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Demonstrating decomposition of hydrogen peroxide using iron(III) nitrate catalyst
I need a way to prove/show that hydrogen peroxide was decomposed through use of catalyst.
I want to ensure that my catalyst: $\ce{Fe(NO3)3}$ or iron(III) nitrate is a catalyst, not a reactant/ consumed during the reaction.

$$\ce{2 H2O2 (aq) ->[Fe(NO3)3 (s)] 2 H2O (l) + O2 (g)}$$

When the reaction is happening, I will introduce a wooden glowing splint over the bubbling reaction and the wooden splint glowing brighter or reigniting will show that oxygen is being produced.

Question:

Now, how do I show that water is produced? Would I just boil the product I get after reaction (image above) and put cobalt chloride paper at the water vapour?

Also, if I were to use the orange solution above again as a catalyst (since it still contains iron(III) nitrate), would hydrogen peroxide decompose again? If so, is there a way to put iron(III) nitrate back to its solid state? or any way to reuse as catalyst?

Edit:
If it is easier to answer, it doesn't have to be with iron(III) nitrate. I have an option to use manganese dioxide, which is another catalyst that I can substitute for iron(III) nitrate. I think it should be OK since it does the exact same reaction.

Detection of oxygen: Detection of $\ce{O2}$ by a glowing splint is a good way to detect the oxygen. Also you could capture the gas by a simple fixture e.g.  and demonstrate the volume change in the receiver. This way you can actually measure moles $\ce{O2}$ produced (by $PV=nRT$) then moles $\ce{H2O2}$ decomposed stoichiometrically by the formula you've written.
Choice of catalyst: Using $\ce{MnO2}$ would be better if you want to make sure you have a catalyst. $\ce{MnO2}$ will not be consumed during the decomposition; I'm not sure about iron(III) nitrate.
Or, detecting change in $\ce{Fe(NO3)3}$ concentration: Addition of a very small concentration of potassium thiocyanate, $\ce{KSCN}$ (say 1/100 of your $\ce{Fe^{+3}}$ concentration) will yield a deep red product, iron thiocyanate ($\ce{Fe(SCN)^{+2}}$). If you have access to a spectrometer you can measure absorbance of the initial solution's product and the product after decomposing $\ce{H2O2}$.
Prove water was produced: The best way I can think of is to measure the (subtle) change of density of your solution before and after decomposition. Using $\ce{MnO2}$ would make this easy because you can remove/filter it as a solid after decomposition and thus measure mass/volume of your solution before and after. $\ce{H2O2}$ and $\ce{H2O}$ have small albeit detectably different densities at RT. $\ce{H2O2}$ is more dense than water so your density should decrease.

How would i remove/filter manganese dioxide as solid from water? Is there any specific equipment? Or wcan i just boil it to remove water?

Just filter paper through a funnel, maybe a vacuum if you want to be thorough.

Would there be a way to prove water for iron(III) nitrate?

You can probably still detect a small decrease in density even with iron(III) nitrate in solution

Yes, apart from potassium iodide which is commonly used as a catalyst in the decomposition of hydrogen peroxide, $\ce{Fe^3+}$ salts, manganese dioxide and nickel hydroxide can be used as a catalysts as alternatives. Since, iron nitrate contains $\ce{Fe^3+}$, it can be used as catalyst.
There are some papers that discuss the use of $\ce{Fe^3+}$ salts as catalyst. Following is the relevant information from the papers:

Paper 1

We should consider the role of the Ferric Chloride ($\ce{FeCl3}$)
  as catalyst in the decomposition reaction of hydrogen peroxide.(...)
  The fact is  that Iron can exist in two different oxidation states,
  $\ce{Fe^2+}$ (Ferrous) and $\ce{Fe^3+}$ (Ferric), allows the catalyst
  to break the reaction into two different redox steps, each of which
  has a lower energy barrier to completion than the uncatalyzed
  reaction:
$$\ce{H2O2(aq) + 2Fe^3+(aq) -> O2(g) + 2 Fe^2+(aq) + 2H+(aq)}$$
  $$\ce{H2O2(aq) + 2 Fe^2+(aq) + 2 H+ (aq) -> 2H2(l) + 2Fe^3+(aq)}$$
Note the first step in the catalyzed reaction involves reduction of
  the Ferric Ion ($\ce{Fe^3+}$) to the Ferrous Ion ($\ce{Fe^2+}$), which
  is then re-oxidized to Ferric Ion ($\ce{Fe^3+}$) in the second step.
  Hence, on net, the catalyst is not consumed during the course of the
  decomposition.

Paper 2

$\ce{Fe^3+}$ ions is actually a homogeneous catalyst. The catalytic
  decomposition of hydrogen peroxide can be essentially explained by two
  different mechanisms based on the mutual redox transition
  Fe(III)/Fe(V) (KREMER-STEIN mechanism) and Fe(III)/Fe(II) (HABER-WEISS
  mechanism), respectively.
According to the mechanism proposed by KREMER and STEIN an
  intermediate oxygen complex of iron with oxidation number +V is
  primarily formed by the reaction of $\ce{Fe^3+}$ with $\ce{H2O2}$.
  This complex reacts with another $\ce{H2O2}$ molecule to water and
  oxygen thereby reforming $\ce{Fe^3+}$.
$$\ce{Fe^3+ + H2O2 <=> [Fe^{III}OOH]^2+ + 2H+ <=> [Fe^{V}O]^3+ + H2O ->[H2O2] Fe^3+ + 2H2O + O2}$$
According to the mechanism proposed by HABER and WEISS the $\ce{Fe^3+}$ ions
  initiate a radical reaction, after which the chain reaction consumes
  the hydrogen peroxide. This mechanism can explain the high reaction
  rate very well.
Chain initiation:  $\ce{Fe^3+ + H2O2 <=> [Fe^{III}OOH]^2+ + 2H+ <=> Fe^2+ + HOO. + H+}$
Chain propagation: $\ce{Fe^2+ + H2O2 -> Fe^3+ + 2OH.}$
  $\ce{Fe^3+ + H2O2 + OH. -> Fe^3+ + HOO. + H2O -> Fe^2+ + H+ + O2 + H2O }$

Answering the update
Yes, manganese dioxide can also be used as a catalyst. From paper 2:

Manganese dioxide is an example for a heterogeneous catalyst.
  The surface of solid manganese dioxide provides a particularly
  favorable environment to catalyze the decomposition, though the
  mechanism is not understood very well. (...)


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Dilution question, "70 wt. % in H2O"?
If I have a solid and want to make an aqueous solution of 70%, how much water do I need? Is it as simple as weighing out 70 $\mathrm{g}$ of solid and adding water until I have 100 $\mathrm{mL}$? Will this solution be 70 wt. % $\ce{H2O}$?

To make a solution that is 70% solute by weight with $\ce{H2O}$ as the solvent, take 70 $\mathrm{g}$ of solute and add 30 $\mathrm{mL}$ of water. The resulting solution will have a mass of 100 $\mathrm{g}$, 70 $\mathrm{g}$ of which is the solute; hence, 70 wt. % in $\ce{H2O}$.

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How to calculate the rate of formation of a product when given the rate law?
The following data are given for the reaction of $\ce{NO} \text{ and } \ce{Cl2}$:

$$\ce{2NO + Cl2 -> 2NOCl}$$
The reaction is second order in $[\ce{NO}]$ and first order in $[\ce{Cl2}]$,
and the  initial rate equals $\pu{1.43E-6 mol L^-1 s^-1}$ at the instant when $[\ce{NO}]_0 = [\ce{Cl2}]_0 = \pu{0.25 mol L^-1}$.

The problem then says:

Calculate the rate of formation of $\ce{NOCl}$ when $[\ce{NO}] = [\ce{Cl2}] = \pu{0.11 mol L^-1}$.

I thought it's just $(2) × (\pu{9.152E-5})× [0.11]^2[0.11]$, which is $\pu{2.44E-7}$,  but the answer provided is half that $(\pu{1.22E-7})$, so I must have gotten the rate equation incorrect.
Is the rate of formation of $\ce{NOCl}$ equal to twice the rate of the reaction, as when I consider this reaction equation: $\ce{2NO + Cl2 -> 2NOCl}$?
Or is the rate of formation of  $\ce{NOCl}$ directly equal to the rate of the reaction, as when I consider this reaction equation: $\ce{NO + 1/2Cl2 -> NOCl}$?
I'm really confused how to tell which of these two approaches to use.

$$r=k \times [NO_2]^2 \times [Cl_2]$$
From the initial condition information =>
$$k=\frac{r}{[NO_2]^2 \times [Cl_2]}$$
Plugging the numbers=>
$$k=\frac{1.43 \times 10^{-6}}{0.25^2 \times 0.25}$$
So, $$k=9.152 \times 10^{-5}$$
For 0.11 condition =>
$$r=9.152 \times 10^{-5} \times 0.11^2 \times 0.11$$
or $$r=1.21 \times 10^{-7}$$

The question asked to determine the rate of formation of NOCl{R(NOCl)}   ?IsR(NOCl) = r , corresponding to the stoichiometric coefficient  of NOCl in this reaction equation :  NO+(1/2)Cl2⟶NOCl  or  R(NOCl) = 2 r   corresponding to the stoichiometric coefficient  of NOCl in this reaction equation 2NO+Cl2⟶2NOCl

My bad, now I understand. You are right, it should be 2r for NOCl production.  The answer provided is wrong.

If we multiply the reaction equation  by 2 , does the  rate of formation of NOCl change ?How much ?how and why ?

How the way of writing the chemical reaction -Duplication the reaction-affect on rate ?

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Why can't a double displacement reaction be a redox reaction?
Why can't a double displacement reaction be a redox reaction?

Oh yes it can, especially when an element is formed.  The reaction
$\ce{NaH + H_2O->NaOH + H_2}$
is a double displacement reaction in which the hydride and hydroxide ions switch partners, and also a redox reaction in which some of the hydrogen changes oxidation states.


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Water-soluble powder for traction on wet floor
Do you know of a substance that can be used for traction on a wet floor and dissolve over time and will not corrode steel?


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Effect of Clemmensen reduction on carboxylic acids
Does the Clemmensen reduction have any effect on the carboxyl group $\ce{-COOH}$ if it is attached to some organic compound? If yes, what does it get reduced to?

You're putting zinc in acid, so there will be a reaction, but maybe not the one you wanted?

I don't think so that there would be removal of carbonyl group because the -cooh group would be in resonance and also clemmensen reduction is preffered for aldehydes and ketones to remove oxygen .
More help would be afyer understanding its mechanism hereMechanism of clemmensen reduction
Also the zinc(amphoteric) might react with the acid giving salt.


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Side chain formation in polymerisation of ethylene
I am having difficulty understanding (or at least need confirmation of my understanding) of this diagram regarding the formation of sidechains of polyethylene. If someone could explain it, that would be great :). Also, what do the rotation symbols mean? Below I include my understanding of the reaction.

My understanding is that the chain terminates with a carbon with an extra bond available. It covalently bonds to a hydrogen from back in the chain, thus forming a complete carbon (connected to the carbon before, and 3 hydrogens).
The group it stole the hydrogen from now becomes a carbon with a missing hydrogen, with an extra bond to fill. Therefore that carbon is connected to the carbons on the left and right of it, and missing one hydrogen on the side. Therefore the group should be CH, since it only has 1 hydrogen left after its other one was stolen. Why does the diagram say CH2?
Then, to the best of my understand, an ethylene group attaches to the side with the missing hydrogen, with the carbon filling the hydrogen's place.  Thus the chain can be continued on the side, indefinitely. In option B: butyl side chain, the hydrogen from another part detaches and attaches to the open carbon on the side chain, terminating it, and forcing the original chain to continue.
In option A: what happens? I don't understand. Are the 2 ethyl side chains never ending? If so, how does that happen? How does the second side chain be produced?


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how to find pressure of a liquid when given volume (1 and 2) and one pressure?
A 5 mL sample of an unknown liquid is placed into a 250 mL flask that has had all of the air removed.  The pressure measured in the flask is found to be 35.4kPa.  In a seperate trial the student puts 214 mL of the same liquid into the same flask at the same conditions.  What will the pressure measure (in kPa) in the second trial?
I've looked through my notes and I don't have a formula or anything in there that I can use volume with. Can someone help me start this question?

Maybe the liquid volume doesn't matter.  Maybe you don't need an equation.  Remember that in an (ideal) gas, pressure and concentration at a given temperature are really just two ways of talking about the same thing.  I hope I didn't just confuse you more, but I think you are making it too hard.

i didnt think of it that way. im definatly overthinking! thanks for your help!


text:
Identifying T1 and T2 in the Claperyon equation
Is there a way to identify T1 and T2 in the two point form of the claperyon equation?
I suspect that the greater temperature is T1 and the smaller temperature is T2.

$T_1$ simply corresponds to $P_1$, same with $T_2$ and $P_2$.
It is not a matter of smaller/greater. $T_1$ is the temperature at $P_1$


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Is the hydrogen atom attached to the benzylic carbon in benzaldehyde acidic?
If benzaldehyde loses hydrogen atom at the benzylic carbon, the negative charge will disperse into the benzene ring via resonance. So shouldn't the conjugate base formed be stable, implying that the hydrogen was acidic?

Its pKa is 14.90, which is not acidic, but is acidic than most non-carboxyl-containing organic compounds.

Yes and no.
What you discuss is only possible if the $\ce{C-H}$ bond is perpendicular to the ring plane. However, since the carbonyl group is planar and better conjugation can be achieved with the $\ce{C=O}$ bond participating in resonance with the benzene ring the predominant configuration of unsubstituted benzaldehyde should be planar, i.e. the $\ce{C-H}$ bond being in the ring plane.
If you introduce steric hindrance, e.g. 2,6-di-tert-butylbenzaldehyde, the chance of the carbonyl group turning out-of-plane is much greater resulting in a better overlap of the $\ce{C-H}$ bond with the ring system and thus a lower $\mathrm{p}K_\mathrm{a}$.

I think I like this answer better than mine.  I'm out of votes for today, but in another 50-odd minutes I'll probably up-vote yours and delete mine.  Nice job.

So, it is solely up to the OP.  I found this http://chemistry.stackexchange.com/help/deleted-answers , that says "Answers can be deleted at any time by their authors, unless the answer has been accepted by the question asker. "

@Suprateek Das, I think was the better answer, good change ;)

Acidity of a weak acid is relative by definition (e.g. Ka).  The wording of your question may not be quite right; you really mean "proton" or $\ce{H+}$ rather than hydrogen atom leaving or we'd be talking about radicals, not ions.  But your statement is correct in that if you had a cyclohexane ring for example rather than a benzene ring, then you would not get the stabilization as you discussed.  Thus, that hydrogen is much more acidic (meaning it leaves as $\ce{H+}$) because of the aromatic functionality than it would be in a compound without such a negative-charge-stabilizing functional group.

There's no mesomeric stabilisation in this case. OP is wrong.

It looks right to me.  But if you can show in an electron-pushing diagram how it's wrong I will consider deleting my answer.

Electron pushing isn't the complete story here - as long as the C=O remains coplanar with the ring, the negative charge will reside in a sp2 orbital that is orthogonal to the pi system of the ring, so it cannot enjoy the stabilisation by delocalisation into the ring.

Good point @orthocresol. I think I may pull this answer in favor of Jan 's. I don't know the ramifications for the OP having already accepted this answer, but I assume it's no problem and I can just delete it.

@airhuff You can’t delete an accepted answer afaik. You’ll have to wait for OP to mark the other answer accepted.

@Jan, thanks for the input.  Since delete still appears to be an available option, I'll try it and see what happens.

Well I guess even without mesomeric, there should be some inductive stabilisation, so I guess your answer is still valid.


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Using the quasi-steady assumption for reaction-diffusion equations
I have a question regarding an equation for changes in chemical concentration. The chemical concentration we are looking at changes due to diffusion, decay and production of the chemical. Since the diffusion happens so much faster than the other two, it is given that we can you a so-called quasi-steady assumption. This then gives the equation:
$$D \nabla^2c-a \cdot c+b\cdot c=0$$
where $a$ and $b$ are rates of decay and production, respectively. $c$ is the chemical concentration and D is the diffusion coefficient.
Is it correct that the reason we only have a derivative expression for the diffusion because it happens so fast relative to the other two, and therefore we can assume the chemical concentration for the two other processes to remain constant (using the quasi steady assumption), or is that not a correct reasoning for why the equation has the simple expression for decay and production, while we need a more complex expression for diffusion?

Read about the diffusion equation (Fick's law). ▽² is a spatial derivative.

Diffusion by nature is associated with concentration by a second-derivate gradient. Decay and production are not necessarily.
I think you're just asking if it's a fundamental thing. It is.

This is an example of a reaction-diffusion system where the participating species both react as well as diffuse. Such systems can be described using partial differential equations.
The mathematical model describes how the system changes with time. So we have $$\frac{\mathrm dC}{\mathrm dt}=f(C)$$
where $C$ is a vector with each element denoting the concentration a species ($C_i$).
At quasi-steady state the rates of some of the species will remain unchanged over time.
$$\frac{\mathrm dC_i}{\mathrm dt}=0$$
The reaction part usually comes from mass action kinetic models (which gives you rates for zeroth/first/second order reactions).
The diffusion part comes from Fick's laws.
$$\frac{\partial C}{\partial t}=D\nabla^2C(x,y,z,t)$$
Here the $\nabla^2$ denotes spatial double derivatives $\large\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}$
which kind of describe how the different species would be distributed in space. The reaction-diffusion equation is a sum of the reaction and the diffusion parts. At quasi-steady state, the concentration of a given species at any given point in the space would remain unchanged over time. This only sets the time derivative to zero, as I previously mentioned.
The time derivative denotes the net accumulation rate of a species. In the absence of diffusion, it will simply be the sum of rates of all reactions leading to formation of the species minus the sum of rates of all reactions causing removal of the species (principle of mass balance).
In the absence of reactions the diffusion terms will describe how the spatial pattern of the species will change over time.
At (quasi) steady state, the time derivative goes to zero which means that all the reaction and diffusion terms balance each other out such that there is no change in the concentration of the corresponding species over time.

Thank you very much for your reply! Does this mean that the expression for production and decay are defined as $a\cdot c$ and $b \cdot c$ also if we did not assume quasi-steady approximation, because the quasi-steady approximation just deals with time, and this equation for decay, production and diffusion is just over space?

@David At (quasi) steady state the time derivative of concentration becomes zero. In this case the time derivative is a sum of reaction rates described by the kinetics and spatial diffusion. In the absence of reactions i.e. when there are only diffusion terms, you can still solve for the steady state spatial distribution. When there is no diffusion then it would mean that the production and degradation reactions rates are equal thereby causing no net accumulation of product. (Note that the time derivative denotes accumulation rate).


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Does the green color represents existence of Copper in it
World is full of green, because of plants. Does it really signifies existence of Copper, in particularly by leafs.
I well understood that Prolific Tamarind Colored Salt has base element Copper. More usage of Tamarind makes us dump as per Siddhas in India.
Would like to know specifically that green color means existence of Copper. Is that completely true. Or by Alkali elements it might be faulty in some situation.

https://en.wikipedia.org/wiki/Chlorophyll

Azurite is a copper mineral and it is blue. Metallic copper is reddish. Cuprite is a copper mineral and it is red. So there are a lot of copper minerals that are not green.

No, there are plenty of compounds, even metal complexes, which are green without containing copper

Die or pigment gets color by right agent which is been in use. Synthesizer gets [or] reveals color by/on physical principles which is not bounds to Chemical Bond, maybe its just a relevant to it.

The green in plants arises not from copper but from chlorophyll.  It is true that many oxidized species of copper are green, but in plants, it is clorophyl as described for example in this Wikipedia page:

"Chlorophyll (also chlorophyl) is any of several closely related green
  pigments found in cyanobacteria and the chloroplasts of algae and
  plants. Its name is derived from the Greek words χλωρός, chloros
  ("green") and φύλλον, phyllon ("leaf"). Chlorophyll is essential in
  photosynthesis, allowing plants to absorb energy from light.
  Chlorophyll absorbs light most strongly in the blue portion of the
  electromagnetic spectrum, followed by the red portion. Conversely, it
  is a poor absorber of green and near-green portions of the spectrum,
  which it reflects, producing the green color of chlorophyll-containing
  tissues."

Does green pigment is green because of Copper?

No, there is actually no copper in chlorophyll, or enough copper in plants to produce the strong green we see.  Chlorophyll is purely organic, with a Mg bound to it.

Oh! ok! By mutation or by laws is there intervention that color formation gets using Copper, because it is pigment.

Mg I agree. But not every green is by Mg, if I am right.

That is correct.  To answer your question, copper plays absolutely no role whatsoever in the "green-ness" of plants.  Nice guess with your knowledge that forms of copper can be green, but as you know plenty other things are green, like the copper-free chlorophyll ;)

Thanks for Agreeing! When it is Organic Mutation plays important role than in In-Organic. So while, other elements can be placed as part of mutation in it, however it is so pure. And as per Material Science nothing is pure so.

Great!  I'm not sure whether I fully answered your question, but if you feel I was helpful please consider accepting the answer (or another one if a better one comes along).

Note that liquid chlorine is also greenish, hence chlorine.

@WYSIWYG, I don't want it on my veggies, but good point ;)

@airhuff  "Oh! ok! By mutation or by laws is there intervention that color formation gets using Copper, because it is pigment." You actually understood this?! I applaud you ;p (+1's answer)

@paracresol, no that one was lost on me.  My "that is correct" comment referred to the immediately previous statement "But not every green is by Mg, if I am right"...I agree that not all green is caused by Mg ;)


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Can ML algorithms be applied to predict the products of a chemical reaction given a set of reactants?
How would the training data look like ? Would it be a set of reactions ? What would be a good algorithm to run on the data ?
Wolfram Alpha does have an interface for chemistry, but does not seem really robust. The other databases for chemical reactions seem to work on a static database.
The ultimate goal is to predict the products given a set of reactants. Any suggestions will be helpful !

Yeah. I initially went to chemistry stack exchange but I thought they might not be familiar with ML. If you suggest, I will post it on chemistry stack exchange.

Your question has nothing to do with ML. You need to ask for whether it makes sense to use the predictors. ML is not magic, if your inputs are non-sense, your model is also bad.

Adding on top of Emre's answer, product formation depends on certain reactive groups say for e.g., $-OH$ (Alcohol) having more chance of releasing a Hydrogen than $-CHO$ (Aldehyde) or a Methyl $-CH_{3}$. Instead of using reactants as categorical variables you should use reactive groups on that, as variables.
For example if you take a vector of say (Chlorine, Alcoholic_group, Amine) to be coded for Para-Amino Phenol, it will be $(0,1,1)$ as chlorine is absent and for $CCl_{4}$, it will be $(1,0,0)$. But remember the first vector will be same for O-Amino Phenol as they are positional isomers. But both have different properties and reactivities. So categorical is kind of wasted here.
In my more deeper and frank opinion, these values should be continuous. You should be kind of looking for a value to quantify these each dimension of the vector coded. Then you can use classifiers from this vectors to find the probability of reaction happening or for product formation as @Emre mentioned.
So, take a simple, small set of molecules and make a vector having fewer dimensions and test it with categorical and continuous values(Give a value to each dimension based on your chemistry gut opinion). That is the better way to find whether your hypothesis/assumptions are reasonable.
EDIT 1:
My answer basically answers whether reaction is feasible or not but does not answer which products are returned. I am sorry for that. But I hope it kind of points to what you can achieve with ML if you want to take your first step.
Hope this helps.

+1 That's precisely the kind of domain expertise I meant. I think the predisposition you refer to can be encoded with a regularizer. Can you quantify how likely one product is to be released relative to another in such cases?

@Emre Yes that can be achieved, but requires quantifying the groups based on reactivity + availability etc. for the molecules involved which I think is a tough job. I don't even know whether there is any quantifiable term for reactive groups :( . That is why I asked him him/her to use chemistry gut knowledge to mark some molecules and test it on a set to see the class probability.

Have a look at this paper
http://pubs.acs.org/journal/doi/full/10.1021/acscentsci.6b00219
(Added to beat character limit)

Let the question run its course here before reposting there. Without knowing much chemistry, I would approach it as a quantized regression problem (i.e., pick a loss function that does not penalize rounding errors). The input is a list of (chemical, coefficient) tuples, and the output is likewise. The order does not matter. The chemicals can be represented as categoricals, and you should also created derived categorical variables for the elements in each chemical. In other words, the input is simply the list of coefficients for the reactants, plus indicators for the derived variables. The goal then is to estimate the coefficients of the resulting products; zero would mean the product is not present.


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How is solid PVA filament (for 3D printing) synthesized?

Polyvinyl alcohol (PVA; also referred to as: PVOH, PVAl) filament--the sacrificial support polymer filament used in 3D printing, how is that made?  All the reactions I've seen that make PVA create a gel-like substance that could be used as a glue or perhaps a film.  But it's not solid.  (See image with beaker.)

The reactions I've seen use an alcohol (ethanol or methanol) and an alkaline catalyst (sodium hydroxide) which are combined with polyvinyl acetate.  Here's the reaction that was used to make the beaker's gel PVA:

Another preparation can be seen in this video.
Is there some kind of coagulation/further reaction I must do to the gel to solidify its structure, or is the trick to do something during the first reaction, such as varying the temperature or pressure?
To clarify, I don't actually need/want it in the typical extruded filament form that would be needed for a 3D printer.  A simple solid blob of the stuff would suffice. I'm very new to materials science and only a novice in chemistry, but I'd love any answers you wonderful folk could provide, even if they are dense scholarly articles. Thank you for your time!

All the reactions I've seen that make PVA (also called PVOH, PVAl), create a gel-like substance that could be used as a glue or perhaps a film.

Yes, the reaction you describe is done in a solvent (water, ethanol, methanol, etc...) so that no acetate groups are unaccessable to reaction in a crystalline phase. A gel is formed because when poly(vinylacetate) is fully hydrolyzed, it is less soluble, typical white glue has about $12\%$ of the acetate groups unhydrolyzed to reduce crystallinity and keep solubility high.  The polyvinyl alcohol in filament is just dried out solid PVA, whereas the gel has water in a polymer matrix. All they have to do is melt and extrude the dry PVA powder into filament.

I don't actually need/want it in the typical extruded filament form that would be needed for a 3D printer.  A simple solid blob of the stuff would suffice.

Side note: you can buy PVA powder online with less mess than self synthesis or depending on your application, you might be able to just use dried out white glue.

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text:
What is the mechanism for the auto-dissociation of nitrogen tetraoxide
How does nitrogen tetraoxide dissociate into nitrosyl ion and nitrate ion?
Is it similar to the auto-dissociation of $\ce{SO2}$ into thionyl ion and sulfite ion by acceptance and donation of oxide ion?
I couldn't find a mechanism for the auto-dissociation of nitrogen tetraoxide.

Do you have any evidence that $$\ce{N2O4 <=> NO+ + NO3-}$$ is actually happening and under which conditions? Maybe then one would be able to also find mechanistic evidence. I would actually be surprised when this would be more feasible than $$\ce{N2O4 <=>2NO2}.$$

Are you interested in the gas-phase dissociation? Liquid-phase dissociation? Dissociation in aqueous media?

Maybe you confused it with N2O5 which is comprised of nitronium and nitrate ion?

Related: http://chemistry.stackexchange.com/questions/58564/are-any-bonds-broken-when-no2-becomes-n2o4

Dinitrogen tetroxide $\ce{N2O4}$ is a colourless liquid composed of planar dimers of nitrogen dioxide $\ce{NO2}$.$^{[1]\ \text{[a]}\ [2]}$

As Martin noted and is to be expected from the structure of $\ce{N2O4}$, the process
$$\ce{N2O4 <=> NO+ + NO3-}$$
is not very likely to happen. Autoionisation of $\ce{NO2}$ is more common$^{[1]}$.
$$\ce{NO2 -> NO2+ +  e^-\\ NO2 +  e^- -> NO2-}$$
Ionisation energy of $\ce{NO2}$ is $9.8\ \mathrm{eV}$ while its electron affinity equals $1.62\ \mathrm{eV}$.$^{[1]}$ Technically you could look at this as heterolytic bond dissociation $\ce{N2O4}$ but that is probably not what is happening. This autoionisation is, in a way, realised for aqueous solutions$^{[2]}$:
$$\ce{\underset{electron transfer}{NO2 + NO2} + H2O -> \underset{really just HNO3}{NO2OH} +  HNO2},$$
$$\ce{\underset{electron transfer}{NO2 + NO2} + 2KOH -> KNO3 + KNO2 +  H2O}.$$
Table 1. Melting and poiling points of $\ce{NO2}$ and $\ce{N2O4}$. Note the "equality" of melting points.$^{[2]\ [3]}$
$$\begin{array}{|c|c|c|}
\hline
\mathbf{Molecule} & \mathbf{mp\ (\deg  \mathrm{C})} &  \mathbf{bp\ (\deg  \mathrm{C})}\\
\hline
\ce{NO2} & -11.2 & 21 \\
\ce{N2O4} & -11.2 & 22 \\
\hline
\end{array}$$
Table 2. Percentage of  $\ce{NO2}$ and $\ce{N2O4}$ at various temperatures (pressure $1 \ \mathrm{atm}$).$^{[2]}$
$$\begin{array}{|c|c|c|c|}
\hline
\mathbf{Molecule} & \text{below}\  \mathrm{mp}\ (\mathrm{\%}) &  \mathbf{40\ \deg  \mathrm{C}\ (\mathrm{\%})} & \text{over}\mathbf{\  150\deg  \mathrm{C}\ (\mathrm{\%})}\\
\hline
\ce{NO2} & 0 & 31 & 100 \\
\ce{N2O4} & 100 & 69 & 0 \\
\hline
\end{array}$$

$\text{[a]}$ To the best of my knowledge, the enthalpy $-55\ \mathrm{kJ}$ applies at $20 \deg \mathrm{C}$ and $1\  \mathrm{atm}$ for both components in the gas phase. NIST gives $-57.12\ \mathrm{kJ}$ at $25 \deg \mathrm{C}$ and $1\  \mathrm{bar}$. See $\ce{N2O4}$ and $\ce{NO2}$.

$[1]$ N. N. Ahmetov. Anorgaaniline keemia. (1974) (pages 375, 376)
$[2]$ H. Karik, Kalle Truus. Elementide keemia. (2003) (pages 424, 426)
$[3]$ Wikipedia:

Nitrogen dioxide
Dinitrogen tetroxide


text:
Chemical plating of glass with copper?
It is commonly known how to plate glass with silver using silver nitrate, ammonia and glucose.
With about the same effort, is there a way to plate glass with copper?

According to Chemical Demonstrations: A Handbook for Teachers of Chemistry (Vol 3, p226), a similar effect can be obtained for copper using copper(II) acetate, ammonia, and hydrazine.

The copper metal forms when hydrazine reduced and ammonia complex of
  copper(II) in basic solution. Hydrazine slowly reduces the complex to
  copper metal. When the surface of the test tube is very clean and wet,
  the copper metal will adhere to it, forming a copper-colored mirror.
The overall reaction that occurs in this demonstration can be
  represented by the equation $\ce{2[Cu(NH3)4]2+_{(aq)} + N2H4_{(aq)} +
> 4OH-_{(aq)} -> 2Cu_{(s)} + N2_{(g)} + 4H2O_{(l)} + 8NH3_{(aq)}}$

Whether or not this can be described as "about the same effort" depends entirely upon the access you have to hydrazine, and to appropriate handling/safety equipment for working with the reagents. While hydrazine in solution is -- if I recall correctly, but I am not an MSDS -- considerably more stable than pure hydrazine, it's still toxic and corrosive (among other things).

Apologies if the Google Books preview for that page is not publicly viewable -- I'm currently on a university library network, so it's possible I have different access permissions; however, I've tried to reproduce enough to at least start to answer the question.

Thanks for the answer. Indeed, getting hydrazine is a little complicated to get if you don't work in chemistry, and is nasty stuff, too. It would be great if there were a way with chemicals easier to get...

Perhaps, hydrazine was picked in this case solely because it gives convenient products. So if you're not worried about impurities, other reducing agents -- perhaps even glucose -- might work?

Yes, it's quite possible.
Remember the Tollen's Reagent Silver mirror test? A Copper mirror can be created using Benedict's reagent but the glass surface must be first treated with Tin chloride sensitizer solution and Palladium chloride activator, when the Benedict's solution is heated with the reducing sugar a bright copper coat will be deposited on the glass

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


text:
Identifying the oxygen atoms that undergo reduction

Identify the oxygen atoms that undergo reduction in the reaction below:
$$\ce{O2 + Cr^3+ -> H2O2 + Cr2O7^2-}$$

I know that the oxygen atoms in $\ce{O2}$ have an oxidation state of $0$.
In the product $\ce{H2O2}$, the oxygen atoms have an oxidation state of $-1$; and in $\ce{Cr2O7^2-}$, oxygen atoms have an oxidation state of $-2$.
Based on how the problem was solved on the answer key it was assumed that all the oxygen atoms from $\ce{O2}$ are reduced to peroxide atoms.
How do we know that there are no oxygen atoms that are oxidized to become chromate oxygens with a $-2$ oxidation state?
I understand that there are other $\ce{H2O}$ oxygen atoms available to provide the chromate oxygens that would change the oxidation state.

It would be nice if we started with a reaction that at least listed all of the component participating. Even nicer would be if that reaction were balanced.

The oxygens in the dichromate ion on the right have -2 oxidation number, the chromium being +6.
http://ch302.cm.utexas.edu/worksheets/basic-redox-wkst-KEY.pdf
https://socratic.org/questions/what-is-the-oxidation-state-of-chromium-in-the-dichromate-ion-cr2o7-2
As you seem to have deduced, the peroxide ions are -1 so all oxygens, as shown, have been reduced.

This does not answer the question. How do you know that the chromate oxygens do not come from dioxygen?


text:
Titration to determine the amount of hydrogen peroxide
I have a mixture of sodium carbonate, sodium bicarbonate, ammonium carbonate, copper (II) sulfate, hydrogen peroxide, and sulfuric acid. I'm titrating potassium permanganate against this mixture to find out the amount of hydrogen peroxide in it. Will anything else in this mixture react with potassium permanganate? If so, is there another better way of determining the amount of hydrogen peroxide in it? Thank you.

I would not use potassium permanganate with sulfuric acid present:

You'll throw off your result due to the competing reaction..and risk a fire depending on the concentration of sulfuric acid.
If you measure the amount of $\ce{O2}$ produced by the decomposition of $\ce{H2O2}$ you can back-calculate how much $\ce{H2O2}$ was present in the sample. I think (with a decent certainty) that none of the species you list here should produce gas besides the peroxide.
That said I would use a catalyst (like $\ce{MnO2}$) for decomposition instead of a potentially reacting species (like potassium permanganate). You will likely get side reactions that consume the catalyst but they don't produce gas. So simply use excess $\ce{MnO2}$.
Measure the volume of $\ce{O2}$ produced (i.e. wait until the gas volume no longer increases) and use the ideal gas law (as a fair approximation) to get moles of $\ce{O2}$ produced. Once you have that, you can stoichiometrically determine moles of $\ce{H2O2}$ in your sample of known volume, thus concentration of $\ce{H2O2}$ (recalling that $2\ce{H2O2} \rightarrow 2\ce{H2O} + \ce{O2}$).
I've done this before with $\ce{MnO2}$ from batteries, washed thoroughly with water. I capture the gas produced in an upside-down grad-cylinder in a tub/beaker of water, e.g.:

P.S. if you manage to / happen to get a pure $\ce{H2O2}$ sample, you can very accurately get concentration (in %) by the following:
$C = \frac{d - 0.9946}{0.00456}$ where $d$ is the density in g/mL at 20deg. C.
That's a standardized function I derived from known and experimental densities of $\ce{H2O2}$


text:
Is aluminium amphoteric?
In my book it says that aluminium can react with water as a base when it's in the form $\ce{[Al(OH)2(OH2)4]^+}$.
And this way as an acid: $\ce{[Al(OH2)_6]^3+}$.
So ... does it depend on with which ion it's accompanied by? Let's say, $\ce{AlCl_3}$ will react as an acid?

Note that these species are in equilibrium. If you meant whether aqueous $\ce{AlCl3}$ is acidic, then sure: the aluminium ion has a large charge density, hence it polarises the water molecules. A base then deprotonates the water ligand. The first deprotonation equilibrium is shifted to the right (pH $< 7$).  Welcome to Chem.SE and see here for help on Mathjax/mhchem.

Yes, you are right. The reaction depends on with which ion it is accompanied with. Note that these species are in equilibrium and present in the solution  and that particular reaction proceeds according to which ion is present in the solution in concentrated form.

Aluminum hydroxide is formed when soluble salt of $\ce{Al^3+}$ i.e
$\ce{[Al(H2O)6]^3+}$ when it is reacted with base.
$$\ce{[Al(H2O)6]^3+(aq) + 3 OH-(aq) → Al(H2O)3(OH)3(s) + 3 H2O(l)}$$
This insoluble metal hydroxide can act as a base, since it can be
redissolved by reacting it with an acid.
As Brønsted base
$$\ce{Al(H2O)3(OH)3(s) + 3 H3O+(aq) → [Al(H2O)6]^3+(aq) + 3 H2O(l)}$$
Alternatively, the metal hydroxide can act as an acid, since it can
react with a base.
Lewis acid
$$\ce{Al(H2O)3(OH)3(s) + OH-(aq) → [Al(H2O)2(OH)4]-(aq) + H2O(l)}$$

Thus, all aluminium compounds are said to be amphoteric.


text:
What kind of 'product' is the 'product' in 'ionic product of water'?
My language has different words for product (=the result of a process) and product (=the sum of multiplied quantities), so I need to know the right meaning to choose the right word.
Which of the meanings above is present in the English term 'ionic product of water' ($K_\mathrm{w}$)?
This definition from http://www.chemguide.co.uk/physical/acidbaseeqia/kw.html seems to suggest it is a sum of multiplied quantities:

The units of $K_\mathrm{w}$: $K_\mathrm{w}$ is found by multiplying two concentration terms together. Each of these has the units of $\mathrm{mol~dm^{-3}}$. Multiplying $\mathrm{mol~dm^{-3}} \times \mathrm{mol~dm^{-3}}$ gives you the units above.

However, the use of the verb produce in this sentence from https://en.wikipedia.org/wiki/Self-ionization_of_water suggests something that is produced as a result of a process:

According to the theories of Svante Arrhenius, this must be due to the presence of ions. The ions are produced by the self-ionization reaction $\ce{H2O + H2O <=> H3O+ + OH-}$

Alternatively, could both meanings apply depending on your perspective?
Thanks in advance for any clarification.

It refers to multiplication.

Per Zhe's comment, in the specific context of the term ionic product of water, the word product is used in the sense of the result of the mathematical operation of multiplication:

The ionic product of water equals the concentration of $\ce{H+}$ times the concentration of $\ce{OH-}$:
  $$K_\mathrm w = \ce{[H+][OH-]}$$

(Strictly, per MaxReid's comment, the ionic product water is of course the product of the activities of $\ce{H+}$ and $\ce{OH-}$, but concentrations are commonly used in routine/non-rigorous calculations as they provide good accuracy for dilute solutions.)
This sense of the word product applies any time one is referencing the mathematical description of the equilibrium among the ionic forms of a given species. This usage also includes the solubility product (again with the above caveat about activities):

The solubility product of silver chloride equals the concentration of $\ce{Ag+}$ times the concentration of $\ce{Cl-}$:
  $$K_\mathrm{sp} = \ce{[Ag+][Cl-]}$$

As an aside, note that the main exception to this terminology is when the expression describing the distribution of species requires division by one or more quantities. In this case, the term to use is quotient, for which I expect little ambiguity exists:

The reaction quotient of the reaction
  $$\ce{A + B->C + D}$$
  is
  $$ Q = {\ce{[C][D]}\over \ce{[A][B]}}$$

However, to be complete: chemistry does also use the word product in the other sense you describe, as in something being produced by a reaction.  Due to this, the word product in the phrase the ionic product(s) of X could also bear this meaning.  For example:

The ionic products of the aqueous decomposition of dinitrogen tetroxide at neutral $\mathrm{pH}$ are nitrite and nitrate ion:
  $$\ce{N2O4(aq) + 2H2O -> NO2-(aq) + NO3-(aq) + 2H3O+}$$

In general, I think the difference boils down to whether one is talking about the ionic distribution of a specific species of interest, as opposed to the transformation that occurs in a particular reaction of interest.
As I noted above, though: In the specific case you are asking about, product is used in the sense of the result obtained from multiplication.

@PrettyHands Well, now that I think about it -- the answer isn't so cut and dried. See edit.

@PrettyHands And, well, now that I think about it some more: for the specific case you're asking about, it is cut and dried. See further edit.

@hBy2Py Isn't it defined as the product of the thermodynamic activities? In this case it should be equivalent, but for completeness sake I'd like to mention this.

@MaxRied Strictly, you are of course correct. I'll edit.

The product $K_w$ is a mathematical construct.
The ion species produced from ionization of water are not mathematical constructs. They are resulting entities (actual, tangible "things").
The word does have two meanings, but the essence in both cases is "result". There isn't really a case when both meanings would apply to the same "result".

Thank you. So in the term 'ionic product of water' specifically, it refers to multiplication?

I think your last paragraph doesn't have the right focus. The gist to me is that just the phrase "ionic product of water" is ambiguous. You have to understand the context in which the phrase is being used to be sure if the mathematical formula, or a chemical species is being discussed. // Needing context is fairly common in English. Look at definition for shoe. https://www.merriam-webster.com/dictionary/shoe

I agree with Max. Yet the phrase 'ionic product of water' is reserved as the well-known $K_w$.


text:
Van't Hoff Equation with changing Enthalpy
Is there a way to derive a modified van 't Hoff equation if we can't make the assumption that $\Delta H_\mathrm{r}$ doesn't change?
We know that $\mathrm{d} \ln K/\mathrm{d}T = \Delta H_\mathrm{r}/(RT^2)$ and $\Delta H_\mathrm{r} = \Delta H_\mathrm{r}^\circ + \int C_p\,\mathrm dT$
Is there a way to integrate the $C_p$ term along with the $T^2$?

There is. Suppose that at 298 K $$\Delta G^○_{298} = -RT\ln{K_{p\,298}} = \Delta H^○_{298} - T\Delta S^○_{298},$$ and if both $\Delta H^○$ and $\Delta S^○$ depend on $T$, $$-RT\ln{K_p} = \Delta H^○_{298} + \int_{298}^T \Delta C_pdT - T\Delta S^○_{298} - T\int_{298}^T \frac{\Delta C_p}{T}dT.$$ Then both sides of it can be divided by $T$ and knowing that $\int udv = uv - \int vdu$, where in our case $v = 1/T$ and $u = \int_{298}^T \Delta C_pdT$, we can "simplify" the last equation: $$-R\ln{K_p} = \frac{\Delta H^○_{298}}{T} - \Delta S^○_{298} - \int_{298}^T \frac{\int_{298}^T \Delta C_pdT}{T^2} dT.$$ I'll not endeavor to differentiate that with respect to $T$, but now you can probably figure out the rest. Since $C_p$ is usually a not-that-complex polynomial of $T$, the math should be rather straightforward.


text:
Which carbon(s) in cyclohexene have the highest electron density?

I think that carbons with highest electron density will be both $\ce C1$ and $\ce C2$ because of the loosely held $\pi$-bond between them, but according to the answer given the carbons with most electron density should be both $\ce C3$ and $\ce C6$.
How is this possible?

:-( The book gives an answer, not necessarily the right one.
Looking at the $\ce{^13C}$ NMR of cyclohexene the carbons at positions 4 and 5, in your diagram, are shifted the least and therefore have the most shielding. Thus those carbon atoms also have the greatest electron density around them.

So, $C1$, $C2$, $C4$, $C5$ all have highest electron density, while $C3$ and $C6$ have the lowest?

Looking at the $\ce{^13C}$ NMR of cyclohexene the carbons at positions 4 and 5, in your diagram, are shifted the least and therefore have the most shielding. c4, c5 > c1, c2 > c3, c6

Actually, I'm a high school student so it is tough for me to understand that 13C NMR of cyclohexene

Ugh. I should have written that (C4 or C5) > (C3 or C6) > (C1 or C2).

Intuitively, I would expect carbons 1/2 to have the highest electron density. MaxW's NMR spectrum does indeed suggest that 1/2 have the least density, then 3/6, with carbons 4/5 being the most electron-dense.
I suppose you could rationalize this conclusion by saying the the CH2 carbons (3/6, 4/5) will suck a bit of density from the attached hydrogens (because C is slightly more electronegative than H), whereas carbons 1/2 only have one H each to pull electron density from, therefore 1/2 have the least electron density. But my gut feels uncomfortable about this.
I attempted to calculate an electrostatic potential map of cyclohexene, but unfortunately I'm not proficient with the software. I think that's what we need here though.
For what's worth, here is a computer-generated model showing that there is more current density around the carbons you labeled 1/2:

(http://pubs.rsc.org/en/content/articlelanding/2012/cs/c2cs35037h)

I appreciate your efforts, but according to MaxW, carbon atoms at 1/2 positions have least electron density, but the model suggests that carbon at 1/2 have highest electron density. What should be the ultimate correct answer?

I think the ultimate answer should come from an electrostatic potential map. I'm not sure if NMR is the most direct way to assess electron density, there may be other factors at play that could be misleading. But I'm not a physical chemist so I'm not sure. I guess I'd take the NMR data over intuition and say carbons 4/5 are most electron-dense. But that's just settling in lieu of the potential map.

This has been bugging me. I think the gist is sloppy wording of the problem.  There is a difference in "electrostatic potential" and "electron density." As chemists we tend to ignore the core electrons which contribute to "electron density" but are not reactive chemically. "Electrostatic potential" is much indicative of chemical reactivity.


text:
Determination of +/- reflections in sigma molecular term symbols
This page, which depicts the molecular term symbols for the $\ce{O2}$ and $\ce{O2-}$ (Edit: Mistaken on $\ce{O2-}$) molecules, perhaps best summarizes the full scope of my questions. In general, I don't understand how a specific electron configuration relates to either +,- or both reflections.

Why does the linked page switch reflections for different spin multiplicities, ie. why does $^3\Sigma^-$ have a minus reflection but $^1\Sigma^+$ a positive reflection?
For $\ce{O2-}$, the left configuration leads to $^3\Sigma^- + ^1\Sigma^+$. Why are two sigma terms needed to describe this configuration?
As a more general question, how accurate is it to use the molecular orbital representation for deriving term symbols? Most sources I have seen describe electron configurations as three degenerate p-orbitals or 5 degenerate d-orbitals and never invoke MO diagrams. However, it seems necessary for deriving both parity and reflections. Is the MO approach always correct? Can the term symbols be derived without using MO diagrams, specifically for the parity and reflections?

Thank you for any advice.

The $ \pm $ superscripts refer to symmetry operations defined in the $D_{\infty h}$ point group.  The operation on the molecular orbital is reflection in a plane containing the nuclei. If the orbital in unchanged then the sign is $+$, if changed then $-$.
The electron configuration for ground state oxygen is
$(1\sigma _g)^2(1\sigma _u)^2(2\sigma _g)^2(2 \sigma _u)^2(3\sigma _g)^2(1\pi _u)^2(1\pi _g)^2(1\pi _g)^1(1\pi _g)^1$
However, the only electrons we need to consider are the two in the $\pi_g$ orbitals. Adding up the individual orbital and then spin angular momenta separately (equivalent to Russell-Saunders coupling in atoms) then combining both gives the term symbols $^3\Sigma, ^1\Delta ,^1\Sigma $. The calculation is done in a similar way to that for atoms and is shown below in the notes.
The subscript g is found because there is an even number of $\pi_g$ electrons.
The superscript $ \pm $ is because any state ( singlet or triplet) must be asymmetrical overall to exchange and as the triplet has a symmetrical spin wavefunction the spatial part is asymmetrical, and vice-versa for the singlet, the superscript for the triplet is $-$. Thus the superscript gives the term symbol $^3\Sigma_g^-$. Hund’s rule dictate that this is the lowest energy level, followed by $^1\Delta$ at $7918 \pu{cm^{-1}}$ and $^1\Sigma_g^+$ at $13195.22 \pu{cm^{-1}}$.
The $\ce{O_2^-}$ has one unpaired electron in a $\pi$ orbital (as does $\ce{O_2^+}$) so the term symbol is $^2\Pi_g$.
As to your third question the overall symmetry should not change when you go from atomic to MO's.
Notes
The individual $m_l$ and $m_s$ are listed below. The values restricted by the Pauli principle are removed.
The $m_l$ values are $\pm 1$ and the $m_s \pm 1/2$. The the table the spins are labelled as $\pm$ only. The total angular momentum is given as $\Omega = \Lambda + \Sigma$ by analogy with $J=L+S$ in atoms.
$$\begin{matrix}
m_{l1} & m_{l2} & m_{s1} & m_{s2} & \Lambda & \Sigma \\
1 & 1 &  + & + &\text{etc.  Pauli forbid'n}\\
\\
1 & 1 &  + & - & 2 & 0 & ^1\Delta\\
1 & 1 &  - & + & 2 & 0 &  .. \\
-1 & -1 &  + & - & -2 & 0 & ..\\
-1 & -1 &  - & + & -2 & 0 & ..\\
\\
1 & -1 &  + & +  & 0 & 1 & ^3\Sigma\\
-1 & 1 &  + & +  & 0 & 1 & ..\\
1 & -1 &  - & -  & 0 & -1 & ..\\
-1 & 1 &  - & -  & 0 & -1 & ..\\
\\
1 & -1 &  + & - & 0 & 0 & ^1\Sigma\\
1 & -1 &  - & + & 0 & 0 & ..\\
\end{matrix}$$


text:
Hydrates and melting points
Basalt melting point: wet vs dry
I had a student ask me this in class the other day; I thought about it, but I still can't work it out.
Many mineral and rock hydrates (take basalt, for instance) have much lower melting points than their anhydrous counterparts.  I could not craft a plausible explanation for this.
I realize that hydrates have different crystal structures due to the presence of water, but I am unsure how this will affect the melting points.  If anything, the possibility of hydrogen bonding would make the melting point higher, no?

Basalt is not a hydrate. :-(

Thanks for the correction, but that still doesn't answer my question.  See the hyperlink I edited in for the image.  It refers to basalt as "wet" or "dry."

Melting point depends on symmetry of the crystal structure. So you need data on structures. If the structure is symmetric the melting point will be higher. This many a times results in data which can't be explained on the basis of things like hydrogen bonding. It also depends on the surface area of the molecule

http://www.masterorganicchemistry.com/2010/07/09/chemical-tetris/
Though this is about organic chemistry these concepts can safely be applied to inorganic.

Melting of crystal hydrates is more like a dissolution in their own water. Then again, basalt (even if wet) is not a hydrate and may be a totally different story.

I suppose you need to click the link to understand the question properly (because I don't really get it); if so, include the relevant portions here.

It's been so long you probably won't even see this, but in case someone else is searching for an answer, here are some interesting answers I've found: (The websites are in French, so I tried to translate the intersting bits)

Igneous rock samples can be melted in the laboratory by heating them at elevated temperatures (1000 ° C to 1300 ° C). If one operates under pressure and in the presence of volatile elements, the melting temperature of the rock and its minerals is much lower (650 ° C to 950 ° C). It is known that magmas contain varying amounts of volatile elements that lower the melting point of many silicates, thus allowing the magma to remain liquid even at temperatures below those required to melt rocks in the laboratory. Indeed, the temperatures that exist in the different levels of the earth's crust are not sufficient to melt rocks in the absence of volatile elements.
  The main volatile elements found in magmas can be commonly observed in the gaseous state at the vents of active volcanoes. It is essentially water vapor, carbon dioxide, hydrogen, chlorine, fluorine (in the form of acids) and many other gases [...].

Taken from http://www.fossiliraptor.be/rochesetphenomenesignes4.htm

In the case of hydrated paragneiss, during the warming of this rock, the melting begins as soon as it reaches the first reaction ((4) Qz + Ab + Gold + H2O ⇄ Liquid), around 650 ° C. Indeed, this gneiss contains all the reagents necessary for this reaction to take place.
  For a dry paragneiss, on the other hand, nothing happens on the curve corresponding to reaction 4; indeed, one of the reagents, water, is lacking. However, when the temperature exceeds 850 ° C, this rock crosses the curve corresponding to reaction 5 ((5) Bt + Ab + Qz + AlS ⇄ Liquid + Cord / Gt + KF, which can be decomposed as follows: (6) Bt + AlS ⇄ Gt / Cord + KF + H2O, then 7) Qz + Ab (Pg) + Gold (KF) + H2O ⇄ Liquid) and the two coupled reactions described above take place: the biotite decomposes to liberate water; therefore, we are in a situation where all the necessary products are present: the merger can take place.
  In other words, a hydrated system will melt (cross its solidus) at a lower temperature than a system without free water: the two reactions correspond respectively to "hydrate solidus" and "dry solidus"!

Taken from: https://planet-terre.ens-lyon.fr/article/reactions-de-fusion.xml#equation4

It sounds like freezing point depression to me. But it's more complicated than that.

The melting point of a pure element or chemical compound is the transition from one phase to another. Wet basalt is different: it can contain up to ~5% H2O, along with 2-3% alkali metal oxides and 45% silica. Under pressure, this could form a glass that flows; a viscous process, facilitated by the water, not a complete dissolution to a homogeneous phase. This glassy phase will flow when hot, but become much stiffer when the water is driven off and so seem to have a higher melting point when dry.


text:
Throwing water into lava
While discussing magma in class with my students, I showed a video of a man throwing water onto molten, running lava.  A large flame ensued.  I realize that the water is being converted into steam and expanding extremely rapidly, but why does this create a small "explosion," if you will?
Surely none of the lava is being "atomized" into the air in small droplets, is it?  But even so, they would not light aflame.

https://www.youtube.com/watch?v=0XgpYXVN-Kk

Are you certain it was a flame and not just lava bubbles with thin enough walls that it appeared more orange than red, thus giving the illusion of a flame?

I think linking the video you are referring to would be potentially helpful in answering this question.

The first thing I think of, is this.

Edited the original to add the hyperlink on the words "a video".  Should be about 1 minute 30 secs into the video.  The thermal decomposition sounds like a plausible explanation, but basaltic magma sits around 1250C at best.  Surely not hot enough to decompose a sufficient enough amount of water?  Or perhaps the initial decomposition and ignition creates more heat, starting a chain reaction?

The liquid looks like it is in a water bottle, but is it water? From that amount of liquid and flame I'd suspect alcohol or gasoline. The numb-numb was putting on a show not a science demonstration.

This is a Steam explosion generally the flame is due to the hydrogen produced when water is superheated so thermal decomposed into $\ce{H2}$ and $\ce{O2}$.
Quoting wikipedia:

Steam explosions are not normally chemical explosions, although a
  number of substances react chemically with steam (for example,
  zirconium and superheated graphite react with steam and air
  respectively to give off hydrogen, which burns violently in air) so
  that chemical explosions and fires may follow.

Theory:  Is the lava hot enough to decompose some of the water into hydrogen and oxygen gas, thus making way for a hydrogen explosion, which releases a lot of energy, which is absorbed by other waters, which decomposes more of them?  It seems like good potential for a chain reaction.


text:
How do orbitals fill, what is the difference between them, and why is this stable?
I know about the Aufbau principle and the order in which electrons fill orbitals, but how is it that atoms are more stable when they have a full valence shell? It seems that any shell can be comprised of many different energy levels(3s,3p,3d,etc) and that it would be kind of arbitrary to say that "since all the orbitals that start with three are full, the atom is stable". Are all the "3" energy levels linked in a special way that differentiates them from 2 or 4? Also, why would an electron even be inclined to join a specific orbital, and do electrons change their spin in order to fit with the pre-existing particle in the orbital.

An electron doesn't care about a specific orbital; it just goes to the lowermost vacant place. Yes, in doing so it might have to change its spin.

What is stable?
In this context, stability is that it neither gains electron to form negative anions nor loses electron to form positive cations. I will only discuss noble gases because they are so exceptionally stable.
It does not lose electrons
This is not unique to noble gases. Elements in group VII such as fluorine and chlorine also do not tend to lose electrons.
This is also not entirely true. Compounds like $\ce{XeF4}$ and $\ce{Cl2O7}$ do exist in which $\ce{Xe}$ and $\ce{Cl}$ bear partial positive charges respectively, although $\ce{Xe}$ is a noble gas.
$\ce{Xe}$, being a noble gas, indeed has a high ionization energy (energy taken to remove an electron) but it is not infinite. This means, as long as we have enough energy, we can make them form compounds.
The ionization energy increases from Group III to Group 0 because there are more protons in the nucleus which exerts a higher pull to the orbitals, which lowers their energy. For example, the ionization energy of helium is twice that of hydrogen, and helium has exactly twice the number of protons than hydrogen.
It does not gain electron
This is because the next orbital has a positive energy. Orbitals with negative energy means that it takes work to free the electrons from the orbitals. Orbitals with positive energy means that no electrons can reside in the orbitals: they are already free.
For example, the electronic arrangement of argon is $1s^2 2s^2 2p^6 3s^2 3p^6$. The next orbital, if it existed, would be $4s$, which has a positive energy. Therefore, its electron affinity (energy released by gaining electrons) is 0. This is evidenced by the ionization energy of the next element, potassium, which is small, suggesting that the extra proton pulls the $4s$ orbital just enough so that it has a slightly negative energy.
Conclusion
Noble gases are stable because they have a full $p$ orbital, so the next electron would need to go to the next orbital which has a positive energy, and the extra protons pulling on the orbital lower the energy of the electrons in the orbital, making it not tend to lose electrons.


text:
Would gas come out of solution in an infinite volume?
Say I have an infinite volume of water with some level of dissolved air in it. I lower the pressure to below the gas' saturation pressure. Does the air come out of solution if there are no interfaces (nuclei) for it to begin? If so, what is the driving force behind the formation of the bubbles? Are there any estimations of the time that it would take to form a bubble of a given size, or an estimation of the equilibrium size of said bubble? Presumably it would be a balance between the surface tension, radius and the water pressure (like $P + 2S/R$)...
To give some motivation, I have a cavitation tunnel where I can apply a pressure field to a large volume of water. In the literature, people speak of homogeneous nucleation where fluids below saturation pressure start to release their gas as bubbles, and I'm struggling to understand how that occurs.

First of all, welcome!  I would try to clarify your question a bit.  For example you seem to be talking about both a hypothetical "infinite volume" and the observation of a real phenomenon in the description.  It may just be me but I'm having a hard time picturing that.  Also you could expand on the equation you present and how you think that might explain the observation of bubbles.  I think it may be an interesting question just in need of some tweaking.

Thanks for the welcome. I added a bit of 'motivation' if that helps.

There will be no balance. Fusion of two bubbles into a larger bubble is always energetically favorable. The equilibrium state is one huge bubble amidst the liquid, which means your situation is essentially non-equilibrium. As for the gist of your question, you are dealing with cavitation.

Thanks @IvanNeretin, I guess I'm wondering where these two bubbles come from in the first place. My understanding is that cavitation is driven by fluids below the vapour pressure - whereas this is taking place above that.

Be it vapor pressure or the pressure of dissolved gas, the phenomenon is pretty much the same. As for nucleation, well, that's a hard problem indeed.

I feel that cavitation is a change in phase while this phenomenon is more a grouping together of gas molecules? The gas does not undergo a phase change to come out of solution does it?

According to this abstract from a paper titled "Why Is It Much Easier To Nucleate Gas Bubbles than Theory Predicts?":

Bubble nucleation from supersaturated gas solutions generally takes
  place at much lower supersaturations than are expected from the
  theory. Furthermore, the same theory predicts that the threshold
  concentration of gas needed to cause nucleation should be essentially
  independent of the gas species used, a finding contradicted by
  experiment. There are two general explanations:  first, that the
  theory is wrong, or second, that there is a previously unidentified
  factor which is influencing the results of the experiments. Given the
  success of the fundamental theory in other areas, the second
  explanation is preferred. The previously unrecognized factor is
  identified here as being the surface activity of the gases which form
  the bubbles.

Although this is discussing conditions of low supersaturation and not subsaturation, I assume the same concepts apply, though it would cost me $40 to get the full article text and find out for sure.
I'm pretty sure what's happening here is that through the random motion of dissolved gas molecules, occasionally enough of them adhere to one another to form the nucleus of a bubble.  Particularly at very small radii, the high surface tension of water dominates otherwise destructive forces, and a micro-bubble begins to form.  But by the definition of saturation with respect to bubble formation, it would seem that you cannot have "macro-scale" bubble formation and growth persisting under conditions of subsaturation as equilibrium is reached.
If I can paraphrase your question as:

"If I lower the gas pressure to below its saturation pressure, does
  the gas come out of solution if there are no interfaces (nuclei) for
  bubbles to form?"

then the answer is yes, regardless of the presence or absence of bubbles.  The dissolved gas molecules will partition at the gas-liquid interface until equilibrium concentrations are achieved in each phase.  If some gas bubble began to form, it would quickly redissolve under conditions of subsaturation with respect to the gas solubility.
Unless I'm missing something about the importance of bubbles in your question, this would pretty clearly be the bottom line answer.  Please comment if I've missed the crux of your question.

Your answer is very helpful. I'll get a hold of that paper and read it myself. I guess the bit that gets me is "through the random motion of dissolved gas molecules, occasionally enough of them adhere to one another to form the nucleus of a bubble". If there was some quantitative analysis of that then I'd be more comfortable.

"it would quickly redissolve under conditions of subsaturation" Is this a 'hard cut-off' phenomena at the saturation pressure or is this a gradual phenomenon?

I think you are mixing kinetics and thermodynamics in a way that is causing you confusion in both your question and comments. For example, when you talk about "where fluids below saturation pressure start to release their gas as bubbles", you are addressing the thermodynamic property of "saturation (implying thermodynamic equilibrium) vapor pressure". My choice of the word quickly may have confused the issue more for you, but when you ask "is this a gradual phenomenon?", then you are talking kinetics. It may help you to remember those concepts are separate from one another.


text:
How to calculate what temperature change in entropy for the reaction equal to the change in entropy for the surroundings given Enthalpy and entropy

A reaction has $\Delta H_\textrm{rxn} = \pu{-107 kJ}$ and $\Delta S_\textrm{rxn} = \pu{285 J/K}$. At what temperature is the change in entropy for the reaction equal to the change in entropy for the surroundings?

My thoughts:
I wasn't really sure if there was an equation that related the entropy and the enthalpy and would like to know if there is such an equation.

You're looking for the equation of Gibbs free energy:
$$\Delta G^\circ =\Delta H^\circ - T\Delta S^\circ$$
Per Wikipedia's Gibbs free energy page:

The equation can be also seen from the perspective of the system taken
together with its surroundings (the rest of the universe). First
assume that the given reaction at constant temperature and pressure is
the only one that is occurring. Then the entropy released or absorbed
by the system equals the entropy that the environment must absorb or
release, respectively. The reaction will only be allowed if the total
entropy change of the universe is zero or positive. This is reflected
in a negative ΔG, and the reaction is called exergonic.

So effectively you're trying to deduce the $T$ that produces $\Delta G = 0$ which is the "tipping point" (then the reaction will occur for $\Delta G < 0$)
So you solve for $T$ in:
$0 = \Delta H - T \Delta S$
Thereafter as $T \rightarrow \infty$ you see that $\Delta G$ keeps decreasing, making the reaction more favorable.
Thus $S_\textrm{system} = S_\textrm{surroundings}$ is a fancy prerequisite for reaction occurrence, which we commonly associate with $\Delta G \le 0$.

When we talk about heat of reaction and change in entropy of reaction, we are talking about going from pure reactants to pure products in some (unspecified reversible) process at constant temperature and pressure.  Since the process is at constant pressure, the heat exchange with the surroundings, according to the first law, is equal to the change in enthalpy.  So,
$$Q_{sys}=\Delta H_\textrm{rxn}$$
In this context, when we talk about heat exchange with the surroundings, all the heat transfer is taken to occur to an ideal constant temperature reservoir.  So the change in entropy of the reservoir is
$$\Delta S_\textrm{res}=\frac{-Q}{T}=-\frac{\Delta H_\textrm{rxn}}{T}$$
If the process is carried out reversibly, the sum of the change in entropy of the system plus the change in entropy of the surroundings must be zero:
$$\Delta S_\textrm{rxn}+\Delta S_{res}=\Delta S_\textrm{rxn}-\frac{\Delta H_\textrm{rxn}}{T}=0$$
(The problem statement must only be referring to the magnitudes of the entropy changes for the system and the surroundings, since clearly they will be opposite in sign.)


text:
Quantitative expression for steric hindrance
Steric hinderance is a major component in determining the feasibility and the rate of a chemical reaction. Wouldn't it be useful to measure it quantitatively then? This would make it easier to compare the property of two molecules. Are there currently ways to measure steric hindrance, or is it not possible for some reason?

Steric hindrance is as much a made up concept as ring strain. You can try to quantify it based on model systems and reference states, but it will never be more than a simple guideline. Even quantum chemistry is not sophisticated enough to correctly account for all effects and then it is almost impossible to separate them.

This paper gives an extremely good introduction to various ways of quantifying steric hindrance: https://www.nature.com/articles/s41598-019-56342-w

There are two ways I know of to measure steric hindrance. Once such quantitative measurement are A values, which give the energetic preference of having the substituent on a substituted cyclohexane ring in the equatorial position versus the axial position. This preference is a result of 1,3-diaxial strain.
Additionally, the Taft equation is a linear free-energy relationship used to measure the steric effects of a substituent on the rate of reaction. The equation is given by:
$${\displaystyle \log \left({\frac {k_{s}}{k_{\ce {CH3}}}}\right)=\rho ^{*}\sigma ^{*}+\delta E_{s}}$$
where $\displaystyle \frac {k_{s}}{k_{\ce {CH3}}}$ is the ratio of the substituents rate of reaction relative to a methyl group, $\sigma^*$ is the polar substituent constant that describes the field and inductive effects of the substituent, $E_s$ is the steric substituent constant, $\rho^*$ is the sensitivity factor for the reaction to polar effects, and $\delta$ is the sensitivity factor for the reaction to steric effects.$^{[1]}$
Though A values are useful as a quick reference in deducing the major products of an unfamiliar reaction, the Taft equation would be used preferentially when doing an in depth mechanistic study of a reaction for publication.

$^{[1]}$ Wikipedia, Taft equation

This shows both the power of organic chemistry and its weakness. The power as it enables empirical estimates to be made by comparison with other data, the weakness in that one measurement is analysed by four empirical parameters. However, there is no alternative, as experimental data on a fundamental level, not rate constants, but observation of transition states is almost entirely lacking, and is so hard to obtain it may never be.  Hard because its difficult to observe one successful event that lasts only a picosecond, or so, in a million failed ones possibly spread over  a microsecond.

There is a full way but it's not easy to calculate.
Steric hindrance is indeed a generic term for a quantifiable phenomenon: electron-electron repulsion, or (much) more broadly, chemical physics.
Electron-electron repulsion can be measured simply/crudely by Coulomb's law:
\begin{align}
E &= \frac{q_1 q_2}{4\pi \epsilon_0 r}, &
F &= \frac{q_1 q_2}{4\pi \epsilon_0 r^2}.
\end{align}
Here, e.g. you can approximate atomic charges with some atomic volume assigned and quantify the forces involved between molecules.
But the full dynamics requires quantum mechanical calculations (ignoring things like relativity and gravity).
Coulomb's energy does appear in QM equations but electron-electron interactions are more complicated than simple point-charges so more terms appear for correlation/exchange energies etc. The general principle is the same as for classical physics though: you calculate total kinetic and potential energy of your system.
The quantitative expression would be the way potential energy changes with intermolecular distance. So you could set up a series of systems with different distances between molecules and see how the energy changes.
One can calculate QM energies with QM software like NWChem, QChem, Orca, Gaussian and others.

How would you separate out steric hindrance from a QC calculation? I think Coulomb's law is the go to for electrical effects, and I would agree that steric effects are just a different way of looking at electric effects, but what would constitute a steric effect in the first place?

In my QM-saturated mind I don't distinguish steric effects from any other effects... Everything is simply the physics of the atoms; specifically the trajectories and charges of particles.
Understandably, known empirical/approximate methods may be a better answer to the question. I think the underlying physics should be not ignored though.
It is fair to say that all physical effects are produced by energy which can be quantified by QC in the case of super-small stuff. Perhaps the beauty of QC is that we lose the need to distinguish "types" of interactions and just quantify them all by E.

I do understand and feel similarly. Unfortunately not everyone does, and for others to understand we must try to speak the same language. Classifications (or rather simplifications) of interactions have helped designing and explaining quite a few things and can provide a powerful tool. Also, distinguishing everything by energy only gets you the difference, quite a trivial point, it is usually the how-much, what, and why, we need to understand molecules and reactions.

While not directly "measuring" steric hindrance but rather "distortion energy" the computational Interaction/Distortion model from K.N. Houk (1, 2) and Bickelhaupt who calls it "Activation Strain" model (3) can give pretty good insight in such things. Especially since steric hindrance is only half (or even less) of the story.
For similar reactions this distortion energy can (!) be related to steric hindrance, but is also inflenced by electronic effects the substituents might have on the reactive center. Over all this model provides pretty good insight if a reaction proceeds fast/slow due to electronic effects or energy needed for distortion.
So how does it work?
We pretty much arbitrarily devide our energy of activation into two parts, the interaction energy and the distortion energy. Distortion energy is the energy needed to "twist" the reactants from ground state geometry into geometry at the transition state structure. For this energy is needed. If you then bring this distorted reactants together they will interact and what you get out here is the interaction energy. Sum of all distortion energies and the interaction energy is the energy of activation.
$$\Delta E^{\ddagger}_\mathrm{act}=\Delta E^{\ddagger}_\mathrm{dist} + \Delta E^{\ddagger}_\mathrm{int}$$
Houk himself explains this very well in this talk.

References:

R. W. Strozier, Pierluigi Caramella, and K. N. Houk. J. Am. Chem. Soc. 1979, 101 (5), 1340–1343.
Daniel H. Ess and K. N. Houk. J. Am. Chem. Soc. 2007, 129 (35), 10646–10647.
Axel Diefenbach and F. Matthias Bickelhaupt. J. Phys. Chem. A 2004, 108 (40), 8460–8466.


text:
Finding threshold frequency of a metal

Find threshold frequency, in $\mathrm{Hz}$, of a metal if it emits an electron with kinetic energy of $2.74 \times 10^{-19}~\mathrm{J}$ when illuminated with light with frequency of $1.49 \times 10^{15}~\mathrm{Hz}$.

I am not sure what to do with this.

Welcome to this site. Note that this site is not for homework help. For [tag:homework] questions, please include your attempt, or else your question might be closed as off-topic.

What have you tried?  Take a shot at it anyway ;)  You'll be much more likely to get help when people see you've put some effort into it.


text:
Does the term "Alkaline" necessarily indicate the presence of an actual alkali?
A month back, @CowperKettle had asked this at The Periodic Table chat room:

Does "alkaline" imply the presence of an actual alkali?

Now I've come across the term "alkaline" on numerous occasions, and pretty much all examples that I know of (in Chemistry) actually do involve the presence of an alkali (often NaOH or KOH).
Wikipedia defines an alkali as:

In chemistry, an alkali is a basic, ionic salt of an alkali metal or alkaline earth metal.

Further, it states:

An alkali also can be defined as a base that dissolves in water.

Now when something is a "base", it could mean a lot of things (An Arrhenius base? A Lewis base? A Bronstedt base?), which doesn't really make the Wikipedia definition any clearer.
Take for example, this case:

...mucosal cells of the small intestine secrete bicarbonate ions into the intestinal lumen to render it alkaline...

I know that was from a "Biology perspective", but that raises the following questions:
1) Is there a universally accepted (I guess that would be IUPAC) definition for an alkali? (Since the Wikipedia definition seems a bit oversimplified)
2) Is it "acceptable" to refer to, say, as suspension of a (Lewis) base in water as "alkaline"? And what about that excerpt from my Bio. text book? Is that use of "alkaline" acceptable as well?

For the first question: the IUPAC Gold Book does not contain the entry "alkaline" or "alkali".

Use of alkaline to refer to basic pH is certainly abundant in Biology. I've taught Anatomy/Phyiology as well as basic (alkaline?) Biology, it's commonly used in textbooks.

Most every English language dictionary definition of alkaline that I've found defines it as the adjective form of alkali and includes something like "having a pH greater than 7".  As pointed out in the comments there is no IUPAC definition, so it's going to be difficult to come up with an authoritative scientific answer.  Here are some interesting items I've found on the word's origin, usage in different contexts, etc.:
Online Enemology Dictionary:
alkali: late 14c., "soda ash," from Medieval Latin alkali, from Arabic al-qaliy "the ashes, burnt ashes" (of saltwort, which abounds in soda due to growing in alkaline soils), from qala "to roast in a pan." Later extended to similar substances, natural or manufactured. The modern chemistry sense is from 1813.
Cambridge Dictionary :
alkali: chemistry - a substance that has the opposite effect or chemical behavior of an acid. adj. alkaline
Biology Online:
alkaline: (Science: chemistry) Having the reactions of an alkali. Origin: L. Alkalinus Relating to or containing an alkali; having a ph greater than 7; alkaline soils derived from chalk or limestone.A substance that has a ph higher than 7.
Merriam-Webster:
alkaline: :  of, relating to, containing, or having the properties of an alkali or alkali metal :  basic; especially, of a solution :  having a pH of more than 7
The almighty Wikipedia, fairly representative of a half-dozen other online definitions:
In chemistry, an alkali (/ˈælkəlaɪ/; from Arabic: al-qaly القلي, القالي , “ashes of the saltwort”) is a basic, ionic salt of an alkali metal or alkaline earth metal chemical element. An alkali also can be defined as a base that dissolves in water. A solution of a soluble base has a pH greater than 7.0. The adjective alkaline is commonly, and alkalescent less often, used in English as a synonym for basic, especially for bases soluble in water. This broad use of the term is likely to have come about because alkalis were the first bases known to obey the Arrhenius definition of a base, and they are still among the most common bases.
I think in the end its usage is simply tied to the jargon and vocabulary of the context in which it is used. And with so many definitions, most of the ways that we hear it used are probably at least partially correct.  And as is so often the case, I think the Wikipedia definition probably comes as close to the best definition as any.

A Lewis base often also reacts as a Brønsted base, hence the sharing of the word "base", but that is not so by definition. Wrong, wrong, wrong.
"Alkaline" means Brønsted base containing alkali metal ions. The statement "An alkali also can be defined as a base that dissolves in water." is utter nonsense. Ammonia? Phenoles? They are NOT alkalis.
To say that a plain ammonia solution is alkaline is sloppy, at least.
Biology: All biologic tissue is full of sodium and potassium ions, so it's hard to argue against the use you mention.


text:
Do amino groups need to be protected during a Clemmensen reduction?
In class, we learned about the Clemmensen reduction of carbonyl groups to the corresponding hydrocarbon. This is all well and good, except that it would also seem that the conditions used in the Clemmensen reduction might also affect other groups, such as the reduction of nitro groups to amino groups.
Generally, would a Clemmensen reduction have any affect on either an aromatic or aliphatic amino group within the compound? If so, I am assuming that a protecting group would then be necessary. I know that nitro groups will be reduced to amino groups, and am also wondering if they would then be further acted upon by the reaction conditions.

What could happen to the amino groups?

I am not exactly sure right now.

I don't have time to write an actual answer right now, but amines are usually tolerated. Secondary and tertiary without issue, primary have some potential issues.


text:
Why does the light catalyzed radical addition of tetrachloromethane to an alkene produce only a single regioisomer?
This reaction always occurs in the following, regiospecific way:
$$\ce{CCl4 + R-CH=CH2 ->[$h\nu$]  R-CHCl-CH2-CCl3}$$
The same should apparently also happen when there is a carbon that has two substituents (e.g. methyl groups) and yield only one major product:
$$\ce{CCl4 + (CH3)2C=CH2 ->[$h\nu$]  (CH3)2CCl-CH2-CCl3}$$
I believe the initiation step goes like this:
$$\ce{CCl4 ->[$h\nu$] CCl3^. + Cl^.}$$
Can't both of these radicals, that form in the initiation step, react with the double bond and form a secondary or tertiary radical carbon? Why does only $\ce{CCl3^.}$ apparently attack the double bond?

Is there any other information missing from this problem or in the wording of it?  I would have expected both of the radicals to react significantly, although I would expect the CCl3 radical to be a less stable intermediate than the Cl radical.

Is it not just a steric effect making the $\ce{CCl3\cdot}$ react more slowly and so the $\ce{Cl\cdot}$ wins out at the more crowded site?

While this reaction is highly regioselective, it is not 100% regioselective. Consider the nearly identical reactions:
$$\begin{align}
\ce{CCl4 + H2C=CH-(CH2)5CH3 ->[(PhCO2)2][90\text{-}105ºC] Cl3C-CH2-CHCl-(CH2)5CH3} \hspace{.63cm} \text{75%}^{[1]}\\
 \ce{CCl4 + H2C=CH-COH-(CH3)2 ->[(PhCO2)2][80ºC] Cl3C-CH2-CHCl-CH(OH)-(CH2)5CH3} \hspace{.63cm} \text{70%}^{[2]}\\
\end{align}
$$
Granted these reactions use radical initiators instead of light, but that wouldn't make the difference between 100% and 70-75% regioselectivity. The reality is that at the start of the reaction, $\ce{.CCl3}$ and $\ce{.Cl}$ are both competing to attack the alkene, and $\ce{.Cl}$ does in fact attack faster. Looking at the activation energies for the abstraction of hydrogen atoms by different radicals, we gain insight into their relative stability:
$$ \small
\begin{array}{lcc}
\hline
\text{Radical} & \ce{CH3-H} & \ce{CH3CH2-H} & \ce{(CH3)2CH-H} & \ce{(CH3)C-H}^{[3]} \\
\hline
\ce{.F} & \text{1-1.5} & \text{<1} & & \text{<1} \\
\ce{.Cl} & \text{3.4} & \text{1.1} \\
\ce{.Br} & \text{17.5} & \text{13.0} & \text{9.5} & \text{6.9} \\
\ce{.CH3} & \text{14.0} & \text{11.6} & \text{9.6} & \text{8.1} \\
\ce{.CF3} & \text{10.9} & \text{8.0} & \text{6.5} & \text{4.9} \\
\ce{.CCl3} & \text{17.9} & \text{14.2} & \text{10.6} & \text{7.7*} \\
\hline
_{^* \text{units are } \mathrm{kcal\ mol^{-1}}}
\end{array}
$$
As this reaction continues, however, chlorine atoms are abstracted from $\ce{CCl4}$, creating the product along with $\ce{.CCl3}$ radicals. Abstraction of $\ce{.CCl3}$ is highly disfavored due to steric effects. Thus, the radical chain reaction is as follows:
$$\ce{CCl4 ->[$h\nu$] .CCl3 + Cl.}$$
$$\ce{.CCl3 + CH2=CH-R ->  Cl3C-CH2-\stackrel{\displaystyle .}{\ce{C}}H-R}$$
$$\ce{Cl3C-CH2-\stackrel{\displaystyle .}{\ce{C}}H-R + CCl4 ->  Cl3C-CH2-CHCl-R + .CCl3}$$
The less than ideal yields of the radical addition of $\ce{CCl4}$ can be attributed to the strength of the $\ce{C-Cl}$ bond (higher activation energies correspond to slower rates of reactions). Because $\ce{CCl4}$ is in competition with the polymerization of the $\ce{.CCl3}$-alkene adduct, $\ce{BrCCl3}$ and $\ce{CBr4}$ are generally preferred for their higher selectivity, owed to the lower strength of the $\ce{C-Br}$ bond. One such example is shown below.
$$\ce{BrCCl3 + H2C=C(C2H5)2 ->[$h \nu$] Cl3C-CH2-CBr(C2H5)2} \hspace{1cm} {\text{91%}^{[4]}}$$

$^{[1]}$M. S. Kharasch, E. W. Jensen, and W. H. Urry, J. Am. Chem. Soc., 69, 1100 (1947).
$^{[2]}$P. D. Klemmensen, H. Kolind-Andersen, H. B. Madsen, and A. Svendsen, J. Org. Chem., 44, 416 (1979).
$^{[3]}$F. A. Carey, R. J. Sundberg, Advanced Organic Chemistry Part A: Structure and Mechanisms, Springer Science, New York, 2007, pg. 1034.
$^{[4]}$M. S. Kharasch and M. Sage, J. Org. Chem., 14, 537 (1949).


text:
Why is carbon dioxide considered a Lewis acid?
$\ce{CO2}$ is considered a Lewis acid. How it is an acid? According to Lewis: “species that accept an electron pair are acids”. But $\ce{CO2}$ can't accept electron pairs because oxygen and carbon both are electron sufficient (have complete octets). If the reason is within a lone pair of electrons on oxygen then ammonia ($\ce{NH3}$), which also has a lone pair of electrons, should be acidic, too. But ammonia basic in nature.
So why can carbon dioxide be considered a Lewis acid?

$\ce{CO2 +H2O <--> H2CO3}$ so the oxygen of the water molecule donates the electrons and is the Lewis Base. Thus $\ce{CO2}$ must be the Lewis acid.

Please see this image taken from this page.

MaxW That is just an experimental observation and gives no insight into why CO2 is a Lewis acid.

If you've learned how to draw resonance structures, they will help you greatly here (and in general). In the attached figure, 1 is the relevant resonance, implying the carbon in $\ce{CO2}$ is partially positive and therefore may be prone to accept electrons (see the resonance hybrid).
In theory, any Lewis base (electron-donor) can react with $\ce{CO2}$ (Lewis base abbreviated B in 2 & 3). We can draw the reaction mechanism using the charged resonance structure (2) or the neutral, most "stable" structure (3). In reality, these are not two discretely different forms but just "snapshots in time" of a true $\ce{CO2}$ molecule, so it doesn't really matter which structure we as humans choose to represent on paper.

I hope you can see how $\ce{CO2}$ is a Lewis acid because it is able to accept electrons. Although in the principal resonance structure you see all atoms have full octets of electrons, the double bonds can easily break, rearranging the electrons. If full octets prevented atoms from receiving electrons, a great percentage of all known reactions would not occur. The key is that the electrons in the octet must have an alternative place to go if new electrons are donated to the molecule (in this case, the electrons migrate to the $\ce{O}$ atom). In more advanced chemistry, we would say $\ce{O}$ acted as an electron sink in this process.
It is interesting to note that any $\ce{C=O}$ structure can react as a Lewis acid in exactly the way your $\ce{CO2}$ molecule did, and this reaction of the $\ce{C=O}$ group (known as a carbonyl functional group) is among the most important patterns in the biochemical reactions occuring in our bodies.

This is an extension. Would you consider carbon dioxide to be a strong electrophile since two oxygens are "pulling" on the electron density of the carbon instead of only one oxygen in the typical carbonyl group?

It's better to stick to relative comparisons like "CO2 is a stronger electrophile than benzaldehyde". This example is claimed here [http://www.chemicalforums.com/index.php?topic=51558.0], and should give you some idea of how strong of a Lewis base CO2 is.

You cannot adequately understand the Lewis acid theory with respect to main group elements if you only draw simple Lewis structures (which, per definition, should have complete octets on all main group elements) and then attempt to find electron pair acceptors. Well, to be perfectly honest, you might get lucky with boron compounds but the game will end there.
It is relatively clear to most people that metal cations can act as Lewis acids: they are positively charged because they lost at least one electron. They also have relatively low-lying orbitals (e.g. the one that lost an electron) with which the Lewis bases can interact. One has to apply more thought to molecules such as $\ce{CO2}$, though.
To analyse a molecule such as $\ce{CO2}$, you should start by thinking about the polarity of the bonds in question. The periodic table or whichever data page you wish to consult will tell us that oxygen is much more polar than carbon; therefore, the bonds are polarised towards oxygen. This can be exemplified by the partial charge formalism as shown below.
$$\ce{\overset{\delta -}{O}=\overset{\delta +}{C}=\overset{\delta -}{O}}\tag{1}$$
This first depiction already tells us that carbon is somewhat ‘electron-lacking’. While that in itself is not yet a proof of Lewis acidity, it is a strong hint. Diving deeper, one would have to consider the molecular orbitals of $\ce{CO2}$. There are two symmetry-equivalent but perpendicular π systems both populated by four electrons; and both can be seen as an allyl anion type of π system. Since it is wholly symmetric, we can draw the following three resonance structures (for clarification, the allyl anion $\ce{C3H5-}$, has also been included in the scheme).
$$\begin{array}{ccccc}\ce{O=C=O &<->& \overset{+}{O}#C-\overset{-}{O} &<->& O=\overset{+}{C}-\overset{-}{O}}\\
\ce{H2\overset{-}{C}-CH=CH2 &<->& H2C=CH-\overset{-}{C}H2 &<->& H2\overset{-}{C}-\overset{+}{C}H-\overset{-}{C}H2}\end{array}\tag{2}$$
In this depiction, I have chosen the left-hand double bond to belong to the perpendicular π system and the right-hand double bond to be the interesting one analogous to the allyl anion. Hence, there will always be one more bond on the left half of carbon dioxide than the allyl anion. The negative charge (i.e. lone pair) present on the leftmost allyl anion carbon corresponds to a lone pair of the leftmost oxygen in carbon dioxide; the lone pair is unmarked as per convention.
In these depictions we can observe a resonance structure wherein carbon dioxide’s carbon has a positive charge and thus an electron sextet. If you look at the actual orbitals in question, the two populated bonding orbitals (one may be called nonbonding) have a high oxygen contribution while the antibonding π3 orbital is a primarily carbon-centred one. This orbital is also the lowest unoccupied molecular orbital (LUMO) and thus the one we are interested in when determining Lewis acidity. We find that it is energetically low-lying and easily accessible and therefore Lewis bases (e.g. hydroxide, ammonia) can easily interact with it.
The Lewis acid-base reaction can also be guessed looking at the resonance structures. Because the Lewis base interacts with the antibonding π3 orbital, it will break one of the $\ce{C=O}$ double bonds leading to the following structure:
$$\ce{\color{cyan}{O}=C=O + \color{blue}{O}H- -> H\color{blue}{O}-C(=\color{cyan}{O})-\overset{-}{O}}\tag{3}$$
I have marked the oxygens in blue and cyan (and black) so they may be distinguished from another.

I love your answer. However, I have a hunch you're going to lose him around "There are two symmetry-equivalent but perpendicular π systems both populated by four electrons; and both can be seen as an allyl anion type of π system."

At the very least your answer certainly helped me see the the issue in a new light

On a lot of these questions I have wondered if we need multiple levels of answers. So high school, undergraduate, graduate and research level answers. ;-)


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Do alcohols react with alkyl halides?
I wonder if alkyl halides react with alcohols or not. I searched on Google but could only find reaction of alcohols with hydrogen halides

Please have a look at the Williamson ether synthesis

Not directly. As I hope you know, alkyl halides are electrophiles, accepting an electron pair at the electron-poor $\alpha$-carbon. Alcohols are nucleophiles, donating an electron pair from the electron-rich oxygen. So your question is a very reasonable one. However, both reagents are only mildly reactive, and the reaction will not proceed at useful rate.
We can deprotonate the alcohol $\ce{ROH}$ to form an alkoxide anion $\ce{RO^-}$. This requires a relatively strong base ($\ce{NaNH2}$ and $\ce{NaH}$ come to mind). The alkoxide is a much stronger nucleophile, and will attack the alkyl halide forming an ether. This usually occurs via the $\mathrm{S_N2}$ mechanism (unless the alkyl halide $\alpha$-C is tertiary). In fact, this reaction has a special name, the Williamson Ether Synthesis:
$$\ce{RO- + RX -> ROR + X-}$$
TL;DR Yes, if the alcohol is activated first.

Can we say that they are less basic than amines because they don't react with alkyl halide to give addition product on these basis?

Alcohols are indeed less basic than the corresponding amines, but no this substitution reaction would not be a good rationale for that. For one thing, this substitution requires the O (or N) to act as a nucleophile, which has subtle but important differences to acting as a base. I would rationalize nitrogen's greater basicity than oxygen by the fact that nitrogen is less electronegative, so it "holds its lone pair more loosely", making it more available to be shared. I think there is a more rigorous explanation using molecular orbitals, if you're interested in digging deeper.

Please note that although in general N is a stronger nucleophile than O and N is a stronger base than O, the nucleophilicity and basicity trends will not always be the same for any given pair of atoms.

And the base had better not be a good nucleophile for carbon.  Putting a Grignard reagent in and having it react directly with the alkyl halide gets ugly.

I am assuming that you want the alcohol to react with the $\ce{OH}$ group and the alkyl halides to react with the $\ce{Cl}$ group.
Both groups like to withdraw electrons, so they tend to be negative, and if we take them away, the resulting ions would be positive so they cannot react.
This leaves us to the possibility of the $\ce{H+}$ being cleaved off from the $\ce{OH}$, producing an alkoxide anion, which can react with the cation resulting from the cleaving off of the halide.
However, to effectively cleave off an $\ce{H+}$, we would need an acidic alcohol, such as phenol, which does react with alkyl halides (taken from here):

In the presence of a base, such as potassium carbonate ($\ce{K2CO3}$), and heat, phenol reacts with an alkyl halide to form an ether.

For example:
$$\ce{\underset{phenol}{PhOH} + \underset{1-chloropropane}{CH3-CH2-CH2Cl} ->[K2CO3] \underset{phenyl propyl ether}{PhOCH2-CH2-CH3} + HCl}$$
where $\ce{Ph}$ denotes the phenyl group.

The reaction proceeds through the ionization of phenol:
$$\ce{\underset{phenol}{PhOH} ->[K2CO3] \underset{phenoxide}{PhO-}}$$
The next step is an SN$2$ step:
$$\ce{\underset{phenoxide}{PhO-} + \underset{1-chloropropane}{CH3-CH2-CH2Cl} ->[K2CO3] \underset{phenyl propyl ether}{PhOCH2-CH2-CH3} + Cl-}$$

For other alcohols, they are generally not acidic, so you would need the corresponding alkoxide by reacting with sodium:
$$\ce{\underset{alcohol}{R1-OH} ->[Na] \underset{alkoxide}{R1-O-} ->[R2Cl] \underset{ether}{R1-O-R2}}$$
This, as the comments have pointed out, is called the Williamson ether synthesis.


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Infrared absorption spectrum of primary amides?
It's my understanding that the infrared absorption spectrum of primary amides beyond having a broad peak also has 2 sharp peaks representing the symmetric and antisymmetric stretch. My question is why isn't there a third peak representing a bending motion where the two hydrogen molecules bend towards each other?

Bending vibrations tend to have quite different frequencies, much lower compared to stretching vibrations. They tend to fall in regions of the spectrum which are more difficult to interpret. This page gives the primary amide N-H bend as 1570-1515 cm-1. So, the bend is indeed there, but the stretch is diagnostically much more useful.