text:
Why are there drastic changes sometimes in radius of isoelectronic ions?
$\ce{S^{2-}}$, $\ce{Cl-}$, $\ce{K+}$, and $\ce{Ca^{2+}}$ each have the same number of electrons. Their effective ionic radii are 182, 181, 138, and 100 picometers, respectively, and the number of protons for each are 16, 17, 19, and 20. My question is, why does the effective ionic radius drastically go from 181 to 138 picometers between chlorine and potassium? Is it really necessarily because they are in different periods or is there something that I am missing here?

You're missing argon.

@Mithoron $\ce{Ar}^0$ isn't an ion.

Obviously it isn't but it has at. num. 18, not K and difference between  Cl and K is 2.

@Mithoron Sure - I'm pretty sure OP's $Z$'s were erroneous. See my edit.

Yeah, his confusion might have partially stem from this.

The whole concept of ionic radius is not sharply defined one. Note, that the same is true for the notion of atomic radius, as well as for any other kind of radius or any other notion of size. In general, the notion of size of an object looses its usual (i.e. classical) meaning in the microscopic realm since objects there do not have well-defined physical boundaries. So from the very beginning one should not expect a perfect trend in a not so well defined property.
Anyway, values of the ionic radii are based on crystallographic data, but what is actually determined in X-ray crystallography is the distance between two ions, not their radii. Then it is basically up to you how to divide this distance into the radii of ions in question. But whatever way the division is done, the ionic radius of an ions is not a fixed, perfectly defined property, since for the same ion, the radius can differ in different crystal lattices due certain variables such as coordination number and electron spin.
For instance, you mentioned the ionic radius of 138 pm for $\ce{K+}$. But this value is for 6-coordinate compounds only while for 8- and 12-coordinate compounds of $\ce{K+}$ the ionic radius is 151 pm and 160 pm respectively. Same story for $\ce{Ca^2+}$: 100 pm, 112 pm, and 135 pm for 6-, 8-, and 12-coordinate compounds respectively.1
Interestingly, for anions there is a little variation in the ionic radius depending on coordination number, so that for many of them only one value independent of coordination number is specified. To be honest I could not come up with a quick explanation why, so it feels like I have to dig into the matter deeper. But I'm quite reluctant to do so, since, as I said in the beginning, the whole concept of ionic radius looks suspicious, so I expect to find just some obscure rule arriving out of the blue which will explain this little variation in the anionic radius relatively to cationic ones.

Slater's rules to the rescue!
Now, I guess, I know which rule can be used to explain this drastic change in ionic radii. Let us examine the application of the good-ol' Slater's rules to the matter. These rules allows us to calculate the effective nuclear charge  $Z_{\mathrm{eff}}$, which is the net positive charge experienced by an electron in a multi-electron atom.
$$
    Z_{\mathrm{eff}} = Z - s \, ,
$$
where $s$ is the screening constant. Slater's rules provide a way to calculate the values for the screening constants, and consequently, to obtain the values of effective nuclear charges.
All the ions $\ce{S^2-}$, $\ce{Cl-}$, $\ce{K+}$, $\ce{Ca^2+}$ are isoelectronic: the electron configuration of them is that of $\ce{Ar}$, i.e. $\mathrm{[Ne] 3s^2 3p^6}$. Thus, we have to compare effective nuclear charges for $\mathrm{3p}$ electrons in these ions. And as a first approximation we could compare effective nuclear charges for $\mathrm{3p}$ electrons in the corresponding atoms. Now without going into the details (listing the actual Slater's rules & applying them) I will just leave here the final numbers, effective nuclear charges for $\mathrm{3p}$ electrons for the corresponding atoms:
$$
\begin{array}{c|cccc}
     & \ce{S} & \ce{Cl} & \ce{K} & \ce{Ca} \\
\hline
Z & 16 & 17 & 19 & 20 \\
Z_{\mathrm{eff}}(\mathrm{3p}) & 5.482 & 6.116 & 7.726 & 8.658 \\
r_{\mathrm{ion}} & 182 & 181 & 138 & 100 \\
\end{array}
$$
Now you can see that despite a relatively small change in the nuclear charge going from $\ce{Cl}$ to $\ce{K}$
$$
   (19-17)/17 * 100 \% \approx 11.8 \% \, ,
$$
there is a rather drastic change in effective nuclear charge
$$
   (7.726-6.116)/6.116 * 100 \% \approx 26.3 \% \, .
$$
And the drastic change in ionic radius is exactly in line with this drastic change in effective nuclear charge
$$
   (138-181)/181 * 100 \% \approx -23.8 \% \, .
$$
To summarize: the effective nuclear charge for $\mathrm{3p}$ electrons increases by about 26.3% when going from $\ce{Cl}$ to $\ce{K}$ and the ionic radius accordingly decreases by 23.8%.
Note that we used effective nuclear charges of the atoms just as the first approximation for that of the ions. For anions an additional electron will occupy an outer orbital and there will be an increased electron-electron repulsion. This will push the electrons further apart and increase the shielding, so that $Z_{\mathrm{eff}}$ for anions will be smaller than for the corresponding atoms. For cations the effect of the removal of an electron is the opposite: when an electron is lost, electron-electron repulsion and therefore, shielding decreases, so that $Z_{\mathrm{eff}}$ for cations will be bigger than for the corresponding atoms. It would not change the above mentioned trend though; in fact, it will make it even more prominent.

1 Ionic radii values are taken from here.

I completely agree with Wildcat's answer.  Excellent stuff. This part, though, got me thinking:

Anyway, values of the ionic radii are based on crystallographic data, but what is actually determined in X-ray crystallography is the distance between two ions, not their radii. Then it is basically up to you how to divide this distance into the radii of ions in question. ...

In particular, I started to wonder: what might quantum chemistry have to say about ionic radii of isoelectronic species, for isolated ions in vacuum away from the influence of crystal packing and counter-ions?
Calculations
To try to answer this, I computed $\ce{S^{2-}}$ through $\ce{Ca^{2+}}$, including $\ce{Ar^0}$, at the RHF/def2-QZVPP level of theory, with the minimal diffuse functions of Zheng et al. added, in ORCA v3.0. I also computed the correlation energies with fully-correlated orbital-optimized CCSD(T), to confirm that the RHF orbitals were a reasonable representation of the systems$\dagger$. As an example, the input I used for neutral argon was$^\ddagger$:
! RHF CCSD(T) ma-def2-QZVPP NOFROZENCORE
! TIGHTSCF
! PAL2

%mdci
   DenMat OrbOpt
   end

%output
   Print[ P_OrbPopMO_L ] 1
   end

* xyz 0 1
  Ar 0 0 0
*

With these data in hand, I then calculated the radial distribution functions (RDFs) of the wavefunctions using MultiWFN. For completeness, the radial distribution function $R(r)$ is defined as:
$$
R(r) = 4\pi r^2 \left|\Psi(r)\right|^2
$$
It bears units of electrons per Angstrom, and represents the number of electrons one can expect to see, on average, within a spherical shell of inner radius $r$ and differential thickness $dr$.  The integral $\int_0^\infty{R(r)dr}$ theoretically should evaluate to the number of electrons in the system (here, $18$), and serves as a quality check on the computed $R(r)$.
RDF Results
The RDF all-space integrals for all five of the systems each fell short of the target value of $18$ by less than $0.06$ electrons $(0.33\%)$, with those for all but $\ce{S^{2-}}$ falling short less than $0.01$ electrons $(0.056\%)$.  Seems pretty solid to me!
The following is a plot of $R(r)$ for the five systems:

For a better view of those inner-shell peaks, the following is the same plot but with a logarithmic $x$-axis$^*$:

The trend of the positions of the local maxima in the RDF$^\#$ is the same for all three peaks: a gradual shifting to larger distances from the nucleus as you go from $\ce{Ca^{2+}}$ to $\ce{S^{2-}}$. Looking at the second figure, it almost seems that the shift is log-linear with $Z$, but a more refined calculation would be necessary to investigate the behavior in detail. At an initial glance, though, this doesn't appear to be a trend quite as dramatic as that noted in the question.
A more subtle feature of the curves (for me, anyways, on my first look at them) is the substantial decrease in the height of each peak as $Z$ is decreased from $Z=20$ for $\ce{Ca^{2+}}$ down to $Z=16$ for $\ce{S^{2-}}$. At the same time, the 'tail' of the distribution shifts substantially outward, especially for the anions $\ce{Cl-}$ and $\ce{S^{2-}}$.  Zooming in the $y$-axis on the plot with the linear $x$-axis confirms that the electron distributions of the anions indeed are quite diffuse (I'll explain the dashed line in a minute):

Ok, fine, but how to quantify the ionic radii from this data?
RDFs and Ionic/van der Waals Radii
The way I ended up going about it was to look at a proxy for the ionic/van der Waals radii of the various ions, in the form of a cutoff value of the RDF function$^\&$.  In other words, what value of $R(r)$ might correspond, if roughly, to these radii?
As the neutral species, argon is the most reliable reference point, since it's really hard to obtain a mass of isolated cations or anions to take measurements from.  WebElements reports argon's vdW radius as $1.88\ \mathring{\mathrm{A}}^{**}$.  Looking at the RDF for $\ce{Ar^0}$, the value of $R(1.88\ \mathring{\mathrm{A}})$ is approximately $0.23\ \frac{e^-}{\mathring{\mathrm{A}}}$.  The radial distances at which $R(r)$ falls to this value for the five species are:
$$
\begin{array}{|c|c|c|}
\hline
\mathrm{\bf{Species}} & \mathbf{r\ @\ R=0.23\ \frac{e^-}{\mathring{\mathrm{A}}}} & \mathrm{\bf{Lit.\ Ionic\ Radius}} \\
\hline
\ce{Ca^{2+}} & 1.44\ \mathring{\mathrm{A}} & 1.00\ \mathring{\mathrm{A}} \\
\hline
\ce{K^{+}} & 1.62\ \mathring{\mathrm{A}}  & 1.38\ \mathring{\mathrm{A}} \\
\hline
\mathbf{\ce{Ar^0}} & \mathbf{1.88\ \mathring{\mathrm{A}}}  & \mathbf{-} \\
\hline
\ce{Cl^{-}} & 2.27\ \mathring{\mathrm{A}} & 1.81\ \mathring{\mathrm{A}}  \\
\hline
\ce{S^{2-}} & 3.08\ \mathring{\mathrm{A}} & 1.82\ \mathring{\mathrm{A}}  \\
\hline
\end{array}
$$
So... my estimates don't compare terribly well to the ionic radius values.  But, recall, my calculations were for isolated ions in vacuum, not in the context of a crystal structure. So, it makes sense that the experimental ionic radii, however determined, would be smaller than those determined in vacuo due to packing effects in the crystal structure.  Further, it also makes sense that the anions would be "squashed down" more when packed into a crystal structure as compared to cations, because the cations are already much more compact than the anions even without external 'constraints.'
Conclusion
Isoelectronic anions can get bigger much faster per-proton-removed than can  isoelectronic cations.  Just how much faster they get bigger depends on the size metric you choose, and in what chemical context and by what method you measure (or compute) that size metric.

$^\dagger$ Aside from a slight shift to high-angular-momentum functions, the results were quite similar. T1 diagnostics were all well below the 0.02 warning threshold.
$^\ddagger$ NOFROZENCORE correlates all electrons. TIGHTSCF requests more stringent convergence criteria. PAL2 makes ORCA run on two cores. %mdci DenMat OrbOpt end enables orbital optimization. The P_OrbPopMO_L part requests a full printout of the final orbital compositions for inspection.
$^*$ Note that the spacing of data points near the nucleus is quite sparse for all systems, and so the curves are only as smooth as they are because I plotted with Excel's 'smoothed line' feature. If this were an actual publication, I would have worked the data up with a much denser grid in the core region.
$^\#$ While it would be tempting to associate the two sets of peaks squished up against the $y$-axis with the $n=1$ and $n=2$ shells, and the broader, smaller set in the vicinity of $0.5\ \mathring{\textrm{A}}\leq r\leq 0.8\ \mathring{\textrm{A}}$ with the $n=3$ shell, in reality electrons from all three shells $1\leq n\leq 3$ contribute to the inner peaks, only those from $n=2$ and $n=3$ contribute to the middle peaks, and only those from $n=3$ contribute to the outermost peaks.
$^\&$ I initially started out looking at the radial integral function of the RDF, $I(r)=\int_0^r{R(x)dx}$, but realized the plain RDF gave better results.
$^{**}$ Of course, there are multiple ways of determining/defining the vdW radius, but let's take that used for WebElements's data as representative.

Very detailed answer: I would double thumbs up this if I could!


text:
Is there an adhesive with an on-and-off switch?
Is there a strong adhesive (can carry about 2 kg in the air) that can be switched on or off, for example with electricity or changing the acidity of the surrounding medium, though I prefer electricity if possible. I have found a few articles on the internet but can't verify them. This is one of them.

I'd suggest looking into research such as this through publications, and not articles written by tech websites. Try looking at ScienceDirect, Journal of American Chemicals Society, and Nature.

source: Stack Exchange

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

sort: votes


text:
Convert grams of water to steam
I'm working through problems in my textbook. How much heat is required to convert 2.55 g of $\ce{H2O}$ at 28 °C to steam at 100 °C?
In simpler problems I've used the fact that $\Delta H_\text{fus} = 6.02\ \mathrm{kJ/mol}$ for $\ce{H2O}$ at a melting point of 0 °C to do these kinds of calculations. But here the temperatures are different. How can I approach this problem? Is there some other conversion factor I should be aware of?

You need the vaporization $\Delta H_\text{vap}$ and you need to use this equation: $$Q = m~c~\Delta T$$
Where $Q$ equals the energy to heat $1\ \mathrm{mol}$ of water up to $\mathrm{100~^\circ C}$. $m$ is the mass of water, $c$ is the specific heat capacity and $\Delta T$ is the difference between the initial and final temperature.
In this stage you have the energy that is used to heat the liquid water up to $\mathrm{100~^\circ C}$ but the water is still liquid and you need steam water.
So you need to calculate the energy for phase transition from liquid to steam and the equation for this is: $$\Delta H_\text{vap} \cdot n = Q$$
$Q$ is the necessary energy for the phase transition, $n$ is the amount of substance and $\Delta H_\text{vap}$ is the energy used for phase change of $1\ \mathrm{mol}$ of the substance.
You also need to convert the mass to amount of substance (or use latent steam heat instead) and to calculate the $Q$ in the two stages.
This is the energy used for heat and change the phase.

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


text:
How to define a new solvent in Gaussian09 to run SCRF?
I want to run computation of my system in sulphuric acid with Gaussian 09; however sulphuric acid is not among the built-in solvents in G09. I know I have to give some parameters for the new solvent I define (including static and dynamic dielectric constants). Here's an example of my input files where I try to redefine a solvent water2.
#p b3lyp/6-31G(2df,p) int(grid=ultrafine)  SCRF(Solvent=Water2,Read)

Comment

0 1

...

stoichiometry=H2O1
solventname=Water2
eps=78.3553
epsinf=1.77556

G09 keeps telling me that "SCRF(Solvent=Water2,Read)" is not correct. I want to know what the correct input will be.

try it with SCRF=(Solvent=Generic,Read), that worked for me

The correct input should be more like (replacing your Water2 with Generic)
#p b3lyp/6-31G(2df,p) int(grid=ultrafine) SCRF(Solvent=Generic,Read)

Comment

0 1
...

stoichiometry=H2O1
solventname=Water2
eps=78.3553
epsinf=1.77556

as it is stated in the Documentation under Additional Input for PCM Calculations.

Note - this only gets your calculation running. Your shown current input section might be insufficient to properly define sulphuric acid.
As Geoff noted, the Gaussian Documentary states an important fact short before showing some $\varepsilon$-values of common solvents that should be taken into account.

We list the ε values here for convenience, but be aware it is only one of many internal parameters used to define solvents. Thus, simply changing the ε value will not define a new solvent properly.

That's probably not sufficient. As described further along in the page you link, it says " We list the $\epsilon$ values here for convenience, but be aware it is only one of many internal parameters used to define solvents. Thus, simply changing the $\epsilon$ value will not define a new solvent properly."

Thanks! I have tested and I've found stoichiometry, static and dynamic dielectric constants are the only parameters you need to run a default SCRF calculation. If you use Solvent=formamide in the beginning and you replace the three parameters with those of water's. you will get exactly the same result with using Solvent=Water.


text:
Pubchem, InChI, SMILES, and uniqueness

PubChem compound 6140 is L-phenylalanine in its neutral (not zwitterionic) form.  According this PubChem, this molecule has the following SMILES and InChI indentifiers:

SMILES: C1=CC=C(C=C1)CC(C(=O)O)N
InChI: InChI=1S/C9H11NO2/c10-8(9(11)12)6-7-4-2-1-3-5-7/h1-5,8H,6,10H2,(H,11,12)/t8-/m0/s1

PubChem compound 6925665 is the zwitterionic form of L-phenylalanine (protonated amine and deprotonated carboxylate).  Pubchem has decided that this species should be called "(2S)-2-azaniumyl-3-phenylpropanoate".  The SMILES and InChI identifiers are:

SMILES: C1=CC=C(C=C1)CC(C(=O)[O-])[NH3+]
Inchi: InChI=1S/C9H11NO2/c10-8(9(11)12)6-7-4-2-1-3-5-7/h1-5,8H,6,10H2,(H,11,12)/t8-/m0/s1

Confusion.  My confusion is why these different compounds have different PubChem entries (CIDs), and different SMILES identifiers...but the same InChI structure.  The different SMILES identifiers each appear to reflect the respective structures displayed by PubChem, but the single InChI identifier given for both compounds seems to reflect only the neutral form.  I even put the InChI into rdkit and converted it back to InChI.  The result was the same and rdkit interpreted this InChI as the neutral (not zwitterionic) species.  What is the reason for this discrepancy between InChI duplicity yet structural uniqueness?

Here are some possibilities:

PubChem is in error.  They should change the InChI for the zwitterionic compound.  (If so , to what?)
PubChem is right but is using definitions of compound and CID that are different than mine.
I have some kind of fundamental misunderstanding of the purpose of InChI, which I had thought would uniquely specify a molecular structure.  But InChI is designed to handle ambiguities like zwitterion vs. neutral.
rdkit interprets InChIs weird.
Something else.

I know this discussion is long gone, but... I've faced a similar struggle. I needed to cluster all compounds having identical first three layers in standard InChI. I made a perl script to create these clusters. In case anyone is interested I can share it.

Please do share!  Also I think your answer will be deleted soon because it is not really an answer.  If you can comment (not answer)  to an existing answer or the original question with a link to the script, thank you!

Unfortunately Pubchem is right, the two structures have the same InChI string and key, since the protonation state is the same in the zwitterion and the neutral form. So the reason for the discrepancy is by design.
I also always thought, InChI was designed for distinguishing between these conformations, but it turns out to just be one of the limitations of the system. The issue is addressed in section 13.2 of the technical FAQ of the InChI trust:

The different protonation states of the same compound will have InChIKeys differing only by the protonation indicator (unless both states have a number of inserted/removed protons greater than 12; in this case the protonation flag will also be the same, ‘A’).
  This is exemplified below by standard InChIKeys as well as standard InChI strings for neutral, zwitterionic, anionic and cationic states of glycine (note that neutral and zwitterionic states do not differ in the total number of protons so they have the same standard InChI/InChIKey):

Great answer.  I had no idea that InChI was designed this way.

@Curt Yeah, i thought the same thing as you and to be frank it does not make any sense to me why it should be that way. Maybe they address this in the second version, but knowing IUPAC they'll take their time *sigh*.

Based on your link, it seems like specifying zwitterions and some tautomers might be possible with a non-standard InChI (InChI=1/... instead of InChI=1S/... that includes a fixed-hydrogen layer...if I figure it out I might post a separate answer, but I never would have known if not for your link so thanks again.

@CurtF. Although the FAQ is still the same, there is a new version of InChI available. However, I doubt they address this issue, but you still might be interested.

InChI is intended to ignore tautomeric forms. As Martin indicates this also means zwitterions are considered identical to the neutral form.
Unlike you and Martin, I'm not sure I see this as a bug, since predicting the most stable tautomer or zwitterion/neutral is a complicated issue.
If you want to keep track of zwitterions, I think SMILES is a better format, since you can specify exactly what you want as far as explicit hydrogens and charges. You'll need to stick to a particular toolkit to create a canonical ordering.
Moreover, there's a complicated relationship between CIDs and InChI / InChI keys.
There are other cases where PubChem will have separate records for compounds that might be "the same" under InChI.

Axial or non-traditional stereochemistry. For example hexihelicene (CID=98863) should have two enantiomers, but the InChI reflects no stereochemistry.
"Extra" stereo centers. PubChem takes depositions in the SD file format, which allows 2D stereochemistry in wedge/hash notation. If you take the actual 3D geometry, you might realize the wedge/hash were used for appearance, not for indicating a stereo center (i.e., multiple CIDs generate the same InChI from the full 3D molecule).

There are also cases where PubChem indicates an InChI key computed from the 2D depiction in the SD file, but there are missing or undefined stereo centers.
So I'd say that because of incomplete stereochemistry and/or inconsistencies in representations, PubChem CIDs will not always match up with "structural uniqueness" and this is by design, both for PubChem and InChI.
The moral of the story, frankly, is that chemistry is complicated and coming up with "perfect" unique identifiers is incredibly hard.

Thanks for the suggestions on SMILES and the discussion of PubChem / InChI is very illuminating.  On whether Inchis treatment of zwitterions/tautomers is a bug or a feature, I agree that predicting the "most stable" one is very complicated, but how is stability related to representing InChI?  What would the rate of interconversion between gaseous CID 6140 and CID 6925665  be in a near vacuum at say 100 or 200 Kelvins?  My guess would be very low, which is why I think of them as separate molecules.  So to me separate InChIs would be good.

@CurtF. Your comment indicates gaseous compounds.. Keep in mind that most organic chemistry is done in solution and what's the rate of interconversion between these compounds in water or another protic solvent? As I said.. this is why it's incredibly hard to come up with "perfect" chemical identifiers.

Oh yeah I don't think anyone is after "perfect".  Just better.  Or at least more consistent.

Martin's answer led me to discover an important extension of InChI that does allow for specification of some tautomer and zwitterion identification.

InChi identifiers that begin with InChI=1S/... are standard InChI.  In standard InChI, the InChI identifier "must be the same for any arrangement of mobile hydrogen atoms", with the quote from Section 6 of the Technical FAQ of the Inchi Trust.
However, non-standard InChIs are also possible.  These start with InChI=1/....  Note the missing S.  In non-standard InChI, there can be an extra layer beginning with /f that is called the fixed hydrogen layer.
Playing around with rdkit (via its Python API) I was able to produce a non-standard InChI that I think corresponds to Pubchem compound 6925665.

from rdkit import Chem
zwitterion_phe_smiles = 'C1=CC=C(C=C1)CC(C(=O)[O-])[NH3+]'
zwitterion_phe_mol = Chem.MolFromSmiles(zwitterion_phe_smiles)

# produces "standard" InChI so not explicitly zwitterionic
print Chem.MolToInchi(zwitterion_phe_mol)

     InChI=1S/C9H11NO2/c10-8(9(11)12)6-7-4-2-1-3-5-7/h1-5,8H,6,10H2,(H,11,12)

# produces "non-standard" InChI with fixed-H layer so zwitterion can be IDed
print Chem.MolToInchi(zwitterion_phe_mol, options='/FixedH')

     InChI=1/C9H11NO2/c10-8(9(11)12)6-7-4-2-1-3-5-7/h1-5,8H,6,10H2,(H,11,12)/f/h10H

# going from respective InChIs to SMILES
## The standard InChI produces neutral SMILES
zwitterion_nonstandard_inchi = 'InChI=1/C9H11NO2/c10-8(9(11)12)6-7-4-2-1-3-5-7/h1-5,8H,6,10H2,(H,11,12)/f/h10H'
standard_inchi = 'InChI=1S/C9H11NO2/c10-8(9(11)12)6-7-4-2-1-3-5-7/h1-5,8H,6,10H2,(H,11,12)'

print Chem.MolToSmiles(Chem.MolFromInchi(zwitterion_nonstandard_inchi))
print Chem.MolToSmiles(Chem.MolFromInchi(standard_inchi))

     [NH3+]C(Cc1ccccc1)C(=O)[O-]
     NC(Cc1ccccc1)C(=O)O

So for zwitterions tautomeric InChI's are possible.  Worryingly, however, some kinds of tautomerism are not handled even by non-standard InChI, again quoting from section 6 of the FAW:

In its current state, InChI recognizes the most common form of H migration (for the full list, see Table 6, Section IVb of the InChI Technical Manual). However, several ways of tautomeric migration that are not supported by default may appear important for some chemists. In particular, these are keto-enol and long-range tautomerisms.

Why Pubchem chose to list standard InChI instead of non-standard InChI is not entirely clear to me.  I suppose it is difficult to figure out programmatically when non-standard InChI would be required.  Ideally I suppose Pubchem would have both standard and non-standard InChIs for each of their molecules, but I'm not sure when/if they will ever make that change.

Basically you're not supposed to use non-standard InChIs in a public-facing service. A standardized InChI is what it is - but you'd need to know the exact flags passed to the code to understand a non-standard one.

PubChem clearly has different ideas about what a "compound" is than standard InChI does, or else the different CIDs wouldn't exist in the first place.  To me using the same InChI for different CIDs obscures this fact.  Plus, there is a widespread conception, which I now realize to be very wrong, that "there is a 1:1 correspondence between every organic chemical structure and a single InChI".  Maybe that's the real issue.

I don't think PubChem obscures the difference. If you search by InChI or InChI key, you'll get multiple CIDs. I agree that many people don't realize that InChI is imperfect and there is not a 1:1 correspondence between a chemical structure and an InChI.

I do think that many people using PubChem don't realize its limitations (e.g., as listed in my answer). That included myself when we started to build PQR.


text:
What causes the difference in heat energy released by combustion for the butanol isomers?
I am currently working on an investigation where I am burning isomers of butanol to test for the heat of combustion values to see which fuel is most efficient. The fuels I am testing are 1-butanol, 2-butanol, 2-methylpropan-1-ol (isobutanol) and 2-methylpopan-2-ol (tert-butyl alcohol).
I have researched that the heat of combustion values for these fuels are as follows:

1-butanol: −2671 kJ/mol
2-butanol: −2661.1 kJ/mol
Isobutanol: −2662.6 kJ/mol
Tert-butyl alcohol: −2644.8 kJ/mol

So evidently, my experimental results should also show similarity in the ordering of greatest to lowest heat of combustion values.
However, I am wondering why isomers of butanol have different heat of combustions. I know it has to do with the shape of the molecules and whatnot, but how does this affect the energy released? What about bond angles?

The heat of combustion (or any reaction) is the energy that is set free when you make the products ($\ce{CO2}$ and $\ce{H2O}$ here, in their respective stochiomeric amounts) out of single atoms (this is the same for all isomers), minus the energy needed to break all bonds in the educts. The latter is different for the different isomers, because the energy in a chemical bond not only depends on the two partners but also on their neighbours (the "shape" of the molecules).
The usual procedure for this calculation is to not go over single atoms but the elements (and compounds) in their most stable form at standard conditions. These energies a called standard enthalpies of formation. There are long tables listing them for all possible compounds.


text:
Organic solutions that corrode metals
Most of the common organic solvents are regarded as noncorrosive, but  stainless steel can be attacked by formic, acetic, and propanoic acids.
Corrosion of stainless steel by organic solvent mixtures, Alice Y. Ku , David H. Freeman, Anal. Chem., 1977, 49 (11), pp 1637–1638
DOI: 10.1021/ac50019a044
Are there some aromatic hydrocarbon that attack stainless steel? Which organic solvents can corrode copper and aluminium?

Picric acid (2,4,6-trinitrophenol) is an aromatic compound has been used as an etchant in the past. Picral is an ethanol solution of picric acid and could etch steel. Look up www.metallographic.com/etchants/etchants.htm

This probably should be a comment.

In fact, I think this can stand well as an answer to that question. It would however, be really nice of you to be more elaborate on it. When was it used as a solution? What references do you have? etc etc.

Do you know any organics that can corrode copper or aluminium?

A lot of the strong organic acids like formic corrode aluminium. https://books.google.co.jp/books?id=iEeiQEeLOmYC&pg=PA156&lpg=PA156&dq=aluminium+corrosion+resistance+in+organic+acids&source=bl&ots=EJldV_Y-Rk&sig=9TLKDJBoelMHddrRFFLOopoeT3s&hl=en&sa=X&ved=0CCgQ6AEwAWoVChMIzuDTof6RxwIVBpKUCh2edA0Z#v=onepage&q=aluminium%20corrosion%20resistance%20in%20organic%20acids&f=false


text:
Can I equalise osmotic pressure using two different solutes?
I understand that there is a pressure gradient between an aqueous solution, e.g. salty water, and, say, pure water due to osmosis. My question is, is it possible to match or almost match, the osmotic pressure between two solutions of different solutes, but using a solution with a different solute. For example, would it be possible to make the osmotic pressure gradient between salty water and some other solution zero, assuming both are at the same temperature and pressure?
I am not a chemist, so purely qualitative answers are helpful.

Just to clarify - the solvent would be the same for both solutions?

Hi Todd,yes, the solvent would be the same (just made a nonsense comment saying the opposite and had to edit it)

Does the isotonic case mentioned here illustrate what I think you're looking for? The example is biological in nature, and is that of a blood cell membrane that is selectively permeable to one of the solutes and the water, but not the other. If so, I'll go ahead and edit the answer I was working on and post it. I want to make sure we're on the right track.

I think that is pretty much the situation I'm thinking of, yes.

Yes, I think so. Based on our comments above, I believe you want to prepare two solutions that are isotonic with each other (have the same concentrations of different solutes) and allow them to communicate across a membrane that is selectively-permeable to water only.
Tonicity is described here as:

Tonicity is a measure of the effective osmotic pressure gradient (as defined by the water potential of the two solutions) of two solutions separated by a semipermeable membrane. In other words, tonicity is the relative concentration of solutions that determine the direction and extent of diffusion. It is commonly used when describing the response of cells immersed in an external solution. Unlike osmotic pressure, tonicity is influenced only by solutes that cannot cross the membrane, as only these exert an effective osmotic pressure. Solutes able to freely cross the membrane do not affect tonicity because they will always be in equal concentrations on both sides of the membrane.

I found this nice explanation with illustrations and one example where it shows that you could prepare a 10 mM solution of albumin and a 10 mM solution of glycine, separated by a water-permeable membrane, and no net flow of water occurs in either direction.

Could this principle somehow be used to desalinate water?

I believe not: you have to get the salts out of the water, and they're the species which create osmotic pressure if they're not able to cross a membrane. Hence the use of reverse osmosis, where the osmotic pressure of the salt water is exceeded by external pressure on the system (at a very high energy cost).


text:
Identifying coordinate bonds
How can we know whether a compound has a coordinate bond or not? I know that I have to draw the Lewis Structure for the compound. But what to do after that? Is there anything to do with formal charges?

Tere really isn't a perfect way to do this other than by observing the structure of the compound.

The first thing needed for a dative bond is a lone electron pair in one atom and an empty orbital in another atom.
This condition is necessary but enough.
There is also a formula that fails in many cases but may sometimes become handy:
Formal Charge = (Oxidiation State of the centeral atom - 8 + Digit of its group number)/2
e.g.: For SO3 it would be: (6-2)/2 Which means that it has 2 co-ordinate bonds.
However for a compound like NH4+ it won't work.
Of course, safest way is to find out how the compound had formed in the first place.
e.g.: If we know that BF3 + F- made BF4-, we would find out that F- had shared with B two of its electrons, filling the empty orbital belonging to B.

After drawing the structure, you can basically check if the shared pair of electrons are coming from the same molecule. If the answer is yes, then the compound contains coordinate covalent bond.

You should seriously improve it

You should also tell the way to identify that shared pair of electrons are coming from same molecule.

Check the formal charge.If the central atoms contains +1 formal chatge then it is donating electron pair to anothe atom.And if it contains -1 formal charge then it is accepting electron pair from other atom

Hmm, it's not useful to oversimplify it like this. What if the atom had been assigned a formal charge more than -1, like many of the sulfur compounds?


text:
What is the transparent film that covers water-soluble dishwasher tablets?
These dishwasher tablets (Finish Powerball Shine) seem to be covered in plastic!

But that can't possibly be true since there's no unwrapping, and they're completely soluble in water.
What is the film made of?

http://www.google.com.au/patents/US5972870

Keep in mind "plastic" encompasses many different chemical species. Many are soluble in water.

The MSDS sheets for the product you've linked to are available via email in the U.K. Here in the U.S., the manufacturer makes them available online. That said, there is not a U.S. product called Finish Powerball Shine, but here is a link for Finish Powerball Quantum Max (Fresh Scent) where it is stated that "Ingredient polyvinyl alcohol" functions as a "Dissolvable Film." You might peruse the other U.S.-hosted Finish products to find a more suitable equivalent.

It is made using Polyvinyl Alcohol (Mowiflex material).

Welcome to [chemistry.se]! Please consider [edit]ing your answer to add some references about this material.

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


text:
Finding rate of reaction from a concentration vs. time graph

A tangent drawn on “conc. of product vs. time” curve of a 1st order reaction makes an angle 30 degrees with the $y$-axis at 20 min. Find the rate after 20 min.

I first tried to find the equation like this:
$$k=\frac{2.303}{t}\log\frac{a}{a-x}$$
Let $\frac{k}{2.303}$ be $c$ to make calculations less messy. Here, $a$ is the initial concentration of reactant.
$$\frac{a}{a-x}=10^{ct}$$
$$x=a\left(1-10^{-ct}\right)$$
$$\frac{\mathrm dx}{\mathrm dt}=a\left(c\cdot\ln10\cdot10^{-ct}\right)=ak10^{\frac{kt}{2.303}}$$
Now, the tangent makes an angle of 30 degrees with concentration axis (or $y$-axis in this case). So the angle made with time axis (or $x$-axis) is 60 degrees. Hence $\frac{\mathrm dx}{\mathrm dt}=\tan\theta=\sqrt{3}$. But the value of initial concentration is not given for finding rate constant. Should I go for a different approach or am I making any conceptual mistake?

I think that your approach is ok. If P is product and R$_0$ initial reactant, the $rate$ and slope of the graph is $dP/dt=R_0.k.exp(-kt)$ which is dimensionally ok. I don't see how you can get the $rate$  $constant$, you are not give enough information. The slope  has units of concentration/time and so taking the tangent is not strictly correct as the tan should be dimensionless.

Let the reactant be $S$ and the product be $P$.  Let $p$ be the dimenionless product concentration, $\frac{P}{S_0}$.  First order kinetics:

$$\frac{P}{S_0} = p = 1- e^{-k t}$$

Slope of first-order kinetics:

$$\frac{dp}{dt} = k e ^{-kt}$$

Slopes and angles.  One confusing thing about this problem is the definition of $\theta$ is different than what you usually see in math class.  It's the angle with respect to the $y$ axis, not the $x$ axis.  As you note, if the $y$ angle is 30 degrees, then the $x$ angle is 60 degrees, meaning the slope is $\tan~60^{\circ} = \sqrt{3}$.
Units problem.  However there is a big units problem in this question.  The slope is $\sqrt{3}$ what?  Gigamolar per nanosecond?  Zeptomolar per eon?  The angle of the tangent curve will change depending on the units used to plot the data.  If you use nanoseconds as the time unit, the angle of the tangent curve will be very flat, but conversely if eons is the time unit, the the angle will be very sharp.  The question does not explicitly say what units are used in the plot.  And the expression for the derivative of a first-order reaction shows that the slope should have units of concentration per time, but $\sqrt{3}$ is dimensionless.  So there is not really enough information to solve the problem.
Dicey assumptions about units.  Let's assume that the time unit is minutes.  Let's further assume that the $y$ axis plots a dimensionless concentration, i.e., $p$. Then the derivative (slope) is $-k e^{-k t}$ and apparently has a value of $\sqrt{3}~\mathbf{{min}^{-1}}$.

The question is asking for the rate, so you are done.  But there are a lot of weird assumptions in this problem.


text:
Does imidazole and hydrochloric acid yield imidazole hydrochloride salt?
I'm asking this because I found a link for imidazole-HCl buffer: , but from what I understand, a buffer is a weak acid or base with its corresponding salt. Would imidazole-HCl buffer be a salt solution, or does it have a weak acid or base as well (if it does, what is the weak acid/base)? Also, I found two different pictures for imidazole hydrochloride:

Which one is correct? Is imidazole-HCl a salt or a buffer?

Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center. Nice question, BTW!

The correct structure of the salt is as shown above. This structure is more stable than the one that has two hydrogens on the same nitrogen - due to electron delocalization.
If you have a water solution of about 50/50 imidazole base and the hydrochloride salt, you get the maximum buffer capacity, due to the fact that both acid and base forms are present, and a small amount of a strong acid or base subsequently added to the mixture will react with these imidazole species, and therefore have little effect on the pH.

It may be worth adding that since the lone pair in the other nitrogen is a part of the pi-cloud (aromatic sextet) and therefore the basicity of this nitrogen would be low.


text:
Where can I obtain research material on the synthesis of polypyrrole?
I'm studying Chemistry at A-level and a large part of the course is a researcher paper that takes most of the year.
Conductive polymers interest me. I'm struggling to find any surfeited researcher on the synthesis of polypyrrole. I not sure if I need to buy a paper or another course of action would be more suitable.

Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center.

Hi Jacob, the trouble when you're first starting out is knowing where to look! Any book on conducting polymers should have the answer, but also look up how you might make bipyrrole. This can then be extended to a polymeric approach. My undergraduate dissertation was on conducting polymers and LED's. It might take me a while, but I can dig out a good reference text as there were no online references when I wrote it!

Thank you! Any text references would be extremely useful. I'm planning on making the polypyrrole electrochemically.


text:
Is there a relationship between viscosity and enthalpy of vaporization?
It would be useful to be able to predict relative evaporation rates of liquids.  For small reduced temperatures the heat of vaporization is relatively constant, but it is not one of the more common properties cited for substances.
Is there any sort of functional relationship between enthalpy of vaporization and common properties like viscosity, melting point, and boiling point?

I don't believe there necessarily is one, and I considered making a plot to confirm that suspicion, but finding the data is going to be a pain since both of those properties vary with temperature. I think you would get substantially different answers with viscosity at the melting point of individual compounds vs. at a fixed temperature across all compounds. I'd be interested if someone did have the data to plot such.

@WilliamKappler: Good point.  Yes, we probably have to specify it at some standardized temperature.  Melting point is a good one.  Or perhaps the "liquid-midpoint": i.e., the temperature halfway between boiling and freezing.

Yes. The phase envelope (or equilibrium curve) of most liquids have a similar shape and can be approximated, to a reasonable degree, by using the Clausius-Clapeyron equation:
$$
\Delta h_v=T(v^V-v^L)\frac{dP^s}{dT}
$$
which requires finding ${dP^s}/{dT}$ from e.g. a vapor-pressure-temperature correlation like the Antoine equation, and also a separate estimate for $\Delta v$ before $\Delta h_v$ can be obtained. This is an exact thermodynamic relationship, but is often simplified by using the ideal gas law for $v^V$ and neglecting the liquid volume $v^L$ i.e. $v^V \gg v^L$, giving
$$
\Delta h_v=\frac{RT^2}{P^s}\frac{dP^s}{dT}
$$
However, it is very important to note that the term $(v^V-v^L)$ becomes increasingly decisive as you go above 1 bar, where the assumption $v^V \gg v^L$ is no longer valid. In such a case, separate correlations for $v^L$ and $v^V$ are required before the correct curvature is obtained. $v^L$, for instance, can be calculated from e.g. a density correlation like the Rackett equation, and $v^v$ from an equation of state like Peng-Robinson.

Wow, so this looks like relatively cutting-edge thermodynamics: Do the Rackett and Peng-Robinson equations really date only from the 1970s?

$\Delta h_v$ is the specific enthalpy of vaporization [J/mol], $v$ the pure component molar volume for the vapor ($V$) and liquid ($L$) phases, while $P^s$ is the saturated vapor pressure ($T$ is in Kelvin).

Cutting-edge? Google the names Rudolf Clausius and Clapeyron... Not so much as cutting-edge as it is still very relevant. New correlations for $P^s$, $v^V$, $v^L$ or any other thermodynamic property are constantly being developed. Most of the time, however, you are only interested in answers that are sufficiently precise and that follow known trends i.e. 5 decimal place accuracy is typically not required (which is why you find many older correlations still in active use).

I'm still trying to digest this.  The differential equation form is needed to find an equilibrium, but before solving for equilibrium I like to be able to find instantaneous values.  So given a T and P what determines a chemical's instantaneous evaporation rate?  Or, put another way, holding the conditions constant and changing the chemical, what do these equations tell us?  For "everyday" T & P it looks like it reduces to saying, "Evaporation rate is proportional to saturated vapor pressure," so Clausius-Clapeyron doesn't actually relate it to other characteristics?

Thermodynamic properties are most definitely influenced by molecular characteristics, which is why different molecules have different saturated vapor pressure curves and normal boiling points (just to name a few). Therefore it is not necessarily correct to say that the Clausius-Clapeyron equation does not relate to other molecular characteristics.

I am not exactly sure what you are trying to get at with this "instantaneous evaporation rate"... If you had a pot of water at $100~^{\circ}\mathrm{C}$, open to the atmosphere, then your evaporation rate should be proportional to the rate of heat that is being added. However, if you wanted to weld a lid onto the pot then things start to get very messy because it becomes a dynamic problem requiring additional levels of complexity involving additional heat and mass transfer equations... [Hopefully this time my latex is right ;)]

That's exactly why I'm interested in the "instantaneous" question: I'm only interested in the differences between chemicals, not in the dynamics of the phenomenon of evaporation.  By analogy: The boiling point of a chemical at standard pressure is a fundamental and universally quoted characteristic, even though in any system as soon as you start adding heat that number is going to change.  Now vapor pressure and vaporization enthalpy are, in my mind, one step removed from boiling temp., but I can't find anything that relates either to boiling point or any other "fundamental" characteristic.

authors: Владимир Владимирович


text:
Why are Silicates solid while carbon dioxide is a gas?
I was under the impression that chemistry almost exclusively involves valence electrons because there isn't enough energy to strip off electrons located closer to the nucleus.
If that is true, and elements in the same period have similar properties because they have the same number of valence electrons, then why is $\ce{SiO2}$ a solid and $\ce{CO2}$ a gas? Surely a difference in mass by a factor of 2.5 can't be that big a deal.
Is it because of a van der Waals force difference, because silicon has extra electrons which result in compounds being formed from it being more symmetrical than those formed from carbon?

CO2 is based on double-bonds, O=C=O. So the heart of the question is: "Why can carbon easily form strong double (and even triple) bonds, while silicon is barely capable of forming weak double bonds?"

Considering that the sublimation point of dry ice is only like $-70{}^{\circ}$ C, a factor of 2.5 of the mass is actually probably plenty to explain this.

Also easy if one is a chemist: Silicium can form $(d-p)\ \pi$ bonds, as it can use its d orbital while carbon can not. Also siliciums radius larger and pi bonds can not be formed efficiently. Long story short: Mainly because of the larger radius of silicium.

Do all elements in a column of the periodic table share the same boiling point? No. So why would you expect them to be the same phase of matter at room temperature?

Above 2230 C, both are gases.

As a side point, given enough pressure carbon dioxide can be rendered macromolecular.

The reason why carbon dioxide is a gas and silicon dioxide is a solid is because their chemical structures are different.
Carbon dioxide is a linear structure with two double bonds between carbon and oxygen. It is a small molecule and non-polar with only weak bonds between the molecules. Hence it is a gas.
Silicon dioxide is not formed of small molecules. It consists of an infinite array of silicons where each silicon is bonded to four separate oxygens (and each oxygen is shared between two silicons). This creates a strong refractory solid (glass and sand are mostly silicon dioxide aka silica). So the same apparent overall formula doesn't describe the actual structure of the compounds at all. But the structures explain the difference in behaviour.
Of course this doesn't explain why silicon prefers to bond with four oxygens when carbon prefers just two. This is not completely simple and results from the relative bond strengths of carbon-oxygen bonds, carbon-oxygen double bonds and the equivalent bonds for silicon and oxygen. The simple version is that silicon oxygen bonds are strong relative to their double-bond equivalents whereas carbon-oxygen double bonds are strong relative to their single bond equivalents. Or, more precisely, if we could make a carbon-oxygen network solid with the equivalent structure to silica, it would tend to fall apart into carbon dioxide. If we could make silicon dioxide molecules, they would react with the release of energy to give silica.
Deeper explanations would need to look at why the relative strengths of double and single bonds turn out that way, but that would get into molecular quantum mechanics and would not be much more useful as an explanation.
The simplest explanation is the fact that the structures are different.

Hi, thanks for your answer! In my book (amazon), it is given that "Si cannot form stable bonds with O, hence, Si has to form a 3d lattice. This is the reason why $\ce{CO2}$ is a gas while $\ce{SiO2}$ is a solid" What would you comment about it? Is it correct or not? Thanks!

@GaurangTandon If your quote is exact, then it is wrong. Si is happy forming stable bonds to oxygen: the issue is that collectively single Si-O bonds are far more stable than double bonds. You can't make a small uncharged molecule with just single Si-O bonds so you end up with a network. The issue is the relative stability of single and double bonds.

If you look at the crystal structure of silicon dioxide ($\ce{SiO2}$), you will see it is comprised of tetrahedra (one silicon surrounded by 4 oxygen atoms) and these tetrahedra are in turn connected to each other, and subsequently form a 2-dimensional network which is large enough to make this substance a solid.
Also, the factor of 2.5 in mass makes a big difference, but it is not relevant to this discussion. I would recommend reading any inorganic chemistry book where such topics are discussed in depth. Van der Vaals forces only play significant roles in long-chain carbohydrates, such as fats, and are not really observed in "inorganic" molecules.
$\ce{CO_2}$ molecules cannot form a crystal in the way $\ce{SiO2}$ units do, and I do not think I have to explain why. Also, it has no dipole moment and no van der Waals forces between the molecules.

Silicon dioxide should be nonpolar by symmetry, too, whether in the same structure as CO${}_{2}$ or no.

It is, but it does not really matter for this question, because as I stated, SiO2 forms large 2-D networks.

Sure, but you bring up the fact that CO${}{2}$ has no dipole moment, which doesn't distinguish it from SiO${}{2}$.  What structure does dry ice take?  I would assume it would be similar to that of solid silicon dioxide.  I really think the mass is the main difference here.

You are very wrong, it is not even a bit similar to the structure of solid CO2. The mass difference is surely not the main difference, every inorganic textbook will show you.

Silicon dioxide is a 3D network not a 2D network.

Van der Waals forces are relevant to everything, even carbon dioxide (they would be why CO2 can be turned into a solid or liquid). They just happen to be irrelevant to the difference between CO2 and SiO2 which is a consequence of the different structures of covalent bonding in the two compounds.

Elements of the answer are also contained in other answers here but this needs some more teasing out. It has to do with the bond energies of C-O, Si-O, C=O, and Si=O. I don't have the numbers at hand, but work it out and it will point you to carbon being the most stable in the form of $\ce{CO2}$ and silicon the most stable as a tetrahedral network (similar to diamond.)
The reason why $\ce{CO2}$ is a gas can easily be explained using intermolecular bonding principles, which you can look up easily.

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


text:
Which element has a greater second electron affinity, sulfur or oxygen?
I found a question asking the above which states that sulfur has a lesser second electron affinity than oxygen. But since the inter-electronic repulsion in sulfur is lesser, shouldn't it be willing to take up another electron more easily than an oxygen ion with a -1 charge?

The most important fact here is that the electron added to $\ce{S-}$ goes into an $n=3$  orbital ($3p$), whereas the one added to $\ce{O-}$ goes into an $n=2$ orbital ($2p$).
The larger $n=3$ orbitals are further from the nucleus and hence experience a weaker attraction to the nucleus and are much higher in energy.  This dependence with quantum number scales approximately $E\sim - n^{-2}$ so is very important [1].
The dependance of orbital energy also scales quadratically with effective nuclear charge, however effective nuclear charges falls significantly compared to the true nuclear charge due to shielding and repulsion from other electrons.  A common first order approximation for the effective nuclear charge is given by Slater's rules [2]
Combining these two approximations we can calculate very approximate orbital energies.
$$
\begin{align}E_{2p}(\ce{O^2+}) & {} = - R_\infty h c \frac{Z_{eff}^2}{n^2} \\
& {} = - R_\infty h c \frac{(8-4.15)^2}{2^2} \\
& {} = - 3.7 \text{ Ry} \\
E_{3p}(\ce{S^2+}) & {} = - R_\infty h c \frac{Z_{eff}^2}{n^2} \\
& {} = - R_\infty h c \frac{(16-11.25)^2}{3^2} \\
& {} = - 2.5 \text{ Ry}
\end{align}
$$
Where 1 Ry is the electronic energy of a hydrogen atom.
As you can see, to first order, the energy of the new electron is much lower when added to $\ce{O-}$ than when added to $\ce{S-}$, leading to a more favourable electron affinity for $\ce{O-}$ than $\ce{S-}$, driven by the fact that the orbital occupied has a lower principal quantum number $n$
[1] - See the energy of a hydrogenic ion - Wikipedia.
[2] - Slater's rules - Wikipedia

Sulfur  has 3 shells while oxygen has 2.This increase in shell reduces the nuclear charge and lessens the second I.P.Further Explanation:
Due to the extra shell, the electron in the outermost shell is held , but not as strongly as an oxygen atom has held its electron. So the second Ionization Enthalpy of sulfur will be less than that of Oxygen,because more energy is required to extract the electron from oxygen atom.

For some reason OP asked about affinity not ionisation.


text:
Does battery self discharge when connected?
I am an electric engineer, and I read about battery self discharge phenomenon. I was wondering whether a battery self discharge rate is the same when it is unconnected (i.e. in storage) and connected (i.e. when it is plugged into a device and being discharged "normally").
In other words, when I calculate battery life time for a product, should I take add the rated self discharge of a battery to the general current calculations or not? And also-is the answer similar to all types of battery?

@santiago, hmmm, I disagree. Self-discharge happens due to a chemical reaction taking place, thus, the question seems to be perfectly on-topic here.

Great question! I studied battery chemistry as part of my graduate work, and I think the answer depends on the mechanism of self-discharge.
I think that by discharging the battery through an external circuit, you will generally cut off the self-discharge mechanisms (depending on the discharge rate), so if you add the discharge rate and the self-discharge rate, you will probably underestimate the battery life. You should also keep in mind that the rate of self-discharge depends on the state-of-charge for the battery, and is generally worse at high state of charge. Thus, even the manufacturer specification for self-discharge rate isn't a constant.

Yep. More than likely, when you connect to an external load the actual cell potential drops somewhat, decreasing the overpotential for the self-discharge mechanisms.  You probably don't eliminate them completely, but odds are their rates are substantially reduced.


text:
What are the products of a reaction between copper sulfate and sodium bicarbonate?
I have been trying for a while to determine what products are produced when copper sulfate reacts with sodium bicarbonate at room temperature and standard pressure. However, while I was looking, I found conflicting information concerning what products are formed. Some sources claim that the products are copper hydroxide and sodium sulfate, while other sources state that the products are copper carbonate and sodium sulfate.
I believe the reaction is:
$$\ce{CuSO4 + 2NaHCO3 -> CuCO3 + CO2 + H2O + Na2SO4}$$
Is this correct?

It's surprising how one can find quality, directly relevant chemistry videos on YouTube. Look here for what seems to be the exact answer to your question.

Copper sulfate reacts with sodium bicarbonate reacts to form basic copper carbonate due to  the great affinity of the $\ce{Cu^2+}$ ion for the hydroxide anion $\ce{OH−}$. From its Wikipedia article:

Basic copper carbonate can also be prepared by reacting aqueous
  solutions of copper(II) sulfate and sodium bicarbonate at ambient
  conditions. Basic copper carbonate precipitates from the solution,
  again with release of carbon dioxide:
$$\ce{2 CuSO4 + 4 NaHCO3 → Cu2(OH)2CO3 + 2 Na2SO4 + 3 CO2 + H2O}$$

Here, the "ambient condition" is very important because if we alter the temperature and/or pressure , the reaction may tend to form copper(II) carbonate or any non-stoichiometric copper compound.
Heating the basic carbonate in an atmosphere of carbon dioxide (produced by the decomposition of silver oxalate at 500 °C) produces copper(II) carbonate but it is itself quite unstable and react with moisture to convert back to basic carbonate.
For advanced reading, you can refer to reference 2 below in the reference which describes Copper/bicarbonate equilibria in solutions of bicarbonate ion in details (which clearly shows formation of basic ion $\ce{CuCO3(OH)^2-}$ in the solution).
Notes and References

http://www.sciencemadness.org/talk/viewthread.php?tid=24056
https://www.sciencedirect.com/science/article/abs/pii/0043135471900492
http://science.wonderhowto.com/how-to/synthesize-copper-ii-carbonate-sodium-bicarbonate-302466/ (Although this site claims to make copper(II) carbonate, it is actually making the basic carbonate. You can clearly see its bluish green color of basic copper carbonate).


text:
Analytical solution of the Schrödinger Equation for AB^+ systems
According to Wikipedia, there are not many problems for which the Schrödinger equation can be solved analytically. [1] The $\ce{H2+}$ ion is probably the most complex molecule that can be treated this way.
As far as I know, every diatomic homonuclear system like this can be solved, e.g., $\ce{U2+}$.
Up to this point the molecules always consisted of two equal atoms which makes me curious about the following:
Is the Schrödinger equation not only solvable analytically for diatomic homonuclear systems $\ce{A2+}$ but also for diatomic heteronuclear systems like $\ce{AB+}$?

Schrödinger equation can be solved analytically as long as there is no electron-electron interaction involved. Thus it is potentially solvable for any one-electron ion system.It is unsolvable even for mere helium atom, obvious it isn't solvable for any diatomic molecules with more than one electron.

Related: How was the quantum three-body problem solved?

Related: Why cannot the Schrödinger equation be solved exactly for systems in which more than two particles interact?

First, one has to be really careful with the words: when it is said that the Schrödinger equation can be solved the dihydrogen cation completely analytically, several remarks has to be made.

There are few different Schrödinger equations out there, so it is better to explicitly mention that one which can be solved analytically for $\ce{H2+}$ is the electronic Schrödinger equation. So, first of all, we are talking about the time-independent Schrödinger equation here, and secondly, we employ the Born-Oppenheimer approximation and solve only the electronic problem. No claim is maid with respect to analytical solutions of the whole time-independent Schrödinger equation without any approximation introduced, as well as regarding analytical solutions of its time-dependent counterpart.
The second remark is about the meaning of the word "analytical", or "analytic". The definitions taken from the Wikipedia are as follows:

An equation is said to have an analytic solution if and only if at
  least one solution can be expressed as an analytic expression.
An analytic expression is a mathematical expression constructed using well-known operations that lend themselves readily to calculation.

These definition seems to be rather vague for me, especially taking into account that,

[..] the set of well-known functions allowed can vary according to context [...]

so we have to be careful relying on some intuitive notions of "analytic expression" and "analytic solution".

Having said so, and in hope that my intuitive notions of "analytic expression" and "analytic solution" are not miles away from the truth,  it is important to note that $\ce{H2+}$, for which analytic solutions are known, is (a special case of) a three-body system. And since there is no claim about analytic solutions for a four-body system, a five-body one, etc., from the beginning we have to restrict ourselves for three-body systems.
From that viewpoint the notation $\ce{AB+}$ used by OP is misleading, because the only three-body chemical system of $\ce{AB+}$ kind is $\ce{H2+}$. Already for the system composed of $\ce{H}$ and $\ce{He}$ the charge has to be +2 since we have to remove 2 electrons so that we end up with just one of them which together with 2 nuclei constitutes a three-body system, and not a four-body one.
Now, what about analytical solutions for $\ce{AB^{n+}}$ systems where $n$ is chosen appropriately so that the system is a three-body one? Well, I think, as for $\ce{H2+}$, there are analytic solutions for electronic Schrödinger equation in the Born-Oppenheimer approximation. The reason is pretty simple in fact: the clamped nuclei model reduces the three-body  problem to a single-body problem of the motion of a single electron within the fixed nuclear potential.
Why there are no related references (at least I could not find one)? It is very likely because chemist has little or no interest in three-body chemical systems of $\ce{AB^{n+}}$ type in general, and specifically, little or no interest in analytic solutions for these systems. Except for the simplest $\ce{H2+}$ system which serves as a primary model for the concept of the one-electron bond.


text:
kinetic reaction rate greater than extent of reaction for equilibrium
I'm comparing a model with chemistry at equilibrium versus a kinetic one.
I take a network of two reactions: methane steam reforming (MSR) and water gas shift (WGS) at 700°C.
I calculated extent of reaction of both reaction, with the equilibrium hypothesis, solving a system of two non-linear equation containing the two $ K _{eq} (T) =f(\lambda_{MSR}, \lambda_{WGS}) $ equations, which give me these results:
$\lambda_{MSR} =  1.4167e-4 [mol/s]$
  $\lambda_{WGS} = -6.6788e-5 [mol/s]$
Then I calculate rates of reaction with an homogeneous and stoichiometric coefficient expression, with Temperature dependence constant (Arrhenius) taken here, I get:
$r _{MSR} = 20.9761e-4 [mol/s/m^3]$
  $r _{WGS} =        0.0 [mol/s/m^3]$ (I have no reactants nor products...)
Now to compare them I have to multiply $r$ for reactor's volume.
Here start my doubt:
taking for sake of simplicity $1[m^3]$ reactor, I get

$r_{MSR}>\lambda_{MSR}$

this what does it mean?
does it have a physical meaning?
what is the meaning of a so large reactor?

My thoughts, my idea, is that equilibrium shows the maximum rate of reaction so kinetic should, or better must, respect it giving me a lower value of reaction rate; but, as you can easily see in my above example, this statement (that I have never read somewhere in any chemistry book, so I can take it as truth) can be superimposed, forced. How do you explain it?

please provide some feedback, I can add some other information if you need it!


text:
What is the favoured geometry of chlorine trifluoride?

Why is the first structure on the left more stable than the second one on the right?
In the first one the angle between the two pair of nonbonding electrons is about 120° which is a lot less than that in the second structure. And it is also very well known that lone pair-lone pair repulsions are worse than bonding pair-bonding pair repulsions. So why is the first structure more stable?

Short Answer
The structure on the left is "preferred" because the structure on the right cannot exist.  You cannot put 4 electrons in a p-orbital.
Detailed Explanation
In $\ce{ClF3}$ the central chlorine atom is roughly $\ce{sp^2}$ hybridized.  This means that we will have 3 $\ce{sp^2}$ orbitals emanating from the chlorine; they will form an equatorial plane and will either contain a lone pair of electrons  or bond to a ligand.  Further in the case of $\ce{sp^2}$ hybridization, we also have 1 unhybridized p-orbital perpendicular to the equatorial plane.
The drawing on the right has two lone pairs placed in one p-orbital.  A p-orbital cannot contain 4 electrons, therefore the drawing on the right does not represent a valid chemical structure.
The question really becomes, does $\ce{ClF3}$ prefer to exist with

2 axial fluorines and 1 equatorial fluorine

or

1 axial fluorine and 2 equatorial fluorines?

We can use Bent's rule to answer this question.  Bent's rule tells us that more p character will be placed in those orbitals directed towards more electronegative substituents (or conversely, more s character will placed in those orbitals directed towards more electropositive substituents). This is because orbitals with more p-character are higher in energy than orbitals with more s-character (because a p-orbital is higher in energy than an s-orbital).  An electronegative substituent will draw electron density away from the central atom.  Consequently there is less electron density to stabilize so the orbital will rehybridize such that it contains more p-character, saving the s-character to stabilize another orbital that contains more electron density.
Since fluorine is very electronegative and an electron pair is very electropositive, applying Bent's rule we see that the two electronegative fluorines will prefer axial positions where the orbital is rich in p-character (in fact, it is roughly 100% p) and the two lone pairs will prefer to occupy two of the s-rich $\ce{sp^2}$ equatorial orbitals.  Therefore, your structure on the left with 2 axial fluorines and 1 equatorial fluorine correctly depicts the preferred arrangement of fluorines and lone pairs in $\ce{ClF3}$.
A final note on the bonding in $\ce{ClF3}$, the two axial fluorines bond with the chlorine p-orbital to form what is referred to as a hypercoordinated or hypervalent or a 3-center, 4-electron bond (3-center because there are 3 atoms $\ce{F-Cl-F}$ involved in the bond and the bond contains 4 electrons, 2 from chlorine and 1 from each of the two fluorines). Hypercoordinated bonding is common in situations where the central atom appears to have more than an octet of electrons.  See these earlier answers for more detail on hypercoordinate bonding.

Why does F replace an axial bond in $\ce{PCl5}$?
How can Iodine bond with 5 Fluorines in Iodine Pentafluoride?
Hypervalency and the octet rule

Why not sp3d hybridised?

I believe the description is meant to show the orbitals containing the electrons as 2 hybridised orbitals, not a single p orbital

Other than in organometallic compounds, d-orbitals are not usually involved in bonding.  P-orbitals are lower in energy and more directed than d-orbitals.

Using the usual "AXE method" all molecules of $\ce{AX3E2}$ type, where

$\ce{A}$ represents the central atom;
$\ce{X}$ represents the number of ligands (atoms) bonded to A;
$\ce{E}$ represents the number of lone electron pairs surrounding the central atom;

are assumed to be T-shaped. I think it does not matter what the central atom $\ce{A}$ and ligands $\ce{X}$ are.

OP is right in that lone pairs repulsion is smaller in the second structure, however, we have to consider repulsions between lone pairs and bonding pairs as well. Yes, generally, lone pair - lone pair (lp-lp) repulsions are greater than lone pair - bonding pair (lp-bp) and bonding pair - bonding pair (bp-bp) ones, but the later could not be just  neglected.
For the first structure we have:

One lp-lp repulsion at 120°;
Two lp-bp repulsions at 120° (one bonding pair in the same plane interacts with two lone pairs);
Four lp-bp repulsions at 90° (two bonding pairs out of plane interacts with two lone pairs);
Two bp-bp repulsions at 90° and one at 180°.

For the second structure we have:

One lp-lp repulsion at 180°;
Six lp-bp repulsions at 90° (three bonding pairs out of plane interacts with two lone pairs);
Three bp-bp repulsions at 120°.

So, in the second structure the lp-lp repulsion is smaller, but the overall lp-bp repulsion is bigger while the overall bp-bp repulsion is basically the same. Well, it is difficult to say a priori, will the bigger lp-bp repulsion outweigh the smaller lp-lp repulsion, but it assumed in VSEPR theory that it will, so the first structure is more favorable.


text:
What does a change in temperature do to the concentration of gases at equilibrium?
I am confused about a specific part of Le Chatelier's Principle. My question revolves around a homework assignment question:

$$\ce{2SO2(g)  + O2(g) <=> 2SO3(g)} \qquad\Delta H^\circ = -198.2\:\mathrm{\frac{kJ}{mol}}$$
Comment on the changes in the concentrations of $\ce{SO2}$, $\ce{O2}$, and $\ce{SO3}$ at equilibrium if we were to increase the temperature of the system.

Now I understand that increasing the temperature of an exothermic process will cause the equilibrium to shift towards the left (in the direction of the endothermic reaction); however, what does the temperature do to the concentrations of the gases?  I think it does nothing, but I am not sure.

You are correct in saying that the increase in temperature causes the equilibrium to shift to the left, in the direction of the endothermic reaction. As the reaction continues in the endothermic direction, the concentrations of $\ce{SO3}$ will decrease in favour of the evolution of more $\ce{SO2}$ and $\ce{O2}$, due to the additional heat available.


text:
How to Convert Sucrose to Glucose and Fructose
If you have some sort of sports drink that contains sucrose, how would you take that sucrose and break it down into its components, glucose and fructose? I know that sucrose + water will break it apart, but how can it still remain together (as sucrose) while in the bottle of sports drink?
I am trying to compare the levels of glucose in various sport drinks. So I'm planning to use Benedict's solution and a spectroscope in order to determine the amount of glucose present in solution (I'm going to add activated charcoal to the solution to remove the colouring from it before placing in the spectroscope).
I will prepare standards by adding 5 grams of glucose to 100 mL of water, 10 grams of glucose to 100 mL water, and so on.
As a high school student I only have limited resources. Any information about my proposed plan or alternatives that are low-cost are appreciated.

Is eating it allowed? That will certainly work. Only issue is that you can't get back the glucose + fructose.

I know this is a stab in the dark and you may or may not have even done this experiment, however, I am in high school and by coincidence happen to be doing this exact experiment. What method did you go for in the end and how did it work out? Did you manage to get the results you wanted? Really appreciate any help!

Sounds like a great experiment!
As you state, a sucrose solution in water eventually hydrolyzes to fructose and glucose, but without a catalyst, the reaction rate is extremely slow; beverages are stable at room temperature for years. However it is hydrolyzed in hours at room temperature by adding sucrase enzyme (secreted in the small intestine), invertase enzyme (derived from yeast) or even a weak acid such as citric acid, tartaric acid, or even dilute hydrochloric acid.
You may need to test whichever of these are available for compatibility with Benedict's solution, though, and you may need to test some of the other ingredients in the drinks to see if they also cause false positive otr negative effects. You might compare the results from the Benedict's test with those from a blood-sugar glucometer (the solutions may need to be diluted in both tests to stay within a measurable range).
BTW, study of yeast fermentation of sucrose is responsible for the name enzyme, meaning leavened with yeast, so one could say invertase is the epitome of an enzyme.
Please share your experimental results here.


text:
Why is microcosmic salt called so?
I really see no relation between the terms microcosmic and the salt's structural formula, $\ce{Na(NH4)HPO4.4H2O}$.
Devoting sufficient time on Google didn't help at all.
Information about the name of this compound is what I'm searching for, thanks in advance for sharing.

From Meriam Webster Dictionary Online
translation of New Latin sal microcosmicus; from the fact that it was originally obtained from human urine- still not  enough  information to discern the link between microcosmicus and urine.

Devoting sufficient time on Google didn't help at all.

Hmmm... I found this PDF in just a few seconds, where we read the following.

The origin of the name microcosmic salt.
Beads of sodium metaphosphate
saturated with some oxides cool to a clear glass. Small crystals grow
in the glass on repeated heating and cooling. A microscope with a 80
to 100 x magnification is needed to inspect these crystals with
various morphologies. Such beads resemble a microcosmos. Some authors
tried to use the effect for analytical purposes in the 19th century,
but these efforts failed. Borax beads sometimes show the same effect.

Microcosmos here is understood (as usual) as a "small world", a "world in miniature".


text:
Calculating Molar Volume from crystal volume
I am trying to calculate molar volume of certain crystal lattices to be used for crystal growth kinetics calculations. The information I have is the crystal volume & crystal lattices. For example the crystal $\ce{Cu41Sn11}$ has the structure cF416 and the information taken from a certain crystal database is as follows:
Atom Formula: Cu41.5Sn10.5
Cell volume:    5.81258 nm3
Cell density (calculated):  8.85 Mg m-3
Now I have tried 2 approaches to it.

First was the wikipedia formula for molar volumes

Vm= NAVcell / Z
molar volume = (Avorgados number x crystal cell volume) / number of formula units in a crystal
This gives the following output for the above mentioned crystal
Molar Volume = 6.022 x 1023 x 5.81258 x 10 -27 / 51
Molar volume = 6.731 x 10 -5

The second one is i figured out the formula

Molar volume = Molar Mass of crystal / Cell Density
So in this case:
Molar mass of crystal $\ce{Cu41Sn11}$ = 41 x 63.546 + 11 x 118.71 = 3911.2 gms (3.9112 Kg)
Molar Volume = 3.9112 / 8850 = 4.41944 x 10 -4
These two approaches yield two different results, so i wanted to make sure where am i making the mistake?
In case you're interested in the crystal database, here it is:

well figured out the answer, I had to take the correct value of Z from the database. The difference in values was because of the incorrect value of Z (which was actually 8) . When you use Z =8 , the two solution yield 'approximately' equal results.


text:
How could a rechargeable battery recharge itself?
I'm an engineering student preparing for my graduation, in one step of my graduation project I used a rechargeable battery(1.2 volt), i was shocked when this battery is starting to recharge it self until it reach 1.04 volt (not fully) for many many cycles and immediately when the voltmeter reads 0 volt from it, when i asked my doctors they told me that maybe its chemical issue.
so is there any explanations please?

It depends on how it was discharged. Most batteries will recover (somewhat) and have a higher voltage after a rapid discharge. When the battery is in use, the internal resistance increases with a high discharge rate. When the battery is removed from the circuit, the internal resistance recovers. The internal resistance has a dependence on ion flow, concentrations, interactions at the electrodes and the redox reaction (there may be others).

As the comment indicated, it is probably equilibrating. When you are discharging the battery, you are causing the materials in the electrodes and the electrolyte to rearrange. The surface of the electrodes are where the reactions are happening, so the products of the discharge reactions tend to accumulate at or near the surfaces of electrodes. This causes the measured potential to go lower.
Once the discharge stops, the contents start to homogenize within the cell bringing the discharge products away from the electrodes and bringing the material from the charged state closer to the electrodes. This in turn increases the measured potential.

thank you very useful . but please how much times this cycle will repeat ?

Not sure what you mean by how much. This will happen every time you discharge or charge with significant current. If you are asking how long this relaxation will take, that depends on a lot of parameters including the SOC, SOH, viscosity of the electrodes, diffusion rate of the electrolyte ions etc...

thank you a lot :) but please can you give me a paper name or a book name that illustrates this that i can study , because i have to mention this in details on my graduation project documentation , i'm an electrical engineering so i don't have a good background about this :(

Your best bet is to read basic electrochemistry books like Bard&Faulkner, particularly chapters explaining concentration gradients and controlled current experiments. I don't know of any specific document applying that to batteries.


text:
How to find the average half life of radioactive nuclide which undergoes two different decays?

Find the average life of a radio nuclide which decays by parallel paths,
  \begin{align}
A  &\rightarrow B\\
2A &\rightarrow B,
\end{align}
  where the decay constants are $\lambda_1 = \pu{0.018 s-1}$ and $\lambda_2 = \pu{0.001 s-1}$, respectively.

I used the formula
$$\frac{1}{t} = \frac{t_1 t_2}{t_1 + t_2,}$$
where $t$ represents the mean life. The equation can be written as:
$$\frac{1}{t} = \lambda_1 + \lambda_2$$
But I am getting the correct answer only if I take $\lambda_2$ as 2 times the given decay constant for the second reaction. Is this because of $\underline{2}A$ on the reactant side instead of $A$?

A source for the exercise would really go a long way as this is a very strange problem.

The first answer and question is misleading, second order reactions do not occur during radioactive decay. I think that you need to consider branching decays such as that of bismuth-212. This will give you a better way of dealing with a real world problem of multiple reactions from the same nuclide.

But I am getting the correct answer only if I take $\lambda_2$ as $2$ times the given decay constant for the second reaction. Is this because of $\underline{2}A$ on the reactant side instead of $A$?

Yes.

The way I tend to approach half-life problems is to recast them as the relevant kinetic differential equations.  Assuming the first-order kinetics as stated in the problem, the contribution of reaction 1 is
$$
\left(\frac{\mathrm{d}[A]}{\mathrm{d}t}\right)_1 = -\lambda_1[A],
$$
and that of reaction 2 is†
$$
\left(\frac{\mathrm{d}[A]}{\mathrm{d}t}\right)_2 = -2\lambda_2[A].
$$
Implicit in the above is the assumption that the first-order decay constants apply to the stoichiometry of the reactions as-written. As you correctly note, that extra factor of $2$ in the differential equation for reaction 2 is because two $A$ nuclei are involved per "unit" of reaction progression.
The total rate of loss of $A$ is the sum of the above:
$$
\frac{\mathrm{d}[A]}{\mathrm{d}t}
= -\lambda_1[A] -2\lambda_2[A]
= -\left(\lambda_1 +  2\lambda_2\right)[A].
$$
This is a straightforward first-order ODE, readily solved for the overall half-life by standard methods.

† While it seems quite unusual to me to have a second-order reaction ($2A\longrightarrow B$) with a first-order rate constant (units of $\mathrm{s}^{-1}$), for the sake of the problem I'm content to roll with it. I'm sure they cast the problem this way so that the calculus was straightforward.
As well, simultaneously having $A \longrightarrow B$ and $2A \longrightarrow B$ processes would seem to violate all kinds of conservation laws--the problem makes no physical sense as a result. Plus, I know of no second-order nuclear process that occurs outside of a particle accelerator. $2A \longrightarrow C$ would have been much less preposterous, even if implausible in the context of low-energy radioactive decay.

I think that both the person asking the question and hBy2Py have both got things rather wrong. Have no fear a nuclear worker is here.
There are some radionuclides such as neptunium-238 which can decay in multiple  reactions.
For this neptunium isotope there are three different modes, alpha, beta- and beta+. We have three rate constants.
Lalpha, Lbeta+ and Lbeta-
The rate constant for the disappearance of the radionuclide is given by summing the rate constants for the three routes to get the total rate constant for radioactive decay.
Lphysical = Lalpha + Lbeta+ + Lbeta-
Things get more complex if the neptunium is inside the body of an animal. Then we have to consider the biological half life. If Tbio is the biological half life for neptunium in the animal then we can write
Kbio = ln(2)/Tbio
Then Ktotal for the rate of loss of the neptunium from the animal (first order reaction) is given by
Ktotal = Kbio + Lalpha + Lbeta- + Lbeta+
If we have a radionuclide whose decay does not branch then the algebra is a bit more simple. Consider Cs137
Then we can write
Teff (effective half life) = 1/((1/Tbio)+(1/Tphys))
Tphy is the phsyical half life which is the radioactive half life.

I am very curious about the particular example of such a real case, as I cannot remember any. It is common to have parallel decays ( E.g. potassium-40 decays by all 3 beta decay modes -- e-, e+, e-capture). But A -> B vs 2A -> B rather smells by mass/energy conservation law breach of a not well thought through made up task.

If we consider a nuclide which has two decay branches, say for arguments sake that 90 % of decays are alpha and 10 % are beta-. Then 1 in 10 of the atoms will undergo beta decay while 90 % undergo alpha. Then the rate constant for the beta decay will be ten times lower than the rate constant for the disappearance of the nuclide

That would be clear - for usual branching. But these branches will have different products, not like for the OP, not speaking about different reagent /product ratio. Unless those 2 reactions are some error. // If the 2 at 2A is a typo and should be just A, I can imagine  $\ce{^{40}K -> ^{40}Ar + e+}$ and $\ce{^{40}K  + e- -> ^{40}Ar}$

I think that the OP dreamed up something which is possible to write maths for but is impossible to happen in real life

Rather, she(?) had some funny (or sad) teacher or textbook.

Maybe, sadly there are some dreadful textbooks out there written by reckless people who write about things which they no next to nothing about. Also some teachers are not too bright, there is also the problem of students getting the wrong end of the stick also

@Poutnik Actually OP is a male and his name is one of the common Indian male names. Coming to the textbook expect, I can confirm that he might be using the school curriculum textbook (which I had to also use, but I never took it seriously). I advertise this all over the site but it seems like some people doesn't take this lightly :(

@NilayGhosh Thanks for the correction. About the textbook, it reminds me the book by Richard Feynman "Surely you are joking, Mr. Feynman!" about his criticism of Brazilian textbooks of Physics.

@Poutnik is right -- this is a homework problem (from the era before the [homework] tag was removed from the site, fwiw). Please don't give OP a hard time about the physical absurdity, it's not their fault.


text:
Modeling of photocatalytic and photoelectric devices on a macroscopic device level
As a computational materials chemist studying photocatalysis, how close to realistic systems can your modeling get?
I have a sort of high-level question about the theoretical modeling of photocatalysis. Most of the computational work on photocatalysis involves DFT calculations looking at band gaps, free energy profiles, and reaction pathways. I have also seen a handful of papers that use reactive MD to probe how the surrounding water molecules get affected by adsorbed species.
As a computational chemist or materials engineer, and you want to take your work closer to a real system, what is the logical next step? Would it be device modeling, where you take inputs from DFT (such as overpotentials), and then try to calculate the I-V curves, for instance? Are there any other multi-scale modeling techniques or coarse-grain methods that can be used?
I guess this question can be extended more generally to different classes of applications, such as $\ce{CO2}$ sensors, other electrochemical reactions, and so on. I'm not really sure what the next step would be to go beyond DFT in order to model more macroscopic length and time scales.

At first sight the question appears overly broad for me.

@Wildcat: I understand, but I'm not really sure about how to make it more specific. I have edited my question a bit though.

Look,
I try to give you a broad answer.
If you want band gaps, you need electronic structure that means DFT!
If you do MD that means classical mechanics means no electronic structure no band gaps.
You want to have solvent effects for your, I presume large system, then DFT is costly, you would be left with QM/MM or sometimes called double resolution.
Reactive forcefields such as Reaxff use some valence bond theory to treat electronic structure which is low level of theory and can not be extended to inorganic compound unless it's parametrized.
Finally all these methods are highly approximate, but for engineering purposes everything is approximate.


text:
What is the cause of hygroscopy?
Hygroscopy is the ability of a substance to attract and hold water molecules from the surrounding environment.
Why are some compounds hygroscopic and others are not? What is it about their structures that allows them to hold water? Is it due to the atoms being arranged in a certain structure due to bonding?
How can you tell if a compound is more hygroscopic than another?

Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center.

I was also wondering this as well. Can ions be hygroscopic, and if so, are ions like $\text{Na}^+$ and $\text{Cl}^- $ be considered hygroscopic?

related http://chemistry.stackexchange.com/questions/33742/what-makes-a-compound-a-dehydrating-agent

Hygroscopy is macroscopic property - not applicable to single ions or molecules.

Ions are hygroscopic because they form very stable, low energy hydrates, which are thermodynamically favorable over water vapor. I.E. salt attracts water vapor because the hydrated salt is lower energy than the water vapor. Other very porous materials are hygrosopic due to capillary action. Water vapor is attracted to the surface initially due to a variety of reasons, then the narrow channels in the porous material cause the water to spread out to lower their energy.

Also, it is worth mentioning that formation of hydrates is a property of compound as a whole, rather than individual ions. See, for example: neither $\ce{NaCl}$ nor $\ce{CaCO_3}$ form hydrates, but both $\ce{Na_2CO_3}$ and $\ce{CaCl_2}$ do, the latter quite eagerly.

Some salts have high tendency to be hydrated, like Li salts. The reason is that their hydration energy is very high, meaning that when they are hydrated, lot of energy is released. Since lower energy is always more stable, such salts tend to show high hygroscopy.


text:
How does the following work in Alkyne?
I have the following reactions from a book, but it is not explained how the different substances come into being. I don't know what happens to $\ce{3H2O (H3O)}$ in the second line.

Pretty much the same thing as in next reaction - do you get the reaction with peroxide?


text:
Do aromatic dienes undergo the Diels-Alder reaction?
Do aromatic dienes undergo Diels-Alder reactions?
My understanding is that the $\pi$ bonds in aromatic compounds are not as localised as they are in other compounds, such as aliphatic dienes. Therefore, the dienophile shouldn't be as willing to go to these compounds. Am I correct?

Yes, you are correct.  Aromatic compounds are stabilized by their resonance energy. This means that the activation energy for reaction of an aromatic compound will typically be greater than that for a similar reaction of a non-aromatic compound.
That said, aromatics will react as the diene component in a Diels-Alder reaction if

the dienophile is extremely reactive (o-benzyne for example)

if the aromatic system is destabilized (for example by steric interactions or structural constraints) so that the resonance stabilization is decreased

or if the aromatic system is large enough that loss of one aromatic ring does not significantly destabilize the entire system.  Anthracene is such an example where the middle ring will react via a Diels-Alder reaction.

(image source)

But then why does benzyne undergo the Diels Alder reaction with Furan? Both of them are aromatic! Is it because benzyne is highly unstable?

Yes, o-benzyne is a high energy species, therefore extremely reactive.

What is this o-benzyne? Is it any different from benzyne or are they just different names for the same thing?

there is ortho-, meta-,  and para-benzyne, depending whether the removed hydrogens are located 1,2, 1,3 or 1,4.

Does benzyne count when answering this question? Since the pi system involved in the cycloaddition is orthogonal to the aromatic pi system...

@PCK  Benzene, paracyclophane and anthracene are used above as examples of aromatic pi systems that participate as the diene componenet in a Diels-Alder reaction.  The in-plane ~ $\ce{sp^2}$ orbitals in o-benzyne are reacting as the the olefinic component of the DA reaction.


text:
Can aluminum and hydrochloric acid be used to generate hydrogen gas?
I am looking to produce hydrogen gas in a small amount with supplies I currently have. Most of the methods I have seen involve reacting zinc or magnesium with $\ce{HCl}$ to obtain the gas, but I do not have zinc or magnesium at the moment.
Is there any reason why aluminum cannot be used just as effectively as zinc? According to the reactivity series, aluminum should be even more reactive than zinc.

Obviously, as has been shown, Aluminum works well.  Is there a reason why it can't be used as effectively as Zinc?  Yes.  1) It tends to take some time to start reacting because of its oxide coating and 2) In my experience it is more reactive so it sometimes is harder to keep cool and you end up with more steam and vapor.  I prefer Zinc because it tends to start easier with HCl in an ice bath, and not accelerate as fast as it gets warmer.  You can scour the Aluminum to get it started.  Iron works too.

Yes. See this video for a demonstration.
The reaction is:
$$\ce{ 2Al (s) + 6 HCl (aq) -> 2AlCl3 (s) + 3H2(g)}$$
You can capture the hydrogen and proceed with your experiments.

Oh yes it worked, I posted my video and process here http://www.volatileinterface.com/generate-hydrogen-gas/

The fact that aluminium may reduce H(+I) to H(0) can be made clear by studying the standard reduction potentials. By definition, the reduction potential
$$E^0_{\ce{H^{+1}} \ \rightarrow \ce{H^{0}}} = 0 V$$
From a regular Latimer diagram, we see that the reduction potential
$$E^0_{\ce{Al^{+3}} \ \rightarrow \ce{Al^{0}}} = -1.676 V$$
Since you are studying the opposite reaction, oxidation of Al, we change sign of the potential. Thus the total $E^0$ for the reaction (at standard conditions) is positive, which indicates a spontaneous reaction (although the kinetics may be slow).
In addition, the precipitation of $\ce{AlCl3}$ drives the reaction toward $\ce{H2}$ production.

Magnesium or calcium would also work well, but they are more expensive. Aluminum is a safe bet.


text:
What is the stochiometric equation of the sucrose and potassium chlorate reaction with sulfuric acid?
I am researching this reaction for my chemistry assessment. Various sources I have dug up state different answers and opinions on what the stoichiometry is and what the products are, sometimes giving no answer at all.
In one source, sulfuric acid is a catalyst that speeds up the reaction while in another source, sulfuric acid is one of the reactants, and is therefore part of the stochiometry. The final products also vary from source to source, and include pure carbon or potassium chloride in addition to carbon dioxide and water.
Can anyone clarify the subject for me?

In this paper [Sugar Dehydration without Sulfuric Acid: No More Choking in the Classroom! T.P. Silverstein and Y. Zhang. J. Chem. Ed. 75 (6) 748-749 (1998)], the authors outline two reactions involving sucrose and sulfuric acid. The first one does not include $\ce{KClO3}$ and utilizes a significant amount of $\ce{H2SO4}$ in order to fuel the dehydration of sucrose: this is not the reaction you're interested in.
The second reaction detailed therein makes use of a very small amount of sulfuric acid to start the reaction (a drop is the commonly-cited amount): the sucrose is oxidized by $\ce{KClO3}$. This is the reaction you want to examine.
The authors give the reaction stoichiometry as:
$$\ce{C12H22O11(s) + 8KClO3(s) \xrightarrow{\ce{H2SO4}} 12CO2(g) + 11H2O(l) + 8KCl(s) + heat}$$
They note that the fumes from the reaction contain $\ce{Cl2, ClO}$, and $\ce{ClO2}$, but the amounts are not quantified. I suspect (but don't know for sure) that the amounts are finite but negligible. And even with the scant amount of sulfuric acid used here, I think you're safe going with the reaction as presented above.


text:
How Many mL of 1.0M HCl Would be Required to Completely Decompose Sucrose into Glucose and Fructose?
I am currently conducting an experiment into the glucose content of certain sports drinks. However, I have hit a roadblock. A majority of the drinks that I intend to test, contain sucrose. I am thinking that by adding $\pu{1.0M~ HCl}$, to say, $\pu{100mL}$ of solution (sports drink), it will decompose the sucrose into its constituents. I'm planning to use Benedict's solution and a spectroscope in order to determine the amount of glucose present in solution (I'm going to add activated charcoal to the solution to remove the coloring from it before placing in the spectroscope). As a high school student I only have limited resources.
So my question is,
Assuming there is 10 grams of sucrose per $\pu{100mL}$ of solution, what volume of $\pu{1.0M~ HCl}$ would be required to completely break apart sucrose into glucose and fructose?
Would I have to use stoichiometry to find the amount?
So far I know I have $\ce{C12H22O11 + HCl}$, What happens to the $\ce{Cl}$ from $\ce{HCl}$ when fructose and glucose is formed?

Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center. Please visit [this page](http://chemistry.stackexchange.com/help/notation " latex="" at="" chemistry.se""="), this page and this one on how to make your future posts better.

Um, you do know what the products are... You even told us what they are.

I understand that fructose and glucose are formed...But what happens to the Cl of the HCl?

The hydrolysis of sucrose is acid-catalysed, so HCl will not be in the chemical equation (it will be listed above the arrow). In theory, this means that you could have little HCl, as it will never be used up and will keep reacting to decompose all of your sucrose. So it isn't so important how much you use, but you should still use enough that it won't take ages to decompose all the molecules. If you're interested you could look up mechanisms for acid catalysis, but you wouldn't be expected to know this at high school level.
To find the chemical equation for the decomposition, you can add up the number of elements of fructose + glucose and compare to sucrose. You will find that two hydrogens and one oxygen is needed, so it is simply a hydrolysis reaction (adding a water molecule).

authors: امنين محمد عبد الكريم