How to derive the Gibbs free energy for an ideal gas? Determine the Gibbs Free Energy for an ideal gas and show that, this energy, and its derivative can be related to the enthalpy of the system via Gibbs-Helmholtz equation
  $$H = -T^2 \left( \frac{\partial (G/T)}{\partial T} \right)$$

So I need to determine the fundamental relation for an ideal gas in the Gibbs free energy representation.

My thoughts:

I know the Gibbs free energy, is defined by
$$G(p,T) = U + pV - TS = H - TS,$$
where $H$ is the enthalpy, then
\begin{align}
\mathrm{d}(U + pV - TS) &=
   V\mathrm{d}p - S\,\mathrm{d}T + \sum{\mu_i\,\mathrm{d}N_i}
   - \sum{X_i\,\mathrm{d}a_i}\\
\mathrm{d}G &=
   V\,\mathrm{d}p - S\,\mathrm{d}T + \sum{\mu_i\,\mathrm{d}N_i
   - \sum{X_i\mathrm{d}a_i}}\\
\therefore \mathrm{d}G &=
   V\,\mathrm{d}p - S\,\mathrm{d}T
\end{align}
Now, if the temperature is constant, and only the pressure change then $\mathrm{d}G = V\,\mathrm{d}p$. It follows that $$G(p)-G(p_0)=\int_{p_0}^{p}{V\,\mathrm{d}p}$$
If $V = \frac{nRT}{p}$, we have
$$G(p) = G(p_0) + nRT \ln\left(\frac{p}{p_0}\right)$$

But, I don't know if this is correct.

The fundamental relation for an Ideal Gas, in the entropy representation is

$$S(U,V,N) =
   NS_0 + NR\ln\left[ \left( \frac{U}{U_0} \right)^c
                      \left( \frac{V}{V_0} \right)
                      \left( \frac{N_0}{N} \right)^{c+1} \right]$$

The Gibbs Free Energy, is the Legendre Transform $G \equiv U[T,p]$

so, $G = U + pV -TS$

First, let's get the equation of state $S(T,V,N) $ from the the fundamental relation. We eliminate $U$ using $U = cNRT$

$$S(T,V,N) =
  NS_0 + NR\ln\left[ \left( \frac{T}{T_0} \right)^c
                     \left( \frac{V}{V_0} \right)
                     \left( \frac{N_0}{N} \right) \right].$$

Also, for an Ideal Gas, $p =\frac{NRT}{V}$

$$G(T,p,N) =
  cNRT - TNS_0
  - NRT\ln\left[ \left( \frac{T}{T_0} \right)^c
                 \left( \frac{p_0}{p} \right)
                 \left( \frac{N_0}{N} \right) \right] + NRT $$

Exercise: Find $G_0 = G(T_0,p_0, N_0)$ and substitute into the fundamental relation derived above to get rid of $S_0$


Using Hydrates as a skin moisturizer I was just in my AP Chemistry class earlier today learning about hydrates and their properties. It occurred to me that a compound with so much water could aid in moisturizing the skin. Is this a valid thought? And if so, could hydrates be irritating to the skin?

Through research, I personally thought Natron may be an option in a solution added to a cream or lotion.

I do not think that hydrates will serve as good moisturizing agents, because they usually do not loose their crystal water unless when being heated. Magnesium sulfate heptahydrate, for example, starts to loose water at temperatures above 50 °C, and complete dehydration of the salt is only achieved at temperatures above 310 °C. Some hydrates can even have a dehydrating effect when they react with moisture to hydrates with higher $\ce{H2O}$ content. Calcium sulfate hemihydrate, for example, reacts with water to form the dihydrate in an exothermic reaction. The stability of hydrates is the reason why many anhydrous salts which are easily hydrated are used as drying agents, for example, sodium sulfate and magnesium sulfate.


Deviation from Raoult's law because of adhesive and cohesive forces A solution made of two or more compounds have adhesive and cohesive forces. Cohesive forces are the attractive forces between the same molecules. Adhesive forces are the attractive forces between different molecules of the solution.

A real solution shows positive deviation from Raoult's law when the cohesive forces are stronger then adhesive forces. It shows negative deviation when the adhesive forces are stronger than cohesive forces.

Why do stronger cohesive forces in real solution make positive deviation and stronger adhesive forces make negative deviation from an ideal solution?

Foreword
This topic is treated in most books and sites so I've tried to explain it to you in a more unusual and friendly way to grasp the concept but maybe the canonicals explanations are more suited for exams and serious conversation.
First step idealize and simplify the Raoult law
You can see Raoult law as the most basic attempt to predict the partial vapour pressure of a multi-components solution. If $A$ is your first component with higher vapour pressure an $B$ is the second component with lower vapour pressure, it seems quite obvious that if you mix an equal amount of $A$ and $B$ you have a vapour pressure equal to the mean of the two.
Back to reality
In fact, the reality is more complex and you have to take into account the force between the molecules of the two components. If you have strong adhesive forces the molecules of the first component near the molecules of the second keep themself more stuck together. So this creates a sort of "net" that doesn't allow the molecules to "escape" from the solution and so the vapour pressure (that in fact quantify the ratio of molecules that "escape" from the bulk) is lower and finally  you have a negative deviation.
But what if the interaction between the molecules of the same component is greater than the interaction between the two components? As you state, this is the case of strong cohesive forces. In this case you have to imagine that the components act as if they were separate.  Molecules of the component A lower the positive interaction between the molecules of the component B interposing them self and avoiding the formation of bonds between molecules of component B and finally increasing the vapour pressure, a positive deviation!
Deep into fantasy
So if you think the molecules as humans exiting from an elevator, humans A are more claustrophobic and so tend to exit more rapidly than humans B when the doors open. If they interact positively with humans B they will keep calm when exiting. If humans B don't find funny humans A they preserve their panic attitude and at the same time separate B humans from each other and rushing out from the elevator pull humans B out more rapidly too!


Does a rotary evaporator need a draft chamber and a vacuum pump? I am using DCM ( dichloromethane) in my research, as a solvent. At the end of the day I have to separate the salute and measure it.
Up until now, I have been using a constant temperature oven, set a about 65 degree Celsius (evaporation temp of DCM). The exhaust DCM is expelled into the atmosphere and the dried salute is then cooled down and measured.

I want to replace the oven with a rotary evaporator. However I have an argument with my advisor. He claims that I require two things:

Aspirator /vacuum pump
Draft Chamber

While I agree that a vacuum pump can be of use, I do not require my solvent to boil, therefore I do not need vacuum.

As for the draft chamber, I completely fail to see its implication.
I believe that since the rotary evaporator is a completely closed system, there is no way DCM vapor could leak, rendering the draft chamber useless.

I think you are taking too dogmatic a position overall.

1) Vacuum also increases the rate of evaporation, without boiling.  This is just practical.

2) Your claim that your system is "completely closed" is over confident. What if you are wrong and you have a leak?  Now you are free to dismiss this, but this is the safety issue your adviser is getting at.

In other words, it would be easy to incorporate your practice with industry best practice.


What Causes there to be a Maximum Amount of Salt which can be Dissolved in Water? When you add salt to water and stir, after a while the salt will no longer dissolve. Why is this?

When a salt dissolves in water, there is a competition between the electrostatic forces that keep the anion and cation together and the ion-dipole forces that solvate the ions.  One could write this equation as:

$$\ce{AB + q H2O \leftrightarrows A(H2O)_{m}+ + B(H2O)_{n}-}$$

where $q=m+n$. Since this is an equilibrium, we can write an equilibrium constant, which is oftentimes referred to as the solubility product: $$\ce{K_{sp}=[A(H2O)_{m}+] * [B(H2O)_{n}^{-}]}$$

Very soluble salts will have a large, but finite, $K_{sp}$.  The magnitude of the solubility product can be rationalized by comparing the energy required to break the electrostatic bonds (the lattice enthalpy) and the energy released by forming solvated ions (solvation enthalpies).  One can use a Born-Haber cycle to predict whether the dissolution of a salt will be endothermic or exothermic, and we would need a similar approach to address entropy changes before we could get a complete, quantitative understanding of the solubility.


Change in inductive effect of a susbstituent when -R group replaces -H atom There are several examples where the negative inductive effect of a substituent gets increased when a hydrogen atom on that substituent is replaced by an alkyl group.

Some particular examples:

$$\ce{-NR3+} > \ce{-NH3+} \qquad \ce{-OR} > \ce{-OH}$$

I don't understand why it should be that way, as an alkyl group shows more positive inductive effect than a hydrogen atom which should, in turn, reduce the negative inductive effect of the substituent.

The answer here gives the reason that a carbon atom is more electronegative than a hydrogen atom and hence it should increase the −I effect (the answer also talks about bond length but I think it's irrelevant to mention that). But wouldn't the inductive effect shown by the whole group be a better parameter than the electronegativity of its primary atom to judge its electron withdrawing/donating ability?

Then there are also examples where replacing hydrogen atom by an alkyl group reduces the negative inductive effect of the substituent, as with $\ce{-CHO}$ and $\ce{-CRO}$ groups. I would usually explain it using the +I effects of $\ce{-R}$ and $\ce{-H}$ groups, but since that logic doesn't hold for the previous case, there should be some other reason for it.

So my question is: How does replacing an H atom in a substituent by an alkyl group affect the inductive effect of the substituent?

P.S. There is a related question, but it is too specific and does not have an answer to my question.

There are several examples where the negative inductive effect of a substituent gets increased when a hydrogen atom on that substituent is replaced by an alkyl group.
Some particular examples:
\begin{align}
\ce{-NR3^+} &> \ce{-NH3^+}& \ce{-OR} &> \ce{-OH}
\end{align}

About comparing $\ce{-NR3^+}$ and $\ce{-NH3^+}$, at first, the electron-donating inductive effect of the methyl groups (+I) makes us feel that the electron-withdrawing inductive effect trend should be the opposite. Neither, could I find any reliable sources with an explanation. So, I came up with a few of my own.
Part I: The Bent's Rule Interpretation (More Reliable)
To analyse the inductive effect, let's take the groups attached to a hydrocarbon chain and then see.

The $\ce{C-H}$ bond orbitals will have more s-character than in the $\ce{N-C}$ bond orbitals because there is more electron-density in the former due to the greater electronegativity of carbon over hydrogen and nitrogen over carbon, which requires the presence of more, lower-energy s-character in them to lower the system's overall energy. Thus, an implication of this would be the increase of p-character in the three $\ce{N-C_{Me}}$ bond-orbitals and the subsequent decrease of p-character in the $\ce{N-C_{chain}}$. This results in an increase of electronegativity of $\ce{N}$, thus the (-I) trend is:
$$\ce{-NR3^+} > \ce{NH3^+}$$
Part II: I don't know what to call this (Need help with this)
Let's consider the examples of phenol and anisole. It is considered that phenol donates more electron density to the benzene ring than anisole because it exists in equilibrium with the phenoxide ion, which is a far-more electron-donating group, despite anisole's $\ce{-OCH3}$ group being a better electron-donating group.
If we apply this to our case, we can assume that $\ce{-NH3^+}$ may be in equilibrium with $\ce{-NH2}$ which makes it a better (+I) group and poorer (-I) group, while $\ce{-NMe3^+}$ can't deprotonate itself and establish an equilibrium like that, making it a better (-I) group.
(somebody please help here)
Part III: Should we really consider $\ce{-CH3}$ as a (+I) group here? (Just thinking, very reliable*)
It is considered that the inductive effect of $\ce{-CH3}$ is mainly attributed to its hyperconjugative effect. However, hyperconjugation requires an empty p orbital carrying a positive charge into which it donates its $\sigma_{\ce{C-H}}$ bond-electron density. However, nitrogen's orbitals are sp³ hybridized here, with no empty orbital carrying the charge as such. So, I think (not really sure*), we should only consider carbon's greater electronegativity over hydrogen, thus making $\ce{-NMe3^+}$ more inductively electron-withdrawing than $\ce{-NH3^+}$. You can also apply this to $\ce{-OR}$ and $\ce{-OH}$.
* Just saw Jan's answer in the middle of my writing process. That makes this a valid point. Great answer, a must check-out.


Exergonic vs Endergonic Reactions Problem
Problem 3, concept check 8.2 from Campbell Biology [1, p. 150]:

Some partygoers wear glow-in-the-dark necklaces that start glowing once they are “activated” by snapping the necklace. This allows two chemicals to react and emit light in the form of chemiluminescence. Is the reaction exergonic or endergonic? Explain.

Answer
Answer key [1, p. A-8]:

The reaction is exergonic because it releases energy—in this case, in the form of light. (This is a nonbiological version of the bioluminescence seen in Figure 8.1.)

From [1, p. 143]:

Figure 8.1 The green glowing spots on the outside of this Brazilian termite mound are larvae of the click beetle, Pyrophorus nyctophanus. These larvae convert the energy stored in organic molecules to light, a process called bioluminescence, which attracts termites that the larvae eat. Bioluminescence and other metabolic activities in a cell are energy transformations that are subject to physical laws.

Question
The answer is an exergonic reaction. What I understand of an exergonic reaction is that not only does it have a negative $ΔG,$ it releases energy. In this case, the energy released is light energy.
My original answer endergonic reaction as for the problem given, to “activate” the necklace, doesn’t it require energy from partygoers to snap the necklace? Is the $ΔG$ value for the light being released much less than the $ΔG$ value of the energy being used to snap the necklace?
Reference

Urry, L. A.; Cain, M. L.; Wasserman, S. A.; Minorsky, P. V.; Orr, R. B.; Campbell, N. A. Campbell Biology, 12th ed.; Pearson: New York, NY, 2020. ISBN 978-0-13-518874-3.

The reaction to consider is the transformation of chemical energy into visible radiation (light) and a little of heat.  A glowing stick consists of at least two compartments which -- on purchase -- a supposed to be kept separate.  The snap only breaks the wall between the two so that chemicals may mix with each other to start chemoluminescence (work principle), a principle similar to a firefly in late summer, too.
To decide if the reaction is exothermic, endothermic; exergonic or endergonic you have to compare the energy of initial state (when the chemicals are separated, no emission of light) with the one after the reaction.  There may be an energetic barrier to trigger the reaction, equally known as activation energy, but in its current form, this detail is not relevant to answer the question.
Take an analogy employing potential energy you might recall from the physics lecture: imagine you ate in a restaurant on a mountain and now walk back home.  It is not relevant that some energy was required to get up from the table, it is only about the comparison between
$$\mbox{state A «being in the restaurant»} \ce{->} \mbox{state B «back in the valley» } + \mbox{energy}$$
Even the pathway, straight downhill, or passing some smaller hills to reach the destination is not important here.  The naming then is, by convention, a formalism.


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

Oxygen/nitrogen ratios vary seasonally and geographically even at sea level on the order of 200 ppm.  Oxygen concentration has also been slowly decreasing in correlation with carbon dioxide increase.

Try using the MSIS-E-90 Atmosphere Model which allow location, date, time and height input and provides composition output.


Why does having equivalent resonance structures give more stability? Although the phenolate ion has more resonance structures (4) compared to acetate ion (2), acetate is more stable because it has two equivalent resonance structures of same energy.

Why does having equivalent resonance structures give more stability? Shouldn't a compound with more resonance structures be more stable?

Preamble
One important thing to know is that what we call "resonance structure" is a byproduct of our chemical notation which can't describe the structure of some compounds effectively using only one chemical structure. Personally I find the old term mesomeric structure more appropriate (meso- Greek mésos in the middle;  -merism from Gk. merismos "dividing, partition") because resonance is already used for a phenomena, a real phenomena, in physics that has nothing to do with resonance in chemistry. Furthermore the compound is not continuously changing it's structure between the possible resonance structures, like the term resonance seems to suggest, but is a "normal steady" compound that simply has an electron distribution that cannot be described with a single chemical structure.
Why does having equivalent resonating structures give more stability?
The resonance energy is due to electron delocalization, so more the electrons are delocalized in the molecule higher is the resonance energy ( this means a decrease in the overall molecule energy). This is clear in the acetate ion - we can draw a resonance hybrid  like this:

Here the electrons are equally distributed (delocalized) between the two oxygen atoms.
However for some compounds the real structure is not always a structure right in the middle between the different resonance structures; the first assumption is true only if the two structures are two equivalent structures. So it can occur that the electrons are not well delocalized between all the atoms hence the resonance energy is lower and so the overall stability.
In the case of phenolate ion I think the first structure in the figure [2] gives a greater contribution to the final structure of the phenolate ion so the electrons are not well delocalized, but if I'm not wrong are mainly in the oxygen atom. Unfortunately  I can't find the energy of the different resonance structures.

Finally, however, I think that you should study case by case if the number of the resonance structures or their contributions are determinant to the overall resonance energy taking into account valency and electronegativity of the atoms involved.


Calculating equilibrium concentrations of silver and halide ions $\pu{500 ml}$ of $\pu{0.01 M}$ $\ce{AgNO3}$ is mixed with $\pu{500 ml}$ of solution containing $\pu{0.1 M}$ $\ce{NaCl}$ and $\pu{0.1 M}$ $\ce{NaBr}$. $K_\mathrm{sp}$ of $\ce{AgCl}$ is $10^{-10}$ and $K_\mathrm{sp}$ of $\ce{AgBr}$ is $5×10^{-13}$. Find $[\ce{Ag+}],$ $[\ce{Cl-}],$ $[\ce{Br-}]$ in equilibrium solution.

After adding all of these molarities will get halved. At equilibrium condition, both $\ce{AgCl}$ and $\ce{AgBr}$ would have precipitated out. So, using the $K_\mathrm{sp}$ expression, we can find maximum amount of $\ce{Ag+}$ required to precipitate both the salts, and this becomes the equilibrium concentration of $\ce{Ag+}$.

Plugging these values in $K_\mathrm{sp}$ of respective salts, we can get the equilibrium concentration of remaining anions.

The $[\ce{Ag+}]$ for precipitation of $\ce{AgCl}$ is $\pu{2E-9},$ and for precipitation of $\ce{AgBr}$ is $\pu{E-11}.$

So, $\ce{AgCl}$ gets precipitated at last.

$$K_\mathrm{sp}(\ce{AgCl}) = [\ce{Ag+}][\ce{Cl-}]$$

Therefore, $[\ce{Cl-}]$ at equilibrium is $\pu{5E-2}.$

$$K_\mathrm{sp}(\ce{AgBr}) = [\ce{Ag+}][\ce{Br-}]$$

Therefore, $[\ce{Br-}]$ at equilibrium is $\pu{2.5E-4}.$
$[\ce{Ag+}]$ at equilibrium is $\pu{2E-9},$ which is already calculated in first step.

However, this doesn't match the answer key. Where am I wrong?

Both chloride and bromide ions are present in 10-fold excess over silver ions. That means that the chloride and bromide concentration in solution will not drop by much (they will remain major species).

The solubility product of AgCl is 200-times higher than that of AgBr. If both AgCl and AgBr precipitate, the chloride solution would be 200-times higher than that of bromide. That is impossible because so little silver ions are available, and bromide and chloride start at equal concentrations. Thus, no silver chloride will precipitate (if this explanation is not convincing, suspend your disbelief until we can check the equilibrium concentrations).

Most of the silver ions will precipitate. As a first approximation, let's say the silver ion drops from 5 mM to zero, so the bromide concentration will drop from 50 mM to 45 mM. Of course, this can't be the equilibrium, so we can dissolve some of the AgBr again.

$$[\ce{Ag+}] = \frac{K}{[\ce{Br-}]} \approx \frac{\pu{1e-11}}{\pu{45e-3}} = \pu{2.2e-10}$$

So the concentration of silver ions is roughly $\pu{2.2e-10 M}$, that of bromide ions roughly $\pu{45 mM}$ and that of chloride ions $\pu{50 mM}$, with no silver chloride precipitating.

  At equilibrium condition, both AgCl and AgBr would have precipitated out.

Now that we have solved the exercise, you can verify that AgCl will not precipitate. The reason is that the silver ion concentration is extremely small due to the presence of 45 mM bromide ions. In fact, if NaCl and AgNO3 had been mixed first, AgCl would have precipitated. However, adding NaBr would have dissolved it again eventually, leading to the same final state.

  However, this doesn't match the answer key. Where am I wrong?

As mentioned in the comments, the set of final concentrations does not make sense from a stoichiometry (charge balance) perspective, and also contradicts the claim that AgCl is saturated. If you assume that both AgBr and AgCl have formed and attempt to solve the problem by brute force, the solutions to the two equations with two unknowns will only yield solutions that are not attainable (negative concentrations, or concentrations higher possible with the available starting materials).


Food which does not produce urea My professor of bioengineering said that all foods produce urea. Do foods exist which does not produce urea?

Thank you very much.

Food doesn't produce urea, your body produces urea from the nitrogen content of the food you eat (mostly comes from proteins). So you can eat e.g honey, which contains minimal, if any, amount of nitrogen.


Order of dipole moments What is the order of dipole moments for $\ce{SO3, SiO2, P2O5,}$ and $\ce{Cl2O7}$? This question is from a book and the answer given is $\ce{Cl2O7 < SO3 < P2O5 < SiO2}$. Can someone explain why? When I drew the Lewis structures and applied VSEPR theory, it seemed that all the given compounds have zero dipole moment.

I wondered about that for a long time, but Ben Norris and BigGenius have the truth of it: the question is about the bond dipole moment, otherwise the answer would make sense. Here's why.
Molecular dipoles

Sulfur trioxide has planar trigonal geometry, and its molecular dipole is zero. Similarly, molecular $\ce{SiO2}$ has the same geometry as $\ce{CO2}$, i.e. linear, and thus its molecular dipole is zero.

$\ce{Cl2O7}$ is bent at its central oxygen atom:

and thus has a non-zero dipole moment. While $\ce{P2O5}$ generally occurs as its “dimer” $\ce{P4O10}$, it is possible to write a molecular structure for $\ce{P2O5}$ which then strongly resembles that of $\ce{Cl2O7}$, being bent at the central O atom. It thus also has non-zero molecular dipole.

Finally, ordering between $\ce{Cl2O7}$ and $\ce{P2O5}$: because the P is less electronegative than Cl, the partial charges borne by the P atoms will be larger than those of the Cl atoms, and thus I expect the molecular dipole of $\ce{P2O5}$ to be greater than that of $\ce{Cl2O7}$.

Bond dipoles
If the question is talking about the $\ce{X–O}$ (or $\ce{X=O}$) bond dipole in the series where X = S, Si, P or Cl, it becomes easier to understand. For one thing, because you're asked about $\ce{X–O}$ bonds where X spans a row of the periodic table, which is typical of this sort of problems. The answer would then be that bond dipole moment get bigger when the electronegativity difference of X and O becomes larger (the sharing of electrons gets more asymmetric). This leads to the answer you gave.


Reference for UV/Vis Spectroscopic Data request I am looking for a complete reference of UV/Vis spectroscopic data for organic materials ( mostly interested in λmax ) .

I have the CRC handbook but it only has data about some organic solvents in water and very limited data on dyes.

Can anyone suggest a comprehensive reference that includes the λmax of a wide variety of molecules in different solvents ?

To know the chemical name, molecular structure, or best: the unique registry number provided by the American Chemical Society (CAS-number) is beneficial in your search as trade names may vary depending on the supplier of the material you are interested in.  The index number of the Colour Index (like C.I. 73000 for indigo) frequently used in the business of dyes potentially is of lesser use here.  In decrease of data coverage:

There are commercial sources like Reaxys (by Elsevier) allowing to search for absorption data.  As this (in part) emerged from a much senior database about chemical reactions, Beilstein (back to 1771) this is a typical database used in Departments of Chemistry at a University or Industry. SciFinder, based on the data indexed by the American Chemical Society equally allows you to find such data -- for both it might be helpful to request assistance / initial training to fully benefit from these sources.  Note: there is so much published data, none of both huge data bases may index / point back to all of the original journal articles, patents, conference proceedings, etc. pp.  Depending on your search strategy, typically you find the data you look for in both data bases--at different places.  Sometimes "The Two" are complementary to each other, sometimes an other (third) search engine finds the data on the net that neither ACS nor Reaxys indexed (yet).

Personally, because of the design "how the numerical data are managed" I would suggest Beilstein as the better choice here: if you find your compound, the data base will list in a table wavelength and molecular absorption coefficient altogther with the bibliographic reference -- often including the solvent of the analysis.  And the thought occurs to me, it is easier to query with Boolean logic, like "find a compound of molecular mass in the range of xyz to x'y'z with an $\epsilon$ greater than 20000", for example.

Less in coverage -- yet worth to point to are the compilations Sadtlers Handbook of ultraviolet spectra (today part in Bio-Rad's "KnowItAll" database) and the handbook of stains, dyes, and indicators by Sigma-Aldrich, too mention some of the printed references usually found in the section of spectroscopy/chemistry in university libraries.  Here you find not only one $\epsilon$ at $\lambda_{\mbox{max}}$, but a complete print of the spectrum.  (Be aware, the elder the sources sometimes use wavenumbers or kilo-Kaysers instead of nanometers to reference the wavelength.)

Taking into account your user profile related to quantum chemistry, perhaps you want to train your quantum chemical model with known experimental data, to evolve from a descriptive to a predictive -- and small molecules may serve you well, too.  If so, take a look at NIST's webbook based on these references.

Last but not least: Take these data with a grain of salt, often there is some limitation: the same compound / material often exhibits a slightly different UV-Vis absorption spectrum if analyzed once in solvent A, and subsequently in solvent B.  Bathochromic / hypsochromic shifts (solvatochromism) may still occur.  Solvent polarity is among the  parameters in this play.


Why is the Kelvin Temperature Scale Based on the Triple Point of Water? Why would the Kelvin temperature scale not just be based on the freezing point of water, which is only .01 degrees Celsius off from it? Is this just a more accurate way of measuring and quantifying the value?

In reality all the "primary" scientific values have changed over time as experimental techniques have improved. Science has evolved to the point where a set of seven "SI base units" (ampere, kelvin, second, meter, kilogram, candela, and mole) have been selected and defined so that better measurements won't cause disruptive changes in scientific data.

So the gist is that the "temperature of ice water" depends on the atmospheric pressure. So it is necessary to define that 0 °C is the temperature of ice water at "standard atmospheric pressure." That means that if the "standard atmospheric pressure" changes, then so does the "temperature of ice water at standard pressure."

However the triple point of water doesn't change and has a unique value. So by defining the triple point of water as exactly 0.01 °C in 1954, then the Celsius scale is anchored as much as the Kelvin scale. By also defining absolute zero as 0 K and −273.15 °C then the temperature scale itself won't change if better ways to measure temperature are devised.

Thus by the time the whole notion of the primary base units was sorted out there was a massive amount of scientific data that had been accumulated using the old definition of the Celsius scale that was based on ice water at a standard atmosphere. Letting the triple point be exactly 0.01 °C kept all of that "old" temperature data "reasonably accurate" since 0.01° is only a 0.01/273.15=+0.0037% change.


Solubility of NaOH in Organic solvent? I am conducting organic reaction in Dimethoxyethane as solvent in the presence of base (NaOH), but the NaOH is not completely soluble ( Sparingly soluble) in Dimethoxyethane even after the addition of excess of solvent. Peoples says that for a reaction, the reactants must be completely soluble in solvent.

Can someone comments about solubility of Bases during organic reaction?

I actually want conduct similar kind of reaction, the paper mentioned that NaOH and organic compound stirred vigorously until to obtained clear solution. In my case my compound is completely soluble in DME but NaOH is poorly soluble.

Your best bet is to publish the procedure. Procedures from pier reviewed journals are generally reproducible (there are exceptions). So if they say "dissolve x grams in y mL" just do it.

NaOH is poorly soluble in dimethoxyethane. This might be sufficient if base is used as a catalyst or if reaction is slowly shifting to product.

Devil is in details. Heat, presence of water and other factors might impact soluility greatly.


Nomenclature of methyl groups in an alkanol I'm having trouble with this question since I'm getting (A), when the answer says its (B).

My logic was that there is a methyl group on the first carbon, and a hydroxide functional group on the second carbon, hence the name 1-methyl-2-pentanol.

Could someone please explain why I am wrong?

Could someone please explain why I am wrong?

Find the longest chain! This determines the base name: hexane.
Find the position of the $\ce{OH}$ substituent. Lower number wins.


Meaning of "Nonyl-phenol 10 mole" In my workplace we are purchasing "Nonylphenol 10 mole" and "Nonylphenol 30 mole" for use as non-ionic surfactants.
At first I assumed, because of the name, that the substance was simply nonylphenol. But further research shows that commercial non-ionic surfactants are actually ethoxylates of nonylphenol. Thus, I'm assuming that what we are buying is in fact ethoxylated nonylphenol, in two different degrees.
Further research is fruitless though. I simply cannot find any source explaining the meaning of "10 mole" and "30 mole" at the end of the name.
Can anyone tell me the actual real chemical formula of "Nonylphenol 10 (or 30) mole", or at least give me an explanation of the name's meaning?

Nonylphenols are a family of closely related organic compounds composed of phenol bearing a 9 carbon-tail (Wikipedia). Following is an example of nonylphenols:

However, "Nonylphenol $n$ mole" is a commercial name for non-ionic surfactants where $n$ stand for the amount of ethelene oxides (EO) in moles in one surfactant molecule. For example, if the non-ionic surfactant is $\ce{R-O-(CH2CH2-O)n-H}$ where $\ce{R}$ is hydrophobic tail of the surfactant, the commercial name "Nonylphenol 10 mole" means $\ce{R-O = 4-C9H17-C6H4-O}$ and $n = 10$ in the structure. An example for a linear chain non-ionic surfactant is pentaethylene glycol monododecyl ether:

where $\ce{R-O = C12H25-O}$ and $n = 5$ (available in Sigma-Aldrich).


How is pumping "Air" into a bicycle make dW (delta work )negative , when in fact there is a increase in pressure and a slight increase in volume? Considering that it's an adiabatic process hence $\mathrm dQ=0$, and $\mathrm dU=\mathrm dW$. And since the work is done on the system then $\mathrm dW$ must be positive.
(The answer which is in the textbook is that $\mathrm dU$ is positive, while $\mathrm dW$ is negative.)

Depending on context and domain dependent convention, the work done is assumed in 2 different, opposite ways, with opposite signs.
Work done on a system $\Delta U = Q + W$, preferred by chemistry
versus
work done by a system $\Delta U = Q - W$, historically preferred by physics. It is used by the former way as well recently.

Speculatively said, physics way was defined during creation of theoretical background for thermal machines. Physicists may had been focusing on work done by machines.
Later chemists came, who invented their way, focusing on energy of systems.

As a prevention, it is always good to explicitly mention which way is used, to prevent confusion.
If applied, your pump is doing positive "chemical" work on the system=tyre, but the system=tyre does negative "physical" work on the pump. The textbook takes the convention of physics.

In an important sign convention, preferred in chemistry, work that adds to the internal energy of the system is counted as positive. On the other hand, for historical reasons, an oft-encountered sign convention, preferred in physics, is to consider work done by the system on its surroundings as positive.

Wikipedia - Work_(thermodynamics)


Do superhydrophobic/hydrophobic materials generally have a lot of friction or only a little friction Also, are there any exceptions that you know of?

Since you are talking about hydrophobicity I am assuming you are interested in liquid-solid friction. If you are interested in solid-solid friction I can say only one thing: hydrophobicity and solid-solid friction are in general unrelated.

So for the liquid-solid case. I am going to give you an answer that gives some insight, but no concrete rules, because, as far as I know, there aren't any.

Superhydrophobic surfaces typically have some surface structure that allows the liquid to 'float' partly on air and partly on top of the surface structure . Friction is related to the amount of contact area where there is a no-slip boundary so that will decrease therefore friction will be lower. A cool application of this effect can be found in this 2013 PNAS article by Karatay et al.

However, if you have a droplet moving, instead of a full liquid stream, then there is a counter-acting effect which is called contact angle hysteresis, which allows a droplet to sustain a difference in contact angle at the front and back of the droplet. This causes a force counter-acting motion of the order of the surface tension of the liquid $\gamma$ and the difference between the contact angles $\Delta \theta$: $F=\gamma l \Delta \theta$, where $l$ is a lengthscale of the order of the droplet radius. This force basically acts as a static friction on the droplet.

The value of $\Delta \theta$ is a complex function of the amount of surface roughness, being 0 for a atomically smooth surface but also approaching 0 for roughness with a small typical lengthscale. Therefore it is hard to make a general statement whether superhydrophobic surfaces will have a lot or a little friction. It all depends on the details of the surface roughness.


Inorganic Chemistry: the order of MO diagram I understand that to create a molecular orbital diagram, the following rules are supposed to be met:

number of A.Os = number of M.Os
Symmetry of the binding orbitals
Similarity in energy

They will eventually create bonding, anti-bonding, and maybe non-bonding orbitals. However, I am having trouble stacking them up in a correct order. For instance, if you refer to the attached diagram below, I don't know how it first starts with 4a1' and goes up to 5a1'. The only clear ones for me are the non-bonding orbitals, which have the same energy as the original group orbitals.

Could you please help me understand this? The coefficients that come in front of letters are also confusing me as well...

Thank you so much!

The numbers that come in front of each molecular orbital irrep are simply referring to the energy order in which the MO's are arranged, with the maximum number (for example, $5a_1'$ for $BF_3$) corresponding to the highest energy MO that has an $a_1'$ symmetry.

For understanding the molecular orbital energy order, you need to explicitly consider the combination of the fluorine ligand group orbitals and the boron AO of the same symmetry. If you do that you'll find $4a_1'$ has perfect bonding interactions between all the ligand p-orbitals involved in $\sigma$-bonding and the boron s orbital (both transform as $A_1'$ in $D_{3h}$). The identity of the p-orbitals would depend on the co-ordinate axis you choose to define as the molecular axis. $3e'$ would arise because of $\sigma$ interactions between the same ligand p-orbitals, this time slightly differently oriented and the boron p-orbitals. The $a_2''$ MO is pi-bonding in nature, so on and so forth.

But without further ado, I'll point you to this excerpt from Purcell and Kotz's Inorganic Chemistry, which is self contained and provides a detailed pictorial description of the various ligand group orbitals very nicely.

Best of luck!


What alkali and alkaline earth metal oxides will turn moist red litmus blue and finally white? How many of the following will turn moist red litmus blue and finally white?
$\ce{Li2O}$, $\ce{KO3}$, $\ce{RbO2}$, $\ce{Cs2O2}$, $\ce{BeO}$, $\ce{MgO}$, $\ce{BaO2}$, $\ce{SrO}$.

Since bleach convert moist red litmus blue and then white, but how we can identify which compound will act as bleach ?

The standard oxides $\ce{Li2O, MgO}$ and $\ce{SrO}$ are water soluble and hydrolyse to form hydroxides and a basic solution. $\ce{BeO}$ is insoluble in water.

$\ce{RbO2}$ is a superoxide which in aqueous solution undergoes disproportionation to $\ce{O2}$ and $\ce{OH-}$ so will give a blue litmus test.

The peroxides $\ce{Cs2O2}$ and $\ce{BaO2}$ react with water to give hydrogen peroxide and the hydroxides, so again a basic solution(source). The hydrogen peroxide produced will bleach the litmus paper.

$\ce{KO3}$(Potassium Ozonide) is highly unstable, decomposes in water to $\ce{O2}$ and $\ce{KOH}$

The oxides which will NOT decolorize litmus are $\ce{Li2O, MgO, BeO}$ and $\ce{SrO}$. The peroxides ($\ce{Cs2O2}$ and $\ce{BaO2}$), the superoxide($\ce{RbO2}$) and the ozonide($\ce{KO3}$) will decolorize litmus.


Properties of stilbene isomers (E)-stilbene may be separated by crystallization from (Z)-stilbene and the side product of the Wittig reaction, triphenylphospine oxide (TPPO), by using absolute ethanol. My question is why? (E)-stilbene and (Z)-stilbene are isomers of each other, why (and how) do they differ in polarity? If both are of low polarity, why is (Z)-stilbene soluble in ethanol, but not (E)-stilbene? Also why is (Z)-stilbene the more polar isomer of the two and has a lower $R_f$ value in TLC?
An initial reasoning of mine was that (Z)-stilbene were more hindered than the (E)-isomer, and thus less stable. Yet, this doesn't make sense to me.

With the $\ce{C=C}$ double bond of the stilbenes as geometric reference, you may think of the phenyl rings as substitutents increasing the electron density of the former.  This creates a small, locally constraint, dipole moment.  Dipoles like dipoles differ from a scalar (like e.g., temperature) that they have a direction, which is why I drew little arrows in the simplified representation of either isomer in the illustration below:

In the case of (E)-stilbene, these vectors oppose each other; for symmetry reasons, they may cancel out each other as if there were no vector present.  This contrasts to the case of (Z)-stilbene where the sum of the two vectors does not vanish; symbolized (the doodle is a concept, not at exact scale) by the dashed arrow: there is a non-zero molecular dipole moment.
The drawing of (Z)-stilbene's configuration however does not state (much) about the molecule's energetically preferred conformation(s).  To lessen intramolecular steric hindrance of the hydrogens of the two phenyl rings, the following spatial arrangement is more likely to happen:

Note, regardless how the phenyl rings rotate around the bond between $\ce{C_{ar}-C_{sp^2}}$, the overall (molecular) dipole moment will not be affected from these local variations.
Equally note that to achieve an efficient recrystallization of the three constituents of your mixture, the isomers of stilbene and TPPO, you use a mixture of ethanol/water heated to reflux then allowed to cool slowly to room temperature.  The addition of water is to attenuate the solution capacity of ethanol for each of the three compounds so that one preferentially crystallizes out.  This is why this purification method is called fractional crystallization.


Newman projection of 3-methylpentane How would you draw the Newman projections of 3-methylpentane, sighting along the C3-C4 bond? I am unsure as to how to position the $\ce{-CH3}$ and $\ce{-CH2CH3}$ groups on the C3.

The Newman projections of 3-methylpentane are:

This is a staggered conformation.
Another one would be

This is also a stggered conformation and is the most stable of conformers.

You can generate other conformers by rotating the C-3 or C-4 atom along the C-3 C-4 bond axis.


Which compound will have the most stable enol form? Which compound will have the most stable enol form, Phloroglucinol or 4-Pyridone?

My Thoughts:

Both in their enol form are aromatic. I am not sure how do I compare their stabilities. The answer given is 4-Pyridone though.

Any help would be appreciated.

The most questions about keto-enol tautomerism can be answered by their NMR spectral data (e.g., evidence of existing enol-form of acetylacetone at room temperature). According to its $\ce{^1H}$-NMR spectral data, it was declared that phloroglucinol, in its neutral form, exists nearly exclusively in its aromatic form (Ref.1). However, their dianion form (e.g., disodium phloroglucinolate) favors keto structures. These results were further analysed and confirmed by kinetic studies (Ref.2).

Wikipedia states that:

  4-Pyridone is an organic compound with the formula $\ce{C5H4NH(O)}$. It is a colorless solid. The compound exists in equilibrium with a minor tautomer, pyridin-4-ol.

Martin - マーチン has given here a nice answer using computational evidence for this preference  of 4-Pyridone (data are from Ref.3).

Based on these, it is clear that in their neutral forms, phloroglucinol has the most stable enol form, existing in predominantly in phenol form (1,3,5-trihydroxybenzene).

References:

R. J. Highet, T. J. Batterham, “The Structure of the Phloroglucinol Dianion,” J. Org. Chem. 1964, 29(2), 475-476 (https://doi.org/10.1021/jo01025a501).
Martin Lohrie, Wilhelm Knoche, “Dissociation and keto-enol tautomerism of phloroglucinol and its anions in aqueous solution," J. Am. Chem. Soc. 1993, 115(3), 919-924 (https://doi.org/10.1021/ja00056a016).
Peter I. Nagy, Giuliano Alagona, Caterina Ghio, “Theoretical Investigation of Tautomeric Equilibria for Isonicotinic Acid, 4-Pyridone, and Acetylacetone in Vacuo and in Solution,”  J. Chem. Theory Comput. 2007, 3(4), 1249–1266 (https://doi.org/10.1021/ct6002252).


Separation of solvent from residue of material. I have a residue of 2kl in which 1.8kl is solvent (EDC) and rest material is residue. First i was thinking to go with normal distillation to recover our solvent. But i heared somewhere that there is some kind of RESIN by which if it got passes solvent will be pass only and residue will collect on another side of that resin. if there is any concept like this please help me.
or there is any other to do?

Dichlorethane has a boiling point of about 84°C - If I would have to separate EDC from a product / residue, I would use a rotovap with water bath configuration. This distillation is a very gentle way to separate solvents from product / residue.

Set the temperature of the water bath slightly higher than the boiling point of the EDC.

If you want to get fancy to make your solvent super pure again after the rotovap distillation, you could perform a fractional distillation on it - however in most cases I think this might not be necessary.


CO-vibration in metal carbonyl I'm struggling to understand the trend of $\ce{CO}$-vibration in isoelectronic metal carbonyls.

$$
\begin{array}{lr|lr}
\hline
\text{Compound} & ν/\pu{cm-1} & \text{Compound} & ν/\pu{cm-1} \\
\hline
\ce{Ni(CO)4} & 2060 & \ce{[Mn(CO)6]+} & 2090 \\
\ce{[Co(CO)4]-} & 1890 & \ce{Cr(CO)6} & 2000 \\
\ce{[Fe(CO)4]^2-} & 1790 & \ce{[V(CO)6]-} & 1860 \\
  & & \ce{[Ti(CO)6]^2-} & 1750 \\
\hline
\end{array}
$$

Since the complexes are isoelectronic, the reason behind the trend should lie in the atomic charge $(\ce{Ni} > \ce{Co} > \ce{Fe}).$ Higher atomic charge means more attraction between metal cation and the ligand electron pair → higher electron density → more π-backbonding → weaker $\ce{C#O}$ bond → lower wavenumber.

But it is not the case here. I'm not sure why. I think it has something to do with the energy level of HOMO.

I think the mistake you're making is this conclusion: "Higher atomic charge. . .  -> higher electron density -> more pi-backbonding". With higher (ie more positive or less negative) atomic charge, there is less electron density donation towards the ligand, because the electron density is pulled to the metal center more strongly. Thus, $\pi$ back-bonding is weaker and C-O bond is stronger.

The more positively charged metal centers do not pull enough electron density from CO in the sigma interaction to be able to donate more back through pi interaction, which seems to be the way you're looking at it.


References to draw 3D molecules with directionality of non-bonding electron pairs and p-orbitals My instructor has been drawing 3D molecules that show the directionality of non-bonding electron pairs and p-orbitals. I've been trying to find references online that show this process, but I'm having a hard time finding anything useful.

Can someone please explain the process or link to a source that does? I've included two pictures of molecules that show what I'm looking for.

As it is already pointed out in the comments, what you are learning for is hybridisation and some basic MO theory. Do not go too far into the MO direction, because you will get into somewhere completely different place, and you will hard time to make the connections.

So aftre you reviewed hybridization: Once you can identify in a molecule the sp1, sp2 and sp3 carbons, you should be able to orient the hybrid orbitals. For example in the case of sp2 carbon of the lower image, it is in the HHCC plane. In case of sp1 there are two, for sp2 one un-hybridized p orbitals. These unhybridized orbitals should be perpendicular to the hybridized ones: perpendicular to the axis of sp1 or the plane of sp2.


Why are copper salts less soluble than zinc salts? Compounds with cations having pseudo-inert gas electron configurations have more covalent character than those that don't, therefore they should be less soluble.

Here are some solubility products:

$$
\begin{array}{lcc}
\hline
\text{Anions}    & \ce{Cu^{2+}}                 & \ce{Zn^{2+}} \\
\hline
\text{Arsenate}  & \color{red}{ \pu{7.95e-36} } & \color{red}{ \pu{2.8e-28} } \\
\text{Hydroxide} & \color{red}{ \pu{4.8e-20} }  & \color{red}{ \pu{3e-17} } \\
\text{Sulfide}   & \color{red}{ \pu{8e-37} }    & \color{red}{ \pu{2e-25} } \\
\text{Phosphate} & \color{red}{ \pu{1.4e-37} }  & \color{red}{ \pu{9e-33} } \\
\hline
\end{array}
$$

Zinc ($\ce{Zn^{2+}}$) has a pseudo-inert gas configuration ($\ce{[Ar] 3d^{10}}$), and its salts are more soluble than those of copper ($\ce{Cu^{2+}}$), which does not ($\ce{[Ar] 3d^{9}}$). Why?

It is because the difference in electronegativity.
The higher the difference in electronegativity between the cation and the anion, the stronger the ionic bond.
Since we are comparing species with the same anions and different cations, we only have to look at the values of the latter ones.

cation    electronegativity   configuration
  Cu             1.9         [Ar] 3d^10 4s^1
  Zn             1.65        [Ar] 3d^10 4s^2

$\ce{Cu}$ is more electronegative so the bonds have less ionic character which makes for stronger bonds.

The same happens in the same period between $\ce{K}$ and $\ce{Ca}$, but here the situation with regard to the electronic configuration is reverse.
Potassium salts are much more soluble than calcium salts, and again it is explained by the electronegativity:

cation    electronegativity   configuration
  K              0.82           [Ar] 4s^1
  Ca             1.0            [Ar] 4s^2


How to convert Excel comments to ChemDraw .cdx? Not sure whether this might be a Superuser or StackOverflow question... but I hope some Chemist can help me with this.
I have an Excel table with cells which contain comments "by Cambridge Soft". I assume that these comments somehow contain the ChemDraw data of the structure. It seems to be ASCII-characters only.

How can I get these comments into a .cdx file?
I tried to save these comments as a text file, renamed it to test.cdx and tried to open it with ChemDraw... no success. I don't have ChemOffice available, which possibly might handle this.
So far, I haven't found any useful information about this on the web. The CambridgeSoft pages are not very informative.
Since these are quite a few structures I don't want to do this manually, but in an automated process. Python would be available. Any hints or ideas?

Because you find these entries in a Excel workbook, it is likely that this spreadsheet was created with Chemdraw's add-in for Excel (company's demo) to embed chemical formulae:

(source).
to provide a bridge between the two worlds to display chemical libraries (promotional example, or manual chapter 13).  To use it, you need both access to Excel as well as ChemDraw.
The (part of the) string shown by you does not look like the typical content of the binary .cdx, nor the human readable .cdxml file which would be understood by openbabel, a program written in the intent to interconvert structure files used in chemistry (it may read .cdx; but read and write .cdxml) on a one-by-one basis (e.g., in a GUI), or programmatically (e.g., C, Python).
Maybe you may identify the author of the spreadsheet.  Or find support by the software's authors (for some years now by PerkinElmer) / add-in maintainers (e.g., on GitHub).  It is not obvious their format were readable by similar products (e.g., ChemAxon's JChem).


Why can a wave-function be written as a product of 1-electron wave-functions? Why can a wavefunction be written as a product of 1-electron wave-functions when making approximations to solve atoms other than hydrogen? (ie. helium)

We used this when solving multielectron systems such as that of helium. Why is this assumption valid or at least reasonable?

To start with your first sentence: "Why can a wavefunction be written as a product of 1-electron wave-functions"? Well, you can write whatever you want to approximate a wavefunction; physics police is not going to come and take you away. But the two important questions are: How useful will the results from your approach be, and how hard is it to work with mathematically? These criteria obviously reduce your number of options quite a bit.

A wavefunction that is constructed as a product of one-electron orbitals definitely has that second point going for it. As you have already experienced yourself, it is straightforward to write down and calculate with. Arguably the most important implementation of this approach is Hartree-Fock theory where the electrons never see each other as distinct point charges in space, only as fuzzy charge distributions that each one has to wade through individually. Hartree-Fock gives you fairly straightforward equations to work out the optimal way to build the independent-particle wavefunction; and while it cannot accurately capture the energetics of interactions among the electrons "out of the box", it's a clean and simple starting point to describe the quantum mechanics of your molecule. And it's often not all too bad, either - for typical well-behaved cases, Hartree-Fock and its one-electron orbitals can already land you within a few percent of the full energy value. The remainder is still way too large for it to be practically useful, but you can reuse this initial wavefunction in more refined algorithms to reduce the error further.


Could I settile for galvanized steel over stainless for salt corrosion resistance? Bear with me, I'll get to the chemistry soon.

I have a float tank that I want to attach a skimmer to.  Unfortunately, the enclosure is weakened by cutting a hole out of the ABS plastic.  The weight of the salt water (1.25SG) to somehow causes a bend in the plastic just enough that getting a good seal isn't possible without additional support.

The float tank consists primary of (pardon the units) 800lbs of $\ce{MgSO4}$ Heptahydrate raised to about 96 degrees Fahrenheit.  The pH is roughly 7.2.

My concern with supporting the ABS plastic is that the Epsom salt water may corrode the metal, if it somehow isn't protected by a liner.  Would I absolutely need to have stainless steel, or could I get by with galvanized (which is much cheaper).

In case it matters, I estimate the volume of water directly pressing on the skimmer opening to be about 9+ gallons (again at 1.25sg).

Thank you for answering non-chemist,
Ben

I'm a chemist, but this seems more like a question for an engineer.
Unfortunately I don't have some references to back up my answer.

If you use galvanized steel you have a protective layer on the outside of the steel. As soon as you handle it for attaching, welding... you destroy/scratch/bruise this top layer and the underlying steel gets in contact with the salt solution which is normally aggressive towards metal. This mostly is right for e.g. chloride salts in a mix with oxygen and I'm not quite sure how corrosive MgSO4 is.
Here is a paper about corrosive currents of steel in Saltwater/cement solutions, but that`s not my speciality: http://www.jmst.org/fileup/PDF/pass543.pdf

Stainless steel is more robust, because it's an alloy of steel and chromium. Therefor it's not just covered with a protective layer but saver as a whole. An I guess it's therefor recommended.

(As a side note, some years ago I had a field trip to a big chemistry company with lots and lots of pipelines. We had some time there with the technical stuff in charge of teh pipelines. They explained, that the most annoying pipelines were the ones with salt water, since it was more aggressive towards the pipes than acids, mineral oils, gases etc.)


Activity of hydrogen ion I am not from chemistry background. Could you tell me in simple words what the activity of hydrogen $a_{\ce{H+}}$ is? It arises in the context that the negative logarithm of the activity of hydrogen gives the pH: $\text{pH} = -\log_{10} a_{\ce{H+}}$.

Building on @Philipp and @chipbuster's answers-
What is Activity?
When we have dilute solutions or small quantities of a gas or other substance, the substance behaves as if the particles are just points, rather than having a size. In these settings, we often use a simplified set of equations.
For example, with the hydrogen ion, we use $\mathrm{pH} = -\log_{10}[\ce{H+}]$ where $[\ce{H+}]$ is the concentration, i.e. the number of particles (in moles) divided by the volume (in litres).
Once you get a lot of these particles, they bang into each other and take up room (think of billiard balls on a pool table or a room full of tennis balls flying around).
In these circumstances, we talk about ‘activity’, which is a way to correct for this. Basically, it is what the concentration or amount appears to be if we were using the simplified laws (the effective concentration). For the hydrogen ion, it is also what the concentration appears to be to physiological systems. In other words, it uses a correction factor for the simplified laws.
Activity, Concentration and Activity Coefficient
The thermodynamic definition of activity is:

$a_{i} = \operatorname{e}^{\frac{\mu_i - \mu_i^⦵ }{RT}}$

Which in low partial pressures and dilute solutions is simplified to:

$a = \gamma\cdot\frac{c}{c^⦵}$ for a solution
$a = \gamma\cdot\frac{p}{p^⦵}$ for a gas

where:

$a$ is the activity
$\gamma$ is the activity coefficient (a constant number based on what the substance is, what the solvent is if there is one, the temperature and other factors)
$c$ is the concentration (for something dissolved); $p$ is the partial pressure (for a gas)
$c^⦵$ is the standard state concentration; $p^⦵$ is the standard state pressure

pH and Activity
The original version of pH as defined by Sørensen is:

$\mathrm{pH} = -\log_{10}[\ce{H+}]$ (hereafter referred to as $\mathrm{p}[\ce{H+}]$)

However, the modern definition that takes into account activity is:

$\mathrm{pH} = -\log_{10}a_{\ce{H+}}$

$\gamma$ is found experimentally. In the case of the hydrogen ion, it is found using electrochemical methods.
Activity of Hydrogen Ion in Dilute Solutions
Standard thermodynamic state for $\ce{H+}$ is considered to be 1 mol/L ($\mathrm{p}[\ce{H+}]$ 0); thus:

$a = \gamma\cdot\frac{c}{c^⦵}$ (as a dilute solution)
$a_{\ce{H+}} = \gamma\cdot\frac{c}{c^⦵} = \gamma\cdot\frac{[\ce{H+}]}{1\ \mathrm{mol/L}} = \gamma\cdot [\ce{H+}]$

Also, in dilute solutions, activity can be approximated/calculated using an extension of the Debye-Hückel equation, the Güntelberg equation, which applies to concentrations of < 0.1 mol/L:

$\log_{10} \gamma_i=\frac{-Az_{i}^{2}\sqrt{I}}{1+\sqrt{I}}$

where:

$A$ and $B$ are constants that depend on solvent and temperature (0.51 and $0.33\times10^8$ respectively for water at 25 °C)
$z$ is the charge of the ion (1 for $H^+$)
$I$ is concentration of the ion

For $\ce{H+}$ in a solution of water at 25 °C this would be:

$\log_{10}\gamma=\frac{-0.51\sqrt {[\ce{H+}]}}{1+\sqrt {[\ce{H+}]}}$

Substituting, we get:

$a_{\ce{H+}}=\gamma[\ce{H+}]$
$\log_{10}a_{\ce{H+}}=\log_{10}[\ce{H+}]+\frac{-0.51\sqrt {[\ce{H+}]}}{1+\sqrt {[\ce{H+}]}}$
$-\log_{10}a_{\ce{H+}}=-\log_{10}[\ce{H+}]+\frac{0.51\sqrt {[\ce{H+}]}}{1+\sqrt {[\ce{H+}]}}$

and so for dilute solutions only:

$\mathrm{pH} = \mathrm{p}[\ce{H+}]+\frac{0.51\sqrt {[\ce{H+}]}}{1+\sqrt {[\ce{H+}]}}$


Calculating the UV/Vis spectrum of perylene I recently started to do computational chemistry and am now trying to write some basic inputs for important calculations. One, which I guess is quite important to me would be to predict UV/Vis spectra of molecules. So I looked for a basic input file on the net, and although my tutor warned me about using DFT for absorption spectra (it's not a topic of the lecture and therefore he wasn't sure what to use either) the first few simple molecules, mostly long conjugated systems and a few azo-compounds gave really good solutions in comparison to literature values.

Now I wanted to predict a perylene in benzene as solvent as well and gave the structure optimization (HF/VDZ) six cycles, which took forever to calculate before I stopped it. Then I used the geometry at each of the six steps to predict the spectrum and compared the results. What surprises me is that the optimization, while becoming better, moves the first absorption peak at ~435 nm further from experiment. From its position and the direction, it seems to better describe the second peak (~410 nm) instead of the first one.

I am also only getting one peak only but for the other molecules, it had been pretty much the same as well, where one peak was quite big and the rest, although being at the right positions were too small to really see.

Now I know about the quantum mechanics and the theoretical backgrounds of UV/Vis spectroscopy but we never really discussed the effects in real spectra. I know from NMR that peaks can split and multiply in experiment, while your theoretical model only contains one signal.

Is there a reason, perhaps due to the big delocalized system why perhaps the bands shift or multiply in some way or why I am not able to see the first band? Or perhaps my calculation shows the first band and it's really far off?

All UV/Vis calculations were done with Orca using RI-BP86/SV(P) and COSMO for the solven.

Here is the experimental spectrum I am referring to:

Please note: Those are my very first calculations and I am still trying to understand when the simple approaches yield good results and when I need to switch to different approaches.

I won't comment on the accuracy or otherwise of you calculation but that does not seem to a problem just now.

In the spectrum you show and in similar ones, the electronic transition energy is at approximately the mid-point of the first absorption and emission peaks, at approx 22800 cm$^{-1}$ in your figure. The different bands you see are due to vibrational transitions from various normal modes 'on top', as it were, of the electronic 0-0 transition. The energy gaps between peaks in the absorption spectrum measure the vibrational frequency in the excited state and vice versa for the fluorescence spectrum. Look up a Jablonski diagram to see generic sketches vibrational levels and transitions.

The number and magnitude of the vibrational transitions depends on the shape of the potential energy surface and is governed by the Franck-Condon factors which are $|\int \psi_e\psi_g|^2$ between the ground (g) and excited state(e).

The width of the lines in solution spectra (such as the one you show) is partly also due to rotational lines  but mainly due to collisions with solvent shifting and broadening energy levels. Consequently if you look up gas/vapour phase spectra at high resolution, such as for benzene as there a many published spectra, you will see many narrow lines corresponding to different vibrational normal modes and if resolution is high enough, even rotational lines.


Cycloalkanes and saturation I thought molecules that have a ring have an added degree of unsaturation, but this explanation has confused me:

How can something like cyclohexane which has degree of unsaturation of 1, be considered to be "saturated"? Are there different meanings of saturation?

Saturated is an old term in organic chemistry, dating back when the atomic structures and electrons did not exist. The meaning of saturated, even today is "chemically combined to the maximum extent, having no ability to combine further." The carbon atoms in cycloalkane are saturated in the sense that all the four valencies of each carbon atom are satisfied by four individual atoms.

The degree of unsaturation is a related concept: It tells us how many rings or double bonds are in the molecule. Don't mix these two terms.


How to calculate the heat of dissolution from a calorimeter experiment? A $\pu{1.50 g}$ sample of $\ce{KCl}$ is added to $\pu{35.0 g}$ $\ce{H2O}$ in a styrofoam cup and stirred until dissolved. The temperature of the solution drops from $24.8$ to $\pu{22.4 ^\circ C}$. Assume that the specific heat and density of the resulting solution are equal to those of water, $\pu{4.18 J g-1 ^\circ C-1}$ and $\pu{1.00 g mL-1}$, respectively, and assume that no heat is lost to the calorimeter itself, nor to the surroundings.
$$\ce{KCl (s) + H2O (l) -> KCl (aq)} \qquad \Delta H = ?$$
a) (2 points) Is the reaction endothermic or exothermic (circle the correct answer)?

Endothermic

b) (4 points) What is the heat of solution of $\ce{KCl}$ expressed in kilojoules per mole of $\ce{KCl}$?

$$q_\mathrm{rxn} = -q_\mathrm{cal}$$
I multiplied the sample $\pu{1.50 g}$ by $\pu{4.18 J} \cdot (-2.4) = \pu{-15.048 J}$
Divided that by $1000 = -0.015048$; thus, $0.015048$
However, my answer seems to be wrong. I know that reaction is endothermic since the temp drops, but I am wondering which values I should be using to correctly determine the "Heat of Solution".

You have multiplied the mass of the sample, 1.50g, by temperature change and heat capacity.

However, the water provides most of the heat for the reaction.

The total mass of the solution is 1.50g + 35.0g = 36.5g.

You should be multiplying  36.5g by the temperature change and heat capacity.

Then, you need to consider how many moles 1.50g KCl is.
Divide the change in enthalpy of the solution by the number of moles of KCl to determine the molar heat of solution of KCl.


Finding the final temperature If $125$ Calories of heat is applied to $60.0\text{g}$ piece of copper at $21.0^\circ\text{C}$, what will the final temperature be? The specific head of copper is $0.0920 \text{ cal/gC}$.

All I have to answer this question is a formula:

$SH = \text{heat/(mass}\times \Delta T)$

The energy amount used for heating up a material can be calculated according to the following equation:
$$ E = c_\text{s}\, m\, \Delta T$$
where $E$ is the energy, $c_\text{s}$ the specific heat capacity, $m$ the mass of the object and $\Delta T$ the temperature difference that is to be achieved.

Solving the above equation for $\Delta T$ we get:
$$ \Delta T = \frac{E}{c_\text{s}\, m}$$

Plugging in the values provided by you, I get $\Delta T = 22.6\;\mathrm{K}$ (or °C, since we're talking about the difference and the scale is the same, it doesn't matter).

Adding this increase in temperature to the initial temperature gives the final answer:
$$ \vartheta_\text{f} = \vartheta_\text{i} + \Delta \vartheta = 21.0\; \mathrm{°C} + 22.6\; \mathrm{°C} = 43.6\; \mathrm{°C}$$

Note the following:

Number of significant digits: Since the data given consistently had three significant digits, I've rounded to the third digit.
Symbol for temperature: I used $\vartheta$ for temperatures in °C and $T$ for temperatures in K. I strongly suggest you follow this example as to never confuse the two.


Non-hardening adhesive I am wondering if any of you know about glue/adhesives that do not harden over time after they are applied or glue/adhesives that take a really long time to harden after application. I need the glue to temporarily hold polymer sheets before they are sewed together.

I think you could use some types of Butylrubber. If I'm not wrong from Isobutylene is derived the Post-it adhesive. However you can find Butyl tape easily in every big hardware store (especially nautical) or online.
EDIT
At this link you can find some more information this is an excerpt:

Tapes and flooring adhesives  Exxon butyl rubber can be used in
adhesive formulations for transparent tape, hot-melt pressure
sensitive adhesives, mastic for pipe wrap tape and vinyl floor tile
adhesives. It can also be used for semiconducting electrical splicing
tape. A rubber splicing tape must be self-fusing; i.e., when wrapped
upon itself, it must fuse to the underlying layer, forming a permanent
bond. The self-fusing is a result of compounding and eliminates the
need for application of a separate adhesive coat. These compounds have
very high tensile strengths and exhibit some degree of recovery or
elasticity. Electrical splicing tapes based on Exxon butyl rubber have
moisture resistance, outstanding aging resistance and good electrical
properties. Polyisobutylene elastomers have long been successfully
used in pressure sensitive adhesives both as primary elastomers and as
tackifiers and modifiers. Exxon chlorobutyl rubber can be used alone
or in blends with other polymers to achieve special properties in
pressure sensitive adhesives.


Confusion between various formulas of Gibbs free energy According to the assumptions of Constant P and T and no non mechanical work, we get
ΔG=ΔH-TΔS
But there is another form of the Gibbs free energy equation which takes Reversible process and no non mechanical work as assumption
dG=Vdp-SdT
If I Use the assumptions constant Pressure and Temperature, according to the above equation we get dG=0,which contradicts the first equation.
What is the mistake I am making? Also, when ΔGsystem=-TΔStotal

The Gibbs free energy of a multicomponent mixture is a function not only of temperature and pressure, but also of the number of moles of the various species in the mixture:  $$G=G(T, P, n_1, n_2. m_3. ...)$$So, $$dG=\frac{\partial G}{\partial T}dT+\frac{\partial G}{\partial P}dP+\frac{\partial G}{\partial n_1}dn_1+\frac{\partial G}{\partial n_2}dn_2+\frac{\partial G}{\partial n_3}dn_3+...$$or
$$dG=-SdT+VdP+\mu_1dn_1+\mu_2dn_2+\mu_3dn_3+...$$ where the $\mu's$ are the partial derivatives of the Gibbs free energy with respect to the number of moles of each of the species, and are referred to as the so-called Chemical Potentials of the various species in the solution.  The numbers of moles of the species in the mixture change as the reaction proceeds, and that causes the Gibbs free energy to change.


Configuration of a water molecule? Water molecules are described as an oxygen atom covalently (ignoring for the monent protonization/deprotonization) bonded to two hydrogen atoms with an angle of ~104° between the hydrogen atoms (like Mickey Mouse). I can understand this with multiple water molecules (the stacking would - for want of better descriptive powers - surround the oxygen's relative negative charge.)

Would a single water molecule retain this configuration? It would seem that in space, it should look like CO2?

The HOH bond angle for an isolated water molecule is accurately known to be 104.5°  (104.52°±0.05° from Rotation‐Vibration Spectra of Deuterated Water Vapor J. Chem. Phys. 24, 1139).

In the liquid phase, values of 105.5° (calculated) and 106° (experimental) are reported in Structural, electronic, and bonding properties of liquid water from first principles J. Chem. Phys. 111, 3572.

The gas phase water dimer (two otherwise isolated water molecules hydrogen bonded to each other) has also been considered and an angle of 104.7° is reported in the above reference.

For ice Ih (common form of ice), values ranging from 106° to 108° have been reported.

Attempting to explain the trend, the intermolecular hydrogen bonding in the liquid and solid phases weakens the intramolecular covalent O-H bonds, and from a VSPER approach causes the electrons of the O-H bond to be more like the lone pair electrons.


What is the reason behind the numbering of nitrogen bases? I am studying nitrogen bases and their structures and roles in nucleotide, polynucleotides and nucleic acids formation. In the structures of the nitrogen bases like Adenine, Guanine, Cytosine and Thymine, the numbering is baffling me. Why are the atoms in these molecules numbered so? Can someone please specify the rules for numbering these four molecules? I am a biologist and I'm not so good at chemistry.

ACDLabs maintains a nice summary of the IUPAC nomenclature rules with examples:

Heterocyclics (pyrimidine): http://www.acdlabs.com/iupac/nomenclature/79/r79_702.htm

Fused heterocycles (purine): http://www.acdlabs.com/iupac/nomenclature/79/r79_960.htm

Briefly, apply the following rules:

Number your rings so that the nitrogens end up with the lowest number combination. Thus pyrimidines have (1,3).
Number your rings so that other functional group have lowest possible numbers. Hence pyrimidine is numbered to the exocyclic carbonyl groups are (2,4).
In a fused system, numbering should prefer (in this order): ring with more nitrogens, rings with other heteroatoms, larger rings, nitrogen atom closer to ring junction.
Number toward the farther ring junction from where you started. Do not number junction positions.

Number 3 is fine and all, but purine seems to have a numbering scheme that does not match these rules, and it is supported by IUPAC. Likely, this represents a historical numbering pattern that predates the IUPAC rules. Everyone was using it and so it was adopted. You will likely just need to memorize it (or look it up when you need to).


Is Bromide anion acting as a base in reaction with ethylamine and methylbromide? What is happening in Step 2 and would you please give the reaction mechanism for it as well?

The bromide anion is acting as a counter anion to balance out the ammonium charge in step 1.

Given sufficient heat, step 2 can occur, with bromide acting as a weak base.  This is likely to produce gaseous produces which would drive the equilibrium as the gasses escape.

Mechanistically, this is possible, but synthetically, it is more likely another base would be added in a typical synthesis.


Is it possible to obtain the amount of air required for a combustion analysis? I'm not very sure if the required amount of oxygen (I assume) can be found by only knowing the amounts of the produced subtances after a combustion.

To illustrate this I found this problem:

  $1.14\,kg$ of octane $C_{8}H_{18}$ is burned with a certain amount of
  air. The resulting products from the combustion are known to have the
  following percentages by volume: $CO_{2} = 41.18\%$ ; $CO = 5.8%$ ;
  $H_2O_{(vapor)} = 52.94\%$. Find the weight of required air in
  kilograms. Assume the composition of the air in this combustion is
  $21\%$ of $O_{2}$ and $79\%$ of $N_{2}$ by volume and its average
  molecular weight is $28.8\,\frac{g}{mol}$.

The alternatives given are as follows:

$\begin{array}{ll}
1.&9.0\,kg\\
2.&16.4\,kg\\
3.&12.5\,kg\\
4.&45.1\,kg\\
5.&120\,kg\\
\end{array}$

What I think to approach this problem was to use the given percentages to get the number of moles using the initial moles of octane.

Using these moles I can obtain the grams of oxygen produced. Since it is given the percentage which is of oxygen in the air I could use this to find the required mass of the air to get that amount of air.

This is shown as follows: (for the sake of brevity I'm ommiting units but they are consistent.)

$\textrm{FW of octane} = 114$

$n_octane=\frac{1140}{114}=10$

Then moles of each product:

For $CO_{2} = 41.18\%$

$\frac{41.18}{100}\times 10 = 4.118$

Then in this amount of $CO_{2}$ there must be these grams of oxygen:

$4.118 \times \frac{32\,g\O}{1\,mol\,CO_{2}}= 131.776 \,g$

For $CO = 5.8\%$

$\frac{5.8}{100}\times 10 = 0.058$

$0.058 \times \frac{16\,g\O}{1\,mol\,CO}= 0.928 \,g$

For $H_2O_{(vapor)} = 52.94\%$.

$\frac{52.94}{100}\times 10 = 5.294$

$5.294 \times \frac{16\,g\O}{1\,mol\,H_2O}= 84.704 \,g$

Adding these together:

$131.776+0.928+84.704=217.408\,g\,O$

Then using the known percentage of oxygen which is $21\%$

$\frac{21}{100}x=217.408$

$x=1035.28 \,g$

Which is approximately 1.035 kg. but this answer doesn't get any close to the specified alternatives. Did I overlooked something. Or could it be that this approach is not applicable?. In this given situation is it possible to find the required mass of air?.

For the given problem there isn't much $\ce{CO}$ created. Given that the question is multiple choice an exact amount isn't needed. The problem only needs to be solve close enough to determine which choice is the answer. So assume complete combustion and then guess that the answer will be about 5% high. If octane, $\ce{C8H18}$, is completely burned with oxygen then the reaction is:

$$\ce{C8H18 + 12.5O2 -> 8CO2 + 9H2O}$$

since the problem states that 1.14 kg of octane was burned, the number of moles is:

$$\dfrac{1140}{114} = 10.0$$

Thus for complete oxidation:

125 moles of oxygen would be needed.
Air is only 21% oxygen so 125/0.21 = 595 "moles" of air are needed.
At 28.8 g/"mole" that would be 17.1 kg of air.

In reality there is about 5% less oxygen required since there is some $\ce{CO}$ produced, so $17.1*0.95 = 16.3\ \pu{ kg}$ is a good guess.

I'll leave finding the exact solution to the OP.


The effect of hyperconjugation on the stability of alkenes with MO theory Hyperconjugation stabilizes carbocations and that makes sense because electrons are given to the empty p orbital.

But how does it stabilize alkenes? Can molecular orbital theory be used to explain it?

Since you are familiar with how carbocations are stabilized via hyperconjugation I will keep this brief: With alkenes, it really is the same principle at work. Instead of the empty p orbital, you have to consider the empty $\pi^{*}$ orbitals of the $\ce{C=C}$ double bond. They interact with the filled high-lying neighboring $\ce{C-H}$ or $\ce{C-C}$ $\ce{\sigma}$ bonds in $\pi$-bond-like manner

and so from those two fragment orbitals you will get two molecular orbitals: one from the in-phase combination of $\pi^{*} (\ce{C=C})$ with $\sigma (\ce{C-X})$, which will be lower in energy then the fragment orbitals, and one from the out-of-phase combination of $\pi^{*} (\ce{C=C})$ with $\sigma (\ce{C-X})$, which will be higher in energy then the fragment orbitals. Since in this interaction there are only two electrons involved, i.e. the ones in $\sigma (\ce{C-X})$, you get a stabilization from the interaction since those two electrons will now occupy the in-phase MO and have thus lowered their energy. The out-of-phase MO is empty and doesn't contribute to the energy of the molecule.


Identifying nucleophiles and electrophiles in an acid-base reaction Is ammonia a nucleophile and hydrogen the electrophile in the Brønsted-Lowry acid-base reaction of acetic acid with ammonia?
$$\ce{CH3COOH + NH3 <=> CH3COO- + NH4+}$$

I say that ammonia is the nucleophile because I drew a curved arrow from the lone pair on the ammonia to the ionizable hydrogen on acetic acid, and I drew a curved arrow from the bonding pair in the hydroxyl group to the oxygen in the hydroxyl group of acetic acid. It seems that ammonia is the nucleophile because it is attacking the hydrogen proton (a bare proton) and that hydrogen is the electrophile because it likes the lone pair on ammonia.
So if the above is right, this would make ammonia a Lewis base and the hydrogen proton a Lewis acid? Or do we say the entire acetic acid molecule is a Lewis acid?

In my opinion, your statements are both right, this is the IUPAC definition [1]:

Lewis acid: A molecular entity (and the corresponding chemical species) that is an electron-pair acceptor and therefore able to react with a Lewis base to form a Lewis adduct, by sharing the electron pair furnished by the Lewis base. For example:
$$\ce{\underset{Lewis acid}{Me3B} + \underset{Lewis base}{:\!NH_\phantom{3}} -> \underset{Lewis adduct}{Me3\overset{-}{B}-\overset{+}{N}H3}}$$

You can see that the author says molecular entity (atom, molecule, ion, ion pair, radical, radical ion, complex, conformer etc., identifiable as a separately distinguishable entity [1]) and the corresponding chemical species (a boarded concept, that in this case I interpret as the combination of different molecular entity), so I think that the exegesis of this definition is that both the acetic acid and the hydron can be called a Lewis acid. And in fact is what usually all the chemists use.
Reference

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.


Why does water volume decrease when salt is added? Why does water volume decrease when salt is added? Our teacher asks us in the class but I don't find any strong reason.

What you are experiencing is the phenomenon of partial molal properties - components of a mixture contribute unequally to the properties of the mixture.

Sometimes the components of a mixture do not contribute linearly to the properties of a mixture. A well-known example is the water-ethanol system, which has negative excess volume. When 5 mL of water and 95 mL of ethanol are mixed, the final volume is less than 100 mL. See the following graph constructed by Wikipedia user Wilfried Cordes:

Similar behavior is occurring in your sodium chloride and water system. Now, as for why these phenomena occur, a lot of it has to do with intermolecular forces. For example, you could consider that the hydrogen-bonding and dipole-dipole interactions between water and ethanol are perhaps stronger than between water and water or ethanol and ethanol. Additionally, water molecules are smaller than ethanol molecules. There is also a thermodynamic consideration. This state of affairs must represent minimum free energy (maximum entropy).

How can you relate the interactions of sodium and chloride ions with water to the observation of negative excess volume?


Comparing O−O bond length in hydrogen peroxide and dioxide(2−) ion Problem

How can we qualitatively compare the bond lengths between $\ce{H2O2}$ and $\ce{O2^2-}?$

Answer

 Bond length in $\ce{O2^2-}$ is slightly larger than in $\ce{H2O2}.$

Attempt
I tried using molecular orbital theory, but since both have peroxide $\ce{O2^2-},$ the differentiating factor must be some sort of repulsion, but I'm not getting it.

The way to tackle this is to look at the lone pair repulsions between the 2 molecules. It is known that $\ce{N-N};$ $\ce{O-O};$ $\ce{F-F}$ single bonds are quite unstable due to lone pair (lp) repulsions.
An evidence of this is that $\ce{N}$ exists in molecular state as $\ce{N2}$ using multiple bonds, so that it's lone pairs do not repel each other, however $\ce{P}$ doesn't do that. It is capable of forming $\ce{P4}$ because in the third period the size of the atoms are large enough to properly distance the lone pairs.
One more evidence of this act is that among diatomic halogen molecules the bond length is in the order $\ce{Cl2} < \ce{Br2} < \ce{F2} < \ce{I2}$ the bond length of $\ce{F2}$ should have been lower (2p–2p overlapping is much stronger hence the bond should have been the shortest). However, due to lp–lp repulsions it is now much longer bond.
With this we can now see that in $\ce{H2O2}$ there is only two lone pairs on each oxygen atom while in $\ce{O2^2-}$ it has three lone pairs on each atom, hence it has increased lp–lp interaction.
As a result of this $\ce{O2^2-}$ tends to elongate (relatively more than $\ce{H2O2})$ its bond to decrease this intense lp–lp interaction, hence the longer bond.


What's the explanation for the chelate effect? I got that multidentate ligands form more stable coordination complexes than monodentate ligands, but why?

The chelate effect is generally agreed to be a thermodynamic effect caused by the change in entropy upon binding of a bidentate ligand.

General description of the chelate effect:

Consider a coordination complex,  [M]L2, where L = a standard monodentate ligand. If we now add a bidentate ligand, two monodentate ligands are released.

The ligand substitution reaction in the forward direction is therefore generally favourable:  ∆S, the entropy term is positive due to two molecules reacting to become three, with this favourable entropy term having the effect of tending to make ∆G negative (negative ∆G is favourable, i.e. the reaction likely to proceed).[*]

Once a bidentate ligand is bound, it then becomes less favourable to go in the other direction (the entropy term would be negative due to entropy decreasing in the system, which in turn would tend to make ∆G positive, and hence the reaction unfavourable).

Example and thermodynamic data:

To give a more concrete example demonstrating the significance of the entropy term, consider the following reaction and associated thermodynamic data (taken from Physical Inorganic Chemistry, S. F. A. Kettle, 1996):

Clearly, the entropy term dominates, causing ∆G to be negative in both cases (hence causing the forward reaction to be favoured in both cases).

[*]: The caveat here is that we're assuming that enthalpy really isn't contributing a whole lot, this is clearly a vast over-simplification, and there are cases where enthalpy becomes important (i.e. when the complex is significantly more stable than the starting materials). We're also ignoring the fact that the act of binding a bidentate ligand itself has some unfavourable entropy, since we're containing the ligand and removing some conformational flexibility


Why does Area of Interface depends on size of particles? The interface is usually a few molecules thick but its area depends on the size of the particles of the bulk phases.

*What is  Area of interface? What is its sense and How is it defined.I mean if we have two bulk faces with significant difference in particle size,and we define surface area for one layer then it is surely not the same as that for the other layer then how that area of interface make sense.

One of the explanation (or i should rather say an example) i found on internet is oil dressing on salads ,where interface is spherical having less surface area.

Now that example is itself confusion because it is mathematically proved that for any given volume sphere has least surface area, then how is that area being less attributed to particle size?

Area of interface is the area of contact between two phases. If one phase is dispersed in another one, it then consists in the sum of the surface for each particle/droplet.

If you consider a spherical particle, its surface is $S=4\pi r^2$ and its volume $V=\frac{4}{3}\pi r^3$. Now consider the ratio $\frac{S}{V}=\frac{3}{r}$, so when the size of a sphere (but it is true for any shape) diminishes, the ratio $\frac{S}{V}$ increases. In other words, if you take an initial volume that you divide in small spheres, the global volume does not change (the sum of all spheres) but the surface increases.


Clemmensen reduction of 1,3 diketones As part of a synthesis I was considering the following reaction:

How selective is the Clemmensen reduction? I know that this reduction favours ketones adjacent to benzene rings but will it also reduce the other ketone, and/or affect any other parts of the molecule. Are there any other methods that would achieve this transformation more effectively.

A Clemmensen reduction of a 1,3-diketone probably isn't the best idea. The reduction is likely to proceed via a ketyl radical.

In the case of a 1,3-diketone, formation of a dihydroxy cyclopropane, which undergoes opening/rearrangement cannot be excluded.

This reaction has for instance been observed for dimedone, where 2,4,4-trimethyl cyclopentanone was formed.

EDIT 1

Which other viable procedures exist?

If you insist on using the given starting material, the following might (!) work:

regioselective (asymmetric) catalytic hydrogenation of the acetophenone-type carbonyl using a 1,2-diamine-substituted Noyori catalyst
Barton-McCombie reaction of the resulting alcohol

If you are rather interested in the product, you might want to rethink the strategy:

The intended product is an arylbutanone, related to the raspberry ketone. A possible retrosynthesis would be:

If you get your hands on the necessary benzaldehyde, an aldol reaction with acetone, followed by catalytic hydrogenation might give the right product.

EDIT 2

5-tert.-butyl-2-hydroxybenzaldehyde is commerically available, but not really cheap.

However, 4-tert.-butylphenol is cheap as dirt. In case you need significant quantities of your final product, you might want to consider the synthesis of the benzaldehyde. The substitution pattern cries for a Reimer-Tiemann reaction.


Difference between radial, planar, angular and spherical nodes What is the difference between them?

I think radial nodes and spherical nodes are the same, and angular and planar nodes are the same.

Reference

Finally, how many spherical nodes are there in a $2p$ orbital?

The wave function $\Psi(r,\theta,\phi)$ of a one-electron (hydrogen-like) atom is seperable as the product of a radial function $R(r)$ and an angular function $Y(\theta,\phi)$

$\Psi(r,\theta,\phi) =  R(r)Y(\theta,\phi)$

If $R(r_1) = 0$, there exists a radial node.  The radial node is a sphere with radius $r_1$. Therefore the terms "radial node" and "spherical node" are the same.

$Y(\theta,\phi)$ is further seperable as

$Y(\theta,\phi) = P(\theta)F(\phi)$

If either $P(\theta)$ or $F(\phi)$ is zero for a given respective angle value, there is an angular node.  However, a angular node is not necessarily a planar node.  An angular node could be a planar node or a conical node.

$F(\phi)$ being zero corresponds to a planar node.

$P(\theta)$ being zero corresponds to either a conical node or a planar node (some think of the planar case as a specical case of conical, with apex angle being 180 degrees)

Overall, there will be $n-1$ nodes.

$l$ nodes will be angular

$n-l-1$ nodes will be radial (spherical)


Is sand a mixture or a compound? I was given an assignment to find out if sand is a mixture or a compound I'm sure it can't be an element. Considering the fact that a mixture contains two or more substances that are not chemically combined and not proportionally fixed and a compound is contained substances that are chemical combined and are fixed proportionally. Sand contains a lot of divided small rocks and mineral particles.

Sand is a mixture. Sand is classified as a heterogeneous mixture because it does not have the same properties, composition and appearance throughout the mixture. A homogeneous mixture has a uniform mix throughout.


The colour of fireworks When the electrons that are excited by the energy of burning drop back into lower energy levels they emit light at the wavelength of colour.

When metals are burnt, metal oxides(some coloured) are formed.

When metals are burnt under high temperature, will they turn to plasma state? If so, should the metal ions be responsible for the colours and sparks?

Which one is the reason for the beautiful colours of fireworks?

The colours of fireworks are caused by the electrons emitting energy when dropping orbital levels. Metal oxides are solids, mainly dull colours (even the yellows and reds are not in the same class as firework colours), and could not cause bright light in the air. The temperature of fireworks is far below that needed to form a plasma.


What is the mechanism for thermal decarboxylation of a generic carboxylic acid (not special cases like beta-ketocarboxylic acid or malonic acids)? I am looking for the mechanism for thermal decarboxylation for any RCOOH, and I am guessing that its possible for the thermal energy to cleave the R-C bond homolytically leaving a R• and •COOH. Then I would guess the R• would add its electron to the H and the O-H bond would undergo homolysis as well leaving the single electrons on the C and now O to form a bond leaving RH and CO2. However I have no idea if this is what actually occurs. Thanks

Under sufficiently alkaline conditions, i.e. when the acid is deprotonated, the carboxylate can undergo an electron transfer reaction with a suitable partner (oxidant). In the course of this process, the carboxylate is oxidized to an acyloxy radical, which subsequently fragments to yield an alkyl (or alkylaryl) radical and carbon dioxide.

$$\ce{R-COO- ->[-e^-] R-COO\cdot -> R\cdot + CO2}$$

A photochemical variant of this reaction was intensively examined in the group of Axel Griesbeck at the University of Cologne, Germany. For further reading, have a look at

Acc. Chem. Res., 2007, 40, 128-140 (DOI)
J. Phys. Chem. A, 2006, 110, 3356-3363 (DOI)
Synlett, 2004, 2347-2350  (DOI)


Receiving milk for working in harmful environment I just got a job in a chemical/biological lab (post-Soviet Country). Chemists here receive milk for "harmfulness of the work" (i .e. their work involves working with all kind of substances/reactives). I'm not a chemist; I do their analysis. The tradition of receiving 0.5 liters of milk a day comes all the way from the Soviet Era. I found this question here: Does milk drinking prevent long-term chemical poisoning?

However, I would like to know if milk is not that great for preventing chemical poisoning (as seen from the accepted answer), why is there still such a tradition of giving milk? And why did this tradition appear in the first place? And do people working in chemistry labs in other countries also get milk? (I'm not sure, but I think in UK they do.)

I know that in India there was a very similar law extant at least back till the 1990's when I had encountered it. Not sure if it is still on the books.

The association with the USSR makes sense: during the cold war a whole bunch of Indian factories were constructed with Soviet aid. And the conventions, technology, etc. came along.

I had heard of several alternative explanations for the milk-law; not sure which (if any) were true. Maybe partly all were.

The Vitamin D or Calcium in milk competing against absorption of Lead or some other nasty element, I forget which the workers were exposed to in chemical factories.
The Milk just contributing to the general health of factory workers in an otherwise grossly malnourished nation (remember, this is cold war era India). Presumably due to the attendant health hazards of chemical factory work, keeping a worker there healthy was more crucial.
Just having the stomach full of a competing innocuous substance (e.g. milk) reduced the propensity of competitive absorption of other nasties in accidents, fugitive exposures etc.


NMR spectrum of plasma treated glucopyranoside This is a 1H NMR spectrum of plasma treated methyl alpha D-glucopyranoside.

Now I'm having problems trying to read the spectra and interpreting if there are any changes to it. What I suspected would happen during plasma treatment is that the methoxy group would break off, leaving it or replacing it with any other possible ROS or RNS. I don't know how to identify the methoxy groups present or to read those peaks to see if there are other possible structural changes.

The usual approach to determining change by NMR following a reaction or procedure is to compare your final spectrum to your starting sample's spectrum, run with identical parameters. That is, look at the spectrum you ran before your plasma treatment and look for changes in peak position and intensity. Within Topspin (which your spectrum here has been processed with) this is very easy to do; you can run a dual display and overlay your spectra on the screen, and even compute a difference spectrum.

Of course, if you are having problems assigning the final spectrum, you should again go back to the spectrum of your starting material and compare to the assignments you made for that. That will make it easy to identify which parts of the molecule have undergone some change. Of course, you might expect that you haven't had complete clean conversion, and there may be a number of products constituting only partial conversion. Otherwise, you are potentially having to characterise a completely new molecule.

You do have the spectrum for your starting material, don't you?


Molecular chirality and optical rotation Why does having molecular chirality result in optical rotation? The dissymetry or chirality of molecules translates to the rotation of plane polarized light, the magnitude and direction depending on the concentration and the nature of the substance. But why does molecular chirality cause the rotation of plane polarized light.
If this rotation is due to the assymetry around the bond, the different molecules are oriented in all possible directions(randomly) and hence the light will fall on them in all different orientations, and should ultimately be undeflected since for every orientation of the molecule, we can reverse the orientation such that the light appears to be falling on the molecule from a direction other than the one for our original molecule.
Moreover, why does inverting the configuration invariably reverses the direction of rotation of light?

since for every orientation of the molecule, we can reverse the orientation such that the light appears to be falling on the molecule from a direction other than the one for our original molecule.

This is false.

Let's take 2-butanol. For this stereoisomer, light is turning clockwise when viewed from the right side (I'm not sure of this, but we can assume).

Here, the disks are there to clarify the direction from which we are viewing the entire setup. Here, the "face" of the disks is to the right so we are viewing it from the right.

By the principle of reversibility of light, if light enters from the other side, we get the reverse path:

Rotate this diagram (by $180^\circ$ around an axis perpendicular to your screen):

Note that the disks are being viewed from the other side (the left) now. But we can flip them:

Take a close look at this. This is the same molecule as the first, just oriented backwards. And looking from the right, it still gives a clockwise twist to the angle.

So while rays of light incident on different orientations of a molecule may differ in the exact value of the angle of rotation, rays of light incident on a reversed (not stereoisomeric, just spatially rotated) molecule have the same direction and value of the angle of rotation and do not cancel.

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