text:
Find number grams of the solute in a dissolution given its density and Molarity

Have a dissolution of $\ce{HNO_3}$ in water. The density is
  $1.42\ce{g/mL}$ with a concentration of $\ce{16M}$. What is the
  concentration in % mass/mass?

Alright. So the % mass/mass is expressed as
$$\frac{\text{grams of } \ce{HNO_3}}{\text{grams of the dissolution}}\cdot100$$
So I need to find the grams of $\ce{HNO_3}$ first. They tell me that the Molarity is
$$16 = \frac{\text{moles of } \ce{HNO_3}}{\text{liters of dissolution}}$$
$\ce{HNO_3}$ has a molar mass of $63$. So to find out how many moles of $\ce{HNO_3}$ are in the dissolution I need to divide the grams of $\ce{HNO_3}$ by $63$... But how am I supposed to find out the grams of $\ce{HNO_3}$?
I was told that the numerator is $63$. But... $63$ is the molar mass of $\ce{HNO_3}$, not the number of moles of $\ce{HNO_3}$ in the dissolution.
Anyway, let's suppose that it is indeed $63$. Therefore
$$16 = \frac{63}{\text{liters of dissolution}} \implies \text{liters of dissolution} = 3.9$$
So there are $3.9$ liters of dissolution? I'm not sure - I feel like I got this all wrong. How do you know there are $63$ moles of $\ce{HNO_3}$?

Using 1L of solution, you have 16M of HNO3, each mole of which weighs 63g. Therefore, in 1L you'd have 63 X 16 = 1008g. Mass of 1L is given as 1420g from the density.
%mass/mass = 1008/1420 = 70% HNO3

Ok, it all makes sense, but what I don't understand is why are you picking $1L$ of solution? Why one and not, say, five liters? Where/how do you get that number? In other words, how do you even know there is one liter.

1L is convenient. Use 2.8L, for the same calculation, HNO3 = 16 X 2.8 X 63 = 2822.4, mass liquid = 2800 X 1.42 = 3976, 2822.4/3976 = 70%, same value.

Well I meant the Molarity part. The Molarity formula in your case is $16 = \frac{63}{1}$, but if you change that $1$ by a $2.8$ instead of getting $1008$ grams you would get 360 grams of $\ce{HNO_3}$.

Actually, the formula is: Total Moles =  [concentration] X volume. So, 2.8L of 16M solution has 44.8 total moles, each with a mass of 63, which leads to a total mass of 2822.4. You have the concentration and volume, but you don't have the total moles, so that's why you are looking for.

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text:
Calculate the molar fraction of a mixture component given only its pressure

You have a mixture of $\ce{He, Ne, Ar}$ gases with a total pressure of $8.40$ atmospheres. $\ce{He}$ has a pressure of $1.50$ and $\ce{Ne}$ has $2.00$. What is the molar fraction of $\ce{Ar}$?

The molar fraction is
$$\frac{\text{moles of Ar}}{\text{total moles of mixture}}$$
However, to know the number of moles of any of these components, I need to know their mass in the mixture. How can I know if all I am given is their pressures?
The answer is $0.583$ and I am not sure how was this concluded. All I can tell is that $\ce{Ar}$ has a pressure of $4.9$ atmospheres by Dalton's law but that's it. By the same law I know that all gases there have the same temperature and volume, but since I don't know either value for any of them it isn't useful at all.

The molar fraction of a gas can be calculated as
$$
x_a=\frac{n_a}{n}= \frac{V_a}{V} = \frac{p_a}{p}
$$
In your case, firstly you need to calculate the partial pressure of Argon. Dalton's law states that in a mixture of non-reacting gases the total pressure exerted equals the sum of the partial pressures of the individual gases.
Thus $p = p_{He} + p_{Ne} + p_{Ar} $, from which you get $ p_{Ar} $ = 4,9 atm.
So the molar fraction of Argon is
$
x_a = \frac{p_a}{p} =\frac{4.90 atm}{8.40 atm} = 0,583
$

Hi! Ok, I understand that the pressure of $\ce{Ar}$ is $4.9$. However, my understanding is that the molar fraction is $\text{moles of Ar} / \text{total moles}$, so I need the number of moles for each substance in the mixture, not the pressures. Or perhaps the molar fraction can also be calculated by $\text{pressure of Ar} / \text{total pressure}$?

As you said it yourself, the molar fraction can also be calculated from pressure data. If I remember correctly, it is so beacuse of the combined gas law.

Can it be calculated with something else? Other than mole and pressure data?

Also from volume data. But in this exercice just from pressure data.

@midi-chlorian this is a homework question that should not to be answered completely as you did because this site is no homework portal. Instead the questioner should be guided to get to the answer themself.

@PH13 Thanks for the remark! I should have paid more attention to the guidelines before answering.

Your mole fraction is $$x_\ce{Ar}=\frac{n_\ce{Ar}}{n_\ce{He}+n_\ce{Ne}+n_\ce{Ar}}$$ and you somehow want to insert your pressures. You can try to convert all moles that you don't know into pressures that you know by using the ideal gas law $$p~V=n~R~T$$ which will make you find that everything you have (1.5, 2.0, 4.9 atm) is all what you need to come up with $x_\ce{Ar}=0.583$.

As your question was answered by midi-chlorian without further explanation, here is why you don't need more than those pressures given to you.
I said that you can use the ideal gas law to find out the amount of substance of each gas. Therefor we need to rearrange it to give $$n=\frac{p~V}{R~T}$$
We know the gas constant is constant ... that's where the name comes from ... and we can assume the temperature to be constant at whatever temperature we want ... it's not important to know the temperature here. On top of that the missing volume of each gas is equal for all three ideal gases, as Avogadro's law says.
This means that not only R and T are constant but that the volume will also not change, which in turn simplifies everything a lot.
$$n=p~\frac{V_m}{R~T}=p~k$$
Let's insert it into the mole fraction (as I told you to do) and see what happens.
$$\begin{align}
x_\ce{Ar}=&~\frac{n_\ce{Ar}}{n_\ce{He}+n_\ce{Ne}+n_\ce{Ar}}\\
=&~\frac{p_\ce{Ar}~k}{p_\ce{He}~k+p_\ce{Ne}~k+p_\ce{Ar}~k}\\=&~\frac{p_\ce{Ar}}{p_\ce{He}+p_\ce{Ne}+p_\ce{Ar}}\cdot\frac{k}{k}\\=&~\frac{p_\ce{Ar}}{p_\ce{He}+p_\ce{Ne}+p_\ce{Ar}}\end{align}$$
And this is the simple reason why you do not need more than your pressures.

But to use the ideal gas law I need the temperature and volume to be able to calculate the number of moles for each gas?

You don't. Just write it down and you'll see.

Hmm... let's try for Ar:
$$4.9 \cdot V = n \cdot 0.0821 \cdot T$$
then
$$\frac{4.9 \cdot V}{0.0821 \cdot T} = n$$
To find the number of moles $n$ I would need $V$ and $T$? (Clearly there's something I don't quite grasp about the usage of this formula sorry!)

Ideal gases do all have the same volume. R is a constant by definition and as the temperature will not change it is also a constant. Now don't enter any numbers, just write down what you get after rearranging the ideal gas law and inserting it into the mole fraction.

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text:
Does bleach (without rinsing) leave residues and can vinegar, baking soda neutralize them?
It's very difficult to get accurate information concerning household bleach residues. Imagine, for instance, that you pour pure (or diluted) bleach onto fabric or synthetic foam and let it dry without rinsing (as a scientific experiment). Some sources say: it will totally "vanish" (with air and/or light), leaving water and salt, and some others say it will leave some toxic compounds behind, such as sodium hydroxide (which is added to the bleach to slow its decomposition).
Question 1: Will the bleach only degrade into water and salt ($\ce{NaCl}$), or will it leave other substances? If so, which ones? If the first case is true (i.e., the degradation products are water and salt only), why does the solution still smell like bleach for a long time?
Question 2: If bleach (a base) does leave toxic residue(s), can we use vinegar (an acid) to neutralize it? $(~\ce{OH- + H+}$ should create $\ce{H2O}$)

White vinegar: I'm confused since there is contradictory information on Wikipedia. Some says vinegar will neutralize harmful residues from bleach, while another source says it "could result in much more caustic and dangerous chemicals, such as hypochlorous acid" being produced.
Will an antioxidant such as vitamin C work as neutralizer? (as suggested by Brian)

It seems that everybody agrees about the use of peroxide, bisulfite, metabisulfite or thiosulfite, and sodium thiosulfate to neutralize bleach residues, but I wonder if more common products such as vinegar or vitamin C can be used instead: What do you think about these two? Any other suggestions?
EDIT: I removed the washing example which was confusing since it implied rinsing + I updated my question since I thought bleach was an acid.
EDIT2:

Household bleach is, in general, a solution containing 3–8% sodium
  hypochlorite and 0.01–0.05% sodium hydroxide; the sodium hydroxide is
  used to slow the decomposition of sodium hypochlorite into sodium
  chloride and sodium chlorate. wiki

Brian and Drmoishe's answer suggest several ways to way to neutralized bleach but what about the sodium Hydroxide residues? (I check on a french bleach, it shows only on the composition: "3.7% of sodium hypochlorite", nothing else)

Question 1
No. Bleach requires something to react with in order for it to be consumed. There will still be bleach residues on the material.
Question 2
I think this question is actually based on a misperception, likely due to the multiple senses of the English word 'neutralize':

neutralize, verb
: to stop (someone or something) from being effective or harmful
: to cause (a chemical) to be neither an acid nor a base

While you are completely correct that adding acid to sodium hypochlorite would neutralize it in the second sense, you would in fact be doing the opposite of neutralizing it in the first sense. Your 'other source' is correct -- adding any acid to sodium hypochlorite will shift the equilibrium toward hypochlorous acid ($\ce{HOCl}$), which in turn equilibrates with molecular chlorine ($\ce{Cl_2}$), which is a nasty character you don't want to mess with$\dagger$.
What you need in order to neutralize it in the first sense, since bleach is an oxidizer, is a reducing agent.  My recommendation would be to, say, crush up a few dye-free vitamin C (ascorbic acid) tablets, dissolve in a bucket of water, and soak the article to be treated for half an hour or so.  Note that I've not ever actually tried this, so I have no idea if it will work.  It might also work to pour such a vitamin C solution into the 'liquid fabric softener' fill area of a washing machine, if it has one.
EDIT: I just came across this Chem.SE question from two years ago, which confirms that ascorbic acid is suitable for scrubbing of chlorine/hypochlorite.
$\dagger$ Note: Technically, since $\ce{Cl2}$ is volatile, soaking the material in an acid like vinegar would probably eventually result in all of the bleach/chlorine coming out of the material, making the actual answer to Question 1 a 'conditional yes.' However, this is a really bad idea from a health/safety perspective.

1/2 Thank you Brian to show me my misconceptions. I finally understood that bleach is a base, but not much more. I thought we could neutralized a base with an acid, but it seems unfortunately more complex than just a bath of OH- vs H+ swimming on it.Your Vitamin C suggest that introducing stuff that contain too much electrons (suc as VitC) would neutralized the bleach compounds which desperately need some. Is that (globally) correct?

2/2 The thing I find weird is that since bleach is a base, (conntains OH-), we should provide more positive charge to neutralized it, and thus acid (which contain H+), but you suggest vitamin C which will provide more electrons... Why?

@GuillaumeCombot See my edit. Acid/base and too-many/too-few electrons are mostly unrelated in general, despite the fact that they both can involve charged species.  You are generally correct -- bleach "has too few electrons" and thus wants to react with stuff that "has about the right number of electrons" or that "has too many electrons." That's what it means when I say that bleach is an oxidizing agent.

Any idea about the sodium Hydroxide residues? (cf edit2)

@GuillaumeCombot There shouldn't be any $\ce{NaOH}$ residues, just $\ce{NaCl}$.  The oxygen in $\ce{ClO-}$ will be transferred to the reducing agent in some fashion in the course of the 'neutralizing' reaction.

After a thorough wash, rinse and dry cycle, little or no bleach should be left in the fabric, but there is possible concern about the effect of chlorine bleach on organic compounds. However, if disinfection is needed, then the infinitesimally small risk of VOC* is outweighed by the elimination of microorganisms.
If any remaining smell of chlorine bothers you, an extra water rinse should be helpful. @Brian suggests a reducing agent, and $\ce{Na2S2O3}$, sodium thiosulfate, is inexpensive and is commonly used to remove residual chlorine for aquaria. However, leaving garments to air-dry or airing them for a day would be as effective.
(*) "Volatile organic compounds"

Ah, thiosulfate, good idea.

EDIT: I removed the washing example which was confusing since it implied rinsing + I updated my question since I thought bleach was an acid.

@drMoishe Thanks for your edit!  Any idea about the sodium Hydroxide residues? (cf edit2)

The concentration of any residue should be minimal, as only tiny amounts of chlorine or thiosulfate are present.  There would probably be more trace minerals in tap water than the reaction products after a final rinse.

@DrMoishePippik  Ok, but the (edited) question specify "no rinsing"...

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text:
A compound that absorbs all visible light
Is there a compound that absorbs all visible light?

Substances which absorb almost all the light falling on them appear black. Therefore you are looking for the blackest known compound.
The record is currently held by Vantablack[1], a substance composed of vertically aligned carbon nanotubes, which absorbs up to 99.965% of visible light incident upon it. As you can see it is really black.

Previously, the record was held by Superblack[2,3], a nickel-phosphorus alloy with an absorbance of up to 99.65%.
For comparison, regular black paint generally absorbs around 97.5% of visible light.

Sources
[1] https://en.wikipedia.org/wiki/Vantablack
[2] https://www.newscientist.com/article/dn3356-mini-craters-key-to-blackest-ever-black
[3] http://pubs.rsc.org/en/Content/ArticleLanding/2002/JM/b204483h#!divAbstract

Making a box with inside walls covered by this material and with small hole. This hole will be almost 100% darkness.

All metals are capable of absorbing photons of any wavelength below hard ultraviolet, as ideally there are allowed electronic transitions of arbitrarily small energy between states in the unfilled valence band. This means metals are theoretically the materials with the widest wavelength range for photon absorption (except for plasmas).
However, this comes to odds with the idea most people have of metals, which is a collection of shiny (highly reflective) grey solids. This is because while they can absorb photons of almost arbitrary wavelength, the absorption probability is not as high as in other materials. As such, a secondary effect can take place where the incoming photons interact with the free electrons in the metal and cause the light to be reflected back, rather than absorbed. One simple way to decrease reflection and hence turn pure metals into black substances is to grind them into a very fine dust or otherwise make a very rough surface on the microscopic level, which causes incident light to bounce around more times when hitting the material, increasing the odds of the light being absorbed.
Furthermore, while we often think of metals as materials which are composed chiefly of atoms from the $s$/$d$/$f$/early $p$ blocks of elements from the periodic table (the "metallic elements"), metals are actually not defined by their composition at all, but by their electronic structure. It is even possible to have metals made out of "non-metals" such as pure carbon, for example (carbon nanotubes of certain chiralities). Thus metallic character is actually likely a contributing factor to the darkness of Vantablack, which bon mentions in his answer.

Good point for mentioning powders. A lot of things appear black if you powder them finely enough.

Any black compound absorbs photons in all the visible spectrum; that is why essentially it appears black to our eyes. So, for instance, iron(II,III) oxide, $\ce{Fe3O4}$ will do so; it is even used as a black pigment. It is very unlikely though that it will absorb each and every photon of any wavelength in the visible spectrum, but the search for such a perfect absorber is meaningless in any case since the very notion of a visible spectrum is not strict.


text:
Getting 3,4-dimethylhex-3-ene from but-2-ene
How can one get 3,4-dimethylhex-3-ene from but-2-ene ?
$\hskip2in$
Using $\ce{Br_2}$ then two equivalents of $\ce{CH_3CH_2MgBr}$, I can get the corresponding alkane (3,4-dimethylhexane) but I don't see how to do it while preserving the double bond...

Are you looking for a synthetic route that produces the trans isomer?  That's the isomer pictured in your diagram.

Convert 2-butene to 2-butanone via hydroboration and oxidation. Couple 2-butanone using the McMurry reaction (TiCl3; Zn/Cu). See this --> https://chemistry.stackexchange.com/questions/108388/ring-contraction-when-cyclohexene-oxide-is-treated-with-methyl-grignard/108402#108402

Your approach, which uses $\ce{Br2}$ to convert 2-butene into 2,3-dibromobutane and react that with ethylmagnesium bromide $\ce{CH3CH2MgBr}$ has two flaws:

This approach removes the alkene and there is no obvious way to get it back (there is a way)
More seriously, Grignard reagents are pretty terrible for $\text{S}_N 2$ reactions. They are very strong bases, so elimination can compete, but more importantly, they can undergo metathesis:

$$\ce{R-Br + R'-MgBr <=> R-MgBr + R'Br}$$
Assuming your approach worked to generate 3,4-dimethylhexane, you could carefully radically brominate it to produce 3,-bromo-3,4-dimethylhexane, and I am sure you can get back to the alkene from there.
A better approach uses two molecules of 2-butene. Do you see how you can break your target molecule up into two 4-carbon pieces? Convert one into an electrophile and the other into a nucleophile. It might take two reactions in each case. If you choose your electrophile correctly (think about what makes a great combo with a Grignard reagents), you will form the $\ce{C-C}$ bond between the two pieces and you will still have a functional group left that you can use as a handle to direct the elimination reaction to regenerate the alkene.

Ben is right that your product is simple dimer of substrate. However, there is much easier approach to conduct dimersisation of butene - it happens easily if you use strongly acidic catalyst, like conc. sulfuric acid or ion-exchange resin, such reaction is used for example in production of isooctane. It's similar to cationic polymerization of alkenes, so parameters have to be carefully adjusted to reduce creation of heavier oligomers.

To follow up on Mithoron's answer and to answer this question, a journal article covers alkylation of isobutane. Its graphical abstract seems to suggest that isobutene is acidified to a tertiary carbocation. Another isobutene adds to that carbocation and combine themselves to be isooctane carbocation at C-4 location. A hydride transfer from a third isobutane molecule to the isooctane carbocation gives the isooctane.
If this analogy is possible imaginarily, then but-2-ene could be acidified to a butane secondary carbocation. Another but-2-ene adds to that carbocation and combine themselves to be 3,4-dimethylhexane secondary carbocation. Internal hydride shifts from the secondary to a tertiary carbocation. Elimination of or returning that acidic hydrogen to the stronger acidic catalyst creates that desired double bond at that location.
Comparing this imaginary case to that of the journal article, hydride shift is one of numerous pathways to product in that abstract but it must occur in this case.

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text:
I need to report MoS2 values from Mo and Sulfur numbers
I'm having a hard time figuring out how to do this and yes, I have the molecular weights, just don't know how to calculate. Been 20 years since college.
EDIT: I'm working with Copper ores. One customer reported Mo% as 5.2, S% as 5.3. The other customer reports $\ce{MoS2}$ values (this particular one is 12.34%) I need to compare these values.
EDIT #2::  I need to report 5.2% Mo and 5.3% Sulfur as ONE value - As "MoS2" if that is possible.

What values? What numbers? Please, clarify a little more the question.

I'm working with Copper ores...  One customer reported Mo% as 5.2, S% as 5.3.  The other customer reports MoS2 values (this particular one is 12.34%)  I need to compare these values.  Thanks

I presume that's % by mass?

And now you want to calculate which perventage of 12.34 % belongs to m% Mo and which percentage belongs to m% S ?

No.  I need to report 5.2% Mo and 5.3% Sulfur as ONE value in terms of MoS2, so that I can compare it to the value that was reported by someone else (The 12.34% MoS2)

Is Mo or S in any other compounds in the sample besides MoS2?

So you want to find the percentage of MoS2 in the sample from the values told by the first customer and check whether it is equal to the value told by second customer?

As the molecular mass of molybdenum is $95.95~\mathrm{g/mol} \approx 96~\mathrm{g/mol}$ and the molecular mass of sulfur is $32.06~\mathrm{g/mol} \approx 32~\mathrm{g/mol}$ one can construct a ratio between both molecular masses that in turn can be solved easily.
$$\frac{96~a}{32~b}=\frac{5.2}{5.3}=0.98\approx1$$
Which gives us after the multiplication with 32/96 on both sides
$$\frac{a}{b}=1\cdot\frac{32}{96}=\frac{1}{3}$$
So customer 1's determined "compound" is not $\ce{MoS2}$ but more $\ce{MoS3}$. Now I would guess that they don't analyze the copper ore for the compound $\ce{MoS2}$ but for both elements $\ce{Mo}$ and $\ce{S}$ separately which is why they came up with $5.2~\%~\ce{Mo}$ and $5.3~\%~\ce{S}$.
In my opinion, you should ask customer 2 to split up it's $12.34~\%~\ce{MoS2}$ into the percentages of each element and compare them with $5.2~\%$ and $5.3~\%$.
Using the ratio for $\ce{MoS2}$ which is $60:40$ one would estimate those $12.34~\%$ to split up into $7.40~\%~\ce{Mo}$ and $4.94~\%~\ce{S}$ or into $6.16~\%~\ce{Mo}$ and $6.18\%~\ce{S}$ if it is more like $\ce{MoS3}$.

Or to answer your question ... probably you can simply add both values to give you $10.5~\%$ which nonetheless will not be equal to $\ce{MoS2}$ based upon the data you are given!


text:
What is the correct nomenclature for diphenyl methane?
Can $\ce{Ph-CH2-Ph}$ be also be named as  as (phenyl methyl) benzene instead of diphenyl methane?
I came to this thinking that $\ce{CH2-Ph}$ can be considered as side chain at side chain.

You could also call it "benzylbenzene", and although I don't think that would be IUPAC approved, it does seem to be in common use.

@CurtF. According to the pubchem database, this is the preferred IUPAC name.

Whoah, thanks for the correction.  I guess I need to bone up on IUPAC rules...

The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-44.1.2.2 Systems composed of rings and chains (exclusive of linear phanes)
Two methods are recognized to name systems composed of rings and chains
  (exclusive of linear phanes).
(1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent. Rings and chains are chosen regardless of their degree of hydrogenation. As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons.
(2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.
(…) For selection of a preferred IUPAC name, see P-52.2.8.

P-52.2.8 Selection between a ring and a chain as parent hydride
Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name. In general nomenclature, a ring or a chain can be the parent hydride (see P-44.1.2.2).

Method (1): “1,1'-methylenedibenzene” (preferred IUPAC name; ring preferred to chain)
Method (2): “diphenylmethane” (treats phenyl groups alike as substituents)
(This example is is explicitly mentioned in the Blue Book.)


text:
Confusion between bases (not alkali) and other -OH containing compounds
Since some bases are not alkaline, we can't check whether they give $\ce{OH-}$ ions or not. Ethanol also has a $\ce{-OH}$ group. Why isn't it a base?

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text:
How can we balance one-half of an oxygen molecule, or any other diatomic, in a chemical reaction?
How can we balance oxygen or any other diatomic molecule with a fractional coefficient in a reaction?  I mean, if we can balance oxygen atoms in an odd number what happens to other oxygen atom, and can this sort of reaction even happen?

The use of fractions is irrelevant, as it's only the ratio that matters. If you don't like fractions you can multiply through by the denominator to get all whole numbers.

Whenever a chemical reaction is written with fractional coefficients, implicitly it's meant to be interpreted as the amount of substance in moles of each species that is involved.  So, it's not really one half of an oxygen molecule reacting, but 0.5 moles of $\ce{O2}$ reacting.
bon's comment is spot-on -- if you are trying to think about a reaction written this way on a molecular level, you need to multiply the reaction through by whatever constant eliminates all the fractions.

Brian you mean in molecular level it is impossible to write reaction in Fraction but if the coefficient numbers mean moles we can use Fraction is that true ? and THANK YOU SO SO MUCH  and please answer my question in Easy English because i am not native speaker

@Yasser Yes, that's exactly right. On the molecular level it makes little sense to write in fractions, but on the 'moles level' it's fine.

Please note that the proper term for "number of moles" is amount of substance. The former would be the same as referring to the mass as "number of kilograms".

@Martin-マーチン How would you rewrite? For the life of me I can't concisely frame how to express that the coefficients are to be multiplied into a one-mole quantity of each species.

Simply substitute number of moles with amount of substance (in moles of you really need to add that).

@Martin-マーチン Hm, indeed, that sounds okay to me. But one wouldn't say "amount of eggs in dozens," right?  I guess 'eggs' is a countable thing, though, while 'substance' as a quantity measure is an uncountable thing... which is probably where the difference comes in.

I actually would say that, or simply number of eggs in dozens, but we're drifting into linguistics here. In this very case, the quantity has a name, the unit has a name, these should be used at the appropriate places, no matter if it were different with other things.

@hBy2Py IUPAC also recommends enplethy for international usage. I have been trying to use it as of late due to brevity and to sound posh.

I see amount of substance as a definition that stress our current understanding. In particular it stresses the nature of matter as substance  , thus giving a physical and chemical identity and assume that matter to be a collection of things.  To date we do not have dark matter as substance. I found amount of matter really distinct from other quantities. It is just a consideration. It has little to do with the specific question


text:
Lemieux-Johnson oxidation on cycles
Consider 5-cyano-2,2-dimethylpentyl 4-methylbenzene-1-sulfonate:

What would it yield by a Lemieux-Johonson oxydation ($\ce{OsO4}$, $\ce{NaIO4}$)?
My first guess would have been that $\ce{OsO4}$ was likely to attack any of the bonds in the cycle (since they are delocalized); but would it stop here, or attack the remaining bonds? Or would something even different happen?

I doubt that it will attack the benzene ring in the same way that it attacks an alkene. The two do not behave much like each other as I'm sure you know.

Yeah, probably nothing would happen.

@Mithoron Something does happen though. I'll try to find the molecules  that were given after and update my question.

Can you oxidize the nitrile?

SciFinder gives zero hits for the compound. It also does not know any reactions from compounds found by ‘substructure search’ and $\ce{OsO4}$. However, if I had to guess something would happen to the nitrile (the only remotely non-inert functional group) since nucleophilic substitutions are not possible. Oh yeah, the Kürti-Czakó is also not helpful.


text:
Are concentrated salt solutions stable over time?
I have to use dilute salt solutions incredibly frequently, and it takes valuable time to make them from pure reagents every time I need them.
Are concentrated salt solutions fairly stable over time?
Does anyone know of any papers describing their ability to maintain static concentrations for long periods?
Edit:
To qualify, I am curious about a wide range of salts. NaCl, ammonium chloride, magnesium sulfate, calcium chloride, potassium phosphate etc...
I am also not especially concerned about analytical precision for my use, but understanding how this could be affected might be useful to others.

Do you mean 'table salt solutions' ?

If you mean sodium chloride in water, I don't think chemical stability is your problem, but rather analytical precision. Contamination and solvent evaporation could throw off your measurements, for example.

Can we mark this as "homework"? The question lacks any display of own thought.

Mark it however you wish. I have my own thoughts about the issue, but I felt sharing them would have guided answers towards my own confirmed suspicions rather than being objective.

Salt solutions generally do not spontaneously change in concentration. However, there are a few factors that could influence this: the first one is that the solvent (water, for example) could evaporate, increasing the concentration of the salt. You can prevent this by sealing your containers. The second one is that, as the temperature decreases, salts could crystallise out of the solution. However, this can be easily fixed by re-warming the solution.

There is more to it than that. Some salts would slowly hydrolyze over time. Some would react with oxygen or other components of air. Some would just up and decompose, even in a sealed container. Some (though none of the mentioned in the question, which are all inorganic) can even support primitive life.


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Is NADPH sensitive and unstable in UV light?
According to the MSDS for NADPH Tetrasodium Salt, it says to store it in a place away from heat and light. Does this include UV light or just visible light? I was told that a pure NADPH stock solution (with pH 7 phosphate buffer as solvent) could be re-used after absorbing UV light at 340 nm, but based on the MSDS it seems that shouldn't be the case. So my question is: could I reuse a sample of pure NADPH stock solution that was exposed to UV for my enzyme activity assay?
datasheets.scbt.com/sc-202725.pdf
https://www.interprise.com.br/produtos/download/90289.pdf

NADPH isn't destroyed by absorbing 340 nm light; if it were, then spectrophotometry wouldn't be a good technique to monitor its concentration.  The primary stability concern with NADPH and NADH is acid-catalyzed decomposition.  A great 1986 article summarizing this phenomenon is available here.
Some highlights from the article:

At 30 °C, the pseudo-first order rate constant for NADPH degradation in aqueous solution goes from 0.5 min$^{-1}$ at pH ~3 to $10^{-5}$ min$^{-1}$ at pH 10.  At pH 7 the rate constant is $10^{-3}$ min$^{-1}$.
The rate is elevated slightly by higher concentrations of phosphate buffer, about 10x higher for 1 molar (!) phosphate vs. extrapolation to infinite-dilution  phosphate.

A secondary pathway for NAD(P)H decomposition is oxidation by atmospheric air, but it is usually pretty minor compared to acid degradation.
So I don't think there will be a big problem with light destroying your NADPH.  You might be able to re-use it, but it depends how long you let it sit in solution, and at what temperature.


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Explanation of steric hindrance and existence of PI5
I have come across a following line in my textbook:

$\ce{PI5}$ does not exist due to steric hindrance.

I searched for the explanation of steric hindrance in Wikipedia but couldn't understand it. Can anyone explain steric hindrance in simpler language?
Wikipedia says:

The existence of $\ce{PI5}$ has been claimed intermittently since the
  early 1900s. The claim is disputed: "The pentaiodide does not exist
  (except perhaps as $\ce{PI3\cdot I2}$, but certainly not as
  $\ce{PI4+I-}$)".

What does this mean?

Interestingly, $\ce{PI4^+}$ does exist. So, the questions are 1)why phosphorous avoids forming more than four bonds (just like nitrogen) and 2) why $\ce{PI4^+}$ (and for that matter $\ce{PBr4^+}$) could be a rather strong oxidizer, strong enough to oxidize corresponding anion.

The simplest explanation is that you can not fit five iodine atoms around a phosphorous atom.
Consider the pentachloride equivalent - this is already quite crowded. Since chlorine atoms have a smaller covalent radius than iodine, you can get five of them around the phosphorous. Of course, all this depends on the bond length of P-I and P-Cl and the size of the phosphorous atom.
Interestingly, there seems to be some suggestion of an uranium pentaiodide compound, for example this PubChem compound. However, there are no references on that page, so presumably this has never been synthesised either?
In general, steric hindrance means geometric conformations of a molecule that are impossible (or difficult to achieve) due to clashes between groups and atoms within the molecule.

Uranium pentaiodide is also hypothetical.

Quite possibly (the reference on that page is from 1968!) however, there is a tantalum compound that has been prepared (although I don't quite see how the shared edge X2Y10 structures are pentagonal bipyramidal, but whatever).

Sterical hindrance does NOT prevent formation of $\ce{CI4}$ (and similarly crouded $\ce{C6I6}$ for that matter), so it is not sufficient to say that $\ce{PI5}$ is too hindered to form

@permeakra Admittedly, I should not have assumed that steric hindrance is truly the main reason for preventing PI5. Are you saying that the carbon-iodine bonds in CI4 are shorter than phosporous-iodine bonds? Presumably CI4 is also tetrahedral, which PI5 could not be.

Hmmm. The bonds in CI4 are 212 pm whereas in PI3 they are 234 pm, so perhaps you are right.

@gilleain both $\ce{CI4}$ and $\ce{C6I6}$ have short I-I contacts well below $\ce{I}$ van-der-waals radius. In fact it seems that in $\ce{C6I6}$ there is enough overlapping to fully delocalize two positive charges over iodines, forming a rather stable $\ce{[C6I6]^{2+}}$

@permeakra Interesting! My point was only that if steric effects were the reason for the non-existence of PI5 then an explanation could be the VDW overlap of the iodines preventing them packing in the required geometry. Basically, I was taking the textbook at its word - always dangerous :)

$\ce{PI5}$ versus $\ce{CI4}$
More is involved than steric hindrance  in preventing the formation of a $\ce{PI5}$ isomer with pentacoordinate phosphorus. There is a subtle antibonding interaction in this $\ce{PI5}$ isomer which does not occur in $\ce{CI4}$.
This answer describes the hidden antibonding ligand-ligand interactions in the "nonbonding" ligand-based orbitals of "expanded octet" molecules such as $\ce{SF6}$. The ligand-ligand antibonding interaction is small enough to minimize its impact on $\ce{SF6}$. But $\ce{OF6}$, where central atom is smaller and the ligands are closer together, is less fortunate.
Similarly in $\ce{PI5}$ there is one ligand-based orbital that becomes antibonding with the ligand-ligand interactions. Occupation of that iodine-based orbital combined with its crowding around the smaller phosphorus atom thus poses a combined electronic-steric hindrance.
With $\ce{CI4}$, however, there is no such ligand-based "hidden antibonding" orbital because there is no formal octet expansion, and without this electronic factor $\ce{CI4}$ can hold together (at least if it's chilled) despite the unfavorable steric condition.*
*This answer mentions the unusual deep red/purple color of this saturated compound. The relative weakness of the carbon-iodine bond, which moves normally ultraviolet electronic transitions into the visible range, explains the color.


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A compound that only absorbs UV-A radiation
As a similar idea to the question A compound that absorbs all visible light, I am curious if there is a compound (or compounds) that absorb UV-A radiation (shown below)?
However, preferable the compound would have relatively good transmission in the UV-B and Visible wavelengths.

Image source

Why would you want to avoid a tan but get burnt? ;)

Well, this question is related to observation of specific wavebands

Yes, there are many such compounds.  "The evolution of UVA protection" is a nice article detailing the progress on UVA blockers.  I'm sure you're aware that  sun care is a big business ($8.7 billion USD worldwide in 2013), so there is no shortage of chemists working in the area. In fact, there is enough (medical and clinical) focus that UVA (315-400 nm) has been subdivided into UVA1 (340-400 nm) and UVA2 (315-340 nm)
Here is a figure from the above link showing the UV spectra of some compounds.  The top part shows UVA absorbers and the bottom half shows broad-spectrum UV absorbers.

and a table with the names of the compounds

BMBM was an early UVA absorber (solid blue line), but was photochemically unstable (dashed blue line), in other words, its blocking power deteriorated with UV exposure time.  More recent UVA absorbers, like DHHB are photostable.
UVA absorbers belong to many different structural classes and contain many different types of functional groups.  Here are few examples.
BMBM

DHHB

TDSA


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Each electron shell can hold up to 2n² electrons. Is there a geometric reason for this?
It seems such a simple formula, that each electron shell can hold in principle two times a square number. I imagine there must be a geometrical reason for this, that's to do with the way electron orbitals "fit" around the nucleus. If this is so, what is it?

$2n^2$ rule is not always true; mind it!

When you say "geometrical reason" it makes me think "if only chemistry was so simple" haha. The reason is quantum mechanical in nature.

@orthocresol http://www.ncbi.nlm.nih.gov/pmc/articles/PMC219917/

@user5764 I don't think so. Can you back your claim up with some examples?

A good insight!
The number of electrons in each orbital are double the odd numbers. The s orbital can hold 2, p holds 6, d holds 10 and f has 14: double 1, 3, 5 and 7. If you add the odd numbers, you get a square number:
1 = 1
1 + 3 = 4
1 + 3 + 5 = 9
1 + 3 + 5 + 7 = 16
The doubling is due to the Pauli exclusion principle; in each orbital, an electron of a specific energy can be in one of two spin states.
It is interesting to see spectral lines split in a magnetic field (Zeeman effect) or an electrostatic field (Stark effect): one line may become 3, 5 or 7 lines. It clearly shows how orbital theory was derived.


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Molar enthalpy vs Enthalpy of formation?
(Note I've refocused a previous unanswered question into a more general question. You can find the original much more specific question here)
There seems to be two main ways of looking at the enthalpy of substances: molar enthalpy and heat of formation. What are the differences/advantages and disadvantages of each and are either one or more or less suitable for different purposes?

Molar enthalpy is the energy released when the substance undergoes combustion. It is essentially the energy required to fragment the substance into its component sub-compounds.
Heat of formation is the theoretical difference between the energy each atom typically has at STP and the total energy that the compound actually has. Essentially, it is the energy that goes into formulating the bonds between the atoms.
Molar enthalpy is generally used a lot more in experimental chemistry because it can be measured directly, and it gives useful values on the energy that can be extracted from a substance.
Heat of formation is preferred in theoretical chemistry, as it is the theoretical maximum amount of energy able to be extracted in perfect conditions.

This confuses me because for example the CHEMKIN thermodynamic data base (http://www3.nd.edu/~powers/ame.60636/kee2000.pdf) for many things that don't seem 'combustible' in that there is no real obvious 'sub-components' for them to break into.

For example, KCl, Na+, O+, O-, O etc.. Many of these things literally can't break down so I don't see how this can be right..

The document refers to these values as: 'Standard State Enthalpies and Entropies at 298 K.'

authors: Валера Сташевский


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Units in modified Arrhenius equation?
The modified Arrhenius equation is used to express the rate constant in a chemical mechanism model I'm working with. The equations is as follows:
$$k_\mathrm{f} = A\times T^b\times\exp\left(-\frac{E_\mathrm{a}}{RT}\right)$$
The paper states that "Units are Moles, cm3, Seconds, K, and Calories/Mole" so what would be the final units of the rate constant? I'm a bit confused due to the presence of the exponential function. When I simply plug the values in as given in the paper the rate becomes huge when multiplied by the molar concentration, is this because it gives molecules per second? To give you and idea for the following reaction:
$$\ce{C2H5OH + OH <=>C2H4OH + H2O}$$
The Arrhenius constants are as follows:
$$A = 1.74E+11$$
$$b = 0.27$$
$$E_a = 600.0$$
I am yet to calculate the reverse reaction using Gibbs Free Energy, is it just equally as large and thus it all cancels out or are the final units really not in $\mathrm{mol\ s^{-1}}$?

The exponential must be dimensionless.  That means if $E_a$ is in units of $\mathrm{cal\cdot mol^{-1}}$, then $R$ must be units of $\mathrm {cal\cdot mol^{-1}\cdot K^{-1}}$.  Make sure you use the right value of $R$ for these units; $R=8.314~ \mathrm{J \cdot mol^{-1}\cdot K^{-1}}$, but $R = 1.987~ \mathrm{cal \cdot mol^{-1}\cdot K^{-1}}$.  Temperature must obviously be in Kelvins.
The example reaction you gave is bimolecular, so I would think $k_f$ is meant to have units of $\mathrm{cm^3 \cdot s^{-1}\cdot mol^{-1}}$.  That way, when you multiply $k_f$ by the molar concentration of both reactants, e.g. ethanol and hydroxyl, you get a reaction rate that is in units of $\mathrm{mol \cdot s^{-1}\cdot cm^{-3}}$.
The remaining terms must therefore combine to give the right units for $k_f$. Since the units of $T^b$ will have units of $\mathrm K^b$, then the units for $A$ will be the units for $k_f$ divided by $\mathrm K^b$, i.e. the units of $A$ are $\mathrm{{cm}^3 \cdot s^{-1}\cdot mol^{-1} \cdot K^{\it-b}}$.

You additionally mentioned multiplying by the "molar" concentration, but the concentration in mol per cm^3 will be 1000-fold less than the concentration in mol per L, so if you really are using molar concentrations, make sure you divide them both by 1000 first.


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Can molecular structure be determined without modern characterization methods?
Molecular composition can be determined in some "primitive" ways (1, 2). What about the molecular structure? Is there a way to know what's bonded to what without spectroscopy etc.?
By "modern" I mean any method that emits and detects particles (photons, electrons, etc.).

Structure elucidation via degradative studies and decomposition chemistry has sadly all but passed from the chemists' toolbox, but this was the method once used to determine molecular structures up until about half a century ago. One of my very favourite examples that I use to teach structure determination by NMR is the story of strychnine. Strychnine (below) is by no means a trivial structure; it is hard enough to solve by 'modern' spectroscopic means. The story of how Robinson and Woodward, standing upon the shoulders of those before them, unravelled the the structure of this alkaloid over a decade and a half is an intriguing story, and one well worth investigating further. (AMBIX, Vol. 48, Part 3, November 2001)

Essentially, the structure is slowly pieced together through a series of degradation reactions. It is a methodology that relies on chemical processes, reactivities and properties only. For instance, it was worked out in the late 19th century that strychnine contained an amide as a lactam through a hydrolysis reaction to give strychnic acid. Were it not a lactam, two separate products would have been formed, rather than the one. Suggestions of the quinoline skeleton came from the reactivity of a dimethyl derivative in a series of chemical tests. Over the next 20 or so years, this was revised and superceded by more likely candidates. So on and so forth over a number of years, each little piece of the puzzle is fitted together, all through the lengthy process of chemical reactions to give smaller easier to identify products. A number of complicated framework structures were proposed, mulled over and discarded for a number of years, and final structure was tested through total synthesis reproduction. This whole process requires an incredibly talented chemist, and it simply makes my mind boggle at the thought of how these guys operated.
And though, it can be argued that UV spectroscopy did play a role in the final jigsaw piece in 1947, the correct structure was proposed by Robinson in 1942, and total synthesis confirmed this in 1954.
This is just one example of determination of molecular structure through degradative or decomposition chemistry, and many complicated structures were determined this way. Thank goodness for high field NMR.

Are you talking about mistakes in evidence or sth 'cause strychnine is lactam and there's no quinoline skeleton...

@Mithoron Sorry - a bit sloppy trying to summarise over 100 years of research and a few hundred papers in just two sentences! I have clarified above for you: the amide was known - it was shown to be a lactam through this reaction. The quinoline (or isoquinoline) skeleton was believed to be correct for many years, and the relative position of the two N atoms caused much debate.  And by quinoline/isoquinoline skeleton, it means a hydrogenated/substituted N-containing 6-membered ring fused to the aromatic ring.


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Does Fluoroborate gets used up in Electroplating
I've been working on Tin electroplating on the side for a while now. I've formulated a mix from my usual electroplating formula template. The mix stubbornly keeps on having the Tin ions form Tin Hydroxide with water. I've tried adding different organics and complexes in hopes of preventing this. But the Hydroxides keep on forming and so the mix does not provide enough Tin ions to allow electroplating at relatively low voltages (at least 16V DC is necessary).
So I finally relented and poured in my "Liquid Tin" http://www.mgchemicals.com/downloads/pdf/specsheets/421.pdf into the mix. Liquid Tin has Stannous Fluoroborate and its the Fluoroborate that keeps the Tin from forming Hydroxides. The effect was really good as plating happens at less than 1V, but then again Liquid Tin is for Electroless plating. After a while it needs higher voltage to electroplate.
So that got me wondering, does the Fluoroborate ion get used up in Electroplating? If there's a danger with Fluorine gas escaping, please also tell me.


text:
Why is sulfur the central atom in the Lewis structure for SO2?
So apparently the Lewis structure of $\ce{SO_2}$ is

It was my understanding that the central atom is the one that is more electronegative. And an atom is more electronegative the closer it is to Fluorine (top-right). Oxygen is definitely closer to fluorine than sulfur is. Then, why is sulfur the central atom?
Also, one thing that has been bugging me for a while: if there are two or more atoms for the central element, how do you make the Lewis structure?

can you give example for "two or more atoms for the central element"?

@Freddy I don't have an example, but I imagine there could be a molecule with two types of elements, and the least electronegative one happens to have two or more atoms. Or is such a thing impossible?

I don't know any such  molecule. But you can check out http://chemwiki.ucdavis.edu/Organic_Chemistry/Fundamentals/Lewis_Structures

@Freddy how about H2O2 (just found it in the book)? Oxygen is the central since Hydrogen can't ever be. But I got two oxygens. Apparently the structure looks like this: http://i.stack.imgur.com/Y39wf.png, but I don't get it: shouldn't the two hydrogens be connected to the same oxygen (because it is the central)?

Let's say left oxygen is A and right oxygen is B. Initially both oxygen(without hydrogen) would have 3 lone pair of electron. Then 1st hydrogen will attach lets say oxygen A. Now oxygen A will have 2 lone pair of electrons and oxygen B have 3 lone pair of electron. So naturally 2nd hydrogen will attract towards oxygen B and not to oxygen A.

Central atom seems like a poor descriptor in this case. Central to me has a profound geometrical aspect.

Being the central atom has nothing to do with electronegativity. Sometimes the central atom is the most electronegative, sometimes it is the least electronegative.

I think you messed up something. Because it is exactly opposite.
"Central part of atom is normally least(not the most) electronegative atom i.e. S in this case.

What about H2O ?

For polyatomic molecule least electronegative polyvalent atom acts as central atom and by using this we wlll get most plausible  structure  (structure having least magnitude of formal charge) for polyatomic molecule.


text:
How can I tell if a molecule is polar or not?

Consider $\ce{PCl3}$, $\ce{CH2Cl2}$, $\ce{HCN}$, and $\ce{SiO2}$. Which of these molecules are polar substances?

I appear to be lacking on the concept of "polar substance". I was just studying about Lewis structures and this question popped up in the book, but I'm not quite sure how is this related.
I think that a chemical bond is polar when one atom is more electronegative than another. But my basic understanding tells me that a chemical bond is non-polar only when the two atoms are the same (like $\ce{F-F}$ or $\ce{O-O}$). But all the above options have different atoms, so the answer would be "none", which is clearly wrong because that's not one of the options.
How can I tell if a molecule is polar or not?

Try drawing the three-dimensional structures of each molecule, including lone pairs.

Bond polarity and molecular polarity are different (though related) concepts. In short the molecular dipole moment is the vector sum of the individual bonds dipole moments, so you have to know the geometry of the molecule to check is it polar or not. See e.g. here.

Strongly related: Why is tetrafluoromethane non-polar and fluoroform polar?

Every sufficiently asymmetric molecule will be polar, but some more than others
The polarity of molecules is related to the polarity of bonds within the molecule, but just having polar bonds is not enough to create a polar molecule.
 Consider, for example, CCl4 and CHCl3. Carbon tetrachloride has 4 fairly polar bonds but they form a regular tetrahedron and the polarity of the individual bonds cancel each other out to leave a molecule which has no net dipole moment and is non-polar. But the chloroform molecule (which swaps a hydrogen for a chlorine) is fairly polar (~1.1 Debye–the unit used to measure dipole moments–usually abbreviated to D) because the symmetry is lower and the three C-Cl bonds now all point away from the C-H bond and their dipoles don't cancel out.
Similarly carbon dioxide is a linear molecule so the dipoles of the (fairly polar) C=O bonds point in exactly the opposite direction and so cancel out. Water, however is bent so the dipoles of the polar O-H bonds  do not cancel out leaving a polar molecule with a significant dipole moment (~1.8 D).
Saturated hydrocarbons are usually thought to be non-polar for most meaningful purposes. But those that lack symmetry still have measurable (but tiny) dipole moments. Propane $\ce {CH3CH2CH3} $ and isobutane $\ce {CH3CH(CH3)CH3} $, for example, both have small but measurable dipole moments (0.081 D and 0.132 D) so are, strictly, polar. In practice, though, these molecules have such low polarity that the effects of their net dipole are dominated by other non-polarity related intermolecular forces and can be though of as non-polar.
But the critical message is that the most fundamental feature that determines polarity is symmetry or the lack of it.
Also note that we are talking about discrete molecules here. One of your initial examples, $\ce {SiO2}$, is a network solid not a discrete molecule and the ideas won't apply.

Definition of dipole moment $$\vec{\mu}=q\times\vec{r}$$
where $q$ is the charge of the two atoms involved, $\vec{r}$ the distance between them.

It is a vector parallel to $\vec{r}$.
It origins in a separation of two equal in magnitud and opposite
charges,
Usually: bigger electronegativity difference ($\chi$) between atoms implies bigger charge ($q$) and so bigger $\mu$. Source.

But both $q$ and $r$ determine the $\mu$-value.
When there are many atoms, and many bonds ($b$) net dipole is obtained by summing up:
$$\mu_{net}=\sum_b\vec{\mu_b}=\sum_b\,q_b\times\vec{r_b}$$
Importantly, to sum up vectors we use bond angles between them. In consequence, molecular geometry is needed.
Molecule geometry
Lewis structures and $\texttt{VSEPR}$ are necessary to predict molecular geometries, and molecular geometry will indicate the angles between dipoles.
One could summarize the previous lines as: $\texttt{VSEPR}$ $\rightarrow$ molecular geometry $\rightarrow$ net dipole moment.
Let's write two examples:

Carbon dioxide

We can see dipoles are on the same line ($x$), and opposite direction. The sum is zero: $\vec{\mu}_{net}=\mu_x - \mu_x = 0$.

Water

In the $\ce{H2O}$ example, there is a net dipole. If angles are measured from $x$ axis, which bisects the bond angle, and contributions are added it yields:
$\vec{\mu}_{net}=2\,q\, r\times cos(52.25)\vec{i}$
Only components on $x$ axis survive. For a detailed explanation see this video.
Next step could be $\ce{SO2}$, which has a net dipole smaller than water. Same analysis gives zero dipole moment for $\ce{BF3}$.

Applications
See this post for simple applications of this knowledge in reality.
*Molecules were drawn with chemfig package.

To find answer to the question that you encountered you can follow these steps
Step 1 : 1st you have to understand what is hybridization and how to find hybridization of molecules , also each hybridization has different structure .
Step 2 : Once you have determined the structure successfully you can use the formula for calculating dipole moment
μ = q*d  and this formula shows dipole moment for a single bond .
A more generalised form :

Here are some examples which can help :

For multiple bonds use concept of vector addition/subtraction.
Remember that polar compounds means  the compounds that have a net dipole moment , but Polar Solvent means that the compound has high dielectric constant .

Hybridisation is not the cause for a molecular structure, it is, - if at all -, the result of it. Better yet, it is a mathematical model to explain orbitals within a molecule. Step 1 therefore does not work. I know that it has often been taught that way, but unfortunately this is wrong.

@Martin-マーチン Yes your are correct Hybridization is a hypothetical concept , but while solving problems. ,like mentioned , it is the easiest way to find whether a molecule has dipole moment or not , Its simply not possible to remember all structures and until the structure is known , we will not be able to find the net dipole moment .

It is wrong to use hybridisation to determine the structure; even if you think it works and is easy. It fails for hyper-coordinate compounds like $\ce{SF6}$, and everything (else) that involves predominantly ionic bonds. Simpler models to estimate structure exist, are equally (un)reliable, and don't rely on wrong theory. Also note that the model you are proposing fails for $\ce{CO}$.

@Martin-マーチン Yes There are many exception I cant diagree . But This is the way I would solve the question that is mentioned.

It has been disproved thoroughly that spd hybridisation is wrong. You are pushing a model that works sometimes by accident, but is wrong all the time. I know it is easy, but that doesn't make it correct, unfortunately. Your very example of $\ce{PCl3}$ is wrong. The bond angles are approximately 90°, there is no hybridisation in this compound. The simple model will betray you, try something more concise and not much more difficult like MO theory, or ligand field theory.

@Martin-マーチン PCl3 has suffieciently electronegative atom attached to it and therefore it is not a part of Dragos Rule , In PH3 or AsH3 where the electronegativity difference is not much , the atom fails to hybridize and therefore the angle is very much lower than 109.28 as in tetrahedral complexes . But PCl3 the hybridisation occurs and angle is about 107° due to lp bp repulsion

I see there is no reason arguing with you; this discussion is already too long. I just wanted to tell you - as a courtesy - why I down-voted your answer. Regarding Drago's rule, please see What is Drago's rule? Does it really exist? Unfortunately the lone pair repulsion is also a post rationalisation of the failing VSEPR model. The bond angle in $\ce{PCl3}$ is about 100°; which means there is some hybridisation, but far from sp³, too. Anyway, I respectfully bow out of this discussion.

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


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The partition coefficient for a chemical species across a membrane
What factors influence the partition factors across a membrane for a chemical species? Wikipedia defines the partition coefficient as:

In the physical sciences, a partition-coefficient (P) or distribution-coefficient (D) is the ratio of concentrations of a compound in a mixture of two immiscible phases at equilibrium.

If we let these "two immiscible phases" be either side of the membrane, In this situation the equilibrium concentrations would be determined by the solution either side of the membrane? If there is water either side of the membrane would they have a P of 1?
I'm not a chemist (or biologist for that matter) so apologies for any misuse of terminology or ambiguity.

Please note that the term partition coefficient is not recommended and should not be used as a synonym for partition constant $K_\mathrm{D}^ \circ$, partition ratio $K_\mathrm{D}$, or distribution ratio $D$.

@Loong Ok. For clarity I am referencing http://www.pharmpress.com/files/docs/MolBiopharmaceutics_sample.pdf page 6 (140 numbered), eq 3.2.6, partition coefficient defined as alpha.

For partitioning across a membrane as you have defined it, you need a system of a solute, two solvents, and a membrane where the solvents do not pass through the membrane and the solute does.

Ok, so either side of the membrane there exists 98% water, so we shall say both solvents are water, then my solute is a large molecule which can pass through the barrier? As it is defined in 3.2.6 in the attached book chapter how would this relate to what you describe?

Partitioning across a membrane is better known as Osmotic pressure. Think of a tube with a "bag" made out of the membrane attached to it. The more concentrated solution goes inside the bag, the less concentrated solution suspends the bag. The liquid from the suspending liquid then diffuses into the bag causes the fluid level to rise against gravity. The height of the rise yields the Osmotic pressure. So it is the solvent that diffuses not the solute.
If the solute diffuses through the membrane then it is just a diffusion problem such that the rate of diffusion is related to pore size. Bigger pores give faster diffusion.
To a chemist a "partition coefficient" would also really seem to apply best to a system at equilibrium. If you're considering drugs in the body then an equilibrium probably isn't obtained. The blood sweeps the drug away to react somewhere else in the body. So the drug is continually transported into the blood.
The other side of this is something like urea in urine and blood. The kidneys actually concentrate urea from the blood plasma into the urine. However this biological process consumes energy. So this sort of process is different than a simple equilibrium.


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What color is solid methane?
Is it blueish, perfect white or some other color? I am interested because some celestial bodies are said to be covered with methane ice.

Are you referring to solid methane or methane clathrates?

@bon to solid methane

Very interesting question, particularly with recent discoveries in mind.

The answer may have to do with light scattering rather than absorption, but I'm not completely sure.

This article appears to suggest that methane ice is colourless. Scientists haven't entirely figured out what color the ice would be, though they suspect it would be colorless, as it is on Earth

gaseous methane is colorless and a property like color depend on electron transfer between molecular or atomic orbitals and the energy of these orbitals is independent to temperature.so methane in solid state(low temperature and high pressure)is still colorless.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the [help]. I do not believe this is true. For example, carbon dioxide in the gas phase is colourless, but as a solid it is shiny white. Or consider water, in its solid state the colour rages from transparent to white to blue depending on the modification (pressure).

water  in solid form can be white but i didn't see blue ice. furthermore after solidification water become turbid instead of transparent. liquid water can pass light through itself  but when it become solid, solid water is turbid it is mean that light can't pass through it,  because ice can't absorb light it reflect all the visible wavelength and appear  white. i have same opinion about solid methane.

I was talking about the appearance, sorry to be imprecise. But here is what I was thinking about: https://en.wikipedia.org/wiki/Blue_ice_(glacial)

Check color of oxygen in different phases - perceived color depends strongly on concentration and in gas it's very low.

Your question is not very specific. So the answer is it depends. Think of water as an analog.

Methane "snow" would be white. (Due to scattering of light).
Solid methane under sufficient pressure would form a clear colorless solid.
Methane "snow" (which would be on surface" of some celestial body) could be any color if contaminated by even small amounts of other hydrocarbons. A reddish color seems to be typical. See ref: http://www.caltech.edu/news/astronomers-find-ice-and-possibly-methane-snow-white-distant-dwarf-planet-1714

Can you provide a reference for your claims?


text:
Selective nitro reduction of poly nitro compounds
Is there any way through which I could selectively reduce one $\ce{NO2}$ group from a dinitro- or trinitro- compound to an amine.
I was reading Solomons and Frhyle and it says that m-dinitrobenzene can be selectively reduced by using $\ce{H2S}$ in aqueous or alcoholic ammonia. It also says that when it is carried out with 2,4-dinitrotoluene it leads to reduction of the 4-nitro group.
So, is there any way to predict which nitro group will be reduced, and/or any mechanism related to this. I have been frantically searching in books and even an internet article would suffice.

Anyone interested in writing an answer for the "mechanism" part of this question may wish to review: HODGSON, H. H. (1946), The Apparent Anomalous Monoreduction of 2:4-Dinitrotoluene explained on the Theory of Hyperconjugation (The Baker-Nathan Effect). Journal of the Society of Dyers and Colourists, 62: 114–115. doi:10.1111/j.1478-4408.1946.tb02402.x AND H H Hodgson J Chem Soc 1944 75 (both cited by Porter in his paper page 457)

Could you give the page number in Solomons and Fryhle?

Is there any way to predict which nitro group will be reduced

After having gone through three pages of Google Scholar and reading several research papers looking for an answer, there is definitely no way to predict the product here for every general case. The most detailed paper on the topic of Zinin reduction by far is Porter, H. K. Organic Reactions 2011, 455–481 (link). Even this paper went as far as to say that (p458):

It is evident that much remains to be learned about the mechanism and that studies using modern techniques and instruments would be desirable.

However, it does give reasonably detailed examples of of selective reduction of substituted dinitrobenzene which are worth mentioning. To begin with:

The positions of the substituents play an important role in the progress of the reaction. ... In polynitro compounds, the relative positions of the groups determine whether reduction gives monoamino or polyamino derivatives and determine the isomer distribution of the monoamino derivatives.

For example: (p459):

In substituted dinitro- and trinitro-benzenes, the least hindered nitro group is preferentially reduced.

In dinitro- and trinitro-phenols and their ethers, a nitro group ortho to hydroxy or alkoxy is preferentially reduced.

This is the most you need to know about this reduction. The other details are more technical, and are attached below are interested readers.

Further detailing:
In the end of the paper, Porter has provided huge tables detailing several (substituted) mono-, di-, tri-nitrobenzene derivatives and their major products on Zinin reduction. So, I studied nearly a hundred different reductions through those tables, and tried to devise other reasonable ways to predict the major product than the two already mentioned above, especially in asymmetrical polynitrobenzene derivatives. I've not listed below mononitrobenzenes since they have only a single nitro group to be reduced. Further, symmetrical polynitrobenzenes (like 2,4,6-trinitromesitylene) have all their nitro groups equivalent, so the product obtained on reduction of either group is same. I've ignored biphenyls and napthalenes in my calculations to keep results simple. Further, exceptional products are not listed here, as they have already been detailed in the paper (p460-p464 under "Side reactions").
The following findings may be used to predict the major product in simple polynitrobenzene derivatives. Keep in mind that exceptional side-reactions do occur, so the prediction may not always be accurate. Yet, a substantially large list of predictable products is obtained as shown below. Examples of compounds are enclosed in parentheses.

The less sterically hindered $\ce{-NO2}$ is reduced in alkyl derivatives of polytnitrobenzene (tables 2 and 3).
Example: 5-t-butyl-2,4,6-trinitro-m-xylene yields 4-t-butyl-3,5-dinitro-2,6-xylidene i.e. a fancy way to say that the $\ce{-NO2}$ group para to the tert-butyl group got reduced.
When a hydroxy or alkoxy group is present, then the ortho nitro group is reduced, even in presence of other halogens on the ring.
(table 4: 2,4-dinitrophenol, 2-chloro-4,6-dinitrophenol, 4-bromo-2,6-dinitrophenol)
If both an alkyl substituent and a hydroxy/alkoxy group are present, and: (table 4)

a single nitro group is ortho to the hydroxy/alkoxy, then that ortho nitro group is reduced
(2,4-dinitroanisole, 2,4-dinitrophenetole, 2,3-dimethoxy-5,6-dinitrotoluene)
no nitro group is ortho/both nitro groups are ortho to the hydroxy/alkoxy group, then the least sterically nitro group hindered is reduced
(2,3-dimethoxy-5,6-dinitrotoluene; 1,2,4-trimethoxy-5,6-dinitrobenzene yields 3,4,6-trimethoxy-2-nitroaniline)

Table 5, 6, 7, 8, 10, 11 do not have any analyseable compounds.
Table 9 lists nitro derivatives of aniline.

preference in reduction is given to the nitro group ortho to the amino group.
(2,4-dinitroaniline, 2,6-dinitroaniline, 3,5-dinitroaniline)
even in presence of other alkoxy or alkyl substituents, the nitro group ortho to the amino group is only reduced.
(4,6-dinitro-o-toluidine, 4,6-dinitro-o-anisidine, 4,6-dinitro-2,5-xylidene)
the ortho nitro group is reduced even in case of N-alkyl anilines, though they are much more bulky.
(N-ethyl-2,4-dinitroaniline, N-methyl-2,4-dinitroaniline)

There are a few cases which I was unable to resolve. The author has not noted them as an exception either. Suggestions are welcome:

Table 4 states that 2,4-dinitroanisole yields 5-nitro-o-anisidine. It is not entirely clear to me why the expected 4-nitro-o-anisidine was not formed. The same yield is observed in 2,4-dinitrophenetole yieldings 5-nitro-o-phenetidine.
Table 7 gives the reduction of 2,4-dinitro-$\alpha$-toluenesulfonic acid as 2-amino-4-nitro-$\alpha$-toluenesulfonic acid, which is not clear why, given the steric constraints at the ortho position.
Table 9 states reduction of 3,5-dinitro-2,4-xylidene gives 2,4-dimethyl-5-nitro-m-phenylenediamine. It is not clear to me why the more sterically hindered nitro group is reduced. It isn't even ortho to the aniline group.

I believe this is a reasonably detailed analysis of the selective reduction of the polynitro compounds.

The Porter review cited claims four examples of isolating phenylhydroxylamines as possible intermediates in the Zinin reduction. My experience is that the reaction mixture is usually heated. The review does not reference Willstatter and Kubli, Chem. Ber., 1908, 41, 1936-40, who prepared the hydroxylamines under ambient conditions. It is unclear to me whether or not the hydroxylamines have been subjected to elevated temperatures to demonstrate their intermediacy in the Zinin procedure.

Source: Maity, Sunil K.; Pradhan, Narayan C.; Patwardhan, Anand V. "Reduction of p-nitrotoluene by aqueous ammonium sulfide: Anion exchange resin as a triphasic catalyst". Chemical Engineering Journal. 141 (1-3): 187–193. doi:10.1016/j.cej.2007.12.013.

The name of the reaction of reducing  di/trinitrobenzenes to di/triaminobenzene is called Zinin reduction. The reaction is carried out using sulfur anion (sulfide, hydrosulfide and polysulfides) or aqueous ammonium sulfide was used in this work which is basically hydrogen sulfide  in aqueous or alcoholic ammonia.
The basic equation of the reduction is given below:-

(a) Reduction by sulfide ions:
  $$\ce{4ArNO2 + 6S^{2−} + 7H2O → 4ArNH2 + 3S2O3^{2−} + 6OH- \label{a}\tag{1}}$$
  (b) Reduction by disulfide ions:
  $$\ce{ArNO2 + S2^{2−} + H2O → ArNH2 + S2O3^{2−} \label{b}\tag{2}}$$
  (c) Reduction by hydrosulfide ions:
  $$\ce{ArNO2 + 3HS− + H2O → ArNH2 + 3S + 3OH- \label{c}\tag{3}}$$
In aqueous ammonium sulfide (hydrogen sulfide in aqueous ammonia), the sulfide ions  and the hydrosulfide ions  remain in equilibrium.  The sulfide ions reduces the nitroarenes according to $\ref{a}$, whereas hydrosulfide ions follows $\ref{b}$  Since the concentration of the sulfide ions relative to hydrosulfide ions is very small, the course of the reduction proceeded
  through $\ref{c}$ predominately.
Although many published works on Zinin reduction exist, the exact
  mechanism of this important reaction is still not clear. The first
  product probably is a nitroso compound, which is rapidly reduced to
  hydroxylamine and then to amine . The rate-determining step is
  considered to be the attack of negative divalent sulfur on the nitro
  group, because no intermediate compounds are observed to be formed
  during the reaction.

Zinin reduction was discovered by a Russian organic chemist Nikolay Zinin in 1842 (Wikipedia) and revieved by Porter (1973; Ref.1). As pointed out by @Gaurang Tandon and @Nilay Ghosh in their answers, this reaction gives the best chance to convert polynitro aromates like 1,3-dinitrobenzene to 3-nitroaniline by selectively reducing one nitro group with sodium sulfide ($\ce{Na2S}$) or other derivatives of $\ce{H2S}$. My only intention here to give few more examples, which may help to deduce a reasonable mechanism for the reaction.
Preparation of 2-Amino-4-nitrophenol has been successfully achieved by selective reduction of 2,4-dinitrophenol using $\ce{Na2S/NH4Cl/aq.NH4OH}$ at $\pu{85 ^{\mathrm{o}}C}$ (Ref.2: It was noted that if the reaction is run at temperatures below $\pu{70 ^{\mathrm{o}}C}$, it is impossible to obtain a pure product even after several recrystallizations):

Interestingly, 4,6-dinitrobenzimidazole has been reduced selectively at 4-nitro group to give 4-amino-6-nitrobenzimidazole using sodium polysulfide (Ref.3). It'''s abstract gives some hints for possible mechanism:

Abstract: 4,6 (5,7)-Dinitro and 5,6-dinitrobenzimidazoles were reduced in aqueous ethyl alcohol by sodium polysulfide to give 4(7)-amino-6(5)-nitrobenzimidazole and 5(6)-amino-6(5)-nitrobenzimidazole. The spectroscopic properties were examined with respect to the effects of the 4- and 5-nitro groups present in the benzimidazole ring. Thus, the preferential reduction could be related to the chelating structure in the studied compounds.

Sodium polysulfide ($\ce{Na2S_n}$) is sodium salt of sulfanes ($\ce{H2S_n}$), which can be easily prepared by heating commercially available $\ce{Na2S\cdot 9H2O}$ with appropriate amount of $\ce{S8}$. For example, If $\pu{50 g}$ of $\ce{Na2S\cdot 9H2O}$ is heated with $\pu{25 g}$ of $\ce{S8}$ to $\pu{100 ^{\mathrm{o}}C}$, an aqueous solution of $\ce{Na2S_{5.5}}$ can be obtained (Ref.4).
Most recent work with 1,3-dinitrobenzene and $\ce{Na2S_n}$ has also been published (Ref.5). In this work, authors have claimed superiority of $\ce{Na2S_n}$ to selectively reduce one nitro group over nitro group selective reducing reagent, $\ce{Zn/NH4Cl}$ in water (Ref.6)), which evidently has given mixture of compounds. They have also showed that $\ce{H2S_n}$ prepared using more $\ce{S8}$ gives higher efficiency in reduction. For example, $\ce{H2S_n}$ prepared with $6:5$ weight ratio of $\ce{Na2S\cdot 9H2O}$ to $\ce{S8}$ gives 57% 3-nitroaniline while  $\ce{H2S_n}$ prepared with $9:2$ weight ratio of $\ce{Na2S\cdot 9H2O}$ to $\ce{S8}$ gives only 30% of expected product.
Based on above analysis, one can conclude that $\ce{Na2S\cdot 9H2O/S8}$ is a superior reagent to selectively reduce one nitro group in poly nitro compounds.

References:

H. K. Porter, “The Zinin Reduction of Nitroarenes,” Organic Reactions 1973, 20(4), 455–481 (https://doi.org/10.1002/0471264180.or020.04).
W. W. Hartman, H. L. Silloway (Checked by H. R. Snyder and J. W. Kneisley), "2-Amino-4-nitrophenol," Organic Syntheses 1945, 25, 5 & Organic Syntheses,  Collective Volume 1955, 3, 82 (DOI:10.15227/orgsyn.025.0005).
S. Dincer, “The preferential reduction of 4,6 (5,7)-dinitro and 5,6-dinitrobenzimidazoles,” Dyes and Pigments 2002, 53(3), 263-266 (https://doi.org/10.1016/S0143-7208(02)00018-9).
R. Steudel, “Chapter 3: Inorganic Polysulfanes $\ce{H2S_n}$ with $n > 1$,” In Topics in Current Chemistry, Volume 231: Elemental Sulfur and Sulfur-Rich Compounds II; R. Steudel, Ed.; Springer-Verlag: Berlin, Germany, 2003, pp. 99–125.
E. F. Lejarazo Gómez , E. Santos Santos, S. Suarez Torres, “Reduction of Nitro Compounds, Through different Reaction Conditions (Combinatory Chemistry),” Journal of Chemistry and Chemical Engineering 2018, 12(2), 74–82 (DOI:10.17265/1934-7375/2018.02.004).
T. Tsukinoki,  H. Tsuzukia, “Organic reaction in water. Part 5. Novel synthesis of anilines by zinc metal-mediated chemoselective reduction of nitroarenes,” Green Chemistry 2001, 3, 37–38 (doi.10.1039/B008219H).


text:
Preparing an amine from an azide
Is there any mechanism available for reduction of alkyl azides to amines.
One of the reactions is conversion by $\ce{Na}$ in ethanol or by $\ce{LiAlH4}$.
But how do these proceed?
$\ce{LiAlH4}$ will produce a hydride ion which attacks where?

Different reagents will have different mechanisms. Are you interested in one reagent in particular?

the ones I have mentioned i.e. Na in ethanol and LAH

I googled it for a while and this was the best thing I found (Source):

This is a tough question, that is probably not known with any certainty.
  Even Jerry March (Adv Organic Chem, 4th ed, page 1219) doesn't commit himself to a mechanism.
You'll have to speculate.   Add a hydride on the N next to the alkyl group, then
  grab a proton on the terminal N.    Add a hydride on the terminal N, then grab a proton on the middle N.  After that, it gets a little sketchy for breaking the N-N bond.
Try a SN2 attack by hydride on the middle N, giving hydrazine and  RNH-, which is a lousy SN2 leaving group, but we're stuck with it.
  Protonation of the RNH- gives the amine and hydrazine.   Or, you could do a SN2 on the N next to the alkyl group, and kick out a NH2NH- anion, which is also a lousy leaving group for the SN2.
  You can see why most books conveniently leave out a mechanism for this reduction.

I don't see the mechanism on that link. Which page are you referring to exactly?

At the current state, this doesn't provide an answer to the question. Am flagging for deletion.

This is a better one, but it's highly encouraged (read: required) that you put the link to the source here.

I'm finding https://answers.yahoo.com/question/index?qid=20100525170905AA9PDiD as a source

If you are quoting verbatim, always mention the source. I will insert it for now since @austinian already found it, but please avoid that in future answers.

@austinian thanks for putting the link; I generally put it, but in this case, I don't know how I forgot...

@Martin-マーチン as I said, I don't do it, but I forgot in this case; maybe because I previously provided not so correct answer,  then edited later. Anyways, thanks for reminding.


text:
Why is tert-Butyl (tBu) often not listed as protecting group for alcohol?
I am not seing tBu listed as alcohol protection group in all of these places:

https://en.wikipedia.org/wiki/Protecting_group#Alcohol_protecting_groups
http://www.synarchive.com/protecting-group/
http://www.organic-reaction.com/organic-synthesis/protecting-groups/ether/

On the other hand it seems to be quite common in peptide synthesis to protect the alcohol of serine, threonine and tyrosine with tBu.
It is listed on the German Wikipedia page: https://de.wikipedia.org/wiki/Schutzgruppe#Alkohole
Can I conclude from this that tBu is very rarely used in non-peptide syntheses? Or is it just plain missing from the English Wikipedia article?

Protecting group selection is a balancing act of ease of preparation and deprotection. In common practice, use of a protecting group adds two steps to a synthetic route. Installation and removing a tert-butyl group requires stronger (acidic) conditions. Of course, it's a rock while in place and inert to nearly all conditions short of strong acid.
Peptide synthesis is particularly amenable to use of tert-butyl protection of hydroxyl groups because (1) all of the suitably protected residues are commercially available and (2) the protecting groups have been designed to be globally deprotected with trifluoroacetic acid. These factors minimize the inherent inefficiency to protecting groups.
The workhorse protecting groups for alcohols are silyl ethers (e.g.trimethylsilyl, triethylsilyl, tertbutyldimethylsilyl, triisopropylsilyl, tertbutyldiphenylsilyl). The silyl ethers are easier to form and cleave than standard ethers and have excellent compatibility with reactions that require alcohol protection.
Specialized (carbon) ethers can be useful as well, including unsubstituted or substituted benzyl ether (removed by oxidative or reductive conditions) and acetals (removed by mild acid treatment).
For a more complete description of protecting groups (including tert-butyl), see the books by Greene and Wuts or Kocienski.


text:
Using vapor mole fraction and pressure to determine liquid mole fraction
I am studying Chemistry by Zumdahl and Zumdahl (9th edition) and I have come across a question in Chapter 11 that I can't figure out the explanation for:
Liquid A has vapor pressure x , and liquid B has vapor pressure y . What is the mole fraction of the liquid mixture if the vapor above the solution is 30% A by moles?
I know that Raoult's Law tells us that the total vapor pressure is χx + (1-χ)y in this case. I also assume that the total vapor must be 0.3x + 0.7y given the other constraint of the question. This would mean the answer should be:
0.3x + 0.7y = χx + (1-χ)y
However this leads to the answer χ = 0.3, whereas the text gives the answer as 0.3y/(0.7x+0.3y)
What have I got wrong? (Also sorry for not formatting it properly but I'm a noob here and I can't figure out the equation formatting).

Welcome to Chemistry.SE. A short introduction to the maths and chemistry notation using MathJax is here, with more information here.

Raoult's law tells us the relationship between the composition of the liquid and the vapor pressure above it:
$$P_A=\chi_A P^\circ_A$$
Where:

$P_A$ is the partial pressure of $A$ in the vapor above the solution
$P^\circ_A$ is the vapor pressure of pure $A$
$\chi_A$ is the mole fraction of $A$ in the solution

For a two component solution $A$ and $B$ (I don't want to use $x$ to avoid confusion with $\chi$) we can then use Dalton's Law to get:
$$P_T=P_A + P_B = \chi_A P^\circ_A +\chi_B P^\circ_B= \chi_A P^\circ_A +(1-\chi_A)P^\circ_B$$
However, from the definition of mole fraction we can derive a relation between the mole fraction in the vapor (from here on out $\phi$ to avoid confusion between vapor and solution, which is what your mistake is).
$$\phi_A=\frac{n_A}{n_T}$$
Where $n$ is the number of moles, and since at constant temperature and volume, $n\propto P$...
$$\phi_A = \frac{n_A}{n_T}=\frac{P_A}{P_T}$$
So, we can rewrite $P_A = \phi_A P_T$ and $P_B=(1-\phi_A)P_T$
We are given $\phi_A=0.3$. Thus $\phi_B=0.7$
Here we go. Substitute $P_A=\chi_A P^\circ_A$ and $P_T=\chi_A P^\circ_A +(1-\chi_A)P^\circ_B$ into $P_A=0.3 P_T$:
$$\chi_A P^\circ_A=0.3\left(\chi_A P^\circ_A +(1-\chi_A)P^\circ_B\right)$$
$$0.7\chi_A P^\circ_A=0.3(1-\chi_A)P^\circ_B$$
$$0.7\chi_A P^\circ_A + 0.3\chi_A P^\circ_B=0.3P^\circ_B$$
I think you can get it from here.

$\kappa_A $ is the mole fraction of A in the solution at equilibrium or initially

IUPAC recommends the lowercase Roman letter $x$ for solution and $y$ for gas, but the lowercase Greek $\chi$ is common for solutions. I have not seen kappa used for mole fraction before. Here is the link to the IUPAC recommendations: https://goldbook.iupac.org/terms/view/A00296


text:
Using a modified Arrhenius plot to find activation energy
I have a question on how to determine activation energy from a plot using the Arrhenius equation. I am working as an assistant with general chemistry II students and their lab is the iodination of acetone. I am familiar with using a plot of $\ln(k)$ vs $1/T$ to get the slope of $E_\mathrm{a}/R$ to solve for $E_\mathrm{a}$. Their lab manual though tells then to plot $\ln(t)$ vs $1/T$ to determine $E_\mathrm{a}$. I was wondering if anyone could shed some light on the derivation of this equation.

I have no clue but as the exponential decay function $N(t)=N_0~e^{-\lambda~t}$ looks very similar to Arrhenius function $k=A~e^{-\frac{E_a}{R~T}}$ there might be some way to either convert one into another or to simply replace the equal terms ...

The use of $t$ only makes sense if $t$ is defined as the time it takes for the reaction to progress a certain relative amount (and if we are talking about first-order reactions).  For example if $t$ is the time it takes for half (or 90% or 99% or whatever, as long as it is specified) the reactant to be consumed, then the idea makes sense.  In that case it would be more helpful and clear to say $t_{1/2}$ (or $t_{90\%}$ or whatever) instead of just $t$.
Assuming $t_{1/2}$ is what is meant by $t$:
Arrhenius equation:
$k = A_0 e^{\frac{-E_a}{RT}}$
$k = A_0 e^{\frac{-E_a}{R}\frac{1}{T}}$
$\ln k = \ln A_0 - \frac{E_a}{R}\frac{1}{T}$
First-order kinetics:
$C = C_0 e^{-k t}$
$\ln \frac{C_0/2}{C_0} = - k t_{1/2}$
$k = \frac{1}{t_{1/2}} \ln 2$
$\ln k = - \ln t_{1/2} + \ln \ln 2 = -\ln{t_{1/2}} + \ln \ln 2$
Combining:
$\ln k = \ln A_0 - \frac{E_a}{R}\frac{1}{T} = -\ln{t_{1/2}} + \ln \ln 2$
$\ln t_{1/2} - \ln \ln 2 = \frac{E_a}{R} T^{-1} - \ln A_0$
$\ln t_{1/2} = \frac{E_a}{R} T^{-1} - \ln A_0 + \ln \ln 2$
So if $t$ means $t_{1/2}$, then a plot of $\ln t_{1/2}$ vs. inverse temperature has a slope of $+\frac{E_a}{R}$, instead of $-\frac{E_a}{R}$ if you plot $\ln k$ vs. inverse temperature.
N.B. If $t$ just means time, then the idea doesn't make any sense, because then instead of $\ln \ln 2$ you would have $\ln \ln \frac{C(t)}{C_0}$, which is a term that depends on time, so instead of being an explicit equation that is clearly a line, you would have an implicit equation, non-linear in $t$, that would need to be solved before the plot was made, and the result would not be linear and would be impossible to use to determine $E_a$.

I believe that is the equation that they are trying to get at in this lab. Although it has no explanation as to what t really is. I assume it's t when 100% of the iodine has been consumed in the reaction. I don't know if it is easier to do the ln(t) vs 1/T or the ln(k) vs 1/T. Personally I preferred the normal Arrhenius equation. That's for helping me understand. Maybe I should ask them to clarify the lab manual.

First, if all you care about is the activation energy, it doesn't matter if it is $t_{1/2}$ or $t_{90%}$ or whatever; the equation shows it only affects the intercept, not the slope.  Second, it can't be $t_{100%}$, as that would imply an infinite intercept!  It must really be $t_{99.99%}$ or similar, corresponding to the highest iodine concentration that the measurement method cannot distinguish from zero.


text:
What are the structures of potassium superoxide and xenon hexafluoride?
I have 2 questions in mind:
1) What is the structure of the superoxides? More specifically, what is the structure of $\ce{KO_2}$?
2) What is the structure of and dipole moment for $\ce{XeF_6}$?

Welcome to chemistry.SE. If you have two questions then please post them as separate questions, not together in the same post. Also please add a bit more detail and your thoughts about the problem to make us sure we aren't doing your homework for you.

@ToddMinehardt I assume you are very familiar with LaTeX syntax ;) We have this neat package called mhchem installed, which makes typesetting chemical formulae much easier. You can have a look at some meta posts to learn more: this one  and this one. Thanks for helping out :D

Xenon hexafluoride's geometry is not easy to be described but it is assumed to be in $C_{3v}$-symmetry - not octahedral! - in gas phase which is supported by electron diffraction measurements and quantum chemical calculations. It crystallizes in at least six to seven different known structures. [1]
As it is further written

Konrad Seppelt, an authority on noble gas and fluorine chemistry, says, "the structure is best described in terms of a mobile electron pair that moves over the faces and edges of the octahedron and thus distorts it in a dynamic manner."

On you question regarding the dipole moment - for the $C_{3v}$-symmetry $\ce{XeF6}$ the dipole moment is calculated  to be $0.46~\mathrm{D}$. [2]
The structure of $\ce{KO2}$ can also be found on wikipedia. There you find the following picture, showing it in it's      body-centered cubic unit cell.

[1] S. Hoyer, T. Emmler, K. Seppelt, J. Fluor. Chem. 2006, 127 (10), 1415-1422
[2] T. D. Crawford, K. W. Springer, H. F. Schaefer III J. Chem. Phys. 1995, 102 (8), 3307-3311

Computed dipole moments for 3 symmetries here gives 0.46 D for $C_{3v}$.

It would be very nice, if you could provide the complete reference when pointing to articles. This way we can be sure that the quotation can be found even if the pages/links change. And I am very surprised about the structure of xenon hexaflouride.

T. D. Crawford, K. W. Springer, and H. E. Schaeffer III. "A contribution to the understanding of the structure of xenon hexafluoride." J. Chem. Phys. 102, 3307 (1995).


text:
How can nitrogen achieve +5 oxidation state?
Nitrogen has no available d-orbitals. Yet, it forms compounds where its oxidation state is +5, like in nitric acid. How?

What makes you think that available d-orbitals are essential for achieving high oxidation states?

Lower group-15 elements could be penta-valent because they have d-orbitals. They help form the hybrid $sp^3d$ orbitals, isn't it?

@AgnishomChattopadhyay No, they don't.

Generally, the simpliest intuition for this  is to imagine that nitrogen looses one electron, becoming isoelectronic to carbon. After forming four bonds with, say, fluorine, we gain $\ce{NF4^+}$ cation where nitrogen is connected to four more electronegative atoms  and have positive charge to begin with, thus +5 oxidation state. In some compounds, say, nitric acid, the electron is not moving too far, instead migrating to nearest oxygen, so nitrate anion is actually isoelectronic to carbonate dianion.

d-orbital hybridisation in expanded octet compounds is a subject of much debate. It has been discussed extensively on this site and the general consensus is that hypervalency provides a better description of the bonding in expanded octet molecules. See here, here and here for some initial reading.

By the way, nitrogen (and any atom type) has d-orbitals. They just aren't typically occupied by electrons.

The concept of oxidation states is just that: A theoretical concept. You assign electronegativity values based on one of the numerous scales available, split the bonds heterolytically according to those numbers, and then count electrons.
As such, it is very easy to make a cation like @permeakra stated in the comments that satisfies the octet rule and still gets to state +5 (the molecular cation in question is $\ce{NF4+}$).
There are no $d$ orbitals necessary (which are also theoretical, but slightly closer to reality) to explain how a +5 state is possible.


text:
The Lewis structure for NCO-
I was having trouble drawing the Lewis structure for $\ce{NCO^-}$. I found this image online:

But why is structure number one preferred over structure number two? Why does carbon prefer to have a triple bond with nitrogen, rather than have a double bond with nitrogen and a double bond with oxygen?

If you count electrons and determine the formal charge on each atom, you find that in structure #1, the negative charge is on the oxygen.  Do the same exercise for structure #2 and you find that the negative charge is on nitrogen.  Since oxygen is more electronegative then nitrogen, the negative charge is more stable when its on the oxygen atom.  Therefore, structure #1 is more stable (preferred) than structure #2.
Structures #3-#5 are even less stable.  They either violate the octet rule (#4, #5) and / or have charges (sometimes multiple) on more than one atom (#3-#5).

When talking about resonance structures, I think stable is the wrong word. I agree with your analysis in terms of contribution. Therefore #1 should have the highest contribution of all.

@Martin  Perhaps this is just semantics, but I disagree with your comment.  Why is one resonance structure a larger/smaller contributor than another - because it is lower/higher energy than another - e.g more/less stable than another.   "It" referring to the resonance structure as if it were an isolated, existing structure, which in reality it is not.

I agree it is a question of semantics. Since individual resonance structures are not real, none of them are stable, only the complete set of structures is. However, the electronic ground state can be described as a linear combination of configurations, which can be chosen to represent individual Lewis structures. From the coefficients of these individual configurations we can see a contribution to the ground state. I just can't bring myself to talk about a stable resonance structure, since I know they are individually not stable in the sense of the definition of the word.


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Maximum number of double bonds a carbon atom can make

What is the maximum number of double bonds a carbon atom can make?

Well, carbon has $4$ valence electrons so... it needs $4$ more to fulfill the octet rule right?
If a double bond yields $4$ electrons, does it mean that the carbon atom can make only one double bond?
OR does it mean it can create up to TWO double bonds (because they would yield $8$ electrons which fulfill the octet rule)?
What about Calcium? I ask because Calcium doesn't follow the Octet rule: it only wants to have 4 electrons, so...

Calcium generally forms fairly ionic compounds so the concept of a double bond isn't necessarily applicable. In regards to carbon, each double bond gives it two more electrons, on top of those that it already has, so I'll let you figure out which answer that leaves you with.

Rather than focus on what it's gaining, try to draw a Lewis structure of carbon connecting to a carbon on each side. Then try changing the connections between them to different numbers and arrangements of bonds and check whether or not the central carbon has a satisfied octet. By doing this systematically you can probably "prove" the answer to yourself.

Well known carbon compounds have two sets of double bonds which agrees with the simple idea from the octet rule
Simple bonding theories like the octet rule often don't work because they overly simplistic (just look at the variety of bonding in sulfur compounds).
It is usually better to look at actual examples of real compounds we know about. In the case of carbon we have a whole class of compounds known as allenes which have a single carbon with two double bonds.
So, from observation alone, we can conclude that a carbon with two double bonds is possible.
The situation with calcium is irrelevant as most compounds will not be covalent but ionic (and counting bonds in ionic compounds is pretty much irrelevant).

[OP ...] carbon has 4 valence electrons so... it needs 4 more to fulfill the octet rule right?

Yes, that is correct.

If a double bond yields 4 electrons, does it mean that the carbon atom can make only one double bond? Or does it mean it can create up to TWO double bonds (because they would yield 8 electrons which fulfill the octet rule)?

The easiest way to do this in your head is to assume that for each bond, one electron "comes" from the carbon atom of interest, and the second "comes" from the bonded atom. So carbon often makes four bonds (which could be two double bonds, as in carbondioxide, in allenes or cumulenes). There are less common cases where carbon has a positive or negative formal charge but it is useful to first become familiar with the common cases (where carbon forms four bonds) before you move on to those.

What about Calcium? I ask because Calcium doesn't follow the Octet rule: it only wants to have 4 electrons, so...

Calcium typically does not form covalent electrons. In many compounds, it occurs as ion with two positive charges. Instead of completing the octet, it forms ions with a completely empty outer shell (so the fourth shell is empty, and the third one is full, with an octet).


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Simple procedure for showing presence of milk proteins in clarified butter (ghee)
By clarifying butter, e.g. by melting butter and filtering it, we rid ourselves of most of the protein in the butter, and are left with mostly fat. However, the separation of protein and fat is not complete. Some people cannot eat milk proteins due to allergic reactions, and cannot therefore use unrefined butter in their cooking.
Is there a simple "home-kitchen" procedure to test whether the filtration sufficiently removed the milk proteins? Any proteins in the ghee can be precipitated by adding acetic acid, but this is not specific to milk proteins.
Does anyone have any ideas?

Is the thinking here that removal of whey protein will also help remove lactose?

No, the sugars are not relevant to those allergic to milk proteins.

A classic test for protein is Biuret solution, using readily available reagents:  sodium hydroxide, copper sulfate and potassium sodium tartrate (Rochelle salt). The test is not specific for milk protein, but what other protein would be in pure butter?
Use caution with the chemicals -- the first listed is quite caustic, and the second is poisonous. BTW, save some leftover Rochelle salt if you'd like to make a piezoelectric crystal.

Are there compunds readily available that binds to milk proteins, while not affecting the taste of the clarified butter? Or that binds to every type of protein. If all protein is lost, then the butter would be safe for those with allergies.

I am not sure that the limit of detection of the Biuret solution is low enough to reliably indicate the absence of an allergen.

Biuret is only good down to ~1 g/dL.  Coomassie blue is commonly used for protein concentrations 2-3 orders of magnitude lower (mg/dL).  Even then, I don't know that  I'd risk it.

@Gossar, point well taken: if the problem that milk protein causes is mild indigestion, it might be worthwhile to test... but if there's chance of a severe allergic reaction, i.e. anaphylactic shock, this would not be a good home experiment! And thanks for mentioning the Bradford (Coomasie blue) test.