text:
Bohr Radius Confusion
I saw online that the radius for each level is defined by the formula:
$$r_n=n^2 r_1$$
Where $r_1$ is approximately $53\ \mathrm{pm}$.
However, on the Barron's AP chemistry book 7th edition, it says "The radii of the other orbits are whole-number multiples of the Bohr radius" (Page 89). It even gives an example of when $n=2$ that the radius is $106\ \mathrm{pm}$.
But that's different from what the formula above says, which when $n=2$ radius should be $212\ \mathrm{pm}$. Is this an error in the book? I'm quite confused by the discrepancy.

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Hmm...looks to me like they're equating the fine-structure constant with the principal quantum number $n$, but that's just from eyeballing the expression you've written. That said, if it is indeed a formula for estimating the Bohr radii for different values of $n$, then it does look like a mistake in the text.

Bohr's theory of the atom precedes quantum theory but does describe the Balmer and other lines in atomic hydrogen.  I've outlined the calculation so that you can see where the Bohr radius comes from.
He  assumed that the electron orbits the nucleus in a circle with a radius r and is determined by the balance of centripetal acceleration and Coulomb attraction towards the nucleus. For an electron of mass $m_e$ and speed $v$ this is
$$\frac{m_ev^2}{r}=\frac{e^2/(4\pi\epsilon _0)}{r^2}$$
 where e is the charge on the electron and $4\pi\epsilon_0$ puts us into SI units, ( $\epsilon_0$ is the permittivity of free space).
The total energy is the sum of kinetic and potential parts $$E=\frac{m_ev^2}{2}-\frac{(e^2/4\pi\epsilon _0)}{r}$$
and using the first equation gives
$$E=\frac{e^2/(4\pi\epsilon _0)}{2r}$$
Bohr had to make the assumption that electrons circling in an orbit do not radiate energy, which is contrary to classical electrodynamics, but he then needed to determine which orbits were allowed. He assumed that the angular momentum is in quantised units of Plancks constant h and a (principal) quantum number n. This gives
$$ m_evr = n\frac{h}{2\pi}$$
which when combined with the first equation gives
$$ r=a_0n^2$$
which is the equation you ask about and $a_0$ is conventionally the symbol for the Bohr radius and is 0.0529 nm.
It is calculated as
$$a_0 = \frac{\hbar^2}{e^2/(4\pi\epsilon_0)m_e}$$
where $\hbar = h/2\pi$.
Re-expressing the  energy using $a_0$ gives
$$E=-\frac{e^2/(4\pi\epsilon_0)}{2a_0}\frac{1}{n^2}$$
which is the famous Bohr equation and explains the energy levels in hydrogen atoms and so the transitions between them.

So does that mean it was an error in the book to say radii of other orbitals are whole number multiples of 53pm?

I don't have the book you refer to but 'yes' it seems so from what you have written in your question.


text:
Organic chemistry - Electrophiles
Can all the positively charged species be called electrophiles?

Perhaps in some sense, but there is a fundamental difference between $\ce{H3C+}$ and $\ce{H4N+}$, for example. The carbocation has an empty (p-) orbital and will form a covalent bond with a nucleophile, like any typical electrophile would. Ammonium on the other hand has a full octet, so the nitrogen cannot accept a new bond and an electron donor would instead bond to a hydrogen, deprotonating the cation. Cations with incomplete octets can be conventional electrophiles (Lewis acids), whereas cations with full octets and hydrogen are better considered Brønsted-Lowry acids (a special case of electrophile).

According to me basically those species of atoms and molecules which can gain a pair of electron and are positively charged or are neutral are called electrophiles. So only those positive species which can gain a pair of electron can be called electrophiles.

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text:
Oxidation product of alcohol

What is the oxidation product of a primary alcohol?
(A) aldehyde (B) alkene (C) ester (D) ketone

My try: So because an alcohol is $\ce{R-OH}$, and oxidation is loss of electrons, I thought that the product would be something like $\ce{R-O}$, which is an ether. However that is not an answer choice. The correct answer in this case is A. Can someone explain how?

An alkene has no oxygen.
An ester has an oxygen between two carbons.
A ketone does not have any oxygen on a primary carbon.

I remember this from school. Don't mind the experts telling you other things, you don't need to know that. You need to oxidize a primary alcohol.
Start with this structure, not with the one you suggested:
$$\ce{R-CH2-OH}$$
Look at the carbon atom, that's the one you oxidize! Right now, it has an oxidation number of $-1$ ($+0$ (doesn't matter) from $\ce{R}$, $2x-1$ from the $\ce{H}$ and $x+1$ from $\ce{O}$). Oxidation means increasing that number. So you need to substitute one hydrogen. Let's check your options:
Aldehyde:
$$\ce{R-CH=O}$$
Perfect. You've got $+1$ now.
Alkene: Doesn't make any sense  here.
Ester:
$$\ce{R-C(=O)-O-R'}$$
Although this is an oxidation, it usually needs a second molecule (which could be the same species) of alcohol.
Ketone:
$$\ce{R-CO-R'}$$
That's an oxidation. Well, let's come back to the fact that you're in school. It's in the question really. You need to oxidize your primary alcohol. Just remember that a ketone is the product of the oxidation of a secondary alcohol, which I hope you can see makes more sense:
$$\ce{R-COH-R'}$$
Btw, don't count on my oxidation numbers. The concept is somewhat useless in actual chemistry, it's been a while since I did that.

Thanks for whoever edited. In my first post I made an ether instead of an ester. Man this was last relevant to me such a long time ago ...

With a weak oxidizing agent, the product would be an aldehyde. Synthetically useful strong oxidants will oxidize the primary alcohol to a carboxylic acid.

Well it depends on how you define "strong oxidant," but the reaction could go so far as to convert all the organic matter to water and carbon dioxide.

Alcohols on oxidation always give aldehyde first and if we are using strong oxidizing agent aldehyde will further get oxidized into carboxylic acids and the final product we get from oxidation of alcohol is carboxylic acid in that case.
Thus alcohol oxidizes to aldehydes and aldehyde oxidizes to carboxylic acids. On using weak oxidizing agent product will be aldehyde.

A primary alcohol either oxidizes to an aldehyde or carboxylic acid, and it depends on two factors: strength of the oxideser and quantity
If the alcohol is in excess, then it is oxidised once to an aldehyde, especially if this aldehyde is distilled as it is formed. Full oxidation can then occur once all of the initial oxidation occurs; this time to a carboxylic acid if there is left over oxidiser and it is heated
Sometimes by using a strong oxidiser such as $\ce{KMnO_4}$ it may directly oxidise to carboxylic acid, or by using a weak oxidiser prevent formation of the carboxylic acid
The way oxidation to aldehydes works is the formation of a double bond between the oxygen on the alcohol and central carbon (carbonyl group), with the oxidser oxygen combining with the hydrogen on the central carbon and the alcohol hydrogen to produce water.
There is a stack exchange post describing the mechanisms of oxidation using  $\ce{KMnO_4}$ at Mechanism of the oxidation of alcohols with KMnO4

I'll bet you I could use half as much KMnO4 as PCC and still yield more carboxylic acid.

The standard procedure is to put more oxidiser to oxidise to carboxylic acid

But the extent of oxidation also depend on the nature of the oxidizing agent itself, not just quantity.

Ok, fair enough, I will edit

Thank you. I think the second to last paragraph gives a misleading sense of the oxidation mechanism. The O that picks up the hydroxyl H will likely not be part of the transition metal oxidant complex itself, but just some added base. It is also not likely the same O that will deprotonate the C. It should also be noted that the whole idea is: deprotonate the hydroxyl, add a transition metal (oxidant) leaving group to the O, then perform E2-like elimination by deprotonating the carbon, forming the C=O pi bond. Keyword: elimination.

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text:
Verify my expected product of two acid/base reactions and does chosen solvent change reaction?
My education in chemistry goes as far as what I research on the internet. I have two different acid/base reactions for which I've tried to predict the resulting product.  Can anyone here verify if I did this right or have an alternative prediction?
Reaction 1: p-toluenesulfonic acid (anhydrous) $\ce{C7H8O3S}$ reacts with potassium hydroxide $\ce{KOH}$. From this I calculated I should expect toluene $\ce{C7H8}$, potassium sulfate $\ce{K2SO4}$ and water $\ce{H2O}$:
$$\ce{C7H8O3S + 2KOH -> C7H8 + K2SO4 + H2O}\tag{1}$$
Reaction 2: p-toluenesulfonic acid (anhydrous) $\ce{C7H8O3S}$ reacts with
sodium bicarbonate $\ce{NaHCO3}$:
$$\ce{C7H8O3S + 2NaHCO3 -> C7H8 + Na2SO4 + 2CO2 + H2O}\tag{2}$$
If correct, then both of these reactions result in toluene. Does anyone think I got it right (or wrong)?

If the p-toluenesulfonic acid is dissolved in hexane, and I wish to neutralize it, neither base is soluble in hexane. If I first dissolved the base in ethanol (anhydrous) which is miscible with hexane (if I'm using the term correctly), will it neutralize the acid in the hexane, and does the fact that hexane and ethanol are involved change the reactions above? Any prediction if the ethanol was 95% instead of anhydrous?
So separating ethanol from hexane is easy if you want to keep the hexane -- just add water and separate. But if I actually have hexane/ethanol/toluene and add water and dispose of the ethanol/water solution, I'm assuming I'm left with a hexane/toluene mix. The toluene has a much higher polarity than hexane, but much lower one than ethanol. Toluene has a boiling point higher than water, and approx. 40 degrees higher than hexane.
Under this circumstance it appears removing toluene from the hexane and solute is not realistic, and would leave me distilling the toluene from my solute at a too high a temperature for the solute ($\pu{111^\circ C}$) whereas the boiling point of hexane is more like $\pu{68^\circ C}$.
If I'm correct, is there any thoughts on how to neutralize the acid in hexane without creating toluene which is not easy to separate from the hexane or am I stuck with toluene as long as I use this particular acid?

You're trying to neutralize tosic (toluenesulfonic) acid in hexane as a post-reaction workup?

Welcome to Chemistry.SE! You seem to assume that the neutralization of p-toluene sulfonic acid with potassium hydroxide (or sodium hydrogencarbonate generates) toluene. This is wrong! The neutralizations  only lead to the potassium (or sodium) salts of the sulfonic acid. With other words, the (most) acidic proton is replaced with a potassium (or sodium) cation. There's no further degradation that would lead to toluene.

Thanks for your responses! -- based upon everyone's comments thus far I had another go at it (see comments in response to AK's answer).

You will not get toluene in either reaction, you will get the potassium and sodium salts of p-toluene sulfonic acid for the respective reactions. This is because 1.) the toluene sulfonic acid is not a salt of toluene and sulfuric acid and 2.) aromatic rings (unless special conditions are applied) undergo electrophilic aromatic substitution. i.e. positive groups cause substitution on the ring, which sodium and potassium are not electrophilic ("electron loving") enough to effect such reaction. If you wanted toluene from these reactions you would have to use a dilute mineral acid in water and the $\ce{H+}$ ions from the acid would cause a substitution, removing the sulfonyl group and yielding toluene and sulfuric acid.
If you really need to neutralize this material, just add the mentioned bases as solids and allow them time to react. Bases don't need to be in solution to react with acids.

Thank you AK and Electronpusher, so much for reminding me how little I know ;) I really appreciate you taking the time! –
I took advice from the two prior answers and rethought a bit and here is what I got.
First instead of looking at the p-Toluenesulfonic acid like C7H8O3S I look at it like CH3 C6H4 SO3 H
So CH3 C6H4 SO3 H + 2KOH
gives me:
CH3C6H4OH  (p-Cresol)
K2SO3 (Potassium Sulfite)
and H2O (Water)
Do I have it right yet?

No, this would require a nucleophilic substitution which will not happen.  You should watch some youtube videos on electrophilic aromatic substitution so that you may understand the underlying principles of substitution on aromatic rings. It is not like most other reactions.

Ok, so, I checked out some videos on KahnAcademy.  Thank you for that.  So basically all that is happening is the neutralization of the acid like initially shown in ElectronPusher's structure.  Basically the H of the OH of the acid gets swapped with the K from the KOH, the H from the acid goes to the OH of the KOH making H2O.  This leaves me with H2O and Potassium p-toluenesulfonate, the latter of which I assume will be a solid precipitate, insoluble in either hexane or ethanol?

Now you have the right idea for the acid-base neutralization. However, the potassium p-toluenesulfonate salt (AKA potassium tosylate), should be soluble in ethanol and maybe somewhat suble in hexane too. See all those C's and H's in the ring? That nonpolar structure is soluble in nonpolar solvents, like hexane. The ionic charges on the sulfonate side would be soluble in polar solvents. This could cause some precipitation in hexane, but ethanol can accommodate both domains and I would expect no precipitate in the alcohol solvent.

How can I predict if, given a solution of hexane and ethanol which contains the dissolved potassium tosylate, if I were to add water to separate, if the potassium tosylate will go with the etoh/h2o or stay in the hexane?  I would guess the polar sulfonate side would create a stronger bond with water than the bond between the non-polar side and the hexane, and perhaps the non-polar side would be satisfied by the alcohol at the same time the polar side is teamed up with the water, i.e. gone from the hexane?

This is more than a simple acid-base neutralization, you are hydrolyzing the $\ce{C-S}$ bond. Is this your intention, to create toluene, or is that a byproduct? Your balanced reactions look correct, but I would recommend you use formulas that show more detail to see what's going on (better yet, structures! see photo). I see no risk in aqueous conditions, since you are hydrolyzing tosic acid anyway. Is there a reason you can't use an organic-soluble base?
Edit: I erroneously went with the OP's assumption that the hydrolysis of the ring would take place. The other comments have reminded me that these conditions are not sufficient for substitution of the aromatic ring.

A sulfonic acid substituent on a benzene ring could, under some conditions, undergo nucleophilic aromatic substitution by hydroxide to give a phenol (in this case, p-cresol). This type of reaction is more favorable with additional electron withdrawing groups on the benzene ring.
See Aromatic Nucleophilic Substitution Reactions. Joseph F. Bunnett and Roland E. Zahler, Chem. Rev., 1951, 49 (2), pp 273–412.


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Why is the uncertainty of a digital instrument always the smallest scale division?
For a mass balance that can give readings to 2 decimal places, the uncertainty is assumed to be $\pm 0.01$. Can someone explain the rationale for this, as I would think it more logical for the uncertainty to be given as $\pm 0.005$.

Well, the value is recorded as a binary value, so back computing the accuracy is not trivial. Presumably $\pm 0.01$ describes the actual faith you should have in the value. No one said it will be correct, but they claim it won't be off by more than 0.01 units.

If I'm following correctly, you're assuming it's like a safety net of sorts? As in the uncertainty is unlikely to ever be as large as +/- 0.01, but it's written that way anyway just in case it's larger than +/- 0.005?

The uncertainty is in some binary digit. Suppose it's $\pm 0.0001_{2}$. That translates to $\pm 0.0625$. But you might round that up to $\pm 0.1$ if your scale only has 1 place after the decimal point. Maybe I should just write this up as an answer.

Whatever the details of the measurement as a binary number translated to decimal, the only sure way to know what the manufacturer of your scales intends by an error of $\pm 001$ is to look at the specifications. If you work in a certified lab the scales should adhere to some standard such as ISO9000.  If not, all you can assume is that the error is literally what you mention.  You could always check by repeated measurements of the same object. Also, you could try measuring in groups as this will reduce the noise introduced by the scales and see if this has any effect.

Migrating my comments to an answer:
The big issue with your line of reasoning is that the transducer in the scale does not produce decimal output. Like all digital devices, the signal is digital (i.e., binary), so the representation of the weight will also be a binary number.
Then you're in a scenario where the uncertainty is something like $\pm 0.0001_{2}$, which is $\pm 0.0625$. In this case, I would say that the error is in the first decimal place after the decimal point.
You can't give an error of $\pm 0.005$, partly for the reason I gave, but also because the error in measurement is not necessary an error in rounding. It's possible that the transducer is temperature sensitive, so within a reasonable range, it might over or under measure by $0.01$ units.


text:
Which crystals can exist permanently (and do not break down) under normal earth like outside conditions?
Hi I am looking for crystals I can use in the world of my novel but I have no idea what types to use or what there are.
So my question is what kind of crystals fit these criteria:

They must be able to exist “permanently” without breaking down in normal earth conditions/atmosphere.
They must be able to grow quite large. (I’m thinking about large hexagonal crystal columns)

I would be very happy if you could list any crystals you know that fist the criteria. With the name I should be able to research further myself and narrow it down.
Thanks in advance.
-
P.s. some of you might have some feeling of déjà vu reading my question as I asked a similar question before but that one was marked as to brought so I am asking it again but trimmed down.
link: How do i create a significantly correct world with large crystal formations in my novel?
P.p.s. if this question is to brought as well please leave a comment explaining how I can make it better because I otherwise really have no idea how to fix it.

Instead of re-asking a question, you might want to edit the original and get it reopened. If the questions are too similar, they might get closed as duplicates.

There are thousands of minerals out there. A good many of them would exist permanently if exposed to elements; they've been through it before. Most of those are crystalline and would grow into large crystals, given the right conditions (which are rarely there, but anyway).

Have a look for 'Cave of the Crystals' in the Naica Mine in Mexico. The crystals there should suffice.

@porphyrin Thanks but they break down (although slowly) when exposed to air so they are not an option. Honestly it was the first thing i thought off when i started thinking about the idea but since they need to be "permanent" so they are not an option.

I would go with quartz, you can even make it basically any color you like. But I would ask that question in a geology forum.

@ TheRWS96 Its possible to grow very large Alum crystals quite easily.

Quartz (silicon dioxide) sounds like a good bet. It is inert, safe to be around, and insoluble in water. Some examples of naturally occurring quartz crystals are really huge - visit the National Museum of Natural History if you are near DC - or any other natural history museum near you. There you will likely see many other naturally occurring minerals of defined chemical composition that form large crystals. Or go online and look for pictures.

for a permanently persisting cristal, I would think of diamond.
For an hexagonal structure, I would think of basaltic organ (bu this is not a cristal)


text:
What kind of string to use for the ice fishing experiment for kids?
Classic ice fishing experiment for kids.
I used nylon because I saw it in a stationery, and apparently that doesn't work because I guess it's too slippery. I tried thread from my mom's sewing kit, but that doesn't work either.
I'm not sure I can find yarn or kite string in the stationery or nearby so please suggest alternatives expected to be found in the average household, supermarket, grocery store or convenience store.
Also how long should we wait? The above video takes only 10 seconds while this video takes about 2 minutes (off screen). What factors influence waiting time? Amount of salt? Type of string?

I've had good results using sewing kit string or thin twine.

While this is an interesting question, I'm voting to close it as off-topic because it has essentially nothing to do with chemistry.

@hBy2Py OP wants to know about the factors influencing waiting time. Surely that's 'on-topic'?

@paracresol I did not take the time to visit the link provided at the start of OP's question ("ice fishing experiment" gives no indication that chemistry is involved, to my mind). Perhaps the information there makes it more of a chemistry question? If so, OP should edit the question to include details of the experiment, and I will gladly retract my close vote.

"Ice fishing" is trying to melt a string into a piece of ice using salt. The ice re-freezes, allow you to pull the ice up with the string. I aggre that "ice fishing" is a very poor description though. Should be fixed to keep open.

@hBy2Py why is having the details in question so important that you 'will' and not 'might' retract? How about we just pretend the details are there? I think not having details makes post less wordy. Why use links in the first place instead of just copying entire webpages? See office fire

@BCLC Because without them there is no obvious relevance to chemistry. In my experience, the term "ice fishing" refers most commonly to 'cutting a hole in a frozen-over lake, and catching fish through it.' With an explanation present, indicating that 'ice fishing' is used in the sense of 'exploiting the freezing point depression induced by salt to penetrate a string into a piece of ice,' the relevance is clear and straightforward. There are still elements to the question that are of uncertain relevance, but they are minor enough for me to retract the close vote.

@hBy2Py okay so that's your experience. Is that necessarily most people's experience? False consensus ? PS Please see edit of previous comment

@BCLC This is from the 'answering' section of the Help Center, but see the "Provide context for links" blurb here. What happens if that YouTube video gets taken down for copyright violations? Your question will then have lost crucial context. Thus, the importance of at least summarizing what you want the reader to get out of the content at the link.

@BCLC "okay so that's your experience. Is that necessarily most people's experience?" Clearly not, since as of this comment only one other person has voted to close. :-)  My close vote and above comments are not meant to tell others what to think; they are to declare what I think. If an insufficient number of people also vote to close within the next couple of days, my close vote will expire, the question will stay open, and I'll be perfectly content with it. That's the SE model for making decisions like this.

Let us continue this discussion in chat.

This resource has an explaination of the behavior, which relies on the salt-dissolved water wicking into the string. This would mean that you need a type of string which will absorb water. It's surprising to me that your nylon string doesn't work (as long as it's thick/low-density enough to absorb water, probably unlike fishing line), is it possible it has a hydrophobic coating?
I would recommend any cotton-based string or twine. Cooking twine and sewing thread is often cotton, if you have access to that. I've had no problems with these.
Edit: I just thought, that you might want to try this not in a cup of water first (like in the link above). You might potentially run into not-string-related problems if your water isn't cold enough, or if your ice gets too submerged when you put the salt on it. It's possible your nylon string is fine.

Hmm. Do you wait long enough for the salt to melt the water on the top of the ice-cube?

idk. I think so? I thijk i resd 10 seconds. I waited 30. I'll try again later

I'm sure you've already thought of this too, then, but my other idea on timing is that you also want to be sure to wait long enough that the water (with salt wicked away) freezes between the string and the ice.

I put ice in a bowl or plate and then put salt over ice and then lightly tapped salt over ice with back of spoon. Then it worked. Any idea why the tapping works? Thanks once again. Merry Christmas, Happy Holidays, and Happy New Year!


text:
Why is the ionic product of water not divided by the concentration of water?
I know the ionic product of water is $[\ce{H+}][\ce{OH-}]$, by following the equation:
$$\ce{H2O <=> H+ + OH-}$$
But actually it's just another way of writing:
$$\ce{2H2O <=> H3O+ + OH-}$$
Since, $K_\mathrm{w}=K_\mathrm{c}[\ce{H2O}]$ and
$$K_\mathrm{c}[\ce{H2O}]= \frac{[\ce{H3O+}][\ce{OH-}]}{[\ce{H2O}]}$$
Why can't the following equation be used? $$K_\mathrm{w}= \frac{[\ce{H3O+}][\ce{OH-}]}{[\ce{H2O}]}$$

This website and references therein (particularly ref 1) may be useful to people attempting to answer.

I am certain this is a duplicate but I cannot seem to locate it...

You yourself state that $\mathrm{K_c [H_2O]}$ is a constant, and then arbitrarily rebrand it as $\mathrm{K_w}$, so why don't you multiply that by the remaining factor of $\mathrm{[H_2O]}$ and rebrand $\mathrm{K_w [H_2O]}$ as another constant, say  $\mathrm{k_w}$? If you want to use $\mathrm{K_w}= \frac{[\ce{H3O+}][\ce{OH-}]}{[\ce{H2O}]}$, then you need to use $\mathrm{K_w=1.81 \times 10^{-16}\ mol\ L^{-1}}$ in ambient conditions. We generally just use $\mathrm{k_w= [\ce{H3O+}][\ce{OH-}]=1.01 \times 10^{-14}\ mol^2\ L^{-2}}$ because it's simpler. Either way is right.

The short answer is: substances in their standard form are not included in equilibria expressions, and most commonly this includes pure liquids and solids.
The long answer is: the equilibrium expression relates all species to a standard activity (this is why they are all unitless). This question and answer detail that well.

Just to hammer this home, notice that for autoionization, we do not write $\mathrm{K}{eq}$. We write $\mathrm{K}{w}$ because it's a different value.

One might want to add, that it is a (usually good) approximation to assume [H2O] as constant. If we go to high acid concentrations or other unusual environments this might not be applicable anymore.

@mcocdawe, I would say that it is usually a good approximation to assume the activity of H2O is a constant. Then, not only do you have to be cautious of having high concentrations of other species, but how much other species interact with and order water, decreasing the "free water" in the system. That might be significant even if the concentration of water stays constant within significant figures.

It is because water is in excess.
At pH = 7 which is assumed to be normal pH of water, though it is generally more acidic due to auto absorption of carbon dioxide gas making acidic HCO3-, we can write find the concentration of water molecules expected to auto-ionize.
Water reacting with itself: H2O + H2O ---> H3O+ + HO-, remember that pH = -log[H3O+], so then:
7 = -log[H3O+]
Rearrange:
10^(-7) M or moles / liter = [H3O+]
which is actually the same about of HO- that has formed from water reacting with itself at pH 7, so appreciably not much H2O has auto ionized.
Some calculations:
Density of water at room temperature is about 1.000 grams / mL
Molecular weight of water is about 18 grams / mole.
Use that to convert to moles / liter by cancellation of units.
(1 grams / mL) * (1000 mL / 1 L) * (1 mole / 18 grams) = 55.56 M H2O
Which H2O is much in excess of the concentration of concentration of hydroxide and hydronium (across the whole pH scale like pH 1 to pH 14), so when you have a reagent in unreacted excess, you ignore it in your equilibrium calculations with respect to chemical kinetics.

I would argue that it's not fundamentally because water is in excess, it's the fact that water is in excess that the activity of water in the system does not change. The fact that the activity does not change, and is thus the same as the standard activity, means that it doesn't appear in the expression. The fundamental issue is the activity of the species.

Because $K_w$ is defined in terms of activity, and then the approximation is made that "activity of water" = 1.
$$\ce{2H2O <=> H3O+ + OH-}$$
$$K_w = \frac{a(\ce{H3O+})a(\ce{OH-})}{[a(\ce{H2O})]^2}$$


text:
Organic problem: acidity of benzene substituents
Ortho nitro benzoic acid is more acidic than para benzoic acid since intermolecular H bonding comes into play in the latter compound. But why is that among para nitro phenol and ortho nitro phenol, para compound is more acidic than that of ortho? There is intermolecular H bonding in para nitro phenol too, right?

Related: Why is meta-nitrophenol less acidic than para-nitrophenol? and Acidity order of nitrophenols

This isn't the answer you were looking for probably, but this is the answer you should take to heart as a chemist.
I want us to look at the difference in free energy changes between the $\mathrm{p}K_\mathrm{a}$'s in question:
\begin{array}{|c|c|}
\hline
\text{Compound} & \mathrm{p}K_\mathrm{a} \\ \hline
\text{2-nitrophenol} & 7.23 \\
\text{4-nitrophenol} & 7.15 \\ \hline
\text{2-nitrobenzoic acid} & 2.16 \\
\text{4-nitrobenzoic acid} & 3.44 \\ \hline
\end{array}
$$K_\mathrm{eq} = \frac{K_\mathrm{a}}{\ce{[H2O]}}$$
$$\Delta \Delta G = -RT\ln K_{\mathrm a1} + RT \ln\ce{[H2O]} - (-RT\ln K_{\mathrm a2} + RT \ln\ce{[H2O]}) = -RT\Delta\ln K_{\mathrm a}$$
$$=-2.303 RT\Delta \log K_{\mathrm a} = 2.303 RT \Delta \mathrm{p}K_\mathrm{a}$$
For $298\,\mathrm{K}$, we can compute the following $\Delta \Delta G$ values:
\begin{array}{|c|c|}
\hline
\text{Series} & |\Delta \Delta G| \\ \hline
x\text{-nitrophenol} & 456\ \mathrm{J}\,\mathrm{mol}^{-1} \\ \hline
x\text{-nitrobenzoic acid} & 7303\ \mathrm{J}\,\mathrm{mol}^{-1} \\ \hline
\end{array}
For reference, the standard entropy of solvation for acetate is $-192\ \mathrm{J}\,\mathrm{mol}^{-1}\,\mathrm{K}^{-1}$, which at $298\ \mathrm{K}$ is equivalent to $-57\,216\ \mathrm{J}\,\mathrm{mol}^{-1}$.
Now, personally, I don't have a great handle on how the solvation of the conjugate bases are going to be relative to the benzoic acids or phenols. But you can see that solvation effects can play a huge role (and these are aqueous dissociation constants).
My personal opinion is that it's not particularly useful to understand where such a small difference comes from, particularly for the phenol series. We're talking about one-tenth of an order of magnitude. Perhaps, as you say, the benzoic acids are different because of intermolecular hydrogen bonding. I'm a bit suspicious since I don't see this acting in a differential way between the two species.
Nevertheless, the main point still stands. Acid-base equilibria in aqueous medium is a non-trivial process to understand because water is a complicated solvent. You can try to rationalize this, but you may well be wrong without very cleverly designed experiments.
And the fact of the matter is that rationalizing the difference in acidity is not that useful. Maybe if you were using linear free energy relationships, but in that case, you know there is a difference, and that's more important.
Again, as I said at the beginning, this is probably not the answer you wanted, but I think you should be trying to understand the limits of what is useful/practical to know.

When I checked the pKas for o- and p-substituted nitrophenols, p-substituted nitrophenol is slightly more acidic (7.1) than o-nitrophenol (7.2).
In general, to judge the relative acidity of substituted phenols, you should first look out for how resonance can stabilize the conjugate base (e.g. the deprotenated molecule). Since the difference in acidity is so slight, you might expect that there shouldn't be big difference in the resonance stabilization, as that would usually result in a larger pKa difference.  This website shows the resonance for o-,p-,and m-substituted nitrophenols. Indeed, In the o-nitrophenol, the electron lone-pair on the deprotenated phenol can only delocalize onto the nitrogen, resulting in N having a +1 charge and both O having -1 charges (as shown in the link above). This is the same for the p-nitrophenol.
However, the reason for the slight difference between the p- and o-substituted nitrophenols is due to the OH hydrogen being able to also hydrogen bond with the the ortho-nitro group. This additional hydrogen bonding interactions makes it harder for the molecule to completely 'let go' of the hydrogen. This shifts the acid-base equilibria towards the unprotenated form of the acid, lowering the acidity/raising the pKa (when compared to the p-substituted compound)

So resonance of phenolate ion in the para form makes it more acidic where as in para nitro benzoic acid , delocalization of electron is not possible?

@user129402 In the o-nitrophenol, the electron lone-pair on the deprotenated phenol can only delocalize onto the nitrogen, resulting in N having a +1 charge and both O having -1 charges (as shown in the link above). This is a higher formal charge than the delocalization shown for the p-nitrophenol. Indeed, when I checked the pKas for o- and p-substituted nitrophenols, p-substituted nitrophenol is more acidic. Why do you think it's the opposite? (I've edited my answer to reflect this--sorry, I couldn't comment previously)

@user129402 Oh sorry, I made a mistake about the relative formal charges: they are actually the same. The slight difference in the acidity does in fact come from hydrogen-bonding interactions in the o-substituted molecule. I've clarified this in my edited answer.


text:
Modern open-source tools for simulation of NMR spectra
I am interested in predicting the NMR spectrum of small organic compounds.  It doesn't matter to me if the prediction is very accurate.  I'll eventually be comparing the prediction to experimental results.
It looks as if several free online tools offer the ability to do such predictions:

NMR DB is a free website that offers a prediction of 1H, 13C, and supposedly even 2D NMR experiments.  Structures are entered via a GUI editor.
VeSPA and PyGamma are Python packages (or Python wrappers around C libraries) that supposedly simulate NMR spectra of molecules.  However, both packages are very old and have not been substantially updated in some time.

My question is, what is the best available (free) computational tool for predicting NMR spectra?  As I mentioned, I understand that predictions are imperfect.  What software, if any, are widely used by practicing chemists for NMR simulation?  What is the best reference to read about such software, including the heuristics and algorithms it uses to simulate spectra?  Is Vespa and/or PyGamma state of the art?  The NMR DB site is nice, but I'd very much prefer an interface that is amenable to scripting, and where structures could be supplied computationally rather than via a GUI interface.
(I'm a mass spectrometrist and biochemical engineer with little training in NMR methods.)

I know that Orca can perform such calculations.  I never used Orca for that.  It is not open source.  I would look for Nwchem as an open source alternative .

I'm not sure about their ability for computing coupling constants.

Thanks!  Both nwchem and orca appear to rely on electronic structure calculations, e.g. DFT or other quantum chemical calculations.  That isn't what VeSPA/PyGamma or NMR DB do, is it?  Is "everyone" aka "the chemist on the street" doing these types of calculations these days?

Yes, Nwchem and Orca rely on electronic structure calculations. NMR DB is very different. I do not know exactly what VeSPA/PyGamma do, but I bet they are different too.
Yes, there are a lot chemist doing this kind of calculations. They are useful mostly to predict spectra or chemical shifts. Take in mind that electronic structure calculations of NMR spectra can be very time consuming as their requires the computation of the expected values of many hamiltonian's derivatives.

Everybody uses the empiric prediction routines integrated in Chemoffice, Mnova, etc., which are based on chemical shift increments, afaik. Very, very few people have a need for accurately simulated spectra. You need to know not only the correct structure but also the dynamics in the system.

@Karl what do you mean for "You need to know not only the correct structure but also the dynamics in the system."?

I say you need not only the geometry and electron structure of the most stable state of your molecule, but also that of every excited state that is reasonably possible at a given temperature (e.g. rotation of methyl or phenyl group, or bihedral angle at any C-C bond), and their conversion rates.

@Karl I'm many years late, but the typical approach to calculating NMR properties is to generate a set of conformers and then take the Boltzmann average over the conformers. So yes you would need some knowledge of dynamics, but you only need local minima (conformers), not everything else along the PES. Also you don't need excited electronic states, which is what "state" usually refers to ;) Although, there is some research which suggests that if you do model the PES properly you can get stupidly accurate results: https://doi.org/10.1021/acs.jctc.5b00856 This could be a whole new question!

In Org Biomol Chem 2016, 14, 3943, Goodman reports the replacement of some "tradition" (expensive) programs with free/open-source ones, as applied to their dp4 method.
Specifically:

Molecular mechanics with tinker (I believe as a replacement for MacroModel), you may or may not want this step depending on what you're doing. The paper is looking at molecules with stereocentres and as such, they appear to have wanted to ensure the NMR values calculated matched the most likely conformation of the molecule experimentally studied.
NMR prediction using NWChem (replacing Jaguar as the gold standard) to do the DFT calculations. There is some mention that this was slower, but again, it would depend on what you're interested in studying.

I think that fulfils your requirement for being able to do the prediction using open-source programs.
As for scripting, there is a mention in the paper of wrapping the whole thing up in a python script (and to my knowledge, there isn't actually a GUI for either piece of the aforementioned programs), so it should be amenable to this.
One thing that it doesn't appear to do, is to graphically plot the resulting data generated (I believe they just compare a list of values to a list of experimental values), but you should be able to plot this in something like MatLab, or, if you play enough, it should be importable into TopSpin via something to turn it into a set of suitably written values.

Since I work in this area now, I feel qualified to comment... you are correct that DP4 only cares about comparing chemical shifts; the scripts developed by the Goodman group don't actually generate a spectrum. Going from a list of shifts to a spectrum is unfortunately a non-trivial problem: see my answer for more info. However, if all one wants are the chemical shifts (and/or couplings), then quantum chemistry software is an excellent choice (as long as computational resources are available).

You can submit the structure to the NMR database website using this link.
Concerning the algorithms, there are references on the nmrdb.org website. For $\ce{^1H}$ prediction we are using Spinus (based on the neural network). For $\ce{^13C}$ we are using NMR shift DB that is based on hose code.
To get an overview of the prediction algorithms you may be interested in our last publication.

Thanks!  Does it work with $\ce{^{13}C}$ SMILES?  http://www.nmrdb.org/service.php?name=nmr-13c-prediction&smiles=CCO gives something reasonable, but http://www.nmrdb.org/service.php?name=nmr-13c-prediction&smiles=[13C]CO doesn't seem to.

The problem with computational chemistry software (ORCA, GAMESS etc.), as described in NotEvans' answer and in various comments, is that they only spit out a list of chemical shifts and coupling constants. For example, this is part of a typical ORCA output:
 --------------
 Nucleus  34H :
 --------------

[...]

Total shielding tensor (ppm):
            28.820         -0.203         1.178
            -2.079         29.083         1.927
             1.392          3.320        24.895

 Diagonalized sT*s matrix:

 sDSO            23.705           26.255           39.984  iso=      29.981
 sPSO            -0.555            2.878           -9.468  iso=      -2.382
        ---------------  ---------------  ---------------
 Total           23.149           29.133           30.516  iso=      27.599

Getting from a list of numbers to something that looks like a spectrum is an entirely different matter altogether. In order to do so, one would have to use these parameters to construct a spin Hamiltonian:
$$\hat{H}_{\text{free}} = \sum_i \Omega_i\hat{I}_{iz} + \sum_{ij} 2\pi J_{ij}\hat{I}_{iz}\hat{I}_{jz}$$
and then perform a quantum mechanical simulation, where a density matrix $\rho$ evolves under this free Hamiltonian, as well as any other Hamiltonians corresponding to pulses, according to the Liouville–von Neumann equation. After the pulse sequence, one can obtain the FID by measuring the expectation value of the operator $\sum_i\hat{I}_{i-}$ at regular intervals corresponding to the NMR dwell time, and then the spectrum by Fourier transforming that.
The best general tool for this is Spinach, which is coded in MATLAB. Spinach itself is open-source, but MATLAB isn't (and the Spinach code does not run on GNU Octave). So, it does not quite meet the criteria in the question, but I thought I should mention it anyway.

Specifically for use with ORCA, Stefan Grimme's group have done a lot of work in this area, see Angew. Chem. Int. Ed. 2017, 56 (46), 14763–14769. There are various Python scripts that are capable of doing this spin Hamiltonian simulation as well as plotting the spectrum. Indeed, it should now be possible to interface with all these directly from ORCA, such that with one ORCA input file you can perform all the steps described in the paper (conformer generation, NMR property calculation, spin Hamiltonian simulation, and plotting). I suggest looking in the ORCA manual for more information.
ORCA is not open-source per se, but it is free to use, which is what I think you're broadly aiming for (I'd rather not get into debates about what "free" software means).

For what it's worth, if you have the programming knowledge required, it's easy enough to write a (e.g.) Python script that is capable of simulating two- or three-spin (maybe four-spin) systems. At the end of the day, an NMR experiment is "just" a bunch of unitary operators acting on density matrices, i.e. lots and lots of matrix multiplication. For example, you could take a look at some of my MATLAB code here which simulates the first FID of a sensitivity-enhanced HSQC experiment. It is fairly trivial to port this to Python, since numpy provides you with pretty much every function you might need.
[Disclaimer: the code will not necessarily stay there forever.]
The only problem is that this scales exponentially with the number of spins, so unless you do some serious optimisation of the code (like that done in Spinach), it becomes intractable very quickly. (It depends on what kind of spectrum you want, too: a simple 1D proton experiment is easy, but if you want a 2D then you need to simulate a number of FIDs, and the complexity increases rapidly if you are using shaped pulses and especially if there is any kind of spatially-dependent element, like pulsed field gradients.)
But if your molecule can be "decomposed" into several spin systems which are small, then the overall spectrum is just the sum of the spectrum of each spin system, so you might be able to get away with simpler code. I've never thought about how you might do that, although my guess is that you would want to construct a Hamiltonian matrix, then pass it to a function which would call itself recursively if the matrix can be written in block-diagonal form.

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text:
What needs to happen to this: Ca5(PO4)3Cl to get this: CaCl2?
$\ce{Ca5(PO4)3Cl}$ is a mineral called Chlorapatite
So lets say I have some chlorapatite in my garage, what do I need to do to it so that I am left with Calcium Chloride $\ce{CaCl2}$?

Probably decompose it.


text:
If Oxygen has a lower electron affinity than Sulfur, then why doesn't sulfur has a lower electron affinity than selenium?
I've read that oxygen has a lower electron affinity (as shown the picture below), because it has a smaller atomic radius than sulfur and thus the electrons experience significant electron-electron repulsion.
However, I was wondering... Why is this not the case for sulfur and selenium or chlorine and bromine? Because of sulfur's and chlorine's smaller atomic radius compared to selenium and bromine, the incoming electron should also experience a slight electron-electron repulsion (just like it does with oxygen), so $\ce{Se}$ and $\ce{Br}$ should have a higher electron affinity.

This is likely due to second period elements' being quite small, so electron-electron repulsion is much more significant than in a third period element. The general trend is that EA is more positive as you move down the periodic table since effective nuclear charge is the same, but the electrons are farther away from the nucleus. Maybe it's easier to consider the second period the anomaly...

Then is the 3rd period bigger because of the d orbitals or something?

Not sure what you mean by bigger. Are you talking about the atoms? They're bigger because the electrons are in higher energy (and thus "bigger") orbitals.

Yes, but I still don't get why the 2nd period atoms are the anomaly. Shouldn't the same apply to the third period atoms since they are also smaller than the 4th period atoms?

For example, Chlorine atom is smaller than Bromine, so why doesn't its electron -electron repulsion repel the incoming atom just like it does for Flourine? What I'm saying is, Chlorine should have a lower EA than Bromine, Bromine should have a lower EA than Iodine, etc for the same reason Flourine has a lower EA than Chlorine, which is because of the electron-repulsion.

See, that's the great thing about chemistry. You have two competing factors. One factor (the small size) is much more significant in period 2. In all other cases, it's the fact that the orbital sizes are larger. It's important to consider that these effects come from slightly different reasons. One is the electron-electron repulsion, which is less as the atom increases in size because of the increased distances, but also because of increased polarizability. The other is simply the decrease in stabilization due to larger (higher energy) orbitals.

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text:
Why are the nitrate salts hygroscopic?
Any metal is never extracted from its nitrate salt because nitrate salts are highly soluble in water, so flow away along with rainwater, or any other source.
But why are the nitrate salts so hygroscopic in nature? Is there any relationship with hydrogen bonding? Or something else is the cause?

maybe hydration enthalpy?

Nitrate salts are no more hygroscopic than other salts. If anything, they are more soluble on the average, but that's another story. This has nothing to do with the reasons for not using them as a source of metals.

Anions like nitrate forms strong hydrogen bond with water. Here water bridges the nitrate oxygen with hydrogen bonds giving it a very stable arrangement.

Are you sure that solvation is driven by enthalpy and not entropy?


text:
Why are radical intermediates are more stable on tertiary carbons?
Is this because there's more electron density? More carbons around means the tertiary carbon "has more electrons around it"?

It’s hyperconjugation and works by the same mechanism as for carbocations.

What exactly is hyperconjugation?

Before thinking about why tertiary radicals are 'stable', you need to have some kind of understanding about the structure of radicals.
Carbon centered radicals adopt similar geometries to carbocations: three coordinate and planar. Whereas carbocations have an empty p-orbital, radicals have a singularly occupied p-orbital. Overall the geometry is usually described as sp2 if we're thinking about hybridisation. (*)

With the structure of the radical now defined, we can think about what gives the tertiary radical such stability, and to do that we need to discuss what hyperconjugation is.
A good place to start (other than a textbook) might be the IUPAC definition from the Gold book:

...The concept of hyperconjugation is [...] applied to carbenium ions and radicals, where the interaction is now between σ-bonds and an unfilled or partially filled π- or p-orbitals...

Essentially, in the sp2 geometry, adjacent C-H bonds are able to interact with the vacant p-orbital, donating a small amount of electron density in order to stabilise the electron deficient p-orbital.
Individual C-H bonds to not exert a particularly strong stabilising effect, however, in a t-butyl cation/radical, there are 9 adjacent C-H bonds, each donating a very small amount of electron density, the sum of which amounts to a net overall stabilisation relative to primary radicals.

Whilst this explains the stability of the tert-butyl radical, there are other mechanisms of stabilisation for radicals which carbocations are not able to participate in. Due to the single electron in the p-orbital, radicals can also be stabilised by resonance, in a similar (but not identical) way to lone pairs or carbanions. A classic example is a radical generated next to a C=O group or C=C group.

Both of these effects discussed (hyper conjugation and resonance) are thermodynamic effects. The stabilisation is lowering the energy, making the radical less reactive.
(*) In the case of radicals, this is a bit misleading, theres a lot of spectroscopic and computational evidence to suggest that radicals deviate from planarity towards a pyramidal structure (the idea of hybridisation and sp2 doesn't hold up well)- especially when electron withdrawing groups such as fluorine are bonded to the radical centre. From the point of view of undergraduate chemistry this is largely irrelevant, however, given that the planar model explains the features of radical reactions that we're usually interest in, as organic chemists rather than theoreticians (EPR studies are one place in which the planar model falls down, the measured values don't correlate well to a completely flat carbon centre).

"there are other mechanisms of stabilisation for radicals which carbocations are not able to participate in." Might not be worth mentioning if you're not going to present an example. It could cause a new student to misunderstand that the resonance you mention immediately after does indeed also apply to carbocations/anions.

There's an example underneath the sentence of the allyl radical. The resonance does indeed apply to carbanions. I feel the sentence is clear.

Oh I see, so you're saying the geometry deformation is a notable stabilizing factor for radicals (which does not apply to the ionic species)? Interesting, that's new to me. Thank you for the clarification.

That's not what Im saying. Perhaps we can continue this discussion in chat

I'm interested in that, but I cannot figure out how to start the chat.

How do you know that carbocations have an empty p-orbital?

@Hamze, thats a whole new question. You might perhaps look in a textbook as any organic chemistry text will explain the structure of a carbocation.....

Radical intermediate's stability is same as that of carbocation's stability. It is stabilised by hyperconjugation and +I effect of alkyl groups.

No, this is not a good generalisation. Unlike carbocations radicals are also stabilised by electron-withdrawing groups such as carbonyls.


text:
Activity of Materials in Equilibrium constant
I learnt that we use activity to define the equilibrium constant . But while calculating the equilibrium constant we plug in their concentration instead of activity .
For a general reversible reaction
$$\ce{aA + bB <=> cC + dD}$$
$$K=\frac{[C]^c [D]^d}{[A]^a [B]^b}$$
We put their concentration instead of activity.
Why is that done ?

While calculating concentrations is trivial (divide the amount of substance by the volume), activity is a complete different story. Activity is usually calculated by multiplying the concentration by a dimensionless number called activity coefficient($\gamma$):
$$a = c \times \gamma$$
The problem here is that $\gamma$ is not only different for every substance, it also varies with temperature, concentration of the substance itself, the diluent used, and so on. If the substance is a salt that dissociates then the calculation gets even more complex.
Since having the values of the activities would be quite a challenge, equilibrium constants are often written using concentrations, which are more readily available. It's not the most correct way of represent it, but it makes your life easier.


text:
Dissolution of lead chloride because of the formation of- Na2PbO2 or [Pb(OH)4]-2?
I read that $\ce{NaOH}$ precipitates lead hydroxide which dissolves with excess hydroxide.
In some sources, I read that lead chloride dissolves due to the formation of complex $\ce{[Pb(OH)4]^2-}$ but my teacher taught that it is because of the formation of $\ce{Na2PbO2}$
I don't think that both are same. I cannot find much information about $\ce{Na2PbO2}$ either.
Which of these two is actually formed?

Lead hydroxide do dissolve in excess of alkali solution forming complex (the solution becomes clear):
$$\ce{Pb(OH)2(s) + 2OH-(aq) -> [Pb(OH)4]^2-(aq)}$$
There can be confusion because some articles(e.g. here) says it lead to formation of sodium plumbate(II) but plumbate is poorly defined as it can considered lead oxyanion or oxides of lead.
Wikipedia further mentions:

In solution, lead(II) hydroxide is a somewhat weak base, forming
lead(II) ion, $\ce{Pb^2+}$ under weakly acidic conditions. This cation
hydrolyzes and under progressively increasing alkaline conditions,
forms $\ce{Pb(OH)+}$, $\ce{Pb(OH)2(aq)}$, $\ce{Pb(OH)3−}$, and other
polynuclear species like $\ce{Pb4(OH)4^4+}$, $\ce{Pb3(OH)4^2+}$,
$\ce{Pb6O(OH)6^4+}$


text:
How to convert chemical rate of reaction into time?
I am very new to chemistry. I am using a simulator for knowing rate of chemical reaction whose unit is $\mathrm{(kmol/m^3)/s}$. I want to convert it into time taken only. I have got concentration of each chemical can I convert it into time.

Perhaps you are looking to express the rate of the reaction in terms of a half life. That is the time it takes for the concentration of a reactant to reach half its original value. Assuming it is a first-order reaction, t½ = 0.693 / k. If your rate constant was expressed in units of concentration per second, the half life will be in seconds.

The first-order rate constant is in units of per-second, not concentration per second (that is zeroth order).

Well, there isn't usually a specific time that a reaction reaches completion or equilibrium. You need to choose an arbitrary limit, such as "when 90% of the reactant has been converted to product".
You can calculate your time from an integrated rate law. This will require you to know what order your reaction is (perhaps CSE can help you there).
For example, for a first-order reaction, the integrated rate law is:
$C_t = C_0e^{-kt}$
Or, in a form more convenient for you:
$t = \frac{-1}{k}ln(\frac{C_t}{C_0})$
where $k$ is the rate constant of the reaction, $C_0$ is the initial concentration, and $C_t$ is the concentration at the time $t$ you are interested in.
If you choose the arbitrary limit I mentioned earlier (convenient for demonstration, but you'd probably want more product formed that 90%), at your "completion" time there would be 10% reactant left. The math would look like:
$\frac{C_t}{C_0} = \frac{10%}{100%} = \frac{1}{10}$
$t = \frac{-1}{k}ln(\frac{1}{10}) = \frac{1}{k}ln(10) \approx \frac{2.303}{k}$
where you have now found the time it takes for the reaction to proceed to your chosen point. Note that if you choose 100% conversion, the remaining reactant concentration would be 0. This gives you $ln(0)$, which is not defined for real numbers. To get a more meaningful understanding, you can take the limit as the reactant concentration approaches zero (the reaction approaching completion), and you will find:
$t_{completion} = \lim_{C_t \to 0}\frac{-1}{k}ln(\frac{C_t}{C_0})$
$= \frac{-1}{k}\lim_{u \to 0}ln(u) = \frac{-1}{k}(-\infty) = \infty$
And thus by our most useful models, it takes literally forever for every single molecule to react. This is true for any reaction order. This is why we cannot easily say when a reaction is "done", and we must pick an arbitrary point.
Note that all the math examples I presented only apply to first-order reactions.
To find the time your reaction "takes" you will have to: 1) choose an arbitrary point to call completion, 2) know the order of the reaction, and 3) know the rate constant of the reaction.
P.S. If you wanted, you could set $\frac{C_t}{C_0} = \frac{1}{2}$ in the equation solved for $t$, and you will derive the half-life formula $t_{1/2} = \frac{ln2}{k}$ mentioned elsewhere.


text:
Why does temperature affect the dissolved oxygen concentration?
I know that an increase in temperature decreases the solubility of oxygen in water, but I don't know why. Could anybody explain?

Have you tried searching a bit on your own? You didn't say what you know so far or where you looked prior to posting here.  Searching google for "oxygen dissolution temperature" yields this as the third result. Give it a try.

@Don_S oh cool, it says it's because dissolving oxygen is exothermic, and according to Le Chatelier's principle, adding heat would favor the endothermic side, so adding heat shifts the equilibrium to the left. Thanks for the link, perfect!

The solubility of a many gases in a variety of liquids is observed to decrease with increase in temperature, for example oxygen, nitrogen and methane in water.  This follows the qualitative predictions of Le Chatelier’s principle, which in thermodynamics is supplanted by the Clapeyron equation. As one of the two phases involved is a gas the Clausius–Clapeyron equation can be used.
This assumes that the molar volume of any gas/vapour is far greater than that of the solution. Assuming that all values correspond to one mole the Clausius–Clapeyron equation is
$$\frac{\mathrm d\ln(p)}{\mathrm dT}=\frac{\Delta H}{RT^2}$$
which in its integrated form between pressure and temperature $p_1, T_1$ to $p_2, T_2$, and by assuming that $\Delta H$ is independent of temperature gives
$$\ln\left( \frac{p_2}{p_1}\right)=- \frac{\Delta H}{R}\left( \frac{1}{T_2} - \frac{1}{T_1}   \right)$$
Despite the approximations involved in its derivation this equation does describe well the measurements of the vapour pressure over many solids and liquids, for example depression of freezing point, elevation of boiling point and solubility of solids and gases in liquids, although the latter is not as frequently described as the other two.
In the elevation of the boiling point the solution at a temperature $T_1$ has a lower vapour pressure $p_1$ than the pure solvent. The solution only boils when the temperature is increased to $T_2$  where its vapour pressure has increased to  $p_2$ (normally 1 atm ), which is the pressure at which the pure solvent boils.  By Raoult’s law  $p_1=p_2x_{\mathrm s}$, where $x_{\mathrm s}$ is the mole fraction of solvent
$\ln(p_2/p_1)=\ln(1/x_s)$
As $x_\mathrm s=1-x_\mathrm t$ where $x_\mathrm t$
is the mole fraction of solute,
$$\ln\left( \frac{1}{1-x_t}\right)= -\frac{\Delta H_\text{vap}}{R}\left( \frac{1}{T_2}  - \frac{1}{T_1}   \right)$$
and as $x_\mathrm t$ is far smaller than unity
$\ln(1-x_t) \approx -x_t -x_t^2/2 \ldots$
from which the second term can be ignored as it is insignificant compared to the first then,
$$x_\mathrm t= \frac{\Delta H_\text{vap}}{R}\left( \frac{1}{T_2}  - \frac{1}{T_1}   \right)$$
Thermodynamically the elevation of boiling point and solubility are very similar but we must swap solute and solvent over so what we formerly called solvent has now to be solute and vice versa.
Considering now solubility, the variation of the mole fraction dissolved $x_\mathrm t$ can be written as
$$x_\mathrm t=\frac{\Delta H_\text{vap}}{R}\left( \frac{1}{T_2}  - \frac{1}{T_\mathrm B}   \right)$$
where what was $T_2$ is now the normal boiling temperature of the liquefied gas. Also $\Delta H_\text{vap}$ (a positive quantity) is now the (molar) heat of vaporisation of the liquefied gas and we approximate this to be the heat of vaporisation from solution.
This last equation indicates that the solubility decreases with increase in temperature and should be the same in all solvents in which it forms ideal solutions.
In the solution a gas molecule has an attractive interaction energy between itself and the solvent. (Experiment shows that $\Delta H_\text{vap}$ is positive thus the heat to dissolve is negative). The solvent molecules also have an attractive interaction between themselves otherwise they would not form a liquid.  Thus for a gas-solvent interaction a solvent-solvent one has to be destroyed. Overall the difference in energy is probably small.  However, there is an entropic effect to consider. Clearly this will have increased when a gas dissolves as now there is another species in the solvent and thus the number of ways the molecules can be arranged has increased and this will have increased the entropy. Together these effect make the free energy negative as expected for a spontaneous process.
As the temperature is increased the molecules gain in average energy and so are more able to overcome the attractive potential that keeps them in solution and thus the vapour pressure of the solvent increases and the solubility of the gas decreases. It is also worth noting that the average kinetic energy $(3/2)kT$ is the same in the solution and in the gas phase, in solution it is merely restricted to a smaller range on motion. Thus the notion that it is the kinetic energy that is acquired that causes a molecule to enter the gas phase in incorrect.

For ideal gases (probably in solution the physics are a little different) but anyway as oxygen is an ideal gas, the equations from physics that you need to keep in mind are ...
kinetic energy = (1 / 2)(mass)(velocity)^2 = (3 / 2)(Boltzmann constant)(Temperature),
so therefore temperature is a measure of the velocity of your molecules, where as temperature increases, so does the velocity of the molecules increase. Thus, all molecules in solution, including gaseous oxygen move more rapidly gaining kinetic energy to escape solvation energy, and so they are more prone to bubble up the surface and escape acqueous solution because oxygen gas is less dense than liquid water.

Unfortunately this answer is not correct either.

Cold water holds more oxygen molecule than hot water because water molecules are closer together in cold water which makes it harder oxygen molecule to escape.
At higher temperature, the oxygen molecule have greater amount of kinetic energy. So the gas molecule in hot water can more easily overce the weak binding forces within the water and escape through the surface.

Unfortunately this answer is not correct


text:
Why chelation increases entropy

It is written here that chelation increases overall entropy?
I thought it should decrease entropy because the ligands are now bonded to the central metal atom forming a ring like structure. According to me entropy must decrease because the ligands can no longer move around so there is a decrease in randomness. Why this logic is not correct?

Because you don't chelate bare cation. It's actually a substitution.

Can you please elaborate??

Read up on Martin's answer here (http://chemistry.stackexchange.com/questions/31005/whats-so-special-about-chelation/31007#31007)

The chelate effect is very simple. Consider the addition of $\ce{Na2H2edta}$ to a random transition metal (e.g. manganese(II)). Your starting complex is a hydrate complex; we shall assume it is $\ce{[Mn(H2O)6]^2+}$. The reaction equation is given below:
$$\ce{[Mn(H2O)6]^2+ + edta^4- <=>> [Mn(edta)]^2- + 6 H2O}\tag{1}$$
You started with two molecular entities, you end up with seven. Since $2<7$, the product side has a higher entropy (more independent particles). The same is true for any chelating ligand that displaces more than one monodentate ligand.

Yes, upon chelation the chelating ligand does lose degrees of freedom concerning bond rotation. However, the number of particles greatly outweighs that.

I think Mithoron has the right idea. Say before chelation your cation is in water, surrounded by its typical hydration sphere of solvent molecules coordinating to it. Then the chelating agent comes along, displacer the water. I'm guessing the driving force for the substitution is enthalpic in origin, the cation-chelator coordinate bonds are more energetically stable/preferable.
Intuitively, which system can be arranged more ways: 1) Bulky chelator molecules strewn about a sea of solvent molecules, and metal cations coordinated by some number of solvent molecules, or 2) that number of solvent molecules freely dispersed in the solution, and the chelators coordinating the metals...?
Since the free chelators themselves require an ordered solvent structure around them, my intuition suggests that performing the chelation and in the bulk solution swapping chelators for solvent molecules would increase entropy, because these newly freed solvent molecules can occupy nearly any position and orientation (whereas previously the free chelators required solvation spheres).
tl;dr Perhaps the entropy increase of dissembling the ordered solvation spheres around the free chelator molecules compensates for your entropy decrease of forming the ordered chelator-cation complex, leading to a net increase in entropy.
I think your situation may be interesting to compare to the rationalization of the hydrophobic effect.


text:
is there any boundary or membrane around atom to recognize its shape?
we know that every Isolated atoms are spherically symmetric.how we know its shape, is there is any membrane covers it as skin of our body?

No boundary. The "shape" may be detected from diffraction experiment. It comes from interaction between atom and external objects.

http://chemistry.stackexchange.com/questions/13766/why-are-atom-spherical-in-shape?rq=1

There is no membrane, no skin, and moreover, no shape and no body. If you ever think of applying any of these words to an atom, then you have a major misconception about the nature of atoms and elementary particles.
An isolated atom is spherically symmetrical in that it reacts the same way no matter which side you approach it from. As for the actual picture of an atom, take a photo of a streetlight in a thick fog from far away, or be content with the image from Wikipedia.

There is no membrane. In principle the electron could be at a very large distance from the nucleus, but the probability decreases fairly quickly as a function of distance.
Edit:
Technically, you could create a boundary surface diagram that draws a "surface" around the atom that represents some cutoff for the probability. Of course, it doesn't mean much because the electron can go right through it. For an atom, this is just spherical, but it is certainly more interesting if you have a molecule.


text:
What is Drago's rule? Does it really exist?
My textbooks states the Drago's rule in inorganic chemistry as follows:

The more electronegative atom prefers the orbital having more $\mathrm{p}$ character, and lone-pairs or less electronegative elements prefer such orbitals, as have more s character.
Now, according to hybridization, $\mathrm{sp^3d = sp^2 + pd}$, $\mathrm{sp^3d^2 = sp + p^2d^2}$, $\mathrm{sp^3d^3 = sp + p^2d^3}.$
So, in $\mathrm{sp^3d}$ more electronegative elements like fluorine go
to axial position, but such elements go to equatorial position in $\mathrm{sp^3d^2}$ or $\mathrm{sp^3d^3}$.
Also, as a result of Drago's effect when atoms of group $14,$ $15$ and$16$ (belonging to third period and above) are attached to atoms having electronegativity less than $2.5$ then it has valence electrons in pure s-orbitals. And apparently no hybridization effect is observed in such cases. That is why in $\ce{NH3}$ the bond angles are $107.8^\circ$ but for $\ce{PH3},$ $\ce{AsH3},$ $\ce{SbH3},$ $\ce{BiH3}$ the bond angles are very
close to $90^\circ$.

I couldn't make much sense out of the above rule. My questions are as follows:

What does the following part mean?

according to hybridization, $\mathrm{sp^3d = sp^2 + pd}$, $\mathrm{sp^3d^2 = sp + p^2d^2}$, $\mathrm{sp^3d^3 = sp + p^2d^3}$

Does it mean that the hybridized orbitals split in this way? If so, why should they split?

Why would the more electronegative atom prefer the orbital with more $\mathrm{p}$ character?

Even if Drago's rule holds, why is the statement in the paragraph beginning with "Also, as a result of Drago's effect when atoms …" true? I could not understand why the same won't occur for atoms with electronegativity $>2.5$.

I could not find much information regarding Drago's rule on the internet. Not even on Wikipedia. One of my acquaintances (who is a chemist) said that no such rule exists. But I didn't get much time to discuss the matter with him.
So does such a rule really exist? And, in that case, how can we answer the above two questions given as bullets  using the existing theories in chemistry?

Sounds rather like https://en.wikipedia.org/wiki/Bent's_rule

Related:http://chemistry.stackexchange.com/questions/14087/why-is-the-bond-angle-h-p-h-smaller-than-h-n-h?noredirect=1&lq=1

Also related:http://chemistry.stackexchange.com/questions/14791/hybridization-in-ph3

@Doraemonドラえもん If the term is textbook specific, you might want to also post the name of the textbook & authors (and edition, if any). There's a chance that that information will help track down the usage. (Also, is this an English language textbook? "Drago's rule" might be a translation error - again, knowing the textbook details may help.)

@R.M. It is not actually a standard textbook which is available worldwide. It is a study material prepared by a coaching institute. I don't think posting the name of the institute will be of much use here. Anyway, I don't think the propositions of the rule are wrong. Some of it is indeed explained by Bent's rule as pointed out by Mithoron. But I could not find the explanation regarding the orbital splitting as proposed by the so-called "Drago's rule". I hope someone can verify whether such a splitting really takes place or not.

@R.M. And yeah, the textbook/study-material is written in English. But I don't think the name of the rule should be much of a problem. The main problem is checking whether all the claims made by the rule are true or not.

That sounds like an extension of Bent's rule shoehorning in some d orbitals. That kind of hybridisation doesn't make much sense as we have often discussed here. I would not waste time on understanding this rule.

@Martin-マーチン "That kind of hybridisation doesn't make much sense as we have often discussed here." Can you give some reference or links for this ? Where has it been discussed before ?

@Doraemonドラえもん In like every question where using of d orbitals in p-block was mentioned? Like http://chemistry.stackexchange.com/questions/32695/is-clh7-possible or http://chemistry.stackexchange.com/a/49844/9961 or...

This might go back to Professor Russell Stephen Drago (1928-1997) https://webster.chemistry.msstate.edu/prof/drago.htm and his textbook of Physical Methods in Inorganic Chemistry from 1965. @EdV any ideas (Drago was born in Turners Falls and received his B.S. from UMass Amherst).

@Karsten Theis Well, I vaguely remember being in the room when Russ Drago and Alan Katritzky visited our (Jim Winefordner’s) lab at the University of Florida at Gainesville: https://chemistry.meta.stackexchange.com/a/4757/79678. I never talked with Drago, but I do not doubt that the rule is named after him.

Going through what you posted, I think ‘Drago’s rule’ (which I never encountered, either at school or at university) gives good predictions but uses a largely terrible set of arguments.
For any atomic system with more than one electron, quantum chemistry predicts the energies of the $\mathrm{s}$ and $\mathrm{p}$ subshells of a shell to differ — while for hydrogen-like systems (one-electron systems) all subshells of a given shell have the same energy. The $\mathrm{s}$ subshell of a certain shell in multi-electron systems has a lower energy than the $\mathrm{p}$ subshell, which in turn has a (far) lower energy than the $\mathrm{d}$ subshell and so on and so forth. Since the $\mathrm{s}$ subshell is determined to be more stable by this method, it is not surprising that the first two electrons of a given shell are added to said $\mathrm{s}$ subshell; the next six electrons are added to $\mathrm{p}$ subshells whereafter a new shell is used.
This immediately explains that it is most stable for any atom to treat the $\mathrm{s}$ electrons of its valence shell as ‘extended core electrons’; i.e. we can predict any configuration that does not touch the $\mathrm{s}$ electrons to be energetically favourable with respect to any configuration that does a priori. (You can consider that part of the quoted statement to be correct.) Therefore, the a priori predicted most stable configurations of certain coordination numbers are:

mono-, di- and tri-coordination: use 1, 2 or 3 $\mathrm{p}$ orbitals to form bonds; keep one lone pair in an $\mathrm{s}$ type orbital. (Predicted bond angle: $90^\circ$.)

tetracoordination (no additional lone pairs): since the former is not possible, hybridise $\mathrm{s}$ and $\mathrm{p}$ orbitals to form $\mathrm{sp^3}$ hybrid orbitals. (Predicted bond angle: $109.5^\circ$.) Actually, in MO terms you should consider the four ligands to supply four group orbitals whose symmetry always fits one of the three central atom’s orbitals, resulting in one lower bond energy and one higher bond energy corresponding to triple-degenerate bonds. The $\mathrm{sp^3}$ description is mathematically equivalent, though.

penta- and hexacoordination; tetracoordination with additioal lone pairs and related: attempt to form as many normal bonds with p orbitals as possible; keep one lone pair in an $\mathrm{s}$ orbital if possible. Use remaining $\mathrm{p}$ orbitals to construct four-electron-3-centre bonds to the remaining atoms. (Predicted bond angles: diverse. $90^\circ$ going from 2e2c bonds to 4e3c bonds; $180^\circ$ between a pair of coordinating atoms contributing to the same 4e3c bond.)

Using the aforementioned 4e3c bonds nicely explains why electronegative atoms prefer to occupy these positions. You should always remember that 4e3c bonds can be described by the following resonance structures (drawn for $\ce{ClF3}$):
$$\ce{F-Cl^+\bond{...}F- <-> F^-\bond{...}Cl^+-F}$$
In this resonance, each outer atom taking part in a 4e3c bond has a formal average charge of $-0.5$; it is generally better to assign charges to electronegative atoms.

The theories I have outlined so far do not adequately predict the bond angles of water, ammonia and related compounds. Indeed, these compounds are remarkably complicated although they seem simple. The gist, however, is that a bond angle of $90^\circ$, while being most favourable on a pure orbital basis, introduces steric strain between the outer atoms which approach each other too closely if the central atom is relatively small. To combat this destabilisation, the bond angle is extended by mixing $\mathrm{s}$-contribution into the bonding orbitals — which in turn means that the lone pair will be less $\mathrm{s}$-like. The ideal bond angle for dicoordinated systems based on sterics is $180^\circ$, showing that the electronic contribution ($90^\circ$) is more important than the steric one ($180^\circ$). Larger central atoms than oxygen and nitrogen, such as phosphorus, antimony, selenium, etc., allow for smaller bond angles since the bond lengths are larger increasing the spacing between the ligands.
The reason for the special behaviour of nitrogen and oxygen is not their high electronegativity but their small size. Chlorine, which has an electronegativity similar to nitrogen, behaves as predicted for large atoms.

It has been commented that you should not invoke $\mathrm{d}$ orbital participation for main group elements. Remember what I said in the second paragraph: a shell’s $\mathrm{d}$ subshell has a higher energy than the next shell’s s subshell. Thus, before any same-shell $\mathrm{d}$ subshell is added to hybridisation, the next shell’s $\mathrm{s}$ subshell should be. Since hybridisations involving two different shells’ $\mathrm{s}$ subshells are not observed, we should reject the possibility of using $\mathrm{d}$ orbitals for hybridisation. Therefore, we can also flat-out reject the central two paragraphs of your quotation.

"...you should not invoke d orbital participation for main group elements". Does that mean all hybridizations like $\mathrm{sp^3d}$ ,$\mathrm {sp^3d^2}$ and $\mathrm{sp^3d^3}$ are invalid for main group elements ? And which elements do you consider as "main group" elements ?

And, in that case, how do you explain the hybridization of compounds like $\ce{PCl5}$ ? If it is not $\ce{sp3d}$ then what is it ? How can you explain its hybridization using only $s$ and $p$ orbitals ?

@dora http://chemistry.stackexchange.com/q/52179/4945 hybridisation simply isn't a good enough concept for these cases.

' Since hybridisations involving two different shells’ s subshells are not observed, ...' No one ever observed any hybridisation. But try mixing 1s and 2s (etc), they have the same symmetry...

@Doraemonドラえもん $\ce{PCl5}$ is $\mathrm{sp^2}$ hybridised. The two axial chlorine atoms form a 4-electron-3-centre bond with phosphorus.

PCl5 is sp3d hybridized

@ShubhangWalavalkar That is incorrect. It is an old theory proposed, I believe, by Pauling that has never managed to die despite the overwhelming evidence presented against it possibly because it was supported by a Nobel laureate. Please see this answer as a starting point. If that doesn’t satisfy you, there are many other posts on this network that can enlighten you.

so according to you 3 center 4 electron bond changes bond angle of 90 degree( 2 center 2 electron ) to 180 ????

@Samardeepsingh Molecules typically don't 'go to 4e3c bonds from 2e2c bonds'. So there is no 'change of bond angle'. Rather, a molecule will either adopt a configuration with two 2e2c bonds or one 4e3c bond, depending on electronic factors. One is predicted to give a 90° bond angle, the other is predicted to give a 180° bond angle. Which is which is up to the chemist to determine.

These rules go back to a publication by Russel S. Drago criticizing the use of the VSEPR model in Introductory Chemistry. As an alternative, he suggests a set of heuristics. First, if there are no lone pairs, use essentially the same rules as VSEPR. How to treat lone pairs is described below:

Here we shall consider central atoms with eight or
less electrons about it. When there are lone pairs on the
central atom, each lone pair counts as a group in the
above scheme if the central atom is a second row element.
If the central element is a third, fourth, etc., row atom
and if the groups attached to the central atom are oxygen
or a halogen, the lone pair also counts as a group. If the
groups attached are less electronegative than bromine
(that is, if almost anything other than a halogen or oxygen
is attached), the lone pair occupies an unhybridized orbital
and does not count as a group. When the s orbital is
used to accommodate a lone pair, only p orbitals remain
for bonding, and the geometry involves an arrangement of the groups at about 90° angles to each other. Some examples
are listed in Table 2.

(Table 2 states that $\ce{H2O, NH3, and PCl3}$ are tetrahedral whereas $\ce{PH3 and H2S}$ use p-orbitals to bond and have ninety degree angles). The claim is that the rule of thumb is better than VSEPR, and not off by more than 4 degrees from the observed structures.
He goes on to say that chemists, when they encounter a molecule they don't know, would draw a Lewis structure and compare it to those of molecules they know rather than trying to predict a structure on paper from first principles. Another quote:

It is deceptive to lead a freshman
into believing anything else. The rules presented above
summarize these analogies and enable you to predict the
geometries of many molecules.

Source: A criticism of the valence shell electron pair repulsion model as a teaching device, RS Drago - Journal of Chemical Education, 1973


text:
Hydrolysis products of chlorides of group 15
$\ce{NCl3}$ on hydrolysis produces $\ce{NH4OH}$ and $\ce{HOCl}$.
$\ce{PCl_3}$ when hydrolyzed produces $\ce{P(OH)3}$ i.e. $\ce{H3PO3}$.
$\ce{AsCl_3}$ when hydrolyzed produces $\ce{As(OH)3}$.
However when $\ce{SbCl3}$ or $\ce{BiCl3}$ are hydrolyzed they produce $\ce{SbOCl}$ and $\ce{BiOCl}$ respectively.
Why are the products so widely varying ? How can we explain the reactions theoretically ?

The first one is because nitrogen is more electronegative than chlorine while all the others are electropositive with respect to chlorine. The antimony and bismuth reactions are probably due to some stability of those compounds that I, unfortunately, don’t know about.

Nitrogen trichloride

Nitrogen trichloride is quite unstable and hydrolyse easily:
$$\ce{NCl3 + 3 H2O → NH3 + 3 HOCl~~~~~~~~(1)}$$
$$\ce{NH3 + H2O -> NH4OH~~~~~~~(2)}$$
$$\ce{NCl3 + 4H2O -> NH4OH + HOCl~~~~~~~(1) +(2)}$$

Phosphorus trichloride

$$\ce{PCl3 + 3H2O → H3PO3 + 3HCl}$$

Arsenic trichloride

$$\ce{AsCl3 + 3 H2O → As(OH)3 + 3 HCl~~~~~~~~(1)}$$
This is the first and main reaction of hydrolysis of arsenic chloride to form arsenous acid. But other species (arsenite ions) like $\ce{[AsO(OH)2]−}$ and $\ce{[AsO2(OH)]^2−}$ also exist in solution and are the conjugated bases.
$$\ce{AsCl3 + 4H2O ⇄ H[As(OH)4] + 3HCl~~~~~~~(2)}$$
$$\ce{AsCl3 + 2H2O ⇄ HAsO2 + 3HCl~~~~~~~~~(3)}$$
And at last aresnic trichloride fully hydrolyse to form arsenic(III) oxide
$$\ce{2AsCl3 + 3H2O → As2O3 + 6HCl~~~~~~~~~~(4)}$$

Antimony trichloride

$$\ce{SbCl3 + H2O → SbOCl + 2HCl~~~~~~~~(1)}$$
With more water it forms $\ce{Sb4O5Cl2}$ which on heating to 460° forms $\ce{Sb8O11Cl2}$.
$$\ce{SbCl3 + H2O ⇄ Sb(OH)Cl2 + HCl~~~~~~~(2)}$$
$$\ce{4SbCl3 + 5H2O ->[50 C] Sb4O5Cl2 + 10HCl~~~~~~~(3)}$$
Final reaction to form antimony trioxide.
$$\ce{2SbCl3 + 3H2O ->[\Delta] Sb2O3 + 6HCl~~~~~~~~~~(4)}$$
Note that speculation on the composition of antimony oxychloride has been raised. Some say it is a mixture of antimony trioxide and antimony trichloride.

Bismuth tricloride

$$\ce{BiCl3 + H2O -> BiOCl + 2HCl~~~~~~~~(1)}$$
Main reaction to form bismuth oxychloride. Further hydrolysis yield and intermediate monohydrate $\ce{BiCl3.H2O}$.
$$\ce{BiCl3 + H2O(vap) ->[50 C] BiCl3.H2O~~~~~~~~(2)}$$
$$\ce{$$BiCl3 + 2H2O → Bi(OH)2Cl + 2HCl~~~~~~~~~(3)}$$
Explaination
Now, you may been be wondering that hydrolysis of nitrogen and phosphorus yield one product but that of arsenic, antimony and bismuth yield 3 or 4 products. This is because of presence of vacant d or f orbitals where electrons can reside easily. For this reason, $\ce{Sb^3+, Bi^3+}$ solvate in aqueous solution to form antimonyl or bismuthyl ($\ce{SbO+, BiO+}$) and other species like $\ce{[Bi6O4(OH)4]^6+, [Bi6(OH)12]^6+ ,  [Bi(H2O)9]^3+}$ and similar antimony ions. Other factors also play role like relativistic effects, lanthanoid contraction, inert pair effect etc.


text:
Synthesis of trans alkene with organoborane
Trans alkenes can be synthesized with organoboranes. The following sequence is represented in Smith's Organic Synthesis

How does the mechanism look like after the first step (hydroboration)?

I think this might be a alkyl shift that results in an $\mathrm{S}_{N}2$ type substitution at an $sp^{2}$ center. Cool. I've only ever seen $\mathrm{S}_{N}2$ type substitution at an $sp^{2}$ center for vinyl iodides. Unfortunately, I don't have my teaching materials on me at work, so I can't find the reference for this for you.
EDIT:

Note the "inversion" at the $sp^{2}$ center.
Greg Fu had this great example of a substitution on a vinyl iodide, like I mentioned above. I can't seem to find this reference, but I just emailed him, so I'll update when he writes me back.

I don’t think it’s a substitution: I would assume addition of methanolate to the boron centre which then induces a migration/insertion of some sort. But I don’t know myself, unfortunately.

Notice that even in that case, you need inversion of the $sp^2$ center. I am in fact proposing the exact same thing as you are.

Oh, you’re talking about a formal inversion! I thought you meant an actual inversion =D

Here's the reference as promised. Somehow my email ended up in Greg Fu's spam folder... :( JACS 1998, 120, 2275


text:
Extracting iron oxide from ceramics waste
In my ceramics class, we use iron oxide in our glazes, but a lot of it goes to waste. We have a bucket full of scrap glaze which contains quite a lot of iron oxide, but is also mixed in with silica, alumina, titanium dioxide, tin oxide, and other substances.
Is there any way to extract the iron oxide or other metal oxides from the scrap bucket?

Assuming the iron oxide is just mixed and has not been fused with the other oxides you could use a magnet to pull it from the other oxides.
If it is fused with the other oxides, you could smelt the iron out with carbon. This will also smelt out the tin, which mixes with iron quite well. If you re-oxidized the iron in a furnace with air the iron will selectively oxidize first leaving tin metal when done. This is likely an impractical and cost-inefficient procedure as iron oxide is incredibly cheap relative to the cost of separation.

According to Alibaba, iron oxide prices are sometimes below US$0.80 per kilogram.  I'm sure the stuff for ceramic glazing is more than that, but even if it's $3 per kilogram, it's still really cheap.

Yes, they are just mixed together not fused. If I was to melt it, all of the ingredients would fuse with the silica, making it impossible to extract. I'll try the magnet. Thank you

There are more than one iron oxide out there, and not all of them are magnetic.

I once was using a silver nitrate and had collected a small amount of silver chloride. At the time, we were operating a lab muffle furnace (for a related reason), so I decided to reduce the $\ce{AgCl2}$ to $\ce{Ag}$. I needed some graphite crucibles and it took 48 - 72 hours of high heat. I collected an amount of silver equal in mass to maybe two quarters (US coins) with a value (at the time) of around \$15 (iirc). Total cost? Oh, probably \$50 to \$100. (Graphite crucibles were consumed, my time, energy costs, and wear and tear on furnace.)
Iron oxide is a LOT less valuable than silver chloride. There is nothing you can do to make any reclamation effort cost effective when you are competing with people digging it out of the ground by the ton.
Of course, most elements can be separated from one another by classical (wet) chemistry - after all, that's how the concept of an 'element' began - elements are the irreducible components of all matter.
Iron is much more reactive than silver - if you don't believe me, go take a look at all the rust surrounding us. Rust is, after all, just one kind of iron oxide. Since it is more reactive, isolating it will be much more expensive than isolating silver. So we have a perfect storm: it's less valuable, and would cost a lot more to process. Titanium is by far more valuable, and yet they use it in paper to make it 'bright' - so generally the only reason to extract metals would be to reduce the cost of disposal. Things like lead or other toxic heavy metals might be best handled by recovery - although generally inclusion into a ceramic is one of the best long-term storage choices, so even then, it's unlikely to be useful. And none of the processes that would be effective would be efficient with the small quantities from a lab.
Just consider all of the $\ce{CO2}$ you'd need to create (either locally, or at the (gas or coal powered) power plant just to heat this scrap to reaction temperature. As I learned long ago, (much to my chagrin), in almost all cases, (small) lab wastes are cheaper to dispose of than to repurpose.

Cost of heating it isn't a problem, it could just go into the kiln to be fired with the rest of the ceramic pieces. I just can't do that because there's silica in the mix. Heating that will make glass, which will make it impossible to extract. I'll need some other way to reclaim it. I'll look into chemical processes

There are ways, of course.   Smelting, for one (reducing the oxides in
a high temperature furnace), and leaching for another (flowing a
solvent through the crushed media).
Oxides and silicates are rather stable, however, and the temperatures, reagents and
solvents that attack them are not tabletop-friendly.   If the scrap was well-sorted, or valuable, or too toxic to dispose of, there might be some
utility in treating it.