text:
Solubility of salts
A saturated solution is prepared at 70 °C containing $\pu{32.0 g}~ \ce{CuSO4.5H2O}$ per 100 g solution. A 335 g sample of this solution is then cool to 0 °C so that  $\ce{CuSO4.5H2O}$ crystallises out. If the concentration of solution of a saturated solution at 0 °C is $\pu{12.5 g}~\ce{CuSO4.5H2O}$ per 100 g solution how much  $\ce{CuSO4.5H2O}$ is crystallised?
I found that in 335 g saturated solution at 70 °C there is $\pu{107.2 g}~ \ce{CuSO4.5H2O}$ and then found that there is $\pu{41.875 g}~\ce{CuSO4.5H2O}$ in 335 g saturated solution at 0 °C . Then I found that $$\pu{107.2 g} - \pu{41.875 g} = \pu{65.325 g}~ \ce{CuSO4.5H2O}$$ is crystallised, but the correct answer should be $\pu{74.47 g}$.

(1) Bless you for writing the whole problem. (2) In the future please add your calculations below the question and format them so that the calcs are easy to follow.

Hello and welcome to Chemistry.SE! As noted above, you did some very good things with your question. At this point, I suggest you take the short [tour] to better familiarize yourself with this site, and in particular take a look at this discussion. Also, it would be much easier to read the question if you would edit it to include the information in your first comment, then just delete the comment. Again, welcome and good luck!

There is a twist in the calculations...
I would encourage you to write out all the details in the steps instead of just equations with a bunch of numbers. It is so easy to get lost in the numbers alone.
GIVEN
(1) For the 70 °C solution.
% $\ce{CuSO4\cdot5H2O}$ = $\dfrac{32.0}{100.0} = 32.0\% $
% $\ce{H2O} = 100 - 32.0 = 68\%$
(2) For the 0 °C solution
% $\ce{CuSO4\cdot5H2O} = 12.5\%$
% $\ce{H2O} = 100 - 12.5 = 87.5\%$
SOLUTION
(1) Calculate the g of $\ce{CuSO4\cdot5H2O}$ and water in 335 g the 70 °C solution.
g $\ce{CuSO4\cdot5H2O} = 0.320 \times 335 = 107.2 $
g $\ce{H2O} = 0.680 \times 335 = 227.8$
(2) Calculate g of solution at 0 °C which has 227.8 g water.
g 0 °C solution = $\dfrac{227.8}{87.5} = 260.3$
(3) Calculate g $\ce{CuSO4\cdot5H2O}$ in 0 °C solution
g $\ce{CuSO4\cdot5H2O} = 0.125\times260.3 = 32.5 $
(4) Calculate g $\ce{CuSO4\cdot5H2O}$ crystals
g $\ce{CuSO4\cdot5H2O}$ crystals $= 107.2 - 32.5 = 74.7$

HINT :
Let me call the mass of the whole solution $\text{S}$. I'll call $\text{L}$ the mass of the liquid and $\text{C}$ the mass of the crystals, so
$$\begin{cases}
\text{A = L + C} \\
\text{A}x_\text{A} = \text{L}x_\text{L} +\text{C}x_\text{C}
\end{cases}$$
So if we assume only $\ce{CuSO4, 5H2O}$ cristalise then $x_\text{C}=1$ and then
$$\text{C}=\text{A}\left(\frac{x_\text{A}-x_\text{L}}{1 - x_\text{L}}\right)$$
Now all you have to do is to find the values of the $x_i$ (which are the mass fraction of $\ce{CuSO4, 5H2O}$ in the considered phase.)


text:
making patinas for copper
Copper nitrate, calcium carbonate and ammonium chloride mixed will patina copper green. I need to know

If calcium carbonate should be a heavy powder or a reagent.
If calcium chloride is a dehydrate or reagent.
If ammonium chloride should be granular or granular reagent.

These mix with water in equal parts. I`m not a chemist, and this is for a sculpture with a recipe from an old metal techniques book.

Hmm... "These mix with water in equal parts." So if 4 kg of solution were being made then 1 kg of calcium carbonate. I can't image that much calcium carbonate would dissolve in 1 kg H2O, 1 kg CaCl2 and 1kg NH4Cl.

When making the solution you might want to heat it to help everything dissolve. Suspect you don't have a lab, but with any reasonable precaution these particular chemicals would be fine to heat on kitchen stove.  There are very few chemicals that I would use in my kitchen.

Could it have been calcium bicarbonate instead of calcium carbonate?

@MaxW, calcium bicarbonate is not a stable solid compound as it readily looses $\ce{CO2}$ and water to form calcium carbonate.

@airhuff - Thanks for the correction. Ammonium chloride has  about half the molecular weight of calcium carbonate. So you must be effectively making calcium bicarbonate in solution. So I'd dissolve ammonium chloride first, then add calcium carbonate, and last add calcium chloride.

Yea, I'm sure it's being made in solution like you say.

Nothing about your process is quantitative in nature, nor does it require high-purity reagents.  So most definitely do not invest in reagent-grade anything.  To answer your questions in order:
1) The "heavy powder" calcium carbonate will be sufficient.  It may be "clumpy", in which case you will want to physically break it up prior to mixing it in with the water.  You should wear rubber gloves and a dust mask if you need to do this.  The only purpose of this compound in your procedure is to raise the pH.
2)  You definitely do not need anhydrous or dried calcium chloride. You are not trying to quantitatively measure out this compound and you are just going to put it in water anyway, so hydration is not an issue for you at all.  The purpose of this reagent (generic term, not referring to any "grade" or purity) is to provide chloride ions, that at high pH will form a layer of cupric chloride, which is green when it absorbs water from the air to form cupric chloride dihydrate. Again, this will be a product of your procedure and the degree of hydration of the calcium chloride is not relevant. As in step one, this compound may be clumpy and if so you should break it up as described there.
3) The phrase "should be granular or granular reagent?" is the key. As stated above, there is no need for reagent grade materials for your process, so if your options are "granular" vs. "granular reagent" then just get the granular, not the granular reagent ;)  The role of this compound is to both raise the pH and to provide chloride for the purpose described in step 2.
Again, it may be tempting to think you will get a better patina by investing in expensive reagents, but for your process that would simply be a waste of your money to buy reagent-grade anything if there is a lower grade alternative.

airhuff, thank you for this information. Its exactly what I was hoping for. Obviously Im no chemist, but with your answers, I can move forward. All of these chemicals are available at Amazon. So because of your generous help, I can purchase and mix like a pro. [???] Again, thank you very much. Ken

Awesome! So glad to help. Note that the way we say thank you here is to accept (and upvote if you like) the answer ;) Cheers...

airhuff, will be glad to vote and show acceptance. How I do Dat.

There is a checkmark to the left of the question, below that big number 2 between the arrows.  If you think the answer was particularly good, it's your choice if you then want to hit the arrow above the 2, changing it to 3 (meaning 2 other people have upvoted it).  And dat's how dat works! Appreciate it.

I meant the checkmark is next to my answer, duh, sorry.

No problem ken, happy to help any time.

source: Stack Exchange

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

sort: votes


text:
Distinguishing atom types
How does one recognize lipophilic points, hydrogen bond acceptors and hydrogen bond donors in organic molecules? For example, provided this SMILES instance
O=C1NC(=O)SC1Cc3ccc(OCCN(c2ncccc2)C)cc3

I know it contains 2 lipophilic points, 5 hydrogen bond acceptors and 1 hydrogen bond donor. How can I determine them?

Can you draw a molecular structure from this SMILES, to begin with? I, for one, can't. Human readability was never the primary objective of this format.

@IvanNeretin, check the updated question, please. This is rosiglitazone (https://en.wikipedia.org/wiki/Rosiglitazone).

Thank you, that's much better. (Not sure if the SMIILES is still needed, BTW.) Now, all hydrogen atoms attached to N or O are hydrogen bond donors, and all N or O atoms with lone pairs are acceptors.

@IvanNeretin, could you clarify the statement about "lone pairs", please? And one more question: do I get it right that lipophilic atoms are all remaining non-ring atoms that are nor H-bond donors, nor H-bond acceptors?

What's unclear about that? As to what constitutes a lipophilic point in your definition, I'm not sure.

To identify hydrogen bond donors: Look for a polarized hydrogen (typically bonded to O, N, S)
Hydrogen bonds form from the interaction of a partially positively-charged hydrogen and a (partially) negatively-charged atom. Thus, to "donate" a hydrogen for hydrogen bonding, you need a polarized hydrogen. Hydrogens are polarized by being bonded to electronegative elements: in organic molecules that will typically be O, N, or S.
To identify hydrogen bond acceptors: Look for atoms with electron lone pairs, or otherwise partially negatively-charged sites on the molecule. (Typically O, N, S, or halides).
To form an interaction with a polarized hydrogen, you need a source of (partial) negative charge, which typically can come from electron lone pairs. The most common species you'll encounter in an organic molecule include O, N, S, or halides (F, Br, Cl, I).

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


text:
Are the various forms of polyethylene polymorphs?
Polyethylene (PE) plastics comprise high density PE, low density PE, cross-linked PE (PEX), linear PE, low density and high density linear PE, etc. Are these all considered polymorphs, or is there another name for them? (Polymorphism refers to crystal structure, so I'm thinking the various forms of polyethylene probably aren't polymorphs.)I guess a more general form of this question is: what is the name for the same basic chemical (polyethylene, in this case) that can form different compounds with different properties based on how the basic molecule branches (isomer doesn't quite seem to fit, either)?

I wouldn't even call them different compounds. See, there is no strict border between them; this, IMHO, is more important than the difference in properties.

Yeah, agreed. But I couldn't think of another term, but each of the different forms of polyethylene have different densities, melting points, etc.

@IvanNeretin, is the term allotrophs maybe more appropriate?

I think I'd personally just say different types of polyethylene. Allotropy refers to a pure element, so probably doesn't work here.

Polymer?  But I thought that just refers to long-chain molecules like polyethylene, polytetrafluoroethylene, polychloroprene, etc.


text:
Why is 1,3-pentadiene favored in the reaction of pent-3-en-2-ol with sulfuric acid?
Let's say we have the following reaction, a simple dehydration reaction.

Why do we get 1,3-pentadiene and not 2,3-pentadiene?

Hydrogenation of 1,trans-3-pentadiene (your top molecule) and the allene, penta-2,3-diene (your bottom molecule), both produce pentane plus heat.  Heat is given off because both of the dienes plus hydrogen is less stable than pentane.  With 1,trans-3-pentadiene, 53.1 kcal/m of heat is given off; with penta-2,3-diene 66.8 kcal/m of heat is given off (data from Frances Maron Fraser and Edward J. Prosen, Journal of Research of the National Bureau of Standards Vol. 54, No. 3, March 1955 Research Paper 2575, p. 143-148). Clearly the allene is of higher energy than the 1,3-diene by about 15.7 kcal/m.
The reaction conditions (strong acid and heat) suggest that an equilibrium mixture of possible products will result.  Around room temperature, each 1.4 kcal/m difference in energy between two isomers will shift the equilibrium between them by a power of 10 ($\ce{\Delta G=-RTlnK}$).  A 15 kcal/m energy difference between your two dienes suggests that the equilibrium will be ~ $\mathrm{10^{10}:1}$ favoring the 1,3-pentadiene. If any of the allene forms, it will rapidly convert to the 1,3-pentadiene.
Side Note: Why are cumulenes generally less stable than 1,3-dienes?

Conjugation which results in resonance stabilization is present in the 1,3-diene but not in the 2,3-diene.
Allenes (in fact, all cumulenes) have considerable strain built into them. If you view a double bond as a two-membered ring, it is easy to see this.  The bond angle in the two-membered ring is severely pinched down.  In turn, the bond angle external to the double bond tries to open up and relieve strain.  This strain relief, can't happen with the central carbon in an allene because it has a double bond on both sides.


text:
Assumptions to be taken when rearranging terms in first law of thermodynamics

Assuming a closed system, show how the first law of thermodynamics.
$$\mathrm dU = \mathrm dQ + \mathrm dW$$
becomes,
$$\mathrm dQ = \frac{C_V}{R}V\,\mathrm dP + \frac{C_p}{R}P\,\mathrm dV$$

I have an answer, but am hesitant to say it is a final answer or even the correct one at that.
I am concerned with the partial derivatives when rearranging this equation. How do certain parts cancel out or how does one approach the problem without them? Is it safe to assume that since you know how the state variables will act in a certain process, that you can use that knowledge to provide a better or more exact answer to the problem?
Attached you will find a picture of the problem that I have done to the best of my ability (which is limited I must say).
How do I make my assumptions?
Do I assume holding one part constant to solve the equation in one way - that is via an adiabatic process, isothermal process - do the same for another path and compare?

This equation can not be applied in general because it can't describe the amount of heat transferred to an incompressible solid or liquid at constant pressure.
For these cases, it would predict that $\mathrm dQ=0$ (which is not necessarily correct).  If it is not generic, it also cannot describe real gases beyond the ideal gas region.  Therefore, the equation must apply exclusively to an ideal gas. The presence of the ideal gas constant $R$ in the equation is a dead giveaway.
For a reversible path,
$$\mathrm dQ=T\mathrm dS=C_v\mathrm dT+P\mathrm dV$$
But, for an ideal gas,
\begin{align}
&\mathrm dT =\frac{P\mathrm dV+V\mathrm dP}{R}\\
&C_v+R = C_p
\end{align}
Therefore,
\begin{align}\mathrm dQ&=C_v\frac{P\mathrm dV+V\mathrm dP}{R}+P\mathrm dV \\ &=\frac{C_v}{R}V\mathrm dP+\frac{(C_v+R)}{R}P\mathrm dV \\
\implies\mathrm dQ&=\frac{C_v}{R}V\mathrm dP+\frac{C_p}{R}P\mathrm dV
\end{align}
This equation applies only to an ideal gas and a reversible path.

I think it might be wrong because when you were differentiating $PV=RT$, you have assumed pressure to be constant and in your derivation, in some places, you have assumed changing pressure.
You have to know what parameter is changing when you differentiate an equation. When you assumed $P$ to be constant you got $P\Delta V = R\Delta T$. If $P$ weren't constant you would have had a completely new formula that would complicate things even more.
I would suggest determining what kind of process is occurring and then manipulating the first law equation accordingly.

I would do a derivation for a single process first then make a generalization for that particular process Because, for example, work done in adiabatic process wont be the same to the work in isothermal process. But the first law equation is general, a factual equation, for the processes that we encounter in our questions.

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


text:
Why is the molecular geometry of a molecule with 5 bonds not uniform?
I'll note that I'm a student in a rather introductory course, and today we learned about molecular geometries.
I'm curious as to why certain geometries (say, $\ce{CH4}$, tetrahedral) are uniform (evenly distributed in a sphere) while other geometries, say "$\ce{SbCl5}$," (5 bonds to the central atom) are less-uniformly distributed - two of the atoms are linear and 3 of them of trigonal planar.
Is this just a simplification for teaching introductory students, or is this how the molecule actually behaves?  I would think that to achieve "equilibrium" from the $\ce{Cl}$ atoms repelling each other, the distribution of the $\ce{Cl}$ atoms would be perfectly normal around the sphere.

Duplicate: A more general version of this question, and one of the two cases asked about in this question.

The short answer, summarizing from the answers here, is that it's not geometrically possible to arrange five points in a mutually equidistant fashion around a sphere.

You could have five ligands equidistant from their nearest neighbors in a regular pentagon.  But that is not generally a favorable structure.

You may want to look into VSEPR theory. To a good first order approximation you can think of electron pairs and external atoms on a central atom as taking up about the same space.
This explains why water's bond angle (104 degrees) is very close to the 109.5 degrees exact equidistant angle associated with methane, or $\ce{AlCl4+}$.
Your example with $\ce{SbCl5}$ is a good example where the theory gets a bit more complicated than simply maximizing distance between steric centers. The bonds to the chlorines are not equal. And yes this is how the chemical would look: see the Wikipedia article on orbital hybridisation.
Now, as your example has 5 identical groups, inter-conversion can occur, where the axial group "switches" with an equatorial group. But at any moment there will be two distinct chemical environments around the compound.

This doesn't explain why a uniform distribution of five groups is impossible.

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


text:
Difference between lattice energy and thermal stability
For an ionic compound, is it always true that the greater the thermal stability, the greater the lattice energy. E.g. for 2 ionic compounds MX and MY, if MX has a higher thermal stability than MY, does MX necessarily have a higher lattice energy?

Related: https://chemistry.stackexchange.com/questions/80448/thermal-stability-of-alkali-metal-hydrides-and-carbonates/80460#80460

1) Lattice energy is the measure of the stability of an ionic solid.
It is the energy required to separate the ions in one mole of a solid ionic compound into gaseous ions.
The weaker the bond between the ions, the easier it will be to break that bond and separate the ions into gaseous ions and the lesser will be the lattice energy.
Therefore, the stronger the bond between the ions, the more is the lattice energy of the ionic compound.
2 )Thermal stability depends upon the amount of heat required to break the ionic compound. Clearly, the stronger the bond between the ions is, the more heat will be required to break the bond and as a result, the thermal stability of the compound increases.
Therefore, stronger the bond, more is the thermal stability of the ionic solid.
In general, we can say that as thermal stability increases, the lattice energy also increases, but they also also depend upon the crystal structure of the solid and other factors.
So, you cannot say with certainty that if the thermal stability is high, then lattice energy will be high too.
Hence, the answer is no. If MX has a higher thermal stability than MY, then it may or may not have higher lattice energy than MY.

That correlation is not true. There's a very simple counter-example: $\ce{MgCO3}$ vs $\ce{BaCO3}$. $\ce{MgCO3}$ has a larger lattice energy ($\ce{Mg^2+}$ smaller than $\ce{Ba^2+}$), but lower thermal stability (decomposes at a lower temperature).

You mean the part where I said one implies the other and vice versa? Yeah, I didn't take into account the crystal structures.

It's not about crystal structure, really. The thing is that you have to consider the products - in this case it forms $\ce{MO}$ and that's very favourable for $\ce{Mg}$ and not as favourable for $\ce{Ba}$. (That's because of, again, lattice energy.)

Yeah, but crystal structures are one factor. Right? So that means that thermal stability and lattice energy don't have a one on one dependency. Are there some other factors as well?

Off the top of my head, I can't think of a specific case where the crystal structure would matter; however, in general yes there is no broad correlation between the two.

Why wouldn't it matter ? Clearly, some crystal structures are more stable than others. So, energy required to break the ionic lattice and separate the ions into its constituent atoms would be obviously high. Okay, could it be because crystal structures would affect both, lattice energy and thermal stability?

See, I'm not saying it doesn't matter, I'm just saying that right now I can't think of an example where it matters. There's probably something out there.

So, you find more gaps in my answer??

I don't know why you're saying that. Essentially I'm telling you that I don't know what else affects the crystal structure.

Oh, well I was not talking about the crystal structures. That discussion is over. I was just asking if there is something more in my answer that needs to be fixed because I previously thought that both thermal stability and lattice energy have a one-on one dependence but when you told me that exception, I had to edit my answer. So, is there anything else?

The factors would probably depend on the precise route of thermal decomposition. Essentially if thermal decomp. is a reaction $\ce{A -> B + C +}\cdots$, to find how favourable this reaction is, you would need to not only look at the stability of the reactant $\ce{A}$ (which is indeed measured by the lattice energy) but also the stability of the products $\ce{B + C + }\cdots$. The exact factors would depend on exactly what $\ce{B}$ and $\ce{C}$ etc. are.

carbonates become more thermally stable down the group

When you supply energy to an ionic compound to separate it into its component ions (e.g. $\ce{MgCO3}$ to $\ce{Mg^2+}$ and $\ce{CO3^2-}$) it is called the lattice energy. The same amount of energy is released when this same ionic compound is formed ($\ce{MgCO3}$).
You can also heat this compound ($\ce{MgCO3}$) until it breaks down to $\ce{MgO}$ and $\ce{CO2}$. And that is called thermal decomposition. I don't know what determines whether the heat supplied will break down the ionic compound or separate it into its component ions but that is the difference between the two processes.


text:
Is an acid (just) a proton donor and if not, what is the correct definition of an acid?
Why is acid called proton donor?
On the internet, I found:

Acids are substances that can donate $\ce{H+}$ ions to bases. Since a hydrogen atom is a proton and one electron, technically an $\ce{H+}$ ion is just a proton. So an acid is a 'proton donor.'

However, my teacher said that it is not fully correct. If this is the case, then what's the correct definition of an acid?

All H+ donors are acids but not all acids are H+ donors. To explain why other substances which didn't donate H+ ion were acids, we needed to revise our definitions of acids. We have three most common definitions of acids-Arrhenius, Bronsted-Lowry and Lewis acids. You can read about them anywhere.

Related: http://chemistry.stackexchange.com/questions/6751

Your teacher said that this is not completely correct as there are (at least) three common definitions for acids and bases (Wiki has a good article for this https://en.wikipedia.org/wiki/Acids_and_Bases):

Arrhenius: In water, acids under this definition protonate the water and form Hydronium (H3O+). Bases in this definition deprotonate the water.
Brønsted–Lowry: This is your proton donor definition. Acids in this definition donate protons to bases.
Lewis: In this definition acids are electron pair acceptors and bases are electron pair donors.

The relationship between these definitions is that as chemistry evolved we ran into more chemicals (acids and bases) with similar reactivities that the previous definitions didn't quite encompass and so chemists broadened the definitions. So, as you go down the list above you will find more and more chemicals that match your new definition of acid or base.

No, you’re misciting Arrhenius’ definitions. An Arrhenius acid is a substance that dissociates in water to give $\ce{H+}$ and an Arrhenius base is a substance that dissociates in water to give $\ce{OH-}$. Ammonia, for example, deprotonates water but does not dissociate to give $\ce{OH-}$ and thus is not an Arrhenius base.

Jan I'm afraid that I cannot agree. In Arrhenius's 1887 paper (I have only read an English translation) "On the Dissociation of Substances Dissolved in Water" he expressly calls out Ammonia as a base and talks about it as an "active molecule" i.e. for every molecule of ammonia in solution two ions are produced (in very dilute solutions). So, what he seemed to care about was the production of either Hydroxide or Hydronium as the definition of acid or base in solution.

Which Arrhenius paper are you referring to? I found one with a matching title (Z. phys. chem. 1887, 1, 631–648. DOI: 10.1515/zpch-1887-0164.) where he does call ammonia a base but not an active molecule. (Having read the German original.)

Acids have different definitions according to different concepts. Acid is a proton donor according to brønsted lowry concept. According to Lewis concept, acids are electron pair acceptors. Now there are some acids like Boric acid which doesn't donate hydrogen by itself but still its an acid. The reason being it accepts electron pair from OH-  and hence obeys Lewis definition. So all acids cannot be explained through brønsted lowry or Arrhenius concept. Here are the three definition of acids:
Arrhenius acids are those which when added to water, increases the concentration of H+ ions in the water.
The Brønsted-Lowry definition is the most widely used definition; unless otherwise specified, acid-base reactions are assumed to involve the transfer of a proton (H+) from an acid to a base.
Lewis acids is a species that accepts a pair of electrons from another species; in other words, it is an electron pair acceptor.


text:
Why don't we measure mass of oxygen used in the calorimeter

The apparatus shown in the diagram can be used to measure the enthalpy change of formation of magnesium oxide (MgO).

List the measurements you would need to make using this apparatus in order to calculate the enthalpy change of formation of magnesium oxide (MgO).

And the mark scheme said:

Measurements needed:

volume/mass/weight of water (in the calorimeter)
initial + final temperature/temperature change/temperature rise (of the water)
mass of Mg (used)/mass MgO
(Not volume/moles/mass of oxygen used)

Why shouldn't I measure the mass of oxygen? Oxygen is also the reactant in its standard state as same as the magnesium element, isn't it?

Hint: what is the heat capacity of the water in the system and that of the oxygen? (Hint to the hint: calculate the moles each of water and oxygen then look up their molar heat capacities on Wikipedia).

Thx:) but i can still measure the mass of Mg used in order to find the MgO mass(mole) produced, right? Can I do the same for oxygen instead of Mg to find the mass(mole) of MgO?

Well, if I understand the comment, yes, you should measure the masses of Mg & MgO gravimetrically. If you had a way of measuring the O in MgO or measuring the amount of $\ce{O2}$ before and after the experiment then you could do that too, in theory. But in reality you do not. The whole point of the hints in my first post was to lead you to an estimation of how much more thermal mass of water you have in your system compared to that of oxygen in your system. Is that of oxygen small enough compared to water to be negligible? Are the Mg and MgO masses even used in that part of your calculations?

@airhuff Sorry>< That's my fault. I thought it was the standard enthalpy change (enthalpy change per mole) instead of enthalpy change of formation. So, the Mg and MgO masses are not necessary. Thank you^^

I'm not sure that I get that explanation, but hey, so longs as you do I guess all is good ;) Do note that in your procedure it says you need the masses of Mg and MgO; they are just not considered as part of the heat capacity of the system. The thermal mass of oxygen is just so small compared to that of the water that it is negligible.


text:
Why is UuX used as a symbol for unnamed elements on the periodic table?
Until recently, the last few elements on the periodic table had the chemical symbols (now they have formal chemical symbols!):
Uub, Uut, Uuq, Uup, Uuh, Uus, and Uuo
Why did they have these temporary symbols? Why the letter U? Is there any reasoning behind the third letter? Why not:
Uua, Uub, Uuc, Uud, Uue, etc...
I was just wondering some of the history of nomenclature and the thought process behind this specific nomenclature.

Related: Why is a temporary name given to an element with an atomic number above 100?

Technically, only one element (119) still uses Uux (specifically, Uue). All others from 110-118 are experimentally known and have assumed proper names, and for 120 or more the second letter becomes different. See ringo's answer for more details..

In 1978, the IUPAC Commission on the Nomenclature of Inorganic Chemistry decided that it would be necessary to have a systematic naming for the elements with atomic number greater than 100, even for those which had not been discovered. The recommendations are as follows:

The name is derived directly from the atomic number of the element using
  the following numerical roots:

$$\require{enclose}
\small
\begin{array}{lcc}
\hline
\text{Number} & \text{Latin Word} & \text{Greek Word} & \text{IUPAC Nomenclature} \\
\hline
\text{0} & \enclose{circle}{\text{nihil}} & \text{mēdén} & \text{nil} \\
\text{1} & \enclose{circle}{\text{ūnus}} & \text{en} & \text{un} \\
\text{2} & \enclose{circle}{\text{duo}} & \enclose{circle}{\text{dúō}} & \text{bi } \\
\text{3} & \text{trēs} & \enclose{circle}{\text{tria}} & \text{tri} \\
\text{4} & \enclose{circle}{\text{quattuor}} & \text{téttara} & \text{quad} \\
\text{5} & \text{quīnque} & \enclose{circle}{\text{pénte}} & \text{pent} \\
\text{6} & \text{sex} & \enclose{circle}{\text{héx}} & \text{hex} \\
\text{7} & \enclose{circle}{\text{septem}} & \text{heptá} & \text{sept} \\
\text{8} & \enclose{circle}{\text{octō}} & \enclose{circle}{\text{oktṓ}} & \text{oct} \\
\text{9} & \text{novem} & \enclose{circle}{\text{ennéa}} & \text{enn} \\
\hline
\end{array}
$$

The roots are put together in the order of the digits which make up the
  atomic number and terminated by 'ium' to spell out the name. The final
  'n' of 'enn' is elided when it occurs before 'nil', and the final 'i' of
  'bi' and of 'tn' when it occurs before 'ium'.
The symbol of the element is composed of the initial letters of the
  numerical roots which make up the name.
The root 'un' is pronounced with a long 'u', to rhyme with 'moon'.
  In the element names each root is to be pronounced separately.$^{[1]}$

Though not in the original text, I went ahead and added the words from which the IUPAC names were derived. Though not obvious, bi- comes directly from di-, which is a result of duo, found in both languages.

$^{[1]}$ Recommendations for the Naming of Elements of Atomic Numbers Greater than 100, Pure Appl. Chem., Vol. 51, No. 2, (1979) 381-384.


text:
Do we get high on "methampetamine" naturally?
First off, I'm new here and I cant tell if this is a chemistry or a biology question. It's both and if I ask on biology forums, they say to ask on chemistry forums and vice versa. So sorry if this question is more geared towards biology.
I am really amazed that our brain produces Phenethylamine any time we do something rewarding. I need to confirm this, because if this is true, I'm high right now... (wait so meth isn't bad?)
Phenethylamine gets biosynthesized in our brain to get converted into dopamine (and also L-DOPA). But the funny thing is that the molecular formula of Phenethylamine is exactly the same as methamphetamine just without the two methyl groups.
From my understanding, if you consume methamphetamine the liver metabolizes methyl groups into hydroxyl preventing it from metabolizing Phenethylamine instantly and making the hydroxyl group a free radical, so the hydroxal group is separated once it reaches the brain, therfore consuming methamphetamine would be 10x more potent then if you consume Phenethylamine. The same way the high from Dextroamphetamine (the chemical used in Adderall which is the same as methamphetamine, just with one methyl group instead of two, making it not as potent) is 5x weaker then methamphetamine.
Consuming pure Phenethylamine would make the liver metabolize 90% of it making the brain receive only around 10% of it. Phenethylamine is naturally occurring and is sold on Amazon as pure powder.
So my question: is Phenethylamine the same as meth, just with 1/10th of a high, considering that is what methamphetamine gets metabolized into?
I didn't mean get high on methamphetamine in the title, but the end product of methamphetamine (apples and oranges)
And does that mean that meth isn't that bad as the media claims it is (besides hepatoxicity)?

Strictly speaking we get naturally high on analogues of methamphetamine just as we suppress pain with natural analogues of NSAIDs and opiates. None of this says that the artificial versions are good for you.

All the drugs I know of are closely related to a molecule we produce naturally or are a molecule that gets produced by our body. This is kind of the point of drugs. But the main issue is the dose. We all have uranium in our bodies, does that mean consuming it isn't bad for you? Or to put it differently: We all have bones, don't we? But putting another bone into your body isn't really considered healthy.
First of all: Chemically, those two molecules of course are not the same. Let's get the obvious out of the way.
I think this is the wrong forum again because really what you are asking is more of a legal/ethical question than a scientific one (both questions). But I'll try to answer the 2nd question "scientifically" with a light esoteric touch: Consuming drugs means messing with a delicate balance in our brains that is so complicated we do not even begin to understand it. I'm not saying don't use drugs, I'm just saying maybe you want it to be not as bad as it really is.


text:
What kind of building materials could be synthesised from basalt?
Building tall cities is made possible by the invention of concrete with steel rebar which has much more tensile strength than any other massively available material that I am aware of. Cement (the most crucial ingredient) is made by processing limestone which is an organic rock formed from corral and skeletal remains.
Rocky "earth like" planets have no limestone (or coal or oil or any other organic material) but they do have allot of basalt. What kind of materials could you manufacture from basalt? Could you make materials that have enough tensile strength to build a city?

Do you have a specific planet and city in mind or do you just want to build a space city and it really doesn't matter how big and where? Humans have been using basalt in certain areas for millenia. If their cities do not meet your needs: Please consider that it might be easier to simply bring the materials to build your city with you on your space ship than to do some fancy basalt stuff. Also, you might get more of an answer in a geology or construction worker forum. Gonna give you an upvote and hope someone tells me why we don't build everything from basalt.

The planet is Tau Ceti E. The city is a metropolis.

I would say that building tall cities is made possible by steel much more so than concrete. At least here on Earth real skyscrapers did not exist until steel allowed architects to get away from the 'massive walls carry the load of the building' paradigm. So, the premise is off. Second, the chemical reactions that limestone allows that result in the concrete curing back in to a solid just aren't going to happen with a basalt-based starting material.

In that case, since the gravity is much higher on that planet - if it even exists - I wouldn't recommend building high skyscrapers.


text:
How many d-electrons are in the outer shell of metals like Cadmium (this is counting the 4d electrons)?
First I know similar questions have been asked about counting d-electrons.
However, my question is about the electrons in the 4d orbital. For example if I have CdS. The cadmium has a charge of +2. So does this affect the number of d-electrons?
Would it be 10 d-electrons as the 2 extra have been removed from the 5s orbital?
Also is the electron configuration of Cd in CdS:
[Kr] 5s2 4d10
or is it:
[Kr] 4d10

Hello and welcome to Chemistry.SE! At this point I recommend that you take the short [tour] to better familiarize yourself with the site. Also, as this appears to be a homework type of question, you should read through this discussion. Regarding your question, which answers do you suspect are correct and why?

Thank you! I suspect that because Cd has a +2 charge, the 5s2 electrons will be lost. So in CdS the electron configuration of Cd is [Kr] 4d10. But I didn't know this was possible... I think i've just made up that I can remove the 5s2 electrons in compounds like this.

Most often it is the valence electron which takes part in formation of ionic compounds, even in transition metals. So Yes $[Kr] 4d10$ is correct.

Without doing some sort of at least half-serious calculations, you cannot really answer the question. For all we know, a cadmium(II) ion could be $[\ce{Kr}]\,\mathrm{4d^{10}}$, $[\ce{Kr}]\,\mathrm{4d^9\,5s^1}$ or $[\ce{Kr}]\,\mathrm{4d^8\,5s^2}$.
However, given the analogies to other transition metals and the general structure of practically all transition metal ions, the first configuration, $[\ce{Kr}]\,\mathrm{4d^{10}}$ is most likely. I can’t recall a case where a transition metal ion had any s electrons in the valence shell before the d electrons were completely populated; i.e. even if you have two s electrons in the neutral atom and ionise only once you would end up with only d electrons in the ion.


text:
What results from the reaction of pent-3-en-2-ol with hydrogen bromide?
I have seen the following reaction in my textbook:

This was quite easy to understand, though I wonder why 2-bromopent-3-ene is not a possible result:

Isn't this how $\ce{-OH^}$ groups and hydrogen halides interact? Any clarification would be appreciated.

It is very rare for a reaction to have no side products, and this reaction is certainly not exceptional in that regard. I'm actually a bit surprised your book chose this reaction to illustrate the regiochemistry of $\ce{HBr}$ addition to an alkene, as this would compete with the formation of an allylic cation, which are generally more stable than alkyl carbocations. Being more stable, these cations will form more quickly:

giving the product you correctly assumed would also be formed. You probably wouldn't obtain this though, as it would continue to react, forming a dibromoalkane (as would the products listed by your book):

Although what ringo said is true, it can be explained quite simply.
Br- is more stable than OH- owing to it's larger size(and deceased electron density) and thus is a better leaving group than OH-
Therefore, OH- has a tendency to stay which results in the reaction you have shown.


text:
Why can't the reaction of grignard reagent with carboxylic acid give a ketone?
I understand that a Grignard reagent can act both as a base and a nucleophile.
It acts as a base in the presence of a proton source (eg: alcohols, amine, water, etc). But when reacting with a ketone or an aldehyde, it acts as a nucleophile and uses a specific reaction mechanism. Why isn't that same reaction mechanism not viable for reaction with carboxylic acid?

Why is this reaction mechanism wrong? Is the formation of a diol not possible due to repulsions or is there some other reason?

Grignard reagent is  a superbase. It would rather take the acid's hydrogen rather than attack electrophilic carbon.

Not sure if I would call a Grignard a superbase, but it will react with the proton in a carboxylic acid rather than reacting as a ncuelophile on the carbonyl carbon

Also worth noting that an organolithium will deprotonate the carboxylic acid, and then attack the carboxylate ion to give a doubly-deprotonated gem-diol, which then reverts to a ketone upon work-up.  This is because organolithium compounds are more nucleophilic than Grignards.  This is an alternative to the answer about using a Grignard + a nitrile.

Acid-base reactions are much faster than nucleophilic attack.

Huy Ngo says it all.  A Grignard reageant will react first with just about any hydrogen that has even a modicum of proton donation capability, before "resorting" to the slower nucleophilic attack.
To get a ketone (in most cases), avoid the dissociable proton by using a nitrile instead, assisted by a copper salt. This forms an intermediate imine which is hydrolysed to a ketone after water/acid workup:
$\ce{CH_3MgI + CH_3COOH -> just a salt}$
$\ce{CH_3MgI + CH_3CN ->[Cu^+] (CH3)2C=N-M ->[H2O] (CH3)2C=O}$

Grignard addition to a nitrile usually needs a copper salt added to get decent yields of ketone.

It is known fact that organolithium compounds react with carboxylic acid to give ketones (e.g., Ref.1). Until recently, however, no report has been published on Grignard reagents on this manner. Now, it is evident that Grignard reagents can also be used to prepare ketones under specific conditions (Ref.2):

References:

Robert. Levine, Marvin J. Karten, “Reactions of carboxylic acids with organolithium compounds,” J. Org. Chem. 1976, 41(7), 1176–1178 (https://doi.org/10.1021/jo00869a020).
Kilian Colas, A. Catarina V. D. dos Santos, Abraham Mendoza, “$\ce{i-Pr2NMgCl•LiCl}$ Enables the Synthesis of Ketones by Direct Addition of Grignard Reagents to Carboxylate Anions,” Org. Lett. 2019, 21(19), 7908–7913 (https://doi.org/10.1021/acs.orglett.9b02899).


text:
Alphabetization rule in case of consecutively placed substituents
This is the compound:

I am stuck on assigning the locants to the given halo substituents. We know that all halogens are of equal priority. So, using first point of difference, we get either an $\ce{I \rightarrow Br \rightarrow Cl}$ path or a $\ce{Cl \rightarrow Br \rightarrow I}$ path. I am confused as to which of the two paths I should choose. I could choose alphabetical order, but the bromine in between is ruining the alphabetical order of C-I
Is there any new theory that I am missing?

Rule P-45.2.2 in the 2013 IUPAC Recommendations reads:

The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes (other than 'hydro/dehydro') to the parent structure.

The term "lowest set of locants" is defined in P-14.3.5; see Loong's answer here. As such the correct set of locants is 1,2,3, not 1,2,5 as was proposed in another answer since "1,2,3" is lower than "1,2,5". (The other answer has since been deleted.) As far as I can tell you have correctly identified this. However, the ambiguity between these two possible numbering schemes is not resolved by P-45.2.2:

In order to decide one first needs to generate the corresponding names for both numbering schemes. Regardless of locant numbering, the three substituents are listed in alphanumerical order (rule P-14.5). Scheme A would give the name "2-bromo-3-chloro-1-iodocyclopentane", whereas Scheme B would give the name "2-bromo-1-chloro-3-iodocyclopentane", since "bromo", "chloro", and "iodo" come in that order alphabetically.
Turning to the next rule P-45.2.3:

The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes to the parent structure (other than 'hydro/dehydro' prefixes) in their order of citation in the name.

Since Scheme A gives a name with the locants cited in the order 2,3,1, whereas Scheme B has the locants cited in the order 2,1,3, Scheme B is used to generate the preferred IUPAC name 2-bromo-1-chloro-3-iodocyclopentane.
As for what is missing in your train of thought: From what I understand of your question, you seem to be thinking that the prefixes must be ordered in increasing locant number i.e. the name generated is "1-X-2-Y-3-Zcyclopentane", and that the locant set must then be chosen such that X, Y, and Z come in alphabetical order. As you already noticed with this example, that is not quite a foolproof rule. The correct way to do it is the other way round: first, order the prefixes by alphabetical order, and then choose the locant set in order to minimise the locants.

Wow! Thanks a lot! Just two more clarifications: (1) "that is not quite a foolproof rule." => could that mean the rule that I was thinking of is in fact wrong and non-existent? is the actual and only rule for alphabetization the one that you've explained? (2) Which software did you use to draw those two molecules? (I've been searching a lot but couldn't decide easily on the best organic molecule sketcher)

@GaurangTandon (1) These are the rules enunciated in Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013. Your rule isn't listed inside there. If you use a different rule, you will end up with non-preferred nomenclature, which is not really a problem in and of itself, but if you want to obtain the preferred IUPAC names then you will need to follow their rules. (2) I use ChemDraw. Costs lots of money. I don't know about the free ones.


text:
What happens to photopolymer chemically when it is cured by uv light?
I'm writing an essay on how 3d printing process effects the structure of the material used and im finding it hard to find what actually happens to the polymer when it is cured. I know that it is made up of monomers, oligomers and photoinitiators and other information would be much appreciated.

Search for Radical Polymerisation. The Wikipedia article has an example of a photochemically initiated reaction.

Curing is simply crosslinking between polymer chains. In the first days of rubber, sulfur was used to crosslink the natural double bonds in the rubber (known as vulcanisation).
Similarly,  in the case of your question, UV light will cause a photoxidation or other photoinitiation process to bond the relevantly, reactive atoms in the chain to others in another polymer chain. The extent of cross-linking will determine if you obtain an elastomer ~1% crosslinking) or a thermoset polymer (definitely >2% crosslinking).


text:
POMs catalysis of benzyl alcohol oxidation, why Ni is the best cation with POM for that catalysis?
Why is nickel the optimal cation for benzyl alcohol oxidation with POMs and hydrogen peroxide?
In this article (Chinese Journal of Chemistry, Volume 30, Issue 2, 2012), they came to a conclusion that Ni gives better conversion then the other metals (Zn, Co, Mn, Cu, Fe) when they used it with Keggin type POM $\ce{Ni3(PW12O40)2}$. See the next note about the active structure:

How would you prove that Nickel is the best?

Welcome to Chemistry! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using LaTeX syntax. If you receive useful answers, consider accepting one.

Do you know that the three results columns are probably %conversion (of starting material,  to be clear) ,  %selectivity and %yield?  The table should then be self explanatory


text:
Why is fluorine more reactive than chlorine?
Why is fluorine more reactive than chlorine despite chlorine having a higher electron affinity?

Why the downvotes? It is a perfectly legitimate question. Now, "more reactive" is the property of molecular fluorine vs molecular chlorine. Higher electron affinity is the property of an atom. To get atoms from molecules, you have to split the latter. And it just turns out that fluorine molecule is awfully easy to split.

Related: Why is fluorine more reactive than iodine despite the weaker I-I bond?

It depends on what reaction you are talking about.

In the context of organic SN1 reactions, the reactivity is $\ce{R-F}$ (least) $\ce{< R-Cl < R-Br < R-I}$ (most). [1] This reflects the leaving group ability, with larger atoms better able to accommodate the extra electron.
In the context of radical homolysis of $\ce{X-X}$ the order of reactivity is $\ce{F2 < Cl2 < Br2 < I2}$ as the sigma bonds become weaker as you move down the period.[2]
In polar aprotic solvents (DMSO) nucleophilicity follows $\ce{F- > Cl- > Br- > I-}$. This order reverses in polar protic solvents due to solvation: J. Chem. Educ., 1970, 47 (6), p 473
Transition metal $\ce{MX_n}$ bond enthalpies have been measured empirically and calculated: J. Chem. Phys., 1957, 26 (6), 1644–1647.. It would be inappropriate to over-generalize, as any discussion here needs to also account for metal oxidation states, HSAB theory, electronic configurations, and ligand-field theory

References:

Clayden, Greeves and Warren Organic Chemistry, p. 348.
Clayden, Greeves and Warren Organic Chemistry, p. 971.

F-F bond is much weaker than Cl-Cl...

There are three things to consider every single time relative reactivity is unknown; atomic radius, shielding, and number of electrons. The reactivity is the halogens ability to gain an electron, so number of electrons already in the atom plays a vital role. Chlorine has more electrons so repels a reacting electron with greater force than fluorine, making it less likely to react.
Fluorine also has fewer electron shells than chlorine, so there are fewer electrons between the positive nucleus and the reacting electron to essentiallly block, or weaken, the electromagnetic attraction. This is shielding.
Lastly, fluorine is much smaller molecule than chlorine, and the shorter distance, or radius, between the nucleus and the electron again makes it more likely to attract the electron and react to gain a noble gas configuration.

Chlorine has more electrons so repels a reacting electron with greater force This doesn't seem to explain it: chlorine also has more protons in the nucleus, which should compensate for higher repulsion.

Reactivity is an elements ability to gain an electron. So the better it is at "stealing" electrons, the more reactive it will be. The thing that makes fluorine so reactive is its electronegativity. Electronegativity is the tendency of an atom to attract a shared pair of electrons. In simple terms, think of this as tug of war, if fluorine went up against chlorine, it would always win the tug of war due to its higher electronegatvity. There is a periodic trend for electronegativity that you can look up. Basically it starts at the bottom left with francium having the lowest electronegativity and goes diagonally left to right all the way up to fluorine having the highest electronegativity. Electronegativity is probably the biggest thing that plays into reactivity. Therefore, since fluorine has a higher electronegatvity than chlorine, fluorine is more reactive.

Fluorine molecule has a small atomic radius with 5 electron in 2p orbital. This means it has a high charge density on its atom. When fluorine molecule breaks up into F- ions it remains highly unstable and just wants to bond with other element. Due to this charge density fluorine has high polarizability which makes it more reactive.

I disagree. High charge density does not make a species more polarisable. Fluorine has high electronegativity and small covalent radius, thus it should be rather non-polarisable.


text:
Proof of Markovnikov's Rule
Is there a proof for Markovnikov's Rule? Specifically, why does the H bond to the carbon with the most hydrogen atoms already bonded to it?

Please see Electrophilic Addition to Unsymmetrical Alkenes on Chemguide

In most reactions that follow markovnikov's rule, a carbocation is formed and rearranged such that it is stabilised to the maximum extent .

The stability usually comes from the inductive effect, that is the electron releasing methyl groups that lead to a better charge distributed system. More is the number of such methyl groups surrounding the carbocation, more will be the stability and more such intermediates will be formed. These stable intermediates will be attacked by the negative reagent. This justifies the Markonikov's rule (at least in cases where carbocations are formed).
Note: This only applies to reactions such as acid catalysed hydration, action of halogen acids on alkenes, etc. In some reactions such as oxymercuration-demercuration no carbocation is formed but it still follows Markonikov's rule and uses a different reaction mechanism.


text:
How P(C2H5)3 acts as a ligand?
How does $\ce{P(C2H5)3}$ acts as a ligand and forms dπ-dπ bonds with transition elements. Why not pπ-dπ bond?
I understand that transition elements have d subshell available to accept electrons. But what about the electrons from phosphorus? From Chemistry Class 12 (NCERT), Part 1, p. 168.:

Another factor which affects the chemistry of nitrogen is the absence of d orbitals in its valence shell. Besides restricting its covalency to four, nitrogen cannot form dπ-pπ bond as the heavier elements can e.g., $\ce{R3P=O}$ or $\ce{R3P=CH2}$ ($\ce{R}$ = alkyl group). Phosphorous and arsenic can form dπ-dπ bond also with transition metals when their compounds like $\ce{P(C2H5)3}$ and $\ce{As(C6H5)3}$ act as ligands.

What do you mean by dπ-dπ? PEt3 acts as a $\sigma$ donor towards metal centres via the lone pair on phosphorus. It's also a $\pi$ acceptor but the acceptor orbital is the $\ce{P-C}$ $\sigma^*$ orbital, not the 3d orbitals of P.

@orthocresol My NCERT textbook says 'Phosphorous and arsenic can form dπ-dπ bonds also with the transition metals when their compounds like P(C2H5)3 and As(C6H5)3 act as ligand.'

@Reeshabh -- That is outdated.  The outer-$d$-orbital bonding theory was invented to force-fit pairwise covalent bonds with observed valences.  Molecular orbital theory which allows delocalized bonds over multiple atoms can explain the valences without need for "outer $d$" irbitals.

As I already mentioned in my comment, phosphines such as $\ce{PEt3}$ can bind to transition metals with partially filled d orbitals as a σ-donor as well as a π-acceptor.
The σ interaction is simply a donation of electrons from the lone pair on phosphorus to empty metal d orbitals.

The π interaction is backdonation of electrons from filled metal d orbitals to an empty orbital on the phosphine. Traditionally this was thought to be the 3d orbitals on phosphorus, hence "dπ–dπ". However, as Martin pointed out in the comments this is a rather outdated mode of thinking. The current accepted view is that the acceptor orbitals are $\ce{P-R}$ σ* orbitals (to be precise, a linear combination of them that has appropriate symmetry to overlap with metal d orbitals).


text:
Reactivity of magnetized materials

Does the magnetisation process affects the reactivity of a magnetic material?
Does the electronegativity change if a metal is magnetised?

These are two entirely different realms.  With good precision, you may consider any influence nonexistent altogether.

Magnetization as a process condition (during synthesis procedure) can affect the crystal structure and nanoparticle size and morphology which in turn can affect accessibility of reactive sites. See the following papers:
Hu, Dawei, Yanmin Wang, and Qi Song. "Weakly magnetic field-assisted synthesis of magnetite nano-particles in oxidative co-precipitation." Particuology 7.5 (2009): 363-367.
Xiao, W., et al. "Magnetic-field-assisted synthesis of magnetite nanoparticles via thermal decomposition and their hyperthermia properties." CrystEngComm 17.19 (2015): 3652-3658.
Li, Yang, et al. "Fe 3 O 4@ PSC nanoparticle clusters with enhanced magnetic properties prepared by alternating-current magnetic field assisted co-precipitation." Colloids and Surfaces A: Physicochemical and Engineering Aspects 520 (2017): 348-354.

+1 for this. I was guessing that there must be some weird material reacted in some weird conditions that would show an effect. // Nanoparticles in general have weird behavior. Too big to be a "free" atom or molecule and too small to be the bulk material.


text:
Is a buffer solution really inactive beyond a range of pKa±1?
I have learnt from several books that the range of a buffer solution is always from $\ce{pH = pKa - 1}$ to $\ce{pH = pKa + 1}$. But is there any solid reason to this and any exception?

Did you derived that expression for buffer solution? Then you will find some reason…

I have read so in several books.

Similar or possibly duplicate questions here, here, and here. The best answer is probably at that second link.

This question is more quantitative than the others, though -- so there may be opportunity for new content in an answer here.

The 10:1 ratio is really just something convenient for the chemists.  The buffer still works outside this range.  True, its effectiveness against base addition decreases with the amount of available acid at high pH, and vice versa for acid addition at low pH.  But if enough capacity remains, you can still live with the buffer even outside the 10:1 range.
Er, actually you do.  Our main blood buffer, based on dissolved carbon dioxide as the acid and bicarbonate as the base, has about 20 times as much bicarbonate as carbon dioxide.  Factors other than what a laboratory chemist calls "optimal" should be given their due.  Metabolic processes commonly produce (weak) acids, and the solubility of carbon dioxide is more limited than that of bicarbonate salts.  Both of these factors make a base-rich buffer desirable.

What I think Fawad was asking hints you can derive the result of +-1 pH around the pKa yourself from the Henderson-Hasselbach (H-H) equation.  BY that equation, the ratio of the anion form of the buffer to the acid salt form of the buffer, is plugged into the equation and solves as pKa+1  if the anion to salt ratio is 10:1, because the log(10)=1. IF the ratio were 1:10, then log(0.1)=-1 giving a pH of pKa-1.     SO that's where the idea that buffers work best in a range +-1 around pKa of the buffer.
PRACTICALLY when the ratio of the buffer's two forms of anion and acid salt are close to 1:1 (e g. 50% each form)  the buffer chemical equilibrium easily absorbs added acid or base by absorbing the donated protons or hydroxyls.  In the H-H equation that solves to a pH equal to the buffer's pKa ,  since log(1)=0.
SO close to the pKa the buffer as its strongest buffering.  PH of +-1 is outer limit for efficient buffering based on the implied ratio of 10:1 FOR the buffer components.  It is theoretical and approximate though.  You should try to plot titration curves of your buffer versus added acid or base.
IF that is not deep enough answer for you, read further into Henderson-Hasselbach, buffering prediction, and titration curves of buffer systems. You might be interested to read about Dr. Good' s buffers too.

Why should it be particularly 10:1, is there any reason for that? - This was my question.


text:
how do protons in the nucleus of an atom overcome repuulsions
Why don't protons present in the nucleus of an atom repel each other being  likely charged  ?

Why, they surely repel, and that with great force. But they also attract.

The repulsive forces that exist between two like charges (protons here) are called electrostatic forces. These are strong, no doubt, when you consider a mass as small as the nucleus.
However, you are missing one more point here. The repulsive forces between two protons in a nuclei are "overpowered" by the attraction that exists between them due to Nuclear forces. These forces only exist between protons when they are present in the nucleus. FREE protons never attract each other as then nuclear forces do not come into play.
     These nuclear forces are said to be short range, although there is mo mathematical variation derived yet. Still, short range means they are VERY VERY strong over short distances, typical of the distance between two protons in a nucleus. So repuslive electrostatic forces become insignificant.


text:
chemical reaction that causes electrolysis of salt water without adding electric current
I am producing hydrogen using house hold items in a capacitor like arrangement in salt water. I assume this is chemical electrolysis. The apparatus produces gas for many weeks that burns orange. Its easily assembled from stuff you would normally find in the trash. Do I have something of interest, or is this common chemistry?

The capacitor is providing electric charge, right?  Electrolysis should occur.

Although what you've done is really cool, from a purely technical perspective you've essentially re-invented the wheel. This is simple hydrolysis in which you use a capacitor as the source of current rather than a battery or some other power supply. Eventually you have to re-charge the capacitor, though again that's an impressive setup if it really maintains a flame for weeks at a time. Maybe you can market it as a candle of sorts?

OP is not using a capacitor.  He is describing the electrode (capacitor plate like) arrangement of his electrolysis apparatus.  He is likely creating O2, H2 and some Cl2 by electrolysis.  This is well researched and large quantities can be dangerous if exploded.


text:
What other molecules can methane plasma transform into?
Imagine a sealed chamber with electrodes or an RF coil to turn gases inside into a plasma. Into the chamber we will put methane. There would be only methane inside, no other gas would be present.
Then the methane will be turned into plasma by the electric current. Will it turn into other stable molecules? If yes, how many? For example will it only turn into ethane and ethane will not turn into anything, so the total number of hydrocarbon molecule types will be 2,or will it dissociate the methane into individual hydrogen and carbon atoms and from this create every possible hydrocarbon in existence given long enough time?

It will dissociate into atoms and then form pretty much every non-strained hydrocarbon you can imagine. Given the huge excess of hydrogen, though, I expect methane to remain the predominant component.


text:
How does NMR spectrometer determine area under each peak?
I am confused about exactly how does NMR machine figure our area under each peak, which then tells us how much protons present in the molecule. Does it have to do something with radio waves being emiitted after proton relaxes? Thanks.

It integrates.

Can u explain in terms of the energy absorbed or given of? It must integrate through some information known.

You got a curve intensity vs. frequency and you can just integrate that.

The NMR integrates the electrical current under the peak.

It's certainly not current. You have a free induction decay plot. The more protons you have of a certain type, the more they contribute to the intensity of that frequency. Once you Fourier transform the signal, you get a stronger intensity for those frequencies.

And how exactly  do you go from more intensity of frequency to having more protons? That was what I was trying to get at from my question.

@Zhe - Back when I was doing NMR the data came out on a strip chart recorder. The integration was literally done by integrating the electrical current under the peak.

@TLo You basically shock the protons into an excited state that causes them to radiate EM radiation over time. The more protons of a given kind, the more energy they can radiate...

@MaxW Ah, sorry. Old school. I've only ever used a continuous scan NMR once in 2000. 60MHz. Didn't even have to cool the magnet...

The modern basic FT 1H NMR experiment involves irradiating and simultaneously exciting a wide bandwidth of proton frequencies. That is to say, all spin populations of the proton nuclei have their spin state reversed with a carefully calibrated Rf pulse that delivers the correct amount of power for the correct duration. As each spin relaxes back to equilibrium, it emits a photon of Rf energy. So, yes your assumption is pretty well correct. This energy packet has a specific frequency (a product of the chemical shift), and a specific intensity - it is a quantum packet of energy. If two nuclei of the same frequency each emit a photon of energy, the intensity of that frequency detected will be twice as large a single nucleus emitting a photon of energy.
Across the entire frequency range, the data collected in the free induction decay is the sum product of all frequencies,  and their intensities. The Fourier transform function then converts this time domain data into the preferred frequency domain spectrum.


text:
How to go from, Iron(II) Sulfate (anhydrous, FeSO4) to, Ferrous Sulfate Heptahydrate (FeSO4.7H2O)
In the lab I need some ferrous sulfate heptahydrate ($\ce{FeSO4.7H2O}$). We have some old anhydrous $\ce{FeSO4}$ from several years ago.
It seems that at normal humidity, $\ce{FeSO4}$ will hydrate to the heptahydrate on its own after a few days at most.
Is this correct? Can I "hydrate" my $\ce{FeSO4}$ just by leaving it out overnight?
Thank you.

I suppose you need that heptahydrate for a more meaningful purpose than making a water solution out of it, right?

According to this Wikipedia article:

Iron(II) sulfate (British English: iron(II) sulphate) or ferrous
  sulfate denotes a range of salts with the formula $\ce{FeSO4·xH2O}$.
  These compounds exist most commonly as the heptahydrate (x = 7) but
  are known for several values of x.

While the anhydrous form appears as white crystals, the article above continues:

Iron(II) sulfate can be found in various states of hydration, and
  several of these forms exist in nature.
  $\ce{FeSO4·H2O}$ (mineral: Szomolnokite, relatively rare)
  $\ce{FeSO4·4H2O}$ (mineral: Rozenite, white, relatively common, may be  dehydratation product of melanterite)
  $\ce{FeSO4·5H2O}$ (mineral: Siderotil, relatively rare)
  $\ce{FeSO4·6H2O}$ (mineral: Ferrohexahydrite, relatively rare)
  $\ce{FeSO4·7H2O}$ (mineral: Melanterite, blue-green, relatively common)

This all suggests that simply letting the anhydrous form sit in air overnight is not a reasonable procedure for the quantitative production of the heptahydrate form.
One qualitative hint that you could get regarding the degree of hydration of your old anhydrous material is the color.  While both the anhydrous and quadrahydrate are white, the heptahydrate is a bluish green color.
You could also measure the melting point of the old material; it should be $\mathrm{60-64^oC}$ if it has hydrated all the way to the heptahydrate.
If you really want to try to convert the old anhydrous material to the heptahydrate, you could expose it to a humidified environment overnight.  Then observer any color changes and measure the melting point. You could even weigh a sample of it then heat to ~$\mathrm{110^oC}$ for a few hours then re-weigh it and calculate the degree of hydration.
That's all a bit crude, but you could try it. I recommend just purchasing some of the heptahydrate compound if possible.


text:
Does higher electron density mean better transport?
In a nano device, do areas with higher electron density necessarily contribute more the to the overall transport compared to areas with lower electron density?

No, insulators also ave fairly high electron density


text:
Possible Water peak in PMMA nmr spectrum

solvent: $\ce{CDCl3}$
I'm unsure as to whether the large sharp peak at ~1.5 on my NMR spectra is water, I was under the impression water gave a broad peak in proton NMR, but may be confusing it with something else.
Can anyone confirm is deny?

Your peak at δ1.6 in $\ce{CDCl3}$ is indeed water. This is typical of the appearance of water in $\ce{CDCl3}$ at the sample concentration you present. Water can sometimes be broad; this depends on the overall sample concentration, acidity of the solution and relative concentration of other labile protons. This looks like a fairly new bottle of $\ce{CDCl3}$.
You will notice that the residual chloroform peak at δ7.26 has $\ce{^{13}C}$ satellites; intensity approximately 0.55% of the parent peak, with a splitting of ~210Hz (so about 105Hz either side of the parent $\ce{^{12}C}$ peak). You will then also see that the water peak, although of similar intensity, does not have any carbon satellites. You could also confirm this by adding a small amount of $\ce{D2O}$, and observing a change in chemical shift for this peak.


text:
Will oxalic acid react with ethanol?
Will oxalic acid, in a saturated solution in ethanol, be able to undergo an esterification reaction with the ethanol? This is assuming no acidifying agents (or anything else, for that matter) are added, and room temperature is also assumed.
I tried to find an equilibrium constant to predict the extent of such a reaction, but I couldn't seem to find anything. This suggests either that the reaction does not take place, or that it is so obscure that nobody has bothered to examine it.

I heard that someone put a lot of effort in finding kinetics of such reaction, but in case of acetic acid. Oxalic is stronger so reaction should be faster, but still not very fast.

Here is an article where they developed a catalyst for the esterification, suggesting it's probably not very fast without a catalysts...for whatever that's worth ;)

it does in presences of dehydrating agent (an agent help in pull water molecule) such as conc. sulphouric acid


text:
At what temperature does tautomerization stop?
The enol and keto forms are said to be tautomers of each other, and the interconversion of the two forms involves the movement of an alpha hydrogen and the shifting of bonding electrons.
At what temperature does tautomerization stop? (E.g I can obtain both the enol and keto form without them interconverting)
To make this simple, let's just assume the molecule of choice is: acetaldehyde
In addition there is no tunneling of the two tautomers. Lastly it is known the the two tautomers of acetaldehyde are stable and do not convert at 11 K with argon matrix trapping. What's the highest temperature that do not interconvert?

Tautomerisation, as an observable process by NMR, is entirely system dependent, and there are many cases where non-exchanging or very slow-exchanging systems are observed at room temperature and higher. As always, equilibrium is both sterically and electronically controlled. Some systems require very low temperatures (liquid nitrogen and liquid helium temperatures) to slow the exchange processes, and of course, going to higher magnetic fields will also improve the case of observing individual exchanging species.
Acetaldehyde is a very small molecule; this would require extremely low temperatures; as you have said 11K allows for observation without interconversion.
A lot of this sounds like a homework question .... you might want to be more specific about what your own research is contributing towards answering your own questions.

Well absolutely there are cases where non-exchanging or very slow-exchanging systems are observed at room temperature and higher. Take for example phenol as an example. What I wanted to know is at what temperature does it truly stop. As for why I mentioned acetaldehyde is because it is the most simplest case. To be honest it could have been any aliphatic aldehyde or ketone. It is easy to say make a generalization and say "at low temperatures", but I already knew that. What makes you think this is a homework answer, it is just pure curiosity.


text:
How to determine the specific heat of aluminum?
I have a question from my AP preparation book that I got wrong and do not understand the explanation:

A student designs an experiment to determine the specific heat of aluminum. The student heats a piece of aluminum with a mass of $\pu{5.86 g}$ to various temperatures, then drops it into a calorimeter containing $\pu{25 mL}$ of water. The following data is gathered during one of the trials:
Initial Temperature of $\ce{Al}$: 109.1
  Initial Temperature of $\ce{H2O}$: 23.2
  Final Temperature of $\ce{Al + H2O}$: 26.8

I correctly calculated the heat gained by the water ($\pu{376.2 J}$), and the specific heat of aluminum ($\pu{0.78 J/g*C}$). However, how do I calculate the enthalpy change for the cooling of aluminum in water? The book's procedure is as follows:
$$\frac{\pu{5.86g}~\ce{Al}}{\pu{26.98g}~\ce{Al}} = \pu{0.217 mol}~\ce{Al}$$
$$\frac{\pu{376 J}}{\pu{0.217 mol}} = \pu{1730 J//mol} = \pu{1.73 kJ//mol}$$
I see that one may calculate specific heat by dividing the heat gained by water by moles of aluminum and converting to kilo-joules per mole in this scenario. However, why is that? What is the reasoning behind this procedure? How would you phrase this procedure to apply to this general type of question?

This is simply conservation of energy. Assume you controlled for loses of heat to, say, the container and atmosphere; using a styrofoam cup and cover is a good way to do this is a student-lab situation. Then, the total heat lost by the metal is equal to the total heat gained by the water.
The last line of calculation in your post is using the total heat lost by the metal, same as that gained by the water, and finding how much to cool/heat 1 kg. This amount applies to the entire temperature difference through which the metal cooled. Convert to J/g, then calculate how many J/g per each degree.

If the water and aluminum are regarded as incompressible, then the internal energy is a function only of temperature.  In addition, the heat capacity at constant pressure is equal to the heat capacity at constant volume.  If the pressure is constant, then the change in enthalpy will be equal to the change in internal energy.  So the change in enthalpy of the aluminum is equal to its heat capacity times its temperature change.  The same goes for the water.  So the total change in internal energy and/or enthalpy for the combination of water and aluminum is zero.


text:
How many decimal places should a volume measurement in a beaker include?
I have a beaker which goes up in $0.5~\mathrm{mL}$ increments and my teacher told me that my measurement $(15.50~\mathrm{mL})$ was incorrect, and instead the answer was $15.5~\mathrm{mL}$. I don't understand. I know that the liquid is between $15.5~\mathrm{mL}$ and $16~\mathrm{mL}$. My estimated digit is $0$ (it looks like it lies on the line), so my answer is $15.50~\mathrm{mL}$. How is it $15.5~\mathrm{mL}$?

When you say $15.50$, it assumes that you have the measurement correct up to 2 decimals. Since you're measuring equipment can do only one decimal, you should report it likewise. So, though numerically $15.50 = 15.5$, they mean different in terms of accuracy of reporting.

Technically your teacher is correct. Since your beaker only has increments of 0.5 mL, you can only report a volume measurement to the nearest 0.5 mL. However, what you did is also a common practice and appropriate in some cases. Your eyeball can certainly interpolate to within a couple tenths or so by estimating the level of the liquid as you described. You can do this as one element of a calculation for example, but in the end you have to round your final result to the level of accuracy that your beaker measures to, since you don't actually know what the uncertainty of your eyeball is ;)

@airhuff - That just isn't true. If you have a burette marked to tenths of a ml  then as you pointed out you can interpolate to 0.01 ml. The error isn't +/- 0.01 but probably something more like +/- 0.02. But that is a lot better precision and accuracy than rounding off to 0.1 ml. // That being said, teachers make the rules in their classes.

Now that I've thought about it -- are you sure that your glassware is a beaker? I don't think I've ever seen a beaker graduated that finely. Can you link to a picture of something similar?

If it's actually not a beaker but a graduated cylinder or other volumetric instrument, see also: Using sigfigs when measuring with an instrument with marks other than powers of ten?

@hBy2Py If that thing really has graduation divisions of 0.5 ml, it's probably a 25 ml graduated cylinder and not a beaker.

@Loong Exactly my suspicion. The 'beak' of a grad cylinder does look a lot like that of a beaker.

It may have to do with the fact that it's a beaker, rather than some other kind of glassware.
First, beakers are poor vessels for volumetric measurements due to their large diameter/height ratios.  You have to add a lot of liquid, relatively speaking, in order to increase the height of the fill level by (say) a millimeter, and that means the uncertainty in the fill volume is large.
As well, beakers are usually not graduated precisely. Thus, even if you very very carefully fill the beaker to exactly the $15.5~\mathrm{mL}$ line with liquid, there's a very real chance that the line doesn't actually correspond exactly to a $\mathit{15.5~mL}$ fill volume.
So, unlike with a piece of glassware specifically made for accurate volumetric measurements, such as the burette mentioned by MaxW in a comment, it is likely not appropriate to assign more significant figures to a beaker measurement than are "available" in the graduation increment.


text:
Enantiomer or not?
My professor said that this compound is an enantiomer but I think it has a plane of symmetry and has no chiral carbons. Am I missing something ?

"it has a plane of symmetry and has no chiral carbons"  you're exactly right

The necessary and sufficient condition for enantiomerism is the chirality or dissymmetry of the molecule as a whole. Presence of chiral carbons isnt a guaranty that the molecule will exhibit enantiomerism. The image must not be super-imposable. This molecule shouldn't show enantiomerism because the image of the molecule looks superimposable to me.

Related (if you want to name this compound): What does lowercase r-s notation mean?

this compound shows enantiomerism or this compound is an enantiomer(of what?) ?

The comments are right.  The molecule shown is not chiral and will not show enantiomerism.
But still, remember that an alkane ring, unlike an aromatic ring, is generally not planar and take that into account when evaluating molecules.  Thus chloromethylbenzenes with a planar ring are never chiral, but chloromethylcyclohexanes will be chiral if the chlorine and methyl group are not on opposite ring carbons.

Since a cyclohexane has a bond angle which is quite close to tetrahedral bond angle, the molecule is not planar. Similarily, p-chlorotoluene would also not be planar.
If you observe the conformers of cyclohexane, you would realize that the chair and boat conformers are non chiral due to chair conformer having an alternating axis of symmetry and the boat conformer having a center of symmetry. But the twist conformer is chiral.
Similarily, if we substituted any two of the hydrogens with a methyl and chlorine group, we would get many more conformers which would be chiral since the new molecules would not have the plane of symmetry or the alternating axis of symmetry which the orginal molecule had. Therefore, I would say that the molecule is chiral though you should confirm it with your professor since drawing the conformers of a molecule is difficult and there can be exceptions.


text:
Why do we still use the 'P' subscript in a thermodynamic derivative such as (dH/dT)P?
Consider the following thermodynamic derivative:
$$
\left(\frac{\partial H}{\partial T}\right)_P
$$
Taking the partial derivative w.r.t. $T$ means that all other variables are held constant. Why do we still write it with a subscript like $P$ or $V$?

I made some pretty substantial edits to your post -- if I've misrepresented anything you wrote, or if you don't like anything I changed, please feel free to edit it back, or to roll back the edits entirely.

In particular, the use of MathJax in titles is discouraged (here on Chem.SE, at least), as it makes it hard to read search results on, e.g., Google.

Long story short, all other variables aren't and can't be held constant.

Because we distinguish f(p,..) and f(V, ...) functions

its a maths thing not related to thermodynamics per se; if you want to differentiate a function of several variables  $f(x, y, z)$ wrt. x then then y, z need to be held constant, hence the curly $\partial$ to indicate a partial derivative and subscripts to indicate what is held constant, $(\partial f / \partial x)_{y,z}$

In thermodynamics, the number of degrees of freedom available to describe a system is given by Gibbs' phase rule:
$$
F = C - P + 2
$$
Here, $C$ is the number of components in the system, $P$ is the number of phases in the system, and $F$ is the number of resulting degrees of freedom.
Consider a pure substance in a closed system.  One can name a wide variety of thermodynamic parameters to describe its state: temperature, pressure, specific volume, internal energy, enthalpy, entropy, etc. However, Gibbs' phase rule says that we can only independently specify a subset of these parameters before we run out of degrees of freedom:
$$
F = 1 - 1 + 2 = 2
$$
So, in any mathematical description of (in your example) the enthalpy, $H$, it can only properly be a function of two independent variables:
$$
H = H\left(X_1, X_2\right)
$$
But: which two variables?
Simply writing the partial derivative $\partial H \over \partial T$ indicates that temperature is to be one of the two available degrees of freedom:
$$
H = H\left(T, X_2\right)
$$
Now, though – what is $X_2$? Well, including the subscript on the differential is a way of explicitly indicating what that other independent variable is:
$$
\left({\partial H \over \partial T}\right)_P \quad \Longrightarrow
\quad H = H\left(T, P\right)
$$
As porphyrin rightly notes in a comment, this applies to multivariate calculus generally, not just the calculus of thermodynamics.

Ok but what if something is a function of more than two variables ?

@mathemather For any given first derivative in any given system, if you have $F$ degrees of freedom, you should specify $F-1$ independent variables as subscripts to the derivative.

It just means one is holding the experimental pressure constant when determining Cp.


text:
Entropy and Enthalpy
∆S (Entropy change) is given by ∆H/T (∆H = Enthalpy change , T= absolute temperature). Then how can the Entropy change for a reaction be negative if the Enthalpy change is positive, and vice versa?

Who says the enthalpy change is always positive?  And what about the equation $\Delta G=\Delta H-T\Delta S$?  What if $\Delta G$ is not zero between the initial state of the reactants and the final state of the products?

You are looking at two different entropy changes. The one computed directly from the enthalpy change is the entropy change of the universe. The entropy change of the system is measured by computing the component due to heat transfer. But internal changes in the density of states can also result in a change in entropy, and this can take place without any transfer of heat.

What do you mean the same way? You just said that they're two different things, surely they're not measured the same way?

"Same way" means $dS = dq/T$ @RobChem

The entropy change you quote is not in general true, but applies only to specific circumstances.  Such an instance is the equilibrium between a solid and liquid or between a liquid and its vapour at a temperature T and pressure P.
In Chemistry it is usual to conduct experiments at constant pressure and temperature, rather than at constant volume, and so the place to start with many thermodynamic calculations is the (Gibbs) free energy, $\Delta G=\Delta H - T\Delta S$.
As the two phases are in equilibrium the free energy change upon vaporisation is $\Delta G_{vap}=\Delta H_{vap}-T\Delta S_{vap} = 0$ and thus
$\Delta S_{vap}=\Delta H_{vap}/T$. Only in the special case when $\Delta G = 0$ does this apply.


text:
Phase description for a substance dissolved in a solvent other than water?
I'm trying to write an equation for which I have $\ce{CuI}$ dissolved in acetonitrile. Usually if you have a salt dissolved in water you can denote that using $\ce{(aq)}$, but is there a notation for substances dissolved in a solvent other than water?

I'd write $\ce{CuI(s) ->[acetonitrile] Cu^+(sol) + I^-(sol)}$

@MaxW That is, if it really is mostly ionized in the said solvent.

In terms of expressing [partition coefficient equations], this link (Wikipedia citing IUPAC, sort of) suggests the following to distinguish an aqueous phase from an organic phase: $\mathrm{(K_D)A = [A]{org} / [A]_{aq}}$.

IUPAC “Green Book” recommends to use $\mathrm{sln}$ for denoting a solution in general [1, p. 54], referring to earlier Recommendations 1981. Appendix No. IV to Manual of Symbols and Terminology for Physicochemical Quantities and Units [2, pp. 1240–1242].
This has been extensively covered in the following posts:

What is the standard way to denote physical states in a chemical reaction? (Meta)
Phase abbreviations for non-aqueous solutions

On the contrary, abbreviation $\mathrm{sol}$ is used to denote a process (typically a sub- or superscript to a symbol for a thermodynamic quantity), not state of matter [1, pp. 59–60]:

2.11.1 Other symbols and conventions in chemical thermodynamics
(i) Symbols used as subscripts to denote a physical chemical process or reaction
These symbols should be printed in Roman (upright) type, without a full stop (period).
$$
\begin{array}{ll}
\ldots \\
\text{solution (of solute in solvent)} & \mathrm{sol} \\
\ldots
\end{array}
$$

Another reason not to use $\mathrm{sol}$ for denoting a state of aggregation is its ambiguity: $\mathrm{sol}$ is often used as an abbreviation for “solid”, as listed in Appendix 10-2 Abbreviations, Acronyms, and Symbols in ACS Style Guide [3, p. 197]

$$
\begin{array}{ll}
\ldots \\
\text{sol} & \text{solid} \\
\text{soln} & \text{solution} \\
\ldots
\end{array}
$$

Also, specialized literature on electrochemistry of non-aqueous solutions uses $\mathrm{sv}$ to denote solvation process separately, see e.g. [4, p. 27]:

$ΔH^\circ_\mathrm{sv},$ $ΔS^\circ_\mathrm{sv},$ $ΔG^\circ_\mathrm{sv}$: Enthalpy, entropy, and Gibbs energy of solvation of the electrolyte

With this obligatory standardization part, the simplest reaction can be written as
$$\ce{CuI(s) ->[CH3CN] Cu^+(sln) + I^-(sln)}$$
However, copper(I) is extremely stable in acetonitrile due to solvation [4, p. 33]:

Interactions by Back-Donation from $\mathrm{d^{10}}$-Cation to Solvent Molecules
Acetonitrile (AN) has relatively small DN and usually solvates rather weakly to
metal ions. However, it solvates very strongly to $\ce{Cu+},$ $\ce{Ag+}$ and $\ce{Au+},$ which are univalent $\mathrm{d^{10}}$-metal ions.
This is because these metal ions have an ability to back-donate their electrons into a π*-antibonding orbital of the $\ce{CN}$ group of AN, as shown by $\ce{CH3C#N\overleftarrow{\ce{:\bond{->}A}}g+}.$
As a result, $\ce{Cu+}$ and $\ce{Ag+}$ in AN are stable and not easily reduced to metal, while the weakly solvated $\ce{Cu^2+}$ is very easily reduced to $\ce{Cu+}$ … , making $\ce{Cu^2+}$ in AN a strong oxidizing agent.

and may theoretically be coordinated by up to five acetonitrile ligands [5] with C.N. 4 being the most common unless there are other ligands.
Copper(I) and iodide don't seem to form a complex in acetonitrile [6, p. 194]:

Also in tetraethyl ammonium iodide and thiocyanate solutions, the dissociation is seemingly complete, … while a slight association is indicated for bromide … and, especially, chloride.

To sum it up, upon dissolving of copper(I) iodide in acetnitrile the formation of tetrakis(acetonitrile-N)copper(I) is the most likely scenario so that the final reaction can be written as
$$\ce{CuI(s) + 4 CH3CN(l) ->[CH3CN] [Cu(CH3CN)4]^+(sln) + I^-(sln)}$$
It should be noted that in a presence of better coordinating ligands $\ce{L}$ composition of the complex may, of course, vary $(\ce{[Cu(CH3CN)_xL_y]^z+}).$
Also, copper(I) iodide may form various coordination polymers based on $\ce{(CuI)_n}$ cluster core [7, 8, 9].
References

IUPAC “Green Book” Quantities, Units, and Symbols in Physical Chemistry, 3rd ed.; Cohen, R. E., Mills, I., Eds.; IUPAC Recommendations; RSC Pub: Cambridge, UK, 2007. ISBN 978-0-85404-433-7.
Cox, J. D. Notation for States and Processes, Significance of the Word Standard in Chemical Thermodynamics, and Remarks on Commonly Tabulated Forms of Thermodynamic Functions. Pure and Applied Chemistry 1982, 54 (6), 1239–1250. https://doi.org/10/d783th. (Free Access)
The ACS Style Guide: Effective Communication of Scientific Information, 3rd ed.; Coghill, A. M., Garson, L. R., Eds.; American Chemical Society; Oxford University Press: Washington, DC; Oxford; New York, 2006. ISBN 978-0-8412-3999-9.
Izutsu, K. Electrochemistry in Nonaqueous Solutions; John Wiley & Sons, 2005. ISBN 978-3-527-60065-6.
Vitale, G.; Valina, A. B.; Huang, H.; Amunugama, R.; Rodgers, M. T. Solvation of Copper Ions by Acetonitrile. Structures and Sequential Binding Energies of $\ce{Cu + (CH3CN)_x},$ $x = 1−5,$ from Collision-Induced Dissociation and Theoretical Studies. J. Phys. Chem. A 2001, 105 (50), 11351–11364. https://doi.org/10/ckvsxw.
Ahrland, S.; Nilsson, K.; Tagesson, B.; Haaland, A.; Schilling, B. E. R.; Seip, R.; Taugbøl, K. Thermodynamics of the Copper(I) Halide and Thiocyanate Complex Formation in Acetonitrile. Acta Chem. Scand. 1983, 37a, 193–201. https://doi.org/10/bqvn5f.
Knorr, M.; Guyon, F.; Khatyr, A.; Strohmann, C.; Allain, M.; Aly, S. M.; Lapprand, A.; Fortin, D.; Harvey, P. D. Construction of $\ce{(CuX)_{2n}}$  Cluster-Containing ($\ce{X} = \ce{Br}, \ce{I}$; $n = 1, 2$) Coordination Polymers Assembled by Dithioethers $\ce{ArS(CH2)_mSAr}$ ($\ce{Ar} = \ce{Ph},$ $\ce{\textit{p-}Tol}$; $m = 3, 5$): Effect of the Spacer Length, Aryl Group, and Metal-to-Ligand Ratio on the Dimensionality, Cluster Nuclearity, and the Luminescence Properties of the Metal–Organic Frameworks. Inorg. Chem. 2012, 51 (18), 9917–9934. https://doi.org/10/f37kdx.
Knorr, M.; Khatyr, A.; Dini Aleo, A.; El Yaagoubi, A.; Strohmann, C.; Kubicki, M. M.; Rousselin, Y.; Aly, S. M.; Fortin, D.; Lapprand, A.; et al. Copper(I) Halides $(\ce{X} = \ce{Br}, \ce{I})$ Coordinated to Bis(Arylthio)Methane Ligands: Aryl Substitution and Halide Effects on the Dimensionality, Cluster Size, and Luminescence Properties of the Coordination Polymers. Crystal Growth & Design 2014, 14 (11), 5373–5387. https://doi.org/10/f6pnff.
Knorr, M.; Bonnot, A.; Lapprand, A.; Khatyr, A.; Strohmann, C.; Kubicki, M. M.; Rousselin, Y.; Harvey, P. D. Reactivity of $\ce{CuI}$ and $\ce{CuBr}$ toward Dialkyl Sulfides RSR: From Discrete Molecular $\ce{Cu4I4S4}$ and $\ce{Cu8I8S6}$ Clusters to Luminescent Copper(I) Coordination Polymers. Inorg. Chem. 2015, 54 (8), 4076–4093. https://doi.org/10/f7cjk2.


text:
Calculating the pH of the endpoint in a titration of weak acid and strong base
Show that the pH at the endpoint in the titration of $\mathrm{0.100\ mol\ dm^{-3}}$ ethanoic acid with $\mathrm{0.100\ mol\ dm^{-3}}$ sodium hydroxide is 8.72. The dissociation constant Ka of ethanoic acid is $\mathrm{1.80\ x\ 10^{-5}\ mol\ dm^{-3}}$.
I managed to obtain the value of Kb using the formula (Ka x Kb = Kw = $\mathrm{10^{-14}}$).
I worked out the chemical equation at the endpoint which is:
$$\ce{CH3COO- + H2O <--> CH3COOH + OH- }$$
However, I am unsure as to what the concentration of $\ce{CH3COO-}$ should be.
I understand that if I find [$\ce{OH-}$] then I would be able to find pOH and by extension pH.

Would it be correct to assume that the concentration of CH3COO- to be the concentration of ethanoic acid that was present at the start of the titration (0.100 mol dm^-3), given that at the equivalence point, all CH3COOH had reacted to form a solution containing only the conjugate base CH3COO- ?

I also have a major problem with the wording of the question. The endpoint is when the end of the titration is detected like when phenolphthalein has changed pink. The equivalence point  is when you have added as many moles of base as there were moles of acid in the solution. So technically the problem as stated is unanswerable. The point of course that we want to make pH(endpoint) $\approx$ pH(equivalence point ) by selecting a good indicator for the titration.

Would it be correct to assume that the concentration of $\ce{CH3COO-}$ to be the concentration of ethanoic acid that was present at the start of the titration ($\mathrm{0.100~mol~dm^{-3}}$), given that at the equivalence point, all $\ce{CH3COOH}$ had reacted to form a solution containing only the conjugate base $\ce{CH3COO-}$?

You can easily rearrange the $\text{K}_\text{a}$ equation to solve for the information that you need to answer that question.
$$\text{K}_\text{a} =\frac{\ce{[H^+][A^-]}}{\ce{[HA]}} $$
or
$$\frac{\ce{[HA]}}{\ce{[A^-]}} = \frac{\ce{[H^+]}}{\text{K}_\text{a}} = 1.06\times10^{-4}$$
Since $1.06\times10^{-4}$ is insignificant compared to 0.100 you can assume that essentially all the HA is converted to $\ce{A^-}$.
NOTE: This does not mean that $\ce{[HA] = 0}$

We can't assume [HA] = 0 because of the hydrolysis that A- undergoes in solution to form back HA. Is that correct?

Exact right. Even if you add pure A- into the solution by dissolving a salt (e.g sodium acetate), then A- undergoes hydrolysis in solution to form some HA (e.g. for sodium acetate you'd get acetic acid).

Realize that this is about significant figures. So if you add 0.100 moles of sodium acetate to a liter of distilled water you can assume that the acetate concentration is 0.100 because little acetic acid is formed. But in math class this wouldn't work because 0.1 means 0.1000000...


text:
Why does the Michaelis-Menten constant decrease in the presence of non-competitive inhibitor?
Non-competitive inhibitors don't bind to the active sites of enzymes. However, in the presence of non-competitive inhibitors, the Michaelis-Menten constant is found to decrease. Why is this so?

The premise of the question is wrong.  In MM kinetics, non-competitive inhibition does not change $K_M$.


text:
Bravais Crystal Lattices and Voids
What is the difference between Hexagonal Close Packing (HCP) structures and the hexagonal crystal Bravais lattice? I've looked up some sites but can't seem to understand them. It would help if you backed up the answer with your specific explanation of Bravais lattices and unit cells.
Edit:
I thought the difference was that the lattice consisted of many unit cells and the unit cell packing could be like hexagonal, cubic or simple. But then I looked at the stoichiometric defects which talked about molecules being in interstitial spaces. But in the unit cell, the spaces are already filled. So how is that possible.

Here are some important definitions:
A space lattice provides the framework with reference to which a crystal structure can be described. A lattice is different from crystal. In fact, a lattice gives rise to crystal when lattice points are replaced by atoms, ions, or molecules.
A 2-D or 3-D lattice is a regular arrangements of points. In order to specify it completely, only a small part of the lattice is described. This figure is known as a unit cell.
In 2-D, there are 5 possible lattices namely, square, rectangle, hexagonal, parallelogram and rhombic.
In 3-D, there are 14 possible lattices, and these lattices are called Bravais lattices (after the French mathematician who first described them) like cubic primitive, hexagonal primitve, etc.
For example,
In a cubic system there are 3 possible Bravais lattices possible namely, primitive, body centered and face centered.
Similarly in hexagonal crystal system there is only one Bravais lattice viz, Primitive.
Crystallographers have been able to divide 32 point groups and 14 space lattices into seven crystal systems and 14 Bravais lattices. Remember that the primitive cells of any two crystals are not the same. Primitive unit cells mean that the lattice point must be at the corners of the unit cell.

HCP is ABA-ABA arrangement of layers in which tetrahedral void of second layer are covered by the third layer.

So each lattice site doesn't represent a site that could be replaced by a unit cell but actually by the constituents(atoms/ions) of the unit cell?

Yes, Unit cell contains many lattice points in/on them and the constituents are placed at those lattice points. Lattice is like a framework, a blueprint but without the constituents. A crystal is the building defined by that blueprint but with the constituents at those lattice points..

Also, what do you mean by point groups?

Search Google for crystallographic point group..You'll find your answer there..

2 brief observations about the answer by Mitchell.

a hexagonal lattice does not exist in 2-dimensions.

By definition,A lattice is an infinite array of points in space, in which each point has identical surroundings to all others. For example consider a 2d cubic lattice; now, for example, choose one of the lattice point (we call it point A) and connect it with an arrow with the point just above and with another arrow with the point on its right. Now move these two arrows as a solid object to another point of the lattice (we call it point B); you will see that the arrow will again connect this other lattice point with other two lattice points, one just above, the other at its right. You find the same thing for both point A and point B, whatever choice you make.
You can do the same job with a hexagonal arrangment of points. Choose again an arbitrary point A in the hexagonal arrangement and plot the two arrows connecting point A to two first neighbouring point. Then again move the two arrows as a solid object from point A to one of the 3 first neighbouring three points. You will see that now arrows will end where no point is present in the space. This is because the 2D hexagonal arrangment of points is actually constituted by two different sublattices, whose symmetries are trigonal.

the different crystal systems are defined by the symmetry operations that are present, not by their metric.

The group of the symmetry operations define the symmetry along the different main crystallographic directions. For example, in the cubic system we have a=b=c and alfa=beta=gamma=90 metric because the symmetry operations that are present impose this.
The tetragonal symmetry simply dictates a=b and alfa=beta=gamma=90; there's no METRIC restriction on the c-axis. The SYMMETRY along the c-axis is different from that along a and b, but in principle it is possible to find a crystal structure that has a tetragonal structure (= symmetry), but a cubic metric, that is a=b=c (and it is still tetragonal!). there are several compounds where this is observed. In conclusion, inequalities strictly hold on symmetry, not on the metric of the crystal structure.

What is the difference between Hexagonal Close Packing (HCP) structures and the hexagonal crystal Bravais lattice?

HCP structures have a lattice that is classified as hexagonal Bravais lattice. There are many other structures that have a hexagonal Bravais lattice but are not HCP structures.
The HCP structure contains a single type of atom closely packed in hexagonal layers, just like the face-centered cubic (FCC) pattern. The two differ in what pattern these layers are arranged (HCP shows an ...ABABAB... pattern while FCC shows an ...ABCABCABC... pattern).
There is a 3D browser showing HCP here: http://lampx.tugraz.at/~hadley/ss1/crystalstructure/structures/hcp/hcp.php.