text:
Why can't some salts be used in the salt bridge of specific voltaic cells?
I had an exam question in reference to a galvanic cell with $\ce{Pb{(s)}}$ as the anode in $\ce{Pb^2+}$ solution and platinum cathode in a solution of  $\ce{Fe^2+}$ and  $\ce{Fe^3+}$:

Identify an appropriate electrolyte to use in the salt bridge.

I gave $\ce{KNO3}$ which was correct. However, the marking criteria specified that sulphate and chloride salts were not usable.
Why are these salts not usable? Is there any way to tell which salts would be usable for other galvanic cells?

The main reason would be that sulphates and chlorides can form precipitates in solution. Potassium nitrate essentially never forms in precipitates in solution by going off the solubility rules. The lead in the cell would form precipitates with the sulphate and chloride, and thus it would be ineffective as a salt bridge.


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Why, in titration, is universal indicator not commonly used?
My teacher said that we always used indicators such as methyl red, methyl orange, litmus, phenolphthalein, etc. rather than universal indicator. I don't understand why, though, would not universal indicator serve the same function, but for essentially all titrations rather than select few?

The thing with universal indicators is that you have 6 different colors or so differentiating certain pH-ranges.
Now if you want to titrate to a certain pH-Value, it is easier to have an indicator like phenolphtalein that changes from colorless to a color at a certain point (phenolphtalein being colorless from 0 to 8 and red for $< 0$).
So you can titrate more accuratly to a certain pH-Value with specialized indicators. On the other hand you can't see where on the pH scale you are.
With an universal indicator you can get a feeling for where you are approximatly, but titrating to a certain value is very difficult.

I'm a chemistry teacher. I quite like using universal indicator.  It usually goes from red/orange to yellow in one drop, then to green in another drop, then to blue/purple after another drop.  We stop at green (neutral). I like it because it's clearly green.
With phenolphthalein we stop when the pale pink persists for approx 30 seconds, but students usually aren't satisfied this and go past the end point by making sure the solutions remains pink by which point it's probably gone past neutral.

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text:
Distinguishing stereoisomers using reagents
Given the following possible reagents:

acetone,
$\ce{MnO2}$,
ozone, or
aluminium propoxide,

which one would distinguish between the cis and trans isomers of cyclopentan-1,2-diol?

The only reagent listed that will distinguish between cis and trans diols is acetone. In the presence of catalytic acid it will form an acetal with the cis diol. This cannot occur with the trans diol as the two OH groups are too far apart. Try drawing the cis and trans diol in 3D.

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text:
Van't Hoff Factor of Sucrose
What is the Van't Hoff factor for sucrose?
According to me it should be 2 as sucrose is a disaccharide.

So what? Salts have factors like 2 or 3 because they dissociate into that many particles. Does sucrose dissociate? No.

Sucrose is a organic molecule. It doesnt show any dissociation in water and hence its van't hoff factor is 1.
Moreover disaccharide is the sugar formed when two monosaccharides (simple sugars) are joined by glycosidic linkage. Glycosidic linkage would never break on simply dissolving it in water. So how are you expecting van't hoff factor to be 2 ?


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Help interpreting IR spectrum
Whilst attempting to create salicylic acid/2-hydroxybenzioc acid, one of our experiments produced a compound with a melting point of over $300^\circ\mathrm{C}$ ( as high as the machine would go) and the IR absorption spectra attached. Can anyone help us identify the compound we have produced?
We used:

$4 \,\mathrm{g}$ of oil of wintergreen
$100 \,\mathrm{mL}$ of $2 \,\mathrm{M}$ aqueous sodium hydroxide solution
Sulfuric acid

The experiment was left for two weeks at which point the crystal pictured had formed.

I second @Ezze, there is something fishy here and the spectral data recorded do not match with how a spectrum of your targeted material should look like:

(Notice that this spectrum was recorded in transmission mode with a pellet of KBr and hence differs slightly from the ATR-FTIR setup you likely deploy [the $\ce{CO2}$ bump]).

In addition, the melting point is far to high compared with literature data.
Recommended action Figure out the error with your spectrometer, record a spectrum of a compound you know about its identity and have access to a reference spectrum.  Then record data of your compound (do not stop at 1000 1/cm, record if possible between 4000 to ca. 500 1/cm, as the displayed reference).
Added after your edit: Could it be sodium sulfate?

I think sodium sulfate nails it....

Yes we ran the experiment three times and the other two times the IR looked like it should and the melting point was in the normal range. The lab tech that ran the IR agreed she had never seen anything like it and our lecturer was also intrigued. The sample has now been destroyed so we have to attempt to work out what we have made with the data we have. Sodium sulfate seems a strong contender thanks @Buttonwood

@JessicaMadden The ester was cleaved-off under basic conditions; yet aq. NaOH naturally will deprotonate both the acid and the phenolic OH and I speculate this disodium salt is much more water soluble, than expected.  As both the pure starting material, as well as the pure product are colourless, perhaps you accidentally confused once the aqueous phase with the organic phase during and extractive workup?  Perhaps you did not add enough acid (both for neutralization + acidification .and. thoroughly mixing .and. check) to the aq. phase prior back-extraction? I myself know both errors, too...

@JessicaMadden As an additional suggestion for future improvement, taking into account that there was a first lab class, and a second one about two weeks later -- maybe used to further purify the product by a slow recrystallization: ATR-FTIR spectroscopy does not consume a lot of sample material, does not require additional chemicals, and is rapidly recorded; just record IR data of your pre-purified product, prior to cleaning the work space. I do so after column chromatography, to confirm "the correct fraction", too -- often quicker, and much less expensive than to collect a H-NMR spectrum.

It probably is the spectrum of sodium sulphate...some things still seem a bit off though.
http://webbook.nist.gov/cgi/cbook.cgi?ID=B6000542&Units=SI&Mask=80#IR-Spec

I do believe something is wrong with this spectrum. I find it highly improbable that a compound would exhibit a single strong absorption peak at low wavenumbers, while showing no other absorbance whatsoever. The fact that there is no overtone of your band also shows me that there was some experimental error when recording this spectrum.


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What is the maximum covalency of oxygen?
My text says it to be 4. I believe it's 2. Is there any compound where covalency of 4 is observed?

Guess they mean coordination compounds.

Can you give me an example?

Basic beryllium acetate will do.

Yeah basic beryllium acetate has four coordinate O and can be thought of as significantly covalent due do the polarisability of Be. There are also numerous three coordinate species containing oxonium ions eg Me3O+

$\ce{H4O^{2+}}$ is also an option.

We know the maximum covalent is at least 3 with such species as  $\ce{H3O+}$.  Also the pyrylium ion (https://en.m.wikipedia.org/wiki/Pyrylium_salt) has an oxygen covalency of 3 by forming a pi bond for the third bond.
Can we get 4-coordination?  As mentioned by @Rob, yes.   @Rob's comment states that basic beryllium acetate appears to have 4-coordinate, covalently bound oxygen.
And it check out.  Basic beryllium acetate, $\ce {Be4O(C2H3O2)6}$, has the four-coordinate oxygen in the center, then four beryllium atoms coordinated to this oxygen at the vertices of a tetrahedron, then an acetate group bridging each edge of the tetrahedron.  In addition to beryllium, zinc forms a similarly structured basic acetate, which can be obtained by heating normal zinc acetate (https://en.wikipedia.org/wiki/Zinc_acetate).
Might the central oxygen be mostly ionically bonded?  Magnesium and heavier alkaline earth metals, which form more strongly ionic bonds with oxygen, are not known to form such a basic acetate structure.  This corroborates the hypothesis that covalent bonding of the central, four-coordinate oxygen is involved.


text:
Is there a convention for breaking chemical formulae across lines?
I found a chat where a user asked for chemical formulae to be broken across lines (with LaTeX/mhchem). I could imagine offering a switch to allow this. But only, if line-breaking within formulae is not completely uncommon.
According to him/her, possible breaking points in $\ce{[nBu4N]2[(UO2)2(\mu-\eta^2:\eta^2-O2)(NO3)2(\mu-Au(CN)2)2]}$ would be
$\ce{[nBu4N]2~\color{red}{|}~[(UO2)2~\color{red}{|}~(\mu-\eta^2:\eta^2-O2)~\color{red}{|}~(NO3)2~\color{red}{|}~(\mu-Au(CN)2)2]}$.
So a rule could be "allow a line-break before (, [ or \{ (which would introduce another possible breakpoint into the example).
Is it common to break long formulae across lines and is doing so sensible?

I think, chemical formulae should not be broken across lines. I can’t remember seeing it happen in large scales anywhere — but it could happen in journals due to their thin column layouts. I’ll try and keep an eye out.

What if the formula is too long to fit on one line?  That looks like the real q.

I remember a couple of books from 1970-80s, where long formulas were split with what would be a $\rm \LaTeX$ \cdot "$\cdot$" notation, I believe. This is an example from Pope, M. T.; Jeannin, Y.; Fournier, M. "Heteropoly and isopoly oxometalates" in Inorganic Chemistry Concepts; Springer: Berlin, 1983; Vol. 8.:

The reason for that, it seems, is that the book was printed in A5 format. Personally, I find an option for mhchem to be able to split formulas across pages very useful.


text:
What does the colon in µ-η2:η2-O2 mean?
In the formula $\ce{[nBu4N]2[(UO2)2(\mu-\eta^2:\eta^2-O2)(NO3)2(\mu-Au(CN)2)2]}$, what does the colon stand for? Is this an official notation?

Can we have a source?

$\mu$ means that the $\ce{O2}$ ligand is bridging two metal centres, i.e. it is bound to two different metal atoms/ions. $\eta^2$ means that the $\ce{O2}$ ligand is acting as a bidentate ligand (i.e. both oxygens are bonded to the metal). $\eta^2\text{:}\eta^2$ means that the $\ce{O2}$ ligand is acting as a bidentate ligand towards both metal centres.

So, it looks something like this (the structure of oxyhemocyanin, bottom): http://web.tock.com/kalee/chem32/spec/Image1.gif There, $\ce{O2}$ (in the middle) is acting as a bidentate ligand towards two copper centres. In your compound I presume it is acting as a bidentate ligand towards two U centres. Now, as to whether it is an "official" notation, IR-2.5.2, IR-10.2.3.1, and IR-10.2.5.1 (notably the text accompanying Example 21 on p 222) in the 2005 Red Book mention this use of colons.

Source: http://chat.stackexchange.com/transcript/message/36064040#36064040

In the Red Book p. 28 defines "Colons are used in names in the following ways." (italics as in the original). Page 222 (examples 21 to 25) has more examples. But these are all for compound names, not formulae. Is it allowed or common to mix both notations?

Good catch. I looked through again and can't find any examples of the use of the colon in a name. However, as far as I can tell it is permissible to use both μ and η in formulae, see IR-4.6.5 (p 67) and Table V (p 259). The colon is not a structural descriptor such as μ and η; it simply serves to delineate two structural descriptors. Just based on this, I would assume that the use of : in formulae is allowed, but I acknowledge it isn't really hard proof.

This notation is mentioned in the IUPAC Nomenclature of Inorganic chemistry, Recommendations 2005 (‘Red Book’):

IR-10.2.5.1 The eta (η) convention
The special nature of the bonding of unsaturated hydrocarbons to metals via their π-electrons
  has led to the development of the ‘hapto’ nomenclature to designate unambiguously the
  unique bonding modes of the compounds so formed. (…) The Greek
  symbol η (eta) provides a topological description by indicating the connectivity between the
  ligand and the central atom. The number of contiguous atoms in the ligand coordinated to the
  metal is indicated by a right superscript numeral, e.g. η3 (‘eta three’ or ‘trihapto’), η4 (‘eta
  four’ or ‘tetrahapto’), η5 (‘eta five’ or ‘pentahapto’), etc.
(…)
If an unsaturated hydrocarbon serves as a bridging ligand, the prefix μ (see Sections
  IR-10.2.3.1 and IR-10.2.3.4) is combined with both η and κ, where necessary. The colon is
  used to separate the locants of the bridging ligand which indicate binding to different metal atoms.
(…)
Examples:
21.

(μ-η2:η2-but-2-yne)bis[(η5-cyclopentadienyl)nickel](Ni—Ni)

(This convention is not used for unsaturated hydrocarbons only, there is also a more complicated example (25) with carbonyl (CO) ligand.)
Use of this notation for formulae does not seem to be explicitly mentioned or “allowed”. If it is not an omission in the rules, I think that such construct could be called a name-formula hybrid, which might useful for detailed description the discussed structure detail in a shortest way possible.
The $\ce{[^{n}Bu4N]2[(UO2)2(\mu-\eta^2:\eta^2-O2)(NO3)2(\mu-Au(CN)2)2]}$ polymeric complex is described in [1], which talks about $\ce{U-O2-U}$ linkage with bidentate peroxo ($\ce{O2^{2-}}$) unit.
References:

Brown, M. L.; Ovens, J. S.; Leznoff, D. B. Dicyanoaurate-Based Heterobimetallic Uranyl Coordination Polymers. Dalton Transactions 2017, 46 (22), 7169–7180.


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Finding change in enthalpy for a reaction with an atom in elemental state using bond energy
For example, consider $\ce{C(s) + O2(g) -> CO2(g)}$.
If we know the bond energies for $\ce{O=O}$ and $\ce{C=O}$, can we determine $\Delta H$, even though there is carbon in the reaction? Can we ignore it, or is there a certain way to compute the energy?

There are two things we need to note with this. For one, you need to know the bond enthalpies if you are going to obtain $\Delta H$, not the bond energies. Second, we can't just use the bond enthalpies in this case because that would ignore that we have $C(s)$ when we need $C(g)$ to make this reaction occur. We have to explicitly include the atomization energy of carbon (the energy needed to break graphite into single carbon atoms).


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Why is sulfur unhybridized in dimethyl sulfide?

In which of the following molecules is the central atom unhybridized:
  (a) $\ce{S(CH3)2}$
  (b) $\ce{SO2}$
  (c) $\ce{SiH4}$
  (b) $\ce{PCl3}$

The answer given is (a), but I can't figure out why this would happen.
According to me $\ce{S(CH3)2}$ should have a hybridization of $\mathrm{sp^3}$ since it has two sigma bonds and two lone pairs. In fact, I can't identify a reason for any of these molecules in which hybridization is absent since the only condition I know for the absence of hybridization is when the central atom has a large size (Drago Rule).
Can anybody help me to understand why the central atom is unhybridized here?

We can immediately discard $\ce{SiH4}$ — a tetrahedral molecule can never be unhybridised.
We can also pretty quickly discard $\ce{SO2}$. The dated description of this molecule included octet expansion which only worked with hybridisation. But even the more modern description employing charge separation ($\ce{\overset{-}{O}-\overset{+}{S}=O}$) requires some hybridisation to allow for all three bonds to happen, and the Wikipedia-stated structure with an angle of $119^\circ$ does not allow non-hybridisation. (Remember that hybridisation is determined by geometry, not the other way around.)
This leaves both $\ce{PCl3}$ and $\ce{Me2S}$ for discussion. And in general, it is possible for both compounds to have an unhybridised central atom.
Contrary to what is often taught at the high-school level, it is typically most favourable for an atom to remain unhybridised when forming bonds. Bonds would then be formed with p orbitals first, which are typically closer in energy to the bonding partner, allowing for a greater stabilisation. p orbitals are also directional (while s orbitals are spherical) meaning that using the p orbital of the bond’s direction a greater overlap can be generated which in turn means a more stable bond. Likewise, a lone pair prefers residing in a more stable orbital and an s orbital is lower in energy than a p orbital. So whenever an atom has the chance, it will form its bonds unhybridised.
However, most ‘simple’ compounds such as water and ammonia feature bond angles far away from the ideal $90^\circ$ of an unhybridised central atom. This is due to a $90^\circ$ bond angle inducing far too much steric strain between the bonding partners (hydrogen atoms in the water and ammonia molecules). The central atom hybridises to such an extent that the destabilising steric strain is minimised while still keeping the bonding orbitals’ p contribution maximum.
Going from water and ammonia to their heavier homologues hydrogen sulfide and phosphane, the bond angles are tightened to almost exactly $90^\circ$. This is due to the greater sizes of the central atoms which have longer bond lengths per se and thus allow the hydrogens to move closer together without additional steric strain. Both hydrogen sulfide and phosphane have one lone pair that resides in a practically pure s orbital.
Moving on from $\ce{H2S}$ and $\ce{PH3}$ to $\ce{Me2S}$ and $\ce{PCl3}$, we are now increasing the size of the substituents. Whether or not the central atom can remain unhybridised depends on how big the substituents are and whether an angle of approximately $90^\circ$ is too small or not. Merely looking at them, one might assume that chlorine is bigger than a methyl group (the methyl group’s hydrogens are pointing away from sulfur at angles of approximately $109^\circ$) and therefore one might be inclined to assume, given the question in its stated version, that $\ce{Me2S}$ feature an unhybridised sulfur while $\ce{PCl3}$’s phosphorus slightly hybridises to reduce steric strain.
A conclusive answer can only be given by looking at published structures, however. For $\ce{PCl3}$, the Wikipedia page reports a bond angle $\angle(\ce{Cl-P-Cl}) \approx 100^\circ$ — this angle is incompatible with non-hybridisation and thus phosphorus is hybridsed.[1] However, performing a literature search for the bond angle of $\ce{Me2S}$ gave me two sources at the top of the list: one claiming a bond angle $\angle(\ce{C-S-C}) = 99.05^\circ$[2] and one claiming a bond angle of $\angle(\ce{C-S-C}) = 98^\circ 52'$[3] — again neither of which being consistent with non-hybridisation.
Given the physical data of two bond angles practically equal (what is $1^\circ$ anyway?), I conclude the question is based on a false premise and can be challenged.

References:
[1]: https://en.wikipedia.org/wiki/Phosphorus_trichloride
[2]: T. Ijima, S. Tsuchiya, M. Kimura, Bull. Chem. Soc. Jpn. 1977, 50, 2564–2567. DOI: 10.1246/bcsj.50.2564.
[3]: L. Pierce, M. Hayashi, J. Chem. Phys. 1961, 35, 479–485. DOI: 10.1063/1.1731956.

Thank you so much...I had no idea that the main cause of hybridization was steric hindrance and not the stability due to hybridization. Is hybridization unfavourable even for molecules with no lone pairs like $\ce{SiH4}$ (keeping steric hindrance aside)?

@OsheenSachdev Molecules like $\ce{SiH4}$ complicated for a different reason. On paper, we do the hybridisation scheme and thereby create four equal bonds. Nature uses molecular orbital theory to arrive at two different bonding orbital energies — but still four identical bonds because that is how quantum mechanics work. If you then go and localise the molecular orbitals, you do indeed arrive at something that can be labelled $\mathrm{sp^3}$ for each bonding silicon orbital.

In those highly symmetric molecules, the solution that gives the highest symmetry is typically observed. Any distortion would require reducing the symmetry which typically means destabilising at least one orbital.

Here dimethyl sulpfide is the correct answer as if you will see in sulphur atom both 3s and 3p are completly filled and in order to become sp hybridised it has to excit it's electrons to the vacant d-orbital. This is only possible if energy difference between 3p and 3d is low. Since carbon has low electronegativity it is unable to bring 3p and 3d close so the energy differece is large enough that excitation cannot take place and the sulphur atom remains unhybridized.
for example- Since fluorine, chlorine, etc are highly electronegative so they bring 3p and 3d close and so excitation is possible and sulphur becomes sp3 hyb.
Hope this answer will help.

you wrote "here", where is here? Did you plan to attach an image?


text:
Hybridization of Oxygen in Orthoboric Acid

What should be the hybridization of the oxygen atom?

A) It should be $\mathrm{sp^2}$ as 1 lone pair must be in the unhybridized orbital for back bonding with boron which has a vacant p-orbital
B)It should be $\mathrm{sp^3}$ (2bp + 2lp)

(In my book it is given as $\mathrm{sp^3}$ but I cannot find a reason to rule out option A. I am not convinced with option B.)

If you have mesomeric structures, for a compound, which suggest different hybridisation, then then you have something in between, not any of them.


text:
A contradiction regarding the reaction coefficient expression in the Nernst equation
I came up with the following question in my NCERT textbook:

Write the Nernst equation and the emf of the following cell at $\pu{298 K}$.
$$\ce{Fe}|\ce{Fe^2+}(\pu{0.001 M})||\ce{H+}(\pu{1 M})|\ce{H2 (g)}(\pu{1 bar})|\ce{Pt (s)}$$
Given: $E^\circ(\ce{Fe^2+}/\ce{Fe}) = \pu{-0.44 V}$.

As per my understanding, the data of activity coefficients must given because the reaction coefficient in terms of concentration is not dimensionless and its logarithm would be meaningless.
$$\ce{Fe + 2H+ -> Fe^2+ + H2}$$
$$Q=\ce{[Fe^2+]/[H+]^2}$$
Secondly, using a unit of concentration other than mol/litre would definitely yield a different value. So I decided to check its solution from a reference book. In the solution, they have used the above expression of molar concentrations.
So, my question is, is this expression correct?

In functions, such as  log, sine, exp etc.   it is implied that all concentration terms are divided by unit concentration, e.g.  $1 \pu{mol dm^{-3}}$  or whatever  concentration  units you use, so that the terms are always dimensionless. Many texts do not mention this.

This is why for equilibrium expressions we use activity and fugacity. Both of these values are dimensionless and are generally quite closely related to the concentration and partial pressure, respectively, for reasonable concentrations.

But does that mean the expression is incorrect?

The expression is fine because you just take the unitless concentration.


text:
How can nucleophilicity be estimated accurately?

Given the following: $\ce{CN-,OH-,SH-,I-}$. Determine  the relative order of nucleophilicity in a protic solvent.

I know that, more the nucleophile is solvated by hydrogen bonding in a protic solvent, the lesser will be its nucleophilicity. Also, solvation depends upon the size of the anionic nucleophile or its charge density (smaller the size greater the solvation).
Again larger the size, greater will be the polarizability of the anionic nucleophile and thus stronger will be its nucleophilicity.
All these indicate that the larger the size, the greater the nucleophilicity will be, and so the relative order of nucleophilicity should be $\ce{I-}>\ce{SH-}>\ce{CN-}>\ce{OH-}$. But the answer is $\ce{SH-}>\ce{CN-}>\ce{I-}>\ce{OH-}$. So where have I gone wrong? How can I determine nucleophilicity accurately? Is there any other factor on which nucleophilicity depends?

http://www.cup.lmu.de/oc/mayr/ReactScalesPoster.pdf

It has been the lifetime research project of Professor i.R. Mayr of the University of Munich (LMU) to determine reactivity scales of nucleophiles. They key points are that any nucleophilic addition’s rate constant can be derived by the equation
$$\lg k_\pu{20^\circ C} = s_\mathrm{N}(N + E)\tag{1}$$
where $E$ is an electrophilicity parameter, $N$ is a nucleophilicity parameter and $s_\mathrm{N}$ is a solvent and nucleophile-dependent additional nucleophilicity factor. A multitude of nucleophiles and electrophiles have been entered into this scheme over the years and a few posters showing nucleophilicity scales have been created (see the webpage).
Mayr’s research could show that nucleophilicity depends on the solvent and on the nucleophile, that it is a kinetic parameter and that it can only be estimated to a certain extent. Simple parameters such as size or even the charge per volume ratio are never enough to explain a nucleophile’s relative nucleophilicity unless all other parameters remain the same.
So your discussion which seems to base itself on size alone is bound to fail.
I wholeheartedly invite you to scrutinise the database data (also linked on the webpage) to attempt to find truly general rules. If you do, do not hesitate to submit them to the academic journal of your choice.


text:
Removing Lysol from oven
My son put Lysol in the oven, how do I make it safe again? I really don't want to blow up my house over a chicken pot pie. I have new recipe that I really want to try out and it does not include bleach, or Lysol.

Lysol uses lots of different mixtures for their cleaning products. Does this product list the active ingredients?

Have a self cleaning oven? // Running that would burn up pretty much any organic compounds. Then wipe oven out with a damp cloth.

Wipe the oven down with water and let that dry. If you have an oven that has a circulation fan, turn just the fan on and let that run for a few minutes. If you want to run the self cleaning cycle after that, you should be fine to do so. The reason I suggest the wipe down first is so you don't have a lot of residual chemicals that can lead to an atmosphere that can support a fire (being above the LEL).


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How to choose the correct name for calcium sulfate based on a calculations of the crystallisation water included?
I have a test review question which i really need help on. It states :

A student heated a sample of the hydrate $\ce{CaSO4 * xH2O}$ in an evaporating dish over a bunsen burner to remove all the water and recorded the following data:
Mass of empty evaporating dish: $\pu{19.73g}$
  Mass of evaporating dish + hydrate: $\pu{21.50g}$
  Mass of evaporating dish + anhydrous $\ce{CaSO4}$: $\pu{20.80g}$
Determine A) the value of $x$ and B) the most likely name/formula of the hydrate.

I already figured out $x$ by getting the mass of water then simplifying the ratio of anhydrous to water.
But the second question which asks to name the formula of the hydrate really confused me. I would appreciate if anyone can guide me how to solve B.

For the hydrate part, you just add [di, tri, quadra, penta, hexa, etc.] to hydrate.  So if x were 2 (I don't know, I didn't calculate it) it would be calcium sulfate dihydrate.

It would indeed be straight forward to answer, if you had included your finding of $x$.

From my calculations
\begin{align}
x &= \frac{M(\ce{CaSO4}) \times \left[m(\ce{CaSO4 * xH2O}) - m(\ce{CaSO4})\right]}{M(\ce{H2O}) \times m(\ce{CaSO4})} =\\
&= \frac{\pu{136.14 g mol-1} \times (\pu{1.77 g} - \pu{1.07 g})}{\pu{18.02 g mol-1} \times \pu{1.07 g}} =\\
&= 4.94,
\end{align}
which gives $\ce{CaSO4* 5 H2O}$, calcium(II) sulfate pentahydrate.

If the value of x turns out to be 2 than the given salt is gypsum or if it turns out to be half then the given salt is plaster of Paris. These are the common names of hydrates of calcium sulphate Otherwise you may use the prefixes.


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How to calculate volume for NaOH and drano max gel (includes dilution and pH, pOH)
I have a question on an assignment that I don't quite understand. It says:

What volume of $\pu{0.0100 mol/L}$ $\ce{NaOH(aq)}$ would have the same number of moles of hydroxide ions as $\pu{100.00 mL}$ of drano max gel?

The pH of drano max gel is $14.7$.
My first step was to switch from pH $14.7$ into mol/L. That turned out to be $\pu{5 mol/L}$ of $\pu{OH}$ concentration. Then I multiplied the $5$ by $\pu{100 mL}$ ($\pu{0.1 L}$) and got $\pu{0.5 mol}$. Now I don't understand how to get the volume needed. Would you use the formula for dilution ($C_1V_1 = C_2V_2$)?

Yes.
If you have for drano $C_1 = \pu{5 mol/L}$ and $V_1 = \pu{0.1 L}$ that is $\pu{0.5 mol}$, and for $\ce{NaOH}$ $C_2 = \pu{0.01 mol/L}$ and $C_2 = x$. Using formula for dilution:
$$C_1V_1 = C_2V_2 \quad \implies \quad V_2 = \frac{C_1V_1}{C_2}$$
General tip: With encountering questions like these, always think about what units you have (mol/L, L etc.) and what units you want to go to. Most of the time, it gets clear when you just look at the units.


text:
The open circuit voltage of a battery: Does it decrease with increase in pH of electrolyte while the voltage across a load increase?

Why does the open circuit voltage of a battery decrease with increase in $\mathrm{pH}$ of electrolyte, but the voltage across a load increase?

I tested a cell which had aluminum sulfate as its anode electrolyte ($\pu{12 g}$ aluminum in $\pu{50 mL}$ water and a $\mathrm{pH}$ of 3.2), and potassium hydroxide as the cathode electrolyte ($\pu{5 g}$ of $\ce{KOH}$ in $\pu{200 g}$ of water, $\mathrm{pH}$ of 12.3). The anode electrode was aluminum foil and the cathode was platinum. The open circuit voltage (OCV) of the cell was $\pu{0.9 V}$, and when connected to a $\pu{1 k\Omega}$ resistor, the voltage was $\pu{0.324 V}$.
When I increased the concentration of $\ce{KOH}$ from $\pu{5 g}$ to $\pu{10 g}$, and all other material concentrations were kept constant, the OCV decreased from $\pu{0.9 V}$ to $\pu{0.7 V}$, but when I attached the $\pu{1 k\Omega}$ resistor, the voltage increased to $\pu{0.42V}$.
I do not understand why increasing $\mathrm{pH}$ is causing the OCV to decrease, but increase the voltage under a load to increase.
I am guessing for the voltage under the load that maybe the higher concentration of $\ce{KOH}$ leads to more ions in solution thus increasing the current flow. But if this assumption/hypothesis were true, shouldn't the OCV also increase?
Can someone please help?

Did you clean/change the aluminum foil in the mean time? It is possible that the aluminum surface is changing/passivating.

i cleaned the aluminum, yet the observation was the same

A solution containing $5 g$ KOH in $200 mL$ water has a concentration equal to $\frac{5 g}{57 g/mol·0.2 L} = 0.438 M$. Its pH should be $13.6$, and not $12.3$ as given in your text....How do you explain this difference ?

Prediction and evaluation of the measured voltage of any galvanic cell involving metallic aluminium is an advanced magic. It just slow and hesitantly pretend to approach theoretical equilibrium values, partly due oxid layer, partly due polarization as it reacts with water. It has also significant memory, affected by its history, so comparison of subsequent measurement with the same meta piece is nearly useless. I advice to use other methals for experiments, like zinc.

Your aluminum half cell has a standard OCV (i.e., at 1 M $H^+$) of 1.676 volts. Ref. 1
You don't have standard conditions, but that doesn't matter; your conditions do not change for this half cell.
The other half cell is a reduction of water, in alkaline solution, with the electrons from the aluminum to produce hydroxide ions.

The CRC Handbook confirms -0.8277 volts for this reaction. I had intended to use a picture from Ref 1 showing the OCV, but it was incorrect by a factor of 10! Just goes to show you that you can't trust it just because it's in print, or online!
The point is, this reaction wants to consume hydroxide ions and generate electrons, but you are running it in reverse by pumping aluminum electrons into it.  When you increase the pH of the half cell, you increase its voltage by giving the half cell more reactant - but then you are fighting it because it is yielding to the aluminum voltage, so your total OCV out decreases. That's why your cell output is so low (that, and the non-standard conditions). But the good thing is that the extra ionic concentration decreases the cell resistance, so it can supply more current to your 1k resistor even with the lower OCV.
Ref 1. https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Voltaic_Cells/The_Cell_Potential

Not sure why the downvote. The same conclusion is reached by considering Le Chatelier’s Law.
This counterintuitive effect is a function of the internal cell resistance and the external load. The initial test gives 324 microamps current and 420 with added KOH. Corresponding internal cell resistances come out to 1777 ohms initially and 666 ohms for the second test. If the external load resistance is raised to 3k ohms or more, the effect disappears.

I have not downvoted. But I must state that the redox potential of Aluminum has no influence on the battery, because Aluminum does not exist in the presence of water. It immediately reacts producing $\ce{Al(OH)3}$ and $\ce{H2}$. Usually this reaction does not occur, because the aluminum metal is entirely covered by a protective waterproof thin and transparent layer of aluminum oxide. But as soon as this layer is destroyed, for example in an electrochemical cell, the metal reacts quickly with water, and it is $\ce{H2}$ that works as anode and not $\ce{Al}$.

@Maurice: Thanks for your interesting comment. But the question involved the concentration increase of KOH in the cathode compartment, which fights whichever anodic process is occurring, yet reduces the cell resistance so the current can increase. Do you think that part reasonable?

Are you sure of your measurements ? Because $\pu{5 g KOH }$ is about $\pu{0.09 mol KOH}$. Dissolved in $0.2$ L water it produces a concentration $\pu{0.45 M}$. The pH of such a solution is $\pu{14 + log 0.45 = 13.65}$. And you say that you measured a $\pu{ pH 12.3}$. This corresponds to a concentration $20$ times lower.

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


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In a galvanic cell from where does the electron in the external circuit come from?
I know that in a galvanic cell anode donates electrons through the external circuit. Is that the same electrons that the electrolyte gives it ?

Electrons don't come with serial numbers. In a way, all electrons in the world are the same electron. Also, I fail to see how is your title related to your question. Anyway, welcome to Chem.SE.

Anode is the terminal where oxidation occurs.
In galvanic cell, anode is the negative terminal while in electrolytic cell anode is the positive terminal.
And the electrons you are talking about are the free electrons which are liberated in the oxidation half after the oxidation.
Consider:  m A + n Bm+  ->  n B + m An+
Actually what happens is in the oxidation half cell, electrons are liberated because of the oxidation reaction and in the reduction half reaction, for the reduction to take place of the B metal, the required electrons are pulled from the oxidation half container (which are liberated during oxidation of atoms A) via wire which joins the two electrodes.
So basically electrons which flows in the external circuit are the electrons liberated from the oxidation of atoms A in the oxidation half.


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How to to test the concentration of melatonin in a strong acid?
If a small tablet of melatonin was coated with a substance to ensure a slow dissolution in water and was then placed in a strong acid, neglecting the limits imposed by maximum saturation, it would slowly but surely dissolve. I would imagine the simplest method for determining exactly how much has dissolved would be by testing its concentration, however the information available on the topic uses methods beyond my capacity to preform as a high school student. Is there any relatively simple method to determine the concentration?

Is there any relatively simple method to determine the concentration?

That depends on your equipment. Having the following at hand would be helpful:

a UV/VIS spectrometer
mortar and pestle
UV-grade ethanol (or methanol)
pure melatonin

How to proceed:

Prepare a stock solution of the pure melatonin in ethanol.
Prepare some dilutions for the expected concentration range.
Measure the UV absorption of all solutions at $\lambda = \pu{277 nm}$.
Draw your calibration curve (absorption vs concentration).
Powderize the tablet using a mortar an a pestle.
Dissolve the powder in a defined volume of ethanol.
Measure the UV absorption of the sample.
Determine the concentration in the sample using your calibration curve.

You have just applied the Lambert-Beer law. Others have done the same and reported the method in Pharmacophore 2014, 5, 252-257 (free PDF).


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Hydration reaction
Currently I'm measuring the rate of completion of a hydration reaction by putting a temperature probe in the middle of a cup full of reagents and measuring the temperature in an isolated box. The main reaction going on in the cup is:
$$\ce{CaSO4\cdot 0.5H2O + 3/2 H2O -> CaSO4*2H2O}$$
Now I'm planning to change from one type of cup to another kind of cup with a different volume and a different shape (that will also be completely filled). Will this influence the temperature profile of the reaction?
Cup 1 (with a volume of 192 ml):

Cup 2 (with a volume of 250 ml):

Hello and welcome to Chemistry.SE. If you have any questions about the site please feel free to visit the [help]. Best of luck!

@Fl.pf. Do you have something to back your answer up? Because I'm not gonna believe you if you simply say yes.

@Fl.pf. I can see that the generated heat would be different for a different volume, but that heat would also be spread over a different volume. Wouldn't this cancel out then?

@Michthan Over volume maybe. But if you change the shape aswell? Also you didn't state if you used the same amount of material for the different volume of cup. Maybe I don't understand exactly what you mean, but I don't think your question is written clearly enough (too much room for interpretation of what you really want to know).

@Fl.pf. I'm leaving work now, but I'll add more specifications tomorrow to make the question more clear.

@Michthan I would still say yes. Since your surface/volume ratio is different in the two cups, you should have different rates of heat dissipation etc?
I'd recommend reading this if you have the access: T. Boddington, P. Gray
Proc. R. Soc. Lond. A 1970 320 71-100

I don't believe there would be a temperature difference unless the concentration of reagents changed too.
If per unit area the same amount of product is created, then energy created by said conversion would be the same, regardless of the total volume.


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What is the difference between metaphosphoric acid and phosphonic acid?
Are metaphosphoric acid and phosphonic acid the same? I have been told that they're same BUT the structures online are different.
Metaphosphoric acid has formula: $\ce{(HPO3)_n}$
phosphonic acid has the formula: $\ce{H3PO3}$
Is this correct?

See this answer: http://chemistry.stackexchange.com/questions/40506/what-does-the-prefix-meta-mean-in-inorganic-chemistry/40517#40517

Metaphosphoric acids general formula is (HPO3)n ; where n denoted number of phosphoric acid units present in the ring with n being greater or equal to 3.
These acids are actually the phosphoric acid units bonded together in rings (cyclic structures), hence forming metaphosphoric acid molecules.
In Metaphosphoric acids, each phosphorus has an oxidation state of +5.
Here I am giving you the example of trimetaphosphoric acid i.e. when n is 3.

Phosphonic acid or Orthophosphorous acid is H3PO3 with the oxidation state of phosphorous being +3. Here is its structure:

Not sure phosphorous is +3.  Hydrogen attached to it might be hydridic?

Yeah, P claims the two electrons from the P-H bond to give 5-2=3.


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Why is 1-bromo-3-methylbut-2-ene named like that and not as 2-(bromomethyl)but-2-ene?
I had this confusion related to nomenclature of halo substituted compounds.
Why is it that 1-bromo-3-methylbut-2-ene is named like that and not as 2-(1-bromomethyl)but-2-ene? why do we prefer the carbon chain including the carbon attached to the bromine and not the chain including the carbon of the methyl group?
Which IUPAC rule was used in making this preference between those two carbon chains of equal length?

Regardless of the correct IUPAC nomenclature, the proposed names “1-bromo-3-methylbut-2-ene” and “2-(bromomethyl)but-2-ene” describe different compounds.

Here's 1-bromo-3-methylbut-2-ene's structure:

You have only two possible chains to choose from: a 3-carbon chain and a 4-carbon chain (doesn't matter which methyl group would be part of that chain, the other one will become a substituent on the chain).
You won't choose the 3-C chain for two reasons:

It is not the longest chain.
It does not contain the double bond in the molecule.

Thus, you are left with a 4-C chain, which necessarily includes the bromine atom.

But why does the numbering start from Br and not the other side if C=C is the functional group here, Br is a substituent.

@Loong can you please help? (See my other comment on this answer)


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Can the decompistion of one carbonate inhibt the decomposition of another carbonate?
If two different carbonates of Group II, are mixed in equal amounts and then heated could one of the carbonates decomposition inhibt the other ones decomposition?
I don't understand how it would because (as far as I am aware) the reaction is not an equilibrium.
However I was told that if:

Equal masses of barium carbonate and magnesium carbonate powders are mixed together, and then heated using a Bunsen burner flame until there is no further change.

then

The residue left after heating reacts with aqueous hydrochloric acid to produce carbon dioxide.

This suggests that not all of the carbonate decomposes. However (from Wikepedia)

Maximum temerature of a Bunsen burner flame is 1560 degrees celcius
Thermal decompistion of barium carbonate starts at 1360 degrees celcius
Thermal decompistion of magnesium chloride starts at 350 degrees celcius and is considered complete above 900 degrees celcius

This suggests that all the carbonate decomposes, however I am skeptical of this (Wikapedia) data.
So is thermal decompistion being inhibted? Or is barium carbonate simply not undergoing full decompistion (Wikapedia data is wrong)?
I couldn't find anything about it but maybe barum carbonate, like magnesium carbonate needs a much higher temperature than stated to fully decompose?

https://en.wikipedia.org/wiki/Bunsen_burner mentions that 1560 is max. for Tirrill burner not bunsen. You simply won't make the mixture hot enough.

Barium carbonate DOES NOT decompose on heating with a bunsen burner.

@Mewton barium carbonate does decompose on heating but the temperature required for its decomposition is not provided by the Bunsen burner


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Confusion in hybridisation of coordination compound
I want to know the configuration and hybridisation of the compound, hexaaquavanadium (III) ion that is $\ce{[V(H2O)6]^{+3}}$.
In this Vanadium is in $+3$ oxidation state its electronic configuration is $\mathrm{[Ar]4s^0 3d^2}$.
So there are $3$ more empty $\mathrm{d}$ orbitals.
Hence the hybridisation should be $\mathrm{d^3sp^2}$.
But it is $\mathrm{d^2sp^3}$ and paramagnetic .
I could not understand how .
Can you explain me?

Related
Basically, it isn't really wise to use hybridization on coordination complexes to begin with. However, the hybridization of $d^2sp^3$ makes sense if you think of it as just describing the shape of the molecule. This should be octahedral, due to its 6 equivalent groups around the metal center, and $d^2sp^3$ is the valence bond description of the hybridization of an octahedron.

Using Ligand Field theory and MO diagrams is probably a better way to understand how the orbitals overlap and where electrons are in the final structure. This diagram is for $\ce{[Ni(H2O)6]^2+}$, but the diagram will look very similar for the $\ce{V}$ complex, just with fewer electrons. (Image source: dx.doi.org/10.1021/jp4100813 | J. Phys. Chem. B 2013, 117, 16512−16521)

So actually it is not $d^2 sp^3$ . We write it as to denote the octahedral shape .

@TrYiSCheM For the most part, I would avoid hybridization with complexes altogether, but yes the hybridizations are commonly used to represent the shape. We can see some of the common associated hybridizations/geometries here.

But actually in which sub shell the ligand donate their electrons

In V what will be filled in empty orbitals.

@TrY iS CheM I'm not sure I understand your question. Based on MO theory, we would say that the electrons from the ligands went into Molecular orbitals of the complex rather than the atomic orbitals of the original vanadium.


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Magnetic behaviour of complex compounds

Which one of the following cyanide complexes would exhibit the lowest value of paramagnetic behaviour?

$\ce{[Cr(CN)6]^{3-}}$
$\ce{[Mn(CN)6]^{3-}}$
$\ce{[Fe(CN)6]^{3-}}$
$\ce{[Co(CN)6]^{3-}}$

Here I determined central atom configuration and it is $\mathrm{3d^3}, \mathrm{3d^4}, \mathrm{3d^5}, \mathrm{3d^6}$ for each option in order. $\ce{CN-}$ is a strong-field ligand so electron will get paired hence option 2 and 4 will have least magnetic behaviour
But only option 4 is correct. Why?

Magnetic behaviour of complex compounds are determined on the basis of magnetic moment, rather the extent of paramagnetism is measured in terms of the magnetic moment μ.
μ = √n(n+2) BM , where n notates the number of unpaired electrons.
First lets give a look towards the number of unpaired electron each of the central metal atom has:

Here the question is asking to determine the complex compound which shows the lowest paramagnetic behaviour. As Co^3+ has no unpaired electron, its μ = 0, hence it is diamagnetic, and hence shows shows lowest paramagnetic behaviour among the above complexes.
And best of luck for jee'17.
Me too appearing the exam.
Hope it helps

why does pairing occur only once, in $\ce{Mn^3+}$? @Yb609

Because 2 vacant d orbitals are required by the CN- .  Basically hybridisation is just a method to explain the molecular geometry. Don't think that molecular geometries are based on hybridisation. Its to true. Hybridisation is just a theory to explain molecular geometries. It has been found experimentally that [Mn(CN)6]^3- shows octahedral geometry and hence to explain that we use the concept of hybridisation and hence we take it as a dirty rule that for coordination number 6, in presence of strong ligand electrons are paired up so as to create 2 vacant d orbitals.

And hence the above rule is a conclusion of what we have experimentally observed.

Oh, thats why this happens, then what about in complexes like $\ce{[V(H2O)6]}$? in case you know, there was a question in the new section, pls do post it there, Im eager to know that too, thanks @Yb609

@QuarkyLittleThing ya i have also seen that question. This same reason works out there as well. That guy who asked that question has a misconception that the molecular structures are given by hybridisation. Hybridisation is not a phenomenon. It is just a concept to explain. You too know that in chemistry experiments are done first and then we give theories (like hybridisation) to explain that.Similarly in [V(H2O)6]^3- case, experimentally it was found to be octahedral low spin complex and hence to explain that we say the hybridisation is d2sp3.

This is because in Co+3 there is no unpaired electrons .i.e all paired up (due to the strong ligand's effect) due to which it will be diamagnetic in nature and hence it will have the least paramagnetic behaviour.


text:
Populations of a two level system
Take the two level system in atomic iodine where the ground state is four fold degenerate and the excited state at $7603 \,\mathrm{cm}^{-1}$ above the ground state (which is taken to have zero energy) and is two fold degenerate from the $^2P_\frac32$ and $^2P_\frac12$ terms arising from the configuration of atomic iodine.
I want to find the temperature at which the ratio of the population of the upper state to the lower state is 1:50.
Using statistical mechanics is it correct to say that this is the case when:
$$
\frac{n_u}{n_l} = \frac{g_{u}\mathrm{exp}(-E_u/k_{\mathrm{B}}T)}{g_{l}\mathrm{exp}(-E_l/k_{\mathrm{B}}T)} = \frac{2\mathrm{exp}(-7603.hc/k_{\mathrm{B}}T)}{4} = \frac{1}{50}
$$
Rearranging for T and substituting in the values of the constants gives me $3401.9 \,\mathrm{K}$ which seems alarmingly high. Where have I gone wrong?

Doesn't seem all that high to me.

Really? Only a 1/50 population for the lowest lying excited state at 3400K? Is my method definitely right?

I didn't check the math, but electronic excitations typically require some thousands K.

What units are using for everything? That's the only thing I could see causing major errors.

The units work alright, I'm fairly sure of that. Perhaps it's correct, I just assumed if done something wrong as it seems such a high temperature to reach such a low level of excitation

We are used to thermal energies and  at room temperature this is small $kT \approx 210 \pu{cm^{-1}}$ so thousands of degrees are needed to reach thousands of wavenumber energies as your calculation correctly  shows.  Photons would normally be used to excite these energy levels and they are very energetic compared to average room temperature $kT$.

I'm getting basically the same answer when I work out the math. I got $T=3398.45\ K$, but we probably just rounded the constants differently.
This is a good lesson in how much thermal energy is carried around at any given time by molecules in the air, as well as the importance of distinguishing between thermal processes and photo-processes.
That is, if we wish to sustain an ensemble of excited particles, there must be sufficient energy such that when we spread the energy out evenly, which is one of the things that this equation implies given that it claims all possible microstates are equally likely, some of the energy must be stored in the form of electronic excitations.
The reason this probably seems like a very high temperature is because we use equipment all the time that relies on electronic excitations and we never have problems with the equipment melting due to it being at thousands of degrees Celsius. That is simply because in these photo-process, meaning the energy is carried in the form of photons, the temperature never is that hot.
That is, if I shoot a photon in, or even a stream of photons at a sample with the transition you are describing, each photon carries a very small amount of energy, and even though there may be quite a lot of them, the energy is quickly dispersed because the excited state is very short-lived, so the photon gets re-emitted, and likely just makes it way out into an infinite reservoir known as the surroundings.
Thus, it is very difficult to maintain a population of excited states thermally, because the system needs to basically be a closed system or else it will eventually just radiate all the energy away.

Another way of thinking about this is that the value of $k_bT$ at room-ish temperature of $298\ K$ (in a familiar unit) is $207\ cm^{-1}$. This makes it clear then why rotationally excited states are frequently occupied at room temperature, because these transitions are on the order of tens of wavenumbers. On the other hand, we often get off free assuming everything is in the vibrational ground state because these transitions are on the order of thousands of wavenumbers.
This is also where the idea of rotational temperature and vibrational temperature come from. These can be understood physically as,

an estimate of the temperature at which thermal energy (of the order of $k_bT$) is comparable to the spacing between rotational [or vibrational] energy levels.

When we make use of excited states via photons, in general how large a fraction of the population is actually excited at any one time? I know in something like a laser or an ICP-MS you're aiming to keep a majority of the relevant population excited, but for most applications the excited fraction is pretty small, isn't it?

@hBy2Py Right that's what I was trying to say. It wasn't a totally pertinent point I suppose, but when exciting using photons, the fraction in an excited state should be quite small. I was just trying to make a distinction between photo excitation and thermal excitation.

Ya and lasers are quite different because you need three states. I wonder if you can make a thermal laser? Seems unlikely and unclear what that even means.

re. number of excited states, if the sample has an optical density of , say, 2 then 99% of light is absorbed so you can work out what fraction of molecules are excited. With a laser its not hard to completely bleach a sample and similarly it is easy to invert the population and make it into a 'gain medium' i.e. a laser amplifier.

, jheindel, I think the answer's really good. I was just trying to point out a minor aspect that you hadn't touched on (as best I could tell).


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What stops intermediates of alcohol hydration and dehydration reacting in the same way?
I am aware they both occur under different conditions and that water attaches in a secondary position in elimination but primary position in hydration.
(Stage 2)
(Stage 3)

But both appear to be the same intermediate, why exactly do different reactions follow this stage?

dehydration requires conc. acid, whereas hydration requires dilute acid, it is mostly the conc. of the reagents used , that determine the path of a reaction, in case of multiple paths possible. The conditions provided, like activation energy through heating or some other such means also play a major role, in such scenarios

This is one reversible reaction, you only change equilibrium with suitable conditions.

As you have spotted, the reaction is reversible and can proceed both ways. Which way it takes depends on the conditions and by implication on Le Chatelier’s principle.
In concentrated $\ce{H2SO4}$, any water will be immediately ‘captured’ by the strongly hygroscopic sulphuric acid. Thus, water is actively drawn out of the equilibrium and the alkene is produced by elimination.
If the reaction is run in aquaeous solution (or at least with water as a cosolvent) and only catalytic quantities of acid, then the nucleophilic attack of water is more likely and the alcohol hydration product is produced.
(In modern synthetic lab-scale chemistry, neither of the reactions are used due to their rather poor selectivity (especially hydration) and harsh conditions (especially elimination). Instead, hydration is typically performed with boranes (anti-Markovnikov) or Markovnikov-selective reagents while elimination is typically achieved by transformation of the alcohol functionality into better leaving groups such as mesylate.)


text:
Do both H (enthalpy) and ∆H, or ∆H alone have a physical significance?
My chemistry teacher said that H (enthalpy) alone has no physical significance it is ∆H or change in enthalpy that has a physical significance. But my book says that both have a significance and H alone represents the total heat content of the body. Thus I am in a confusion, please help me.

related http://chemistry.stackexchange.com/questions/19541/why-it-is-not-possible-to-find-absolute-value-of-internal-energy

Offcourse H and ∆H have physical significance. H represents the enthalpy of a system. Enthalpy refers to the sum of the internal energy of a system plus the product of the system's pressure and volume.
The ∆ symbol is used to represent change. Therefore, ∆H represents the change in enthalpy of a system in the reaction. So your book is completely correct.

That being said, the teacher is right too (in a way). We pretty much never know absolute enthalpy of anything. Arguably, it may not even be well-defined.

@IvanNeretin I completely agree with  your point.

@IvanNeretin Ya thats true. Agreed with you


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In Glucose, how will you prove the presence of the following?

One carbonyl group
5 hydroxyl groups
One primary alcholic group

1.  Reaction with $\ce{HCN}$, or with hydroxlyamine ($\ce{NH2OH}$)
Glucose reacts with $\ce{HCN}$ to give a cyanohydrin, or with a schiffs base to give an oxime. This shows the presence of a carbonyl group.
2. Reaction with acetic anhydride $\ce{(CH3CO)2O}$
Glucose gives a penta ester with acetic anhydride, revealing that it has 5    hydroxyl groups.
3. Reaction with Nitric acid
On oxidation with Nitric acid ($\ce{HNO3}$),glucose as well as gluconic acid both yeild a dicarboxylic acid (saccharic acid). This indicates the presence of a primary alcoholic group in glucose.

To prove the presence of $\ce{-CHO}$ group:
When glucose reacts with hydroxylamine, oxime is formed and cyanohydrins are formed on the addition of hydrogen cyanide to it. This reaction can confirm the presence of carbonyl group in glucose. And also on the reaction of glucose with a mild oxidising agent like bromine water, the glucose gets oxidised to carboxylic acid containing six carbon atoms. This indicates that the carbonyl group is present (as an aldehyde group).
To prove the presence of $\ce{-OH}$ groups:
The presence of $\ce{-OH}$ group is confirmed after the acetylation of glucose with acetic anhydride, which gives glucose pentaacetate.
Hope this helps

The $\ce{OH}$ groups react with acetic anhydride to give acetates. Thus, I took the advantage to correct acetic acid to acetic anhydride. Please switch it back if you meant acetic acid.


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Why is boric acid written as B(OH)3 and not H3BO3?
Why is boric acid written as $\ce{B(OH)3}$ and not just $\ce{H3BO3}$, like it should be for an acid?

I've seen it written both ways. If anything, the second way seems to be more common.

yeah, i did agree in the question, its written both ways, and not just as B(OH)3, but why does this happen only for boric acid? being written in the form of a base @IvanNeretin

@QuarkyLittleThing Writing H3PO4 as PO(OH)3 is actually weird. Same applies for H2SO4. For H2SO4 writing it as SO2(OH)2  looks very confusing and rather i would prefer writhing it as H2SO4. I think its just a way of writing. If there is any chemical or structural reason behind that, then I am eager to know.

This is mostly because boric acid is commonly believed to have different acidity mechanism. While common acids generally dissociates
$\ce{HNO3 <=> H+ + NO3-}$
boric acid is commonly believed to associate with water
$\ce{H2O + B(OH)3 <=> H+ + [B(OH)4]-}$
this type of acidity is rather common for metal hydroxides, hence the way of writing.
For that matter, not every hydrogen should be written at the beginning of a formula of an acid, but only actually acidic ones. So, $\ce{H2SO4}$, but $\ce{HPH2O2}$

Even the convention of writing acidic hydrogens at the beginning is not always followed, and it is all too easy to suppose that  $\ce{HCOOH + NaOH}$ will give ... $\ce{NaCOOH}$.  Oy!

And given that the second $\mathrm pK_\mathrm a$ of sulfuric acid is a modest $\sim 2$, one could also envision it being written as $\ce{HSO3OH}$. Interestingly, in the early development of chemistry apparently it was thought simply to be an adduct of $\ce{SO3}$ and $\ce{H2O}$!


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Naming this oxo-carboxylic acid
On a textbook question, the answer key revealed this structure to be 3-methyl-2-oxopentanoic acid:

I found this strange, as the carbon from the left (#3) is methyl, but the oxo prefix is on the right for the $\ce{C=O}$. Why is this?

Why does it not say 3-methyl-5-oxopentanoic acid or 4-methyl-2-oxopentanoic acid? Is the nomenclature correct?

Yes it is correct

Why does it not say 3-methyl-5-oxopentanoic acid or 4-methyl-2-oxopentanoic acid? Please explain

Why would it be either of those? The methyl and oxo groups are right next to each other.

@ringo how is it correct?

@ringo methyl (ch3) isn't right next to oxo (C=O)

It is obviously incorrect. CH3 from either sides is situated at #3.  And how is oxo at 5?

Starting from the group with the highest priority, the carboxylic acid group, we label the longest carbon chain:

This molecule might otherwise be named a ketone, but since the carboxylic acid group has a higher priority, it is treated as another substituent (oxo), located at the 2 position. The methyl group is attached to the 3 position, hence the name 3-methyl-2-oxopentanoic acid.


text:
Does 30% hydrogen peroxide eat away at platinum?
if I soaked a platinum dish with 30% hydrogen peroxide, will it eventually eat away at it?

Nope. On the contrary, hydrogen peroxide will decompose pretty quickly.

Yes, as @Ivan said, hydrogen peroxide will not corrode platinum, instead hydrogen peroxide will get decomposed.
Catalytic decomposition of hydrogen peroxide by metals and alloys of the platinum group:

The catalytic decomposition of dilute aqueous solutions of hydrogen
  peroxide has been studied on a number of Group VIII metals and binary
  alloys, including palladium-gold. The kinetics were measured at 27 °
  as a function of catalyst composition and pH of the solution. In
  neutral solution, the activity of the individual metals of Group VIII
  was generally greater than those of Group VIII2, the order of specific
  activity (per unit surface area) decreasing in the sequence $\ce{Pt > Os >
 Ir > Pd > Ru > Rh}$. (...)

Other references:

http://ac.els-cdn.com/S0926337306001561/1-s2.0-S0926337306001561-main.pdf?_tid=a114888e-1c29-11e7-9622-00000aacb360&acdnat=1491635258_a3e20984dda9eaaf07e78f4814383b96
http://pubs.acs.org/doi/abs/10.1021/ed080p788
http://jcsp.org.pk/ArticleUpload/2541-11513-1-CE.pdf


text:
Surface area: what is the difference?
What are the differences between:

van der waals surface area
solvent accessible surface area
topographical polar surface area

What is the significance of these differences, and what are the differences in how each of the three is used?

Well, you should elaborate. IMO question isn't bad, but it gathered downvotes.

@Mithoron Attempted an edit. idk, if you don't like what I changed, please feel free to edit further, or roll my edits back.

A Van-der-Waals surface Area (VdW) is a surface constructed by spheres around atoms with the radius of each sphere being the van-der-Waals radius of the respective atom.
The Solvent accessible surface area (SAS) is an surface normally being constructed by "rolling" a sphere the size of the vdw-radius of the solvent molecule around the solute and taking the middle of the solvent sphere as the surface of the solute.

Picture is taken from the original GEPOL publication of Pascual-Ahuir and Silla J. Comp. Chem., 1990, Vol. 11, No. 9, 1047-1060
So the difference is just the size. Both have the same questionable physical meaning. They are used for example in continuum solvation models like COSMO/PCM.
I used the SAS for Frozen-Density-Embedding-models to close unphysical holes (in PCM the surface constitutes an infinite continuum of solvent, so there would be an infinite volume of solvent between two subsystem molecules) between molecules of different subsystems in a QM/QM/Cont. multilayer method. Nonetheless, they are both only viable for qualitative results.
For the PSA, just read the wikipedia article.
The PSA is only the surface of the polar atoms of a molecule and it is used in medicine to calculate permeability of cell membranes of molecules in the body.


text:
Why doesn't nitric oxide react with water?
Why doesn't $\ce{NO}$ (nitric oxide) react with water? I know it is a neutral oxide, but I'm really curious as to why that is. Why will $\ce{NO}$ not react with $\ce{H2O}$?

Why would it? Normally we explain why things happen, and not why they don't.

Think about it, oxides dissolve in water to give their corresponding acids or bases. Here, what acid would you get, and how stable would its conjugate base ? i believe now you have your answer

@IvanNeretin what is "normal"?

Although nitric oxide is thermodynamically unstable, it is kinetically stable as its decomposition rate is very slow[ref] and so it is considered stable and won't normally react to anything.
So, nitric oxide wouldn't normally hydrolyze easily. But in the presence of air, it forms nitrous acid $(\ce{HONO})$:
$$\ce{4NO + O2 + 2 H2O -> 4 HNO2}$$
However, nitrous acid is very unstable and quickly disproportionates:
$$
\begin{align}
\ce{2 HNO2 &-> NO2 + NO + H2O &(cold dil. soln.)}\\
\ce{3 HNO2 &-> HNO3 + 2 NO + H2O &(warm conc. soln.)}
\end{align}$$
Since $\ce{HNO2}$ is required to prepare diazonium salts (a versatile compound in organic chemistry), it is prepared in-situ by reaction of $\ce{NaNO2}$ and $\ce{HCl}$. Due to its instability, it is prepared for one-time use and is not stored for further preparations.

Reference: J. Phys. Chem., 1958, 62 (3), pp 360–361 https://doi.org/10.1021/j150561a030

I have also heard about a disproportion reaction which happens with a productivity of around $0,01 \%$ as,
$$\ce{4NO + H2O -> N2O + 2HNO2}$$

Hi thanks for your contribution. Though in such cases, it's much better to cite your source as well. A research paper or a famous textbook generally works. Thanks!

My chemistry teacher, a professor at my university told me about this reaction a long long time ago. Since then the only source where I could find this reaction was a Russian website which also contains a vast database of known chemical reactions. There is no original source, there's a mistake (HNO3 instead of HNO2) and the only note is "almost no reaction". Sorry, it's a really interesting reaction and I keep wondering about its mechanism but I just fail to find any papers or books, at least on the Internet.

No worries, and I appreciate your response. But in this case, I think it's rather better if you instead ask a new question, looking for proper citation for the reaction. We support [tag:reference-request] as well, so it wouldn't be a problem :-)


text:
Calculation of empirical formula from elemental analysis

A compound contains 40.0% Carbon, 6.7% Hydrogen, and 53.3% Nitrogen (by mass). Calculate the empirical formula.

I'm getting $\ce{C7H14N8}$, but my professor says it is $\ce{CH2N}$.
Which is the correct answer?
When brought to moles, 40 g C is 3.33 moles of C, 6.7 g of H is 6.63 moles of H, and 53.3 g of N is 3.80 moles of N.
Divided by the lowest mole count (3.33), you get one carbon, two hydrogen, and 1.14 nitrogen. I am multiplying all by 7, to get $\ce{C7H14N8}$.
He is saying 1.14 is 1, basically, and that is can all be simplified as it is, no need to multiply, and your final answer is $\ce{CH2N}$, marking my answer half off. It's not worth going to battle for, but I want to know who is really right, because I feel my answer is better.
So the real question is, with N, should you get it closer to a whole number, or leave "as is"? Thanks.

Could you include your calculation? It will probably clarify the situation.

The professor's answer can't be right. The carbon to nitrogen mass ratio ought to be 6:7 if the mole ratio is 1:1.

Incidentally - "homework" is kind of a misnomer, because around here it doesn't mean literally homework. See here for more info. Anyway, with the extra detail you added, your question is fine, so don't worry about it.

These would be some terrible analytical results for a compound of nothing but C, H and N. The carbon value in particular would fail any audit if the known compound were 42.9% C and the reported value was 40.0%. Nitrogen gets a fail too, though you don't expect quite the accuracy for N as you do for C. Finicky H gets by on this one, but overall, if the emperical formula is CH2N, these are aweful results. What a terrible problem for an exam.

A sad Ph.D. student had been working on the synthesis of a natural product, with empirical formula $\ce{CH2N}$, for several years. He finally succeeded in synthesising it - or so he thought - because his NMR data seemed to fit the target structure. Excited about his breakthrough (in fact, he was so excited that he didn't even bother taking a mass spectrum!), he sent a sample for elemental analysis, expecting to get some results along the lines of 42.8% carbon, 7.2% hydrogen, and 50.0% nitrogen.
Not long after this, the results came back, and he discovered that his compound inexplicably had 40.0% carbon, 6.7% hydrogen, and 53.3% nitrogen. "That's not what I expected," he thought to himself - "I wonder if I can publish it and try to gloss over the data."
Upon checking the Author Guidelines for the Journal of Organic Chemistry (section 2.2.2), he found that they were actually pretty strict.

Found values for carbon, hydrogen, and nitrogen should be within 0.4% of the [calculated] values for the proposed formula.

Thus was the story of the sad Ph.D. student who had to go back to the drawing board. (Thankfully, he had synthesised enough of the immediate precursor and by varying the conditions, he eventually succeeded in making his target.)

TL;DR Rounding down 1.14 to 1 is a gross error and won't ever get published in a reputable journal. Considering that this is not even a real-life scenario, there's absolutely no reason why standards should be lowered.

Exactly. As an analytical guy, that's what drove me nuts about this terrible test/homework problem. (see my comment ripping the analytical data on OP if interested). Big +1 for clever presentation!


text:
Charging Paint either negatively or positively to repel paint with the same charge
Is it possible to give paint a negative or positive charge.  Once dry, if you tried to paint on the surface with the same charged paint, would it just run down and refuse to stick?

In general, it is not possible to give anything a charge.

But what about this:  https://en.wikipedia.org/wiki/Electrostatic_coating

I am curious about the possibility for Anti-Graffiti purposes.  If it's possible to charge the paint a municipality uses to cover graffiti so that it repels spray paint in the same manner that two negative poles of a magnet repel each other.


text:
How to obtain pure alcohol from mixture of alcohol & carbonyl? (Grignard experiment)
If one discovers that after finishing up a Grignard reaction that their alcohol product has been contaminated with the carbonyl starting material, how could they separate the alcohol from the carbonyl compound to obtain a purified alcohol product?
I was thinking that they could recrystallize the contaminated product using an alcohol as a solvent, but is there a better option for separating these two compounds?

I rather doubt that there is a better option in general. Other possible methods are

distillation
To give an example, acetone has a boiling point of 56 °C, while tert-butanol boils at 82 °C.
column chromatography on silica

But in many cases, your suggestion to separate the compounds by crystallization is a good idea. This is in particular true when your product has a high boiling point and/or you have to purify large amounts of substance.


text:
Positive electrode potential of copper
My book (NCERT India) says that the electrode potential of metals is decided by the following factors:

Enthalpy of atomisation
Ionisation enthalpy
Hydration enthalpy

It says that the first two require energy while the latter releases energy. That's all good and I can follow till there. But then it provided this table of thermochemical data for 3d series metals (Table 8.4 page 218):

It says that the electrode potential for $\ce{Cu^2+/Cu^1+}$ is positive because the hydration enthalpy does not balance the enthalpy of atomisation and the ionisation enthalpy - which requires energy. By the same logic, simply summing up 1 & 2 (sum of the first and second ionisation enthalpies) & 3 would give a positive value. How is it feasible for any of the there elements to have a negative electrode potential?

"simply summing up 1 & 2 (sum of the first and second ionisation enthalpies)"Your point 1 was atomization enthalpy instead...

The potential as it is given there is vs. standard hydrogen electrode(SHE) which has an arbitrary assigned value of 0.0000V. Through the choice of another reference, it is very easy to make all of those numbers positive or negative. The absolute value of the half cell potential can only be useful in comparison to others. So you could make arguments like Cu has a more positive potential compared to Ni(by 0.59V based on the above numbers), but that is it.


text:
Is cycloocta-1,3 diene planar?
I know that cyclooctatetraene is non-planar because of

Angle strain and
Antiaromaticity if it was planar.

But what about cycloocta-1,3,5 triene or cycloocta-1,3 diene. It doesn't risk being antiaromatic, but there still is angle strain. Would it lose its planarity and hence inhibit resonance?
In general, when is the angle strain enough, that releasing the angle strain but also inhibiting resonance at the same time lead to a net increase in stability?

Guess those would be somewhat non-planar, but not enough so to break the conjugation completely. Moreover, I expect the conjugated part to be more planar than the rest.

That's what I'm guessing too, but I posted this question so that someone could confirm/debunk this thought.

I think the more dominating factor for the non-planarity of cyclooctatetraene is antiaromaticity, rather than angle strain. So in case of cycloocta-1,3,5 triene and cycloocta-1,3 diene though angle strain may be a destabilizing factor but the other one i.e. antiaromaticity is absent, on the other hand they have a conjugated system of 6 and 4 carbons respectively which gives them stability to some extent. So I agree to Ivan Netetin's comment that the conjugated part will be more planar than the rest.

Closely related, but not a duplicate: Why is cyclooctatetraene non planar but the cyclooctatetraenide anion planar?

Short answer: there's some twisting of the conjugated structure, particularly in the diene
I ran initial calculations for these using B3LYP density functional theory. If anything this method has a slight tendency for extra delocalization.
I'll follow up later with other methods, but I suspect the general answer will be the same.
Cycloocta-1,3,5-triene
Honestly, I expected the delocalization to be retained in the central conjugated section (albeit not perfectly flat) with some twisting on the outer double bonds.
Indeed, the dihedral angle around that central C=C is ~6.9°, but it's not that different around the other CC=CC bonds (~3° and 5°, respectively).
The whole C=CC=CC=C is fairly co-planar, with the remaining two carbons profoundly out of plane, likely relieving ring strain.
It's clear there's some delocalization because the "single" bonds between the double bonds have bond lengths ~1.46Å, smaller than usual.
Thus, in the triene, there's slight non-planarity to the conjugated section, but it's fairly delocalized.

Cycloocta-1,3-diene
Interestingly, the result is a bit different for the diene. While the C=C bonds are planar (i.e., dihedral angle ~2°) the middle C-C bond is strongly non-planar (i.e., dihedral angle ~38.4°). Moreover, it's 1.475Å in length, longer than the comparable bonds in the triene.
Here, I think the delocalization of two double bonds isn't enough to balance the ring strain.

With more saturated carbon atoms and their hydrogen atoms sticking out above and below the "plane" of the ring, the diene could be more affected by conformal strain than the triene.  That disfavors planar bond geometry.

Agreed. There's several things going on that make the diene different than the triene.


text:
Why do some reactions need to be maintained at a particular temperature for prolonged periods of time?
Some reactions require that temperatures of 70–100 °C to be maintained for periods of time as long as 24 hours. Why is this so?

Quite simply, reactions happen faster at higher temperatures (because who likes waiting), and some may not proceed at all until heated. From a statistical viewpoint, you are giving more reactant molecules more energy, which thereby increases the number of collisions, which in turn increases the number of successful collisions (those which produce product). A typical way to attain such conditions is to place a reaction under reflux.

While I think both of the other answers are good, I want to comment on two practical constraints that yield long reaction times:

Solvent choice - We perform a lot of chemical reactions in solvents. This allows easy mixing and usually facilitates separations and purification steps after reaction. On the other hand, if some reaction requires a high activation energy barrier, we cannot always simply raise the temperature or most practical solvents will evaporate (e.g., we can't run most reactions at 300-400 °C even if it would minimize reaction times).
Human timescale - Even if some reactions might complete sooner, it's practically much easier to run a reaction for ~24 hours - you set up the reaction in the morning and then it's done the next day. It's neither safe, nor practical for someone to come in at 2 AM to shut down a reaction.

From a fundamental standpoint, diffusion/transport issues limit reaction times and thermal energy increases reaction rates but practical concerns also affect reaction times.

Long hold times are needed for one of two reasons:
(1) Mass Diffusion limitation - For example, in a drying reaction moisture needs time to diffuse through a sample to the surface before evaporating.
(2) Temperature limitation - For example, in a conventional drying reaction, the part of a sample nearest to a heating element heats first. The heat needs time to travel through a sample without getting any part of the sample too hot.  If part of the sample gets too hot an undesirable reaction like combustion may occur.


text:
Dipole moment of CH2Cl2 and CHCl3
Why does $\ce{CH2Cl2}$ have a higher dipole moment than $\ce{CHCl3}$?
My argument being that the $\ce{CCl3-}$ ion being rather stable, can generate a good amount of polarity in the bond, moreover, shouldnt the dipoles of both $\ce{Cl}$ atoms cancel out?

The dipoles of both Cl atoms aren't pointing in the exactly opposite directions, hence they cancel out only partially.

In that case, wouldnt the addition of one more such dipole add to the net resultant vector @Ivan Neretin

The said dipole is not quite aligned with the sum of the first two, so the resulting change could be either way. As it happens, it is actually in favor of CH2Cl2. If you are familiar with vectors and 3D geometry, you can check it yourself.

Don't use mhchem in question titles.

As the dipolar moment is a vectorial property, you have to take into account both the number of chlorine atoms around the carbon atom, as well as their relative arrangement -- among them, as well in respect to the carbon atom.
Lacking access to experimentally determined data about the molecular geometries,you may sketch the molecules in question, for example with Avogadro, and optimize the geometry (here, MMFF94 was deployed). You are literally able "to walk around" the molecule and its dipolar moment.

You easily will recognize a feature in common -- the similar direction of the resulting dipolar moment.  Simultaneously, you will recognize the different angle defined by Cl-C-Cl, too.  However, beside the angle, distances like C-Cl and C-H are affected by the different steric demand and electronic properties, too; equally influencing the overall dipole moment.  Hence the molecule of dichloromethane is more polar than the one of chloroform.
Sidenote: a force field like MMFF94 is a computationally cheap approach to optimize molecular structures.  There are better methods, but more costly; for the level of the question addressed, however, I thought it already provides the insight necessary here.

In $\ce{CHCl3}$ the dipole moment of the $\ce{C-Cl}$ bond is towards $\ce{Cl}$. Since it has a tetrahedral geometry and the dipole moment is a vector quantity, the vector sum of all dipole moments would try to cancel out. As they are in the outward direction, they will cancel to some extent.
While in $\ce{CH2Cl2}$, the $\ce{C-H}$ bond has the dipole towards $\ce{C}$ and there are only two $\ce{C-Cl}$ bonds. So they wouldn't cancel out.
And talking about the $\ce{CCl3-}$ ion. The carbanion would form only in certain (extreme) conditions.

Sorry. My mistake

How exactly would 3 dipole vectors oriented like a tetrahedron (3 vectors with a common origin give a tetrahedron) cancel out each other

PS. 4 vectors with common origin give tetrahedron. And I didn't say they completely cancel out.they try to

Im sorry, but we need just 3, and point accepted, that they only try to, then why dont they result in a dipole moment larget than what would arise due to 2 vectors trying to cancel each other

I can't attach the image, but search on Google tetrahedral geometry. You would see that it requires 4.  Also the angle between any two bonds is approx 109° so if you would try to do vector sum of three vectors mutually aligned at 109° and two at 109°, the first case would give lesser resultant. And the increase in H also contributes to more dipole

Understood the answer to the question,P.S: i still beilieve you havent got me, although i understood that youre trying to keep a central point and project 4 vectors from it. Try it this way-Keep a central point an project 3 vectors from it, youll see a pyramid like figure, technically, that is a tetrahedron, even i was confused at first, but once you visualize what i said, you'll get it @Ava

Actually the thing about three vectors is that it is associated with trigonal planar. If you dont place them in the same plane yeah a pyramid would be formed.          So I hadn't been able to understand you in this aspect only right? @Supernova


text:
What is the ground state electron configuration of isoprene?
Also the electron configuration of isoprene at its singlet state and also that of the triplet state?

The ground state is a singlet, when excited by UV light a $\pi\pi^*$ singlet excited state can be produced. (Singlet has electron spin paired with one electron in $\pi$ HOMO orbital one in LUMO [or higher orbital for other excited states]). The singlet may then form a triplet or decay to the ground state by emission/ internal conversion/reaction as per normal. In some polyenes the lowest excited state may be symmetry forbidden from the ground state, you will have to look into the details for each molecule.


text:
how can I prepare 5-nitro-2-indanone?
I want to make 5-hydroxy-2-indanone, so I need make 5-nitro-2-indanone first.But I can find any documents for reference. please give me some suggestion, thanks

I would not limit my options looking for the nitro-substituted precursor while other strategies might be available too. Related: http://chemistry.stackexchange.com/questions/65040

Drawing on Klaus's comment and reference to the preparation of 2-indanone, I suggest starting with the commercially available 3,4-dimethyl anisole. Bromination under radical conditions (non-polar solvent, AIBN, arc lamp) with NBS should give the bis-bromomethyl compound. Reaction with the anion of dithiane with 2eq of strong base should give the dithiane of 5-methoxy-2-indanone. Deprotection with boron tribromide reveals the OH and dithiane removal by acid hydrolysis or Hg++ gives 5-hydroxy-2-indanone.

From personal experience, 3,4-dimethylanisole may be activated enough to under electrophilic bromination even under nonpolar/radical conditions.

it is nice for me to get some information about my question from your kind. but do you have any documents for me to refer to? because I need some documents in detail to start our work. please!! thanks  in advance ,friends!


text:
Why is the H-H Bond Dissociation Enthalpy so high?
"The H-H Bond Dissociation Enthalpy is the highest for a single bond between 2 atoms of any element."

The bond dissociation enthalpy is the energy needed to break one mole of the bond to give separated atoms - everything being in the gas state.
For covalent bonds, bond energies and bondlengths depend on many factors:

electron afinities,
sizes of atoms involved in the bond,
differences in their electronegativity,
and the overall structure of the molecule.

There is a general trend in that the shorter the bondlength, the higher the bond energy.
Hydrogen has a relatively small atom and short bondlength hence very strong strong attractions between the atoms which will consenquently increase its bond dissociation ethalpy.
Credits

Bond enthalpy (Bond energy)
Inorganic chemistry bond enthalpy of Hydrogen]2


text:
Why does rice become sticky after cooking?
Rice when not wet isn't sticky at all. But when you soak it in water for some time it gets sticky.
What's the chemical change that occurs in rice that makes it sticky?

For the most part, starch may be responsible for the stickiness. See here

You should make that an answer @Don_S, I was about to say that when I saw your comment.

Sorry, I was in a bit of a hurry. Go ahead @Stian, post an answer and you can use the reference in my comment, if you like (obviously, it's not actually mine...).

Starch is what makes rice sticky. Starch surfaces, when dry, are closely packed together and they do not stick to each other. If they get moist, especially if left to soak, or boiled, the starch will "unwind" as it absorbs water and become sticky. Depending on the source of starch and the type, you can have cleavage of starch to sugar and it will become even more sticky. You can read this pdf for more information
Funfact, you can make your own glue with flour and water as an example.

I do not take responsibility for any mess or any unwanted results if for instance you decide to glue the broken baby seat with home made glue. =)

Well, you mean starch is stored in granule and when it is soaked the granules open up and become sticky?

See also this for a review of the molecular aspects of starch-water interactions (especially section 5).


text:
Can decarboxylation of carboxylic acids be effected by bases other than soda lime?
What about other bases such as Group 1 or Group 2 hydroxides? Does it have to do something with the mechanism?
I don't understand the mechanism either:

What is NaOH supposed to with the acetate ion? I thought the acetate ion would be produced when NaOH reacted with acetic acid after which the reaction would proceed as follows:

Decarboxylation can be effected by pretty much any base strong enough to produce the anion plus a high enough temperature. I've used DBU to effect an aromatic decarboxylation.
The role of the $\ce{NaOH}$ is this reaction is to capture the $\ce{CO2}$ as sodium carbonate.

So does NaOH act as a protonic acid over here (by donating H+ to the alkyle anion) and then capture the CO2 as Na2CO3? If yes, then isn't that quite odd that NaOH donates a proton?

Draw out the whole equation. 2NaOH + CO2 gives Na2CO3 + H2O which provides your proton source

Yes, I understand.


text:
Difference between Sodium Bicarbonate and Sodium Carbonate reacting with HCl
In lab, we performed the Solvay process where we had to produce sodium carbonate as the final product. There is a certain step where we put something in the oven and after this, we expect that the product is sodium bicarbonate ($\ce{NaHCO3}$). This product does not get named in the instructions; it is simply referred to as "dry product". After this step, we were asked to add 2 drops of $\ce{HCl}$ to a small amount of the dry product in a watch glass. Finally, we should repeat the watch glass step using $\ce{NaHCO3}$ and $\ce{Na2CO3}$.
After doing these 3 experiments, I observed the same thing (bubbles). I think that the reactions were:
$$\ce{NaHCO3 + HCl -> NaCl +  CO2 + H2O}$$
and for the third one (since I added $\ce{HCl}$ drop-wise to $\ce{Na2CO3}$, $\ce{Na2CO3}$ was in excess):
$$\ce{Na2CO3 + HCl -> NaHCO3 + NaCl}$$
and then:
$$\ce{Na2CO3 + HCl -> NaCl + CO2 + H2O}$$
I believe the purpose of the 3 experiments was to identify that the dry product was $\ce{NaHCO3}$, but after adding $\ce{HCl}$, I observed exactly the same thing (bubbles) for both known reagents. So I'm confused as to what should be my answer.

Or the reactions should be:  Na2CO3(s) -> 2Na+(aq) + CO3-2(aq)    ,
(2) CO3-2(aq) + H+ <--> HCO3-(aq)                                 ,
(3) HCO3-(aq) + H+ <--> H2CO3(l) ---> H2O(l) + CO2(g) ?

Your 3rd equation is not balanced. You will need 2 moles of hydrochloric acid to get the same products and that's how you can separate sodium carbonate and sodium bicarbonate. NaHCO3 needs one mole of HCl whereas Na2CO3 needs two moles. Also the amount of NaCl produced in Na2CO3 is more than that of NaHCO3.

It seems to me that the purpose of the 3 experiments with the HCl was not to identify the dry product. You can just write the reactions you observed, that was probably the point.


text:
Is vapour pressure of a liquid solution constant at a given temperature, no matter the size of closed container and amount of liquid taken?
Does vapour pressure of a liquid solution depend on the size of closed container, amount of solution taken, given that, temperature is kept constant? Or will it change, if some extra gas is added to the container at constant temperature?

In a fixed container assuming that the liquid and vapour are in equilibrium then vapour pressure is determined by the temperature. It does not matter what the size or shape of the container is. Adding inert gas only increased the total pressure.

The equilibrium vapor pressure of a compound at a given temperature is an intrinsic property of the compound, meaning it does not depend on variables like volume of the container, quantity of the compound, etc.  Other examples of intrinsic properties include density and melting point. None of these intrinsic properties depends on the quantity of the material or it's physical surroundings.

The simple answer is that in a thermodynamically closed system composed of a condensed phase (either solid or liquid) and a vapor phase of a chemical compound (or element), then only temperature matters - at equilibrium. This means vapor pressure may not be the same as partial pressure - even at equilibrium - even if the temperature is the same. That is; despite Dalton's Law of Partial Pressures, the partial pressure of a compound does depend on the other components present in the gas phase. This effect is usually so slight, when dealing with pressures around STP, that it can be ignored for a good approximation and Dalton's Law can be assumed to hold. You ask if the amount of material matters. Well, only if it isn't enough to provide for any condensed phase to exist after equilibrium has been established. That is, yes a drop of water in a volume of a cubic meter at 150°C will only exist in the vapor phase, and it's pressure will not correspond to the vapor pressure of steam at 150°C. The partial pressure of a compound depends on temperature, gravitational gradient, surface curvature, and the interactions of the compound with the other materials present; but it is a good approximation to assume it depends only on temperature (assuming enough of the compound is present to have some of it in a condensed phase). There's also another complication, which is surface absorption (onto the walls of the container), but again, this can generally be ignored.

Vapor pressure is the pressure exerted by the vapors when in dynamic equilibrium with its condensed phase at a given temperature.
It is the maximum value of partial pressure at a given temperature. So if pressure is increased after this value then some vapors will condense and if pressure is decreased then more liquid will vaporise until the vapor pressure is reached.
Vapor pressure doesn't depend on the amount of the liquid, volume of vessel, shape/size of container, area of the surface, etc.
It only depends on the nature of the liquid and the temperature.
Let's compare vapor pressure of ethanol and ethylene glycol. Since intermolecular forces of attraction are less in ethanol as compared to glycol (More H-bonding in glycol), molecules of ethanol will find it easier to escape into gas phase and hence vapor pressure of ethanol is more than that of glycol. In short, vapor pressure is inversely proportional to forces of attraction.
Now let's see it's relation with temperature. As the temperature increases kinetic energy of the molecules increases and therefore vapor pressure increases. The relation is clearly described by the claussius clapeyron equation:
ln(P°) = -∆H/RT + c
Where P° is the vapor pressure
∆H is the enthalpy of vaporisation
R is universal gas constant
c is the constant of integration (check out the derivation)
d(lnP°)/dT = ∆H/RT²
By variable separable method of solving differential equations we obtain the above solution.
Using claussius clapeyron equation vapor pressure can be easily compared at two different temperatures.
Changing the composition will also change the vapor pressure. It depends on the nature of the solute. That can be given by Raoult's law for ideal solutions and the deviations in case of non ideal ones.
Hope this helps!

YES, the vapor pressure will change if gas is added to a closed system.
Vapor Pressure is Temperature and Pressure dependent. So adding gas to a closed system increases the pressure, in turn changing vapor pressure.


text:
Is the reaction of hydronium and thisulfate an acid-base reaction?
Is this reaction
$$\ce{2H3O^{+}(aq) + S_{2}O_{3}^{2−}{(aq)} -> 3H_{2}O(l) + SO_{2}(g)  + 1/8S8(s)}
$$
an acid-base reaction?

It starts with the protonation of the anion, so that part is acid-base, but then it continues as a redox reaction.


text:
Titration of caffeine using Iodine
I was trying to do an investigation on the variation of caffeine content with temperature. I used 6 Lipton yellow label tea bags (black tea) to brew the tea solution.
Next, I perform the extraction of caffeine using standard procedures (separate by dichloromethane and evaporate). I was able to extract ~80mg of caffeine in 70-degree celsius water. I redissolve the powder into water to a 100 mL volumetric flask. I then pipette 25mL of the caffeine and added excess $\ce{I2}$ and $\ce{H2SO4}$. The chemical equation is given by:
$$\ce{C8H10N4O2 + 2 I2 + KI + H2SO4 → C8H10N4O2 .HI .I4 + KHSO4~~~(1)}$$
$$\ce{I2 + 2 Na2S2O3 → 2NaI + Na2S4O6 ~~~(2)}$$
Yet, I was unable to get any precipitation. I carried on with the titration and found that some Iodine did react in the (1). In fact, I did get a value of caffeine close to my theoretic result. So why didn't I get any precipitation?

This seems to be based on a protocol mentioned in https://chemistry.stackexchange.com/questions/62208/how-to-quantify-caffeine-content-in-coffee/85907#85907


text:
How do native sulfur crystals form?
I was visiting an elements exhibition in the Ulster Museum (Northern Ireland) and there were these native sulfur crystals on display.

My question is how do sulfur crystals like these form in nature? Does the sulfur precipitate out of a solution? Are they the product of a chemical reaction?
I am confused because of sulfur's negligible solubility in water and I couldn't imagine any other way such perfect crystals could form.
I am really interested in making such crystals myself but as I have said, I don't know by what process or under what conditions this crystallization of sulfur occurs.

Nature makes huge crystals out of many things; there is no guarantee that you can reproduce those even in a well-equipped lab. Negligible solubility over a million years may pretty well yield something worthwhile. Then again, maybe it was formed in an entirely different way. Slow cooling of molten sulfur, for example - why not? And by slow, I mean really, really slow. No, I mean even slower than that.

Hint: by slow temperature lowering of molten sulphur, near volcanos, for example (https://en.wikipedia.org/wiki/Sulfur#Natural_occurrence).

Crystals like the lower specimen, with large yellow crystals, couldn't come from cooling molten sulfur. Rather they would be made from an aqueous solution where there was a very low concentration of elemental sulfur depositing out.

I used to dissolve sulfur in carbondisulfide and acetone, and some pretty cool little crystals would form if a tube of it dried out. By "pretty cool" I don't mean anything like what's in that photo though.

If you want to grow any sort of crystals an Alum is good to thing try as they easily form very large crystals.

Tl;DR: Wikipedia mentions some of the ways in which elemental sulfur is produced naturally:

Through volcanic emissions, including emissions from hydrothermal vents. Some of them found near hot springs and near volcanic deposits. Also, lakes of molten sulfur up to ~200 m in diameter have been found on the sea floor due to submarine volcanoes.
As minerals: sulfide minerals, such as pyrite (iron sulfide), cinnabar (mercury sulfide), galena (lead sulfide), sphalerite (zinc sulfide), and stibnite (antimony sulfide); and the sulfate minerals, such as gypsum (calcium sulfate), alunite (potassium aluminum sulfate), and barite (barium sulfate).
Action of anaerobic bacteria on sulfate/sulfide minerals.

Long answer:
I found a paper1 which discussed #3 point in details. It said that although sulfide minerals can be oxidized to zero-valent sulfur by sulfate reducing bacteria in presence of molecular oxygen, it may pose the following problems:

exposure to oxygen would drastically decrease growth of microbes thereby slowing down sulfide production
on geologic timescales, excess supply with oxygen would convert
sulfide into sulfate rather than native sulfur
enormous amounts of oxygenated water would be needed to produce sulfur in large amounts

So, in order to overcome these problems, scientists proposed four possible mechanisms as a means to form native sulfur:

a modified sulfate reduction process that produces sulfur compounds
with an intermediate oxidation state,
coupling of sulfide oxidation to methanogenesis that utilizes
methylated compounds, acetate or carbon dioxide
ammonium oxidation coupled to sulfate reduction
sulfur comproportionation of sulfate and sulfide.

These reactions are found to be thermodynamically favorable and especially useful in environments containing dissolved sulfide. This provide evidence that microbial species functioning in such environments can produce native sulfur in absence of light and external oxidants such as $\ce{O2}$, nitrate, and metal oxides.
Reference

Labrado, A. L., Brunner, B., Bernasconi, S. M., & Peckmann, J. (2019). Formation of Large Native Sulfur Deposits Does Not Require Molecular Oxygen. Frontiers in microbiology, 10, 24. https://doi.org/10.3389/fmicb.2019.00024

Related: Which allotropes of sulfur exist naturally?