text:
Is NaCl crystal always going to have even number of atoms?
I just read:

In $\ce{NaCl}$, atoms don't aggregate so as to form discrete molecules but they are held together in a network structure. The ratio of the $\ce{Na}$ to $\ce{Cl}$ atoms in a sodium chloride crystal is $1:1$.

Here I have a $\ce{NaCl}$ crystal. Assume that I can break this crystal in all possible ways. So that I have its two pieces every time with a different mass ratio. Would these pieces always contain even number of atoms of $\ce{Na}$ and $\ce{Cl}$. Or, is this ratio going to be same for all conditions?
Suppose I have a $\ce{NaCl}$ crystal of mass $a$. I break it into two pieces of masses $x$ and $y$ such that $x,y<a$ and $x+y=a$. $a$ is a constant by its definition whereas $x,y$ are variables. Then, is the following true?
$$R(x)=R(y)=1:1$$
Where $R(x)$ and $R(y)$ denotes the ratio of $\ce{Na}$ to $\ce{Cl}$ atoms for all values of $x$ and $y$ respectively.

So that's what happens when mathematician ask about chemistry ;) 1:1 means that there's no significant inherent tendency for non-stoichiometry. Macroscopic crystals are never ideal.

:) I don't know what you mean by ideality of  crystal.

They are never pure and undefected

What if I assume it ideal? Then, there must be an answer to my question.

@Mak would you like to add why you removed your answer?

I removed the answer for now since I need to elaborate on some stuff and address your issues comprehensively, hopefully.

Scratch that @Sufyan, Jan is hopefully writing an answer to your question.

Tl;dr: Theoretically yes, practically no.

Needless to say this is a very mathematical question that almost calls for a philosophical answer. The key point to consider is usually the crystal’s surface. Here, there are most defects and the least ideal structure since there is at least one row of atoms which are lacking partners on one side. But I’ll keep that out of my reasoning for a second.
Consider an ideal crystal and a Subtle Knife. You can use the knife’s almost infinitely thin blade to cut exactly between two sheets of atoms. The initial crystal had a $1:1$ distribution, every single sheet has a $1:1$ distribution and thus so will your two bits.
Then consider cutting one of the fragments again, but three atom rows away from where you cut it before. If our ideal crystal is infinitely large, we still get a $1:1$ distribution. Now continue to consider cutting the fragments into smaller fragments until you arrive at a theoretical body of $3 \cdot 3$ size. This cube has to violate the $1:1$ condition. It will be charged depending on which atom type is present in the corners. (The same is true for any other cube of an odd side length when measured in atoms.) However, we have a charged particle (consisting of nine atoms) now. It will attrackt particles of the other charge and pull them towards it. In the long run, one of the following may happen:

An ion of the opposite charge is drawn towards the crystal and attaches to it or inserts into it.
An ion of the extraneous charge is removed.

Both will make us regain our $1:1$ crystal — but it is no longer ideal. However, see below for the evolution of a more realistic view.

Now let’s consider a slightly more realistic case. We no longer have an infinitely large ideal crystal but one with a finite size. Let’s have a cube of even atom length just for simplicity ($1:1$ ratio). And because we want to see our cube, let’s assume a macroscopic size, e.g. a cubic centimetre. Now cut it with your Subtle Knife so that the even side length is split into to cuboids with one odd length. One must have a $+1$ charge, the other must have $-1$. Thus, neither have $1:1$.
One could now say that maybe a chloride ion ‘jumps’ to the other side or something, again creating a non-ideal crystal. Or maybe moisture from the air supplies one side with a proton and the other with a hydroxide. But we could also say that we have a charge of $+1$ and one of $-1$ each delocalised over some $10^{20}$ atoms — no big deal if you think of it.

And let’s go a step further and consider a real crystal. It’s going to have defects, impurities, non-aligned domains and a surface. If it ever had a mathematically exact $1:1$ ratio, then that was purely by chance. We can still delocalise the effective charge over billions of billions of atoms. We would need an extensive charge in a very small area to cause a problem at all.
Even worse if our crystal is put into a saturated $\ce{NaCl}$ solution. (Saturated, because I don’t want it to dissolve.) All the time ions will be entering and leaving solvation in a non-coordinated manner. Both random in time and random in location. The numeric ratio of atoms will change all the time. If we consider a very sub-saturated solution, of course there will be more ions dissolving into solution than recrystallising. But also, those that end up in solution also dissociate from each other. Thus, there is nothing actually keeping an atom of a charge of one (or two, or three, see other metals) close to its counter-ion — charge dissociation is not as bad as it seems on the paper.
So real world crystals will only have the ideal stoichiometry in a short-lived transition state between being too positive and being too negative — and being philosophical I could say that that almost be like a person’s mood.

authors: אתי איבנבוים


text:
Oxidation state of metal ion in unit cell

If a cubic unit cell for a metal nitride has nitride ions at all face and corner positions, and metal ions at all edge positions, what is the oxidation state for the metal ion in the nitride?

I think that if something is in a face position, there is half of that something present in the unit cell. So if there are nitride ions at all face positions (6) then there must be 3 total nitride ions. And at the corners, there are eighths of nitride ions. So given 8 corner positions, there must be an additional nitride ion formed by the fragments at the corners.
This gives us a total of 4 nitride ions, or a negative 12 charge that must be balanced by the metal ions at the "edge" positions. But what are edge positions? If I know what it is, then I can visualize how many metal ions there are in the unit cell.
Are the edge positions directly halfway between the corner positions and butted up against the face positions? So is there one edge position per edge of the unit cell? This gives us 12 edge positions, and each fragment is a fourth of an ion, so we have 3 metal ions total. So the oxidation state of the metal ion is +4?

Edge positions are those at the centre of an edge of the cube; i.e. in the middle between two closest corner positions. As such, there are 12 edge positions and each belongs to 4 unit cells giving us 3 metal ions.
The way the nitride positions are described is basically a verbose way of saying cubic close-packed or face-centred cubic. Therefore, we can redescribe the cation positions as ‘all octahedral voids except for the one in the centre of the unit cell’. Since there are 4 atoms per unit cell in ccp and 4 octahedral voids, that again gives us a $3:4$ ratio and thus $\ce{M3^{IV}N4}$.

The edge atoms actually occupy the the center of the edge, In the first 2 figures, red atoms show the positions of the cation.
The last figure gives a clear idea as to how many cubes actually share one edge. 4 cubes share 1 edge, so effectively, each edge atom contributes 1/4 to the unit cell.
There are 12 edges in total, which means the effective contribution is 12/4=3. Since it is given 3*(oxidation state of metal ion)=12 for neutrality of molecule, we get the O.S. to be +4.

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text:
Regiochemistry and stereochemistry in synthesis of alcohols from cycloalkenes through oxymercuration and hydroboration
I have been posting several practice final questions for my organic exam next Tuesday. I have worked through all the questions, but my prof does not post the answers. I would greatly appreciate your feedback.
Draw full structures, including any relevant stereochemistry of the organic products expected in each of the following reactions:

One reaction uses a syn addition, and the other uses oxymercuration. Question: for the oxymercuration, this is anti-Markovnikov because the hydrogen doesn't end up on the most substituted carbon, right?

Markovnikov addition means that the hydrogen ends up on the less substituted carbon. That means oxymercuration gives the Markovnikov product. Go back and think about the simple addition of HBr. I feel like you could have checked a textbook first or the Internet though. Most textbooks I know would have said exactly the same thing that I did: hydroboration is anti-Markovnikov and oxymercuration is Markovnikov.

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text:
Do orbital models derive from the solution of the stationary Schrödinger equation?
I know that hydrogen-like atomic orbitals can be seen as a solution of the stationary Schrödinger equation - but what about atoms with more electrons? As I am a pupil and quite uneducated about this subject, I would be very grateful for an answer that is easy to understand.

Go read about the orbital approximation.

There is no answer that is easy to understand for an uninitiated since any real explanation of the orbital model requires a deep immersion into the world of quantum chemistry.

I know that hydrogen-like atomic orbitals can be seen as a solution of
  the stationary Schrödinger equation - but what about atoms with more electrons?

Orbitals are one-electron wave functions, i.e. they are wave functions that describe the state of only one electron. For a hydrogen-like atom, which is a one-electrons system, if treating its nucleus as a classical positively charged point particle, the wave function is indeed an orbital, since there is just one quantum particle out there and it is the single electron. So, right, orbitals are the solutions of the (non-relativistic) time-independent Schrödinger equation if the nucleus is assumed to be a classical point particle.
For any many-electron atom, though, even when its nucleus is though of as a classical point particle, the wave function describes more than one electron, so by the very definition it is not an orbital. It can be written in terms of the one-electron wave functions (orbitals) as their product, but this can be done only approximately due to interactions between the electrons. It is in this sense we say that orbitals aren't real.
Note though that, strictly speaking, even for a hydrogen-like atom orbitals aren't real since they arise in the non-relativistic treatment which is also an approximation. The reality is even more weird than its non-relativistic description, but for atoms from the first and second rows of the periodic table the non-relativistic description is usually a good one in a sense that it introduces relatively small errors that can be neglected for the most of the chemical purposes. The orbital approximation on the other hand might be the source of much larger errors. For instance, for a $\ce{He}$ atom in its ground electronic state the orbital approximation introduces the error of the $10^{-2} E_{\mathrm{h}}$ order while the relativistic correction is about two orders of the magnitude smaller, $10^{-4} E_{\mathrm{h}}$.

Let's take a look at the Hamiltonian for Hydrogen atom with one electron:
$\hat{H}_{H} = T_e + T_p + V_{ep}$
You need for each particle (electron and proton) an expression for the kinetic energy ($T$) and a expression for their interactions as potential energy ($V$). For an additional electron you have to add additional expressions:
$\hat{H}_{H^-} = T_{e_1} + T_{e_2} + T_p + V_{e_1p} + V_{e_2p} + V_{e_1e_2}$
Each expression increases the difficulty for solving the Schrödinger Equation. Thus there are no other element orbitals that can described analytically exact. But there are a lot of good numerical approaches.


text:
Transforming a cis double-bond to a trans double-bond using any mechanisms of my choice
For my final, I need to be able to a trans-double bond from a cis double bond. Here is what I drew:

I essentially created a triple bond and oxidized

Yes that is fine. But converting a triple bond to a double bond is reduction, not oxidation.

Heating it also would work, but that's probably not what the boring syllabuses want from you. :)

Adding HBr and subsequent elimination should also produce a trans double bond as the main product.

I like the bromination step.  Since you have geminal halogens you can use magnesium shavings to elimnate them to re-form your alkene which should give you a trans maolecule.


text:
Does ethylene glycol sink in water?
I’m a little concerned that there is a far stronger mix of antifreeze in my car's coolant system than I am aware. I check the antifreeze level in my car (using a refractometer) when the engine is cold (after being left overnight).
During the night, can the antifreeze settle to the bottom of the cooling system, thus leaving a weak mixture of water and antifreeze within the coolant reservoir up top, where I take a sample with the refractometer?

No. The only way for any separation to occur would be if some of the coolant froze. The frozen part would be rich in water, the liquid part rich in ethylene glycol.

Ethylene glycol and water are miscible in all proportions: https://en.wikipedia.org/wiki/Ethylene_glycol. That said, if you were to add antifreeze directly to the coolant overflow bottle, as is normally recommended, it would take a few restarts and cool-downs of the engine before it mixed with that in circulation.

Ethylene glycol is completely miscible with water in all proportions (as the commenters point out). Therefore, once mixed, the glycol will not separate from the water, ever.
The only exception is if it gets so cold the mixture starts to freeze. Freezing will cause some degree of separation as the water will solidify first. But if it is that cold, separation will be the least of your problems.

In a car radiator below the freezing temperature of water, the water part increasingly turns to slush. The glycol just prevents the mix from freezing hard enough to expand and crack engine blocks, etc.
But, ice is lighter than straight water and I haven't seen that mentioned or addressed here. Ice/slush rises in the system and then might be subsequently melted back into water the following day when the mixture is checked with a refractometer.
So. The glycol doesn't sink as was suspected, but ice/slush rises and then turns into water.
Similarly, separation is the greater of your problems because the slush will not pass through a small tube radiator and it can interfere with the thermostat and water pump bypass. If you fired up an engine for a short period of time, you can cause a log jam of slush in one side of a radiator tank. If the side tank has the radiator cap, I can see getting weak test results the following day when the slush melts.

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text:
Is mixing ammonia and bleach spray in the same parcel of air considerably dangerous to health?
I've read a few articles advising not to mix bleach and ammonia in the same room because it makes chloramine gas. But my boyfriend and I are having an argument - he thinks that since those articles don't specifically say that mixing them in the same room (in our case, our small, poorly-ventilated bathroom) is bad, the only situation in which we shouldn't do this is spraying them directly on the same surface.
I don't see any reason why the chemical reactions producing chloramine gas wouldn't occur if they're mixing in the same room, from our bleach cleaner and glass cleaner spray bottles, though they would be less concentrated. However, he would only trust an answer from a chemistry expert. Please advise!

What do you mean by "mixing in the same room" exactly?

First bleach and ammonia will yield chlorine gas. // If you clean the mirror with the ammonia spray and the sink with the bleach spray then you should be fine. However you certainly should not spray both cleaners on the same surface at the same time.

MaxW, that is exactly what I was wondering.

Gerhard, by mixing I mean I am spraying them on surfaces that are close to each other, around the same time, and in a closed and unventilated room, so the sprays are mixing in the air.

And by the way Max W, why wouldn't chlorine gas be formed by mixing them in the air alone? Is there something about the sprays mixing in the air that would make inhibit the reaction?

Using a spray might leave a tiny, tiny amount of droplets in the air which could combine with a second spray. However  the amount of liquid left in the air will be so small that it wouldn't be a problem. You'd smell the chlorine if it formed. So just clean with one spray, wipe it up, then use the second spray and different rag/paper to clean up second spray.

However the amount of liquid left in the air will be so small that it wouldn't be a problem.[citation needed]

In a poorly ventilated room, I'd certainly avoid using both at the same time. Ammonia is very volatile and will usually "find" the only slightly less volatile chlorine gas - or perhaps the bleach itself. -Chemist. PS: A poorly vented bathroom is in itself a bad idea. ;)

This information is only for academic and educational purpose only. Please never ever try to mix ammonia and bleach.
Mixing bleach and ammonia is extremely dangerous, since many toxic vapors (nitrogen-chlorine species) will be produced. The primary toxic chemical formed by the reaction is chloramine vapor and hydrazine. (very, very toxic).
The following is the list of chemical species produced by the reaction:

$\ce{NH3}$ = ammonia
$\ce{HCl}$ = hydrochloric acid
$\ce{NaOCl}$ = sodium hypochlorite
$\ce{Cl2}$ = chlorine gas
$\ce{NH2Cl}$ = chloramine
$\ce{N2H4}$ = hydrazine
$\ce{NaCl}$ = sodium chloride or salt (non-toxic)
$\ce{H2O}$ = water (non-toxic)

The bleach decomposes to form hydrochloric acid:
$$\ce{NaOCl → NaOH + HOCl}$$
$$\ce{HOCl → HCl + [O]}$$
And then the ammonia and chlorine gas react to form chloramine, which is released as a vapor:
$$\ce{NaOCl + 2HCl → Cl2 + NaCl + H2O}$$
$$\ce{2NH3 + Cl2 → 2NH2Cl}$$
If ammonia is present in excess, toxic and potentially explosive liquid hydrazine may be formed.
$$\ce{2NH3 + NaOCl → N2H4 + NaCl + H2O}$$
References:

http://chemistry.about.com/od/toxicchemicals/a/Mixing-Bleach-And-Ammonia.htm
http://home.onehowto.com/article/what-happens-when-you-mix-bleach-and-ammonia-7512.html

Not sure about hydrochloric acid (not mutually stable with ammonia), I would have put ammonium chloride instead.  But products $4, 5, 6$ are nasty enough.

My story is simpler than other answers.  When I was in graduate school one of the cleaning guys told me he was knocked out mixing ammonia and bleach in his cleaning solution.
Bleach itself is best stored (and used) well separated from other household chemicals.  Ammonia, acid, organic solvents  -- all can react with bleach to give chlorine gas and other dangerous stuff.


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H2O2 + NaCl + H(+) reaction?
So I found these reduction potentials online:
$$\begin{alignat}{2}
\ce{H2O2 + 2H+ + 2e- \;&<=> 2H2O}\qquad &&E=1.78\ \mathrm V\\
\ce{2Cl- \;&<=> Cl2 + 2e-}\qquad &&E=-1.36\ \mathrm V
\end{alignat}$$
If the net reduction potential is positive will chlorine gas be spontaneously formed when salt and hydrogen peroxide are mixed in a acidic solution?

Exactly what is an acidic base...? Anyway, theoretically, yes. However kinetically the reaction may not be favoured.

Not necessarily.  If you get the pH right and the stoichiometry absolutetly 1:1 then you might form $\ce{Cl2}$.  However your reaction will compete with other reaction and you will more likely oxidize the chlorine into chlorate ions ($\ce{ClO-}$, $\ce{ ClO2-}$, $\ce{ClO3-}$, or $\ce{ClO4-}$).

If you want to form chlorine I suggest calcium hypochlorite and hydrochloric acid or potassium permanganate and hydrochloric acid.


text:
How do I find the rate constant in this situation?
Say I have two kinds of molecules $A$ and $B$.
The molecules of type $B$ are small and bind onto molecules of type $A$.
(Any number of them may do so, but in practice this number is $\leq 8$.)
We say a molecule of type $A$ is of type $A'$ if it has at least one copy of $B$ bound to it.
More specifically, we say a molecule of type $A$ is of type $A^k$ if it has exactly $k$ copies of $B$ bound to it.
We can easily detect experimentally when a molecule is of type $A'$, but not so easily when a molecule is specifically type $A^3$, for example.
However, it was done at one point, and the researchers gave us a way (given a concentration of $A'$) to determine the individual concentrations of $A^k$'s. (Poisson distribution)
I would like to consider the dissociation reaction $$A^k \rightarrow A^{k-1} + B$$ where $k \geq 1$ and $A^0$ is considered to be $A$.
Because the molecules of $B$ are small, I can assume that the rate of this reaction is independent of $k$ and that the molecules of $B$ dissociate independently of any other molecules of $B$ attached to the same copy of $A$.
My question is: how can I find a constant $c$ so that the rate of the reaction is $r = c[A']$?

Is $A'$ the concentration $A$ molecules with at least one $B$ bound, or is it the concentration of $A$ molecules with precisely one $B$ bound?

In my notation $A^1$ would be the molecules with exactly one $B$ bound.  Essentially $A'$ is the collection of all the molecules of type $A^1, A^2, A^3$ etc.

You've written the dissociation reaction as irreversible.  Is that what you intend?  In that case every reaction is unimolecular so the decomposition kinetics will be a series of exponentials


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Potassium vs sodium bicarbonate in fire fighting

Why is it that potassium bicarbonate reduces flashback as opposed to sodium bicarbonate when fighting fires?
From what I understand, high heating of bicarbonate salts decomposes them into carbon dioxide, which helps suffocate the fire, and the corresponding metal hydroxide. Could it be potassium hydroxide that is interacting differently with the fire versus sodium hydroxide that causes this issue?

I would guess from the observed that the sodium hydroxide sinks more easily to the ground of a puddle of previously burning oil, thereby leaving a open surface that is still somewhat hot and can easily get reignited once the carbon dioxide has been blow away.

@Karl why would sodium hydroxide sink more easily to the ground?

I've no idea. It just seems logical that a reignition could take place easily only if the surface wasn't covered.

To answer this question, firstly we must understand both how fires are started, and how they are perpetuated. It is common knowledge that in order for fire to exist there must be heat, fuel, and an oxidizing agent (usually oxygen). Once started, a fire is then self perpetuated by a radical chain reaction. Removing any of these three ingredients or interrupting the radical chain reaction can be effective ways of extinguishing a fire. Both sodium and potassium bicarbonate are effective at performing several of these actions, though there are slight differences between them.

Differences between $\ce{NaHCO3}$ and $\ce{KHCO3}$

It is widely believed that fireextinguishing powders can function as both energy-absorbing materials and solid surfaces on which free radicals can be destroyed. Heat may be absorbed by the heat capacity of the solid [$C_p(\ce{KHCO3})=90.05\:\mathrm{J\:mol^{-1}\:K^{-1}}$ vs. $C_p(\ce{NaHCO3})=87.61\:\mathrm{J\:mol^{-1}\:K^{-1}}$], the heat of fusion, the heat capacity of the liquid, heat of dissociation from breaking of chemical bonds [$\Delta H_\mathrm{decomp}(\ce{KHCO3})=1063.28\:\mathrm{kJ\:mol^{-1}}$ vs. $\Delta H_\mathrm{decomp}(\ce{NaHCO3})=995.72\:\mathrm{kJ\:mol^{-1}}$], and heat of vaporization. All of these contribute to the total endothermicity of the fireextinguishing powder.
From a chemical aspect, it has been found that potassium salts are more effective than sodium salts, and iodide anions are more effective than chloride anions. Presumably, there is a catalytic path for destruction of free radicals, such as $\ce{H}$, $\ce{O}$, and $\ce{OH}$, utilizing the potassium in the salts. It must be remembered that any powder that has a chemical fireextinguishing capability will also have a heat-absorbing (endothermic) capability.$^{[1]}$

The Chain Radical Mechanism

Combustion of hydrocarbons is thought to be initiated by hydrogen atom abstraction (not proton abstraction) from the fuel to oxygen, to give a hydroperoxide radical ($\ce{HOO}$). This reacts further to give hydroperoxides, which break up to give hydroxyl radicals. There are a great variety of these processes that produce fuel radicals and oxidizing radicals. Oxidizing species include singlet oxygen, hydroxyl, monatomic oxygen, and hydroperoxyl. Such intermediates are short-lived and cannot be isolated.$^{[2]}$

The Chain Radical Breaking Mechanism

The mechanism by which chemical inhibition occurs when certain powders are added to fuel-air flames is not completely understood. Solid particles may scavenge chain propagating species by surface adsorption or reaction; or alternatively, the solids may vaporize in the combustion zone to produce gaseous products that react with active species via homogeneous gas reactions.
Most investigators seem to favor the latter mechanism, although direct experimental verification is very lacking. If homogeneous gas reactions are, in fact, responsible for the inhibition, then for cases in which potassium salts are the inhibiting agents, the chemical reactions
$$\ce{KOH(g) + H->K(g) + H2O}\:\:\:\:\:\Delta H = -33.2\:\mathrm{kcal\:mol^{-1}}$$
  $\hspace{78 mm}$and/or
$$\ce{KOH(g) + OH -> KO(g) + H2O}\:\:\:\:\:\Delta H = -1.7\:\mathrm{kcal\:mol^{-1}}^{[3]}$$

[1] Fire Extinguishing Powders
[2] Oxygen Fueled Combustion Reaction Mechanism
[3] Flame Inhibition by Potassium Compounds

Potassium bicarbonate is preferred over sodium bicarbonate for possible reasons such as:

Counterflow diffusion flame: $\ce{KHCO3}$ is approximately 2.5 times
more effective than $\ce{NaHCO3}$ on a mass basis in extinguishing the flame. This may be attributed to (perhaps) a difference in enthalpies of decomposition of the two compounds (I am not including them here).
the particles effectiveness probably increases with surface area. This includes heat transfer, radiation, decomposition/vaporization. and surface catalysis. Therefore small particles are expected to be
more effective at fire extinguishing.
In Sheinson, R.S., "Fire Suppression by Fine Solid Aerosol", they found that the effectiveness of the agent increased gradually as the particle diameter decreased until reaching a critical diameter. At this diameter there was a dramatic increase in the the suppression efficiency and the effectiveness remained constant for all particles below the limit.The limit sizes were reported as 16 micro and 22 micro for $\ce{NaHCO3}$ and $\ce{KHCO3}$, respectively.

You can refer to:
Evaluation of Bicarbonate Powders as Fire Suppressants
Fire Extinguishing Powders


text:
How to calculate the oxidation number of sulfur in H2SO5?
I am in class 11th and I am having trouble solving the question.

Calculate the oxidation number of sulphur in $\ce{H2SO5}$?
  (The answer is given as as $+6$.)

$\ce{H2SO5}$ exists it is named as "peroxy sulfuric acid".
How tried it:
\begin{align}
2(+1) + 1(x) + 5(-2) &= 0\\
2 + x - 10 &= 0\\
x&=\pm8
\end{align}

https://en.wikipedia.org/wiki/Peroxymonosulfuric_acid It has oxygen on higher ox state as all peroxyacids have

@hackwarewright Also called Caro's acid.

Structures are more important than chemical formulae while finding oxidation numbers of individual atoms.

Note that an oxidation number of $+8$ on $\ce{S}$ would require to open up the 2p shell, which does not happen for sulfur.

$\ce {H_2SO_5}$ has an oxygen-oxygen bond. This means that two of the five oxygen atoms have an oxidation number of $-1$. Same case as in $\ce{H2O2}$.
$$2\cdot (+1) + 1\cdot(x) + 3\cdot(-2) + 2\cdot (-1) = 0\\2 + x - 6 - 2 = 0\\
x = +6$$
You have 3 oxygens with oxidation number $-2$ and 2 oxygen atoms with $-1$.

$\ce{H2SO5}$ has the Lewis structure shown below:

It is possible to assign the oxidation # of each atom by considering the electronegativities of the two atoms involved in each bond and assigning the bonding electrons to the more electronegative atom in each case. Oxygen atoms 3 and 4 are bonded to each other, so the bonding electrons are assigned one to each atom.
After assigning bonding electrons to the more electronegative atom in each bond, splitting the O-O bonding electrons and assigning lone pair electrons to the atom they are on, the oxidation # of each atom is found by the following formula:
$$ oxidation~number = group~number~of~element - assigned~electrons~in~the~structure$$
e.g. Oxygen 1: group # 6 (for oxygen) - assigned electrons 8 = -2 oxidation #
using these rules, the assigned oxidation #'s are:
Both H's: +1 each
Oxygens 1,2: -2 each
Oxygens 3,4: -1 each
S: +6
Note: there are no formal charges in the Lewis structure shown. There is another contributing form with single (dative) bonds to the two oxygens (# 2). This does not change the assigned oxidation numbers, but it does put 1- formal charges on the oxygens numbered 2, and a 2+ formal charge on the S.

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text:
Reaction of moist egg shell powder and acid?
So I performed an experiment to investigate the effect of fluoride ions on the rate of reaction between egg shells (calcium carbonate) and acid.
When doing this I had a dry sample of crushed egg shells and a sample previously soaked in distilled water which had dried for 3 days but was still slightly moist in a 'gloopy' texture.
When I reacted the moist egg shells with hydrochloric acid they made a frothy mixture of bubbles where the carbon dioxide was trapped in bubbles while the dry egg shells fizzed and the carbon dioxide was not trapped in bubbles.
What could be a reason for the differences in the way the carbon dioxide was trapped?
(membranes were removed from both samples)

I don't think anything chemically has occured. It's likely that the membrane of the dry eggs became rigid and brittle and thus fractured into pieces when the bubbles formed where as the moist membranes would be more deformable allowing them to trap $\ce{CO2}$ bubbles.

hi thank you so much but both samples of these egg shells had their membranes removed which is why i am so confused but i think this might answer my other question about membranes so thank you

It sounds like something chemically occurred if he reacted the egg shells with HCL and gas formed?


text:
Dissociation of benzoic acid
Benzoic acid is a weak acid, so it does not completely dissociate in water. I know that the dissociation equation of benzoic acid is:
$$\ce{C6H5CO2H <=> H+ + C6H5CO2-}$$
The question I'm having trouble with is:

Would the dissociated form be more or less likely to dissolve in water?

Is this basically asking if the ions on the right are more likely to dissolve when compared to $\ce{C6H5CO2H}$?
If this is what the question is asking, I think yes, the dissociated form will be more likely to dissolve as they are very polar when compared to $\ce{C6H5CO2H}$ and they can be surrounded by polar water molecules. But I'm not sure.

Solubility of benzoic acid(mw 122.12):

1.7 g/L (0 °C) - 0.00139 moles/100 ml
2.7 g/L (18 °C)
3.44 g/L (25 °C)
5.51 g/L (40 °C)
21.45 g/L (75 °C)
56.31 g/L (100 °C) - 0.04611 moles/100 ml

Solubility of sodium benzoate (mw 144.10):

62.69 g/100 mL (0 °C) - 0.4350 moles/100 ml
62.78 g/100 mL (15 °C)
62.87 g/100 mL (30 °C)
71.11 g/100 mL (100 °C) - 0.4935 moles/100 ml

So yes, sodium benzoate is more soluble than benzoic acid, but consider calcium benzoate which is less soluble.
Calcium benzoate (mw 282.31):

2.32 g/100 mL (0 °C) - 0.00822 moles/100 ml
2.72 g/100 mL (20 °C)
8.7 g/100 mL (100 °C) - 0.03081 moles/100 ml

So the "real" answer is - it depends...

Interesting that unionized methyl benzoate is soluble 2.75 g/L (https://foodb.ca/compounds/FDB012198). It would seem that the benzoate ion is relatively compatible with water, but even more attracted to calcium dication. Solubility of unionized organic molecules (like alcohols, https://www.solubilityofthings.com/water/alcohols) decreases markedly with more than 4 carbons. Linear alkanoic acids, liquids up to nonanoic, are more polar, but drop off in solubility at hexanoic (https://en.wikipedia.org/wiki/Hexanoic_acid) as melting point increases.

The question i'm having trouble with is:

Would the dissociated form be more or less likely to dissolve in
    water?

Is this basically asking if the ions on the right are more likely to dissolve when compared to $\ce{C6H5CO2H}$?

Yes, that is the gist of that question.
It basically doesn’t only touch benzoic acid or medium-chain fatty acids which are both much better soluble than the corresponding aldehydes or alcohols. It is also leading the way to things like the structure of amino acids. Take glycine. In organic solvents, it is best described as $\ce{H2N-CH2-COOH}$ while in water it assumes the form $\ce{^{+}H3N-CH2-COO-}$.


text:
How does a reaction exhibit kinetics with a non-integer reaction order?
What does it mean when the order of reaction isn't an integer, e.g., a reaction order of 2.43. Not with something acting as a catalyst but rather as a inhibitor. Why does the inhibitor produce an unusual reaction order? Specifically fluoride ions slowing down the reaction between calcium carbonate and acid by forming calcium fluoride.

Never heard of the phrase "decimal point orders."  Reacting calcium carbonate with hydrofluoric acid would be slow since calcium fluoride is insoluble in aqueous solution. Hence the calcium fluoride would coat the calcium carbonate.

I guess I meant a fractional order with respect to an inhibitor? I've heard of the term broken order reaction but I've no definite source to reference

Semi-related answer.

Can you provide a reference for the example reaction and data you are describing?


text:
Melting point of alkaline earth metal carbonates
The melting point of carbonates of group 2 elements is (From Wikipedia)

$\ce{BeCO3}$ : 54
$\ce{MgCO3}$ : 540
$\ce{CaCO3}$ : 1339
$\ce{SrCO3}$ : 1494
$\ce{BaCO3}$ : 811

A similar trend is followed by alkali metal carbonates (from Wikipedia). But in the inorganic chemistry book by Shriver and Atkins, this is given

Is there a difference between decomposition temperature and melting point? Or is the values given in the book wrong?

Melting and decomposition are different things, as you guessed in your question.
Melting:
Something can melt, and then re-solidify, undergoing a physical but not chemical change.
A simple example would be ice, but it is also true for many chemicals. For example tert-butanol is a solid at room temperature (at least in a well air-conditioned lab), but melts readily at slightly elevated temperatures without any degradation. Dimethylsulfoxide is another example of a chemical which often solidifies in a cool lab, but can be melted back down with only the heat of your hand.
Decomposition:
Decomposition is a different thing, whereby the material undergoes a chemical change (usually in addition to physical change). This is often seen in laboratory settings when trying to measure the melting point of molecules with high melting points, such that at the temperatures needed to melt the solid, side reactions are often observed, destroying the material. For this reason, in the literature, you will see melting points reported as  mp. (decomp.), i.e. the temperature at which your molecule decomposed rather than melted.

So barium carbonate melts before decomposing?

The data you present certainly suggests so, which is entirely possible. You could try looking up the sources cited by Wikipedia and Atkins' to find the conditions used to measure the melting and decomposition points, as you can only really compare values if they were measured under the same conditions (for instance, if they were both measured by heating in the open atmosphere, or under inert atmosphere, or under vacuum and extrapolated back to melting/decomposition values at atmospheric pressure).


text:
Enthalpies of formation of alkali metal halides
Shriver and Atkins's Inorganic Chemistry has this graph:

How can I explain the different direction of the trends for fluorides and chlorides? I know that the decreasing lattice energy (as M goes from Li to Cs) is one factor that needs to be taken into account, but the trend in lattice energy should be similar for all the metals.

The equation for the formation of $\ce{MX(s)}$ from its constituent elements is:
$$\ce{M(s) + 1/2 X2(g) -> MX (s)}$$
In general, $\ce{X2}$ could be liquid ($\ce{X} = \ce{Br}$) or solid ($\ce{X} = \ce{I}$). However, since your question is asking about the fluorides and chlorides, we can just limit it to the gaseous state. The principles that will be described can be applied just as easily to the bromides and iodides without any problem.
For any question that asks about trends in thermodynamics, the best way to go about solving it is always to draw a cycle. This will allow you to identify which terms contribute to the change in the trend, and from there rationalise the observed trend. So, without further ado:
$$\large\require{AMScd}\begin{CD}
\ce{M (s) + 1/2 X2 (g)} @>\color{blue}{\Delta_\mathrm{f}H(\ce{MX})}>> \ce{MX (s)} \\
@V\color{red}{\Delta_\mathrm{atm}H(\ce{M})}VV  @AA\color{red}{-\Delta_\mathrm{L}H(\ce{MX})}A \\
\color{red}{\ce{M (g) + 1/2 X2 (g)}} @. \color{red}{\ce{M+ (g) + X- (g)}} \\
@V\color{red}{I_1(\ce{M})}VV @AA\color{red}{E_\mathrm{a}(\ce{X})}A \\
\color{red}{\ce{M+ (g) + e- + 1/2 X2 (g)}} @>\color{red}{\frac{1}{2}B(\ce{X-X})}>> \color{red}{\ce{M+ (g) + e- + X (g)}}
\end{CD}$$
By Hess's law, the enthalpy change for the blue path must be equal to the enthalpy change for the red path. This follows from the fact that enthalpy, $H$, is a state function. So:
$$\Delta_\mathrm{f}H(\ce{MX}) = \Delta_\mathrm{atm}H(\ce{M}) + I_1(\ce{M}) + \frac{1}{2}B(\ce{X-X}) + E_\mathrm{a}(\ce{X}) - \Delta_\mathrm{L}H(\ce{MX})$$
Note the sign convention I am using for the electron affinity and lattice enthalpy (these seem to be defined differently in different places). The lattice enthalpy used here is always positive (endothermic).
Clearly, the terms $\Delta_\mathrm{atm}H(\ce{M})$ and $I_1(\ce{M})$ only depend on the identity of the metal $\ce{M}$. Therefore, it is helpful to lump them together and consider them as one term, called $\Delta H_\ce{M}$. Similarly, the next two terms only depend on the halide $\ce{X}$ and we can collect them together and call it another term, $\Delta H_\ce{X}$. It is by no means necessary to do this; however, it saves some clutter in the expressions:
$$\Delta H_\mathrm{tot} = \Delta H_\ce{M} + \Delta H_\ce{X} + \Delta H_\ce{MX}$$
with
$$\begin{align}
\Delta H_\mathrm{tot} &= \Delta_\mathrm{f}H(\ce{MX}) \\
\Delta H_\ce{M} &= \Delta_\mathrm{atm}H(\ce{M}) + I_1(\ce{M}) \\
\Delta H_\ce{X} &= \frac{1}{2}B(\ce{X-X}) + E_\mathrm{a}(\ce{X}) \\
\Delta H_\ce{MX} &= -\Delta_\mathrm{L}H(\ce{MX})
\end{align}$$

The fluorides
As shown in your graph, the trend for the metal fluorides is that going from $\ce{LiF}$ to $\ce{CsF}$, $-\Delta_\mathrm{f}H(\ce{MF})$ decreases, i.e. $\Delta_\mathrm{f}H(\ce{MF}) = \Delta H_\mathrm{tot}$ increases (it becomes less negative). Here, the halide ion $\ce{X-}$ is not changing, so $\Delta H_\ce{X}$ is a constant and we only need to consider the other two terms:

Going from $\ce{Li}$ to $\ce{Cs}$, the atomisation enthalpy decreases and the first ionisation energy decreases. That means that $\Delta H_\ce{M}$ decreases.
Going from $\ce{Li}$ to $\ce{Cs}$, the radius of the metal ion increases, and so the lattice enthalpy decreases. This means that $\Delta H_\ce{MX}$ (which is the negative of the lattice enthalpy) increases.

So, there are two opposing factors. The first factor says that $\Delta H_\mathrm{tot}$ should decrease, and the other says that it should increase. Since the observed trend is that $\Delta H_\mathrm{tot}$ increases, this means that the second factor is greater than the first. That is to say, the increase in $\Delta H_\ce{MX}$ outweighs the decrease in $\Delta H_\ce{M}$, leading to an overall increase in $\Delta H_\mathrm{tot}$.

The chlorides
The analysis here - in terms of the individual trends in $\Delta H_\ce{M}$ and $\Delta H_\ce{MX}$ - is exactly the same. However, this time, the overall trend is reversed: $\Delta H_\mathrm{tot}$ decreases. That can only mean that, when $\ce{X} = \ce{Cl}$, the decrease in $\Delta H_\ce{M}$ now outweighs the increase in $\Delta H_\ce{MX}$. Why is that so?
First, we note that $\Delta H_\ce{M}$ is entirely dependent on the identity of the metal ion. It is independent of the identity of the halide. That means that the decrease in $\Delta H_\ce{M}$ when $\ce{X} = \ce{F}$ must be exactly the same when $\ce{X} = \ce{Cl}$ - there is absolutely no reason for it to change.
Logically, since the decrease in $\Delta H_\ce{M}$ is not changing in magnitude, it must mean that the increase in $\Delta H_\ce{MX}$ must have become smaller in the chlorides. Otherwise, there is no way for the decrease in $\Delta H_\ce{M}$ to suddenly outweigh the increase in $\Delta H_\ce{MX}$. And we can explain this very easily using what we know about lattice energies, because that's exactly what $\Delta H_\ce{MX}$ is: a lattice energy (with a minus sign).
The lattice energy of an ionic salt can be estimated using the Kapustinskii equation. For a metal halide, this reduces to the equation;
$$\Delta_\mathrm{L}H \propto \frac{1}{r(\ce{M+}) + r(\ce{X-})}$$
Clearly, going from $\ce{LiX}$ to $\ce{CsX}$, $r(\ce{M+})$ increases and $\Delta_\mathrm{L}H$ decreases. The difference is that when $\ce{X} = \ce{F}$, $r(\ce{X-})$ is smaller than when $\ce{X} = \ce{Cl}$. Mathematically, his means that in the fluorides, the lattice enthalpy is more sensitive to changes in $r(\ce{M+})$. This makes sense - if we went the other way and took the limit of $r(\ce{X-}) \to \infty$, the lattice enthalpy would not depend on $r(\ce{M+})$ at all.
That means that the decrease in $\Delta_\mathrm{L}H$ - which is equivalent to the increase in $\Delta H_\ce{MX}$ - must be larger in the fluorides, and smaller in the chlorides. That is why in the chlorides, the trend in $\Delta H_\ce{M}$ overrules the trend in $\Delta H_\ce{MX}$, and the reversal of the overall trend is observed.

Ending words
How would you know all of this if you were not given the graph? Well, the truth is, without looking up values of thermochemical radii, ionisation enthalpies, etc., there is no real way to know. The fact that the overall trend is reversed in the chlorides means that the variations in $\Delta H_\ce{M}$ and $\Delta H_\ce{MX}$ must be pretty similar in magnitude. However, if you are given data and asked to rationalise it, you can definitely do so by constructing a diagram and relating the variation in the various energetic terms to variation in physical properties that you are familiar with (ionisation energy, ionic radii, and so on).


text:
Polarity of the phosphate ion?
I understand that phosphate is bonded covalently, however, it is characterized as a polyatomic ion. This is making me confused on whether it's polar or non-polar.

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http://chemistry.stackexchange.com/questions/8667/determining-the-lowest-rank-non-vanishing-multipole-moment

There are multiple different things to consider for phosphate $\ce{PO4^3-}$:

The bonds themselves. Each of the four $\ce{P-O}$ single bonds is polarised towards the oxygen. Each bond is polar.
The molecule as a whole. It is of near-perfect tetrahedral shape thus having the point group $T_\mathrm{d}$. This point group includes the inversion symmetry operation $i$. Therefore, macroscopic properties of a vectorial type (like the overall dipole moment which is a vector more or less pointing from one pole to the other) must, by definition, be $0$. Thus, the entire molecule is not a dipole.
The molecule’s total charge. It has an overall charge of $-3$. This charge is located on the surrounding oxygens but the entire thing can be simplified from far away to a ball with the charge in the centre, i.e. the phosphorus atom. We would consider this molecular ion a monopole.

Located entirely in centre? Really?

Symmetry, my dear @Mith; symmetry.

It' symmetrically on oxygens, however not exactly 3 - there's partial charge on P but definitely not such high.

@Mith I reworded. Better?

It's not monopole - if sth hasn't dipole moment it still can quadrupole moment

@Mithoron I admit defeat when it comes to any $n$-pole with $n > 2$. Mind noting that in an answer of your’s so I can learn? ;)

A poly-atomic ion is a group of atoms that act as a single ion, it has no bearing on the bonding with-in the molecule.  The oxygen-phosphorous bonds are covalent but polar, the molecule its-self is charged, but lacks a direction across the molecule of the charge making phosphate non-polar.  The molecule its-self is free to bond covalent or ionicly depending on what it reacts with.


text:
Which preposition to use for the reducing agent?
Which expression is correct when we use the prepositions with or by for indicating that the reduction is carried out using $\ce{NaBH4}$ as a reductor:

It is reduced to sodium phenylselenolate with $\ce{NaBH4}$ in alkaline ethanol.

or:

It is reduced to sodium phenylselenolate by $\ce{NaBH4}$ in alkaline ethanol.

Hi and welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site. I improved the formatting of your post; among others by including MathJax. For more information on how to do so yourself, check out the [help], this meta-post or this one.

Very funny, but there's  SE site for learning English, you know?

@Mithoron Well yes, but this is a rather specialistic chemical context. It could be that normal language requires preposition $x$ while chemistry speak requires preposition $y$. Thus I feel that it is OK to post the question here.

Either is fine.  In the most literal interpretation of the statement I would say by (as it does the actual action of reduction), however I personally would likely use with, to a scientist there is little difference though.  If you were filing a patent, the term might might matter a bit more, which a quick survey of the patent literature suggests with.  Again though this is minutia and nit-picky.
I would suggest phrasing your sentence more as:

...is reduced to sodium phenylselenolate with a solution of NaBH4 in alkaline ethanol .

perfect, you are right , thanks for these clear explanations @Jan,@A.K

I don't think they're exchangeable, and you give the right example. "By" leaves no doubt about who the acting party is. "A is reduced by B", or "A is reduced by addition of B". If you say "A is reduced with B", I'm inclined to ask if there is also a "C". ;-)


text:
Unpaired electrons
Am I right in saying this: If there is an unpaired electron, then there is no neutral charge in that atom and a negative charge is given since the electron doesn't cancel out, since there is no pair. So does that negative atom carry that unpaired negative charge? And does that atom/element, like cobalt attract things because of the negative unpaired electron and how it attracts a proton from a different atom? Someone please make it clear to me.

Welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site. Visit the [help] for any unanswered questions remaining. Your first statement is incorrect, btw. A lot of atoms and even stable molecules have unpaired electrons: $\ce{O2, NO, NO2, \dots}$

Pairs of electrons do not cancel out each others charge. They are both negative charged. The electron charge in atoms is compensated with the protons of the atom. So your statement is completely false.

Pairing is about spin not charge.

Your statement is unfortunately wrong, as the electrons all carry a negative charge (that you could say is balanced by the positive charge of the protons in the atom nucleus). Single atoms of an element are neutral but can have unpaired electrons.
Atoms with paired charges can be negatively charged, most common ions like Cl- have no unpaired electrons and the addition of the electron that formed the final pair added the negative charge.
Electron pairing is about spin, electrons with opposite spins pair together to form a much more stable arrangement releasing energy.


text:
Which chemical should I use to wash soot precipitate from cast iron?
The soot deposition is quite strong (almost like graphite) and my iron object has a complicated shape with many grooves.
Which chemical would dissolve the carbon without oxidizing the iron?

There is not practical ways to do this. Technically, carbon dioxide at higher temperatures should work. However, simply blowing it in oxidizing flame should not significantly oxidize cast iron. Possibly, bathing it in molten sodium nitrate may work, though likely give the iron strange look.

Depends on your resources and how delicate the object is (is it an antique?).

You can put it in a furnace or self-cleaning oven to burn-off the soot. Determining the correct temperature for this will be an art.
You could put it in a furnace under a hydrogen atmosphere.  the hydrogen will react with the carbon to form hydrocarbon gases and it definitely won't oxidize the iron.  You can also use carbon dioxide which will for carbon monoxide with the soot.  You may not have this sort of equipment though.
You can clean the object in an ultrasonicating bath. These are widely available and the best part is that you will not require chemicals to remove the soot.  This method may take a while but it is the most gentle.

For #2, what should be the temperature in the furnace with a hydrogen atmosphere?


text:
What is the relationship between something magnetic and unpaired electrons?
I don't get how some elements are magnetic. Can someone please explain the relationship between something magnetic and unpaired electrons?
Why is cobalt magnetic?

Pretty sure hw questions need to have an "attempt" so I'll give you some suggestions to look at and maybe you could incorporate them in your question so we can answer it. You might want to look at the concepts of "ferromagnetism", "paramagnetism" and "diamagnetism". It has to do with the electron spin. After reading it, you'll probably get it. Cobalt is considered ferromagnetic. Also, have you seen a trend in magnetism (hint hint, d and p). Good luck!

There are five forms of magnetic behavior. Ferromagnetism is when a material is attracted to a magnetic field and can spontaneously hold a magnetic field alone (the typical form of magnetism).  Paramagnetism is when a material is attracted to a magnetic field but cannot spontaneously hold a magnetic field alone (the attraction may be too weak to observe practically).  Diamagnetism is a material that expels magnetic fields (essentially non-magnetic). there is also ferrimagnetic and anti-ferromagnetic but though these are interesting they not incredibly relevant to the answer.
Electrons form the magnetic fields of materials.  If a metal's atom have unpaired electrons the metal can attract a magnetic field.  If all of the electrons are paired in opposite spin orbitals then the spins cancel each other out and no response to a magnetic field is observed.
Cobalt has unpaired electrons making it at minimum demonstrate paramagnetic behavior. It is however able to hold a magnetic field making it ferromagnetic.  Most metals have unpaired electron and are paramagnetic (for example: aluminum, zirconium, titanium, magnesium and potassium)  Ferromagnetic metals become paramagnetic at elevated temperatures and inversely many paramagnetic metals become ferromagnetic at low temperatures.


text:
Cis and trans substituents on rings different stereodescriptors
The attached two compounds are diastereomers. They can be differentiated by cis/trans descriptors. Is there another way of describing them as cis/trans seems to be outdated?

EDIT: How to determine the r/s configuration of the pseudochiral C atoms in this case? The usual rule is that R precedes S, but here two identical groups are bounded to the same pseudochiral atom. How to assign r/s in this case?

Related: IUPAC name for 1,2,3-trichlorocyclopropane?

Edit: strongly related to your previous question How to determine the configuration of C atom bounded to two groups with same configuration? and Yolo123's question that was suggested to you Number of diastereomer pairs of 1,3-dichloro-1,2,3-triphenylpropane where the nomenclature was explained in detail by Jerrepierre. The same situation applies here.

Although the cis/trans nomenclature might seem 'outdated', it is still by far the easiest way to describe disubstituted ring systems, and is commonplace in both conversation between organic chemists, and in the organic literature when such relationships are introduced into a molecule.
As these ring systems possess some symmetry, the (R) and (S) descriptions commonly used are not applicable, but (r) and (s) may be assigned to carbons ('pseudo asymmetric'), based on the IUPAC 2013 recommendations for describing achiral cyclic compounds.
I've included the assignments for the two molecules you presented in the image below. In essence (taken from March's Organic Chemistry):

The small letters used for the psuedoasymmetric center are assigned using established rules. [...] The descriptors “r” and “s” are used to denote such centers; they are assigned in accordance with Sequence Rule 5, taking into consideration that “(R)” has precedence over “(S)” in the order of priority.

This additional example, taken directly from the IUPAC blue book 2013 may clarify the reasoning behind assignment of (r) and (s):

Its worth noting that this nomenclature only applies when you can find symmetry, in many cases this is non-existent, and normal (R) and (S) nomenclature can be applied.

Yes.  I think, perhaps the resident IUPAC expert will correct me, but I think this situation arises because the molecules aren't chiral (their mirror images are imposable).So whereas if we had something (R,R) and flipped one of the centres we'd get (S,S), this isn't the case here because the two centres are basically labelled with respect to one another (i.e. the label given to one of them depends on the other one, and vice versa.). I added an additional example before I read your comment which may help clarify.

I don't understand how do you assign r or s (R precedes S).

I edited my question.

Could you perhaps edit to show how you've tried to assign r and s so I can see where you might be going wrong?

Never mind, I understood it now. I had problems with the phantom C atoms.


text:
Why don't electrons have the same spin even when in same orbit?
Do electrons not have the same spin because that is just how it was found or does it contribute to another factor?

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. This appears to be a homework question, please share your thoughts and attempts towards the solution. It is really hard to judge your background and how deep an answer would have to go. Please provide more information [edit]ing your post.

The Pauli Exclusion Principle says no two electrons can have the same quantum numbers, and there are roughly two ways to interpret this.
First, an 'orbital' may host two electrons so long as they have opposite spins.
Second, an 'orbital' may only host one electron, but there is a second orbital that is geometrically and locationally identical which hosts an electron of the opposite spin of the first orbital.
Why?  It is written into the fabric of the universe.  That is how the math works out.  Experiments have strongly supported Quantum Mechanics.


text:
Reversablity of dehydration
Aqueous magnesium bicarbonate when evaporated decomposes into magnesium carbonate. Is that reaction reversible by dissolving the carbonate in soda water?

What was that carbolic supposed to mean?

Isn't carbolic the old way of saying phenol (or phenoxide in this case, possibly).

yes should read carbonic acid

Yes, in fact this process is similar to the processes in lime stone that form cave stalagmite and stalactites but with calcium carbonates.
$$\ce{M(HCO3)2 <-> MCO3 + H2O + CO2 }$$
$$\ce{M} = (\ce{Mg},\ce{Ca})$$
Though I should note the process in not called dehydration.


text:
What chemical compounds used in qualitative chemistry can be distinguished by smell?
I will be competing in a contest and one part of it is qualitative analysis. And a fully legit way to help yourself during it is smelling compounds (to just get a theory to prove, you have to give a proof mixing given compounds, smelling is just to help, it is no proof). Oh, and I have a good nose, I think that much better than people usually have.
I mean by that I can quite easily distinguish solutions of e.g.:

Iodium (in $\ce{KI}$, 0.05 mol/l)
$\ce{Fe^3+}$ jones (0.05 mol/l)
Thymol blue (0,02 mol/l)
Sodium thiosulfate (0,05 mol/l)

or distinguish sugar from salt (kitchen ones).
And I have a access to most of the compounds used in analytical chemistry.
And my question is to write a list of compounds that should be able to be distinguished by smell. I also mean the most obvious things like ammonia. I don't mean just water solutions. I would really appreciate you being specific like "Acetate salts' water solutions, smelling like acetate acid" rather than "Acetate(s)". Thanks.

This is a terrible idea, chemists stopped using smell and taste as a means of chemical analysis many years ago.
Of course, when working with compounds, you occasionally catch a smell of something (DMS being an unpleasant example), but purposefully smelling compounds is dangerous and potentially fatal if you're presented with some unknown compound  for which you haven't looked up the safety data form.
Even simple chemicals that you list, such as ammonia and acetic acid will quite happily destroy the linings of your nose and respiratory tract, leaving you unable to smell for the near future. In more severe examples, the chemicals could leave you with permanent damage.
There are many many tools available for qualitative and quantitive analysis in the lab today, make use of them.


text:
How to calculate the heat released when sodium hydroxide is dissolved in hydrochloric acid solution?
How to calculate the heat released when sodium hydroxide is dissolved in hydrochloric acid solution? Here is the data I gathered from a lab experiment:

$50\ \mathrm{mL}$ of hydrochloric acid solution
$2.00\ \mathrm g$ of sodium hydroxide (in solid form)
Initial temperature is $24\ \mathrm{^\circ C}$
Final temperature is $44\ \mathrm{^\circ C}$

The chemical equation of the reaction is:
$$\ce{NaOH(s) +H+(aq) + Cl- (aq) -> Na+(aq) +Cl- (aq) + H2O(l)}$$
This is the ONLY information I can use and I cannot search up anything online.
I calculated:
$$
\begin{align}
q&=mc\Delta T\\
q&=\left(50\ \mathrm g\right)\left(4.18\ \mathrm{J/(g\ ^\circ C)}\right)\left(20\ \mathrm{^\circ C}\right)\\
q&=4180\ \mathrm J
\end{align}
$$
So calculating the heat released by the reaction, I assumed that the mass is $50\ \mathrm g$ since hydrochloric acid is a solution of hydrogen chloride in WATER and the reaction produces water as well. Therefore, since the density of water is $1\ \mathrm{g/mL}$, $50\ \mathrm{mL}$ must be approximately $50\ \mathrm g$. Also, the specific heat capacity must be similar to water as well which would be about $4.18\ \mathrm{J/(g\ ^\circ C)}$.
So my question is: Is my assumption correct? If not, why?

I am pretty sure in reality your assumptions are valid. However at a school level, this is the correct and expected way to do this question. But some teachers may insist that you use 52 grams instead of 50 grams since you also add 2 grams of sodium hydroxide. But that depends on the teacher.

If that's all the information you have (no concentration or mass or density for the HCl solution, no mass for the final solution), and you weren't instructed to obtain any of that information, then what you did (assuming 50g and heat capacity of water) is ok for an introductory lab class.
There will be many sources of error:
The density of HCl solution is greater than water, so the initial solution will be more than 50g.
The heat capacity of HCl solution is less than that of water.
Without knowing the concentration of the initial solution, it is unknown if the final solution is NaCl solution, a solution of HCl and NaCl, or a solution of NaCl and NaOH.
The heat capacity of the final solution will be less than that of water.
The final solution will be somewhat more than 52g.
Just dissolving NaOH in water is exothermic. It can't be assumed that all the NaOH reacts without knowing the concentration of HCl.

This answer would have been very useful if it came earlier, but thanks anyways for the useful information. I will take note of it for future lab experiments.


text:
Find the velocity constant K given the speed

This reaction is of first order in [A]. Consider that

What is the velocity constant $K$ here?

So let's see... The velocity equation is
$$V = K \cdot [A]^n\cdot[B]^m$$
Assuming that it is bimolecular... But... we don't know if it is bimolecular! All they tell me is that there is some reaction with a molecule $A$ involved.
Anyway, they also tell me that the reaction is of first order... on [A]. It is my understanding that this is called a partial order, and the global order is not necessarily the same, as it would depend on the involvement of any other molecule (which we don't know about). Therefore, it is not necessarily true that the equation of velocity is
$$V = K \cdot [A]^n\cdot[B]^m$$
So how do I even approach this kind of problem?
I can't even calculate the general velocity of the reaction, because I don't know the coefficients of the reagents. Heck, I don't even know how many reagents and products are there!
(I know this is an extremely simple exercise and the answer is also quite basic, but I just fail to find an example of exactly this scenario anywhere).

You are really confusing yourself. The problem is very simple. First of all, remove the fact that the rate of reaction depends on all reactants from your mind. It is not the case. Sometimes, even though multiple reactants are involved, only some of them actually matter when we need to find the rate. Let me explain it without chemistry.
Consider a factory where a car is made. The first level makes the engine in 1 hour. The second level makes the tyres in 1 hour. The third level makes the rest of the parts in 1 hour. The fourth level puts them all together in 1 day. Now, even though all the 4 levels are involved in the process, it is the slowest level (level IV) that determines the rate of the process because the rest of the levels can keep sending the parts, but the finished car won't exit the factory unless level IV puts them all together. This is the case with chemical reactions too! The slowest step decides the order and rate of the reaction.
Now, back to the problem. Since they have given that the reaction follows first order kinetics in $[A]$, the rate law can be written as $R=k[A]$. After a little bit of integration and other stuff, we have the equation, $k=\frac{ln(2)}{t _{1/2}}$ , where $t _{1/2}$ is the half-life, i.e. time taken for concentration (actually, it's no. of molecules) of reactant to become half its original value. In this problem, $t _{1/2}$=10.0 seconds. Substitute and find the answer.

It's a first order reaction in $\ce{[A]}$:
$$-\frac{d\ce{[A]}}{dt}=k\ce{[A]}$$
By integrating the above equation, we get:
$$\ln\frac{\ce{[A_0]}}{\ce{[A]}}=kt$$
Where $\ce{[A_0]}$ is the concentration  of $\ce{[A]}$ at $t=0$
So, you plot $(\ln\frac{\ce{[A_0]}}{\ce{[A]}})$ as function of $t$. You get a line which passes through the origin, with a slope equals to $k$.

Another good approach is to use the method I described here: Finding [A] when looking for the velocity constant
Try it and see what you think.  It doesn't give the exact answer, but, for this set of data, it gives an excellent approximation.


text:
How to find out pH of aquarium and apply it when adding HCl?
I'm trying to rescue a neglected fish tank. I've been trying to figure out how much extra hydrochloric acid to add to my water change for my fish tank to soften the water a little and reduce the pH a little, so it peaks at about 7.6 at the end of the day. Though I don't I want the KH to drop below about 3 dKH. Also note that I don't want to change the pH by more than 0.1 per water change.
I would like to know how to measure/calculate/estimate the pH and how to apply that knowledge to find out the resulting balance between pH and carbonate hardness of the fish tank after some of its buffer has been used. See my spreadsheet for my calculations so far.
To make things easier, let us assume that

cows are spherical
there is only pure water, $\ce{CaCO3}$, and $\ce{HCl}$ being mixed.
dissolved oxygen and $\ce{CO2}$ levels equalise with the atmosphere
other chemicals in the tank can be used if it is convenient.
the label on the pool acid is accurate even after being opened a few times.
my measuring is accurate.

All of the info below is for additional context. My main question is: How do I calculate the balance between the $\ce{CaCO3}$ dissolved in water and the resulting KH and pH balance when adding $\ce{HCl}$?
Edit: lets be specific and simplified and start with

a 9L bucket of tap water
with 3dKH (0.0005349585373 mol/L) $\ce{CaCO3}$
and starts with a pH of 7.2 or 0.00000006309573445 mol/L of H+
and add 1ml of HCl at 1.7 mol/L

Does that mean that for the bucket of water before adding the acid
$${
pKa = 7.2 - log_{10}(\frac{0.0005349585373}{0.00000006309573445}) = 3.271679877
}$$
If so, how can this info be used to predict how much of the $\ce{CaCO3}$ is used when the HCl is added and what the new pH will be in the bucket after equilibrium is reached? Nevermind the significant digit issue.

So far the things I have and can measure are

API freshwter master test kit to test for pH, Ammonia, Nitrite, and Nitrate.
API KH&GH test kit to measure degrees carbonate and general hardness.
pool acid with 370g/L $\ce{HCl}$ on its label ( 10.15 mole/L )
Tap water with

pH 7.2
3 dKH
4 dGH

science products aquarium science chlorine neutralizer
Fish Tank

Tank volume 55L
18 months old
filter is sponge and mini K1 bio media before a 700 L per hour pump going to a spray bar at the top of the tank, that looks a lot like this one
substrate is shovel of dirt from back yard covered with aquarium sand
some wood pieces that have been in there for about 3 months.
planted with java swords and java moss
17 pacific blue eyed fish
only using mechanical and biological filter. No charcoal in use.
pH 7.8 at the end of the day. Possibly as low as 7.6 in the morning before the grow lights turn on. (Nitrification of ammonia reduces pH, and photosynthesis increases pH)
5 dKH
20 dGH
Ammonia less than 0.25 ppm
Nitrite less than 0.2 5ppm
Nitrate 5 to 20 ppm depending on how I hold the test to the colour chart

Water change size is 1 bucket (9L)

I've mixed 50 ml of the pool acid in to 250 ml of tap water to make it non-fuming to be easier to work with, and am calling this "diluted" acid. So far I've only been adding 1 to 2 ml of the diluted acid to the water change, because there are fish in the tank and I want to take it slow. I use a 1 ml glass eye dropper with 0.25 ml markings on it to add the diluted acid to the bucket of water.
Seeing that this amount of added acid is not changing much, I've been trying to figure out how much more I need to add to have the desired effect. After doing some initial calculations in a spreadsheet I see that 2 ml of my diluted acid is not enough to use up the buffer in the 9 L bucket of tap water let alone have much of an effect on the fish tank.
My problem is figuring out the balance of amount of carbonate buffer remaining and the resulting pH level.
this CE question and wiki seems to be most of the info I need, but I seem to be missing some info to fill in the equations.
Update for even more context: For the benefit of those who are concerned about the procedure I am following and for those who want even more detail, this is what I do when the water change includes acid:

Test the tank water for pH and KH morning and evening to determine the current min and max values. Remember fish, bacteria and photosynthesis all affect pH. Also test for ammonia, nitrate, and when in the mood also test nitrites and GH. Though since this is a well established tank I often skip the nitrite test.
start filling the 9L bucket with tap water.
(sometimes) test pH, KH, and GH of water fresh out of the tap. But I'm lucky and my tap water is fairly consistent quality. Though I suppose this means they treat the heck out of it.
while filling the bucket I add science products aquarium science chlorine neutralizer according to the directions on the bottle.
while still filling the bucket I have been adding 1 to 2ml of the diluted (1.7mol/L) HCl with a glass eye dropper with 0.25ml markings with about 5mm between each marking.
(most times) once the bucket is full, I test the bucket water for pH, and KH
I leave the bucket to rest for at least a few hours, or overnight to give it time to reach equilibrium.
Once the bucket has rested, I test the bucket water for pH, KH, and sometimes GH. This is a very important step, that I take extra care to perform correctly.
if the bucket water pH is below 6.2 I would discard that water and start again. I would consider a pH of 6 ok except this is the minimum reading on my test kit, and a pH below 6, even well below 6, is indistinguishable from a pH of 6 with my test kit.
if the fresh water is safe, I siphon enough water and muck out of the tank so 9L can be added (yes this is a 2nd bucket), and tip that into the garden bed.
slowly pour the bucket of fresh water into the tank. This takes at least 3 minutes because if I pour any faster I disturb the sand substrate too much or am in danger of dislodging one of the plants. This is due to the downward flow pattern of pouring. By this time the filter pump has pushed almost 35L through itself (700L per hour), maintaining a good stirring rate in the tank.
test tank water for pH, and KH.
wait a few hours and test pH and KH again.
the following day test pH and KH morning and night

Because this is a fairly labour intensive process I don't add acid every time I do a water change. When not adding acid in the water change I skip nearly all of the steps except the chlorine neutraliser.

You really need someone with significant aquarium experience rather than a chemist. Perhaps the Biology section would be a better fit since you don't want to kill the fish or plants. // It isn't clear how you're measuring the pH nor what you want the final pH to be.

While I think this is an excellent researched question, I also fear that you might not reach the audience you need here. If you'd like to have your question migrated to maybe [biology.se], let me know.

Stop! Don't add $\ce{HCl}$ to the aquarium with live fish in it! I don't know whether your jurisdiction prohibits cruel treatment of animals, but... well, please, just don't do that.

@IvanNeretin do you have any info on why I should not add HCl to my water change water?

@MaxW I've updated the Q to have the extra info. Also I'm not after a step by step instruction manual on fish keeping. I realise the simplifications I've specified will result in the calculated result will be different than if I actually followed the formula's advice. My next few water changes will likely be 2.5ml, then 2.75ml, then 3ml to actually measure reality with all of its complexity. Unless Ivan can convince me that I shouldn't.

@Martin-マーチン no thank you. I've purposefully specified some significant simplifications to make this a suitable question for a chemist. The extra info of this being for a fish tank is just for context.

@BeowulfNode42 Because it's way too harsh. It's like heating an apartment (with people inside) using a flamethrower. Also, it's because of the nature of typical titration curve (and remember, your cows are not quite spherical, so you don't really know that buffer capacity). You add some acid, nothing changes, you add some more, and more, and then suddenly one extra portion drives it all the way into acidic pH and kills everything inside.

@IvanNeretin That is why you add the acid to the water change bucket, mix, measure, wait and then measure again the water in the bucket before adding to the tank. So it is like using a flame thrower to heat a bucket of water with a thermometer in it so you can put the bucket in a room with people in it. If there is a problem with the bucket of water you either fix it or discard it, and don't add it to the tank/room until it measures ok.

Do you take out a 9L bucket, discard that water, and and 9L of fresh water from tap? Is tap water chlorinated?

@MaxW Yes I remove water from the tank first then add 9L of fresh water from the tap. Yes it is chlorinated. Yes I use a chlorine neutraliser. Yes I have updated the question to show that.

I think you're removing 9 liters of tank water, and replacing it with 9 liters of unchlorinated tap water.
So:
(1) Remove 9 liters of water from tank into bucket.
(2) add enough HCl to get 9L of water in bucket to the desired pH.
(3) Discard the water in bucket removed from tank.
(4) Fill bucket with tap water & adjust that to desired pH.
(5) Add 5 times amount of acid used in step 2 to bucket of water from (4). (Bucket is 9 liters, but tank is 55. 55-9= 46. 46/9=5.1).
(6) Add water from bucket back to tank 1 liter at a time giving tank a couple of minutes or so to mix.
The first time that you do this it would be prudent to check pH after 3 liters, 6 liters and the whole bucket has been added just to be sure that the tank is acting as predicted.
Notes:
(A) The fish, plants and gunk at the bottom of the aquarium aren't figured into how much acid to add. So the pH after a couple of hours (a day?) will be probably be less than the the pH change observed in the bucket in step 2.
(B) I'd guess that one the pH gets shifted to the desired point that less acid would maintain that level, and that the acid needed would be fairly constant.
(C) I'm not sure how often you remove 9 liters of water from tank. If you do it once a week, then you might want to spread adding acid over several days. So take 2 liters of water out of tank and add as much acid as used in step 1, then dump that water back into the tank. Do acid addition 5 times over a week, measuring pH each time before and after 10 minutes.
(D) record all data about measurements to develop a record of how your tank behaves. Don't rely on memory.

Good idea about testing the tank water throughout the addition of the fresh water with acid. Though I was actually looking for some chemistry equations to fix up my spreadsheet calculations, not fish keeping advice. Since a pH of 7.8 is still within the realms of ok, I am prepared to wait a week or more between acid additions even if the desired results take months, so adding acid between water changes is not required. ps. it is a 55L tank and 55-9=46L that the tank is drained down to during the water change, not 36L.

@BeowulfNode42 - I don't think that you can really use any equations to extrapolate from the current tank pH. I really think you need to titrate the "sample" from the tank (9 liters of water to be discarded) and extrapolate from that. Note (D) was really anticipating that you'd eventually get to the desired pH and that the HCl additions would become fairly constant. In any kind of science experiment a good scientist writes stuff down... // PS - I'll fix tank size.

Tank size fixed. // I'm a chemist and don't know anything about aquariums. The chemistry that I outlined is sound. The extra pH checks were thrown in since I'm a nervous Nellie and didn't want to give advice that kills your fish. I'd really think that trying to change pH 0.1 units at a time would be within the tolerance of the fish. How much acid the soil in the tank bottom will neutralize is the big mystery.

Re-reading your answer I see that it should provide a way of achieving the desired results, but there are a couple of issues. 1) It still doesn't teach me any of the formulas involved or how to apply them. Which is what my question is about. 2) It also still leaves me guessing, with no idea of a starting point, of how much HCl to add to the bucket of tank water and will leave me testing over and over again. This wouldn't be such an issue if I had a pH meter that provided a continuous readout.

@BeowulfNode42 (1) I'm not sure that any "formulas" apply. I'm not sure what gunk is in the water or at what concentration. (2) To cut down pH tests on bucket of water removed you could add 1 ml acid and test, then 2 ml and test, then 4 ml and test, and so on.  If you change the pH in the bucket by 0.1 pH units, then 5 times as much acid should give about a 0.1 pH change in tank. If you change the pH in the bucket by 0.2 pH units, then 2.5 times as much acid should give about a 0.1 pH change in tank. (pH change isn't really linear, but for such a small change it should almost be...)

Nevermind the tank or what is in the bottom, or even what strange things are in the tap water. Can we stick to the simplifying assumptions and start with the bucket of fresh water? Surely there is some way to at least estimate what would happen in an ideal situation, perhaps with a table of known properties for the chemicals involved to help fill in any blanks.

@BeowulfNode42 - (1) The tap water should be fairly consistent. So if it takes 3 ml HCl to get to pH 7 (or whatever desired pH) then it should always take that much. In the tap water carbonate is likely the only anion.. (2) For the tank, I'd guess at least urea, ammonia, and carbonate. But there is also algae, bacteria and who knows what else floating around in the tank. // All in all I'd expect that typically there would be "consistency" in both the tank and the tap water. So if 8 ml of acid changed the 9L bucket of tank water by 0.1 pH unit, then 8 ml of acid should work the next time too.

If you need to check the pH of your aquarium, you can use pH test paper. Its results can be seen immediately, which is convenient for you to carry out the next step. Sincerely recommend this pH test paper 0-14https://www.hawachlab.com/ph-test-strips/universal-ph-test-strips/

You are seriously over-thinking it. Get some liquid indicator ( methyl blue ?). Measure the pH. Likely no correction will be needed. Natural waters generally have buffering. When I was a boy in college I calculated how much nitric I needed for my 50 gal ; put in 10 times the amount and hardly changed p
H ( Lake Michigan water). Only time pH can be a problem is with rain/distilled water . I did accidently kill fish with rain water . Otherwise fairly successful with aquariums for 70 years.


text:
Finding [A] when looking for the velocity constant

Some reaction of first order on $[A]$ has the data:

What is the velocity constant for this reaction? The units are
  $s^{-1}$.

Well, since the units are $s^{-1}$, then the reaction must be of global order one. Meaning that $A$ is the only reagent, yes?
Anyway, the velocity equation is
$$V = k \cdot [A]^1$$
Let's pick the first experiment...
$$\left (-\frac{0.40 - 1.60}{10} \right ) = k \cdot [A]$$
Huh. I don't actually know how to find $[A]$. Should I add $1.6 + 0.40$? Subtract them maybe? Or pick just one?
I tried all three of them and still couldn't get the answer. The options were

$0.030$
$3.1x10^{-3}$
$0.013$
$3.0$
$0.14$

dah... $\ce{[A]}$ is in the table. You're trying to solve for $k$.

@MaxW then it's none of the answers, as I mentioned.

You should read Wikipedia article https://en.wikipedia.org/wiki/Rate_equation Your use of "velocity constant" is weird terminology.

Actually you should really just read a textbook. The questions you have posted are very common examples and should be covered in pretty much any general chemistry textbook or books tailored for the IB / A level syllabus (or its equivalent in the US). Right now, I get the feeling that you do not understand the topic very well and I think you would probably help yourself more by reading the topic instead of trying to do questions.

For a first-order reaction, the half-life is constant and is given by the equation
$$t_{1/2} = \frac{\ln 2}{k}$$
In $10~\mathrm{s}$, two half-lives have passed (the concentration of $\ce{A}$ drops to a quarter), so $t_{1/2} = 5~\mathrm{s}$. Accordingly,
$$\begin{align}
k &= \frac{\ln 2}{5~\mathrm{s}} \\
&= 0.1386~\mathrm{s^{-1}}
\end{align}$$

A way to get an approximate value of k is to divide by the average value of [A] over the time interval.  In your example, it would be 1 (over the first time interval).  So you would get an approximate value of 0.12 for k.  This approximate value is closest to the 0.14 in your multiple choices.

If you use the second time interval, you would get $0.03$, which would be wrong. I imagine it was just a coincidence that the first interval worked.

No way.  if you use the 2nd time interval, you get 0.03/0.25 = 0.12, the exact same result.  The 0.25 is obtained as (0.4+0.1)/2


text:
Problems with a lithium battery
I have a lithium battery that is separated from any device and is producing a whitish-gray smoke every time the wires move. Sometimes it makes hissing noises.
Is there a way to solve these problems?

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. It seems to me that this battery is broken and should be disposed of in a secure manner. Other than that I cannot see how this question relates to chemistry, so it might get closed as off topic.

It may be a side reaction or short circuit

You probably want to drain its charge and put it into recycling.  It's probably a fire hazard at this point, and the damage that a fire from it could cause will cost more than a new one.


text:
What explains the very high density of osmium and iridium?
Iridium with a density of $22.56$ g cm-3 & osmium with $22.59$ g cm-3. Data from their respective Wikipedia pages.
Only these two naturally occurring elements have such a high density. I wish to know why. Also If possible the factors affecting the density of any/every elemental substance.

Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center.

Well, by definition, at least one element will be the densest. So you are asking why particular elements adopt a close-packed crystal structure (various reasons) and have heavy nuclei. I'm not sure that there is a 'simple' explanation for that.

The reasons behind osmium's density are explained here


text:
Bond strength by Similar size or Overlap?
Which of the sp³-sp³ or sp³-sp² forms a more stronger bond.. As we know that former one will make weaker bond as both are larger orbitals but in latter one.. One is bigger and other smaller.. So latter one would be a stronger bond..
But if we see by other thing tbat similar Orbitals make more effective bond.. Then sp3-sp2 should make a weaker bond.. So which of the two would actually make up for a stronger bond?

Possbily a dupe of http://chemistry.stackexchange.com/q/42212/7475


text:
First order differential equation for transmembrane diffusion
I want to model transmembrane diffusion using a first order reversible differential equation. The diffusion rate is known from literature. There are two volumes $V_{\mathrm{intra}}$ and $V_{\mathrm{extra}}$. The former is $10^{12}$ times smaller than the latter. $c_{\mathrm{intra0}}=0~\mathrm{mol~L^{-1}}$, and $c_{\mathrm{extra0}}=x~\mathrm{mol~L^{-1}}$, with $x>0$.
Thus far I am using this:
$$\frac{\mathrm{d}c_{\mathrm{intra}}}{\mathrm{d}t}=k_{\mathrm{diff}}\cdot c_{\mathrm{extra}} - k_{\mathrm{diff}} \cdot c_{\mathrm{intra}}$$
However, this does not account for the fact, that $V_{\mathrm{extra}}$ is much larger than $V_{\mathrm{intra}}$. I expect $c_{\mathrm{extra}}$ to be virtually constant. So far the the simulation results in the equilibrium being at $c_{\mathrm{extra0}}/2$.
Can someone suggest a method to account for the volume discrepancy?

Edit:
From the outlined scenario above, I expect in REALITY, my $c_{\mathrm{extra}}$ to not noticeably change from its initial value due to the large difference between $V_{\mathrm{intra}}$ and $V_{\mathrm{extra}}$.
However, in the simulation I don't want to set $c_{\mathrm{extra}}$ as constant, because I want to apply a 'extracellular' decay rate later on. Basically, I want to be able to modulate the extracellular concentration over time and see how the concentration within the cell changes.
This is why I want to account for the volume. Maybe, one needs an entirely different approach for that?

It's hard to see how, from your differential equation, $c_{intra}$ at equilibrium could be $cextra0/2$.  At equilibrium, your equation tells us that, at equilibrium, $c_{intra}=cextra0$.  I suspect an issue with the mathematics.

@ChesterMiller I never had a course on differential equations, so this might well be. My train of thought: Forward and reverse reaction rate are the same, and $c_{intra0}=0$, hence the equilibrium of $c_{extra}/2$.
This is also the result when simulating it in MATLAB

This is not the solution if $c_{extra}$ does not change from its initial value of $c_{extra0}$

Yes.  You need an entirely different approach for that.  You want to specify $c_{extra}(t)$ (i.e., as a function of time t) and you want of determine the response $c_{intra}(t)$ as a function of time t, correct?  We can provide the general solution for that if this is what you want.

@ChesterMiller Yes, exactly

The solution to your differential equation, subject to the prescribed initial conditions is $$c_{intra}=c_{extra0}(1-e^{-k_{diff}t})\tag{1}$$ under the constraint that $c_{extra}$ does not change from its initial value of $c_{extra0}$.  The true material balance constraint is $$V_{intra}c_{intra}+V_{extra}c_{extra}=V_{extra}c_{extra0}$$
which would yield a result virtually indistinguishable from Eqn. 1.
CASE WHERE $c_{extra}$ IS NOT CONSTANT
OK.  Here is the solution for the case where $c_{extra}$ is a function of time $c_{extra}(t)$, and $c_{intra}=c_{intra}(0)$ at time t = 0:
$$c_{intra}(t)=c_{intra}(0)e^{-k_{diff}t}+k_{diff}\int_0^t{e^{-k_{diff}\tau}c_{extra}(t-\tau)d\tau}\tag{2}$$
where $\tau$ is a dummy variable of integration and where $c_{extra}(t-\tau)$ represents the value of the concentration $c_{extra}$ at time $t-\tau$.  This solution reduces to the previous result if $c_{intra}(0)=0$ and $c_{extra}$ is constant.
So, to find the value of $c_{intra}$ at time t, you need to use Eqn. 2 and perform the indicated integration (usually numerically).

Agree. OP: use Bernoulli method to integrate your eq. analytically and obtain (1). At $t\rightarrow \infty $, $c_{intra}=c_{extra}=c_{extra0}$. The second eq. above is mole balance (no change of moles, because no reaction, change in T and P).

I think I didn't explain myself precisely enough. I don't want  $c_{extra}$ to be constrained to its initial value. I will update the original post so that I can hopefully clarify my question. Sorry for the inconvenience.

Did you see the addition I edited into my answer for the case in which  the external concentration is not constant?


text:
Calculating the physical properties of isotopes
I am trying to determine the theoretical physical properties (namely melting and boiling temperatures) of isotopic substances.  Harold Urey used Debye thermodynamics (which is quite dense) in his 1946 paper to calculate the physical properties of isotopic substances. Most of the equations appear to be for exchange rates of the isotopes within a given system.  I would like to know which equations are available to determine isotope effects on physical properties of isotopologues and the assumptions that go with them.

Besides being 'dense' what is the problem with Urey's work?

He doesn't explain the calculations used for calculating physical properties, but rather selective enrichment of isotopes in various compounds.

It seems you were rather asking about isotopologues not isotopomers.

@ mithoron You are correct.

The Debye frequency of a crystal is a theoretical maximum frequency of vibration for the atoms that make up the crystal.

It seems safe to assume melting will be occurring at the Debye frequency.
It is quite possible that the authors of your paper used the Lindemann model, which was published in 1910, in conjunction with Debye frequency, which is unique to each topoisomer, to predict the melting point using the following equation of Lindemann:
$$ T_m = \cfrac{4\pi^2 m \nu^2 c^2 a^2}{k_B} $$
$$m = \mathrm{atomic\ mass}$$ $$ v = \mathrm{frequency} $$   $$ k_B=\mathrm{Boltzmann\ constant}$$ $$T=\mathrm{temperature}\ \left(^\circ \mathrm K\right)$$ $$c = \mathrm{Lindemann\ constant}$$   $$a = \mathrm{atomic\ spacing}$$
Different diastereomers have different resonant frequencies which can be found through either Debye calculations or IR and Raman spectroscopy.
Substituting in the Debye expression for frequency (I copied this from Wikipedia; I didn't actually derive it myself) gives:
$$ T_m = \cfrac{2\pi m (\theta^2_D k_B)c^2 a^2}{\hbar^2} $$
Where $\theta^2_D$ is the Debye temperature and $\hbar$ is Planck's constant.
Another way of doing it is by using a hard sphere model in combination with Debye/Lindemann (PDF link).


text:
Visualizing PDB file in VMD, how to upload molecules?
Problem: When I load PDB-file of my protein of interest in VMD, nothing happens. The molecules is uploaded, but when I push "D" the screens with with big "VMD" letter freezes instead of presenting the structure of the protein in the PDB file. This is what is looks like:

Any Ideas what causes this? Should the PDB file be in a specific folder? I am a complete novice at this, so I wont understand very complicated answers.
I have downloaded the mac version of VMD version 1.9.2, I have Mac OS X version 10.11.2.  Here is the PDB file that I want to open.

I am getting it perfect in my windows version of VMD.

I have tried to uninstall the program and reinstall it, I do suspect its something wrong with my installation - I just can't figure out what, because I am just clicking on the download link then installing automatically like mac does for you - there is nothing I can really "mess-up". So I don't understand what is wrong. However - I will try some more, thanks.

Perhaps there is some issue with the "Gatekeeper" feature of newer versions of the Mac OS? I upgraded my desktop recently and VMD 1.9.2 now has some dialogs about "unknown developers" when it runs.

I have also experience some problems with other applications after installing the new version of Mac OS, so this is a likely cause of the problem. Perhaps time will resolve it.

I found a shortcut, and that is if I go to (click on):

file
New Molecule
Enter in the "browser" window the direct PDB accession code "3WD5" (for my protein) then VMD retrieved the protein structure form the PDB database on its own, and is showing it as it should in the visualization window.

However, this does still not resolve why it will not load a downloaded PDB file, but I suspect its most likely due to some problem with downloading of the files or installing the program its self that has gone wrong.


text:
Lithium compounds with good solubility in organic solvents?
I want to dissolve enriched $\ce{^6Li}$ into a liquid scintillator (made almost purely of di-isopropyl-naphthalene, mixed isomers, with a tiny bit of fluors and wavelength shifters) for enhanced neutron detection and background rejection. My group's previous attempts at this have been to use Lithium-6 Chloride with surfactants, but this negatively affects the attenuation length of the scintillation light passing through the scintillator (since the droplets of the microemulsion are essentially giant scattering surfaces).
Having had some O-Chem and P-Chem as an undergrad I know some organolithium compounds are soluble in organic solvents, but those are eat the container walls nasty. I know that lithium bromide has some limited solubility in organic solvents, but I need to get to at least 0.1%, preferably 0.5% atom doping/fraction, so I need something that dissolves well. I know that lithium perchlorate is soluble in many organic solvents, but the explosive nature of the mixtures it makes with those solvents kills that idea.
This leads to my question:
Are there any lithium compounds that are not caustic, nor explosive, but are soluble in greasy organic liquids (like di-isopropyl-naphthalene)?

LSC cocktails for the detection of neutrons usually contain boron-10 instead of lithium-6. Did you consider using such a cocktail?

We are looking at that as well, and 10B is nice because of the much larger neutron capture cross-sections and the greater ease of finding compounds that will dissolve. However, 10B, when it breaks up into a 7Li and 4He on capture, releases much less light than 6Li breaking up into a 3H and 4He. Therefore it is harder to separate the capture light from background. 10B is being examined, but since this is for a neutrino measurement, we need to make sure that we can separate the signal from noise well to give better precision. If you have a 10B compound suggestion it would be welcome.

I would suggest lithium iodide as it is very soluble in many polar organic solvents or a lithium alkyl carboxylate like lithium stearate if a non-polar solvent is required. Cyclopentadienyllithium would also be a good option for your specific solvent though more expensive.

How reactive is Cyclopentadienyllithium? Would it attack acryllic (what the container walls are made from), 2,5-Diphenyloxazole (fluor), or 1,4-bis(5-phenyloxazol-2-yl) benzene (wavelength shifter)?

I do not forsee any reaction of cyclopentadienyllithium with the aforementioned materials.

Also, the lithium stearate might be interesting, though with that atom fractions above 1.7% are impossible, but it is worth a look. Assuming it is not so large that when dissolved it acts as a scattering center.

Lithium Benzoate may be a viable option

Sodium and lithium Cp (cyclopentadienide) salts are very sensitive to oxygen, so I think that LiCp is a poor choice. Also I worry about it reacting with the chemicals in the LSC cocktail.

Under inert conditions you may try dissolving pure metallic lithium. Lithium forms naphtalenide salt , I don't see why di-isopropyl-naphthalene should be any different.
Another option would be use of 12-crawn-4, which make a lot of salts lipo-soluble. Presence of TMEDA (tetramethylethilendiamine) should affect solubility of many lithium salts as well, though I would focus on salts like benzoate or perchlorate.

How reactive is that naphtalenide salt? Also, I had read that lithium perchlorate is can form explosive mixtures with organic solvents, which would be a big problem in our lab.

It does react with air and water. However, I beleave, it may be prepared in a sealed vial by careful heating of lithium and a freshly distilled DIPN (beware, some gases may develop).  It does not explode, but may self-ignite on air. It is a very strong reducing agent| Yes, it can. But it is a lithium salt with very weak crystal lattice. Though I think, nitrate should work just as well.  But none of Li salts are likely to dissolve in DIPN without a helper , like TMEDA or DME

The napthelenide salt seems a bit too scary for this (we have certain conditions on what is in the detector cell because of where we want to put it) same for the perchlorate. However, the crown ether for solubility might be attractive, and the TMEDA may also be worth a look. Thanks for your suggestions.

Could you elaborate about temperature conditions?

The LS will sit at room temperature. So anything in it needs to be stable at that temperature, including not attacking the fluors (2,5-Diphenyloxazole), wavelength shifter (1,4-bis(5-phenyloxazol-2-yl) benzene) or walls of the container (acryllic).

Conside lithium benzoate, lithium croconate, lithium tosilate and lithium acetate. none of them is likely to dissolve without a helper. In case of aromatic solvent, you may need an aromatic helper, such as ortho-dimethoxybenzene

Actually, also consider https://en.wikipedia.org/wiki/Lithium_hexafluorophosphate in presence of DME

Again I am sure that the lithium compound suggetsed is too sensitive to oxygen and will be too reducing, the wavelength shifters in many LSC cocktails are things like PPO which is an oxazole. I worry that the lithium napthalide will reduce this thus stopping the LSC cocktail from working. I think that the crown ether would be a much better choice.

For a boron compound I would like to suggest that you either blend boric acid with glycerol or some other polyalcohol. Then I would add a little alkali (sodium hydroxide not potassium hydroxide). I would then mix this with a LSC cocktail designed for high ionic strength samples such as urine or sea water.


text:
hydrated halide structure
What is the structure of $\ce{LiCl.2H2O}$?
I think when $\ce{LiCl}$ is put in aqueous medium, it will break into $\ce{Li+}$ and $\ce{Cl-}$. Then $\ce{H2O}$ will give LP to Li and hydrate it. But why 2 $\ce{H2O}$ molecules? 2 means 2 coordinate bonds? Octet is not filled? Then why does this happen?

Crystal structures in general are not the kind of thing you can deduce through logic alone. There is no simple answer to "why are there $2\ce{H2O}$". Also, $\ce{LiCl}$ has other hydrates as well.

So it's like an ionic bond between Li+ ans Cl- and 2 coordinate bonds by water to Li+?

So we don't know and have no way of knowing what is it like, unless we look up the actual structure in a database.

Google: Water of crystallization

The radius ratio of $\ce{Li+}$ and $\ce{Cl-}$ is very small, this means the size of $\ce{Li+}$ is so small than $\ce{Cl-}$ that the crystal lattice of $\ce{LiCl}$ becomes unstable. So to make the size of $\ce{Li+}$ bigger $\ce{H2O}$ molecules surround the caption bigger and make the structure stable.