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How to prepare elemental gallium from a mixture of gallium and aluminum nitrate?
How would one go about preparing and purifying elemental gallium from a mixture of aluminum(III) nitrate and gallium(III) nitrate?

If your mixture contains aluminum nitrate and gallium nitrate, then you need to separate the aluminum nitrate first. This patent deals with separation of various nitrates from a mixture.
When you have pure gallium nitrate, decompose it to gallium oxide. Then reduce gallium oxide to gallium using a reducing agent like magnesium. (Note: hydrogen reduces gallium(III) oxide to gallium suboxide: $\ce{Ga2O3 + 2 H2 → Ga2O + 2 H2O}$)
You will find this youtube video of help.

Thank you, this is exactly what I was looking for! That YouTube video was also quite helpful in learning how gallium aluminum amalgams are purified.

@ComputerCraftr glad i could help. Alternatively, you can also check this video.. It shows how gallium can be extracted from gallium-aluminum alloy.


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Decomposition through thermolysis, electrolysis, photolysis
What decides whether a compound will dissociate (break-up) under thermal decomposition or photolysis or electrolysis i.e if I am provided with a compound and I'm asked to dissociate it then how will I get to know that it would dissociate under thermal decomposition or photolysis or electrolysis.
Please try to provide an easy-to-understand as I'm in just 10th grade

The most general answer is that every compound will breakup if enough energy can be added to it.  In a simple diatomic thermal energy will increase the vibrational (and rotational) energy until the bonds breaks. Electromagnetic radiation (uv, vis , ir) can also do this, as well as forming excited electronic states that  decompose.

You should analyse the chemical bindings in your compound, e.g. ion bindings will easily break-up in water, stronger bindings may require You to use heating or electric current to dissociate the molecule. Also if you know the structure of the compound, look for some similar structures on the internet, the more similar the structures, the more similar properties they have, so similar factors will dissociate them.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


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Can titrant and analyte be swapped?
I just did a titration experiment at school, our teacher told us multiple times to add the solution of known concentration into the solution of unknown concentration. I am wondering if we can do it the other way, if not, why?

From what I know, you can add the solution of unknown to the solution of known concentration. The only thing that matters is you have a known volume of the solution to which you're adding the other solution.
For example, suppose we're titrating $\pu{0.15 M}$ $\ce{Na2CO3}$ with an unknown concentration of $\ce{HCl}$, if we need to add $\pu{say 15 mL}$ $\ce{Na2CO3}$ to a known volume of say $\pu{30 mL}$ of $\ce{HCl}$ to reach the endpoint, we can calculate the concentration of the HCl solution.
$$\ce{Na2CO3 + 2HCl -> 2NaCl + H2O + CO2}$$
$n(\ce{Na2CO3}) = \pu{0.15 mol/L} \cdot \pu{0.015 L}$
$n(\ce{Na2CO3}) = \pu{0.00225 mol}$ (i.e. the known concentration times 15 mL change to L)
We know that 2 moles of $\ce{HCl}$ will react with 1 mole of $\ce{Na2CO3}$ to form the products on the right hand side of the equation. So,
$n(\ce{HCl}) = 2 \cdot n(\ce{Na2CO3})$
which we calculated above using the relation of moles as product of concentration times volume. So
$n(\ce{HCl}) = \pu{0.0045 mol}$
Since we know that we had $\pu{30 mL}$ of a $\ce{HCl}$ solution with $\pu{0.0045 mol}$ of $\ce{HCl}$ in it,
$[\ce{HCl}] = \frac{\pu{0.0045 mol}}{\pu{0.03 L}}$
$[\ce{HCl}] = \pu{0.15 mol/L}$
So, we had a $\pu{0.15 M}$ solution of $\ce{HCl}$.
If we're now adding $\ce{HCl}$ to $\pu{15 mL}$ $\ce{Na2CO3}$ and it takes $\pu{30 mL}$ of $\ce{HCl}$ to reach the endpoint, we can use similar calculations to once again find the concentration of $\ce{HCl}$ (and get the same answer).
It doesn't matter whether you add the known solution to the unknown one, the only thing you need to know is the volume of the thing you're adding to and how much of the other one you need to add to reach the endpoint.
Think of it like this,
We have 4 variables so to speak;
- concentration of the known solution (we know)

concentration of the unknown (what we want to work out)
volume of the known solution (either adding or added to) which we measure
volume of the unknown solution (either adding or added to)which we also measure

So we know 3 things no matter what we add what to and these 3 values are enough to work out the concentration of the unknown solution. You should clarify with your teacher what she meant. She may have said always add the acid to the base because if we place the base in the burette it may react with the glass, $\ce{SiO2}$ which acts as an acid and that might degrade up the precision for future use.

For future reference, let me suggest to include \mhchem while formatting your answers; functions enabled on ChemSE are compiled here: https://chemistry.meta.stackexchange.com/questions/3044/hidden-points-of-editing-you-probably-didnt-know/3045#3045  As a second recommendation, I suggest to keep carrying (within reason) the units in the equations, which is eased with \pu{}.

Thanks for your suggestions I appreciate it!
I'm not too familiar with formatting thanks for directing me to that page.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


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Is Be2 diamagnetic?
From  molecular orbital theory, it has no unpaired electrons but can we conclude it is diamagnetic considering the fact that it does not exist?

Technically exists: 1, 2, 3 Don't think they managed to measure its magnetic moment though. (Disclaimer: I didn't read through)

Diberyllium is thought to be non-existant. This website explains how diberyllium cannot exist in brief and simple language. According to it:

A simple MO diagram explains why $\ce{Be2}$ must be indefinitely unstable.
  When the two beryllium 2s orbitals interact they split in two energy
  levels - one bonding (lower energy) and one antibonding (higher
  energy) - and when the four electrons of both Beryllium atoms are divided over these two new molecular orbitals(symmetric and antisymmetric, or g and u combinations of the 1s orbitals) ,both levels get fully
  occupied and there is no net energy gain and therefore no bond.

The compound does not exist but that doesn't mean its MO diagram cannot be made. If you look at the MO diagram taken from this site:

You can see that electrons in each energy levels are paired and hence diamagnetic. From that site:

Actually, if a quantum mechanical calculation is performed (at the
  ωB97XD/6-311G(d,p) level), the bond length emerges as 2.81Å and a
  vibrational wavenumber of 167 $\ce{cm^{-1}}$ is predicted. Despite the zero bond
  order, a weak bond IS predicted, and this is the van der
  Waals or dispersion bond.

You can also check the links @orthocresol sent for more information about diberyllium.

2s-2p mixing means combining the s and  orbitals together with the same plane lying and associated with the same energy which leads to the energy change and the bond order respectively.
And From the MOT concept Be2 doesn’t exists as its Border is 0 and in case of para or dia it is diamagnetic.


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How do molecules know how to behave when a disturbance occurs in an equilibrium?
I am well aware of Le Chatelier's principle, but what causes molecules to behave in such a way, my guess is that it all has to do with rates of reactions and frequency of collisions.
When at a constant volume, the addition of an inert gas causes no change in the equilibrium, since the concentration of each species will remain constant, I understand that there will be the same amount of molecules per unit volume as before the addition, however wouldn't the rate or probability of collision increase since now each unit of volume has less space in which the molecules can move about due to the addition of the inert gas in that unit of volume, hence if a reaction had more moles in the forward then the reverse the probability of the forward occurring would be greater, causing a forward shift?

The inert gas does not affect the equilibrium if the volume is help constant because it is present on both the reactant and product sides of the equation and cancels out. As the partial pressures are important increasing volume will decrease these and more reaction towards side with greater number of moles.

The molecules don't know, they aren't alive in any way for that matter!
Here's an easy way to see it (note that I'm avoiding a mathematical approach to give an intuitive reason):
Consider a hypothetical reaction where $\ce{A}$ and $\ce{B}$ combine and gives $\ce{C}$:
$$\ce{A + B <=> C}$$
Let's say initially, there's an equilibrium. The rate of combination of $\ce{A}$ with $\ce{B}$ is the same as the rate of dissociation of $\ce{C}$. Let's say we added an excess of $\ce{B}$. If there's lots of $\ce{B}$ around, molecules of $\ce{A}$ have more chances of bumping into molecules of $\ce{B}$. This pushes the equilibrium forward, until there's enough $\ce{C}$ to counteract the amount of $\ce{A}$ and $\ce{B}$ present. A similar argument can be given if $\ce{A}$ was in excess than in $\ce{B}$.
You could now think about what happens if we removed some amount of $\ce{C}$. In such a case, the rate dissociation of the limited amount of $\ce{C}$ in the vessel would not balance with the rate of combination of $\ce{A}$ and $\ce{B}$, and thus equilibrium would again flow towards the products.
Now, lets see what happens if you add an inert gas $\ce{D}$ to that mixture. In such a case, $\ce{C}$ will continue to dissociate in it's usual rate, but $\ce{A}$ and $\ce{B}$ cannot combine with each other as fast because they may bump into $\ce{D}$ instead of each other. Thus equilibrium shifts towards the reactants.
I hope this makes sense. Leave a comment if you're not understanding any part of this.

Your answer makes alot of sense ...however I still have trouble understanding equillibrium shifts concerning addition of inert gases

@LM26 Edited the answer

Well imagine that it is already in a gaseous environment. When you add an inert gas, all you're doing is replacing that 'air' environment with  an inert gas.

In that case, your not removing air are you? Then the $\ce{A}$ and $\ce{B}$ would bump into air and inert gas as well.

Note that adding inert gas does not change equilibrium unless the volume is also changed.


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IUPAC naming molecules?
Can someone help me name these 3 molecules (they may be only hypothetical)?

I have only guessed at the order of functional groups

How did you arrive at your suggested names? Have you tried searching for the IUPAC priority rules?   If you search chem.SE for posts written by @Loong there are many worked through examples of this kind of thing that may allow you to answer your own question.

Please see our guidelines on how to ask nomenclature questions.

You have already found out that the carboxylic acid group is the principal characteristic group, which is expressed at the end of a name by means of a suffix or a class name.
You have also found the correct names for the parent structures butanoic acid and but-3-enoic acid, respectively.
Furthermore, you have correctly started the numbering at the carboxylic acid groups since low locants are given first to the principal characteristic group; thus, the numbering starts at the carboxylic acid groups.
The remaining problem is the order of citation of substituent prefixes. Simple prefixes are arranged alphabetically disregarding any multiplicative prefixes. Any multiplicative prefixes are inserted later and do not alter the alphabetical order.
The corresponding actual wording of the nomenclature rules (taken from Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)) reads as follows.

P-14.5 ALPHANUMERICAL ORDER
Alphanumerical order has been commonly called ‘alphabetical order’. As these ordering principles do involve ordering both letters and numbers, in a strict sense, it is best called ‘alphanumerical order’ in order to convey the message that both letters and numbers are involved
Alphanumerical order is used to establish the order of citation of detachable substituent prefixes (not the detachable saturation prefixes, hydro and dehydro), and the numbering of a chain, ring, or ring system when a choice is possible.
(…)
P-14.5.1 Simple prefixes (i.e., those describing atoms and unsubstituted substituents) are arranged alphabetically; multiplicative prefixes, if necessary, are then inserted and do not alter the alphabetical order already established.

Therefore, the names for the compounds that are given in the question are 4-bromo-3-methylbutanoic acid, 4-bromo-4-hydroxy-3-methylbutanoic acid, and (3*Z*)-4-bromo-4-hydroxy-3-methylbut-3-enoic acid.


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Why do electronegative atoms depress coupling constants?
Related to this previous question about detection of small couplings between protons.

In a recent conversation with our labs NMR technician, it was stated that coupling constants decrease in the presence of electronegative atoms:

...and you have a lot of electronegative atoms around, so thats going to make those coupling constants drop off significantly...
Source: 'Personal communications', aka the labs NMR-tech

Thinking about it, this is something I've always known but possibly never thought about to any great deal, despite the fact it shows up in even the most basic organic chemistry textbooks:

...electronegative atoms, which decrease 2J and 3J couplings between protons
Source: Organic Chemistry, Clayden & Warren

Digging a little deeper, this idea comes up a lot in NMR textbooks too, and a few trends are discussed, for instance in Günther's NMR spectroscopy:

For substituted ethanes the relation between the change of electronegativity, caused by the replacement of a hydrogen atom with a group X, and the coupling constant is given by:
3J = 9.41 – 0.80 (EX – EH).
Source: NMR spectroscopy, Günther

What I can't find, in any great detail, is why the coupling constant depends on electronegativity the way it does.
From what I've read, it has something to do with the polarisability of the bond, but again, its maybe not completely intuitive how this effects coupling between two protons separated by 2/3 bonds.

One possibility is that the bonds to electronegative atoms have more p character, so the remaining bonds are shorter and have more s character, than in the case of the unsubstituted compound. But intuitively, two atoms that are farther apart should have weaker coupling, so perhaps there is another factor at play.

I don't know where my Oxford Chemistry Primer NMR book is, but I usually consult it for these types of questions first...

@Zhe. I feel slightly offended :P  In all genuine seriousness, I've looked in a few books (see references in question) and haven't found anything that convinces me.  There is an 'MO' type explanation involving bond polarisation, but its not terribly thorough in explaining why this affects the J value.

:P I was trying to say (apparently failing at it) that the primer is an excellent reference that I've found useful. If I knew where my copy was and had it on hand, I would've looked it up for you (assuming it was there). Unfortunately, it's either packed away with the rest of the chem books in storage, or I donated it to MIT.

@Zhe it isn't in there, unfortunately. I checked inside when I saw this question - surprisingly there is a lack of detail in many textbooks as to why this is observed...

NotEvans, your statement about bond polarization reminds me that spin-spin coupling is actually mediated by electrons. Is it reasonable then to suggest that removing electron density via a electronegative atom would decrease the electron density available to mediate the coupling? I do believe I remember electron mediation detail from the OCP primer. :)

@NotEvans. Try looking at "Relationship Between Proton-Proton NMR
Coupling Constants and Substituent
Electronegativities" MAGNETIC RESONANCE IN CHEMISTRY, VOL. 27, 556-563 (1989) and references cited therein.  I'll try to write answer if no one else does, but I'm not thinking very clearly right now.

@DavePhD. Thanks for the reference. I don't have online access to the back-issues, but will see if I can get the library to dig out the old paper copies from the stacks tomorrow!

@NotEvans. I have access to it, but it cites to older works by Pachler that I don't have like http://www.sciencedirect.com/science/article/pii/S004040200192410X

Let's set some starting parameters so we are hopefully working from the same page. Interestingly, the summary that an increase in electronegative substitution leads to a decrease in coupling constant is not strictly correct per se.
For geminal coupling, electronegative substitution at the alpha position leads to a more positive coupling constant. H-H Geminal couplings are negative, and hence we observe a coupling that becomes smaller in magnitude. (There are a few examples where they actually go the other way, but still become more positive).
 On the other hand, H-F geminal couplings are positive, and so an increase in 2JHF coupling is observed with an increase in electronegative substitution.
Electronegative substitution at the beta position has the opposite effect, leading to a more negative coupling; this actually gives a coupling that is larger in magnitude. Effects from substituents at the beta position are usually smaller than for the alpha position, but the effect is nevertheless opposite.
For vicinal couplings, electronegative substitution leads to a reduction in magnitude of the coupling constant.
The magnitude of scalar spin coupling depends on a few parameters:

the angles between overlapping bonds between coupled nuclei
electronegativity of substituents
bond length

The traditional accounts for the mechanism of spin-spin coupling might go something like...
In the simplest terms, spin coupling comes about because the magnetic moment of nucleus A interacts with the electrons surrounding it, and these electrons produce a local magnetic field at the neighbouring nucleus B. Spin coupling is dominated by the Fermi contact over a short number of bonds. The Fermi contact describes the magnetic interaction of an electron within the nucleus. Only s-orbitals can contribute electron density at the nucleus (all other orbitals have nodes at the nucleus), and so bonding to entities that will promote s-character of molecular will strongly influence the coupling constants.
The more stable orientation of for the nucleus and electron are antiparallel (because they have opposite sign magnetogyric ratios), and for electrons in a bond also to be antiparallel. For the most stable arrangement we get something like (lower case = electron spin, upper case = nuclear spin, u=up, d=down, - are bonds):
Ud-uDu-dUd-uDu-.....
Essentially, the observable nuclear transitions depend on the molecular orbitals, and how well the molecular orbitals stabilise the antiparallel, or spin-polarised configuration.  There are some comprehensive explanations of the molecular orbital model given in Gunther, as you have already encountered, and first published by Pople and Bothner-by. You will also find an excellent account, as always, on Reich's NMR page.
A coupling constant is positive, by definition, if it stabilises the anti-parallel electron spins, as shown above. So, a 1J will be positive, 2J negative (usually) and 3J positive. If the probability of finding an electron in the up state is the same as for the down state, the coupling constant will be zero. For a simple 2 nucleus, 2 electron bond system, if there is an up-electron at one nucleus, then there is a very high chance that there will be a down electron at the other nucleus, and so the coupling will be large. For 2 nuclei across multiple bonds, the spin state of an electron at nucleus A will have a negligible effect on the spin state of electron at nucleus B, and so the coupling will be very small.
...Pause for a breath....
So, to the crux of your question...
If a substitution element can help stabilise the anti-parallel arrangement of the electron spins, by reducing the energy gap for the HOMO-LUMO of the molecular orbitals then the coupling constant will become more positive. Electron-withdrawing groups serve to reduce the energy gap between these orbitals (arguments based on symmetry-allowed transitions - read the above references). σ-acceptors will reduce the LUMO, while π-donors raise the HOMO; both have the same outcome of making the coupling more positive.
This explains the geminal trend, but what about vicinal? There are some theories about alternating polarisation along the bond pathway, so that stabilising the anti-parallel alignment in one molecular orbital will destabilise the adjoining orbitals, hence the trend towards reducing J. It all gets a bit vague about here, and not very convincing. There was some good work by Altona et al in the 80s/90s (see for example) extending the Karplus equation to account for electronegativities, and improved Karplus equations were published, but nothing universal. At this point, we begin to accept that the ability to describe the physical nature of what is going on here is not really going to work using an emperical model.
More recent work by Cremera and Gräfenstein, for example, pretty well put these old 'textbook' explanations back in their boxes, and describe the mechanisms for spin coupling as composites of 4 contributing factors: Fermi contact, spin dipole, diamagnetic spin orbit and paramagnetic spin orbit. It all gets very Hamiltonian early on, but essentially there are co-dependent interactions of all these terms in the electronics. Fermi contact coupling is propagated over multiple bonds and occurs through space via a molecular orbital vector involving the tails of the orbitals, and these tail interactions are strongly influenced by the substituents. An electronegative substituent on a carbon will increase the local Fermi contact, but lead to a decrease in the tail orbital overlap, leading to a reduction of vicinal coupling, for instance.
Some very useful references for you might include:

Phys. Chem. Chem. Phys., 2007, 9, 2791–2816. Calculation and analysis of NMR spin–spin coupling constants
Annu. Rep. NMR Spectrosc. 2000, 41, 55–184. Advances in theoretical and physical aspects of spin-spin coupling constants


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Corrosion rate of a metal in a liquid in relation to the size of electric current
I am trying to determine the formula for calculating the corrosion rate of a metal in a liquid in relation to the size of electric current that is passed between 2 metal plates. For example: we have 2 aluminium plates in parallel in distilled water at 5 mm distance. The plates are 10 mm thick. The plates are connected to 12V 5A power source. Is it possible to calculate the corrosion rate of the plates in relation to the current?

This appears to be what we used to call a homework question. You might amend your query to include what your thoughts are on the answer.

It is certainly possible to calculate the corrosion rate, with the caveat that the current that actually passes through the cell must be measured. Using a 12 V 5 A power source does not mean that 5 A is the current that will go through the cell, only that the current will not exceed 5 A.

It is more of a home project type of thing. I realize that the current we want is the one that passes between the plates, however I was hoping this can be calculated in advance given the stated parameters (update: I realize you would need to know the resistance of the liquid). Ok, lets assume the current passed between them is 3A. I am not looking for actual calculations. I am more after the formula it self or a pointer where I can find one (or deduce it). The example was just to help to illustrate what kind of set-up I am talking about.

It is possible to calculate the corrosion rate, assuming the current is measured and you can assume all of the current is going into oxidizing the aluminum (which is probably not a safe assumption in most cases), simply based on the amount of charge that is passed.
Charge is the number of electrons being passed and current is charge per unit time.
Therefore, if you know the current, you can calculate the rate of reaction.
If we take one form of Faraday's law of electrolysis:
$$n = \frac{It}{zF}$$ where n is the number of moles of product produced, I is current, t is time, z is the number of electrons transferred per reaction and F is the faraday constant (number of coulombs per mole of electrons), we can simply divide both sides by t to get the number of moles per unit time, i.e. the rate:
$$\frac{n}{t} = \frac{I}{zF}$$
Assuming a 3-electron process ($\ce{Al -> Al^{3+} +3e-}$, for example) with 3 A of current, this works out to around 10 mmol/s (you could convert this to mass or volume using Al's molar mass and density, if you like).
The real trouble in doing something like this is that determining the current a priori is not trivial.
In a cell like the one you describe, many factors need to be taken into account.
Here are a few off the top of my head:

The area of the electrodes in the solution; more area generally means more current.
The rate of the reaction depends on the electron transfer rate, as well as things like mass transport, if oxygen or the like needs to get to the surface. N.B. the limiting reaction may actually be at the cathode.
Not all of the current will go into causing the corrosion of the aluminum, some will be lost to producing heat, for instance.
Depending on the voltage, other electrochemical reactions might occur
The current will probably change significantly over time. Starting in a very resistive solution like distilled water, the current will be low, but if ions are being produced by the electrodes, the solution will become more conductive over time. (if it's very pure water, this can be very slow)
I'm not sure if it works in distilled water, but you may also end up with something akin to the process of anodizing aluminum, where oxygen being produced at the anode reacts with the aluminum ions and forming an aluminum oxide layer. Depending on the conditions, this can passivate the electrode and essentially make it non-conductive.

If you are interested in some of the analytical techniques used to examine these properties experimentally, this primer from Gamry is quite nice.
In short, it's not really practical to calculate a corrosion rate in advance, especially with just a normal power supply where you can't compensate for the effects of solution resistance and the like, though it may be possible to tune the conditions of your cell such that the current reaches some consistent state over a period of time. (e.g. anodizing aluminum is often done in acidic conditions to dissolve some of the forming oxide layer, to avoid passivating the electrode)

Thanks. This is exactly what I needed. When I did my research online I had hard time figuring out exactly what to look for but this is gives me a good starting point.  It seems I forgot a lot of stuff from my school days.


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Is 4-chlorotoluene also an IUPAC name for 1-chloro-4-methylbenzene?
I think that toluene is a common name for 1-methylbenzene. And in IUPAC nomenclature, we must provide formal prefixes and suffixes along with correct numbering so the only correct option must be 1-chloro-4 methylbenzene.

I should have searched before answering, but: Possible duplicate of Is 3,5-diethyltoluene an accepted IUPAC name?

The use of 'toluene' as a parent hydrocarbon in nomenclature is described in Rule P-22.1.3 in Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013.

These names [toluene, xylene, and mesitylene] are retained due to a long and well established tradition. Toluene and xylene are preferred IUPAC names, but are not freely substitutable; toluene is substitutable under certain conditions, but only for general nomenclature. (emphasis mine)

"General nomenclature" refers to any name generated by the use of IUPAC principles, rules, and conventions, whereas the "preferred IUPAC name" (PIN) refers to a name that is preferred among two or more names generated from two or more IUPAC recommendations. The introduction P-10 reads:

Any name other than a preferred IUPAC name, as long as it is unambiguous and follows the principles of the IUPAC recommendations herein, is acceptable as a 'general IUPAC name", in the context of a 'general IUPAC nomenclature'.

As stated in P-22.1.3, in the construction of a PIN for the molecule below, the use of 'toluene' as the parent hydrocarbon is not permitted; the PIN is 1-chloro-4-methylbenzene.

Regarding the substitution of toluene in general nomenclature, P-15.1.8.2.2 lists the characteristic groups that can be used in general nomenclature:

[...] halides $\ce{-Br, -Cl, -F, -I}$, pseudohalides $\ce{-N3, -NCO}$ (and chalcogen analogues), $\ce{-NC}$, substituent groups derived from the halogen oxo acids $\ce{-ClO, -ClO2, -ClO3}$ (similarly for groups in which $\ce{Cl}$ is replaced by $\ce{Br}$ or $\ce{I}$), $\ce{-NO2}$ and $\ce{-NO}$, and $\ce{-OR}$ ($\ce{R}$ = alkyl groups), and chalcogen analogues, and $\ce{-SO-R}$ and $\ce{-SO2-R}$, and $\ce{Se}$ and $\ce{Te}$ analogues.

Since chloride falls under this list, the name 4-chlorotoluene is permissible in general nomenclature, but it is not the preferred IUPAC name.

Note that there is no space between the prefix 'chloro' and the parent hydride name 'toluene'.
Also note that there is no such thing as '1-toluene'; in the name '4-chlorotoluene', for example, the methyl group of toluene is automatically assigned the locant '1' (see figure below from the Recommendations). Therefore, the name '4-chloro-1-toluene' is incorrect.

The question says "the  iupac name(s) of the compound is/are" so does it refer to pin or general name

Sorry but I can't interpret ambiguous wording for you.


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How to determine cis or trans for an alkene?
I was doing some practice problems for an exam this week, and I came across a molecule that I swear is trans, but the book says it doesn't exhibit any stereoismerism. I got every other question right (or at least figured out why I got it wrong), but I can't figure this one out.
Here's a picture of the molecule:

As you can see, it has two butyl groups extending to the left and right of the central pi bond, doesn't that make it trans?

In the example you've given, both groups on the right hand side of the alkene are equivalent (n-butyl) meaning its impossible to tell them apart (and hence you can't assign a priority to either one.

The situation would be different if we were considering two of the same groups bonded to different sides of the alkene.

In the example above we can now say that the two butyl groups are either cis or trans to one another, since both sides of the alkene can be distinguished (on each side of the alkene, we can assign a group of higher priority).  Note that E/Z may be more appropriate here.

And note that cyclohexyl has a higher priority than n-butyl.


text:
Coordination Chemistry: Predict colors (simple model, old)
I remember many years ago, when I prepared for a talk on coordination chemistry I found a PDF with a theory about how to predict the colors of simple coordination compounds. I remember the name of the person, who came up with this theory was a Nordic name, like Jorgenson. It sounded pretty similar to that at least but I cannot find it.
Basically, this was a quite empirical route. The metals had different numbers for each oxidation state and the ligands as well and in the end you had to either multiply or add them together (I don't really remember) but you came up with an energy which was sometimes close to the experimental value and sometimes completely wrong.
I know this is not really a good concept to use but I am more interested in the theory itself and would like to add it to a recent summary on coordination chemistry I am writing right now. Perhaps somebody has heard of this theory before.

Heh, I guess I didn't see here any "theory identification" earlier.

Not sure if this is useful. But I was just reading a paper on Hard Soft Acid Base Theory and "Jorgensen" was mentioned for the idea of chemical symbiosis between metal ions and ligands. Hope this is the "Jorgensen" you are looking for. Although nothing was mentioned about the method of predicting colours you are talking about. Refer to https://en.wikipedia.org/wiki/Symbiosis_(chemical) for more details.


text:
What is solubility of silver oxide in NaOH and KOH
What is solubility of silver oxides ($\ce{AgO}$ and $\ce{Ag2O}$) in $\ce{NaOH}$ and $\ce{KOH}$?
How much of silver oxide can dissolve in 1 liter of $\ce{NaOH}$ or $\ce{KOH}$?

Silver(I) oxide is soluble in alkali solution and may contain corresponding dihydroxoargentate(II)  in solution or form silver hydroxide. This can be used to describe the amphoteric character of silver(I) oxide as it also reacts with acid to form silver chloride.
$$\ce{Ag2O + 2NaOH <=>[H2O] 2Na[Ag(OH)]}$$
Source
$$\ce{Ag2O + 2KOH <=>[H2O] 2K[Ag(OH)2]}$$
Source
This paper discuss the solubility of silver(I) oxide in water, alkali and alkaline salt solution and thus form silver hydroxide and thus describe the amphoteric nature.
This link describes how much silver(I) and silver(II) oxide is soluble in alkali solution and very much answers your question.

The solubility of $\ce{Ag2O}$ and $\ce{AgO}$ in alkaline electrolytes
  has been studied with a polarographic technique, using a rotating
  platinum electrode. Only amonovalent (and no divalent) silver species
  could be detected in solutions which had been in intimate contact
  with $\ce{AgO}$ powder over prolonged periods of time.
Quantitative measurements of the solubility
  of $\ce{Ag2O}$ in $\ce{KOH}$ solutions ranging from 1–14 moles per
  liter, were carried out utilizing a potentiometric titration method.
  The solubility has a maximum at about 6N $\ce{KOH}$ where it reaches a
  value of $\ce{4.8 \times 10^{−4} N}$.

Silver(II) oxide is not as stable and readily available as silver(I) oxide. In fact, it is made by oxidizing silver(I) oxide by ozone on alkali solution.

The oxidation by ozone of a suspension of silver or silver oxide in an
  aqueous solution of sodium hydroxide is described. It has been shown
  that the oxidation proceeds in two steps:
$$\ce{AgO3→Ag2OO3→AgO}$$
The experimental results are in good agreement with a mechanism of
  dissolution and precipitation. The silver (II) oxide obtained has
  remarkable properties of stability in alkaline solution and of
  reducibility to metallic silver. These special properties are probably
  due to the large size of the particles.

I really, really doubt you want to know what you are asking. Perhaps you don't understand that both (Na or K)OH are solids at STP and attempting to dissolve silver oxide into them would be extremely difficult.  If they were heated to well above their melting point (318° & 360°C, respectively), then you could mix the powdered oxides in, but then I would expect silver oxide would not dissolve in the molten liquids, but rather react.  However, be warned, this is a speculation on my part. In other words, if I'm correct, at STP there is no "solution" of Ag1,2O in (Na or K)OH. (STP = standard (laboratory) temperature & (air) pressure). If you meant an aqueous solution of the hydroxides, and yet didn't specify their concentrations, then again you need to learn some chemical fundamentals. The exact nature of a mixture of the oxides and the hydroxides at room temperature? IDK. In fact, by AgO, do you mean Ag4O4? which is a mixed (+1, +3) oxide? In the aqueous state, the mixed oxide wouldn't exist, making the question quite problematic on several different levels.


text:
Methods of splitting carbon dioxide into elements
Is it possible to split $\ce{CO2}$ into C and O? If so, what are some methods and their efficacies?
This question originated because I was wondering if it was possible, and if so, how, to convert ice from the ice caps on Mars (made up of frozen $\ce{H2O}$ and $\ce{CO2}$) into oxygen.

I would go for water ice, we have found plenty of it on Mars.

Of course you can split CO2 up. But it takes energy to do it. So, what is your real question?

What kind of scale are you thinking about? In other words, are you thinking in terms of converting Martian CO2 to O2 and carbon on a global scale, or for supplying a future outpost with oxygen? The former is far beyond our technological capabilities, while the latter could be doable. Also note you don't have to go to the poles, as the atmosphere (what there is of it) is mostly CO2. The biggest problem though, no matter the chemical path for getting there, will be the massive energy requirements. This again underscores that whether this is a reasonable possibility or not depends on the scale.

The best way on the planet would be to use an electrochemical way, as there will be power (solar/nuclear) but no many other chemicals.  Failing that, and probably better, would be to genetically engineer plants here on earth to operate in the cold and at low light levels.

I assume you meant convert $\ce{CO2}$ into $\ce{O2}$.
Lithium peroxide ($\ce{Li2O2}$) is a substance that does this. It absorbs $\ce{CO2}$ and converts itself into $\ce{Li2CO3}$ and $\ce{O2}$:
$$\ce{2Li2O2 + 2CO2 -> 2Li2CO3 + O2}$$
It was used as an air purifier in the spacecraft to scrub carbon dioxide. It offers it's bonus of releasing oxygen as well as being non-hygroscopic, so it won't remove the moisture content in the air.
But at any day, you would need a lot of lithium peroxide if you want to make Mars habitable.
Edit: The partial pressure of carbon dioxide in the Martian atmosphere is $\pu{575.8Pa}$. Using the ideal gas law, and temperature of Mars is $218K$, we get amount of substance of $\ce{CO2}$ as $\pu{0.32mol/m^3}$.
Such an amount would require $\pu{14.65g}$ of lithium peroxide per cubic meter of martian atmosphere. According to SigmaAldrich, that costs about ₹21.22 (33 US cents) per cubic meter.

Plants do a pretty good job of using solar energy to convert CO2 to O2. But the intensity of sunlight on Mars is weak and you would need a very well-insulated greenhouse to do this sustainably. I would start with Antarctic algae and see where that takes us.

Nooo! This is very much wrong, and a common misconception. Plants do not convert $\ce{CO2}$ into $\ce{O2}$. In fact the oxygen released comes from water, not carbon dioxide.  Plants use the sunlight energy to split water into hydrogen and oxygen. The hydrogen goes into making glucose along with carbon dioxide. There's very little water on Mars anyway.

@PrittBalagopal That's true but I don't think it really matters for the question. The OP seems to be interested in how to make oxygen and whether it comes from $\ce{CO2}$ or water doesn't really matter.

@iad22agp Exactly, people think that CO2 is broken down in photosynthesis into free oxygen when this is not the case at all.  The carbon-oxygen bond in CO2 is very strong and it takes much energy at high temperatures to break the bond no matter what enzymes are used.  The free oxygen from photosynthesis actually comes from the break down of water.  The hydrogen-oxygen bond is much weaker and the small amount of energy from sun light with the help of some enzymes allows the breakdown of water into free oxygen and the hydrogen is then combined with the CO2 to make sugars,


text:
Is 1,2-dihydronaphthalene aromatic?
According to Hückel's rule a compound is aromatic if it has conjugation throughout the compound and the number of conjugated electrons is $4n+2$. This compound does not satisfy both:

The benzene ring itself is aromatic, though. Is the compound considered aromatic because of this?

Hückel's rules cannot be applied to this compound, it is not a monocycle.

Hückel's rules are a very concise set of very strict rules. Most aromatic compounds do not comply with these.

Hückel (4n + 2) rule
  Monocyclic planar (or almost planar) systems of trigonally (or sometimes digonally) hybridized atoms that contain (4n + 2) π-electrons (where n is a non-negative integer) will exhibit aromatic character. The rule is generally limited to n = 0 – 5. This rule is derived from the Hückel MO calculation on planar monocyclic conjugated hydrocarbons (CH)m where m is an integer equal to or greater than 3 according to which (4n + 2) π-electrons are contained in a closed-shell system. Examples of systems that obey the Hückel rule include:

cyclopropenyl cation (m = 3, n = 0)
cyclopentadienyl anion (m = 5, n = 1)
benzene (m = 6, n = 1)

Systems containing 4n π-electrons (such as cyclobutadiene and the cyclopentadienyl cation) are 'antiaromatic'.
  From IUPAC Gold Book, DOI: 10.1351/goldbook.H02867

Based on this it is actually quite difficult to assess aromaticity to more complex molecules. Aromatic molecules that do not comply with Hückel's rules are for example naphthalene, thiophene, or pyridine, among many more. A more suitable definition, which also makes quite clear how fuzzy the whole subject is:

aromatic

In the traditional sense, 'having a chemistry typified by benzene'.
A cyclically conjugated molecular entity with a stability (due to delocalization) significantly greater than that of a hypothetical localized structure (e.g. Kekulé structure) is said to possess aromatic character. If the structure is of higher energy (less stable) than such a hypothetical classical structure, the molecular entity is 'antiaromatic'. The most widely used method for determining aromaticity is the observation of diatropicity in the 1HNMR spectrum.
  See also: Hückel (4n + 2) rule, Möbius aromaticity
The terms aromatic and antiaromatic have been extended to describe the stabilization or destabilization of transition states of pericyclic reactions The hypothetical reference structure is here less clearly defined, and use of the term is based on application of the Hückel (4n + 2) rule and on consideration of the topology of orbital overlap in the transition state. Reactions of molecules in the ground state involving antiaromatic transition states proceed, if at all, much less easily than those involving aromatic transition states.

From IUPAC Gold Book, DOI: 10.1351/goldbook.A00441

While we tend to say a molecule is aromatic, that is hardly ever the case. Think about substituted benzene molecules; while the phenyl ring retains aromatic character, the side chains are not necessarily aromatic. When describing the molecule it is better to refrain from absolutes.
Within this framework it is possible to identify molecular entities that posses aromatic character and refer to these. Strictly speaking the answer that 1,2-dihydronaphthalene is aromatic is therefore incorrect. Better is to say that because of the phenyl subunit, 1,2-dihydronaphthalene possesses some aromatic character. In this particular case you could even argue that because of the additional double bond, the  main reactivity of this molecule is more similar to alkenes than it is to aromatic compounds.
Aromatic character is sometimes like a unicorn: it's hard to describe in unambiguous terms, but when you see it, you'd probably know what it is.
There is another definition of relevance in the IUPAC Gold Book, which actually gives a clearer, more systematic aproach for determining aromatic properties, and can be seen as an extension of the above:

aromaticity
  The concept of spatial and electronic structure of cyclic molecular systems displaying the effects of cyclic electron delocalization which provide for their enhanced thermodynamic stability (relative to acyclic structural analogues) and tendency to retain the structural type in the course of chemical transformations. A quantitative assessment of the degree of aromaticity is given by the value of the resonance energy. It may also be evaluated by the energies of relevant isodesmic and homodesmotic reactions. Along with energetic criteria of aromaticity, important and complementary are also a structural criterion (the lesser the alternation of bond lengths in the rings, the greater is the aromaticity of the molecule) and a magnetic criterion (existence of the diamagnetic ring current induced in a conjugated cyclic molecule by an external magnetic field and manifested by an exaltation and anisotropy of magnetic susceptibility). Although originally introduced for characterization of peculiar properties of cyclic conjugated hydrocarbons and their ions, the concept of aromaticity has been extended to their homoderivatives (see homoaromaticity), conjugated heterocyclic compounds (heteroaromaticity), saturated cyclic compounds (σ-aromaticity) as well as to three-dimensional organic and organometallic compounds (three-dimensional aromaticity). A common feature of the electronic structure inherent in all aromatic molecules is the close nature of their valence electron shells, i.e., double electron occupation of all bonding MOs with all antibonding and delocalized nonbonding MOs unfilled. The notion of aromaticity is applied also to transition states.
  From IUPAC Gold Book, DOI: 10.1351/goldbook.A00442

One of the key elements is the thermodynamic stability and the tendency of retaining the structural type during the course of a reaction. This certainly reflects the benzene-like aspect of the earlier definition.
The last part comes directly from the original (4n + 2) criterion, but significantly weakens it, as it is not reduced to the very symmetry restricted number Hückel derived from the hydrocarbon cycles.
For simple compounds like thiophene it is easy to construct an approximate MO diagram (if you for further simplification ignore the σ-sub-structure). And the same can be done more complex molecules like 1,2-dihydronaphthalene.
In general I am not a big fan of oversimplifying complicated situations; questions that are asking for absolute statements are always oversimplified. In case of aromaticity it is tempting to reduce it to a simple electron-count, which therefore still prevails in textbooks although horribly incomplete; but it is also dangerous, because it neglects a lot of the underlying chemistry.
Aromaticity is a beast; fuzzy definitions, or criteria, the ongoing debate among scientists only show that you need to be open minded and accept that many things are not as easy as it seems at first glance.
For the time being you can help yourself by identifying molecular entities, or moieties that are derived from (simpler) aromatic compounds and check whether a conjugated π-system was destroyed while creating them. As one of the key features you can be almost certain, that aromaticity is retained in all but the most extreme cases.

By your logic, would benzaldehyde be non-aromatic because it shows reactions of aldehydes? The non-benzenoid ring can be broken into two and they can be treated as two substituents, right?

@Yashas Yes and no; benzaldehyde is still aromatic, at least the phenyl ring. It's also an aldehyde. It's just not that simple, and it would be wrong regarding the one without the other. Certain reactions can target one functionality, while others target the second.

Benzaldehyde certainly is "aromatic" in the sense that it has an aroma. Hence, the term aromatic. While certain amino acids are "essential", essential oils are not. They are fragrant, however.


text:
Is 1,2-dihydronaphthalene "partially aromatic"?
I've learned that if any one ring of a structure like the below is aromatic, the whole compound is aromatic.

But one of my friends suggested that such compounds cannot be aromatic and gave the following resonance structures as the reason:

I feel that the non-aromatic/anti-aromatic resonance structures wouldn't make significant contribution. Is my reasoning correct?

Antiaromatic species are unstable. No molecule would attempt to take up an antiaromatic contributor.

I edited your post to make the picture more readable. I used Web Chemdoodle for this purpose. See my guide on how to format chemical compounds and mechanisms. I used Microsoft Paint to write "Aromatic" and "2e" and stuff.

I think I address this in the answer to the very similar (possibly duplicate) question: Is 1,2-dihydronaphthalene aromatic?

Aromaticity is a quite difficult concept but in this case we can answer the question by simply looking at similar compounds and experimental data.
What you got here is a styrol derivative, the pi-system (which we need to look at if we are talking about aromaticity) is the same. So if this compound is not aromatic then styrol shouldn't be either.
But styrol is in fact aromatic (or better: the benzene ring in styrol is). It reacts like an aromatic compound should, it got the right shift in NMR, the bond lengths are in accordance with aromaticity and so is the UV/VIS spectrum.
So a conjugated double bond clearly doesn't make a benzene (significantly) less aromatic.

According to Huckel Rule the compound is aromatic only if the conjugation is spreaded all over the structure and the compound is flat or nearly so, so that the sideways overlapping of pi orbitals can take place for conjugation and there should be conjugation of (4n+2)pi electrons on the whole compound and as can be seen from your provided structure the second(right hand sided) ring does have conjugation on the whole structure.

So, is the compound aromatic or not?

I don't think so!

The way you describe the rule implies that substituted benzenes would not be aromatic as electrons are not delocalised over the substituents. That's just wrong. This compound contains an aromatic ring.

This answer is plain wrong.

As the question asked the whole compound is not aromatic but can be said because of the presence of the benzene ring. I actually know this question which is from the IIT-JEE examination paper!

@VaibhavDixit if a chemist says that a compound is aromatic it means that there's at least one aromatic structure in it.

OK sir! Thanks for correction. I appreciate that!

@VaibhavDixit You realise this answer doesn't comply with the official key.

For a compound to be aromatic or anti aromatic there should be conjugation throughout the peripheral , this compound lacks conjugation throughout the peripheral so it falls under non-aromatic category.

Huckel baba fails many time for prediction of aromaticity of a compound, and the chart for checking if the compound is aromatic or not goes like this: 1- we check if compound is cyclic and presence of conjugation , 2- planarity of molecule,3- 4n+2 pi electrons .There are many examples in which some of double which can conjugate with the ring doesn't conjugate and resides localized on two carbons just to maintain aromaticity  like Pyrene ,in this exmple also the double bond should be localised on those two cabon atoms and not participating in resonance and then only  it only it can be aromatic

@KushalTripathi I don't know which "chart" you are speaking of but you should only count the pi electrons in the conjugated ring system, not every pi-system in the molecule and also not pi-systems conjugated to the cyclic one. If we do that here we got a 6-membered ring with presence of conjugation and 6 pi-electrons, which fulfills the hückel rule. And for the second part: I don't nderstand what you mean at all, that's wild talks. And also Pyrene is completly aromatic.

@DSVA yes , Pyrene is completely aromatic . I was relating this molecule to the Pyrene because in pyrene also the total no of pi electrons are 4n and not 4n+2 . But the pi bond at the center of pyrene is not considered to be conjugating with other pi bonds of molecule, so we do not count that pi electron and that's why we say it has 4n+2 pi electrons, simply in this example double bond in the non benzenoid ring should not be disturbing the conjugation of the benzene ring and electrons of benzene are only counted. It is an aromatic compound ,my answer is wrong and I was explaining this only

@KushalTripathi the hückel rule doesn't work for fused polycyclic compounds, in fact it does only work correcly for monocyclic systems. The middle bond in Pyrene is in conjugation with the others...

@DVSA It may be wrong whether the middle pi bond conjugate with rest of the double bonds, but to the best of my knowledge that bond is not considered as pi bond and this is why we say that no. of non pi electrons in pyrene is 2. You can confirm this from the fact that reaction of pyrene with Br2/CCl4 gives the addition reaction at the middle pi bond.


text:
What is 'space'?
What is in the space between atoms?
I understand that molecules are constantly being formed from collisions and such, but what I do not understand, is, on a tiny level, within the level of the atom, what is in between? 'Air' is not the answer, what if we were talking about the molecules and therefore the atoms between air! What's there! Is it matter? Is it just 'nothingness?' Is it subatomic particles?
Similarly, what is space? For example, in outer space, there is ideally a vacuum. What is in a vacuum, if not atoms? What is in space, if not atoms?
I have had these questions for as long as I could comprehend 'open space.'
:)

It's basically just vacuum. Nothing more.

In that case, what's in a vacuum?

Literally nothing at all

As  @fyrepenguin said, nothing

Well, there are electromagnetic and gravitational fields and such. Even in deep space there these things as well as particles of matter (mainly H and He), cosmic radiation, dark matter, etc. But I think you are asking what lies between any two things, which is nothingness. Probably a better question for physics.se or astronomy.se, depending on your angle.

I'm voting to close this question as off-topic because it is about physics

nothing is not true. In the vacuum of space, there are always particles coming in and out of existence.

https://chemistry.stackexchange.com/questions/28981/what-is-there-in-an-atom-between-the-nucleus-and-electrons

From a chemists point of view, nothing. There are no particles or molecules, there are only (sometimes) photons carrying the electromagnetic signal between molecules. You could, even say that virtual particles constantly fill (and leave) this space, provided some shielding from the electromagnetic radiation (however I believe this usually happens inside atoms, between orbitals). You can also say there are gravitational fields between particles/inside the atom too, but they aren't actual particles (at least ones that we have not found).
From a physicists point of view, there's everything. What we perceive as "empty space" is just a blank piece of paper waiting to be written on. Over on the Physics SE a question similar to this has been answered. Essentially this "empty" space is full of potential energy, fields waiting to be simulated (typically by energy) to create particles.

Our best understanding is that the Universe is composed of fields. So, I guess you could look at "space" as being a 3-d volume slice of the 4-dimensional space-time manifold which does not contain matter (although matter is "just" another field, it is one of the most localized fields (by localized I mean has a well defined (and measurable) position)). Sub-atomic particles are ...hmmm...embodiments of their respective fields - it doesn't really make sense to logically separate the particle from the field. Virtual particles exist in empty space, or looking at it another way: no space is empty. We also believe space possesses an energy density (a constant energy density), but that's a bit more difficult to pin down. To answer your more philosophical question: what is in a vacuum if not matter, I guess I'd say it's "potential" - not in the sense of potential energy (although there is that) but in the sense of an ability to form virtual particles (as long as they obey the laws of physics (quantum mechanics)). Unfortunately, the reality is that not only can you not really separate space and time, but you can't separate space-time from energy. For instance, a photon experiences no time and the space in the direction of a (relativistic) particles velocity is compressed - the idea that these slices are fixed or constant in dimension is inconsistent with Relativity. So, in a sense, space is subjective: it depends on the observer. In artsy-craftsy language: space is the canvas upon which nature draws.


text:
Minimum heating temperature for dehydration of cobalt chloride
I found online several sources that state 100 °C for a rapid dehydration of cobalt chloride in its various forms. I'm wondering if that's a minimum, or would it be possible to dehydrate it at a much lower temperature, provided ambient humidity is controlled.

The minimum I could find is 110 °C (Chemical forums) although not verified. The same is written in this link although there are two temperature, I am assuming the hexahydrate start to lose water above 110 °C.

mp 86 °C; loses $\ce{6H2O}$ at 110 °C
[...]
Anhydrous cobalt chloride is blue in color. It can be prepared by
dehydration of the hexahydrate either at 150–160 °C in vacuo.

Also, you can find this paper useful. The temperature mentioned here is 200 °C.

One of the "various forms" of cobalt chloride is Drierite, a mixture of cobalt chloride and calcium sulfate which is a common dessicant. It is blue when active and pink when hydrated (Ref 1). "Drierite can be regenerated by spreading the granules in a tray and heating them in an oven at about 425°F (220°C) for 1 to 1.5 hours."
A more informative website (Ref 2) explains; "The temperature at which DRIERITE desiccants are regenerated is crucial in restoring DRIERITE to its original condition. Absorbed moisture is water of hydration and is chemically bound to the calcium sulfate of DRIERITE. Temperatures in the range of 400° - 450° F are required to break these bonds and release absorbed moisture. Lower temperatures, regardless of heating time, will not regenerate DRIERITE unless applied under vacuum (26" Hg, 325° F or 28" Hg, 275° F). Care should be taken not to overheat DRIERITE Desiccants. High temperatures can alter the crystal structure and render the desiccants permanently inactive."
There are three well-defined hydrates of CaSO$_4$: the dihydrate, the hemihydrate (Plaster of Paris) and the anhydrous material, which can be especially useful (i.e., reactive) when not quite totally anhydrous. Wikipedia explains (Ref 3): "On heating to 180 °C (356 °F), the nearly water-free form, called γ-anhydrite (CaSO4·nH2O where n = 0 to 0.05) is produced. γ-Anhydrite reacts slowly with water to return to the dihydrate state, a property exploited in some commercial desiccants. On heating above 250 °C, the completely anhydrous form called β-anhydrite or "natural" anhydrite is formed. Natural anhydrite does not react with water, even over geological timescales, unless very finely ground." Apparently, some porosity remains in the anhydrite until all the water is gone, and then the lattice is closed up tight.
Although the times and temperatures mentioned above apply to CaSO$_4$ rather than to CoCl$_2$, and stray from the actual question, they indicate that a mixture including CoCl$_2$ performs its function (color change vs dessicant capability) usefully when the properties of the actual dessicant are taken into account.
Ref 1. https://secure.drierite.com/catalog3/faq.cfm
Ref 2. https://www.foxscientific.com/PDF/Drierite.pdf
Ref 3. https://en.wikipedia.org/wiki/Calcium_sulfate

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


text:
Is 4-methylchlorobenzene an acceptable IUPAC name for this compound?

The answer given for this is A and B
My doubt is if toluene is an accepted IUPAC name then why not chlorobenzene?(If that is the reason why the answer is given wrong)
Or else is there any other reason why the option C is wrong?

Related, but not duplicate: Is 4-chlorotoluene also an IUPAC name for 1-chloro-4-methylbenzene?

In the generation of a preferred IUPAC name (PIN), the locant '1' may only be omitted under certain circumstances, as stipulated by Rule P-14.3.4.2 in Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013.

(c) in monosubstituted homogeneous monocyclic rings, e.g. cyclohexanethiol (PIN) or bromobenzene (PIN) ...

Since 1-chloro-4-methylbenzene is not a monosubstituted ring, it does not fall under this category. Therefore, in the preferred IUPAC name, the locant '1' must be cited to avoid any ambiguity.

The omission of the locant '1' when there is no ambiguity is widespread, and is permissible within general nomenclature (refer to Loong's answer here). In this case, the omission of the locant '1' from the name "4-methyl-1-chlorobenzene" is likely to be unambiguous. However, note that the name "4-methyl-1-chlorobenzene" is not in accord with IUPAC nomenclature to begin with. Therefore, the abbreviated name "4-methylchlorobenzene" is probably not acceptable, either.

P-14.5.1 Simple prefixes (i.e. those describing atoms and unsubstituted substituents) are arranged alphabetically...

Since 'chloro' comes before 'methyl' in alphabetical order, the order of citation of substituents must be chloro followed by methyl; hence the correct name must be of the form x-chloro-y-methylbenzene.

What about TNT then? Should it be 1-methyl-2,4,6-trinitrobenzene?

@Mockingbird No, it should be 2-methyl-1,3,5-trinitrobenzene because the locant set (1,2,3,5) is considered to be lower than (1,2,4,6), as outlined in rules P-14.3.5 and P-45.2.2 (see here for the wording).


text:
Naming a molecule with both alcohol and aldehyde functional groups
How would you name this molecule as it has an alcohol and aldehyde group. The aldehyde surely has the higher priority but if I start numbering there, I'm not sure how to write alcohol.

Best try I have is 3-methylpent-2-ene-1-hydroxy-2-al.

I'm guessing the parent chain starts from the aldehyde right? I think this is too advanced for my level to be honest

Excellent. You found out the parent chain. What are the substituents now?

I'm not sure what you call the CH2OH but others are methyl and a c=c bond

$\ce{-CH2OH}$ is a complex substituent. You would call it "hydroxymethyl".

Related: How to name molecules with a hydroxyl group in a side chain?

You have correctly identified the principal characteristic group as well as the principal chain. Thus, the parent structure of the compound is pent-2-enal. You also correctly included the ‘3-methyl’ substituent.
The missing $\ce{-OH}$ group cannot be expressed as suffix ‘ol’ since it is neither the principal characteristic group nor situated on the principal chain. Therefore, it has to be indicated by the prefix ‘hydroxy’. The corresponding rule in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows.

P-63.1.4 Substitutive nomenclature, prefix mode
Hydroxy groups are indicated by the prefix ‘hydroxy’ when:
(1) a group having priority for citation as the principal characteristic group is present; or
(2) a hydroxy group cannot be denoted by a suffix.

In this case, the hydroxy group is placed on a methyl side chain, which results in the compound substituent group ‘hydroxymethyl’.

P-29.4 COMPOUND SUBSTITUENT GROUPS
P-29.4.1 Compound substituted substituent groups
A compound substituted substituent group is formed by substituting one or more simple substituents into another simple substituent that is considered as the principal chain. The choice of the principal chain is fully discussed in Section P-44.3. The first criterion to be applied is that the principal chain is the longest chain; it will be applied in the following examples of acyclic compound substituent groups.
(…)

Parentheses are used around such compound prefixes in order to avoid ambiguity.

P-16.5.1 Parentheses (also called curves or round brackets)
P-16.5.1.1 Parentheses are used around compound (see P-29.1.2) and complex (see P-29.1.3) prefixes; after the multiplicative prefixes ‘bis’, ‘tris’, etc.; to enclose a multiplied parent compound consisting of a parent hydride and a substituent suffix or a multiplied substituent prefix even though preceded by a basic numerical prefix, such as ‘di’, ‘tri’, etc.

Therefore, the complete systematic name for the compound that is given in the question is (2Z)-2-(hydroxymethyl)-3-methylpent-2-enal. (Also note the stereodiscriptor for the double bond.)


text:
Why are neutralisations involving weak acids and bases less exothermic than those involving only strong acids and bases.?
I read somewhere that for example the neutralisation reaction between sodium hydroxide and acetic acid is less exothermic than those of sodium hydroxide with hydrochloric and nitric acid because some energy is needed to cause the weak acid (acetic acid) to completely ionise. Is this true?

Think carefully. What is the main reason why a neutralization reaction is exothermic?

Correct me if I'm wrong please, but I'd say that most neutralisations simply involve the reaction between a hydronium ion and a hydroxide ion to form two water molecules which is an exothermic process. But, I still don't see what the difference is between the neutralisation of HCl with NaOH than that of CH3COOH with NaOH.

You're on the right track. Acetic acid is a weak acid. It furnishes lesser $\ce{H3O+}$ in solution compared to a typical strong acid. So, less hydronium ions, less combination of hydronium and hydroxide, and less energy released.

I still don't understand why the fact that a weak acid does not ionise completely is an explanation as to why it is neutralised less exothermically. If we have equimolar solutions of HCl and CH3COOH both of which are monoprotic, won't we still need an equal number of moles of NaOH to neutralise both?

I think the answer is closer to what was stated in the original question: "some energy is needed to cause the weak acid (acetic acid) to completely ionise". Stated differently, less energy is released from making acetate ion from acetic acid than from making chloride ion from hydrochloric acid (or water from hydronium ion).
Despite the lesser tendency of acetic acid to ionize, the overall stoichiometry of the two reactions is the same--as pointed out in a comment by the OP.


text:
Finding ions in water samples
So for a school chemistry assignment I was given 5 water samples and a bunch of different reagents to help me find out what was in the solutions. I'm  stuck on two.
The first one had a pH of 7 and only reacted with silver nitrate and formed a white precipitate. the second had a pH of 14 and turned brown when silver nitrate was added to it. Does anyone know of any chemicals that would do this?

The white precipitate is pretty obvious what it could be, for the brown one, see my answer below. Also welcome to SE.

@Pete's answers correctly determined what is the salt in the water sample which react with silver nitrate to form the white and brown  precipitate respectively. My answer just describe the reactions.
The first one is very much likely to be a chloride salt as it would react with silver nitrate to form silver chloride. Silver chloride has a prominent white color and it is not at all soluble in water.
$$\ce{AgNO3 + Cl- → AgCl + NO3-}$$
It could have been a sulfate salt as silver sulfate is white but is somewhat soluble in water.
The second one is a hydroxide salt.
$$\ce{2AgNO3 + 2OH- → Ag2O + 2NO3^{-} + H2O}$$
Silver oxide is brown in color. It is somewhat soluble in water and actually form dihydroxoargentate ion instead of silver hydroxide.

It is slightly soluble in water due to the formation of the ion
  $\ce{[Ag(OH)2]^{−}}$ and possibly related hydrolysis products.

Note that silver hydroxide is not an actual compound. It is just like a solution of silver oxide in water just like ammonium hydroxide(solution of ammonia in water).

It's Silver(I) Hydroxide
The solution was very alkaline, as evident by it's pH value of 14. Although  $\ce{AgOH}$ is sparingly soluble, with a $K_{\text{sp}}$ of $\pu{6.8×10^{-9}}$, under alkaline conditions it precipitates as a brown solid.
This is a picture of precipitating silver hydroxide. Is it what you saw as well?

The first one contains chloride, likely. AgCl is a white solid.
The second one contains some sort of hydroxide. Likely sodium hydroxide. If you expect there is only 1 salt in each sample, then you can stop there.


text:
Separating a powdered mixture of Zinc Oxide and Silver Oxide
I have some powders that contain silver, $\ce{Ag2O, ZnO}$ and maybe some $\ce{AgO}$
Can I wash powders with $\ce{NaOH}$ to remove $\ce{ZnO}$ (as Sodium zincate) without any silver or silver oxides loss?

I think you can do that.

Your second suggest that you also want to save the silver oxides apart from zinc oxide. Sodium hydroxide won't help because it also reacts with silver oxides. So, you need a weak acid to selectively react with zinc oxide but neither silver oxide or silver. Hydrobromic acid will work because it reacts with zinc oxide to form zinc bromide which is highly soluble in water. leaving silver oxide and silver untouched. Silver bromide is also expected to form but hydrobromic acid is not strong enough to do that. Instead, if an alkali bromide salt was used, silver bromide would definitely form. From wikipedia article of zinc bromide:

$\ce{ZnBr2 · 2H2O}$ is prepared by treating zinc oxide or zinc metal
  with hydrobromic acid.
$$\ce{ZnO + 2 HBr + H2O → ZnBr2 · 2H2O}$$
The dihydrate can be dehydrated by passing it over hot $\ce{CO2}$

Thanks but I don't want to  loss silver oxides too! You say washing with NaOH can dissolve silver oxides ?

@user2779537 Your original question was to dissolve zinc and silver oxides leaving silver untouched for which NaOH method is a plausible way. Now, you are saying that you also want to save the silver oxide too for which my answer become obsolete. You should have mentioned before. Now I have to delete my answer and start again from scratch.

Yes I am sorry, Also I said want to dissolve just zinc. I want to remove zinc without loss of silver and silver compounds

zinc phosphate is insoluble in water so does it precipitate in solution ? If yes, it mixes with silver again!

@user2779537 see edited answer.


text:
Saponification value of a fat/oil

The saponification value of a fat was determined by adding $\pu{9.0 g}$ of the fat to flask containing $\pu{50.0 cm3}$ of $\pu{1.00 M}$ ethanolic potassium hydroxide. The remaining solution was titrated with $\pu{0.50 M}$ hydrochloric acid.
  Calculate the saponification value of the fat, if $\pu{15.0 cm3}$ of the $\pu{0.50 M}$ hydrochloric acid were required.

My teacher has left notes to take $\pu{15 cm3}$ of $\pu{0.5 M}$ $\ce{HCl}$ as $\pu{7.5 cm3}$ of $\pu{1 M}$ $\ce{HCl}$, which I understand. What I don't understand is why the following line is followed by $\pu{42.5 cm3}$ of $\pu{1 M}$ $\ce{HCl}$ have reacted. How do you go from $7.5$ to $42.5$?
I've calculated that the saponificaiton value is $\pu{264 mg}$ using $\pu{0.0425 mol}$.

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


text:
Will water allow salts to be dissolved in methanol (that are otherwise poorly soluble)?
If we make a mixture of water, methanol and a salt like $\ce{FeCl2}$, $\ce{NaCl}$ etc, will the ability of water and methanol to mix allow $\ce{FeCl2}$ and $\ce{NaCl}$ to be a part of this mixture and form a solution as opposed to when $\ce{NaCl}$ and $\ce{FeCl2}$ are just added to methanol in which they are poorly soluble?
Or will we see layers forming (which I don't see happening because of the strong tendency for water and methanol to mix)? Or will we just see a methanol-water layer with some salt dissolved in it and some salt left at the bottom (even though they are soluble in water)?
Thanks in advance

This is a highly complicated question, and the answer can't be derived by logic alone.
Maybe the liquid phase will remain mixed, and just dissolve some salt (more than would dissolve in methanol, but less than would in pure water). Or maybe it will separate into water-rich and alcohol-rich layers, with nearly all dissolved salt staying in the former. This may be considered an extreme case of salting out.
I know for sure that separation does occur in water/ethanol/$\ce{CaCl2}$ system, even though ethanol by itself is miscible with water in any ratio, as is methanol. Then again, $\ce{CaCl2}$ is much better soluble in water than $\ce{NaCl}$, so the latter might not be strong enough to produce the same effect.
The bottom line is that you can't tell for sure until you try it. Chemistry is an experimental science, after all.


text:
Two different containers full up with pure water and salt water in osmosis. What happens?
I would like to understand what happens when there are two containers which are connected together and separated by only semipermeable membrane. Then in one container I fill up (all the way to the top) pure water and in the second container I put water with a lot of salt ($\ce{NaCl}$). Now In container 2 it said to be an high osmotic pressure.
Then my question is: Is the water in container 1 will drain out / will be pushed out by the osmotic pressure in container 2? (see sections a,b,c below).
■  My question popped up after reading the following things in a book called "RBS - physiology" (6t edition p. 5):

Osmosis is the flow of water across a semipermeable membrane from a
  solution with low solute concentration to a solution with high solute
  concentration.

Example of osmosis:

a. Solutions 1 and 2 are separated by a semipermeable membrane.
  Solution 1 contains a solute that is too large to cross the membrane.
  Solution 2 is pure water. The presence of the solute in solution 1
  produces an osmotic pressure.
b. The osmotic pressure difference across the membrane causes water to
  flow from solution 2 (which has no solute and the lower osmotic
  pressure) to solution 1 (which has the solute and the higher osmotic
  pressure).
c. With time, the volume of solution 1 increases and the volume of solution 2 decreases.

@Crafter0800 please don't use MathJax in titles, see https://chemistry.meta.stackexchange.com/questions/261/is-mathjax-in-titles-a-problem-and-why

In your described scenario, there will be osmotic flow in an attempt to equalize the chemical potential of the solvent (water in this case, assuming the membrane is only permeable to water). The effect of the osmotic flow will be that pure water from container 1 will flow into container 2. Since this is a closed system, one will observe the liquid level rise in container 2 and fall in container 1, and the amount of displacement will be a function of how different the chemical potential of the water is between container 1 and container 2. There is no classical force that will push or pull the liquid from one container to another, osmotic flow is an instance of diffusion which is governed by the concentration gradient that is present.

Thank you 1+ What do you mean by "closed system"? My case, is for example two cups (glasses) connected and between them a membrane permeable to water. Now the two cups are open from above. And if I understood you well, you are talking about hermetically closed container (which is an interesting by itself) while I'm talking about open system. Isn't it?

Closed system meaning that there are no energy or mass flows into and out of the system you have described, nothing special like a hermetically sealed container.

Then basically to answer to my question is that the water will not flow out of the container because it needs energy. Isn't it? If it is, I don't really understand why if the containers were filled with half of their space, some of the water had flown from side to side. Does it not need the same energy as it would needed when they were full up?

Water will flow from container 1 to container 2, the driving force (i.e. energy) for this flow is the concentration gradient of the salt between the two containers.

So it cannot be the driving force to push out the water in the case of the full up containers?

Even with the containers completely full, water will still flow to equalize the chemical potentials, maybe enough water will flow and container 1 will overflow, that all depends on the actual concentrations and calculated osmotic pressures. The only reason water would not flow in the manner described would be if there was additional pressure applied to the higher concentrated side to cancel out the energy from the concentration gradient, which is the basis for reverse osmosis.

Let us continue this discussion in chat.

Indeed when one speaks of a flow from one compartment to the other, this is not true: water exchanges in the two directions, but the equilibrium is when there is a net flow, as water molecules are more attracted by salt than from nothing.


text:
Is that correct to say that osmosis happens with water only?
Is that correct to say that osmosis happens with water only (based on the definition below) or it can happen with other liquids as well?
I've read the following definition in a book called "BRS - physiology" (6th edition, p.5):

"Osmosis is the flow of water across a semipermeable membrane from a
  solution with low solute concentration to a solution with high solute
  concentration."

Maybe helpful: arXiv:physics/0305011 [physics.gen-ph]

"BRS - physiology" -> https://www.amazon.com/Physiology-Board-Review-Linda-Costanzo/dp/0781798760 ? Attribution to used sources is important. We prefer having human-readable citations and (if possible) stable links. If you would like to learn more, see meta.chemistry.se.

Almost:) "RBS - physiology" (6th edition) p. 5

No, it is not correct because it happens with other liquids (solvents) also.
Taking Wikipedia's definition:

Osmosis is the spontaneous net movement of solvent molecules through a
  semi-permeable membrane into a region of higher solute concentration,
  in the direction that tends to equalize the solute concentrations on
  the two sides.

Only as extra information: osmosis occurs until chemical potentials of every substance on each side of the membrane are equal. As the substance cannot move through the membrane, the liquid (solvent) do.
According to the definition, the book isn't correct, because it says with water only, but take into account that in biology the solvent is always water. So, as an internal definition of biology (or physiology) maybe is not wrong.

I'm pretty sure by definition, osmosis refers to the diffusion of water across a semipermeable membrane along a concentration gradient (ie from where water is in high concentration to where it is in relatively lower concentration). Diffusion refers to the movement of any particular substance along a concentration gradient, not necessarily through a semipermeable membrane (but can be). So yes, it is correct to say that osmosis refers to a process that is specific to water.

IUPAC doesn't has a definition for osmosis itself but for whatever reason "reverse osmosis" is defined in the gold book as "Liquid-phase pressure-driven separation process in which applied transmembrane pressure causes selective movement of solvent against its osmotic pressure difference.". If reverse osmosis is defined with just solvent and not water, why should osmosis only include water?

Biologists, physiologists, biochemists, etc. define it with water ONLY, but other disciplines (chemistry, physics, engineering, etc.) will include other solvents as well.  So, it really depends on who your audience is.

@Dr.J. Nonsense. Why on earth should they use a different definition? Just because their samples usually are water-based?


text:
Can formaldehyde be stored at a higher temperature than room temperature?
I have studied that formaldehyde stored at temperature lower than room temperature convert into 1,3,5-oxane. So, it's not stored at colder temperatures.
But my question is - Can a similar degrading reaction happen if methanal is stored at temperature higher than room temperature?

http://pubs.acs.org/doi/abs/10.1021/j100260a032

Trimerization of formaldehyde can happen at high temperatures, but appear to require catalysts. There is a patent for synthesizing 1,3,5-trioxide at high temperature. This patent uses phosphoric acid catalysts.
According to the patent:

The process of the invention is advantageously carried out at reaction temperatures of from 80 to 160° C., preferably from 90 to 145° C., at an inlet partial pressure of the formaldehyde of from 0.5 to 5 bar absolute, preferably from 0.5 to 2 bar, and at an inlet concentration of formaldehyde of from 1 to 100% by volume, preferably from 20 to 100% by volume

There is also an early study which explored the kinetics of the reaction but for a different catalyst. In "Vapour-phase trimerization of formaldehyde to trioxane catalysed by 1-vanado-11-molybdophosphoric acid" by Kern and Emig, they use a molybdophosphoric acid catalyst. They used a tube reactor and varied the temperature from
102.5 to 110° C. According to their study:

The
  conversion
  observed
  increases
  extremely
  with
  decreasing
  temperature.

However, they comment that

the
  temperature
  dependence
  cannot
  be
  attributed
  to
  thermodynamic
  effects
  only.
  It
  has
  to
  be
  related
  to
  the
  properties
  of
  the
  heteropoly
  compound.

Another study is "Synthesis of Trioxane Using Heteropolyacids as
Catalysts" by Mazamoto et al. Their process uses heteropolyacids as catalysts and they were able to obtain a conversion % of 22.9 at 110° C. Where
$$\text{Conversion %}=\frac{3[\text{trioxane}]+2[\text{methyl formate}]+[\text{methylal}]}{[\text{feed formaldehyde}]}\cdot 100$$


text:
Acidities of (dimethylamino)acetophenone and (diphenylamino)acetophenone
How would you compare the acidities of these two compounds?

The deprotonation is obviously alpha to the carbonyl/amine. I think that the diphenyl substituent will be more electron withdrawing than the di methyl, just because the lone pair of nitrogen will have conjugation with the benzene rings, vs methyl groups electron donating onto N.

Yeah, you're thinking in the right way! In the first structure, the hydrogen atom will not have a tendency to leave since the negative charge on the  corresponding conjugate base is destabilized due to lone pair of nitrogen atom(the N atom intensifies the negative charge on the  carbanion formed by donating its electrons in the p orbital of carbanion). Whereas in the second structure the lone pair of N atom is already dispersed in the two phenyl rings so the lone pairs have very less tendency of going into the p orbital of carbanion. Therefore, the second structure is more acidic since its conjugate base is more stable( or less destabilized) than the first structure


text:
Determine from pKa if acid or base
I'm studying medicine, not chemistry, but I hope you can help me anyway.
I just had to solve a question where it is given that the drug Propranolol has a $\text{p}K_{\text{a}}$ of 9.5 and the un-ionized form is non-polar.
Is there any way to say if propranolol is an acid or base just from the $\text{p}K_{\text{a}}$?
Hope to hear from you, and sorry if it is a stupid question.

It's not a pKa of propranolol, but its pKbH+ - pKa of its conjugated acid.

Probably, but it only says "pka = 9,5". So if i don't have the information about propranolol being a base, how can i know?

pKa of coniugated are often not properly distinguished, though luck kid...

Hi @Nanna, to answer your specific question, if all they tell you is 'Propranolol has a ${pKa}$ of 9.5 and the un-ionized form is non-polar', and you don't know anything else about it, then no, you can't tell if it's an acid or a base. Un-ionized molecules are generally non-polar, or at least less polar than ionized ones: for an acid the ionized form is an anion; for an amine, the ionized form is the protonated amine. Doesn't help, does it? I suspect what they actually meant to tell you was that in the acid-base reaction the loss of a proton produced a less polar species. Then it's a base.

You're welcome. And don't be afraid of asking questions, they are never stupid if you're someone who wants to learn. In fact, it's the exam/test question that is badly formulated, as it does not give you sufficient information to answer, or actually even misleads you.

Based on the structure, this drug is weakly basic. It has an amine moiety, which if protonated, would likely have a $\mathrm{p}K_{\mathrm{a}}$ value of around 9.5.

I know you can say a lot about a drug based on the structure - but my exam is without any forms of 'aid' - I can't just look it up on the internet, so I have to use the information I am given. :-)

At a pKa of 9.5 for a drug, you can probably assume an anime as a first guess.

@Zhe That must be one hell of a drug to "assume" anime from:D

No, you cannot say because the terminology is, unfortunately,  used interchangeably.  For example, phenol has a pKa around 10. That means it is weakly acidic. However, amine pKa's are frequently referred to as being around 9-11,  when in fact those numbers are the pKa of the conjugate acid. The pKa of secondary amines for example, is in the 20's as a very strong base is required to deprotonate them.
So in your example, I would wager the pKa is of the salt form, not the neutral molecule.

We are taught (assuming the $\ce{pK}$ values are about aqueous solutions)
\begin{align}
\mathrm{p}K_\mathrm{b} + \mathrm{p}K_\mathrm{a} &= 14&
\text{so} &&
\mathrm{p}K_\mathrm{b} &\approx 14 - \mathrm{p}K_\mathrm{a}.
\end{align}
In this case, $\mathrm{p}K_\mathrm{b} \approx 3.5$, which is not a strong base ($\mathrm{p}K_\mathrm{b} < \approx 2-3$) but it's close. In medical literature a "drug" (i.e. a medicine) is a family of chemical compounds (or mixtures of compounds) where the exact chemical structure is often ignored. Propranolol may refer to the nonpolar compound, or one of several salts (such as Propranolol.HCl). There's no way to know without knowledge of how the drug is actually provided.
Anyone studying medicine should also study $\mathrm{p}K$ values, you may have done that in high school. As far as identifying the organic chemical groups which may ionize, the good news is that there are only a few, so you can learn them. At least, only a few at physiological pH.
Knowing that a group can ionize isn't as good as knowing it does ionize, but it's probably the best you can do from basic chemical knowledge. Experience will help some, but generally you'll have to rely on the literature since the exact electronic (and perhaps steric) effects that contribute are complex and difficult for the non-expert to analyze.

I have removed the unscientific banter at the beginning, which came off a bit rude. I invite you to review or be nice policy and act accordingly.


text:
Why is 2-pyridone so much more acidic than a regular amide?
I am comparing the structures of 2-pyridone and a regular (cyclic) amide-what is the main factor contributing to the increased acidity of 2-pyridone compared to a normal amide. For pyridone, I thought the proton was orthogonal to the pi-system, so no delocalisation of the negative charge could take place, so I am confused how it is more acidic?

The main factor is aromaticity.
Let's look at the main mesomeric forms of DMAc and 2-pyridone.

For DMAc the true structure is somewhere in the middle. We know the C-N bond cannot rotate freely but the barrier is quite low. If we look at structure 2 we can imagine that this structure contributes a lot to the acidity of the amide, since we got a protonated imine like structure, while on the left it's more like an amine.
For 2-pyridone the right structure (4) is aromatic, so it is quite stable. In this case this mesomeric structure will contribute to the real structure much more, which means it is closer to the right structure than in case of DMAc. But in structure 4 we basically got a protonated pyridine. Which is very acidic. It also tautomerizes to 2-Hydroxypyridine, in that case we have a phenol like structure, which is quite acidic too.


text:
Why isn't Indigo soluble in water?
Why isn't Indigo soluble in water? I mean, it has 2 H-N groups, shouldn't it then be able to hydrogen bond with water?

True, Indigo can form hydrogen bonds, but with what? Is it water? Nope.
It makes hydrogen bonds with itself.

This makes the molecule incapable of bonding with water. Not only that, but the molecule itself is quite symmetric, with oppositely oriented polar bonds, that cancel out each others' dipole moments. Notice the molecule's similarity to an ace of diamonds:

This type of symmetry is called center of symmetry as it is about it's center we can observe the symmetry.
So we got it, it can't associate itself with water, and it's nonpolar. What do we expect? It doesn't dissolve in water.

Yes, and the lone pair of Nitrogen is interacting with $\pi$ system of the ring, so it diminished more the nucleophilicity(i am  not sure if it is a correct word). Nice reasoning about the dipolar moment. It can be done "rigurously" by symmetry theory also.

Thank you  for the elaborate answer. I was aware of it's symmetry, it's just the hydrogen bonding that puzzled me, thank you.


text:
Dissociation of zinc hydride
How can we determine if zinc hydride is a ionic compound? If it is a ionic compound then why doesn't it dissociate when put in water?

Zinc(II) hydride is not at all an ionic compound rather the elements are arranged in a one-dimensional network bonded covalently.  From wikipedia article of zinc(II) hydride:

New evidence suggests that in zinc(II) hydride, elements form a
  one-dimensional network (polymer), being connected by covalent bonds.(...)Solid zinc(II) hydride is the irreversible
  autopolymerisation product of the molecular form, and the molecular
  form cannot be isolated in concentration.
Molecular zinc hydride in the gas phase was found to be linear with a
  Zn-H bond length of 153.5 pm. Quantum chemical calculations predict
  the molecular form to exist in a doubly hydrogen-bridged, dimeric
  groundstate, with little or no formational energy barrier.The dimer
  can be described as di-μ-hydrido-bis(hydridozinc), according to IUPAC
  additive nomenclature

As it is not an ionic compound, it does not dissociate in aqueous solution, rather it gets hydrolyzed slowly. If you want to decompose into component element, just heat it at 90℃.

Zinc(II) hydride slowly decomposes to metallic zinc and hydrogen at
  room temperature, with decomposition becoming rapid if it is heated
  above 90°C.
$$\ce{ZnH2 → H2 + Zn(0)}$$
It is readily oxidised and is sensitive to both air and moisture;
  being hydrolysed slowly by water .


text:
How does a salt bridge in a galvanic cell neutralize each half cell?
I am not exactly sure what purpose the salt bridge has.

In the above image, I understand that electrons flow from $\ce{Zn}$ through the wire to $\ce{Cu}$ due to the redox reaction. However, I do not understand how the Salt bridge neutralizes the charges in each half cell.
image from https://en.wikipedia.org/wiki/Galvanic_cell

Hello and Welcome to Chemistry SE. This question, I'm afraid is too broad and would require an entire book or chapter of a book to explain it. This question is a bad fit for SE. Consider referring a book, and come back if you have any conceptual doubts in it. Happy learning.

I just need a very simple answer for a high school Chemistry class. Nothing too specific.

Then, the first paragraph of the English wikipedia about the topic (https://en.wikipedia.org/wiki/Galvanic_cell) -- where the original figure appears, too -- condenses the definition to the bare minimum.  Equally, some keywords mentioned are introduced there with a hyperlink; and it lists additional references / textbooks, too. For future reference (and not only for the purpose of addressing questions and providing answers on ChemSE), it is good practice to quote sources accessed, which includes to cite the origin of data and figures used.

The galvanic cell use the theromdynamic preference of the mixture of $\ce{(Zn^{2+}|Cu)}$ over $\ce{(Zn|Cu^{2+})}$ (which results in the transfer of electrons from zinc to copper(II) ions) and forces this process to occur through a wire.
The salt bridge completes the circuit without allowing the zinc and copper(II) ions to mix directly. If they did, they would react by touching and no electrons would pass through the wire.

If you don't want the Zn and Cu to touch, why can't you completely leave out the salt bridge?

Have you tried using a battery where you didn't complete the circuit?

I know that you have to complete a circuit, but I'm not completely sure as to of what purpose the salt serves. Why doesn't it work if you just have Zn and Cu(II) connected to each other?

The solutions will immediately build up excess charge due to the flow of electrons. You need some ions to shift also to balance out the charge distribution. It can't be the reacting ions, so you use a salt bridge to mediate the ion flow.


text:
If A is a stronger acid than B then is B necessarily a stronger base than A?
It seems like this must be true. So if it is could someone please prove this?

I'd say it's not true in general. Unless in the 'stronger base' part of your sentence you actually mean the conjugate anions of each acid. Then it's true by definition: if the forward reaction (acid going to proton+anion) is more favourable, the backward reaction (conjugated base catching a proton and forming the acid) must be less favourable.

No I don't mean the conjugates. As an example, water is a stronger acid than ammonia but ammonia is a stronger base than water.

Then no, I confirm what I stated earlier, it's not a general rule.

BTW,  interesting to see how some people feel completely unashamed securing a nice +5 by just copying someone else's pre-existing comment and posting it as an original answer.

@user6376297 I certainly didn't mean to copy your comment as an answer and I'm sorry it seems that way. It's hard to see how an answer could be anything else, however, since the answer to the question is basically "yes" or "no". You'll also notice that my answer was meant to copy the form of OP's comment. I'm not even giving an explanation. Only a counterexample.

In general this is not true.
For instance, water is a stronger acid than ethane and is also a stronger base than ethane.


text:
Reversible Sulfonation of Aniline?
If I have aniline in acid and fuming $\ce{SO3}$, will it reversibly sufonate? I am worried that the nitrogen will be protonated and reduce the nucleophilicity of benzene ring to the point where sulfonation just isn't possible and instead you just get a salt? Or is it possible that you only get mono-sulfonation?
Here is my reaction scheme:

See this question. Quoting: "Sulfonation [of aniline] is a reversible reaction. (However) The p-isomer, being more stable, does not get desulfonated easily. Hence, p-sulfanilic acid is the major product."

It sulfonates on N first and then rearranges on heating to give sulfonation in the 4 position. JACS 68 (1946) 969


text:
Is it safe to lyophilize/Freeze-dry a chelated compound?
I have a compound that consists of a metal ion chelated by a linear organic structure with assorted moieties in aqueous medium. I'd like to dehydrate this mixture fully to get solid product, but heating is out of question due to unwanted repercussions on aforementioned "assorted moieties". If I lyophilize the compound, will it remain chelated or is there a risk of the metal ion unbinding?

I lyophilized my sample and redissolved a portion of it and confirmed chelation by verifying the presence of the absorption peak of the Lanthanide I was chelating and using Beer's Law.
Thanks for your help.

I don't think this can be answered with the amount of information provided.

Lyophilization is regarded safer as normal drying because it is performed at considerably lower temperature; however in order to achieve this, you have to lower the pressure above the condensed phase, too.  Naturally volatile compounds (including some organic compounds solid at room temperature) may sublime under these conditions, too; which may be hold for hours, even days to remove the water gradually.  This simultaneous process may lower the yield of isolated product.
Other light organic molecules (like methanol, ethanol, THF) equally may leave the vessel, too.  The complexation may be affected if leaving molecules were involved as part of the no-further specified chelate, too.

This can easily be done by using rotary evaporator, water can be removed if you have strong rotary pump from the metal complex. The chemistry of metal complex would not be disturbed and metal will remain intact with your chelator.
Change in the pH or changing the condition may would decrease the stability of your metal complex.
If you don't have access to rotary evaporator having strong pump, Simply freeze dry your sample using dry ice and then use freeze dryer to remove water from sample and let it dry overnight.
Lyophilization is much safer technique and mostly used to remove water from heat sensitive organic compounds.
Sincerely ,