text:
If you are given ΔH for this reaction which occurs at constant pressure, what additional information you need to determine ΔE?
Reaction: $\ce{2Al + 3Cl2 → 2AlCl3}$
Answer: We must have either temperature, or the values of P and ΔV in order to calculate ΔE from ΔΗ.
Since ΔH is given, we can use the formula ΔE=ΔH-PΔV to calculate change in internal energy, and only pressure and ΔV are required to calculate ΔΕ.
So my question is, why do we need temperature in this question?

Because:
1) your question assumes that the volumes of the solids ($\ce{Al}$ and $\ce{AlCl3}$) are negligible compared to the volume of the $\ce{Cl2}$ gas. This means that $\Delta V \approx V(\ce{Cl2})$.
2) your question assumes that chlorine gas behaves ideally. This means that $V(\ce{Cl2}) = \frac{nRT}{p}$.
In the end, it means you only need two of the three variables ($p$, $T$, $V$) in order to obtain all the information you want.
I would however be very careful about using molar quantities, in particular $V_m$ (since $p$ and $T$ are intensive properties), if the question is asking you to find $\Delta H$ and $\Delta E$ in terms of $\text{kJ mol}^{-1}$.


text:
Enthalpy Change
If I mixed $25.0\ \mathrm{cm^3}$ of $0.350\ \mathrm{mol\space dm^{-3}}$ sodium hydroxide solution with $25.0\ \mathrm{cm^3}$ of $0.350\ \mathrm{mol \space dm^{-3}}$ hydrochloric acid. The temperature rose by $\mathrm{2.50^oC}$. Assume that no heat was lost to the surroundings.
What is the molar enthalpy change for the reaction?
How do I know the mass of water?

Assume the density of the resultant solution is the same as that of water.

Regarding your question about the mass of water, lets calculate how much water is produced by the reaction:
$\pu{25.0 cm^3} = \pu{0.025 dm^3}$
$\pu{0.025 dm^3} \cdot \pu{0.350 mol dm^-3} = \pu{0.00875 mol}~\ce{H2O}$
$\pu{0.00875 mol}~\ce{H2O} \cdot \pu{18 g mol^{-1}} = \pu{0.16 g}~\ce{H2O}$
As this only represents a fraction of a percent of the total water mass (50 g), we can pretty safely neglect any water mass contributed from the neutralization reaction. This is justified as the problem  instructs us to make assumptions that would introduce similar errors, like assuming no heat is lost to the surroundings (i.e. we're ignoring the calorimeter constant).
From here the problem is straight forward. Because this is a 1:1 molar neutralization reaction, we can just calculate the heat of the reaction, $q$, then divide by the moles of water produced by the neutralization.
So first we calculate the total heat of the reaction from:
$$q=mC_g\Delta T$$
where:
$q=$ heat of the reaction
$m=$ mass of the solution (water) $= \pu{50.0 g}$
$C_g=$ specific heat of the solution (water) $=\pu{4.18 J g^-1 K^-1}$
$\Delta T=$ temperature increase for the reaction $=\pu{2.50 K}$
$$q = \pu{50.0 g} \cdot \pu{4.18 J g^{-1} K^{-1}} \cdot \pu{2.50 K} = \pu{523 J}$$
Then to calculate the molar heat of the reaction we just divide by the number of moles of water produced as calculated earlier:
$$\pu{523 J}/\pu{0.00875 mol} = \pu{59.8 kJ mol^-1}$$

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text:
QMOT | Orbital mixing in methyl
The following picture is from E. V. Anslyn Physical Organic Chemistry and shows constructing of MO of methyl using QMOT. MOs $\textbf{E'}$ and $\textbf{D'}$ are formed by mixing $\textbf{D}$ and $\textbf{E}$. I can't quite imagine how the mixing goes here.

Can you imagine the mixing of s and p atomic orbitals into sp? It's pretty much the same.

Yes thank you, that would illustrate the polarized p lobes, right? But in the mixing that leads to E', the 3 1s obitals meet out of phase. Why the resulting orbitals have these 3 1s orbital in "brown phase" and are not cancelled?

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Separation of 2 compounds ( Powder )
Suppose I have a mixture of 2 pure binary compounds, a white powder. What is the best method to separate the 2 compounds?
What I have thought of so far is to dissolve powder in DI water and carry out filtration if one compound is soluble and the other is not. However, I have not been able to think of any other methods to separate the 2 compounds if both are either water-soluble or water-insoluble. I considered fractional distillation but the compounds may become anhydrous. Furthermore, the boiling points of the compounds are not given and they could end up being too close for fractional distillation to work.
Is there any efficient, and preferably simple, method to achieve the above-mentioned desired result?
Edit: Is there any general method of approach for separating such 2 compounds as there is for separating water-soluble and water-insoluble compounds?[Filtration]  Or does it have to be achieved purely by trial and error of various methods?

I think, for the most part, designing a separation procedure depends on you knowing the chemical identity of the two compounds. Like if you have AgCl and AgBr (not exactly a white powder, but I'm just trying to make a general point) you can separate them by addition of ammonia. The key is to find a reagent that reacts differently with both solids. Water is a possibility (e.g. Na2CO3 and AgCl), acid is a possibility (e.g. Ca(OH)2 and BaSO4). But if you don't know what the solids are, you're not going to know what that reagent is.

Could also vary the volume and temperature used for dissolving the solids. Cold di water extractions? Or whatever solvents you are using That's if you don't want other information of your unknowns. Without solubility info, if it's weak acid/base, would be doing lot of trial and error and quantitative analysis just to see if any the solvents you may find are reactive. Maybe try using a spectrometer to get better idea of the compounds?

Your question is about as broad as chemistry itself. Like you said, you may try dissolving them in water, if one is soluble and the other is not. If it's not working with water, chances are it will work with ethanol, or acetone, or... see, there are many solvents on the shelf. No luck? Maybe by now you've found something that dissolves them both, so you may try fractional crystallization. Still no luck? Maybe the compounds differ in density or selective adhesion of something, so flotation might help. Still no luck? Chromatography, by any chance? No? Distillation, like you said (provided they both would survive until boiling point), or maybe sublimation (good for $\ce{NH4Cl + NaCl}$)? No? Well, there is a multitude of chemical processes; some of them might change (that is, destroy) one compound or both, but probably in such a way that they can be regenerated later.
And if everything else fails, you can always look at the mixture under a microscope and separate the crystals manually.

"And if everything else fails, you can always task an undergraduate with looking at the mixture under a microscope and separating the crystals manually," you mean? ;-)

If the powdered compounds are of different density you can separate them based on density in a liquid that does not dissolve them. For example, if you want to separate caffeine (density=1.36 g/cm3) from lidocaine HCl (density=1.20 g/cm3) you can tune the density of you separation medium to be between the two compounds in density. The more dense compound will sediment and the less dense will float achieving separation. A good hydrophobic separation medium in this case could be a mixture of tetrachloroethylene (density=1.62 g/cm3) and hexane (density=0.661 g/cm3)-tune the ratio of the two to tune the density of the separation medium. If the compounds do not dissolve in water you can tune the density of water solutions for used as separation mediums using compounds such as sodium chloride, or zinc bromide or similar to achieve high density solutions if necessary. In this case collect your separated fraction and rinse it with DI water to get rid of the salts.

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If a conductivity meter is calibrated with aqueous standards, can it be used to measure the conductivity of an organic-based solvent system?
Trying to figure out if the calibration solutions for a conductivity probe/meter must be of a similar nature to the system which will be measured. I can only find aqueous calibration solutions online, and the solvent systems that I need to monitor are ethanol and methanol based. If I use the aqueous calibration solutions, will my measurements in the organic systems be accurate?

I don't know of any fundamental theoretical reason why you would need to match solvents.  However, it seems like a very easy empirical test!  Make some lithium acetate solutions of known concentration in water, in methanol, and in ethanol, and then compare the results!


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Reaction of Cu in hydrochloric acid bath with exposure to air over several months?
I assume that copper will be slowly oxidized by oxygen, dissolved in water, forming copper(I) oxide (as stated here):
$$\ce{4 \overset{\pm0}{Cu} + O2 -> 2 \overset{+I}{Cu}_2O}$$
Reaction rate might be affected by the hydrochloric acid (cf. here)
And some of the copper could actually be converted to $\ce{CuCl2}$ (cf. here):
$$\ce{2HCl (aq) + \overset{\pm0}{Cu}(s) -> \overset{+II}{Cu}Cl2(aq) + H2(g)}$$
But what will happen to $\ce{Cu2O}$?

Will it degrade to $\ce{CuO}$, which then forms $\ce{CuCl2}$? That would mean that one should find higher concentrations of $\ce{CuCl2}$ in the solution, as time goes on.
Will it not react any further but instead get solved in the hydrochloric acid solution, because it is soluable in acids?
Or will something else happen?

Note: The concentration of hydrochloric acid solution is around 10 %.
Related Questions:
Copper Doesn't React with Hydrochloric acid
Copper and hydrochloric acid

in presence of CuCl2 in excess of concentrated HCl copper will slowly dissolve, forming $\ce{[CuCl2]^{2-}}$ anion, which is readily oxidized by air.  So, yes, copper will slowly dissolve in hydrochloric acid bath, especially if some 'seed' copper chloride is present. The process is slow, though.

also -- oxidation is typically a surface-reaction unless you have a very powerful oxidizer. Cu in air will form an oxide on the surface slowly. Water will speed it up some, HCl even more, and H2O2 (~30%) with HCl will oxidize very fast -- effectively dissolving the Cu and leaving CuCl2 dissolved in water.

In the presence of excess hydrochloric acid, copper oxides are not the immediate reaction product of copper metal and oxygen (e.g. from the air).  The air does oxidize the copper...
$$\ce{Cu + 1/2 O2  + 2 H+ -> Cu^{2+} + H2O}$$
... but the final oxidation product is copper(II) ion, not copper(I).  (Copper(I) is formed initially but is more prone to air oxidation than copper metal is, so it is relatively quickly converted to copper(II).)  Additionally, in the presence of chloride anions, the copper(II) ions are preferentially complexed by chloride.  The low concentration of hydroxide means that hydroxide complexes are not favored, and chloride is a better ligand than water for copper.
The actual complexes formed vary in their stoichiometry and their color.  $\ce{CuCl2(aq)}$, $\ce{CuCl3^{-}(aq)}$ and $\ce{CuCl4^{2-}(aq)}$ are all possible:
$$\ce{Cu^{2+} + n Cl- -> CuCl_n^{(n-2)-}(aq)}$$
$$n \in \{2, 3, 4\}$$
Nurdrage made a nice video showing how the reaction of $\ce{HCl}$, air, and copper metal leads to copper chloride.  It's the first of the three methods he presents for making copper chloride.  As he notes, the reaction is very slow.
$%edit$

Thanks! Do you think that this reaction could, after enough time, lead to a orange layer on the copper surface, when the same solution is exposed to additional "fresh" copper regularly? Because this is what I was observing after approximately one month.

Well, after some time, the $\ce{HCl}$ will be used up, so technically the solution would turn into water. It sounds possible. But I never tried. @AstronAUT

That would mean, that this orange powder I observed could be Cu2O (red) mixed with CuO (black), which form in the absence of HCl, right?


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Water extraction of ethanol from petrol
We have now 10% ethanol mixed petrol. Seller says that if this petrol comes in contact even 10 ml of water, all ethanol which is mixed in the petrol will be extracted by water, is it true?

Hi and welcome to Chemistry.stackexchange.com. Take your time to take a [tour] of the site and visit the [help] to find out, how it works. Please also take your time to use proper punctuation. (The space is after the full stop/comma/question mark.) ‘Conversion’ is the wrong word, by the way.

There is an element of truth to it. You might want to search for the partition coefficient of ethanol in fuel/water systems. Or just check it yourself on a small amount. Chemistry is an experimental science, they say.

Water and ethanol are highly "miscible"; a science term that means virtually any amount of water is fully soluble in ethanol, and vice-versa.  The next thing to consider is that ethanol and gasoline are quite miscible. Think of E-10 (10% ethanol) all the way to the E-85 some petrol engines are designed to function well on (with  some loss in power output).
However, water is very poorly soluble in pure gasoline.
So adding when water to E-10 the water mixes well with the ethanol, but the H2O/ethanol mix looses solubility rapidly as more water is added.
In theory, if you set up a separation flask, basically a clear glass vessel with a petcock at the very bottom, filled it at room temp with a gallon of E-10 and added ~3 tablespoons of water, shook it all up then let it settle to separate, you could drain away the bottom (denser) layer of liquid, which would be the water/ethanol mix, an leave you with ~0.9 gallons of relatively alcohol-free petrol.
My final remark would be to ask why anyone would wish to do this as I can think of no reason in the world why someone would bother.

Definitely not gallon vs 10ml, maybe 100ml vs 10ml

Maybe the concern is that if you get a small amount of water in your gas tank (e,g, by condensation) it will "extract" the ethanol from your gas tank and create a sizable "aqueous" phase - that could cause problems in your engine or gas line. But I think those "dry-gas" products like STP (mostly methanol) may help to dissolve the water that collected in the bottom of the tank into the gasoline so it will not longer clog your gas line.

A reason to consider doing this is with smaller engines like a lawnmower . Ethanol impurities in the fuel can cause a viscous residue in the carby and lines that can build up and cause problems. Because the ethanol is so hygroscopic it attracts moisture from the air which can end up contaminating the fuel also.

I'm a chemist and my Mom's car had this problem. I can agree that it was caused by water in the fuel, but the exact amount that would cause a problem seems fuzzy. There are a couple of factors. One is temperature and the other the presence of various stabilizers which may, or may not, be added to the gas. But it does seem to be caused by a small amount of water. This paper has a ballpark of about 0.5% at 60 degrees F. http://www3.epa.gov/otaq/regs/fuels/rfg/waterphs.pdf


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Understanding the energy produced from oxidizing 1 kg of lithium
I am reading the following paper in preperation for a research project on the theoretical energy density of batteries: Lithium - Air Battery: Promise and Challenges and on the second page they describe the following calculation:

The oxidation of 1 kg of lithium metal releases 11,680 Wh/kg

I was wondering specifically how this number was obtained. I think that this would be the right chemical equation to use for this type of approximation but I am not sure:
$$\ce{4Li (s) + O2 (g) -> 2Li2O}$$
I am wondering if anyone can explain to me how this number was produced.

Nice question! Have a look here and here, both of which indicates that the reaction in question is $2\ce{Li^{+} + 2e^{-} + O2 \leftrightharpoons Li2O2}$.

Ahhh thank you! That certainly makes a lot more sense. So using that I could say that 1 kg of $Li^{+}$ is 144.1 mol of $Li^{+} $ Which would produce 144.1 mol of electrons right? So I guess where I am confused is how to go from there to the Wh/kg which is a unit of energy. I guess you would have to know the voltage of the cell that the electrons would be crossing and use that potential difference to try to get to energy?

Ok So I did a few calculations that get me close to the number in the paper and I am wondering if this is the right direction. 144.1 mol of electrons is $8.7 \times 10^{25}$ electrons. Assuming that the cell for this lithium ion battery is 3.2 volts then it would take 3.2eV to bring one electron through this potential. So 3.2 eV * $8.7 \times 10^{25}$ electrons = $2.8 \times 10^{26}$. Then converting eV to Wh I see that 1 eV = $4.45 \times 10^{-23}$. This is 12388.8 Wh/kg. Perhaps they assume that they are working with a lower voltage cell to get the lower number? Does this look right?

Ok I think that this is about right. It at least kind of convinces me for the moment... I am coming from a physics background so this electrochemistry stuff has had my head spinning.

I think you're on the right track, if only because you get an answer that isn't crazy and I think your process is sound. That said, I am pretty out of my depths with electrochemistry and, more specifically, batteries (google "specific energy" for more on what they're reporting, for instance). Hopefully some more knowledgeable folks will weigh in here.


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Do all conversion factors have an infinite number of significant figures?
I believe conversion factors have an infinite number of significant figures. They should be exact values; but not all of them are, are they?
I looked at a pdf document (via the Internet Archive) and it agreed with the fact that all conversion factors have an infinite number of significant figures.
I know this is true with a conversion factor such as $\frac{\pu{1 cal}}{\pu{g ^\circ C}}$, because this is how a calorie is defined. Similarly, it's the same for $\frac{\pu{1 kg}}{\pu{1000 g}}$, $\frac{\pu{4 quarts}}{\pu{gal}}$, etc.. We know these are exact, and in dimensional analysis, if you multiplied by the first conversion factor, you wouldn't end up with one significant figure because of the single significant figure in $\pu{1 cal}$.
However, what is the case with measured conversions, such as $\frac{\pu{4.184 J}}{\pu{cal}}$ or $\frac{\pu{1.60934 km}}{\pu{mi}}$? These were measured using significant figures, so it would make sense to round with significant figures when multiplying (dimensional analysis), but the above link says otherwise.
What should I do, especially with the below type of conversions?

Your intuition is correct. There are two types of conversion factor.
One kind of conversion factor is based on definitions. These conversions are exact (infinite significant figures).
In addition to your examples of the specific heat of water (in calories), the kilogram/gram conversion, and the quart/gallon conversion, other conversions are exact, including:

The speed of light - $c=299,792,458\ \mathrm{m/s}$ the length of the meter is then derived from this constant.
The price of gasoline/petrol.
The inch is defined as exactly $25.4\ \mathrm{mm}$, and thus the conversions from all imperial to metric lengths (including your  are exact as well.
The specific heat of water in joules - the relationship between the calorie and the joule is defined exactly as $\mathrm{1\ cal = 4.184\ J}$

Note that most conversions of this type are unit conversions. Many units have been defined exactly in terms of SI-units.
Other conversion factors are based on measurements. One common example is atomic mass. Atomic masses are valuable conversion factors that are based on measurements. However, some unit conversions live in this category.
The electronvolt is based on the measurement of the elementary charge of the electron (which is not exact). Thus the electronvolt is $\mathrm{1\ eV=1.6021766208(98)\times 10^{−19}\ J}$.

To clarify, how should I round with measurement-based conversions?

@jlam55555 Most measurement-based conversion factors have a functionally-infinite number of significant figures for the purposes of routine chemistry work. That being said, to be rigorous, one would have to track down the original source of the measurements (or a reputable secondary source such as Perry's Chem Eng Handbook or the CRC Handbook), and see how many sigfigs were reported there. That would then become the number of sigfigs to use in figuring out the final rounding, same as if it were a measurement you made yourself.

Thanks to the new SI, the electron volt is now exactly 1.602176634×10−19 J. Atomic weights are still empirically measured, though

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text:
Recovery and Disposal of Unknown Copper Compound
About a month ago, I ran a small electrolysis experiment in an old fish bowl with a $\ce{MgSO4}$ electrolyte and copper electrodes. It ran for about 3-4 hours.
Now, a month later, I am left with about 40% of the initial volume of water and a blue copper compound on the bottom of the bowl. I believe this is $\ce{Cu(OH)2}$. I've provided a picture.
This has been sitting outside for the entire month under a little greenhouse type thing (but air can still get at it). As winter is coming, I was hoping to collect what I've formed and dispose of it properly. So
1) Could it be $\ce{Cu(OH)2}$?
2) Is there a way I can get the solids out? Or should I just pour everything into a plastic container for disposal?

Unlikely. Copper hydroxide adsorbs carbon dioxide from air rather
fast, forming basic carbonates. if you want to test, try to boil it.
Copper hydroxyde looses water under mild heating, forming black
copper oxide.
Most likely it is either basic copper carbonate or one of basic
copper sulfates or a mix of both. I highly doubt that it is a pure
copper sulfate, as Mg had to go somewhere.
If you want to be a responsible citizen (which is not actually much
a point with relatively safe copper salts), add some soda, leave
overnight and then filter off the precipitate with filter paper.
If no filter paper is present, cotton white, toilet paper, and
similar should work, just be careful to not tear them. The
precipitate then may be heated until black and disposed any way
suitable.

It's probably copper sulfate ($\ce{CuSO4}$) residue (the electrolysis of magnesium sulphate with copper electrodes would yield a decent amount of copper and sulphate ions which would precipitate in absence of solvent i.e. water). A good way to test if it's $\ce{Cu(OH)2}$ would be to use a pH strip in a dissolved solution of your blue salt, and see if the pH strip indicates base (blue color). If it's decently neutral, you likely have copper sulfate.
Copper sulfate is used in stump remover as a common household product. It's fine to dispose of in a plastic bottle etc.
You can also dry it (for example by heating) and see if it turns white. The anhydrous form of $\ce{CuSO4}$ is white. Copper hydroxide wouldn't change from it's blue-green color upon heating.
P.S. nice moth.

This does not look like copper sulfate to me at all, nor does it resemble pure hydroxide (both should be way brighter). Some form of basic carbonate, maybe. Also, heating copper hydroxide does not leave it unchanged - it turns to oxide (black).

It indeed looks more like copper carbonate https://en.wikipedia.org/wiki/Basic_copper_carbonate . Copper sulfate has a more intense blue color! I think you can dispose it down the drain. If you really don't know, visit your local high school. They probably have a heavy metal waste container craving for some copper ;-)

I'd say both of you are right, by appearance. That leaves the question of how the OP got such a substantial amount of carbonate in the sample in the first place, considering he used a sulphate salt to begin with.


text:
How are subshells and shells arranged in an atom?
If $\mathrm{2s}$ and $\mathrm{2p}$ are in the second energy level, and $\mathrm{3s}$, $\mathrm{3p}$, and $\mathrm{3d}$ are in the next (3rd) energy level, how are these subshells arranged in space in relation to one another? Particularly, could this be pictured as the $\mathrm{2s}$ and $\mathrm{2p}$ orbitals being relatively the same size (although different shapes) and superimposed over them being the $\mathrm{3s}$, $\mathrm{3p}$, and $\mathrm{3d}$ orbitals all of a relatively larger size than orbitals of the 2nd energy level?

Everything is superimposed on everything. Orbitals are regions of space - they aren't containers that physically exist. (More properly, orbitals are quantum mechanical wavefunctions, but that's a story for another day)

But in general the size of the orbital does increase with principal quantum number (2 or 3 in your question).

@orthocresol You should write this up into a full answer.

I will start off by offering an apology for anything I might have misunderstood - if this is the case, just tell me in the comments below.

If 2s and 2p are in the second energy level, and 3s, 3p, and 3d are in the next (3rd) energy level

This is, in general, not the best choice of words. The 2s and 2p orbitals are only degenerate, meaning that they have the same energy, in a hydrogenic atom: a species which contains one nucleus and one electron. The hydrogen atom itself satisfies this requirement. Other hydrogenic species include $\ce{He+}$, $\ce{Li^2+}$, $\ce{Be^3+}$, ...
For any species that has more than one electron, the 2s and 2p orbitals have different energies (see this question, although it's a bit of a read) and therefore it would not be correct to refer to them belonging to the same "energy level".
What you can say is that 2s and 2p orbitals belong to the same principal quantum shell. All this means is that the number in front of them - called the principal quantum number - is the same, in this case, 2.

How are these subshells arranged in space in relation to one another?

Orbitals are not actually containers, or tangible objects that occupy space. They are mathematical functions obtained by the solution of the Schrodinger equation. The pictures of orbitals that you are probably familiar with are just pictorial representations of these functions. Without going into too much detail, one quantity that is used in the functions is the radial distance between the electron and the nucleus, denoted $r$. $r = 0$ corresponds to the nucleus's point in space. What this means is that all orbitals - regardless of their shape or energy - are centred at the nucleus. The $\text{s}$ orbitals are spherical in nature, and the centre of the sphere corresponds to the nucleus. And the same goes for $\text{p}$ orbitals, $\text{d}$ orbitals, and so on.
Therefore, to put it very succinctly, everything is superimposed on everything. You could think of it as shining different colours and shapes of light onto a screen - the projected images can all simultaneously exist.

3s, 3p, and 3d orbitals all of a relatively larger size

You're right about this - in general, orbitals with a larger principal quantum number tend to be larger, as long as we restrict the discussion to the same atom. For example, the $\text{3s}$ orbital is larger than the $\text{2s}$ orbital, which is in turn larger than the $\text{1s}$ orbital.
Now since I did say that orbitals are mathematical functions, this does raise the issue of how you define an orbital to be larger than another. If you are interested, you can read up about the radial distribution function, although I shall not expand on it as that would really be beyond the scope of this question.

I think it should be pointed out that because of penetration the suborbitals fill (with some exceptions) in a particular order known as the Aufbau principle. https://en.wikipedia.org/wiki/Aufbau_principle


text:
Medicine and chemical density related
How are medicine densities calculated?For example what is the density of ascorbic acid 0f 250mg a pill?

I'm not exactly sure what's your problem.

I agree with @Mithoron. It is unclear what you are asking.

Ascorbic Acid will have a density of 1.69 g/cm³ (I couldn't find under what conditions, however). If you are asking how to calculate the density of a mixture (be it homogeneous or heterogeneous), there are several chemical engineering books that deal with the subject matter. However, it doesn't make sense to ask about the density of ascorbic acid in a 250 mg pill. You could say that it is the mass of the ascorbic acid divided by the volume of the pill. But then, the "density" loses its meaning. Why? because it is possible to make any material have some density that we want by varying the volume of the container and the mass of the molecule/atom/compound (we could place one atom of any element in a container of humongous volume such that the density is near zero). Density is a description of how molecules/atoms/ions attract each other through intermolecular forces (with the pressure/temperature specified).


text:
Transition metal and complexation
When can we say that such metal forms a complex?
Are $\ce{Pd(OAc)2}$ and $\ce{Pd(PPh3)2Cl2}$ complexes?

Yes, they are. Most so called salts are actually polymeric complexes.

Both of these compounds are coordination complexes. The label complex is preferred by many chemists because despite the appearance of some of these compounds as simple ionic compounds, they often display much more complex bonding motifs.
For example in $\ce{Pd(PPh3)2Cl2}$, there are two neutral ligands, $\ce{PPh3}$ that interact with the metal center through dative bonds. There is also evidence that the $\ce{Pd-Cl}$ bond is more covalent than ionic, since palladium forms a complex ion $\ce{PdCl4^-}$ with chlorine.
Palladium acetate man seem like a more clear cut case until you look at the crystal structure. Each acetate anion bridges two palladium, centers in a polymeric chain:

It is worth noting that IUPAC does not recommend using the term complex to describe these compounds given that complex can refer to a number of other, more loosely bound, species. IUPAC prefers coordination entity, though I have never heard a practicing chemist or seen a textbook use that phrase.


text:
How to remove Ethanethiol from household natural gas?
Household natural gas is injected with a little bit of Ethanethiol to make it a distinctive odor for safety purposes.
But, if it is used to drive cars (*), its odor is sensible and it is annoying.
Is there a way to remove this from the household gas (which is mainly the mix of 90% methane and 10% ethane) on a home budget scale?
My first idea is to have some type of chemicaly which could be inserted into the empty bottle.
(*) In most countries, this can be done legally, although it has to follow relatively strict safety and tax regulations.

After burning in the engine, there should be little or no odor (except perhaps  faint odor from sulfur dioxide), particularly in countries where cars have catalytic converters. Before burning, the odorant is a useful warning of leaks.

@DrMoishePippik With thiole in the fuel, your catalytic converter is dead after twenty kilometres.

Just an idea: You can pass the gas through a bleach solution that can oxidize the thiol. I have no idea how effective this can be since the human smell is extremely sensitive to ethanethiol (to the scale of parts per billion). One possible set-up (very common in organic labs) is shown in this picture. The end Buchner flask should contain bleach and not HCl. The other conical is a safety measure in case you have suck back of the bleach solution.

+1 but the real probelm with the question is that nearly any technique is going to be pretty impractical for processing significant amounts of gas in any reasonable home setup.  But since you got the ball rolling, let me add my only slightly less impractical "solution" of bubbling the ethanethiol-contaminated gas through liquid mercury to soak up the thiols, forming mercaptides.

To add to practicality issues @peterh has to be prepared in case he fails, he might get a visit from the fire brigade (because of the bad odour).

@CurtF. Organic compounds with mercury and sulfur? Hm-hm, afaik contacting them would cause much worser problems as a little bit of bad odor.

@K_P I won't do anything which could be dangerous. As I know, the odor is only bad if the gas-driven car was constructed badly, but in this case it is really dangerous as well. Thus, the solution is: if the gas car smells, it must be repaired, and not the ethanethiol should be removed. But, the bleach-solution had been surely worked, thus I can accept it.

@peterh So exciting that you tried it and you had good results! If you could give some details how you did it by answering your question, that would be great (I'd upvote it  for sure) . Let me add, I was worried that maybe your gas having caught moisture after passing through the aqueous bleach soltn would not be suitable as fuel for an engine. In that case you could add a bottle with a drying agent. But seems to work without it. Also, the fire brigade call has really happened in more than one instances in chemistry departments where a single, even small-scale, experiment was the culprit.


text:
The Krieger-Li-Iafrate Approximation
I am having trouble understanding what KLI approximation is. Could someone explain it to me in an intuitive/simple way?

Could you tell a bit more about it?

Maybe they can

@Hexacoordinate-C Fell free to summarize that poster in an answer:)

The optimized effective potential (OEP) is a natural connection between local density-functional theory and many-body perturbation theory, which is derived from the Sham-Schlüter equation. Converging the full set of OEP equations is quite a challenge. Therefore an approximation was made in order to reduce the integral equations to an analytically solvable one via a dominant orbital approximation, called The Krieger-Li-Iafrate (KLI) approximation.
$$\sum_{i,j\neq i}\frac{\phi_j^*(\mathbf{r})\phi_i^*(\mathbf{r})}{\epsilon_i-\epsilon_r}\left(f_i \langle \phi_i|v_x|\phi_j\rangle - \sum_{k, \alpha}\frac{d_{ik}^{\alpha}d_{kj}^{\alpha}}{2}(\epsilon_i-\epsilon_k)\left[\frac{f_i(1-f_k)}{\epsilon_i-\epsilon_k+\epsilon_{\alpha}}+\frac{f_k(1-f_i)}{\epsilon_i-\epsilon_k+\epsilon_{\alpha}}\right]\right) + c.c.$$
$$=\frac{1}{2}\sum_{\alpha,i,k}\omega_a\frac{\left|d_{ik}^{\alpha}\right|^2 f_i(1-f_k)}{(\epsilon_i-\epsilon_k-\epsilon_{\alpha})^2}\left(\left|\phi_k(\mathbf{r})\right|^2-\left|\phi_i(\mathbf{r})\right|^2\right)$$
Source: Maybe they can see comments above.

Hope this helps, I didn't want to go deeper because you asked for a simple and intuitive explanation. But I am aware that the formula isn't very simple :P

Seriously? Direct copy-paste from a poster found by another user without referencing neither the user, nor the poster, nor the original paper? And formula as an image when there is MathJax up and running? And what is what in this equation? How it can be used and when?

@andselisk better than nothing. I am just surprised that no one used and extracted the information from that poster, which I ended up reading all and having interest for it. There are not very good sources on the internet about this topic. So I tried to do my best without changing a lot the wording because it should stay as it is. But it would always be great to have an expert talk about this (which I highly doubt that there is).

So, you don't even want to edit this answer to improve it in a slightest?

I'm on it, don't worry.

Latex Edition done. What else?


text:
What is the use/function of lithium chloride as an excipient in sofradex eardrops?
Certain drugs have lithium chloride as an excipient, eg. Caprilate, NeyDop N (Revitorgan-Dilutionen N Nr 97), NeyDop N (Revitorgan-Lingual Nr 97), Nisita.
An excipient is not supposed to have a pharmacological effect, but it seems like lithium chloride is an inhibitor of various enzymes (http://www.ncbi.nlm.nih.gov/pubmed/15107743).

Probably the main reason $\ce{LiCl}$ is used in liquid to soften ear wax is that $\ce{LiCl}$ is quite hygroscopic and pharmacologicically inactive under the conditions used (small amount used to retain moisture).
As you state, taken internally, in contact with the eyes, or left in contact with skin for extended time, there are some biological effects, but I'd not be concerned about those with a tested commercial product.


text:
Electrolysis of auric chloride solvent with different metals in anode and cathode
Question:

Write down possible reaction in anode and cathode for electrolysis of $\ce{AuCl3}$ with solid $\ce{Au}$ electrode in anode and solid $\ce{Ag}$ electrode in cathode.

As far, I could write the anode reaction. In the anode, there are 3 competitors ($\ce{Au}$ electrode/$\ce{Cl-}$/$\ce{H2O}$), so the final reaction would be
$$\ce{2Cl- -> Cl2 + 2e-}$$
But I couldn't figure the cathode reaction because there are 3 competitors in cathode ($\ce{Ag}$ electrode/$\ce{Au3+}$/$\ce{H2O}$) that I have no idea what I need to do with the $\ce{Ag}$ here.
Some sources said $\ce{Ag}$  is an inert metal. How the reaction for cathode would be?

How is Ag a competitor at cathode?


text:
What is the pKa of the second ionizable group?

One compound contains an amino group with $\mathrm{p}K_\mathrm{a}$ of $8.8$, and one other ionizable group with a $\mathrm{p}K_\mathrm{a}$ of $5$ to $7$. To $100\ \mathrm{mL}$ of a $0.2\ \mathrm{M}$ solution of this compound at $\mathrm{pH}\ 8.2$ was added $40\ \mathrm{mL}$ of a solution of $0.2\ \mathrm{M}\ \ce{HCl}$ and the $\mathrm{pH}$ changed to $6.2$. What is the $\mathrm{p}K_\mathrm{a}$ of the second ionizable group?

I am a little confused because I am not sure what to do when the $\mathrm{pH}$ does not equal the $\mathrm{p}K_\mathrm{a}$ of the amino group. I'm thinking since the known $\mathrm{p}K_\mathrm{a}$ of $8.8$ is highest at that point there should be no free $\ce{H+}$.

Since you're given that the maximum $\mathrm{p}K_\mathrm{a}$ of the second group is $7$, you can assume that the amino group will fully protonate before the second group starts accepting electrons.  (Perhaps this is what you meant by "at that point there should be no free $\ce{H+}$"?)
Therefore, from Henderson-Hasselbalch$^\ddagger$, you can calculate the fraction of the amino group that is protonated based on the starting $\mathrm{pH}$ and the amino group $\mathrm{p}K_\mathrm{a}$. (Hint: Yes, the amino group should be mostly, but not entirely, protonated.)
It's straightforward to then calculate:

The total number of moles of acid added
The number of moles of not-yet-protonated amino groups before adding the acid

which is equal to

The number of moles of acid consumed to protonate the rest of the amino groups

and, therefore, you can find

The number of moles of acid left over to protonate a portion of the second ionizable group in the molecules.

Having calculated #4, you can then calculate what fraction of the second ionizable group is protonated by the leftover acid.  This feeds back into Henderson-Hasselbalch, from which you can calculate the second $\mathrm{p}K_\mathrm{a}$ value since you are told the value of the final $\mathrm{pH}$.
$^\ddagger\,\mathrm{pH} = \mathrm p K_\mathrm a +\log\left(\frac{\ce{[A-]}}{[\ce{HA}]}\right) = \mathrm pK_\mathrm a + \log \left(\frac{[\mathrm{A_{total}} ] - [\ce{HA}]}{\ce{[HA]}}\right) = \mathrm p K_\mathrm a + \log\left(\frac{1}{f_\mathrm{protonated}}-1\right)$

THIS IS THE SOLUTION:
1. HH Equation:
1a.) 8.2=8.8+log(B/A)
1b.) solving for ratio of B:A= (B/A)=.251
1c.) %B= (.251/.351)*100%= 20%, thus
      %A= 80%
(.351 is the TOTAL amount of moles for acid and base) it comes from the equation (x/(x+1))
2.Amount Reacted:
KNOWN: .1L, .2M of compound
2a.) .1L*.2M= .02 moles
By using percentages, we have
2b.) .8*.02= .004 moles of base
       .2*.02= .016 moles of acid
KNOWN: .04L, .2M of HCl
2c.) .04L* .2M= .008 moles HCl
.004 moles of base will get protonated by .004 moles of H+ from HCl
This forms .004 moles of acid
Adding the original acid concentration gives .02 moles of acid
There are still .004 moles of H+ left
This can protonate the acid
This can occur because this is an ionizable group with pKa from 5-7
See top of thread to see original problem
Thus, .004 moles of the H+ protonates .004 of the .02 moles of the original acid
This is your new Base to Acid ratio for the HH eq.
(.016/.004)
The .016 comes from .02mol original acid - .004 mol H+
The .004 on the denominator comes from .004 mol formation of the new acid
3. PLUG INTO HH Eq.
3a.) 6.2 (given)=pKa + log(.016/.004)
3b.) Solve for pKa

Thus the pKa of the second ionizable group is 5.6

Hopefully this is easy to understand

pH=pKa + log ([A]/[HA])
pH goes to 6.2
8.2=8.8 + log([A]/[HA])
8.2-8.8 = log ([A]/[HA])
the difference is negative but since it represents just "the difference" we can use positive.
.06= log ([A]/[HA])
so using the new pH,
6.2= pKa + .06
6.2-.06= pKa
pKa= 5.6

At minimum, this doesn't seem to account properly for the differing volumes of solution used.

If my mental math is correct, this gives a correct numerical result, but the approach is not generally valid.  It only works in this case by the happenstance of the particular numerical values (concentrations, volumes, dissociation constants, etc.) used in the problem statement.


text:
Calculating the pH
I have a question about a pH calculation which is the following:
Calculate the pH of $\pu{E-8 mol/L}~\ce{HCl}$ solution in water.
($\ce{HCl}$ is a strong acid which completely ionizes in water)
So what I had done:

completely ionizes in water so: $\pu{E-8 mol}~\ce{H+}$ is added to water.
Water equillibrium says: $\ce{2H2O <=> H3O+ + OH-}$
$K_\mathrm w = \ce{[H3O+]} \times \ce{[OH- ]}$
therefore I assume $\ce{[H3O+]} = 10^{-7}$, because  $(10^{-7}\times 10^{-7} = 10^{-14})$
$10^{-7} + 10^{-8} = 1.1\times10^{-7}\,\frac{\mathrm{mol}\,\ce{H3O+}}{\mathrm{L}}$
$\mathrm{pH} = -\log(\mathrm{ans}) = 6.96$

However, after I checked the answer it says the $\mathrm{pH}$ is supposed to be $6.98$ instead of $6.96$.
What they did was: $10^{-8} + \pu{9.51E-8}$
Where does this $9.51\times10^{-8}$ come from?

We can not in this case neglect water auto-dissociation reaction when determining the pH of the solution.
The equation of the electro-neutrality of the solution:
$$\ce{[H3O+]}=\ce{[Cl- ]  + [OH- ]}$$
$$\ce{[H3O+]} =c+ \frac{K_\mathrm w}{\ce{[H3O+]}}$$
where $c$ is the concentration of the strong acid.
By arranging the above equation, we get a second order equation:
$$\ce{[H3O+]}^2 -c\ce{[H3O+]}- K_\mathrm w=0$$
If we solve this equation (and take only the positive solution), we get:
$$\ce{[H3O+]}=\frac{10^{-8}+ \sqrt {10^{-16}+ 4 \times 10^{-14}}}{2}$$
By taking the minus of the decimal logarithm
$\mathrm{pH}=6.98$

I'm sorry, could you elaborate the first step? I do not understand the first 2 equations.

@user21398 The first equation is called a charge-balance equation and says that, since the system as a whole must be electrically neutral, the total positive charge in the system must be equal to the total negative charge in the system. The second equation is just replacing $[\ce{Cl-}]$ with the concentration of the strong acid (since it dissociates fully) and replacing $[\ce{OH-}]$ with the expression involving $K_{\text{w}}$.


text:
Nucleophilic attack by bromine?
According this scheme the bromine act as a nucleophile and attacks pyrimidinium generating Br+ species? I always thought that Br2 acts solely as electrophile with Br- as by product.

I agree it doesn't look right. Perhaps there is an initiation step where $\mathrm{Br^-}$ is created for example by attack of $\ce{SO_3^-}$ on bromine.

@K_P Only there's no $\ce{SO3-}$ in there, only $\ce{SO3}$ (gas) and $\ce{HSO4-}$.

@Jan I meant from the pyridinium suflonate that is formed but it was an example. One could think other ways to create some $\ce{Br^-}$ to initiate the reaction. And in any case it is just a suggestion.


text:
How can I dissolve copper but not zinc?
I have a zinc-body garden spray head with a brass garden hose thread stuck inside.  This is actually an irritatingly common problem.
Which got me wondering: What chemical means exist to dissolve copper but not zinc?
(Going the other way is easy: just dipping in HCl acid dissolves the zinc but doesn't touch the copper.)
Boiling in lime sulfur will oxidize a thin layer of copper, but I don't think that will be sufficient to extract the hose thread.
Tryggvi Emilsson supposedly used KCN and $\ce{K2S2O8}$ to dissolve copper without touching zinc, but claims it was a very tricky process to get right.
Is there a clever way to extract the copper using electrolysis?$%edit$

Your main problem is that zinc is much more reactive than copper, so it is very hard to come up with something that creates some copper ion but not a zinc ion.

Alternatively, try heating to separate the parts. The thermal expansion coefficient for zinc is about $\alpha=30\times10^{-6}\ \mathrm K^{-1}$, whereas the value for brass is about $\alpha=19\times10^{-6}\ \mathrm K^{-1}$.

If it is worth the trouble, you might saw three cuts through the copper hose fitting (which will also make notches in the threads of the zinc female fitting). You can then try to pry the sections of the copper male fitting out. However, the odds are that the zinc is corroded and bonded to the copper. Next time, you might coat the threads with grease or soft wax to discourage electrolytic corrosion. It's not a chemistry answer, but much safer than using cyanides.

@DrMoishePippik - Indeed; good practical answer.  That is how I generally solve this sort of problem.  I'm just a little surprised there's no reliable way to attack copper without hurting zinc.   (Or maybe cyanides are reliable and the authors of that process just wanted to discourage people from "casually" attempting it?  Because of course I keep a good stock of KCN at home ready to open up every time I get copper stuck to zinc ;)

Brass is an alloy of copper and zinc. you dissolve the alloy and may possibly precipitate zinc. Of  course it will not be in the from of a spray fitting. We very rarely saw dezincification of brass in the refinery;  never did figure out exactly how it occurred.

H2S reacts with the copper in Monel metal over the nickel but the metals are intimately mixed. The best suggestion is probably the careful application of heat and /or cold. If corrosion is the cause soaking in a strong ammonia solution might dissolve the interface corrosion.


text:
Reason for isotropy in amorphous substances
I'd be thankful if someone explains why amorphous substances are said to be isotropic and crystalline substances anisotropic

Not a bad question, but if you can clarify a little bit (narrow the scope), it will help you get a better answer. For example, some crystalline solids are isotropic. And there are probably corner cases (in any event) in which the statements don't apply in re: one or the other being isotropic or anisotropic.

Related: How can diamond have a fixed refractive index?


text:
How to purify volatile haloalkanes prepared by Darzens halogenation?
We know that the Darzens method of substitution works on $S_n i$ and is considered the best method in terms of yield of product. Also the side products are $\ce{SO_2}$ and $\ce{HCl}$ so the equilibrium shifts in the forward direction. But we know the lower haloalkanes are gases so if the molecular masses of the halides are comparable with that of the side products then how can we purify them since we cannot use Grahm's Law of Diffuson.

Sir @Jan I have tried to improve by posting a link. I m just an beginner. Please help

At least I understood what the method is supposed to be, and ron took care of the tags (thanks!) I'm still voting to close as unclear because I don't know about your specific case.

If you bubble the product gas stream through water that will remove the $\ce{HCl}$.  There is probably a similar process that will remove the $\ce{SO2}$, perhaps bubbling the gas stream through a tube of  $\ce{CaCO3}$.

If they're low boiling, aren't you just going to purify by distillation anyway?

@Zhe But Both of the haloalkanes are volatile and the boiling points are also comparable.

Then you can't separate them...


text:
Calculating partial pressures from mixture of gases

A gas is enclosed in a container at $35~\mathrm{m^3\,mol^{-1}}$ and has the following composition at $49\,\mathrm{^\circ C}$
$\ce{N_2} \ 2\,\% $
$\ce{CH_4} \ 79 \,\%$
$\ce{C_2H_6} \ 19 \,\%$
Assuming an ideal gas, calculate the partial pressure of each component (in kPa).

My working so far:
I chose $100 \,\mathrm{mol}$ because I've been given percentages and it's a nice number to work with, and I multiplied this with $35 \,\mathrm{m^3\, mol^{-1}}$ to get $3500 \,\mathrm{m^3}$
I used $P=\frac{nRT}{V}$ to calculate an overall pressure but got $76\, \mathrm{Pa}$ which seems way too low. Have I done anything wrong?

$35,\mathrm{\frac{m^3}{mol}}$ is, in fact, in the question?

$35~\mathrm{m^3,mol^{-1}}$ means that 1 mole of gas occupies 35 $\mathrm{m^3}$ of volume. What is the volume of 1 mole of gas under normal pressure and how it compares to that?

First calculate the total pressure from the ideal gas law, explicitly keeping track of units to avoid confusion and mistakes:
$$\text{Ideal Gas Law:}~~~~{P=\frac{n}{v}*R*T}$$
where:
$\frac{n}{v}=\pu{\frac{1}{35}\frac{mol}{m^3}}$
$R=\pu{8.31\frac{m^3~Pa}{mol~K}}$
$T=\pu{322K}$
Then plug the values in to the ideal gas equation to get the total pressure:
$$P_{total}=\pu{\frac{1}{35}\frac{mol}{m^3}*8.31\frac{m^3~Pa}{mol~K}*322~K}$$
$${P_{total} =\pu{76.5Pa}}$$
Then just multiply the total pressure by the fraction of each gas as stated in the problem:
${P_{\ce{N2}} = \pu{76.5Pa * 0.02 = 1.5 Pa}}$
${P_{\ce{CH4}} = \pu{76.5Pa * 0.79 = 60 Pa}}$
${P_{\ce{C2H6}} = \pu{76.5Pa * 0.19 = 15 Pa}}$


text:
Is it possible to make deuterated organic compounds?
Organic compounds always have a $\ce{C-H}$ bond. But can we make compounds having $\ce{C-D}$ bonds where  D is the deuterium part like $\ce{CD4}$, $\ce{C2D5CHO}$, $\ce{C6D6}$  and other complex organic compounds? Overall, can we make organic compounds containing deuterium? Also, can we consider those compounds organic? Are the compounds stable? Give some example of preparatory reaction of those compounds.

Um, I haven't heard of "deteurium bonds" to date. And don't forget to use Mathjax for chemical formulae formatting!

Yes, we can, and they can be quite useful probes of chemical and biological processes. Yes, they are still considered organic.

Just to give an example - $\ce{CDCl3}$ is widely used as an NMR solvent.

@orthocresol As is $\ce{C6D6}$, which the question specifically asks for ;)

@orthocresol, What is a NMR solvent?

@NilayGhosh A solvent used for NMR. I'm going to quote from the Wikipedia page on proton NMR - "Simple NMR spectra are recorded in solution, and solvent protons must not be allowed to interfere. Deuterated solvents especially for use in NMR are preferred, e.g. deuterated water, D2O, deuterated acetone, (CD3)2CO, deuterated methanol, CD3OD, deuterated dimethyl sulfoxide, (CD3)2SO, and deuterated chloroform, CDCl3."

Deuterium occurs at around the 0.02% level in nature (natural abundance level).  So all compounds that contain hydrogen will already contain a low level concentration of the deuterated isomer(s).  So in fact, deuterated compounds do exist in nature.
Despite this low level of natural deuterium incorporation, the deuterium can still be detected by various spectroscopic techniques (mass spectrometry and nmr for example).
Further, organic compounds can be synthesized with deuterium in specific locations. These deuterated compounds are still considered organic compounds. Since deuterium is a stable nucleus (e.g. does not undergo radioactive decay), the compounds containing deuterium are also stable.  Because the $\ce{C-D}$ bond is a bit shorter and therefore a bit stronger than the corresponding $\ce{C-H}$ bond, there is an isotope effect in reactions where these bonds are broken. Because the $\ce{C-H}$ bond is weaker, it will break up to 7 times faster than the corresponding $\ce{C-D}$ bond. This can have a significant effect on biological processes.
$\ce{D2O}$ and $\ce{D2}$ gas are relatively common and inexpensive materials that can be used to introduce deuterium into other molecules. Here are 2 examples involving the use of these materials for deuterium incorporation.  In the first example $\ce{D2O}$ is used as the source of deuterium, in the second example deuterium gas is used to introduce the deuterium.  If you google "organic preparation of deuterated compounds" you'll find many more examples.

image source

image source
Specifically deuterated compounds are widely used in many areas of research and industry.

Is halogen substitution always used? What about preparing deuterated acetylene C2D2 from CaC2 and heavy water?

@Rich Yes, that would be another (of many) useful approach.


text:
Significance of sign of standard reduction potential
I am very confused with the signs of standard reduction potentials.
For example the standard reduction potential of Zinc is −0.76 V and that of copper is +0.34 V.
I cannot understand the  significance of the signs.
As far as I know,if the value is negative then the reaction is non-spontaneous(from the relation between Gibbs Free Energy change and Electrode potential).But if the reaction is non-spontaneous, then why are we classifying it as an electrochemical cell?
Should'nt it be classified as an electrolytic cell?
(Sorry if that sounded incorrect but my basics are weak)

What reaction are you talking about and why is that important? Also, if the reaction isn't spontaneous, then the reverse reaction is, so the cell will still produce some current, only in another direction. Why wouldn't we call it an electrochemical cell?

@IvanNeretin I am talking about a galvanic cell.I am not talking about any reaction in specific.I am just asking about the signs of the Electrode potentials.

Well, then your understanding is about right: the sign of potential tells us whether certain reaction will be spontaneous or not. What's confusing about that?

There are several questions here so let me start by defining the different types of cells and standard reduction potentials.
Definition of Electrochemical (Galvanic) cell: A cell that converts chemical energy into electrical energy. In these cells a redox reaction creates electrons that can do work.
Definition of Electrolytic cell: A cell that converts electrical energy into chemical energy. In these cells the electrical energy source provides the electrons to perform a reaction.
Standard reduction potential: The tendency for a chemical species to be reduced and is measured in volts at STP. The more positive the potential is the more likely it will be reduced.
Now, here are your two reduction equations for $\ce{Zn}$ and $\ce{Cu}$:
$$\begin{alignat}{2}
\ce{Zn^2+(aq) + 2e- &-> Zn}\qquad&{-}0.76\ \mathrm V\\
\ce{Cu^2+(aq) + 2e- &-> Cu}\qquad&{+}0.34\ \mathrm V
\end{alignat}$$
The more positive the potential the more favorable the reaction as it is written will be. Remember that  $\Delta G = -nFE^\circ$ and that when $\Delta G$ is positive, the reaction is non-spontaneous, and when $\Delta G$ is negative, the reaction is spontaneous.  Positive values of $E^\circ$ will lead to negative values of $\Delta G$ and vice versa.
So, the reduction of $\ce{Cu^2+}$ to form $\ce{Cu}$ is more favorable than the reduction of $\ce{Zn^2+}$ to form $\ce{Zn}$. This means that  $\ce{Cu^2+}$ is a better oxidant than $\ce{Zn^2+}$. For an electrochemical cell, the cell potential can be calculated by the following equation:
$$E^\circ_\text{cell}=E^\circ_\text{cathode}-E^\circ_\text{anode}$$
For a working electrochemical cell we need $E^\circ_\text{cell}$ to be positive. I would use $\ce{Zn}$ as the anode (oxidation) and $\ce{Cu^2+}$ as the cathode to give an $E^\circ_\text{cell}$ of $+1.10\ \mathrm V$.

thanks for the edit Ivan wasn't sure how the arrows worked


text:
When expressing a quantity, do the non-significant figures determine whether you round up the last significant figure?
I had thought that the non significant figures determined whether the least significant figure was to be rounded up. For example, given the number $9.1145$, if the number is to be expressed as three significant figures, then, I had thought that, it would be expressed as $9.12$.
However, I noticed that in one of my chemistry professor's calculations in which the quantities were expressed as significant to three figures, she expressed $8.125 \times 10^2$ as $813$. In another calculation she expressed $9.1145 \times 10^{-3}$ as $0.00911$. However, it seems to me that $0.00912$ is the correct expression of $9.1145 \times 10^{-3}$, as the $5$ requires rounding up the $4$, which requires rounding up the least significant $1$ to $2$.
Is there a rule for rounding that my professor is following that I don't know?
(Note: this isn't a homework question, but the homework tag seemed to be the most appropriate one.)

Never round twice. If that was the rule, then 0.46 could be rounded up to 1, which is clearly not the case.

Relevant meta discussion

The international standard ISO 80000 Quantities and units – Part 1: General includes the normative Annex B Rounding of numbers, which stipulates:

Rounding in more than one stage (…) may lead to errors. Therefore, the rounding shall always be carried out in only one step.
Example: 12.254 should be rounded to 12.3 and not first to 12.25 and then to 12.2.

(Note that, in the given example, the ISO standard follows the round half to even rule in which the even multiple is selected as the rounded number. This is the default rounding mode for floating-point computation. Anyway, the rule “rounding shall always be carried out in only one step” is independent of the used tie-breaking rule.)


text:
Why is quickly cooling a a hot pan bad for it?
I've heard that really quickly cooling a hot pan is bad and reduces the lifetime of the pan. I'm wondering, from a chemical perspective, why this would be true. My thought is that as a pan is heated up, there may be slight rearrangements in the lattice of the metal, which, if allowed to cool, would likely reset into the correct portion of the lattice. On the other hand, if quickly cooled, perhaps these atoms don't make it back into the correct place, and this warps the pan.
As a secondary question to this idea of warping a pan, why would reheating the pan not cause it to return to the original shape? Seemingly, that original shape is a lower energy structure of the lattice, so with enough energy to overcome barriers, why would it not "unwarp"?

Specifically it depends on material the pan is made of. Just like how heated glass can crack once immediately expose to a source of the opposite temperature. For example, a cast iron skillet has the potential to crack when at a high temperature it is exposed to a much lower temperature. To answer your second question, the reason why reheating the pan wouldn't return it to its original shape is first, you would have melt the metals to reform it and second of all if a crack has occurred in the situation, it means the elastic limit has been reached. The elastic limit is the greatest stress something can undergo and return to its original shape. Beyond the elastic limit, it is impossible to return something to its original shape. I hope this helps!


text:
Using Le Chatelier's Principle to explain why percent dissociation of acids/bases increase with dilution
$$\ce{HA(aq) +H2O(l) <=>H3O+(aq) +A- (aq)}$$
Say the acid solution is currently in equilibrium. How would diluting the acid by adding water change the equilibrium position? Le Chatelier's Principle says pure liquids like water do not affect the equilibrium position. Using the equilibrium constant, it is easy to show the reaction should shift towards the right, but how can we show that with Le Chatelier's Principle?

Le Chatelier's Prinicple also says that the system will adjust itself based on a change in concentration.
Concentration of an acid is (normally) expressed in moles/liter. If you were to increase the amount of solvent (water) which is on the reactant side of the equation, you would be diluting the acid solution that is in equilibrium and causing the system to adjust accordingly.
As far as the last portion of your request is concerned I am slightly confused. Le Chatelier's Prinicple is built into the equilibrium expression. If I wanted to show how the system were to react based on the addition of more reactants/products I would just use the equilibrium expression for the system.
Hope this helps, D

So when the water is added, the concentrations of HA, H+, and A-, will all decrease, so we can't really use Le Chatelier's  Principle right? But using the equilibrium constant will always work.

Le Chatelier's  Principle can still be used to get an idea of how the reaction is going to adjust according to what is added. The equilibrium constant is a mathematical way of determining exactly how it will adjust.

As concentration of weak acid decreases, a dilution is occurring, thus an increase in water. So equilibrium shifts right to make up more products to make up for increase in h20. Products/reactants. More products = bigger numerator = increase percent dissociation.


text:
Fractional vs Simple distillation
After performing both simple and fractional distillations for a 50/50 mixture of methyl acetate/ethanol, i discovered that my total recovery (in mass) of substances after distillation was lower for the fractional distillation than for the simple distillation. Is this to be expected, or would it be because of an error on my behalf? (To clarify, my initial mass for F. distillation was 17.39g and my recovered mass was 14.95g. My initial mass for S. distillation was 16.54g and my recovered mass was 16.04g.) What would be some reasons in explaining this result?

I'll give you so much for now (because I'm about to leave): It is expected. I'll let you find out why (or leave that explanation to others).

In practice distillation is not 100% efficient: You never get a perfect separation of a mixture.  But fractional distillation is more efficient, so you would expect the product to be of higher purity than that produced by simple distillation.
E.g., if you have a 50/50 mixture by mass of two liquids, you would expect a higher concentration of the higher-boiling-point liquid at the end of a fractional distillation apparatus, which means a smaller fraction of the original mass.


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Do you refer to a bonded atom as an "ion"?
Take H2O, would you call the H a hydrogen ion? Or just a hydrogen atom?

Answer: Hydrogen atom.
Remember, the Hydrogen in this case is polar covalently bonded to the Oxygen. The Oxygen has not completely stripped the Hydrogen of it electron but it DOES have primary ownership.
Take a look at NaOH. In this case you would consider the Sodium portion of this to be an ion. Because the Oxygen has completely stripped the Na of its electron so the two are now held together by opposite charges. [Na+][OH-]
At least that's how i like to think about it.
All the best,
D


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Ice + steam → water
When 50 g of ice at 0 ℃ added to 10 g of steam at 100 ℃, 60 g of water is formed. What is the temperature of water?
Answer: 40.2 ℃.
I'm not asking for the workout of this question. I just want to know how can I reach to this answer, preferably with explanation.  Thanks.

Well, first you'd like to transform your steam to water and see how many Joules you get from that (you'll have to look up the specific heat of vaporization of water in a textbook). Then melt the ice with that heat (here you'll need the specific heat of fusion) and heat the resulting water with the leftovers, if any (here you'll need the specific heat capacity of water). And finally, just mix together two amounts of water at different temperatures.
Upd. OK, I mostly agree with what you are doing, except for the last operation. You check the mass of ice you can melt with the heat obtained from the condensation, and found it to be 67.66 g. Good, but we don't have that much. We only have 50 g. What does that mean? It means you can't spend all that heat on melting. When you're done with melting, there is some leftover amount of heat. Now we use it to heat the water we obtained by melting. Can you do that?

Ivan what do you mean by ' heating the resulting water with the leftovers'?? Which is which? And why do we need to heat the 2 of them? And you've mentioned to ' mix together two amounts of water at different temperatures'. Meaning?? BTW I've did the first 2 parts and I've posted it on my question. I'm not sure whether I did it correctly. Can you have a look at it?

Wait a minute, I'll edit my answer.


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Does milk drinking prevent long-term chemical poisoning?
I have heard some rumors that drinking milk prevents chemical poisoning. I have done a bit research and some sources confirm that.

Corrosive Poisons
The best first aid is to dilute the poison as quickly as possible. For acids or alkalis (bases), give the patient water or preferably milk or ice cream — one (1) cup for victims under five (5) years; or one (1) to two (2) glasses for patients over five (5) years. Milk or ice cream is better than water because it dilutes and helps neutralize the poison. Water only dilutes the poison. Cornell University

In Case of Poison Ingestion: Drink Milk?
As @Jan pointed out in the case of chemical poisoning the paramedics are immediately called.
So my question is, does milk helps chemical poisoning, e.g., for people that daily work with chemicals or are working in chemical facilities (working with or be in the presence of $\ce{HCl}$, $\ce{HF}$, $\ce{NaOH}$, $\ce{HNO3}$, $\ce{H3PO4}$, etc.)? Or what would be beneficial? And I mean long-term exposure and not immediately threat to life or health.

Great question. I would surmise that it primarily helps with ingested poisons and will have little effect on gaseous poisons or poisons that absorb through the skin.

Milk can act as a buffer due to all the proteins dissolved. Water cannot. I think that's the gist. Whatever you do, in case of chemical poisoning immediately call the paramedics! No, seriously, immediately!

If your question is just about the general exposure (i.e. not drinking, not pouring litres of something over your skin/in your eyes/down your nostrils) then the answer is: This level of exposure is not considered toxic, no actions are taken. It is assumed that the body can handle the chemicals.

That can be true, but when you work in a chemical factory, limits are sometimes exceeded. Or you can accidentally inhaled some vapours. Then some preventive actions should be made,  like drinking milk?

Everything that strengthens your bodys immune system will help in some way. So better eat enough fresh vegetables and fruits. And for the same reason it should be better to drink human milk before any other milk.

The high calcium content of milk is supposed to help in case of excess oral fluoride intake. $\ce{CaF2}$ is largely insoluble. But you have a few minutes at best, after that most of the HF has wandered through your stomach lining.

This post only concerns chronic poisoning, i.e. exposure to a harmful chemical in low amounts for an extended period of time. The answer is not applicable to acute poisoning; in that case a doctor should immediately be called.
Basically there are three types of chemicals that have a harmfulness threshold to which one can be exposed over extended periods of time.
Type one includes substances such as mercury that are primarily aggregated in the body rather than excreted. In rare cases, drinking milk may enhance excretion thereby decontaminating the body. I haven’t heard of any such examples, though, and I would assume it to be a very rare case so neglegible.
Type two includes those substances that the body can actually metabolise to harmless products. This would include most protein poisons. However, the metabolism shouldn’t be enhanced by drinking milk so it will likely have a non-effect.
Type three includes those that, while they aren’t metabolised, just don’t do enough damage at low concentrations. That would basically include your list of acidic and basic substances but also non-aggregating metals at low concentrations. For these, excretion is not significantly enhanced or reduced by drinking milk so again no effect is expected.
Comparing the acute exposure to acids to the chronic one, in the former giving a person milk to drink is actively encouraged. This is because milk contains a lot of proteins that can act as buffers, increasing the body’s buffering potential and reducing the effects of the acid. The body does have more than enough buffering capabilities to cope with chronic cases though, so even there regularly drinking milk would not help.
Tl;dr: Milk is a great drink but cannot act as a type of preventive antidote.

Can the same be said for soy juice a.k.a. soy milk?

@Martin-マーチン I don’t know much about soy milk or hazelnut milk or any other of the milk-ersatzes, so I can’t tell you that, sorry.

Drinking milk won't help of course if you're breathing in carbon monoxide. So drinking anything only can help if you ingested the poison.
Whether or not milk would help more than an equal amount of water would depend on the poison ingested. It is possible that the particular poison could be actually neutralized, or it is possible that it would dissolve in the fat, thus allowing more time for other measures such as pumping the stomach out.
So if milk actually neutralizes the poison, then drinking milk would certainly help long term exposure. However if milk just dilutes the poison then over time the long term exposure would reach some steady state.
For the strong acids and alkalies that you listed chronic exposure to low amounts will do no harm. In other words for those compounds the harm will be done in minutes, and not build up in the body over years of exposure.
Another caveat here is that ingestion is the most preventable exposure. In any chemistry lab eating or drinking is against the rules.

Thanks for the answer. You are right, most long therm poisoning will be due to inhalation. Can you provide some source to your statement:
For the strong acids and alkalies that you listed chronic exposure to low amounts will do no harm. In other words for those compounds the harm will be done in minutes, and not build up in the body over years of exposure.

Milk has proportion of fats and For chemicals with high partition coefficient like organochlorines it follows that a certain proportion will dissolve in the fats and taken up by the body. Milk in this case will not be useful as antidote


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Qualitative Analysis of Different Cations
So, we were taught in lab how to identify a couple of cations and anions through elimination and confirmatory tests. I was wondering, however—how would this work if the test solution were made up of multiple ions?
For example, calcium and zinc would form a white precipitate, respectively, if you add $\ce{NaOH}$. In excess, zinc would dissolve; calcium would not. $\ce{Fe^3+}$ would form a red precipitate.
In a test solution with these three in it, I would assume that there would be red precipitate and white precipitate if you add $\ce{NaOH}$, is that right? And if you add in excess, only some of the white precipitate would dissolve … or something like that. How would you effectively go about the process of identifying the ions?

I'm pretty sure you would have learnt about more different possibilities of identifying the cations in solution (or not). Specifically for $\ce{Ca^2+}$, there is a qualitative analysis method that works independently of all other ions being there or not there. Are you, by any chance, in a practical course that has a nickname something like 'ion lottery'?

Eh... this is just how qualitative analysis is taught at a more basic level. Precipitation with NaOH or NH3. Obviously it's rather flawed. @OP you're right this is a rudimentary (and therefore poor) way of identifying ions. People write textbooks on these stuff. Do a google search for "Vogel inorganic QA" if you're interested.


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What other intermoelcular forces can exist between neutral molecules?

What kind of intermolecular forces can exist between neutral molecules?

My answer was London dispersion.
My logic was this: London dispersion is the only thing weaker than dipole-dipole, and as far as I am concerned two neutral molecules can't be attraced by dipole-dipole forces because you need a positive end on one and a negative one on the other: but there cannot be because, well, both are neutral, no?

What are the intermolecular forces present in $\ce{CH3CH2NH2}$?

My answer was also London dispersion because it appears like this molecule is neutral.

Both answers are wrong. It appears I don't properly understand intermolecular forces. Why are my answers wrong?

It sounds like you are confusing polarity with formal charge. Polarity is an uneven distribution of electron density within a molecule. This uneven distribution can occur for several reasons, but the most common is difference in electronegativity between atoms.
Consider the $\ce{O-H}$ bond. The electronegativy of an oxygen atom is 3.44 and that of a hydrogen atom is 2.20. The difference (1.24) is quite large given the scale commonly only goes to 4.0. Thus, the $\ce{O-H}$ bond is polarized toward the oxygen atom. The oxygen atom has a stronger pull on the electron pair than the hydrogen atom. More electron density is around the oxygen atom than the hydrogen atom. Or, if you prefer the other interpretations, the electron pair is closer to the oxygen atom or spends more time around the oxygen atom.
Regardless, there is an uneven sharing of electrons, and the oxygen atom, has more. The oxygen atom thus can be said to have a partial negative charge to represent its overabundance of electron density. This partial charge is not a full formal charge, but rather a shorthand notation to help us remember that the oxygen atom has more than an even share of the electron density in the bond. We represent these partial charges with $\mathrm{\delta^{-}}$. Consider the notation on a water molecule:

Because of these uneven distribution of electrons, water (which is a neutral molecule) has a permanent dipole. These dipoles attract each other.
The type of force that exists between ions is called the ionic interaction.

This can be a bit tricky when you are first starting your journey into the world of intermolecular forces. What I like to look for first are what (if any) heteroatoms are present when the backbone of the molecule is composed of carbon atoms.
In your case there is a nitrogen present and it is attached to 2 hydrogen atoms. This is a perfect example of a molecule that will exhibit hydrogen bonding as an intermolecular force. Hydrogen bonding, at least in my opinion, is the easiest to pick out. N, O, and F atoms bonded to Hydrogen are the only species in which this attractive force between molecules is observed. Now if you were to remove that $\ce{NH2}$ group and just had $\ce{CH3CH3}$ you would be correct. The only intermolecular force between the molecules would be London forces.


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What is the role of dichloromethane in the ozonolysis / Criegee oxidation?
Why is $\ce{CH2Cl2}$ used in ozonolysis of alkenes and alkynes? What role does it play in the reaction? Is dichloromethane a common solvent for organic reactions and used for its inertness, or is it particularly used in ozonolysis reactions? If it is the latter case, how does it enhance the ozonolysis reaction?

Dichloromethane is often used as a 1:1 cosolvent to facilitate timely cleavage of the ozonide.

All of my ozonolyses to date were in dichloromethan …

Is $\ce{Ch2Cl2}$ used as an indicator for the complete reaction?

CH2Cl2 is a nice solvent as it boils at 42 C, making it very easy to remove, and doesn't produce peroxides. This makes it excellent for many reactions. Unfortunately, it requires quite uncommon CaH2 for deep drying.

Sometimes the solvent choice is very important to enhance or decrease reactivity of either of the reactants.
Dichloromethane is a widely used organic solvent. Chemists sometimes say tongue-in-cheek that it can dissolve almost anything organic since it is rather mid-range in polarity. Other than that, it is also non-nucleophilic, non-electrophilic and generally does not take part in many reactions. (There are certain exceptions, however!) This means that it is a standard solvent choice for many reactions and the ozonolysis is no exception. Most reactions require a solvent to allow all reactants to get into close proximity of each other for the reaction to occur at all — not many reactions can also be conducted in solid state.
The standard mechanism for ozonolysis — the Criegee mechanism — is given for solvents such as dichloromethane, hexanes or any others that do not participate in the reaction in any way (that are inert). However, when using methanol or other alcohols the mechanism must be modified as also highlighted by Criegee in the original mechanism publication. I have included the mechanism for ozonolysis both in presence and in absence of methanol in the scheme below.

The resulting peroxy-methyl-hemiacetal is much more stable than the carbonyl oxide and thus various side reactions, e.g. formation of Staudinger ozonides with remote carbonyl groups are inhibited. Thus, in the case of the ozonolysis, adding methanol decreases the reactivity of the ozone in solution. Conduction the reaction in dichloromethane should be considered the standard case — unless one wants to say that it enhanced ozone’s reactivity.
These details can also be found on organic-chemistry.org and this page from Michigan State.

Dichloromethane (DCM) is used in ozonolysis for inertness and it also gives ready access to easy work up through aqueous washing, from which it is easily separated normally containing the product(s).I've not seen any evidence for it participating in the reaction.
Alcohols react to form alkylhydroperoxides post ozonolysis.  Toluene and acetonitrile can be ozonolysed, for example. Ethers and esters are normally fairly inert. Water can help with oxidative work-up.
DCM is very good at solubilising compounds and doesn't interfere with the reaction if further reduction or oxidation are needed. DCM is actually dangerous when used with secondary or primary amines as they accumulate and can cause unexpected thermal events


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What is the mechanism for reduction of alkylmercuric compounds by hydrides?
What is the mechanism of the reductive elimination of mercury metal using hydrides that is carried out after oxymercuration of the alkene? I have come across a free radical mechanism, but that suggests loss of the stereospecificity of the product, while the reaction is stereospecific in reality. Also, why isn't an acid sufficient to remove the mercury as it is with alkynes? Why is a reducing agent necessary at all?

The reducing agent NaBH4 is used for this reaction because it is highly selective. If H2SO4 were used in place of NaBH4 in this reaction the molecule would be dehydrated and we would probably see some HgSO4 form as well. The main thing to remember here is that a reducing agent has much more selectivity than an acid solution.

Do organomercury compounds in general give alkanes with acids (save the case when alcohol groups are also present in the compound, or any such groups which might also react with the acid)? Grignard's reagents are also organometallics, and they do.


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Electrons flow difference between circuit and electroplating process by considering battery's positive and negative side
Electrons flow  from the negative end to the positive end of a battery (via a circuit) but in electroplating process electrons flow from the battery's positive end  to the battery's negative end. Why this happens ?
Here I want to know about the electrons flow difference in battery's positive side  to the battery's negative side between circuit and electroplating system.

Here in the first figure by considering battery's positive and negative side electrons flow from positive side to the negative side.

But in this second figure by considering battery's positive and negative side electrons flow is opposite from the first figure.

Hi and welcome to chemistry.stackexchange.com. Feel free to take a [tour] and visit the [help] for any questions on how this site works. I frankly fail to understand your question, because the electrons are still flowing from negative to positive. Remember that cathode and anode are names for the electrodes of reduction and oxidation respectively and do not reflect a charge.

Does the duplicate not answer your question? If so, please explain why. If you have a new question, your chances are better if you post it using the 'ask question' button.


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Regioselectivity in electrophilic aromatic substitution of benzofuran and indole
Why does benzofuran react preferentially at position 2 when nitrated or formylated, while indole reacts preferentially at position 3 (source: Joule and Mills, Heterocyclic Chemistry 5th ed., p 371)?

In electrophilic substitution of aromatic systems sigma complexes are typically formed as reaction intermediates.  Let's look at the sigma complexes formed in the reaction of benzofuran and indole with an electrophile ($\ce{NO2+}$) at the 2 and 3 positions and compare their stability.

With both benzofuran and indole attack at the 2 position produces a sigma complex where the positive charge is nicely stabilized by the benzene ring (1a and 1b) . The stabilization is analogous to that found in a benzyl carbocation.
Attack at the 3 position produces a sigma complex where the positive charge can be stabilized by the lone pair of electrons on the adjacent heteroatom (resonance structures 3a and 3b), but because nitrogen is less electronegative than oxygen we would expect resonance structure 3b to be a bigger contributor than resonance structure 3a. Therefore we would expect the sigma complex produced upon attack at the 3 position to be more stable in the case of indole than benzofuran.
The overall result would then be to produce more of the 3 isomer in the case of indole as compared to benzofuran.


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If no oxygen is present, will cut avocados stop browning?
Oxygen makes cut avocados brown, or at least it is necessary to facilitate the rapid browning reaction we see within 24 hours of cutting the avocados.
If oxygen was completely removed from the air surrounding the cut avocado, will the browning stop?
If a small fraction of oxygen remained, will the browning slow? In other words, is the rate of browning proportional to the percentage of oxygen in the surrounding air?
I never studied chemistry so my understanding comes from here:
http://www.compoundchem.com/2014/08/03/why-do-avocados-turn-brown-the-chemistry-of-avocados/

Hi and welcome to chemistry.stackexchange.com. Feel free to take the [tour] and visit the [help] for any questions about the site.

Though I'm sure that it's possible for them to brown due to substances other than oxygen, in general, yes, if you remove oxygen from the cut surfaces of an avocado, it will not turn brown, and yes, if you reduce the quantity of oxygen, browning will slow.  The same works for apples, bananas, and pretty much any other fruit that browns upon exposure to air.
It's not particularly practical for intact avocados, but one method for preventing guacamole from browning is to put plastic wrap right against the surface of the dip, preventing air contact.  Additionally, if you look at packaged guacamole, in addition to the browning resistance caused by acidity, the amount of oxygen in any given container is limited to a few small bubbles.  The oxygen in those bubbles is quickly consumed and the remaining gas does not oxidize the guacamole further.
You could conceivably test this with a vacuum chamber and keep normal slices of avocado green for a prolonged time.  You also could use a food vacuum sealer that bags and evacuates food for long term storage, though ripe avocados are delicate enough that they will probably be crushed into paste by the process.

When I used to work as a chef I've soaked avocado segments in orange juice to stop browning , it may be the ascorbic acid in the juice that does this.

Acidity of orange juice is mostly due to citric acid.


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Stereoisomers of Hexachlorocyclohexane
What is the basis of naming the various isomers of hexachlorocyclohexane $\alpha$, $\beta$, $\gamma$, etc.? Here is a list of these isomers. Is it just random naming, or is it according to some convention?

I didn’t find anything definite. However, I found a PDF from the State’s Institution for Protection of the Enviroment in Baden-Württemberg (Landesanstalt für Umweltschutz) which gives me two or three relevant clues. Along with my intuition and a chat with a labmate, I conclude:
There is nothing systematic or conventionalised in the names of the isomers. They were likely named after the order in which they were isolated or their structures were determined.
Clues I used:

The $\unicode[Times]{x3B6}$, $\unicode[Times]{x3B7}$ and $\unicode[Times]{x3B8}$ isomers were synthesised last.
$\unicode[Times]{x3B1}$, $\unicode[Times]{x3B2}$ $\unicode[Times]{x3B3}$ and $\unicode[Times]{x3B4}$ were characterised first by a guy called Teunis van der Linden, after whom the active insecticide $\unicode[Times]{x3B3}$-hexachlorocyclohexane was named lindane.
Thereby concluding that $\unicode[Times]{x3B5}$ was discovered after $\unicode[Times]{x3B1}$ to $\unicode[Times]{x3B4}$.


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Unusual Friedel–Crafts alkylation with pivaloyl chloride
In class we were told of an anomalous Friedel–Crafts reaction with a tertiary acid chloride — $\ce{(CH3)3CCOCl}$, or pivaloyl chloride. When this reacts with benzene in the presence of $\ce{AlCl3}$, instead of effecting the normal acylation, it instead breaks down into the tert-butyl carbocation, which then alkylates benzene to form tert-butylbenzene.
I could find no satisfactory explanation for this, however, except for the fact that a tertiary carbocation is a very stable carbocation — and nor did I find any mention of this reaction on searching. Could anyone please detail this further?

As many reasons already mentioned in previous answers. I also have one reason in mind. I think we can also say that as CO is released, with the evolution of gas , entropy of the reaction increases which makes it more favourable to occur.

You were on right path; tert-butyl carbocation is quite stable$\ldots$ so stable that the acylium cation, which normally reacts as an electrophile itself, instead decomposes via decarbonylation (loss of stable carbon monoxide molecule).

The t-butyl carbocation isn't as stable as the acylium cation (which is stabilised by resonance), but the difference is small enough that with the help of the non-reversibility of CO loss (which is removed from the mixture as a gas) and reduction of the steric strain (which is lower in t-butyl than in acylium), it is enough to make it the main reaction route instead of a side route.

Is there any other similar reaction which you know of?

@Charles https://en.wikipedia.org/wiki/Hayashi_rearrangement, https://en.wikipedia.org/wiki/Wagner%E2%80%93Meerwein_rearrangement

Does the rearrangement in the question asked also have a name?

It's not a rearrangement but F-C alkylation, only using unusual reagent.

What is the major product? Once tert-butyl benzene is formed, the ring is more activated towards substitution

@schrodinger_16 I'd expect some 1,4-bis(tert-butyl)benzene, but tert-butyl benzene should be main product if C6H6 is used in excess.

@Mithoron Why is acylium cation more stable than tert-butyl cation, exactly?

@YUSUFHASAN https://chemistry.stackexchange.com/questions/41162/why-is-the-acylium-ion-stable

Only one example or do all such tertiary compounds giving stable carbonation give similar reaction?

@user137644 You mean if there's another tertiary group instead of t-butyl, there's gonna be similar effect? Yeah, it could be more or less pronounced, depending on bulky it is and how well is cation stabilised.


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What is the mechanism by which lactones are hydrolyzed during a Wolff-Kishner reaction?
I've been away from orgo for a few years but lately someone asked me a question about lactones during a Wolff-Kishner reduction.  I understand the basic mechanism of a Wolff-Kishner but I cannot figure out how a lactone gets hydrolyzed.  Can anyone help me with that step?
Here is a link to the basic mechanism that mentions lactones.

This is a general feature of the hydrazine reagent that is employed in Wolff-Kishner reductions. When attacking an ester’s (includes lactone’s) carbonyl function, the generated tetrahedric intermediate can break down to form a hydrazide as shown in the scheme below.

The carbonyl function is not reduced but the ester (or lactone) is hydrazinolysed.
One could assume that the displace alcoholate could reattack the carbonyl group to regenerate the ester/lactone. However, apart from hydrazine being much more nucleophilic than alcoholates, the resulting hydrazide is also stabilised much like an amide, and is therefore less reactive.


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Calculation of overall rate constant of a parallel reaction
Consider a parallel reaction
$\ce{A->B}$ (rate constant for this reaction is $k_1$ and it follows first order kinectics),
$\ce{2A->C}$ (rate constant for this reaction is $k_2$ and it follows first order kinetics).
Then we can write
$$\mathrm dA/\mathrm dt = \mathrm dA_1/\mathrm dt + \mathrm dA_2/\mathrm dt$$
(where $\mathrm dA_1$ is the amount of $\ce A$ going through the first path and $\mathrm dA_2$ is the amoubt of $\ce A$ going through 2nd path),
$$\mathrm dA/\mathrm dt = k_1[\ce A] + 2k_2[\ce A]$$
Then to calculate the $k$ where $k$ is the rate constant for the overall reaction we write
$$k[\ce A] = k_1[\ce A] + 2k_2[\ce A]\tag i$$
And so $k$ comes out to be $k_1+2k_2$.
My doubt is that to write $-\mathrm d[\ce A]/\mathrm dt = k[\ce A]$ did we assume that in the overall reaction the coefficient of $\ce A$ in the overall reaction to be 1.
Taking a general case
$$\ce{$a$A->products}$$
we can write
$$-\mathrm d[\ce A]/\mathrm dt=ak[\ce A]$$
But according to the equation $\text{(i)}$ we took $a=1$. Why?
If suppose $\ce A$ is a radioactive substance and we have to calculate the mean life of $\ce A$ then we require to know the overall rate constant, so foe that what should be the value of $a$?

Your question is poorly defined. Are both reactions $\ce{A -> B}$ and $\ce{2A -> C}$ first-order with respect to $\ce{A}$, or are they elementary steps (which means you are missing a power of two)? The second equation is missing a minus sign since the LHS is negative and the RHS is positive. Lastly, the stoichiometric coefficient of $\ce{A}$ does not affect the order of the reaction wrt $\ce{A}$. If your "overall reaction" (which btw isn't even defined) has a rate law of $k_1[\ce{A}] + k_2[\ce{A}]^2$ then that is just how it is. You cannot simply label it as "first-order" or "second-order".

Both the parallel reactions are of first order type

In the second equation the lhs is negative and so is the rhs as d[A]1/dt and d[A]2/dt are also negative as they are also decreasing.It is not the order which I am confused about,it is the overall rate constant which I am concerned as if A is a radioactive substance and we have to calculate its mean life we require the overall rate constant.

Related Finding average half life of radioactive nuclide which undergoes two different decays

The key to answering these kinds of riddles is to think about units.  Rates of reaction have units of $\mathrm{\frac{mol}{L\cdot s}}$, but that' isn't fully specified.  You need to know moles of what.
In your case, the confusion is stemming from confusion of what the "moles" in $\mathrm{\frac{mol}{L\cdot s}}$ refers to.
The answer can either be moles of "reaction", or moles of A.  The other key thing to realize is that you can write "the" reaction however you want.  The value of the "rate constant" $k$ depends on how you define the rate.  If for the overall reaction you write:
$$\ce{3A -> \varnothing}$$
...and if the "rate" you are calculating is the rate of reaction, then there are three moles of A lost per mole of reaction, so
$${dA \over dt}=-3k_{\alpha}\ce{[A]}$$
If instead you decide "the" reaction is
$$\ce{A -> \varnothing}$$
then you can rate the rate of disappearance of A as
$${dA \over dt}=-k_{\beta}\ce{[A]}$$
The choice is up to you and it is arbitrary.  You just have to keep track of the moles of what the rate laws you write refer to.

Then to calculate the k where k is the rate constant for the overall reaction we write [[Equation (i)]]

You need to think carefully about what you meant by the words "the overall reaction".  When you write equation 1, you are adding together rates of loss of A, i.e. the units are $\mathrm{\frac{mol~of~A}{L\cdot s}}$.  The only reaction you can write that has the same rate is one where the coefficient of A is 1.
"The" half-life for radioactive elements usually refers to units where units are $\mathrm{\frac{mol~of~element}{s}}$.  The fact that the unit is $\mathrm{\frac{mol~of~element}{s}}$ and not $\mathrm{\frac{mol~of~reaction}{s}}$ means that if we wanted to write the corresponding chemical reaction, the stoichiometric coefficient had better be 1.$%edit$


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Filling up of electrons
Is it necessary that filling up orbitals start from spin-up orientation only? And also, is it mandatory to fill the electrons with one kind of spin or they can be filled with spin up and spin down(as shown in the photofor clarifying my doubt)

You either have them all spin up or all spin down. Not a mixture of up and down. Whether it's all spin up or all spin down is not important (refer to the link in Mith's comment).

Is there any proof? I am asking this just to make sure that I don't miss anything.

also http://chemistry.stackexchange.com/questions/15967/do-electrons-only-fill-spin-up-first-or-could-it-start-filling-down-spins-f


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Predicting isotope patterns in mass spectrometry

Imagine you have a mixture of $\ce{^1H}$, $\ce{^2H}$ and $\ce{^3H}$ (hydrogen, deuterium and tritium) present as diatomic molecules and that the numbers of the atoms of the three species are the same. Sketch the mass spectrum.

The book's answer is

For peaks of mass= 2, 3, 4, 5, ratio of heights = 1:2:3:2:1.

I’m not getting why the answer is this?
Why there is no peaks of mass 1, is it because $\ce{H2}$ consisting of $\ce{^1H}$ can break in mass spectrometer?

Try thinking of the simpler case, only hydrogen and deuterium. Every atom in every $\ce{H2}$ molecule has identical chances of being either hydrogen or deuterium. What would you expect?

Related: Mass Spectrum and Molecular Ion peaks

why there is no peaks of mass 1, because H2 consisting of hydrogen-1 can break in mass spectrometer.

This means they are doing it under the conditions when the molecules wouldn't break.

I can't find where they mention that in the question?

Yes, albeit indirectly: you have to conclude that from the absence of peak with mass 1.

How they mentioned indirectly?

Wait, I misunderstood it: the peaks are in the answer. OK, this means they didn't.

The first step is to write all the possible combinations of H atoms: $$\mathrm{H^1H^1~~ H^1H^2~~ H^1H^3 ~~H^2H^1 ~~H^2H^2~~ H^2H^3~~ H^3H^1~~ H^3H^2 ~~H^3H^3}$$ Now it is important to note that each of these parings have an equal probability of forming and exist in equal quantities in the mixture. Their molar mass are respectively: $$2, 3, 4, 3, 4, 5, 4, 5, 6$$ So there is $1$ way to make $2$, $2$ ways to make $3$, $3$ ways to make $4$, $2$ ways to make $5$ and only $1$ way to make $6$. Since each way has an equal probability of occurring, then the ratio of the molar masses should be in the ratio -  $~1:2:3:2:1$.
Hence for peaks of mass = $2, 3, 4, 5, 6$ ratio of heights = $1:2:3:2:1$
The reason why there is no peak for mass $1$, I am not too sure. However I think it is because if the hydrogen molecule was to break up, it would form a proton and a H radical. Both of these are extremely reactive and unstable and will probably just get lost in the machine and never actually get recorded. This could make sense as the mass spectra for $\ce{HCl}$ or $\ce{HBr}$ doesn't include a peak for mass $1$. Also in mass spectra for hydrocarbons, such as heptane, the smallest mass peak recorded is usually 29 which is the ethyl group. Therefore any thing lower than that (such as the methyl group) is probably too unstable.

The question is not very good because (i) most mass spectrometers that chemists are going to come across don't measure below 5 or 10 Daltons and thus are not useful for measuring hydrogen; (ii) tritium in equimolar abundance to deuterium or protium would produce a dangerously radioactive material, unsafe to handle; and (iii) mass spectrometry actually measures mass to charge ratios, not mass, and it is unclear what ions would be derived from the hydrogen.  All that said, your answer to this not so great question is very good.

The question is not very good indeed. It's a 100% imaginary question and has nothing to do with real MS measurement. 1, The main purpose of Mass Spectroscopy(MS) is to identify compound by measuring the mass to charge ratios. It measures either $[M+H]^+$ or $[M-H]^-$, in positive mode or negative mode, respectively. M refers to the exact mass of the compound. For a organic molecule, it is usually several hundred Daltons. 2, Each isotope of an element has certain natural abundance percentage affecting peak intensity.

@Dejian15 This is incorrect, it depends on the ionisation method. ESI is known to give association complexes such as $\ce{[M + H]+}$ while EI would give $\ce{[M]+}$.

@Jan Thanks for the comment. I should have pointed it out. This is a great website.