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Sulfuric acid fumes from cleaning shower drain
This morning a workman came by to my newly rented property to fix a few things, among which a slightly blocked bathtub drain.
He let some mild water run continuously and he poured down a chemical, and it wasn't before I saw a huge vapor coming out and an intense boiling-like sound, I realized something was wrong. Basically he let down an entire bottle of >95% sulfuric acid down the drain just like that, without even wearing safety goggles or gloves. A pungent odor emanated through the whole flat and I had to open every window (he hadn't advised to open them beforehand).
Now, I don't have much experience with chemicals of this sort, so here are my questions:

Did I inhale toxic fumes?
What should I do to make the bathroom/shower usable again? I noticed that the mirror and several water tiles that are not in the close proximity of the drain but are on the way to the nearest window opening, show condensation streaks, such as those left by drops of water that run down. They were definitely not there before. Is that condensed sulfuric acid vapor? Do I need to clean with a sodium bicarbonate solution?
Needless to say, I also had towels, toothbrushes, and other items in the room, do I need to take any precaution before using them (cleaning, rinsing, etc?).

In short, how do I clean up this mess?

If the workman did use concentrated sulphuric acid (which is not the best choice for drain cleaning) then the best solution if to flush the drain with large quantities of water and wash the surfaces with water.
Sulphuric acid is not toxic but is very corrosive when concentrated. But it is highly water soluble and not harmful when it is dilute enough. So dilution with water will solve whatever problems it seems to cause. If it got splashed around, wash everything thoroughly and there will be no further problem. If you use enough water then it won't even be necessary to neutralise the result as any remaining acid will be swept through the drain and highly diluted by the flow of water.

He wasn’t wrong to be using Sulfuric acid, he just didn’t know what he was doing. Sulfuric acid is sold for cleaning pipes, but you won’t find it easily because it is one of the most dangerous ways to clear a drain.
The boiling-like sound was the acid reacting with compounds in the drain (hair, grease, soap, certain metals, etc.). During the reaction various gases would have been released causing the mixture to bubble and fizz.
There is likely a cocktail of chemicals in the clogged drain so it can be difficult to say exactly what gases will be produced, but the important ones are Hydrogen, Sulfur dioxide, and possibly some Hydrogen sulfide. Hydrogen sulfide is very poisonous. It smells like rotten eggs, but Sulfur dioxide has a similar smell. Hydrogen sulfide in low concentrations isn’t instantly deadly, but it can kill you slowly. In high concentrations one or two breaths of Hydrogen sulfide can be deadly. It is a popular chemical for committing suicide. But that will likely not be the main gas produced. The main point here is that forced air ventilation (a fan) should be setup to blow fumes out a window BEFORE you start using it.
Concentrated Sulfuric acid also reacts with water and releases a lot of heat (this is a whole separate dangerous property aside from it being a strong acid). The mixture in the drain can easily become hot enough to boil the water and when contained in a drain pipe it can erupt a geyser of hot water and acid straight out of the drain. Acid burns are unpleasant. Hot acid burns will send you to the hospital for skin grafts. What can take hours to react with acid at room temperature can happen in just a few seconds with hot acid.
But, yes, Sulfuric acid is used to clear drains, but it is an older method used by people who know how to handle it. Your plumber was clearly not one of these people. Sulfuric acid will instantly strip the nickel finish off the drain ring. Nickel is probably the most common finish on modern bathroom hardware. After contact with Sulfuric acid the fixtures will be black, gray, or yellow gold if it strips down to the brass plating.
Once the Sulfuric acid is drained away and the house airs out there will be no lingering vapors or chemicals to worry about. It’s nasty stuff but it doesn’t linger. If any Sulfuric acid splashed on fabrics or paper you will find the material burned away looking exactly like a cigarette burn. The end result of fire and Sulfuric acid is the same — black carbon.

I doubt that the workman was a professional.
Sulfuric acid has a high boiling point (338 °C) much higher than water so I don't think it would evaporate back out of the drain unless it created a reaction hot enough in the drain for it to sublime.
Without knowing what was in the drain to begin there could be many chemicals that could have been produced by pouring the h2so4 down the drain many of which could evaporate back out the drain, including compounds with higher boiling points than water can leave with the water because there attracted to and soluble in the steam, because this occurs some of the acid most likely evaporated  back out of the drain.
Tap water will neutralize the acid but if it was 95 % which is extremely strong it would take a lot. Taking into account the acid would have been neutralized slightly by the gunk in the drain a few buckets of a weak bicarbonate solution should be OK.  Don't make a strong bicarbonate solution or it'll foam up. It is not good practice to neutralize a strong acid with a strong base or other way around. Sodium carbonate is classes as a weak base and sulfuric acid is a strong acid so if you use a few buckets of a weak bicarbonate solution it should work.
As for cleaning the residues, if it really smells chemically bad like your saying I'd get a professional to clean it up holding the guy that did it responsibly. Don't mess with chemicals if you don't know what they are you might get sick. You may have breathed in toxic fumes it sounds like it. If you feel sick or are worried call the hospital. I'd definitely change toothbrush and towels and anything that goes in or on your body all of which should have been removed from the room prior to using a chemical that strong in the household.
Consider this, the sulfuric acid for car batteries is about 45 % and even that would not be practical to pour down a drain especially if there's metal components or metal pipes. If there is metal involved I'd say a lot of hydrogen would've been created. Hydrogen sulfide may have been produced which has an extremely pungent smell.

Thanks for your reply. Now it doesn't smell that bad, after I left for about 10 hours leaving all the windows open. This is the stuff that he used: (meltmsds)[http://www.markvitow.com/external/commerce/1/productfiles/safety/meltmsds.pdf]. As for the water that would neutralize the acid, how much would it take? And how much and how strong should the bicarbonate solution be? I'm a bit worried that calling a professional cleaner the odds of any professional cleaner not knowing what's going on and using a bleach-based cleaner are quite high and that would make the situation worse.

I don't know exactly how much tap water it would take but a few buckets . Bicarbonate will work better. I think if you used 1/4 cup of bicarbonate in a 9 L bucket of water a few times it should be fine.

I was just reading your link. That guy was an idiot . That chemical is not for use in a home kitchen or bathroom that's an industrial drain cleaner I'd sue him for endangering your life. Have a read of the safety precautions or more importantly the conditions to avoid that are stated in the msds.

Thank you for your response. Suing sounds like too much effort, but I will at least I will report it to his company and my landlord who arranged the work. I just can't believe how he could turn such a simple plumbing job into a chemical mess.

He's an idiot , not the first , definitely won't be the last! It's definitely overkill for a bathroom drain! And as I said read the bit in the msds where it explains how and where not to use the product. If it really is 95% H2SO4 it may have damaged the drain system...

It also explains in the msds how to clean up a spill of the product . Have a look I'm pretty sure it recommends bicarbonate. Acid spill kits usually are bicarbonates or carbonates. The water solution will be fine to clean the droplets too.

Thanks again for your help! I added a few images of the condensation marks I was talking about to the post, would you tell me what they could be the result of in your opinion? I tried in a small part rubbing some away using normal dry paper towels and they come off easily (at least not visible anymore in that part).

I disagree with the beginning of your answer. Sulphuric acid in water will create enough heat for the water to boil, which it will, and which will take sulphuric acid into the environment with it. Other than that a better formatting would also help your answer.

I said some times compounds with higher boiling than water can leave with the steam.  I agree it could be formatted better. Thanks for feedback @ Jan :) i haven't written any science pages for years now so my terminology is a bit slang. I also don't have a computer. How do you suggest i reformat?

I said to at the start it could evaporate if it created reaction hot enough to sublime the acid which it could evaporate back out the drain, which it may have by the sounds of it .

@ AL just from the photo I couldn't identify what the residues are and if I tried I'm afraid I would be criticized for trying by the feedback I've got! I hope I could help, good luck cleaning up mate ...

As I wasn't there I honestly don't know how hot the drain got!

Sulfuric acid seems like an odd choice to clean a drain. Typically drain cleaners are strong alkali mixtures.
Concentrated sulfuric acid ionizes on contact with water with a very exothermic reaction. It could easily boil or splatter up from the drain.
Rinsing with water would be sufficient to clean up any mess. To be on the safe side, I'd wash linens in bathroom before using again.

This is what he used: link

I believe you. Just seems odd to me. The worry would be droplets of sulfuric acid which might have gotten blown out of the drain. The droplets of sulfuric acid won't evaporate.

What does that imply? How should I clean up those droplets? Thank you.

Rinse tub/shower with water.

Water will only dilute the sulfuric acid, not neutralise it. Also the trap may also contain residual acid, so make sure plenty of litres / gallons of water go through it. Avoid bleaching products until this occurs as otherwise chlorine can be released from the NaClO

Tap water contains minerals that will neutralize the acid . Check this with a ph indicator. Due to the fact it would take quite a bit of tap water dilute sodium carbonate is a better option.

Geeze, this isn't about a gallon of sulfuric acid but just what may be a few drops. Since concentrated sulfuric acid was already put in the drain, just rinsing tub/shower with water will easily wash away any slight contamination.

I actually think it was 500 or 750 ml bottle. The guy felt quite proud mentioning he had used the whole bottle.


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Is it possible to recycle an alkaline battery?
Is it possible to recycle an alkaline battery?
The application I am thinking about is that in 3rd world countries often access to western markets is limited. You can't just pop into a store and buy batteries. Would it be possible for a small factory to be in these places and the locals could bring their batteries in for recycling?
The reactions for an alkaline battery are the conversion of zinc to zinc oxide and MnO2 to Mn2O3. So, to recycle a battery of this type you would have to reverse these reactions. Is that easy to do?

related http://chemistry.stackexchange.com/questions/883/can-i-use-one-alkaline-battery-to-recharge-another

Is it possible to recycle an alkaline battery?

Yes, in theory. Everyready has 4% recycled materiel in some of its batteries.

The application I am thinking about is that in 3rd world countries often access to western markets is limited. You can't just pop into a store and buy batteries. Would it be possible for a small factory to be in these places and the locals could bring their batteries in for recycling?

No. The chemistry and processing involved would be far to complicated for some sort of rudimentary factory. There is also the economic factor. Even with a well educated work force such a factory would be very expensive to run and need a massive number of used batteries to recycle.
Part of the expense that recycling would offset is the cost of disposing of the used batteries. For example NiCd batteries in a landfill are bad news because of the toxicity of cadmium. Third world countries generally have much poorer environmental regulations than the developed countries.

To answer the additional question...

The reactions for an alkaline battery are the conversion of zinc to zinc oxide and MnO2 to Mn2O3. So, to recycle a battery of this type you would have to reverse these reactions. Is that easy to do?

No it isn't easy. You can't just mechanically separate the zinc component of an used battery from the manganese component. You'd have do something like dissolve the mess with acid then separate and process the two elements.

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Why is peanut butter viscous?
Why is it that when you chop peanuts, a highly viscous paste is created?
Im assuming that the reason that chopped peanuts dont become just dust is because we cannot chop finely enough, but what causes the chopped particles to adhere to each other?

Well, that's mostly fat, so...

To expand on @Mithoron's comment - the peanut is not a homogeneous substance. Instead, it is comprised of many compounds. Therefore, it can not be neatly subdivided to infinity. The peanut oils behave such that they will come out and make the slurry that we call peanut butter.

Peanuts contain peanut oil. When you crush the peanut you also compress some of the peanut oil out out of the nut particles.

Ok so you now have some oil/solid mixture, so what?

Do the experiment. Try mixing some vegetable oil with flour. The flour doesn't dissolve, you just get a paste.

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Physical Meaning of the Difference between MO and VB Wave Function of H2?
What would be the physical meaning of the difference between the MO wave function and VB wave function of $\ce{H2}$?

What do you mean by physical meaning? Molecular orbital theory and valence bond theory are two approaches to the electronic structure of a molecule. Carried out to infinity with the same prerequisites they are identical.

I mean if you took the actual wave functions of MO and of VB for H2 and subtracted them mathematically, you'll end up with a mathematical expression which represents some sort of physical meaning. Assuming not to infinity, what would this difference represent?

Both theories are mathematical representations of the same thing. If there are any differences they are due to approximations or incompleteness and they do not have any meaning whatsoever.

http://chemistry.stackexchange.com/a/38144/186

I'll give a somewhat simplistic answer. VB assumes that all bonding is between only two atoms. MO bonding considers bonding as a molecular phenomenon.
In $\ce{H2}$ there isn't really much difference. However consider benzene, $\ce{C6}\ce{H6}$. VB theory would have three double carbon bonds and three single carbon bonds which isn't really true. So VB bonding then hybridizes over the six carbon atom ring to predict the aromatic character of the molecule. However in MO theory the aromatic character is fundamental to the model.
From wikipedia

"In particular, valence bond theory correctly predicts the dissociation of homonuclear diatomic molecules into separate atoms, while simple molecular orbital theory predicts dissociation into a mixture of atoms and ions. For example the molecular orbital function for dihydrogen is an equal mixture of the covalent and ionic valence bond structures and so predicts incorrectly that the molecule would dissociate into an equal mixture of hydrogen atoms and hydrogen positive and negative ions."

VB and MO will predict the same structure as the wave functions are essentially the same, if the description of complete. Differences do not carry any physical meaning and this is what the question is about.

Also hybridization is part of VB not MO theory and VB theory does not predict alternating single and double bonds for benzene, it makes use of the concept of resonance to suggest a structure that is essentially the same as that suggested by MO.

@Bon - You're right about hybridization being part of Vb theory. I fixed that part.

@Martin - "VB wave function" seems odd to me. (Granted my chemistry degree was earned when the four elements were air, water, earth and fire.) To me VB theory is more about the linear combination of atomic orbitals.

Well, the aromatic character of benzene is not exactly "fundamental" to MO theory - it is actually derived from the model. You can draw MO diagrams for the $\pi$ system of benzene without actually knowing that is it aromatic, and figure out from the diagrams that it is aromatic - which is the origin of Huckel's famous $(4n+2)$ rule... I mean, it's probably really just how you look at it. If you can get your hands on it, see Ian Fleming's Molecular Orbitals and Organic Chemical Reactions for a description.

@orthocresol - I was thinking that this was a high school level question. But  the answer from Wikipedia indicates a higher  level of  understanding would be required.


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Why don't substances have a condensation point?
All substances have a freezing point at which they transition from a liquid to a solid. They all also have a melting point at which they transition from a solid to a liquid. However, as far as I can tell a substance only has a boiling point at which it transitions from a liquid into a gas. There is no analogous concept of a "condensation point" at which a gas transitions into a liquid. Why not?

Melting point and freezing point really are the same thing and so are boiling point and condensation point. See here.

A "condensation point" is sort of a misdirection as to the physical process involved. "Point" is really a stand-in for "temperature."
Consider water in air. Let's say that the temp is 20 degrees C, but the "dew point" is 10 degrees C. So if the temperature drops to 10 degrees C then the air is saturated with water and and a further drop in temperature will result in the formation of dew. So being past the dew point is dependent on the amount of water in the air and the temperature, not just the temperature alone.

Gases (such as $\ce{O2}$, $\ce{Br2}$, $\ce{N2}$, $\ce{F2}$) can be condensed in a chamber under specific conditions ( things like humidity, pressure, temperature) but the conditions rely on more than just temperature or one variable that could be called a condensation point.


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What are examples for state functions?

List some examples of state functions. Check all that apply:

internal energy
work
volume
pressure

I know that a state function is independent of the path taken, so I think work and volume are state functions. Am I correct?

No.  Work is not a state function.  To be more precise about the definition of a state function, the change in a state function from thermodynamic equilibrium state A to thermodynamic equilibrium state B is unique and independent of any and all process paths that take the system from state A to state B.  In the case of work, the amount of work required to get from state A to state B depends on the process path, and varies for example with the amount of heat added or removed during the process or if the process is reversible or not.

Work and heat are major examples that are not state functions. Major examples of state functions are internal energy, entropy, enthalpy, free energy, and Heimholtz energy.

Let's say you go to Stonehenge.
You see a big rock there.
You can observe the state of the rock.
But how much work was done to bring the rock there?
How far away did the rock come from? Was it a downhill or uphill journey?  Was the stone dragged, moved with wheels, or pulled on a sled over snow?
So you can know the exact state of the rock, but not how much work was done to bring it to that state.
Therefore work is not a state function. The others given in the exercise, i.e. internal energy, pressure, volume, temperature are all state functions.

Examples of state function include (1) internal energy, (3) volume, and (4) pressure.
This is because a state function is a value in which is dependent on the state of that particular system. The value of a change in a state function is always the difference between the final and initial values. It is important to note that the value depends only on the state of the system, not on how the system arrived at that state.

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Entropy of mixing
$$S=k_\mathrm B\ln\mathit\Omega$$
For the mixing of two types of molecules:
$$\mathit\Omega=\frac{N!}{N_1!N_2!}$$
Therefore, subbing in and using Stirling’s approximation ($\ln{N!}=N\ln N-N$):
$$\Delta S = k_\mathrm B(N\ln N-N_1\ln N_1-N_2\ln N_2)$$
However, I know that the result should be:
$$\Delta S =-k_\mathrm B(N_1\ln x_1+N_2\ln x_2)$$
How do I get there? Have I done something wrong?

You haven't done anything wrong, just bear in mind that $N = N_1 + N_2$ and see whether you can simplify your expression further.

Thank you. However, I only get the number of moles not the mole fraction in the logarithms. i.e i get $\ln \frac {N_i}{N}$

Find an expression for the mole fraction in terms of the symbols you already have (which are basically $N$, $N_1$, and $N_2$), then see whether you can manipulate the equation to get the natural logarithm of that.

I'm sorry, I am really stuck. Would you mind just telling me? Or giving me a large clue? Thank you so much

The mole fraction $x_1$ is $n_1/n = N_1/N$ (since each mole is exactly $N_\mathrm{A}$ particles). Likewise for $x_2$.
$$\begin{align}
\Delta S &= k_{\mathrm{B}}(N \ln N - N_1 \ln N_1 - N_2 \ln N_2) \\
&= k_{\mathrm{B}}(N_1 \ln N + N_2 \ln N - N_1 \ln N_1 - N_2 \ln N_2) \\
&= k_{\mathrm{B}}[N_1(\ln N - \ln N_1) + N_2(\ln N - \ln N_2)] \\
&= k_{\mathrm{B}}\left[N_1 \ln{\left(\frac{N}{N_1}\right)} + N_2 \ln{\left(\frac{N}{N_2}\right)}\right] \\
&= -k_{\mathrm{B}}\left[N_1 \ln{\left(\frac{N_1}{N}\right)} + N_2 \ln{\left(\frac{N_2}{N}\right)}\right] \\
&= -k_{\mathrm{B}}(N_1 \ln x_1 + N_2 \ln x_2)
\end{align}$$
If you want, you can go further by noting that:
$$\begin{align}
n_1 &= \frac{N_1}{N_\mathrm{A}} & n_2 &= \frac{N_2}{N_\mathrm{A}} & k_{\mathrm{B}} &= \frac{R}{N_\mathrm{A}}
\end{align}$$
which gives
$$\Delta S = -R(n_1 \ln x_1 + n_2 \ln x_2)$$
or
$$\Delta S = -nR(x_1 \ln x_1 + x_2 \ln x_2)$$
whichever you prefer (they're of course all the same).


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is my reasoning for enantiomers/meso compounds correct?

I think a and b are enantiomers because a has R configuration but b has S configuration.
I think c and d are identical compounds which makes them a meso compound. If you just turn it 180 degrees they look the same to me. Am I right or is there something wrong with what I'm saying?

Related: Confusion on enantiomers and meso compounds

Seems about right to me.


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is this the proper identification of this cyclic molecule?

Attempt:
(1S,3S)-1,3-Dichlorocyclohexane
I'm just not sure about the nomenclature for R and S configurations (our teacher was a little unclear). I gave it my best shot after looking through some textbooks but would like just a quick confirmation or someone to tell me if it's wrong.
In addition, if I exchanged the positions of the Cl and H groups on each side with each other would I get an enantiomer? ((1R,3R)-1,3-Dichlorocyclohexane)

Correct and yes.

It's correct, as already mentioned in a comment.
Some might find more or less helpful to redraw the structure to:

Your additional question can be tricky, understood a different way. It's possible to exchange H and Cl positions, or rather axial to equatorial bonds and the other way around – by switching to the other boat conformation, i.e. without changing the configuration:

(The two structures are identical, one of them can be rotated to superimpose the other.)

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Group 13 elements reactivity with halogens
All the trihalides of the Group 13 elements are known, but $\ce{TlI3}$ is not. Why is this so?

https://en.wikipedia.org/wiki/Thallium_triiodide answers it...

This is due to the Inert Pair Effect. As we move down the 13th group, addition of d orbitals and f orbitals are taking place. Since the d orbitals and f orbitals are away from the nucleus, they have poor sheilding power or capability and thus the s orbitals of the outermost shell are attracted towards the nucleai, making it difficult for s orbital electrons to leave the orbitals and only pairing of the 1 electron in p orbital is possible. Thus +1 valency of Tl is more stable than +3 as it is not energetically favored. This,ofcourse, does not mean that formation TlI3 is not possible. It's just that it requires more energy.

There's also relativistic effects.

You should also explain why only iodide is unable to bring out the +3 oxidation state.

@orthocresol Isnt that due to size of iodine?

Indirectly, yes, but just saying that doesn't explain anything


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Phase subscript for supercritical reagent?
When writing a chemical reaction, one can use subscripts to denote the phase of a reagent, e.g. $\ce{SiO2 (s)}$, $\ce{H2O (l)}$, $\ce{F2 (g)}$ or $\ce{Na^+ (aq)}$. However, I need to write a reaction involving a reagent that is above its critical point. What is the subscript for a supercritical reagent?

I don't think there are simple subscripts for more exotic phases.

@Mithoron How should I notate it then?

Just a note on the notation itself: the phases should not be set as subscripts. It is still very common, but the recommendations are now to put it in line, I.e. $\ce{H2O(l)}$.

The accepted if not standardised notation is Sc or sc - I've seen both.  Capital "S" followed by a lower case "c", e. g.  ScCO2 is the one I've seen most often in chemistry related papers.
EDIT - apologies, this is a prefix and not a subscript. Hopefully you get the point

Could you add example? I doubt it should be written the way you did.

Try http://www.nottingham.ac.uk/supercritical/scwater.html or http://www.sciencedirect.com/science/article/pii/S0040403900019171?via%3Dihub for examples used by supercritical fluid practioners

I think the proper answer is (g), the same as a gas.  After all, the critical point of hydrogen is -240 °C and 2.24 atm.  But no one refers to a compressed cylinder of hydrogen gas as "supercritical hydrogen".  It's just a compressed gas.
After all, supercritical fluids meet the formal definition of a gas: they have neither a defined shape nor a defined volume.  (Liquids have no defined shape but do have a defined volume.)  The critical transition is the loss of stability of the condensed (liquid) phase; in contrast, both before and after the critical point there is a a well-defined gas phase.
Usually when people say "supercritical" they mean "above the critical point, but not by much" or something similar.  It's a not-well-defined term. I'd avoid using a special abbreviation for it, since supercritical fluids really are just gases.


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Using CIP rules in a bicyclic compound with carbonyl subsituent
In bicyclo[2.2.1]heptan-2-one, arrange the numbered groups according to CIP rules:

Group 1 will have highest priority because oxygen atom is attached to 1. (It can be considered as two oxygen atoms).
Group 4 will have least priority due to least atomic number among the numbered groups.
But I am confused about order of priority of group 2 and 3. Both have same atoms. (So I will check the next atom which is also same.)

Group 2 has -CH2-C(CHH); group 3 has -CH2-C(CCH) where (XYZ) means that the carbon is bonded to XYZ atoms. I mean, you already used this rule to determine that 1 was the highest, didn't you. 1 is -C(OOC) whereas 2 and 3 are both -C(CHH). Since O > C in atomic number 1 > 2,3. Same goes for 2 and 3 you just need to go one carbon further.

OK. So 3 has more priority.

Yup it does, which gives you an S configuration, I think.

Also note that the given compound has a second chirality center at the other bridgehead.

@orthocresol: It's an S-configuration if the methylene 3 is above the plane of the 6-membered ring. I presume that your "I think" considers this point.

@user55119 yes, an important point. I must admit, I have no recollection of what I was thinking at that time. :-)


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Better solvent for KCl than water?
In my experiment, I used an atomizer to generate $\ce{KCl}$ aerosols. I saw that the nozzle that was dipped into the 10% wt $\ce{KCl}$ solution got partially clogged after an hour. I then had to disassemble the nozzle in order to clean it, which I did by rinsing it with water and drying it with a stream of air.
 Is there any way to avoid the clogging?

This seems like more of an engineering problem than "chemistry problem." I'm not sure how long you want the process to run, nor how critical the 10% salt concentration is.
I'd try adding a pure water bubbler in front of the salt solution to saturate the air being blown through with water.

I'd agree with @MaxW that this is mostly an engineering problem.
Your best bet for increasing solubility would be to increase temperature; I don't think you'll find a solvent better than water for KCl, given how cheap/available it is.

In general, increasing velocity of the particles will increase solubility (as Kinetic Energy increases). You may consider putting your solution on a stir plate or air-bubbler like @MaxW suggested.
If you're really looking for other solvents that may work better for your setup, stick to something highly polar.
Figures reference: http://pubs.acs.org.ezproxy.lib.usf.edu/doi/abs/10.1021/je049922y

It looks to me like the poster has a glass frit bubbler. I'm conjecturing that the back and forth motion of the fluid results in part of the solution becoming "trapped" in the frit by bubbles, and then drying out. The bubbles then insulate the solid salt preventing it from contacting the bulk of the solution. So I think the simple solution is to hydrate the gas before bubbling it through the salt solution.

Thanks, looks it might be the case and hydrating the gas is one solution. For now, in my experiments i've cleaned the stainless steel nozzle by dipping into to the boiling water. Apparently, this has fixed the problem.


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Is every substance in equilibrium with its vapours?
In one of my previous questions, I got an answer from user Jan:

The smell of a compound as we perceive it usually depends on which receptors in the nose will bind the molecule (or not).
These receptor, being proteins, are usually rather specific for certain shapes, not for bonding patterns. So chlorobenzene might fit into similar receptors as toluene which would lead to at least partly a similar smell, but that is it.

This means that a substance must be in the vapour phase in order to smell it, which brings me to my question:

Do all substance (at least those which have a characteristic smell) maintain an equilibrium with their vapours (if the substance is in closed container)?

But if it's true for those solids, then it must be true for all other solids, as well as nonvolatile liquids that do not have a characteristic smell to humans. Is this not the case? Also, if this substance is placed in open air, will it evaporate totally?

You can maintain equilibrium only if vapour is contained.

The solid or liquid is at equilibrium with its vapor at the air interface.  But, typically, the partial pressure of the substance in the bulk room  air will be less than the equilibrium vapor pressure at the interface (except, as Mithoron points out, if the room is sealed).  As the substance diffuses from the interface into the bulk room air, its partial pressure decreases with distance from the interface.  The pressure gradient provides the driving force for the diffusion.

Yes, everything is partly vapourised, even high-boiling (or -subliming) solids. Equilibrium can, of course, only be reached if the air does not exchange. So put a smelly cheese in your fridge and come back two days later; the smelly cheese will be at vapour pressure equilibrium with the fridge air. Remove the cheese but keep the door closed and the smell will stay for a while.

If they were maintaining an equilibrium with their vapour you wouldn't be smelling it.
In theory any equilibrium in the case of vapor pressure occurs in a closed container, in reality a vapor pressure equilibrium is rarely if ever reached.

authors: حامد السروري


text:
Validating a method to meet accuracy criteria using precision and accuracy?
If a method has a precision of 10% and an accuracy of 95 to 105% is required, can this method be validated to meet accuracy criteria? Explain your rationale.
This question was posed by my professor and I'm struggling to come up with an answer.
I understand that precision refers to whether your method gives consistent repeatable results and that accuracy refers to how close your results are to the value you should actually obtain. However, I don't understand how a method's precision and accuracy could validate a method to meet accuracy criteria as the question asks.

I highly recommend Eurachem's guides on method validation. See this page for an excellent guide: https://www.eurachem.org/index.php/publications/guides/mv

Possibly, but the rational is a bit convoluted.
First, as you noted there is a fundamental difference between precision and accuracy. You explained the difference correctly.
Now let's think of a hypothetical example. Suppose that you have a pan balance in the lab and an analytical balance down the hall in a balance room.
So you are going to use the analytical balance (which is accurate and precise) to validate the pan balance.
You weigh a rock specimen on the pan balance and the weights have a precision of 10%. This means +/- 10%. So to get a better precision, you weigh the rock multiple times. Now the average weight will have a better precision than a single weighing. The average will be smaller by a factor of 1/sqrt(n). So 4 weighings will give you +/- 5%.
So you get a bunch of rocks that span whatever weight range that you would use. You weigh the rocks on the pan balance and on the analytical balance (which is assume to give the "true" weight). Now you can establish a calibration curve for the pan balance.
The calibration curve is going to have some error too, so you'll probably have to adjust how many repeat measurements and how many different samples you need to get the desired accuracy. But it should be possible in this example. The real twist here is how you "validate" the calibration.


text:
Why is starch insoluble in ethanol?
I understand that starch will precipitate from an aqueous solution in the presence of alcohol. I also understand that molar mass effects solubility and starch has a relatively high molar mass. There is also an issue regarding the polarity of the solute and solvent.
Is the insolubility of starch (which has the form $\ce{(C6H10O5)$_n$}$) in ethanol only due to the fact that starch is polar and ethanol is only slightly polar?

It’s certainly somewhat soluble although a lot less than in water.

Due to the fact that starch (amylum) is a polysaccharide , a long chain compound consisting of glucose monomers it has poor solubilty in most solvents.
Some starches are water soluble because of the hydroxyl groups involved in the polysaccharide chain, some like the branched form amylopectin are more insoluble.
Starches form granules in solution with cold water, the granules break down as the water is heated and can be solubilized.
When ethanol is added, which is less polar than water, it lowers the solubility and potential hydrogen bonds of the starch , because the water is more attracted to the alcohol than the starch, the starch is dehydrated by the alcohol and precipitates.
So while polarity and temperature are variables to starches poor solubility, the critical variable in this case is the starches ability to hydrogen bond with the solvent.

Joel has the right answer though it is somewhat hidden. Think of having the starch dissolved in water and the adding lots of ethanol. As  Joel then says "When ethanol is added, which is less polar than water, it lowers the solubility and potential hydrogen bonds of the starch because the water is more attracted to the alcohol than the starch, the starch is dehydrated by the alcohol and precipitates." So the answer to the original question is  Yes. The question is a little off focus though because the critical property  isn't really polarity, but hydrogen bonding capability.

I improved the answer I hope it's ok.


text:
How can I tell if a liquid is (not) emitting light?
I found an image of Nuka Cola Quantim, a soft drink from the Fallout video game series, which depicts the bottle as glowing, but it doesn't feel quite real:

Then I also saw this video of a chemiluminescent reaction of luminol:
https://youtu.be/TJj7t2mrpxc
Experienced chemists and/or photo editors might tell right away which is real and which is fake. What hints should I look for when determining if a liquid itself is glowing or is being lit by a different light source?

unfortunately there is no "luminescence" tag and I don't know what else to pick

I doubt it's on-topic here, maybe on Skeptics SE would be.

I personally have no problem with having the question, but I agree, that it is not really on topic since the question is more about the picture than the chemistry which might be behind.

Tricky indeed, though it may be a useful question to discuss if fluorescence can be observed any easy way and distinguish from reflections.

I'm voting to close this question as off-topic because this appears to be more related to photography and editing graphical images, and is probably better suited for an audience that specializes in producing/manipulating/enhancing images.

The bright line of light at the 'shoulder' of the bottle looks more like a reflection of a light source above. This looks more like the product of 3-D rendering software, of a pale blue liquid having a bright light cast on it from a line source.

In a photograph, a light source is always brighter than any surface it is illuminating.
Therefore, if the liquid is the actual light source, the liquid should be significantly brighter than the rest of the bottle. See for example this picture by David Mülheims CC BY-SA 2.5, via Wikimedia Commons:

In the given picture, however, the parts above and below the water level have a similar brightness.


text:
Is there a better way balancing reaction equations than systematic trial and error?
I was introduced to the concept of balancing chemical equations in school and I did some practice examples, which were easy and took me about 2 minutes per question. I used the systematic trial and error method and I don't really like it.
I'm worried that my teacher will give us a more difficult question, so I'm looking for a method to solve chemical equations such as
$$\ce{C3H8 + O2 -> CO2 + H2O}.$$
I have tried matrices, but I don't really understand them, as I am not in grade 12 yet. If you know matrices, can you explain how I can balance the chemical equation with aforementioned method (thoroughly)? I would also be grateful if you have any other useful and quick methods which work for pretty much any high school chemical equation balancing problem.

I really wouldn't bother learning matrices just for balancing at this early stage. You can just do these by observation, where you 'juggle' and make sure that you always have the same number of each element on either side. If you do insist on learning matrices just to do this (which I advise against), you can start here

When using matrices, you are really just solving a system of equations. You can use any technique you know to solve the system of equations that a chemical equation represents. Using matrices is only one technique that may be useful for complicated equations or if you have a calculator/MATLAB on hand. In the given example, it would be easier to solve it directly with guess and check.
Writing the system of equations
In your example
$$\ce{C3H8 + O2 -> CO2 + H2O},$$
write in some variables as coefficients:
$$\ce{a C3H8 + b O2 -> c CO2 + d H2O}.$$
Now, we can write out a system of equations based on the stoichiometry of each element on either side of the equation. In other words, write equations that express the number of each element on the left hand side to the number of each element on the right hand side:
\begin{align}
3a &= c\\
8a &= 2d &\Longleftrightarrow&& 4a &= d\\
2b &= 2c + d
\end{align}
Finally, we have to choose a value for one variable to substitute in. This will be the fourth and final equation. We want integer coefficients here, so we will choose a substitution that gives us all integer coefficients. Otherwise, just multiply your coefficients at the end so that they are all integers. Here, I'll let $a = 1$.
Solving the system
Substitution
Substituting the first two equations into the third gives:
\begin{align}
2b &= 6a + 4a = 10a &\Longleftrightarrow&& b &= 5a
\end{align}
\begin{align}
a &= 1\\
b &= 5\\
c &= 3\\
d &= 4
\end{align}
And the balanced equation is
$$\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}.$$
Using matrices
To answer the question though, let's do it with matrices. I won't go through the whole explanation of solving the system but I'll show you how to build the matrices and get an answer. Remember, we are just representing the same system of equations from before, but with matrices now!
We need to build two matrices, $A$ and $b$. Reorganize the equations we wrote previously so that all of the variables are on the left hand side and a constant is on the right hand side. Also, notice I added in our equation let $a = 1$:
\begin{align}
1a + 0b + 0c + 0d &= 1\\
3a + 0b -1c + 0d &= 0\\
0a + 2b -2c - 1d &= 0\\
4a + 0b + 0c - 1d &= 0
\end{align}
These equations are written like this:
\begin{align}
A &=
\begin{bmatrix}
1 & 0 & 0 & 0\\
3 & 0 & -1 & 0\\
0 & 2 & -2 & -1\\
4 & 0 & 0 & -1\\
\end{bmatrix}\\
b &=
\begin{bmatrix}
1\\
0\\
0\\
0\\
\end{bmatrix}
\end{align}
Notice that $b$ is just a column of the constants on the right hand sides of the equations, in the same order as they were given. The order is important here!
There are a few ways to solve this manually or in a calculator/MATLAB. If you have a TI-8x calculator or similar, you could use its rref function (this transforms an augmented matrix into reduced row echelon form, I'm sure google could help you out) or you could multiply $A^{-1}b = x$. In MATLAB, just do: A\b. This gave me:
$$
x =
\begin{bmatrix}
1 \\ 5 \\ 3 \\ 4
\end{bmatrix}
$$
Each row in this final matrix $x$ is the value for $a$, $b$, $c$, $d$ in that order, giving the balanced chemical equation
$$\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}.$$

@aabmass How would I be able to solve this (matrices) manually?

@ImagineDragons Using Gauss-Jordan elimination would be simple for you. I can't explain in a comment, but I'm sure you can find great examples online. However, it boils down to the elimination method for solving a system of equations. You basically multiply/add/subtract rows with each other like you would for elimination.

Always start balancing by equating the number of atoms of elements which appear only once. For example, in the equation you have mentioned, oxygen appears in both the compounds, whereas, hydrogen and carbon appear only once. So, balance $\ce{H}\;(=8)$ and $\ce{C}\,(=3)$. Now, we have
$$\ce{C3H8 + O2 -> 3CO2 + 4H2O}.$$
The equation isn't balanced yet. Now we need to balance the only remaining element, $\ce{O}$.
We have $\ce{2 O}$ atoms on the left and $10$ on the right hand side, balancing these numbers we get:
$$\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}.$$
Just follow these steps and you can balance almost all chemical equations.

We have the unbalanced equation:
$$\ce{C3H8 + O2 -> CO2 + H2O}$$
You don't really need to solve a system of equations or matrices to balance such chemical equations.
First, observe which elements only occur once on each side of the equation:
We can see that $\ce{C}$ and $\ce{H}$ only occur once on each side, so let's begin by trying to balance these.
There are three $\ce{C}$ on the left and one on the right. So we will add $3$ as a coefficient to $\ce{CO2}$:
\begin{align}
\Rightarrow&& \ce{C3H8 + O2 &-> 3CO2 + H2O}
\end{align}
There are eight $\ce{H}$ on the left and two on the right. So we will add $4$ as a coefficient to $\ce{H2O}$:
\begin{align}
\Rightarrow&& \ce{C3H8 + O2 &-> 3CO2 + 4H2O}
\end{align}
It is useful to leave the $\ce{O2}$ on the left until the end because we can add any coefficient to it to balance with the right side.
Now, on the right side we have a total of ten $\ce{O2}$. So we will add $5$ as a coefficient to $\ce{O2}$:
\begin{align}
\Rightarrow&& \ce{C3H8 + 5O2 &-> 3CO2 + 4H2O}
\end{align}
We now have a balanced chemical equation.


text:
How to calculate and denote small uncertainties?
I am writing a report and I'm stuck on determining how to write the uncertainty.
I have this:
$$1.0 \times \left( 0.1000 \pm 0.0005 \right).$$
If I multiply the values, I need to keep two significant figures, but then my uncertainty would be beyond the correct amount of decimal points. What would I do in this case?

See here for an explanation why $\pm$ is discouraged by several bodies.

When a number is given without any further information, it is generally interpreted so that the last digit is rounded.
Therefore, the first given value $a=1.0$ is generally assumed to represent a value between $0.95$ and $1.05$, or $a=\left(1.00\pm0.05\right)$. Thus, the assumed uncertainty of $a$ is $u(a)=0.05$.
Conversely, the second given value $b=\left(0.1000\pm0.0005\right)$, where the assumed uncertainty of $b$ is $u(b)=0.0005$, might simply be written as $b=0.100$.
When multiplying numerical values, according to the simplified rules of significance arithmetic, the result is rounded to the number of significant digits in the factor with the least significant figures. Therefore, when you multiply $a=1.0$ (two significant digits) by $b=0.100$ (three significant digits), the result is $\boldsymbol{y=a\cdot b=0.10}$ (rounded to two significant digits).
This result may be estimated to represent a value between $0.095$ and $0.105$, or $y=\left(0.100\pm0.005\right)$; i.e. the assumed uncertainty of $y$ is $u(y)=0.005$.

By way of comparison, a generally better result for the uncertainty $u(y)$ may be obtained by using the rules of propagation of uncertainty (or propagation of error) as follows.
When calculating the product of $a$ and $b$
$$y=a\cdot b$$
and if we assume that the uncertainties of $a$ and $b$ are not correlated, the relative uncertainty of the result $y$ may be estimated as
$$\begin{align}
\left(\frac{u(y)}{y}\right)^2 &= \left(\frac{u(a)}{a}\right)^2 + \left(\frac{u(b)}{b}\right)^2 \\[6pt]
\frac{u(y)}{y} &= \sqrt{\left(\frac{u(a)}{a}\right)^2 + \left(\frac{u(b)}{b}\right)^2}
\end{align}$$
Thus, the absolute uncertainty of $y$ is
$$\begin{align}
u(y) &= \sqrt{\left(\frac{u(a)}{a}\right)^2 + \left(\frac{u(b)}{b}\right)^2} \cdot y \\[6pt]
&= \sqrt{\left(\frac{u(a)}{a}\right)^2 + \left(\frac{u(b)}{b}\right)^2} \cdot a \cdot b\\[6pt]
&= \sqrt{\left(\frac{0.05}{1.00}\right)^2 + \left(\frac{0.0005}{0.1000}\right)^2} \times 1.00 \times 0.1000\\[6pt]
&= 0.005024938\\[6pt]
&\approx 0.005
\end{align}$$
The same result may be obtained using the general formula
$$\begin{align}
u(y) &= \sqrt{\left(\frac{\partial y}{\partial a}\right)^2 u^2(a)+\left(\frac{\partial y}{\partial b}\right)^2 u^2(b)} \\[6pt]
&= \sqrt{b^2\cdot u^2(a)+ a^2\cdot u^2(b)} \\[6pt]
&= \sqrt{(0.1000)^2\times (0.05)^2+ (1.00)^2\times (0.0005)^2} \\[6pt]
&= 0.005024938\\[6pt]
&\approx 0.005
\end{align}$$
Thus, in this case, the result that is obtained by using the simplified rules of significance arithmetic is a good approximation of the result that is obtained by using the rules of propagation of uncertainty.
(Note that the numerical values of the estimate $y$ and its uncertainty $u(y)$ should not be given with an excessive number of digits, although in some cases it may be necessary to retain additional digits to avoid round-off errors in subsequent calculations.)


text:
why maximum (w/w) percent concentration of commercial acids is same to their molar mass?
i want to know why maximum (w/w) percent concentration of commercial acids is same to their molar mass, for example for HCl the maxmim concentration is 37 and it's molar mass is 36.5 and maximum concentration of H2SO4 is 98%(w/w) and it's molar mass is 98.
thanks alot!

I think this is just a coincidence. As you may have already noticed, the molar mass is not exactly the same as the maximum concentration.

It's a coincidence as HBr is 48% (MW ~81), phosphoric acid is 85% (~98), hydrofluoric acid 97% (MW ~20).Look here for more http://www.csudh.edu/oliver/chemdata/acid-str.htm

Bon. We must have posted at the same time. Apologies


text:
Protons and the Heisenberg principle
The Heisenberg uncertainty principle implies that it is not possible to calculate the position and velocity of a body at the same time accurately. Electrons follow this principle since their orbitals are just probabilities where they may occur. But in the case of protons, we are kind of certain about their position in the atom. So shouldn't they have a very, very, very high velocity to balance the certainty in position?

You are using the Heisenberg uncertainty principle to relate the uncertainty in position $x$ to the uncertainty in velocity $v$.
However, the quantitative version of this principle actually is
$$\Delta x\cdot\Delta p\geqslant\tfrac12\hbar $$
where $\Delta x$ is the uncertainty in position $x$ and $\Delta p$ is the uncertainty in momentum $p$.
Certainly, since
$$p=m\cdot v$$
or
$$v=\frac pm$$
where $m$ is mass and $v$ is velocity, you may also use the uncertainty principle to compare the uncertainty in position $x$ to the uncertainty in velocity $v$.
However, the mass of a proton is $m_\mathrm p=1.672\,621\,923\,69(51)\times10^{-27}\ \mathrm{kg}$[CODATA2018], whereas the mass of an electron is only $m_\mathrm e=9.109\,383\,7015(28)\times10^{-31}\ \mathrm{kg}$[CODATA2018]; i.e. a proton is over a thousand times heavier than an electron.
Therefore, a given uncertainty in position $x$ and the corresponding uncertainty in momentum $p$ result in a smaller uncertainty in velocity $v$ for the proton.

But in the case of protons, we are kind of certain about their
  position in the atom.

Well, yeah, kind of certain. The very notion of molecular geometry arises in the Born-Oppenheimer approximation. Nuclei are much heavier than electrons so that when solving the electronic Schrödinger equation they can be assumed to be stationary. This clearly violates the uncertainty principle since we precisely know both position (whatever it is) and momentum (zero by assumption) of each and every nucleus simultaneously. But the thing is that when we solve the electronic Schrödinger equation nuclei are treated as classical particles, so everything is fine. Kind of fine, I mean.

So shouldn't they have a very, very, very high velocity to balance the
  certainty in position?

No, on the contrary, nuclei are assumed to be perfectly stationary classical particles. That's the trick. And this is, of course, an approximate description of the reality which usually works fine, though.

"so that when solving the electronic Schrödinger equation they can be assumed to be stationary" There is no need for that, and the usual (at least in physics texts) solution involves the canonical transformation into CoM coordinates using the reduced mass. What the OP described as the motion of the electron is actually the motion of the difference in position. Motion of the proton only makes up about .06% of that, but it is there. This has the effect of allowing you to predict the spectral differences between protium, deuterium and tritium from the changes in the reduced mass.

@dmckee, in chemistry, there is a need to assume at first that nuclei are stationary, since otherwise the very notion of molecular geometry is undefined. So, yeah, we do CoM separation as usual, but that is not yet chemistry, it still physics. To get into chemistry you have to introduce the BO (or adiabatic) approximation in which at the first step (electronic SE) nuclei are treated as stationary classical particles. And this is the step where orbitals arise, so electrons are still quantum particles, but not nuclei.

Hmmm ... that's what I get for stepping out of my area of expertise. But it seems that you could still define the molecular geometry in terms of the CoM's of the atomic systems. Or are there surprises lurking in the details?

@dmckee, well, true, for an atomic system you don't need to invoke the BO approximation, but still even in atoms the nucleus is treated classically in chemistry. Except for some very rear cases when the quantum description is unavoidable, such as, for instance, the accurate studies involving the proton transfer: it is known that quantum tunneling is important there. But usually we treat nuclei as classical particles.

Physically, of course, nuclei are not classical particles. But what I wanted to say, is that orbitals (mentioned by OP) arise after it is already assumed that nuclei are classical particles, so at this stage there is no sense anymore to apply the uncertainty principle to them. Thus, there is no paradox out there, just an approximation.


text:
What is the physical quantity/magnitude of pH?
I have a database to classify variables/parameters in water resources as follows:
Variable: Precipitation (P); Quantity: Length; Units of measurement: mm (millimeters).
Following the example, what would be the properly quantity of variable pH?
Variable: Potential Hydrogen (pH); Quantity: ?; Unit of measurement: pHU (pH unit, unit less, dimensionless).
Maybe concentration or activity?
[I know what pH is, It is not necessery explain, I'm looking for the most appropriate word to describe its quantity]

My question is about quantity not unit.

If you don't want to know what pH is, and you don't want to know what it's units are, then it's not clear to me what it is that you do want to know.

@SimonW know the most appropriate physical quantity, if possible, to pH.

@Artur_Indio surely, using that definition, pH is a physical quantity, of various substances. So the answer to your question is the same as the answer to "what is pH"? I think you need to (a) think carefully about how to ask the question, as at present it's very unclear; (b) most probably, ask it over at chemistry.stackexchange.

I disagree with the answer that says "pH measures acidity or basicity" (no matter what is in Wikipedia).
pH is the negative of the log of the hydrogen ion (H+) activity. Activity is a thermodynamic concept that is related to but not the same as concentration. Activity is similar to chemical potential. The same concentration of hydrogen ions will have different activities, depending on what else is in solution. In weak solutions, the activity coefficient that relates the concentration to the activity depends mainly on the charges of the other ions in solution and not on the particular ion species. But at higher amounts of other species, the particular chemicals and chemical interactions become increasingly important.
On the other hand, acidity is the ability to provide hydrogen ions to a reaction and considers not only the hydrogen activity but also the amount of acid-base reaction that can occur. You may have heard of pH 'buffers'. These are solutions that are designed to have a balance of chemical species so that the pH is very stable, even if small amounts of acid or base are added or if the ionic strength of the solution changes. If you add a bit of base to the buffer solution, the solution reequilibrates to provide acid to react with the base. You have to add a fair bit of base before the buffer capacity is used up and the pH changes very much. Then, with further addition, the pH will change a very quickly until it approaches the pH of the base that is being added. So a pH buffer has high acidity and alkalinity, keeping the pH stable.
In groundwater geochemistry, acidity is measured by reacting the solution with strong base until the acid is essentially used up. This is done by titration either to a fixed endpoint pH or to the inflection point in the curve of base added vs pH curve.
Bottom line, pH doesn't measure acidity. In terms of the original question, I would say:
Variable: pH, Quantity: negative log of the hydrogen ion activity, Units: dimensionless pH units.

I agree with everything except for the last two words (which should be removed imho).

I left in 'pH units' because it is often used. One problem with dimensionless numbers in equations (IMO) is that it can be unclear where they came from. For example it is sometimes hard to tell if someone is using a weight ratio or a volume ratio. In the case of pH, you could say that it is based on the ratio of the activity of H+ to the activity of H+ in the standard state (in this case, at infinite dilution, and defined as 1).  But yes, pH units is not strictly correct but is descriptive.

pH measures the activity (for practical purposes, this is the same as concentration) of H+ cations in an aqueous solution (i.e water).
At standard room temperature and pressure, and at conditions not too far from it, a value of 7 means that the water is neutral. Lower values means acidic (lemon juice, sulfuric acid, rain water). Higher values mean basic (ocean water, lime water).
It is unit less, therefore the term "quantity", as I understand from your question, does not mean much.

So, "concentration" seems a good word to be the physical quantity/magnitude of pH.

pH measures acidity or basicity: https://en.wikipedia.org/wiki/Acidity_function.
There are several acidity measuring scales and pH is one of them. Another example is the Hammett acidity function.

So, Quantity: acidity or basicity, Variable: pH, Unit Measument: pH units?

Quantity: acidity or basicity, Variable: pH, Unit Measument: pH units (or better, unitless)

authors: 04 ก้องภพ เจริญพันธ์


text:
Why don't we have hyperconjugation with C-C bonds?
C-C bonds are weaker than C-H bonds.So why is hyperconjugation with C-C bonds not possible?

How so? Could you provide an example?


text:
How many grams of sodium sulfate and water in 100ml of 0.5M solution

Calculate how many grams of sodium sulfate ($\ce{Na2SO4}$) and water ($\ce{H2O}$) should be used to prepare $\pu{370 g}$ of 15% solution of this salt. How many mL of this 15% solution should be taken to prepare $\pu{100 mL}$ of $\pu{0.5 M}$ solution of $\ce{Na2SO4}$?

My calculations:
$\pu{370 g}$ = 85% of water, 15% of sodium. So we need $\pu{370 g} \times 0.85 = \pu{314.5 g}$ of $\ce{H2O}$ and $\pu{370 g} \times 0.15 = \pu{55.5 g}$ of sodium sulfate.
We need $\pu{100 mL}$ of $\pu{0.5 M}$ solution.
The amount of sodium sulfate we need:
$$
\begin{array}{cc}
\pu{1 L} & -\quad \pu{0.5 mol} \\
\pu{100 mL} & -\quad \pu{0.05 mol}
\end{array}
$$
$$m = \pu{0.05 mol} \times \pu{142 g/mol} = \pu{7.1 g}$$
To get $\pu{7.1 g}$ of sodium sulfate we need
$$
\begin{array}{ll}
\pu{100g} & -\quad \pu{15 g} \\
x & -\quad \pu{7.1 g}
\end{array}
$$
$x = \pu{47.3 g}$ of solution that's over $\pu{50 mL}$ of it.
But the answer states we need $\pu{3.75 mL}$ of solution.
Could you tell me where my reasoning is wrong or it is a problem with the answer?

I agree with your mass calculations, and the mass of sodium sulfate required as well.  You should need the 47.3g you've calculated as well.  To find a volume, you'd either need to know the density of the solution or make some (not insignificant) assumptions about the solution's density.  In any case, the answer you were given is incorrect.

The density of a 15 % aqueous solution of sodium sulfate at 20 °C is about 1.14 g/ml.

As Jason indicated your calculations are corrected.

370 grams of 15% solution of sodium sulfate does have 314.5g $\ce{H2O}$ and 55.5g $\ce{Na2SO4}$.
Your calculation for grams of solution are correct. 47.3g of solution is required.

However the density would be expected to be somewhat above 1.00 g/ml not less. So a little less than 47.3 ml of solution would be required. However the density of the solution is nowhere near 10 g/ml so the book answer is horribly wrong.
To solve for ml of solution, I'd have just stated an assumption that the density of the solution was 1.00 g/ml since you weren't given that information. There is absolutely no way to calculate the density given the information in the problem.


text:
How do I find the rate constant when I only have product concentrations?
I have done an experiment on the hydrolysis of aspirin, which is pseudo-first order. I need to find the rate constant, but I only have the final ($0.0278\mathrm{~M}$) and initial ($0\mathrm{~M}$) concentrations of the product (salicylic acid), the volume of the solution ($0.06\mathrm{~L}$), the change in time ($24$ hours), and the temperature ($25\mathrm{~^\circ C}$). Is there a way to find the rate constant of this reaction? My main problem is that I do not know the initial or the final reactant concentrations; I cannot simply subtract the concentration of product formed from initial reactant concentration to find the final reactant concentration, as I do not have either reactant concentration.

What's the relationship between the concentration of your product and the concentration of your reactant? I won't tell you any more. Figure it out yourself. It's important.

The reaction is Acetylsalicylic Acid + Water to Salicylic Acid + Acetic Acid

pseudo first order

You're not answering the question

what do you mean by relationship

An equation. Not a chemical equation. A mathematical equation relating the concentration of acetylsalicylic acid to the concentration of salicylic acid. The chemical equation will help you get there.

Try and figure it out yourself first. If you're really stuck look at this previous question http://chemistry.stackexchange.com/questions/32346/ I have to go off and won't be around to answer comments.

My main problem is that I do not know the initial or the final reactant concentrations; I cannot simply subtract the concentration of product formed from initial reactant concentration to find the final reactant concentration, as I do not have either reactant concentration.

Is this a real experiment in the lab, or is it just a question where you're given some kind of data?

This is a real experiment

Ok, sorry... Well, if you don't have data points along the way, it's going to be hard. Normally, in a kinetics experiment, you take samples from the reaction vessel at set intervals and measure the concentration of product at that point in time. This allows you to plot a graph of $\ln [\ce{X}]$ against $t$ and therefore find the rate constant.. However, if it is a real experiment, you should at least know the initial reactant concentration. I mean, you're the one who put a certain mass of the reactant in a certain amount of solvent.

Having two data points (initial and final) is not the end of the world, although it is not very ideal. You can still plot $\ln [\ce{X}]$ against $t$ (according to the integrated rate law) and you will have 2 data points - one at $t = 0$ and one at $t = 86400\mathrm{~s}$. Find the gradient and then find $k$. And when I say $\ce{X}$ I mean aspirin.

I used low dose Aspirin tablets with 81 mg of Acetylsalicylic Acid in them; however, due to error in the spectrophotometric evaluation of Salicylic Acid with Iron (III) Chloride, I achieved higher concentrations of Salicylic Acid than my initial reactant concentration, and so I was trying to determine a way to go backwards; to use my products to find the reactant concentrations.

The thing is I do not know the X; I know the Salicylic Acid concentration changes

Can I use the integrated rate law with products?

No... not when it's first-order and when one of your data points is 0. For the simple reason that you can't take the logarithm of 0.

Yeah that is why I am trying to find an alternative quantitative way to solve this.


text:
How to distinguish between a coordinate bond and a covalent bond?
I’m confused as I am unable to find a completely trusted way of finding coordinately bonded compounds from a bunch of other compounds.

Please don’t use all-caps in titles (or in the body or anywhere else, for all that matters) Thanks @Wild kitty for being quicker in editing ;)

Thnx @Wild kitty for being quicker in editing ;)

Coordinate bonds are covalent bonds and don't really differ in practice from non-coordinate ones. Only difference is source of electrons.

Coordinate bonds can be distinguished from (other or ‘normal’) covalent ones experimentally.
Let’s look at the bond creation first as Newbie suggested. Typically, one would use a Lewis base and a Lewis acid to create a coordinate bond between them.
$$\ce{NH3 + BH3 -> H3N\bond{->}BH3}\tag{1}$$
In contrast, one would usually consider a normal covalent bond to be created from two radicals:
$$\ce{H3C^{.} + {}^{.}CH3 -> H3C-CH3}\tag{2}$$
But we can also create covalent $\ce{C-C}$ bonds by using a carbanion source and a carbocation source (slightly different example, because both ions are actually known and react with each other):
$$\ce{(H3C)3C+ + {}^{-}CN -> (H3C)3C-CN}\tag{3}$$
Thus, we can’t really tell a difference here. The difference in fact lies in the way a bond dissociates if the bond dissociation enthalpy $D = \Delta_\mathrm{D} H^0$ is provided. Standard covalent bonds will dissociate homolytically by the following equation:
$$\ce{H3C-CH3 -> H3C^. + ^.CH3}\tag{4}$$
A coordinate bond will, however, dissociate heterolytically thus liberation the original Lewis acid and base:
$$\ce{H3N\bond{->}BH3 -> NH3 + BH3}\tag{5}$$
In cases such as the ammonium ion ($\ce{NH4+}$) or say $\ce{[AlCl4]-}$ this means we have a certain number of normal bonds and another certain number of coordinate bonds (three and one, respectively in both examples). Although they are all between the same atoms, we can distinguish if we remember that neither ion can exist by itself, there will always be a corresponding counterion. Thus, if we consider $\ce{NH4Cl}$ or $\ce{Na[AlCl4]}$, we realise how the first bond is broken differently:
$$\begin{align}\ce{NH4Cl &-> NH3 + HCl}\tag{6}\\[0.6em]
\ce{Na[AlCl4] &-> NaCl + AlCl3}\tag{7}\end{align}$$
And all further bonds are cleaved homolytically. We gain two different bond dissociation energies, one corresponding to the coordinate bond and one corresponding to the remaining bonds. (Note that salts such as $\ce{AlCl3}$ or $\ce{NaCl}$ are cleaved homolytically by heating in the gas phase — flame-colouring would not happen if we had $\ce{Na+}$ in the flame.)

Firstly, you seem to be talking about "covalent bond" and "dative covalent bond" as two distinct types of bonds, which is not the case as one is a subset of the other (specifically "dative covalent" is a subset of "covalent").

Secondly, you also seem to misinterpret "covalent bond" as the pure covalent case (i.e. the bond is not polarised; the electronegativity difference between the two atoms involved is negligble).

Consider the case of the hydrohalic acids. They have been cited in many literature as "covalent substances". By your definition of homolytic fission, the bonds in these chemical species would not be "covalent" because they often dissociate to give ions. By your definition, wouldn't these bonds be considered "coordinate" and not "covalent".

In conclusion, your answer is problematic. These clarify your answer regarding the points I have raised.

@TanYongBoon The fallacy of your penultimate comment: I explicitly said upon addition of the bond dissociation enthalpy. This is not the case for deprotonation reactions; the bond is not broken by the bond dissociation enthalpy but by a certain reaction mechanism which involves breaking and reforming of bonds.

Also, you seem to be seeing covalent bonds as something ‘pure’ or somehow different from polarised covalent bonds (or ionic bonds). They are not; all bonds are some variation of a wave function. All we can do is pick out certain cases and attempt to interpret them with our models. Argument in point: even the ionic NaCl bond dissociates homolytically (as mentioned at the end of the answer) much like any covalent bond, proving it is much closer to a covalent bond than to a dative or coordinate bond.

I completely understand the different types of covalent bonds. My point of contention is your view of "covalent" as the "pure covalent" type, based on how you have defined it in your answer. I cannot be certain but I do not think that all covalent bonds would homolytically cleave upon addition of the bond enthalpy. Do you have any evidence to support your point?

Regarding the fallacy you have pointed out in your first comment, I admit that I have overlooked that phrase. However, the fact remains that you clearly have distinguished the "covalent bond" and the "coordinate bond" as two distinct types of bonds in your answer.

If you are not convinced, consider the following argument derived from your answer: Conclusion 1 (C1): Coordinate bonds heterolytically dissociate (upon addition of BDE). Conclusion 2 (C2): Covalent bonds homolytically dissociate (upon addition of BDE). However, let's now add in the general understanding of coordinate bonds as covalent bonds: Premise 1 (P1): Coordinate bonds are dative bonds. Conclusion 3 (C3: Based on C2 and P1): Coordinate bonds dissociate homolytically. As can be seen, there is a contradiction between C1 and C3 contradict when we factor in P1.

@TanYongBoon Ah, that’s where you’re coming from. Note that a coordiante (covalent) bond or dative bond or whatever you may call it is a bond very similar to an ordinary covalent bond but which dissociates heterolytically as outlined in my answer. This is a subset of covalent bonds that behaves differently from all other covalent (and ionic) bonds. Unfortunately, no word exists to the best of my knowledge to describe covalent non-dative bonds. Until it is created, I’ll have to live with the ambiguity of the term covalent.

@TanYongBoon A quick and dirty reference for dissociation energies and that they refer to homolytical dissociation is the BDE Wikipedia page.

My contention is not in the terminology. Even after clarifying that the "covalent bond" in your answer is the "non-dative covalent bond", your answer remains problematic. Consider covalent substances like hydrogen fluoride do they dissociate to give a hydrogen radical and a fluorine radical upon addition of BDE or will it more likely give hydrogen and fluoride ions.

By your definition, do we then say the bond in hydrogen fluoride is a coordinate bond because it certainly would not homolytically cleave? If you agree with me on this, then your answer is indeed problematic.

In my opinion, you seem to be distinguishing the difference between a polarised bond and a pure covalent bond in your answer, and not the difference between a coordinate covalent bond and a non-coordinate covalent bond.

@TanYongBoon It seems to me that in lieu of experimental evidence you immediately assume that addition of the BDE to a compound such as hydrogen fluoride would result in heterolytic bond cleavage. That is not the case; e.g. irradiating $\ce{HF}$ molecules with light of the appropriate wavelength will generate hydrogen radicals and fluorine radicals.

All right, you are right in saying that I am making an assumption here regarding hydrogen fluoride. Therefore, any conclusions derived from that example may be false. So let's not talk  about that. Since I cannot be sure of a single non-coordinate covalent molecule like HF, how can you be sure that all non-coordinate covalent bonds would dissociate heterolytically? I am very sorry but I remain firm in my stance and am unconvinced by your answer.

@TanYongBoon I’ve noticed. I cannot convince you. Maybe reading up in the textbook of your choice convinces you =)

How about this? Which source(s) did you get this piece of information from?

@TanYongBoon Comments are not for prolonged discussions.

The main (and probably the only) way is to apply the definition. In other way, you need to understand what the valence structures of the examined compound's ingredients are and based on this information to determine how they have bonded.
If you struggle to implement this analysis directly, try to work step by step, starting with thinking out a synthetic reaction, e.g.:
$$\ce{A + B -> AB}$$
that produces the examined compound $\ce{AB}$. The least you can conclude from knowing $\ce{A}$ and $\ce{B}$ is whether any of them provide elecro-deficient (or ionic) constituents of $\ce{AB}$, which, if so, is a clue that there may be coordinate bonding "in a close proximity" to the bond $\ce{A - B}$ …
Example: In ammonium hydroxide:
$$\ce{NH3 + H2O <=> NH4+ + OH-}$$
the hydroxide part is an anion, so it has a surplus of electrons. By knowing that $\ce{OH-}$ came from $\ce{H2O}$, it becomes obvious that the additional electron has came from one of the hydrogen atoms in $\ce{NH4+}$, hence, the bond between $\ce{N}$ and this particular $\ce{H+}$ is coordinate (both electrons come from the nitrogen atom $\ce{N}$).


text:
Is this problem doable with the information given?
Can I solve this problem without knowing any starting amounts?
An element $\ce{X}$ forms iodide ($\ce{XI3}$) and chloride ($\ce{XCl3}$). The iodide is quantitatively converted to the chloride when it is heated in a stream of chlorine:
$$\ce{2 XI3 + 3Cl2 -> 2XCl3 + 3I2}$$
Calculate the atomic mass of $\ce{X}$. Identify element $\ce{X}$.

Hi and welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site. I improved the formatting of your post using MathJax. More information on how to do so yourself can be found in the [help] and this meta post.

Also, the problem is not solveable without the masses of reactant and product.


text:
Calcium chloride hexahydrate has separated in jar - is it beyond saving?
I work in a lab where I'm in charge of ordering new supplies. A graduate student brought up a $500~\mathrm{g}$ amber jar of $\ce{CaCl2.6H2O}$ and pointed out that it had separated into a clear liquid (on top) and a white crystal (on bottom). The cap seemed to be on normally and there is no expiration date on the jar.
Before I decide if I need to order more, I am wondering:

What happened to the $\ce{CaCl2}$?
Can it be saved?

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. While it may seem strange at first we do not use salutations in our Questions and Answers. Please use the upvote button to say thanks, or leave a comment how the post can be improved. I assume that the packaging was not sealed? It would further be good if you include your thoughts on the matter.

It likely just drew water from the atmosphere.

Can't see why recrystallization wouldn't help it

Calcium chloride is incredibly hygroscopic (it sucks in water), this property makes it incredibly useful (for instance when used in a desiccator or in a drying tube to prevent water from getting into a reaction).
The wet $\ce{CaCl2}$ can be turned back into a solid  by heating (ideally under vacuum), however you can end up with a mixture of different compounds (for instance heating very wet $\ce{CaCl2}$ has been known to give $\ce{Ca(OH)2}$).  It can also be recrystallised from ethanol to give the di-hydrate, which in turn may be dehydrated by heating, however fresh $\ce{CaCl2}$ is very inexpensive, and so I would suggest buying a new bottle, and ensuring it is sealed well after use to prevent it taking up moisture from the atmosphere.

Note that OP’s jar was apparantly labelled $\ce{CaCl2 . 6 H2O}$

"purification of laboratory chemicals" only gives details of anhydrous $\ce{CaCl2}$ or the di-hydrate. I assume that the hexahydrate is the lowest grade available as a solid. In any case, the advice holds: recrystallisation may work, but its probably cheaper and easier to buy new material. Heating is an option but may lead to degradation into a mixture.

As others have said, $\ce{CaCl2}$ is absolutely spiffing in drawing water from where-ever. You might have the cap firmly on the jar but humidity probably found its way in, just because the screwing of the cap was not enough to completely seal away the outer atmosphere. The relative humidity in that jar will be very low so entropy will draw ever more water inside. It is not that uncommon for $\ce{CaCl2}$ to end up in the middle of a puddle of water drawn.
It depends on what you want to do with the stuff whether you can still use it.

If you just need calcium chloride and don’t care about specifics: Go for it!
If you just need hydrated calcium chloride: Go for it!
If you need to weigh a semi-exact amount of hydrated calcium chloride: Filtration should be enough.
If you want to do some real sensitive analysis that requires exact stoichiometric amounts of $\ce{CaCl2}$ you need to get a new one or recrystallise to a defined hydrate.
If you wanted to use $\ce{CaCl2}$ as a drying agent: Why the heck are you even considering the hydrate?

Then again, the stuff is cheap, so if you feel safer, buy a new one. Don’t chuck the old one, though, it can most certainly be put to good use!


text:
Does trans 1,3-dichlorocyclobutane have zero dipole moment?
Does this compound have zero dipole moment?

I think it has a non-zero dipole moment. If I assume the ring to be planar, then the dipole moments of each $\ce{C-Cl}$ bond cancel out. (One is upward and the other is downward.)
But if I take the actual puckered structure of cyclobutane, one $\ce{C-Cl}$ bond is axial and the other is equatorial and hence they won't cancel out and the compound has a non-zero dipole moment.
But the answer given in my book says the compound has zero dipole moment.

Is that possible when one Cl is at axial position and one Cl is in equatorial position?

Should I consider ring flip? Because flipping switches chlorine from axial to equatorial and equatorial to axial.

Is ring inversion in our IITJEE syllabus ? There are so many things at conflict here .

Yes. There or not there, it is indirectly present in questions

Cyclobutane and its substituted derivatives readily undergo a ring flip (or ring inversion) as pictured below.

The barrier to ring flipping is very low, around 1.5 kcal/mole, so at room temperature the flipping process is very rapid.  The lowest energy conformation of cyclobutane exists in a puckered geometry as depicted in $\ce{A}$ and $\ce{B}$. In the ring flipping process, the molecule passes through a transition state where the cyclobutane ring is planar.
Only molecules that belong to symmetry classes (point groups)

$\ce{C_{n}}$ (the molecule only contains a $\ce{C_{n}}$ axis)
$\ce{C_{nv}}$ (the molecule contains a $\ce{C_{n}}$ axis and a $\ce{\sigma}_{v}$ plane)
$\ce{C_{s}}$ (the molecule only has a plane of symmetry)

can have a dipole moment.
Conformers $\ce{A}$ and $\ce{B}$ both have $\ce{C_{s}}$ symmetry (the only symmetry element is a plane that bisects the ring and contains the two cyclobutane carbons bearing the substituents) and therefore do have a dipole moment. However, the dipole moments of conformers $\ce{A}$ and $\ce{B}$ are equal and opposite, so when flipping is rapid the dipole moment averages out to zero.
Therefore at room temperature, where flipping is rapid, the molecule has no measurable dipole moment. If you cooled the system down to a very low temperature where conformers $\ce{A}$ and $\ce{B}$ were not rapidly interconverting, then you could measure a non-zero dipole moment.

So cyclobutane can be assumed to be planar? Can I apply the same logic for cyclohexane?

No, cyclobutane cannot be assumed to be planar, but it passes through a planar transition state when ring inverting. Here's an example of why you can't assume cyclobutane (or cyclohexane) is  planar.  As you noted, the two substituents are different in the puckered form.  One substituent is axial and one equatorial, therefore they will have different reactivities.  If you treated cyclobutane as planar you would not have axial and equatorial positions, both positions would be equivalent and you would expect equivalent reactivities, which is not correct.

Is it strictly correct to say that when flipping is rapid the molecule has zero dipole moment?  Logically, it seems that it should have an oscillating dipole whose time average is zero (would it be stochastically oscillating?).  So the ensemble average and the time average would both be zero but the instantaneous dipole would not.

@JasonB What you write is correct, even when the molecule is inverting rapidly it does have an instantaneous dipole moment. I attempted to make that point when I wrote, "when flipping is rapid the dipole moment averages out to zero." I've now changed the wording in the last paragraph to say things a bit better.  If you'd like to further modify the answer, please feel free to do so.  In a situation like this I think it might be common usage to say that the molecule has zero or no dipole moment. It seems akin to the situation where a chiral conformation is in equilibrium with an achiral

conformation. In this case we might say that the molecule is achiral, even though it may be in the chiral conformation at any given instant.


text:
Irreversible process (Isothermal and Adiabatic)
How exactly would we define an irreversible change for an isothermal process? Why can we say what the external pressure will remain constant? Please give an example.
Also, why can we not define both reversible and irreversible processes for adiabatic process simultaneously?
$$\Delta E_\mathrm{rev} = \Delta W_\mathrm{rev}$$
$$\Delta E_\mathrm{irrev} = \Delta W_\mathrm{irrev}$$
I got that if these both take me to the same point then $\Delta E$ will be same, but why must $W_\mathrm{rev}$ and $W_\mathrm{irrev}$ be different?

If you have an adiabatic reversible path that takes you from state 1 to state 2, any irreversible path from state 1 to state 2 will require removal of heat, so it won't be adiabatic.  This is a consequence of the 2nd law of thermodynamics, captured by the Clausius Inequality.


text:
Radiationless energy transfer: how to arrive at the equation that give R0 between two chromophores?
I have a homework question where I should calculate  $R_0$ of an energy transfer between two chromophores. However, I don't understand how my professor arrived at the equation that we should apply (which is important to understand for the exam).

In a radiationless energy transfer experiment (RET) it was observed that the efficiency of transfer was $0.0185$ at $\pu{5 nm}$ distance between the two chromophores. Estimate $R_0$ for the two chromophores, assuming the energy transfer efficiency is $0.7$ at short distances.

In our lecture we are given the formula:
$$[E_\mathrm{eff}] = \frac{R_0^6}{r^6 + R_0^6}$$
I understand that $E_\mathrm{eff}$ is the efficiency of the transfer, and $r$ is the non-variable distance between the two chromophores. However, I am not able to solve for $R_0$. I tried to ask my professor but he thinks faster than I do and skips steps. So I was not able to follow completely, but he somehow mathematically solved the equation to (where $Eff_\mathrm{max}$ is maximum efficiency of energy transfer):
$$r = R_0 \sqrt[6]{\frac{E_\mathrm{eff}}{E_\mathrm{eff,max}}}$$
When I apply his formula I "plug in" the given numbers and solve for $R_0$ as following:
\begin{align}
R_0 &= \frac{r}{\sqrt[6]{\frac{E_\mathrm{eff}}{E_\mathrm{eff,max}}}}\\
R_0 &= \frac{\pu{5 nm}}{\sqrt[6]{\frac{0.0185}{1}}}\\
R_0 &= \pu{9.72 nm}
\end{align}
I would like to understand how he arrived at the final formula that I apply. Is it correct to solve for $R_0$ like I have shown above?

I think I found the answer to my own question through some help from some classmates. What the professor did, was to solve the equation mathematically by converting $R_0^6$ to $1$.
So first, the efficiency of the transfer $E_\mathrm{eff}$ is actually a ratio of the actual transfer efficiency divided by the maximum efficient (which we should assume is "one", unless we are told otherwise.
The next step is a bit hard to explain, as I am not that good in math, but I will try my best. What I understand is that you can "gather" the two $R_0^6$ by multiplying both the numerator and denominator by what we should think of as the number one, but we actually divide by $R_0^6$ and in this way we don’t have to include $R_0^6$ that we divide by on both sides of the equal sign - as it is technically the number one (mathematically allowed), as following:
$$\frac{E_\mathrm{eff}}{E_\mathrm{eff,max}} = \frac{\frac{R_0^6}{R_0^6}}{\frac{R_0^6}{R_0^6}+ \frac{r^6}{R_0^6}}$$
We then get as following:
$$\frac{E_\mathrm{eff}}{E_\mathrm{eff,max}} = \frac{1}{1 + \frac{r^6}{R_0^6}}$$
So by applying some math and assumptions we can solve the equation like this, and what I have explained in my question should therefore be correct.

As you have r = 5nm and the efficiency at the same distance using your first equation only $R_0$ is unknown. The 0.7 bit seems to be a red herring!


text:
How to calculate the radius of gyration (Rg) and the molar mass from light scattering data (Rθ and θ/degree)?
I have a homework problem that I don’t have a solution for, so I would like understand how to solve it and to know if my suggested way of solving it is on the right track. The problem is as following:

A polysaccharide was analyzed by light scattering at several angles ($\theta$), and the following data were at hand:

the system constant $K = 2 \times 10^{-8}$,
the concentration is $2.5~\mathrm{mg/mL}$,
the effective wavelength $\lambda = 400~\mathrm{nm}$,
the second virial coefficient is assumed to be $0$ (meaning that no correction is needed for the concentration dependence).

Calculate the molar mass and the radius of gyration for the polysaccharide.

We are also given the following information, where I think $R_\theta$ is the measured Rayleigh ratio, and $\theta$ is the scattering angle in degrees:

\begin{array}{rr}
R_{\theta}& \theta\\\hline
5 & 60\\
4 & 90\\
3.33&120
\end{array}

From the lectures I was given the following equation to find the molar mass and $R_\mathrm{g}$:
$$
\frac{K \times C}{R_\theta}
  = \frac{1}{M} \left( 1 + \frac{16\pi^2}{3\lambda} R_\mathrm{g} \times
     \sin \left(\frac{\theta}{2}\right)^2 \right)
$$
I am unsure if the way I think is correct — and I don’t really understand all the steps. But here is what I think I have to do, or at least how to starts:

extrapolate to zero concentration to account for large molecules (polysaccharide) and non-ideal solution. To do so I think I have to plot $\frac{K \times C}{R_\theta}$ against $\sin\left(\frac{\theta}{2}\right)^2$.
From this the plot I get the linear equation $y = ax + b$ which I can use to:

Find molar mass because: $y - \text{intercept} = \frac{1}{M}$
Find $R_\mathrm{g}$ because: $\frac{\text{slope}}{y - \text{intercept}} = R_\mathrm{g}$

I read somewhere that I should make some sort of a second graph from the first one, in order to remove the $\frac{1}{M}$ part of the equation, and thus, I can somehow calculate $R_\mathrm{g}$, but I don't understand this.

My Questions:

Am I generally on the right track here?
What units do I get the molecular mass in from the above equation? Is it KDa?
I think I get $R_\mathrm{g}$ value in nm?


text:
Dipole moments of pyrrole and furan
Why do pyrrole and furan have dipoles oriented in different directions?

I found some description here http://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/12/oc/vlu_organik/heterocyclen/arom_gessaet_heterocyclen.vlu/Page/vsc/en/ch/12/oc/heterocyclen/fuenfaromat/fuenfring_aromat.vscml.html

Pyrrole's major contributor has double bond to nitrogen, furan's looks like on your picture.

Both pyrrole and furan have a lone pair of electrons in a p-orbital, this lone pair is extensively delocalized into the conjugated pi framework to create an aromatic 6 pi electron system.
Where pyrrole and furan significantly differ is that,

in pyrrole there is an $\ce{N-H}$ bond lying in the plane of the ring and directed away from the ring
whereas in furan, there is a full lone pair of electrons in roughly the same position.

The localized lone pair of electrons pointing away from the ring has a very significant effect on the dipole vector and is enough to cause the observed reversal in dipole moment direction between furan and pyrrole.

It is due to more than just the presence of the lone pair on the oxygen. The oxygen is more electronegative than nitrogen therefore although both furan and pyrrole have the inductive effect of electron withdrawing, this effect is greater in furan because of the greater electronegativity.
So the dipole moment of furan faces towards the oxygen, a result of greater inductive effects than resonance effects. Whilst the dipole moment of pyrrole faces away from the nitrogen as a result of greater resonance effects

This may be the same as other answers, but viewed a different way.
Imagine a strong base were to remove the (most acidic) proton attached to nitrogen in pyrrole, giving a structure isoelectronic and isostructural with furan.  With the lone pair on the most electronegative atom in both cases we would then expect similar dipole moments (the charge on our deprotonated pyrrole would exist independently of the dipole moment).  But the actual pyrrole has that proton attached, at what would be the negative end of the furan-like dipole.
So it is the extra hydrogen atom in pyrrole, which would effectively be a proton in a comparison with furan, that reverses the dipole in the former.