helium has to be a triplet, the most simplest case would be still singlet.


text:
How to determine concentration of an unknown solution using Beer's Law?
What if the concentration of an unknown is higher than the range of concentration used when you were doing an Absorption v. Concentration graph? How can you figure out the concentration of an unknown solution, basically, if it falls above the range? I have literally no idea how to go about doing it.

Fit a line to the standard measurements and extrapolate. There's no guarantee, of course, that the absorbance vs concentration response is linear beyond your standards.

... or dilute the sample and remeasure the spectrum.

This can be an issue in many calibration experiments, and falls into the broader and extremely important field of "design of experiments" within the field of chemometrics. Clearly, your range of calibration data is not quite sufficient for your application, but you have several options available to you, each with pros and cons, some of which have been mentioned in the comments

Go back to the calibration curve, and take some standard measurements at higher/lower concentrations than you did. Pros: this will be the most accurate method, and will help you to avoid this problem with future measurements. Cons: takes time and requires more measurements.
As commented, extrapolate from your model. Pros: extremely quick. Cons: likely to be inaccurate unless you are 100% sure Beer's Law holds for your sample at all concentrations. There is also no way of estimating the errors.
As commented, dilute your sample with a known quantity of solvent and remeasure. Then multiply your answer by the dilution factor. Pros: Fairly accurate, takes less time than performing more measurements for your model. Cons: Depending upon your lab skills and precision of your equipment, errors could start to propagate and become larger.

Good luck!


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Denaturing of Proteins and Nucleic Acids - Effect of Temperature (Heating vs Cooling)
Heating a protein/nucleic acid will disrupt inter-molecular and intra-molecular forces in the tertiary structure. It will also interfere with the shape of the active site if the protein.
Does this apply with cooling also?

Complimentary nucleic acids are held together mainly by hydrogen bonding, so I'm not sure if they are able to be cold denatured, but cold denaturation is a well known phenomenon for proteins. Here is a decent review of the processes involved. In short, most proteins' tertiary structure is dominated by hydrophobic interactions and the energy barrier to unfolding a protein is generally quite small. The stabilizing effect of the hydrophobic interactions decreases with temperature and at a certain point, is insufficient to keep the protein folded. For many proteins, the cold denaturation temperature is below 0 °C, so it's difficult to study. Generally, studies on these proteins are done by adding chemical denaturants or changing the pressure of the system to shift this temperature higher, having the risk of conflating this deliberate denaturation with the temperature effect. Of course, not all proteins are stabilized this way and things like metallothioneins probably don't denature in this way as their tertiary structure is maintained by coordinating metal ions.