text:
Why do Zr, Sn, and Ti enolates form syn aldols selectively?
In Organic Chemistry (2nd ed.) by Clayden, Greeves, and Warren, it is stated that (p 869)

[...] the enolates of some metals [Sn(II), Zr, Ti] give syn aldols regardless of enolate geometry.

There are also literature examples of syn-selective Zr aldol reactions, e.g. Tetrahedron Lett. 1980, 21 (48), 4607–4610.
If they were to obey a simple Zimmerman–Traxler chair-like transition state model, as is applicable for Li or B enolates, then (Z)-enolates should give syn aldols and (E)-enolates anti aldols. How does the mechanism or stereochemical model differ for these enolates?

Would you consider splitting this into three questions? I'd be keen to research and answer it, but I also think there's enough content to make one question per metal. Essentially I don't want to write an essay for one question ;)

For most aldol reactions, the relative 1,2-diastereoselectivity is a consequence of the enolate geometry.*
Both tin and titanium enolates are easily formed, and have been shown to exist predominantly in the (Z) configuration.
When reacting a cyclic (Zimmerman Traxler) TS, its commonly observed that (Z) enolates afford 1,2-syn aldol adducts, due to the orientation of the enolate   (the (Z) enolate always has an axial R-group).
The following examples are instructive. In both cases (tin or titanium), we first form a (Z) enolate using the conditions shown. When the aldehyde (R-CHO) is introduced, chelation using the metal organises the Zimmerman Traxler TS. The lowest energy TS possible is the one in which the aldehyde is arranged with its bulky R group equatorial, however the enolate cannot put the methyl group (in this case) equatorial, as the enolate geometry enforces its axial positioning. Following both transition states through, we get the 1,2-syn aldol adduct in both cases.#

Image taken from Andy Myers' CHEM115 Handout: Stereoselective, directed aldol reaction

The description given in Carey is fundamentally incorrect for the reactions shown (even if it does predict the correct product). Looking at the references he gives, they're all fairly old, before more recent experimental and computation studies have validated the cyclic TS model for these kinds of metal enolates.

*: Mukaiyama aldol reactions do not conform to this, since they occur via a fundamentally different transition state (an open TS) to the Zimmerman Traxler model commonly found for other aldol reactions.
#: In this case, we also get enantioselectivity via the use of a chiral auxiliary, and depending on the metal used the enantioselectivity can be switched around due to chelation or lack thereof between the auxiliary and the metal

Is there any reason why (Z) is favoured?

@orthocresol. I was actually thinking of posting that as a question. I have never found a good model for why tin gives (Z).  Lithium obeys the Ireland model, Boron has well defined models of its own. Tin, not so much!

@orthocresol Just had a nice chat with NotEvans., but if memory serves me, the oxaziridinone's side chain creates a lot of strain for the (E)-enolate one the transition metal is coordinated to both carbonyl groups. I think that was one of the goals of the design for the oxaziridinone directing group.

I've already realized that Carey, while being an excellent book, is pretty much outdated. By the way, Myers lectures are very interesting.

In Carey, Sundberg Advanced Organic Chemistry it is stated:

This preference arises from avoidance of gauche interaction of the
  aldehyde group and the enolate beta-substituent. The syn
  stereoselectivity indicates that reaction occurs through an open TS.

The relevant conformations adopted by the two molecules in respect to each other, just before the reaction takes place are attached. The lower one, which leads to the anti aldol, has the energy increasing gauche interaction (the R groups may differ):

Therefore is doesn't matter whether the enolate configuration is Z or E, the syn aldol will be obtained stereoselectively.


text:
Calculating the pH of a highly dilute solution of HCl
For relatively high concentrations of $\ce{HCl}$, I usually just assume that $[\ce{H+}] = [\ce{HCl}]$, because $\ce{HCl}$ is a strong acid and is completely ionized in solution. By taking the negative logarithm of the concentration of $\ce{[H+]}$, the pH comes out more or less correctly.
However, at very low concentrations, this method doesn't work out so well, because water has an amount of auto-ionization, which is described by $K_\mathrm{w} = 10^{-14}$. My old method was just to say that
$$[\ce{H+}] = [\ce{HCl}] + \pu{10^-7 mol dm-3},$$
to include the auto-ionization of water in my calculations. I can't really describe why, but I believe this method to be wrong, because the number $\ce{10^{-7}}$ comes from water just by itself, not with $\ce{HCl}$ added to it.
Then I came across this website, which implies that the above method is an OK method, but mentions that "However, that more complex technique will not be discussed here.". What is that 'more complex technique', and how could I use it?

At first you have to think of where the protons come from. That is as you mentioned it, hydrochloric acid (for every chloride ion in solution there has to be a proton) and water (for every hydroxide ion from the autoprotolysis there also has to be a proton). $$\ce{[H+] = [OH-] + [Cl-]}$$
The concentration for the hydroxide ions can be derived from the autoprotolysis of water, as I've said before.
$$k_\mathrm w = \ce{[H+]~[OH-]}
\Rightarrow \ce{[OH-]}=\frac{k_\mathrm w}{\ce{[H+]}}$$
Using some funny math, one can find out, that the concentration of the chloride ions in solution can be described the following eqation. Recognizing shortly after that the hard way was not necessary yields to the simplification that hydrochloric acid is completely dissociated.
$$\ce{[Cl-]} = \frac{\ce{[HCl]0}~k_\mathrm a}{\ce{[H+]} + k_\mathrm a} \xrightarrow{k_\mathrm a \gg \ce{[H+]~~}} \frac{\ce{[HCl]0}~k_\mathrm a}{k_\mathrm a} = \ce{[HCl]0}~(=c_0)$$
This changes the first equation to be like
$$\ce{[H+]} = \frac{k_\mathrm w}{\ce{[H+]}} + c_0$$
Rearranging everything gives you a second order equation in $\ce{[H+]}~(=x)$
$$x^2-c_0~x-k_\mathrm w = 0$$
Wherefrom you get two solutions (from which only the positive one makes sense):
$$x = \frac{c_0}{2} + \sqrt{\frac{c_0^2}{4}+k_\mathrm w}$$
Let's compare this solution to your mentioned version. The following graph shows you the difference between both assumptions as $$\operatorname{abs}(f_1(c)-f_2(c))$$
$\hskip3.5cm$
A maximum difference of 0.1 pH between the easier and the "more complex" version is acceptable.

The gist is that assuming that any HCl added will completely dissociate in less than 1 molar solution is absolutely reasonable. If $x$ moles of HCl are added the the autoionization equilibrium is shifted such that  $$x = \ce{[H^+] -[OH^-]}$$

Would anyone be able to show where the expression for $$\ce{[Cl-]=\frac{[HCl]_0K_a}{[H+] + K_a}}$$ comes from?

@pH13 why is it that k_a >>$ [H^+]$? I mean how we are sure regarding [H+] to be coming out to be very less than ka

@H.Linkhorn Solve $k_a [HCl] - [H^+][Cl^-] = 0$ and $[HCl] + [Cl^-] = [HCl]_0$ together for $[Cl^-]$

@ProblemDestroyer to replace activities with concentrations is only ok for diluted solutions and when you look up the k_a for HCl, you will find it is much higher than the acceptable concentrations within this approximation.

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text:
What advantages does the mole have over counting large numbers using SI prefixes?
I'm trying to understand why chemists use the mole unit instead of just counting and using SI prfixes to simplify the big numbers with units such as zetta- or yotta-molecules (yotta=10^24).
Here's what I've found so far, but it doesn't seem to be enough to make moles particularly important:
Reasons for moles:

You can easily approximate the number of atoms from the mass of a sample because the mass number of an isotope approximately equals the mass of 1 mole of atoms in grams.
It's a historical convention that would be too expensive or uncomfortable to change now.
You can use millimole and micromole instead of having to learn several more obscure SI prefixes like exa and peta. EDIT this reason is added from Jan's answer
Shorter descriptions, eg "mole of carbon" vs "yotta atom of carbon" (from matt_black's answer).

Reasons against moles:

For accurate calculations (beyond about 3 significant figures), the relationship between mass number and mass in the first reason above breaks down. So this is a risky thing to use.
It's a additional concept and set of facts that chemists have to spend effort learning but which describes human convention rather than nature.
We need extra conversion factors such as 1/mol which people often neglect and Faraday's constant used in Q = n(e-) x F [1] which wouldn't exist without moles.

Non-reasons for moles:

You can easily calculate the number of atoms from the mass of a sample because the atomic mass of an element equals the mass of 1 mole of atoms in grams. This relationship only exists because of the special units commonly used for atomic mass. If periodic tables listed atomic mass in grams (or perhaps yoctograms), then we could do the same calculations just as easily without moles.
In practice, we can't measure the number of molecules so we have to measure mass or volume instead and should therefore count them using a unit that's defined in terms of mass. I don't think the precise details of how a unit is defined matter for practical purposes. If you have 1 litre of an ideal gas, you still have to do calculations to find out many moles it contains just as you would to find how many molecules it contains. There is even a proposal [2] for SI to redefine the mole to be independent of the mass of any substance, indicating that keeping the definition isn't very important.
Chemists would make mistakes with those big numbers. They'd use "yotta" in the same way they use "mole", not doing calculations with the actual number it represents and not being at risk of other types of error. (from matt_black's answer) However it would be more complex replacing millimole with zetta because you'd have to remember that a zetta-atom is a milli-yotta-atom.

Here's a similar question but it's mixed in with the idea of measuring number rather than mass or volume -
Why do people still use the mole (unit) in chemistry?
This other similar question mainly addresses the uncommonness of quantities as big as 1 mole in other areas of life -
The mole is used extensively in chemistry, why not elsewhere?
[1] http://www.ausetute.com.au/faradayl.html
[2] https://en.wikipedia.org/wiki/Proposed_redefinition_of_SI_base_units

Your point about the accuracy doesn't make a lot of sense. Many elements have a fractional mass (Chlorine 35.5, Boron 10.8, ...), so for any practical use (particle physicist don't use moles), your "integer mass number" approach breaks immediatly.

Yep, that's what I mean by the 1st reason against. It's a bit like saying "10N is the weight of 1kg". Enough to get a quick feel for something without looking up tables or getting out a calculator. However, so far, this seems to be the only substantial reason for using moles!

Chemists don't memorise the number of neutrons (it's of no use chemically), they know the ordinal number (=protons) and the average  molar mass.

The unit "moles" is just extremely convenient. One simple additional convention in at least a thousand more complex concepts your learn until you get your degree (or hundreds of concepts for a good high school diploma).

Karl - what makes it more convenient than are simpler unit like in my question? If chemists don't memorize mass numbers then even this most important reason disappears, along with any value in Jan's suggestion of an imperial mole.

Mass numbers of isotopes (except deuterium) are of no concern to chemists. What is so hard to understand about not wanting to work with SI prefactors like zetta and yotta?

Let us continue this discussion in chat.

Notwithstanding switching costs, I also think that memorizing SI prefixes is a non-argument, this is what engineering notation is for. I personally find obscure SI prefixes as cumbersome as moles.

Of course, it’s convention and has been so for a good century. And there is no real reason why this convention (and not a different one) happened in the first place — it is entirely conceiveable to define an ‘imperial mole’ so that the atomic mass of an element is equivalent to the same mass in ounces. 1 imperial mole of carbon atoms would then be twelve ounces of carbon atoms.
So there are basically two questions in this:

Why replace a large number with a unit just meaning ‘this certain large number’?
Why is the presently defined mole a good choice for this unit?

The thing about large numbers is that they are large. Almost everybody who grew up in a metric country can name at least three SI-prefixes: kilo, centi and milli. Thanks to IT, many people now also know mega and giga (and maybe tera), even if they don’t use metric units at home. But tera only gets you to $10^{12}$. We need some $10^{21}$ for moles.
I often work with milli- or micromoles of substances in my research. In plain numbers, that’s $10^{20}$ or $10^{17}$ — I don’t know those prefixes and thus would have to learn an entirely new subset. With the mole, everything one uses in the lab will nicely fall into something between nano and kilo.
It also helps to have a single unit there. Molar mass is expressed in grams per mole, concentrations in moles per litre and many more. If there were no unit, it would be simple grams per 1 or 1 per litre — precisely the reason why some people prefer to use rad or some other way to show radians rather than just writing the number. If the unit is there, you’re unlikely to forget it, you know if your calculations are good and more. If there was no name for this unit, it would have to be invented.
So if the mole didn’t exist, it should be invented for simplicity.

The good thing about the size definition of the mole is, as noted above, that it brings everything into one general range. Whether it’s mass, volume, concentration or amount, every unit is going to be prefixed by only a small subset of the SI-prefixes: kilo, milli, micro, maybe nano. Thankfully, those are the ones that are most used in everyday life, too (excluding nano and maybe micro).
It doesn’t really matter where one ends up. If the mole had originally been defined imperially,[1] that would be fine, I wouldn’t have memorised $12.01\,\frac{\mathrm{g}}{\mathrm{mol}}$ for carbon but $340.48\,\frac{\mathrm{g}}{\mathrm{mol}}$. That would create significantly larger molar masses, but that shouldn’t be a problem it should only mean that nano is more prevalent.

The thing about the definition of the unit is that it doesn’t bother $99.9\,\%$ of the scientists working with the unit. Seconds are defined (by SI) according to a number of transitions occuring in some weird isotope that I wouldn’t even know how to measure. I would explain a second by saying it’s the 86400th fraction of a day if someone asked me. Same meaning, different exactness. If the moles are soon defined by mere counting rather than weighing atoms then so be it. Nothing will change for me in practice. (Maybe the fourth digit of a molar mass but I don’t really count those.) So as long as a new definition doesn’t break anything, we’re good to carry on.

Notes:
[1]: With imperially, I meant to assume the following definition using imperial units:

One mole is the number of atoms in exactly $12.00~\mathrm{oz}$ of atoms of the carbon isotope $\ce{^{12}C}$.

Exactly. High cohesion, low coupling.

I just want to point out that since 2019, the connection with carbon 12 is now only an approximation because Avogadro's number has been redefined to be an exact arbitrary integer. So it wouldn't make sense to have an imperial mole anymore.

@user1318499 Not wrong, but if we had had an imperial mole then Avogadro’s number would simply be a different value meaning it would still be pretty much the same.

I think the biggest reason it's still used is because there's no escaping our need of Avogadro's constant, $N_\mathrm A=6.022\times10^{23}\ \mathrm{mol^{-1}}$. Even if we were to use some metric prefixes meaning $10^{24}$, etc. we would still need Avogadro's constant because it is used in the definition of other physical constants.
For instance, Boltzmann's constant, $$k_\mathrm B=\frac{R}{N_\mathrm A}$$
And, Faraday's constant, $$F=N_\mathrm A\cdot e$$
where $-e$ is the elementary charge of an electron.
Or, the number of particles in an ideal gas, $$N=\frac{p\cdot V\cdot  N_\mathrm A}{R\cdot T}$$
All of those constants have to stay the same for many of the formulations of Chemistry to be correct, so we would have stumbled across this magic number one way or another.
While it might be true (and I don't know this) that the mole was not defined with any of that in mind, it's really quite convenient if you ask me to have a mole be both a unit and a physical constant which everyone knows.
It's particularly useful because you don't have to memorize what the value of $k_\mathrm B$ is, you just remember the equation and figure it out using a calculator. After all, every knows Avogadro's number and the ideal gas constant.

I think these are non-reasons. Without moles, we would have no need for the gas constant R, we'd just use Boltzmann's constant instead. The question already identifies Faraday's constant as redundant for this reason. Without moles, we would just use e instead of N_A * e. You've highlighted that there's this whole parallel system of constants and formulas that exist just to accommodate the mole. n=P.N_A / (R.T) can also be written as n=P/(k_b.T).

The thing about the mole is that it simplifies chemistry terminology in a way that can't be avoided if you want to talk about chemical reactions.
A mole is a count of the things involved in a reaction, not necessarily a count of the atoms involved. So a mole of oxygen gas contains two moles of oxygen atoms. A mole of a protein contains hundreds of moles of amino acid resides and, well, an awful lot of moles of atoms. When thinking about this we don't usually have to worry about the number of units of the thing we are discussing.
If we didn't have the idea of the mole, we would have have to use much longer descriptions every time we talked about chemical entities or chemical reactions to be clear what it was we were actually counting. A mole of carbon atoms would have be described as 602 Zetta carbon atoms: a mole of carbon is shorter. Everything would have its own name. And we'd have to use a lot of exa and zetta SI prefixes which could get awkward in calculations.
The value of the mole doesn't matter that much most of the time when thinking about reactions: only the ratio. That it is an very large number is mostly unimportant and the calculations chemists mostly do use molar ratios where they only need to calculate using atomic or molecular mass for the components in the reaction. Including the actual size of the mole in these calculations would introduce very large and unnecessary numbers that would end up cancelling out in the small number of calculations where nobody made an error and would give completely incorrect answers in all the others where people got confused by all the extra prefixes or digits.
Tha actual size of the mole is a little arbitrary, but has the advantage it is the sort of scale chemistry is often done on. A mole of water is ~18g or about 18mL of water. You can do a reaction with that scale on your tabletop. And you don't need to count or even remember Avogadro's constant to do it or any of the calculations associated with it. 6.02*1023 molecules of water is a much less convenient unit. I could happily talk about fully electrolysing a mole of water to produce a mole of hydrogen and half a mole of oxygen with no mention of Avogadro, no big-number SI units.
PS The mole is truly inconveniently large for some objects not on the atomic scale. A mole of moles (the furry burrowing mammal) would make up a sphere about the same size as the moon.

You second paragraph about "count of things" applies to SI prefixes too. A "zetta protein" would be a number of protein molecules, just like a "mole of protein" is. You 3rd paragraph about "use much longer descriptions" could be a disadvantage. Is "a mole of hydrogen" referring to atoms or molecules? You need context to know the meaning. Your 5th paragraph about "The actual size" applies equally well to SI prefixes so I think it should be deleted for clarify.

Moles are simply a built in conversion factor that keeps us from either having to use amus as our macroscopic unit of mass, or grams or kgs as our microscopic unit of mass.
So the basic advantage is that if you want to find the gram mass of $10^{24}$ oxygen molecules in standard units you would have to use Avogadro's number to do it.
Grams were already entrenched in our units of force, energy, pressure, volume and even length (a meter is equivalent the cube root of the volume of $10^6$ grams of water at what 298K?).  You can derive a meter, joule, Newton, second or K from a gram of  water  in the Earth's gravitational field. Moles are not a fundamental concept in chemistry, they are a practical concept, but the ability to move between non-conforming unit bases is fundamental as long as we live in a universe where not every scale of matter is a precise power of ten up from the last one.
In chemistry, by the way, the amu is effectively/practically close to being the "quantum" unit of mass.  We either use the fundamental quantum unit of mass in chemistry when describing molecules and dispense with our units of force, energy, power, pressure, volume, and length, or we ignore the chemical quantum of mass, or we find a way to move from one to another.

I have a suggestion that your 4th paragraph isn't helpful and perhaps could be deleted. I think the "effectively quantum" property is covered in the first bullet point of the question. Karl's comments on the question say that this  property isn't useful. And using macroscopic units for atoms is already the main question.

Actually, the "quantum" point does make sense, because in organic chemistry, basically everything is H,C,N,O, and using integer mass numbers is correct to <1%. The argument also holds for Li, Na, K, P, S, F, Br, I. Chlorine is the only exception in the common main group elements.

I am a chemist and I am still trying to understand why SI prefixes are not used by chemists.
I am familiar with all the SI prefixes, but even if I was not familiar, I would prefer to learn a pair of universal prefixes rather than deal with the mole and all the issues, inconsistencies, and headaches it brings to us:

I have to learn SI prefixes the same. I.e. the mole is redundant.
I have to learn two mole concepts: the chemical mole and the thermodynamic mole [1].
I have to learn three definitions of mole in the SI. The original definition in the SI (1971), the amended definition (1980), and the new definition proposed for the future SI [2]. I have to learn the differences and check what one is being used in a given reference. For instance, the definition of mole in the current SI gives a uncertainty to the Avogadro constant, the proposal for the new SI sets a exact value for this constant and moves the uncertainty to $M(^{12}C)$.
Then I have to deal with "correction factors" to compensate for the
incompatibility between the redefinitions of mole and kilogram [3].
I have to learn an abstruse name for the physical quantity that
supposedly is measured in moles, "amount of substance", like if a
mol of electrons is a substance. And then I have to deal with the
proposals to change that physical quantity to "chemical amount",
"stoichiometric amount", or "enplethy" [2].
I have to deal with the ontological confusion between counting and measuring [4].
I have to deal with redundant equations such as $pV=nRT$ and $pV=Nk_BT$.

REFERENCES
[1] Failures of the global measurement system. Part 1: the case
of chemistry. Gary Price. Accred Qual Assur (2010) 15:421–427.
[2] What is a Mole? Old Concepts and New. Jack Lorimer. In CHEMISTRY International January-February 2010.
[3] Comments on recent proposals for redefining the mole and kilogram. B. P. Leonard. Metrologia 47 (2010) L5–L8.
[4] The Mole is Not an Ordinary Measurement Unit. Ingvar Johansson
Accred Qual Assur (2011) 16:467–470.
EDIT
I want to add the words that a synthetic chemist wrote in a letter to C&EN:

In theory, we can dispense with the mole (C&EN, Aug. 4, page 32). The mole, molarity, and molality were adopted as convenient ways to express relative amounts and concentrations of substances when we didn’t know the actual mass of atoms and molecules—when the very existence of atoms and molecules was in dispute.
Now we know, which has enabled us to come up with the number $6.022 × 10^{23}$, give or take, and requires us to explain to students—who question why anyone would specify such a ridiculous number— that it wasn’t chosen but is experimentally derived. Now that we know the actual mass of atoms, ions, and molecules, we can describe the actual number of entities directly—for example, an acid as $1.5 \; \mathrm{YH^{+}}$ per liter ($\mathrm{Y} = \mathrm{yotta} = 10^{24}$), which is about $2.5 \; \mathrm{mol \, H^{+}/L}$. We could rid ourselves of an unnecessary and confusing-to-beginners concept, but of course it will never happen.

I agree with his viewpoint, except maybe with his "it will never happen". Time will say.

I think it’s safe to assume that over $90~%$ of the users here are chemists. As a chemist, I fail to grasp your points one to seven, although not grasping the difficulty of two to four could be due to me being an organic chemist, not a physical one. In order: 1) In my life I have not yet experienced an SI prefix larger than tera ($10^{12}$). I need thousands or millions of ‘tera’ to reach the range of a mole, so I would be learning new prefixes just for that. 2) Nobody in organic or inorganic chemistry even knows these exist.

learning the definitions is not required, you just need to understand how to apply them. Whether the molar mass of carbon or the Avogadro constant have an uncertainty is fully irrelevant because their uncertainties are way below any other lab uncertainties. 4) The correction factors average out to $1$ for practically all lab work. 5) I don’t find amount abstruse at all. And I can safely ignore weird renamings much like I can mostly ignore the unit psi if it pops up. 6) … seriously …? 7) Technically, all equations are redundant. In this case, you just forgot the transformation

factor $k_\mathrm{B} N_\mathrm{A} = R$ which links the two. In the same vein, there are numerous ‘redundant’ equations for energy.

I know you are a chemist, you started your post stating at. I was just reminding you that it’s safe to say practically all power users here are chemists. I don’t think I need to go through your answers to me comments because they don’t offer new information. Except that I rebuke 3 for being wrong and that I see the problem in 5) even less.

In a world where many chemists still rely on kcal/mol, I'm afraid your vision of abandoning the mole will be difficult to pursue.

Why oh why don't we just abandon kcal and just use the kJ?  :(

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text:
Name the following coordination complex
What are the IUPAC rules when naming a coordination complex like:

$$\ce{[Pt(NH3)4][Pt(NH3)Cl3]}$$

Now according to my textbook, it is named as tetraammineplatinum(II) amminetrichloridoplatinate(II). Both the $\ce{Pt}$ atoms have the oxidation state +2 .There are no charges assigned to the two complex ions in the formula, so, my assumption is that the entire complex is neutral and thus the oxidation states for $\ce{Pt}$ should be assigned accordingly. However, on assigning the oxidation states as mentioned in the answer the assumption is proven wrong. I'm confused about the mentioned oxidation numbers.

How do you calculate the oxidation numbers of the metal atoms in such a complex?
Shouldn't the charge on each complex ion be mentioned in the formula if the oxidation states given in the answer are correct?

This is a salt of two complex ions. Consider it as two separate complexes, happened to be in same salt: one is cationic $\ce{[Pt(NH3)_4]^{2+}}$ and another is anionic $\ce{[Pt(NH3)Cl3]^-}$. However, it seems that it should have two anions, with formula $\ce{[Pt(NH3)_4] [Pt(NH3)Cl3]2}$

That's exactly what I was thinking, but I asked an expert who told me that the given formula was correct. He was kinda' unsure though.

He is incorrect in this particular regard. Howevere, there is $\ce{[PtCl4]^{2-}}$ anion, wich could  ... messed with his line of though.

Your textbook contains an error. Either the formula is written incorrectly or the name is. Since according to comments the name is correct, it must be an incorrect formula. Let’s break it down:
If the formula is correct
We have two complex ions that together create a doubly-complex salt. According to the rules of (pseudo-) binary ionic compounds, one ion must be positively charged and the other negatively. By convention, the cation is written first so it is safe to assume that $\ce{[Pt(NH3)4]^{?}}$ be the cation — which fits in well with chemical intuition because $\ce{NH3}$ is a neutral ligand and platinum is a metal.
Therefore, $\ce{[Pt(NH3)Cl3]^{?}}$ must be the anion — very nice, because whe have chlorido ligands that are commonly negative. Since every chlorido ligand adds $-1$ charge to the complex, the anion can be anything from $-1$ to $-3$ with decreasing probability.
The complex needs to be a $1\,:\,1$ binary complex, so the negative charge needs to be equivalent to the positive charge. From here onwards it gets difficult. We have the following possibilities:

$\ce{[Pt^{I}(NH3)4]^{+} [Pt^{II}(NH3)Cl3]^-}$
$\ce{[Pt^{II}(NH3)4]^2+ [Pt^{I}(NH3)Cl3]^2-}$
$\ce{[Pt^{III}(NH3)4]^3+ [Pt^{0}(NH3)Cl3]^3-}$

They all differ in the assumed oxidative states of platinum. We cannot decide a priori which one of these three is correct, although the last one is by far less probable. Technically, we can even assume further diverging oxidation states but they get proportionally less probable. Only if the charge were clearly stated could we distinguish between the cases. For the sake of consistency, here are the names:

tetraammineplatinum(I) amminetrichloridoplatinate(II)
tetraammineplatinum(II) amminetrichloridoplatinate(I)
tetraammineplatinum(III) amminetrichloridoplatinate(0)

If the name is correct
Then we can logically build up the resulting salt from the name. Let’s start with the cation:

It is platinum(II), so we get $\ce{Pt^2+}$
adding four ammine ligands gives us $\ce{[Pt(NH3)4]^2+}$

For the anion:

It is still platinum(II) (although rendered as platinate, so we need to end up with an anion): $\ce{Pt^2+}$
We have one ammine ligand: $\ce{[Pt(NH3)]^2+}$
and three chlorido ($\ce{Cl-}$) ligands: $\ce{[Pt(NH3)Cl3]-}$

Since we have a cation with $+2$ charge and an anion with $-1$ charge, we need two anions per cation, giving us the final formula:
$$\ce{[Pt(NH3)4][Pt(NH3)Cl3]2}$$
So what is correct?
Going by chemical intuition (platinum(II) being a well-known $\mathrm{d}^8$ system that forms square-planar complexes), I would say that the name is correct and the formula isn’t. This also has the implication that they messed up the question by missing a subscripted 2 (rather likely) rather than messing up the answer by getting a Roman numeral wrong (not quite as likely).
Hence there is nothing miraculous or confusing about assigning oxidation states to metals in complexes — and the IUPAC even formally defines oxidation states for promiscuous ligands such as nitrosyl (which is always considered to be a radical neutral ligand, never $\ce{NO^+}$ or $\ce{NO-}$) allowing for a clean-cut naming of complex compounds without knowing the chemistry involved.
The charge does not have to be mentioned in doubly-complex salts for any component. However, it always helps students to give them and in non-trivial cases (like the assumed correct formula) it might be vital to know the overall charges. If there are no charges written, though, the final product has to be assumed neutral.


text:
Mass percentage of potassium chlorate in decomposition reaction

When heated, a $3.40\ \mathrm g$ sample mixture of $\ce{KClO3}$ decomposes and produces $685\ \mathrm{cm^3}$ $\ce{O2}$ gas at $30\ \mathrm{^\circ C}$ and $101\ \mathrm{kPa}$. What is the mass percentage of $\ce{KClO3}$ in the mixture?

What I've done so far is write out a balanced equation which is:
$$\ce{2KClO3 -> 3O2 + 2KCl}$$
Where do I go from here?

Generally, it is preferable to write down the complete quantity equation before plugging in the numbers. If necessary, perform all mathematical operations on quantities symbolically. Do not plug in the numbers until you have only one equation for the desired result. In the following, we use this method to calculate the amount of oxygen $n_{\ce{O2}}$, the mass of potassium chlorate $m_{\ce{KClO3}}$, and the mass fraction of potassium chlorate $w_{\ce{KClO3}}$ directly from the parameter values given in the question (although only the last calculation is actually required to answer the question).
Amount of oxygen
For simplicity’s sake, you may use the ideal gas law to estimate the amount of oxygen $n_{\ce{O2}}$ as follows. The volume $V=685\ \mathrm{cm^3}=0.000\,685\ \mathrm{m^3}$, the temperature $T=30\ \mathrm{^\circ C}=303.15\ \mathrm K$, and the pressure $p=101\ \mathrm{kPa}=101\,000\ \mathrm{Pa}$ are given in the question. The molar gas constant is $R=8.314\,459\,8(48)\ \mathrm{J\ mol^{-1}\ K^{-1}}$.
$$\begin{align}
p\cdot V&=n_{\ce{O2}}\cdot R\cdot T\tag1\\[6pt]
n_{\ce{O2}}&=\frac{p\cdot V}{R\cdot T}\tag2\\[6pt]
&=\frac{101\,000\ \mathrm{Pa}\times0.000\,685\ \mathrm{m^3}}{8.314\,459\,8\ \mathrm{J\ mol^{-1}\ K^{-1}}\times303.15\ \mathrm K}\\[6pt]
&=0.027\,448\,609\ \mathrm{mol}\\[6pt]
&\approx0.027\,4\ \mathrm{mol}
\end{align}$$
By way of comparison, the amount of oxygen calculated using standard reference data for oxygen instead of using the ideal gas law yields $n_{\ce{O2}}=0.027\,47\ \mathrm{mol}$. Considering the estimated uncertainty of the result obtained by using the ideal gas law, the difference is not significant. Hence, the approximation using the ideal gas law is acceptable in this case.
Mass of potassium chlorate
The chemical equation
$$\ce{2KClO3 -> 3O2 + 2KCl}\tag7$$
indicates that $2\ \mathrm{mol}$ of potassium chlorate yield $3\ \mathrm{mol}$ of oxygen. Therefore,
$$n_{\ce{KClO3}}=\frac23\cdot n_{\ce{O2}}\tag8$$
The molar mass of potassium chlorate is $M_{\ce{KClO3}}=122.55\ \mathrm{g\ mol^{-1}}$. Since the molar mass $M_{\ce{KClO3}}$ is defined as
$$M_{\ce{KClO3}}=\frac {m_{\ce{KClO3}}}{n_{\ce{KClO3}}}\tag9$$
the mass of potassium chlorate $m_{\ce{KClO3}}$ is
$$\begin{align}
m_{\ce{KClO3}}&=M_{\ce{KClO3}}\cdot n_{\ce{KClO3}}\tag{10}\\[6pt]
&=M_{\ce{KClO3}}\cdot \frac23\cdot n_{\ce{O2}}\tag{11}\\[6pt]
&=M_{\ce{KClO3}}\cdot \frac23\cdot\frac{p\cdot V}{R\cdot T}\tag{12}\\[6pt]
&=122.55\ \mathrm{g\ mol^{-1}}\times\frac23\times\frac{101\,000\ \mathrm{Pa}\times0.000\,685\ \mathrm{m^3}}{8.314\,459\,8\ \mathrm{J\ mol^{-1}\ K^{-1}}\times303.15\ \mathrm K}\\[6pt]
&=2.242\,551\,324\ \mathrm g\\[6pt]
&\approx2.24\ \mathrm g
\end{align}$$
Mass fraction of potassium chlorate
The mass fraction of potassium chlorate $w_{\ce{KClO3}}$ is defined as
$$w_{\ce{KClO3}}=\frac{m_{\ce{KClO3}}}{m}\tag{13a}$$
where $m=3.40\ \mathrm g$ is the total mass of the sample. Therefore,
$$\begin{align}
w_{\ce{KClO3}}&=m_{\ce{KClO3}}\cdot\frac1m\tag{13b}\\[6pt]
&=M_{\ce{KClO3}}\cdot \frac23\cdot\frac{p\cdot V}{R\cdot T}\cdot\frac1m\tag{14}\\[6pt]
&=122.55\ \mathrm{g\ mol^{-1}}\times\frac23\times\frac{101\,000\ \mathrm{Pa}\times0.000\,685\ \mathrm{m^3}}{8.314\,459\,8\ \mathrm{J\ mol^{-1}\ K^{-1}}\times303.15\ \mathrm K}\times\frac1{3.40\ \mathrm g}\\[6pt]
&=0.659\,573\,919\\[6pt]
&\approx0.660=66.0\ \%
\end{align}$$


text:
Are long oxygen molecules possible?
Are large oxygen containing molecules possible? Either large rings, or chains with hydrogen atoms at the ends. Like this:
$\ce{H-O-O-O-O-O-O-O-O-O-$\cdots$-O-O-O-O-O-H}$

Related: Why does sulphur, but not oxygen, catenate?

See also https://en.wikipedia.org/wiki/Allotropes_of_oxygen https://en.wikipedia.org/wiki/Tetraoxygen https://en.wikipedia.org/wiki/Solid_oxygen

Yes, they are possible; several have even been detected.  They all have the general formula $\ce{H2O_{n}}$ and belong to the class of hydrogen polyoxides.
Of course, the $n=1$ member of the series is water $(\ce{H2O})$, and the $n=2$ member is hydrogen peroxide $(\ce{H2O2})$.
The $n=3$ member is trioxidane $(\ce{H2O3})$, and it has the following structure:

It can be prepared in a number of ways as outlined in the linked article, such as the reaction of ozone and hydrogen peroxide. Trioxidane is also an active antimicrobial ingredient. Trioxidane has a half-life around 16 minutes in organic solvents, but only milliseconds in water.
Hydrogen tetroxide ($n=4$, $\ce{H2O4}$) is also a known compound as is hydrogen pentoxide ($n=5$, $\ce{H2O5}$).  Hydrogen pentoxide is formed as a byproduct in the preparation of trioxidane.
This abstract of a computational study provides bond dissociation energies for the hydrogen polyoxides. Starting from water, they steadily decrease (the $\ce{O-O}$ bonds become weaker) to $\ce{H2O6}$ and then start to increase again.
Abstracts from other computational studies dealing with the expected structures and stabilities of these molecules can be found here and here.

A recent paper in the Angewandte claims to even have synthesised pure $\ce{H2O3}$ as a white crystalline solid. — G. Strle, J. Cerkovnik, Angew. Chem. Int. Ed. 2015, 54, 9917.

Wow!  Hopefully no (in)flammables were around ... .


text:
Which are the appropriate washing solvents for gas chromatography equipment?
I work in a laboratory where gas chromatography is used extensively. We handle terpenes and terpenoids (alcohols, ketones, and acetates derived from terpenes), organic solvents (acetone, acetonitrile, ethyl acetate, hexane, and toluene), acetic acid, and water.
I want to know which are the most appropriate solvents to use for washing the equipment or which criteria should I use to decide among the existing solvents. Now, we are using acetone and ethanol, but sometimes it seems that they do not work for some terpenoids (the syringe sticks and sometimes the equipment crashes). Any suggestions about this?

What does the manual for you system say?


text:
What angle limits should apply to Graham condensers?
I have seen Graham condensers set at an angle (perhaps as low as 30 degrees from horizontal).  I guessed this wouldn't be a problem because siphoning would clear the condensate.  But I recently saw an argument that Graham condensers should only be used at near-vertical angles.  Thinking harder this makes sense to me, because as soon as enough liquid condenses in the first loop to form a trap the condenser can't work on any other vapor.  Absent some vacuum on the receiving end the only way it could clear itself is if there happened to be enough vapor to eventually close a loop and start the siphon effect, or if enough pressure built up on the boiling end to push the liquid through which would be bad!
But I still see diagrams and photos of Graham condensers set at shallow angles.  Here's an example from the first few Google image results:

So when and why, if ever, is it considered a good practice to run a Graham condenser in a non-vertical orientation?

I have no idea why there are so many photos showing Graham condensers in a non-vertical position.  Maybe just because they look so cool and photos tend to be landscape.  The only time that might be safe is if there were a vacuum pulling to the collection side.
Graham condensers should be used vertically, precisely for the reasons you stated: Particularly because otherwise they will bubble, bump, and potentially explode.

One might add that for non-vertical use, the plain old Liebig-type is the only suitable choice. I wonder why anyone calls this Graham-thing a condenser, it's just a cooler.

As far as I can tell, the main reason that graham condensers are shown in near horizontal position is because the setup required to put it properly in a vertical position looks odd to the untrained eye, and requires quite a tall support. The boiling flask, heat source, and so on are elevated as high or higher than the top of the condenser.
The Graham is really only useful as a condenser in the vertical position, and even then, it will tend to "spit" as condensate traps gas above it. To my knowledge it is almost exclusively used in vacuum distillation, for that reason. In spite of numerous drawbacks, it does put a great deal of coolant around the coil. As mentioned, it would be downright daft to use this as a reflux column.
Why this very mediocre piece of glassware should be so common (about half of all distillation apparatus kits seem come with these), is beyond me. Even a plain Liebig condenser is of greater value for any common use that a graham might be used for (vacuum distillation aside). The technique for making graham condensers (to outward appearances) seems like it would be extremely similar to that for making coil condensers, which are far, far more useful, both as reflux columns and distillation condensers, so why the Graham would be so much cheaper is beyond me.

Your reasoning seems sound. For myself, I wonder why anyone would want to use this type of condenser at all. It has a very small bore which reduces the effective surface area. There are a number of other (better) types of condensers to choose from. Was in the lab for 20 years and never found a reason to use this type of condenser.

Maybe they're only used for photos (because they look so neat)?  The small bore does seem problematic, but in simple terms of effective surface area my guess is the coiling actually maximizes that for a given condenser length.

I came here through the VLQ queue. While I believe this answer has enough quality not to be treated with the trash we deal with in this queue, I think you should be a little bit more elaborate. Shrugs Some might think this is better as a comment.

Perhaps this answer could be improved by explaining when and why you use the other condenser types?  Sort of a process of elimination?  (Because it didn't take long to convince myself that a small-diameter but proportionally longer (coiled) bore has the maximum surface area as compared to other liquid-cooled glass condensers of the same length of the Liebig or Allihn style.  Spiral cold-fingers like a Dimroth, on the other hand, would appear to be yet more efficient, though a Friedrichs would appear to have the same perils as I cite for the Graham.)

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


text:
Why does phosphorus give a free electron in silicon doping?
Extrinsic semiconductors are created by doping an intrinsic semiconductor; typically silicon is doped with phosphorus to create free electrons or boron to create "holes".
In the case of phosphorus doping, phosphorus takes the place of a silicon atom in the silicon lattice, and its fifth electron becomes a free electron.
Here is what I am struggling to understand with my basic knowledge. Phosphorus has the configuration $\ce{[Ne] 3s^2 3p_x^1 3p_y^1 3p_z^1}$.
This page gives an description of the promotion of a 3s electron to a 3d shell, allowing five bonds to form in the case of $\ce{PCl5}$. I would assume that a similar process results in four bonds forming in the silicon lattice and leaving a free electron.
But that same page later on mentions a 2007 paper which essentially says that that model is not accurate, because such a state is less stable than other ways of modeling the bond (assuming I'm understanding correctly. The problem is described here, I have not read the actual paper.
What causes phosphorus to form four bonds with silicon and leave a free electron rather than, say, forming three bonds and leaving a silicon with an unpaired electron in the lattice, or bonding with another "free" electron? Is it promoting an electron to form the fourth bond and freeing the last?
In short: Why does phosphorus form four Si bonds and leave a free electron rather than forming three Si bonds when doping silicon?

First, don't forget that the phosphorus is sitting on a diamond cubic silicon lattice site, which pretty much defines the symmetry of the potential around it - the P cannot change that, but must respond to it. So, you should not think in terms of atomic orbitals.
Second, the fifth electron is not initially a free electron - it occupies a state that happens to be in the band gap.  Now, that state is really close to the conduction band, only 45 meV below it at 300 K, so at room temperature there is a really good chance that it will jump into and roam about the conduction band.  Nor is that the only state in the gap for P - there is another state 72 meV above the valence band (i.e. it is slightly easier to ionize a 4-fold coordinated P than a Si).
Why doesn't the P only bond with 3 Si? Well, that would leave one unhappy Si atom, and one not much happier P atom.  Again, this is occurring in a crystal, not an isolated molecule. The P will occupy the lowest free energy state. Note that there are other impurities that preferentially occupy various interstitial sites. It is just that P can be coerced into pretending to be pretty much a Si atom - peer pressure, perhaps...


text:
Solubility of Gold in low concentrations
When preparing ICP-MS calibration standards for trace metal (~100 ppt) analysis, the values for Gold do not remain stable for more than a day. The standards are prepared using commercially available 10 ppm solutions in Nitric Acid with traces of HF.
A known amount of the Gold standard is added to 2 mL of 70% ultra pure Nitric Acid, an Indium Internal Standard and diluted to a pre-determined volume with ultra pure deionized water. The samples being analyzed are in 49% hydrofluoric acid to which has been added 2 mL of 70% ultra pure Nitric Acid, an Indium Internal Standard and diluted with an equal volume of ultra pure deionized water.
Would preparing matrix-matched calibration standards solve this problem?

Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center. || To other users: Please note that this isn't HW, as the question seems to demonstrate the effort  and is more conceptual that a normal homework question.

Whenever you prepare very dilute working standards, it's good practice to do a stability study to see how long you can store them. There are many factors that can influence the stability of these solutions. The choice of container can make a big difference. Dilute metal solutions should be stored in polyethylene or polypropylene containers, not glass. Another well-known problem is photoreduction. Gold solutions should be stored in opaque containers and be exposed to light as little as possible. Some solutions may also be sensitive to atmospheric exposure. (though I don't think it's an issue for gold)
I don't have experience with working with gold in hydrofluoric acid, but such a large change in such a short period of time in the absence of the above factors may indicate some actual reactivity with your matrix, for which matrix-matching might help, though you may be better off reformulating your solutions if that's the case. It may also be an option to just make fresh working standards very frequently or increase storage life through refrigeration.

Bingo.  Storing standards and samples for metals analysis in glass containers is a huge problem when doing metals analysis.


text:
When do one have to actually calculate the net charge of an amino acid?
When the pH and the pKa levels differ by 2 or more you don't need to calculate the net charge. What you do then is to just count the negative and positive charge and add them up.
When pH=pKa you also don't need to calculate, because you get half of the different types.
BUT what about if we want to know the net charge for an amino acid where one of the groups have an pKa value of 4,25 , when the pH is
a) 4,0  , b)5 , c) 5,5 and d)6
Do you have to calculate it for all of these? Can you say that since a) is pretty close to the pKa we consider them the same, or that since d) is pretty close to being 2 units more, we can just say that and make it easier? And is there even really a point in calculating for c) when it's more than 1 pH unit over?
My question is basically: where do I draw the line?

You draw the line where the error is acceptable for your application. Without knowing what you want to know the charge for, it's hard to give much more of an answer than that. N.B. many software packages like MarvinSketch will easily calculate the average charge and even give you species distributions for different pH values.

Your question hardly makes sense to others, because you provide your thinking and a fragment of the info in the question, and there's no sign of any real question. Please see What is the XY problem?, and edit your question to include a real question.


text:
Why does water spills on our shirt change its colour?
While drinking water, some of it spilled on my shirt and as usual, as has happened to many of us, the colour darkened. So why does this happen? Is it due to some chemical effect or physical one?

For anyone interested, this question was asked and answered on Physics.SE by another user before.

This is not so much a change (physical and especially not chemical), it is more about perception of the diffuse reflection and refraction of light. According to Australia's ABC Science page Darker when wet:

When your clothes are wet, all those air gaps (between the fibres) get filled with water. So when light hits a wet patch it's moving in and out of water and fibres, not air and fibres. The refractive index of fabric is a lot closer to that of water than it is to air, so the light doesn't change speed quite as dramatically going between water and fibres. And that means it bends at a smaller angle when it goes from one to the other.

This is summarised in the diagram below:

The information on the linked page is from Professor David Jamieson from the School of Physics at The University of Melbourne, Australia.
Another explanation according to Dr. Christopher Baird of the website Science Questions with Surprising Answers is that we perceive a darkening due to the fabric becoming more transparent - an explanation from the website:

Spill some water on your white shirt and the situation changes. In the
  spot that is wet, the water fills all the little pockets in the fabric
  that used to be filled with air. In terms of its interaction with
  light, water behaves very similar to fiber material. Adding water to
  fabric effectively removes the multiple reflecting surfaces (at least
  as far as the light is concerned) and turns it back optically to a
  solid chunk of material, which is transparent.

Essentially, what we are perceiving when the fabric appears to darken, is due to the colour of what is underneath - view the cloth with a light source behind it, it appears 'lighter', due to being more transparent.


text:
Mixing 1 cubic meter of HCl and NaOH: Temperature rise of a metal container?
Say I (perfectly) mix $1\ \mathrm{m^3}$ of 1 M $\ce{HCl}$ with $1\ \mathrm{m^3}$ of 1 M $\ce{NaOH}$ at standard air pressure and room temperature within a metal vessel. Heat of neutralization for strong acids and bases is $-55.8\ \mathrm{kJ/mol}$ (is that actually just the bond enthalpy of an $\ce{H-OH}$ bond?).
So the $2\ \mathrm{m^3}$ of $\ce{H2O}$ amount to around $2 \cdot 10^6\ \mathrm{mol}$ of water ($M(\ce{H2O}) \overset{\wedge}{=} 20\ \mathrm{g\ mol^{-1}}$). Therefore, $\Delta_\mathrm n H$ is around $2 \cdot 10^6 \times 55.8\ \mathrm{kJ/mol}$, around $110 \cdot 10^6\ \mathrm{kJ}$.
I would now like to calculate the temperature rise.
The heat released is $Q = m c_p \Delta T$, where $m$ is the mass of the solution ($2000\ \mathrm{kg}$), $c_p$ is the specific heat capacity of the solution and $\Delta T$ ist the change in temperature. But that's exactly what I'm trying to find. So unless I know $Q$, I can't find $\Delta T$. Therfore, how do I find $Q$?

Please don't use MathJax in titles due to searchability issues and URL slug issue. (I don't think the compound $\ce{CeHCl}$ exists)

You have 1 mole/liter of each reactant.  How many liters of each solution do you have?  That gives you the number of moles of reactants. They will create way less than $2\cdot 10^6$ moles of water-you were counting the water that was already there.  Once you have the right number of moles of reactant, that is the number of moles of water produced.  Multiplying by your heat per mole gives $Q$.  Then divide by $mc_p$ to get $\Delta T$

Which theory tells me that here in this case $H$ is just another symbol for $Q$?

They are defined differently, but in this problem you are supposed to recognize that they are the same. $H$ is the energy released by one class of reactions.  You have claimed that this reaction is typical of the class and it is the only reaction going on. In that case it is the equal to $Q$.

As Ross has said, $\Delta H$ is the same as $q$. This is true only under three conditions:

Pressure is constant;
Only p-V work is performed;
The process occurs in a closed system.

The First Law of Thermodynamics states that $\mathrm{d}U = \mathrm{d}q + \mathrm{d}w$. The expression for p-V work is $\mathrm{d}w = -p\,\mathrm{d}V$, giving $\mathrm{d}U = dq - p\,\mathrm{d}V$.
Enthalpy, $H$, is defined as $H = U + pV$. At constant pressure, this means that $$\begin{align}
\mathrm{d}H &= \mathrm{d}U + p\,\mathrm{d}V + V\,\mathrm{d}p \\
&= \mathrm{d}U + p\,\mathrm{d}V \quad\quad \mathrm{(since}\,\mathrm{d}p = 0) \\
&= \mathrm{d}q - p\,\mathrm{d}V + p\,\mathrm{d}V \\
&= \mathrm{d}q
\end{align}
$$
Integrating over the path of interest, you would find that $\Delta H = q$.

Apart from that, I think you would also find that the number of moles of water produced is wrong. You will have to calculate it stoichiometrically, using the equation:
$$\ce{HCl + NaOH -> H2O + NaCl}$$

2 cubic meters of water aren't produced in the reaction, but they come from the acid and base solutions, so I do not see why you are multiplying 2 cubic meters of water with the enthalpy of neutralization. By multiplying the amount of water actually produced during neutralisation with the enthalpy of neutralisation, you get Q, the amount of heat released, so the temperature is the only unknown.


text:
Formalin fallen out of solution
Is there a way to bring back into solution a bottle of formalin that has crystallized? The bottle has been stored for several years at room temperature.
Thanks in advance.

I don't have the knowledge to answer this, but if the answerer feels this lacks some information, I'd be happy if they comment.

The 'crystals' are polyformaldehyde, more commonly called paraformaldehyde.
Heat it up, with some base catalyst.


text:
Do only metals form ions?
Do nonmetals form ions? I am asking this because I was reading something and I came across the phrase "metal ions," which left me to wonder whether or not there are nonmetal ions as well.

Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center. || There are a lot of nonmetal ions. The book may wanted to separate the two because the metals usually form cations.

In short, yes. Single-element ions are common for both metals and non-metals. Some simple examples: $\ce{F^{-}, Cl^{-}, I^{-}, S^2-}$ Multi-element non-metal cations are a bit more common: $\ce{NH4+}$ and $\ce{H3O+}$ being the obvious examples and there are plenty of anions like $\ce{CN-}$, $\ce{SO4^{2-}}$, $\ce{NO3-}$, etc. Things like proteins and polypeptides are also frequently ions in solution and may be positively or negatively charged depending on the pH.
Simply due to their electronegativities, all the single element non-metal ions are anions in solution, but in the gas phase, all bets are off and you can knock off an electron to get a cation, (or have negatively charged metal ions) so long as it doesn't bump into anything else.

Ammonium nitrate is a fertilizer that is composed of nonmetal ions.
There are myriad other examples.

Ions are classified into cation  and anion, cation is a positive ion and anion a negative ion. Cations are mostly from metal nature while anions are mostly from non-metals, so yeah there are non-metal ions!


text:
Why is CO practically nonpolar?
This question was in my book. According to me CO should be polar as it should have a dipole moment.
But I found that the $\sigma$-electron drift from C to O is almost nullified by the $\pi$-electron drift from O to C. What is about the $\pi$-electron drift towards the Carbon atom?
In Google it is given CO is polar, but my book says it is practically nonpolar ... now I am confused.

CO is polar only. it is proved by its activity to ir and microwave rays,

@user19485 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]. At the moment this read more like a comment than an actual answer - could you elaborate a little more. With a bit more rep, you will be able to post comments on any question/answer. For now I have converted your answer to a comment.

Since the difference in the polarity caused by sigma drift and pi drift is quite small, it is practically non polar. But, chemically, it is polar in nature, as you see in the answers below.

Well, it's just like you said: $\sigma$ electrons drift from C to O, because O is more electronegative, and $\pi$ electrons drift in the opposite direction, because... Look at it this way: C and O are both sp, each with a lone pair facing away from the other atom. They form a $\sigma$ bond and a $\pi$ bond, too. Now they want to form another $\pi$ bond, but C got no more electrons to use, so O has to throw in two of his own. That's the drift we are talking about.
As it happens, the two effects almost negate each other. The resulting dipole moment is so insignificant, I don't even remember which way it is.

This is not true. All orbitals except for the HOMO, including the two pi orbitals, are polarised towards the oxygen. The lone pair of carbon has a larger dipole moment than all of the other orbitals combined due to its size and location.

I never said they aren't. Surely all orbitals are polarized towards O. Now, on one of these orbitals sit two electrons that were his in the first place. They belonged to O, and now they are somewhere between C and O - admittedly, closer to O, but still in between. That's the drift I'm talking about.
As for the lone pair of carbon, isn't it matched by that of oxygen?

It's not like I'm inventing a bicycle here. Look: https://en.wikipedia.org/wiki/Carbon_monoxide#Bonding_and_dipole_moment

Preliminaries: I am using the wrong (but still common) notation of the dipole moment. Please see the question about the direction of the dipole moment.

The reason why carbon monoxide is often referred to a being practically nonpolar is because of its very small dipole moment.
\begin{align}
\ce{{}^{\ominus}\!:C#O:^{\oplus}} &&
\text{Dipole:}~|\mathbf{q}|=0.11~\mathrm{D} &&
\text{Direction:}~\longleftarrow\hspace{-1ex}+
\end{align}
Unfortunately the explanation of the bonding in carbon monoxide is far from simple, and it is perfectly understandable that you are confused. The dipole moment can be very well explained with molecular orbital theory, which I have already written about.
I will include a brief explanation at this point as I saw that the explanation on Wikipedia is from my point of view incomplete (not to say wrong).
Let's first state the obvious: Oxygen is more electro-negative than carbon. Therefore one can expect that all bonding orbitals are polarised towards the oxygen. That is of course only true when we are staying in the Lewis bonding picture. Based on this one would expect that the lone pairs cancel each other and since all orbitals of the triple bond are polarised to the oxygen, the dipole moment should also point in that direction.
This is obviously wrong, since the experiment tells us something different. There are ten valence electron in this molecule. They all occupy orbitals with bonding character, i.e. there is no nodal plane perpendicular to the bonding axis between the elements. The low lying orbitals are all polarised towards oxygen, because of electronegativity. The HOMO on the other hand is strongly polarised towards the carbon and is very large. In addition the nuclei contribute a dipole moment because of their asymmetry of charge.  With less positive charge on the carbon nucleus than on the oxygen nucleus, the dipole component of the nuclear charge contribution also has its negative end on carbon.  The combination of HOMO and nuclear dipoles in the reverse direction is larger than the combination of all the other occupied orbitals.

It is important to understand that it is often not possible to know about the properties of a compound simply by looking at a structure on paper. Very often the simple models we love and use break down easily for small molecules. A bonding situation has to be explained as is and not in terms of what would have been when we look at individual atoms and how does it change.

‘They all occupy orbitals with bonding character, i.e. there is no nodal plane perpendicular to the bonding axis between the elements.’ I don’t know for certain, but in the picture you showed in the other post I see MO number 4 which looks antibonding to me …

@Jan It is not, the nodal plane 'curves' through the oxygen. You can only see that when you would cut through the molecule. The same applies for the home, but it's the carbon in this case.

@Aneek when it comes to carbon monoxide, there are no simple answers. The accepted one is wrong, but very easy to comprehend. This one is harder to understand, but correct. You can choose.

There is actually another dipole contribution at work ... from the nuclei.  Carbon has six protons there, oxygen has eight, thus the charge distribution from the nuclei is the same as a pair of +7 charges plus a dipole of +/-1 charges separated by the internuclear distance.  This latter dipole, like the net dipole  from the lone pairs, is negative towards carbon.

@martin check my edit.  We need to add the nuclear contribution because it is bigger than the net result.

@oscar I really need to think about this. In the calculations I have linked a am quite certain, that this is already included for each calculation (it is a static field in a way). But I don't currently have the time to do that now. For the time being, I don't really disagree with the fact, so I let the edit live. But I need to think about the double counting so I might (big might) be coming back on that.... Thanks for the input though..

Let me know how it turns out.  If it were up to me I would be rendering the nuclei separately.  Easier to understand the physics that way.


text:
Aromatic Elimination Addition
Chlorobenzene will react with hydroxide to give phenol. But will the reaction on Chlorobenzene also work with methoxide, to give anisole?

Note that it takes brute force or good catalysts to turn chlorobenzene into phenol. SciFinder gave me a few results (10 % of what I got for phenol) of turning chlorobenzene into anisole with heat and catalysts. None from the early 20th century though, so either they didn't boil it in methanol or it didn't work directly.

Yes, It will.
It's a Nucleophilic Substitution reaction,
Chlorobenzene + Sodium Methoxide --> (Under extreme conditions ) Anisole
Extreme conditions because Aryl halides don't go Nucleophilic Substitution easily.

Without other electron withdrawing groups on the ring I'm not so sure that it is a nucleophilic substitution reaction.  The reaction could proceed through an aryne intermediate.

Actually my chemistry prof told me that it will go thu Nu Substitution, I am just sharing knowledge here


text:
Spontaneity of reversible reactions

If a reaction has positive $\Delta G$ then reverse reaction has negative $\Delta G$. How can both occur simultaneously, e.g., in a reversible reaction such as dimerisation of $\ce{NO2}$ to form $\ce{N2O4}$? Do forward and backward reactions occur spontaneously? How can both directions be spontaneous at the same time?

I read in some book that  Haber's process to form ammonia is spontaneous at a temperature below $170\ \mathrm{^\circ C}$ (the author proved this using $\Delta G < 0$ for spontaneous process). This means that above this temperature $\ce{N2}$ and $\ce{H2}$ shouldn’t combine to form $\ce{NH3}$. But industrially Haber’s process is performed at temperature around $400\ \mathrm{^\circ C}$. Isn't this a contradiction?

I just want to be really picky about your use of the symbol $\Delta G$. The symbol $\Delta G$ can be used at any stage of the reaction and in fact, gives no information about the final position of equilibrium. The correct symbol to use is $\Delta G^\circ$. (Actually it's not a circle, it should be the standard state symbol, but whatever.) The former varies as the reaction proceeds; the latter is a constant, defined to be equal to $\Delta G$ when $Q = 1$, that determines the position of equilibrium.

Refer to this question for more info - http://chemistry.stackexchange.com/questions/28644/what-is-the-difference-between-%CE%94g-and-%CE%94g-with-an-o-on-top

If a reaction is not reversible, spontaneity depends on free energy being positive or negative. If a reaction is reversible, it can be spontaneous in both direction until equilibrium is reached.

You seem to be mixing up thermodynamic and kinetic considerations. All reactions occur spontaneously all the time, and their backward reactions too. If left alone for long enough, a system will eventually reach the point where the forward and backward reactions go at the same rate (and it's not like they both stop altogether!) - that's an equilibrium. Where that equilibrium will be is judged by $\Delta G$; if it is negative, then the equilibrium will be closer to the products.
As for ammonia, that's where the kinetics kicks in. At room temperature $\Delta G$ is in our favor, but the equilibrium would take forever to reach. We raise the temperature so as to make the reaction faster, but that also shifts the equilibrium in the wrong direction. So we raise the pressure, too. This does the trick: now the reaction goes the way we need, and at reasonable rate.

Great explanation but I just want to be picky (same thing with the question text) - the correct symbol in this context is $\Delta G^\circ$ and not $\Delta G$.

The sign of $\Delta G^\circ$ is not an indication of whether a reaction will occur spontaneously.  For a reversible reaction, if you start with a mixture of only reactants, the reaction will proceed to produce products until the system reaches equilibrium.  If you start with a mixture of only the products (i.e., the products of the forward reaction), the reverse reaction will proceed to produce the reactants (of the forward reaction) until the system reaches equilibrium.

$\Delta G^⦵$ is the free energy change, for the reaction as written, of the reaction of the reactants and products in their standard states to their concentrations at equilibrium. The reaction written one way will be the negative of the reverse.
$\Delta G$ is the free energy change for the actual concentrations involved. At equilibrium $\Delta G = 0,$ the activities are the equilibrium activities, and $\Delta G^⦵ = -RT\ln K_\mathrm{eq}$.
As Chet Miller states, if there are only reactants, regardless of the reaction direction $\Delta G$ will be negative and the reaction will proceed to equilibrium. The extent of the reaction is determined by the equilibrium constant. Of course, the reactants must be properly mixed, have sufficient activation energy or catalysis, and there must be a possible mechanism.

Please familiarize yourself with recommended notations for standard state and refrain from using ALL CAPS for emphasis. Structure your answer with paragraphs for readability. Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.


text:
Relationship between normality and molarity
I found this formula for calculating the normality from the molarity:
$$\text{Normality} = n \cdot \text{Molarity}$$
with
$$n = \frac{\text{Molecular weight}}{\text{Equivalent weight}}$$
But in one book I found that if equivalent weight is equal to half of the molecular weight then
$$2N = M$$
Under the given condition, I am confused if the above statement is true.
Am I following a factually correct book for solving problems on volumetric analysis?

The normality of a solution is the molarity of the solution multiplied by the number of equivalents per mole:
$$\mathrm{N} = \mathrm{{eq\over mol}}\cdot\mathrm{M}= \mathrm{{eq\over mol}}\cdot\mathrm{{mol\over L}}=\mathrm{{eq\over L}}$$
Example: Calculate the normality of a 1 $\mathrm{M}$ solution of $\ce{H2SO4}$. There are 2 equivalents per mole, from the reaction
$$\ce{H2SO4 -> 2H+ + SO4^{2-}}$$
which yields 2 moles of $\ce{H+}$ for every 1 mole of $\ce{H2SO4}$.
Using the relation given above, we find that a 1 $\mathrm{M}$ solution of $\ce{H2SO4}$ is 2 $\mathrm{N}$:
$$\mathrm{{2\;eq\over mol}}\cdot\mathrm{1\;M}=2\;\mathrm{N}$$
One resource you might find useful in order to check your math is Sigma-Aldrich's Normality & Molarity Calculator. Additionally, there are details relevant to the definitions of molarity and normality that you might find clearer that the ones you've conveyed in your question.


text:
Why does the Birch reaction not yield fully saturated products?
Why isn't benzene reduced all the way down to cyclohexane when performing a Birch reaction? (Why is it not reduced further?)
Similarly for the reaction with sodium dissolved in liquid ammonia, why isn't an alkyne reduced fully to an alkane?

If you look at the mechanism for the Birch Reduction, you will see that the mechanism wouldn't work with only 2 double bonds. As for the metal reactions, they do go fully to alkanes, unless it is a poisoned catalyst like in lindlars.

Without going in a lot of detail a first order simple answer is that 1,4-cyclohexadiene doesn't give the Birch reaction and so the initial reaction from benzene stops there. Also what is "disolv. metal reaction" other than a Birch reaction?

@K_P Sodium metal in liquid ammonia reduces alkynes to trans alkenes

@CognisMantis yes it would. You could add an electron into any of the remaining double bonds (and if they were next to each other, they could even stabilise by resonance).

The mechanism of the Birch reaction includes single-electron transfers from the sodium dissolved in ammonia to the aromatic compound. These transferred electrons will attempt to occupy the lowest possible orbitals.
In the case of benzene, we have a delocalised $\unicode[Times]{x3C0}$-system. Delocalisation can be understood here to mean higher occupied orbitals and lower unoccupied orbitals. Therefore, the electron will find an orbital at a lower energy and it will gladly hop in to create the radical-anion intermediat.
Once this is protonated, the remaining radical system can still be seen as a delocalised pentadienyl radical. Thus, the energy level a new electron will occupy is still rather low and another anion can be created. Further protonation yields the product of the Birch reaction, a cyclohexa-1,4-diene — most importantly, a non-conjugated system. This can be regarded as a pair of isolated double bonds.
The energy of the unoccupied $\unicode[Times]{x3C0}^*$ orbitals in an alkene seems to be just too unfavourable for a free electron to occupy them. I don’t have the facilities to do calculations (and neither am I firm enough to be able to do them) so I can’t provide you with values to compare, but it has to be the case otherwise the reaction would continue. If it were not so, there would be no reason for the reaction to stop.

The radical anion after the first step of the Birch reduction needs to be conjugated with (=stabilised by) another pi-system, otherwise it simply will not be formed from a pair of sp2 carbon atoms.


text:
Reading enthalpy of mixing graph
I'm struggling to interpret enthalpy of mixing data. Given this plot,  from this paper, say I want to calculate the heat released from mixing 0.29 mol of 7.7% ethanol with 0.71 mol of 28.7% ethanol to get 1 mol of 22.6% ethanol.
Is it correct to say $\Delta h_m = 753 - (0.29*605 + 0.71*672) = 100 J/mol$?
Or, in other words, can I subtract the molar weighted enthalpies of mixing at the initial states from the enthalpy of mixing at the final state to get the change of enthalpy of mixing from that process?
It seems trivial, but something about it feels weird.

Consider it this way: You un-mix each of the two solutions, meaning you must undo the enthalpy of mixing for each. Now remix the combined solutions, getting out the enthalpy of mixing for the whole thing.  And, watch you signs...


text:
Electric field required to split water into oxygen and hydrogen
Others sources typically say that the minimum voltage required to split water into hydrogen and oxygen is 1.229 V. Overvoltage is typically required in a practical device due to inefficiency leading to heat production.
However, suppose that two electrodes are separated by a large distance.  This will result in a very low electric field strength (generally, electric field equals $V/d$ where $d$ is the distance of separation).  Is there a threshold electric field level required for electrolysis to take place?

Pardon? A. I'm not doing homework.  B. I am asking a serious question.  Yes, maybe I am stupid but that has never stopped me from asking and this seemed like the appropriate place to do it.  My apologies for asking a question that any high school student should be able to answer.

When it comes to separating charge and many other things, the electric field strength is the important characteristic. Yet, with splitting water, the only thing that is ever mentioned is the voltage required.  It seems to me that there would be an electric field threshold for splitting the water, but nothing seems to ever be said about this.

This is nothing to do with the "electric field strength" between the electrodes (in vacuum or with solution, whatever), because this is not a process when you literally tear apart the molecule by this field. It is a chemical reaction, and equilibrium with some reaction barrier and the potential required to overcome the free energy difference between reactant(s) and product(s).

In part, there is no abrupt cutoff because the gradient of the field is not spread evenly across the electrolyte: the gradient is steep at the interface between electrodes and the electrolyte, where gas evolution actually takes place, not in the body of the electrolyte. Note where bubbles arise in this video.
However, if the conductivity of the electrolyte is low, whether due to electrode separation or low ionization, then electrolysis proceeds more slowly, perhaps imperceptibly, though there is no complete cutoff point.

Thank you for the answer.  The thin film in which the gasses are produced is exactly what I am working on.  Gas production is a nuisance that I want to avoid or minimize, so understanding the dynamics of gas production is important in my work.  I can describe the electric field strength as a function of position and was hoping that keeping the value under a threshold would solve the problem.  The film effect that you cite seems to be the basis for people ignoring the role that electric field strength plays in the process.


text:
Can the hydroxyl group of 4-hydroxybenzoic acid react with phosphorus pentachloride?
I learned that, a phenol will not react (or react very slowly) with $\ce{PCl5}$ due to its stabilized structure.
How about the $\ce{-OH}$ group on the benzene ring of 4-hydroxybenzoic acid? Will that $\ce{-OH}$ group react with $\ce{PCl5}$? If the answer is yes, is it anything to do with the $\ce{-COOH}$ group?

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Carboxylic group would react.

a phenol will not react (or react very slowly) with $\ce{PCl5}$

That's not true. Phenol and $\ce{PCl5}$ do react, producing mixture of compounds of formula $\ce{PCl_n(C6H5O)_{5-n}}$ (with $n > 1$ ). It is even possible to produce $\ce{(C6H5O)5P}$![1]
However, unlike in alcohols, in phenols:

the carbon–oxygen bond is much stronger, having a bond order of more than one and being formed with $\mathrm{sp^2}$ hybridized orbitals
the formation of the corresponding transition state for nucleophilic substitution would destroy the aromaticity of the benzene ring.

Thus, the formation of chlorobenzene does not proceed at a significant rate.
It is still possible to replace the phenol group by a chloride atom in some cases. For example, it is possible to prepare picrylchloride (1,3,5-trinitrochlorobenzene) by reaction of pyridine picrate and $\ce{PCl5}$, since three nitro groups greatly increase the reactivity of the phenyl fragment in nucleophilic aromatic substitution reactions.[2] However, as a general rule, benzene derivatives are not active in nucleophilic substitution except if there are a lot of strong π-electron withdrawing groups in the ring.
As for 4-hydroxybenzoic acid, my guess is that it should react as both phenol (as above) and benzoic acid (which forms benzoyl chloride).

Addition
While I was unable to find precise information on para-hydroxybenzoic acid, there is some info on ortho-hydroxybenzoic (salicylic acid). $\ce{PCl5}$ replaces the $\ce{OH}$ fragment of the carboxylic acid group with chlorine. The phosphorus oxochloride then reacts with the phenolic group, producing an $\ce{O-POCl2}$ fragment. The information, however, does not look absolutely trustworthy. Still, assuming it is correct and extrapolating it onto para-hydroxybenzoic acid, I'd say that the product should be $\ce{Cl-C(O)-C6H4-O-P(O)Cl2}$.

References
[1]: Ramirez, F.; Bigler, A. J.; Smith, C. P. Pentaphenoxyphosphorane. J. Am. Chem. Soc. 1968, 90 (13), 3507–3511. DOI: 10.1021/ja01015a038.
[2]: Boyer, R.; Spencer, E. Y.; Wright, G. F. Can. J. Chem. 1946, 24b (5), 200–203. DOI: 10.1139/cjr46b-025.


text:
Frequency of Mpemba Effect
What are the chances of warm water freezing faster than cold water? (How many percent?)
Does it have to do with container size or something in it?

Why do you believe this has anything to do with chance?  Have you read, e.g., the wikipedia page?

so what do you say? which one of them always happens? warm or cold?

The phenomenon ("warm water freezing faster than cold water") is not probabilistic.  Rather, it either occurs or it doesn't depending on the precise definitions and control of all relevant conditions.
From the wikipedia page:

A reviewer for Physics World writes, "Even if the Mpemba effect is
  real — if hot water can sometimes freeze more quickly than cold — it
  is not clear whether the explanation would be trivial or
  illuminating." He pointed out that investigations of the phenomenon
  need to control a large number of initial parameters (including type
  and initial temperature of the water, dissolved gas and other
  impurities, and size, shape and material of the container, and
  temperature of the refrigerator) and need to settle on a particular
  method of establishing the time of freezing, all of which might affect
  the presence or absence of the Mpemba effect.

This article explains the randomness of the phenomenon:New Explanation. Initially-hot water has lost much of its ordered clustering (higher entropy) and, if the cooling time is sufficiently short, this will not be fully re-attained before freezing. Experiments on the low-density water around macromolecules have shown that such clustering processes may take some time.Entropy reduction curves function of temperature S=f(T) appear retardation (lagging) relative to entropy growth curves.At any temperature point T the entropy S=mclnT during cooling is more than this during heating.The water after was heated and recooled at the initial temperature,has more entropy than before it was heated.This means that molecules have now the same average kinetic energy,but thermal motion before heating was more oriented by the structure mentioned above.Recooling random collisions are more possible leading to faster temperature’s reduction.

Please don not post verbatim the same answer to multiple questions, instead make us moderators aware, that the questions are identical witn a custom flag.


text:
Find the Structural Formula of Hydrocarbon Compound X
There is a specific compound $\ce{C3H6O}$ that exists as hydrocarbon compound X. It does not react with metallic sodium, bromine does not lose it's color when applied to the compound, and does not give the silver mirror reaction when ammonium silver nitrate, in fact it doesn't react.
I've realized since it does not react with sodium, it cannot have a hydroxyl functional group, and that since bromine does not give the discoloration reaction, the Carbons of the compound cannot have double bonds. And obviously, if it does not give the silver mirror reaction, than it cannot be an aldehyde or a carboxyl acid. However, the answer that the book gave was $\ce{CH3COCH3}$. Am I doing something wrong here?

It's the acetone. You are doing well.

The test for an alkene is decolorization of bromine WITHOUT evolution of HBr, acetone would decolorize Br2 and evolve HBr [the reaction is initially slower].  The book loses trust when it states C3H6O is a hydrocarbon Please check on Tollen's reagent what is it a test for?

Your reasoning is correct. Are you confused because acetone is not a hydrocarbon? That has to be a mistake in the wording of the question. Another common mistake is to forget that there are some reactions that aldehydes can do that ketones are inert to.
There are seven constitutional isomers for $\ce{C3H6O}$, not counting the enols, which are the minor tautomers of acetone and propanal:

Acetone - a ketone
Propanal - an aldehyde
Cyclopropanol - an alcohol
Allyl alcohol - an alkene and an alcohol
Methoxyethene - an alkene and an ether
Oxacyclobutane - a cyclic ether
Propylene oxide - an epoxide

Your reasoning is correct based on the chemical tests:

Sodium reacts with alcohols - a negative sodium test means no alcohol functional group is present, eliminating allyl alcohol (and the enols if you consider them).
Bromine reacts with alkenes and alkynes (and a small number of other functional groups, like phenols) - a negative bromine test means no alkene functional group is present, eliminating allyl alcohol, methoxyethene, and the enols.
Ammoniacal silver nitrate (Tollens' reagent) reacts with aldehydes - a negative Tollens test means no aldehyde function group is present, eliminating propanal.

The results of the qualitative test leave you with three possibilities, acetone, oxacyclobutane, and propylene oxide. With the evidence you have, you might not be able to distinguish between the three.
An iodoform test would confirm or eliminate acetone.
A test using silver nitrate and periodic acid would confirm or eliminate propylene oxide.


text:
Misconception? "That gas is heavier so it sinks."
I've heard people say many times that one gas is heavier than the other so the heavier gas sinks. Now this must certainly be true to a certain extent (I think of gas used in WWI that sunk into trenches). However, this idea of the weight of a gas being a driving force seems unusual because, in my experience, entropy is the dominating motive for particles in the gas phase. A good example of this is that gas phase clusters are studied extensively in the literature due to how unusual they are.
So, why would some gases "sink" and give up their entropic gain for what doesn't seem to be much of an energy minimization? Could the answer be simply that this concentration of heavier gases is quite short lived and they soon float away to their entropy filled wonderland?
I would also give as an example the fact that CO2 certainly gets up high in the atmosphere quite well and that is more than twice the "weight" of other particles in the air. So what's different about that than about the examples given previously?

I think you're right with your answer.

Without significant mixing, diffusion takes a long time to mix gases.  Our understanding of entropy tells us that we will indeed finish with mixed layers, but that doesn't give us a time frame for that mixing, only an outcome.
Given a slow but steady production of a dense gas, layers absolutely will form due to density differences.  There are plenty of places where this can easily be seen, from your WWI example to the ability to pour carbon dioxide or hexane vapor into a container.  They're not stable over the long term, but they do exist for substantial periods of time, days in some cases.
If confined by an open container, huge areas can experience this effect, as at Lake Nyos.  Heck, even small temperature based density differences drive atmospheric layering that has major impacts on weather and pollution.
It's not an enormous effect, but you may be underestimating the strength of the forces involved here.  The same pressure differential that forces a helium balloon upward acts on an open volume of helium as well, and more strongly due to the missing mass of the balloon.  A body of a gas like $\ce{NO2}$ has about as much force pushing it downward (in air) than an equal volume of helium has pushing it upward.  Without outside mixing to cause it to disperse, it will happily form a short term layer until diffusion carries it away.

No diffusion does not take a long time.  You can smell a fart across the room in seconds.    Thermal layers is a different thing.

@Frisbee Diffusion is fairly slow.  The reason you can smell a fart quickly is the presence of large scale air currents in the room.  Sure, "fairly slow" is qualitative, but for an easy example showing that it takes a significant amount of time for gases to diffuse through less than a meter of air, look at the famous demonstration of the reaction between ammonia and hydrochloric acid in gas phase.   That's just to get the two gases in contact, not to fully mix the contents of the tube.

https://www.youtube.com/watch?v=1PJTq2xQiQ0


text:
The pentagonal pyramidal structure of XeOF5-
Why does the $\ce{XeF6}$ molecule have an octahedral structure while $\ce{XeOF5-}$ has a pentagonal pyramidal structure, despite both molecules having 7 electron pairs? I suppose the extra electron on oxygen would distort the molecule somewhat, but this seems like an extreme outcome.

Related: What is the molecular structure of xenon hexafluoride?

$\ce{XeF6}$ is not octahedral, but distorted (or capped octahedral), one of the possible arrangement for seven electron pairs.
Wikipedia provides the following, quoted from a paper by Konrad Seppelt in Accounts of Chemical Research[1]:

...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...

The molecular geometry of $\ce{XeF6}$ is thus not truly an octahedron, nor is it static. In fact, the referenced article by Seppelt indicates that $\ce{XeF6}$ is only monomeric in the gas phase. In solution and the solid phase it is tetrameric $\ce{Xe4F24}$ clusters. The solid phase appears to be more like clusters of $\ce{[XeF5+ F^{-}]}$ with the fluoride ions bridging two $\ce{XeF5+}$ centers. Thus, in the solid, every $\ce{Xe}$ atom has 7 bonds to fluorine, with two that are more ionic. The structure for this scenario looks like the pentagonal pyramid.
Reference:

Seppelt, Konrad (June 1979). "Recent Developments in the Chemistry of Some Electronegative Elements". Accounts of Chemical Research 12 (6): 211–216. doi:10.1021/ar50138a004

In XeF6 monomeric structure where will the lone pairs be situated ? In the equatorial plane or axial plane...?

@SubhadipPal - see the block quote in my answer. The lone pair is mobile (or perhaps delocalized).

Can u explain the structure by VSEPR thoery..?

@SubhadipPal - VSEPR is a model based on classical mechanics that survives into the quantum mechanics age. VSEPR survives because it is a useful model, but it has its limits. This compound is one of its limits.


text:
Is there a way to make long delay ink?
I would like to make some invisible ink that apears by itself after some time, preferably at least a couple of days. I want to write a dedication in a book with it, so it would probably be fine if the colour apears after half an hour of air exposure. Is this even possible using easy to get ingredients?

Some organic molecule with colored oxidation product and low oxidation rate, dissolved on solvent, but cant think none in particular.

You might experiment with fresh-ground apples, which contain polyphenol oxidase and catechol oxidase, enzymes that turn brown in air (a tiny amount of copper or ferrous salt is also present, acting as a catalyst).
For that matter, dilute ferrous salts are often almost colorless (pale green) but darken to brown on exposure to air. (Digging in a podzol, soil of a pine forest, shows that color change.) So, you could extract a bit of liquid from greenish podzool and load your pen or marker with that.


text:
Green color substance forming during electrolysis
A green-colored substance formed during electrolysis, wherein the electrolyte was magnesium sulfate and minerals in tap water.
I did not have any leads to stick into the water, so I just put a 9 volt Duracel battery in the solution. The battery didn't leak.
White flakes formed after 1 minute before green substance at 10 minutes.
So, what is the green stuff and what are the white flakes?

The green material is probably an iron(II) salt, corroded from what I assume is the steel composing the positive pole of the battery.  If it has a gelatinous appearance, and slowly turned orange-ish over time, my guess would be it's $\ce{Fe(OH)2}$, with any orange color deriving from slow oxidation to various iron(III) species such as the oxide-hydroxides.
Per the above link, the white flakes are possibly that same iron(II) hydroxide, prior to reaction with $\ce{O2}$ to form the green color.  Alternatively, it could be $\ce{Mg(OH)2}$, or a related magnesium compound, formed at the cathode (negative terminal) from the hydroxide ion produced by reductive water electrolysis:
$$
\ce{2 H2O + 2 e- -> H2 + 2 OH-}
$$
A photograph would be tremendously helpful in facilitating identification.


text:
Carbidopa : zwitterion
The pKa of carbidopa is 7-8 of the base hydrazin. Is this because of the zwitter form with the acid? Why is the basicity so low? If it can form a zwitter ion wouldnt  the pKa be higher?
https://en.wikipedia.org/wiki/Carbidopa

Where did you find such value? Check this: http://www.drugbank.ca/drugs/DB00190


text:
Relative strength of metal hydrides as reducing agents
Question:
Out of $CH_4, CaH_2, FeCl_2$ and $BeH_2$, which compound would produce the most strongly reducing aqueous solution?
Thoughts:
Out of the hydrides, I would say that $CaH_2$ stands a good chance at being the strongest reducing agent, since the Ca-H bond energy is lowest (allowing the extremely reducing hydride to be released more readily into solution). The only rationale I can think up for this is the relative sizes of the other ion in the hydrides (Ca>Be>C) which implies that bond length is greatest in the calcium hydride (thus bond energy is lower). I am unsure as to how I should tackle the chloride compound. Much assistance is needed.

You are correct that calcium hydride is good reducing agent. It is also a strong base and reacts with water, rendering its aqueous solutions non-reducing:$$\ce{CaH2 + 2H2O -> Ca(OH)2 + 2H2}$$

@BenNorris just about any hydride is a good reducing agent. But is it the most? For some reason I think that beryllium hydride might be better but it's not based on anything.


text:
Does the aluminium chloride assay (Ordonez et al.2006) for total flavonoid content work on coloured extracts?
It is said that after applying aluminium chloride assay, the extracts will turn yellow due to formation of Al-Flavonoid complexes. So I was wondering if the assay still works if the extracts are already yellow.
For example, the yellow extracts are applied aluminium chloride assay and determined for total Flavonoid content using colorimetric method.
Then, would the result show the TFC (total flavonoid content in those yellow extracts?

Depending on the intensity of the yellow color (specifically, the degree of absorption of light at the wavelength of your measurement) you might be able to subtract out the yellow from the extraction in order to differentiate between that and the color arising from the aluminium chloride assay.
In other words, just measure the absorbance of your material before the addition of your reagents, and subtract that value from your final absorbance reading.
Alternatively, since this is a matrix interference issue, you might do well using the method of standard additions. This is described in this Wikipedia article:

The method of standard addition is a type of quantitative analysis approach often used in analytical chemistry whereby the standard is added directly to the aliquots of analyzed sample. This method is used in situations where sample matrix also contributes to the analytical signal, a situation known as the matrix effect, thus making it impossible to compare the analytical signal between sample and standard using the traditional calibration curve approach.

This article gives a reasonable outline of the procedure that should be sufficient should you decide to try this method.


text:
Why don't gases escape Earth's atmosphere?
Some gases are lighter than others and rise. Why don't they continue going up, leave the atmosphere, and then enter outer space?

Related: http://physics.stackexchange.com/q/202138/17547

Related: http://earthscience.stackexchange.com/questions/736/

They did. All the helium ad hydrogen gone from the atmosphere and keep escaping nowadays, too.

The atmosphere actually loses gases to outer space.
The average velocity $\bar v$ of gas molecules is determined by temperature $T$.
However, not all the molecules travel with the same velocity.
The probability of finding a molecule with a velocity near $v$ is described by the Maxwell distribution of speeds
$$\begin{align}
f{\left(v\right)}&=4\pi\sqrt{{\left(\frac m{2\pi kT}\right)}^3}v^2\exp\left(-\frac{m{v^2}}{2kT}\right)\\[6pt]
&=4\pi\sqrt{{\left(\frac M{2\pi RT}\right)}^3}v^2\exp\left(-\frac{M{v^2}}{2RT}\right)
\end{align}$$
where $m$ is the mass of the molecule, $k$ is the Boltzmann constant, $M$ is the molar mass of the gas, and $R$ is the molar gas constant.
Individual molecules may reach escape velocity $v_\mathrm e$ and thus be able to leave the atmosphere.
Escape velocity is the minimum velocity that is sufficient for an object to escape from the gravitational attraction of a massive body. For a planet, the escape velocity may be estimated by using the formula
$${v_\mathrm e}=\sqrt{\frac{2Gm_\text{planet}}r}$$
where $G$ is the gravitational constant, $m_\text{planet}$ is the mass of the planet, and $r$ is the distance from the centre of mass of the planet.
Therefore, atmospheric escape depends on the mass of the planet, the temperature of the atmosphere, and the molar mass of the gas.

There is also the strength of both the solar wind and the magnetic field to take into account

By my calculation, the escape velocity of a particle about 6400 km from the center of the earth is around 11,000 m/s. That seems unreasonably fast for any particle to ever escape which I suppose is good proof of your answer, but does a particle ever get moving that fast while in the atmosphere?

@jheindel Yes, escape velocity on Earth is 11.2 km/s (Mars: 5.0 km/s; Moon: 2.4 km/s). The average speed of a hydrogen atom (M = 1 g/mol) at T = 1000 K is about 5 km/s. (Note that the temperature in Earth’s exosphere can be well above 1000 K.) Particles from the high-speed tail of the Maxwell distribution may actually reach escape velocity. Including other processes, Earth loses hydrogen at a rate of 3 kg/s (helium: 50 g/s). See also D. C. Catling, K. J. Zahnle: The Planetary Air Leak, Scientific American, May 2009, 26

Actually some do. There is a problem of helium escaping the earth's atmosphere.
https://en.wikipedia.org/wiki/Atmospheric_escape
There are many other sources about this very problematic occurrence.

Lighter gases don't go up after they've mixed with the surrounding atmosphere. Buoyancy (like with a balloon filled with $\ce{He}$ rising or $\ce{CO_2}$ from a leak in a line dropping to the floor and filling the room) occurs only with bulk masses.
You need extreme g fields/gradients, like in a ultracentrifuge, to separate gases by weight.
There is a certain net flux of light compounds upwards, because e.g. $\ce{He}$ (from radioactive decay) comes into the atmosphere at the ground and escapes into space from the upper layers (see other answer).

Above the turbopause (about 100km) in the heterosphere atmospheric gasses stratify according to molecular weight.

Argument taken, but 100 km is already "space" ;-), with mean free path lenghts in the range of centimetres. A good amount of the hydrogen up there comes from photolysis of water, i think.