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If copper has started to corrode, can it stop?
I've recently obtained a couple of electronics units which have suffered corrosion on their copper inductor coils. They were operated in conditions of intense humidity/heat and were placed near the sea. This has resulted in the complete failure in one of these units, where the inductor has come apart due to being eaten away by corrosion, and the other coil shows noticeable amounts of corrosion (patina present).
If I were to operate the latter device in a more forgiving environment (ie: putting a filter in front of it in an air conditioned room) will the copper continue to corrode regardless or will the oxide layer stop spreading further?

Unlike rust on iron, patina naturally protects underlying copper, and stops further corrosion directly underneath.

The previous answers doesn't take into account the environment:

They were operated in conditions of intense humidity/heat and were placed near the sea.

Near the sea means that there are certainly a lot of $\ce{Cl-}$ ions in your metal. Chloride ions are a very dangerous species for metals conservation they tends to depassivate the metal forming complex ions. In the case of copper occurs this series of reactions:
$$\ce{Cu^{+} + Cl^{-} <=> CuCl}$$
$$\ce{Cu^{+} + 2Cl^{-} <=> CuCl2- + Cl- <=> CuCl3^{2-} + Cl- <=>CuCl4^{3-} }$$
That lead to different copper corrosion product. (e.g. atacamite $\ce{Cu2(OH)3Cl}$, malachite $\ce{CuCO3.Cu(OH)2}$ (Did you see some blue-green pigment?), Cupric hydroxide $\ce{Cu(OH)2}$.
These are all dynamic equilibria so your copper with dry-wet, hot-cold cycle will continuously object of degradation until chloride ions are present.
Use distilled water with cotton to clean the salts deposit, dry quickly where you clean.
Reference:

The Open Corrosion Journal, 2009, 2, 71-76
Corrosion Science 46 (2004) 109–135

Can you tell me, what $\ce{Cu(COH)3}$ is?

@PH13: I fixed the formula for malachite.

Copper doesn't need a protective layer for atmospheric corrosion- its inherently inert. It might not be a patina rather sulphide pollution from sea air caused simple corrosion and so it won't affected in other environments. Yes rust is very weakly protective- only slightly physically protective as long as its not broken but rust is rather weak and non-adhesive.


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Experiment to show that air contains about 20% oxygen
I remember that in my primary classed I had an experiment in science to show air contains 20% oxygen. The experiment involved taking a trough of water, lighting a candle in the middle of the trough and inverting a glass jar over the candle so that air supply for the candle is cut of. When candle extinguishes, water level in the glass jar rises. I have seen this happen.
My question is why should the water raise in the jar, since burning of candle should produce an equivalent amount of carbon dioxide, so the molar amount of gas in jar is constant? Also the heat from the candle should expand the gasses in the jar, water level should go down I think.

This is not a good experiment because it makes a bunch of assumptions about combustion products.  The water should condense, the carbon dioxide will be less than the oxygen into the combustion, but some of that will dissolve in the water.  A better test might be to freeze out the nitrogen and see what is left.

@OlinLathrop The oxygen condenses before the nitrogen (at about 90 K).

The paraffin wax is an alkane ($\ce{C_nH_{2n+2}}$); its combustion reaction is (complete combustion):
$$\ce{2C_nH_{2n+2} + $(3n+1)$ O2 -> $2(n+1)$ H2O + $2n$ CO2}$$
or (incomplete combustion):
$$\ce{2C_nH_{2n+2} + $(2n+1)$ O2 -> $2(n+1)$ H2O + $2n$ CO}$$
In the two cases the quantity (in moles) of $\ce{O2}$ is greater than the quantity of $\ce{CO2}$ or $\ce{CO}$.
Moreover the gaseous $\ce{H2O}$ becomes liquid.

You mean that the rise in water level is due to the water produced during combustion?

It's due to 2 reasons: first, the quantity of gases present at the end of the reaction ($CO_2$ and $CO$) are less than the initial quantity of $O_2$ and so the pressure of the gas decreases and the water rises. Second, a small amount of $H_2O$ is added to the liquid water already present but I think this effect is very small compared to the first one.

@Gautam Moreover, I read now the following "It can not be that the oxygen depletion is responsible for the water raising, because the water does not rise immediately. The water rises only after the candle dims. If gas would be going away, this would lead to a steady rise of the water level, not the rapid rise at the end, when the candle goes out".

@Gautam It is not "only" the oxygen depletion which matters. There are two effects which matter: the chemical process of the burning as well as a physical process from the temperature change. These effects cancel each other initially. Since these effect hide each other partially, they are more difficult to detect. (source: http://www.math.harvard.edu/~knill/pedagogy/waterexperiment/index.html)

I would also expect that the heat of the flame of the candle will heat up the gas and expand, so some of the gas will escape through the bottom. So when the gas cools down again, the pressure will drop and water will be sucked up.

@fibonatic Look at the third source of error in http://www.math.harvard.edu/~knill/pedagogy/waterexperiment/index.html

One problem with using combustion reactions to determine the percent of oxygen in air is the additional products, namely $\ce{CO2}$ and water vapor.  If performing a water displacement reaction, these gases add to the uncertainty and imprecision of this method.
Water displacement by a combustion reaction can, however, be used if an appropriate combustion reaction is chosen.  A more reasonable experiment to determine the percent oxygen in air is to use the oxidation of iron $$\ce{4Fe(s) + 3O2(g) -> 2Fe2O3(s)}$$ The benefits of this reaction are (a) only one product is formed and (b) the product is a solid so it will not influence the water displacement in a significant way.
This experiment has been reported in the Journal of Chemical Education and a brief search of the web found an adaptation of this experiment if you don't have access to the journal.  Briefly, one measures the volume of a test tube, adds a known amount of steel wool into the tube and inverts the tube in a beaker of water.  The oxidation of the iron is catalyzed by first rinsing it in a dilute acid (vinegar) solution.  Knowing the density of iron one can calculate the volume of the test tube occupied by the solid and then determine the volume of air before and after the reaction is completed.
There are a number of possible errors that can be introduced into this experiment, but unlike the combustion of a hydrocarbon, these errors are much more manageable for an undergraduate or high school chemistry experiment.  Most significantly, the reaction occurs near room temperature so the temperature change of the gas can be ignored.

I have done this demonstration in class to talk about gas laws, but not to talk about the composition of the gas.
What is happening is the candle inside the glass is obviously hot and the flame with heat up the surrounding air.  As temperature of a gas increases, the volume of the gas increases.  This is why you might see some bubbles coming out of the bottom of the glass if you don't have a good seal.  The candle will continue to burn until the oxygen is consumed from the air.  Then when the candle goes out the temperature of the air will decrease.  As the temperature decreases the gas will contract, which causes a vacuum inside of the glass (or an area of low pressure).  Water will move from high pressure to low pressure because the pressure of the air is higher on the water outside of the glass than the pressure on the water inside of the glass.

I doubt that the CO2 dissolves quickly enough. My gut feeling is that this experiment is not working as suggested.
I think it is a physical, not chemical, effect.
When you put the glass above the candle, there is already a column of hot air. When that air cools, the water rises.
It would be a worthwhile experiment to ignite the candle electrically.
A better experiment would use steel wool, like bobthechemist suggested.
Btw. it is very easy to electrically ignite steel wool.

This is better off as a comment under the main post IMHO.

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Peltier Bismuth-Telluride toxicity
From what I understand, the working agent inside a peltier cooler is Bismuth-Telluride. I've read a toxicity data-sheet regarding Bismuth-Telluride here.
If I were to use a peltier cooler inside a sealed cooling system, for example a thermoelectric cooler, will any groceries stored in it be "infused" with Bismuth-Telluride?

Yes there are little cooling boxes, for camping, for example, where one of the walls (partially) is made of a sheet of a food-safe-metal, back-to-back to one side of a Peltier element as a cooling device.  This works fine, for example if attached to the car's battery.  Less efficient (by rate of heat transfer) may be an separation of the Peltier element viz. the inner of the cooling box by glass or food-safe plastic.
Of course, a complete immersion of the Peltier device into a cooler or cooling system won't work, as necessarily one of the flat sides of the element is going to become cool, but the other going to be warm.

Thanks for your reply. The cooler I have has a 12 volt fan situated in the lid of the cooler with the cool side of a peltier element next to it to circulate the air inside of the cooler (it looks like the heat created on the other side is dispersed with a heatsink) – Doesn't that establish a direct contact between the food and the Bismuth-Telluride? How much Bismuth-Telluride is actually given off from a bare-metals peltier element?

I do think, by accident your description inversed "cool side" and "warm side".  Because as you rightly state, the "warm side" serves as a "heat sink", the "cool side" of the Peltier element really is used to cool the inner of the cool box.  Already in light of this perspective, it makes sense to separate the "warm side" from the goods to be cooled; the presence of the fan (on the "warm side") even increases this need.  Yet if there such a separation, then I assume the likelihood of BiTe-particles, suspended in air, to be spilled into the inner of the coolbox by the fan, to be little.

If you visit wikipedia's entry regarding Thermoelectric cooling the right-hand figure "Peltier element schematic" places the single elements in stacks, and these on top of (I quote:) "ceramic plates".  Provided i) absence of cracks and gaps in these plates, and 2) a tight connection of these plates with the surrounding wall/frame of the cooling box, I assume this decreases well the possibility of BiTe-particles to enter the inner of the cool box.

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Is there an efficient and effective way to clean oil from floor?
I've recently had several jerry-cans of oil (awaiting disposal) topple over in the attached garage, and several rooms following influx of water following heavy rain. After the water level receded the floor is left with a thin layer of burnt oil.
Various sites on the net recommend using sawdust, baking soda etc. I'm talking close to 800sq.ft of area split into several levels with a step between different levels.
The question I have to ask, therefore, must be obvious now. How does one clean large volumes of burnt oil spilt distributed by water over cement, and stone floors?
EDIT: Nature giveth, and Nature taketh away. Another cloudburst, this time over 100mm of rain in a couple of hours, and muddy waters found their way indoors again. Now the waters have receded, and almost all the oil is replaced with a thin layer of mud. Hopefully washing with water will get rid of the most of it.

http://en.wikipedia.org/wiki/Detergent

Assuming you only want chemistry-related answers, I'd recommend any type of detergent.
If you lift that restriction, however, then I'd recommend pressure washers.

Erm. Would a pressure hose/detergent be suitable for such a large area?

If it's a very large amount of oil and area, I'd recommend mopping up the bulk of it using something like this: http://www.eco-absorb.com/ or http://www.clear-passage.com/absorb.html or http://www.oildri.net/Gran_QuickSorb.html (NB I have no idea if these particular products any good, but it came up when I did a google search)
Afterwards, you can use a pressure washer and/or detergent for the rest.


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Different reactions of clean and rusty iron nails in vinegar (acetic acid)
I have submerged a iron nail in vinegar twice recently. What is puzzling me is the different results each time. The first time I submerged it, it was covered in rust. When I put it in the solution of vinegar, I got a red precipitate dissolved in the solution that I believe was Iron(II) and Iron(III) acetate. However, when I put the newly cleaned nail in ANOTHER bath of the same concentration of vinegar, I did get some bubbles but no red precipitate that was dissolved in the solution. It's only been about 2 days, though. However, there was a black precipitate on the surface of the nail, but I do not believe it is iron acetate because of its color.
I am confused as to the results because Iron is higher than hydrogen on the activity series, so it SHOULD be reacting with the vinegar to give me that red precipitate and hydrogen gas. I added some hydrogen peroxide to the solution to oxidize the iron, and the solution is getting a reddish tint to it.
Why do you think I am not getting that reddish precipitate when I put in the cleaned iron nail in the new vinegar solution? Thank you!

In your first experiment, the rust ($\ce{Fe2O3. xH2O}$) probably reacted with vinegar to form Iron (III) acetate which makes the solution reddish in color.
In your second experiment, the iron on the clean nail surface would have reacted with vinegar to form Iron (II) acetate. There are two main reasons why the solution might appear to have no change:

Usually, Iron (II) acetate is prepared using concentrated acetic acid and scrap iron (or ferrous oxide/hydroxide). Since vinegar is extremely dilute (around 5-8% acetic acid solution in water), the reaction would be very slow and the conc. of Iron (II) acetate formed will be very low.
Iron (II) acetate happens to be very soluble in water and has a very faint light green color in solution which is not easily detectable in lower concentrations.

Then again, when you added $\ce{H2O2}$ to the second solution, you mentioned that a reddish color appeared. This shows that the dissolved $\ce{Fe^2+}$ ions are oxidized to $\ce{Fe^3+}$. The $\ce{Fe^3+}$ formed will react with the acetate ions to form red Iron (III) acetate.

What are the gas bubbles?

When you dip a rusty iron nail in vinegar solution, the vinegar reacts with $\ce{Fe2O3.nH2O}$ (rusty iron or oxidized iron).
Whereas in case of clean nail, it directly reacts with iron.
Your product is probably different due to this reason only, if I understood your question right. ;)
Edit : Chemical equation for pure iron in vinegar -
$\ce{Fe(s) + 2CH3COOH(aq) -> Fe(CH3COO)2(aq) + H2(g)}$
Chemical equation for rusty iron in vinegar -
$\ce{Fe2O3 + 8CH3COOH(aq) -> 4Fe(CH3COO)2(aq) + O2 + 4H2O}$ (Not sure about this equation.)

So then what do you think the products of the reaction are? My thinking is that they are the same, but the clean nail just takes longer. However, I am not sure and that's why I asked.

@ashu Your 2 reaction is wrong so it was good that you weren't sure ;)

So, at first the bubbles from my rusted jackplane are oxygen, and after all the Fe3O4 is consumed, the iron is reducing the hydrogen ions in the solution?

There is no O2 formed in the dissolution of iron oxide.

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Crystallization of solutions and liquids
The process of crystallization of a liquid or any solute occurs when its concentration exceeds the maximum concentration which the solvent can carry or when the temperature drops below the freezing temperature in cases of liquids. Then why is it that these materials require a small "nuclei" or a jagged edge to start forming crystal on, why cant it form crystal anywhere within the bulk of the matter. (eg cloud formation from condensation of water)

Crystallization is rightly called by you a process.  Simplified, there are two steps to be considered: Nucleation and crystal growth.  For the sake of simplification, let me consider the crystallization of a molecular compound in absence of ionic charges.

Nucleation refers to the initial stage, where once isolated molecules of the dissolved substance assemble to a first crystal germ. This is energetically hindered because these particles of especially small size have unfavourably high ratio of particle surface energy per molecule in these just formed particle (or perhaps better: aggregates). This energetic drawback has to be overcompensated. In addition, molecules loose degrees of freedom accessible in solution, at first: dynamic freedom, and secondly, depending how strictly rules are followed to build the crystal, a number of orientations in space otherwise available and accessible--In a perfect crystal, the structural close environment of one molecule is going to be repeated for all molecules, the very next ones and for their next ones, over ranges of Angstroms to Centimetres.

Once such a germ is formed,

Crystal growth more or less refers to simply attachment of additional molecules according to these "(energetically justified) previously determined rules" to yield a bigger entity.  The need of oversaturation of the solute is less prominent than during nucleation, yet still existent to energetically favour this process. In this description, crystal growth is again simplified, too.  It may happen, in the (sub)process of crystal growth, conditions are met that allow crystallisations in realisations even more energetically favourable than at early stages of crystal growth so that a compound preferentially forms a polymorph A initially, then ones in polymorph B. And often a crystallization does not yield only one (perhaps even single)crystal, rather than multiple ones.  So nucleation may occur in one spot of the solution while simultaneously crystal growth already takes place on germs already formed in other parts of the solution.

Complementary, think this way: it is every time easy to dissolve a perfect crystal (already considering entropy), but easily difficult to bring molecules back into a perfectly ordered crystal.
The science and art (like craftsmanship) of crystallization is of high importance in our daily life, spanning areas of electronic devices (semiconductors) till pharmaceutics. Journals like Crystal Growth & Design or CrystEngComm are examples concentrating on this subject.

But why does the crystallization process become energetically favourable when a nuclei or a 'seed' is added allowing the particles to crystallize on, so that the process starts at saturation level itself?


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How to remove Sn from Sn60Cu40 alloy?
I have practical question.
How can I chemically remove $\ce{Sn}$ from $\ce{Sn60Cu40}$ alloy?
I want to get something like copper sponge this way.
Something like nanoporous gold produced from $\ce{Au50Ag50}$ for heat exchanger, but cheaper.

Welcome to Chemistry.SE! I have edited your question to make it look better for our site. First, I removed the quotation marks from your title - they are unnecessary. Second, I used our MathJax plugin to make your chemical formula look like a chemical formula. You can learn to do this for yourself. Start by checking out the [faq].

@Kamil see http://chemistry.stackexchange.com/help/notation specifically

You can remove tin from bronze (copper-tin alloys) and leave behind the copper. However, doing so may not leave you with copper in an easy to use form.
Tin is a less reactive metal than copper. Tin reacts with all acids, like $\ce{HCl}$, while copper only reacts with oxidizing acids, like $\ce{HNO3}$.
$$\ce{Sn(s) +2H+(aq) -> Sn^{2+}(aq) + H2 (g)^ }$$
$$\ce{Cu(s) + 2H+ (aq) -> No \ Reaction}$$
$$\ce{Cu(s) + 4H+(aq) + 2NO3- (aq) -> Cu^{2+} (aq) + 2H2O (l) + 2NO2 (g) ^}$$
Soaking your bronze in acid (even vinegar might work) will eventually dissolve the tin away from the copper. Heating, or using a stronger acid like the $\ce{HCl}$ in muriatic acid, will likely increase the rate. However, since the composition of the alloy is uniform throughout, the copper will likely come out of solution as a fine powder.

Well... Powder will not satisfy me. Please take a look at my updated question.

It is possible you would get the sponge this way. It might still be worth a try.

I think, that Sn/Cu ratio is not 50/50 and it may not work, but I have to try :)

Selective corrosion would definitely produce a copper sponge but I don't what chemicals selective removes tin in a tin/copper alloy.

https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uact=8&ved=0CCAQFjAA&url=http%3A%2F%2Fwww.ams.tuke.sk%2Fdata%2Fams_online%2F2004%2Fnumber4%2Fmag09%2Fmag09.pdf&ei=LeXiVK3mLoigNrCsg7gI&usg=AFQjCNGTjKIuTiRAh3LXqbx4CRpPQbWMsQ&sig2=9bY2yapC1xJ9bD8QA5zwbg&bvm=bv.85970519,d.eXY

Well You need some kind of crystal structure even for sponge. Tin and copper forms an alloy and they have formed crystal structure together. So solving bronze in acid will destroy that structure, witch hold the peace as whole. Sponge can be formed from sintered peace if it is formed from zinc and copper powder. Then solving zinc whit acid will not destroy coppers crystal structure.


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Difference between exothermic and exergonic
In High School I learned that an exothermic reactions releases energy, while an endothermic reaction needs energy to occur. Now I learned that there is a separate, somewhat similar classification scheme of exergonic and endergonic reactions.
What is the difference between these two classification schemes? Are exothermic reactions always exergonic, and if not, can you give me an example?

The classifications endothermic and exothermic refer to transfer of heat $q$ or changes in enthalpy $\Delta_\mathrm{R} H$. The classifications endergonic and exergonic refer to changes in free energy (usually the Gibbs Free Energy) $\Delta_\mathrm{R} G$.

If reactions are characterized and balanced by solely by heat transfer (or change in enthalpy), then you're going to use reaction enthalpy $\Delta{}_{\mathrm{R}}H$.
Then there are three cases to distinguish:

$\Delta{}_{\mathrm{R}}H < 0$, an exothermic reaction that releases heat to the surroundings (temperature increases)
$\Delta{}_{\mathrm{R}}H = 0$, no net exchange of heat
$\Delta{}_{\mathrm{R}}H > 0$, an endothermic reaction that absorbs heat from the surroundings (temperature decreases)

In 1876, Thomson and Berthelot described this driving force in a principle regarding affinities of reactions. According to them, only exothermic reactions were possible.
Yet how would you explain, for example, wet cloths being suspended on a cloth-line -- dry, even during cold winter? Thanks to works by von Helmholtz,  van't Hoff, Boltzmann (and others) we may do. Entropy $S$, depending on the number of accessible realisations of the reactants ("describing the degree of order") necessarily is to be taken into account, too.
These two contribute to the maximum work a reaction may produce, described by the Gibbs free energy $G$. This is of particular importance considering reactions with gases, because the number of accessible realisations of the reactants ("degree or order") may change ($\Delta_\mathrm{R} S$ may be large).  For a given reaction, the change in reaction Gibbs free energy is $\Delta{}_{\mathrm{R}}G = \Delta{}_{\mathrm{R}}H - T\Delta{}_{\mathrm{}R}S$.
Then there are three cases to distinguish:

$\Delta{}_{\mathrm{R}}G < 0$, an exergonic reaction, "running voluntarily" from the left to the right side of the reaction equation (react is spontaneous as written)
$\Delta{}_{\mathrm{R}}G = 0$, the state of thermodynamic equilibrium, i.e. on a macroscopic level, there is no net reaction or
$\Delta{}_{\mathrm{R}}G > 0$, an endergonic reaction, which either needs energy input from outside to run from the left to the right side of the reaction equation or otherwise runs backwards, from the right to the left side (reaction is spontaneous in the reverse direction)

Reactions may be classified according to reaction enthalpy, reaction entropy, free reaction enthalpy -- even simultaneously -- always favouring an exergonic reaction:

Example, combustion of propane with oxygen, $\ce{5 O2 + C3H8 -> 4H2O + 3CO2}$. Since both heat dissipation ($\Delta_{\mathrm{R}}H < 0$, exothermic) and increase of the number of particles ($\Delta_{\mathrm{R}}S > 0$) favour the reaction, it is an exergonic reaction ($\Delta_{\mathrm{R}}G < 0$).
Example, reaction of dioxygen to ozone, $\ce{3 O2 -> 2 O3}$. This is an endergonic reaction ($\Delta_{\mathrm{R}}G > 0$), because the number of molecules decreases ($\Delta_{\mathrm{R}}S < 0$) and simultaneously it is endothermic ($\Delta_{\mathrm{R}}H > 0$), too.
Water gas reaction, where water vapour is guided over solid carbon $\ce{H2O + C <=> CO + H2}$. Only at temperatures $T$ yielding an entropic contribution $T \cdot \Delta_{\mathrm{R}}S > \Delta_{\mathrm{R}}H$, an endothermic reaction may become exergonic.
Reaction of hydrogen and oxygen to yield water vapour, $\ce{2 H2 + O2 -> 2 H2O}$.  This is an exothermic reaction ($\Delta_{\mathrm{R}}H < 0$) with decreasing number of particles ($\Delta_{\mathrm{R}}S < 0$). Only at temperatures at or below $T$ with $|T \cdot \Delta_{\mathrm{R}}S| < |\Delta_{\mathrm{R}}H|$ there is a macroscopic reaction.  In other words, while the reaction works fine at room temperature, at high temperatures (e.g. 6000 K), this reaction does not run.

After all, please keep in mind this is about thermodynamics, and not kinetics. There are also indications of spontaneity of a reaction.

So they're just synonyms for spontaneous and nonspontaneous?

@user3932000 No, they are not synonyms for spontaneous, or nonspontaneous.  They assess the energy difference, comparing the energy state of the starting material(s) with the one of the product(s).

Then are they two ways of expressing the same states? Exergonic/endergonic when describing energy differences, and spontaneous/nonspontaneous when describing reaction thermodynamics.

Both exergonic and exothermic reactions release energy, however, the energies released have different meanings as follows:

Exothermic reaction

Energy released is just called energy
Energy of reactants is greater than that of products
Energy of the reaction system decreases relative to that of the    surounding, i.e. the surrounding becomes hotter.

Exergonic reaction

Energy released, has a special name called Gibbs energy or Gibbs free   energy
Energy reactants is greater than that of the products
It has nothing to do with how hot or cold reactants become. Has a more chemical meaning - it relates to the spontaneity of the reaction; thus it always means that a reaction is feasible, i.e. reaction will always happen.

In summary, whereas, an exergonic reaction means that a reaction is spontaneous, an exothermic reaction has nothing to do with spontaneity, but that an energy is released to the surrounding.

In Exothermic and Endothermic reactions we are mostly talking about the changes in potential energy, these changes tend to manifest themselves as the flow of heat under constant pressure conditions circa the first law of thermodynamics. When we measure Enthalpy we are measuring the energy involved in the formation/breaking of chemical bonds in a particular reaction.

This is a very useful metric for predicting what compounds will form under certain conditions and the TOTAL potential energy changes however.. the 2nd law of thermodynamics tells us that we cant use ALL of the energy in a chemical reaction to do work, only a small amount of it. So we had to come up with Endergonic and Exergonic to explain how changes in GIBBS FREE ENERGY works with a chemical reaction

TLDR: Exo/Endotehrmic we are measuring changes in potential energy states
cant use all potential energy to get work done
gotta measure energy we can use for work as energonic and exergonic

For a exothermic reaction, $\Delta H\lt0$. For a exergonic reaction constraint is (from Gibbs-Helmholtz eqn): $\Delta G\lt0 \Rightarrow \Delta H-T\Delta S\lt0 \Rightarrow \Delta H\lt T\Delta S$
Hence, even if $\Delta H>0$ (endothermic reaction), a reaction can be exergonic provided it follows the constraint for it ($\Delta H\lt T\Delta S$; high temperature or greater no. of degree of freedom).
So there is no such imposition that a reaction has to be exothermic if it is exergonic or vice versa.

Please edit your answer - as written, it's incomplete. See this style guide for how to typeset your posts.

Yes, all exergonic reactions are exothermic. Consider a reaction occurring spontaneously we know that energy would be released i.e. '$\ce{\Delta H}$ is negative' (since a reaction or process absorbing energy makes it non-spontaneous) and according to the second law of thermodynamics, entropy(or disorder) of the system must increase.
Negative $\ce{\Delta H}$ and increasing, positive entropy together make $\ce{\Delta G}$ negative according to the equation: $\ce{\Delta G = \Delta H~ - ~T\Delta S}$ (where $\ce{\Delta }$ = change; G= Gibb's free energy; H= enthalpy; T= Thermodynamic temperature and S= entropy).
Therefore, if enthalpy change is negative and change in free energy is negative, they are both (respectively) exothermic and exergonic.
The same applies for endothermic and endergonic.

Your first sentence is incorrect. See here for a spontaneous (i.e. exergonic) yet endothermic reaction. Examples are not so common because at low temperatures the entropic factor often turns out to be small, so free energy changes are mostly influenced by enthalpy changes.


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Open flame combustion in bizarro world
Humans typically achieve combustion by starting a chemical reaction of a fuel (often a carbon $\ce{C}$ containing fluid like methane $\ce{CH4}$), which is is brought into an an oxidizer (a gas containing $\ce{O}$, like air). For example, when you turn on a lighter.

What would happen if the situation were turned around?
We have an open space entirely filled with fuel and we let in some oxygen-containing fluid. How would the combustion look?

I know the (adiabatic?) temperature maximum of a flame will be achieved at conditions which are slightly on the fat side. But I have no idea how, for example, the production of unburned $\ce{C}$'s, soot production and so on, would look.

To bizzaro world humans, I guess the oxygen would be a "fuel" and the fuel would be their "oxygen". :P +1 Nice question.

@kaliaden: Turns out the word fuel comes from fire and oxygen relates to acid. I don't know enough chemistry to image what it would amount to switching the reactivity and possibilities of combinations of air to that of carbon based gases.

tl;dr: You are likely to see more soot, $\ce{CO}$, $\ce{H2}$, less of $\ce{CO2}$ and same amount of $\ce{H2O}$.
There are different type of combustion that you can talk about. For example when you mix fuel and oxidizer before burning them, it is referred to as premixed system. For example a gas cooking range, gas turbine engines, IC engines, etc. On the other hand if you do mix the fuel and oxidizer you get what's called a diffusion system. For example a candle flame, flame thrower, apartment fires, etc. Typically premixed systems will produce a blue flame, whereas diffusion systems will produce yellow-orangish flame.
It is important to distinguish between a premixed and diffusion flame system because they will behave differently. A lighter produces a diffusion flame. For this answer we will focus on diffusion flame.
An important characteristic of a diffusion flame is that it burns where there is just the right amount of fuel and oxidizer. That means irrespective of whether there is fuel entering oxidizer or oxidizer entering fuel, the flame will only be present where there is just the right amount of fuel and oxidizer. So for example for a methane-oxygen system at the flame we get, $$\ce{CH4 + 2O2 -> CO2 + 2H2O}$$
So in both the cases we would expect to see same amount of $\ce{CO2}$ and $\ce{H2O}$ begin produced when same amount of fuel is burnt. That doesn't sound interesting at all. Plus our ideal situation ignored the effect of heat produced and its effect on the surrounding.
What happens when we take this in account?
Soot
Well in most methane systems it is very difficult to produce soot, so let us consider something like propane $(\ce{C3H8})$ or higher order hydrocarbon (anything with more carbon atoms), which is more likely to soot. Soot is produced when unburned $\ce{C}$-atoms combine to form larger molecules, which combine to form particles. How soot is produced is still an active research area. But we do know that most soot is produced at relatively low temperatures (compared to the flame temperature) in fuel rich regions.
In a candle/lighter flame the soot is produced near the wick, but most of it is oxidized as it reaches the flame. Only some close to the center escapes from the tip because there is not enough oxidizer there to burn it all. This is the main reason diffusion flame appear yellow.
Now imagine your lighter which starts off with oxidizer in the center, there won't be any soot particle inside the flame. So the flame will appear bluish. But the heat produced from the flame may be enough to break (pyrolyse) the fuel in the surrounding region. This will cause soot to be produced. Also because there is no reason for the soot to approach the flame it will not get oxidized easily and you will be left with a lot more soot.
Unburned fuel
Next consider other unburned combustion products. When you burn methane in oxygen you produce more than just carbon dioxide and water. The unbalanced reaction can written as:
$$\ce{CH4 + O2 -> CO2 + H2O + CO + O2 + H2 + C + ...}$$
The unburned $\ce{C}$ forms the soot. A side effect is production of $\ce{H2}$. So now not only will you see more soot, there will be excess hydrogen too. Also in case of higher hydrocarbon fuels you might see some pyrolyzed fuel which is yet to form soot or get oxidized fully.
In a premixed flame system when you have lot of fuel lying around you are more likely to produce $\ce{CO}$ rather than $\ce{CO2}$. A simple reason being you first need to produce $\ce{CO}$ before you get $\ce{CO2}$. A side effect of this is that the temperature will be lower. This is because most heat is released when $\ce{CO}$ is oxidized to $\ce{CO2}$.
So what happens in a diffusion flame system?
As mentioned earlier in diffusion flame systems, the flame is around where there is just the right amount of fuel and oxidizer. The products formed at the flame diffuse out (into the fuel and oxidizer) and since there is scarcity of oxygen, $\ce{CO2}$ will get reduced to $\ce{CO}$ i.e. you will see increase in $\ce{CO}$ production and decrease in amount of $\ce{CO2}$.
Side note: Saturn's moon Titan has an atmosphere which primarily consists of nitrogen (98.4%). The next abundant gas is methane (1.4%) followed by hydrogen (0.1-0.2%). So it's not a bizarre world after all.

I'll leave the science fiction/fantasy to more creative contributors, but I think your conditions can be replicated in our not-so-bizzaro world (although those over at the Chemistry World Blog might be able to convince you that our world is a bit bizzare).
If you returned from watching some videos, consider the following chemical equation:
$$\ce{CH4 + 2O2 -> 2H2O + CO2}$$
This is clearly a combustion reaction with two reagents: a fuel, $\ce{CH4}$, and an oxidant, $\ce{O2}$.  It is important to emphasize that the reaction requires both reactants in order to go to completion. If we look closely at the stoichiometry, we require two molecules of oxygen for every one molecule of methane.  The question posed was

If we have an open space filled with fuel, and a little amount of oxygen is added, what would it look like?

Well, perhaps it would look like this:

Here, the blue dots represent $\ce{CH4}$ and the red dots represent $\ce{O2}$.  There is an excess amount of methane (fuel) for the amount of oxygen (oxidant) in this box.  In General Chemistry class we refer to this as a limiting reagent problem.  The oxygen is a limiting reagent, and the reaction will only go on so long as there is a sufficient amount of both reagents.  When all the oxygen is used up, we have this:

Here the orange dot represents $\ce{CO2}$ and the green dots represent $\ce{H2O}$.  Nothing else is likely to happen at this point.  So getting back to your question, what would this world look like?  Well it would look a little wetter and a little hotter.

Are you therefore implying there will be absolutely no differences if you switched the fuel and oxidiser?

@MarchHo Nope, not implying that at all.  The chemistry is described by the chemical reaction that can happen and the amounts of the reagents provided to the system.

How then is your image meaningful? If I switched blue and red I get the exact same image.

@MarchHo If by switching the blue to red you mean "what if we have an excess of oxygen?" then the lower image will contain two orange dots and four green dots because the stoichiometry of the reaction indicates that four water molecules and two carbon dioxide molecules will be produced.

To add to the previous answers, for a real-world example look no further than your average rocket engine.  Especially hydrogen/oxygen engines are run fuel-rich; at stoichiometric rato the chamber temperature would be too high and the combustion gases too reactive for the engine to survive.

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


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Does variations in air pressure affect the melting point of PLA and ABS while 3D Printing?
Tight temperature controls are needed to ensure that the filament used in fabrication does not begin to "ooze". In places where the air pressure rises and drops during the course of the day there may be more or less ooze taking place. Controlling ooze is paramount to obtain better print quality.
Question: Given Boyle's law, shouldn't the melting point of PLA and other plastics used in the Fused Filament Fabrication 3D printing methods change based on the ambient air pressure of the machine's location? Should air pressure be taken into account when calculating how much heat to apply to the print nozzles? Does humidity affect PLA's melting point?

Boyle's law refers to properties of an idealized gas, where the product of pressure and volume remains constant.  Taking into account your description, as well from filament-based 3D printers seen so far, my answer is no, Boyle's law does not apply here.
If running by extrusion, polylactic acid (PLA) is heated to overpass melt, however this liquid is not heated to become a gas. Temperature control clearly is of importance, because the viscosity of the polymer depends on it.  And viscosity co-influences the precision of details you may print, actually down to the scale of less as a tenth of a millimetre.  Without experience to run one 3D print in at sea level, and the same again during an intercontinental flight, I assume it is more likely, machine control takes into account the pressure difference between ambient air and polymer melt, and not (yet) the (in comparison) lesser variations of air pressure.
The chemical structure of PLA shown in Wikipedia's article shows a relatively low number of functional groups susceptible to water, relatively low to the inner backbone of the polymer (in the structural scheme, this is written in the square brackets).  The article refers to typical molecular weights of 130 kDa when PLA is produced.  These two let me say, humidity does not play an important role.


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speeding up etching process of a copper board
Background:
I use Sodium persulfate to etch a copper board to manufacture home made Printed Circuit Board (PCB).
On the etching powder's plastic, it is written that:

1.5-2dl of etch powder in 1 L of hot water or corresponding volumes

So I add 2 cup powder for 10 cup hot water. In the first time usage it works perfect and it etches the board (double layer 8x16 cm FP4 board) in 20-30 min. However, when I use it for the second time (in another day), the solution works tooo slow. The last time I waited 13 hours for board to be ready. As the board stayed too long on the enchant, the acid ate away many traces on the PCB.
I add extra sodium persulfate (around 2 cup again) when I use the solution second time.
I use microwave (900 W) to warm the enchant again. I have not measured the temperature of the acid but I guess it might be around 40-60 C and I don't have any temperature stabilization so that the solution get colds in 5-10 min again and I repeat the same procedure maybe 6-8 times per process.
Question
Why it takes too long to etch the copper when I use the solution second time although I add the same amount (2 cups) of sodium persulfate to the solution?
What can I do to speed up the etching process apart from mixing the solution?

Only what you don't want: prepare a fresh solution each time as the disolved copper is making things slower. You can't use another reagent? I know of ferric chloride and the mix of hidrochloric (or muriatic) acid/peroxide.


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When measuring avidity should you use the ligand concentration or the nanoparticle concentration?
Looking at avidity between a ligand-receptor, you're looking at an enhancement of the $K_d$ compared to a lone ligand.
Is it more appropriate to compare the $K_d$ using the concentration of the "free" ligand or the concentration of the free complex?
Using the ligand is odd at the lower regimes since the ligand efficency is fairly low but using the nanoparticle changes how you have to calculate the effective concentration. I've seen papers that do both and I'm curious which is most appropriate/correct

What, precisely, do you mean by avidity? Your question seems to be asking about cooperative binding (multiple ligand-macromolecule events at discrete sites) rather than avidity (sum of interactions of a single ligand-macromolecule across a contiguous binding interface).

In the case where there is not cooperativity between binding events, the Scatchard Equation is the generalized solution to multiple ligand binding to a macromolecule. The Scatchard plot uses the free ligand concentration.
There are multiple models used to address cooperative binding. The Hill Equation is the most frequently encountered in biochemistry literature:
$$ \theta = \frac{[L]^n}{K_d+[L]^n} $$
Where $[L]$ is the concentration of free ligand, $\theta$ is the fraction of bound ligand-receptor complex, $K_d$ is the apparent dissociation constant, and $n$ is the Hill coefficient, describing cooperativity.


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Sigma and pi symbols in molecular orbital theory
In molecular orbitals diagram, sigma ($\sigma$) and pi ($\pi$) symbols are used. What do these symbols really mean?

These symbols are for bonding orbitals. A covalent bond is formed when two orbitals in different atoms hybridize. $\sigma$ and $\pi$ refer to two different configurations in which this can happen.
In a $\sigma$ bond, two $s$ orbitals or two $p$ orbitals oriented towards each other join into one orbital, with the electron density concentrated between the two atoms.
In a $\pi$ bond, two $p$ orbitals oriented perpendicular to the bond axis merge, and, in the resulting orbital, electron density is concentrated above and below the bond axis.

so, pi and sigma symbols are actually referring to pi bond and sigma bond, right?

Technically they're for ABMOs as well, when used with an asterisk

@MuhammadRafique Yep. Though in the case of sigma-star and pi-star, it's not exactly a bond. It's better to look at them as sigma-bond-like and pi-bond-like orbitals, as bonds don't exactly correspond to orbitals like in VSEPR here. It's similar, but not the same.


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Is it safe to condense water into a liquid nitrogen trap?
I'm trying to remove water from an imidazolium salt (i.e. dry the salt).  However I want to know if it is safe to remove the water under vacuum.  Is it safe to condense water into a liquid nitrogen trap?  Wouldn't the water freeze in the trap since the freezing temperature of water is 0 degrees Celcius?  Just curious b/c I was informed that it was ok to do it this way.

I suppose it depends on how much water!

Im not really sure the exact amount.  However, based off of the peak that I observed in the NMR spectra, it shouldnt be more than 15 mL of water.

The water does freeze in the trap so it'll work as long as it's not more than the trap's capacity.

One use of a liquid nitrogen cold trap is to collect organic and water vapors that are removed from a system under vacuum.  In this case, the vapors are collected as ices (solids).  See this reference which has some useful information on Schlenk line safety and use.
Where you want to be cautious is when your collection is finished.  You do not want to keep a liquid nitrogen cold trap exposed to the atmosphere because oxygen will condense and cause an explosion hazard.  Typical procedures would include shutting of the vacuum pump, dismantling of the vacuum apparatus and allowing the trap to warm to room temperature (in a fume hood if organic vapors were collected).
Some additional references might be helpful, including this one that not only describes proper Schlenk line technique but also provides some good references; and you can see how liquid nitrogen traps are used for several industrial purposes here and here.

Thanks for the response bob.  I will check out the references you gave.


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Wave function for particle in a infinite well located at -L and +L
The wave function for a particle in a infinite potential well located at $0$ and $L$ is $$\psi_n=\sqrt{\frac2L}\sin\frac{n\pi x}L$$
I want to find our wave function for the same situation where the walls are located at $-L$ and $+L$
I can write the Schrödinger equation, but I am unable to apply the boundary conditions, so I am unable to find the wave function.
How can I find the wave function in case of walls at $-L$ and $+L$?

I will answer this question by copying a passage out of the very good QM book bei Shankar (see here) which I can recommend wholeheartedly to anyone trying to learn quantum mechanics. I considered writing up a solution myself but since the passage answers your question exactly (ok, not quite exactly since the calculation is performed for the infinite wells being located at $-\frac{L}{2}$ and $\frac{L}{2}$ but that doesn't change anything important) it would have been a waste of time and effort to do so. If you have any questions about the derivation I'm happy to help you.

It is very simple, there is no need to solve it again since you already know the solution of infinite square well. The idea is very simple, the trap looks the same, but wider now $2L$. So first you should extend it by a factor of two and then move it back to the center, it is given by the rescaling and translation:
$$L \mapsto 2L \tag{1}$$
$$x \mapsto x-L \tag{2}$$
So the solution is:
$$\psi_{n}=\sqrt{\frac{1}{L}}\sin\left(\frac{n\pi x}{2L}-\frac{n\pi}{2}\right) \tag{3}$$
$$E_{n}=\frac{n^2\pi^2 \hbar^2}{8mL^2}$$
As expected. Note that $\sin(\pi/2-x)=\cos(x)$ so that you can split it into odd and even series, but the Eq (3) given above is more succinct.

I think it useful to note that there is a lowest possible energy level.  The uncertainty principle, knowing the distance between the walls, allows us to compute the minimum momentum the particle can have.  Since $E = p^2/2m$, you have the lowest energy.  The rest is as hwlau has it.


text:
Which compound reacts most slowly?
I'm trying to figure out why the answer is C. Can anyone explain to me why that is the answer?

What have you tried? What do you already know? I'm closing this by our homework policy (note: it applies to homework-like questions too), but if you can edit that in I'll be happy to reopen it. :)

Nitration is a general class of chemical process for the introduction of a nitro  group into an organic chemical compound.

For introduction of Nitro group, the compound must have either partial charge or proper charge (ions) or must form a radical to form the bond between Nitro group and compund. Compound III (option C) is more stable and ease of charge generation or free radical generation will be less due to less difference in electronegavities of atoms.
     Due to less electronegativity difference, the shared pair of electron contributes equally to the electron density of both atoms. If electronegativity difference is higher, ease of Charge generation is higher, eg. $\ce{HCL}$ (both have Partial charges, $H^{\delta+}$ & $Cl^{\delta-}$).
This is the only reason for which the reaction will be slowest for compound C.

Please use backticks only  for code.

Electrophilic aromatic substitution, of which nitration is one specific example, is often a kinetically controlled process, i.e., the distribution of products is determined by the energies of the transition states and intermediates of the respective reaction pathways available. Under conditions where thermodynamic equilibrium cannot be established, the predominant product will be the one with the lowest activation energy for the rate-determining step, and more stable transition states and intermediates imply lower energy of activation. This is also the product that forms most rapidly.
(Incidentally, even under thermodynamic conditions where equilibrium can be established, it's frequently the case that the major pathway will involve the initial kinetic product in equilibrium with the thermodynamic product. If I remember correctly, this is the case with, e.g., the isomerization that occurs with certain Friedel-Crafts polyalkylations.)
Given that, in order to answer your question, you need to examine the transition states of the nitration process for each of your reactants. If you consider the mechanism shown on the Wikipedia page for electrophilic aromatic substitutions, you'll see that the cationic intermediate of the reaction is additionally stabilized when electron-donating groups are present. In your specific question, molecules I and II are both very strongly electron-donating by means of $\pi$-conjugation, which is nicely illustrated by the resonance structures one could draw for those molecules. Molecule IV is stabilized by a similar effect, although it's slightly less efficient due to the lone pair on the nitrogen also being conjugated to the electron-withdrawing carbonyl group. Molecule V is very weakly donating, but by different means. In the case of that molecule, it donates electrons by induction, and by hyperconjugation of the vacant $p$ orbital with a $\sigma$ carbon-hydrogen bond. Because the energies of $p$ orbitals and bonding $\sigma$ molecular orbitals are vastly different, and because the orbital overlap is incomplete and sensitive to conformation, the stabilization provided by the methyl substituent is comparatively minute.
As an aside, I recently became aware of another, somewhat orthogonal explanation for the reactivity of substituted aromatics, namely the frontier orbital theory approach. Fleming, in his excellent book on molecular orbital theory, explains (by analogy with the benzyl anion) that conjugation of the benzene ring with an electron-donating substituent disrupts the normal degeneracy of the $\psi_2$ and $\psi_3$ molecular orbitals, and yields a non-bonding HOMO with nodes at alternating atoms. The locations of those nodes results in relatively large coefficients on the atoms in the ortho and para positions, and very low (idealized to zero, but experimentally, of course, non-zero) coefficients on the atoms in the meta positions. I suppose this can serve as a complementary theoretical justification for the observed kinetics and perhaps regioselectivity of these types of reactions.


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Would cooling of a battery using cooler affect its life time?
My smartphone gets over heated where I use mobile data for long period. Last day I found a cooler of an old cpu, the thing silver in colour to absorb heat(with eventually gives out heat as it has more surface area on the other side). I used to keep my phone above it and it gets cooled. So my question is whether it will cause any problem for the battery if I tries to cool it using the same? I am asking this because there are a lot of thermodynamics involved in batteries and would it cause any problem to my battery life?

Let's see what I know. Not really sure whether this is a terrific answer to your question, but it's what I have.
Through knowledge of using laptops, I know that lithium-ion batteries will decrease in performance when overheated. This probably applies to most redox reaction batteries for the reason that the reaction proceeds more quickly when temperature is increased, and as a consequence, kinetic energy of the molecules are increased.
Increasing the temperature of the battery also increases internal resistivity, which is somewhat more of a physics question. The electrons collide with the material that they are flowing through much more frequently in a higher temperature situation as opposed to a lower temperature situation. This doesn't mean, however, that you can dip your phone in LN2 and have it work forever. There's obvious physical limitations to everything.
What you're doing to your phone (a.k.a. cooling it) shouldn't have a problem at all on your phone's performance, mainly because you're doing to it what others do to their laptops - placing it on a cooling pad.


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Can anyone tell me the mechanism behind this reaction?
$\ce{1-octene -> octan-1-ol}$
The reaction occurs in the presence of $\ce{ROOR}$ and $\ce{HBr}$.
There is no information about temperature.

Welcome to Chemistry Beta! Please see our homework policy Try to include an indication of your attempts to solve this, maybe a partial mechanism. I can reopen this once edited. Thank you.

I have only trivial knowledge of organic chemistry, but here goes:
Assuming ROOR represents a peroxide, we have homolytic cleavage of ROOR to form $\ce RO^{.} and ^{.}OR$. Afterwards, these radicals interact with HBr to produce ROH and a Br radical. The ROH then is protonated on the alcohol group to make a better leaving group, and the $\ce {H2O}$ is cleaved, leaving a radical R group, which is then attacked by a Br radical. Water is then added to the alkene oct-1-ene to form octan-1-ol.

HBr will add to the octene and the product will be 1-bromo-octane. This in turn can be hydrolyzed in a second (and separate!) reaction with water to give 1-octanol. R-O-O-R acts a a radical starter only, the addition of HBr is a radical chain reaction.


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What does lowercase r-s notation mean?
I came across a naming convention which I haven't seen before. I let ChemDraw name the following compound for me and got a name containing lowercase "r" and "s" configurations.

Can someone tell me what this means and something about the actual convention behind it? I think it must have something to do with rings and maybe with some kind of "pseudo-chirality" due to a certain conformation of the ethyl group which is attached to the ring, but I'm only guessing.

For examples and methods for r/s assignments, see http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/isomers/RS14272/achiral.html

This notation is used to designate so-called "pseudoasymmetric carbons." This occurs in instances in which precisely two of the groups bonded to a tetrahedral carbon are structurally indistinguishable in terms of connectivity (i.e., the same atoms, bonded with the same multiplicity, and in the same order moving outward from the carbon whose absolute configuration is being assigned), but contain chiral centers which have opposite configurations. See the IUPAC Gold Book entry on pseudo-asymmetric carbons.
As for the molecule in your example, it is named incorrectly. There is an inherent symmetry present in 1,4-disubstituted cyclohexanes that prevents them from being chiral (except in cases where the substituents on the ring themselves contain chiral centers, which methyl and ethyl groups do not), and without any chiral centers in the molecule, there obviously can be no pseudoasymmetric carbons either. I would add, however, that cis/trans isomerism is, of course, possible for 1,4-disubstituted cyclohexanes, but that's entirely different from the property of chirality. The only stereochemical designator that could reasonably be added to the molecule from your example is "cis-".

The name (1s,4s)-1-ethyl-4-methylcyclohexane is correct. This is not a bug in ChemDraw. Note that the presence of stereogenic units such as pseudoasymmetric centers does not require the molecule to be chiral. The absolute configuration of achiral cyclic molecules is specified by CIP stereodescriptors, too.

The name (1s,4s)-1-ethyl-4-methylcyclohexane is correct; it is even the preferred IUPAC name (PIN). It refers to the cis isomer of 1-ethyl-4-methylcyclohexane.

The preferred IUPAC name for the corresponding trans isomer is (1r,4r)-1-ethyl-4-methylcyclohexane.

In general nomenclature, according to Subsection P-93.5.1.2 of the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the stereodescriptors 'cis' and 'trans' may be used to denote the relative configuration of substituted monocyclic compounds.

P-93.5.1.2 Relative configuration: the stereodescriptors 'cis' and 'trans'
The stereodescriptors 'cis' and 'trans' are used to show the relationship between two ligands (atoms or groups) attached to separate atoms that are contained in a ring or a ring system. The two ligands are said to be located 'cis' to each other if they lie on the same side of a plane. If they are on opposite sides, their relative position is described as 'trans'. The appropriate reference plane of a ring or ring system (the ring being in a conformation, real or assumed, without reentrant angles at the two substituted atoms) is the mean plane of the ring(s). The stereodescriptors denote relative configuration; the absolute configuration must be described by CIP stereodescriptors such as 'R' and 'S'.
(…)

In preferred IUPAC names, all stereogenic units shall be specified (unless an omission is explicitly allowed). The presence of stereogenic units does not require the molecule to be chiral.
According to Subsection P-93.5.1.1.1, the absolute configuration in substituted monocyclic compounds shall be described by Cahn-Ingold-Prelog (CIP) stereodescriptors.

P-93.5.1.1.1 Absolute configuration
In substituted monocyclic compounds the absolute configuration is specified in preferred IUPAC names by the CIP stereodescriptors such as 'R', 'S', 'r', and 's'.

This rule explicitly applies to achiral cyclic compounds, too.

P-93.5.1.1.2 Achiral cyclic compounds
The configuration of achiral cyclic molecules is also specified by CIP stereodescriptors in preferred IUPAC names.
(…)

Therefore, this rule also applies to (1s,4s)-1-ethyl-4-methylcyclohexane, which is given in the question, and its isomer (1r,4r)-1-ethyl-4-methylcyclohexane. (The examples mentioned in the Blue Book include the similar compounds (1s,4s)-cyclohexane-1,4-diol, (1r,4r)-cyclohexane-1,4-diol, and (1r,4r)-4-chloro-4-methylcyclohexan-1-ol.)

The lower-case stereodescriptors 'r' and 's' are CIP stereodescriptors like 'R' and 'S'. The stereodescriptor 'r' or 's' is chosen by applying the CIP sequence rules as for the stereodescriptors 'R' and 'S'. However, the stereodescriptors 'r' and 's' are used to indicate the absolute configuration of pseudoasymmetric centers, whereas 'R' and 'S' are used to indicate the absolute configuration of real chirality centers.

P-91.2.1.1 Cahn-Ingold-Prelog (CIP) stereodescriptors
Some stereodescriptors described in the Cahn-Ingold-Prelog (CIP) priority system, called 'CIP stereodescriptors', are recommended to specify the configuration of organic compounds, (…), and in the nomenclature of natural products (…). The following stereodescriptors are used as preferred stereodescriptors (…):
(a) 'R' and 'S', to designate the absolute configuration of tetracoordinate (quadriligant) chirality centers;
(b) 'r' and 's', to designate the absolute configuration of pseudoasymmetric centers;
(…)

The 'pseudoasymmetric' is defined in Subsection P-92.1.1 (d) as follows.

Stereogenic units are called pseudoasymmetric (center, axis or plane) when they have distinguishable ligands 'a', 'b', 'c', 'd', two and only two of which are nonsuperposable mirror images of each other (enantiomorphic). (…). The 'r/s' (…) stereodescriptors describing a pseudoasymmetric stereogenic unit are invariant on reflection in a mirror (for example 'r' remains 'r', and 's' remains 's'), but are reversed by the exchange of any two ligands ('r' becomes 's', and 's' becomes 'r'). Lower case stereodescriptors are used to describe pseudoasymmetric stereogenic units.

For example, (2R,3r,4S)-2,3,4-trichloropentane contains two chirality centers (labeled 'R' and 'S', respectively) and one pseudoasymmetric center (labeled 'r').

The two 1-chloroethyl ligands of the pseudoasymmetric center are nonsuperposable mirror images of each other. On reflection of the entire molecule in a mirror, the configurations of the chirality centers are reversed ('R' becomes 'S', and 'S' becomes 'R'), but the configuration of the pseudoasymmetric center is invariant ('r' remains 'r').
The same principle applies to (1s,4s)-1-ethyl-4-methylcyclohexane, which is given in the question.

The two halves of the 1,4-disubstituted cyclohexane are nonsuperposable mirror images of each other. Therefore, the compound contains two pseudoasymmetric centers, which are invariant on reflection in a mirror ('1s,4s' remains '1s,4s').

Although this case is explicitly mentioned in the Blue Book, it may not be obvious how to choose the stereodescriptors 'r' or 's' for 1,4-disubstituted cyclohexanes. The ranking is a complex process and requires the use of auxiliary descriptors.
Generally, the stereodescriptors 'r' and 's' are assigned by applying the Cahn-Ingold-Prelog (CIP) Priority System as for the stereodescriptors 'R' and 'S'. The order of precedence is reached by the construction of a hierarchical digraph and the application of the Sequence Rules.
The constitutional formula of a cyclic molecule is transformed into an acyclic digraph according to the methodology that is described in Subsection P-92.1.4.3 and which has been recommended by Prelog and Helmchen.
In order to obtain the acyclic digraph of a stereogenic unit, the didentate ligand (i.e. the ring) is transformed into two monodentate ligands by leaving intact in each case one bond with the core of the stereogenic unit and cleaving the other bond. At the end of each of the two branches thus obtained, a duplicate atom of the core is attached. (Note that this methodology is similar to the case of multiple bonds.) For example, for the pseudoasymmetric stereogenic unit at the atom numbered '1' of (1s,4s)-1-ethyl-4-methylcyclohexane:

where $\ce{(C)}$ is a duplicate representation of the core of the pseudoasymmetric stereogenic unit.
As the digraph is acyclic, it is now possible to specify the stereodescriptor 'R' or 'S' for the stereogenic unit at the atom numbered '4', which is carrying the methyl group, in each branch. These are auxiliary descriptors that are different in each branch.

Since the auxiliary descriptors are different, they allow the branches to be ranked using Sequence Rule 5, which states that 'R' has priority over 'S'.

P-92.1.3.5 Sequence Rule 5
An atom or group with descriptor 'R', 'M', and 'seqCis' has priority over its enantiomorph 'S', 'P' or 'seq Trans'.

Therefore, the configuration of the pseudoasymmetric stereogenic unit at the atom numbered '1' of (1s,4s)-1-ethyl-4-methylcyclohexane can be determined to be 's'.

In the same way, the configuration of the pseudoasymmetric stereogenic unit at the atom numbered '4' of (1s,4s)-1-ethyl-4-methylcyclohexane can be determined as 's'.

Therefore, the complete configuration of (1s,4s)-1-ethyl-4-methylcyclohexane, which is given in the question, is indeed (1s,4s).

ChemDraw16 confirms this, but Marvin 18.3 generates this: "(1s,4r)‐1‐ethyl‐4‐methylcyclohexane". I can't confirm that anywhere online, places like this https://pubchem.ncbi.nlm.nih.gov/compound/1-Ethyl-4-methylcyclohexane don't use r/s descriptors in any of the 63 synonyms they list and it seems like the IUPAC name should be "cis-1-Ethyl-4-methylcyclohexane". My confusion in what you wrote here is why it's valid to assign an R/S to an achiral site after sketching it as an acyclic digraph. I don't have easy access to the Blue book but the Gold book seems to require chirality in the ligands.

I did some more reading about "auxiliary descriptors" here: http://onlinelibrary.wiley.com/doi/10.1002/anie.198205671/abstract
(see their 1,3-cyclobutanediol example) and that really helped me understand how pseudo-asymmetric centers are identified.

"The two halves of the 1,4-disubstituted cyclohexane are nonsuperposable mirror images of each other." I have some trouble understanding this. I have two doubts: (1) Would 1,4-dimethylcyclohexane need the s / r stereodescriptors? I think NO because it's superimposable on its two halves (2) Is the non-superimposability of the two halves of 1-ethyl-4-methylcyclohexane because of the ethyl group at one end? The ethyl has one extra sp3 carbon, which the methyl does not. Is this the reason?


text:
Confusion in unit cells of crystal system
Unit cells are divided into two main types

Primitive
Non-primitive

Primitive includes simple cubic lattice whereas non-primitive includes fcc bcc end centered..
Among the seven types of crystal systems orthorhomic crystal system posses both primitive as well as non-primitive unit cells i.e. face centred body centred rnd centred and simple cubic but the cubic system posses primitive , body centred and face centred arrangements only.
My question is if face centred is possible in cubic system why is not end centred type of unit cell arrangement possible in cubic system if there is no sort of hindrance in vacant space as in orthorhombic which posses all types of unit cell arrangements.

I'm a bit confused. What's going on here? Essence of the cubic system is equivalence of a, b and c axes, namely, presence of 3-fold axis in the body-diagonal direction of the cube. There's no chance of "base-" centred cubic.

Actually there is. You should do some research before answer questions since most of the time a reference is needed. http://www.ndt-ed.org/EducationResources/CommunityCollege/Materials/Structure/metallic_structures.htm and http://en.wikipedia.org/wiki/Cubic_crystal_system

First, a note: the classification of crystal systems, or the reduction of possible lattice types into primitive Bravais lattices, has nothing to do with “space available” or “hindrance”. It is purely a mathematical property having to do with the symmetry of the lattices.
To answer your question, the base-centered cubic lattice is not a Bravais lattice, because it is equivalent to a simple tetragonal lattice:

i got what you had to say about the space thing but i am still confused as. to why cubic systems do not possess end centred arrangement if face centred is possible?

@AanalDesai it is possible to have a base centered cubic system, but it is not a minimal representation of the system. Just like the unit cell must be the smallest cell that can reproduce the full periodic system by a set of translation, the Bravais lattices are by definition the smallest lattices. Imagine a base-centered cubic system (drawing on the left): you can represent this particular lattice as “base-centered cubic”, but you can also view it as a “simple tetragonal” lattice… and the second representation features a smaller unit cell, hence it is the irreducible one.

I had one more query, Wouldn't a face centred simple cubic be an in-centred tetragonal then..?


text:
Removing titanium from tantalum crucible
I'm trying to remove titanium filling from a tantalum crucible for a high-temperature effusion cell.  I've managed to get a majority of the material out with tweezers, but some remains and is difficult to remove.  I understand that sulfuric acid will corrode titanium, but not tantalum, and was considering this as a method for extraction.
If anyone has some advice on a method to (safely) remove $\ce{Ti}$ from the $\ce{Ta}$ crucible without destroying it, I would very much appreciate it.  I am certainly not a chemist and could use some help.

in our lab we use chromic acid to remove by products from crucible (not sure its Ta)..you may also try sonication it may remove Ti

Straight hydrochloric acid- Say 20% is virtually  inert to Tantalum- but significantly corrosive to titanium. If you heat it to boiling point of hydrochloric- it will eat through the titanium.
http://www.parrinst.com/wp-content/uploads/downloads/2011/07/Parr_Titanium-Corrosion-Info.pdf
http://www.tantaline.com/Tantalum-Corrosion-71.aspx

Titanium and Tantalum are quite nonreactive, so using 'safe' acids, such as diluted acetic acid won't make any difference. That's why using strong acids such as (sometimes has to be hot ) sulfuric acid will do it. If you are looking for not having to work with sulfuric acid, you could always use hydrogen iodide: that is even stronger than sulfuric acid...
It's documented here that:

Tantalum appear not to be attacked by many acids at room temperature
  but does dissolve in  hydrofluoric acid, $\ce{HF}$, or oleum (a solution of
  sulfur trioxide, $\ce{SO3}$, in sulfuric   acid, $\ce{H2SO4}$, also known as
  fuming sulfuric acid).
Reaction of tantalum with water
Tantalum does not react with water under normal conditions. The
  surface of tantalum   metal is protected by a thin oxide layer.

Then I read for titanium (Source):

Reaction of titanium with water
Titanium metal is coated with an oxide layer that usually renders it
  inactive. However, titanium will react with steam form the
  dioxide, titanium(IV) oxide, $\ce{TiO2}$,   and hydrogen, $\ce{H2}$.
$\ce{Ti(s) + 2H2O(g) → TiO2(s) + 2H2(g)}$

So Tantalum doesn't react with steam so you might be able to use steam for that separation as seen as $\ce{TiO2}$ is white and tantalum not.
Source : http://caldersgr9science.wikispaces.com/73+Tantalum


text:
Which of mono-, di-, triphenylmethane is the most acidic?

Question
How would you compare the acidity of the most acidic proton in each of the following molecules?

diphenylmethane
toluene
triphenylmethane

The answer says 3 is the most acidic, followed by 1, and finally 2 is the least acidic. Can someone please tell me how to determine this? I know the benzene group is electron donating group, but am not quite sure how to correlate this knowledge with determining the relative acidities.

You should consider the stability of carbanion that is left behind, after removing $\ce{H+}$.   One could make arguments for resonance stabilization of each of the carbanions. You could "push" the minus charge around the phenyl rings, and there is a general rule that the more "good" resonance structures that one can draw, the more stable the molecule.
It turns out that the tertiary carbanion in 2) is flanked by three phenyl groups (thus many good resonance structures compared to 1) and 3) ).


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How to store molecule data using SMILES when building a toy chemistry model
I'm currently building a toy chemistry model based around Gillespie's algorithm. Starting from a food set of {H,O,N,C} I let my system evolve, I have been storing my molecules as strings in SMILES format.
For instance:
[O] reacts with [H] --> [O] ([H])
[O] ([H]) reacts with [H] --> [O] ([H]) ([H])
Using square brackets I've been able to keep the molecules in my system consistent with their SMILES interpretation.
However this system of storage is only allowing me to construct linear molecules.
How can I create, for instance, a carbon ring, so that it is interpreted in SMILES as only a ring of carbon atoms, without hydrogen atoms?
For instance C1CCCCC1 gives http://www.chemicalize.org/structure/#!mol=C1CCCCC1&source=calculate
How would I generate 6 carbon atoms in a ring without the 12 hydrogen?

There is no such molecule as a six-membered carbon ring without hydrogens, there have to be some substituents there. To remove the hydrogens you would have to assume triple bonds, and those are linear and you couldn't form a ring.

See this link for decent documentation on the SMILES format.  SMILES can be drawn and written in both hydrogen-suppressed and hydrogen-complete formats.  Regarding @MadScientist's comment, are you trying to get your algorithm to evolve the unrealistic carbon-ring-without-hydrogens or are you more concerned with the visualization of your evolved molecules?

As you may have guessed, I almost no background in chemistry whatsoever. Why can't this molecule exist? How does this form, does a hydrocarbon fold in on itself?
@bobthechemist I'm trying to get my algorithm to be capable of evolving rings of atoms, I'm planning on adjusting the physical realism later.

@Freeman The answer to your question probably can't be contained in the 600 characters allowed for a comment, and is beyond the scope of the question you asked.  Nonetheless, I suggest looking at this online textbook with particular attention to chapter 9, which describes chemical bonding.  (You might need to skim chapters 2 and 3 beforehand).


text:
Is this relation in case of a mixture of ideal gases valid?
I came across the following relation in a book (written as a fact, with no proof), for a mixture of ideal gases both at the same pressure and temperature.  The final volume of the mixture of gases after mixing is:
$$ V_\text{mix} = \frac {M_1V_1 + M_2V_2}{M_1+M_2} $$
where $M_1$, $M_2$ are the molar masses of the individual gases and $V_1$, $V_2$ are the respective volumes.  I could not figure out its proof.  Is this relation valid at all?
Please do not confuse with a similar looking relation,with $ M $ corresponding to molarity, in case of a solution mixture.

The relation appears to be incorrect.
For ideal gas law states:
$$ pV = nRT = \frac{m}{M} RT $$
This can be expressed as:
$$ VM = m \left(\frac{RT}{p}\right) $$
Now for mixture of two ideal gases at same pressure $(p)$ and temperature $(T)$:
$$ m_\text{mix} = m_1 + m_2 \tag{mass conservation}$$
$$ m_\text{mix} \left(\frac{RT}{p}\right) = m_1 \left(\frac{RT}{p}\right) + m_2 \left(\frac{RT}{p}\right) $$
$$ V_\text{mix} M_\text{mix} = V_1 M_1 + V_2 M_2 $$
$$ V_\text{mix} = \frac{V_1 M_1 + V_2 M_2}{M_\text{mix}} \tag{1}$$
This is not same as the mentioned relation,
$$ V_\text{mix} = \frac {M_1V_1 + M_2V_2}{M_1+M_2} \tag{2}$$
because this require $ M_\text{mix} = M_1 + M_2 $. Which is not correct!
To understand why, imagine mixting two gases such that $ M_1 = M_2 $ and $ V_1 = V_2 $, then equation (2) gives $ V_\text{mix} = V_1 = V_2 $ instead of the correct final volume of $ V_\text{mix} = 2 V_1 = 2 V_2 $

I think the relation should have read:
$$ M_\text{mix} = \frac { M_1 V_1 + M_2 V_2 }{ V_1 + V_2 } $$
which follows from equation (1) and Amagat's law which states $V_\text{mix} = V_1 + V_2$.

This relation is not valid if the gases are assumed to be ideal. see Amagat's law of partial volumes
Moreover, if the gases are non ideal, the relations between the volumes of the combining gases and the final resultant volume is very complex and the given relation is too simple and includes too less variables to even begin describing the actual non-ideal gas mixing.
It, however unlikely, might represent a certain special case that can only be judged from the context wherein the aforementioned equation was taken from.


text:
What does an endpoint saturated signal mean. How does it relate to equilibrium constant?
I have had a question that has been bugging me for a couple of years in my graduate school career, and I was wondering if perhaps someone could help explain it to me. Given an endpoint assay like this one I've shown below (an ELISA assay, but I assume it could work for any endpoint), how does the endpoint signal (the signal at the highest concentration) relate to binding? And how does $EC_{50}$ (the concentration between the maximum and minimum signal) relate to the equilibrium constant?

For years, biologists have used the $EC_{50}$ to give a rough approximation for the equilibrium binding constant $K_d$. However, it always bothered me that all curves features on the graph could produce nearly the same $EC_{50}$, but with different levels of maximum absorbance. For instance two of the curves are saturated at around 1.2 and the others are around 2.5.
Do all of these curves have roughly the same $K_d$, or is it worrisome that we don't get equivalence signal for the maximum concentrations?

$EC_{50}$ is concentration of a bioactive molecule which gives half the maximum response. You should note that the response is bound by a maximum which means this dose-response curve follows a saturation kinetics.
Saturation kinetics can be modelled in many ways and Hill and Michaelis-Menten kinetics are the most commonly used models in biochemistry. Again, a point to note is that many mathematical functions/models do not explain the underlying mechanism. They just mimic the behaviour of the system.
Lets assume that the antibody-response follows a two-step kinetics like Michaelis-Menten.
$$\ce{Ab + Ag <=>[k_a][k_d] AbAg ->[k_{res}] Response }$$
Lets call $R_{max}$ as the maximum response rate. Like in the case of MM kinetics, the response rate $R$ would be (see the wiki page for the derivation):
$$R=R_{max}\frac{[Ag]}{K_m+[Ag]}$$
Here $K_m$ is the concentration of antibody which would give half the maximum response rate (half response rate times times a sampling time would give you the $EC_{50}$).
Now, what $K_m$ really is depends on the approximation of the model:

Equilibrium model: this assumes that the binding and unbinding reactions are much faster than the response reaction and are hence at equilibrium (wrt response dynamics). In this case $K_m$ would be same as the dissociation constant.
Quasi Steady State model: this assumes that enzyme-substrate complex is at a quasi steady state (QSS) i.e. it doesn't change with time. In the case of enzyme substrate reactions, the substrate gets consumed and so the QSS would result from both the binding-unbinding reactions and conversion reaction. In this case $K_m=\dfrac{k_d+k_{res}}{k_a}$. However in this case the antibody is not consumed to give the response. So QSS would be equivalent to equilibrium approximation.

This would also apply for the Hill kinetics in which response can be described as:
$$R=R_{max}\frac{[Ag]^n}{{K_m}^n+[Ag]^n}$$
where $n$ is an integer which represents co-operativity in binding i.e. one bound antibody enhances the binding of the next (total number of binding sites is the upper bound of $n$).
Note that equilibrium model may not be applicable for drugs that can get consumed in a reaction. Also, not all drug induced reaction would follow this simplified model. There are other models for co-operative binding too.
Therefore, before concluding anything about kinetics from an empirical measure like $EC_{50}$, we need to understand (or at least assume) the underlying model.


text:
How to measure amount of oxygen (or carbon dioxide) in air?
I'm not chemist, and I'm looking for easiest way to measure how much oxygen plant in closed container will produce per hour [can measure over time for increased accuracy].
My first idea was to measure percentage of oxygen in air with automotive lambda sensor.
However I don't expect good precision of measurement like this, and this is a little complicated (lambda sensor needs high temperature, so I need to heat it up).
I wonder if there is some chemical way to perform measurement like this.

Possible duplicate of this question?  Different application, I agree.

Thats pretty hard to do... Produced water will condensate everywhere... And its pretty hard to light it up in closed jar (but not impossible). Im wondering about purchasing oxygen sensor for $70 and making electronic circuit for it. I thought there is simple reaction with some powder that will absorb oxygen in few minutes and change its mass or something.

If easy is important, then purchasing a sensor is the way to go.  DIY solutions probably require a bit of Chemistry knowledge and materials (e.g. platinum) that might end up costing as much as a commercial product.

I edited my question. Maybe measuring carbon oxide is easier?

Unlikely, the amounts of gases produced are very small.  You might benefit from a web search of "how to measure plant respiration" all of the non-high-tech methods use either water traps or commercial probes.

@bobthechemist Thanks! "Plant respiration"... It did not I think to ask google like this.

This is untested, but I think it might work. You could light some fuel on fire inside the closed container and see how much of it burns (just put the cup of fuel on a kitchen scale) before it dies out and than back calculate how much oxygen was present. Preferably you should take a fuel that burns easily to make sure that the total combustion reaction is the only reaction that occurs. Also if you want to repeat the experiment every hour or so, you want to be sure that the fuel does not evaporate (substantially) because that might cause the entire container to light up when you try to burn the fuel.
I am not really sure which 'household' fuel would be a good option, but maybe one of the other people on this site have suggestions?!
You can probably use one of those candle lighters and stick the long side through a small hole in your container and than seal the rest of the hole to be able to light a fire inside, without having air flowing in or out.
Just be careful not to set the plant (or anything other than the fuel) on fire!

I think some alcohol would be right choice, however I think this will be very imprecise measurement. Other idea is propane mini-torch with constant pressure. Time of burning in constant flow should be proportional to amount of oxygen...

Other idea - "alcohol candle" made of probe, plug (to reduce evaporation) and some kind of candlewick in little hole in plug.

@Kamil - what are your main worries in the imprecision? local depletion of oxygen? or something else?

Im worried, that amout of burned fuel (alcohol) will be very small. I can minimize local depletion by some fan if convection from flame will be not enough (in case of big container, like terrarium or something).


text:
How to determine a Reynolds number for mixture of 2 liquids
I'm experimenting with synthesis in microreactors and I'm using 2 reactants. I'd like to estimate the Reynolds number of the flow containing two miscible fluids. I have the values of their properties (density, viscosity, flowrate).
Do I just average both property values for both liquids or is there a more intricate relation?
Thank you very much in advance for your solutions and suggestions!

As a very rough estimate, you could do that and just weight them by the fluids' respective concentrations.
Depending on how ideally the two fluids mix, you may want to look up or calculate the actual properties of the particular mixture you're working with. If you can't find the values in the literature but have access to physical property estimation software such as Aspen, you could calculate these values yourself.

You are assuming that both density and viscosity are additive.  Any evidence this is true?

No, in fact it will be entirely untrue unless the two substances mix ideally. But as mentioned, it's a good place to start as a ballpark estimate.

Bob thank you very much for your explanation on how to tackle the problem. The thing is that the fluids are miscible, but they also react, so it would be almost imposible to conduct an experiment having an ideal mixture. For my estimation of Re the relation you suggested (Arrhenius equation) will be good enough.
Thank you once again for all the help!

The Reynolds Number is the ratio of inertial to viscous forces and is calculated according to this equation: $$Re = \frac{\rho v L}{\mu}$$ where $\rho$ is the density, $\mu$ is the dynamic viscosity, $v$ is the fluid velocity and $L$ is a characteristic length that depends on the flow being modeled. When estimating the Reynolds number of a mixture, the term $v L$ would not change but we need to make the assumption that the term $\frac{\rho}{\mu}$ of the mixture can be derived from the properties of the pure components.  This assumption will be valid if

The components have similar structure
The components are non-polar
The components are not associated
The difference in the kinematic viscosities is small (~15 cP)

Some of these conditions could be relaxed if one species is in significant excess.
Densities are additive so long as the volume of the mixture equals the sum of the volumes of the pure solutions.  (Search for volume weighted densities for more information.)  Viscosities of mixtures is a much harder thing to estimate.  One method is the Grunberg-Nissan equation: $$Log\eta_m=x_1Log\eta_1+x_2Log\eta_2+x_1x_2d$$ Where $\eta_i$ is the dynamic viscosity of component $i$, $x_i$ is the mole fraction of component $i$ and $d$ is an interaction coefficient.  If you assume $d=0$ then you have the Arrhenius equation for the viscosity of a mixture.
So, at the end of the day, if you state you are working under the assumptions that volume weighted densities equal mass weighted densities and use the Arrhenius equation to determine the viscosity of the mixture then you can average the weighted properties of the pure components to get an estimate of the mixture.
Just a word of caution under these assumptions you should take the mass weighted average of the densities but the mole weighted average of the viscosities.
I relied on this reference for much of this answer.  It will require a subscription to the service to access the PDF, but some of it can be found on google books.


text:
Gold nanoparticles in a water/oil environment
I am currently investigating the assembly of gold nanoparticles in a mixture of water and oil. The thing is the following: If we have gold nanoparticles (35nm diameter) dispersed in water and we put some oil on it. After that we shake it, then the gold will be kept between the layer between water and oil and now I was wondering why this happens.
So why is it not possible that the gold nanoparticles stay solely in the water or go over in the oil completely? Or it could be possible that the gold particles agglomerate, why does this not happen? Or maybe it does happen and I do not know this.
Does somebody have some background information on this topic?

It is caused by the wettability of the surface of the nanoparticles. For naked particles this will be determined by the properties of gold but, as Daniel mentioned, it is likely that there is a capping agent, in which case the wettability will be governed by the properties of the capping agent. From this point on I will just call this the solid.
Collecting at the interface
Depending on the surface energies between the solid and water (sw), water and oil (wo) and solid and oil (so) you can have 2 situations both with their own extreme case. We can understand which of the three by looking at the famous Young's equation$^1$ :
$$\tag{1} \gamma_{so}-\gamma_{sw}=\gamma_{wo} \cos \theta $$
where $\theta$ is called the contact angle and in this case I have defined it through the water phase as shown in the picture below.

If the surface energies are such that a high contact angle occurs then the solid particles will be mostly on the oil side of the interface (situation 1). But given the surface energies it is still beneficial (energetically) to have some contact with the water, hence the angle. If the surface energies are such that $\cos \theta < -1 $ it means that there is no longer a contact angle and it is preferable for the solid particles to be completely submerged in the oil.
Situation 2 is exactly the opposite, the solid 'likes' the water resulting in a low contact angle. Here again an extreme case can occur in which the equation requires $\cos \theta > 1$ which means that the particles will fully submerge in the water.
Because many liquid-liquid-solid combinations have at least some contact angle (i.e. not 0 and not 180) the particles tend to collect at the interface of the two fluids.
Agglomeration
The reasoning for whether agglomeration will occur is in fact similar. Here again you will look at the surface energies and determine whether it is energetically favorable to have reduce the solid-liquid area by making solid-solid contact or whether it is favorable to have the solid in contact with the liquid. Capping agents are typically designed to repel themselves thus making sure that agglomeration doesn't occur (because it is usually not desired)

1: note that the original essay doesn't contain the equation in mathematical form, only in wording

I believe this would depend more on the capping agent (citrate in Turkevich method, TOAB in Brust method etc. see Colloidal gold) capping agents are amphiphilic and will therefore concentrate at the oil-water interface.


text:
Can you determine whether a compound can form given particular elements?
Say you are given a set of particular elements, such as Dy, C, I, Er, P. Is it possible to determine whether a compound exists with all of the given elements and no more?
If not, is it possible to find this out if given the number of a particular element, such as; two hydrogen and one oxygen where you could come to the conclusion that $\ce{H2O}$ is possible.

are you sure about Dy, Er?

Answer on first question is no, on second is yes.

Consider this short list C, H, O, N this elements can produce unlimited number of compounds.
If you have short formula like C6H6 and you assuming only covalent bonds in molecule, define valence 4 for Carbon and 1 for Hydrogen you can make mathematical calculation according to graph theory and you will get 217 structures. But most of them doesn't exist due to bonds strain and intersection. If you want to obtain existing molecules you can filter this 217 models using for instance quantum mechanics calculation.

There's also the rule of thirteen and the index of hydrogen deficiency that can narrow down the list algebraically without resorting to quantum mechanics.


text:
Why do batteries specifically vent *hydrogen* in the event of abuse?
When a battery is subject to overcharging or overdischarging (including attempts to recharge a primary cell), it may vent hydrogen.  It seems that every type of electrochemical battery I've read about or used specifically emits hydrogen under these circumstances.
Why is hydrogen emitted in all cases, and not some other gas or chemical? What role does hydrogen play in electrochemical batteries that other chemicals can't?

Lead storage batteries use an electrolyte of sulfuric acid and water. On overcharging you are electrolysing the water with emission of $\ce{H2}$ in one electrode and $\ce{O2}$ in the other.
Other batteries use molecules with H and Li. The idea is to use light elements so you can accumulate more charge for the weight.


text:
Z* effective charge and Ionization Energy
I'm trying to figure out the patterns for Ionization Energies. I am familiar with the periodic trend, however things become quite different when we hit the 1st I.E. For example, Na has an I.E(1) of 495.8 kJ while its second I.E. rockets up to 4562 kJ while the atoms towards the right are much lower than this. The trend says that the I.E. increase up and right of the periodic table, which is not the case here.
My point is, in order to get a better estimate, would it be safe to say that the Zeff charge and atom size relate to the I.E?
Example: Effective Charges
Mg = 2
Al = 1
S = 4
Si = 2
Na = 1
We see that the strongest pull towards the charged center would be Al and Na in this case. However, having Al being the smaller atom would require more energy to remove the electron from its valence shell.
Now in the case of first Ionization Energy we have:
Mg = 3
Al = 2
S = 5
Si = 3
Na = 2
In this case, Na now has reduced its size due to the fact it jumped from n=3 to n=2 level and Al also reduced in size but is still a bigger atom than Na due to our trend.
My question is, is this approach fairly accurate or should I be looking somewhere else?

Using the common idiom: "full subshells are stable"
It is a little more compact to use the incorrect explanation and correct it than to explain in terms of correct. Also, when you hear this incorrect explanation, you will understand what is meant.
Yes, size is a factor (thus the "up" part- IE is higher in lower periods), but the observed anomaly in second-ionization energies in $\ce{Na}$ and $\ce{Al}$ is better related to removal of an electron from a full subshell.
$\ce{Na}$ has an electron configuration of $\ce{[Ne] 3s^1}.$ The first ionization gives $\ce{Na}$ makes it $\ce{[Ne]}$, the same as a noble gas- full 2p subshell. Because full subshells are stable, it takes a lot of energy to remove the first electron from it. $\ce{Al}$ is analogous; the second ionization means removing an electron from a full subshell.
The same is observed in the first ionization energies (in $kJ/mol$):
$$\ce{Na}:495$$$$\ce{Mg}:737$$$$\ce{Al}:577$$
Ionization of $\ce{Mg}$ removes an electron from a full $\ce{3s}$ subshell, so it is a little higher than the trend.
There are many factors that contribute to exact values, but these are difficult to predict, which is why they taught a simple version in your class.
Correcting the myth
While it is common to refer to full subshells as being more stable, it isn't really what is going on. In reality, the next electron added in destabilized. This occurs because of shielding. Outer electrons feel repulsion from inner electrons, thus outer electrons are easier to remove than inner ones. The saying that "half-filled subshells are stable" is also used. Again, it is that the next added electron is destabilized, but in this case it is because the energy needed to spin pair the two electrons.

I'm glad you didn't provide the explanation based on full/half-full shells alone without qualifying it and providing a correct explanation. I think the pedagogical value of that explanation is next to nil, and it may even be harmful if students find it satisfying and hence don't probe more deeply for more rigorous and less arbitrary rationalizations.

$Z_{eff}$ is certainly a major factor in determining ionization energies, however atomic and ionic radius probably shouldn't be viewed as having a direct causative relationship to ionization energy. It's more correct to say that ionization energy and atomic/ionic radius have some of the same underpinnings (namely, $Z_{eff}$ and various electron-electron interactions). Because many of the same underlying forces are at work, there is some correlation (i.e., ionization energy increases in the same direction as atomic radius decreases, certain specific anomalies aside), but don't confuse that correlation with causation.
As one moves down a group on the periodic table, $Z_{eff}$ obviously remains constant, however the number of core electrons shielding the nucleus increases, and consequently the valence level electrons become increasingly energetic as their distance from the nucleus grows. This fact largely accounts for the increase in atomic radius and the decrease in ionization energy that occurs as one moves down along any given group on the periodic table.
Conversely, as one moves right along a period, the number of core electrons remains constant, while the $Z_{eff}$ increases. It's reasonable to expect, therefore, that valence electrons will experience a stronger electrostatic attraction to the nucleus as one moves right along a period, causing both an increase in ionization energy and a decrease in atomic radius. The ionization energy trend mostly conforms to that expectation, with the notable exceptions of transitioning from group IIA to group IIIA, and from group VA to group VIA, where ionization energy (perhaps unexpectedly) drops. To explain these exceptions, compare the electron configurations:

When removing the first group IIIA valence electron, it is being removed from a $p$ orbital, while the first group IIA valence electron would be removed from an $s$ orbital. Electrons in $p$ orbitals are somewhat more energetic due to the nuclear charge being partially shielded by the electrons of the preceding $s$ orbital (in addition to more complex quantum mechanical effects), hence they are easier to remove.
The first group VIA electron to be removed is paired with another electron in the same $p$ orbital, while all $p$ orbital electrons of group VA elements are unpaired (in accordance with Hund's rule). Electron pairing causes some mutual electron-electron repulsion, making these electrons more energetic, resulting in a drop in ionization energy for group VI by comparison to group V.

As you move further down the periodic table, the contributions of electrons in $d$ and $f$ orbitals become significant, the energy gaps between subsequent principal energy levels get narrower, and the ionization energy trend becomes more strictly linear for main group elements (specifically, the exceptions I described above no longer apply once you reach principal energy level five).

"As one moves down a group on the periodic table, Zeff obviously remains constant" - unless you are using Clementi's rules.  Come to think of it, don't Slater's rules give you different values for Al and Ga?

@bobthechemist, I was referring to the most rudimentary calculation by taking the difference between protons and core electrons, which I thought would be sufficient for explaining the trend at the gen. chem level without introducing additional complexity. I'm aware of Slater's rules, but this is the first I've ever heard of Clementi's rules, so thanks for that and I'll investigate further.