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Is there any chemicals that I can add to soy bean wax
I recently decided to make homemade candles in my spare time since all 6 kids have left the house.
After much research and trial and error and reading many articles, I have found that Soy wax does not release fragrance oil well on its own, and needs an additive/chemical of some sort, but being that I am not a chemistry kinda gal, I don't want to buy certain acids or whatever and risk blowing up my kitchen, or making deadly fumes just because I am not aware of what chemicals can or can't go together.
You can add additives such as Vybar to certain waxes such as paraffin and it will help the color retention, and a few other things, but I am wanting something in the raw form that I can add to the soy wax before it solidifies so when the wick is lit, the fragrance will smell. Soy wax alone has a reputation of not releasing fragrance without using an additive/acid/compound to it. I just don't want to mix something that would be, or could be dangerous.
Any suggestions would be greatly appreciated..

Then why use this kind of wax in the first place?

Because they say it is safer for you to breathe then the toxic's that come from paraffin wax. And of course it helps support the farmers. Yet, you can't add a bunch of fragrance, or it will rise to the top and basically drown out the candle, and still won't smell it!

What about beeswax?

Beeswax is a quite a bit higher in price, and alters the scent. So pumpkin won't really smell like pumpkin, because of the wax type. However if you use  paraffin, the old fashion wax you will see the black smoke rise on the glass/container up the walls or just black smoke in the air.  I know there are additives I can throw into soy wax just like when ppl use paraffin wax they use vybar and its an additive to promote various things, I just was looking for something like stearic acid to force the soy wax release the fragrance when the wick is lit. Thank you tho!


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What happens when pyridine N-oxide reacts with dichlorocarbene?

It seems that the combination of sodium methoxide (a base) and chloroform will generate dichlorocarbene, $\ce{CCl2}$, by α-elimination.
What happens after that, though? It is an ortho formylation like in the Reimer–Tiemann reaction? Or does the nucleophilic oxygen simply attack the carbene?

Dichlorocarbene has been shown to deoxygenate pyridine N-oxide producing phosgene as an initial byproduct. In this case, since methoxide is present, dimethyl carbonate is ultimately formed.[1]
The proposed mechanism involves nucleophilic attack on the carbene by the oxygen, followed by an elimination:

(1) Schweizer, E. E.; O'Neill, G. J. A Novel Deoxygenation Method for Pyridine N-Oxide. J. Org. Chem. 1963, 28 (9), 2460–2461. DOI: 10.1021/jo01044a516.

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Do proteins have a smell?
Proteins are heavy and not so volatile. Does it mean that pure proteins do not have a smell?

Sounds about right. Then again, proteins are hard to make pure.

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What is the geometry of the intermediate during an SN2 reaction?
Suppose we have 2-bromo butane. The carbon connecting to the bromine will have the following valence electron configuration:
$$\underset{2s}{[\uparrow]}\underset{2p}{[\uparrow \vert \uparrow \vert \uparrow]} $$
This hybridizes to an sp3 configuration which yields a trigonal pyramidal configuration.
Suppose in an SN1 reaction the bromine comes off. Then the connecting carbon will have a valence electron configuration of
$$\underset{2s}{[\uparrow]}\underset{2p}{[\uparrow \vert \uparrow \vert\;]} $$
which yields an sp2 or trigonal planar configuration.
Suppose we instead have an SN2 reaction such as with NaOH. Then the intermediate would be a carbanion that has a carbon that has 5 bonds.
$$ \underset{2s}{[\uparrow ]} \underset{2p}{[\uparrow \vert \uparrow \vert \uparrow]} \underset{3s}{[\uparrow]} $$
would be the wrong configuration but I'm having trouble understanding what the right one would be.
A trigonal bipyramidal molecular geometry seems to me like it'd fit the shape of what the carbanion would be but the numbers don't seem to add up to me.
In a trigonal bipyramidal geometry there needs to be two $sp$ orbitals two point up and down.
I learned from How are the hybrid orbitals of sulfur hexafluoride shaped? that higher orbitals are usually only slightly involved so the resulting bonds would only have a small d characteristic. For example, instead of a $sp^2d$ hybrid something like a $sp^{2.75}d^{0.25}$ characteristic is more likely.
Would the rest of the bonds have something like a $sp^{2.67}d^{0.33}$ characteristic?
According to molecular orbital theory what would the shape of that carbanion be and how would the groups attached to it be arranged?

Molecular orbitals theory works fine, but it is incompatible with hybrid orbitals. The transition states involves a 3-center-4-electron bond.


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What are the half-equations of manganese and oxygen in the following redox reaction?
The equation is $$\ce{2Mn^2+(aq) + 4OH^-(aq) + O2(aq) -> 2MnO2(s) + 2H2O(l)}$$
I know that the Manganese (II) gets oxidized to Manganese (IV), but I'm not sure about the other half. I know oxygen gets reduced, but using simply $\ce{O2 +2e^-->2O-}$ doesn't work because the equations won't balance out then.
By the way, this chemical equaltion is used in the Winkler titration.

No wonder it won't balance. What is the oxidation state of O in the products?

The reaction doesn't look right for what occurs in the Winkler titration

@Tyberius There are different views for what the first equation is in the winkler method. Wikipedia mentions this.

The half-equations should look like this:
$$\ce{2Mn^2+(aq) \to 2Mn^4+(aq) + 4e^-}$$
and
$$\ce{4OH^-(aq) + O2(aq) + 4e^- -> 4O- + 2H2O(l)}$$

Let us consider these half equations:
$$\ce{Mn^2+(aq) + 4OH^-   \to MnO_2(aq) + 2e^- + 2H2O}$$
and
$$\ce{ 2H2O(l) + O2(aq) + 4e^- -> 4OH- }$$
By multiplying the first equation by two and adding it to the second one, we get the overall reaction.

Are these actually half equations though? Is this one correct or is the first answer correct?

Both answers are correct. However, I prefer mine, because they give directly the overall reaction . The first half equation shows the oxidation of Mn (II) ion into manganese oxide (IV), while the second half equation shows the reduction of two oxygen atoms into ion hydroxide (with the oxidation number -II). Please notice that in the second half equation, the other two atoms of oxygen in the left side keep their oxidation number (-II) as hydroxide ions


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How does one understand the product of a Diels Alder reaction?
am having trouble making sense of products of diels alder reaction. Like I understand the stereochemistry and everything but what I am confused about is how come textbooks don’t give a variety of products that is possible in a diels alder reaction. Like, I know that pericyclic reaction is all about the symmetry of lobes of dienphile and diene. That is exactly how the difference between antafacial and superfacial come in, in that in antarafacial since the lobes are not in symmetry, you have to twist the molecule to get symmetry. So in the same sense, I have realized that when u do diels alder reaction depending on the way you organize the lobes of the diene and dienophile you can get so many possible products. However textbooks don’t really mention this phenomenon at all. I have shown what I mean by different lobe symmetry in the picture below. Can anyone tell me what is going on or am I overthinking this or what?

You are best served by looking up in a textbook "regioselectivity" and "stereoselectivity" of "Diels-Alder reactions". Any good organic chem textbook (e.g. Clayden) will have a section on it.

I think the answer to the first question is that it does not serve beginning students to know the great variety of Diels-Alder reactions.

Okay. Orthocresol, yes, sometimes, when there are certain substituents that make preference for one product over another, this question of lobe symmetry and diferent products does not matter. But when I was thinking of this question, I was thinking in terms of those substituents that don't really make this preference.

The diene would be too strained to undergo an antarafacial reaction. Was that your question?


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Concentration of gas in water: What is the time derivative of concentration toward steady state?
The Henry's law gives the steady state concentration for a gas in water. But it does not tell about the speed for this to happen if the concentration is not in steady state. What is the time derivative of concentration of oxygen in gas?

The transient diffusion equation for the oxygen concentration is $$\frac{\partial C}{\partial t}=D\frac{\partial^2 C}{\partial x^2}$$where D is the diffusion coefficient.  Henry's law applies at the interface.


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If there are multiple bases in solution and a strong acid is added, will both bases react?
For example, if there is one mole of carbonate ions ($\ce{CO3 ^2-}$) and sulfate ions ($\ce{SO4 ^2-}$) in a solution, and one mole of hydrochloric acid ($\ce{HCl}$) is added, will the one mole of hydrochloric acid react just with the sulfate ions as that is stronger, forming hydrogen sulfate ions?
Or will the acid react with both the ions forming hydrogen carbonate ions ($\ce{HCO3-}$) and hydrogen sulfate ions ($\ce{HSO4-}$)?

The key concept you need to understand is that acid-base reactions are relatively fast, so you'll do whatever the equilibrium for the mixture dictates.

Sulfate ions are much weaker than carbonate ions.  Based on my answer below this implies that even if the acid reacts with both bases, the protons will ultimately end up almost all on the carbonate (or on water molecules from any  carbonic acid that forms and decomposes before we can get to full equilibrium).

Both based will indeed react with the acid.  But wait, there's more.  When the bases react to form their conjugare acids, the conjugate acid from one base could react with the other base.  When one base is strong and the other weak, this secondary reaction will favor the direction where the strong base gets protonated.  So you see the strong base gets more protons while the weak base tends to be pushed back to the basic form.
What if both bases are similar in strength?  Then we apply equilibrium thermodynamics.  Say hydrochloric acid is added to a solution containing both sodium acetate and sodium nitrite.  The equilibrium between each of the latter two ions is governed by the appropriate acid dissociation constants:
$\ce {HC2H3O2 = H+ + (C2H3O2)-} $  $K_{a1}={\ce {[H+][(C2H3O2)-]}}/{\ce {HC2H3O2}}$
$\ce {HNO2 = H+ + (NO2)-} $  $K_{a2}={\ce {[H+][(NO2)-]}}/{\ce {HNO2}}$
To get the equilibrium that relates the competing weak acids and bases directly, just divide to eliminate the hydrogen ion concentration; in this case
${\ce {[HNO2][(C2H3O2)-]}}/{\ce {[(NO2)-][HC2H3O2]}=K_{a1}/K_{a2}}$
So your acid dissociation constants aren't just about acid dissociation.  They also govern how a limited supply of protons gets distributed between competing bases.

Though your title question would be difficult to answer for a general case, your example is pretty straight forward. Carbonate is a much stronger base than sulfate, so you would convert a mole of carbonate to a mole of bicarbonate before any significant protonation of sulfate would occur. And in the general case, it will be the stronger base that is protonated first, but the picture can get complex with bases of similar basicity and such.


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Finding theoretical mechanism of sodium thiosulphate and hydrochloric acid
In class we conducted an experiment where we timed reactions between sodium thiosulphate and hydrochloric acid. We kept the concentration of the $\ce{HCl}$ constant, while changing the concentrations of sodium thiosulphate.
We then graphed this data to discover there is a linear relationship between the concentration of thiosulphate and 1/time, indicating the it is a first order reaction. This means in the rate determining step there will be one thiosulphate ion.
In the analysis question it asks you to come up with a theoretical mechanism. I'm having alot of trouble with this because the first order relationship found on the graph indicates there is 1 thiosulphate ion in the rate determining step, but the overall equation also has one thiosulphate ion.
Can anyone help me out, I have tried multiple methods but none seem to add up to the overall equation.
$$\ce{2 HCl (aq) + Na2S2O3 (aq) -> 2 NaCl (aq) + SO2 (g) + S (s) + H2O (l)}$$

I think you are running into problems partly because the reaction isn't first order and partly because the real mechanism is fairly complex. It looks like this lab came from Flinn Scientific and they suggest a pretty convoluted mechanism for the reaction.

You can format mathematical and chemical expressions on Chemistry.SE using MathJax; this post contains further details.


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Accelerated stability testing for chemicals: Shelf life / Expiry Date
How does one use the results of Accelerated stability testing for chemicals to determine the typical Shelf life  and Expiry Date?
e.g. Suppose the typical storage is going to be at 25 C and the expected shelf life is 2 years (as printed on the labels or the best-before date).
Now in the interest of time suppose we test at a higher temperature how do we extrapolate those numbers to the typical shelf life at room temperature.
Any standard protocols for this? Also what's a reasonable high temperature to use for the accelerated tests. I am dealing with a fairly temperature stable & nonreactive material here. 60 C?
Stability testing seems quite common in the Pharma industry etc. so I'm sure there's norms out there.
Note that I'm not dealing with a drug / medicine / food etc. This is strictly for an intermediate which will not be consumed directly by the end user.

What kind of intermediate is it? Packaging, excipient, processing additive (removed later)? Have you checked for FDA and USP guidance docs (or whatever applicable govt. body or pharmacopoeia is applicable where you live)? ASTM has documentation of accelerated aging, but I think it is mainly for materials.

@J.Ari Thanks. It's a molecule that goes through 3-4 more processing steps to lead to the API.

The molecule itself does not have a USP / FDA standard monograph. My two big questions are what temperature to test it at and how to extrapolate the accelerated testing results to typical temperatures.

Your accelerated test temperature will likely be a function of what the final product (or some representative intermediate hold state) must be kept at. If everything has to be kept cold (for ease we will say cold = not room temp), then studying at room temperature could be sufficient. I'm less sure about room temperature stable products. I would propose going to worst case storage temp of the product. This article below may help you figure out the extrapolation.
http://www.biopharminternational.com/assessing-shelf-life-using-real-time-and-accelerated-stability-tests?id=&sk=&date=&pageID=4

@J.Ari Sorry, I should have been clearer: Sure studying at Room Temp. would be sufficient. The accelerated testing is only to save time. e.g. We believe 2 years is a reasonable shelf life based on what we know. But we don't want to wait for two years for test results. So we raise T and extrapolate results to room temperature. That, I believe, is the motivation behind accelerated testing.

So you want to find out how long to test at accelerated conditions to mimic storing at room temp for 2 years, is that correct?

Let us continue this discussion in chat.

Yes, there are guidelines. For example from ICH and WHO for pharma, and from CIPAC for pesticides. Some can be downloaded freely.
Yes, there are guidelines. For example, from ICH and WHO for Pharma, and from CIPAC for pesticides. Some can be downloaded freely.
Accelerated stability tests are generally not accepted as replacements for full-scale stability tests, but can provide an indication of the formation of undesirable by-products etc.
As a rule of thumb, a very general rule, chemical reaction rates in-/decrease by a factor of 2 for every 10C the temperature changes. According to this rule, if the test temperature is raised from 20C to 30C, the test period could theoretically be halved. However, if you look through the temperature/time combinations in the guidelines, you will find this relationship hard to find :-).

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Dess–Martin oxidation work up
I'm doing an oxidation of an alcohol with Dess–Martin periodinane which works really well but having trouble with the work up. Doing this reaction with about 5 g of alcohol so the reaction ends up using about 17 g of DMP. As the reaction progresses the DMP byproduct crashes out. And working up turns it into a gun that is quite difficult to remove
Running reaction in DCM and work up with bicarb. Even after filtration DMP is still present. Separates fine on a column but with how much there is it seems to be trappings some of my product.  I've read that a thiosulphate wash is also used in these reactions, but I don't think this will be useful to get rid of the gum.
Any suggestions?

It looks like you are not the first to get gummed up by the mono-acetoxy iodinane byproduct of this reaction. A seemingly simple solution is given in this article about converting an alcohol to ketone via DMP:

The mono-acetoxy iodinane by-pdt can be converted to easier to remove by-pdts during workup by treatment with aq $\ce{NaOH}$ or $\ce{Na2S2O3}$.

This is not a procedure I have experience with specifically, so I hope this a do-able solution for your problem as I understand it.  Good luck regardless.

The answer to your question really depends on your aldehyde, and how sensitive it is (for instance if it's chiral at the alpha position then you don't want to be doing aqueous work ups with anything basic).
1. Filtration
Filtration is by far the most convenient work up for a DMP as the filtrate is often clean enough to telescope through into another reaction.
Firstly, if your reaction isn't already buffered with sodium bicarbonate, add a spatula tip before trying to filter, it quenches out any acetic acid which will not be trapped by the column.
Secondly, add something like ether or hexanes— the DMP related stuff isn't soluble in these, so they'll just sit as solids on the top of your column compared to if you elute with a tonne of dichloromethane for example.
As you've found, it can be a little difficult as the amount of solid gets quite large and traps product. Using a sintered funnel under vacuum makes it far easier, as does using a wide but thin plug rather than a long column.
2. Aqueous workup
I often avoid this one because if I'm doing a DMP it usually means I'm making a very sensitive aldehyde and don't wish to have it sitting around removing the last traces of water afterward.
As AirHuff notes, aqueous thiosulfate washes are very effective. As previously, I'd suggest initially quenching with sodium bicarbonate (a solution this time), then extracting with ether, followed by a final thiosulfate wash, drying, and evaporation.
3. Avoid DMP entirely
As a reagent DMP is great for doing small scale reactions as you essentially add a scoop of reagent and solvent in an open flask and it's done in ten minutes.
On a large scale however the procedure generates a lot of solid byproducts, which can, as you've discovered be difficult to remove. There is also a huge cost associated with DMP, even when prepared in house. For the material you have, you'd be better served doing a different oxidation.

Thanks! I tried again today. Thiosulphate wash definitely made it more crystalline and easier to filter than the previous gum.

After a few more repeats of this turns out the base washes are essential. If your compound is base-stable then washing with higher molarity base works much better, e.g. 4M KOH

A base wash would indeed help. Though if your aldehyde is stable to strong base, I would question why you're bothering with DMP at all ...


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What might serve as an initial starting photocatalyst for this large water-splitting solar simulator?
Question: What might serve as an initial starting photocatalyst for this large water-splitting solar simulator? Surely there must have been some planned experiments!

The Gizmodo article Insane Light System Blasts the Energy of 10,000 Suns and Phys.org's Let there be light: German scientists test 'artificial sun' both describe the large, powerful concentrated light source called Synlight which has been built by the German Aerospace Center (DLR).
This large (~350 kW!) solar simulator - built from 149 high intensity xenon arc lamps, all collimated/focused to a single 20x20cm area - will be used to test and develop materials and catalysts that will allow photochemical splitting of water to yield hydrogen gas, to be scaled up later and used with a solar concentrator instead of electrically powered lamps. Hydrogen could be used as fuel directly, or to synthesize hydrocarbon fuels when combined with carbon dioxide or monoxide.
Water is transparent over most of the sun's useful spectrum, and this is for the obvious reason that the wavelengths reaching the ground are those not already absorbed in the atmosphere. Thus a photocatalyst will be necessary for a high yield of hydrogen gas from concentrated solar light.
From the Gizmodo article:

When the array is focused onto a single spot, it heats metal to 1,475 degrees Fahrenheit (800 degrees Celsius), which is then sprayed with water vapor. The metal reacts with the oxygen in water, and hydrogen remains. With further heating, the oxygen is once again separated from the metal.

There are certainly easier ways to heat a metal plate than building a giant solar simulator, so I assume the point is to provide an intense source of photons with a somewhat similar spectrum to sunlight as a test platform for various catalysts. But the quote suggests the surface will react with the oxygen released, and need to then be cycled to higher temperature to release it. This does not sound like a very practical photocatalytic process.
So far I have not found anything clearer about what kinds of catalysts are likely to be investigated, but there must be some potential candidates. Based on known photocatalytic water splitting mechanisms, or if someone can find a discussion by DLR, what would be a starting point, or proof-of-concept photocatalyst that might work here?

below: Images from here, photo credit: DLR (top), DPA (bottom). The shape of each reflector is likely to be close to an ellipsoid with the xenon arc at one focus and the sample near the other. This way a ~100 cm diameter reflector can illuminate a much smaller 20x20cm sample.
[

One possible candidate: http://www.sciencedirect.com/science/article/pii/S1364032114009265

@Tyberius Thanks! That looks like a very helpful review of this subject, exactly what I need to read.

According to wiki, the cerium (IV) oxide - cerium (III) oxide cycle is "a two step thermochemical water splitting process based on cerium (IV) oxide and cerium (III) oxide for hydrogen production." Maybe this helps.

possibly helpful; Phys.org's Water-splitting module a source of perpetual energy

Efficient light-harvesting relies on two primary factors: an efficient absorption of solar radiation, and an effective means for charge separation. The absorption of solar radiation by a photovoltaic material induces the creation of an exciton, a quasiparticle comprised of an excited electron that is Coulombically attracted to the positive electron-hole formed in its excitation. The formation of an exciton is essentially the same as simply exciting an electron, but when this motif can move through a bulk material it is often useful to frame it this way. The higher efficiency of a photovoltaic, the more excitons that are generated. The better the charge separation, the longer the excitons last and thus the more likely that their potential energy can be converted into work.
So long as the photocatalyst can 1) absorb light in the solar spectrum 2) stabilize excitons long enough for them to reach water molecules and 3) incubate excitons with energies high enough to activate water-splitting, then it would be a suitable candidate for testing with the Synlight.
The characteristics of a good photocatalyst for water-splitting would optimize the above traits (maximize efficiency of light absorption to maximize exciton formation, maximize charge separation as to increase exciton lifespan, and maximize excitons of sufficient energy to ignite the water-splitting process). There are many many candidates, but some promising candidates in the literature in re water-splitting are cerium dioxide heterostructures (CeO$_2$/___).
Cerium dioxide has promising properties for photovoltaics: cerium dioxide has good charge mobility and oxygen-species mobility due to the high reversibility of the cerium ion between its Ce$^{3+}$ and Ce$^{4+}$ oxidation states; the material is generally non-toxic (it is used in photocatalysis currently for cleaning of wastewater at wastewater treatment plants); and it is generally resistant to corrosion. It has the issue, however, that it is only efficient at absorbing in the ultraviolet range of the solar spectrum; this is due to the material's large band gap, the energy separation between ground-state electrons and excited-state electrons.
In order to ameliorate this, most research on CeO$_2$ photovoltaics focuses on perturbing the crystal structure, surface shape/topology, and purity of the material to lower the band gap and increase efficiency across more of the spectrum. Doping the material with other semiconductors or charge-transferring materials such as cadmium sulfides (CdS), zinc oxides (ZnO), and silver phosphates (Ag$_3$PO$_4$) has led to many shifts in efficiency across the spectrum. Changing the morphology of the structure—pure or doped—has profound changes as well, such as the creation of CeO$_2$/CdS nanoparticles for photovoltaics. Here is a link to a 2021 review article related to this subject by Mekonnen, An Overview on the Photocatalytic Degradation of Organic Pollutants in the Presence of Cerium Oxide (CeO2) Based Nanoparticles: A Review if you would like to learn more.
A more complex photocatalyst that could take two excitons of lower energy and combine them into an exciton of higher energy would also be ideal, as it would be able to access parts of the solar spectrum below the threshold of water-splitting activation—this combination process is called up-conversion, and it is highly technical.

Vitamin B-12 is a photocatalyst, see, for example, https://pubs.rsc.org/en/content/articlelanding/2018/ra/c7ra13037f#!divAbstract and also https://www.ncbi.nlm.nih.gov/pubmed/29271445 , but it has not been applied to water splitting!
What has been used is TiO2, see this source to split H2O into H2 and O2.
Mechanics: The action of light on TiO2 creates electrons (e-) and electron holes (h+). Some reactions:
$\ce{H+ + e- -> .H}$
$\ce{.H + .H -> H2}$
$\ce{OH- + h+ -> .HO}$
$\ce{.HO -> H+ + .O-}$
$\ce{.O- + h+ -> .O}$
$\ce{.O + .O -> O2}$
One of the less desirable reaction is:
$\ce{.H + .HO -> H2O}$

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Hydrogen bonds: 1 donor with 2 acceptors at the same time possible (or vice versa)?
Can 1 donor (eg N-H) facilitate 2 H-bonds with 2 different acceptors (eg 2x O) at the same time (assuming that both are within range and angle constraint)? i.e. building an "h-bond triangle"?
Is it possible the other way round: 2 donors and 1 acceptor at the same time?

Yes, both are possible and called bifurcated hydrogen bonds.


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How does SO2 have 2 π bonds?

The hybridization of $S$ in $SO2$ is sp2 which means it has only one unhybridized p orbital to form a $\pi$ bond. How does it form two?
Could it be one electron from a $p$ orbital getting promoted to a $d$ orbital leading to one $d-p$ and one $p-p$ $\pi$ bond. Is this even possible or are $\pi$ bonds only between p orbitals? Could the electrons possibly be delocalized in the $d-p$ and $p-p$ $\pi$ bonds?

Yes, you are right. One electron from 3p orbital is unpaired and excited to a 3d orbital. The hybridization of sulfur atom is sp2 hence a lone pair and two bond pairs(due to sigma bonding) reside in these hybrid orbitals.
The unpaired electrons are 3p and 3d hybridized orbitals are used in pi bonding with oxygen's unhybridized 2p orbitals. Hence two pi bonds are formed which are p-p pi and d-p pi bonds. Because of delocalisation these bonds are identical and have properties intermediate to the two different pi bonds.
To understand how the electron is able to make a jump with high energy difference, you will have to understand that d orbitals usually participate in bonding only when central atom is joined to many highly electronegative atoms. Because of polarization, there is a slight positive charge on the sulfur atom due to which orbital contract. This contraction is very very significant in d orbitals. Because of this electrons can be excited to the less energetic and tightly held d orbital. Also, the bonds formed will have a greater extent of overlap because d orbitals are not that diffused anymore. This is the reason that PF5 exists but PH5 does not.
Also, note that we cannot use dz^2 orbital for pi bonds as its lobes have the same phase and hence net overlap during pi bond will become zero.

It's complete non-sense and you're saying that years after "d-orbital hypervalency" was disproved.


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How to determine a poisonous or healthful plant?
I'm not a chemist or scientist, however i am interested in figuring out if the plants around me:
can provide any nutritional value
Can be eaten in salads or poisonous
I don't have any idea of the path to take to get my answers.
Going back to school isnt an option.
Please advise on the path to take to know these things.
Thank you

Library: there are pocket books and bigger references that help you identify a plant, then you can search literature if it is edible or not. Just for the record: many edible wild plant is mildly toxic.

i would advise you to not risk taking random plants, and eating it based on others opinion of what it looks like, rather, ask someone who knows such stuff, like a botanist. Or try giving pictures and asking for help at the cooking stack exchange

Ironically, the best answer is that the local elders and medicine men/women are the best sources. If you live in an advanced civilization, then you're probably out of luck unless that lore has been written down. Now, there's no guarantee that it's correct or complete--we don't just go around testing random plants on people.

I'm voting to close this question as off-topic because this is about botany and not chemistry.

There are a few books on this topic available on Amazon. They might be useful, depending on where you live.
A Field Guide to Edible Wild Plants: Eastern and central North America (Peterson Field Guides) Sep 1, 1999
by Lee Allen Peterson and Roger Tory Peterson
https://www.amazon.com/Field-Guide-Edible-Wild-Plants/dp/039592622X/ref=sr_1_1?ie=UTF8&qid=1490368083&sr=8-1&keywords=edible+plants
Edible Wild Plants: A North American Field Guide to Over 200 Natural Foods Apr 7, 2009
by Thomas Elias and Peter Dykeman
https://www.amazon.com/Edible-Wild-Plants-American-Natural/dp/1402767153/ref=sr_1_2?ie=UTF8&qid=1490368083&sr=8-2&keywords=edible+plants
Edible Wild Plants: A Folding Pocket Guide to Familiar North American Species Dec 26, 2014
by James Kavanagh and Waterford Press
https://www.amazon.com/Edible-Wild-Plants-Familiar-Naturalist-ebook/dp/B00USBMCE8/ref=sr_1_3?ie=UTF8&qid=1490368083&sr=8-3&keywords=edible+plants
Bushcraft 101: A Field Guide to the Art of Wilderness Survival Sep 1, 2014
by Dave Canterbury
https://www.amazon.com/Bushcraft-101-Field-Wilderness-Survival/dp/1440579776/ref=sr_1_4?ie=UTF8&qid=1490368083&sr=8-4&keywords=edible+plants
Identifying and Harvesting Edible and Medicinal Plants in Wild (and Not So Wild) Places May 20, 1994
by Steve Brill and Evelyn Dean
https://www.amazon.com/Identifying-Harvesting-Edible-Medicinal-Plants/dp/0688114253/ref=sr_1_5?ie=UTF8&qid=1490368083&sr=8-5&keywords=edible+plants

I don't think there is a scientific approach that would be useful here.
I advise you to buy a book on wild plants that can be eaten, maybe look for a book specific to your region.
Identifying every plant, then checking if you can eat it would take a long time, so the faster approach would be to see what is edible, and then look for the plant.

Ok, thanks guys for all your responses.  I thought I needed a scientist to help me with this issue so that's why I posted here. But your responses have helped to steer me in a more accurate direction.


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Colligative properties and sizes of particles
Why do colligative properties only depend on the number of particles and not the size?
*My argument being that since the root cause of these properties is the occurrence of solute particles on the surface, reducing the vapour pressure of the solvent, shouldn't bigger molecules, such as urea, give a much more appreciable deviation in properties, as compared to smaller ones like NaF?

Colligative properties depend upon the "relative number" of solute particles and not on the number of solute particles.

relative? i beg your pardon, am i missing something? @Cotton

Consider two solutions-A and B. Say the number of solute molecules in A is 5 and number of solvent molecules is 10000. In B, the number of solute molecules is 5 and number of solvent molecules is 10. In which solution do you think relative lowering of vapour pressure will be more?

Wait, colligative properties were applicable for dilute solutions, right? or is it valid for conc. solutions too? @Cotton

Of course they are valid for both dilute and concentrated solutions.

You are confused probably because the formule given in your textbook are valid for dilute solutions. But they are just special cases. Colligative properties are applicable for both dilute and concentrated solutions.

Please. PLEASE  pay attention to the tags you're creating. You've been creating a lot of typo-y and wrong tags lately, and they've been a pain to clean. Don't edit, and take a second look at the tags on your questions.

Colligative properties From Wikipedia

In chemistry, colligative properties are properties of solutions that depend on the ratio of the number of solute particles to the number of solvent molecules in a solution, and not on the nature of the chemical species present. The number ratio can be related to the various units for concentration of solutions. The assumption that solution properties are independent of the nature of solute particles is only exact for ideal solutions, and is approximate for dilute real solutions. In other words, colligative properties are a set of solution properties that can be reasonably approximated by assuming that the solution is ideal.

Emphasis in above text is mine.
So it the assumption that colligative properties are independent of particle size, not the reality. The assumption is very useful, but not perfect.

Vapor pressure reduction is not truly explained as a mass transport phenomena where solute particles at the surface of the solvent prevent solvent molecules from entering the vapor phase. Since vapor pressure depression is dependent on solute concentration, which is uniform throughout the solution, the proper explanation is the increased entropy of the solution as compared to the pure solvent. The entropy increase is not dependent on the size of the solute, but the relative quantity of solute to solvent molecules (expressed as the concentration of the solute).
http://www1.lsbu.ac.uk/water/colligative_properties.html


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How does increasing ionic strength affect the pH of solution?
I'm having trouble interpreting the data for an experiment. With increasing ionic strength, I found lower solubility but also lower pH of the solution. What is the effect of the ionic strength on the pH? I'm looking at the solubility of boric acid and changing ionic strength by varying concentrations of NaCl solutions.

Do you mind to provide additional details on the experiment?

The effect of ionic strength on pH will depend heavily on what your solution is to start with, and what you're using to adjust the ionic strength.

I'm looking at the solubility of Boric acid and changing ionic strength by varying concentrations of NaCl solutions.


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Why do tealights burn violently when nearly burnt out
I'm not entirely sure if this is the right place for this question, but I decided that it's better than physics.se (and I had no other ideas). I hope this is a good match, otherwise please move the question where it would belong. :)
I often use a mildly scented tealight in my apartment to cover the lingering smell of food. I usually light them using a matchstick, and sometimes I would toss the matchsticks into the tealight after lighting it up.
Usually, the matchstick would burn out after a short while, and for most of the life of the tealight nothing interesting happens. However, when the tealight is very close to burning out with the remains of the matchstick inside, it starts burning very violently, with the flame forming fantastic shapes (it looks like the whole remaining volume of the wax is burning).
Once the flame was so violent it broke the heat resistant glass container I kept it in (I was careless and I let the side of the cup touch a thinner part of the glass, rather than just the bottom).
Why does this happen?

There are two factors that probably matter here. One is that, for most of the life of the tea light, only a small amount of wax is liquid and actually burning. Most of the flame is liquid wax burning in the wick with a small pool of liquid wax below the wick. But later, when most of the solid wax has gone all the wax will be liquid. The second is that the matchstick can act as a second wick.
When the wick is still vertical (and held there by the presence of solid wax) the area of melting is kept small as the flame is vertical and the area of melting is constrained. Later, the wick may fall over as there isn't enough solid wax left. Also the heat can spread wider as the volume of wax decreases as it is also conducted more effectively through the aluminium container (wax isn't a good conductor of heat). The net result is more liquid wax near the end of the candles life.
Then two other factors may intervene. If the wick falls over, there is now a larger area of wick where burning can happen and the whole wick (not just the tip) can burn. Sometimes the matchstick will now also act as a wick, increasing the area of flame. Both these effects make more flame and more heat in a sort of runaway reaction which might also cause enough heat to reach the flash point of the wax leading to an even bigger flame. Even if the flash point isn't reached, the flame from a bigger wick area and a match acting as a wick can be a lot bigger than the flame form a normal vertical wick.

"Even if the flash point isn't reached ..." — This is anyway not a question of the flash point; this one will quite surely not be reached. It is only a question of the kindling point who will surely be reached.

@Zeus I have no idea what a kindling point is.

Sorry, I see now that you are right (I was struggling with translations). "Flash point" is correct.

When only about 3 mm of wax is left, the wax in the tea light becomes so hot that its flash point is reached, and it begins to burn independently of the wig as a yellow/orange flame, just as is well known to happen in a more bluish version with alcohol at room temperature. This effect has principally nothing to do with the matchstick, which has been thrown in (as mentioned in the question).
Given that the tea light is already burning independently of the wig, if one would now extinguish the flame by covering the vessel for a second, one could see that an extremely strong cloud of white fumes evaporates (which should not be inhaled!): this is inflammable gaseous wax, which may be torched again even if one removes the wick beforehand with tweezers.
In the case of a still full or half-full wax-vessel, it is not possible that the temperature of the wax raises to the flash point because the flame is then burning essentially above or on level with the rim of the vessel, so that the vessel and the wax are heated up much less. Additionally, when only few wax is left, this heating up is much more effective because the heat may then focus on less wax. (The heating effect is stronger if the wax is in an aluminium vessel, because metal reflects the heat back into the wax; however, the stronger effect is helpful to elucidate the real causal coherences.)

This sounds good, but I've never seen a "clean" tealight burn nearly as violently as the ones with matchsticks thrown into them. It looks like a small, violent bonfire. Perhaps the temperature is so high it reaches the kindling point of (larger than normal parts of) the matchsticks, so they burn more than normal, and that's the reason?

@tomasz, Did you ever utilize tea lights with aluminium vessel? — For I could not observe any difference whether with or without matchsticks. It was also a bonfire of about 4 or 5 centimeters height. What do you mean by "larger than normal parts"? Extra long matchsticks?

No, I mean the temperature might be high enough that some pieces whose kindling point is not reached normally (when you just light the matchstick) actually start burning and fuel the fire. Or maybe it's just high enough to achieve complete combustion. About aluminium vessel - I have never seen any other kind.


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Which is more reactive among alkane, alkene and alkyne?
Since there are two pi bonds in alkynes - are they more reactive? Is it correct? I heard that acetylene is less reactive than ethylene.

Reactivity depends on the process. What type of reaction do you want to compare them undergoing?

http://chemistry.stackexchange.com/questions/48181/reactivity-of-alkanes-alkenes-and-alkynes

The reactivity order depends on the process(as mentioned in the comment).
1) In acid-base reactions, alkynes are the most reactive followed by alkenes and alkanes. This is due to the stability of conjugate base of alkyne by sp hybridized carbon atom.
2)In electrophilic substitution, alkynes are less reactive than alkenes. It is because of relative stability of carbocation formed in the case of alkenes.
3)In case of hydrogenation reactions, alkynes are more reactive than alkenes.

The reactivity is:
Alkynes > Alkenes > Alkanes.
Alkynes have two pi-bonds between two (or more) carbon atoms, along with a sp-sp hybridised orbital bonding (sigma bond). The pi-bonds can be easily broken to free the valence shell electrons for combining with other atoms. Hence they are the most reactive.
Alkenes have one pi-bond between two (or more) carbon atoms, along with a sp2-sp2 hybridised orbital bonding (sigma bond). The pi-bond can be easily broken to free the valence shell electrons for combining with other atoms, but as the number of pi bonds is lesser than Alkynes,  they are less reactive.
Alkanes have no pi-bonds between carbon atoms. They only have a sp3-sp3 hybridised orbital bonding (sigma bond). For Alkynes to react, first the sigma bonds with Hydrogen (s-sp3 sigma bonds) have to be broken, and only then can other atoms combine. This entails more energy, so Alkanes are the least reactive among the three.
This also explains why Alkynes and Alkenes undergo addition reactions, while Alkanes only undergo substitution reaction.

It also shows why acetylene is much much more reactive than ethene or ethane.

Gave a +1. Would you consider to be a private beta user on Materials? https://area51.stackexchange.com/proposals/122958/materials-modeling?referrer=ZjcxNmMxZjA3ZGFkNWJiMjRhZGQxNjAzY2E0ZTAxYzYzOTc2Yjk4YTFiYWVlZDA2YjlmYWI3NTE0OTNmOGI3OBKIibVWqhq0HqTbmyaNJVtUCXoNz7CCexf2SvvYcVjQ0 It's for the design of new materials: solar cells, photovoltaics, bio-materials, superconductors for quantum computing, new materials for better batteries, energy storage materials, molecular magnets, energy conversion materials, organic LEDS for screens, stronger and lighter materials for cars, planes, spaceships, etc.


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Chemical hardener to cast breastmilk into gemstones
I'm looking for a chemical hardener that I can add to breastmilk to turn it into a stone. I have seen others use one but they will not share what they are using. In their video they add about 1ml of secret solvent to 15ml of breastmilk and heat it on low heat for a couple minutes. It hardens completely within a few days. Any help would be greatly appreciated.

Please have a look at this very detailled and illustrated blog post from 2011. They seemingly have tried different methods and finally decided to use a polyester clear casting resin. And btw, happy welcome to Chemistry.SE!

Referring to real milk (bovine, human breast milk; but not coconut milk) contains casein which may be polymerized.  You may end up with edible cheese, or green adhesives, or gemstone-like galalith, for example.
Frankly, I have no hands-on experience to experiment with breast milk.  And I recommend to refrain from experimenting with formaline at home.  However, the procedure "Sculpted Science: Turn Milk into Plastic!" in the open access section Bring Science to Home by the Scientific American, which triggers the process just by vinegar and gentle heat, sounds much safer and may be worth to follow.


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Why several of isomers of octane such as 2,3-dimethylhexane and 2,4-dimethylhexane doesn't have a value for melting point from any databases?
This is a very rough data (as shown below) I have collected from CRC Handbook of Chemistry and Physics 93rd edition in black and ChemSpider in blue. As you can see, I am not quite done, but I was wondering why several of the isomers of octane such as 2,3-dimethylhexane and 2,4-dimethylhexane doesn't have a value for melting point from both databases? (I have tried searching up the melting point for each on a reliable database and have come up with no value given. Wikipedia does give a melting point, but I am very cautious about using that data as it does not a source to prove it.

I recommend to include more than one (secondary) reference and to widen the querry -- if there is access -- to general chemistry data bases like Chemical Abstracts (by the American Chemical Society), or Reaxys (by Elsevier), and Springer Materials, too.
In addition, do not stick to one chemical name, and equally include in your querry unique identifiers like the CAS registry number -- applicable regardless how the organic nomenclature is applied (including databases in other languages) -- rather than one chemical name.  By doing so, you may rapidly find addtional data, for example among chemical suppliers like Sigma-Aldrich, that lists 2,3,4-trimethylpentane (CAS-RN 565-75-3) in its catalog with a melting point of −110 °C and a boiling point of 113-114 °C.
Beside electronic ressources, consider additional ones available in print, like volumes in Landolt-Bornstein dedicated to condensed matter (now included in Springer Materials).


text:
What does the plus-minus sign with a value in parentheses mean?
I always see the plus-minus sign with a value in parentheses after a physical property is described such as melting and boiling point, like this:
$$\text{Melting point }(\pm 1~^\circ\mathrm{C})$$
What does it mean?

Related: What does a number in brackets after another number mean? I.e. 21(1)cm-1

Where is that snippet from? Context helps.

If you search up "CHEMISTRY INVESTIGATION FACTORS EFFECTING THE BOILING AND MELTING POINTS IN ORGANIC HOMOLOGOUS SERIES." It's the pdf.

It's simply describing the absolute error of the relevant quantities.
So if this precedes a table of melting points, you can expect the true value to vary by 1 degree, e.g.
$\ce{H2O}: 0.0 \pm 1~^{\circ}\mathrm{C}$
means the melting point is between $-1$ and $+1~^{\circ}\mathrm{C}$.

I suggest to adhere to the less ambiguous, proper notation, that is either i) value and incertainty enclosed in parentheses, followed by the unit like (1.5 ± 0.3) kg.  Or ii), without a parentheses like 1.5(3) kg.  To state a measurement like 0.0 ± 1 ºC  however rises questions.  As one useful reference here, consult  BPIM guide JCGM 100:2008, section 7.2.2. -- A document in consent with IUPAC, IUPAP, ISO, and others; equally worth to read for a general picture, too.  (Source: http://www.bipm.org/en/publications/guides/gum.html)

The errors you quote is the absolute error as described in the answer by @khaverim, however, what this means is not clear. This is because what the error measures is often not defined in tables or papers. Normally it is assumed to be one standard deviation but could be two standard deviations, or if there are only a few measurements might be just the range (largest-smallest value).
If the error is one standard deviation it means that by chance only $68$ % of the time will a measurement fall within the range, (and so by chance $32$ % outside the range), if two std. dev. then this increases to $95$ % so it matters quite a lot.
Recall also that the error can only be know if the true value of the quantity being measured is also known. Thus, although called an 'absolute' error it is only an estimate of the the error just as the mean of quantity is an estimate of its true value. Clearly as the number of measurements increases these estimates get better but the estimate of the error only increases as $\sqrt N$ for N measurements so it is easy to reach a point of diminishing returns.
[ Sometimes the Relative Error is quoted and this is (error)/(true value). The Relative Standard Deviation is the ratio (calculated std. dev. )/ (calculated mean value). When multiplied by $10^6$ these relative values become part per million, p.p.m. ]


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How can I calculate the number of branches in an alkane isomer?
I am so confused. So my textbook says that

2-Methylheptane has 1 branch
2,2-Dimethylhexane has 2 branches

However, when I get to:

2,4-Dimethylpentane
2,2,3-Trimethylpentane
2,2,3,3-Tetramethylbutane

I have no clue how to count the number of branches from the structural formula. My textbook assumes you'd understand how to calculate the number of branches after looking at the example. Can someone please help me understand?

Each substituent is a branch off the linear chain, assuming it's named correctly.

Each instance of a comma-delimited number in the prefix is directly indicating a branch on the "backbone" or base-chain. It's quite literally the number of numbers.
The most complicated example you have can be dissected like this:
$2,2,3,3-$ tells us to anticipate four branching groups from the base-chain on the 2nd, 2nd, 3rd, and 3rd carbons.
$Tetra$ tells us that all four branching groups are identical (since there are four total)
$methyl$ tells us the name of the branches. The $methyl$ group is a $\ce{-CH3}$ attached to the backbone.
$butane$ tells us what the backbone is; in this case, a 4-carbon ($but-$) fully-saturated carbohydrate ($-ane$).
So the molecule will look like this, with the 4 branches indicated:

Thanks, but when looking at 3-ethylpentane, from the "literally the number of numbers" it would be 1 branch but when looking at the structural formula it looks like 2 in 1 branch? Is that whole branch just 1?

Ethyl is one branch. It's just a larger one than methyl.


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What causes the "Gd break" in the trend of lanthanide-EDTA formation constants?
Smith and Martell obtained a series of data for the binding of trivalent lanthanide ions, $\ce{Ln^3+}$, with various carboxylic acid ligands (amongst them the well-known EDTA).1 A graph of the formation constants is attached ($K_\mathrm{f} = [\ce{Ln(edta)-}]/[\ce{Ln^3+}][\ce{edta^4-}]$):

Due to the contracted nature of the 4f orbitals, the lanthanide ions characteristically do not exhibit covalency in metal-ligand bonding, which is strongly reflected in their chemistry. Therefore, the almost linear increase in $K_\mathrm{f}$ going from $\ce{La^3+}$ to $\ce{Lu^3+}$ is entirely to be expected. The ionic radius decreases going across the 4f block, and consequently, the interactions with the ligand (primarily electrostatic in nature) become stronger.
However, as the title suggests, I'm interested in the slight blip at gadolinium ($\ce{Gd^3+}$). The value of $K_\mathrm{f}$ seems to be lower than would be expected, and the astute reader will notice that this applies not just to the EDTA complexes but also for some of the other ligands investigated.
Is there a reason for this anomaly?
There is a review by Moeller et al. from 1965,2 which wrote:

For all ligands which have been studied, the gadolinium complex is less stable than would be expected from the simple electrostatic model. This behavior has been called the “gadolinium break” and cannot be explained, as was originally attempted, by assuming a steric effect, since it is still apparent in ligands for which there should be no steric interference.

Of course, this is more than half a century ago. I would be grateful if anybody could provide either a sound theoretical explanation for why the Gd formation constants should be smaller or a newer reference that explains this effect.

References:

Smith, R. M.; Martell, A.E. Critical stability constants, enthalpies and entropies for the formation of metal complexes of aminopolycarboxylic acids and carboxylic acids. Sci. Total Environ. 1987, 64 (1–2), 125–147. DOI: 10.1016/0048-9697(87)90127-6.
Moeller, T.; Martin, D. F.; Thompson, L. C.; Ferrús, R.; Feistel, G. R.; Randall, W. J. The Coordination Chemistry of Yttrium and the Rare Earth Metal Ions. Chem. Rev. 1965, 65 (1), 1–50. DOI: 10.1021/cr60233a001.

Two questions.  1)  Can you correct the 1987 DOI?  The given link fails.  2)  Have you considered crystal field theory?  It seems to fit the pattern quoted here.

Thanks, I have fixed it. 2) Yes, CF effects would produce the pattern shown here with a minimum at $\mathrm{f^7}$. However I'm not entirely convinced that CF effects would play a significant role in the chemistry of the lanthanides. I'm working on the premise that the interaction is electrostatic - although I would certainly not mind being proven wrong.

I'm also now curious about the transitions at dysprosium: why does malonate ~plateau and, especially, why does DTPA  fully turn over??

@hBy2Py I didn't read ref 1 fully (this was originally something I saw in an exam question and I just wanted a reference for the data), but my first guess is that it could be to do with coordination numbers. The authors listed $\Delta H$ and $\Delta S$ values inside there, too, so you may find an explanation inside there.

, ortho, I agree -- DTPA has a seventh ligand site, so it could well be just a steric effect there -- when the cations get small enough, that seventh ligating position can't bind effectively and that frustrated ligand messes with the free energy. Not sure of the malonate, though -- why it'd be level, instead of decreasing, if it were a steric effect there, too.

If you consider the solvent extraction of lanthanides by reagents such as H-DEHPA, then to a first approximation the distribution ratio is given by the equation.
$$\ce{DLn}= k[\ce{H-DEHPA}]^3[\ce{H+}]^{-3}$$
If you plot k as a function of atomic number for DEHPA dissolved in most organic solvents (solvent extraction term is diluent) then the value goes up as a function of Z. But the line has bumps on it.
The reason is that the extraction reaction is
$$\ce{Ln(H2O)_n^3+(aq) + 3H-DEHPA(org) -> [Ln(DEHPA)3](org) + H^+(aq)}$$
As you go from one side of the lanthanide group to the other the binding of the lanthanide to the anionic DEHPA ligands is likely to become stronger, but the value of $n$ changes in several steps. This will give bumps on the graph of $K$ for both the extraction of the lanthanides and also for their binding to ligands such as EDTA.
As the lanthanide shrinks as it becomes smaller then the hydrated lanthanide $$[Ln(H2O)n]^3+$$ will change in stability, if we assume that for a moment we have a ten coordinate (decaaquo) complex which has a metal atom which is slightly too big for the ten waters, what will happen is that as it shrinks it will go through an optimum size and then become too small for the ten waters. We will then have a complex with nine waters.
Consider the chemical reactions of a trivalent f-block metal with a ligand such as a EDTA tetraanion or something neutral such as a pair of BTBPs. We will have
$$EDTA^(4-) + [Ln(H_2O)_n]^(3+) --> [Ln(EDTA)(H_2O)_q]^- + (n-q)H_2O$$
When I checked the cambidge database Ce,Pr, Eu, Gd, Dy, Ho in the solid state has q = 3.
For Yb and Er q = 2
So as n and q do not change at the same time going along the lanthanide series the issue of the number of waters will add some bumps to the graph of stability constant as the stability constant relates to the delta G of the complexation reaction.
If we are changing suddenly the number of waters formed by the reaction then the entropy change (Delta S) of the reaction will change suddenly. This will introduce some sudden changes into the graph of binding constant. You might also want to look at a paper by Mike Hudson, Mike Drew, Mark Foreman and Karren Kennedy. In this paper they form the coordiantion complexes of all the non radioactive lanthanides with a ligand (not EDTA) and then you can compare the strucutres. When I last looked I noticed that the La, Ce and Lu complexes were slightly different to those in the middle.
You might help yourself if you were to compare all the crystal strcutures of the lanthanide EDTA complexes, choose only those which are merely hydrated.
M.G.B.Drew, M.R.StJ.Foreman, M.J.Hudson, K.F.Kennedy, Inorganica Chimica Acta, 2004, 357, 4102, DOI: 10.1016/j.ica.2004.06.032


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Conductivity of electrolyte solution
I found a relationship online between the conductivity of electrolyte and current.
$$i = F \sum_i z_i N_i = F^2 \left( \sum_i z_i^2 u_i c_i \right)\nabla\phi = -\kappa\nabla\phi \qquad \kappa = F^2 \sum_i z_i^2 u_i c_i$$
This equation is hard for me to follow.
I get what most of the variables mean and I am guessing that $\Delta\phi$ is the potential difference. However, I thought that the conductivity was also related to temperature, which I do not see in the equation.
It has been a struggle for me to find results on the internet to relate conductivity and current together. Can someone help me and tell me if this equation is right.
Also, this equation is true for dilute solutions, what will the equation be for concentrated solutions (solutions near $\pu{1 M}$)?
Also, how would you calculate the potential gradient?

As it stands, your question is a little broad and you are better served by breaking it up into multiple queries. That said, all your problems will be solved by visiting the Wikipedia page on Debye-Huckel theory, or others you might find with a search on those terms (pardon my lack of umlaut on the "u", apparently HTML and TeX are not working in this box).

@Todd o.o Mathjax works in comments.


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Silanization of glass cover slips
I was trying to silanize the glass coverslips using a protocol mentioned in some papers that say to make up a solution of 595µl N-trimethoxysilylpropyl-N,N,N-trimethylammonium chloride (TCMP-TAC) and 216µl of vinyltrimethoxysilane (VTS) in 300ml ultrapure water and incubate overnight at 65ºC.
However, I don't know about the concentrations of each of the components of the mixture containing TCMP-TAC and VTS individually. Gelest provides TCMP-TAC in 50 % methanol and VTS as a solid. How should I go about making the individual components?


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Buffer and strong acid
If a weak acid is added to a neutral solution, it creates equal amounts of  $\ce{H3O+}$ and conjugate base.
So what then what happens to the original $\ce{H3O+}$ created by the weak acid when a strong acid is added to a solution.
From what I understand, some of the newly created $\ce{H3O+}$ from the strong acid will react with the conjugate base of the weak acid, preventing the pH from getting lower. But as I stated above, what about the original $\ce{H3O+}$ from the weak acid? For every molecule of the conjugate base that reacts with a $\ce{H3O+}$ from the strong acid, there must have been a $\ce{H3O+}$ created from the weak acid. So how is pH actually prevented from being lowered?

I have read your question several times to figure out what your problem is.
Below, I will try to give you a simple explanation without any fancy mathematics or complex equations.
Now, suppose we dissolve the weak acid "A" in water. The acid will react with water and form hydrogen ions and the corresponding base "B" of the weak acid. We further suppose that the pH of this solution is 3 at equilibrium.
We have the following equilibrium:
$$\ce{A + H2O <--> H3O+ + B}$$
The concentrations are supposed to be low, so we can state an equilibrium constant:
$K = [\ce{H3O+}]*[\text B]/[\text A]$
Now, we add a strong acid to this solution. What will happen?
The pH of the solution will drop. Suppose it will drop to pH = 2. We will get an immediate response from the equilibrium reaction of the weak acid system. Since the amount of hydrogen ions have increased in the solution by the addition of the strong acid, the equilibrium will be pushed to the left.
Pushed to the left? What does it mean?
Well, the new situation implies that the weak acid cannot give off  the same amount of hydrogen ions as it did at pH = 3. pH is now 2 and we have a completely new situation.
Now I will target on what I think is your main problem.
When you add a strong acid to the weak acid solution, the pH will drop.
The equilibrium of the weak acid will then reset into a new equilibrium determined by the actual pH and the equilibrium constant of the system.
At pH = 3, hydrogen ions were formed from the weak acid. These (or other) hydrogen ions will be forced to react with the corresponding base B, giving more A at pH = 2.
Thus, the net effect is that the hydrogen ion concentration related to the strong acid will be in principle unchanged.


text:
What is A in the following equation
This equation describes the total current in bulk solution in a battery caused by migration in the electrolyte.
Then the source says what A is: In the bulk solution (away from the electrode), concentration gradients are generally small, and the total current is carried mainly by migration. All charged species contribute. For species j in the bulk region of a linear mass transfer system having a cross sectional area, $A$...
$$i = \sum_j i_j = \frac{FA\,\Delta E}{l} \sum_j |z_j| u_j C_j$$
I am confused if $A$ is the electrode area or the solution cross sectional area.
I will describe what the rest of the variables are

The $i$ is the current in the bulk solution.
F is Faraday's Constant
∆E is the change in potential (i.e. Voltage measurement of the battery)
l is the distance of how far the electrodes are separated (or at least that is what I think)
z is the charge of species j
u is mobility of species j
C is the concentration of species j

Can you show us where this equation is from?

just giving some random equation doesnt seem such a good idea, could you describe its origin, and/or define what it gets you? ie what is in the LHS of the eqn?

sorry for that. I edited the question to hopefully make it more understandable

$A $ is the cross-sectional area in solution. The  "bulk" is the relatively homogeneous "mess" of charged particles and solvent -- $A $ is not the area of the electrode.
What this equation tells you in terms of $A $, is that, the larger the area through which some number of charges are passing in a given moment, the greater the possible current, because more area means more room for more charges that can pass through in a moment. If you squeeze the area you start to make a bottleneck wherein less charged particles can pass in a given time. (This is why copper wires with larger diameter (or, smaller US gauge) have higher current capacity).
This makes sense, right? Current is measured as charge per unit time. The implication there, is that it is motion of charge through some point (1D) or cross sectional area (2D, $A $ here) per unit time.

So when you say A is the cross-sectional area in solution, do you mean if I had a KOH solution in a beaker, then A would be the cross-section area of the beaker in which the solution is being held?

Not quite. It's the  cross sectional area of the actual solvent (water) between electrodes. Not the beaker.

okay so if I have a basic galvanic cell setup with two cylindrical beakers that contain the electrolytes connected by a salt bridge, can I approximate the A as the cross-sectional area of the beaker? If not, how can I find the area between the electrodes?

The salt bridge is the bottleneck. That is where current will be determined.

sorry, but what do you mean when you say the salt bridge will be the bottleneck?

also, why will the current be determined there and not at the electrolyte?

nvm, so if I use a paper towel electrolyte, the A would be the cross-sectional area of the paper towel?

The charged particles can only travel as fast as the slowest area from one electrode to the other.  You may consider getting an ammeter and trying experiments to compare currents where the only variable is the area of your salt bridge.

so the area i could use is the area of the paper towel in the equation above? Also, how do I know what the concentration of the electrolyte species will be   in the salt bridge?

I recommend you move the conversation to chat. Comment section isn't made for extended discussion.

Let us continue this discussion in chat.


text:
Need for hybridization
I have learned hybridisation as a concept created to explain the same $\ce{C-H}$ bond lengths in $\ce{CH4}$.
I found it easy to find the type of hybridisation in different molecules, but my problem is how to find whether an atom in a molecule under goes hybridisation or not.
For example, $\ce{N}$ goes under hybridisation in $\ce{NH3}$, but there are some examples where $\ce{N}$ won't hybridise (not sure abt $\ce{N2}$, if you may suggest). In $\ce{ClF3}$, $\ce{Cl}$ is hybridised but in $\ce{PCl5}$ $\ce{Cl}$ is not.
I want the reason for undergoing hybridisation.

Hybridization occurs in our heads, rather than in atoms.

Then when should we think about it.

The thing is, orbitals and everything concerned with that is an oversimplified model to explain a lot of chemistry rather okayish. If you want to get a better understanding of how it might actually work, you need to start with quantum chemistry.

As it has already been said hybridisation is a theory we use so as to explain things (e.g. why the bonds of methane are the same) that otherwise do not make sense. In general you need to use each time the theory that best describes the result. Other times you'll use the valence bond theory or other times you'll use the molecular orbitals theory.

If you really want to, you could consider Cl in PCl5 to be sp3 hybridized with 3 of those sp3 orbitals occupied by lone pairs

Thanks i actually wanted to find hybridization of N in -C triplebond N

That would be sp.

In general, you should never assume a need for hybridisation unless you have a bonding situation (such as a tetrahedron or related tetravalent structures) which cannot be explained without hybridisation.
As soon as you have a lone pair on any atom, it is more favourable to have this lone pair in an unhybridised s orbital rather than in a hybridised $\mathrm{sp}^n$ orbital. This can be seen for example in phosphane, hydrogen sulphide, etc. in which practically unhybridised central atoms are possible and thus the bond angles show pure p orbitals taking part in the $\ce{A-H}$ bonds.
If you consider very small central atoms (e.g. nitrogen in ammonia) or very large ligands (e.g. chlorine in $\ce{PCl3}$), then having a perfectly unhybridised central atom — meaning a bonding angle of $90^\circ$ — would introduce too much steric strain between the ligands. Thus, the overall system is more favourable if the central atom’s bonding orbitals are partially hybridised, i.e. if the bond angles are enlargened slightly to reduce steric stress. The final bond angle and thus the final central atom’s configuration is a carefully balanced compromise between a hybridisation with a p contribution as high as possible and the ligands separated maximally.

By the way, chlorine is practically never hybridised and it is most certainly unhybridised in $\ce{ClF3}$. Rather, this molecule should be understood as having:

a single $\ce{Cl-F}$ σ bond using a p orbital both of chlorine and of fluorine

a four-electron-three-centre bond using a p orbital of each of the three atoms that form the linear $\ce{F\bond{...}Cl\bond{...}F}$ fragment (i.e. a p orbital of chlorine). This bond can be understood with the following resonance structures:
$$\ce{F-\overset{+}{Cl}\bond{...}\overset{-}{F} <-> \overset{-}{F}\bond{...}\overset{+}{Cl}-F}$$

one chlorine lone pair in a p orbital perpendicular to all $\ce{C-F}$ bonds

one chlorine lone pair in an s-type orbital.

authors: Αντώνιος Κελεσίδης


text:
Properties of metallic hydrogen
Given the possibility that metallic hydrogen may have recently been produced, and the massive media bloviation that followed, I am curious about some of the "properties" that I heard from multiple media outlets.

Does metallic hydrogen decompose to mono-atomic hydrogen, then recombine to diatomic hydrogen?
How much energy is produced by the decomposition?
How certain are we that it will be stable(ish) at STP?

The Harvard Gazette had a story on metallic hydrogen. One of the authors is quoted with "One prediction that’s very important is metallic hydrogen is predicted to be meta-stable...”. This sounds like an interesting perspective, but at this point, it seems to lack experimental proof.

Great link @KlausWarzecha . The research group did claim they made shiny, electrically conductive metallic hydrogen though. However, Wikipedia's metallic hydrogen page pointed out that "Shortly after the claim was published in Science, Nature's news division published an article stating that some other physicists regarded the result with skepticism. Recently, prominent members of the high pressure research community have criticised the claimed results,  questioning the claimed pressures or the presence of hydrogen at the pressures claimed." So, it is definitely controversial!

Regarding stability, like @KlausWarzecha said, it's predicted to be meta-stable. The grand claim is that it will be a room-temp solid (the Harvard group wasn't sure if they had liquid metal or solid metal) superconductor, also having enormous energy density, as you would imagine. Unfortunately, according to this Wikipedia article: "In February 2017, it was reported that the sample of claimed metallic hydrogen was lost, after the diamond anvils it was contained between broke." So for now, it sounds like it's more "meta" than "stable" ;)

@airhuff Glad that you had a look at the more recent Wkipedia article, which I didn't do. A detailled comment on the initial article is freely available at arxiv.org. Your closing statement is right on the button :)

I don't know if anyone expects metallic hydrogen to be metastable. Lots of people sure would love that to be true, since it's a very energy dense fuel and could be a room-temperature superconductor, but metastability is more just wishful thinking than a predicted property.


text:
Color change of potassium permanganate during a titration with Fe(II)
Is it possible for potassium permanganate to have color change (in a titration) from colorless to pink? If so, then in what circumstances (particularly for a titration)? I have been always been taught that color change for potassium permanganate is purple/pink to colorless. I have scoured the whole internet searching for an example and I couldn't find one.
The problem is a question in a recent A level paper which asked about the color change of $\ce{KMnO4}$ with an $\ce{Fe^2+}$ solution. The answer was colorless to pink. The examiner report further mentioned that many students wrote purple/pink to colorless, however it was marked incorrect.
Can someone explain the ambiguity here?

"Potassium manganate", without further disambiguation, usually refers to Mn(VI) $\ce{K2MnO4}$. You need to use either "potassium permanganate" (much more common) or "potassium manganate(VII)".

An aqueous solution of potassium permanganate, $\ce{KMnO4}$ is of purple colour, regardless if the solution is neutral, acidic, or basic.  Potassium permanganate is a relative powerful oxidizer. (As a look-up in a table of standard electrode potentials may tell.) If one considers acidic conditions $\ce{KMnO4}$ may oxidize material, like
$$\ce{Fe^{2+} -> Fe^{3+} + e-}$$
while being reduced from oxidation state +7 to oxidation state +2 in the form of $\ce{Mn^{2+}}$.  Aqueous solutions of, say $\ce{MnSO4}$, however are almost colourless.  On the other hand, $\ce{Fe^{2+}}$ alone is not a strong oxidizer, under normal conditions incapable to generate permanganate.  That is why this analysis is widely used (application note, example 1, example 2 in lab classes and beyond).
The end point of the redox titration is reached if all $\ce{Fe^{2+}}$ is oxidized, so that under addition of additional $\ce{KMnO4}$, your analyte solution remains purple.

But in the provided example of $\rm KMnO_4$ and $\rm Fe^{2+}$, it is not possible for the colour change to be from colourless to pink right (assuming no additional $\rm KMnO_4$ is added)?

@Faiq You have to add an extra bit of $\ce{KMnO4}$, otherwise you can't tell when the titration is done.

By observing for the colourless phase. The moment the colour disappears is my endpoint

@FaiqRaees  During the titration, $\ce{KMnO4}$ is consumed, hence the purple colour of this added reagent solution, added to you analyte of $\ce{Fe^{2+}}$, vanishes.  Once all the $\ce{Fe^{2+}}$ is consumed, the reaction with $\ce{KMnO4}$ is discontinued; hence, further addition of $\ce{KMnO4}$ will now yield the persistent purple colouration.


text:
Solubility of epinephrine in water
Do you know how can I dissolve epinephrine in water? because it is insoluble in ordinary situation. When I decreased the pH by $\ce{HCl}$ it really worked but I need pH=7 to continue my experiment. Can I add $\ce{HCl}$ and dissolve the EPI and later using buffer to reach pH=7 and in this way the EPI wont be precipitate?

Hello and welcome to Chemistry.SE. If you have any questions about the site you can take the short [tour] or visit the [help]. It might me easier to help you if you stated your end goal, or at least what you need to do after it's dissolved.

What the context of aiming for a highly concentrated aqueous solution of epinephrine / adrenaline?  What would be a sufficient concentration?  Would -- assuming a chemical reaction is intended -- in sight of a low reaction rate a steady of epinephrine in low concentration be sufficient?
While this lists a typical concentration of saturation in water of about 180 mg/L, other conditions are provided, too.  The list equally includes a statement like "soluble in acetic acid" -- so perhaps you do not need to decrease the pH value down to 1 (by HCl), and maybe you were able to obtain a sufficient concentration of epinephrine in a mixture of water / acetic acid, too?
Are there restrictions, imposed by the intended application / compatibility  toward (biological?) materials to stay near neutral conditions?  One way to reply to this and to tailor solubility of pharmaceutical active ingredients is to refrain from their form as free acids or bases, and to use them as salt.  In this sight, would be epinephrine tartrate, or an acetate, an alternative option?

Thank you so much for your consideration. the final concentration of Epinephrine is 23 mg/100mL. I can not use it as a salt and I have the biological measurement purpose.

@ghazi If your department has access to a database like Reaxys (by Elsevier) or Chemical Abstracts / scifinder scholar (by the American Chemical Society) to mention just two large ones relevant in (organic) chemistry, you may consider a substructure search.  Draw / insert the molecular structure of epinephrine / adrenaline, and intentionally allow the database to search for literature known salts.  Some of them will be more "inorganic", other more "organic" and maybe not interfering with your biological experiment intended.

I need to use florescence so I have to have the clear solution. maybe Acetic acid is a better idea.


text:
Hydration enthalpy of cations
Can someone provide reliable data on the solvation / hydration enthalpies of $\ce{Ti^{4+}}$ and $\ce{Cr^{3+}}$ ions, or explain which is more stable qualitatively? According to this article1
CFSE in $\ce{Cr^{3+}}$ is around 59.7 kcal = 249.7848 kJ
and according to the abstract of an article by Uudsemaa and Tamm,2 the estimated hydration energy of $\ce{Ti^{4+}}$ is around 7800 kJ/mol
Is this data correct? can someone explain qualitatively as well?
References
(1) http://hydration%20energy%20of%20Cr3+%20value
(2) Uudsema, M.; Tamm, T. Calculations of hydrated titanium ion complexes: structure and influence of the first two coordination spheres. Chem. Phys. Lett. 2001, 342 (5–6), 667–672. DOI: 10.1016/S0009-2614(01)00617-0.

Your first link is broken. Anyway CFSE is merely one small contribution to hydration enthalpy (which are ion-dipole interactions). You also need to explain very clearly what you mean by "stability", because that is not a very clear-cut concept.

Logically one would expect the hydration enthalpy of Ti(4+) (smaller, more highly charged cation) to be much more exothermic than that of Cr(3+). However there's a reason why it's an "estimated" hydration energy and there's a reason why it's a computational study. Ti(4+) doesn't exist in water, probably not even as an aquo complex. The paper writes "The Ti(4+) cations have too large charge to be stable as simple hydrates, and usually exist in hydrolysed forms, e.g. Ti(OH)2 or [Ti(OH)3]+." Reference given is Holleman-Wiberg.

In order to help you out on this question, it would be good if you could include human readable citations in your post.

More the charge density more the hydration energy...You can find the stability this way...


text:
(Why) is it difficult to liquidize the polluting exhaust of a vehicle?
Considering the hazardous effects of car and other vehicle exhaust when released into the atmosphere, I began wondering why we are not dominantly preventing the exhaust from entering into the atmosphere, for example by liquidizing it and storing it either in the driving vehicle or elsewhere?
My question is: why is it so difficult to liquidize the exhaust, in order to prevent it from entering the atmosphere? Or isn't it difficult at all perhaps?
Please note that it is not forbidden to theorize about adding any equipment and or chemical reactants to the vehicle.
P.S.: I would be equally interested in just storing the exhaust of a vehicle inside the vehicle by means of compression, rather than liquidization.

It would be an extraordinarily energy intensive process, more so than running the vehicle.  Assuming you are trying to trap CO2, you would have to cool the hot exhaust to well below 100C, then store it for transport to recycle. Since burning a kg of gas produces 3 kg CO2, that means your storage tank is 3x the size of your gas tank. There are better ways to reduce emissions ;)

@airhuff What about adding another gas/reactant to the CO2, so that, after the reaction, the gas could more easily be compressed or liquidized and stored?

If you could somehow bubble the exhaust through a strong base, that would absorb the CO2. The problems with that are that it's messy and dangerous to work with and you would need a lot of base. Although you could just say these are engineering problems that could be overcome...I don't want to discourage you from this line of thinking. One way to do this, rather than putting such a device in each vehicle, would be to centralize the collection of CO2 on a large scale at a power plant that makes electricity for everyone's electric cars. That has it's own advantages and disadvantages though.

Here is something you might find interesting. This is an article about recovering the other major combustion product, water. It's a proposal for military vehicles in the desert to recover drinkable water from vehicle exhaust. Water would be much easier to collect than CO2 and this probably isn't relevant, but I thought you might find it interesting anyway.


text:
Determining rate constants for parallel reactions from experimental data
I have a question regarding the rates of parallel reactions. I have gathered experimental data for the overall reaction (i.e. conversions and selectivities and specific times) but due to the two reactions running parallel to each other for this reaction means I can't use a simple integrated rate plot.
Say for
$$
\ce{A ->[$\large k_1$] B} \\
\ce{A ->[$\large k_2$] C}
$$
By substitution and integration, I have found that:
$$
[A](t) = [A](0) \mathrm{exp}(-(k_1+k_2)t)
$$
and then the following equations can be derived in terms of the products formed:
$$
\frac{d[B]}{dt} = k_1 [A](0) \mathrm{exp}(-(k_1+k_2)t)
$$
which gives
$$
[B](t) = \frac{k_1[A](0)}{k_1+k_2}(1-\mathrm{exp}(-(k_1+k_2)t)
$$
when integrated. Also,
$$
[C](t) = \frac{k_2[A](0)}{k_1+k_2}(1-\mathrm{exp}(-(k_1+k_2)t)
$$
is the equation for the other parallel reaction.
My initial thoughts were to plot the graphs (as I know the values for $[C]$, $[B]$, and $[A]$) and form some simultaneous equations, but this is an exponential plot and so i'm not sure how this would work.. or if I could even plot it, as I don't know $k_1$ and $k_2$ - that is what I need to determine.

To clarify, you know the values of A and either B or C for some value of $t$ other than zero?

One way you could get $k_1$ and $k_2$ is to first find their sum using your equation for $[A](t)$ at a known $t$.
Once you have done that, you can solve for $k_1$ using your equation for $[B](t)$ at a known $t$, since you now have the denominator of your first factor. You could alternatively find $k_2$ using $[C](t)$.
Once you have $k_1$, then $k_2$ is easily found from the sum.

Yes that is how I have done this.. well sort of - I found the total k from equation [A]t by a first order rate plot, and then I used [B]t/[C]t = k1/k2, and used an average of the concentration ratios (as I had a few values for different t values) and then solved the two equations to get the values! Thanks though :D

As B and C always appear with the same rate constant ($k_1+k_2) as A disappears. The only way to get individual rate constants is to look at the relative proportions (i.e. yields) of B and C produced.


text:
What is the difference between strontium-86 and strontium-90?
I know that strontium ($\ce{Sr^90})$ has a higher mass number than its counterpart, but strontium ($\ce{Sr^86}$) is non-radioactive and not deadly to humans.
What makes $\ce{Sr^90}$ so different that it is cancerous and lethal to humans? Is it just the chemical makeup of the element that makes the difference?

wikipedia "While $\ce{^90Sr}$ (half-life 28.90 years) [$\beta^-$ emitter] has been used similarly, it is also an isotope of concern in fallout from nuclear weapons and nuclear accidents due to its production as a fission product. Its presence in bones can cause bone cancer, cancer of nearby tissues, and leukemia." https://en.wikipedia.org/wiki/Strontium See also https://en.wikipedia.org/wiki/Isotopes_of_strontium

Related: http://chemistry.stackexchange.com/questions/20146/what-makes-elements-radioactive

The big deal is that Sr is similar enough chemically to Ca to be taken up by the body. This means that the energetic betas will damage living tissue. If external to the body, or not taken up by the body, beta emitters are less of an issue (depending on the exact energy of the beta, of course).

4 neutrons$%and a MathJax comment$

For all practical purposes, Sr90 is chemically identical to Sr86. However, the extra neutrons crammed into Sr90 make it radioactive.
As mentioned in the comments above, alkaline earth metals (e.g. Ca, Sr, Ba) behave chemically in a similar fashion, and tend to deposit in bone like calcium does. This keeps the radioactive Sr90 in contact with the rapidly-growing hematopoietic material in bone, increasing the risk of cancers such as leukemia.