Does standard electrode potential depend upon temperature? Does standard reduction potential depend on temperature?

I am confused because my institute gives the definition for standard reduction potential as the potential that is measured at standard state which defines as measured in atmospheric pressure at $\pu{298 K}$, and when the concentration of all the substances is unity. But as far as my understanding goes standard states can be defined at any temperature and not just $\pu{298 K}$.

Also, as its directly related to $\Delta G^\circ = -nFE_\mathrm{cell} = \Delta H^\circ - T\Delta S^\circ$, this points toward it being dependent on temperature. Also, when someone says standard state, does that put restriction on temperature being $\pu{298 K}$??

Yes, the standard reduction potential does depend on temperature. The definition of the standard reduction potential is stated in Ref.1 as:

A standard electrode potential $E^\circ$ is defines as the potential (in Volts, $\pu{V}$) of a half-reaction relative to a reference electrode at a specific temperature, all chemicals being at their standard states at unit activity. These states may be arbitrarily defined as pure crystalline solids, pure liquids, ideal gases at one atmosphere fugacity ($\pu{101325 \times 10^{5} Pa}$), and ideal solutes at unit molality. The most common temperature for the tabulation of standard electrode potentials, as for their thermodynamic data, is $\pu{25 ^\circ C}$ ($\pu{298.15 K}$).

Accordingly, tabulated data is chosen to be at $\pu{298.15 K}$ for the convenience. This is also beneficial that the most common reference electrode is the Standard Hydrogen Electrode (SHE) for the solvent water:
$$\ce{2H+_{(aq)} + 2e- <=> H2_{(g)}} \tag1$$
$E^\circ_{(\ce{H+/H2})}$ for the above half-reaction is been assigned as zero volts at all temperatures. Extended work over $\pu{298.15 K}$ by De Béthune and coworkers shown that the temperature dependence of $E^\circ$ is approximately linear between $\pu{273.15 K}$ and $\pu{373.15 K}$ according to the following equation (Ref.2-Ref.4):
$$E_T = E^\circ_{298} + (T - 298.15)\left(\frac{dE^\circ}{dT}\right)_{298} \tag2$$
References:

Steven G. Bratsch, “Standard Electrode Potentials and Temperature Coefficients in Water at $\pu{298.15 K}$,” Journal of Physical and Chemical Reference Data 1989, 18(1), 1-21 (https://doi.org/10.1063/1.555839).
A. J. De Béthune, T. S. Licht, N. Swendeman, “The Temperature Coefficients of Electrode Potentials: The Isothermal and Thermal Coefficients—The Standard Ionic Entropy of Electrochemical Transport of the Hydrogen Ion,” Journal of The Electrochemical Society 1959, 106(7), 616 (doi:10.1149/1.2427448).
G. R. Salvi, A. J. De Béthune, “The Temperature Coefficients of Electrode Potentials: II. The Second Isothermal Temperature Coefficient,” Journal of The Electrochemical Society 1961, 108(7), 672 (doi: 10.1149/1.2428187).
André Jacques De Béthune, Nancy A. Swendeman Loud, In Standard Aqueous Electrode Potentials and Temperature Coefficients at $\pu{25 ^\circ C}$; C. A. Hampel: Skokie, ILL, 1964.


Types of concentration units I would like to ask about a concentration unit.
What does it exactly mean when stock concentration of something is,
for example, 5X?

Thank you in advance!

The “X” factor simply indicates that the solution is in a concentrated form that must
usually be diluted to a “1X” concentration for use. For example, a 5X concentrated solution must
be diluted 5-fold, while a 100X concentrated solution must be diluted 100-fold. The dilutions
are usually done using water.

See more  here under part 3:

http://csmbio.csm.jmu.edu/biology/courses/bio480_580/mblab/SolutionPrepReview.html


Acetophenone reacted with LDA and diethyl carbonate. Why do I need HCl? I'm working on a practice problem that asks for the product of acetophenone and LDA reacted with diethyl carbonate. After this, it says step 2 is HCl. After I react the first step, I get the correct answer to the problem - ethyl benzoylacetate. I'm asking why the problem says that HCl is added even when everything is neutral in the product I got and it's correct.

If you draw what you have carefully, without omitting any charges or adding magically appearing protons, notice that the actual product you initially get is either the ethanolate addition product of ethyl benzoylacetate, a mixture of ethyl benzoylacetate and lithium ethanolate or the lithium enolate anion of ethyl benzoylacetate and ethanol. In any case, you still have an anionic product.

For isolation and purification, you want to get rid of that negative charge (initially introduced by the strongest base you added: LDA). This is called quenching and typically performed with weak acids. I personally would use ammonium chloride for this type of reaction, but carefully adding dilute $\ce{HCl}$ will do the trick just as well. You should use $\ce{HCl}$’s proton to remove any remaining negative charges and eliminate $\ce{LiCl}$.


Can I use the hypotenuse of the pythagorean theorem to compute the work? A friend of mine told me that she used the hypotenuse of the pythagorean theorem to compute the work in graph below and the to compute the coefficient of efficiecy in this thermodynamics triangle cycle. What do you tnink about?

The work would be area of triangle, which is no way related to the hypotenuse (other than $\small H^2=P_0^2+V_0^2$)$$W=\frac12P_0V_0$$
Heat is :
$$Q_{in:A-C}=(C_V+R/2)\left(3\frac{P_0V_0}{R}\right)+C_V\left(2\frac{P_0V_0}{R}\right)$$
Efficiency is :
$$\eta=\frac{W}{Q_1}=\frac{\frac12P_0V_0}{(C_V+R/2)\left(3\frac{P_0V_0}{nR}\right)+C_V\left(2\frac{P_0V_0}{R}\right)}=\frac12.\frac{R}{3C_V+3R/2+2C_V}$$
$$\eta=\frac{R}{10C_V+3R}=\frac{R}{5fR+3R}=\frac1{5f+3}$$
which for  a monoatomic,diatomic and polyatomic(linear/non-linear) are:
$$\begin{array}{|c|c|}\text{Monoatomic}&1/18\\
\text{Diatomic}&1/28\\
\text{Polyatomic-Linear}&1/28\\
\text{Polyatomic-Non-Linear}&1/33\\
\end{array}$$


1H-NMR spectrum for furfural I was supposed to identify this compound by knowing its molecular formula and the following spectra:

$1\mathrm{H}\ (\mathrm{s},\ 9.67\,\mathrm{ppm})\\
1\mathrm{H}\ (\mathrm{quartet},\ 7.73\,\mathrm{ppm})\\
1\mathrm{H}\ (\mathrm{quartet},\ 7.30\,\mathrm{ppm})\\
1\mathrm{H}\ (\mathrm{quartet},\ 6.63\,\mathrm{ppm})$

My question is: how come we get quartets for the protons on the furan ring? I can see it getting doublets and triplets, but it seems impossible for them to couple with the aldehyde proton (which, in itself, doesn't couple with any of them).

They are not coupling with the aldehyde proton. Because the three protons around the ring are locked in such close proximity to one another, they will all couple each other. Let us consider these protons:

$\hspace{6cm}$

As by the $(n+1)$ rule of multiplicity, a proton will couple with each nearby, chemically unique proton and be split into $(n+1)$ peaks. These split signals can then also be split into $(n+1)$ more peaks by other nearby, chemically unique proton. Since all three protons are in the same region of space around the ring, they all couple each other, splitting each proton's signal into $(1+1)(1+1)=4$ peaks, making each signal a doublet of doublets (a quartet).

If you'd like a detailed analysis of the NMR of furfanal, you can read about it here.


The Boiling or Evaporation of Diethyl Ether When we heat water, we see bubbles appear as the boiling point approaches. The water then "boils" vigorously as it is converted from the liquid phase to a gas.

I realised recently that during an ether extraction of a product, a lot of ether had evaporated, as was evident from the mass balance. The ether vapours were visible, but no "boiling", or fluid motion that we tend to associate with boiling, was observed. The boiling point of ether is $34.6\ \mathrm{^\circ C}$. The ambient temperature then was around $35\ \mathrm{^\circ C}$.

Question: Why do we not see ether "boiling" near its boiling point?

Let me also put forth my thought process: The most apparent difference between the two cases of boiling water and boiling/evaporating ether is the location of the heat source. The ether was kept in a beaker on a stable platform, so it could exchange heat through the upper surface and the walls of the beaker. Heating water from below might create pockets of 'hotter' and hence less dense water that might lead to the "boiling". In the case of ether, the only possible explanation I found believable was that the walls of the beaker were really bad conductors of heat, as we would have seen some fluid flow in the ether case as well.

We also see puddles of water evaporating off. In that case, the heat exchange is possible only from above. Considering no disturbances, there is no visible fluid motion in the puddle. I felt this was because the rate of evaporation was too low. In the case of ether, it is already at its boiling point. This leads me to believe that in a restricted environment as a beaker, when the ether changes phase, the vapour accumulates near the surface as it is denser than air (almost 2.5 times as much). This further slows down the transfer of heat. Taking this into account, I find it very hard to believe that about 5 g of ether was lost in a mere span of a bit less than 10 minutes.

Is my reasoning right, or is it flawed? Please do tell.

Notes & References:

Diethyl Ether - Wikipedia
weather.com and personal experience
Modified Ideal Gas Law: $\rho=\frac{pM}{RT}$, i.e. $\rho\propto M$, taking $M_\text{air}\approx 28.84\ \mathrm{\frac{g}{mol}}$

The issue here, as eluded to in your discussion, is that near the boiling point the rate of heat being removed from the ether (via rapid evaporation) was greater than the rate of heat being transferred into the ether, so that the temperature of the ether never reached it's boiling point. In fact, the ether was likely several degrees below it's boiling point due to the evaporative cooling.

The rate of heat transfer into the solution via conduction from the surrounding air is a function of the difference in temperature between the ether and the air, which in this case was very small. If the ether had been placed in an oven of a sufficient temperature, the ether would have boiled.

The mass loss that was observed is not at all unexpected, as the temperature of the ether was just below it's boiling point. Imagine how quickly water would evaporate from a beaker if the surrounding air temperature was $\pu{99^oC}$ (and assuming we're at sea level so the water doesn't boil). The vapor pressure of water, and ether, is only slightly less at a few degrees below it's boiling point than it is right at it's boiling point. (Of course the physical action of boiling also allows for more rapid evaporation, so long as sufficient heat is applied to keep up with the evaporative cooling).

Summary, TL;DR:
As the ether was close to it's boiling point, it was evaporating rapidly, resulting in rapid mass loss. The evaporative cooling near the boiling point happens at a rate to greater than the conductive heat transfer from the surrounding air, so that it never quite reaches it's boiling point, and actually cools to several degrees below it's boiling point.


Graphene in Lithium-ions batteries: why? I have recently been reading about the amazing discovery of a super-material: Graphene. This one carbon-layer chemical is not only one of the most resistant materials ever known, but also one of the best heat and electric conductors.
At some point, the article mentioned that the implementation of graphene in batteries - such as the lithium ion battery - would revolutionize the industry. I then found in a closet the dusty chemistry book we used in school, and I took a look at the electrochemistry section (I still remember I had enjoyed it back then).
Now that I understand - or rather believe to understand - again how a Lithium-ion battery works, I can't figure out how graphene should contribute to its improvement. Apparently, both electrodes are made of materials that can incorporate lithium ions into their crystal structure and release them easily. One would, therefore, typically employ a lithium metal oxide such as $\text{LiCoO}_2$ as cathode and graphite as anode. (By the way, what could you use as electrolyte?)
Anyhow, as I read in this paper, graphene is usually somehow incorporated to the cathode... Notwithstanding, I don't understand how this should improve the conventional lithum batteries. What are its key properties in this process?
Please consider that I am by no means a professional chemist, but rather chemistry enthusiast who has the knowledge of high-school. Thanks in advance.

Lithium has a standard redox potential equal to -$3.04$ V. Coupled with any compound at the cathode, it could hardly produce cells yielding more than $4$ Volts, because part of the available energy is needed and lost to extract the external electron of the Lithium atom. The performance of a cell built with a Lithium anode could be more important if the Lithium was already ionized. That's what is done with graphene. When introducing Lithium atom in a graphene structure, its electron is not so strongly attracted by the Lithium nucleus. Lithium inserted between graphene layers behaves as if it was already partly ionized. That is why Lithium in graphene produces an anode which can deliver about $5$ Volts if coupled with a convenient cathode made of Lithium and Cobalt(III) oxides. Graphene plus Lithium may be considered as a mixture of Lithium ions $\ce{Li+}$ and "nearly free" electrons delocalized in the graphene structure.


Interpretation of H as T goes to absolute zero Euler's identity for the enthalpy states that (for a single component and phase)

$ H = TS + \mu N$

What are the implications of this expression in the context of the third law of thermodynamics (consider a "a perfect crystalline substance")? My textbooks describe the behaviour of $\Delta S$, $S$ and $\Delta G - \Delta H$ but not of $\Delta H$ or $H$.

Is it safe to assume that as $C_p$ and $S$ go to zero, so too does $\Delta H$ for any process at 0 K?

Short answer: No. There's no reason to expect that this should be the case.

My standard reference for classical thermodynamics is Callen's Thermodynamics and an Introduction to Thermostatistics, 2nd ed. Ch. 11 is relevant here.

Planck's statement of the third law is, according to Callen, "The $T=0$ isotherm coincides with the $S=0$ adiabat." In addition to the theoretical ramifications of having an isotherm and an adiabat coincide, the Planck statement also assigns to this adiabat $S=0$. This is physically grounded in statistical mechanics.

Let $\Delta$ represent the change in a quantity of a reversible isothermal process. Then, at zero temperature:

$S = 0$, by the third law.
$\Delta S = 0,$ again by the third law.
$\Delta G-\Delta H = T\Delta S = 0,$ because $T \to 0$.
$$\frac{\text{d}\Delta H}{\text{d}T}-\frac{\text{d}\Delta G}{\text{d}T} = \lim_{T\to0}\frac{\Delta H-\Delta G}{T} = \lim_{T\to0}\Delta S = 0,$$ once again by the third law. (This shows why the Gibbs free energy and the enthalpy are essentially equivalent thermodynamic potentials at low temperatures!)

But there is no reason to expect $\Delta H = \Delta G = 0$. Regardless of temperature, some energy is still required to effect a change in the system.


Fluorescein synthesis What is the mechanism of the fluorescein synthesis performed by von Baeyer?

This reaction is considered a modified Friedel-crafts acylation. The reaction is formally a series of electrophilic aromatic substitutions, with hydroxyl leaving groups picking up protons (1), however the Lewis acid catalysis obviously plays a role in the mechanism. I think it increases the electrophilicity of the phthalic anhydride in the first step, and facilitates the elimination of the hydroxyl groups. For the mechanism without the involvement of the zinc(II) chloride, see (1). Here is a mechanism emphasizing the role the Lewis acid might play in catalyzing the reaction:

This reaction works in a similar way to the Nencki reaction (2)
In order to get the form of the molecule shown as the product of your reaction, the $\mathrm{pH}$ needs to be increased in order to open the lactone ring.(3) This is fluorescein-A, the form of the molecule that actually fluoresces.

References:

Jag Mohan, In Organic Analytical Chemistry: Theory and Practice; Alpha Science International Ltd.: Pangbourne, England, 2003, p. 497 (ISBN: 1-84265-119-6).
M. Nencki, N. Sieber, "Ueber die Verbindungen der ein‐ und zweibasischen Fettsäuren mit Phenolen," Journal für Praktische Chemie 1881, 23(1), 147-156 (https://doi.org/10.1002/prac.18810230111).
Clifton J. Stephenson, Ken D. Shimizu, "A fluorescent diastereoselective molecular sensor for 1,2-aminoalcohols based on the rhodamine B lactone–zwitterion equilibrium," Org. Biomol. Chem 2010, 8(5), 1027-1032 (https://doi.org/10.1039/B918823A).


Turn Alpha Helix into Beta Pleat Beta sheets have the peptide backbone parallel whilst the alpha helix has a curled backbone.
But the thing I can't get around my head is why would the same protein always fold into one of these shapes?
It looks to me like I could grab an alpha helix and just straighten it out to form a beta pleat. I can't see anything that wouldn't work - so if I could turn alpha into beta, then what ensures that these proteins will fold the same way each time they're made? I think there must be something that prevents me from turning alpha into beta but I'm not sure what.

In the alpha helix structure as well as the beta pleated structure, the atoms are arranged in such a way that the number of hydrogen bonds is maximum. H bonding is a stabilizing factor, and consequently, any molecule (or polymer) would like to maximize their number. This answers your question as to why the protein would fold into one of these shapes.
Coming to why alpha and beta structures are not easily interconvertible, it is once again because of hydrogen bonding. The H bonds between H and electronegative atoms like oxygen and nitrogen are difficult to break, and hence we cannot simply smash an alpha helix into a beta pleated structure, since in doing so, we would inevitably have to break some hydrogen bonds. Thus, whether a protein exists in alpha or beta form depends on how it was formed.


How to differentiate hydrous and anhydrous compound? For an experiment, my teacher gave me containers of hydrates and their corresponding anhydrous compounds, but I have a feeling that for one of the compounds she switched the hydrate and anhydrous label by accident. How can I check if this is true?

There is no general answer to this question.
Heating is probably the best approach. It is not the only one. For some compounds, heating the hydrated variety produces an emission of water vapor, and nothing else, like $\ce{CuSO4·5H2O}$ or $\ce{Na2CO3·10H2O}$. For others compounds, it may produce a total decomposition like $\ce{AlCl3·6H2O}$, which is transformed into $\ce{Al(OH)3 + HCl}$ at high temperature. Considering the corresponding anhydrous compounds, $\ce{CuSO4}$ is decomposed into $\ce{CuO + SO2 + O2}$ at $650$°C. $\ce{Na2CO3}$ resists decomposition up to $1200$°C. $\ce{AlCl3}$ sublimes at $177$°C. You see that there is no general rule.
You may also check the behaviour of your hydrated compound with substances that react violently with water like thionyl chloride $\ce{SOCl2}$. if there is a reaction, it is probably a hydrate. But it may also be an anhydrous compound that react with $\ce{SOCl2}$, a metal for example, or an organic acid.
If you want a better answer, please give us the names of the particular compounds you are interested in.


Enthalpy change and equilibrium I don't understand why enthalpy change won't change at an equilibrium?

The reason I ask this is because of this question:

At first I thought the answer is C, but I realised that the amount depends on where the position of equilibrium lies. The correct answer is D, but doesn't enthalpy change (delta H) also depend on where the position of equilibrium lies?

The answer should be D. Because of the principle of microscopic reversibility, the forward enthalpy is opposite to the reverse enthalpy. This also follows from the conservation of energy if you will.

In this StackExchange question, there is a nice picture given of a potential energy surface. There you can readily see that the forward reaction enthalpy is opposite to the backward reaction enthalpy.


Solubility of Gold in low concentrations When preparing ICP-MS calibration standards for trace metal (~100 ppt) analysis, the values for Gold do not remain stable for more than a day. The standards are prepared using commercially available 10 ppm solutions in Nitric Acid with traces of HF.

A known amount of the Gold standard is added to 2 mL of 70% ultra pure Nitric Acid, an Indium Internal Standard and diluted to a pre-determined volume with ultra pure deionized water. The samples being analyzed are in 49% hydrofluoric acid to which has been added 2 mL of 70% ultra pure Nitric Acid, an Indium Internal Standard and diluted with an equal volume of ultra pure deionized water.

Would preparing matrix-matched calibration standards solve this problem?

Whenever you prepare very dilute working standards, it's good practice to do a stability study to see how long you can store them. There are many factors that can influence the stability of these solutions. The choice of container can make a big difference. Dilute metal solutions should be stored in polyethylene or polypropylene containers, not glass. Another well-known problem is photoreduction. Gold solutions should be stored in opaque containers and be exposed to light as little as possible. Some solutions may also be sensitive to atmospheric exposure. (though I don't think it's an issue for gold)

I don't have experience with working with gold in hydrofluoric acid, but such a large change in such a short period of time in the absence of the above factors may indicate some actual reactivity with your matrix, for which matrix-matching might help, though you may be better off reformulating your solutions if that's the case. It may also be an option to just make fresh working standards very frequently or increase storage life through refrigeration.


Minimizing Catalyst Residue in Vegetable Oil Hydrogenation We are using a nickel catalyst in our hydrogenation process to help convert vegetable oil to vegetable wax. The issue is ensuring all additives are successfully removed once hydrogenation is complete.

There are numerous catalysts that can be used to assist this process.

The question is: what catalyst is effective for hydrogenation but also extremely efficient to filter back out of the finished wax? Or, is there a cleaner way to hydrogenate oil such that almost no residue remains?

Nickel is usually a good catalyst for this sort of hydrogenation as it's cheap and effective. As to residues, I'm assuming that alludes to residual catalyst; there are a few ways to mitigate against this. Initially, ensure you have minimal fines in the catalyst. The easiest way is to ask your supplier to ensure this. If it's not that sort of process and you just use off the shelf material, then settle your catalyst in water, give it a mix and remove supernatant after a few seconds. Fines will suspend for longer and can be removed in this way. Large particle size material will sink faster. By using the larger particles, you make it easier to remove the catalyst by filtration later. The definitive check is residue on ignition of the final samples. The downside to nickel in batch (as opposed to continuous or flow)  is that it's very dense and needs the right agitator type and speed to get it suspended effectively. However,  this is not normally an issue unless you have inappropriately sized the agitator and type e.g. anchor, to the vessel size and geometry. It may not be worth using precious metal (PM) catalysts for this sort of transformation, but they are easier to suspend when supported on carbon. The advantage of PM catalysts is lower loading.
In all references to catalysts above, they are of the heterogeneous type.


Colour of polytetrafluorethylene Why are teflon coated dishes black or dark grey in colour when teflon as such is white (and the underlying aluminum is not black either)?

Note: I was employed at DuPont in the 1990's, and my understanding of Teflon(R) cookware manufacturing is based on proximity to product research and development and not from hands-on experience with PTFE formulations.

Teflon(R) polytetrafluoroethene coated materials are manufactured in several steps.  The PTFE coating is specially formulated as a waterborne dispersion containing surfactants and pigments.
Versions of Teflon(R) coatings sold in the early 1990's used a primer/topcoat system to improve adhesion of the polymer film to the substrate.  These formulated coatings are applied to the cookware and the pan is fired at high temperatures to form the PTFE film.
Coatings can be applied to both grit-blasted and smooth substrates.
The color(s) observed on PTFE coated cookware results from interactions of the pigments, polymers and surfactants during the high-temperature curing process.


Colour of Nickel thiocyanate What is the colour of $\ce{C2N2NiS2}$ (Nickel thiocyanate)?

In most of the web sites it only molecular weight,formula and other specifications but its colour is missing every where.

After bit research i got finally one link where they mentioned the colour. The other name of(Synonym) of Nickel thiocyanate is "Nickel Thiocyanate Greenpowder" which suggest it is Green in colour.

Reference:ChemicalBook


How do I figure out the hybridization of a particular atom in a molecule? I'm learning how to apply the VSEPR theory to Lewis structures and in my homework, I'm being asked to provide the hybridization of the central atom in each Lewis structure I've drawn.
I've drawn out the Lewis structure for all the required compounds and figured out the arrangements of the electron regions, and figured out the shape of each molecule. I'm being asked to figure out the hybridization of the central atom of various molecules.
I found a sample question with all the answers filled out:
$\ce{NH3}$
It is $\mathrm{sp^3}$ hybridized.
Where does this come from? I understand how to figure out the standard orbitals for an atom, but I'm lost with hybridization.
My textbook uses $\ce{CH4}$ as an example.
Carbon has $\mathrm{2s^2 \,2p^2}$, but in this molecule, it has four $\mathrm{sp^3}$. I understand the purpose of four (there are four hydrogens), but where did the "3" in $\mathrm{sp^3}$ come from?
How would I figure out something more complicated like $\ce{H2CO}$?

If you can assign the total electron geometry (geometry of all electron domains, not just bonding domains) on the central atom using VSEPR, then you can always automatically assign hybridization. Hybridization was invented to make quantum mechanical bonding theories work better with known empirical geometries. If you know one, then you always know the other.

Linear - $\ce{sp}$ - the hybridization of one $\ce{s}$ and one $\ce{p}$ orbital produce two hybrid orbitals oriented $180^\circ$ apart.
Trigonal planar - $\ce{sp^2}$ - the hybridization of one $\ce{s}$ and two $\ce{p}$ orbitals produce three hybrid orbitals oriented $120^\circ$ from each other all in the same plane.
Tetrahedral - $\ce{sp^3}$ - the hybridization of one $\ce{s}$ and three $\ce{p}$ orbitals produce four hybrid orbitals oriented toward the points of a regular tetrahedron, $109.5^\circ$ apart.
Trigonal bipyramidal - $\ce{dsp^3}$ or $\ce{sp^3d}$ - the hybridization of one $\ce{s}$, three $\ce{p}$, and one $\ce{d}$ orbitals produce five hybrid orbitals oriented in this weird shape: three equatorial hybrid orbitals oriented $120^\circ$ from each other all in the same plane and two axial orbitals oriented $180^\circ$ apart, orthogonal to the equatorial orbitals.
Octahedral - $\ce{d^2sp^3}$ or $\ce{sp^3d^2}$ - the hybridization of one $\ce{s}$, three $\ce{p}$, and two $\ce{d}$ orbitals produce six hybrid orbitals oriented toward the points of a regular octahedron $90^\circ$ apart.

I assume you haven't learned any of the geometries above steric number 6 (since they are rare), but they each correspond to a specific hybridization also.

  $\ce{NH3}$

For $\ce{NH3}$, which category does it fit in above? Remember to count the lone pair as an electron domain for determining total electron geometry. Since the sample question says $\ce{NH3}$ is $\ce{sp^3}$, then $\ce{NH3}$ must be tetrahedral. Make sure you can figure out how $\ce{NH3}$ has tetrahedral electron geometry.

  For $\ce{H2CO}$

Start by drawing the Lewis structure. The least electronegative atom that is not a hydrogen goes in the center (unless you have been given structural arrangement).
Determine the number of electron domains on the central atom.
Determine the electron geometry using VSEPR. Correlate the geometry with the hybridization.
Practice until you can do this quickly.


Is cyclohexane‐1,3‐dicarboxylic acid a correct IUPAC name? I did the numbering by taking all the functional groups in principal chain (scheme A), but my teacher did it differently and proposed the name cyclohexane‐1,3‐dicarboxylic acid (scheme B):

The principal chain in my teacher's answer is shorter. Also, he did not put any $\ce{-COOH}$ in the principal chain which he could have done.
Where am I wrong?

The answer to this question is actually pretty interesting, I think.  @user55119 's comment that the principal chain is the cyclohexane ring is correct -- but it seems like you understand that that's what the teacher is saying, and you're asking, "why?"  The best short answer, too, is imho in your comments: @MaxW 's "naming organic compounds is just screwy" is the bottom-line answer, because there are so many crazy special cases to deal with that it sometimes seems like sorcery to come up with them.  But if one is willing to dig into it a little, 99% of them do actually have consistent reasoning behind them -- even if coming up with that reasoning can feel pretty byzantine right up to the point where you get it.  This can be particularly true for cyclic compounds, which can in some cases even have more than one valid IUPAC name.
In this particular case, the reason your answer is incorrect is because what you defined with your numbering isn't a chain.
Per IUPAC, a chain is a sequence of linked units that is bounded by precisely two "boundary units" -- either a branch point or an "end unit", which is itself defined by IUPAC as a unit* of the macromolecule that is bound to only one other unit.
Now, this clearly causes problems with cyclic compounds, so the IUPAC has some caveats included within the definition of a chain -- one being that a cyclic macromolecule with no end groups is also a chain.  So, a chain is either linear and has two boundary groups terminating it, or it's strictly cyclical and includes no end units.  That means a cycloalkane -- including the cyclohexane in your example -- is a valid chain.
Your numbering, however, contains one boundary group.  The unit at position 1 is a valid boundary group, as it is connected to only one other group.  The unit at position 7, however, is not a boundary group; it is connected to a total of two other units (#6 and #2), and worse, it's a non-boundary that you can't continue counting from either since you've already counted the other molecule it attaches to!
"Maybe it solves the problem if I start from the unit I called #3, and leave the hanging methyl group to be the last part of the chain instead of starting from it -- so the chain would start at current #3 and end at current #1, 3-4-5-6-7-2-1."  (You may want to be looking at your illustration while you read this; it'll probably make the counting I am referring to clearer than just imagining it.)  No, that also doesn't work; it obviously that just puts the problem at the other end -- it still has only one boundary unit, because while the unit at the end of that new "chain" attempt, #1, is only connected to one other, the first unit in the "chain" (#3) is connected to two units, not one, so is not a boundary unit.  Still stuck!
And of course, you can try counting from your #2 instead of your #1 or #3, but then you go all the way around and never have a way to include that methyl group hanging off of #2 -- you'd have to "step" onto #2 again, effectively counting it twice, to then "step" onto the tail CH.  (I wish I could make a video of this "pointing" at each unit in the count; it would come out much clearer, I think!)
So in effect, really the only way you have to count the components of the macromolecule and get a coherent chain out of it is to start on #2 and end on the connection between #7 and #2 -- meaning you end up with the most practical solution being exactly the caveat that IUPAC settled on in specifying that a cyclic alkene can itself also be counted as a chain.
From there, I imagine you can figure out the rest: the cyclohexane as a valid chain is also the longest chain,  you start numbering from the functional group, etc.  Hope that helped explain the teacher's answer!

Edit/addendum: Some in the comment have asked for authoritative sourcing of the above.  Good point, and thank you for bringing to my attention that I only stated what I used for sourcing but neither completely cited nor offered links to said sourcing.  First post here in the chemistry StackExchange; please forgive my error.
IUPAC's authoritative reference work for nomenclature is their so-called Gold Book -- the Compendium of Chemical Terminology.

IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific
Publications, Oxford (1997). Online version (2019-) created by S. J.
Chalk. ISBN 0-9678550-9-8. https://doi.org/10.1351/goldbook.

I cite various definitions from within the Gold Book, the following two being of particular importance:
IUPAC - chain (Note 2). https://goldbook.iupac.org/terms/view/C00946
IUPAC - end-group. https://goldbook.iupac.org/terms/view/E02092
To address one of the comments in particular: it should be noted that nowhere in the Gold Book is there an authoritative definition of the term "ring."  That would make claim that "according to IUPAC a ring cannot be a chain," already dubious based on the definition I cited above, even less likely to be the case -- they certainly aren't going to "prohibit" something they don't even define.
The complaint about the definition of "chain" referring to "polymers" as if it were a notation of exclusivity and implying that the Gold Book's definition somehow contradicts or does not apply to the Blue Book should be pretty handily refuted with a bit more reading in both the Blue and Gold books. The Blue Book states it doesn't deal with preferred naming of polymers, but does still define the systematic naming of organic structures whether polymeric or not.  (Blue Book P-11, p.3)  The Gold Book defines "polymer," very simply and with no caveats, as "a substance composed of macromolecules."  Since "chain" and "polymer" thus both use the term "macromolecules," one would naturally be led to read that definition as well, where along with the (hopefully) uncontroversial definition of being a high-molecular mass chain of repeated low-molecular mass components, one might stumble onto Note 2:

"If a part or the whole of the molecule has a high relative molecular mass and essentially comprises the multiple repetition of units derived, actually or conceptually, from molecules of low relative molecular mass, it may be described as either macromolecular or polymeric, or by polymer used adjectivally."

No clearer acknowledgement of the lack of some "hard line" apparently perceived by some between polymers and "regular" organic macromolecules should be necessary.
When it comes to systematic naming, especially, polymers aren't a "special class" of substance which are somehow privy to different definitions and procedures -- indeed, the Blue Book tells us that's precisely the opposite of the case.  So, for those strictly systematic purposes (which is the entire point of the question, and should be in most introductory academic settings because the point isn't to have students memorize some esoteric rules but instead to teach them something about the structure of molecules and their function within a larger compound) - the definition of "chain" given applies just fine. In fact, the definition of "polymer" is expressly defined in the Gold Book, again IUPAC's authoritative text, such that it could apply to any simple alkane with a side branch for the purposes of that text.  Rather than second-guess clearly written prose, I'd rather assume the working group knew what it was doing there by leaving it vague and allowing common usage to define terms that are inherently ambiguous.  (Also see: What is the difference between Alkanes and Polymers?)  They seem more aware that strict legalism doesn't serve their purposes than most of the practitioners of their work do.
And for those of you who skipped this addendum entirely: good for you.  As interesting as some of the contentions below are, it seems generally agreed by all that none actually change the correct answer to the original question asked anyway; whatever your take on the semantic issues around the Blue and Gold (and Purple!) books that took up half the comments, they're just that -- semantics.  One can think that an answer is based on a definition that is "irrelevant" due to its source, which I find an odd and not particularly useful criticism (but hey, that's just me); but thinking then that said definition is "incorrect" is false logic.  If someone who has never seen a math book in their life believes 2+2=4 because some hot dog vendor on a street corner told them so, it doesn't make 2+2 equal anything other than 4.  Fortunately for them, just because their source has "no relevance" doesn't mean that they can't have a grasp of basic math.


Tricky exam question The mark scheme says carbon dioxide is produced from carbonate + acid. I know hydrogen can't be produced because both the reactions are endothermic.
But where did the acid come from? Can someone give me an example of Solid B and D which produces a mixture and when added together it gives carbon dioxide

The obvious choice for the source of carbonate is sodium bicarbonate (UPDATE: sodium bicarbonate is not soluble enough per other answer, but sodium carbonate is a possibility), which has a positive enthalpy of solution even at high concentrations, so let's call that B.

The acid is a little trickier. If you look up the standard enthalpy of solution of most acids, you'll find negative values (meaning the temp would go up when they are added to water). However, these values refer to infinite dilution in neutral water. Once you've added a little bit and the pH has dropped, the reaction $\ce{H+ + HO- -> H2O}$, which has a large negative enthalpy, essentially stops happening and the enthalpy of solution becomes positive. This is easy to tell by looking at the maximum solubility as a function of temperature. If it increases, then the enthalpy change of solution at saturation is positive. So at the high concentrations in the problem (4 g per 30 mL), many acids that are solid at room temp will cause a temperature drop when dissolved. Citric acid is one example.

A solution of citric acid is certainly acidic enough that it will cause rapidly production of carbon dioxide gas when mixed with a solution of sodium carbonate.


Why does the exchange interaction in Hartree-Fock theory lower the total energy? In Hartree-Fock theory, the expectation value of the total energy can be written as

$$E = \langle\Psi| H |\Psi\rangle = \sum_{a} \langle a| h |a \rangle + \frac{1}{2}\sum_{ab} \big( [aa |bb] - [ab|ba] \big),$$

where the first summation contains the single electron kinetic energy and the interaction energy of the external field. The second summation contains a classical Coulombic term plus an exchange term. The exchange term is introduced by the antisymmetrization of the wavefunction (through a Slater determinant) and takes the form of

$$[ab|ba] = \int \mathrm { d } \mathbf { r } _ { 1 } \mathrm { d } \mathbf { r } _ { 2 } \psi _ { a } ^ { * } \left( \mathbf { r } _ { 1 } \right) \psi _ { b } \left( \mathbf { r } _ { 1 } \right) \frac { 1 } { r _ { 12 } } \psi _ { a } ^ { * } \left( \mathbf { r } _ { 2 } \right) \psi _ { b } \left( \mathbf { r } _ { 2 } \right).$$

This integral is always positive, and so the exchange contribution to the total energy is always negative. Can anyone explain why the exchange contribution to the total energy is negative? In Fermi-Dirac statistics, we can see that the effect of Pauli exclusion of indistinguishable Fermions is an effective outward pressure. Hartree-Fock theory seems to suggest that the exchange interaction lowers the energy of the system, and therefore stabilizes it. Shouldn't the effect of exchange, therefore, be an effective inward pressure?

Can anyone explain why the exchange contribution to the total energy is negative?

I find it misleading that exchange interaction is treated as something that changes total energy of the system. This lowering of energy is actually due to the Hartree-Fock scheme being in principle inexact, and is not really specific to indistinguishable particles.

Let me elaborate. Consider a hypothetical system of distinguishable particles with exactly the same masses, charges and spins (and all their other properties entering the Schrödinger's equation). Let these parameters be equal to those of an electron. Now, if you solve the Schrödinger's equation for this system in the electromagnetic field of an atomic nucleus, you'll get some (spinor-valued) wavefunctions $\Psi_n$, where $n$ enumerates the states. These wavefunctions won't in general be antisymmetric.

But the symmetry of the Hamiltonian, namely its invariance with respect to exchange of a pair of particles (since all their relevant properties are the same), tells us that there's a large degeneracy. If you linearly combine the degenerate states corresponding to the same energy so as to get symmetric $\Psi_k^{\mathrm s}$ and antisymmetric $\Psi_k^{\mathrm a}$ linear combinations, these will still be solutions of the Schrödinger's equation we started with.

If you throw away the symmetric solutions and only leave the antisymmetric ones, these will actually be the solutions for the case where our particles are indistinguishable. You see? We've just arrived at exchange interaction between indistinguishable particles, without any change in energy. What we got is just a possible increase in ground state energy due to losing a bunch of states (and thus renaming one of the formerly excited states to the ground one).

Now, why do we get lowering of states in inexact schemes when we introduce such antisymmetrization? That's directly related (at least for the antisymmetrized ground state) to the variational principle, which tells us that an approximation of a ground state will always have mean energy $E_{\text{approx}}$ (which is the Rayleigh quotient) obeying

$$E_{\text{approx}}\ge E_0,$$

where $E_0$ is the exact ground state energy.

Before introducing the Slater determinant, the trial wavefunction was smooth and generally nonzero at the loci of electron-electron collisions. But the exact (even non-symmetric) eigenfunction should have a local minimum (a cusp, see here for an example) or a (smooth) zero at such loci due to the Coulomb repulsion. After antisymmetrization, many of such loci of collisions will get zeros. This will make the approximation better, and the mean energy lower, closer to the exact eigenenergy.

If instead of Hartree-Fock method we used some other approximation that had begun with exact treatment of electron collisions, sacrificing something other, then such antisymmetrization might not lead to any improvement in estimation of eigenenergies (limit case being very precise numerical treatment yet disregarding Pauli principle, which is basically equivalent to an exact solution for distinguishable particles).


Does any element have nutritional (caloric) value? Is there any single element (e.g. possibly carbon) that can be eaten, which has a nutritional value such that it provides calories?

I found on Wikipedia (dietary element) that there are certain elements that our bodies require.  I suppose oxygen could count as such a nutritional element, although we breathe it instead of eat it.

Interesting question! You may get a better answer, but I believe the answer is no.

Calories are units of heat and represent energy available to the body. The form of fuels that can be (biochemically) "burned" by the body are hydrocarbons; i.e. molecules of carbon and water, like sugars. Carbon alone I do not believe can work.

The only other single element fuel is hydrogen, and again I do not think this is biologically available to the body. By the way, oxygen is required as the oxidizer in any combustion reaction, and we definitely need oxygen (from breathing) to convert the food we eat to energy. But oxygen is not a fuel: it is the oxidizer that is reduced during combustion.


Why can zwitterions be difficult detect by HPLC-MS? I understand that to separate molecules by HPLC (or SFC) you commonly adjust the pH of the mobile phase to get preferable charges (i.e. negative charges) on the molecules - so that they interact with the column and are therefore well separated and thus give a large distinct peak.

Im not completely sure I understand what happens to zwitterions like amino acids when you try to separate them. Is it correct to think that they randomly interact with the column and thus "spread out"resulting in a chromatogram with a lot of noise? Why is it that the connected MS also has problems detecting them (should it not continuously detect the molecule)?

Example of amino acids that could be difficult to detect:

Zwitterionic amino acids are routinely detected by HPLC and by MS.  The only challenge is to find an appropriate stationary phase.  The most common stationary phases for this type of separation would include octadecyl-derivatized silica (ODS, also known as C18) and a HILIC stationary phase.  Many column vendors make columns explicitly marketed for separation of amino acids.

Examples:

Sciex markets an LC-MS method using a C18 column.
Agilent markets an LC-MS method using a HILIC column.
A decently-cited paper from 2016 describes a 3-minute LC-MS method for amino acid separation using HILIC columns.

What needs to happen is that analytes need to reversibly equilibrate between the stationary and the mobile phase via a single "mode".  If there are multiple modes by which analytes can interact with the column, then peak shapes will be highly distorted and/or flattened into noise.  An example of this would be if the analyte can both interact hydrophobically with the C18 moiety of ODS and via Lewis acid / Lewis base interactions with bare silanols (Si~OH) groups on the surface of exposed, underivatized silica.  These days, column manufacturers go through great lengths to insure that very few if any exposed silanols remain exposed on derivatized silica resin used for analytical LC columns.  Zwitterions are really no different than other analytes in this regard.

At the MS, zwitterions will need to be transferred to the gas phase as charged ions for detection.  But this isn't really a problem either, at least for e.g. electrospray mass spectrometry.


Dealing with different units (psi) in ideal gas equation My assignment is as follows:

A sample of air is trapped in a container at $\pu{16.2 psi}$. If the initial temperature is increased from $\pu{10 ^\circ C}$ to $\pu{32 ^\circ C}$, what is the final pressure?

I am really confused as to what $\pu{psi}$ is and how am I supposed to convert it to figure out the final pressure.

In the given experiment, there is no indication that the volume of the container changes and/or a reaction takes place. Both the volume and the amount of substance can therefore assumed to be constant.
Let's assume that the enclosed air can be treated as an ideal gas.
Under these conditions, the pressure is proportional to the absolute temperature (in Kelvin):
\begin{align}
p &\propto T & \frac{p_1}{p_2} &= \frac{T_1}{T_2}
\end{align}
This relation is known as Amonton's Law.

I am really confused as to what psi is and how am I supposed to convert it to figure out the final pressure.

Using the equation above, you would only have to convert the temperatures to Kelvin and could give the final pressure as $\pu{psi}$ too.
If you insist on conversion to Pascal ($\pu{Pa}$) as your favourite SI unit for pressure, remember that a pressure is defined as a force acting on a surface (area):
$$\pu{1 Pa} = \pu{1 N//m^2} = \pu{1 kg//m s^2}$$
The $\pu{psi}$ is a still rather common unit and it stands for pounds per square inch.
This is double weird, since both the force and the area are expressed in non-SI units.
Note that the pound is not the German Pfund ($= \pu{0.500 kg}$), but the angloamerican pound with $\pu{1 lb} = \pu{0.453 kg}$.
The area part in $\pu{psi}$ is given in square inches with $\pu{1 in} = \pu{0.0254 m}$.
In summary, $\pu{1 psi} \approx \pu{6895 Pa}$.


Preferred Lewis structure for sulphate anion and XeO4 In a section reviewing Lewis structure, the textbook Atkins' Physical Chemistry, 10th Edition, by Atkins, Paula, and Keeler, gives the following illustration of hypervalence and octet expansion:

I have two questions regarding this illustration:

With regards to the structure of the $\ce{SO4^{2-}}$ ion, notice that the lowest energy structure (1b) has the double covalent bonds placed at angles that are closer than had the double bonds been placed at opposing sides of the sulfur atom. I'm confused as to why this is supposedly the lowest energy state, since one would expect the lowest energy state to be when the two double bonds are at farthest angles from each other (which would be when they are on opposing sides of the Sulphur atom in the diagram), since this would be the case when there is the most distance between the repulsive negative charges of the electrons in the two double covalent bonds? Based on my research, it seems that what I am referring to here would resemble a linear molecular geometry (although, I am obviously not saying that the molecule would actually have a linear molecular geometry), where the we can imagine the red atom in the following image as being the sulfur atom, and the two white atoms as being the oxygen atoms that have the double covalent bond:

My next question is with regards to the $\ce{XeO4}$ molecule. It seems to me that the arrangement of all of the free electron pairs on the oxygen atoms would be in a more energetically favourable state (the molecule would be in a lower energy state) if they were farther away (at greater angles) from the double covalent bonds? It seems to me that their current arrangement would make sense in a scenario where there were an additional 2 free electrons on the oxygens, as is the case for two of the oxygen atoms of structure 1b, but since these oxygen atoms only possess 6 electrons instead of 8 (as is the case for two of the oxygen atoms of $\ce{SO4^{2-}}$), then wouldn't it be more energetically favourable to have the electron pairs at greater angles from the double covalent bonds? Based on some research, it seems that the geometry that I am referring to is a trigonal planar molecule geometry, as shown here (although, I am just referring to the Lewis structure of the molecule -- not the 3-dimensional geometry). In the following image from the same Wikipedia article, imagine the red atom as being the oxygen, and one of the white atoms as being the double covalent bond, and the other two white atoms as being the lone pairs of electrons, so that you have a $60^\circ$ angle between them:

I would greatly appreciate it if people could please take the time to clarify this.

You are reasoning as if the ion $\ce{SO4^2-}$ is planar. It is not the case. It has a tetrahedral geometry. And what you see in Atkins' book is a projection of the tetrahedron on a horizontal plane. So the angles between the bonds are not 90°, but larger. It is the same for $\ce{XeO4}$. This reality will solve all your questions simultaneously.


What are the differences between the various household surfactants? In a typical household, the following surfactants are found, among others:

Soap / handwash
Shower gel
Shampoo
Dishwashing agent
Washing powder / liquid

From my understanding, the role of all these is to lower the surface tension between water and fatty acids, enabling one to "wash" things.

Are they all the same then, and is the only difference between them marketing? Can I wash my hair with dishwashing agent, or pour shower gel into my washing machine?

Can I wash my hair with dishwashing agent [...]!

If it's an agent for dishwashing machines: Please DON'T!

Cleaning in dishwashing machines is performed in alkaline medium.

The tabs usually contain (among others):

alkaline compounds, such as sodium carbonate and sodium hydroxide
bleaching agents, such as sodium perborate and sodium percarbonate
pentasodium triphosphate (to keep calcium and magnesium in solution)

The solution of these tabs is not meant to get in contact with skin and eyes.


With which 2 of these 5 chemicals would Chlorine not react? N, NaOH, O2, NaBr, Fe A question in a Russian EGE test asks to choose 2 chemicals out of 5 with which Chlorine does not react:

Nitrogen - I checked and it reacts
$\ce{KOH}$ -- reacts
$\ce{O2}$ -- reacts
$\ce{NaBr}$ -- reacts
$\ce{Fe}$ -- reacts

I'm at a loss. Which two should I pick then? Maybe Fe and Nitrogen, because they would be the least likely to react?

Here's the question:

$\ce{NaOH}$

$$\ce{2NaOH + Cl2 → NaCl + NaClO + H2O}$$

  Reaction between chlorine and sodium hydroxide to form sodium
  chloride, sodium hypochlorite and water. The reaction takes place in a
  cold dilute solution. (Chemiday 1)

$$\ce{3Cl2 + 6NaOH → NaClO3 + 5NaCl + 3H2O}$$

  Reaction interaction chlorine and sodium hydroxide to form sodium
  chlorate, sodium chloride and water. The reaction proceeds in a
  concentrated solution with heating.(chemiday 2)

$\ce{NaBr}$

$$\ce{2NaBr + Cl2 → 2NaCl + Br2}$$

  Reaction between sodium bromide and chlorine to form bromine chloride
  and sodium. Hot - sodium bromide. The industrial method of producing
  bromine.(chemiday 3)

$\ce{Fe}$

$$\ce{2Fe + 3Cl2 ->[\Delta] 2FeCl3}$$

  The reaction of chlorine and ferric chloride reaction with the
  formation of iron (III). The reaction proceeds at a temperature above
  250 C °.(chemiday 4)

But I couldn't find any reaction of chlorine with neither nitrogen or oxygen. Guess what, chlorine doesn't have sufficient energy to break the triple bond of nitrogen or double bond of oxygen.


How are van der Waals parameters determined experimentally? If the ideal gas law $PV=nRT$ is not enough, as a second approximation the van der Waals equation

$$\left(P+a\frac{n^2}{V}\right)\left(V-nb\right)=nRT$$

is applied. The constants $a$ and $b$ are van der Waals constants which take into account some of the attraction and repulsion  that the ideal gas law does not.

Solving for $P$,

$$P=\frac{nRT}{V-nb}-a\frac{n^2}{V^2}$$

and plugging it into the definition formula of compression factor $Z$ yields
$$Z=\frac{V_{m,real}}{(RT)/P}=\frac{V_{real}}{(nRT)/P}=\frac{V_{real}}{V-nb}-a\frac{nV_{real}}{RTV^2}.$$

This result, supposedly, helps in the measuring of constants $a$ and $b$. But how is it really done? What are the more rigorous definitions of the constants?

At the critical point, the first and second derivatives of pressure with respect to volume are zero.

This provides two equations to solve for the two unknowns (a and b) in terms of the critical temperature and critical volume (or pressure).

See Critical Constants of the van der Waals Gas


Projection Operator Method on Cl4+ I am not quite sure how to start on 1b.We are given 2 diagrams but I am not sure how to use the second one in the question, or if we have to. I got the point group for Cl4+ to be (D2h) however, but after that I do not know how to proceed.

To form a linear combination, a 'victim' vector is chosen and operated on with each symmetry operation in the point group, $C_2, \sigma(xz)$, etc. to find what vector this turns into. This vector's name (the basis function) is then multiplied with the character from the point group corresponding to that symmetry operation and symmetry species. Finally, the names are added together by moving from one symmetry operation to the next along the top row of the point group. We expect the result to be the sum or difference of the displacement vectors, e.g.  $2v_1 + v_2 - v_3 \cdots$ and so forth. The same formula or algorithm is used if the combinations of atomic orbitals that make up molecular orbitals are sought as for vibrational normal modes..

The equation to make a linear combination of vector names $L_M$ is

$$\displaystyle L_M=\frac{d}{h}\sum_{j=1\cdots h}c_jS_j(v)$$

where $M$ is the symmetry species label, $A_g, B_{3g}$, and so forth, $h$ is the order of the group, $d$ the dimension of the irreducible representation, and the sum is over all the classes. Because the resulting vector will be normalized $d/h$ can be ignored. One further point is important. In the $C_{3V}$ point group, for example, there is a heading $2C_3$ as there are two members in this class, and this must be split in the summation into $C_3^+$ and $C_3^-$. This is because moving a vector $120^ \text{o}$ to the right, say, will turn it into a different vector than turning $120^\text{o}$ to the left, similarly, any mirror planes must be separated out. The character for symmetry species $M$ is $c_j$ and $S_j(v)$ is the effect that operator $S$ in column $j$ of the point group has on our victim vector $v$.

The result of this operation is to produce a vector whose name is recorded. For example, if $S$ is the identity then the result is $E_1(v) = v$, other operations may leave $v$ unchanged or change it into another vector. Finally, note that if the reduced representation of the vibrations or molecular orbitals contains two or more symmetry species of the same type, e.g. $2A_1$, then two or more different victim vectors will have to be chosen to obtain all the linear combinations.

The first step is therefore to choose vectors to represent, say $\pi$ orbitals and operate on them to form the reduced representation and so find the only those symmetry species that can be present then use the formula above and record the names of the total result  produced for each symmetry species in turn, e.g.  $2\pi_1-\pi_2$ etc. Using the nodes produced you can guestimate the order of energy levels.

(You can find details of forming a reduced representation here Understanding group theory easily and quickly)


Inorganic chemistry for condensed matter physicists I'm a theoretical condensed matter physicist, on a field of magnetism research. I wonder if any of you could recommend a concise and well-written textbook on inorganic chemistry to learn concepts of the ligand field theory, coordination of transition metal complexes and their magnetic states, etc.

During my work, I just look up the periodic table and see the electronic configurations, but the oxidization of the transition metal complexes depends on ligands, and it's frustrating as I don't seem to understand a general principle.

I'm quite familiar with solid state physics, and a bit of quantum chemistry. It would be great if you could recommend me textbooks or materials on inorganic chemistry for "solid state physics" or "material science".

Any text book suitable for an advanced undergraduate course in inorganic chemistry should cover these topics. Which text may be best for this purpose will be a matter of opinion, so I am not going to recommend a specific one. Links to specific books/editions will risk being out of date for future readers of this question and answer. However, some author combinations have been active in writing text books in this field for many years. I would look for a recent edition of the book by either:

Meissler and Tarr (as of this writing, the most recent edition is the 5th edition by Meissler, Fischer and Tarr, published by Pearson, 2013)
Shriver and Atkins (as of this writing, the most recent edition is the 6th edition by Shriver, Weller, Overton, Rourke, and Armstrong, published by WH Freeman, 2014)

Just be careful since the phrase "inorganic chemistry" in some uses refers to the introductory course sequence in chemistry, what is more commonly called "general chemistry".


Is the amount of substance of acid and base always equal at the equivalence point? Does the amount of substance of acid always have to equal the amount of substance of base at the equivalence point? I know it "works" for monoprotic acids and monobasic bases, but what if we had a diprotic acid and monobasic base?

We speak in this case of multiple equivalence points.

As a concrete example, if we are titrating $\ce{H2SO4}$ with $\ce{NaOH}$, then the first equivalence point occurs when all the $\ce{H2SO4}$ has reacted to form $\ce{HSO4-}$, and the second equivalence point when all the $\ce{HSO4-}$ has further reacted to form $\ce{SO4^2-}$. If you started with 1 mole of acid in solution, then adding 1 mole of base gets you to the first equivalence point, and another mole (2 moles total) gets you to the second equivalence point. In the same vein, one might consider triprotic acids, which have three equivalence points, and so on.

Incidentally, an interesting case to think about might be that of a titration between a diprotic acid and a dibasic base.


Beer's Law: Are the graphs of concentration/time and absorbance/time supposed to look very similar? For my lab, I'm trying to figure out if the reaction between NaOH and crystal violet is a zeroth, first, or second reaction. My slope is 5.714*10^4 and y-intercept 0.008016. In logger pro, after finishing collecting my absorbance/time of NaOH and crystal violet, I tried getting the concentration of crystal violet by using that equation. When graphed, it looks exactly the same as my original graph, just on different numbers. I'm confused by this because when I've tried graphing ln(concentration) and 1/concentration, the graph returned no straight lines. I'm wondering if I'm doing something wrong.

Well, since Beer's Law says

$A = \epsilon⋅l⋅c$

And $A$ hence is directly proportional to $c$, as Beer's Law actually represents a line equation.


Why are plastics lipophilic? My wife prepares lunch for me every morning. After eating it on my desk, I typically wash the lunchbox and utensils in the company's kitchen.
I have noticed that it is very difficult to remove the oiliness from the plastic containers (think Tupperware, but also harder plastics that look like polycarbonate): I need to rub the detergent-soaked sponge extensively and rinse with very hot water, several times. This does not happen with glasses, porcelain dishes or metal utensils like forks.

In other words, all the plastic containers I have come in contact with seem to be quite lipophilic.

Is this because edible oils and plastic are chemically "similar" (i.e., both contain long-ish chains of hydrocarbons), or is there another reason that I am missing?

First, many plastics are somewhat porous. Since plastics are permeable to water vapor and oxygen, coffee and other food products are sometimes packed in two-layer bags of polyethylene (for strength) laminated to aluminum foil (to reduce permeability).

Second, lipids and other non-polar substances can dissolve in and and migrate through common plastics used in food containers, such as polyethylene and polypropylene, as you and @Alchimista note. This is why carotene, lycopene and other chemicals in tomatoes eventually stain poly containers permanently: they've dissolved into the plastic.

In a similar way, mercury dissolves into gold (e.g. jewelry) in a matter of seconds, leaving a silvery stain. That stain then disappears in a week or two as the mercury migrates throughout the object.

The reverse phenomena also occurs, where substances in plastic packaging can migrate out of the packaging.

As the Bard might have put it, "Will all the water in the ocean wash this food from my cans?"


What is the principle of “scrubbing” in liquid extraction or solvent extraction? There are many steps in solvent extraction:

Extraction
Scrubbing
Back extraction

I understand the 1st and the 3rd.
What is the principle of scrubbing?
Is it just washing the organic phase?
How undesirable species get removed from organic phase?

"Scrubbing" in this context is the treatment of the organic phase to remove impurity metals. This is as opposed to "back extraction" which is to remove the metal of interest from the organic phase.


Chemical function of Mg and Conjugated bonds in Chlorophyll This question has 3 parts, each related to the title.

In biochem, we were shown the chemical structure of chlorophyll.  The light-absorbing head has many carbon-rings with alternating single and double bonds (conjugation).  We were taught that this arrangement of bonds is what allows the molecule to absorb light energy.

Question 1: I understand that pi-bonds are generally better at absorbing photons than single bonds, but why do conjugated bonds increase this ability?

Question 2: Also, this molecule does not seem to be have any resonance structures needed for delocalized bonds and thus does not have the light absorbing capabilities like that graphene/graphite or benzene rings.  (If we could engineer an artificial chlorophyll molecule would this increase its efficiency?)

The head of the molecule also contains a Mg ions that is coordinate bonded to four surrounding nitrogen atoms.

Question 3:  If the alternating double and single bonds are responsible for the absorption of light, what role does this metal complex serve in the molecule?  Other pigments like carotenoid don't have this metal complex and they're still able to absorb light.

but why do conjugated bonds increase this ability

Adsorption of light results in excitation of an electron, moving from low-level (usually bonding or non-bonding) to a higher (usually antibonding) orbital. (for understanding of what bonding, non-bonding and antiboding orbitals are, see here)

Generally, when two orbitals of same energy partially overlap, two more are formed: one with slightly higher energy and one with slightly lower. When orbitals of conjugated bonds interact, energy of one bonding orbital goes up and of one antibonding orbital goes does, narrowing the gap. When the gap is withing or closely near ~1.6-3.2 eV it can participate in visible light adsorption.

  this molecule does not seem to be have any resonance structures

It does, see carefully.

  what role does this metal complex serve in the molecule

The relevant metal-free molecule do exist in nature. It has, however, different color. $Mg$ ion seems to adjust the gap mentioned above. For a more clear understanding of the matter you might look for data on phtalocianin pigments, that may be easily obtained with different central atom in a very similar structure.


Changes in which factors affect both the rate and the rate constant of a first order reaction? Changes in which factors affect both the rate and the rate
  constant of a first order reaction?

  I. temperature

  II. concentration

  (A) I only (B) II only
  (C) Both I and II (D) Neither I nor II

I would think that the answer would be (C), both temperature and concentration.

Concentration would obviously affect it (from the rate expression), and the temperature would also affect it because the Arrhenius equation for the rate constant is $$
k = A e^{-E_a/(R T)}
$$

and thus temperature would have an influence on the rate constant, and since the rate constant is part of the rate equation, it would also affect the rate of the reaction itself.

However, the given answer is

  A

I don't understand why this is.

If you see the rate equation for the Arrhenius equation:
$$
k = A e^{-E_a/(R T)}
$$
you can see that the rate constant increases for an increase in temperature (and as activation energy increases, the rate constant decreases.

Now, since the rate constant increases, this implies that the rate of reaction increases with temperature.

Regarding concentration: you are correct in that concentration affects the rate of reaction. However, a rate constant does not change according to concentration. You can observe this in the Arrhenius equation, where A is independent of the concentration of the substrate. The fact that the rate increases does not imply the rate constant changes.

Possibly more clearly: it is implied that increasing the rate constant will increase the rate of reaction. However, increasing the rate does not increase the rate constant.

An increase in temperature increases the rate constant and hence the rate.

An increase in concentration increases the rate but not the rate constant.


Water evaporation from calorimeter giving less exothermic enthalpy value? To find the enthalpy change of combustion for an alcohol, you use a calorimeter filled with water and heat it by burning an alcohol.

I was told that usually experimental values for the enthalpy change of combustion are less exothermic than data book values, because of many reasons (eg- heat loss, incomplete combustion etc.)
One of the reasons why the experimental value may be less exothermic is because water has evaporated from the calorimeter.

I don't understand how water evaporating from the calorimeter would make the enthalpy change less exothermic. If the water evaporates, then the mass of water left in the calorimeter is lower, so it would heat up faster, so wouldn't this give a more exothermic result?

First, we have some things to consider and they are:

the specific heat of water;
the latent heat of vaporization of water;
how the terms mentioned above are compared to each other.

The specific heat of water is $\pu{4200 J kg-1 K-1}$ and it means that it takes $\pu{4200 kJ}$ amount of heat to increase the temperature of $\pu{1 kg}$ water by $\pu{1 K}.$ On the other hand, the amount of heat needed to turn $\pu{1 kg}$ water at $\pu{373 K}$ into $\pu{1 kg}$ water vapor is $\pu{2268000 J kg-1},$ i.e. the latent heat of vaporization of water is $\pu{2268000 J kg-1}.$

Now, if we consider two pieces of data above, it's clearly visible that it takes much more heat to vaporize water than it takes to increase the temperature of the water (given the same mass). As it's this heat which absorbs heat from the reaction materials (mainly products), the high latent heat of vaporization makes the reaction much less exothermic than increasing the temperature of the water (since it requires more heat).

So, reducing the exothermic characteristic of the reaction depends more on the rate of vaporization than it does on the increase in temperature of the water. As a result of this, despite the fact that the amount of water available to absorb heat via increasing its temperature decreases as more and more water is vaporized, the amount of heat absorbed by vaporization is more than enough to compensate for this. Thus, the reaction becomes less exothermic than expected with the vaporization of water.


How does chlorine in water evaporate? Drinking water (e.g. from your home's tap) is spiked with chlorine, for an amount of just some ppm.
Purpose of this operation is to kill bacteria, for our health.

  Background: chlorine in water is potentially dangerous for plants, or better, their soil.

A common solution to get rid of chlorine in tap water is leave it in an open pot all-night-long. In this way chlorine evaporates.

Now, in my mind, chlorine in dissolved in water as chloride ion ($\ce{Cl-}$). Just like salt in water.
And just salt in water, if you leave a pot of that mix in a room, after some time water evaporates completely and salt remains on the bottom of the pot.

Question is, how can chlorine evaporate from water?

Chlorine and Free Chlorine

Chlorine is the term commonly used to describe the addition of chlorine gas, sodium hypochlorite (i.e., bleach), or calcium hypochlorite (i.e., bleach powder) to water to form free chlorine. Free Chlorine is the sum of hypochlorous acid ($\ce{HOCl}$) and hypochlorite ($\ce{OCl-}$); a monoprotic acid with a $\ce{pK_a} = 7.54$.

Chlorine gas ($\ce{Cl2}$) does not exist in water (except highly acidic) as it hydrolyzes to $\ce{HOCl}$. Chloride ($\ce{Cl-}$) forms from the redox reaction of $\ce{HOCl}$/$\ce{OCl-}$, but it is not reactive.

Both $\ce{HOCl}$ and $\ce{OCl-}$ are volatile, with $\ce{HOCl}$ being $\approx\!5\!\times$ greater (Henry's solubility constant being $H = 0.076$ and $0.013$, respectively).$^{[\text{citation needed}]}$ When you open a pitcher of water containing free chlorine to the atmosphere, the free chlorine comes into equilibrium with the atmosphere and thus it volatilizes. As such, you do not lose "$100\%$ or all" of the free chlorine, but enough is removed that the concentrations drops below your taste threshold.

Side Note

If the water contains dissolved organic carbon and organic nitrogen (e.g., natural organic matter, pathogens), then the free chlorine could be reacting while it is sitting and thus you could be losing free chlorine this way as well. Further you can also add combined chlorine (e.g., monochloramine) to water in place of free chlorine. This also forms from free chlorine reaction with nitrogen compounds in water (e.g., urine) and is the common "pool smell" observed in swimming pools that are not properly chlorinated.


How does NMO reform Osmium tetroxide after reaction with an alkene? After Osmium tetroxide adds to a double bond and is then hydrolysed the following species is formed:

Because Osmium tetroxide is toxic, NMO is used to transform the above back to Osmium tetroxide. NMO is:

What is the mechanism for this?

N-methyl-morpholine-N-oxide (NMO) is used to reoxidize and regenerate osmium tetroxide ($\ce{OsO4}$) in what is known as the "Upjohn Dihydroxylation" procedure.

Mechanistically, the redox occurs when a lone pair of electrons on the reduced $\ce{Os(VI)O3}$ are used to form a bond to the morpholine N-oxygen; the electrons from the former $\ce{N-O}$ bond are, in turn, transferred to the morpholine nitrogen.  The $\ce{OsO3}$ gains an oxygen and is oxidized back to osmium (VIII) tetroxide and the NMO is reduced to N-methylmorpholine

image source

Osmium tetroxide is both toxic and expensive. Using the Upjohn procedure allows the reaction to be run with a catalytic amount of osmium teroxide and 1 equivalent of NMO.


Does limonene react with calcium hydroxide or acetic acid? I can't find the results of too many experiments with Limonene. Which has many uses (e.g. making soap).

I'm curious of the results of combining Limonene and Calcium Hydroxide (Lime) in one experiment and another experiment combining Limonene and White Vinegar.

What is the result of these equations? Are the products potentially dangerous?

Could they result in combustion?

C10H16 + CH3COOH → ?

C10H16 + CaH2O2 → ?

I can't find the results of too many experiments with Limonene.

Actually, there are quite a lot and the feature article Limonene: a versatile chemical of the bioeconomy, Chem. Commun., 2014, 50, 15288-15296 by Rosaria Ciriminna, Monica Lomeli-Rodriguez, Piera Demma Carà, Jose A. Lopez-Sanchez and Mario Pagliaro provides an interesting insight:

d-limonene (1) is available in large quantities as a side product in the production of orange juice
it is widely used as a bio-based solvent to  replace petroleum-based solvents for various purposes, such as

general household cleaners
degreasing of metals
outdoor maintenance cleaners

it is used as an insecticide

In order to to prove comsumer-friendliness of d-limonene and its degradation products, a lot of studies on the reaction of d-limonene with oxygen, leading to endoperoxides, hydroperoxides, and the monoterpene ketone carvone (2), have been performed.

In the ongoing BioProMo project, the bacterial conversion of d-limonene to the carboxylic acid 3 is examined. The latter is a natural preservative.

  I'm curious of the results of combining Limonene and Calcium Hydroxide (Lime) in one experiment and another experiment combining Limonene and White Vinegar.

To my information, these combinations aren't used in industrial or academic research, but in another context.

As outlined above, d-limonene is widely used as a natural replacement for other organic solvents, which aren't easily available outside science and/or industry. As a consequence, d-limonene as a solvent has also found its way into recipes for the clandestine extraction of controlled substances from plant material.

Some of these recipes seem to promote the combination of d-limonene, calcium hydroxide, and white vinegar (as a source of acetic acid) to extract N,N-dimethyltryptamine (DMT) as its acetate from the bark of Mimosa tenuiflora.

While this seems technically possible and reactions between d-limonene, the base, and the acetic acid are not to be expected, the posession of the extraction product is apparently illegal in a variety of countries.


Does a calibration curve of absorbance vs concentration need to pass exactly through the origin? If you are making a calibration curve for $\ce{CuSO4}$ where you plot absorbance against concentration and the curve deviates slightly from the origin, does it make the calibration curve useless?

Good answers, but it is almost never a good idea to force the intercept to zero. Unless there is a clearly justifiable and compelling reason to do so, it is better to let the data set 'vote' for itself and not impose a preference/bias/mandate. Among other things, forcing the intercept to zero changes the standard error of the slope, which is an important statistic.

Below is a screenshot of the result of using Excel's Data Analysis tool on a set of linear calibration data I collected in the past:

As shown, the intercept is -0.0541 and its standard error is 0.0352, so the t test statistic of the intercept, defined as intercept/(standard error of the intercept) = -1.537. The negative sign is irrelevant for present purposes, but the P-value is 0.264.

The null hypothesis for the intercept is that the true intercept, which is unknown, is zero. The alternative hypothesis (aka research hypothesis) is that the true intercept is either 1) not equal to zero or 2) not equal to zero, but in a known direction. The former is the 2-tailed option and the latter is the 1-tailed option. For intercepts, the 2-tailed option is most common and this would almost certainly be the case for a Beer's law plot.

The P-value of 0.264 is much higher than the usual 0.05 value and this means that it would be quite risky to reject the null hypothesis based on the OLS analysis. In fact, the risk of mistakenly rejecting the null hypothesis, even though it was true, would be 26.4%. The confidence in the alternative hypothesis would be 100% - 26.4% = 73.6%, which is too little confidence for most people. Typically, you want P ≤ 0.05. Otherwise, it is too risky to 'jump ship' for the alternative hypothesis.

What does it mean? Well, the data resulted in a P-value of 26.4%, so the null hypothesis cannot prudently be rejected. But it is also not accepted: it still might be wrong, but the data did not show that. Replacing the intercept by zero would mean that a decision was made that the true intercept was zero, even though the analysis proved no such thing. It is not a good idea to pretend that the data showed what it did not show. Bottom line: the intercept should not be forced to zero unless there is a clear and compelling reasion to do so.


In acidic solution, what does ClO₃⁻ reduce into? In acidic solution, what does $\ce{ClO3-}$ reduce into and, most importantly, why? It's $$\ce{6e- +ClO_{3}^- +6H^+ -> Cl^- + 3 H_2O}.$$

Now, with the gift of foresight, why was the equation for its oxidation not instead $$\ce {2e^- +ClO_{3}^- +2H^+ \rightarrow Cl^- + O_2 + H_2O}?$$
Obviously it balances and charge is conserved. However, the latter equation would seem more likely: less charges have to be pulled apart (so less work has to be done to move these charges), and the bond energy of bonds in a water molecule is almost twice that in an oxygen molecule?

Where is there a flaw in my logic and does there exist a rough set of rules to determine how things will split up under oxidation or reduction?

The first equation is correct because $\ce{ClO3-}$ is reduced to $\ce{Cl-}$ and oxygen maintains its -II oxidation state in both educts and products. As chlorine changes its oxidation state from +V in chlorate to -I in chloride, $6 \ \mathrm{e^-}$ are required for this reduction.

In your second equation, chlorate is reduced to chloride as well; however, only $2 \ \mathrm{e^-}$ are explicitly supplied. The remaining electrons would have to come from the concurrent oxidation of oxygen from -II in $\ce{ClO3-}$ to 0 in $\ce{O2}$, and this is unlikely to happen, as oxygen is more electronegative than chlorine.


Most common material in the universe I was told that 'water' was the most common material in the universe and I disputed that claim. I was met with the argument that hydrogen and oxygen were the most common atoms in the world so water was as well. If this is true, then how come we can't find water anywhere in outer space? It must be the collection of 2 hydrogen atoms and 1 oxygen atom that is so rare then?

The Big Bang started with about 3/4 hydrogen atoms and 1/4 helium atoms.  Add a pinch of lithium and a whisper of boron.  Supernovae since then have released a lot of helium.  Heavier elements are still rare.

"The most common [molecular] material in the universe" must survive photolysis, radiolysis, and reaction.  Water is poor candidate.

http://www2.astro.psu.edu/users/alex/astro497_7.pdf
What is out there.  CO is the marker.
http://physics.nist.gov/cgi-bin/micro/table5/start.pl
Molecules in space, isotopically and by
http://science.gsfc.nasa.gov/691/cosmicice/interstellar.html
http://www.astro.uni-koeln.de/cdms/molecules
atom count


Polyethylene terephthalate reaction with tap water I have searched the internet and I found that reusing PET no.1 is dangerous. Why it is dangerous and what is the equation of the reaction between it and water tap ? And why it is not danger to when it is used for the first time in water bottles but it is danger when reusing it ?

Thanks!

Links I have found useful:

http://greenliving.nationalgeographic.com/plastic-drink-bottles-safe-reuse-drinking-water-3110.html

http://www.plasticsinfo.org/Main-Menu/MicrowaveFood/Need-to-Know/Plastic-Bev-Bottles/The-Safety-of-Polyethylene-Terephthalate-PET.html

It is probably not a chemical reaction that's the problem, it's that the container should be cleaned before it's refilled, and an unclean container might or will have bacteria.  While you're thinking 'refilled' might be in a matter if minutes, someone labeling the bottle is thinking about it sitting on a shelf with the cap off for months before reuse.  At that point there could be any number of unhealthy contaminants.


Common ion effect math problem The question was:

  If you are given a $1$ Litre solution of $\ce{CaF_2}$where there is $2.0 \times 10^{-5}$ M $\ce{Ca^{2+}}$. How many mililitres of $0.1$ M $\ce{NaF}$ would you need to precipitate $\ce{CaF_2}$? The solubility product of $\ce{CaF_2}$ is $1.7\times10^{-10}$ .

The answer given is

   956 ml

What I attempted:
I thought that if $x$ Litres of  $\ce{NaF}$ is necessary to precipitate $\ce{CaF_2}$; then after the mix, the ionic product($K_{ip}$) of $\ce{CaF_2}$ will exceed the solubity product($K_{sp}$). $$K_{ip} > K_{sp}$$
$$\ce{[Ca^2+][F-]^2} > K_{sp}$$  and from this equation I tried to solve the value of $x$.

Here's what I tried in details:-

step 1) Write down the dissociation  reaction of $\ce{CaF_2}$

$\ce{CaF_2 <=> Ca^{2+} + 2F-}$

and from the solubility product formula $K_{sp} = \ce{[Ca^2+][F-]^2}$ I calculated the concentration of $\ce{[F-]}$ which I got was $2.91\times10^{-3}$.

step 2) Then I again calculated the concentration of $\ce{Ca^2+}$ and $\ce{F-}$ after adding  $x$ Litres, using the formula $\ce{S_1\times V_1 = S_2\times V_2}$, because the volume now was not $1$ Litre, but $(1+x)$ Litres.

I got $$\ce{[Ca^2+] = \frac{2\times 10^{-5}}{(1+x)}}M$$ $$\ce{[F-] = ( \frac{0.1x}{1+x} + \frac{2.9\times 10^{-3}}{1+x})}M$$

step 3) Then as I stated before, the condition of precipitation would be as follows
$$K_{ip} > K_{sp}$$
$$\ce{[Ca^2+][F-]^2} > K_{sp}$$

So, from the equation above, I calculated the value of $x$ which I got was $1.8$ Litres (approximately) which was nowhere near the given answer in the book. How to solve this problem?

First the question doesn't really make sense. $\ce{CaF_2}$ is virtually insoluble, so a "solution of $\ce{CaF_2}$" is a really odd phrase.

Skipping this nit-picking, the reactants are 1 L of a $2.0 \times 10^{-5}$ M $\ce{Ca^{2+}}$ and a $0.1$ M $\ce{F^-}$ solution  and $\ce{CaF_2}$ is to be precipitated.

This is a sloppy use of significant figures since volume of 1L and the molarity of the $\ce{F^-}$ solution are only given to one significant figure.

Ignoring the significant figure problem too, let's work backwards from the given answer of 946 ml.

For the initial solution we can solve for the moles of $\ce{Ca^{2+}}$ present.

$\text{moles }\ce{Ca^{2+}} = 1 \rm{L} * 2.0 \times 10^{-5}\frac{moles}{L} = 2.0 \times 10^{-5}\text{ moles}$

Now if we add 946 ml of 0.1 M NaF, then the moles of $\ce{F^-}$ in solution is 0.0946 moles which vastly more than the calcium. So we can assume the precipitated $\ce{CaF2}$ has a negligible effect on the $[\ce{F^-}]$, and calculate the final concentration of $[\ce{F^-}]$ as:

$\ce{[F^-]}_{\text{final}} = \dfrac{0.0946}{1.946} = 0.0486 \text{M}$

Now from the $\rm{K}_{\text{sp}}$ we have

$\ce{[Ca^{2+}]}_{\text{final}} = \dfrac{\rm{K}_{\text{sp}}} {[\ce{F^-}]^2}= \dfrac{1.7 \times 10^{-10}}{0.0486^2} = 7.20 \times 10^{-8}$

$\text{moles }\ce{Ca^{2+}} \text{ final}= (7.20 \times 10^{-8})(1.946) = 1.40 \times 10^{-7}$

The initial moles of $\ce{Ca^{2+}}$ is $2.0 \times 10^{-5}$ which is +/- $0.05 \times 10^{-5}$. Thus so far as I am concerned the solution is a "magic number" and there is no way to get from the problem statement to that value. In other words why not $2.00 \times 10^{-7}$ or $3.00 \times 10^{-8}$ for the final $\ce{Ca^{2+}}$ concentration?!?

Thus you have to assume that you're going to precipitate some fraction of the $\ce{Ca^{2+}}$. So are you going to precipitate 99%, 99.9%, 99.95% or what? There just isn't anything in the problem statement on which you can pin a value.

Just out of curiosity the % $\ce{Ca^{2+}}$ recovered is:

% $\ce{Ca^{2+}}$ recovered = $(1-\dfrac{1.40 \times 10^{-7}}{2\times 10^{-5}}) \times 100\% = 99.3 \%$

Note that we can also assume that the original solution is saturated in respect to $\ce{CaF2}$. As shown in OP's post, this leads to a concentration of $2.91 \times 10^{-3} \rm{M}$ for $\ce{F^-}$ and given the 1L volume, $2.91 \times 10^{-3} \rm{moles}$ of $\ce{F^-}$. Adding that to the number of moles from the 946 ml of $\ce{F^-}$ solution gives a slightly different value for the total moles of $\ce{F^-}$:

total moles of $\ce{F^-}$ = $0.0946 + 0.00291 = 0.0975$

$\ce{[F^-]}_{\text{final}} = \dfrac{0.0975}{1.946} = 0.0501 \text{M}$

So the magic value for the textbook solution is to assume that: $\ce{[F^-]}_{\text{final}}= 0.0500$

$\ce{[Ca^{2+}]}_{\text{final}} = \dfrac{\rm{K}_{\text{sp}}} {[\ce{F^-}]^2}= \dfrac{1.7 \times 10^{-10}}{0.0501^2} = 6.77 \times 10^{-8}$

$\text{moles }\ce{Ca^{2+}} \text{ final}= (6.77 \times 10^{-8})(1.946) = 1.32 \times 10^{-7}$

% $\ce{Ca^{2+}}$ recovered = $(1-\dfrac{1.32 \times 10^{-7}}{2\times 10^{-5}}) \times 100\% = 99.34 \%$

So we get a slightly different value, but it still seems like a magic number of some sort.

Now if I were to write the problem I would have stated:

  You are given $1.00$ Litre of solution which is saturated with  $\ce{CaF_2}$ and has a $\ce{Ca^{2+}}$ concentration of $2.00 \times 10^{-5}$ M. How many mililitres of $0.100$ M $\ce{NaF}$ would you need to recover 99% of the $\ce{Ca^{2+}}$ as a precipitate of $\ce{CaF_2}$? The solubility product of $\ce{CaF_2}$ is $1.7\times10^{-10}$ .

Curious I played with this some more using the assumption of the $\ce{F^-}$ saturated starting solution. Obviously if you add vast amounts of $\ce{F^-}$ solution then any $\ce{CaF_2}$ which formed would dissolve agin. So there is some amount of $\ce{F^-}$ solution which precipitates the maximum amount of $\ce{CaF_2}$.

The max % $\ce{Ca^{2+}}$ recovery = 99.44% with 1913 ml of $\ce{F^-}$ solution. To +/- 0.01% any volume between about 1400 ml and 2500 ml would yield effectively the same result.


What wavelength of light is released when an electron drops an energy level in complexes? I know that the color of transition metals is related to the splitting of the d-orbitals. Only a certain wavelength is absorbed in the promotion of electron to a higher d-orbital and therefore the remaining colors are interpreted as a certain color. For example in copper(II) sulphate the red wavelength is absorbed and therefore it the complex appears to be cyan. But shouldn't the same amount of energy be released once the electron drops back to its previous state and releases the energy is gained? What wavelength would this drop correspond to?

Not necessarily. The drop may occur gradually so that the colour is "blinded out" by the nonadsorbed colours. The drop may also occur in stages, so that several low energy emissions occur. Some are outside visible range and others are too weak or go unnoticed.


Nucleoside/nucleobase chemical synthesis I'm interested in the chemical synthesis of nucleoside phosphoramidites, but all the procedures I can find have "start with a protected nucleoside" as the first step. Searches for chemical synthesis of nucleosides all seem to find process that start with existing nucleobases, or in vivo processes, or speculative origin-of-life processes. Searching for chemical synthesis of nucleobases ONLY seems to find speculative origin-of-life processes.

What I'm looking for (just out of curiosity) are processes for chemical synthesis of nucleosides from a non-biological basis (not using pre-existing nucleobases) as used in typical commercial/industrial supply of nucleoside phosphoramidites. I'm starting to wonder if they even are chemically synthesized, or if they are instead harvested in some way, but it's hard to find a smoking gun that outright says it one way or the other without access to whitepapers anymore (no more university account).

Does anybody know?

Finally found what I was looking for, especially paired with Mithoron's comments.

"Currently, purine nucleotides are obtained at industrial level mainly by microbial fermentation".

https://microbialcellfactories.biomedcentral.com/articles/10.1186/s12934-015-0234-4

Ledesma-Amaro, Rodrigo & Jiménez, Alberto & Santos, María de los Ángeles & Revuelta, José. (2013). Biotechnological production of feed nucleotides by microbial strain improvement. Process Biochemistry. 48. 1263-1270. 10.1016/j.procbio.2013.06.025.


Reaction of borax with NaOH How does borax  react with NaOH?
My textbook says that sodium peroxoborate is formed. I couldn't find the reaction anywhere else.
Can anyone verify as well as explain it?

The following reaction schemes are adapted from [1, pp. 76–77].
Amorphous borax $\ce{Na2B4O7}$ forms tetrahydroxoborate in concentrated sodium hydroxide solution:

$$\ce{Na2B4O7 + 7 H2O + 2 NaOH → 4 Na[B(OH)4]}$$

Upon fusion sodium metaborate is formed:

$$\ce{Na2B4O7 + 2 NaOH →[\pu{700-750 °C}] 4 NaBO2 + H2O}$$

In order to form peroxoborate, a peroxide should be introduced in the first place, e.g. by using hydrogen peroxide (see also Preparation of Sodium Peroxoborate):

$$\ce{Na2B4O7 + 2 NaOH + 4 H2O2 + 11 H2O → 2 Na2[B2(O2)2(OH)4] * 6H2O}$$

References

R. A. Lidin, V. A. Molochko, and L. L. Andreeva, Reactivity of Inorganic Substances, 3rd ed.; Khimia: Moscow, 2000. (in Russian)


Nucleophillic addition reaction in the presence of an acid catalyst on a molecule with an alkene This is a solution for a question in my textbook.

In the presence of an acid catalyst, am I wrong in my understanding that the acid will protonate the alcohol, and then do an electrophillic addition reaction onto the alkene (in addition to the nucleophillic addition reaction from the alcohol)? If so, was it omitted from the solution because it is not a major product (if it's not a major product, why)?

The acid will protonate the O of the carbonyl carbon, so a positive charge is generated. The oxygen of the alcohol with the free pair of electrons will act as nucleophile. So the alcohol is added to the cycle and a pair of electrons of the double bond of the carbonyl will compensate the positive charge generated. As the medium is acid, in the second step the hydroxyl group is protonated, so it is generated a water molecule, the alcohol attacks again at the same carbon and the water is released to the medium.

This is why this process is used to protect carbonyl groups. The reaction is reversible so if you add water, the reaction direction is reversed and you get the keto form again.


Heat when dissolving solutes in water Why do certain solutes dissolved in water release heat and other solutes absorb heat?

To understand this phenomenon first we must point out that there are very different solid substances and solvents. Different combinations of them lead to different interactions and properties.

Anyway for the regular case - ionic or soluble compound in water - there is at least two steps to realize the solution:

Edited Answer:

Break the solid structure (eg. the crystal lattice). This step generally absorbs energy (endothermic, see lattice energy);
Solvate the compound. Because there are interactions between matter, the solvent might organize itself around the compound (solvation shell). Those molecules involved are not any more available, they are tied to the compound, they are not any more free water. This step generally releases energy (exothermic, see solvation energy). You can think about this step as a reaction of the solvent to recover form the intrusion by coating the intruder with one or several layers of its molecules. Then it 'feels' if there were only water. You will find details of the complex solvation process in good Physical Chemistry book.

Succeeding these two unit steps leads to dissolution. Solubility is the thermodynamic limit beyond that you cannot dissolve further molecule of a substance. Off course this is related with every substance present in the solution and available free water. Solubility are generally expressed for pure substance only.

When you realize your solution, speaking of heat exchanged with the environment, you experienced the balance between those two reactions. This is why the heat balance can lead to a global endothermic or exothermic reaction.