text:
What is the chemistry behind the reaction of KNO₃, K₂CO₃ and Sulfur?
I'm preparing a chemistry show where I end the show with yellow powder bang.
The powder is a mixture of:
$$\ce{K2CO3 + KNO3 + S} $$
My question is, what's the reaction between these chemicals?

Not sure about this, but the nitrate might oxidise the sulphur to sulphur dioxide or trioxide.

It seems that nobody really knows about the role of the carbonate, but there's a nice video in the Free Range Chemistry series of the RSC.

It looks like a variation on a gunpowder mixture which is usually sulfur, KNO3 and charcoal

Yellow powder is said to be unpredictable, so it is probably better to use a different bang.

Here's just a guess, but I do have some confidence that I'm right:
$\ce{KNO3 + S -> K2SO4 + K2SO3 + K2S + K2O + SO2 + N2}$
$\ce{K2CO3 ->[\Delta] K2O + CO2}$
The first, unbalanced equation is the reaction between potassium nitrate and sulfur, producing quite some gas as product, and a variety of potassium compounds left behind, with the composition depending on your formulation of the powder. Of course there might be some sulfur trioxide produced, but the major oxide of sulfur should be the dioxide. In the second equation is the thermal decomposition of potassium carbonate releasing carbon dioxide. This is the part that I'm not very sure of, but it may be to release more gas to give your reaction more puff. However, IIRC the decomposition requires a very high temperature. In contrast, bicarbonate has a much lower decomposition temperature and also release more gas(carbon dioxide and water), but the products would likely be absorbed by potassium oxide to form hydroxide and carbonate, so  bicarbonate might not help much.

Most likely,
a mixture of potassium polysulfides is formed first. Possibly, a further more complex reaction takes place between the nitrate ion, carbonate ion and sulfide that forms small amounts of tetrasulfur tetranitride, disulfur dinitride and sulfites, evolving nitrogen and sulfur dioxide. The remaining solids are clearly (just look at their possible structure) more reactive than simple polysulfides and ought to be responsible for the final high speed decomposition of the mixture.
On a side note, the less costly way to determine if sulfur nitrides are actually present would be to pass the sample under a Raman spectrometer.


text:
Does tetramine copper dissolve cellulose?
I saw a Youtube video which showed that cotton (cellulose) dissolves in deep blue solution of tetramine copper made by adding 20% ammonia in copper carbonate salt. I tried to replicate the process using $\ce{CuSO4}$ and copper chloride. None worked. Where have I gone wrong?

As @iad22agp states, it is not sufficiently alkaline. Schweizer's reagent is tetraamminediaquacopper dihydroxide, $\ce{[Cu(NH3)4(H2O)2](OH)2}$. See Schweizer's reagent.
That is not Doktor Schweizer, I presume.

You might want to check the pH - you would have a less basic solution using copper sulfate or chloride than you would get from copper carbonate, and that could be less effective in dissolving the cotton. Also, your particular sample of cotton might behave differently from others, such as that used in the video.

There should be no anions exept OH- present.

made by adding 20% ammonia in copper carbonate salt.

is wrong, using CuSO4 is even wronger.
You have to precipitate Cu(OH)2 and wash the
precipitate thoroughly, a very tedious operation.
According to a IG Farbenindustrie Patent You
throw copper pieces in  > 25 % Ammonia
(cut from electrical wire) and percolate
with oxygen or air. When using air, you have to
maintain temperature under 8 °C. This needs days
until "ready" but has the advantage that the
reagent is pure without cleaning operations.

Is there any reason this is formatted like a cinquain?

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text:
3rd analytical group of cations
If a mixture of Mn$^{2+}$, Zn$^{2+}$, Co$^{2+}$, Ni$^{2+}$ and Al$^{3+}$ salts is dissolved in water, a white dim solution is obtained. I guess that the aluminium and zinc ions hydrolysed and the respective hydroxides are formed. Am I right, or maybe one of them won't hydrolyze so much that the solution becomes dim or maybe some other cation could have hydrolysed, too?

I'm quite sure that aluminium will produce a white colour due to formation of hydroxide. Manganese, zinc, cobalt hydroxides are probably soluble as they are precipitated in the 4th group as sulphides. Not sure about zinc.

Okay, I just need a confirmation whether zinc will produce a white hydroxide, too.

Just looked at the wikipedia page for zinc hydroxide. Zinc hydroxide will form a white precipitate but will dissolve in excess acid or base due to its amphoteric character.

Aluminium and zinc will produce a white precipitate due to formation of hydroxide. Both these hydroxides dissolve in excess acid or alkali due to their amphoteric character.
Manganese, nickel, cobalt hydroxides are probably soluble as they are precipitated in the 4th group as sulphides.

You mean nickel instead of zinc in the second paragraph?

@Marko I did mean nickel. Updated answer. Thanks for pointing out the error.

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text:
Lambert Beer's law deviations
Can somebody explain as concrete as possible, how the change of temperature and the change of pH  leads to deviations of this law?

I don't understand what you mean. Lambert Beer law  apply for fixed temperature and pH.

A compound S follows the Lambert-Beer law if the absorbance $E_\lambda$ at a particular wavelength $\lambda$ is proportional to the concentration $c$ of S.
\[E_\lambda = \left(\frac{I_0}{I} \right) = \epsilon\cdot c \cdot d \]
A "deviation" is observed when the real concentration $c$ is different from the expected/assumed.
This may happen if

S is an acid or a base:

variation of the pH leads to (de)protonation and forms another species with a completely different absorption spectrum, or at least a different $\epsilon$ at the wavelength under observation

S undergoes a reaction to S' (with different optical properties) at elevated temperatures (or change of pH)
Decrease of temperature or change of pH leads to the formation of a turbid solution from which light is scattered into the detector


text:
Why is it that reactions that cause an increase of temperature are exothermic?
I've been taught that if you have two aqueous reactants that react together and produce heat it is exothermic. Shouldn't that be endothermic? Did my professor misspeak?
If it's gaining heat, that means it's pulling heat from the outside. The opposite of what happens when water freezes and releases heat (which is actually exothermic).

Also, is the "break down of for during digestion" endothermic or exothermic?
It requires energy to break bonds, so I think it's endothermic, but it also releases heat supposedly.

When you burn something, does the thing actually become cooler and the area around it become hotter? Or does it all become hotter?


text:
what are possible isomers of (meso-stilbenediamine)(isobutylenediamine) palladium (II)?
It can form a square planar and a tetrahedral complex but within square planar or within tetrahedral can there be isomers possible?

The use of the term "meso" in the compound name indicates that both phenyl groups are on the same side of the molecule, "cis" to one another, not "trans".
Here are the tetrahedral and square planar representations.

Note: I used $\ce{Pt}$ instead of $\ce{Pd}$ as the central metal, but it makes no difference in terms of isomers.
The tetrahedral molecule has a plane of symmetry containing the "top" 5-membered ring, bisecting the 2 methyl groups on that ring and then bisecting the "bottom" 5-membered ring with the 2 "meso" phenyl groups. Since the tetrahedral molecule contains a plane of symmetry it is achiral and has no isomers.
In the square planar form, there is no symmetry element.  Therefore the square planar molecule is chiral and exists as an enantiomeric pair of isomers.


text:
How to prevent Tin ions from precipitating as Tin(II,IV) Hydroxides
Alright, so Tin ions in most of it's salts have the propensity to form Tin(II,IV) Hydroxides and then precipitating out of aqueous solutions. What can I throw in the solutions to prevent this? Using Tin Fluoroborate doesn't precipitate Hydroxides, but I don't have access Fluoroborates. I really don't want very acidic solutions, either, for safety purposes.
Is there any complexes out there that does the job?

A pH of about 3 or 4 should work as far as I remember from my students time. Do you consider this to be too acidic?

as long as not strongly acidic enough to readily oxidize zinc metal, but i don't know how low the pH has to be to oxidize zinc.

how about ketones and saccharides?

@Badadeeboop do you like chelants?


text:
Chemistry of Stripper Poles: How much weight can different materials support?
I'm looking to create a light up dancing pole. This requires a translucent material that is strong enough to support the weight and force of a human being, even when it's hollowed out to so LEDs can go inside.  The material must also be reasonably affordable, so something like diamond is out of the question.
I've discovered a couple materials during my research (please recommend any I'm leaving out):

acrylic (I know there are acrylic poles out there, but I'm not sure
if they're able to support a human when hollowed out)
polycarbonate

To take the example of polycarbonate, how much weight can it support as a function of its thickness/diameter?
[Also, if the editors feel like this is an inappropriate question I'd just like to point out that the weight that different materials can support is a very appropriate question here.  My intent is not to scandalize, only to build a dance pole and learn some Chemistry in the process.  Thank you]

I'm inclined to flag the question as off-topic. A question on the pros and cons of sex toy surface materials (silicone vs PVC, etc.) woud be more on-topic. One aspect related to chemistry is the possible update of body fluids by the pole material - changing performers change act differently "intense". The major part, however, is pure mechanics (and material science): bending moment of a tube of particular length that cannot exceed a certain diameter: the girls typically don't have hands the size of toilet lids ;) It's therefore not astonishing that the tubes are typically made of steel.

I think that a good (programmable) stage lighting, using high-power leds, will probably lead to better results (and higher tips).

Anyway: Welcome to Chemistry.SE! And yes, I upvoted your question because the idea is funny and not scandalizing at all!

I'm voting to close this question as off-topic because it is more related to mechanics and engineering.

Hmm, I think there is some chemistry in this, but it's way more appropriate in engineering.SE.

I'm fine with moving the question to Engineering


text:
Need of hyperconjugation?
Recently I read about Hyperconjugation. But i have a doubt.

What is the need of hyperconjugation in this structure when there is charge separation resulting in lesser stable structure.

delocalization of electrons takes place through overlap between sigma bond orbital and pi-bond orbital or p- orbitals which dominates unstability due to charge seperation and causes overall stability

What is the need of hyper conjugation in this structure when there is
  charge separation resulting in lesser stable structure.

I think we'd all agree that the charge separated structure you've drawn is a resonance structure.  Charge has been separated so it may well be higher in energy than the first (neutral) resonance structure.  Nonetheless, to whatever extent the charge separated resonance structure contributes to the description of propene - even if it is just a small amount - it will stabilize propene.
Since there are 3 hydrogens on the methyl group you can draw three identical resonance structures, this will increase the importance of the structure.  Note too the charge separated resonance structures involve an allyl carbanion (resonance stabilized) and a proton; two very stable structures - this will further increase the weight, or importance, of the hyperconjugated structures.
These hyperconjugated resonance structures are really showing us that an $\ce{sp^3}$ carbon can donate electron density to an $\ce{sp^2}$ hybridized carbon.  Electron movement in this direction lowers the energy of the electron because it moves the electron from an orbital that is 25% s-character to an orbital that is 33% s-character.  The more s-character in an orbital, the lower the energy of the orbital (an s orbital is lower in energy than a p orbital).  As we stabilize the electron, we stabilize the molecule.
This hyperconjugation argument comes in handy.  It can be used to explain

why double bonds with more alkyl groups are more stable than double bonds with fewer alkyl groups (ref1, ref2,)
why carbocation stability follows the order: tertiary > secondary > primary (ref1, ref2)
why alkyl groups are ortho-para directing in electrophilic aromatic substitution (ref)


text:
Why member ring is favoured in chelates?
why chelates which form 3 or 4 member rings or 6 or 7 member rings are as stable as 5 member ring chelates?
for example if we measure stability of chelates with oxygen donors, say oxalate, malonate, succinate etc oxalate chelated with metal shoes more stability. why malonate , succinate as chelating ligands are less stable?

Chelating increases the stability , probably due to more change in positive entropy

In its general form, your statement on the stability of these complexes is wrong!
$\ce{Be(II)}$-malonato complexes are more stable than the corresponding oxalato complexes, cf. Preparative, potentiometric and NMR studies of the interaction of beryllium(II) with oxalate and malonate.

no sir i was not trying to say it was universal, i saw in books that said 5 member ring forming chelates are relatively more stable than 6 member or 7 member or 4 member ring forming chelates.


text:
Are all naturally occuring reactions exothermic?
If a reaction is endothermic, then the reaction must take energy from the environment to occur (as far as I know), and if this is the case, then to me the reaction seems energetically unfavorable, as there is only so much energy you can get from the environment. Does this means all naturally occurring reactions, (that don't just happen once in a blue moon but at a steady rate) are exothermic? This seems pretty silly, so if not, why not?

What is a "natural" reaction, which reactions are "unnatural"?

It's not necessary that all naturally occurring reactions are exothermic. The definition of Gibbs free energy is:
$$\Delta G = \Delta H - T\Delta S$$
A spontaneous reaction needs to have $\Delta G<0$, and therefore for spontaneous endothermic reactions to occur we only need that $\Delta H < T\Delta S$.

Some examples should be good. Suggestion: enthalpy of dissolution of ammonium nitrate, decomposition of carbonates (e.g. in such natural settings as volcanic heat), and the endothermic reactions in equilibriums such as $\ce{NO2}$ (a very good example since the temperature has a great effect on the color of the gas)

Does this means all naturally occurring reactions, (that don't just
  happen once in a blue moon but at a steady rate) are exothermic?

No, not all naturally occurring reactions are exothermic ($\Delta H > 0$) or exergonic ($\Delta G > 0$).  For example, the reaction fundamental to our existence, photosynthesis, where carbon dioxide is converted to oxygen and sugar is both endothermic and endergonic (see the figure).
$$\ce{n CO2 + 2n H2O + photons → (CH2O)_n + n O2 + n H2O}$$
The energy required to surmount the barrier is provided by photons (radiation).
Numerous other biochemical reactions are also endothermic and endergonic. This reference mentions protein synthesis and the $\ce{Na+/K+}$ pump which drives nerve conduction and muscle contraction.

An actual reaction is given here.  Strongly basic barium hydroxide octahydrate reacts with a weakly acidic ammonium salt, given ammonia which becomes gaseous and so absorbs heat while increasing entropy.  You can actually freeze water underneath the beaker with this reaction.
https://youtu.be/bVBjc8eH1Co
Thermichemical data indicate that the reactions of ammonium chloride with calcium and magnesium hydroxides are endothermic too.  But the limited solubility of the lighter alkaline earth hydroxides means you don't get as good a visual effect.


text:
Partial pressure and chemical equilibirum

Suppose I am given following reaction in a vessel:
  $$\ce{NH4HS(s) <=> NH3(g) + H2S(g)}$$
  $P_\ce{NH3} = P_\ce{H2S} = \pu{1 atm}$ at equilibrium; if volume is doubled at this instant, then what would be $P_\ce{NH3}$ at new equilibrium?

I know that as soon as volume is doubled, pressure would drop to half at this instant, and $K_\mathrm{c}$ will have the same value at new equilibrium, but how to actually use this to calculate the partial pressure of ammonia at new equilibrium?

No, at new equilibrium $K_c$ would actually is the same value. This constant stands for reaction between these two compounds no matter the pressure, it is only affected by temperature.
As ammonium hydrosulfide is a solid, the reaction is only relevant at the surface of the crystals, and the "concentration" of ammonium hydrosulfide at the relevant areas will remain constant. Therefore same partial pressure of both ammonia and hydrogen sulfide is needed to maintain the equilibrium. Hence, at equilibrium, the partial pressure of both ammonia and hydrogen sulfide should be the same as the initial partial pressure.

can you help me write it mathematically ? would it be $(P/2)^2=1$ ?

@Sigma nah, just ${P_1 \over P_2} = 1$ or $P_1 = P_2$

$P_1$=$P_2$=???

@Sigma $$P_1 = P_2 = 1 atm$$


text:
What would follow in the series sigma, pi and delta bonds?
I realise, that this question is a stretch, but I was wondering, how would a bonding orbital be called if it was formed from two $f_{x(x^2−3y^2)}$ or $f_{y(3x^2−y^2)}$ orbitals. Have there been any suggestions on this, was it anywhere proposed or discussed? I am not arguing about the existence of such a thing, but as a thought experiment, how would it be called? Phi, $\varphi$?
I am also throwing in a reference request here, since I am interested in any literature related to the topic.
Some background:

A $\sigma$ orbital is an orbital, that has no nodal plane between the bonding partners. It is $C_\infty$ symmetric with respect to the bonding axis (in first approximation - of course $d_{xy}$ orbitals can align in a $\sigma$ fashion, too, but for the sake of the argument, let's not consider this.) The point group of this orbital can further be assumed as $D_\mathrm{\infty h}$. It shares therefore most features of an $s$ orbital, hence $s$igma, $\sigma$.
The most prominent example for this is probably the dihydrogen, $\ce{H2}$, molecule.
A $\pi$ orbital has one nodal plane and the bonding axis of the involved atoms is part of this plane. It is therefore asymmetric, $C_1$, with respect to this plane. However, it is $C_\mathrm{s}$ symmetric with respect to the plane that is perpendicular, also sharing the bonding axis. The orbital has further the point group $C_\mathrm{2v}$. It has therefore most features of a $p$ orbital, hence $p$i, $\pi$.
The most prominent examples for $\pi$ bonds are probably ethylene, $\ce{C2H4}$, and acetylene, $\ce{C2H2}$.
A $\delta$ orbital has two nodal planes between the bonding partners. These are perpendicular to each other. The bond orbital is asymmetric with respect to both of them. As a result, it is $C_\mathrm{s}$ symmetric with respect to the bisecting mirror planes. The point group of this orbital is $D_\mathrm{2h}$. Therefore it has shares features with a $d$ orbital, hence $d$elta, $\delta$.
One example is the $\ce{[Re2Cl8]^{2-}}$ ion, see Wikipedia's quadruple bond.
I imagine two $f$ orbitals aligning with the bonding axis being simultaneously a $C_3$ symmetry axis, giving it an overall $D_\mathrm{3h}$ point group. Since there is no greek letter starting with an $f$, I think the logical thing would be to choose $\phi$ as it sounds a like. (I cannot be the only one thinking that.)

The following graphic should somewhat demonstrate what I mean:

I have to admit, that I have not done much research about this. It was more like an impulsive question for me to ask.

I think $\phi$ would be the correct name (unless you want digamma $\digamma$), and following the pattern, I would imagine three nodal planes. Other than that, I expect the $\phi$-bond exists only in theoretical models.

Wikipedia has a tiny section mentioning the possibility of $f$-orbital overlap, indeed forming a phi bond.

tl;dr The next in the series is called φ bond. There is even a tiny Wikipedia article about it.

Nicolau pointed me to the Wikipedia article, that had at the time a tiny section about the φ symmetry of the bond. Ben also kindly agreed with my naming proposition. I'd like to back up just a little bit an quote one sentence of this article:

The type of interaction between atomic orbitals can be further categorized by the molecular-orbital symmetry labels σ (sigma), π (pi), δ (delta), φ (phi), γ (gamma) etc. paralleling the symmetry of the atomic orbitals s, p, d, f and g. The number of nodal planes containing the internuclear axis between the atoms concerned is zero for σ MOs, one for π, two for δ, etc.

Basically this was what I was already thinking, but since the section is quite tiny overlooked it accordingly. But since this is a reference request, I'll go a little deeper.
The concept of the bond is actually quite old and I was able to track it down to at least 1984, where Bruce E. Bursten and Geoffrey A. Ozin "demonstrate[d] for the first time the existance of φ bonds between metal atoms."[1] A little further back, 1978, N. Rösch and A. Streitwieser already performed SCF-Xα calculations on thorocene and uranocene, where they came across f orbital contributions in bonding. Yet, they did not coin the phrase φ bond.[2] Following that, there have been more theoretical studies on the diuranium molecule, with increasing level of theory.[3] One of the more recent studies is by Laura Gagliardi and Björn O. Roos, which is available free of charge (see below).[4] Xin Wu and Xin Lu also studied the phenomenon in endohedral matallofullerenes, i.e. diuranium in a Buckminster fullerene.[5]
It is quite safe to assume, that there are more studies about this research topic or in a more general aspect the research on molecules with high bond orders. For this purpose I will include a couple of good reads.[6]
References and further reading

Bruce E. Bursten, Geoffrey A. Ozin, Inorg. Chem., 1984, 23 (18), 2910–2911.
(a) Rösch, N., Streitwieser, A., Jr., J. Organomet. Chem., 1978, 115, 195-200. (b) Rösch, N., Streitwieser, A., Jr., J. Am. Chem. Soc. 1983, 105, 7237-40.
Melanie Pepper, Bruce E. Bursten, J. Am. Chem. Soc., 1990, 112 (21),  7803–7804.
Laura Gagliardi, Björn O. Roos, Nature, 2005, 433, 848-851. Available free at Archive ouverte UNIGE.
Xin Wu, Xin Lu, J. Am. Chem. Soc., 2007, 129 (7), 2171–2177.
Without order: (a) Gernot Frenking, Nature, 2007, 446, 276-277.
(b) Gernot Frenking, Science, 2005, 310, 796-797.
(c) Gabriel Merino, Kelling J. Donald, Jason S. D'Acchioli, Roald Hoffmann, J. Am. Chem. Soc., 2007, 129 (49), 15295–15302.
(d) Fernando Ruipérez, Gabriel Merino, Jesus M. Ugalde, Ivan Infante, Inorg. Chem., 2013, 52 (6), 2838–2843.
(e) Jia Zhou, Jason L. Sonnenberg, H. Bernhard Schlegel, Inorg. Chem., 2010, 49 (14), 6545–6551.
(f) Awal Noor, Rhett Kempe, Inorg. Cimica Acta, 2015, 424, 75-82.
(g) Frank R. Wagner, Awal Noor, Rhett Kempe, Nature Chemistry, 2009, 1, 529-536.
(h) Björn O. Roos, Antonio C. Borin, Laura Gagliardi, Angew. Chem. Int. Ed., 2007, 46 (9). 1469-1472.


text:
Percentage ionic character when electronegativity is given

What is the ionic character of a bond, $\ce{A-B}$, in terms of the electronegativities of $\ce{A}$ and $\ce{B}$ ($\chi_\ce{A}$ and $\chi_\ce{B}$)?

I have been taught that the percentage ionic character is:
$$ \frac{\text{observed value of ionic character}}{\text{calculated value of character}} $$
but I can't understand how electronegativity is used here. I couldn't find anything on the internet either.

It is naive to assume that electronegativity alone determines ionic character.  If you trust the usual electronegativity criteria then metal hydrides can never be mostly iinuc, but in reality all alkali metals and most alkaline earth metals do just that.  Poor covalent overlap between diffuse metal orbitals and the compact hydrogen 1s orbital has something to do with it.

Linus Pauling proposed an empirical relationship which relates the percent ionic character in a bond to the electronegativity difference $\Delta \chi$.
Percent ionic character $= (1-e^{-(\Delta \chi/2)^2} )\times 100$
But I'd like to correct the definition of percent ionic character in your question using dipole moment  $\mu$ (not Observed value of ionic character):
Percent ionic character = $\Large\frac{\mu_{\text{observed}}} {\mu_{\text{calculated}  }}$ $\times 100 \%$
Where $\mu_{\text{calculated}}$ is calculated assuming a 100% ionic bond.
For more details please see this page.

$$\text{% of ionic character} = 16\times ∆\mathrm{EN} + 3.5\times (∆\mathrm{EN})^2$$
where $∆\mathrm{EN}$ is electronegativity difference. For example, in $\ce{H-F}$ $∆\mathrm{EN} = 2$:
$$
\begin{align}
\text{% of ionic character} &= 16\times 2 + 3.5\times 2^2 \\
  &= 32 + 14 \\
  &= 46~(\%)
\end{align}
$$

I don't understand how you obtained $Δ\mathrm{EN} = 2$. $Δ\mathrm{EN} = χ(\ce{F}) - χ(\ce{H}) = 3.98 - 2.20 = 1.78$ ($χ$ is Pauling's EN, not the oxidation state).

Linus Pauling has introduced the idea of defining the percent ionic character possessed by a chemical bond. A covalent bond with equal sharing of the charge density has 0% ionic character, while a perfect ionic bond would have 100% ionic
character. The ionic character of a bond increases with the difference in electronegativity between the cation and the anion. He has established a quantitative ionicity scale for molecules and crystals based on electronegativity difference (Ref.1 & 2):
$$I_{ionicity} = 1 - e^{-\frac14(\chi_A - \chi_B)^2}$$
where $\chi_A$ and $\chi_B$ are electronegativities of atoms $\ce{A}$ and $\ce{B}$, respectively of the diatomic molecule $\ce{A-B}$. I have found the correct representation of the equation in Ref. 2 (hopefully) to avoid the confusion raised by two different answers within this question (I do not have access to Ref.1 to confirm the original equation). Following is the relation ship between some diatomic molecules:

References:

Linus Pauling, In The Nature of the Chemical Bonds and the Structure of Molecules and Crystals: An Introduction to Modern Structural Chemistry; Third Edition; Cornell University Press: Ithaca, NY, 1960 (ISBN-10: 0-8014-0333-2).
P. Ramakrishnan, "Electronegativity: A Force or Energy," International Journal of Trend in Scientific Research and Development 2019, 3(4), 665-685 (Unique Paper ID: IJTSRD23864; ISSN: 2456-6470).

Electro negativity  decides how the bond A-B  is polarized- electron cloud is more attracted towards more electronegative atom. It's assumed that polarization beyond a certain limit, leads to ionic bond. The electronegativity difference serves as a measure of percentage at which a bond  is ionic.Roughly speaking, electro negativity difference of 1.7 is equivalent to 50 ℅ ionic character;.(calculated ionic character in your question ) Thus, ionic character of a given compound is 50% ×∆ (E.N)/1.7

I agree on most of Yomen Atassi's answer, however I would like to "correct/clarify" the formula of the percent ionic character. The following is the one cited by him, which is the one also reported in several sites in the internet:
Percent ionic character = (1 − e^(−(Δχ/2)2))×100
However applying this equation to real cases results in incoherent percentages (negative values larger than 100).
Checking the source of Pauling's equation (Pauling L. The nature of the chemical bond. 3rd ed. 1960. Pag. 98), I realized that the mathematical expression is different. This is the one cited in the book:
Percent ionic character = (1 − e^(−1/4(Δχ^2)))×100
The last expression provides consistent values which are in agreement with values reported by Pauling and other authors.
If you want to apply Pauling's equation for the calculation of "percent of ionic character" (using electronegativity differences, Δχ), consider to use the later equation.

The expressions are in fact the same.  You or the references mistyped the first formula, the second 2 is meant to be an exponent and should be preceded by ^.

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text:
How can ionic substances have an infinitely-repeating lattice structure?
The title's pretty much it. At school, we're taught that ionic substances are composed of an infinitely-repeating lattice, with atoms at fixed angles from other atoms. How is this possible when I can get an individual grain of salt? Clearly if there's an individual grain, it can't be infinitely repeating.

What happens near the edges?

Edge effects and defects are indeed important in some cases, particularly on the nanoscale.

To be technically accurate, it's "practically" infinite.
It's not infinite since as you mention a grain of salt (or even a large crystal) has some finite size.
You have to remember the actual atomic dimensions $10^{-10} \:\mathrm{m}$. So while it's not truly infinite, there are an incredibly large number of atoms in a crystal.

So it's essentially infinite in the middle, but what happens around the edges? Are there sodium ions with only three chloride ions attached to them, rather than the four it'd have with it in the middle?

Yes, the only limitation is that the whole crystal is not charged. So as you said, there will be "there sodium ions with only three chloride ions attached to them". Not only that, there will be also here chloride ions with only three sodium ions attached to them.

Side comment, http://www.physlink.com/Education/AskExperts/ae342.cfm estimated number of atoms in grain of salt to be 10E18. Therefore each edge is about cubic root of it long, about one million. Big, but not incredibly. But the surface effects do not reach deeper than ~5 layers, so the bulk of crystal is real crystal. (See history of http://en.wikipedia.org/wiki/X-ray_crystallography )

The difference between surface and interior properties is particularly notable in topological insulators.


text:
Abbreviations in electronic configurations
I am teaching my students how to write the electronic configuration of the elements of the periodic table.
For example, following the rules on how to fill the orbitals, the electronic configuration of iron is
$$\ce{Fe}:~\mathrm{(1s)^2(2s)^2(2p)^6(3s)^2(3p)^6(4s)^2(3d)^6}.$$
Thus, its abbreviated form is
$$\ce{Fe}:~[\ce{Ar}]~\mathrm{(4s)^2(3d)^6}.$$
However, I has not been easy to find the explanation on why in any periodic table it is written as
$$[\ce{Ar}]~\mathrm{(3d)^6(4s)^2}$$
instead of
$$[\ce{Ar}]~\mathrm{(4s)^2(3d)^6}.$$
Same case for many other elements. Why is that?

I have seen it written both ways.  However, the '3d' orbitals in this case are going to be lower in energy than the 4s and therefore it is written in a way that indicates increasing energy from left to right.

I would recommend following the convention used by NIST which is to list the subshells in order of principle quantum number $(n),$ and in order of angular momentum quantum number $(\ell)$ for a given $n.$
See also ELECTRONIC STRUCTURE OF THE ELEMENTS (PDF).
The question refers to “rules on how to fill the orbitals”, but I don't think there are any rules, just memorization aids.
See The low-lying level structure of atomic lawrencium $(Z = 103):$ energies and absorption rates for an sample of how difficult it is to calculate the ground state configuration of an atom.  The only rule is quantum electrodynamics.
Also, the configurations are really just designations or leading configurations, with the actually ground state being a mix of configuration.  See for example the above lawrencium reference or A critical compilation of energy levels and spectral lines of neutral boron, where it explained “Configuration and
term labels have little physical meaning for highly mixed levels”. Boron ground state is $95\%$ $\mathrm{(1s)^2(2s)^2(2p)}$ and $4\%$ $\mathrm{(1s)^2(2p)^3}.$  Lawrencium is calculated to be $86\%$ $\mathrm{(7s)^2(7p)}$ and $6\%$ $\mathrm{(6d)(7s)(7p)}.$

I guess, from a teaching perspective, listing the sub-shells order of increasing energy.  My goal in teaching general chemistry is to help students make connections between quantum theory, the periodic table, and chemical reactivity.  I teach my students to write electron configurations using the periodic table as a guide.  However, I do tell them about exceptions and why they happen, but I don't want them to be more concerned about them.  They can opt to take physical chemistry if they want to delve deeper.

@TamiM. so would you have your students switch the order of 3d and 4s when you go from Fe to Fe+ since the 4s becomes higher energy, or do you always go by the neutral atom energy level order?

I would probably accept either way.  Honestly, I would accept either valid way of writing electron configurations, even for neutral atoms.  But, when I teach general chemistry, I do teach them to use the periodic table to write electron configurations.  If I were teaching a more advanced course, I would go much more in depth about mixed configurations.   As it is, I go into more detail about these things then many general chemistry lecturers (I can't help it, I am a physical chemist).  As I said before, I just try to focus on connecting theory to laws and chemical reactivity.

In some textbooks that I have read, I came across electron configurations that listed the 3d electrons before the 4s electrons.  I believe it is to keep the electrons with the same principal quantum number $(n)$ together.
For example, $[\ce{Ar}]~\mathrm{(3d)^6(4s)^2}$ in full would be $\mathrm{(1s)^2(2s)^2(2p)^6(3s)^2(3p)^6(3d)^6(4s)^2}.$  When I teach general chemistry, I opt to list the sub-shells in order of increasing energy instead, since it correlates with the structure of the periodic table.

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


text:
Are there any STP liquids not containing large amounts of hydrogen and oxygen that are safe for human contact?
I'm working on a science fiction short story and was wondering if there were any liquids that would be safe to bathe in while traveling on a space craft with artificial (spin-based) gravity.  In such a context hydrogen and oxygen are valuable resources not to be wasted.  While bath water could be recycled, it is not available for other uses while someone is sitting in it, so I was wondering if there are other liquids at Standard Temperature Pressure that would be safe to bathe in (potentially ingesting small amounts).
I thought of using carbon bucky-ball fullerenes in a fluid but the viscosity looks to be much too high (essentially tar-like).  I know there are waterless chemicals that can be used for cleaning (e.g. DryBath), the purpose of this is more to have a hot-tub to relax in than to disinfect and clean.  If the addition of some hydrogen and oxygen (either as H2O or in other compounds) can improve the viscosity or is necessary to form the liquid, that is acceptable as long as less than 10% of the liquid is comprised of hydrogen and oxygen.  The purpose is to save water and oxygen for air while still being able to soak in a hot-tub.
Unfortunately most oils contain a substantial amount of hydrogen and some oxygen, so they don't qualify unless the amount of carbon offers substantial savings in volume.  I'd prefer something without any hydrogen or oxygen if such a chemical exists.
This is my first post here and I'm unfamiliar with the tags, so feel free to edit or add any you feel are appropriate.

This is a very interesting question! I can see oxygen being a potentially rationed element, but hydrogen is the most common element in the universe. Oxygen is the third most common element in the universe. With the right technology, it should be easy to mine hydrogen from gas giants. We could mine water from comets and other Kuiper belt objects

I know hydrogen is abundant (and even exists in space in low quantities), but for the purposes of the story, hydrogen is also a fuel source and so water (and thus hydrogen) is still rationed for habitats in addition to oxygen for both water and air.  The space craft is not in a position to harvest resources from gas giants, asteroids, or planets.

If you want to relax your muscles similarly to a hot tub, without insisting on getting clean, you don't need a liquid. Granulated material of size and composition which doesn't allow it to cling to skin is a good replacement, as well as simply donning a suit with an active heating system. I have back problems and lying on a warmed wax pad (closed up, no skin contact with the wax) is just as good as taking a hot bath.

This is a good suggestion, though I am still most interested in a liquid rather than a solid or gel, since I specifically asked for a liquid that could be a water substitute.

Assuming abundant electricity, water is a safe and compact way to transport large amounts of hydrogen and oxygen. You just need electricity to split them apart again. So water may not need to be as constrained a resource as you imagine.

Several posters have mentioned various approaches to getting or using water on a spacecraft. While that is an interesting topic, it is not what I'm asking and is off topic for this question (feel free to start one on water, hydrogen, and oxygen use in space travel). I'm essentially asking, given that water for whatever reason is not avaliable in large quantities on a space habitat, what are safe chemical alternatives that can be used in the same way we use large quantities of water on humans for purposes other than hydration?

Perfluorocarbons would be a possibility, $\ce{(-CF2-)_{n}}$, they contain only carbon and fluorine and are relatively low cost and widely available in large quantities.  They are inert (e.g. non-reactive) and depending upon chain length and structure, you could pick one that can be warmed up nicely to a bath temperature.  They are also used industrially in degreasing operations, so they might have some cleaning ability as well.
Some perflurocarbons are actually being investigated as human blood substitutes - they can transport and release oxygen to tissues.  Perfluorodecalin is a prime example. As shown below, it is basically a naphthalene ring in which all of the hydrogens and double bonds have been replaced with fluorine.

Ingesting small amounts would not be a problem. In fact, these same materials have been used in liquid breathing experiments.
There are a number of other scientific aspects of these perfluorocarbons that could play into science fiction.  For example, the small size of the molecules (compared to say, haemoglobin) allows it to travel to smaller places that haemoglobin cannot access due to its larger size. Perhaps the perfluorcarbon could also carry small drugs to relax, rejuvenate, repair body parts, or enhance sensory experiences during the bath. Since these perfluorocarbons act as oxygen transport agents they could also be worked into the story as a bath medium that also scavenges "loose" oxygen.

Wow, perfluorocarbons have even been used in liquid breathing research.  This is a really interesting possibility (+1).  I'll wait a while to see if others have any good ideas while I do some more reading, but this is excellent.

Thanks everyone for the excellent ideas and comments.  I've decided to accept this answer, as I feel it comes the closest to what I asked for in the question out of all the suggestions made in comments and Yomen's answer.  Most suggestions other than this one involved gels or solids.  This is the only answer that supplied an actual liquid, and @ron also suggested ways to incorporate other properties of perfluorocarbons into a scifi setting.

I think we should propose something that uses some water. We cannot forget that our body is compatible with water as a solvent. The chemistry in our body is based on the chemistry of aqueous solutions.
On the other hand, we have to take into account the  large quantities of solvents, other than water to be transported to the spacecraft for astronauts baths. The water used for my proposed solution is a secondary product of the fuel cells  actually used to generate electricity in the spacecraft. The fuel for the fuel cells are oxygen and hydrogen. And as @Ben Norris added hydrogen is the most common element in the universe and oxygen is the third most common element in the universe. So, with   the right technology, we should have them on the space craft.
I would like to suggest to use "Bentonite based superabsorbent polymer composite". It's a kind of hydrogels that absorbs water and loaded up to 50% wt/wt with a clay, bentonite (aluminium phyllosilicate). This composite is taken to space as powder. When astronauts  want to take a bath, they can add some (water+fragrance) to the powder that swells and forms a gel.
The value of bentonite as stabilizing and rheological agents is due to the colloidal structure in water. Each smectite particle is composed of thousands of submicroscopic platelets stacked in sandwich fashion with a layer of water between each. A single platelet is one nanometer thick and up to several hundred nanometers across. The faces of these platelets carry a negative charge, while edges have a slightly positive charge.
Bentonite is largely used in cosmetics, it is bio-compatible and has excellent rheological properties to get perfect emulsions and optimize flow properties. On the other hand, the composite I'm suggesting has been formulated and tailored to absorb moderate quantities of water  and fragrance that has a soothing properties to relax astronauts.

The question I have about this approach is: How much water would be necessary for such a gel to be used as a bath.  Remember that the primary purpose in this case is to relax and soak, not get clean (though that is a potential side benefit).  For astronauts in real life what you are suggesting makes excellent sense, but I'm dealing with a futuristic fictional context where people living on a water constrained habitat would like to soak.  I'm not opposed to gels, as that would be similar to a mud-bath in some respects, but it shouldn't be more than 10% water.

Bentonite is largely used in cosmetics, it is bio-compatible and has excellent rheological properties to get perfect emulsions and optimize flow properties. On the other hand, the composite I'm suggesting has been formulated and tailored to absorb small quantities of water 10% wt/wt + fragrance that has a soothing properties to relax astronauts.

Please, take into account the quantities of solvents, other than water to be transported to the spacecraft for astronauts baths.

Point taken that we don't want to carry material that is only useful for baths, although the medical benefits of perfluorocarbons (as a blood substitute and breathable fluid) might justify their inclusion for space travel.  Nevertheless, a Bentonite hydrogel as you've outlined is a valid choice that meets the requirements I outlined (+1).


text:
How chemically homogeneous is air?
By this, I mean if I buy compressed dry air from somebody like British Oxygen how much variation in composition might I expect between batches? For example, is Argon evenly distributed globally, by height, by geography etc? Is $\ce{CO2}$ removed? Are contaminants filtered out eg $\ce{NOx}$?

Oxygen/nitrogen ratios vary seasonally and geographically even at sea level on the order of 200 ppm.  Oxygen concentration has also been slowly decreasing in correlation with carbon dioxide increase.
Try using the MSIS-E-90 Atmosphere Model which allow location, date, time and height input and provides composition output.

Thanks. I was thinking of the possibility of replacing Argon with air in a calibration experiment, but it now looks like "air" is not as pure as I would like.


text:
Paper making alternative
Is there any paper making alternative? What is there in trees that makes it the only thing from which paper can be made? Do we even need paper anymore with all electronic devices and stylus', etc

Paper mostly consists of cellulose, form which wood is a good source.
Another interesting source is the hemp plant and it seems that hemp was indeed the major source in paper manufacturing up to the 19th century.
Nowadays, recycling of paper probably is the most reasonable way to reduce the wood use.

Do we even need paper anymore with all electronic devices and stylus'

Yes, for different reasons:

Think in how long old books and handwritings have lasted. Compare that with the durability of magnetic (tapes, hard disks) or optical recordings (CD, DVD, Bluray, etc.)
Supposed that the ability to read doesn't get lost and a document is written in a well-defined alphabet, you do not need any additional technology to access the information. Who can read data from a 9 track tape nowadays (the technology is only fifty years old) or an "old" MFM hard drive? 8" or 5.25" floppy disks, anyone? ZipDrives, supposed they survived the "click-od-death"?
Most people do not use electronic signatures for documents. Papers are signed, sealed, etc. Consequently, written contracts, bills, etc. are considered more forge-proof than electronic documents. Storing them is therefore mandatory in business.

"" that hemp was indeed the major source in paper manufacturing up to the 19th century""  No, not at all. Before groundwood pulp and chemical wood pulp were invented, paper was made from cotton and linen rags. That was the basis for the "Lumpensammler"-trade :=) High value papers (money etc) are still made from rags and cotton linters.Hemp with its long and very coarse fibers  was outside 19th century technology for papermaking. A lot of green propaganda nonsense is told about hemp today, I guess to facilitate access to cannabis.

I forgot: just before the great oil crisis in 1978 3M invented synthetical pulp (microscopic PP fibers from a proces without spinning) When I started to work in a pulp company in 1978 they were just recovering from that threat :=)

Another thing I forgot: textile fibers and hemp were and are much too expensive to produce standard paper.


text:
Why does the graph of compressibility factor(z) vs pressure increase after reaching the minimum?
I understand why it decreases for most of the gases but i cannot understand why it increases after reaching a certain minimum.

After a certain amount of pressure, the molecules start repelling each other and this leads to a rise in compressibility factor.
At high pressures, the van der Wall constant $b$ which represents the finite volume occupied by the gas molecules plays a greater role than the van der Wall constant $a$ which represents the attractive forces between the gas molecules.


text:
How can you achieve negative pressure?
Pressure is the magnitude (absolute value) of the force over the area, so how can that result in a negative value for pressure as would happen in certain circumstances?

Have a look at this Veritasium video discussing how trees pull water from the soil to their leaves. Derek briefly mentions the mechanics of negative pressure in solids and liquids.

@Nicolau How is the air removed from the capillaries in spring? Cut a twig from a tree or wine and couple a manometer to it. You will see, that there is pressure from below.

The term “negative pressure” is often used in engineering to refer to a situation in which an enclosed volume has lower pressure than its surroundings. If the a region is surrounded by more pressurized area around it would cause substances to flow inwards and thus the term negative pressure.

Pressure is not an absolute value. It might be helpful to know where you got that idea from. I can theoretically completely evacuate a chamber so that one side of a pressure transducer has "zero" pressure and then claim that what I'm measuring (on the other side) is the "absolute pressure". The problem with that is that I can't completely evacuate any chamber, nor can I prove that once I've reduced the measurable gas (or liquid) present to below my detection limits that there isn't some kind of "aether" present.
OTOH, we can approach zero pressure "in the limit" and we can, via other experiments, show that there is no "aether". The Casimir Effect between two plates placed close together is an example of the existence of pressure in the absence of "everything". So, for practical purposes we can claim knowledge of absolute pressure values, even though all measured values are relative.
In the USA, relative gauge readings are often recorded as having units of psig or pounds per square inch gauge pressure, which means the pressure excludes ambient air pressure (which depends on altitude, weather, and temperature (among other things)), other gauges have units of psia (pounds per square inch absolute) which are based on a perfect vacuum being the zero point. It is important when using any pressure gauge to know what the value is relative to (to ambient air pressure, to a standard atmosphere (and which standard atmosphere!), or to a 'full' vacuum).
Another way to put this is that if we assume we can obtain a vacuum, then we can measure relative to it and claim the result is "absolute". While reasonable and quite solidly based on our understanding of the nature of our physical world, we can not prove it is true without making underlying assumptions about how nature influences its measurements.


text:
Finding identities of diatomics from Spectroscopy and Mass Spec Data
Let's say I have a sample of 2 (assumingly different) diatomics A and B.
Through spectroscopy I found the data below:
For molecule A:
$$\begin{align}
B&\approx2.17690\ \mathrm{cm^{−1}}\\
D&\approx4.79000\times10^{-5}\ \mathrm{cm^{−1}}\\
I&=1.28590\times10^{-46}\ \mathrm{kg\ m^2}\\
\tilde ν_e&=928.15\ \mathrm{cm^{−1}}
\end{align}$$
For molecule B:
$$\begin{align}
B&\approx1.51695\ \mathrm{cm^{−1}}\\
D&\approx7.1049\times10^{-6}\ \mathrm{cm^{−1}}\\
I&=1.84533\times10^{-46}\ \mathrm{kg\ m^2}\\
\tilde ν_e&=1401.86\ \mathrm{cm^{−1}}
\end{align}$$
Through Mass Spec, I found 2 strong spectral lines at $30.0077086\ \mathrm{g\ mol^{-1}}$ and $41.9766928\ \mathrm{g\ mol^{-1}}$. How do I deduce which is which and further, how do I determine the bond length. [FYI, I'm thinking the former mass is the molecule NO and the latter NaF just by guessing]
I'm thinking that the heavier molecule will have the larger inertia value. And from the inertia equation $I=\mu R^2$ I'll derive the bond length $R$. but how do I find the $\mu$ from the actually mass of the molecule from the mass spec data?

Is this problem conceptual, based on hypothetical data?

Through [m]ass [s]pec, I found 2 strong spectral lines at $30.0077086\ \mathrm{g\ mol^{-1}}$ and $41.9766928\ \mathrm{g\ mol^{-1}}$.

These mass-to-charge ratios or $m/z$ values are implausibly precise.  The significant figures you show imply a precision of about 30 parts-per-billion (ppb) mass accuracy.  That is well outside the specification of any commercial instrument I am aware of.
Also, if you are right about the first ion being, $\ce{NO}$, the value is inaccurate.  The neutral $\ce{^14N^16O}$ molecule has a mass of 29.99799 Da.  The positively charged ion formed by electron loss from the neutral species would have a mass of 29.9974 Da.  This latter value is off from the value you quote by about 341 parts-per-million.  So if the mass spectrometry data really is precise to within 30 parts-per-billion, and it is accurate, then the molecule can't be $\ce{NO}$.
In fact, with mass spectrometry data that precise (and accurate), we shouldn't need any other data to identify the diatomics, other than the fact that they are diatomics.  We can add the mass of every possible nuclide to every other possible nuclide and see which give the right mass.
Here's how to do that using the R packages ecipex and dplyr.
require(ecipex)
require(dplyr)
nistiso %>%
    mutate(foo = 1) %>%
    left_join(., ., by='foo') %>%
    filter(mass.x > mass.y) %>%
    mutate(diatomic_mass = mass.x + mass.y) %>%
    filter(diatomic_mass < 30.00771, diatomic_mass > 30.00770)

The result is that only the $\ce{^11B^19F}$ diatomic molecule has the requisite mass. Even if you loosen the mass tolerances to allow for e.g. for electron loss or gain, this is still the only possible diatomic with the right mass.
You can repeat this exercise for the other data point, and find that $\ce{^40Ca^2H}$ is the only possible ion consistent with the data.  ($\ce{^40Ar^2H}$ is close in mass but is more than one electron's mass too light.)
I'll leave it to you to decide whether these ions are chemically meaningful or possible for you, and if they are consistent with the rotational spectra.


text:
How does 3dCarbon CLIP printer work?
Recently there has been announced a new 3d printing technology which uses UV light, from lasers or a projector, to harden slices of epoxy resin so that the printed object rises out of a pool of epoxy.
There is some information about an oxygen boundary and UV light, can someone explain the chemistry of the process?
https://www.youtube.com/watch?v=UpH1zhUQY0c

I have seen the video, and wonder if the claim "can use all polymers, elastomers" is true, because you need a "handle" on the molecule to make it uv curable. In other words, if I would claim to use a certain polymer it would have been drastically modified,or not?

Is a polymer drastically modified by putting a UV handle molecule on it? interesting question why don't you ask how that process works? there are uv cureable versions of polyester and rubbers, i don't know what they do.  This printer if it becomes available for homes will have a mechanical transfer to a drying and cleaning area, otherwise it wil be unmarketable, to have dripping models coming out of a box at home!

Actually, the technology isn't that new. It seems however, that the setups are becoming more affordable in the hobby market.
The technology is know as stereolithography and is a form of light-induced polymerization (UV curing).
The liquid monomers used have a substituent that can be activated using UV light. Out of practical considerations, UV is more convenient: polymerization of the resin will not occur under natural light.
In principle, the procedure is comparable to UV curing in dental repair. There, a lamp that irradiates the whole area is used: you want a fast hardening all over the tooth.
In 3D printing, this is obviously different! The light source is a laser with a narrow beam: you only want to irradiate a very small volume element at a time.
You don't want a rapid polymerization either: only the volume element irradiated should react!
The laser beam is focussed at the boundary of the resin and different volume elements are addressed by moving the focus in the $xy$ plane.
Rather than moving the (heavy) laser itself, the beam is "moved" using a mirror optic. This allows for a very fast adressing of reaction points within the plane.

UPDATE
In the example linked, in the question and a comment, however, the setup is different from laser-based STL printers.
Irradiation from the bottom of of the resin tank through an oxygen-permeable membrane indicates indeed that the UV curing proceeds in the presence of oxygen. Anyway, if the light source would be just a lamp, one would nevertheless achieve homogenous curing of the whole layer.
But the manufacturer doesn't use the term lamp or light source. Instead, they write about a projector and that's probably what they do:
They are not illuminating the whole layer, but are projecting masks for every slice of the object to be printed.

Ich habe mich auch maßlos über diese dreise "Neuheit" geärgert.

AKA = i am also immeasurably annoyed by this Dreise "novelty" . Yes Georg it seems that the media is promoting this form of 3d printer as new. perhaps it's the first one that can print as fast and that has been filmed printing in one single movement?

I am curious about how they manage to focus the UV light into a point rather tan a line, and what the oxygen layer is inside the epoxy, if they inject oxygen through the base in some way.

@ufomorace there's a misunderstanding. there is no oxygen layer inside the resin, there's just air on the top of the basin.

there is an oxygen permeable layer: http://www.gizmag.com/clip-carbon3d-3d-printing/36606/pictures#4   The process relies on the right balance of light, which triggers curing, and oxygen, which inhibits it

@ufomorace Interesting! I've tried to work that into the update of my answer.

There is an article on the website slashdot with probably alot of conversations regarding this news story... just found a page that said: With a free radical system, polymerization stops as soon as the light is eliminated.  Free radical systems are also subject to oxygen inhibition, which means that oxygen in the air prevents the molecules at the surface from polymerizing, leaving a wet or tacky surface.... here is the full information - its freely available. https://justpaste.it/k1tn

What's new about the system isn't the hardware; stereolithography has been around for ages, but the chemistry is the interesting part (enough for an article in Science). Normal SLA is done iteratively, debonding the part from the window and applying new resin for each layer between exposures. By using this "dead zone", the part doesn't stick to the window and new resin is drawn under it as it is raised, allowing continuous exposure. And they do use a projector, with a DLP chip.

DLP chip is fascinating. It must be a very strong ray of UV that they shine through the teflon AF 2400 film to have a oxygen permeable film that can stay straight with epoxy on it. perhaps the oxygen zone is under pressure and the teflon is only 1mm thin. The balance of that probably makes a size limitation on the prints. There is a great printer that can print giant parts in a vat of epoxy by shining at the top layer, and indeed it works in steps as it sweeps over the top layer, whereas this has infinite z resolution. / framerate z resolution.


text:
"Alkanes must be vaporised before they will burn" - what does this mean?
What does this sentence mean - I assumed that burning vaporised something? If something is vaporised, isn't that turning it into a gas? How can you burn a gas? This is a sentence in my AS Chemistry book and I am obviously missing something.
"Alkanes must be vaporised before they will burn"

Burning an alkane means reacting it with oxygen (in air). The larger the contact surface between the alkane phase and the air, the better the reaction. That's why cars have carburators: there's no larger "contact surface" than between the molecules in a gas mixture.


text:
How to rationalise the basicity order of phenylenediamine?
Why is the basicity of para- > meta- > aniline > ortho-
for phenylenediamine?
($\mathrm pK_\mathrm a$: ortho: 9.53, meta: 9.12, para: 7.96) I tried deducing it using resonance and inductive effects but got stuck.

related http://chemistry.stackexchange.com/questions/7683/ortho-effect-in-substituted-aromatic-acids-and-bases?rq=1

I am not being able to incorporate the ortho effect concept!Let me know please.

Isn't the basicity trend listed backwards? The para-analog has the lowest pKa for the conjugate acid (strongest acid), which means that the base is the weakest?

in phenylene-diamine shouldn't the basicity order be p>aniline>o>m since -NH2 is not much of a bulky group and thus doesn't exhibit ortho effect. SO with ortho effect out of the picture -NH2 is very strong +M group and effect of +M>-I (for -NH2). In ortho due to hydrogen bonding its basicity is reduced. Meta position only -I. so by that reasoning para is most basic followed by aniline followed by ortho and then meta....So p>aniline>o>m....Am i going wrong anywhere?

As Mithoron pointed out in a comment, we had a similar question convering the cases of 2-methylaniline (o-toluidine) and 2-methylbenzoic acid (o-toluic acid) here.
In the case of 1,2-diaminobenzene (o-phenylenediamine), it is conceivable that hydrogen bonding between the two amino groups reduces the electron density on one of the nitrogen atoms, thereby leading to a slight increase on the other. As a result, o-phenylenediamine is significantly less basic than the m- and p-isomer, but a tad more basic than aniline.

Klaus's answer is nice, but is omits one important detail:
The hydrogen bonding would necessarily require that the lone pair of electrons go 'out of conjugation' with the ring system, and hence there can be no donation of electron density via pi-effects as in the para-phenylenediamine case.


text:
Why is water more acidic than alkanols?
Why is $\ce{H-OH}$ more acidic than $\ce{CH3CH2-OH}$? A common answer to this question is because of the $+I$-effect of $\ce{CH3}$ the negative charge at $\ce{O}$ increases making it less acidic. But if we compare electronegativity difference, the difference between $\ce{O}$ and $\ce{H}$ is more than that between $\ce{O}$ and $\ce{C}$ so there should be even more negative charge at $\ce{O}$ in case of $\ce{H2O}$. In this way $\ce{H2O}$ should be less acidic than $\ce{CH3CH2-OH}$. This is also the reason that $+M$ (electron donating ability) of $\ce{-OH}$ (which is having more negative charge at $\ce{O}$) is greater than the $+M$ of $\ce{-OR}$.

What data are you using to support this? According to Evans' pKa table methanol is slightly more acidic than water in water and notably more acidic in DMSO http://evans.harvard.edu/pdf/evans_pKa_table.pdf

Yes, that's an exception. Edited the question.

\begin{array} {|l|r|r|}
\hline
  &\mathrm{pK_a~in~water} &\mathrm{pK_a~in~DMSO}\\
\hline
\ce{H2O} &15.74 &31.4\\
\hline
 \ce{CH3CH2OH} &16 &29.8 \\
\hline
\end{array}
As we can see from the $pK_a$ data water is slightly more acidic than ethanol in water but in DMSO ethanol is notably more acidic. In DMSO hydrogen bonding is minimal so the acidity is primarily dependent on the acid and not the solvent it is in.
However, in water the hydroxide ion is hugely stabilised by hydrogen bonding (the ethoxide is as well but not nearly to the same extent) so this accounts for the greater acidity of water in this situation.
http://research.chem.psu.edu/brpgroup/pKa_compilation.pdf
http://www.chem.wisc.edu/areas/reich/pkatable/index.htm

Is solvation the only reason? How does inductive effect plays it role? And how will you explain ethanol being more acidic in DMSO?


text:
Does sorbitol dissolve in ethanol?
I know Sorbitol, $\ce{C_6H_14O_6}$ $$\ce {C(OH)H_2 (C(OH)H)_4C(OH)H_2}$$is soluble in water however I did not have much luck finding out whether or not it is soluble in ethanol. My guess is that it isn't however I am not so sure.
Is Sorbitol soluble in ethanol?

Did you see this?

The case is very similar as in http://chemistry.stackexchange.com/questions/19598/is-glycerol-glycerin-soluble-in-ethanol

Hallo Klaus, das ist ja toll! Bei den Grünen ist die Messung von ein paar Löslichkeiten also Wissenschaft?

@Georg I see that with less sarcasm ;) True, the measurements won't qualify for a publication in J. Phys. Chem., but sometimes it's exactly the fundamental data one is looking for. No fancy interpretation, just the boring data, a well-described, reproducible setup and a careful experimenter. This is even more true if a survey covers a whole series of compounds under identical conditions (which admittedly isn't the case here).

Sorbitol's solubility in ethanol seems to be very temperature dependent according to the Merck Index$^{[1]}$

Freely sol in water (up to 83%); sol in methanol, isopropanol, butanol, cyclohexanol, phenol, acetone, acetic acid, dimethylformamide, pyridine, acetamide solutions; practically insoluble in most other org solvents; quite soluble in hot alcohol, sparingly soluble in cold alcohol

1) O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Whitehouse Station, NJ: Merck and Co., Inc., 2006., p. 1498


text:
Difference between continuously overlapping p orbitals and conjugated pi bonds
Is there any difference between the phrase "continuously overlapping p-orbitals" and "conjugated pi bonds" when referring to aromatic compounds? I've heard both used and I wasn't sure if they mean the same thing.

Strictly speaking p-orbitals can overlap to form a sigma-bond if the touch end-to-end

Interesting question!
To have continuously overlapping p-orbitals, the p-orbitals must be arranged in a cyclic loop.  The p-orbitals in benzene are continuously overlapping.  The p-orbitals in benzene are also conjugated pi bonds.

image source
In buta-1,3-diene the p-orbitals form conjugated pi bonds, however, they are not continuously overlapping - there are two "end" carbons that are not connected with each other, there is no continuous loop.

image source
Let's take it a step further:
Continuously overlapping p-orbitals are not always aromatic.
Example 1: Cyclobuta-1,3-diene (the square planar structure, $\ce{D_{4h}}$) has 4 continuously overlapping p-orbitals but it is not aromatic, it is antiaromatic.  For a continuously overlapping system of p-orbitals to be aromatic, it must contain 4n+2 pi electrons.

image source
Example 2: Have you heard of the Möbius strip? The mathematical (topological) concept has application in chemistry.  Imagine a large cyclic loop of p-orbitals where each p-orbital is twisted a few degrees from the preceeding p-orbital.  If the loop is large enough, then by the time we come back to the first p-orbital we have twisted the last p-orbital by 180°; there is a phase inversion (node) between the first and last p-orbital.  In a Möbius system the p-orbitals are continuous, but not continuously overlapping!  Möbius systems are aromatic if they contain 4n pi electrons.

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Continuous pi systems with an even number (0, 2, 4...) of nodes (cyclobutadiene, benzene, etc) are called Hückel systems in order to differentiate then from Möbius systems where there are an odd number of nodes (1, 3, 5...).
Lest you think the Möbius concept is purely theoretical take a look at these two molecules

image source
Isn't this what makes chemistry fascinating!


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IUPAC nomenclature of alicyclic compound with an equal number of cyclic and acyclic carbons

In this compound there are three cyclic and three acyclic carbons. Even the substituent comes on the 2nd position in both cases. So, should the main chain be the cyclic one, since there is a rule that cyclic has a priority over acyclic?

According to current IUPAC recommendations (2013), for the preferred IUPAC name, a ring has seniority over a chain. Therefore, the given structure is a propylcyclopropane. (However, the IUPAC recommendations allow that the context may favour the chain.)
The two prefixes of cyclopropane (chloro- and 2-bromopropyl-) are arranged alphabetically. Thus, the name is 1-(2-bromopropyl)-2-chlorocyclopropane.


text:
How to calculate the bond energy of C-Cl?
$$\ce{C(s) + 2 Cl2 (g)  -> CCl4 (g)}$$
\begin{align}
\Delta H^\circ_\mathrm{f} (\ce{C(g)}) &= 121~\mathrm{kJ/mol}\\
\Delta H^\circ_\mathrm{f} (\ce{Cl(g)}) &= 716~\mathrm{kJ/mol}\\
\Delta H^\circ_\mathrm{f} (\ce{CCl4(g)}) &= -106.48~\mathrm{kJ/mol}\\
\end{align}
Basically I already answered this, but I was a bit confused about the result, cause I don't know if there is a bond that is negative, cause am familiar with the energy given to "dissociation" of a bond or the "formation" of it , but here it's given the energy to the formation of chlorine and carbon, I don't know if that's the same as the formation of the bond or not!
\begin{align}
\Delta H^\circ_\mathrm{f}(\ce{C-Cl}) &= \frac{\Delta H^\circ_\mathrm{f}(\ce{C(g)}) + 4 \Delta H^\circ_\mathrm{f}(\ce{Cl(g)}) - \Delta H^\circ_\mathrm{f}(\ce{CCl_4(g))}}{4} \\
\Delta H^\circ_\mathrm{f}(\ce{C-Cl}) &= \frac{121 + 4 \cdot 716 + 106.48}{4} \\
\Delta H^\circ_\mathrm{f}(\ce{C-Cl}) &= 772.87~\mathrm{kJ/mol}
\end{align}

Should I take the $\triangle H^\circ_\mathrm{f}(\ce{Cl(g)})$ same as $\triangle H^\circ_f(\ce{Cl-Cl})$? In that case I'm going to multiply it by $2$, not $4$.
Should the  $\Delta H^\circ_\mathrm{f}(\ce{C-Cl})$ be prefixed by a minus $-$ or a plus $+$, cause I don't know if it's an energy that is needed to formulate $\ce{CCl4}$? I'm not saying that the enthalpy of the bond should be negative, but the direction.

Using your Given data
$$2\times\Delta{}H^\circ_\mathrm{f} (\ce{Cl(g)}) +\Delta{}H^\circ_\mathrm{f} (\ce{C(g)}) -4\times\Delta H_{C-Cl}=\Delta{}H^\circ_\mathrm{f} (\ce{CCl4 (g)})$$

Hello Sir , thanks For Answering , Can you explain to me why You multiplied the $\triangle H^°_f(Cl(g))$ by $(2)$ , & Why You Put a $(-)$ Before The $\Delta H_{C-Cl}$

$\delta H$ depends on stoichiometry and I subtracted enthalpy of bond formation because that is the "lost" energy that is missing on RHS of equation .When bond is formed then energy is lost from LHS side into heat. Implying RHS has less energy than LHS so I just balanced out total energy.

that's seems Logical now , but still why   $2\times\Delta{}H^\circ_\mathrm{f} (\ce{Cl(g)})$ .
can we say That  $2\times\Delta{}H^\circ_\mathrm{f} (\ce{Cl(g)})$ $= 1\triangle{}H^°_{d}{}_{(Cl-Cl)} $ , & thank you Sir for your time .

You are taking account of energy here so $2$ is there because you have formed 2 moles of chlorine gas.

but it says $2{} Cl_2$ That's Gonna Give Us $4 Cl$ if that's possible , because it says $\Delta{}H^\circ_\mathrm{f} (\ce{Cl(g)})$

notice the word "moles"

I agree with MxH. You must put a 4 in front of $Cl$ or a 2 in from of $Cl_2$. I don't believe Heisenberg is looking at the number of moles correctly.

The bond dissociation enthalpy* $D$ is the enthalpy change for conversion of bonded into dissociated molecules following homolytic cleavage of the bond in question:
$$\ce{A-B (g)->A(g) + B(g)},\qquad D(\ce{A-B})=\Delta H^\circ_f (\ce{A(g)})+\Delta H^\circ_f (\ce{B(g)})-\Delta H^\circ_f (\ce{A-B(g)})$$
This is the property usually discussed as representing the "strength" of a bond, and is positive because the process of breaking a bond requires an input of heat (is endothermic). Breaking stronger bonds requires a greater positive enthalpy change.
The enthalpy change for formation of the bond by the dissociated molecular constituents is negative the dissociation enthalpy:
$$\Delta H^\circ (\ce{A-B})=-D (\ce{A-B})$$
and is a negative quantity because heat is released when bonds are formed (the process is exothermic).
The enthalpy of formation of a $\ce{Cl-Cl}$ bond is the enthalpy change for the following process:
$$\ce{2Cl(g)-> Cl2(g)}$$
and therefore can be expressed as
$$\Delta H^\circ (\ce{Cl-Cl})=\Delta H^\circ_f (\ce{Cl-Cl(g)})-2\Delta H^\circ_f (\ce{Cl(g)})$$
and since $\Delta H^\circ_f (\ce{Cl2(g)})=0$,
$$\Delta H^\circ (\ce{Cl-Cl})=-2\Delta H^\circ_f (\ce{Cl(g)})=-D(\ce{Cl-Cl})$$
In the process
$$\ce{C(g) + 4 Cl (g)  -> CCl4 (g)}$$
four $\ce{C-Cl}$ bonds are formed, so the average heat of formation of one bond is
$$\Delta H^\circ(\ce{C-Cl}) = \frac{1}{4}\left(\Delta H^\circ_\mathrm{f}(\ce{CCl_4(g))}-\Delta H^\circ_\mathrm{f}(\ce{C(g)}) - 4 \Delta H^\circ_\mathrm{f}(\ce{Cl(g)}) \right) $$
which is a negative quantity (exothermic).
Note again that the corresponding bond dissociation enthalpy is then a positive quantity.

*Sources (from IUPAC Gold Book website):

PAC, 1990, 62, 2167. (Glossary of atmospheric chemistry terms (Recommendations 1990)) on page 2176.
PAC, 1994, 66, 1077. (Glossary of terms used in physical organic chemistry (IUPAC Recommendations 1994)) on page 1089.


text:
Is there a common name to refer to the groups 13 and 14?
The group 15, 16, and 17 are called the pnictogens, chalcogens, and halogens respectively. Is there a name for the groups 13 and 14 as well?

Heh. It seems that the name for group 13 and 14 elements is... drums adding the wait for the effect ...Carbon and Boron groups! This guy(s) seems to use all the other names for the groups, except that they don't use any names for C and B groups. Eh, maybe charcoalgens or something for the Carbon group?

It seems that the carbon group is known as either the tetrels or the crystallogens; while the boron group is known as triels or icosagens. These names are very rarely used.
Source: Wikipedia articles on Carbon family (also, crystallogen) and Boron family. Although other (seemingly) respectable sites use them.


text:
How to determine the bond order using the Lewis structure?
What are the steps to follow in order to determine which molecule has a higher bond order by drawing the Lewis structure? For example, with $\ce{CO}$ and $\ce{CO3^{2-}}$?

Determining the bond order starting from a Lewis structure is a task that can range from very easy to rather difficult. Luckily, most cases you will encounter are the easy ones. The first step should always be to draw out your molecules. For carbon monoxide and carbonate, this is what you should initially arrive at:

The second step is to check if you are missing any resonance structures. This is most obvious in carbonate whose three oxygens are all equal and whose bonds can be pushed around as shown below to give these additional two resonance structures:

However, also carbon monoxide can be drawn in a different resonance structure. These two resonance structures are obviously not equal since one equips all atoms with an octet while the second leaves carbon with a sextet:

Therefore, in a third step, you must weight the resonance structures. As stated above, the three of carbonate are all equal, we must consider them contributing the same to the overall structure. Those of carbon monoxide, however, are not equal; the triple-bonded structure will contribute far more than the double-bonded structure.
Depending on the level of your exam, you may be expected to completely ignore the lesser resonance structure (introductory level) or to assume weightings for both structures (advanced level) — or, anything in-between. Here, let us assume that the double-bonded resonance structure contributes almost nothing to the final result so we may ignore it (which is close enough to the truth).
Having done all this preparative work, we are now able to take a look at the actual bond whose bond order we want to determine. For each resonance structure:

Count the number of electron pairs in one specific bond

Multiply that with a weighting fraction

Sum up the values obtained in this way
$\displaystyle \text{B. O.} = \sum_i \frac{n_i(\ce{e-})}2 \times x_i$

For carbon monoxide, of which we have a single non-negligible resonance structure, we have 3 electron pairs in the resonance structure which contribute 1 (or the entirety) to the final structure so our bond order is 3.
For carbonate, we have three resonance structures that we must consider separately and we must multiply the individual structure’s electron pair count with $\frac13$ because each contributes a third to the overall picture. If we consider the upwards-pointing $\ce{C\bond{...}O}$ bond of the second figure, from the first resonance structure we have a contribution of $\frac23$, from the second we have $\frac13$ and from the third we also have $\frac13$. Adding these three values we have an overall bond order of $\mathbf{\frac43}$.

Finally, although this is already well into the advanced level I wish to point out why the weighting step is important. Consider the structure of a carboxylic ester as shown below. At first sight, this might look like ‘half a carbonate’, which would mean that both shown resonance structures are equal, contribute $0.5$ to the overall picture and therefore both bond orders would be $1.5$. However, this is only the case if we are talking about the carboxylate anion, not about the ester. In the ester’s case, as can be seen below, one of the resonance structures includes charge separation. Therefore, we must weight the non-charge separated resonance structure stronger than the charge separated one. We might choose weighting factors of $0.75$ and $0.25$ (without any additional data the choice of factors is entirely arbitrary!). This leads us to two different bond orders for the two $\ce{C-O}$ bonds, one being something like $1.75$ the other corresponding to $1.25$.

It is important to note that in the ester’s case we cannot ignore the contribution of the second, minor resonance structure since comparing the reactivity of an ester to that of a ketone shows a clear difference (the ester being less reactive, i.e. the double bond not being a full double bond). Likewise, replacing an ester e.g. with a thioester or selenoester further reduces the contribution of the second resonance structure closer to where it may safely be ignored again.

Your CO resonance structures look odd without the four more electrons O.o

Firstly draw lewis structure, then count the total number of bonds which is equal to $4$ here. Finally, count the number of bond groups between individual atoms, which is $3$.
The bond order will be equal to $\frac{4}{3}$.

And what about CO ?

Draw lewis structure and repeat the steps , its not that hard.


text:
Make pure phthalocyanines from phthalocyanine dyes
We need a pure copper phthalocyanine for our experiments. Of course, we can buy it from Sigma Aldrich, but this means we'll have to wait a while. In local area we can buy only dyes which are about 98% purity, which is not enough. Is there any fast and easy method to purify it in lab?

Did you consider sublimation in vacuum? This article: Journal of Materials Science, 1982, 17, 2781-2791 might give some directions.

We have no vacuum equipment in our lab.

Purifying phthalocyanines is a pain because even the substituted ones aren't very soluble in common solvents (e.g., for chromatography or recrystallization). I highly recommend sublimation. You don't need high vacuum. Do you have any way to provide reduced pressure?

Unfortunately, we don't have any vacuum equipment. But we have a furnace (not designed for vacuum).

I doubt You will get some higher purity from Sigma Aldrich. Do they garantee some purity?

Don't know. On the site it is written that "99%" and "sublimed grade", but vhat are the digits after comma is unknown. There are papers where this product was used.


text:
Silicon degassing with negative pressure
I am planning to use polydimethylsiloxane (PDMS) to create phantom for mimicking tissue for imaging. For this, I plan on using SYLGARD® 184, which is a viscous liquid. When adding other components to it, the air bubbles will be introduced. I have learned that one way (available to me) is using the negative pressure/ vacuum pump to remove bubbles / de-gas the solution. Does any one knows what negative pressure is sufficient to remove the air from the gel-like solution?


text:
GAMESS and Gaussian : what is the maximum efficient number of supported nodes?
I have looked at GAMESS and Gaussian manuals but can not find the maximum number of efficient nodes. As I tried GAMESS on our supercomputer, I can add as many CPUs as I want but it really does not Improve the calculation time and efficacy . On the other hand just adding maximum number of CPUs is just waste of CPU-Hour time and resources.
Is there any criteria for selection of best number of CPUs for optimum resource usage ?

as a general rule: no hyperthreading 'cores'. If 6+ cores/node, the last physical core usually may be left unused with little to no penalty to performance. However! when you request cores from two nodes, you add a tight bottleneck and latency for memory updates. For smaller systems it is not usually worth it (latency is so large that it imposes enough delays to consume all CPU time), and for systems with large memory use it is prudent to use every trick to reduce inter-node communications.

TL:DR benchmark it. There is no way to say for sure without benchmarking. And the answer is machine-dependent anyway.

@permeakra, sure, that is the only way. But I'm also pretty sure many of us already did benchmarks, so why not share and collect the results in one place?

@Wildcat The answer is machine-dependent. It depends not only on CPU used, but also on memory used (both latency and bandwidth) and network parameters (mostly bandwidth), amount of memory actually used  and hdd i/o.  Consequently, there is little point in sharing.

@permeakra The answer is not only machine-, but problem dependent, too. The size of the molecule, and what actual algorithms are used are very important details. Different parts of the software scales different way: it is no  the same if you integrate all the time or you are busy with diagonalization.

Is there any criteria for selection of best number of CPUs for optimum
  resource usage?

The criteria is basically the same as for choosing the level of theory: benchmarking. Parallel scaling depends on many factors: the level of theory, the size of the molecule, the machine architecture, etc., so there is no golden rule.
Though from my experience I can say that Gaussian 09 single point energy, geometry optimisation and frequencies calculations:

scale well for HF & DFT up to 16 physical cores;
scale well for MP2 up to 8 physical cores;
almost do not scale for CI & CC.

Also note that Gaussian 09 is about 2 times faster for almost any kind of calculations than the previous version (Gaussian 03) and doesn't suffer from some awkwardly strange memory effects (see e.g. here).

That's great !  What about GAMESS? Do you have experience on it?

@Aug, suddenly, I do not have a lot of experience with GAMESS, but for my systems (small organic anions as well as big organometallis complexes) when doing small number of DFT geometry optimizations & frequencies calculations I saw only small speed-up going from 16 to 32 cores. So, presumably, GAMESS is somewhere close to Gaussian.

@Aug, out of free packages, NWChem is usually advertised as extremely well parallelized. I have almost zero experience with it, but you can give it a try. ;-)


text:
When the intermolecular forces in a liquid increase, does the equilibrium vapor pressure of the liquid at a given temperature increase or decrease?
I assumed it would increase, but I think I am wrong. I know that when the forces are strong, the vapor pressure would decrease. Does this same principle apply to the IM forces increasing?

http://www.chem1.com/acad/webtext/states/changes.html

Yeah your hunch is right. Higher the intermolecular forces between the liquid particles, harder it is for it to escape into the vapor phase, ie., you need more energy to convert it from liquid to the vapor
phase, in other words, higher its boiling point.
If it is harder for a liquid to escape into vapor, because it is held back into the liquid by the neighboring particles attraction, you have less vapor and hence low vapor pressure.
So in nutshell, if the intermolecular forces are increased, the equilibrium vapor pressure of the liquid will decrease.


text:
Relative energies between reactants and products
I've been asked to draw a relative energy sketch of the reactants and the two products. The reaction being examined was adding copper acetate monohydrate to glycine in two different fashions to create the two isomers cis-bis(glycinato)copper(II) monohydrate and trans-bis(glycinato)copper(II) monohydrate. I found that the cis-bis(glycinato)copper(II) monohydrate was  thermodynamically stable and trans-bis(glycinato)copper(II) monohydrate was kinetically favored since when adding heat to trans-bis(glycinato)copper(II) monohydrate it changed to the same color as cis-bis(glycinato)copper(II) monohydrate therefore cis-bis(glycinato)copper(II) monohydrate has a lower energy relative to trans-bis(glycinato)copper(II) monohydrate. The part where I'm stuck is where the reactant energy is relative to these two. I know it must be lower than the intermediate trans-bis(glycinato)copper(II) monohydrate, but where the energy level of the reactants compared to the thermodynamically stable cis-bis(glycinato)copper(II) monohydrate has me stumped, whether it is higher or lower.
Cheers

That's interesting; I just did this experiment (I'm actually writing my lab report up now) and for me the trans-bis(glycinato)copper(II) monohydrate was the thermodynamically favoured product and the cis-bis(glycinato)copper(II) monohydrate the kinetically favoured. As far as the reactant energy levels are concerned, I attached a photo of a rough graph my lab instructor gave us that illustrates the energy levels for the two pathways to the final thermodynamically favoured product. I hope that helps!!


text:
What is the enthalpy change for the following equation?
The equation we need it to change to
\[\ce{FeO + CO ->Fe + CO2}\]
The three equations and delta heat:
\[
\begin{align*}
\ce{3Fe2O3 + CO &-> 2Fe3O4 + CO2} &\Delta H=-47.0\ \mathrm{kJ}\\
\ce{Fe2O3 + 3CO &-> 2Fe + 3CO2} &\Delta H=-25.0\ \mathrm{kJ}\\
\ce{Fe3O4 + CO &-> 3FeO + CO2} &\Delta H= 19.0\ \mathrm{kJ}
\end{align*}
\]
What is the enthalpy change and how to combine the 3 equations to make the above equation? (Please show work)
Thank you,
     Tiffany

Please show your efforts  into solving the problem. What have you tried that didn't get answers?

[…] how to combine the 3 equations […]

Equations is the keyword!
This is just an example on how to solve a system of linear equations, disguised as a chemistry problem.
If the calculation with $\ce{FeO}$, $\ce{CO}$, $\ce{Fe}$, $\ce{CO2}$, etc. feels strange, just translate them to $x_1$, $x_2$, $x_3$, $x_4$, etc.
The question for the enthalpy of the first reaction therefore is equivalent to finding the solution for
\[y = x_3 + x_4 - x_1 - x_2 \]
How to work in the other equations?
\[\ce{3Fe2O3 + CO -> 2Fe3O4 + CO2} \quad\quad\Delta H=-47.0\ \mathrm{kJ}\]
You have the reactants, the products and the enthalpy: translate that to a mathematical equation following the naming scheme introduced above:
\[2x_6 + x4 - x_2 - 3x_5 = -47.0\]
Do the same for the other two equations, rearrange and try to solve the equation system.


text:
Ionic equilibrium equation
What form is the ionic equalibrium equation? I have problem asking for it, but I'be done so many different equalibrium equations that I'm starting to get mixed up on which one is called what... Which one is it?

An ionic equilibrium equation is simply one that involves ions.
Example : $\ce {HA <=> H^+ + A^-}$


text:
Iodine color change in different complexes
Why does the color of iodine differ when it forms a charge transfer complex with benzene (pi donor) - pink and ethanol (electron pair donor) - brown? Does the HOMO-LUMO energy differ, if so how?

The formation of charge transfer complexes between iodine and aromatic hydrocarbons according to
\[\ce{Ar + I2 <=> Ar\cdot I2}\] is a long known phenomenon and has been described by Benesi and Hildebrand in A Spectrophotometric Investigation of the Interaction of Iodine with Aromatic Hydrocarbons. The following data for the absorption maxima of iodine in the VIS and UV range (wavelength in nm) in different solvents are taken from that article.
\begin{array}{lrrr}
\mathbf{solvent} & \lambda_{max}(\textrm{VIS}) & \lambda_{max}(\textrm{UV})\\
\hline
\textrm{benzene} & 500 & 297\\
\textrm{toluene} & 497 & 306\\
\textrm{p-xylene} & 495 & 315\\
\textrm{mesitylene} & 490 & 333\\
\hline
\textrm{trifluormethylbenzene} & 512\\
\textrm{n-heptane} & 520 & \\
\end{array}
Note that the effect is stronger in the UV than in the VIS range and the shift correlates with the (combined) $+I$ effects of the methyl groups, while a $\ce{CF3}$ substituent disfavours then formation of a charge transfer complex.
Ethanol, on the other hand, is a protic and polar solvent, nice to stabilize charged species.
One might argue that under these conditions, other species than in the case of aromatic solvents exist - the solvation of triiodide $\ce{I3-}$
 is certainly possible here.
In addition to the coordination of iodine by ethanol - we cannot exclude interaction with several molecules of ethanol:
\[\ce{n\ EtOH +I2 <=> I2\cdot(EtOH)_{n}}\]
it is conceivable that disproportionation of iodine and solvation of the resulting charged species might play a role here:
$$\ce{ I2\cdot(EtOH)_{n} + I2 <=>[\ce{EtOH}] I+\cdot(EtOH)_{n} + I3-} $$

So, the triodide causes the brown color in ethanol. Would the situation be the same with a different electorn pair donor?

There are very many charge transfer complexes and not just involving iodine. They have been know for at least 100 years. In normal (photon) absorption to form an excited state and so give a solution a colour, an electron is promoted from a bonding to non-bonding orbital in the chromophore molecule.
In charge transfer complexes two species are involved, D & A and are close to one another, effectively touching, say benzene and tetracyanoethylene. The lowest excited state now is one in which an electron is transferred from one of the pair to the other during the absorption process. Thus the lowest state above the ground state is the pair D$^+$A$^-$. The charge transfer absorption is the colour produced in this transition.


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Converting 26.62 NOx µg/L into mM (millimolars)?
I have to convert  a whole lot of nutrient data : $\ce{NH_4}$, $\ce{PO_4}$ and $\ce{NO_x}$ data that is in µg/L into mM (millimolars).
Could someone give me an example for 26.62 $\ce{NO_x}$ µg/L? or 7.73 $\ce{PO_4}$ µg/L? or 14.39 $\ce{NH_4}$ µg/L?
I have a large data set in Excel to convert and send to someone and want to get it right. Any help would be really appreciated.

I would make it easy for yourself and convert your µg/L to g/L. You can then divide it by the formula weight to obtain the molarity (in M). After this it is just a simple conversion to mM.
I suppose in Excel you can create a new column and use a formula something like this:
=((CELL/106)/Formula weight)*1000
So for your example of phosphate:
=((0.00000773)/94.97)*1000
=0.00008139 mM // 8.139x10-5 mM

The hard part is what formula weight to use for the NOx.  I suspect that the "μg" of NOx really means of μg of nitrogen as NOx.  If I'm right, then the NOx formula weight is 14, but this would be an important detail to confirm about the OPs data set.

Thanks Curt, I did what you said and sent the spreadsheet away to have rechecked and all was spot on. The molar mass NOx was 14, PO4 was 31 and NH4 was 14 so I guess they're not taking the 'O' mass into account? Thanks so much for your help. Paul.


text:
NMR: Magnetic and chemical equivalence in benzene
According to my textbook, all the protons in $\ce{^{12}C_6{}^{1}H_6}$ are magnetically equivalent, but those in $\ce{^{13}C_6{}^{1}H_6}$ are not.
I understand that for two nuclei to be magnetically equivalent, they need to have equivalent coupling to all other nuclei in the molecule. I do not understand this with respect to benzene as I expect Hc to couple differently to Ha than Hb to Ha and thus for them to be magnetically inequivalent irrespective of the isotope.

All of the protons in $\ce{^{12}C_6{}^{1}H_6}$ benzene are equivalent by symmetry, therefore they are chemical shift equivalent.
You wrote, "I understand that for two nuclei to be magnetically equivalent, they need to have equivalent coupling to all other nuclei in the molecule."
I think the correct statement would be, for two nuclei to be magnetically equivalent, they need to have equivalent coupling to all other non-chemical shift equivalent nuclei in the molecule.  If there are no other non-chemical shift equivalent nuclei in the molecule, then the chemical shift equivalent nuclei are also magnetically equivalent.
In the case of $\ce{^{12}C_6{}^{1}H_6}$ benzene, all of the protons are equivalent by symmetry, therefore they are chemically equivalent and since there are no "different" nuclei (non-chemical shift equivalent nuclei) for them to couple with, they are also magnetically equivalent.
In the case of $\ce{^{13}C_6{}^{1}H_6}$ benzene all of the protons are equivalent by symmetry so they are all chemical shift equivalent.  Since all of the carbons are $\ce{^{13}C}$ (spin=1/2), there are now "different" (the $\ce{^{13}C}$ nuclei are non-chemical shift equivalent) nuclei for the chemical shift equivalent protons to couple with; therefore the protons will be magnetically non-equivalent.  The spectrum is now an AA'A'A"A"A"'XX'X'X"X"X"' (A=proton, X=carbon-13) spectrum.  Since the protons are not magnetically equivalent (at least some of) the couplings will be observable.

In the case of 13-C benzene, the protons will couple to the spin 1/2 13-C nuclei surely ?

Yes, that's why I mentioned the sidebands, but I was thinking natural abundance, I've edited the answer accordingly.


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Why are transition metal hydroxides insoluble?
From what I understand, adding $\ce{NaOH}$ to a solution containing transition metal ions gives transition metal hydroxide precipitates. Why is $\ce{NaOH}$ soluble but transition metal hydroxides insoluble? I have a feeling it has something to do with ligands and complexes formed, which I am familiar with.

Why is Al(OH)3 (and some more hydroxides from main groups) insoluble?

The rule of thumb for hydroxide salts is actually that all hydroxides are insoluble save those of the alkali metals and the heavier alkaline earth metals (namely $\ce{Ca^{2+}}$, $\ce{Sr}^{2+}$, and $\ce{Ba}^{2+}$). There are many resources you can look up to check the solubility rules of certain ions, this one, for example.
Before researching this, I suspected that the solubility of the salts with a certain anion might be determined by the cation's control over its valence electrons, which is measurable through ionization energy. Sure enough, when you compare the first ionization energies of various metals, you find that those with the lowest first ionization energies are those that are soluble hydroxides.
Alkali Metals
The salts of all alkali metals are completely soluble because they have relatively little control over the outermost electron orbiting them, in addition to this being the only electron they must lose in order to attain an octet structure.
Alkaline Earth Metals
For the alkaline earth metals, the complete outermost $\mathrm{s}$ orbital must be lost in order to attain an octet structure. Because the orbital is complete, and among other reasons, the first ionization will require more energy than the alkali metal in the same period. In addition to this, the second electron must then also be lost. This means that overall, it requires much more energy to ionize this group than the alkali metals. This explains the insolubility of $\ce{Be(OH)2}$ and $\ce{Mg(OH)2}$.
The solubility of the hydroxides of the metals lower in the alkaline earth metal group can be explained by the shielding effect, which effectively says that the ability of a positive nucleus to attract electrons decreases as the amount of orbitals surrounding the nucleus decrease.
Transition Metals
The insolubility of the hydroxides of the transition metals can be explained both by the increasingly large ionization energies that result from successively removing electrons (or negatively charged hydroxide molecules in this case) from the metal cation, and the greater effective nuclear charge on the electrons across a period.

there is quite a set of transition metal oxides soluble in water.

Solubility depends on complex interactions between many factors, but ionization and shielding aren't amongst them. IE and shielding have to do with how easily or likely you form the ionic compound at first place, and not how an ionic compound behaves after you make it.
In general solubility depends on......
i) lattice enthalpy of the ionic compound
ii) solution enthalpy of the ionic compound
iii) entropy (disorderedness) changes of both the solvent and the solute
In general group 1 metal compounds have low lattice enthalpies, because the ionic bonds formed by 1+ ions are relatively weak.
Also, for dissolving ionic compounds in general, solvent molecules will experience a decrease in entropy as solvation shells of eg water molecules surround the ions in a particular orientation in order to form ion-dipole attractions. This effect is not too significant for group 1 ionic compounds.
I'm not saying the above gives a complete story, but both factors above can both help explain the relatively high solubility of group-1 ionic compounds in general.

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Anode and Cathode Understanding
I am very much confused: in the cell (Zn, Cu in $\ce {H2SO4}$), Cu is positively charged and is anode, where in this cell (Zn in $\ce {ZnSO4}$ and Cu in $\ce {CuSO4}$), Cu is positively charged but is Cathode. So how can I define it, where both are electric or voltaic cell. Please help me. Please see these picture for better understanding.

reference:A Text Book Of Electrical Technology,Volume 1,B.L.Theraja

A related question: http://chemistry.stackexchange.com/questions/26300/hydrogen-fuel-cell-why-do-the-h-ions-move-through-the-electrolyte

The copper electrode is never the anode (in this cell). It is mislabeled if it is called an anode. The copper ions are reduced at the cathode. The cathode has a negative potential in relation to the anode.

Here Cu plate is cathode and Zn plate is anode, it's written wrong in it. In the first case (Zn in  $\ce{ZnSO4}$ and Cu in $\ce{CuSO4}$) the electrodes are connected through an external circuit (You see that bulb glowing?). But what this article is trying to say is that in the cell (Zn, Cu in $\ce{H2SO4}$), if you leave the solution as it is without connecting externally, Zn having higher tendency to lose electrons, will lose electrons and these will appear on its plate surface making it negatively charged, the $\ce{Zn^2+}$ thus formed will combine with $\ce{ SO4^2-}$ and form $\ce{ZnSO4}$. The remaining two $\ce{H^+}$ ions will more towards copper plate where they will combine with the electrons on its plate and make it positive charged, releasing $\ce{H2}$. Now if a wire is connected across this setup then the electrons on Zn surface will leave it and go towards positive charged Cu plate.