text:
Generating density matrix of a wavefunction from common quantum chemistry software
I am looking for density matrixes for a couple of closed shell and open-shell systems, and I wonder if there is a trivial way to make e.g.  Gaussian just generate an output for me. I may be wrong but most quantum chemistry software seems to generate the density matrix when checking SCF convergence, so it would make sense if there were an option to save it.

I'm sure there is a nicer way to do this, but I haven't found it yet. For Gaussian, IOp(6/8=3) will print the density matrix verbosely in the output file.


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Basicity of nitrogen in amide versus basicity of nitrogen in aniline
I was wondering that how can we compare the basicity of nitrogen in amide to the nitrogen in aniline as in both structures the lone pair on nitrogen is in conjugation as shown below,

Can someone help me?

Which orbital will be lower in energy and hence more easier to approach and break- the C=O π* or C=C π*?

source: Stack Exchange

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Reaction of urea and thiourea with nitrous acid
I was told that urea on reaction with nitrous acid gives nitrogen gas, carbon dioxide and water. While on the other hand, thiourea on reaction with nitrous acid gives $\ce{H+}$, thiocyanate ion and water.
I tried looking for the mechanism and thought that this would be related:

I carried out the above procedure on both the nitrogens of urea and was able to obtain the products of the first reaction enlisted above. Now, my questions are:
• Isn't the nucleophilicity of the lone pairs of urea's nitrogens negligible, and any reaction should proceed via attack by oxygen's lone pairs? (I thought that this was similar to amidines and amides)
• Why are the products of the reactions with urea and thiourea different, as the carbonyl oxygen, or sulphur doesn't seem involved in the reaction according to me
• More importantly, is this even the correct mechanism?

It is similar as in amidines and amides and it doesn't matter. Any attacks on O are unproductive, so effect of conjugation is rather kinetical.

As implied in your link, the reaction of urea and nitrous acid will follow the mechanism you posted, repeating a second time with the nitrogen on the remaining amide. As for why the nitrogen is more nucleophilic, I can speculate, but I don't want to be wrong.
The difference for thiourea is that sulfur can also act as a nucleophile instead of the nitrogen in the thioamide. The product is dependent on the pH. You can read more about it here. Here is the mechanism:

That's a thiourea that I squeezed in on the far right as a product from the strong acid. Sorry for the mess.

What I've  so far is that Thiourea is a stronger nucleophile (Sulfur is less electronegative than oxygen, thus more nucleophilic).  Thiourea reacts at the site of the sulfur whereas urea reacts the the amine sites.

Thanks for the idea, but it would be great help if you could provide a mechanism scheme

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How do flame tests give off color if they only emit the colors when the electrons return to ground state?
Since we are constantly holding the element in the fire and the elements only give off the color when the electrons emit the energy that was supplied and return to ground state, why do we only see the colors when in the fire and don't see the colors when we take the supplied energy ( i.e., the fire) away?

Because in you sample there are enough atoms able to absorb energy and release it as light. It's the relaxation to ground that emits not the ground itself. "The colour is the energy not the ground"

Related: Color imparted to flame and color of coordination compounds

The flame area is a violent place due to high temperature favoring ionization. The color appears due to the dynamic process of the formation of short-lived species formed briefly in the flame. As we are burning hydrocarbons, they get split into unstable radicals easily loosing unpaired electrons. In the area of the flame enriched with electrons metal ions from the probe such as $\ce{Na+}$ are reduced to the metal atoms:
$$\ce{Na+ + e- -> Na^0}$$
which are emitting yellow light in a form of observable doublet of spectral lines ($λ_1 = \pu{589.0 nm}$, $λ_2 = \pu{589.5 nm}$) arising from the electronic transitions $\ce{3s^1 -> 3p^1}$. Note that it is the metal atom, not the ion, that is responsible for the emission of light of the given color.
So, as long as we keep the test probe (metal salt) in the flame, we can observe spectral lines of the given element as both conditions are fulfilled (energy supply (heat) and reducing atmosphere (free electrons from the fuel)). By taking the sample out of the flame, electron transitions are no longer possible as there is neither metal atoms capable of any, nor enough energy to promote the electron to the higher level.

$3s^1$ to $3p^1$ emission?  I thought the energy release was the other way around.

Is reducing atmosphere really a requirement?

@OscarLanzi Yes, you are right, maybe I badly formulated it. I meant that $\ce{3s^1 -> 3p^1}$ transition precedes the energy release which occurs due to $\ce{3p^1 -> 3s^1}$.

@aventurin Not always, I just took sodium salt with sodium ion without any electrons on the outer shell to transfer in the first place, so one needs to reduce it first in order to observe any light emission.

Electrons are pushed into excited states and fall back to the ground state all the time while their atoms are in the flame.  They don't wait in the excited state until they're out of the flame.


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Standard electrode potential
What is meant by the following:

Electrode potential for $\ce{M^3+/M}$

Is it standard oxidation potential (SOP) or reduction potential (SRP)? What does a high positive or negative value for SRP and SOP tells us about the reactivity of the element?

This is standard reduction potential or standard electrode potential $E^\circ$. More positive values refer to the strongly oxidizing agents, and more negative ones – to the strongly reducing agents. In order to convert SOP into SRP, reverse the sign. Examples and quotes from Atkins' Elements of Physical Chemistry [1, p. 209]:

$$
\begin{align}
\ce{F2 + 2 e- &-> 2 F-} \quad &E^\circ = \pu{+2.87 V}\\
\ce{Li + e- &-> Li+} \quad &E^\circ = \pu{-3.05 V}
\end{align}$$
The standard potential, $E$ (Ox/Red), of a couple Ox/Red is then measured by constructing a cell in which the couple of interest forms the right-hand electrode and the standard hydrogen electrode is on the left.
[...]
A brief comment. Standard potentials are also called standard electrode potentials and standard reduction potentials. If in an older source of data you come across a ‘standard oxidation potential’, reverse its sign and use it as a standard
  reduction potential.

References

Atkins, P. W.; De Paula, J. Elements of Physical Chemistry, 5th ed.; Oxford University Press: Oxford ; New York, 2009. ISBN 978-1-4292-1813-9.


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Exact meaning of constant volume heat capacity
From Wikipedia:
$$
\left(\frac{\partial U}{\partial T}\right)_V = \left(\frac{\partial Q}{\partial T}\right)_V = C_V,
$$
$C_V$ is what known to be constant volume heat capacity. I don't really get the exact meaning of 'constant volume'. Does it mean that (a) the whole process takes place under a constant volume, i.e, it can be any volume as long as the volume does not change during the process, the volume is maintained at its initial value throughout the process, or (b) the $C$ is measured under a fixed constant volume $V$?
If (a) is correct, are the rates of change of $Q$ with respect to $T$ same at, for example, $V = \pu{1 dm3}$ and $V = \pu{2 dm3}$? We are not considering only perfect gases.
If (b) is correct, what is the value of $V$ for standard $C_V$?

@Aditya Garg For a real gas, Cv depends on specific volume.

(a) is correct. (b) is wrong. $U$ depends on $V$ only in the combination $v=V/N$. $v$ is the specific volume, an intensive quantity that does not depend on the size of the system.
We can derive this from first principles. Take $U = U(T,V,N)$. Because $U$ is extensive, $$U(T,\lambda V,\lambda N) = \lambda U(T,V,N) \quad \Longleftrightarrow \quad U(T,V,N) = \lambda^{-1}U(T,\lambda V,\lambda N),$$ by taking $\lambda = N^{-1}$, we can write this expression as $$U(T,V,N) = NU(T,V/N,1) =: Nu(T,v),$$ which produces the desired functional relationship.

You need to put consideration of any process out of your mind. $C_V$ and $U$ are the functions of state that depend only on the characteristics of the particular thermodynamic equilibrium state that the system is in. How it gets from one thermodynamic equilibrium state to another is irrelevant.
If you have two closely neighboring thermodynamic equilibrium states that are at the same volume, the difference in internal energy between these two states is $\mathrm dU = C_V\,\mathrm dT$. In the case of an ideal gas (where internal energy is independent of volume), even if the two neighboring states differ by both $\mathrm dT$ and $\mathrm dV,$ the difference between these two states is still $\mathrm dU = C_V\mathrm dT$.
Furthermore, I should add that because $Q$ is a function of process path, it should not really be part of the definition of $C_V.$ So, Wikipedia is wrong to include this. It only confuses the student.
As far as the title question is concerned, the exact definition of molar $C_{V,\mathrm m}$ is:
$$C_{V,\mathrm m} = \frac{1}{n}\left(\frac{\partial U}{\partial T}\right)_{V,n}$$


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Why did we multiply equilibrium constant with concentration of pure water and called it the water product?
Equilibrium constant is calculated by taking active masses of pure liquids as unity. But while calculating ionic product of water we could have simply defined the equilibrium constant as water constant.

I'm a bit confused. Are you asking why $$K_w = \ce{[H^+][OH^-]}$$ instead of $$K_w = \dfrac{\ce{[H^+][OH^-]}}{\ce{[H2O]}}$$

If I understand your question correctly, reading about equilibrium constant of water on Wikipedia should answer your question. If there is something else down the rabbit hole that's not clear to you, then you probably should re-formulate the question or add some more details.


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"Don't pour oil into lye, pour lye into oil"
I was making soap last weekend and I got distracted, and thence I put too much lye into my oils blend. Basically, the whole soap paste is unusable to make soap because the cure will never be enough to neutralize all the lye contained in the soap paste.
I wanted to recalculate the lye quantity and add the quantity of oil needed to salvage my soap paste but I remembered a piece of advice I read on the recipe: never pour oil into lye, always pour lye into the oils.
Why is that?
At first I thought it was because of similar reasons to why one shouldn't pour water into acid (as these two substances react strongly when put together and the hydration reaction may provoke projections). It has been a while since my high school chemistry lessons but I think that a similar thing happens with water and basis. However, oil has pretty much nothing in common with water when it comes to chemistry. Another reason may be that what a user from a soap making community calls the "splash potential":

You are more likely to splash whatever is being poured into than what is being poured. You do not want lye water splashing out onto you and your surroundings.

This sounds more plausible but it isn't a scientific explanation. So, is there a scientific reason behind the restriction: "do not pour oil into lye"?

I would say that it's more or less the splash potential thing as oil is non polar. https://chemistry.stackexchange.com/questions/893/why-first-water-and-then-acid.

Olive oil, while not as reactive as water in this setting - is still the dilute solution and you should add the concentrate to the diluant and not vice versa. It is just common practice. Without ever doing saponification I do think it is easier to keep them mixed that way as well, with oil being so oily and all.

Lye is denser and tends to pour through the lighter oil, improving mixing (and possibly cutting down on the splattering of lye).  The same is true of concentrated mineral acids added to water.
In any case the hazards are minimized by also adding slowly and with stirring.


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pH change with temperature
I understand that the pH of solutions decreases as the temperature increases but is there a reason as to why some solution's pH decreases more than others even if the temperature change is the same?

Briefly, as pH relies on the equilibrium concentrations of $\ce{H3O+}$ and $\ce{H+}$, the question really boils down to how temperature affects these equilibria for the given compounds. For quantitative analysis there is van't Hoff equation.

The pH of a solution is dependent on acid/base equilibria of the substances in the solution. Each of these equilibria will be affected differently by temperature changes. If reactions are endothermic, the equilibrium will shift towards the products upon increasing temperature, otherwise towards the reactants. How much the equilibrium shifts depends on how exo/endothermic the reaction is.
One of the equilibria is the autodissociation of water, which is endothermic. Increasing the temperature will increase product (H+ and OH-) concentration in pure water. Per definition, the pH will decrease as the H+ concentration (or activity) increases.
More complex solutions will either show an increase or a decrease in pH when temperature is raised. For example, some buffers are notorious for changing pH a lot while others are pretty stable, see this table.


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Heat of formation method
For enthalpy calculations why is it that in some cases the "heat of formation method" is used? I know that $ΔH_f$ represents the heat required to form the compound at standard temperature and pressure ($25^oC$, $1$ atm) but why is it used?
For example, in $Ex.1$

NO(g, $300^oC$):  $H$ = $\int (Cp)_{NO}dT$
  where $T_1 = 25^oC$ and $T_2 = 300^oC$

The enthalpy of nitrogen monoxide is simply the integral from the initial temperature to the final temperature.
However, in $Ex.2$

$CH_4$($150^oC$):  $H$ = ($ΔH_f)_{CH_4}$ +$\int (Cp)_{CH4}dT$
  where $T_1 = 25^oC$ and $T_2 = 150^oC$

The enthalpy of methane is the heat of formation + the sensible heat from the initial temperature to the final temperature.
What tells me when I should add the heat of formation?

Unless ΔHf is zero, you should use the second equation. ΔHf would be zero for elements, but not for NO ΔHf.

@KarstenTheis If I should always use the second equation unless the heat of formation is zero, then why didn't the book include it for NO?

You're confusing different enthalpies (don't worry, it seems to be rather common). Let's stick with NO as an example. The standard enthalpy of formation, $\Delta_fH^\circ$, is defined as the enthalpy of the reaction $\ce{0.5N2 + 0.5O2 = NO}$ at any given temperature. That is, if you need $\Delta_fH^\circ(298\textrm{ K})$, or $\Delta H^\circ_1$, your reaction is $$\ce{0.5N2(gas, 298) + 0.5O2(gas, 298) = NO(gas, 298)}, (1)$$ and for $\Delta_fH^\circ (T)$, or $\Delta H^\circ_2$, provided that the gases in question do not liquify at $T$, the formation reaction is $$\ce{0.5N2(gas, T) + 0.5O2(gas, T) = NO(gas, T)}. (2)$$
Remember that $\Delta_fH^\circ$ for the elements in their stable states is zero at any temperature.
Now let's say that we have to calculate $\Delta_fH^\circ(T)$ using the known value of $\Delta_fH^\circ(298\textrm{ K})$. For that we need to know the heats of three other processes: $$\ce{N2(gas, 298) = N2(gas, T)}, (3)$$ $$\ce{O2(gas, 298) = O2(gas, T)}, (4)$$ $$\ce{NO(gas, 298) = NO(gas, T)}. (5)$$
The enthalpies of $(3)$, $(4)$ and $(5)$ are basically the enthalpies of heating from $298$ to $T$. They are often called "enthalpy increments" and are defined as $\Delta_{298}^TH^\circ=\int_{298}^T C_p\,dT$ . Note that the enthalpy increments can be written between any two temperatures, not just from 298 K.
It is clear that when we add $(1)$ and $(5)$, and subtract half the $(3)$ and $(4)$, we'll get $(2)$. The same with enthalpies: $\Delta_fH^\circ(T) = \Delta_fH^\circ(298\textrm{ K}) + \int_{298}^T \Delta C_p\,dT$, where $\Delta C_p$ is the difference between $C_p$ for products and reactants of reaction $(2)$ with stoichiometric corresponding coefficients (1.0 and 0.5 in our case) taken into account.
Returning to your examples: in the first one, you are asked to find the enthalpy increment for $\ce{NO}$, which is just the amount of heat required to, well, heat it up. In the second example, you are asked to find the enthalpy of formation of $\ce{CH4}$, and I hope that my example above shows you the way of doing it. Also, if you copied the formulae from the textbook right, there's $\Delta$ missing from under the integral (should be $\int \Delta C_p(\ce{CH4})\,dT$).
A rule of thumb: when in doubt about the enthalpies of any process, write down the process thoroughly, with temperatures and phases of all reactants and products. This way it becomes much clearer. It is also useful to remember that in the real world there is no such thing as "The Enthalpy": what one has to deal with is always some change, some $\Delta$.


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What is the temperature dependency of the van-der-Waals constants?
What is the effect of the temperature on the van-der-Waals constants $a$ and $b$?
I have checked on the effect of pressure and volume on the van-der-Waals constants, but I've not been able to get the effect of temperature.

See, they are constants; they are assumed not to depend on anything.

$a$ and $b$ are supposed to be constants. But, they in fact aren't, so your question is in principle valid.
vdW is not a very good model to describe $V_m(p,T)$ of any actual non-ideal gas over a larger pressure or temperature range, and if you take a really good model (those have 6 or 8 parameters), you can just mathematically derive $a(p,T)$ and $b(p,T)$.
However I doubt there is a general trend for a and b with increasing temperature or pressure. I'm sure you could find one for similar (e.g. mono- or diatomic, elementar) gases.
vdW is useful (just like virial coefficents and other models with only two or three parameters) to accurately model $p,V,T$ in a limited range without a large computational effort, e.g. for process control.
(The value of the vdW theory is imo an educational one: It gives a convincing reason why real gases are non-ideal, and correctly describes the trends.)


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Hydrogenation of 1-hexene with Raney nickel
For my high school senior chemistry project, I need to create and perform a lab. I've chosen to do the hydrogenation of 1-hexene with the catalyst of Raney nickel.
I'm struggling because I'm not exactly sure how to perform a hydrogenation. Some websites tell me I need nitrogen gas, some don't. I'm most likely performing a batch hydrogenation under atmospheric conditions.
Any advice on the actual procedure of the hydrogenation would be helpful.

Raney nickel, depending on the formulation, can be quite flammable and dangerous to work with, so given your experience, you may want to reconsider what you're designing here...

I've seen several fires caused by Raney Nickel waste, but then Pd on C can also cause fires.

You need an inert gas supply (Nitrogen), vacuum and a stirrer. Google "Balloon Hydrogenation" for detailed instructions.

What equipment have you got available to you? If you're going to do this, I hope it's in the school lab. I would consider the possibility of transfer hydrogenation using a hydrogen donor rather than hydrogen itself, given your limitations. Or, reconsider to use a homogeneous catalyst but these are usually much more expensive.

I am doing the lab in my school, but my main hurdle is that we don't have a vacuum. Is there any way to get around that? I can buy one but I need one that isn't absurdly expensive.

There is a safety issue here which has been pointed out several times. If you are planning to do this (and honestly, I would seriously suggest not using RaNi if you don't need to), please pay attention to your risk assessment and make sure you are supervised by somebody at all times.

I am going to be supervised my chemistry teacher as well as a lab coordinator at all times, as well as filling out a MSDS sheet

@AlexLee water aspirators are cheap ways to introduce vacuum, but depending on what gets sucked into the water stream, you don't want anything too nasty

To meet your brief of a) performing a hydrogenation and b) using Raney nickel, but performed in the safest manner, I would use aluminium-nickel alloy.
This is cheaper than sponge/Raney nickel and easier to handle i.e
 It's not pyrophoric until activated.
I would advise the use of nitrogen, but not a sweep as it can be detrimental to the procedure I'll outline.
I won't give molar details, but work on about 20% of alloy by mass vs your substrate to start with. I won't quote solvents either as I don't know if you need to isolate your hexane or not, nor how you intend to monitor the reaction.
If you just need to do the reaction, go for a less volatile solvent.
To perform the hydrogenation itself, slowly and in a controlled manner with constant monitoring of temperature,  add 4M NaOH.  This reacts with the aluminium in the alloy to make Raney nickel.  At the same time, hydrogen is generated which the nickel can adsorb and hydrogenate your substrate with.
Bear in mind that the NaOH solution introduces a lot of water,  so if your solvent is immiscible with water, your stirring needs to be all the better.
You could use solvents that mix with a lot of water e.g. 1-butanol can contain up to ~20% water, but from which the bulk aqueous layer can be separated if needs be.
So long as the NaOH solution is added slowly enough, you can control the temperature.  One way to manage is this is to only introduce 25% of your solution to be added at one time. That way if you accidentally add it all, it shouldn't go too crazy and boil/eject the contents uncontrollably.
Once complete, you can decant off the liquid but ensure you leave water behind to cover the catalyst.  Rinse and decant until the pH is 8 or below. Then you have some options:
Dispose of the wet catalyst into bags that go off site from your school for disposal,  using the appropriate school procedures or,
Use dilute (1M) HCl to gradually deactivate the nickel. Again, add very slowly over hours if necessary.  Then follow appropriate waste disposal guidelines for the nickel solution.


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Comparing collision energy formats in the NIST 17 MS/MS library
In the NIST 17 MS/MS library, collision energy can be represented as a positive integer (i.e. 1,2,3,...), a percentage (i.e. 35%), or a floating point number (i.e. 7.3 eV). How can I compare these different collision energy settings? Is it possible to convert them all to eV, for example?


text:
Why is my sodium acetate not resulting in any crystals?
My son received a chemistry set for Christmas and we've been working through the experiments over the last few weeks.
One of them involves boiling white vinegar and adding baking soda to produce sodium acetate. After boiling this down it is supposed to result in some crystals forming on the pan. However, after trying it a couple times we haven't been successful in producing any crystals. What are we doing wrong?
Our process is:

Add 4 cups of distilled vinegar to a pan. Boil it on the stove.
Stir in 4 tablespoons of baking soda. Do this 1 tablespoon at a time, stirring gently to dissolve each spoonful before adding more.
Keep at a high boil until it is no longer fizzing.
Reduce to a medium heat and boil until only 1 cup remains. This usually takes about an hour.

At this point our booklet says that crystals should be forming around the outside of the pan. However, both times we have no crystals - only a liquid that looks like weak tea.

You obviously haven't reached the necessary concentration of sodium acetate to cause precipitation. Boil some more, and try letting the solution cool. You want the crystals to form slowly so that you get big crystals.

"only a liquid that looks like weak tea" Just to double check: Are you using clear Distilled White vinegar or White Wine vinegar?  If you're not using clear distilled white vinegar, you may be introducing other compounds from the vinegar that inhibit crystal growth.  (I know you said distilled vinegar in your process, but I've done far, far sillier things, so just double checking the ingredient list.)

@Van It's distilled white vinegar.

@MaxW I took the latest batch off the heat to cool at room temperature. The result is (for lack of a better word) brown goo.

I was going to suggest that you super saturated the system, but that doesn't seem very likely if you have brown goo. You could try to add a single grain of table salt to see if that does anything (I doubt it).

Hmm... well, is it brown goo or is it weak tea? Those sound very different...

@indigochild that you're getting brown goo or weak tea is making me think that you have or are producing caramelized sugars.  As you are reducing down the mixture, does the discoloration start along the edge of the pan and move in, or does it start more along the bottom / middle?  Also, how long are you keeping the solution at high heat?  Lastly, what brand of vinegar are you using?

see top comment here: https://chemistry.stackexchange.com/questions/76139/recrystallizing-sodium-acetate-from-solution?rq=1

From my experience it wouldn't hurt to filter the hot solution before crystallization. Also, if the solution is viscous, it is going to take much longer to for the crystals to grow and their quality is most likely going to be poor.

The solubility of sodium acetate is interesting (that means confusing!). Sodium acetate exists in two forms: anhydrous and trihydrate. When compounds exist in two forms like this, the solubilities, given in grams per 100 mL H20, are predictable: the hydrated form is much more soluble, because it is only partially the compound, and the other part is extra water, the solvent.
However, sodium acetate displays a different behavior:

https://en.wikipedia.org/wiki/Sodium_acetate
Sometimes different hydration stages of compounds have different temperature ranges of stability and solubility, but the behavior of sodium acetate is unique. The precipitate that forms so impressively upon cooling is probably the trihydrate, because the amount of sodium acetate in solution according to the recipe does not seem (in my quick estimate) to exceed the solubility of the anhydrous compound, and so the solution produced might be metastable. For some length of time. But, we may be able to initiate crystallization by scratching the glass container, or stirring it, or allowing a drop to dry completely to produce some solid, then using that as a seed for the rest of the solution. Or freezing it.
What I have done is to heat some (pure) sodium acetate to form a concentrated solution (calculated to be supersaturated at room temp), then cooled it: no ppt. Then I touched the surface of the solution with my fingertip and BOOM! Crystallization proceeded like a shot, producing a mass of crystals that filled the container and solidified the former liquid mass. (Did I say it was supersaturated?) I was surprised at the mass of crystals compared to the amount of material I had added to the water at the beginning, but did not think about it till now.
It turns out that the solubility curves of trihydrate and anhydrous sodium acetate are quite separated:

https://www.google.com/search?q=solubility+of+sodium+acetate&client=safari&rls=en&source=lnms&tbm=isch&sa=X&ved=0ahUKEwjt-un3m4XlAhXrY98KHQExDiEQ_AUIESgB&biw=1004&bih=846#imgdii=qbOI-c2mtjJb3M:&imgrc=lRafmM5w0nsXhM:
The formation of trihydrate at low temperatures must be a difficult process, and the brown goo obtained here might signify some crystallization inhibitor is present. Maybe a slight excess of NaHCO3 or HCH3CO2 is holding things up - you could try adding a drop of the vinegar, or a few grains of NaHCO3 to see if that changed anything. But If you have a glass rod, scratch the side of the glass container - that works almost all the time.

What you are trying to prepare is a "super-saturated" solution. How fast and how much of a substance will dissolve in a solvent often depends on temperature. For example, you can make a lot more sugar dissolve in a glass of hot water than a glass of cold water. A solution is super-saturated when you dissolve the maximum amount of solute at a high temperature then let the solution cool. That solution now has more solvent dissolved in it that it is supposed to at the cooler temperature. Often, this sets things up for crystals to form (the solution expels the overabundance of solvent via crystal-formation).
Anyway, I'd try taking what you already have, heating it back up again. Add more of the baking soda if you still have some. If not, don't worry about it. Let the solution simmer longer than the directions call for, then let it cool down slowly. Sometimes crystals take a couple of days to form.

You said you are a pan to heat the mixture. The vinegar maybe reacting with the metal in the pan and causing the discoloration. I would suggest using a glass beaker in the pan to create a double boiler.


text:
How can benzaldehyde have a pKa of 14.9?
There are numerous websites out there which claim that the pKa of benzaldehyde – C6H5CHO – is 14.90. (Just do a Google search for benzaldehyde pka to see what I mean.) This doesn't make sense chemically, as there are no protons in benzaldehyde which would be so acidic.
However, it doesn't seem to have been made up. The CRC Handbook of Chemistry and Physics lists the same value of 14.90 for the pKa of benzaldehyde, at $\pu{25 ^\circ C}$ in aqueous solution. But I can't find this information in the primary literature.
If this value refers to the acidity of the hydrate, does it include the equilibrium constant for hydrate formation? i.e. is this Ka the equilibrium constant for the reaction:
$$\ce{PhCHO + 2H2O <=> H3O+ + PhCH(OH)O-}$$
or is it just
$$\ce{PhCH(OH)2 + H2O <=> H3O+ + PhCH(OH)O-}?$$
I would appreciate some kind of definitive evidence. Using chemical reasoning and intuition is great, but if this value is to be of any use to anybody, then we need to know exactly what it describes.

Is it possible this is a value for the hydrate?

@DennisCao, indeed, that's my primary suspicion: it might be analogous to the "pKa" of CO2.

I think this must be for the deprotonation of the hydrate.

Not that it's the definitive source, but the book "Dictionary of Food Compounds" (see Google Books preview) suggests it is indeed the hydrate.

Just for some additional context, it does appear to be possible to abstract the aldehyde proton from benzaldehyde directly (as opposed to the hydrate), though it is indeed a much weaker acid. This reference indicates that LiTMP in toluene/hexane at 0 °C performs the deprotonation, but the resulting conjugate base understandably immediately condenses with additional benzaldehyde, eventually forming benzyl benzoate on aqueous workup.

Significant amount of geminal diol of benzaldehyde exists in an aqueous solution of benzaldehyde at 25 °C because $\mathrm{p}K_{\text{hyd}} = 2$ (Ref. 1)

The $\mathrm{p}K_{\mathrm a}$ of benzyl alcohol is listed as 15.40 (Wikipedia). Thus, one can reasonably assume that the given value of $\mathrm{p}K_{\mathrm a}$ 14.9 represents a composite equilibrium constant for the hydration of benzaldehyde and dissociation of the geminal diol thus formed.
Late addition to the answer:
In his paper, "Acidity constants of benzimidazolium ketone and pyridinium aldehyde hydrates" (Ref.2), Terence C. Owen states that:

It is known that the acidity constants of gem-diols typically are about 2.5 units lower than those of the corresponding monohydric alcohols.

When do some extensive literature survey, one can find few examples to backup that statement. For example, $\mathrm{p}K_{\mathrm a}$ of methanol is reported as 15.7 while that of formaldehyde hydrate is 13.3, between which $\Delta\mathrm{p}K_{\mathrm a} = 2.5$ (Ref.2). Interestingly, $\mathrm{p}K_{\mathrm a}$ of 1,1,1,3,3,3-hexafluoropropan-2-ol is reported as 9.22 while that of hexafluoroacetone hydrate is 6.45 where $\Delta\mathrm{p}K_{\mathrm a} = 2.77$ (Ref.3). However, $\mathrm{p}K_{\mathrm a}$ of 2,2,2-trifluoroethanol is reported as 12.37 (Ref.4) while that of 2,2,2-trifluoroethanal hydrate is 10.05 (Ref.3) where the difference is < 2.5 ($\Delta\mathrm{p}K_{\mathrm a} = 2.33$).
Thus, we can conclude that the $\mathrm{p}K_{\mathrm a}$ of benzaldehyde is derived from its hydrate (gem-diol).
Also see: Yoshiro Ogata and Atsushi Kawasaki, In The Chemistry of Carbonyl
Group, Volume 2; Jacob Zabicky, Ed.; John Wiley & Sons Ltd.: New York, NY, 1970, Chapter 1: Equilibrium additions to carbonyl compounds, pp 1–69 (https://doi.org/10.1002/9780470771228.ch1).

Late Edit:
It is also noteworthy that $\mathrm{p}K_{\mathrm a1}$ of the cyclic hydrate of phthalaldehyde is reported to be $11.50 \pm 0.10$ at $\pu{25 ^\circ C}$ (Ref. 5):

A bit of further research I did backs this up (several sources also mention the hydration equilibrium constant, e.g. http://www.nrcresearchpress.com/doi/pdf/10.1139/v79-084), and I think it is plausible that the gem-diol is more acidic than benzyl alcohol (inductive withdrawal). However, I'm still holding out for a more detailed confirmation of what exactly this 14.9 entails, specifically whether it includes the hydration equilibrium or not.

This is a good reference. Accordingly, $pK_a$ of hydrated Phthalaldehyde is suggested as 11.6.

Note that if you combine the 2 orders of magnitude from the hydration equilibrium and the 2.5 orders of magnitude from the gem-diol expected acidity constant, you would expect that the pKa is about 0.5 units smaller, which is actually spot on.

I like that you're still working on this answer. ;) Another paper further suggests that the pKa of benzaldehyde hydrate is 12.0 (observed) or 12.43 (calculated). Together with the hydration equilibrium constant we arrive at a pKa of ~14 to 14.5 for benzaldehyde. That is probably convincing enough for me so I will accept (I upvoted long ago of course!)... I am guessing that this 14.9 number was obtained just by direct pH / conductivity / etc. measurements without any regard for the underlying equilibria...

In a way, we don't have to suppose the existence of a geninal diol. We get the same effect if we assume that the aldehyde acts similarly to boric acid, simply adding a small amount of hydroxide ions from water/a basic solution to its own molecule.

But since benzaldehyde is much less acidic than water, is its pka even measured in water though? Is it possible whereby a nonaqueous solvent was used to dissolve the benzaldehyde, measure its acidity and report its pka relative to water based on those measurements and other data? See here: https://chemistry.stackexchange.com/questions/28022/how-is-the-pka-of-extremely-weak-acids-determined

Wait there's no pka of the 2nd compound of the 1st image you posted (gem-diol of benzaldehyde)? Benzyl alcohol is a different compound.


text:
Explanation of the strength of phosphorus–oxygen bond
When discussing the Wittig reaction, Clayden's Organic Chemistry (2nd ed.) cites the strength of the P=O bond formed in triphenylphosphine oxide as a driver of the reaction through enthalpy:

The P=O bond, with its bond energy of 575 kJ mol−1, is one of the strongest double bonds in chemistry, and the Wittig reaction is irreversible and is driven forward by the formation of this P=O bond. No need here for the careful control of an equilibrium necessary when making acetals or imines.

According to this table of bond dissociation energies, the value for the P=O bond is similar to the bond energy values for bonds such as the Be–F bond (577 kJ mol−1) and the H–F bond (568.6 kJ mol−1).
However, taking into account the large differences in the p-orbital sizes of phosphorus and oxygen atoms, and the sizes of the phosphorus and oxygen atoms, I fail to understand the similarity in the strength of the P=O and other bonds listed above.

The oxygen atom is larger than the hydrogen, fluorine, while the phosphorus atom is larger than hydrogen, fluorine, and beryllium. Thus, I would imagine that the sigma orbital overlap would be weaker due to a larger distance of interaction in P=O than in the H–F and Be–F bonds.
Due to the differences in the sizes of the p-orbitals of oxygen and
phosphorus, I would naturally infer that the p-orbital interactions and, thus, the pi bond in P=O would be weak.

Are there other factors that I am failing to consider that strengthen the phosphorus–oxygen double bond?
References
Clayden, J., Greeves, N., Warren, S. Organic chemistry, 2nd ed.; Oxford University Press: New York, 2012.

(+1) I'll throw it out there that it's not a true double bond. If you discount d-orbital involvement in bonding, then it's better represented as $\ce{P+-O-}$. Unfortunately, I don't have anything useful to add beyond that.

I also believe that this representation of the bond also agrees with the weak p-orbital interactions in the bond, as stated in the question.

The d-orbitals of phosphorus help delocalise the O electrons as well, right?

I guess that since the bond primarily exists as $\ce{P+-O-}$, then the interaction between both the atoms would consist of a sigma bond and an electrostatic potential energy between the opposite charges similar to an ionic bond. This might be responsible for the higher bond energy

The phosphorus oxygen double bond is a relatively short bond. The oxygen p-orbital lone pairs are donated to the phosphorus antibonding orbitals further stabilizing and strengthening the P-O bond. This idea is supported by molecular orbital theory and ab inito calculations. Consequently, this suggests there is no d-orbital involvement in the phosphorus(V) atom.


text:
Why are reversible reactions spontaneous?
If a reversible chemical reaction is spontaneous in one direction, shouldn't it be non spontaneous $(ΔG > 0)$ in the other? In other words for the reaction:
$$\ce{A + B ⇌ C + D}$$
if
$$\ce{A + B -> C + D}$$
has $ΔG < 0$ and thus proceeds spontaneously, doesn't this imply that
$$\ce{C + D -> A + B}$$
would have $ΔG > 0$ and thus proceed non spontaneously?

A negative $\Delta G^0$ implying that a reaction will proceed spontaneously or a positive $\Delta G^0$ implying that a reaction will not proceed spontaneously is just a "rule of thumb" to give you a feel for what to expect.  But all a negative $\Delta G^0$ really means is that the equilibrium constant for the reaction is large (and formation of the products is highly favored), and a positive $\Delta G^0$ means that the equilibrium constant for the reaction is small (and formation of the products is not highly favored).  However, if you start out with only reactants present initially, irrespective of the sign of $\Delta G^0$, the reaction will proceed spontaneously.  However, if $\Delta G^0$ is negative, at final equilibrium, the extent of the reaction will be greater.

Remember not to confuse the standard free energy change (measured/computed for the standard states and concentrations) with the real free energy change.
In order to obtain the $\Delta G$ for the reaction, you should apply:
$$ \Delta G = \Delta G ^{\varnothing} + RT \ln Q$$
Then, a reaction which is nonspontaneous in one context may be spontaneous in another context where the concentrations of participating species are different.

Probably nitpicking, but shouldn't the upper symbol at $ΔG$ be circle $\circ$ or a plimsoll ⦵? An empty set symbol ∅ looks somewhat confusing to me.

Probably? \varnothing is the only one I ever remember...

That's what the compose key is for:)


text:
Is the negative excess volume of ethanol/water the cause of my math error?
I have a binary mixture, say ethanol at $90\%wt$ ethanol, $10\%wt$ water.  According to Perry's Chemical Engineers' Handbook 9th Ed., that mixture has a density of $0.81797g/cc$ (at $20^{\circ}C$).  So my $1L$ of mixture should have a mass of $817.97g$.
Now, I want to create this mixture.  I know the total mass and the mass %'s, so I just multiply them out: we find that we need $736.17g$ ethanol and $81.80g$ water.  Ok, so far so good.  However, say all of my balances are broken, so I only have graduated cylinders to work with.  Easy, just convert those masses back to volumes using their respective pure substance densities: $0.78934g/cc$ for ethanol and $0.99823g/cc$ for water.
However, doing this tells me that I should have $932.6mL$ of ethanol and $81.9mL$ of water, which adds up to a total volume of $1014.6mL$ of binary mixture, so somehow I magicked $14.6mL$ of liquid out of thin air using this conversion algorithm.
Is this due to the negative excess volume of the ethanol/water mixture upon mixing?  That's my current guess, but for the life of me I'm having a hard time figuring out the math here.  It seems like the change in volume should be much smaller: $10\%wt$ water in ethanol is about $10\%mol$ and there's about $4.5mol$ of water in the mixture, so based on this dataset it seems like there should only be a difference of about $3mL$.  Even with all the rounding I just did, it seems like this can't be the only error source.
UPDATE: That last step was where my error was.  For some reason, I calculated the excess volume based on only the number of moles of water, not the total moles of the mixture.  When using the total volume of the mixture (as one should), and a new set of tables (since the old ones can no longer be accessed 4 years later), the total volume effect comes out to -14.4mL, which is nearly exactly what my calculations predicted.  Thanks to Anger Density for finally narrowing down the error, and do check out Ananta's answer for a thorough breakdown of this kind of calculation.
Is there something wrong with the logic of my algorithm here?  Can anyone point me to the reason I'm seeing this numerical error whenever I run this calculation?

Hmmmm.  I see 0.81942 in a different reference for 90 wt.% ethanol.

@JonCuster Ahh, is it hard copy or could you link it?

https://wissen.science-and-fun.de/chemistry/chemistry/density-tables/ethanol-water-mixtures/ for online, and 0.8194 in my old CRC...

@JonCuster Interesting! I usually trust both CRC and Perry's quite a lot, I wonder why the disparity. I'm at the gym, I'll look into it when I get home, or maybe when I get back to work tomorrow

My CRC has 0.8180 as the 'relative density at 20C, kg/l' while 0.8194 is the 'specific gravity at 20C'.

@JonCuster I've been using SPG, since the density of water at 20C is not actually 1 g/cc. iirc Perry's calls this "Density at 20C," while Wikipedia calls it "density relative to 4C water at 20C," but those are equivalent statements

I think your calculations are exactly right. Because of the negative excess volume of ethanol-water solutions, you will mix a larger total volume of separate ethanol and water to make your 90% weight solution of ethanol.
In fact, the excess volume is "built in" to the density measurement listed in Perry's. Let's calculate the required excess volume to reproduce this density. Using your already calculated values, I get:
$n_{total}=n_{ethanol}+n_{water}=16.0+4.5=20.5$ moles total solution
$x_{ethanol}=\frac{n_{ethanol}}{n_{total}}=\frac{16.0}{20.5}=0.78$
$v_{excess}=\frac{\Delta v}{n_{total}}=\frac{\pu{-14.6mL}}{\pu{20.5mol}}=\pu{-0.71mL/mol}$
Comparing this calculated value to the closest one I can find in the tables you referenced, it looks like they list -0.678 ml/mol for an ethanol mole fraction of 0.7805. Seeing as the DDBST data are for 25 degrees Celsius and yours are at 20, I would call the agreement excellent!

Interesting!  This makes it seem like I might have made a mistake not in my original calculation, but in my assessment of how large the excess volume effect should be.  I estimated 3mL when I originally posted this question a few years ago (vs the ~14mL actual), but I may have messed up a conversion between mass and moles or some similar thing.  I'm not in a situation to work through the math right now, but I'll try to remember to look at this during the workday tomorrow and check the math!

I just redid all the math and you're totally right.  Working it through the way I did, there are indeed about 20.5 total moles in the solution, and with an excess volume of mixing of -0.7mL/mol (which I found here since I can no longer access the tables I linked years ago) this comes out to -14.4mL excess volume, nearly exactly what I found!  Accepted.

As additional followup: I can't reconstruct exactly what I did back when I first posted this, since I can't get at the tables.  However, based on the values from the new set of tables I linked in the last comment, it looks like I originally (for some reason) calculated excess volume based off only the 4.5 moles of water.  I have no idea why I did that, but it gives approximately -3mL excess volume, which is the number that caused my initial confusion.  I will update the original question to highlight where my error was.

First Things First
Ethanol and water do form a solution with a negative excess volume, like you correctly said. This is pertinent to answering your question. For further reading, refer to this Wikipedia page on Excess Properties.
Evaluating Your Answer Step by Step
Step 1: Calculation of Mass of Solution.
Since density of the solution is given. This calculation is accurate.
$$
m_\text{solution} = \rho\times V = \pu{0.81797 g cm^3} \times \pu{1000 cm^3} = \pu{817.97 g}
$$
Step 2: Calculating Weights of Ethanol and Water
Since weight ratios are given, this calculation also seems straightforward.
$$
m_\text{ethanol} = \pu{0.9} \times m_\text{solution} = \pu{0.9\times 817.97 g = 7.4\times 10^{3} g}\\
m_\text{water} = \pu{0.1} \times m_\text{solution} = \pu{0.1\times 817.97 g = 8.2\times 10^{1} g}
$$
Step 3: Calculating Volumes of Ethanol and Water
This is where your problem lies. The thing is, when ethanol and water are mixed together, these don't mix ideally. That is, $V_\text{ethanol}+V_\text{water} \neq V_\text{solution}$. Actually,
$$
V_\text{ethanol}+V_\text{water} > V_\text{solution}
$$
Thus, you will need more volumes of ethanol and water than required by calculations using mass.
Remember: mass and volume analysis often contradict due this deviation. This is why some scientists, especially physicists, prefer molalilty as opposed to molarity.
The Fix
Start with volume, we are required to make $\pu{1000 cm^3}$ of ethanol-water mixture. Assume $V_\text{ethanol}$ and $V_\text{water}$ are required to make this solution. Now, We know that
$$
\begin{align}
&V_\text{ethanol}+V_\text{water} > V_\text{solution}\\
\implies &=\dfrac{m_\text{ethanol}}{\rho_\text{ethanol}}+\dfrac{m_\text{water}}{\rho_\text{water}} > \dfrac{m_\text{solution}}{\rho_\text{solution}}
\end{align}
$$
Here you can use the exact values.
$$
\dfrac{m_\text{ethanol}}{\pu{0.78934 g cm^{-3}}} + \dfrac{m_\text{water}}{\pu{0.99823 g cm^{-3}}} > \dfrac{\pu{817.97 g}}{\pu{0.81797 g cm^{-3}}}
$$
This is an equation of the type
$$
\dfrac{x}{\pu{0.78934 g cm^{-3}}} + \dfrac{y}{\pu{0.99823 g cm^{-3}}} > \dfrac{\pu{817.97 g}}{\pu{0.81797 g cm^{-3}}} \tag{1}
$$
The Problem with the Fix
Unfortunately, based on theoretical analysis, this is as far as I can get. I can help you with a strategy to find the exact masses. Since it is given that
$$
\dfrac{m_\text{ethanol}}{m_\text{water}} = 9 \tag{2}
$$
Try different values of $m_\text{ethanol}$ and $m_\text{ethanol}$ following Eqs. (1) and (2). Unfortunately, without a quantitative relation between $m_\text{ethanol}$ and $m_\text{water}$ in Eq. (1), this is as far as we can get. An (informed) trial and error seems like your best option here.
Fixing the Problem with the Fix
Start from the ideal case (the dotted line in the graph) and follow Eq. (2) (dark solid line in the graph), that is Eq. (2), which in terms of volume given below.
$$\dfrac{m_\text{ethanol}}{m_\text{water}} = 9 \text{
 or } \dfrac{\pu{0.78934} \times V_\text{ethanol}}{\pu{0.99823} \times V_\text{water}} = 9
$$
This means you need to follow the dark solid line (eq2) in the green-shaded region.

until you get the desired result. That is,
$$
V_\text{solution} = \pu{1000 cm^{3}}
$$
Note: Since other answers have verified densities, I am assuming these are correct

And so: "Is the negative excess volume of ethanol/water the cause of the OP's math error?"

@ToddMinehardt not only are OP's calculations somewhat inaccurate (since OP reasoned it with "negative volume of mixing"), but all the answers are also wrong suggesting you need a higher than required volume of ethanol and water because of "negative volume of mixing," because ethanol and water show "positive deviation from ideal mixing behavior." Remember, we do need to prepare a $\pu{1000 cm^{3}}$ of solution using only graduated cylinders and the answer I posted provides a well-reasoned method to do just that.

This answer is not correct. The statements "Ethanol and water do form a solution with a negative excess volume..." and "you will need less volumes of ethanol and water than required by calculations using mass" are incompatible.

@AngerDensity thank you for pointing out the discrepancy. I have made the correction.

I appreciate the effort you put into this answer, and to revise it to make sure it is accurate.
I don't think it answers my exact original question, since I was aware of the concept of excess volume but the effect seemed to be of the wrong magnitude.  However I think that for anyone finding this Q&A in the future it's very valuable.  I saw you comment elsewhere that you may extend it to the general case, and I hope you do!  Perhaps one of us could also find a new table of values for excess volume in the EtOH/H2O system, the one I linked 4 years ago seems to have expired.

@realityChemist the answer is a practical method to find volumes of $\ce{EtOH}$ and $H2O$ required to make a desired volume of $\ce{EtOH}/\ce{H2O}$ solution using only graduated cylinders (assuming one has no excess to datasets or instruments such as mass balance) based only on the concept of $V^\text{E}$, $\rho_\text{solution}$, $\rho_\ce{H2O}$, and $\rho_\ce{EtOH}$, latter two of which are available on commercial containers of $\ce{H2O}$ and $\ce{EtOH}$, respectively.

@realityChemist nonetheless, I appreciate the feedback, I should have (re)formulated your question more precicely and answered that more accurately. Thank you for your encouragement and consideration.

When you mix ethanol and water they take up less space due to the way the molecules fit together, so you will have what is called a negative volume of mixing.  In this case, the 932.6 mL of ethanol and the 81.9 mL of water will give you a 100 mL solution of 90% ethanol.

Presumably you mean a 1 liter solution (1000mL)?

Ethanol and water show positive deviation from ideal mixing behaviour. That is, $\mathrm{V_\text{actual}}>\mathrm{V_\text{ethanol}}+\mathrm{V_\text{water}}$.

@Ananta Exactly the opposite, V_act < V_eth + V_water. You may confuse it with positive deviation from Raoult law, where mixture has higher then ideal vapour pressure.

@Poutnik Oh! I see it now. Think I will be deleting this answer then. Still, does the analysis stand correct for solutions with positive excess volumes?

@ananta I think the answer could be  easily tweaked  to address the general case of nonzero excess mixing volume, both signs simultaneously.

@Poutnik for now, I have fixed the answer, I will make another post for a general analysis dealing with wider range of excess volume.


text:
What is the name of this chemical LiB2PO4H7?
I was recently looking at a chemical equation calculator that balances equations for you.

I came across a reaction that I am unsure about, it goes like this:
B2H6 + LiH + PO4 = LiB2PO4H7
I was researching the name of the product of this reaction for a while, but couldn't find a name for the chemical. I ask you guys this, if this chemical actually exists, what is the name of it?

Where did you find this reaction? It's very strange.

I found it on a balancing site.https://www.webqc.org/balance.php

I don't see the equation on the site. Do you mean you input the chemicals yourself into the calculator, and asked it to balance to see what comes out?

Yes, just input the equation and it seems to balance just fine.

These websites would accurately balance any BS you feed them with, they don't care about chemistry, only math. What are the reaction conditions (solutions/molten salts/hydrothermal/...)? Where does $\ce{PO4^3-}$ come from? If it's phosphoric acid it doesn't make much sense as both diborane and lithium hydride would first react quite violently with water ending up in boric acid and lithium hydroxide. Also, the product seems to be an anion, $\ce{LiB2PO4H7-}$, which again doesn't make much sense.

I wouldn't really consider it a duplicate because this question focuses on what the name of this specific compound is. That question asks how to name ionic compounds in general.

@WitherFang136 arguably the point of having the more general question is to address the large number of specific questions asked on this topic. Since the naming follows general rules, the general question addresses all the specific ones.

There is an important concept to be understood underlying this question, something all new chemistry students eventually learn: all sensible chemical reactions will balance, but not all equations that balance make chemical sense.
Balancing is purely mathematical manipulation (solving a system of linear equations), and proper balancing is necessary but not sufficient to represent real chemistry.
As a simple example, $\ce{200 H2 + O2 -> 2 H200O}$ balances just fine, but doesn't actually mean anything chemically. Here are a few more examples of proposed chemical reactions which balance correctly, but don't actually represent real chemistry due to more subtle arguments.
The crux of the problem is that it is very easy to write a program which can do mathematical balancing of chemical reactions, but it is exceptionally difficult to write a program which can tell in general whether the reaction actually happens! Chemists have spent the past several hundred years tabulating enormous amounts of information (e.g. physical chemistry) and making qualitative descriptions of chemical behaviour (e.g. atomic models, Lewis dot structures, etc.) so we can make this problem tractable after "getting a feel for it".
Okay, back to your reaction. In your particular case, things are made worse by the fact that one of your reagents, $\ce{PO4}$, already doesn't exist in any reasonable conditions. Because this is such an unusual chemical, it's hard to predict exactly how it would react based on the rules-of-thumb we know, as most have been developed to understand more "normal" chemistry. What would most likely happen is that it would tear electrons out of just about anything, such as the hydride ions in $\ce{LiH}$. The products could end up as a mixture of $\ce{Li3PO4}$, $\ce{LiBH4}$, $\ce{H2}$ and possibly left over reagents, among others.
The closest chemically valid entity to $\ce{PO4}$ is the phosphate anion $\ce{PO4^3-}$. The calculator you used doesn't understand this by itself, so you would have to write "PO4{3-}" instead of "PO4" in the input field. If you had done this, since $\ce{LiH}$ and $\ce{B2H6}$ are true, neutral compounds, then any combination of all the above must have a negative charge too (conservation of charge). Nevertheless, I would still expect the online calculator to fail. My basic guess is that the most favourable outcome would be the formation of $\ce{LiBH4}$, with other less important side-reactions.

And, of course, the website linked in the question happily confirms that $\ce{200 H2 + O2 -> 2 H200O}$ is the balanced equation for the synthesis of Nicolau's reagent from hydrogen and oxygen.

@DavidRicherby Yes, indeed, the prototypical example of dictacoordinate oxygen. With two extra lone pairs.

I've never stumbled upon a "balancing" website that would be checking the correctness of the reaction. This one doesn't even seem to take oxidation state of any element into account as the charges are not balanced (-3 from $\ce{PO4^3-}$ on the left and -1 from $\ce{LiB2PO4H7^-}$ on the right).
Also, as it's been already ruled out in other answer, these precursors would unlikely result in any borophosphates. However, there is at least one compound characterized that consists of $\ce{H,Li,B,O,P}$ only:

$\ce{Li[B3PO6(OH)3]}$, catena-[monoboro-mono-dihydrogendiboratemonohydrogenphosphate]. Hydrothermal synthesis from $\ce{LiOH · 2 H2O}$, $\ce{P2O5}$ and $\ce{B2O3}$, conc. solution in $\ce{HCl}$ at $433~\mathrm{K}$ [1].

There is also a few dozens of lithium borophosphates with addenda metals, to name a few:

$\ce{LiCd(H2O)2[BP2O8] · H2O}$, lithium cadmium diaqua catena-[monoborodiphosphate]- monohydrate. Hydrothermal synthesis from $\ce{CdCl2 · 2.5 H2O}$, $\ce{LiOH}$, $\ce{H3BO3}$ and $85\%$ $\ce{H3PO4}$ in deionized water at $443~\mathrm{K}$ [2].
$\ce{Li3V2[BP3O12(OH)][HPO4]}$, trilithium divanadium(III) borophosphate hydrogenphosphate. Hydrothermal synthesis from $\ce{H3BO3}$, $\ce{VCl3}$, $\ce{LiH2PO4}$, $\ce{LiCl}$ in deionized water at $553~\mathrm{K}$ [3].
$\ce{LiCu2[BP2O8(OH)2]}$, lithium dicopper hihydroxoboro-bis(phosphate(V)). Hydrothermal synthesis from $\ce{H3BO3}$, $\ce{Cu(OAc)2·H2O}$, $\ce{LiH2PO4}$  in deionized water at $473~\mathrm{K}$ [4].

As you can see, all methods use less volatile precursors and hydrothermal conditions. Feel free to practice with writing and balancing chemical reactions for these real syntheses.
References

Hauf, C.; Kniep, R. Crystal Structure of Lithium Catena-[Monoboro-Mono-Dihydrogendiboratemonohydrogenphosphate), $\ce{Li[B3PO6(OH)3]}$. Zeitschrift für Kristallographie - New Crystal Structures 1997, 212 (1), 313–314. https://doi.org/10.1524/ncrs.1997.212.1.313. (Open Access)
Ge, M.-H.; Mi, J.-X.; Huangm, Y.-X.; Zhao, J.-T.; Kniep, R. Crystal Structure of Lithium Cadmium Diaqua Catena-[Monoborodiphosphate]- Monohydrate, $\ce{LiCd(H2O)2[BP2O8] · H2O}$. Zeitschrift für Kristallographie - New Crystal Structures 2003, 218 (JG), 295–296. https://doi.org/10.1524/ncrs.2003.218.jg.295. (Open Access)
Lin, Z.-S.; Hoffmann, S.; Huang, Y.-X.; Prots, Y.; Zhao, J.-T.; Kniep, R. Crystal Structure of Trilithium Divanadium(III) Borophosphate Hydrogenphosphate, $\ce{Li3V2[BP3O12(OH)][HPO4]}$. Zeitschrift für Kristallographie - New Crystal Structures 2014, 225 (1), 3–4. https://doi.org/10.1524/ncrs.2010.0002. (Open Access)
Yang, M.; Li, X.; Yu, J.; Zhu, J.; Liu, X.; Chen, G.; Yan, Y. $\ce{LiCu2[BP2O8(OH)2]}$: A Chiral Open-Framework Copper Borophosphate via Spontaneous Asymmetrical Crystallization. Dalton Trans. 2013, 42 (18), 6298–6301. https://doi.org/10.1039/C3DT50591J.

Why are you saying "1 H on the left"? I see 6 + 1 + 0 which is 7.

@kasperd Because I should've paid more attention when I wrote the answer. Must've been thinking about something else. Thank you, I'm going to fix it.

This ion is most likely not the correct product, like the answers say, especially since it is not.  However, IUPAC has methods on how to name compounds like these.  Take a look at this page in the IUPAC red book.
However, if you were to just name the compound, ignoring the fact that its charge does not balance, you would get something along the lines of:
Lithium diboron phosphate heptahydride.


text:
Why do people commonly refer to Na2HPO4 as Sodium Hydrogen Phosphate, rather than Disodium Hydrogen Phosphate?
First of all, I am aware of this question.  The purpose of my question is to understand why people seem to break the rules that IUPAC has set (See example five on page 76 of the book), especially since this is not a compound that has a historic name, like $\ce{NH3}$ being ammonia or $\ce{N3-}$ being azide.  This compound has two species of cations, with sodium having a 2:1 ratio to hydrogen.  According to example five on page 76, compounds that have multiple cations are supposed to give prefixes to every compound in an ion that has a quantity greater than one.  Why do we not do this for sodium hydrogen phosphate?

"A person is smart. People are dumb, panicky dangerous animals and you know it." — Agent Kay, in Men in Black (1997).

Strongly related: https://chemistry.stackexchange.com/questions/64417/why-is-potassium-phosphate-kh2po4-in-this-reaction?r=SearchResults

That situatuation is different, since the name of KH2PO4 , potassium dihydrogen phosphate is actually correct, whereas calling K2HPO4 potassium hydrogen phosphate is incorrect.  Likewise, the name of potassium monohydrogen phosphate is incorrect, since cations with a quantity of one do not get prefixes.

Well, turns out that I needed to look at this a little differently.  I looked at calcium hydrogen carbonate and then realized that hydrogen is part of the polyatomic ion.  It's a bit harder to recognize when there are no parenthesis, like in the case of sodium hydrogen phosphate, but the same thing applies there.  Therefore, the name, sodium hydrogen phosphate is correct because there are still only two distinct ions in the compound.


text:
Order of reactivity of nucleophilic addition at carbonyl group with Grignard reagent

I thought the order would be $1 > 3 > 2$ as 1 is aldehyde and 2, 3 are ketones. As generally aldehydes are more reactive, I placed it first. I chose 3 over 2 as two alkyl groups would make cleaving of $\ce{C=O}$ difficult in 2.
But the answer is given $1 > 2 > 3$, even though phenyl is a bulky group and alkyl groups are electron releasing.
How to approach these type of questions?

All the compounds are planar, steric hinderance might have played a more significant role if the attack of nucleophile was in the plane. Nucleophilic addition at the carbonyl group takes place at an angle of 107° with the plane.!

Addition of grignard reagents at the carbonyl group are primarily governed by electronic effects. More the positive charge at the carbonyl carbon more will be the rate of reaction.
The two phenyl groups reduce the $\delta +$ most significantly by resonance, next the dimethyl ketone by two methyl groups which reduce the positive charge by hyper conjugation and inductive effects.
The order can be finalized as:
$$1>2>3$$

References
Clayden, J.; Greeves, N.; Warren, S. Organic Chemistry, 2nd ed.; Oxford UP: Oxford, U.K., 2012; pp. 130


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Hydrolysis under basic medium

The first compound is said to be hydrolyzed more easily than the 2nd compound below under Alkaline medium . How is this possible ?
I thought due to resonance, -NO2 acquires negative charge which faces repulsion from the solvation effect by basic medium. But it will be relatively easier for oxygen in -CH3O( in 2nd compound) to donate its electrons to have it in resonance with benzene. But this contradicts with answer. Where am I going wrong?

Going by the mechanism of hydrolysis under basic medium, the lone pair of hydroxide ion will attack on C=O carbon.
Due to the negative mesomeric (-M/-R) effect and negative inductive effect of the nitro group, the C=O will act as a more electrophilic substrate as compared to the one which is has ether substituent,
(which due to its +M and +I effect will reduce the electrophilic character of the C=O ),
will be easier to hydrolyse.
The effect of solvation on substituents is less likely to tell which compound will hydrolyse easily as it is not the Rate Determining Step of the reaction. So although the observation is not wrong, its still not the deciding factor.

Moreover, since the molecule is completely conjugated(in fact cross conjugated), so only a partial negative charge would exist on the nitro group in the resonance hybrid, so it's hydration will be quite less to consider

Perhaps in alkaline water, the attack on the carbonyl carbon would be by OH- ion.


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Molecular dynamics after modeling a protein loop. Necessary?
Firstly, I am postign this question to both chemistry and physics stacks exchange as I am not sure to which it belongs and to use the benefits from both sites.
I have modeled missing loops in a PDB structure using chimera. Should I run MD or energy minimization on the output file?
Thank you in advance.

What properties of the protein do you want to calculate?

Of course you should. It's optimisation of structure. @IanBush It's not calculating properties, but complementation of crystallographic data  with computational methods.

To expand on Mithoron's comment, it's absolutely necessary to equilibrate (or at least minimize the energy) of your system once you've introduced an appreciable perturbation of any sort (in this case, you've appreciably changed the system itself!). To give you an idea, some lesser changes that would still justify equilibration are: changes in solvation, temperature, pressure, protonation of amino acids, oxidation state of cofactors, etc.
Note: by equilibration, I mean letting your system relax in the presence of surroundings. By minimization, I am referring to minimizing the total energy of the protein (explicitly solvated, unless you're doing some odd gas-phase or implicit solvation) in the absence of temperature and pressure (0 K, no thermostat or barostat).


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What is the difference between expanded structural formula and Lewis dot structure?
I am teaching myself the basics of chemistry and I have done Lewis dot structure before. Right now I am going into carbon chemistry which often use structure formulas to represent compounds.
There is this formula called the expanded structure. From what I can see, the expanded formula is just the same as the Lewis except we use the lines to represent bonds etc.
What is the difference between them and do we have to assign the formal charges in the expanded structure formula? Also, do we show the dots (valence electrons) in the expanded formula?

Yes. Extended molecular structures replace dots with line emphasising electron pairs that are shared between two atoms. In other words each line does represent a bond. Most often the EMS omit electrons that are not directly involved in bonds. As such the depiction of electrons between C and O in, e.g., a carbonyl group is simply sketched as C=O. However, whenever the presence of lone pairs is relevant to the discussion, those are explicitly shown.
In case in which the structure does not correspond to a single Lewis structure, the most important one is depicted and the situation is considered to be known to the reader. An archetype example is benzene. It can be sketched as having three double bonds but their delocalization is assumed. Those six delocalized electrons can be depicted as an inner ring, an expedient which is more evocative but must be abandoned when fine details are to be discussed (mechanisms of reaction or mesomeric effects). Moreover a circle in aromatic rings doesn't always represent six electrons.
Finally you can find depiction of delocalized electrons as variously dotted lines enveloping the group of atoms over which they do delocalize, for instance a kind of "cloud" around the two oxigens of a carboxylate anion. Their meaning is usually clear.
To the second question: on the light of the above yes, formal charge are to be assigned. They are usually important while discussing the relative stability of limiting forms which enter the description of the actual hybrid of resonance.
https://en.m.wikipedia.org/wiki/Lewis_structure. Particularly the subsections Resonance and Usage and limitations.
Edit: For sake of clarity I must add that unpaired electrons are always depicted. Indeed they are always of chemical relevance!

The "expanded structure" is the expansion of condensed structural formula. For example, the expanded structure of the condensed structure $\ce{CH3CH2CH3}$ is given below along with the lewis structure

Basically, the lewis structure has been simplified by representing the two electron covalent by a dash $—$ instead of dots. Lewis structure represents valence electrons or electrons in the outermost electron shell including the lone pairs, but the other structure does not represent the lone pairs if any.


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Why is carbon monoxide a monodentate ligand
Carbon monoxide has lone pairs on both the carbon and the oxygen atoms. However, I have never seen any complex where the oxygen atom or both the oxygen and the carbon atoms are the donors.
I know that the much lower electronegativity of carbon makes it a much stronger donor. However, this difference in electronegativity exists in the cyanide ion too, albeit a little less. However, cyanide ion is ambidentate.
My question basically is :
Why is carbon monoxide not an ambidentate ligand? Further, why is it not bidentate?

It has a lot of modes of coordination. It can't be bidentate because even when it binds with both atoms then it's either bridging or "hapto" mode.

It is possible for carbon monoxide to coordinate through oxygen, as reported here.  However, there is indeed a strong tendency for carbon monoxide to coordinate through carbon, essentially due to the electronegativity difference noted by the OP.  Cyanide is more likely to coordinate through N than carbon monoxide through O, simply because the electronegativity difference is not so large.  Even cyanide requires special circumstances to get it to coordinate through nitrogen, often involving the carbon being "preoccupied" with another atom -- just like carbon monoxide in the referenced article.


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Why are catabolic reactions exergonic and anabolic reactions endergonic?
Catabolism breaks apart large molecules and anabolism creates large molecules. Catabolism is supposed to be exergonic, while anabolism is supposed to be energonic. Id on't really understand this.
By definition a chemical bond is at an energy minimum. So if catabolism is taking apart large molecules with many bonds, which are at an energy minimum, how is this supposed to release energy (be exergonic) if the bonds are already at an energy minimum?
By contrast why is it that anabolism, which is putting creating a large molecule (which is a minimum energy state) from many atoms is enderognic and thus requires energy? I mean you're creating a minimum energy state, why would this require energy?

Besides the biological aspects consider that a stable molecule is at an energy minimum when isolated.  If you put it an environment than you must consider the global situation and how the energy in that system can get redistributed to attain an even lower energy state.  " So if catabolism is taking apart large molecules with many bonds, which are at an energy minimum, how is this supposed to release energy (be exergonic) if the bonds are already at an energy minimum?" because the fragments are at an even lower minimum.

In most chemical reactions, including those of catabolism and anabolism, some bonds break and others are formed. For example, if you hydrolize a biopolymer such as a polysaccharide or protein into its building blocks, you break bonds in the polymer and in water, but form new bonds between parts of the water and parts of the polymer.
Making a polymer from building blocks sometimes is endergonic (like some anabolic reactions) and sometimes is exergonic (like making polyethylene from ethylene). There is no inherent rule that says more complex molecules have higher or lower Gibbs free energy than the small molecules they are made from.
You could say there are three reasons why we eat food: To directly use the substances we ingest (e.g. vitamins), to use the substances we ingest to make new substances we need (e.g. amino acids in proteins), and to oxidize substances (in humans with $\ce{O_2}$ as the terminal electron acceptor) and harvest the Gibbs free energy. We choose food accordingly (e.g. we don't eat sand). Also, we don't eat Grignard reagents or other highly reactive substances because they are usually not available and we are not equipped to handle very exothermic reactions in our bodies. So the food we eat is fairly stable, and often it requires Gibbs free energy to drive biosynthetic reactions (explaining why many anabolic reactions are endergonic)
Molecules that can be hydrolyzed in the digestive tract using exergonic reactions are good as food. It is difficult to run endergonic reactions in the digestive tract (the reactants often used to drive endergonic reactions, most notably ATP, are found inside cells, not in the stomach or intestine). For those reasons, any large molecules we digest can be broken down to smaller molecules in exergonic reactions (these smaller molecules then enter cells, where further reactions could be exergonic or endergonic).
Specifically for hydrolysis (breaking biopolymers into building blocks) and condensation reactions (making biopolymers from building blocks), the hydrolysis reactions tend to be exergonic and the condensation reactions tend to be endergonic. This might be related to the high concentration of water throughout the body.

When we say a bonded substance is more stable, it means that it has less energy than a the total of its atoms freely floating around. Part of the total energy from the composite are stored in chemical bonds, which makes the resulting product has lower energy.
When a catabolic reaction occurs, the large molecule is broken into smaller molecules with the breaking of chemical bonds that held them together. The bond is broken and energy stored in the bonds are released. That is why for most of the times, a catabolic reaction is exergonic.
When an anabolic reaction occurs, smaller molecules or atoms form a large molecule. They store a surplus energy in the form of chemical bonds, making the resulting product lower in energy and more stable. Initially the reactants are good to be themselves, but when more energy is added to them, they would become unstable and therefore seek a way to store the incoming energy. Therefore it seems to us that they are taking in energy, or endergonic.


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Why is the methyl chloride and sodium hydroxide reaction irreversible?
Is the reaction of methyl chloride and sodium hydroxide (aq) irreversible simply due to the relative stability of sodium hydroxide and sodium chloride? Or is there more to it than that?

Which reaction and under what conditions?

Think about what the product of the reaction is going to be, assuming that the $\ce NaOH$ is aq (as stated) and dilute.


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Depicting molecules by Natta projection
Does anyone know the meaning of these two bonds in stereochemistry? These bonds are provided by the $\mathrm{\LaTeX}$ chemfig package. The creator of the chemfig package describes them as "hollow cram bonds".

Cram bonds are synonymous to the "wedged" bonds or, in this case, to "hollow wedged" bonds, which IUPAC [1, p. 1910] describes as follows (emphasis mine):

There are a variety of obscure conventions for depicting arbitrary mixtures of diastereoisomers. The most common (but still obscure) of these conventions says that bold and hashed wedges should be used for absolute configuration, unwedged bold and unwedged hashed lines represent relative configuration and racemic character, while hollow wedges and dashed lines represent relative configuration only.

A bit more context is given in [2] (bond type 4):

Central to the correct perception of a stereounit’s topography is the stylized generation of a vanishing point, realized by the graphic stereodescriptor, or
  stereobond, commonly called a “wedge” as illustrated in Figure 1 as 1a and 1b. Three such wedge symbols as in 3, 4, and 6 are in use, each of them capable of portraying the ligand Z in 1 and 2 as being located in the front (3 and 4) or behind (6) the picture plane. In an apparent search for further simplification, line symbols 5, 7, and 8 were invented as wedge substitutes.
  Lacking differentiation of their termini, line-type symbols appear, a priori, to be inherently incapable of serving such a purpose; nevertheless, they provide a useful function in the context of a convention. Accordingly, the stereogenic unit to which the line-type bond symbol is attached is considered to lie in the picture plane so that the symbol printed in bold (5) reveals its ligand Z in front of that plane, whereas the hashed symbols (7 and 8) are intended to show that the ligand is behind it. In other words, symbol 5 portrays the stereocenter at the vanishing point and hence functions such as 3 and 4.

Figure 1. Perspective drawings of stereocenters with commonly used stereobonds.

References

Brecher, J. Graphical Representation of Stereochemical Configuration (IUPAC Recommendations 2006). Pure and Applied Chemistry 2009, 78 (10), 1897–1970. https://doi.org/10.1351/pac200678101897.
Maehr, H. Graphic Representation of Configuration in Two-Dimensional Space. Current Conventions, Clarifications, and Proposed Extensions. J. Chem. Inf. Comput. Sci. 2002, 42 (4), 894–902. https://doi.org/10.1021/ci025518w.


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Number of products on dehydrobromination
Question:

The total number of alkene products possible from the dehydrobromination of 3-bromo-3-cyclopentylhexane using alcohol KOH is :

This question was asked here before but I am getting more products than those mentioned there
Dehydrobromination of 3-bromo-3-cyclopentylhexane and number of possible alkene products
My doubt is, why doesn't Hydride shift and ring expansion take place here? I would expect the Elimination reaction to be E1 as Br is a good leaving group and it is 3° halide. So there should be a carbocation intermediate on C3.
There is an alpha hydrogen in the cyclopentyl ring wrt this carbocation. Why won't hydride shift or ring expansion into cyclohexane (more stable ring) take place here?

The bromide is tertiary. There are no cations involved. There are five alkenes. There are stereoisomers as your link suggests.

Yes it is a 3° halide(fixed the typo). So are you suggesting that E2 would take place? Why wouldn't E1 take place?

With a strong base, such as KOH, present E2 predominates over E1. E2 is kinetically controlled and given sufficient time, the alkenes from an E1 will be equilibrated by the liberated HBr to a thermodynamic mixture of alkenes. The E2 mixture of alkenes is usually different from the E1 mixture.


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Superscript and subscript together after the same atom
In the chemical formula of some ion compounds, there is the necessity to write the charge in the superscript and the numbers of atoms in the subscript, but together at the right of the same symbol like in the PO43− ion:

Does IUPAC provide universal guidelines saying that I have to typewrite the two numbers in a stacked or staggered way?

https://chemistry.meta.stackexchange.com/questions/2803/formulae-with-charge-and-number-of-atoms-e-g-sulfate-ion-arent-displayed-pro?r=SearchResults

From IUPAC Green Book [1, p. 51]:

In writing the formula for a complex ion, spacing for charge number can be added (staggered arrangement), as well as parentheses: $\ce{SO4^2-}$, $\ce{(SO4)^2-}$. The staggered arrangement is now recommended.

Also, as you are apparently a $\mathrm{\LaTeX}$-user, the same convention is used by default in chemformula package. From the manual:

If a compound does not start with a sub- or superscript and there is both a sub- and a superscript, the superscript is shifted additionally by a length determined from the option charge-hshift = {❬dim❭}, also see page 13f.
The second point follows IUPAC’s recommendations.

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.

International and various national standards (e.g. ISO 80000, DIN 1338) do not agree with the typographic disaster of the staggered notation that was introduced in Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book). The standardized notation still is aligned: $\mathrm{PO_4^{3-}}$ or $\mathrm{(PO_4)^{3-}}$.
Nevertheless, the current IUPAC recommendations as well as the ACS Style Guide stagger the subscript and superscript: $\mathrm{PO_4{}^{3-}}$


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how could I make a file conversion from .spc (thermo fisher) in .cdf (Shimadzu) format file?
how could I make a file conversion from .spc (thermo fisher)  in .cdf (Shimadzu) format file?

I'm voting to close this question as off-topic because it is not about Chemistry but about conversion of file formats.

OpenChrom (open source) should handle both SPC and CDF out of the box, but I haven't used it for years. From what I remember both formats are proprietary (?) and there might be some quirks when using a third-party converter.

Related: https://chemistry.stackexchange.com/questions/51058/how-to-convert-spc-files-to-ascii-without-original-spectroscopy-software


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What explains the stark white colour of my iodometric titration analyte solution?
During my Advanced Higher (secondary school) investigation, I reacted pure copper with concentrated nitric acid to acquire a copper nitrate solution. This was neutralised with excess sodium carbonate and brought back to a clear solution with ethanoic acid.
I then reacted the solution with potassium iodide in a conical flask before titrating it with sodium thiosulfate.:
$\ce{2Cu^2+(aq) + 4I^-(aq) → 2CuI(s) + I_2(aq)}$
$\ce{I_2(aq) + I^-(aq) ⇌ I_3^-(aq)}$
$\ce{I_3^-(aq) + 2S_2O_3^2-(aq) → S_4O_6^2-(aq) + 3I^-(aq)}$
This analyte solution was at first brown, but when titrated it turned a pale coffee colour. When starch was added just before the end point, it turned a dark grey-brown colour, as expected - the iodine forming a complex with starch. https://en.wikipedia.org/wiki/Iodine_test
But when all of the iodine had reacted with the sodium thiosulfate, I was left with a stark white solution. I tested to make sure no iodine was left over by adding in another drop of starch - no reaction. So, this seems to be the endpoint.
Copper iodide seems the likely explanation, but the wikipedia page shows that it is mostly insoluble in water. It is soluble in iodine solutions but all the iodine is gone, presumably.
Does it take time for the copper iodide to precipitate? I quickly rinsed out the conical flask. There was some white staining on the sides which I think is copper iodide.

You probably have the solution of potassium iodide in excess for the oxidation. $\ce{CuI}$ is somewhat soluble in in iodide solutions (not in iodine solution), due to common ion effect (see Wikipedia for solubility).

@MathewMahindaratne Yes, that makes sense! I can't believe I forgot about the excess potassium iodide. It's the small things that always end up being the key. Can you post your comment as an answer so I can accept it?

I agree with one of answers provided that what you obtained at the end point is a colloidal solution of copper(I) iodide. Since $\ce{CuI}$ is insoluble in water (0.0042% w/v), the stark white color (probably milky) solution is nano particles of the solid suspended in $\ce{KI}$ solution, assuming you have used significantly excess $\ce{KI}$ for the reaction. Thus your solution should be in following equilibrium:
$$\ce{CuI + I- <=> [CuI2]-}$$
Wikipedia states that:

$\ce{CuI}$ is poorly soluble in water (0.00042 g/L at 25 °C), but it dissolves in the presence of $\ce{NaI}$ or $\ce{KI}$ to give the linear anion [$\ce{[CuI2]-}$. Dilution of such solutions with water reprecipitates $\ce{CuI}$. This dissolution–precipitation process is employed to purify $\ce{CuI}$, affording colorless samples.

Thus, your intention is to isolate solid $\ce{CuI}$, you must add extra $\ce{KI}$ while moderate heating your solution until you receive a clear solution. Let the solution to cool to room temperature, and add adequate amount of pure water (about 4 times by volume) to cause the precipitation. Upon standing for about $\pu{1 h}$, $\ce{CuI}$ will precipitate.
A typical procedure is given below in literature (Ref. 1):

Commercial copper(I) iodide or a sample that has become discolored ($\pu{3.80g}$, $\pu{0.02 mole}$) placed in a 250-mL Erlenmeyer flask is dissolved with magnetic stirring and moderate heating by the addition of a minimum volume (about $\pu{125 mL}$) of $\pu{ 3.5 M}$ potassium iodide (about $\pu{3 hr}$ is required). Activated charcoal (e.g., Matheson, Coleman, and Bell "Darco" (3-60) (about $\pu{5 g}$) is added to the resulting solution, which varies in color from light tan to dark brown, depending upon the extent of contamination. The mixture is stirred magnetically with moderate heating until the supernatant solution becomes colorless, at which time stirring is stopped and the charcoal is allowed to settle, indicating that the contaminating iodine has been completely adsorbed (about $\pu{10 min}$). The mixture is suction-filtered through a 150-mL (60-mm i.d.) sintered-glass funnel (medium porosity), and the filtrate is diluted with sufficient water (about $\pu{500 mL}$) to cause reprecipitation of the copper(I) iodide upon standing for about one hr.

Important: Vigorous heating in air causes $\ce{CuI}$ to decomposes to $\ce{CuO}$.
Ref. 1: Inorganic Syntheses, Volume 22; Smith L Holt, Jr., Ed.; Inorganic Syntheses, Inc.: New York, NY, 1983, Chapter 2: Transition Metal Complexes and Compounds (sub Title: 20. Purification of Copper (I) Iodide; submitted by G. B. Kauffman & L. Y. Fang; Checked by N. Vishwanathan & G. Townsend), pp 102–103 (https://doi.org/10.1002/9780470132531.ch20).

You’re correct that this is just a suspension of copper(I) iodide. If CuI were soluble in water, the solution would be clear; the white colour is a sign of (very small) particles of the solid suspended in solution.


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What happens if the electron in the lowest energy level of an atom loses its energy?
What happens if the electron in the lowest energy level of an atom loses its energy? Excited states spontaneously give of energy and transition to a lower energy state. Why can't an electron in the lowest energy level spontaneously lose energy and fall into the nucleus? Wouldn't that be an even lower level as it maximizes the coulombic attraction between electron and nucleus?

It just does not do so, by definition of a ground state.

That's like asking what happens if you fall off the floor of your room.

Maybe with the edits, you can reopen the question. It is one that has been studied historically before quantum physics was established, and it elicits some important concepts (in a classical view, the electron would lose energy; there are no electrons at rest; Heisenberg's uncertainty principle; zero-point energy).

By definition of lowest energy level, you cannot go any lower!
As you already suggested, by classical mechanics one could imagine a situation where the electron is even closer to the nucleus (until it falls into it), than the lowest energy level (= quantum mechanical ground state) actually has. But quantum mechanics does not allow this. This is one of the key aspects of quantum mechanics.
This also relates to having no energy level between the ground state and the first excited state. Energy levels are discrete or quantized.
Again in classical mechanics, where energies are continuous, we could describe some energy level in between. But quantum mechanics simply does not allow this.
The difference between what would classically be possible and what quantum mechanics allows for, is commonly also known as zero point energy. (And anyone who tries to sell you on getting some "free energy" out of this is wrong, because the whole point of the term zero point energy is, that you cannot go any lower than that.)
Why does quantum mechanics say there is a lowest energy level?
It does so, because it needs to represents reality. Remember that quantum mechanics is not the ultimate truth, but just a theory developed to give a working description of reality. To do so, it needs to assume a couple of things.
Let's assume for a moment there would be no lower limit and an electron could go arbitrarily close to a nucleus, as classical mechanics allows it to do. This would  make atoms unstable. The electron would constantly emit light (as the Maxwells equations for a moving charge require) and spiral into the nucleus.
However this conflicts with reality, as it would mean: no stable atoms, no stable molecule, no chemistry, no biology, no humans, nobody to ask that question in the first place. Additionally, it could not explain why emission spectra look the way they do.
In order to reflect these experimental facts, quantum mechanics can only be a viable theory if it asserts energy levels are discrete and there is a lowest possible energy level.
Whether there is some deeper explanation where this discretized nature is coming from is unkown. A theory going beyond quantum mechanics might explain it, but this is an area of ongoing research in the physics community. But chemistry is not really concerned with this, as quantum mechanics is all it needs.


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Quaternary Ammonium salts and peroxy acid
I came across this question in my test:

This compound

is treated with $\ce{CH_3CH_2I}$ and then with $\ce{H_2SO_5}$ and heated.
  The product is:
  A)

B)
  $\ce{CH_2CH_2}$
  and two other impossible options.

I first don't understand the role of $\ce{H_2SO_5}$ in this. I know about Hoffmann elimination of quaternary ammonium hydroxides but in that we use silver oxide to form the hydroxide from the iodide. And then heating is done for elimination.
Secondly the answer given is option B but I am not able to understand why is that. The second example in this page says otherwise Quaternary ammonium salts
Does this have anything to do with the $\ce{H_2SO_5}$?
My third doubt is a general one. What happens in general when different alkyl groups are attached to the the nitrogen in a quaternary ammonium hydroxide. Which one undergoes elimination first? Are there any rules to determine that?

@NilayGhosh How is that question related to this one

I think H2SO5 is used to oxidize Iodide so it could go away for sure and cleanly. R3NI might sublimate, but (R3N)(HSO4) likely cannot, and HI is a very strong acid unlikely to be displaced. In classical example Ag2O is used which isn't cheap and the products are distilled off. In this option hydrosulfate remains in reaction flask, which might be useful.

As for elimination, apparently Hoffman elimination favors products eliminating hydrogen from less stericaly hindered position. Usually it is used to judge the product from same radical as Hoffman elimination is usually used with trimetialalkyl salts, but there is no reason to not apply it between radicals as well.

The first question is how much EtI is the starting material treated with? If it is only 1eq then I think the product is the tertiary and the persulfate is oxidising that to the N-oxide then eliminating by a Cope elimination

@Waylander it is in excess

I still think the question is about a Cope elimination. That is quite a hindered amine to try and quaternise with EtI.

I believe this is an example of a Cope elimination mechanism here
The EtI is making the tertiary amine. The persulfate is taking the tertiary amine to the N-oxide. Heating this produces ethene and hydroxylamine of the starting amine.
This is consistent with the Cope as "The sterically demanding amine oxide function reacts preferentially with the more easily accessible hydrogens"


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What would be the intermolecular forces between the molecules of methanetetrol?
Methanetetrol has the following chemical formula: $\ce{CO4H4}$. It is also referred to as orthocarbonic acid.
Since the molecule has a molecular geometry of a tetrahedron, similar to that of a molecule of methane, I would assume that the dipoles at the bonds cancel each other out. Hence, making the molecule non-polar with no dipole or charge separation overall. Thereby, the intermolecular force between the molecules of methanetetrol should be London dispersion forces.
However, I've been told that the intermolecular force between the methanetetrol molecules is hydrogen bonding since the four $\ce{OH}$; hydroxyl groups in the molecule are capable of making 4 hydrogen bonds with 4 subsequent methanetetrol molecules.
Hence, my question is how can methanetetrol have hydrogen bondings as the intermolecular force between its molecules if the molecules do not have a dipole; or so I believe, i.e. what do you think is the intermolecular force between methanetetrol molecules?

You do not need a molecular dipole. The polarization of an individual bond could suffice. According to Wikipedia the molecule remains hypothetical! so somebody told you a hypothetical fact.

This molecule is very polar, despite having zero dipole moment. Besides, it does not exist.

Let's try to guess a reasonable structure for $\ce{C(OH)_4}$. The oxygen atoms and the central carbon would probably form a tetrahedron, therefore, those bond dipoles would basically cancel (as in e.g. $\ce{CCl_4}$). However, the carbon-oxygen-hydrogen angle would probably not be 180°. Whether the hydroxyl group would rotate freely about the carbon-oxygen bond or there would be some intramolecular hydrogen bonding preventing that, I would assume that intermolecular hydrogen bonding could still occur. (Which would be the first step towards eliminating water and forming $\ce{CO_2}$.)

Maybe it would be better to discuss the question whether non-polar molecules can make hydrogen bonds using stable molecules as example, such as 1,2 ethanediol (ethylene glycol). On paper in the most stable conformation, ethylene glycol looks non-polar (2-fold rotational symmetry). However, if we compare its properties to those of ethanol (a polar molecule capable of making hydrogen bonds), it is more viscous, it mixes with water at any ratio (is miscible with water) like ethanol, and its dipole moment is higher than that of ethanol.
There are many such molecules, see for example the discussion on hydroquinone.
On the other hand, carbon dioxide has a dipole moment of zero and does not form hydrogen bonds. In a study by Lam et al (2014) that combines X-ray absorption and computational methods, the authors conclude that

neutral CO2 acts like a typical hydrophobic solute, inducing stronger hydrogen bonds among the adjacent waters


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Acid catalysed dehydration of unsymmetric alcohols
How to find the major product of acid catalysed dehydration of unsymmetric alcohols. For example, in the dehydration of butan-2-ol using conc. H2SO4, which is the major product? But-1-ene or But-2-ene?
And, in the dehydration of 1-methylcyclohexanol, the major product is 1-methylcyclohexene and not methylenecyclohexane. But how did we work it out??

See Zaitsev's rule

Zaitsev's rule clearly addresses your query. Or you could look at it this way. The product with more hyper-conjugation is more stable. Hence but-2-ene is more stable than but-1-ene. Always see hyper-conjugation for more stable, major product.


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Detecting alkaloids - Dragendorff's reagent
I am writing a book and as part of the plot one of the characters has a plant that contains alkaloids in tincture form. They want to do a test to detect for the alkaloids so they use the Dragendorff's reagent test. Quote below:

Bindweed contains alkaloids. The Dragendorff's reagent is a simple test to detect for them. They had all I needed in the chemistry lab... bismuth nitrate, hydrochloric acid, potassium iodide. When I added milk to the solution it turned orange confirming the presence of alkaloids. I did a control test too and nothing happened.

Would this technically work?

Here is how you make Dragendorff's: this prep uses acetic acid rather than hydrochloric https://communities.acs.org/thread/4183  I've seen other preps that use tartaric acid. It is possible to get false positives from other plant extracts that have a basic centre  https://www.researchgate.net/post/How_does_the_Dragendorff_reagent_work_for_detecting_Alkaloids

Thank you for coming back! I need to add that in my book the bindweed plant has been made into a tincture, so dissolved in alcohol, and then a few drops of it are added to a glass of milk. It is this milk that is being tested. I don't know if the Dragendorff's test would still pick up the alkaloids of the bindweed in this form?

It should if the alkaloids have extracted into the alcohol which I think they would. You should have them run the blind of testing milk without the tincture to see if that tests positive - (also an author)