text:
Will adding up protons and electrons (without neutrons) create a new element?
If protons have a positive charge and electrons have a negative charge,
can we add up several protons and electrons together to create a new element, without adding neutrons to hold the nuclei together?

Helium-2 has been detected--but the half life is so short that it's not even known.  Likewise Lithium-3.  Beryllium-4 hasn't even been detected.  I see no point in trying to walk further up the table.

Yes and no.
Elements are defined by the number of protons only. It does not matter if (say) a carbon nucleus has six or seven (or eight) neutrons, they will all react the same.*
With that, to create new elements, you would need to get up to some 115 or so protons fused together. But there is a reason for neutrons: all the positively charged protons in the nucleus repel each other electrostatically so neutrons are there to stabilise the nucleus — you can think of it as cushioning protons apart. For each element there is a set number of neutrons that will create a stable nucleus. E.g. for hydrogen (one proton) it is zero or one, for oxygen it is eight, for carbon six or seven, for tin it can be $62, 64, 65, 66, 67, 68, 69, 70, 72$ or $74$.
Nuclei with other neutron numbers decay radioactively on half-life scales from femtoseconds to millions of years. I would expect no-neutron nuclei (non-hydrogen) to be on the very short edge of that spectrum (although I’m no radiochemist).

*: There are reactions in certain systems that depend a lot on hydrogen being $\ce{^1H}$ rather than $\ce{^2H}$ or $\ce{D}$ because they rely on the tunneling effect whose probability decreases with mass (I think by the factor $m^2$). That’s also why one shouldn’t drink $\ce{D2O}$

The comments section is not for extended discussion, therefore I have moved them into a seperate chatroom. I also imported the conversation from the periodic table to there.

Just for fun, is there any way to extrapolate what the half-life of, say, ununennium-119 would be?  (I know it would be based on ridiculous and unrealistic assumptions, but we might get a hilariously small number.)

The Belt of Stability chart: http://chemwiki.ucdavis.edu/Physical_Chemistry/Nuclear_Chemistry/Nuclear_Stability_and_Magic_Numbers sum it very well.

@NateEldredge It wouldn't form in the first place so its half life is undefined.

Aren't all reaction rates dependent on the particular isotope (thus dependent on the number of neutrons)? - with the relative difference being highest for deuterium?

@PeterMortensen Yes, but the difference is neglegible, usually. Only in the proton/deuterium case does it usually actually make a difference.

Jan's answer is correct. I will try to fill in a few details about why neutrons are essential to creating stable nuclei.
All stable isotopes excepting Hydrogen-1 have neutrons in their nuclei.  Hydrogen, for example has two stable isotopes: The first simply has a proton with no neutrons in the nucleus, while the second, often called Deuterium has a proton and one neutron.
Oxygen has three stable isotopes (O-16, O-17 and O-18), and by stable, I mean that they NEVER decay.  The reason they never decay is that they cannot transmute (decay) to any other configuration of protons and neutrons without existing in a higher energy state.  Since they don't have the energy, the protons and neutrons that make up those nuclei are stuck in that configuration (until something gives them the necessary energy to go to a higher energy state).
Now, as to your question: Neutrons play an important role in stabilizing the nucleus.  Protons and neutrons respond to the Strong Nuclear Force (SNF) and are attracted to each other by that force. Protons, with charge, ALSO respond to the electrostatic force.  Since all protons are positively charged, the more of them present in a nucleus, the larger the force that is trying to rip the nucleus apart.
Neutrons, in a nucleus bring added SNF to the nucleus without adding any repulsive electrostatic force. Thus they play an important role in creating stable nuclei. They increase the tendency of a nucleus to stick together.  Thus all elements beyond Hydrogen have some neutrons in their nuclei.
And that explains why we don't find atoms with large numbers of protons and very few neutrons.  By the time we get to Uranium, we find that nature needed more than one neutron for each new proton in order to create stable nuclei.  U-238 is the most abundant form of Uranium and it has 92 protons and 146 neutrons.
But that leaves us with one other important dilemma.  If neutrons attract each other, and there is NO electrostatic repulsive force between them, one could reason that we should find atoms with large numbers of neutrons and very few protons. Why not Hydrogen 99?  1 proton and 98 neutrons would seem to be extremely stable, yet it is not found in nature. In fact, if you add even one more neutron to Deuterium, you find that it (Tritium) decays with about a 12.32 year half-life.
The answer, is that there exists yet another force besides the SNF and the electrostatic force.  It is called the Weak Nuclear Force because it is weaker than the SNF.
For an introduction to the Weak Nuclear Force I would go here:
https://en.wikipedia.org/wiki/Weak_interaction
So, we need neutrons to make atoms stable.  Electrons cannot exist inside the nucleus due to the Uncertainty principle, but that's another issue entirely.

One additional note: Neutrons on their own are not stable, they decay into a proton, an electron and a neutrino. This is the WNF in action.  The same things happen if you have a nucleus with too many neutrons.  Having the right number of protons and neutrons together stops this somehow. (Your Wikipedia link explains this, but I think it should be said directly)

I'll focus on the nucleus only. As far as I'm aware, there is little to no variation in radioactive decay lifetimes as a function of ion charge state; that is to say that the number of electrons surrounding a nucleus has little to no impact on whether the nucleus will decay.  I'm open to corrections from measurements at, say, the Hollifield facility at Oak Ridge.
The 1-hydrogen nucleus is stable - one proton, no neutrons.
The 4-lithium nucleus is listed as having a decay time of 8E-23 seconds per the Chart of the Nucleides. So, three protons and one neutron is really, really unstable. A 3-lithium nucleus has never been observed. So, three protons with no neutrons is not stable.
The 2-He nucleus has been put forth as the mechanism for the observed 'exotic 2-proton decay' of even-Z nuclei. However, a recent paper on 2-proton 19-Mg decay says 'modeling the decay as the tunneling of a diproton ($^{2}$He nucleus) typically yielded lifetimes several orders of magnitude too short.' Thus, it would appear that the 2-He nucleus, two protons and no neutrons, which has never been observed alone, is not the object leading to observed 2-proton decays.  Thus, while it has been postulated, it does not seem to occur.
A no-neutron universe would then consist of hydrogen, and hydrogen alone.

Thank you for elaborating , so the 2-He is actually a good example for what i was asking, even if it has a short lifespan , does it contain 2 electrons too ?

Even if it would be created in 2-proton decay those protons would get unbound in time similar to that for 4-lithium no electron would "even notice that" :)

@Mithoron Ah, i see your point there :) so the question is how did the researchers ever "notice that" .

Ever notice what? That 2-He was postulated was to try and account for the observed 2-proton decay via one particle leaving the nucleus (solar to an alpha), and then decaying into the two observed protons. You seem very keen still to link electrons with nuclei more closely than chemistry allows or nuclear physics requires.

@JonCuster Yes, because i thought they are entangled , maybe because i think all chemical bonds have to do with electrostatics ,i mixed up the ability of a single atom to exist without electrons as being a cause of instability\stability.

While in almost all cases the electronic structure/surroundings of an atom has negligible influence on its radioactivity, there are reported cases showing an effect. Here's an interesting read on the subject.

Also whether the hydrogen-1 nucleus is stable is something of an open question in physics.  Certain theories predict it to be unstable with a half-life of >$10^{31}$ years, only >21 orders of magnitude larger than the current age of the universe.  All attempts to observe proton decay have never found any, but it's a bit of an analytical challenge to measure something so rare.

Without getting into the specifics of nuclear forces, there is a simple way to think about this.
What matter for chemistry is the number of protons in the nucleus of the atom as this determines the nature of the electron orbitals the atom will have which is what chemistry is about.
So, if you could persuade electrons to joint the nucleus and that led to a proton turning into an neutron then you would indeed have a new element. Whether that would be a stable element depends on all sorts of complicated stuff involving nuclear stability (and the strong and weak nuclear forces I'm trying to avoid dealing with). Besides, we already mostly know what nuclei are stable or not as nuclear physics has explored the possibilities extensively.
Moreover, this process actually happens, though it isn't common. There is a rare form of radioactive decay for proton rich nuclei where the nucleus "captures" and electron transforming a proton to a neutron and thereby creating a different element. An example would be rubidium-83 (37 protons, 46 neutrons) decaying to krypton-83 (36 protons, 47 neutrons). Wikipedia lists more examples.
So the answer is: yes you can add electrons to a nucleus to create a new element. But the fact that it isn't common suggests it is hard. And this reflects both the instability of most of the products you could get (even if it could happen the products would be too unstable to see in most cases) and the difficulty of the reaction mechanism (there is a big energy barrier preventing it happening for most nuclei).

Re "What matter for chemistry is the number of protons": So we would be OK drinking heavy water instead of regular water?

@PeterMortensen I wouldn't advise drinking heavy water, though it can reverse some of the effects of drunkenness. Hydrogen is the one major exception to the "only protons matter" rule as doubling the nuclear mass makes a significant difference to hydrogen chemistry. Otherwise it is a good rule of, thumb.


text:
Density of wooden materials
Would all wooden materials have the same density? Why or why not? (Examples may help). I have the following solution to the question: No, not all wooden materials would have the same density. Due to the makeup and properties of the components found inside certain wooden materials.

You might want to migrate this to the DIY/Home Improvement SE site, where some folks with hands-on experience can address your question.

You might find this table of Wood Densities helpful. For example, note the density of balsa and that of Lignum Vitae. In fact, Lignum Vitae is so dense that it has been used for bearings on ships and even nuclear sub!

No, the sieve tubes(things that carry water and food around the plant) are different sizes depending on type of plant. Check out this video, it shows the difference in the the size of the tubes. https://www.youtube.com/watch?v=L6t2AZubF8U

source: Stack Exchange

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

sort: votes


text:
What is purpose of sulfuric acid in iodine extraction?
I am trying to extract iodine from kelp following this guide here.  It calls for sulfuric acid, but I don't see what this does do help the reaction.  I thought the reaction was simply oxidizing iodide to iodine with hydrogen peroxide, so I don't understand the purpose of the acid.  I do not have any sulfuric acid, and it is expensive to purchase, I do have a large supply of sodium bisulfate.  Could this be used as a substitute to the sulfuric acid if all that is needed is a low pH.

‘It is expensive to purchase’ — that is a lie. It is cheaper than (laboratory) sand.

The acid is not expensive, but the shipping costs are very high for strong acid concentrations

Have you seen this experiment in the Journal of Chemical Education? Revisiting History: Encountering Iodine Then and Now. A General Chemistry Laboratory To Observe Iodine from Seaweed. This article has a simpler procedure, if your experiment is for teaching purposes.
https://pubs.acs.org/doi/abs/10.1021/ed086p206

Reading the guide, you are supposed to use $2\,\mathrm{ml}$ of $1\,\mathrm{M}$ sulphuric acid in $20\,\mathrm{ml}$ of water which will put you in the range of $\mathrm{pH}\, 1$.
It is important due to the equation shown in trb456’s answer that there is acid present, since $\ce{H2O2}$ is usually only an oxidising agent under acidic conditions. You can see why if you deduce the partial equations of both oxidation and reduction to get the complete redox equation.
It doesn’t matter what the counterion to oxonium ($\ce{H3O+}$ is. So any acid that is strong enough will do the trick. Sulphuric acid is one of the cheapest choices, but dilute hydrochloric acid will do the same trick. $\ce{NaHSO4}$ (which I hope is the correct chemical; I never understood the bi- nomenclature and prefer hydrogen sulphate) may work since it is still pretty acidic, but you would need higher concentrations. Acetic acid will probably not work.
Nitric or phosphoric acid should also do the trick, in case you accidentally have them standing around on the shelf. Whatever the acid you use, if you have pH papers, just use one to check the pH. If it is around $1$, you’re good.
Don’t worry about oxidising $\ce{Cl-}$ to $\ce{Cl2}$ with excess $\ce{H2O2}$. For one, the concentration of chlorine will be too low to cause any harm (unless you decide to stick your nose in the beaker and take a deep breath or drink the whole thing). Also, the seaweed itself contains enough chloride ions that this would be a background reaction anyway. Stay in a well-ventilated place (as you always should when experimenting) and you should be fine.

You need a strong acid to provide hydrogen (technically hydronium) ions, which is true for most aqueous oxidation reactions:
$\ce{2I- +H2O2 +2H+ -> I2 +2H2O}$
Unfortunately, bisulfate ion appears to be a weaker acid than water $(K_{\mathrm{a}} =  10^{-2})$, so it may not work unless you use a very concentrated amount to obtain enough hydrogen ions. Try it!
A fairly cheap alternative could be hydrochloric acid, which may be available in hardware stores as muriatic acid, and is stronger than sulfuric acid. But read the MSDS, as hydrochloric acid fumes and is nasty stuff.
Best of luck, it’s a cool experiment!

I thought that sodium bisulfate was a strong acid as it has a pKa of 1.99.  I wanted to avoid using HCl, because I am worried I will also oxidize Cl- to Cl gas with the H2O2.

@popgalop: Oxidizing $\ce{Cl-}$ to $\ce{Cl2}$ will not happen because the reaction of $\ce{H2O2}$ with chlorine is much slower than with iodine. As I noted, the Ka of bisulfate is weaker than water.

But the Ka is still higher than water, because water has a Ka of 10^-14 and bisulfate has a Ka of 10−2

According to wikipedia a solution of NaHSO4 can have a ph less than 1.

That's not a big deal anyway. True, you can't push all iodide into the protonated form, but you don't really have to.

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


text:
What is the significance of the "Standard Temperature" in Standard Enthalpies of Formation tables?
In my data book, there's a list of common compounds and their molar enthalpies of formation -- at 298.15 K. What's the meaning of this given temperature value (is it the final temperature of the compound in question?) and how does it affect the molar enthalpy of formation of a substance?
What, for example, does it mean when the molar enthalpy of formation of H2O(g) at 298.15 K is given as -241.8 kJ/mol, even though water should be in a liquid state at 298.15 K? And what would the molar enthalpy of formation of H2O(g) be at, for instance, 500 K?

Seems you forgot about water vapor.

Sorry, I still don't understand. Isn't water vapour H2O(g)?

Yes, so there's no problem with it at 25C

So is the standard temperature listed in these molar enthalpy tables the temperature of the system upon losing/gaining the given amount of enthalpy?

It's simply enthalpy in this temp. you have to pick temp. so...  Oh whatever, I need to go to sleep

Most enthalpies (or themodynamic observables whatever their type) are not constant across the temperature range — not even across the range of one phase. So the enthalpy of vapourisation of water is different a $273\,\mathrm{K}$ from what it is at $295\,\mathrm{K}$ or $350\,\mathrm{K}$.
Of course, standard conditions have been defined. However, the definition of standard conditions does not explicitly include temperature! Likely because some wanted $273\,\mathrm{K}$, others wanted $295\,\mathrm{K}$. So the temperature at which a certain entropy is valid must always be specified.
Some books put a ‘standard temperature’ at the top and call that their specification.

See also Notation for states and processes, significance of the word standard in chemical thermodynamics, and remarks on commonly tabulated forms of thermodynamic functions (IUPAC Recommendations 1981): “The above definitions of standard states make no reference to fixed temperature. Hence, it is possible to have an infinite number of standard states of a substance as the temperature varies. (…)”


text:
Name of the substance on which water cannot stay?
There is a show on discovery, 'You have been warned.'. In that show, once I remember seeing an unusual substance, when, on water was poured in it, even on a perfectly straight (non-slanting) table, water just didn't stay on it. It "rushed" in random directions and then when it was not in contact of the substance, it stopped. I don't remember the episode, nor the substance.
What could this substance be, or if they're a class of compounds, what are their properties?
Can anybody also  point out that episode?

What you are talking about is ultrahydrophobic coatings. They can be made of different materials; personally, I am familiar with ZnO/PS nano-composites. I urge you to read more into it. Specifically, do look up the Lotus effect. I am sure you will be able to find all sorts of examples.

Have you seen that eqisode?

You may also want to consider researching lipophilicity. Although lipophilicity is not really synonymous with hydrophobicity because there are exceptions.

Thank you @getafix! Found this https://www.youtube.com/watch?v=IPM8OR6W6WE.

Can I move it to television stack exchange?


text:
Is Alkanol an official part of chemical Nomenclature?
There are several references to alkanols here and here but neither of these sources are too reliable. The IUPAC gold book seems not to recognise the term and a quick google search reveals no really good sources.Whilst still not a super reliable source the fact there are no results on Wikipedia but simply a redirect to alcohol also furthers concerns and so the question is: Is the term Alkanol a 'real', 'proper', and 'official' term in chemistry?

IUPAC gold book doesn't recognize it, that's a pretty solid answer. I've never heard anyone say 'alkanol' before either, I've only heard people refer to these as simply alcohols. However, there are people who still use the word 'ethylene' to describe the compound 'ethene' so it could just happen to be some common language I haven't been exposed to.
To be clear, no it's not official, but yes it might be real terminology used by some organic chemists.

The term ‘alkanol’ is widely used in the literature; for example, a search on ScienceDirect yields several thousand results. It is true though that ‘alkanol’ is not included in the systematic IUPAC nomenclature.
You are right that the IUPAC Compendium of Chemical Terminology (Gold Book), Version 2.3.3 (2014-02-24) does not include the word ‘alkanol’. However, the Gold Book is merely a collection of terminology definitions found in other IUPAC publications. It is not a sufficient source for nomenclature rules. Nevertheless, class names of organic compounds are usually included in the Gold Book.
The current rules for nomenclature of organic chemistry can be found in Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book). In many cases, the Blue Book uses the class names of organic compound given in the Glossary of class names of organic compounds and reactivity intermediates based on structure (IUPAC Recommendations 1995). Neither of them includes the word ‘alkanol’.
According to current IUPAC recommendations, the class name for compounds in which a hydroxy group $\ce{-OH}$ is attached to a saturated carbon atom is ‘alcohols’.
Alcohols can be named in several ways. Preferred IUPAC names of simple alcohols are generated by attaching the suffix ‘ol’ to the name of the parent hydride, with elision of the final letter ‘e’ in the parent hydride, if present (methanol, ethanol, propan-1-ol, etc.).
In general nomenclature it is also possible (and particularly relevant in the older literature) to generate the name by linking a substituent groups (a radical) to the class name ‘alcohol’ (methyl alcohol, ethyl alcohol, propyl alcohol, etc.).
Since ‘alkyl group’ is a valid class name, it is in principle possible to use the historical rule and generate the generic name ‘alkyl alcohol’. The term ‘alkanol’ might be considered the analogous generic name in imitation of the currently preferred method using the suffix ‘ol’:
methyl alcohol → methanol
ethyl alcohol → ethanol
alkyl alcohol → alkanol
However, such analogies can be deceptive. The analogous conversion of the (not preferred but correct) name ‘isopropyl alcohol’ would yield the incorrect name ‘isopropanol’. (IUPAC recommends that the name ‘isopropanol’ should be abandoned since there is no hydrocarbon ‘isopropane’).

If it is not given in the Gold Book, then it is not an official part of nomenclature.
However, at the same time the equation alkanol = simple alcohol is incorrect. Alkanol would be used only for completely saturated alcohol, while for exaple allylic alcohol (prop-2-en-1-ol) or but-2-en-1,4-diol are clearly very simple alcohols but not alkanols.
Much like alkanol also similar terms as alkenol, alkanone, alkenone etc. are in use.
You can expect every organic chemist to understand the term correctly after about two semesters’ organic chemistry courses.


text:
Converting hydrogen to methane
I am in need of small amount of pure methane for laboratory purpose.
I have hydrogen generator in the lab.
Is there a possibility of generating methane from hydrogen ?
May be make hydrogen flow over heated graphite ? ( I am newbie and just trying to get the idea...)
Any help in this direction is helpful. thanks.

In laboratory, methane is made by heating sodium acetate with soda-lime.

I'd just buy a lecture bottle of methane of the required purity -  the impurities would be known and upper bounds given by the vendor.

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


text:
Need for enthalpy of vaporization in calculating bond dissociation enthalpy
In our chemical bonding class we were given the following problem:

I am confused by how the question says that you need to consider the heat of vaporization/sublimation for HBr and HI.

Bond dissociation enthalpy is defined as "the energy needed to break one mole of the bond to give separated atoms - everything being in the gas state."

So if you can find the heat of formation of that molecule in the gas phase, why would you need anything else?
My attempt:

The above answer obtained is not the one provided by the professor so I assume I'm missing something here. The answer provided is-

364 kJ/mol

Well, I think that you have to look up the actual way bond dissociation enthalpy is defined in your textbook/lecture notes, since it is a not quite standard term. IUPAC defines only bond-dissociation energy but not enthalpy.

bond-dissociation energy, $D$
The enthalpy (per mole) required to break a given bond of some
specific molecular entity by homolysis, e.g. for $\ce{CH4 -> .CH3 + H.}$, symbolized as $D(\ce{CH3−H})$ (cf. heterolytic bond dissociation energy).

Many resources on the Internet claim that the IUPAC definition of the bond-dissociation energy refers to $\pu{0 K}$ (see e.g. Wikipedia, UC Davis ChemWiki) and that it differs from bond dissociation enthalpy, which refers to the enthalpy change at room temperature ($\ce{298K}$).
Wikipedia

The bond-dissociation energy is sometime also called the
bond-dissociation enthalpy (or bond enthalpy), but these terms are not
strictly equivalent, as they refer to the above reaction enthalpy at
standard conditions, and differ from $D_0$ by about $\pu{1.5 kcal mol-1} \pu{(6
kJ mol-1)}$ in the case of a bond to hydrogen in a large organic molecule.

UC Davis ChemWiki

Officially, the IUPAC definition of bond dissociation energy refers to
the energy change that occurs at $\pu{0 K}$, and the symbol is $D_0$. However,
it is commonly referred to as BDE, the bond dissociation energy, and
it is generally used, albeit imprecisely, interchangeably with the
bond dissociation enthalpy, which generally refers to the enthalpy
change at room temperature ($\pu{298K}$). Although there are technically
differences between BDEs at $\pu{0 K}$ and $\pu{298 K}$, those difference are not
large and generally do not affect interpretations of chemical
processes.

If we follow this definitions, then yeah, at standard conditions $\ce{HBr}$ is liquid so we have to add up $\pu{12 kJ mol-1}$ for its enthalpy of vaporization, and (almost) get the desired $\pu{364 kJ mol-1}$. The small discrepancy is probably due to slightly wrong value of enthalpy of vaporization which I just quickly found somewhere.

Great thank you for your explanation. I believe you're correct as this yields the professor's answer

I think he wants it at 0 K which would make it a solid. If you take the heat of fusion and vaporization into account, the correct answer is given. Thanks again

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


text:
Cheap silocone casting resin (1 or 2 component resin)?
I searched in many places and on Youtube and I found ways
to make silicon mold putty from cheap silicone using ordinary
household stuff and kitchen stuff.
But nowhere is a recipe on making silicon casting resin out of the cheapest everyday silicone.
Clearly, casting resin has a much better usage spectrum than molding putty and it is the way molds should be created. Ethylene-Acetate dissolves silicone and results in very long
hardening times (weeks or months) and I suspect that it also
could make the molds shrink a lot while hardening. Any ideas?
I am not dependent on kitchen materials and I can buy some
chemicals like Ethylene-Acetate in the local painter's store.

The problem with attempting this is the nature of the cheap silicone.  Most store bought silicone is already polymerized before use.  For example, the silicone caulk which can be purchased in home improvement areas is a finished product which  has been infiltrated with solvent.  These caulks contain no catalysts to facilitate reactions.  Any curing is the result of solvent drying from the caulk and solidifying the resin.
If you were to attempt to further dissolve one of these materials with something like ethyl acetate (see here for a simplified list of solvents), it probably would not go well.  All you would be doing is further impregnating the silicone with solvent.  While this would decrease the viscosity, it would increase your drying time.  Shrinking would occur during your casting, and I expect that you would also experience a very messy result, no matter the quantity of release agent you applied.
Consider a very simple silicone (aka polysiloxane).  Two silanol molecules react by hydrolysis to form a disiloxane and a water molecule.  In order to reverse the process and obtain your starting material, you would need to put the water back into the disiloxane to cleave it.  This is energetically expensive.
Even the full polymerization process is not a spontaneous one.  When you purchase a silicone casting resin from a supplier, you are actually buying a Part A composed of n-siloxane (where n is a small enough number to be a liquid) and a Part B composed of crosslinker and catalyst (this is often silicic acid and tin because they are inexpensive).  So what you are actually doing is taking a short chain polymer and crosslinking it into a solid mass.
Adding silicic acid and tin to a store-bought silicone product such as caulk would probably cause a reaction.  The result would probably be a very hard rubber with incompatible properties for casting.  However, I have not attempted this.  Furthermore, I am not sure silicic acid would be easily accessible to a home chemist.
Despite how fun it would be to synthesize your own silicone casting resin at home from everyday chemicals, it probably isn't feasible.  There are some inexpensive options out there for resin casting, and I encourage you to go with those.

many thanks this is a good explanation!
to everybody: I seek inexpensive casting resin vendors,
please add links or names here!
thanks a lot, Gerald

Glad I could help. I didn't want to put any sort of brand support in the answer.  My current favorite is OOMOO 30 from Smooth-On.  For molding projects that only needed a temporary rubber, I used Cast-A-Mold 25T.  I found both through Amazon, although I have been making purchases directly from Smooth-On lately.  For kitchen chemistry, I recommend something like the OOMOO which is mixed 1:1 by volume rather than the comparatively more complicated 10:1 by mass ratio.


text:
Can linear SOCl2 exist?
Specifically, instead of its normal structure, this Lewis structure seems to be valid as a constitutional isomer:
$\ce{Cl-O-S-Cl}$
So what prevents this from existing, or if it already exists how is it made?

Anything with Cl-O bond is a strong oxidizer, sulfur in the oxidation state +2 is a reducing agent, so the thing will pretty quickly transform into common $\ce{SOCl2}$.

Just for clarification: You do mean ‘linear’ $\ce{Cl-O-S-Cl}$ as opposed to trigonal pyramidal $\ce{O-SCl2}$, correct?

Not for long. Sulfur is in this case a reducing agent (love to give electrons away). And Oxygen is an odixidizing agent. It's just not stable or favorable. But sure you can make it.

Ivan's comment is better than this answer - you need to back up claims about possibility of creation in the answer

Not quite. I failed to add a discrediting note like "...should this thing exist, it will..."

@Ivan Neretin Yes, it looks like it would be better, but still...

I highly doubt that it could exist. The co-linear arrangement, $C_{\infty\mathrm{v}}$, of $\ce{Cl-O-S-Cl}$ is a second order saddle point on the DF-BP86+D3(BJ)/def2-SVP level of theory on the $S=0$ potential energy surface. It has two imaginary bending modes, $\nu_1=371.39\mathrm{i}$, $\nu_2=108.23\mathrm{i}$. (Please note that they are actually perpendicular to each other.)

That basically means, that distortion along these modes will lead to an arrangement of the nuclei which is lower in energy. Since this is a second order saddle point, this arrangement can itself only be a transition state (first order saddle point), which also cannot be considered stable. (Compare also here.)
Therefore distortion among $\nu_1$ would probably lead to $\ce{Cl-O}$ and $\ce{S-Cl}$ fragmentation. While distortion along $\nu_2$ might preserve the $\ce{S-O}$ bond, fragmenting the chlorine atoms/ chloride ions - but that is pure speculation of a hypothetical case.

Let's try a very tempting ad-hoc explanation of why this is the case. In that case let us assume all atoms are $\mathrm{sp}$ hybridised and the bonding axis is $z$. The Lewis structure that seems to be valid has therefore three lone pairs ($\mathrm{sp}$, $\mathrm{p}_x$, $\mathrm{p}_y$) at each chlorine and two lone pairs ($\mathrm{p}_x$, $\mathrm{p}_y$) at the oxygen and sulfur. The three σ bonds are formed by the $\mathrm{sp}$ orbitals.
That would result in four lone pairs that are co-aligned in the same plane. Overlap is inevitable, and in the most simplest terms of VSEPR theory, we would call that lone pair repulsion. (Please note that for strictly VSEPR theory we would have to assume tetrahedral hybridisation.)
In other words, same spin electrons have to avoid occupying the same space and since the orbitals are so close together this will rise the energy a lot. You can probably call it Pauli repulsion.
The MO picture is too messy, so that I won't show it here. Basically most of the occupied orbitals are anti-bonding with respect to one of the bond. The HOMO is of π symmetry and anti-bonding with respect to every bond.

This is a good answer, but unfortunately I wasn't really interested in sp hybridized linear, I meant it in the "straight chain alkane" linear sense where the oxygen and sulfur are likely sp3 hybridized.

@Kainui you mean in a zig-zag fashion? Well then it is not linear, is it?

It's not linear in that symmetry-phys chem sense, but in sense of org chem, i.e. 'linear alkane chain', it is somehow "linear".

Yeah exactly @ph13 , I mean it seems kind of funny to imagine an oxygen in a truly linear state, I can't think of a single case where oxygen is truly 'linear' in that sense lol.


text:
Colorless nature of carbonyl complexes
Researching the coloration of transition metal complexes one day, I came across this reference on Wikipedia 1:

I am not sure why this is so. Take $\ce{Mo(CO)_6}$, or any other carbonyl complex of the metals in group 6. If we have a strong $\pi$ acceptor like carbonyl, and hence a low-spin $d^6$ system, would we not get an orbital splitting pattern like this:

The excitation of an electron to one of the empty $e_g$ orbitals seems feasible to me. Why can't one of those paired electrons absorb radiation to exhibit colour for the complex?
Reference
1 https://en.wikipedia.org/wiki/Metal_carbonyl

Remember that going from t2g to eg in an octahedral complex is Laporte-forbidden, too.

Why can't one of those paired electrons absorb radiation to exhibit colour for the complex?

They can, they do and they color. The problem is, the corresponding transition requires so much energye, that the adsorption is well above upper range of visible spectrum in most cases. Linked article provides us with 3 optical spectra of $\ce{Cr}$, $\ce{Mo}$ and $\ce{W}$ carbonyls. For all adsoprtion mostly happens above 300 nm, with only small bits between 300 and 400 nm. Visible spectrum is between ~400 and ~700 nm. Since only bit of upper blue part of the spectrum is adsorbed, only a pale yellow coloring is observed.
This, of course, if you want so badly to apply crystal field theory to carbonyls, which does not work well in the case.

Because the transition need too much energy. If orbital relaxation is neglected, the energy required to excite an electron from the $j^{th}$ (occupied) orbital into the $i^{th}$ (unoccupied) orbital is
$\displaystyle \Delta E_{ij}$   $\textstyle =$    $\displaystyle (\epsilon_{i} - \epsilon_{j})$
    $\textstyle -$    $\displaystyle \int \rho_i({\bf r}) \rho_j({\bf r}') v_{\rm E}({\bf r}-{\bf r}') \,
{\rm d}{\bf r} \, {\rm d}{\bf r}'$
    $\textstyle +$    $\displaystyle \delta_{s_is_j}
\int \phi_i^*({\bf r}) \phi_j^*({\bf r}') \phi_j({\bf r}) \phi_i({\bf r}') f({\bf r}-{\bf r}') \, {\rm d}{\bf r} \, {\rm d}{\bf r}'$
    $\textstyle +$    $\displaystyle \frac{1}{2} \int \rho_i({\bf r}) \rho_i({\bf r}')
\left[ v_{\rm E...
...bf r}-{\bf r}')-f({\bf r}-{\bf r}')\right] \, {\rm d}{\bf r} \, {\rm d}{\bf r}'$
    $\textstyle +$    $\displaystyle \frac{1}{2} \int
\rho_j({\bf r}) \rho_j({\bf r}') \left[ v_{\rm E...
...r}-{\bf r}')-f({\bf r}-{\bf r}')\right] \, {\rm d}{\bf r} \, {\rm d}{\bf r}'\;,$
where $\rho_k = \vert\phi_k\vert^2$ is the charge density from the $k^{th}$ orbital.
So you need a lot of energy for the transition and because $E=hc/\lambda$, the wavelength will be so low that you can't see it. Therfor no colour.


text:
Cell voltage required for refining copper
If you put Zinc or Iron solids into aqueous copper (II) sulphate solutions, copper solids form spontaneously due to the fact, that Zinc/Iron oxides into ions more willingly than Copper does.
And if you put Silver bars into the same solution, hoping to get copper solids forming on them,
(yes, electroplating copper on silver)
you'd need a theoretical minimum of $\rm 0.46\,V$, as that is the difference between the standard reduction potential of $\ce{Ag}$ and $\ce{Cu}$ ions.
Our question is this.
While purifying/electrorefining copper, how much voltage is required to make it happen?
If we simply think in standard potential dimensions, we wouldn't need to do anything. And yet, common sense tells me that surely wouldn't be right.
So, how much, and why?

There is no universal answer for any refining/purifying process. It depends on what you're starting with, and what impurities are present.


text:
What is difference between intramolecular redox and disproportionation redox
I can not understand difference between the intramolecular redox reaction and the Disproportionation redox reaction. Both involve the same molecule as substrate, but what is difference?

Disproportionation is a redox reaction where one element works both ways, with part of it being oxidized and another part reduced. It does not have to be intramolecular in the strict sense, since the oxidizing and reducing agents do not necessarily come from the same molecule (though they come from similar molecules, of course). Think of the well-known reaction:
$$\ce{3KClO -> KClO3 + 2KCl}$$
It is certainly a disproportionation, but one may hardly call it intramolecular; there is not much going on within one molecule (or rather within one ion, for all of these are ionic compounds). Chlorine oxidizes chlorine; now, one ion contains one atom of chlorine, so to oxidize anything it needs to find another ion in the first place.
Intramolecular redox reaction occurs when we have the oxidizing and reducing agents within one molecule. They may or may not be of the same element. In the former case it is going to be a disproportionation, but not in the latter. Think of TNT:
$$\ce{2 C6H2(NO2)3CH3 → 3 N2 + 5 H2O + 7 CO + 7 C}$$
See what's going on? Basically, N oxidizes C. They are indeed from the same molecule, so this may be intramolecular, but they are not the same element, hence this is not a disproportionation.

You're sure TNT decomposition is intramolecular?

Not really. It's a pretty violent thing, with a lot of molecules breaking up and reacting all together. I should have said that this is but an example of a reaction that can happen intramolecularly. Come to think of it, redox reactions with proven and definitive intramolecular character are relatively rare. The Wolff–Kishner reduction  may be a valid example.

Slight disagreement: any elimination reaction with a cyclic transition state (e.g. selenoxide pyrolysis, ester pyrolysis) is an intramolecular redox reaction although the only redox active elements are typically carbon. There are probably quite a few more of that type.

That's right, so they are not that rare, after all.

DISPROPORTIONATION DEFINITION:

Disproportionation is a chemical reaction, typically a redox reaction, where a molecule is transformed into two or more dissimilar products. In a redox reaction, the species is simultaneously oxidized and reduced to form at least two different products.

Disproportionation reactions follow the form:
$$\ce{2A -> A' + A"}$$
where $\ce{A}$, $\ce{A'}$, and $\ce{A"}$ are all different chemical species.
The reverse reaction of disproportionation is called comproportionation.
Examples: Hydrogen peroxide converting into water and oxygen is a disproportionation reaction.
$$\ce{2 H2O2 -> H2O + O2}$$
Water dissociating into $\ce{H3O+}$ and $\ce{OH-}$ is an example of a disproportionation reaction that is not a redox reaction.
In intramolecular redox reaction, the molecule of a single substance undergoes oxidation-reaction by the process decomposition.
e.g. Thermal decomposition of ammonium nitrite  into $\ce{N2}$ and $\ce{H2O}$:
                      $$\ce{NH4NO2 -> N2 + 2H2O}$$
Here, the oxidation number of nitrogen in $\ce{NH4+}$ ion changes from -3 to zero (oxidation), and that of nitrogen in $\ce{NO2-}$ ion changes from +3 to zero (reduction) in the formation of $\ce{N2}$.
Another example is the thermal decomposition of $\ce{KClO3}$ into $\ce{KCl}$ and $\ce{O2}$. In this case, chlorine is reduced to -1 state and oxygen is oxidised to zero state.

I suppose your example for intramolecular redox is better than the one given in the previous answer.

@AaronJohnSabu I don’t; ammonium nitrite (apart from not existing due to immediately decomposing) is an ionic compound composed of ammonium molecular ions and nitrite molecular ions. The ammonium and nitrite ions do not form common molecules thus the redox reaction is not intramolecular.

In intramolecular redox a single molecule is the substrate.
But in case of disproportionation a single element is both reduced and oxidised.The element might be a part of a molecule with other elements.

Could you expand on this a little more? Usually when answering a question that has other answers, you want to make sure your answer is sufficiently different from the others or clarifies some point that other answers missed. Welcome to Chem SE!

Intramolecular reactions are those in which different elements (or species), which are from the same molecule, undergo oxidation and reduction.
Disproportionation reactions are those in which the same element undergoes both oxidation and reduction to give different products.


text:
Do entropy changes have an equivalent of bond enthalpy?
Is it possible to calculate entropy changes in a way similar to enthalpies from bond enthalpies? For example, to work out the enthalpy change of a reaction you can get the sum of all bonds in the products and minus the sum of all bonds in the reactants to get a reasonable approximation of the enthalpy change of reaction. Does such a thing exist for entropy?
Thanks!

Please note that you are calculating the $\Delta U$ with your bond breaking/forming method. If there is low volume work done, then you can approximate:
$$
\Delta U \approx \Delta H
$$
For the entropy: Yes and no. I have no idea for liquids but qualitatively I would guess that it is not possible to have a quick and dirty method as you wish for this phase. And even for gases and solids it is quite a long work.
For gases you can calculate the entropy for every component itself and then add/substract.
Translational (Sackur-Tetrode):
$$
S_\text{trans}= R \ln \left( (2\pi m k_B T)^\frac{3}{2} \frac{V_m}{N_lh^3} \exp(5/2) \right)
$$
Rotational:
$$
S_\text{rot}=R\left(\ln\left(\frac T\Theta_\text{rot} \right) +1 \right)
$$
Vibrational:
$$
S_\text{vib}=R \left(\frac{\Theta_\text{vib}}{T} \frac{1}{\exp\left( \frac{\Theta_\text{vib}}{T}\right)-1} - \ln \left(1-\exp\left( \frac{-\Theta_\text{vib}}{T}\right) \right)\right)
$$
For solids you could calculate the heat capacity $C_p(T)$ with the Debye modell and then integrate from 0K.
I am really sorry, but I do not think, that there are easier methods.


text:
Difference between 'homogeneous mixture of several chemical compounds' & 'non-homogeneous system'
I just started reading Thermodynamics by Enrico Fermi; he started defining state of a system for homogeneous mixture of several chemical compounds & then non-homogeneous system as:

I thought non-homogeneous system to be a mixture of several homogeneous chemical compounds separated by boundaries. But Fermi seems to have done it otherwise; he has defined the state separately for homogeneous mixture of several chemical compounds & non-homogeneous system that is, mixture of several homogeneous chemical compounds. I am not getting how they are different at all. Can anyone please explain how they are different? What's the difference between  homogeneous mixture of several chemical compounds & non-homogeneous system?

You are right that normally a non-homogeneous system would be one where there is boundary between two internally homogeneous phases (like milk which is basically oil and water mixed at a very fine scale).
But Fermi as also allowing for the possibility that there may be only one phase where the concentration of its components varies continuously in space (if there is more than one phase the variation is discontinuous by definition).
It certainly isn't a common situation if the phase in in equilibrium where we would expect things like diffusion to create a uniform concentration of things over time, at least in liquids and gases). But it is very common in the real world where equilibrium is rarely reached (imagine the state of a poorly stirred cup of tea where the concentrations of tea-stuff will be larger near the tea bag than far away from the tea bag).
It is also very common in places where it is hard to get to equilibrium which is often true in solids (I don't know whether this is what Fermi was thinking of). Pyrex glass, for example, is apparently a single phase but the structure is deliberately made different near the surface than far away from the surface to create tension in the surface layers that protect the glass from thermal shock. But the properties vary continuously.
The examples I can think of may not be relevant to what Fermi was saying as they mostly are metastable and won't be at equilibrium which is often what thermodynamics restricts itself to worrying about.


text:
Demonstrational smelting iron
I'm planning an instructional demonstration of smelting. It seems you may be able to take rust (hematite, $\mathrm{Fe_2O_3}$) and carbon (perhaps from charcoal briquettes), place this in a crucible of ample size, light the carbon on fire, and the formed iron may melt and accumulate on the bottom of the crucible.
Yes, iron burns, everyone knows you can light steel wool.  I'm hoping that the interface above the iron will become encrusted with enough slag and ash that there won't be much of the formed iron burning, and what's specific about steel wool versus trying to light a spoon on fire is that the steel wool has a very high surface-to-mass ratio, enabling it to be heated well as well as provide ample mass transfer area.
What are the problems with this demonstration, and what kinds of solutions exist for this?  For example, I foresee that I need to use lots of extra carbon to make sure that any iron formed doesn't reoxidize.

Although the mechanism doesn't involve carbon, the widely-used thermite reaction does give you iron metal from hematite (aluminum metal is oxidized in the process), which is typically collected in sand until it cools. Are you open to something like this, or does it have to involve carbon?

I'm hoping that this would be done like in a blast furnace.  I'm not so interested in using the Thermite process.   I'm looking more for the "blooming" process described on the page "Smelting" on Wikipedia.

I shouldn't have said "blast furnace" as much as "in a historical fashion."

The biggest problem you will have is getting a high enough temperature, not the oxidation of the iron.
Putting the crucible in a charcoal furnace won't produce a high enough temperature to cause the reaction or the liquefy the iron: you need to have a crucible that can be driven to a much higher temperature. Normally this is done by blowing air (or, if you have it, oxygen) into the furnace. In practice what is often done is to mix the iron oxide and an excess of carbon in a single vessel: the carbon burns to create the temperature and also acts as the reducing agent to create the iron. This still requires a source to force air into the mixture. This appears to be how ancient craftsmen did the reaction, kicking off the iron age.
For an account of how to do this see this blog describing a modern recreation of some of those ancient techniques.

Thank you, this is very interesting.  I like his tower.  I think it may be too large of a scale.  I am thinking of proceeding with in a slightly different way.  I will post something on the tube when the results are interesting.


text:
What are the volumetric thermal expansion coefficients for butyl acrylate and methyl methacrylate monomers?
We have some tanks of butyl acrylate and methyl methacrylate monomers, and the level indicators are giving incorrect readings.  The control systems engineer who set up that system suspects that the density values used in programming the level indicators were wrong, so I measured the density of the monomers by hand today.  However, the monomer samples I had were at around 70-72 Fahrenheit, whereas the tanks are usually closer to 52 Fahrenheit.  I'm trying to find the volumetric thermal expansion coefficients for these liquids so I can correct my measurements, but I'm unable to find them.  I've checked online and in the CRC Handbook of Chemistry and Physics (which lists coefficients for some organics, but not these).
I know that the expansion coefficients are temperature dependent, but if I can get even a single value (at, say, 20 Celsius, which seems to be a fairly standard place to measure this for organics) I should be able to get my estimate close enough to be able to judge if this is the source of error in the level indicators or not.

At what temperature did you measure the density? The CRC Handbook gives a density of 0.8936 at 68F for butyl methacrylate. Over such a small temperature range like this you could just extrapolate/interpolate the density based on 2 values, so long as they are at least several degrees or so apart. For example, the density of methyl methacrylate is 0.945 at 68F, and 0.934 at 77F. If we just did a linear extrapolation to 53F, we would calculate a density at that temperature of 0.965. Extrapolation can get you in trouble though. Do you have a density measurement at a lower temperature?

I should have measured density at a lower temperature, yeah.  It would have been easy enough to measure the density in a cool water bath or something (or heck, just using the literature values for the density at a different temperature) and interpolate in that range to get a rough estimate.  These were early days in my career though, so my first instinct was to use one of the things I learned in school and I ended up over-complicating it.  I can't remember now how this issue ended up getting solved.


text:
CuCl via electrolysis of HCl
I need a solution of CuCl in HCl.  Since CuCl reagent is such a pain to source and work with (because it oxidizes so quickly) I was thinking that I could create the solution simply via electrolysis of copper electrodes in HCl.
Is anyone aware of research on this electrolytic reaction?  I don't have time right now to look for side reactions or to play with parameters (solution strength, voltage, etc.).
(Unfortunately I can't find any research specific to this, because CuCl-HCl electrolysis is a popular method of generating hydrogen, so all searches return papers on that.)


text:
Does "parts" mean molar or mass?
Researching old (like century-old) papers it is common for chemists to describe preparations in terms of parts – e.g., "1 part water, 2 parts sulfuric acid."
Is there a reliable convention on whether this means molar parts or mass parts?  I.e., in the above example is the formula 1:2 parts by mass or by molecules?

There is none. Each time, you have to work your way back to the point where the authors specify which parts they are talking about (or where it becomes clear from the context).

Also, don't forget the possibility of parts by volume. Y'know, to make it that much less ambiguous.


text:
Difference between dissociation energy and dissociation temperature for water?
I was trying to find the temperature at which water dissociates into hydrogen and oxygen, and so I came across this Wikipedia page. It states that it takes $\pu{493.4 kJ/mol} + \pu{424.4 kJ/mol}$ to dissociate the bonds in a water molecule which gives a total of about $\pu{920 kJ/mol}$ or about $\pu{51 MJ/kg}$. So I tried to calculate the temperature that would give that huge amount of energy using the heat capacity of water vapor. I took $\pu{3.3 kJ/kgK}$ for water vapor as an average specific heat, and so by simply dividing $\pu{51 MJ}$ by $\pu{3.3 kJ}$, I ended up with about $\pu{15,000 K}$.
But on this other Wikipedia page, I found: "at $\pu{2200 ^\circ C}$ about three percent of all $\ce{H2O}$ molecules are dissociated into various combinations of hydrogen and oxygen atoms. At the very high temperature of $\pu{3000 ^\circ C}$ more than half of the water molecules are decomposed".
So now I am very confused. If $\pu{3000 ^\circ C}$ is enough to dissociate more than half of the water molecules, then it should be expected that almost all the molecules should dissociate at around $\pu{4000 ^\circ C}$ or so and not the $\pu{15,000 K}$ I calculated earlier.
So am I missing something here? Or is there an explanation to this?

One key is an assumption about various degrees of dissociation.

It states that it takes 493.4 kJ/mol + 424.4 kJ/mol to dissociate the bonds in a water molecule which gives a total of about 920 kJ/mol or about 51 MJ/kg.

That sounds like the energy required for $\ce{H2O <=> 2H + O}$.

But on this other Wikipedia page, I found: "at 2200 °C about three percent of all H2O molecules are dissociated into various combinations of hydrogen and oxygen atoms.

Here it sounds like they are considering less extreme dissociations than complete dispersal to individual atoms. For example, $\ce{H2O <=> H + OH}$ and perhaps even $\ce{H2O <=> H2 + O}$ or $\ce{H2O <=> 2H + \frac{1}{2}O2}$ and related equilibria may be happening.
In effect, if just one OH bond breaks in water, that only requires 493.4 kJ/mol and leads to the formation of species such as $\ce{OH}$ and $\ce{H}$ atoms.  Those species have a different heat capacity than $\ce{H2O}$ vapor.  The heat capacity page you linked to says:

The values above apply to undissociated states. At high temperatures above 1500 K dissociation becomes appreciable and pressure is a significant variable.

In the calculation you are doing you are not including a specific pressure variable either.  Heat capacity could also be a function of pressure.  Presumably the data you linked to is all for 1 atm of pressure, although it didn't really say.  Likewise, the claim that at 2200 °C, 3% of water is dissociated must also be pressure-dependent.  For which pressure is it true?
The final complexity is the entropic term.    The three dissociated atoms have 9 translational degrees of freedom but intact water only has three.
These pressure and entropy complexities matter because $\Delta G$ is what determines equilibrium, and $\Delta G = \Delta U + \Delta (PV) - T \Delta S$.  The bond dissociation energies you quote represent the $\Delta U$ term, so the calculation you are doing is implicitly assuming that $\Delta U \gg \Delta (PV) - T \Delta S$.
I think that your calculation is admirable in that it comes so close to the right answer while still neglecting all these complexities!

Quick caveat: the bond dissociation energies are closer to $\Delta H$ than $\Delta U$.  So I mis-spoke a tad above.  But they are "standard" enthalpies, meaning they probably assume that water and its dissociation products are ideal gases.  Thus a real-gas correction is still required.  That said, the $T \Delta S$ term is probably driving most of the difference between your calculated temperature and the observed temperature for equilibration.


text:
Does CDCl3 also Knock People Out?
I know that chloroform can be used to make people unconscious, but would that also work with deuterated chloroform, $\ce{CDCl3}$?
My instinct is "yes", but I'm not sure I'm willing to try this experiment on anyone.

My instinct says the same.

In almost all respects, deuterium behaves like hydrogen, chemically. You can safely drink pure $\ce{D2O}$ in reasonable quantities (though not enough to replace more than ~25% of the $\ce{H2O}$ of the body). In fact, prokaryotes can continue to grow in ~100% deuterated media, albeit more slowly. It does prevent mitosis in eukaryotes in large quantities, though.
Given that similarity, there is no reason to assume any different behavior between deuterated chloroform and that with just protium. But don't knock yourself out trying the experiment, as chloroform is hepatotoxic.


text:
Is there any molecule visualizing software for Python?
I am looking for a free and open-source software written or wrapped for Python that would simply allow me to generate bunch of figures (in eps, jpg, or pdf format) from a set of xyz files, e.g. water clusters of different sizes.
I am aware of PyMol, but it seems a bit complicated to use (it has its own script language) for such a simple task and I guess it does not recognize xyz files, so I'd first need to convert to pdb format, etc.

Just take a look here. There are a lot of relevant Python for Chemists. Or you could just consider to use LaTeX/Tikz for the job?

How do your XYZ-file looks like anyway? And where do you get it from? If you have an example of how the file looks like and what the result should look like, maybe I can help you better (in Python or in Tikz). Maybe this is what you looked for?

Thanks for the tips, I updated the question with a link to wiki for xyz format. I think rdKit is good for 2D images only, but I will check more.

Well It's PyMol for a reason - you can use Python scripts http://pymolwiki.org/index.php/Python

Can I use it like import pymol pymol.read('xyz') etc.?

@Armut Try to take a closer look to this here. Tomorrow I will try to figure more out about it. PS! Armut means "poor" on German ;-)

http://www.pymolwiki.org/index.php/Get_Coordinates_II

Thanks @Mithoron, the last time I checked all pymol tutorials and examples are written in its own scripting language, so I had the feeling that python API is not well developed or supported.

There's GaussSum, which I believe is written in python http://gausssum.sourceforge.net/

Thanks @JanJensen, while reading about GaussSum here, I found about QMForge, which is more close to what I am looking for.

great, glad to hear it!

In case you really need it has to be written in python probably you want to try Atomic Simulation Environment (ASE). It has a GUI that "allows users to visualize, manipulate, and render molecular systems and atoms objects".
Let me add (even when you are asking explicitly for codes written in Python), just in case that what you need is to create images from .xyz files (regardless of the language), Jmol probably will suits you, as it reads .xyz files straightforwardly. For saving your image once it is loaded, you will need to open a console in Jmol and type:
write image 1200 720 jpg 90 "filename.jpg"
where:

1200 and 720 are the width and height of your image (choose the ones you want to)
90 is the quality of the compression


text:
How is it possible to dissolve zinc oxide safely for skin use, preferably without acid?
Zinc oxide comes as a powder that is insoluble in water but soluble in most acids.
I once received a clear liquid containing zinc oxide from an Italian pharmacy as a deodorant.
What could they have used to make the $\ce{ZnO}$ soluble and safe for use on the skin?

I think they just made nanoparticles that appear to be dissolved. When ZnO is dissolved, then you would talk about ions, then you would probably not call it zinc oxide any more.

This is the information I was actually looking for. If you post your comment as an answer I will award you the bounty.

As I previously stated in a comment, I think that the zinc oxide is not actually dissolved. That would just leave ions which you would hardly consider zinc oxide any more. $$\ce{ZnO + 2HCl <=> Zn^2+ (aq) + 2Cl- (aq) + H2O}$$
Instead I think that there are nanoparticles in use, that appear to be dissolved. In deodorant or other skin products it is likely to have quite a low concentration.
An admittedly short search turned up a patent by Marc Paye, in which is stated:[1]

[...]
  (f) 0.01% to 1%, more preferably 0.03% to 0.5% of Nanoparticulates of zinc oxide which functions as an anti-irritant system for anionic surfactants,
  [...]

Or from another patent of the same company:[2]

[...]
  (c) 0.5-10% (particularly 0.5-7.0, and more particularly 0.5-5.0%) of a small particle size zinc oxide, particularly a micronized zinc oxide or a nanoparticle size zinc oxide having a particle size in the range of 20 nanometers-200 microns;
  [...]

I also found, but have not completely read an article that confirmed my theory.[3]

As an ingredient in dry deodorants to reduce wetness under the arm, ZnO can be used between 0.05 and 10% by weight with average particle size in the range of 0.02–200 microns. ZnO may be used to provide a pH range desirable for deodorants designed for use on sensitive skins [174].

At least in the last article you should also find some information about how to synthesise these particles. Spoiler: It's nothing you would try at home.

Paye, M.; Zinc oxide containing surfactant solution. US6774096 B1, 2004.
Hall-Puzio, P.A.; Gale, A.E.V.; Brahms, J.C.F.; Deodorant with small particle zinc oxide. US 6358499 B2, 2002.
Moezzi, A.; McDonagh, A. M.; Cortie, M.B.; Chem. Eng. J., 2012, 185–186, 1-22..

Thank you for gathering all of this information. Very well put together.

ZnO ia an inorganic compound that is oil soluble. It does react with some fatty acids which are responsible for bad odor, thus the use of it in some deodorants (articles below).

http://journal.scconline.org/pdf/cc1989/cc040n06/p00335-p00346.pdf
http://onlinelibrary.wiley.com/doi/10.1111/j.1365-2133.1990.tb06265.x/abstract;jsessionid=75CB8A5514245763CE30DC30CFE27443.f03t02

The fact the it is used on many baby powder products (for rushes) shows that it is safe on the skin. By making a quick search you can find that ZnO particles are within the range of 70 nm to 100 nm (http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2714038/).
These values show that the ZnO particles do not actually penetrate the skin. They usually also say coated or uncoated ZnO. This is done in order to make ZnO less reactive (coated ones).

Another quick search lead me to this site which has very informative stuff about ZnO and how to make your won deodorant.

Thank you for the thorough answer it helped me realize I was wrong in my thinking of using an acid. Sorry if my question was not clear, I was looking for how to make a clear liquid with ZnO. I had already found the article describing how to make a deodorant paste which is easy to find.

Actually, seems like zinc oxide is NOT oil soluble. At least according to what people say on this post


text:
How to get meniscus down to the calibration mark with a volumetric pipette
I am having trouble taking aliquots using the 10 ml volumetric pipette.
After rinsing, I draw the solution up past the calibration mark and stopper the pipette with my index finger. Then I slowly rotate the pipette and try to let a tiny bit of air inside to get the meniscus down to the calibration mark. But every time, the solution just shoots down, or goes just below the calibration mark.
How can I improve my technique?

How do you get the liquid into the pipette? Do you suck it in with your mouth? That would be really dangerous

If you have to stopper the pipette with your finger, it just takes practice. I believe the proper way to do it is to just roll your finger to the side very slowly, but I've heard several strategies including rotating the pipette like you said, and rocking/sliding your finger back and forth.
It is a lot easier if you have access to pipette bulbs like these, which let you squeeze the side arm to expel liquid smoothly:

These are easy to use too - you just roll it down to expel the liquid:

Most labs should have either of these, especially university ones. Maybe less common in high schools.


text:
Volume of products - stoichiometry

$\rm 20\,cm^3$ of ethyne, $\ce{C2H2}$, is reacted with $\rm 500 \,cm^3$ of oxygen.
  The equation for the reaction is:
$$\ce{2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(l)}$$
What is the total volume of gas remaining at the end of the reaction?
(all volumes are measured at room temperature and pressure)

My effort: I think the answer should be $510\,\mathrm{cm^3}$ because the volume of the leftover oxygen is $450\,\mathrm{cm^3}$ and the volume of the products is $60\,\mathrm{cm^3}$ but the answer is $490\,\mathrm{cm^3}$. Can someone please explain why?

490 cc seems to be the correct answer. Please show your work, and I would be happy to help you find your mistake.

Because they count water as liquid, that's why.

I think you failed to notice that one of your products, i.e water is a liquid, and thus will not contribute to the total gas volume at the end of the reaction.

Here's a breakdown: First, you identify that ethyne is the limiting reagent and would be consumed completely. Second, you correctly calculated the volume of oxygen left that is $450\ \mathrm{cm^3}$. $20\ \mathrm{cm^3}$ of ethyne will completely react to produce $40\ \mathrm{cm^3}$ of carbon dioxide.
Thus, the total gas volume when the reaction reaches completion is $490\ \mathrm{cm^3}$.


text:
Structures for NO+ (Nitrosonium)
There are two possible structures for Nitrosonium :

In the first structure, there is a +1 Formal charge on Oxygen, whereas,
In the second structure, there is a +1 Formal charge on Nitrogen.

Q. Which is a better lewis structure?
Q. Are they resonating structures of NO+?
Wikipedia gives the first structure only, so i think that is the only correct structure.
I think structure in which less electronegative atom has a positive formal charge is better, so the first structure is better.
Q. Is the above reasoning correct?

Which is a better Lewis structure?

There is no such thing as a better or a worse Lewis structure. They should be properly referred to as major or minor resonance contributors/forms.
I looked up a textbook to check the rules of determining which resonance structure contributes more. The order is:

Structures that satisfy the octet rule are greater contributors than structures which do not satisfy the octet rule.
As long as the octet rule is not exceeded for second-row elements, the contributing structure with the greater number of covalent bonds contributes more.
When two or more structures satisfy the octet rule with the same number of covalent bonds, the major contributor is the one with the smallest separation of oppositely charged atoms.
Among structural formulas that satisfy the octet rule with the same number of covalent bonds, and in which one or more atoms bear a formal charge, the major contributor is the one in which the negative charge resides on the most electronegative atom.

In the first structure, all atoms possess a complete octet. But in the second structure, nitrogen only has six electrons around it: consequently, by rule 1, the first resonance structure is the major contributor.

Are they resonating structures of $\ce{NO+}$?

Not exactly.
The actual structure of the nitrosonium cation is somewhere in between the two forms. This means that the N–O bond is somewhere between a triple bond and a double bond. However, since we have established that the first form contributes more to the actual structure, it is going to be closer to a triple bond than a double bond. You could think of it as a 2.9-bond. Molecular orbital theory tells us that the actual bond order in $\ce{NO+}$ is 3, but that's an entirely separate story for another day.
The cation does not resonate between two distinct structures. Let me cite another example. Benzene also has two different resonance structures, both of which contribute equally (since they are identical to each other). Does that mean that the C-C bonds flip between being a single bond and being a double bond? No. All the C-C bonds in benzene are 1.5-bonds, at all times. The common usage of alternating single and double bonds (the Kekule structure) is just us being lazy since many mechanisms are easier to draw that way. Similarly, the $\ce{NO+}$ cation does not flip between the two resonance structures, which represent two extreme ends of a spectrum of possible structures. The actual structure is always in the middle of the spectrum. Which side the actual structure leans towards, is determined by which resonance form contributes more.
At risk of sounding like a broken record - The actual structure is somewhere in between the two resonance forms. It does not oscillate between two different resonance forms. This is a very important concept to understand!

I think the structure in which the less electronegative atom has a positive formal charge is better, so the first structure is better. Is the above reasoning correct?

Yes and no. There are several methods of measuring electronegativity, and in all of them, nitrogen is less electronegative than oxygen. By your reasoning, this means it would be more favourable to place the positive formal charge on nitrogen, i.e. resonance form 2. So your reasoning is a little bit off there.
On the other hand, you would be correct if there wasn't a difference in the number of bonds between the two resonance forms. As I mentioned at the top of the post, the number of bonds takes precedence over the electronegativity issue.

In case of double dative bonds even this rule can be broken

For the sake of completeness: What is resonance, and are resonance structures real?


text:
Molar heat of reaction for water in a neutralization reaction
I have a simple lab problem:
$30\ \mathrm{mL}$ of $1\ \mathrm{mol/L}$ $\ce{H2SO4}$ reacts with $50\ \mathrm{mL}$ of $1\ \mathrm{mol/L}$ $\ce{NaOH}$ in a neutralization reaction as per the following equation:
$$\ce{H2SO4 + 2NaOH -> 2H2O + Na2SO4}$$
Initial temperature of reactants was $22.0\ \mathrm{^\circ C}$ and final solution temperature was $30.5\ \mathrm{^\circ C}$.
What I CAN solve is the molar heat of reaction for $\ce{NaOH}$, which I found to be $-56.8\ \mathrm{kJ/mol}$. I found that by using:  $\mathrm dHm = Q/n = mc\left(\mathrm dT\right) / n$, where $n$ = moles of $\ce{NaOH}$.
$n = 1\ \mathrm{mol/L} \times 0.05\ \mathrm L = 0.05\ \mathrm{mol}$
Then $\mathrm dHm = \left[ \left( 50\ \mathrm{ml} + 30\ \mathrm{ml} \right) \times 4.19\ \mathrm{J/\left(g\ ^\circ C\right)} \times 8.5\ \mathrm{^\circ C} \right] / 0.05\ \mathrm{mol} = -56.8\ \mathrm{kJ/mol}$ since the reaction is exothermic.
The trickier part is when it asks me what the molar heat of reaction for water is in this neutralization reaction. I know the general enthalpy is just $mc\left(\mathrm dT\right) = 2.84\ \mathrm{kJ}$. But then what do I do with this value to find the molar heat of neutralization for water? I have to divide by the number of mols of water as per the equation above, but I don't know which number to plug in? I need explanation, please.


text:
Naming Ionic Compounds with Transition Metals
Look at $\ce{FeO2}$.  We know that $\ce{O2}$
is $-2$ then we must have $-4$ for $\ce{Fe}$
So $\ce{Fe}$ must have $+4$ charge so then is it $\ce{Fe(IV)}$ ?

What is your question?

i found that there is only Fe with  2+ and 3+ charge so how do i got 4+ thanx

try using $Fe_2O_3$ instead.

$\ce{FeO2}$ does not exist.

@bon Apparently it does. At least, if your laboratory is in the lower mantle. This "peroxide" may even contain $\ce{O2^{3-}}$ instead of the usual perocide ion.

As stated in the Wikipedia article for iron:

Like other group 8 elements, iron exists in a wide range of oxidation states, −2 to +6, although +2 and +3 are the most common.

Thus, assigning $\ce{FeO2}$ as an iron(IV) compound is not something that is impossible, but it is unlikely. There are only a few compounds with that are iron(IV).
A more likely scenario is that $\ce{FeO2}$ may contain other anions of oxygen:

Oxide $\ce{O^2-}$
Superoxide $\ce{O2^-}$
Peroxide $\ce{O2^2-}$
Ozonide $\ce{O3^-}$

While oxide is the most common anion, there is another on the list above that is reasonably common. Which is it? If you have that anion, what is your oxidation number for iron?

I'm afraid iron(II) peroxide is is even more unlikely.

However OP may ask about hemoglobin with oxygen bound to Fe

FeO2 may be formed under high pressure and even exist in Earth's mantle. This answer references an article suggesting it actually cobtains iron(III)  and $\ce{O2^{3-}}$.


text:
Chemical Test for tin
How can I distinguish whether a metal that I have is tin ($\ce{Sn}$)? I am specifically looking for a chemical reaction that will happen only with tin and not with other metals such as $\ce{Al,~Cu,~Fe,~\ldots}$. Do such reactions exist? What are they?

I think this question would be incomplete without mentioning what metals the sample could be made of, i.e. all of the metals.

@inɒzɘmɒЯ.A.M It could be made of: $\ce{Al,~Ti,~V,~Fe,~Co,~Ni,~Cu,~Zn,~Sn}$.

Beside determining the volumic density (tin: 7.31 g/ccm, aluminum: 2.7 g/ccm, etc. pp) and as asked here on chemistry SE, there is a quick pre-test.  In German it is known as "Leuchtprobe", literally "illumination probe".  Be reminded to protect yourself properly (safety googles, fume hood, gloves).
In a first instance, you have to reduce the tin compound.  For this, a little of your solid under investigation is finely mixed together with some grains of zinc (Zn). In a porcelain dish, to this mixuture is added 5 mL of 20 m% aqueous HCl.  Await this reaction ceased.  Zn will produce hydrogen, and because it is freshly created (statu nascendi), it is of superiour reducing power compared to bottled hydrogen gas, and the solution is cooled to room temperature.
Then dip a cool test tube, partially filled with cold water, into this solution.  While still wetted by the solution, rapidly  place the test tube into the upper part of the non-luminous flame of a Bunsen burner.  Blue fluorescence is an almost specific evidence for tin (caused by $\ce{SnCl2}$, and stannanes), already at low concentrations.
Presence of arsenic may hamper this fluorescence.  Niob-compounds yield some similar fluorescence, too.  Beside using tin-free chemicals and equipment to be used -- and prior to the test of your unknown -- do at least once the test in known absence of tin, and once with a sample with tin to learn how to discern the two instances.

One can better detect the fluroescence by adding potassium permanganate into the test tube to create a dark background. (Or so I heard; I let others do that test in the ion lottery practical course.)

I agree with this comment; a test tube partially filled with a cold aqueous potassium permanganate solution increases the contrast compared to a test tube simply filled with blank water. Own practical experience in the lab (not only during the initial (in)organic Marcha analítica separation process

but as TA, too) that a test tube hold too long unshaken eventually leads to a bumping solution, including a stain on cloth and skin.

Is there any other analytical method to recognize tin?

My initial answer assumed you look for a qualitative test with as little as possible techniqual equipment and chemicals.  To go further depends on the sample matrix and often, there are reliable procedures to do this.  The U.S. american NIST, for example, developped digestion standards how to proceed properly, like this more senior reference (http://nvlpubs.nist.gov/nistpubs/jres/21/jresv21n1p95_A1b.pdf).   Atomic absorption spectroscopy (AAS) would represent an alternative (https://en.wikipedia.org/wiki/Atomic_absorption_spectroscopy), tin being listed in the table "GFAA method development".

By coincidence, AAS was a recent answered question, too: http://chemistry.stackexchange.com/questions/30892/atomizers-in-aas

Don't look at the fluorescence too long, because the water in the tube locally overheats within ten seconds: splash!

@Karl And safety glasses/goggles.  And chemically speaking, to perform the test at least once with a blank (no tin) and once with presence of tin as each measurement is a comparison against a reference. The illumination in the lab (fluorescent tube / LED lumiophore / day light) can influence how well it is seen, too.

We always put some KMnO4 into the tube with cold water, so it's practically black. Makes it much easier to see the fluorescense. And gives brown stains everywhere in case of spillage. And yea, safety glasses! ;)


text:
Can calcium hydroxide + sulphuric acid be a way of preparing insoluble calcium sulphate?
To prepare a soluble salt, we can react an acid with an insoluble base (hydroxide or oxide).
To prepare an insoluble salt, we can use precipitation method which is by mixing two solutions of soluble compound.
However, can I prepare an insoluble salt by reacting an acid with an insoluble base (hydroxide or oxide)?
For example,
 1. Calcium hydroxide + Sulphuric acid
 2. Lead (II) hydroxide + Sulphuric acid
 3. Lead (II) hydroxide + Hydrochloric acid

No.
The outside layer of the calcium hydroxide reacts with the sulfuric acid to form the calcium sulfate that you want. All good, until you realise that the newly-formed calcium sulfate is now coating the rest of the calcium hydroxide that didn't get to react. In the end you just get a big mess.
From a practical standpoint, even if your method worked, it's just much easier to just mix two solutions together and filter. No need to stir stuff and wait for insoluble powders to dissolve. On top of that, you're dissolving a white powder in order to make another white powder - how would you know when your reaction has finished?
Conclusion: Use calcium chloride and sodium sulfate.


text:
How to calculate the number of water molecules of a hydrate based on mass loss upon heating?

A $\pu{4.250 g}$ sample of $\ce{Na2SO4.nH2O}$ the sample loses $\pu{2.388 g}$ upon heating. What is $n$ for this hydrate?

Do I first determine the mass of the water than the amount of substance of the sample and than the whole number ratio?

Determine what the $\pu{2.388 g}$ is.

 It's the amount of water lost from the crystal structure, assuming the sample was heated enough for this purpose.

What does 'mole' mean? Isn't it just an indicator of the number of atoms/ions/molecules? So, what would be the relation between the quantity sodium sulfate and water molecules?

 Yes, 'mole' is an indicator of quantity. Take a look at the formula: $\ce{Na2SO4.nH2O}$. What does that mean? It means in a unit, there are $n$ water molecules trapped inside the crystal framework, which consists of sodium and sulfate ions.

How do you proceed into finding the amount of substance (in mole)?

 You need to somehow 'remove' any unknown variables to calculate the amount of substance, so you need to remove $n$. Try subtracting the amount of water from the weight of the whole sample:
$$m(\ce{Na2SO4.nH2O}) - m(\ce{H2O}) = m(\ce{Na2SO4})\\ \pu{4.250 g} - \pu{2.388 g} = \pu{1.862 g}$$

How many grams of anhydrous $\ce{Na2SO4}$ are there? What is the formula to convert that into moles? How many moles of $\ce{Na2SO4}$ are there?

 The formula is $\displaystyle \mathfrak{n}=\frac{m}{M}$ where $m$ is mass, $M$ is molar mass and $\mathfrak{n}$ is the amount of substance. Thus $\displaystyle \mathfrak{n}=\frac{\text{mass of anhydrous sample}}{\text{molar mass of sodium sulfate}} =\frac{1.862}{142} = \pu{0.0131 mol}$.

You found how many moles of sodium sulfate are there. Isn't that going to be the same for water? Plug into the formula; and find $n$.

 We know that for in any unit of Gluber salt, there are 2 sodium ions, one sulfate ion, and $n$ molecules of water. Thus, it's safe to assume that $\mathfrak{n}$ for water is $n$ times that of what we found for sodium sulfate.
$$\begin{align}0.0131\times n &=\frac{2.388}{18} \\ &= 0.132\bar{6} \\ \Rightarrow n&=10\end{align}$$


text:
What is the name of the compound Sc(HCO2)3?
I would like to know what the name of the compound $\ce{Sc(HCO2)3}$ is. I've been told that it is not scandium hydrogen carbonite and that it is not scandium formate. Any hints or suggestions about how I can solve this problem?

Note that I’ve never heard carbonite before.

@Jan It's very good for freezing people in!

@Jan In Star Wars: The Empire Strikes Back, Han Solo is frozen in carbonite (and then subsequently defrosted in the next film).

This says scandium formate, both taxonomically and structurally.

Sc(HCO2)3 is scandium formate, or scandium methanoate. It cannot be anything else !

The formula you have is definitely scandium formate. There is one of four reasons why someone (like your instructor or your online homework service) might tell you that scandium formate is wrong.

Someone wants you to call this compound scandium(III) formate because scandium can have multiple oxidation states.
Someone wants you to call this compound scandium methanoate (or scandium(III) methanoate) since methanoate is another name for the formate ion.
You are correct and your instructor is wrong (or your assignment software has the wrong answer keyed in).
There is a typo in the formula you were given.

Could be scandium bicarbonate with a typo.

Why it is not scandium bicarbonate?

@Aneek - The bicarbonate anion is $\ce{HCO3^-}$. The formula given in the question is $\ce{Sc(HCO2)3}$.

If an error did occur, and maybe recurring, what, if any, implications should a student draw? This serious issue, in my opinion, is not considered. See my answer below for possible need to check sources and actual background of my personal experience.

@BenNorris can scandium really show multiple oxidation states? AFAIK it only shows a +3 state


text:
What type of glass are Pasteur pipettes made of?
What type of glass are Pasteur pipettes made of? And at what temperature do they need to be heated to create a micro-pycnometer?

This page from the glass shop in the Department of Chemistry and Biochemistry at the University of Delaware lists the properties of several common glasses that are used in laboratory equipment.
The two most common types of glass for less expensive applications are soda lime glass and borosilicate glass. The anneal point (or glass transition temperature) can be thought of as the temperature above which the glass can start to have some molecular motion, but it is not a liquid. The soften point is probably what you are looking for.
$$
\begin{array}{lcc}
\hline
\text{Glass} & \mathrm{Anneal\ Point\ ^\circ C} & \mathrm{Soften\ Point\ ^\circ C} \\
\hline
\text{Soda Lime} & 545 & 724 \\
\text{Borosilicate} & 560 & 821 \\
\hline
\end{array}$$
Soda lime glass is cheaper, and is generally used when you want something very inexpensive or disposal. Borosilicate glass has a lower thermal expansion coefficient so it is less likely to break. If you want to know which type of glass your pipets are made from , look at the wide end from an angle. If there is a slight greenish cast to the curve of the glass, you have soda-lime glass. Or you could look at the box they came in.


text:
Why grain boundary is reactive to acid?
I have read about grain boundary etching. They use acid to make grain boundary visible for light microscope. I know that grain boundary is reactive to acid, but why? Thanks.

A grain boundary is an area with some amount of disorder, strain, and unterminated bonds. This gives some acids the opening they need. See also dislocation highlighting etches.

I think this is due to the disordered atomic arrangement at the grain boundary which implies higher  internal energy of the constituting atoms and consequently more reactivity (attack by acid, corrosion,...).


text:
Why does oxidized copper become green?
Why does oxidized copper turn green? I am sorry if this question is too basic, but am really stumped here. It would be helpful if one could provide a research paper or website with your answer.

Copper oxide is not green, but green verdigris, basic copper carbonate (or acetate) forms on copper exposed to air and moisture.
Heating copper in air can form a near-transparent film of copper oxide, which can exhibit all the colors of the spectrum due to interference of light. The same effect cause a "rainbow" in oil slicks on water.

I don't think this answers the question, just says how not why. I am imagining some kind of answer with d-orbitals.

The question is a misstatement. Read it

Huh, I still don't see it, it sounds like someone's asking for an explanation for the color change of copper when it's oxidized. Oxidized does not mean "gains oxygen".   I think Aaron is trying to understand why copper carbonate and/or copper acetate are green.

Copper when is exposed to air, it gets oxidized and a tarnish of green color is formed on it.As Cu is a transition metal, it possesses color according to the "crystal field theory".
So, Cu reflects green color when it comes in contact with oxygen.


text:
Why do electrons need to be paired?
I’d read that it takes energy to place electrons in pairs, just less than it would take to place two electrons in different subshells. I get that. What I don’t get is, then, why would electrons want to fill their orbitals with electron pairs at all? Wouldn’t it better to just fill them up with single electrons, as per Hund’s rule, and then leave them like that? Would that not be more energetically favorable? Why do atoms insist of having full valence shells, then?

Wouldn’t it better to just fill them up with single electrons, as per
  Hund’s rule, and then leave them like that? Would that not be more
  energetically favorable?

You have to consider two different scenarios here.
Case 1. Orbitals of different energies.
Consider a two electron system, say, a $\ce{He}$ atom, and let us concentrate on $\mathrm{1s}$ and $\mathrm{2s}$ orbitals exclusively. Naively, one could think of the total energy of the system being simply the sum of orbital energies. Then for $\mathrm{1s^2}$ configuration we get $E_{\mathrm{1s^2}} = \varepsilon_{\mathrm{1s}} + \varepsilon_{\mathrm{1s}}$, while for $\mathrm{1s^1 2s^1}$ configuration we get $E_{\mathrm{1s^1 2s^1}} = \varepsilon_{\mathrm{1s}} + \varepsilon_{\mathrm{2s}}$. Now clearly, since $\varepsilon_{\mathrm{1s}} < \varepsilon_{\mathrm{2s}}$, we could expect that $E_{\mathrm{1s^2}} < E_{\mathrm{1s^1 2s^1}}$, i.e. that  $\mathrm{1s^2}$ configuration is indeed more energetically favourable than $\mathrm{1s^1 2s^1}$ one.
This is a bit naive way of looking at things since the total electronic energy is not equal to the sum of orbital energies. As you mentioned, pairing the electrons up takes some energy, so that $E_{\mathrm{1s^2}}$ is clearly not just the sum of $\varepsilon_{\mathrm{1s}}$ and $\varepsilon_{\mathrm{1s}}$, but this amount of energy is relatively small compared with the energy gain you get by having both electrons in the lower $\mathrm{1s}$ orbital. In other words, energy gain from placing both electrons in energetically more favourable $\mathrm{1s}$ orbital overcompensates for having to pair them up during the placement.
How do we know that? Well, we know it from the quantum theory, or, to be more precise, from its application to a chemical systems. Thus, a solid understanding of quantum mechanics & quantum chemistry is required for a quantitive discussion, but for now you can be certain that electrons fill the lowest available energy levels before filling higher levels. This is the principle known as the Aufbau principle which follows from quantum theory and generally holds.
Case 2. Orbitals of same energy.
Consider a $\ce{C}$ atom now and let us have a look at two electrons in  $\mathrm{2p}$ orbitals. You could populate the same, say, $\mathrm{2p_x}$ orbital by a pair of electrons, or you could let the electrons populate two different orbitals, say, $\mathrm{2p_x}$ and $\mathrm{2p_y}$. Naively, we could expect the total electronic energy for both configurations to be equal to each other, since $\varepsilon_{\mathrm{2p_x}} = \varepsilon_{\mathrm{2p_y}}$, so that,
$$
  E_{\mathrm{2p_x^2}}
  = \varepsilon_{\mathrm{2p_x}} + \varepsilon_{\mathrm{2p_x}}
  = \varepsilon_{\mathrm{2p_x}} + \varepsilon_{\mathrm{2p_y}}
  = E_{\mathrm{2p_x^1 2p_y^1}} \, ,
$$
but this is again "not quite right", since pairing the electrons up takes some energy. And this time there is no any gain to compensate for this effect since $\varepsilon_{\mathrm{2p_x}} = \varepsilon_{\mathrm{2p_y}}$, and as a result $\mathrm{2p_x^1 2p_y^1}$ configuration is energetically more favourable than $\mathrm{2p_x^2}$ one.

In terms of real world examples, here are two chemical species whose ground states have unpaired electrons in separate orbitals of the same energy: oxygen and cyclobutadiene. Oxygen, reactive as it is, is stable enough to constitute 20% of the atmosphere, whereas cyclobutadiene rapidly dimerizes or reacts with other species present.
These two examples conform to Hund's rule, and to case 2 in Wildcat's answer, above. Pairing the electrons in the same orbital would place them in closer proximity (hence higher energy) than placing them in two different orbitals where they remain unpaired. A further consequence of having unpaired electrons is that it makes the molecules paramagnetic. This state of having two unpaired electrons is called a "triplet" state since there are three possible spin states for the two electrons (up-up, up-down, and down-down) that can be observed spectroscopically in the presence of a magnetic field.

This seems more like a comment than an answer. Perhaps you could expand on why these molecules have unpaired electrons.


text:
What are the effects of the repulsive forces between electron on the potential energy graph for the dihydrogen molecule?

This diagram is from my textbook. It states that as long atoms get closer, the energy decreases due to the attractive forces between the electrons and the nuclei. When the nuclei get too close, the nuclei’s repulsive forces drives the energy upward. However, what of the repulsive forces between the electrons? As the atoms get closer and closer from infinity, where are the effects of the increasing repulsive forces between the electrons?

First of all, you have to understand that the potential energy depicted on the graph in your question already accounts for repulsive forces between the electrons. This potential energy is the sum of two terms, namely, the sum of electronic energy $E_{\mathrm{e}}$ and nuclear repulsion energy $V_{\mathrm{nn}}$, where the first term contains contributions from kinetic energy of electrons motion $T_{\mathrm{e}}$, potential energy of their interactions with the nuclei $V_{\mathrm{en}}$, as well as potential energy of their interactions with each other $V_{\mathrm{ee}}$. This sum of electronic energy $E_{\mathrm{e}}$ and nuclear repulsion energy $V_{\mathrm{nn}}$ plays a role of potential energy for the motion of the nuclei, hence its name.
All these components of the potential energy, $E_{\mathrm{e}} = T_{\mathrm{e}} + V_{\mathrm{en}} + V_{\mathrm{ee}}$ and $V_{\mathrm{nn}}$, behave differently with respect to internuclear separation. For instance, as you have mentioned, the last term $V_{\mathrm{nn}}$ grows when nuclei get closer to each other. The same has to be true for $V_{\mathrm{ee}}$ term since the closer the nuclei are the closer are the electrons. However this effects of increasing the energy could be overcompensated by a decrease in $V_{\mathrm{en}}$ term thanks to attractive nature of the corresponding forces between nuclei and electrons.
When a decrease in $V_{\mathrm{en}}$ overcompensates an increase in $V_{\mathrm{nn}} + V_{\mathrm{ee}}$ the bond is formed. Not that an overall decrease in potential energy is accompanied by an increase in kinetic energy in accordance with the virial theorem. But the increase in kinetic energy is only half of the lowering in the potential energy, so that the total energy goes down when a bond is formed.
Here is the graph of all these terms for $\ce{H2}$ molecule obtained from quantum chemistry calculations1:

Wilson, C. W. & Goddard, W. A. The role of kinetic energy in chemical binding. Theor. Chim. Acta 26, 195–210 (1972). PDF from caltech.edu.

Very nice answer. I have just one point I'd like to discuss: In your last sentence you state: However this effect of increasing the energy could be compensated [...] by a decrease in $T_{\mathrm{e}}$.. But wouldn't the kinetic energy of the electrons rise as the atoms come closer together? My argument for that would be as follows: When the $\ce{H}$ atoms come closer the orbitals get, sort of, compressed. This compression leads to a higher gradient of the wave function near and inbetween the nuclei and since the kinetic energy of the electrons is proportional to said gradient it rises.

The argument about the proportionality between the gradient of the wave function and the kinetic energy of the electrons I discussed in the last paragraph of this answer of mine in more detail.

@Philipp, yeah, you're perfectly right. My mistake.

Very interesting links.


text:
Concept of Bomb caloriemeters
Considering the example where $0.3212\ \mathrm g$ of glucose (molar mass = 180) being burnt in bomb calorimeter of heat capacity $641\ \mathrm{J\ K^{-1}}$: the temperature rises by $7.793\ \mathrm K$. Is the heat (say '$E$') obtained taking $E=641 \times 7.793 \times 180/0.3212$ the internal energy or heat of combustion? Why? Answer explained in detail will be highly appreciated!
What is the fundamental difference when we completely burn fumaric acid in a bomb calorimeter? Why do we take the heat released as the internal energy? IS it because volume is constant?
Reference: Elements of Physical Chemistry (Paula, Atkins) Q.no:3.21 and 3.22

Let me list some equations first: $$ w = -p\Delta V,\qquad\Delta U = q + w,\qquad \Delta H = \Delta U + \Delta (pV)$$
where $w$ is work, $U$ is internal energy, $q$ is heat transfer and $H$ is enthalpy
Now, when you are trying to calculate the internal energy, enthalpy or anything else, you should first note what values are constant and what values change through out the reaction. For example, if I burnt a sample of glucose in an open beaker, the volume would obviously change, but the pressure would remain constant. This is because the beaker is open, which means that pressure exerted on the products and reactants is the external atmospheric pressure which is always remains a constant value. Noting the fact that pressure is constant is important as it allows us to simply the equations that I have listed above. $$q = \Delta U - w = \Delta U + p\Delta V$$ $$\Delta H = \Delta U + p\Delta V = q$$
Therefore, by realising that pressure is constant, we are able to state the $\Delta H = q$
Now lets consider the case of burning glucose in a bomb calorimeter. Here is a diagram of one below:

So lets consider what values remain constant and what values change throughout the combustion. As you should be able to see, the bomb calorimeter has a fixed volume (it can't expand or contract), hence $\Delta V = 0$. Since the bomb calorimeter is a closed object, the pressure on the system is the internal pressure, not the atmospheric pressure which is a constant. This internal pressure will change as the reaction proceeds (in this case it will increase as gases are produced from the combustion). Therefore $\Delta p \neq 0$.
Now, using the data from above, we can obtain the following expressions: $$w = 0,\qquad\Delta U = q,\qquad\Delta H = q + V\Delta p$$
So when you do the calculation: $641\ \mathrm{J\ K^{-1}}\times7.793\ \mathrm K$ you are finding the total heat transfer ($q$) from the system to the surroundings (the bomb calorimeter). Now form the above information we know that the total heat transfer is also equal to change in internal energy, therefore: $$641\ \mathrm{J\ K^{-1}}\times7.793\ \mathrm K = \Delta U$$
I think you were trying to find the molar internal energy for the reaction, which the calculation will be then: $$\Delta U_\mathrm m = \frac{641\ \mathrm{J\ K^{-1}}\times7.793\ \mathrm K\times180\ \mathrm{g\ mol^{-1}}}{0.3212\ \mathrm g}$$


text:
Determining k value of a Reaction
You and a co-worker have developed a molecule that has shown potential as cobra anti-venom ($\ce{AV}$). This anti-venom works by binding to the venom ($\ce{V}$), thereby rendering it non-toxic. This reaction can be described by the rate law, $\text{rate} = k\left[\ce{AV}\right]^1\left[\ce{V}\right]^1$.
You have been given the following data from your co-worker: $\left[\ce{V}\right]_0=0.20\ \mathrm M$, $\left[\ce{AV}\right]_0=1.0\times10^{-4}\ \mathrm M$. A plot of $\ln\left[\ce{AV}\right]$ versus $t$ (s) gives a straight line with a slope of $-0.32\ \mathrm s^{-1}$. What is the value of the rate constant ($k$) for this reaction?
My logic tends to fail me on this question. So to start off, the overall  reaction order is 2, thus the equation is: $${A}^{-1}=kt+{a}^{-1}$$ However, we're only given a $k$ value for the law: $\text{rate}=k\left[\ce{AV}\right]$. I'm lost as to how to proceed next. Given the k we can calculate the initial rate of $\left[\ce{AV}\right]$ but how is this going to help me find the $k$ for $rate=k\left[\ce{AV}\right]^1\left[\ce{V}\right]^1$? I suspect something tiny that I missed in this problem, any directions or hints would be appreciated!

You can regard the reaction to be pseudo-first order, as the concentration of $\left[\ce{V}\right]$ is much higher than the concentration of $\left[\ce{AV}\right]$. Therefore, $k\cdot\left[\ce{V}\right]_0=\text{const.}=k_{2}$. The rate law of this reaction would change to $r=k_{2}\cdot\left[\ce{AV}\right]$. Integrating this would lead to the expression: $\ln\left[\ce{AV}\right]=\ln\left[\ce{AV}\right]_0-k_{2}\cdot t$. The dependence of $\ln\left[\ce{AV}\right]$ on time has the slope $-k_{2}=-0.32\ \mathrm s^{-1}$, so you can easily calculate $k$ from $k_{2}$ and $\left[\ce{V}\right]_0$.


text:
Will the ions in salt bridge be used up?
Ions in salt bridge move to the electrolytes of two half cells to balance the charges.
But will the ions in salt bridge be used up?
If it will, is the electron flow then stop?

The ions in the salt bridge are meant as charge carriers, and using a suitable diaphragm intends to keep the mass transport pretty low.
Nethertheless, some of the electrolyte (like $\ce{KCl}$, $\ce{KNO3}$, $\ce{Na2SO4}$ or $\ce{Bu4N+ ClO4-}$ to mention a few) does enter the other half cell.  This is why, as one criterion, the electrolyte in the salt bridge is choosen to not interfer with the electrochemical reaction you want to characterize (analysis like cyclic voltammetry) or execute (electrochemical synthesis).  Likewise for solvents used in cyclic voltammetry, there is an electrochemical window, the intervall of electrode potentials where it is safe to use them so that they are not oxidized, or reduced, respectively.
And if the electrolyte is consumed and charge carriers are absent, provided the electrical potential applied is not altered, yes, the electrical current drops to 0.