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Prepare 1 liter of 28% NH3(aq) solution from 1 gal of %10 NH3(aq)
I have one gallon of commercial, surfactant free, ammonium hydroxide solution that I would like to use to prepare a concentrated solution.
I have two options as I see it, and I am still getting into the chemistry side (amateur chemistry), but am pushing ahead and want to do this even without the understanding. I am hoping the solutions will provide adequate explanation as to what is being done, and why, which will help me learn to solve problems like this in the future. Chemistry newbie, here.
Solubility at room temperature is to be assumed.
I see that I have two options.
1) Prepare a receiving flask with cold distilled water (likely in an ice bath) and then prepare a distillation flask with the %10 ammonium hydroxide solution. Slightly heat the distillation flask to the point where the ammonia gas begins to off-gas, which is funneled into the receiving flask and dissolve.
2) Prepare a receiving flask and distillation flask both with an amount of the 10% ammonium hydroxide solution. As above, gently heat the distillation flask to off-gas the ammonia into the receiving flask where it should dissolve, and increase the concentration beyond 10%.
Since I would like a 1 liter solution containing 28% NH3(aq), what volume of cold distilled water should be placed in the receiving flask, which I ask because I assume the solvation of NH3 into water will cause the volume to increase. If I placed 1 liter of water into the flask initially, then this would result in more than 1 liter of solution. This also depends on the amount of ammonia that needs to be dissolved to reach my desired concentration, which varies from option #1 to #2.
So what I want to know is,
Q1) How to determine the volume of a solution given a specific concentration of NH3(aq). This will allow me to account for the increase in volume when adding ammonia to an existing solution, in order to reach a desired concentration.
Q2) How do I know when I have reach %28 NH3(aq) concentration?
Q3) A procedure leading from 10% to 28%, and any advice on more technically correct approaches to this problem. Is it a problem that should be considering molarity, etc.?
Q4) Any quick background info that I can look up to help me understand this problem would be welcome!
Bottom line: Prepare a 1 liter solution of 28% NH3(aq) from 1 gallon of 10% NH3(aq), and please provide an explanation.
Thank you for any help. I am a software programmer, and have delved into he world of chemistry as a hobby, and am getting confused by terms like molarity, normality, and both of those versus concentration, etc. It's maddening to say the least.

Why are you using archaic "NH4OH"? It's Simply NH3 water solution.

I am learning chemistry here. Thank you for the input. I see it referenced as NH4OH, but I think that is what it exists as in an equilibrium in the solution. And archaic? Not sure how that is archaic, as I would think "liquor ammonia" would be the archaic term. Just answer my question, please. I try to be as accurate as possible in my question/description.

NH3 is a weak base so degree of dissotiation is small - writing NH4OH implies total dissotiation, but it was written like that in old books.

Interesting. I knew NH3 was a weak base, and only partially ionized in solution, but could you point me to a place where I can read about these implications? I didn't know that NH4OH actually meant anything other than a molecule formation. I now see where your comment was coming from, because I wasn't sure how to take it exactly.

NH4OH isn't archaic at all. Aqueous ammonia is sold as ammonium hydroxide: Sigma-Aldrich

Disclaimer: ammonia is extremely hazardous including flammable in some vapor concentrations.

How do I know when I have reach %28 NH4OH concentration?

You can measure the specific gravity and compare to the table on page 2 of
this document:
Next, it is important to understand that if you heat/boil/distill the 10% solution the vapor coming off is not 100% ammonia, it is a mixture of ammonia and water, but with increased ammonia concentration.
To predict the percent ammonia in the vapor phase, you need to look at the dew point line in a table like Fig. 1.2 here, but that one is for 16 bars or Fig. 2 of Ganesh et al. Journal of Mechanical Engineering Research Vol. 3. (1), pp. 25-39 that shows many pressures including 1 bar.
As the original solution is heated, the system progresses toward the upper left in Fig. 1.2/Fig.2, the liquid phase along the bubble point line and the vapor phase along the dew point line.  The concentration of ammonia in both phases progressively decreases.  Eventually, the concentration of ammonia in the vapor phase drops below the concentration of ammonia in the original solution.
From Fig. 2 of Ganesh et al. it can be seen that at 1 bar starting from 10 weight% ammonia liquid phase, the initial vapor phase concentration would be close to 28% ammonia.
This shows that even if you added no water to the condensate, there is no way you can get 1 liter of 28% ammonia from 1 gallon of 10% ammonia in one simple distillation step.
Only through fractional distillation could you get that much 28% ammonia from that little 10% ammonia.
Also, practically speaking, it will be difficult to avoid lossing ammonia vapors to the environment and safety needs to be considered.

I mean a 28% NH3(aq) solution, where I believe by weight, 28% of its weight consists of dissolved NH3(g). I realize there are NH4+ and OH- ions in solution, but I think I was trying to calculate the amount of actual NH4OH in solution, rather than the dissolved NH3(g), which is partly where my confusion comes from.

I got the idea of gently heating a NH3(aq) solution below the boiling point of water to produce NH3(g) here: https://www.flinnsci.com/media/620542/91231A.pdf

My questions are more with how do I predict how much water I will need to arrive at 1 liter of 28% NH3(aq) as I gas it with NH3(g), because of the volume expansion. You answered my question about determining the NH3(aq) concentration, but I was also thinking about weighing it and determining completion based on weight gained.

How do you calculate the volume of a liquid when a gas has become dissolved within it, as in the case here? 100g of water has a volume of approx. 100ml, but what about when I dissolve 100g of NH3(g) into it. What effect on volume does that have?

I just checked out the MSDS, and I must point out it is for Anhydrous Ammonia, not gaseous wet ammonia. I think, but might be wrong, but there is a stark difference. Yes, both dangerous in their own ways, but one won't kill you just by being around it.

@JeffreySchultz 1 liter of 28% ammonia is 902 grams at 60 degrees F from the specfic gravity table in the answer.  This means it has 649 grams of water.  So if you add 253 grams of ammonia to 649 of water you have 1 liter of 28% solution.

Thank you, Dave. I understand the gist of what you are saying. In my own time I am currently working through a university-level Chemistry book that covers inorganic and organic chemistry, which was given to me by my wife's uncle because of my interest in chemistry. He used to be a chemist (Yale graduate) with the EPA. However, he lives in a different state, which makes it difficult to learn anything from him.

I have been trying to prepare standard solutions of various bases and acids, and this was attempt at procuring an ammonia stock. I know there are other ways of generating ammonia gas, but was hoping to just reuse an existing ammonia solution. I will likely follow Flinn's direction with a very cold receiving flask of water, with a 10% solution as the ammonia source, and just see what I get using a hydrometer. I will take safety precautions when doing so, and have a dedicated space for this. Thank you! https://www.youtube.com/watch?v=4epBHiBbbmU


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How does heat and/or boiling remove vitamin C?
The www is full of crackpot claims about how various food treatments alter their nutrition content and it makes finding actual science very difficult.
I am specifically interested in the vitamin C content of orange juice.
From an answer I received in Biology SE (referencing Hodgins et al. 2002), pasteurisation of orange juice is unlikely to reduce the vitamin C content by more than 20%. This tallies with what I have been able to find on the reduction of vitamin C in milk by pasteurisation.
My question is twofold:
(i) What are the processes at work which remove some vitamin C during pasteurisation? Is it just the heat? What is the mechanism?
(ii) If I were to boil orange juice for 10 minutes or so, would that remove much more vitamin C and if so, why?

Rob, see this.

@MARamezani Thanks. That is useful. It looks like the half life of the vitamin C in this study is of order hours at temperatures of 70-90 Celsius.

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Is atmospheric pressure acting only on the contents or also on the container?

$22~\mathrm{g}$ of dry ice is placed in an an empty $600~\mathrm{ml}$ closed vessel at $298~\mathrm{K}$. Find the final pressure inside the vessel, if all $\ce{CO2}$ gets evaporated?

Now by applying $pV = n\mathcal{R}T$, the value of $p$ comes out to be $20.4~\mathrm{atm}$, but in the solution shown, it adds $1~\mathrm{atm}$ i.e atmospheric pressure pressure, so the answer comes out be $21.4~\mathrm{atm}$. Now why do we need to add atm pressure, the question tells us to find the total pressure inside the vessel, isn't the atmospheric pressure acting only on the container, and not on its contents?

A quick idea I had was that after placing the dry ice in the container and sealing it, t = 0, the container also contains air at 1 atm. So the $\ce{CO2}$-pressure is 20.4 atm. and the air is 1 atm. Total pressure: 21.4. Otherwise I can't quickly think of an answer.

According to the way this question is worded, adding 1 atmosphere would be incorrect. The reason is that they specify an empty container - if you have a rigid container and it is truly empty (high vacuum), then the absolute pressure would be zero, and the gauge pressure would be -1 atm (see this wikipedia article for reference to these terms).
Once the $\ce{CO2}$ sublimates, the final absolute pressure would be given by $PV=nRT$ - there is no need to add atmospheric pressure. To find the gauge pressure, you would actually subtract one atmosphere.
It is possible that the question wants you to assume that "empty" means "filled with air at 1 atm," but that is a guess and I would argue that it's not safe to assume that based on the way the question is written.

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How to calculate the vapor pressure of water?

$48~\mathrm{l}$ of dry $\ce{N2}$ passes through $36~\mathrm{g}$ $\ce{H2O}$ at $300~\mathrm{K}$, and this results in a water loss of $1.20~\mathrm{g}$. The vapor pressure of water is?

The solution says, that water loss is observed because of escape of water molecules with $\ce{N2}$ gas. Then it says that water vapours occupy the volume of $\ce{N2}$, i.e  $48~\mathrm{L}$, this part I am not able to understand. Why would water occupy the same volume as that of $\ce{N2}$, it isn't mentioned that these two are in some container of volume of $48~\mathrm{L}$. What is the logic behind this? Then the solution proceeds, by applying $pV=\frac{m\mathcal{R}T}{M}$.

It doesn't matter if there is a container.  I'm picturing bubbling the nitrogen through the water, for example the water is in a flask, and there is a hose going down into the water, maybe a frit at the end of the hose.
The important assumption is that nitrogen reaches equillibrium with the water, which is a big assumption.
But adding 1.2 grams of water would increase the total volume by about 1 L.  I would use the total volume to answer the question, not 48L, but it will only change the answer by about 2%.

True, they should have asked for partial pressure, not vapor pressure so that assumption wouldn't be necessary.

This is an example of using partial pressures for a mixture of ideal gases.
We are making a big assumption - that water is an ideal gas - but once we make that assumption, the way to proceed is to treat each gas in the mixture as if it were by itself in the container. As you pointed out, it isn't very obvious from the way the question is worded that we would assume that both would occupy 48 L, but, since the volume of a gas depends either on the container volume or on temperature and pressure when there is no container, it does make sense. The idea is that whatever is constraining the $\ce{N2}$ to 48 L would also constrain the $\ce{H2O}$ to the same volume, since it is a mixture of gases.
Since the partial pressure equals the vapor pressure, the solution is given by calculating the pressure of this mass of water under these conditions.

Thanks again , i did get what you are saying , but i think , we need to make another assumption that , the weight of N2 gas taken= weight of water displaced , only then will they occupy equal volume .

No, for ideal gases you don't need to make that assumption. And I am pretty sure that that assumption would never hold - I can't think of a situation where a gas would displace an equal mass of water while bubbling through. Maybe you are trying to bring in buoyancy? That would make sense for incompressible fluids and for solids, but not for gases.


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Gibbs energy and Entropy
In Gibbs energy equation:
$$\Delta G=\Delta H-T\,\Delta S$$
At constant pressure,
$$Q=\Delta H\tag{1}$$
and we know that $T \Delta S=Q$ (reversible)
so finally
$$T\,\Delta S=Q\tag 2$$
Putting values of Equation $\text{(1)}$ and $\text{(2)}$ in Gibbs energy equation ,
we should always get $\Delta G=0$ for a reversible reaction. ..
Why isn't it so?
I don't know how to search it on Google; if anyone could give me a link or otherwise explain it?

Reason 1:
while $T\,\mathrm dS=\mathrm dQ$ for a reversible process, this does not mean $T\Delta S = Q$
Reason 2:
The term "reversible reaction" is unrelated to "reversible process".
Lame Reason 3:
Because your "always" statement does not repeat the "constant pressure" requirement assumed for $Q=\Delta H$

@Loong This answer does not address the misconception that heat is transferred between systems that have to be specified.

The problem is that you are equating "reversible reaction" with "reversible process."
What you are seeing when you plug equations 2 and 1 into the Gibbs Free Energy equation is that at equilibrium,
$\Delta G = 0$
The reason that this works is because a reversible process is one that progresses by very small movements away from equilibrium states - when we say a process is "reversible" what we really mean is that we are making an approximation that allows us to ignore entropy generation. When you make this approximation and apply it to the Gibbs Free Energy equation, you are forcing it to stay at equilibrium - hence, $\Delta G = 0$.
I think your confusion is coming from the similar names of "reversible reaction" and "reversible process." A reversible process is one in which the entropy generation term is zero, but a reversible reaction is one in which the forward and reverse rates of reaction result in a mixture of reactants and products at equilibrium.
In other words, a reversible reaction can be in an equilibrium ($\Delta G = 0$) or non-equilibrium state ($\Delta G \neq 0$), but a reversible process is always in an equlibrium state.

Even at equilibrium, $\Delta_r H$ and $\Delta_r S$ are typically not zero, so the OP's statements are incorrect even at equilibrium.

You are missing the subscripts. Heat transfer only makes sense if you define a system and its surrounding. Your equations become:
$$\Delta G_\text{sys}=\Delta H_\text{sys}-T\,\Delta S_\text{sys}$$
At constant pressure,
$$q = \Delta H_\text{sys}\tag{1}$$
and we know that $T \Delta S=q$ (reversible)
so finally
$$T\,\Delta S_\text{surr}=-q\tag 2$$
Putting values of Equation $\text{(1)}$ and $\text{(2)}$ in Gibbs energy equation ,
we get
$$\Delta G_\text{sys} = -T\,\Delta S_\text{surr}-T\,\Delta S_\text{sys} = -T\,\Delta S_\text{univ}$$
So the reaction will proceed if $\Delta G_\text{sys}$ is negative or $\Delta S_\text{univ}$ is positive. This is how the textbooks tell it.

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Why are PTFE and FEP named "ethylene" when they don't have double bonds?
PTFE (Polytetrafluoroethylene) and FEP (Fluorinated ethylene propylene) have the following chemical formulas:

It's been years since I have taken organic chemistry, but I do remember alkenes have double bonds and alkanes have single bonds.  Neither of these compounds have double bonds.
Why do these compound names contain "-ene" in them, instead of "Polytetrafluoroethane" or something?

I don't know how in blazes you found that; I searched for 30 minutes using those keywords!  Thanks for the link.

By the way, I give you a warm welcome to chemistry.SE by marking your question as duplicate! :)

What is SE shorthand for?  Also, how can I mark the question as duplicate?

Technically not duplicate because monomers are different even though topic is adequately covered.

@user171828 SE = Stack Exchange

Polymers follow a slightly different nomenclature system than you are used to in organic chemistry. Instead of naming the polymer after the monomers as they are, we name them after the monomers as they originally were. In other words, a polymer made from ethylene monomers is called polyethylene. Since PTFE and FEP are made from fluorinated ethylene and propylene, we just put "poly" in front of the monomer names. This is despite the fact that after the reaction, the bond order is reduced by one.

This is correct thomij. If you are going to think that way (about the product macromolecule instead of the unit structures that made them, why wouldn't you call them difluoromethylene and call PET, polymethylene; or as you are saying polymethane? If you inspect it after they are single, not double, carbons connected. Since nothing but seemingly arbitrary brackets makes them two carbon units in their chain elongated states?

The compounds are polymers. The unit structure of PTFE is tetrafluoroethene. PTFE is named in accordance with the subunits from which they are polymerized. You can apply the same logic to the second one (you apply the same logic to all polymers). There are no common names for polymers just abbreviations. IUPAC rules do have polymer rules too, so you will necessarily see them adhering to the IUPAC rules. It is, however, a matter of the unit structure in brackets, not the monomer that determines the name.


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Reaction between lead dioxide and nitric acid
A wikipedia article suggests that lead dioxide is produced by this reaction:
$$\ce{Pb3O4 + 4 HNO3 → PbO2 + 2 Pb(NO3)2 + 2 H2O}$$
The same article also suggests that lead dioxide reacts with nitric acid in the following way:
$$\ce{2 PbO2 + 4 HNO3 → 2 Pb(NO3)2 + 2 H2O + O2}$$
I am unable to understand why then, in the first reaction, $\ce{PbO2}$ doesn't further react with the residual nitric acid. Is it because the quantity of the acid is regulated? Or is there some error in the reactions?

Exact statements are:

Because of the instability of its $\ce {Pb^4+}$ cation, lead dioxide reacts with warm acids, converting to the more stable $\ce {Pb^2+}$ state and liberating oxygen
  $$\ce{2 PbO2 + 4 HNO3 → 2 Pb(NO3)2 + 2 H2O + O2}$$

and

Lead dioxide is produced commercially by several methods, which include ... or reacting $\ce {Pb3O4}$ with dilute nitric acid:
  $$\ce{Pb3O4 + 4 HNO3 → PbO2 + 2 Pb(NO3)2 + 2 H2O}$$

Since nitric acid in warmer in first reaction, the following reaction occurs easily, which is actually a decomposition reaction:

When heated, lead(II) nitrate crystals decompose to lead(II) oxide, oxygen and nitrogen dioxide.
  $$\ce{Pb(NO3)2->PbO +NO2 +O2}$$

Hence the oxygen might be liberated with decomposition-cum-oxidation reaction.

You need to consider the reaction conditions.
The Wikipedia Lead Dioxide article is saying that in dilute nitric acid the first reaction occurs, while in warm nitric acid the second reaction occurs.
So the first reaction would only proceed to the second reaction under harsher conditions, heat and more concentrated acid.


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Is there a name for the opposite reaction to the dissolution?
When atmospheric $\ce{CO2}$ reacts with water to form $\ce{H2CO3}$, this is called dissolution, isn't it? What term would you use for the opposite reaction when it occurs at atmospheric pressure (e.g. due to temperature change)? For a solid it would be precipitation, but here it is for a gas.

Good question! But it is maybe worth noting that the process of gaseous carbon dioxide dissolving in water and the reversible reaction of carbon dioxide with water to form carbonic acid are two separate processes.

Effervescence if there are bubbles and
Degasification more generally speaking.
There is also the term outgassing, but that term is broader than just gas coming out of liquid solution.

thank you for the link. But is effervescence also valid at atmospheric pressure or is it only for high pressure cases (carbonation/effervescence).

The word can be used at atmospheric pressure, but usually requires bubbles.  If the gas is going to the vapor phase without bubbles I wouldn't use that word.

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Is table salt sticking to my laptop?
I was reviewing first posts here, and eating a juicy and salty (table salt) tomato at the same time. Suddenly, I realized that a few table salt crystals have been poured down on my mousepad.
I got curious and I rotated the laptop $90°$. I witnessed that the salt is somehow resisting to fall! Not sticked (very firmly attached) though, and the mousepad wasn't wet and didn't contain any possible chemicals that could cause the "attaching";
except that apparently the mousepad is "soaked" with the sebum produced by the tip of my fingers. Sebum, however, is a mixture of non-polar lipids such as fatty acids, triglycerides, sterols, sterol esters, wax esters and squalene. (See here also)
There can be another reason which lies in the electrical field that my laptop produces, and this could affect a compound like table salt that is primarily ionic. (No clue how, am I missing an easy chemistry concept?)
So, is there any possible "mild" (as in invisible) reaction, either chemical or physical, as a reason of the phenomenon? Or is this behavior because of the composition of table salt?
Note that I'm certain about the existence of such thing, though research showed nothing relevant in the Net.

Did you say the salt crystals fell off a tomato you were eating? Any chance moisture was from the tomato, perhaps adhered to the crystal, which subsequently 'stuck' to your mousepad?

No @theo. it was the salt shaker's fault :). I'm 97.85% (:D) sure the salt crystals hadn't been in contact with the moisture of any kind. Seriously, they weren't or hadn't been in contact.

the reason could be anything? can you reproduce the proble? or can you give to the experts for checkup?

More than likely the case of your laptop became charged and was attracting the salt. Yes salt crystals can be coerced by an electrostatic field.
Here is a post on a teaching site that actually uses salt (and pepper) to demonstrate electrostatic attraction.
https://www.teachervision.com/electricity/lesson-plan/5787.html


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I have 100 mg of a proteinase K lyophilized powder and I need to make it to a working concentration of 25 mg/mL
So I know that I need to add $4~\mathrm{mL}$ of solution to this powder, which will give me a concentration of the enzyme at $25\ \mathrm{mg/mL}$. What I do not understand is how to decide how much of each of these "ingredients" (below) to add. I have $\ce{TrisHCl}$ and $\ce{CaCl2}$ in granular form ($M = 157\ \mathrm{g/mol}$ and $111\ \mathrm{g/mol}$, respectively) and 100 % glycerol solution. Will I have to make a large batch of Tris and $\ce{CaCl2}$ just so I can actually weigh out the needed amount?
The specifications for the working solution are:
$10\ \mathrm{mM}\ \ce{Tris}$
$1\ \mathrm{mM}\ \ce{CaCl2}$
30 % glycerol
Final $\mathrm{pH} \approx 8$ (use $\ce{HCl}$ if necessary)
MilliQ $\ce{H2O}$ to volume

Simply adding 4 mL of solution to 100 mg of enzyme is not an accurate way of preparing this solution. I am going to describe what to do from an analytical chemist's point of view. If you don't need this level of precision, you could do this following the same steps without volumetric glassware.
The first thing you need to do is decide what volume of your 25 mg/mL solution you need. Ideally, you will have a volumetric flask corresponding to this volume. For now, let's go with the 4 mL that you mentioned.
Next you will need to make a stock solution of your buffer. This will include all of the components except for the enzyme:

Decide what volume of buffer to make. You will need to make at least enough for your working solution, and again you will want to use a volume for which you have a volumetric flask. Also, you want to be weighing out reasonable amounts of your reagents (because weighing out <10 mg of something can really suck). Let's assume you're making 1 L. You're reagents are cheap anyway and the larger scale will make everything easier.
Calculate how much $\ce{TrisHCl}$ and $\ce{CaCl2}$ you will need to obtain the desired concentrations in 1 L of solvent.
Weight out the $\ce{TrisHCl}$ and $\ce{CaCl2}$ and add it to your volumetric flask.
Add 300 mL glycerol to your flask using a graduated cylinder. 30% glycerol probably refers to v/v.
Fill nearly to volume with water. You need everything to be in solution so you can adjust the pH, but you also need to leave some room to work with to do the actual adjustment. I'd shoot for approximately 80% full.
Adjust the pH by adding concentrated $\ce{HCl}$ dropwise.
Fill to volume with water and thoroughly mix the solution.

Now that you have your buffer solution, you just need to weigh out the appropriate amount of your enzyme and dissolve it in your buffer. From the example above, you would weigh out 100 mg of the enzyme (don't just assume you have 100 mg, even if the bottle says so) and dissolve it in your buffer using a 4 mL vol. flask.*
*
If you don't have a 4 mL vol. flask and you actually need all 100 mg of your enzyme in this solution, you will need to mess around with dilutions. Start by by creating a 50 mg/mL (2 mL vol. flask) or 100 mg/mL (1 mL vol. flask) solution.

To get the correct concentration of $CaCl_2$, you would weigh just under 1 mg. That's difficult to do. I would weigh 50 mg of $CaCl_2$ and dissolve into 50 mL, and use 1 mL of that solution. Also, pipettes are very useful for small volumes.

@LDC3 Yeah, the buffer prep should really be scaled up. I didn't actually do the math for the concentrations, but it's pretty obvious once you think about it.


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Why is O2 enough to form a mole of Oxygen?
I understand that this is the most basic knowledge of moles, however I'm still unsure - according to easy research, $\ce{O2}$ forms a mole of Oxygen. As a mole is $6.022*10^{23}$, exactly what on the periodic table can we use to find out that it's 2 Oxygen in one mole? Wouldn't it be $6.022*10^{23}$, instead of 2?

Welcome to chemistry.SE! You can visit the help center for questions you might have about our community. Or, you can just start by taking a little tour.||By the way, I really don't get your question. What has 2 in $\ce{O2}$ to do with the Avogadro number? I think you've misunderstood the concept of mole.

I'm very familiar with the stack exchange websites, however thanks for trying. Regarding the comment - That's exactly what I'm wondering. I've found One mole of oxygen gas, which has the formula O2. - Found at this link

$\rm O_2$ is the most common form of oxygen---the periodic table won't really tell you that.
Avogadro's number tells you how many particles there are in a mole.
There are $6.022\times 10^{23}$ O atoms in a mole of O atoms.
There are $6.022\times 10^{23}$ $\rm O_2$ molecules in a mole of $\rm O_2$.
Since you have 2 oxygen atoms in one $\rm O_2$ molecule, there are
$2\times 6.022\times 10^{23}$ O atoms in a mole of $\rm O_2$.
Do you see the difference?

A 'mole' is not short for a 'molecule'.
The 'mole' is a specific quantity (number of objects) defined by Avogadro's constant $N_A=6.022\times10^{23}mol^{-1}$. So a 'mole' of oxygen molecules has $6.022\times10^{23}$ molecules.
The '2' in $O_2$ means there are two oxygen atoms in an oxygen molecule.


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Obtaining ammonium perchlorate via double displacement reaction
I was tasked with figuring out how to make some $\ce{NH4ClO4}$(ammonium perchlorate). I found that one way of making it was to react potassium perchlorate and ammonium chloride in a double displacement reaction. according to:
$\ce{KClO4 + NH4Cl -> NH4ClO4 + KCl}$
Then I got to work on finding ways of making the necessary reactants without having to buy them by doing additional reactions with easier to find chemicals. After some work I came up with these two reactions:
$\ce{KCl + NH4NO2 -> NH4Cl + KNO2}$ (to make the ammonium cloride)
and
$\ce{KCl + HClO4 -> KClO4 + HCl}$ (to make the potassium perchlorate)
My question is (the newbie part) how should I now go about these reactions? Should I be mixing the reactants together in an aqueous solution? In water? Should I be mixing them into an ionic solution in an electrolytic cell? Or hell do I just mix them together as a granulated form and apply a little heat? How do I get these reactions to occur?

The Wikipedia article for ammonium perchlorate gives a hint at how this can be done. The hint is that ammonium perchlorate is curiously less soluble than sodium perchlorate, and thus with the right concentrations, you would get ammonium perchlorate to precipitate by mixing solutions of sodium perchlorate and and, say, ammonium chloride.
$$\ce{NH4Cl(aq) + NaClO4(aq) -> NH4ClO4(s) + NaCl(aq)}$$
That being said, the best advice that can be given on how to make ammonium perchlorate is

Don't. Especially do not use heat!

Ammonium perchlorate is fairly dangerous explosive that can decompose with not much provocation into a whole heck of a lot of gas for just a little bit of solid.
$$\ce{2NH4ClO4(s) -> N2(g) + Cl2(g) + O2(g) + 4H2O(g)}$$
That this decomposition produces toxic chlorine gas should only make you want to prepare ammonium perchlorate less.
Safely producing, storing, and transporting this compound is going to require a lot of specialized equipment and training (which is expensive). The danger here is not of the type that may lead to funny smells and dizziness in poorly ventilated rooms. The hazards associated with the compound involve loss of life or limb and serious structural damage if you look at it funny. Don't. For more information into the sort of destructive power ammonium perchlorate has, consider that this is the rocket fuel that got the space shuttle into low earth orbit. This stuff has the power to level industrial complexes with the forces of an earthquake.

Thanks for feedback. As to the safety, I am only researching and outlining a method from start to finish which I have to present to the engineering club. We will be making it together (yes our sponsors are chem majors) for use in rocket engines as part of the project. Everything will be handled safely but thanks for your concern.

This is not something that an undergraduate chemistry student will be prepared to do safely. This is the type of substance that has laws attached to it.


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How dangerous are these soil organic carbon digestion methods?
I have several ground up soil samples that I need to get fractions from. The fractions are based on the Century Soil Model. Briefly, the fractions relate to the approximate time frame it takes for an organic carbon pool to be metabolized in the soil. Fast pools ≈1 year, slow pools ≈10+ years, etc.
I have looked at several papers to try to get an idea of the chemical methods used for digesting the soil samples to get at the more stable carbon pools. For more easily digested organic carbon pools, hot water seems to work, but for the middling and more recalcitrant pools, other reagents are used.
Some methods mentioned include:

dissolve in $\pu{0.1 M}~\ce{NaOH} + \pu{0.1 M} ~\ce{Na4P2O7}$
dissolve in $\ce{HCl}$
reflux with $\pu{6 M} ~\ce{HCl}$
"stepwise chemical digestion with cold and hot $\ce{H2SO4}$
(95 - 98% concentration)"
oxidation with $\ce{KMnO4}$ ($33\text{ or }\pu{333 mM}$)  (this one I see mentioned most frequently)

I am not a chemist. I took the required chemistry classes for my degree and passed, but that's years ago now. When I did do chemistry labs, my results hardly ever came out as they were supposed to, so my skill in this field is minimal. Besides these methods not working, and thus wasting time and money, my real concern is how potentially dangerous these chemicals are. Would someone be so kind as to give me an indication of how risky these chemicals might be? We have a functioning fume hood, will buy PPE, and be sure there are safety procedures in place, but I'm still wondering how much I should worry if a flask breaks or something spills.

$\pu{0.1 M}~\ce{NaOH}$ is potentially corrosive when in contact with metals or human skin (see here), so I advise precautions such as wearing protective standard gloves and keep metals away from feasible damage.
$\ce{HCl}$ and $\ce{H2SO4}$, especially when they're hot and are in higher concentrations, are very corrosive acids that are dangerous in contact with many materials. They can also release noxious gases in the reactions, but their weakness is glass. Due to reasons irrelevant, these acids can't corrode glass. I recommend extreme precaution. You can work under a "glassy" protection and as Yomen mentions, under a fume hood. Gloves are very important here. Here's what Caltech thinks about precautions using $\ce{HCl}$ and here's what Berekley does about sulfuric acid.
$\ce{KMNO4}$ is a very strong oxidizer. It may cause fire in contact with different materials (we had a volcano experiment in elementary school where we mixed it with glycerine and the result was very exothermic), and it has biohazards such as kidney damage. Here it's explained more comprehensively.

You must be careful when working with hot and concentrated acid solutions as they are corrosive. Otherwise, the chemicals you mentioned are common ones.
Because you are not used to manipulate chemicals, I advise you to wear goggles and protective gloves while preparing solutions. Work under fume hood when refluxing $\ce{HCl}$.


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Collection of gas over water
When you collect a gas over water, you have to account for the mixed in water vapor by subtracting the room pressure from water vapor pressure. However, why do you not do something similar for volume, where you subtract the total measured volume from the volume of water?
How can $PV = nRT$ still hold when the pressure you are using is only for the collected gas whereas the volume is for the gas + water vapor volume?

You are right! There is an approximation here in only considering the vapor pressure of the gas. But let's consider the value of vapor pressure of water at $298\ce{K}$:  it's  $13.85\,\ce{mm Hg}$. So, if you neglect the vapor pressure of water, the relative uncertainty is  $$\frac{13.85}{760}\times100=1.83\%$$
As you can see, it's negligible.
P.S. To calculate the vapor pressure of water, I used the first equation in http://en.wikipedia.org/wiki/Vapour_pressure_of_water

Maybe I am misinterpreting your answer, but that is not what I asked - I know that you need to subtract out the vapor pressure of water from the room pressure to correct for the pressure contributed  by water vapor, but I want to know why you also do not subtract out the volume contributed by water vapor from the total gas volume.

Any gas occupies the available volume for it. So, for your question water vapor and the gas occupy the same volume. What is different is the partial pressure and the number of moles of each one.


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What is the effect of oxygen on soap foam?
I've been wondering the effect of $\ce{O2}$ on soap foam in a closed, test room. The room contains a candle and soap foam.
That is to say: if the room has lower and decreasing $\ce{O2}$ in its air, does the foam is effected (i.e., the soap produces lower foams)?
PS: I have very basic knowledge in Chemistry, and, excuse me if I didn't formalize the question in a good manner.

Hmmm. I don't know, but I'd say that if you significantly lowered the percentage of $\rm O_2$ in the air, the average molecular weight of the molecules in the bubble would drop (there'd be more low-molecular weight nitrogen and less higher-molecular weight oxygen), so the rate at which gas would diffuse out of the bubble would increase, and the foam would flatten faster.
Try blowing a foam with nitrogen, and compare it with the results of blowing a foam with air or oxygen---if I'm right, the foam blown with nitrogen will flatten a bit faster.
There'll be a bit more $\rm CO_2$ in the air from the candle, and that would diffuse more slowly out of the bubble (in fact, $\rm CO_2$ is used as a "blowing agent" to make foams, partially for that reason). But I'd think in this case that the small amount of $\rm CO_2$ produced by the candle would dissolve in the liquid, with gas dissolution rates enhanced by the lower surface tension and higher surface area of the foam.

"...foam would flatten faster.", by flatten you mean blowing-up faster?

I mean the bubbles in the foam would probably shrink faster, because lower-molecular weight gas would leak out of them more quickly.


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Is it true that under certain conditions, Mg can reduce SiO2?
Is it true that under certain conditions, Mg can reduce $\ce{SiO2}$ and the latter the former? What are those conditions?

http://meta.chemistry.stackexchange.com/q/261/4945 and http://meta.chemistry.stackexchange.com/questions/253/should-we-revisit-the-url-slug-issue

Take a look at the link of the question you've asked. And take a look at this meta post.

Yes it is true.
Conditions are temperature in the 650 to 850 degrees C range.
There is a cool looking youtube video of the reaction, not that that makes it true.
For a more serious discussion see Production and Purification of Silicon by Magnesiothermic Reduction of Silica Fume
and Ordered Mesoporous Silicon through Magnesium Reduction of Polymer Templated Silica Thin Films

Yes, it is true below $\pu{1683 K}$, the melting point of silicon, the $\Delta_\mathrm{f}G^\circ$ curve for the formation of $\ce{SiO2}$ lies above the $\Delta_\mathrm{f}G^\circ$ curve for $\ce{MgO}$, so, at  temperature below $\pu{1683 K}$, $\ce{Mg}$ can reduce $\ce{SiO2}$ to $\ce{Si}$.
On other hand, above $\pu{1683 K}$ the $\Delta_\mathrm{r}G^\circ$ curve for $\ce{MgO}$ lies above for $\ce{SiO2}$, hence at temperatures above $\pu{1683 K}$, $\ce{Si}$ can reduce $\ce{MgO}$ to $\ce{Mg}.$


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Although thermodynamically feasible, in practice, magnesium metal is not used for the reduction of alumina in the metallurgy of auminium. Why?
Everyone knows:
$$\ce{Mg +Al2O3->MgO +Al}$$
This reaction is thermodynamically feasible.
Use can we this for reduction of Alumina. And produce Aluminium. Then why don't we not use this why?

If you compare the magnesium price (about 2$ per pound)
to the aluminum price (about $0.80 per pound)
you can see it is unprofitable.
Even considering the elements' relative atomic weight, you would still be losing money if you did that.

not everyone uses '$' or 'pound'

@ADG regardless of the currency used, the fact remains that magnesium is more expensive than aluminium.

@santiago Dave already mentioned "Even considering the elements' relative atomic weight, you would still be losing money if you did that."

@ADG, my comment was in reference to your comment obviously


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Would there be a difference if deuterium is embedded instead of protium (regular hydrogen) in acids?
So instead of regular hydrogen, it would be a deuterium (still a Hydrogen). For example, instead of $\ce{HCl}$ it would be $\ce{DCl}$ where D is a deuterium.

Yes, there would be a difference. The acids would be weaker.  Why? Because the heavier isotope lowers the zero-point energy (the lowest possible energy) in the acid molecule, making the acid's bond to the D stronger than it would have been to the H. Acids in $\rm D_2O$ are weaker than they are in ordinary water, too.
Any reaction that involves dissociation of these acids will run more slowly, too. This is called "the kinetic isotope effect". You can read more about that here: http://en.wikipedia.org/wiki/Kinetic_isotope_effect
You might want to ask further questions: why is the zero point energy lower for a bond to D than a bond to H? How exactly does the lower zero point energy make the bond stronger? How exactly does it affect the rates of reactions that include acid dissociation?
Ask away!
If you'd like to see how substituting a D for an H in acids affects the acid dissociation constant for particular acids, here's a good reference:
R. A. Robinson, M. Paabo, R. G. Bates, "Deuterium Isotope Effect on the Dissociation of Weak Acids in Water and Deuterium Oxide", J. Res. Nat. Bureau. Standards, 73A(3), 1969, pp 299-308.

That's very interesting - in polymer science we often assume that deuterated species don't affect the chemistry, but from that reference it looks like the pKa can change by about 0.5 - pretty significant for weak acids!

I added a link to that reference - it's hosted at NIST, so you don't need journal access to read it

Ha, I just found it online myself and added the link, too. You guys are quick!

What about the zero point energy of O-H in $\ce{H3O+}$ relative to O-D in $\ce{D3O+}$ ? Why isn't that equally and oppositely important?

@DavePhD, it will be oppositely but not equally important, because the change in reduced mass on isotopic substitution there isn't as great as it is in the acid, right? Assuming the acid anion has a larger mass than water.

@FredSenese I guess it would depend on relative bond strengths (hydronium versus the acid) and relative reduced masses.  I just meant equally important to consider, not that the zero point energies themselves were equal.  For HF or NH3 the reduced masses would be similar to water.  Complicated to extend to polyatomics.


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How to determine number of alpha hydrogens in a cyclic compound?
How do we determine the number of alpha hydrogens in a compound? Where should we start the naming from?
For example, bicyclo[4.2.0]oct-7-ene and bicyclo[4.2.0]oct-3-ene:

Compare the stabilities of the following alkenes

In both these structures, I need to determine their stabilities based on their hyperconjugating structures. I need to determine the number of alpha hydrogens it has, to determine the number of hyperconjugating structures it makes. Where do I start?
I am aware that stability is directly proportional to the number of hyperconjugating structures. More the number of alpha hydrogens, greater is the number of hyperconjugating structures, and hence greater the stability.

The discussion with alpha hydrogens would have worked given structures like hex-3-ene and 2,5-dimethylhex-3-ene.

The relative stability of your two compounds is not determined by arguments based on hyperconjugation.  Rather the relative stabilities have to do with ring strain.  Placing a double bond in smaller and smaller rings increases the angle strain in the ring system.  A double bond, being roughly $\ce{sp^2}$ hybridized, would prefer to have bond angles around 120°.  This becomes difficult in 3- and 4-membered rings where the internuclear angle is 60° and 90° respectively.  Therefore molecules containing double bonds in 3- and 4-membered rings will be more strained than molecules with the double bond in a larger ring.
Edit: Response to OP's comment
No, the given solution is not the correct answer to the problem.  It is correct that structure B has more alpha (alpha to the double bond) hydrogens (4) than structure A (only 2 alpha hydrogens).  It is also true that hyperconjugation can be used to explain why alkyl groups attached to a double bond act to stabilize a double bond (see this earlier answer for an explanation), and that the double bond in B is more stabilized by hyperconjugation than the double bond in A.  However, hyperconjugation has a much smaller effect on the stability of these molecules than the ring strain effect.


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Organic chemistry - alkanes step up reaction
Well I was studying organic chemistry, and I came across this reaction for increasing the length of the carbon chain. Upon asking my teacher about the following reaction, she said she was unsure about it. Is this a valid reaction? And if it is, can you please provide the name of the reagent and the actual mechanism?

I think the reagent you're talking about is ketene (R=H in the figure below).  It is a linear molecule (the central carbon is $\ce{sp}$ hybridized) and very reactive.

When mixed with an alkane and heated it dimerizes to form diketene rather than react with the alkane.

However, when irradiated with light, ketene will produce the reactive carbene "methlylene" ($\ce{CH2:}$).  When this photolysis is performed in the presence of an alkane, the methylene will insert itself into the various $\ce{C-H}$ bonds like you've drawn.  However the reaction is not synthetically useful for a number of reasons.  For example, the $\ce{C-H}$ insertion reaction is very indiscriminate and a variety of products will be formed if different $\ce{C-H}$ bonds exist in the molecule.

Oh, so it is not the best way to increase the length of the carbon chain? Then what is a better reaction to do so? Is there a way to only increase the chain one at a time?

@Gummybears  That's right, the ketene method is not a synthetically useful way to extend a carbon chain.  Starting with an alkane, there is no good way to extend the chain by one carbon at a time.  You could do something like free radical bromination of the alkane, separate (ugh!) all of the resulting isomers, then form a Grignard reagent and react it with carbon dioxide to generate the one carbon longer carboxylic acid.  In the lab, you would actually just buy the alkyl bromide and skip the messy bromination step.  BTW, if the above answer was helpful, please mark it as accepted, thanks!

Well I really don't want it for a lab process. Hmm... then what is the best way to increase the length of the carbon chain of an alkane?

@Gummybears  If you have to start with the alkane (like on a test), then free radical bromination (as described in my earlier comment) is probably the way to go.  I'm not aware of any method that can selectively target a specific C-H bond in an alkane where many different C-H bonds might exist.  C-H bond activation is an active area of current research.  There are some organometallic reagents that can select certain C-H bonds in non-alkanes. You might want to post this as a new question.

I would describe the reaction as a C-H insertion reaction of the carbene generated in situ from ketene. The reaction starts with an initial release of carbon monoxide to generate the highly electrophilic 6 electrons carbene species, which is then involved in a C-H insertion process. The C-H insertion can proceed through two mechanism: via singlet state or triplet state of the carbene. The carbene is a very electrophilic sp2 hybridized species with 6 electrons: a lone pair and two pairs in bond orbitals (with H atoms). Singlet carbenes have the lone pair of electrons in a nonbonding sp2 orbital and have an empty p orbital, while triplet carbenes have two unpaired electrons, one in a sp2 and the other in a p orbital. C-H insertion mediated by singlet carbenes is believed to proceed through a concerted mechanism, with the side-on approach of the carbene which provide constructive orbitals overlapping. Triplet carbene insertions should follow a two-step radical pathway, but it is generally accepted that the reaction proceed through singlet state carbenes (supported by retention of stereochemistry observed when if the C-H bond is at a stereogenic center).


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Hydrogen cyanide - a Lewis acid?
I am told that $\ce{HCN}$ is a Lewis acid.
From what I understand a Lewis acid possesses (an) empty orbital that is accessible to attacking nucleophiles.
Does $\ce{HCN}$ have any empty orbital(s)? So how can it behave as a Lewis acid? I know prototypical Lewis acids are for example boron trifluoride, and these have an empty p-orbital.

You don't have to say that it is specifically a Lewis acid. $CN$ bond, as nitrogen is more electronegative, and due to the presence of triple bond, causes the carbon to become partially positively charged. It thus attracts the bonded pair of electrons with the hydrogen, and just like in hyperconjugation, an $H^+$ ion is released. In fact, this hydrogen is even more acidic than alpha hydrogens.

Could you say in what reaction it would act as Lewis acid? It's quite relative.

HCN has a low empty orbital in the form of the antibonding sigma orbital H-CN.

Low-lying, empty antibonding orbitals will suffice.  The reaction of carbon dioxide and water to form carbonic acid serves as a parallel example.  Water donates one of the lone pairs on oxygen towards bond formation, water is the Lewis base in this case.  Carbon dioxide has empty, low-lying $\ce{\pi}$* orbitals that serve to accept these electrons; carbon dioxide is a Lewis acid in this example.


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Why does heptane have an odour
I read that alkanes are 'colourless and odourless'. Heptane is an alkane. Why does heptane smell like heptane? Impurities?

Alkanes as molecules cannot interact with most receptors, as receptors usually bind to polar or charged groups and to lesser extent to double bonds. This makes light alkanes in small concentrations indeed odorless and colorless. However, alkanes may dissolve in lipid membranes, changing their properties. In addition, yes, alkanes usually are impure. Likely, the most common impurities are alkenes, that also can produce further impurities on contact with air under UV.  || not an answer, as I'm not an expert on odor theory.

Some higher-molecular weight alkanes definitely have an odor, but they aren't as intense as most of the compound types you'll find stinking up an organic lab.

"The odor threshold for hexane is 130 parts per million (ppm), with a faint peculiar odor reported." ---http://www.epa.gov/ttnatw01/hlthef/hexane.html
"[Heptane is a] colorless liquid with a mild gasoline-like odor." ---http://www.cdc.gov/niosh/docs/81-123/pdfs/0312.pdf
"Alkanes have relatively high odor detection threshold (37 to 50 mg/m$^3$ for dodecane) as compared to other VOC groups such as sulphides (0.0003 to 0.16 mg/m$^3$  for
dimethylsulphide), phenolics (0.022 to 4 mg/m$^3$  for phenol), and nitrogen heterocycles (0.0004 to 0.0008 mg/m$^3$  for skatole) (O'Neill and Phillips, 1992)" ---http://www.bae.uky.edu/jbicudo/odorVOC_02a.pdf

According to the following paper, there are apparently at least two olfactory receptors  in rats that respond to alkanes; one accepts alkanes with areas around 3.5 nm$^2$, and another  responds to hydrocarbons with areas of about 5 nm$^2$. The authors note that response to cycloalkanes is higher than for normal alkanes, so the shape matters too.

"Concanavalin A reveals olfactory receptors which discriminate between alkane odorants on the basis of size." ---http://www.ncbi.nlm.nih.gov/pubmed/2803264

Just wanted to add that gasoline in cars is mainly composed of alkanes from 6 to 12 or 14 Carbons. If I'm wrong please correct me.

gasoline is never pure, it usually has alkene impurities, and sometimes special antidetonation additive.

@permeakra, yes, the smell of gasoline comes from many stinky non-alkanes, so I'll retract that as an example. Nevertheless, I have no problem smelling pure heptanes or hexanes, do you?

@FredSenese When I'm not breathing in vapors directly from the bottle, I do have such problem. Comparing to such solvents as pyridine, ethyl acetate and acetone, hexane (I have not worked with heptane, though) virtually has no odor. On the other hand, even $\ce{CO2}$ in high concentrations produces some feeling.


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Sacrificial Anodes and Pit Corrosion
I have a slab of aluminum and one part of the slab will be exposed to a very corrosive environment. If I put a sacrificial anode elsewhere on the slab will the section that is exposed still be protected from pit corrosion? I guess I'm unsure because the pit corrosion is a result of a localized reaction and if an anode is added will it prevent the local reaction from taking place.

What this 'corrosive environment' would be?

Its essentially a small puddle of slightly acidic fluid.

Generally speaking, there is no protection unless the sacrificial anode is immersed in the electrolyte. It might be better to anodize and seal the aluminum, which should protect it against most acids. There are many aluminum anti-corrosion treatments, but in any case the area under attack would have to be covered. See http://en.wikipedia.org/wiki/Anodizing.


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Is it still possible for an amateur, a hobbyist, or a science buff to make relevant discoveries in chemistry today?
Consider astronomy. Hobbyists and amateurs are today still able to make small to moderate scientific contributions to their field. Is the same true in chemistry?
I think there might be at least two restricting factors:

The equipment needed to make scientific discoveries is far too expensive (e.g. hundreds of thousands of dollars/euros/whatever) for an amateur to obtain
Not having a bachelor's + master's degree (so being ready for a PhD) is too limiting ("too much to study alone").

If there is no "correct" answer for the field of chemistry in general, please post an answer per bigger sub-field (e.g. organic chemistry).

I like the question very much. No reason to close it in my opinion. It could be too broad, but that is no restriction for a short summarizing objective informative and (widely) applicable answer. +1

The damn meth cooks have complicated home chemistry. Try explaining the lab in your basement to the cops.

If you consider all the topics that can earn a Chemistry Nobel price, e.g. biochemistry, materials science, there are a lot of unexplored field where tinkers and hackers can have fun and somewhat meaningful projects.

No wonder why every good question is off-topic on S.E.! Can't we make a few exceptions?

I think it is a nice question, with a good intent. Does chemistry.SE have the option for a community wiki?

No, impossible. The rare genius who could become a chemist (physicist, etc.) without direct contact to academia will at some point, before he can go beyond undergrad knowledge, be found and snatched away by academia. Our planet is too small for that to not happen today.

It's conceivable that somewhere out there, someone with a completely fresh point of view could have an annus mirabilis in an area like theoretical chemistry, working with just a paper and pencil (or a laptop) and home-brewed equipment and resources.  Yes, for most of us, all of the low-hanging fruit is gone---but for a real giant, the whole tree might be full of low-hanging fruit.
Most of us aren't giants, but as Newton said, we can stand on the shoulders those who are. Doing that requires some ability to search the extant literature, even for "small to moderate contributions".
In addition to the obstacles you've listed above, there is a third impediment: paywalls.
There was a time when you could walk into a university library and sit down with the Science Citation Index and do a good job of finding research related to yours. Nowadays, you need a faculty or student university ID to do such a search and retrieve the papers you need. Without an ID at a good university, our amateur, hobbyist, or science buff will  hit paywall after paywall. Amateurs might have to pay half a day's wages or more for the pdf of a single paper!
Paywalls are not only a major obstacle for amateurs---they discourage professionals in the third world, and they shut out young people who are beginning their careers outside of university.
Consider Jack Andraka's story, here: http://blogs.plos.org/thestudentblog/2013/02/18/why-science-journal-paywalls-have-to-go/

The UMCP library website visitor section says "Databases:  On campus, anyone can access the databases without restriction." http://www.lib.umd.edu/about/visitors

Nice to know for amateurs in Maryland. It also says that "Off campus, only currently registered UMD students and currently employed faculty and staff can access the site-licensed databases."

I think the libraries that are part of the federal depository library system have to allow public access, so there should be similar access throughout out the USA, I just picked that one because of your location

"Anyone can visit a Federal depository library and will have access to all collections for free" http://www.gpo.gov/libraries/?utm_source=rss&utm_medium=rss&utm_campaign=federal-depository-library-program

I didn't know that, thanks---I'll have to actually try it without my ID.

I have to say, though, even with my ID at my school, I still hit plenty of paywalls.

Thankfully since this answer was made the situation has improved and we now have sites that allow you to circumvent these paywalls

Eric Betzig, one of this year's Nobel laureates in chemistry, did much of the work for which he won the prize (superresolved fluorescence microscopy) in his living room .

He does have a PhD in physics, and OP seems to be saying not even a bachleor's degree

If he did the work on his own time and not as part of an institution, then wouldn't that mean he is an amateur? But not a "rank" amateur.

I would consider someone who got at BA from Caltech, PhD from Cornell, then worked for Bell Labs, then was a VP of R & D, and then worked in the living room of his colleague from Bell Labs to be self employed, definitely not an amateur.

Back in the early days of chemistry, it was all done by "amateurs". And, as far as what costly equipment is needed--go back and look at an old pre 1900 textbook like Fownes' -- it is amazing how much chemistry they got right without NMR, IR, MS, HPLC, GC, etc.

I'm a mathematician by training, and I think math is a decent analogy to your question. The answer is that yes, amateurs could certainly produce new insights, but this rather low probability. The reason is that humanity has been applying the scientific method to all sciences for a long time, resulting in many new results. But this also means that the "low hanging fruit" in all sciences have been heavily picked. It's simply the case that most of the remaining problems are hard in any scientific discipline. You may still be able to find some non-researched problem, but this is increasingly unlikely. And if it is non-researched, it is likely very hard. So this does not rule out amateurs, but it does set a high bar.

With experimental chemistry, simple the financial constraints of running a lab can be a serious issue. Math is not an experimental science, and you don't need a lab, only paper and pencil, literally.

What about the Mpemba effect? Does that count as chemistry? Maybe it's more physics or chemical engineering.
Erasto Mpemba fits the definition of a total amateur. He was in  high school back then.
OTOH, 1963 was half a century ago now.
At least in Chemical Engineering, a fair bit of innovation happens still by amateurs. Ok, not total amateurs, but plant operators, entrepreneurs, businessman, scrap dealer, contractors, foreman,  et cetra. i.e. someone outside the conventional BS-MS-PhD academic system.


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Choosing a balance for home chemistry
A good balance is an important piece of equipment for chemistry. For home chemistry, the price of the balance is unfortunately limiting ..
How should one go about picking a balance (for home chemistry; general guidelines are welcome too)? Specifically, I'm thinking whether to buy a model with 0.01g precision (up to 200g), or a one with 0.001g precision (up to 20g) (the latter is a so called "diamond balance" but probably does the job). Which one would serve me better?

Home chemistry can be difficult sometimes due to the expense of the essential working materials and instruments required. As a mathematician all I need is a sharp pen (which is sometimes still quite difficult to obtain!) and blank paper. I can give you the following general (personal) advice regarding purchasing chemistry lab equipment:

Determine your budget and the needs for the experiments you want to perform. Do you really need a scale that has to weight with an accuracy of 0.001g or is 0.01g sufficient (or even 0.1g)? Ask yourself what you REALLY need for an experiment BEFORE searching for instruments. If you do it in the converse order you often buy something way more fancier (and pricier) than what you actually needed in the first place.
Look at specialized chemistry equipment supply websites to orientate on what is available and in what price range certain products are (think of scales, chemicals, glassware, etc.)
A lot of stuff can be bought incredibly cheap on websites where you not expect (at least I did not) people to sell lab supplies/instruments. Think of ebay.com and amazon.com as examples. They sometimes list items at a fifth of the usual price that would have to be paid at a specialized store.
Take a look at local non-specialized stores like the construction market (you know, the place where they sell all kinds of construction material). They sometimes hide hidden (cheap) gems! As an example I was able to obtain a vacuum source (membrane pump) for < 100\$, which would have cost me easily > 400\$ at a chemical supply website.

About your particular question: I really recommend a 0.01g precision 200g (or 600g) scale. It brings the benefit of having relatively accurate weighting for even small samples, while still being able to weight bulk mass. Furthermore, they are most common in the lab/home environment and hence relatively cheap in comparison with professional analytical scales. If you plan to do a lot of microscale experiments (yielding a product < 200 mg) the analytical scale (precision of 0.001g) may suit your needs better.
I wish you much success, think and buy wisely!

Let me first say that @Jori's answer is excellent. I would only add two things.
1) A range of equipment is usually desirable if it is also affordable. So measures across all 3 accuracies (0.1, 0.01, and 0.001) is what you want ideally, particularly given that precision tends to decrease with absolute mass.
2) There is a trend towards green chemistry, which emphasizes minimal reactants, products, and waste whenever possible. Thus, there will be a trend towards microchemistry where small amounts of reactants are the norm, and this will necessitate higher precision balances (i.e 0.001g precision).


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What is the mechanism by which sugars glycate proteins?
I was looking at why it's ok for glucose to circulate through the blood while fructose is prevented from doing so by the liver.  Fructose glycates proteins more readily than glucose.  This is the opposite of what I expected as glucose is an aldose while fructose is a ketose.  It seems reasonable to expect that the aldose would be more reactive than the ketose in many reactions.  What is the mechanism of glycation of proteins?

What is the source for the idea that fructose is more active at glycation than glucose?  I found an in vitro paper that shows the opposite.

One mechanism of glycation is the Maillard Reaction.
The mechanism starts with nucleophilic attack of the sugar carbonyl group by a side chain amino group from the protein.
The Milliard Reaction occurs in the human body and has been implicated in various adverse health conditions.
The cornea, lens and retina are known to be damaged by this reaction.
See The Maillard reaction in the human body. The main discoveries and factors that affect glycation Pathologie Biologie, Volume 58, Issue 3, Pages 214-219 for a recent review.


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The standard for drinking water is 1300 mg/l and the first area has a value of 837 µg
I am trying to figure out the hazard quotient of copper compared to the standard. The standard for drinking water is 1300 mg/l and the first area has a value of 837 µg. How do I convert this?
I use the formula hq (hazard quotient) = contaminant (c) divided by standard (s).
or $$HQ= \frac{C}{S}$$ I am having problems with the conversion.

Welcome to chemistry.se! See the [help] for questions about formatting and take the [tour] to learn about the scope of the page. Unfortunately I have no idea what you are talking about, so I cannot help you., sorry.

I don't know what this standard or the "first area" are, but the hazard quotient is simply the ratio of potential exposure to the maximum exposure without adverse effects. You can't take the ratio of the two values you have because 1300 mg/L is a concentration and 837 µg is an absolute quantity. You need to first decide how much water your potential exposure will involve and multiply that by 1300 mg/L.
Incidentally, 1300 mg/L of copper shouldn't be any kind of standard level in drinking water. That's more like a standard solution for AAS or ICP, is definitely bright blue, and probably contains several percent of nitric acid.

the US EPA drinking water standard is 1300 micrograms per L, so he/she is off by a factor of 1000. http://water.epa.gov/drink/contaminants/

Yes, that seems more reasonable.


text:
Required HLB on the Griffin Scale
What is the required HLB on the Griffin Scale of 94% water and 6% mineral oil for producing a stable micro-emulsion?

The required HLB doesn't depend on the concentration of water. The required HLB of mineral oil is ca. from 9 to 12. You have to find out the required HLB of your mineral oil at first.


text:
Why don't we use an alkyl halide directly in Gabriel Phthalimide Synthesis?
In Gabriel Phthalimide Synthesis, why do we use $\ce{KOH}$ in first step rather than directly adding an alkyl halide?

It is used to make the conjugate base of the imide which is far more nucleophilic than the imide itself, but not too much to attack another molecule of the alkyl halide.

The goal is to make a primary alkylamine without causing polyalkylation. If you do not activate the pthalimide nitrogen with a base, you have a highly delocaled lone pair on the nitrogen. At best, you get an Sn1 reaction, however that will not give a primary amine as loss of the halide (if you can force it to go) will undergo Wagner-Meerwein rearrangement (hydride shift) to give a secondary or tertiary amine.
This is the point of the Gabriel phthalimide synthesis.


text:
How can the chlorophyll concentration of plants be measured?
Is there a way to measure the chlorophyll content of plants (from e.g. leaves) with basic chemistry equipment (at home)?

To measure the total chlorophyll from leaves, probably not, since you need access to methyl isobutyl ether or pyridine.
But you can separate the different pigments in the leaf. From The Naked Scientist: Cut a rectangle 1 cm wide and long enough to stand up in a small jar from a coffee filter. Place your leaf over the top of the strip and roll the coin across the strip about 2 cm from the bottom. Now add a little nail varnish remover to the bottom of your jar, half a cm is plenty. Suspend your strip of paper so the bottom end is sticking a few mm into the acetone. you can hold it up by folding the top end of the strip over forming a hook which will rest over the lip of the jar. Wait a few minutes, with any luck you should see an interesting effect. Try the experiment with a strongly colored leaf like a purple one.


text:
Increasing the length of a carbon chain in an alkane
What would be the best reaction to increase the carbon chain length in an alkane? Preferably, it would be by one at a time; however, if that is not possible, just generally increasing the total length of the carbon chain (such as producing pentane from propane) would be fine.
Moreover, is there a way that I can add a methyl group to a compound in the form $\ce{R-CH3}$?

Adding alkyl groups to alkanes cannot be done cleanly in a one-step process.  You can do things like add Gridnard reagents to carbonyls and then reduce the resulting product, but these are 2-3 step reactions.

One method could be metathesis of alkanes. There is a nice paper about it published by J. M. Basset et al., SCIENCE 1997, 276, 99-102, "Metathesis of Alkanes Catalyzed by Silica-Supported Transition Metal Hydrides" and for sure more recent ones have been published since then.

Abstract
  The silica-supported transition metal hydrides $\ce{(≡Si-O-Si≡)(≡Si-O-)2Ta-H}$ and $\ce{(≡Si-O-)xM-H (M, chromium or tungsten)}$ catalyze the metathesis reaction of linear or branched alkanes into the next higher and lower alkanes at moderate temperature ($\pu{25^\circ}$ to $\pu{200^\circ C}$). With $\ce{(≡Si-O-Si≡)(≡Si-O-)2Ta-H}$, ethane was transformed at room temperature into an equimolar mixture of propane and methane. Higher and lower homologs were obtained from propane, butane, and pentane as well as from branched alkanes such as isobutane and isopentane. The mechanism of the step leading to carbon-carbon bond cleavage and formation likely involves a four-centered transition state between a tantalum-alkyl intermediate and a carbon-carbon bond of a second molecule of alkane.

Homologation methods exist, but typically not for unbranched alkanes.
A typical example is the Arndt-Eistert homologation of carboxylic acids.
For simple unbranched alkanes, stepwise chain elongation is not a technical relevant process. On the contrary, petrochemical processes reach for the opposite (formation of smaller alkanes) by thermal or catalytical cracking.
Formation of long chain alkanes, although not via subsequent step up reactions, is possible by catalytic oxidative dimerization of terminal alkynes (Glaser, Hay or Eglinton coupling) and subsequent hydrogenation or by Kolbe dimerization of carboxylic acids.
Branched alkanes, such as triptane (2,2,3-trimethylbutane), which are interesting as 'anti-knocking' fuel additives, are another league. Here, the source of the methyl group to be transfered is methanol, catalysts are either zeolites or indium(III)-iodide ($\ce{InI3}$) (DOI).

What about Wurtz reaction? That can be used for increasing the chain length.

@Gummybears The intermolecular Wurtz reaction can't be used for homologation. At least, not in a sane way.

@Gummybears You describe an intermolecular reaction! An intramolecular would take place between two centres of the same molecule.

Oops, sorry. Yes I got confused. But why can't it be used to increase the chain length?

@Gummybears Methylbromide is a gas under standard conditions, which makes life less easy. In addition, consider the mixture of A-Br and B-Br in the presence of sodium. This leads to A-A + A-B + B-B. More trouble in separation. In summary, this is not a recommended method.

Hmmmmm... So basically, there is no viable reaction to increase the length of the carbon chain, or add on a methyl group to an organic compound?

@Gummybears There are lots of methods, some of them suggested in my answer or the anwers of others. But your prerequisite was to use an alkane as a starting material. Typically, you' use these as unpolar and mostly inert solvents, but not as starting materials ;)

Yes, I wanted to use an alkane to add on to. However, in all of the methods provided, the same issue occurs, that a variety of products are obtained. And these products are difficult to separate.

Carbenes, like $\ce{CH2}$, are known to insert into everything, including, in case of absence of better substrate, carbon-hydrogen bonds. However, the process is not clean, as carbene will insert into any suitable position it can find, with no selectivity. Carbenes are usually produced by decomposition of precursors, usually ketene ($\ce{CH2=C=O}$) or diazomethane ($\ce{CH2=N=N}$).
In case there is some group in the compounds, it is often possible to selectively substitute hydrogen, activated by the group. For example, many electronegative groups, like carbonyl group, activate nearby hydrogenes, making them quite acidic by organic chemistry standards. Produced anion can react with equivalent of carbocation. Carbonyl group may be later eliminated by Wolff–Kishner reduction (heating with hydrazine and alkaly metal hydroxide)
However, there is no general way to increase chain of alkanes cleanly and selectively.

What about Wurtz reaction? And for the carbene example you gave, the carbene group would want to add to an $\alpha$ carbon I'm supposing. For example, if we have prop-1-ene, the carbene would add on to create a branched alkane?

@Gummybears Wurtz works with haloalkanes, and is... messy (i.e. unselective) and requires dangerous reagents. Carbenes insert literally into everything, but prefer double/triple bonds, including cojugated into aromatic systems. It is possible to SOME extent brominate tertiary position in alkane and then exchange halogene to selected alkyl group using dialkylcuprates, but it is not a general route. Alkanes are known as chemical corpses for a reason.

They're known as chemical corpses? Well that's something new. And by the way, what is diakylcuprate?

@Gummybears http://en.wikipedia.org/wiki/Corey%E2%80%93House_synthesis\

A question: When we add lithium metal to the alkane, is there a position where it will selectively join? Hopefully the end of the chain? Or is there a possibility that with this method we get a mixture of isomers?

@Gummybears you don't add lithium to alkane, you have to constract alkyl halide first. In case one needs to construct a linear alkane chain, one probably should start with fatty acid with close number of carbons, then add required amount of carbons, and than eliminates leftover groups There are numerous synthetic methods for the task, but they rely on reactivity of other compounds than alkanes.


text:
Bonds within complex anions
Many groups of bound atoms complexively bearing a (negative) charge are called complex anions. Examples are $\ce{NO3-}$, $\ce{CO3^2-}$, $\ce{BO3^2-}$, $\ce{CrO4^2-}$, $\ce{Cr2O7^2-}$, $\ce{MoO4^2-}$, $\ce{WO4^2-}$, $\ce{PO4^3-}$, $\ce{AsO4^3-}$, $\ce{VO4^3-}$ and the several, sometimes polymeric, complex anions occurring in silicate and borate minerals.
I apologise if I am asking something trivial, but I cannot find a source, except for silicates, for which I read that the $\ce{Si-O}$ bonds are basically covalent, explaining whether the bonds within complex anions are covalent in general.
Are they always covalent? If they are not covalent in general, are they in the examples I have written?

All these anions have polarised covalent bonds.

The notion of the mentioned anions being "complex anions is that they don't dissociate if dissolved in water. Hence they have covalent bonds not ionic ones.

All complex anions you noted that include only non-metal atoms are always considered covalent. Except for a few extreme examples including fluorine and hydrogen, non-metals only form predominantly covalent bonds between themselves.
It gets a little more tricky when considering those anions whose central atom is a transition metal. In this case at first approximation, we can consider the anion a coordination compound and the coordinate bonds within coordination compounds are typically considered more covalent. Moving on to a second approximation clarifies the picture: the formal oxido-ligands are in fact very strongly attached to the metal and do not undergo ligand substitution. This means that they are even more covalent than other coordination complexes.
Therefore, it is typically safe to consider the bonds within complex anions to be predominantly covalent.

"Electronegativity" is a measure of the power of an atom in a molecule to attract shared electrons to itself. If you look up the electronegativities for the atoms on either end of a bond, there's a quick and dirty way to decide for yourself how much covalent character the bonds have.
Calculate the electronegativity difference between the central atom and oxygen. The smaller the difference, the more covalent character the bond has. For example, for Si-O, Si and O have Pauling electronegativities of 1.8 and 3.5, respectively, so the electronegativity difference is 1.7.
Some textbooks set (rather arbitrary) thresholds for electronegativity differences to help general chemistry students classify bonds as non-polar covalent, polar covalent, or ionic. Pauling's original text says that if the electronegativity difference is 1.7, the bond has about 50% ionic character, so you can say an Si-O bond is about 50% ionic, 50% covalent. If the electronegativity difference is greater than 1.7, you can call it "predominantly ionic"; if it's less than 1.7, you can call it "predominantly covalent".

Therefore, if i correctly understand, the only ionicly bound complex anions would be those with Cr-O and V-O, having chromium and vanadium respectively electronegativity 1.66 and 1.63 versus the 1.8 of oxygen. Then we observe that $\text{CrO}_4$, $\text{Cr}_2\text{O}_7$ and $\text{VO}_4$ "get some electrons more" as a whole because they are anions. Thank you very much!!!

You're welcome! But oxygen's Pauling electronegativity is 3.5 (a typo in your note?) And it isn't black and white covalent or ionic, if you have an electronegativity difference of 1.8, that means about 55% ionic, 45% covalent.

Yes, a typo in my comment, I should have said "1.8 of silicon" and, if an element is less electronegative than Si, its bonds with O are more ionic than covalent. Thank you again!!!

@Fred Senese I'm afraid that your answer is too long or too short, but definitely misleading - atoms in different oxidation states exhibit different "effective"  electronegativities. Also Si has "standard" Pauling  electronegativity 1.90 and oxygen 3.44.

@Mithoron, well, there are many different electronegativity scales, refinements, and corrections that could be used. I thought it best to answer the question at the genchem level, and I used Pauling's (original) electronegativity values in my answer, and I followed his approach as well.


text:
Prevent crystal growth by adding a chemical to a solution?
If I have a solution of dissolved $\ce{NaCl}$ or sugar (or some other ionic compound), is there a chemical I could add to it that would prevent crystal growth if I boil it down? I'm looking for it to remain as a very fine powder or to keep the crystals as small as possible.

Crystal growth isn't that easy to have- fine (mm) powders is normal with precipitation. If you seeking smaller than that, that is hard because the powders aggregate to larger particles.

Spray it as a fine mist into a drying chamber - the solvent will evaporate and you are left with a powder.
No chemical involved though,  but still process chemistry.

If you decrease the solvent volume, the dissolved material will start to precipitate.
Solubility most often increases with the temperature of the solvent.
Slow decrease of the temperature or slow evaporation of the solvent at constant temperature mostly leads to slow crystallization, starting from a small(er) number of nucleation centres and typically results in the formation of less but larger, well-formed crystals.

Rapid decrease of the temperature with additional scratching the surface of the crystallisation vessel (beaker, etc.) with a glas rod, stirring, or sonification will typically result in precipitation of the solute in the form of smaller particles.


text:
Good choice for heat-conductive coil submerged in sea water
I know only as much about chemistry as I can remember from a year in college.  So, please speak slowly and use short words.
I'm trying to decide on an appropriate material for a heat-transfer coil that is meant to remain submerged for extended periods in salt water, within a few feet of other metal parts.
I'm designing a solar still that will use a cooling/condensing coil to take advantage of the heat differential between the solar-heated evaporation chamber and nearby (cooler) water temperature.  It will be mounted on a sailboat.  The coiling coil will probably be about 12-18 inches long and 2-3 inches diameter.
Impulse was to use copper, but I'm not sure how this will react in a salt-water environment over time.  I know that sea life will stay off of it, since copper is the active element in many bottom paints designed to minimize subsurface growth.
Any help is appreciated.  Maybe the answer is to use a sacrificial anode on the coil?  Or, maybe some other material would be more suitable for long-term use?

My advice is to use copper and protect it with a sacrificial anode.  Copper would be the choice in a non corrosive environment and if you can protect it go for it.
However, if the coils will contact parts of the boat that are important than the coils should be aluminum.  Copper would cause the other metal parts to corrode faster.
Why not use rubber or plastic, something like the pipes used in drip irrigation?

I've leaned away from rubber/plastic for two reasons.  First, those materials are way less conductive, with thermal conductivity numbers in the range of about 0.2 W/(m K), as compared to copper at around 400 W/(m K).  But, secondarily, such a surface would need to be coated with a protective paint to reduce slime and barnacle adherence.  And, such a coating would further reduce thermal conductivity.Would zinc be a good choice for an anode?  Or would that be the opposite of a good choice?  Zinc is used elsewhere to protect steel parts.

Take a look at a table of standard reduction potentials.  They are written as an ion being reduced to metal.  You want the most negative number for an anode, the more positive number for what is being protected.  Copper is .34 (volts), zinc -.76, aluminum is -1.71.  Aluminum can protect zinc and zinc can protect copper.  It sounds like protecting copper is your best option.

have you thought about using cupro-nickle, which is used exensively in marine engineering. I am aware it is expensive, but further research should reveal its suitability. I hope that you have found this helpfull.

Wow.  You're not kidding about it being expensive.  Looks like somewhere around $100 USD for what I would need.  Though, it may save money in the long run.  Thanks for the tip!


text:
Possible products from anion analyses
What are possible products from these two anion analyses?

The nitrite test, reacting diphenylamine with a nitrite:

$$
\ce{(C6H5)2NH + NO2- -> ?}
$$

The sulphite test, reacting potassium permanganate with a sulphite:

$$
\ce{KMnO4 + SO3^{2-} -> ?}
$$

Under acidic conditions, a blue colour is observed. This results from the formation of N,N'-diphenylbenzidine. In the course of the reaction, diphenylamine is oxidized to a quinoid radical which subsequently adds to another molecule of diphenylamine.
Under slightly acidic conditions, bleaching of the solution is observed. Sulfite is oxidized to sulfate while $\ce{MnO4-}$ is reduced to $\ce{Mn^{2+}}$.

No...i want the full reaction

@NeilRoy With other words: Do my homework. No, because... NO!

Well can you please give me the reaction. It is not homework ... it is given in my practical chemistry book but no reactions were given with it! So please help


text:
Hybridization - Mixing one atomic s orbital with 2p orbitals
Explain why three ${sp^2}$ hybrid orbitals result when one s atomic orbital is blended with two p atomic orbitals.

When some number of atomic orbitals are mixed together to produce hybrid atomic  orbitals, the same number of hybrid atomic orbitals must result.  So, when N atomic orbitals are mixed together, N hybrid atomic orbitals are produced.  In the case you describe, mixing one s atomic orbital with two p atomic orbitals means that three hybrid atomic orbitals must result.
These three hybrid orbitals will each contain 33.333% s character and 66.666% p character because that is the ratio of the one s orbital to two p orbitals that we are mixing together.  We can describe such an orbital as $\ce{s^{33.333}p^{66.666}}$, but if we reduce that notation to the lowest common denominator we get $\ce{s^1p^2}$, or the commonly used description $\ce{sp^2}$.  This simply means that this hybrid orbital is composed of one part s character to two parts p character.
To summarize if we mix 3 atomic orbitals, one s and two p orbitals, then 3 $\ce{sp^2}$ atomic hybrid orbitals will result.


text:
X[Y|Z] formulas
I often find formulas of minerals where some element symbols are grouped within square brackets and separated by a vertical bar. An example can be brochantite, whose formula I find expressed as $\ce{Cu_4[(OH)_6|SO_4]}$.
I wonder what does such a notation means, and I have made some conjectures, although I find no reference on line or in books. I suppose that the square brackets are used to enclose the set of the anions present in the compound and the vertical bar is used to separates different anions, while the cations remain outside the square brackets. That is just an intuitive supposition. Is my supposition correct? If it is not, what does the described notation mean?

I'm not sure but wikipedia has just written a normal formula for Brochantite:  $\ce{Cu4SO4(OH)6}$. Could you give me some reference of where you saw that formula?

@MARamezani Thank you very much for the comment! I've found this particular formula in an Italian language book: A. Mottana, R. Crespi, G. Liborio, "Minerali e rocce" ('Minerals and rocks'), but I've found similar notations in several other books of earth sciences.

Maybe, and I say maybe, they wanted to demonstrate something like, for example, brochantite has $\ce{OH-}, \ce{SO4^{2-}}$ and copper.

@MARamezani Thank you again! I've just found other similar formulas in Mottana-Crespi-Liborio's text: Cu[UO$_2$|PO$_4$]$_2\cdot$12H$_2$O for torbernite, K$_2$[(UO$_2$)|V$_2$O$_8$]$\cdot$3H$_2$O for carnotite... but UO$_2$ is cation UO$_2^{2+}$, not an anion. Therefore my initial supposition that the brackets contain anions was wrong...

Now that I think about it, I see that two brackets like this ][, if not awkward, would've looked way messier than a | . Maybe it was the author's or the publisher's attempt to make chemical formulas look a little bit neater.

@Self-teachingDavide Since $UO$ exists as a compound, then $UO_2^{2-}$ can also exist. http://webelements.com/compounds/uranium/uranium_oxide.html

The IUPAC rules for punctuation marks in chemical formulas don't mention [|], so that notation must be peculiar to geology. They need different notation; chemists write  formulas for molecules, while geologists focus more on complicated superstructures built from smaller structures that can be arranged in different ways, even within the same superstructure.
This paper describes the structure of brochantite as layers and chains of copper(II) ions octahedrally coordinated with hydroxides. The octahedra in different layers and chains  are crosslinked by sulfate ions.
Perhaps the notation $\rm [(OH)_6|SO_4]$ is an attempt to indicate this arrangement. The brackets denote the electronegative part of the structure, with what comes before the vertical bar representing directly coordinated ions, with what comes  after representing ions that aren't directly coordinated.
I'm speculating, though. Sorry. I would have put this into a comment, but it's too long and I wanted to give you the links to think about.


text:
How can I increase the signal intensity and precision?
How can I increase the signal intensity and precision when use HPLC or LC-MS/MS ? How Also prevent Contamination and Carryover during Analysis

Welcome to chemistry.SE! You can visit the help center if you had any questions about the policies of this site. It might help you get answers. Or, you can just take a tour.

This is any extremely broad question and you could probably write a book to answer it. Also, strategies for optimizing an HPLC or LCMS method strongly depend on what you are analyzing, the method that you are using, and the specific problems you are seeing. If you have more specific questions, you might get some answers.


text:
Is it possible to calculate atomic radius with electron configuration?
I need to know whether is it possible to calculate the atomic radius according to the number of electrons and electron configuration. Or is there any way to calculate the atomic radius using common characteristics which an atom has?

You can find the effective nuclear charge ($Z^$) using Slater's rule and then plug it into Bohr's formula ($r = \dfrac{53 n^2}{Z^} \pu{pm}$). This value is quite close to the covalent radius.

There's no measuring the radius of a single atom, mainly because electrons are around the nuclei: We can't define a radius for any single atom. Take a loom at the uncertainty principle:

Introduced first in 1927, by the German physicist Werner Heisenberg, it states that the more precisely the position of some particle is determined, the less precisely its momentum can be known, and vice versa. The formal inequality relating the standard deviation of position $\sigma _x$ and the standard deviation of momentum $\sigma _p$ was derived by Earle Hesse Kennard later that year and by Hermann Weyl in 1928: $$\sigma _x \sigma _p \geq \frac{\hbar}{2}$$ Wikipedia

Thus, if you want to measure an atom's radius, you have to measure the distance between two nuclei, and then divide it by two. This results in different types of radius defined for atoms. The most common are covalent radius, ionic radius, and Van der Waals radius. Radii trends are what that are mostly studied in undergrad chemistry, rather than the how of measuring them with advanced techniques.
Chemguide is pretty informative when it gets to teaching the trends:
chemguide: atomic radius

Can you explain the "Calculated" column on the Wikipedia data page to the atomic radii of the elements. What are they calculating, and how? Obviously, judging from the empirical column, the calculated radii are WAY off on most of the elements, so their calculations were suspect, but what were they calculating?

@Jimmy The source in Wikipedia used a quantum chemical model to 'solve' the Schroedinger equation for atoms with a certain number of electrons. Simply put, they used a special method to solve a complicated equation using computers to predict how electrons behave around atoms. The error is high because the paper is from 1967. The methods used have since improved considerably to decrease that error.


text:
Is toothpaste solid or liquid?
My teacher didn't answer this properly:

Is toothpaste solid or liquid?

You can't say toothpaste is a solid because solid material have a fixed shape but toothpaste doesn't. However, you can't say it's a liquid because liquids flow easily but toothpaste needs a certain force to push it out of the tube. So is it a solid or liquid? And are there any other example just like toothpaste?

Though the definitions might seem similar, "solid" and "liquid" (especially "liquid") have crude meanings for anyone. I gave up using them a while ago. :)

just out of curiosity, what "improper" answer did your teacher give you?

Its a colloid. Intermediate between solid and liquid state

She says it is plasma

@Simon-Nail-It That certainly is an improper answer. It may not be obvious what toothpaste is, but it's definitely not a plasma.

@iamnotmaynard  His teacher probably did not mean the state-of-matter or phase Plasma, but rather the archaic/colloquial term (more often "Plasm") for a highly viscous/collodial liquid.

As in "Blood Plasma".

May be semi-solid, like butter.

Toothpaste is what is called a non-newtonian fluid, more specifically toothpaste is a Bingham plastic.  This means that the viscosity of the fluid is linearly dependent on the shear stress, but with an offset called the yield stress (see figure below). This yield stress is what makes it hard to say whether it is liquid or solid. The fact that toothpaste is viscous alone is not sufficient to explain this, because water is also viscous, but doesn't behave like a solid (unless frozen, but that's another phenomenon).

What the yield stress does is the following. Below a certain shear threshold the fluid responds as if it were a solid, as you can see happening when you have put toothpaste on your toothbrush, it just sits there without flowing away. A highly viscous but newtonian fluid would flow away (although slowly as pointed out by @ron in his comment to the answer of @freddy).
Now if you put sufficient shear stress on the toothpaste, when you squeeze the tube of paste, it will start flowing and respond as a liquid.
Other examples, as mentioned in the Wikipedia link in my first sentence, are e.g. mayonnaise and mustard. Another example is silly putty.

I tried explaining this to the TSA, they didn't buy it.

@AnubianNoob What is a TSA? Some sort of teaching assistant?

And what exactly didn't they buy about it? Just curious :)

@Michiel, TSA is the American border agency. There are rules dictating the volume of liquid you're allowed to take on a plane with you.

@Trengot it is not a border agency, but a transportation authority. Any mass transit system (planes, trains, boats) has TSA agents at security check points, not just those entering or exiting the borders of the nation.

@TylerH, Thanks, I stand corrected (although it was close enough to explain the joke).

Mayonnaise is a liquid? I thought that emulsions are outside of the pure solid/liquid/gas scale.

@Trengot Sure, I just wanted to make the distinction in case the definition you gave is used elsewhere :-)

I just have to post this non-newtonian fluid video here, sorry: https://www.youtube.com/watch?v=3zoTKXXNQIU

Here's a genchem-level answer for a genchem-level question about the classification of matter:
Toothpaste is a sol: a stable suspension of tiny solid particles in a liquid. When the toothpaste dries out you can see what the solid part alone looks like.
Mixtures with more than one phase often have interesting properties and behaviors that the components alone don't have; the other answers to your question touch on some of these.
Other examples of sols are paints and solid-pigment inks. Again, you can see the solid part when they dry.

Here's the boring answer: Toothpaste is a mixture of some solids and some liquids.
The question "is it solid or liquid?" makes sense when you're talking about a substance or a mixture with a single phase—that is, a substance or mixture that's pretty much completely uniform throughout space. Examples of single-phase materials include pure water (a liquid), a chunk of copper (a solid), and sugar water (which is a liquid—including the sugar).
But some materials consist of multiple phases, and in this case, we have to ask the question "is it solid or liquid?" separately for each phase. Examples of multiple-phase materials include ice water (a mixture containing both a solid (ice) and a liquid (water)), vinaigrette (a mixture of oil (a liquid) and vinegar (another liquid)), and toothpaste (a mixture of several kinds of solids and liquids).

It is Viscous.
Viscous mean "having a thick, sticky consistency between solid and liquid"(dictionary meaning)
There are many more examples like tomato ketchup, honey, wax, toothpaste, etc.

To know more check out Wikipedia

It is all about viscosity and the timescale of the measurement. On a timescale of a microsecond, toothpaste seems pretty solid; on a time scale of centuries, glass seems somewhat liquid.

Where did you find this definition of viscous? Because I believe it is wrong. The physics definition of viscosity is the degree of resistance to gradual deformation by a shear stress, thus for a liquid to be viscous means it has a resistance to deformation, but it doesn't mean anything in the context of a solid.

@Michiel I got it from Google. Just type "viscous meaning" It's normal dictionary meaning. Click here

@Freddy well, apparently Google picked the definition from a dodgy dictionary, because any physicist will tell you that this definition is wrong.

"glass seems somewhat liquid" - not at room temperature: http://www.cmog.org/article/does-glass-flow As @Michiel explains, viscosity is not sufficient to explain what is going on.

@Michiel I agree with "non-newtonian fluid." On thinking more I realize we cant use viscous as a state. Sorry for that :(

So, it is a viscous liquid?

Toothpaste is a colloid. just like dust particles suspended in water form suspension, colloids are much finer particles suspended in a medium. for example jelly, it is solid particles suspended finely in a liquid....and foam; that is gas suspended in a liquid. As such is deodorant spray that is liquid suspended in air(gas)

Actually, toothpaste is both. I'm no chemist, but I am pretty sure that it can be correctly classified as a semisolid, which means exactly what you'd think. Semisolids have properties of both solids and liquids. Slime would be another example of a semisolid.

Toothpaste is a colloid, more specifically a type of liquid dispersed in a solid dispersing medium. The solid component is much more in proportion than the liquid one, so the paste is quite thick. You can't really call it a solid or a liquid, its a combination of both. As Michiel says, it may be considered as a non-Newtonian fluid. If you have seen a toothpase, you could observe these:
Has no definite shape (liquid property)
Cannot flow (solid property)
Slight tendency to decrease surface area (liquid property)
So you can see that it has properties similar to both solids and liquids. Its a colloid of a liquid in solid.