Can the structure of an azide ion contain a coordinate bond? Can the structure of an azide ion ($\ce{N3-}$) be $\ce{N#N\bond{->}N^-}$?

I know the actual structure but was asking whether the single bond can be shifted to a N to make it a triple bond and a coordinate bond with other Nitrogen.

This is not a duplicate of Hybridization in azide ion? because I am asking whether it can have a coordinate bond or not while the other question asks what the hybridization of centre nitrogen in $\ce{N3-}$ is.

You can’t really tell a dative bond from a ‘normal’ covalent bond until you break the bond. A dative bond will then dissociate heterolyticly while a covalent bond will dissociate homolyticly.

Just using that simplified way of putting it, you could say that there are many reactions of azides wherein $\ce{N2}$ is liberated leaving a nitrene or related structure; just one of many examples would be the Curtius rearrangement. However, that is not what is meant in the definition above; the definition talks about supplying the bond dissociation energy.

The BDE is not typically supplied in chemical reactions except in coordination compound ligand exchange and radical generation. If you were to apply it to $\ce{N3-}$, I predict homolytic dissociation. Thus, the bond between the nitrogens is more of a typical single bond.

Of course, aside from the chemical implications of dative versus covalent bond you may just have decided to use an arrow notation rather than formal charges. Technically nobody can stop you but still you shouldn’t call it a true dative bond.


Effect of Temperature on pH of Water The $K_\mathrm w$ is a function of temperature. It is $10^{-14}$ at $25\ \mathrm{^\circ C}$. When the temperature is $50\ \mathrm{^\circ C}$, the $K_\mathrm w$ can be calculated to be somewhere around $10^{-12}$ using Vant Hoff's equation, but this is so weird.

MY ATTEMPT

If we will solve for $K_\mathrm w=10^{-12}$, we get $\rm pH=6$. Will the water become acidic due to increase in temperature?

TL;DR
$\rm pH=6$ is the neutral pH at $\rm 50\,^{\circ} C.$
Long Version
Actually there is nothing wrong with your calculations. It seems that you are confused about the definition of a solution being acidic. If something is acidic, this means that the concentration of $\ce{H+}$ ions (technically $\ce{H3O+}$ ions) is greater than the concentration of $\ce{OH-}$ ions. At room temperature (25 degrees Celsius) a solution is neutral if its pH is 7. This means that for a solution to acidic, its pH must be lower than 7 at 25 degrees.
When we increase the temperature to 50 degrees, the pH of a neutral solution is actually 6, not 7. You can check this as if you were find the total concentration of $\ce{H+}$ ions and $\ce{OH-}$ ions, you will find that they both equal $10^{-6}$. Since there concentration is equal, the solution must be neutral. So for a solution to be acidic at a temperature of 50 degrees, its pH must be lower than 6, not 7.
Explanation
Now you might wonder what the reason for this. Consider the following equation which is the auto-disassociation of water $$\ce{2H2O + heat \leftrightharpoons H3O+ + OH-}$$ As you can see, this reaction is actually endothermic. This means that when you increase the temperature, due to Le Chatelier's principle, the equilibrium will shift to the right; hence the concentration of $\ce{H3O+}$ and $\ce{OH-}$ ions will increase. This is indicated by the increase in the $\rm K_w$ value, as more products are formed.
Therefore since there is a greater of concentration of $\ce{H3O+}$, a solution which is neutral will have more $\ce{H3O+}$ ions at 50 degrees than it would when it is 25 degrees. Therefore its pH will be lower than 7 despite being neutral.


How does rubber toughening work? I’m doing a research project which involves rubber toughening - the particular example I'm looking at is the toughening of SAN polymer (styrene-acrylonitrile) with polybutadiene to make ABS. What I'm looking for is a description of why the process makes the material tougher, whereas everything I can find online seems to be research about the effects of different factors on the toughness. If anyone could provide a source that would be great.

I think this Wikipedia article answers your question quite directly:

  "This important property [rubber toughening] is entirely due to the
  small rubbery domains that aid in arresting crack propagation, thereby
  increasing toughness/energy needed to break the material."

The same article gives an example using the materials sited in your post:

  "Well-known rubber toughened polymers include high impact polystyrene
  (HIPS), rubber-toughened nylon, and acrylonitrile butadiene styrene
  (ABS). In HIPS, polybutadiene domains are dispersed in a polystyrene
  matrix; in rubber-toughened nylon, typically cis-polyisoprene is
  dispersed in a Nylon matrix; in ABS, polybutadiene domains are again
  used, but for both polystyrene and acrylonitrile."


Difference between a dimethyl and ethyl substituent groups I'm teaching myself organic chemistry and am struggling with differentiating between dimethyl and ethyl substituent groups.
Example: tert-butyl (1,1-dimethylethyl)

This chapter follows the one on bond-line structures, and I realize one of the rules for making a valid bond-line structure for straight-chain alkanes involves zig-zagging carbon. But I am confused on how to represent the substituent groups - is the ethyl group the V, and the methyl groups are each of the branches from the middle carbon atom of the substituent group?

If an organic compound has dimethyl in its name than that means that it has 2 methyl groups. If they are on the same carbon they form a V in the bond line structure. An ethyl group on the other hand  forms a 2 carbon unbranched zigzag in the bond line structure instead of the V that a dimethyl group forms in the bond line structure.

For the ethyl group it is standard on all bond line structures to represent it as a 2 carbon unbranched zigzag. For the dimethyl group it is standard to represent it as a V in all bond line structures.

As to how we know this it is because an ethyl group coming off of a carbon looks like propane if you just look at that carbon and the ethyl group. In fact some people in the early years of organic chemistry might have called propane ethylmethane before the prop prefix was used for alkanes. However if this was the case some of them would call propane dimethylmethane. This was because they were looking at different carbons as the central atom. If you look from one of the ends it looks like an ethyl group sticking off of methane. If you look at the center it looks like a dimethyl group sticking off of methane.


How to calculate the remaining mass in the mother solution after multiple extractions? During an extraction, a solute ($\ce{S}$) will partition itself between the two solvent phases with a partition coefficient ($K$) defined as:

$$\ce{S(solvent 1) <=> S(solvent 2)}$$

$$K = \frac{[\ce{S}]_2}{[\ce{S}]_1}$$

Solute $\ce{S}$ is found to have a partition coefficient of 15 between solvent 1 and solvent 2; i.e., with 15 times as much in solvent 2. Suppose $\pu{150 mL}$ of a $\pu{0.02 M}$ solution of $\ce{S}$ in solvent 1 is extracted with \ce{\pu{50 mL}} of solvent 2. What fraction of $\ce{S}$ remains in solvent 1? If three $\pu{50 mL}$ extractions are performed, what fraction of $\ce{S}$ remains in solvent 1?

In old lecture notes of mine I found this formula

$ g_i = g\left(\frac{V_1}{V_1 + V_2K_2}\right)^n$

where $g_i$ is the remnant mass of a product initially dissolved in volume $V_1$, to be $n$-times extracted by volume $V_2$ with a given Nernst dissipation coefficient $K = \frac{a_2}{a_1}$.  Here, $g$ refers to the mass of the product initially dissolved in $V_1$.

Applied to your case, assuming a molar mass of 180 g/mol, this would read like 0.54 g initially dissolved in you 150 mL solution 1 (3 mmol).  After a single extraction with 50 mL of 2, only 90 mg were remaining (0.5 mmol) in $V_1$, after the third time 2.5 mg (about 14 umol).

I speculate a textbook like P. W. Atkins' Physical Chemistry may contain an example to confirm such a calculation.


Isoelectric pH of colloids? How did we derive the relation between the isoelectric pH and the dissociation constants for the amino and carboxyl groups on zwitterionic amino acids.

$$pI=\frac {pK_{a1}+pK_{a2}}{2}$$
where $$\ce{NH3^+-R2(COO^{-})<=>NH3^+-R2(COOH)-K_{a1}}$$
$$\ce{NH3^+-R2(COO^{-})<=>NH2-R2(COO^{-})-K_{a2}}$$

For more complicated cases (charged side groups) and their isoelectric pH, here is one site which just posts the equations without any derivation. Besides this, there is no reference I could find for this derivation.

Consider the equilibria in dilute aqueous solution:

$$\ce{H2A+ \rightleftharpoons HA + H+} \ \ \ \ \ \ \ K_{a_1}=\frac{[HA][H^+]}{[H_2A^+]}$$

$$\ce{HA \rightleftharpoons A^- + H+} \ \ \ \ \ \ \ K_{a_2}=\frac{[A^-][H^+]}{[HA]}$$

At the isoelectric point:

$$[H_2A^+]=[A^-]$$

Using the expressions for $K_{a_1}$ and $K_{a_2}$, we find:

$$\frac{{[HA]}[H^+]}{K_{a_1}}=\frac{K_{a_2}[HA]}{[H^+]}$$

$$[H^+]^2 = K_{a_1}K_{a_2}$$

$$pH_{isoelectric}=\frac{pK_{a_1}+pK_{a_2}}{2}$$

See also this section on Wikipedia.


Does a double bond or a triple bond have a higher priority in nomenclature? I have this structural formula:

Should it be named pent-1-en-4-yne or pent-1-yn-4-ene?

The relevant section in the 2013 IUPAC Recommendations (Blue Book) reads:

  P-31.1.1 [...] Locants as low as possible are given to multiple bonds as a set, even though this may at times give 'yne' endings lower locants than 'ene' endings. If a choice remains, preference for low locants is given to the double bonds. [...]

In this case, the two possible numbering schemes lead to the same locant set of (1,4).

However, according to the italicised portion of the quoted text, the double bond is considered to have the higher priority, and the name would correctly be pent-1-en-4-yne.

Note that the final "-e" from "-ene" is dropped if the following suffix begins with a vowel or "y," and is retained if the suffix begins with a
consonant. Additionally, spaces are only used in very specific cases, like in the carboxylic acid ending "-oic acid." Otherwise, dashes are used to separate different functional groups and numbers in the name.


What determines whether a step in a reaction is reversible? What makes a step in a reaction not reversible? I'm learning all these mechanisms and having to memorize which steps are reversible. Is there a way to figure out whether it's reversible or not?

There's no way of figuring that out "from scratch", it is a lot of experience and knowledge. There is no general answer to this but there are some steps which are basically always seen as reversible or irreversible*.

protonation/deprotonation:
unless an extremly strong acid/base (like BuLi) is involved those are always in some kind of equilibrium and are reversible.

exchanging a good leaving group with a bad one:
That can be seen as irreversible.

gas evolution:
as @LinearChristmas pointed out: that can be seen as irreversible (if you are working in solution)

in general: producing a stable structure:
for example aromatization, that's irreversible

*theoretically every reaction is reversible, but if the equilibrium is far on one side we call it irreversible.

For an explanation what makes a reaction reversible see: Is every chemical reaction in equilibrium?


Why is the dipole moment of acetone higher than ethanol? It is true that acetone is less polar than ethanol.
I thought the dipole moment is proportional to polarity, and if so, the dipole moment of acetone should be lower than ethanol.
But the dipole moment of acetone is higher than ethanol.
Why is it higher?

Dipole moment is just one of aspects of solvent polarity, as is well known in HPLC separation  analytical technique where the following effects take part:

Dipole moment interaction, permanent and induced
The above for the whole molecule and for particular bonds
Electron pair donor
Electron pair acceptor (including $\ce{H+}$ donation)

Depending on solute molecules specifics, the solvent polarity order is variable, as various molecules have different responses to above interactions.

See also Using solvent triangle in HPLC


Why does liquid water form when we exhale on a mirror? In the descriptions below, I always assume external pressure to be constant at 1 atm, the condition where daily observations are made.

1) When I exhale on a mirror, liquid water forms on the mirror. That's condensation. Obviously, the temperature of mirror must be < 100 °C, so water vapor condensing on mirror makes sense.  However, in that case why do we have water vapor in our breath when our body temperature is also < 100 °C in the first place?

2) One reason for (1) may be like 'water vapors in air is in equilibrium with liquid water', so some water vapors can exist although T < 100 °C. If this is correct, based on this reasoning, then why doesn't ice exist at T > 0 °C? [p = 1 atm only]

3) If liquid water can evaporate into gas at T < 100 °C, then why doesn't ice turn into liquid at T < 0 °C? (I never use the term 'melt' here, just like evaporate≠boil) (both have hydrogen bonds, intermolecular forces should be the same?)

4) When I exhale on a wall, no water droplets form. Why do water droplets form on mirror but not on wall (just an example)? The mirror must have the same temperature as that of the wall, both of them must have achieved thermal equilibrium with their surroundings a long time ago.

Why do we have water vapors when our body temperature is also <100°C in the first place?

At normal pressure, water boils at 100°C, meaning that bubbles of pure steam form under water. At lower temperatures, water molecules reversibly move from the liquid to the gas phase and back. The higher the temperature, the higher the vapor pressure, and the higher the equilibrium concentration (partial pressure) of water in air. On the geological scale, there is no equilibrium, and we experience different temperatures, different humidity (related to partial pressure of water in air), and different pressures depending on location.

  Why does ice not exist at T>0°C

Ice, as a pure solid, and water, as a pure liquid, have defined concentration at given pressure and temperature (the equilibrium constant expression for melting includes neither liquid nor solid water, it is simply 1). Above the melting temperature, all the ice melts, there is no equilibrium. This is different from the liquid:gas equilibrium, which exists at temperatures below the boiling point, with lower and lower concentrations of the gas as the temperature drops (here, the equilibrium constant expression included the concentration or partial pressure of water vapor).

  If liquid water can evaporate into gas at T<100°C, then why not ice turns into liquid at T<0°C?

Again, this has to do with pure liquids and pure solids having a constant (or nearly constant) concentration. If you add salt to the liquid, however, ice will turn into liquid below the freezing point (effectively lowering the concentration of water in the liquid). Also, the surface of the ice melts at lower temperature than the bulk, so even for pure water, there can be liquid at temperatures below the bulk freezing point.

  When I exhale on a wall, no water droplets form.

I would do an experiment and check. I think the water droplets are easier to see on a mirror. Try a grand piano (i.e a smooth surface painted with shiny black paint), you might see the water droplets there as well. Or take a long hot shower and check whether water droplets form on surfaces other than a mirror.


What is a dry electron? I was reading about water photolysis which produces a positive water radical and a dry electron, or an excited water $\ce{H2O^*}$ through the following equation

$$\ce{
H2O ->[$hv$] H2O^{.+} + e_{dry}^-, or~ H2O^*
}$$
(as seen in this link)

By opposition I have also seen $\ce{e_{aq}^-}$ (as seen in pp. 2 of this link).

I can understand the "positive water" and even the "excited water", but what is a "dry electron"? And what is an $\ce{e_{aq}^-}$ (wet electron? I am unsure of how to translate the term I had learned to english). Why not simply say "free electron"?

A "dry" electron is an electron released from the event of an ionization. Depend on source the electron can either be defined as:

  "The $e^-$ is a 'dry electron' that still retains some of the excess
  kinetic energy acquired from the ionizing event."$\ce{^{[1]}}$

or,

  "The 'dry-electron' (defined as an electron before solvation)
  ..."$\ce{^{[2]}}$

A dry electron is thus, an electron that is not assiciated with any atoms, molecules or solutions.

Reference

$\ce{[1]}$ Thormod Henriksen, David H. Maillie; Radiation and Health; page 157

$\ce{[2]}$ J.D. Simon; Ultrafast Dynamics of Chemical Systems; page 144


Conformation identifying cis and trans How to identify cis and trans in conformers of cyclo hexane.i know that there are two positions axial and equatorial but using this data how to identify cis or trans? Also why are functional groups in equatorial positions more stable?

Draw the cycle as if it were planar (in fact it is not, but just imagine for a moment), with one hydrogen at each carbon pointing upwards with respect to the plane, and the other downwards. Now, if we introduce two substituents to different carbons on the same side of the plane, that would be the cis isomer, otherwise trans.

As for the stability of equatorial conformers, that's purely for sterical reasons. Draw the cycle, this time in its real geometry, and see what happens if we have two bulky groups in equatorial positions on the same side. Clearly they will be unpleasantly close to each other.


Looking for a toxic substance that cannot exist at room temperature I am writing a short story, and am trying to find a gas with the following properties:

It must have a very distinctive odor
It must be impossible for it to exist at room temperature under normal conditions
It must be toxic to humans

How large a pool of possible substances are there that meet these requirements? Is there a particular family or class of gases where I should be focusing my attention?

Some additional clarification (apologies for not including this when I first wrote the question):

The scenario I am writing is Cthulhu-based. The idea is that this substance is discovered where it shouldn't be able to exist. This is where the second criteria comes from. They will collect a sample of this stuff, analyse it and realise that its mere existence is impossible in the circumstances they found it. A completely hypothetical example might be a substance that spontaneously combusts in earth's atmosphere, or that immediately reacts in some other way with a gas present in air. This would mean that it cannot exist where the protagonists discover it, yet it does.

The toxicity is just intended to add some jeopardy to the situation

As Ivan Neretin already explained in his comment, you can't have it all.

Anyway, cyanogen, $\ce{N#C-C#N}$ would be a nice candidate. It is colourless, but highly toxic (H330). If it's cold enough, that is below -21 °C, you can even have a pond of it ;-)

In addition, it adds a historical space touch to the story. Back in 1910, when the return of Halley's comet was expected, cyanogen was spectroscopically identified in the comet's tail and the New York Times titled COMET'S POISONEOUS TAIL ;-).


Balancing redox equations with oxygens in multiple oxidation states How do you balance this using the ion-electron method;

$\ce{CrO5 + H2SO4 -> Cr2(SO4)3 + H2O + O2}$

The oxygen has multiple oxidation states in $\ce{CrO5}$, and none of the sites I looked this up on dealt with that.

This is a rather unusual case of what is discussed in answers like this one, where we circumvent problems with multiple atoms being oxidized or reduced by considering whole compounds as oxidizing or reducing agents.

Here, the whole-compound redox-active material is $\ce{CrO5}$, and as in peroxide disproportionations generally this is both an oxidizing agent and a reducing agent.  We thereby render

$\ce{CrO5 -> Cr^{3+} + (5/2) O2 + 3 e^-}$

$\ce{CrO5 + 10H^+ + 7 e^- -> Cr^{3+} + 5 H2O}$

We then apply the usual method of multiplying the first reaction by $7$ and the second one by $3$ to balance the electrons leading to

$\ce{10 CrO5 + 30 H^+ -> 10 Cr^{3+} + (35/2) O2 + 15 H2O}$

and reducing to "lowest (whole number) terms"

$\ce{4 CrO5 + 12 H^+ -> 4 Cr^{3+} + 7 O2 + 6 H2O}$

All that remains now is to add the spectator ions if desired, and we're done.


Most contributing resonance structure of phosphoric acid I am uncertain about the structure of $\ce {H_3PO_4}$ because different sources show different ones.

The structure shown here is:

while Wikipedia gives this one:

Arguments in favour of the 2nd structure:

Structure having filled octet for second row elements (C, N, O, F) are stable.
Structure having minimum number of formal charges and maximum number of bonds.
Structure in which there is minimal charge separation while keeping the formal charges closer together.

Arguments supporting the first structure:

Structure in which negative charge appears on the most electronegative atom should be chosen.
I feel that this could be more contributing because in the first structure sulphur has an expanded octet while $d$ orbitals are of higher energy so it's more favourable to form bonds without using them.
Structure having filled octet a for second row elements ($\ce {C, N, O, F}$) are stable. (followed by both structures)

$\ce {H_3PO_4}$ is different from phosphate ion.None of the answers in the linked question directly state what would be the more contributing structure. They focus on the hybridisation of P in phosphate.

Both structures are not the main contributing structures. In fact, the first structure is outright wrong and should be marked as an error as the electrons of phosphorus do not add up and the non-protonated oxygen is implied to be a radical.
However, the correct most contributing structure can be derived from the first structure simply by adding formal charges: a formal positive charge on phosphorus and a formal negative charge on the non-protonated oxygen. This structure satisfies the stability conditions well:

All main group elements have octets (dublets for hydrogen)
The second proposal violates this most important principle

The number of formal charges is minimalised
We cannot go any lower without violating the more important first point

The type of formal charges matches the electronegativity difference
And also we cannot distribute them differently

The distance between formal charges is minimal
But again we have no choice.


Strength identification in conjugate acid-base pairing I'm a student in highschool AP chemistry. I'm confused about a certain problem in my chem textbook. In summary, I am given a species and asked to identify its basicity as strong, weak, or negligible. I must then give its conjugate acid and identify likewise for its acidity. (in the context of Brønsted-Lowry theory)

I understood all parts of the problem except for $\mbox{O}^{2-}$, to which I assigned the conjugate acid as $\mbox{OH}^{-}$. It was correct thus far. My guess that they were both weak was wrong, however. According to the solutions page, they're both strong!

How is $\mbox{O}^{2-}$ a strong base? How is $\mbox{OH}^{-}$ a strong acid? To my knowledge, neither are found in the canonical strong acid/base lists, and neither seem to satisfy the "completely ionize in water" definition.

$\ce{H2O}$ is a weak acid.  Abstract a proton to give $\ce{HO^{-}}$, a strong base.  Abstract a proton from that to give $\ce{O^{2-}}$, a very strong base.  See a species as a whole, up and down for a property.

Try it with ammonia.  Feel free to wonder why $\ce{Li3N}$ is red.  Look what a very strong base can do,

$\ce{Li3N + 2 H2 → LiNH2 + 2 LiH}$ (reversible at 270°C)


How does de Broglie actually prove Bohr's postulates? We know Bohr said that the angular momentum of an electron is an integral multiple of $nh/(2π).$ And in de Broglie's wave equation, he said the circumference of the path of the electron traveling as a way is $n$ times the wavelength, which is equal to $2πr.$

But in Bohr's model $n$ is the principal quantum number, while the $n$ in de Broglie's equation is the number of wavelengths in the one revolution. I don't understand how they are equal.

de Brogile explains why orbitals are quantised

Strictly speaking de Brogile doesn't prove Bohr's postulates which are mostly wrong. But he did provide an explanation for the most important of Bohr's ideas: electron orbitals are quantised.

Bohr's whole model starts with the classical idea that electrons "orbit" a nucleus. But this has several problems not least of which is why don't they just radiate away their energy and fall into the nucleus (which is what accelerating charges should do in classical electromagnetism). Bohr made the extra assumption that the electrons around a nucleus could only "orbit" in a small number of fixed paths or orbitals which explained the actual observations. He didn't have any good reason why this was true.

de Brogile provided a reason. According to his postulates things like electrons are not just classical particles but have some of the characteristics of waves. and the thing about waves is that they interfere with each other. Electrons orbiting a nucleus can only do so if their wave-like property fits into a specific orbit (the way the vibrations in a guitar string produce only specific notes made from standing waves of integer divisions of the string length). Imagine integer numbers of electron waves around a nucleus and you are a third of the way of getting the picture (electrons orbit in 3D so the compatible vibrations are spherical harmonics in 3D space not the easier to picture vibrations of a 1D guitar string).

de Brogile, in short, produced a justification for the Bohr assumption that electron orbits had to be quantised.

Those pictures you see of electron "orbitals" are, in essence, pictures of the possible "waves" electrons can form when orbiting a nucleus.

All the above simplifies things a bit but the key ideas were from de Brogile and were more fully mathematicised by Dirac and Schroedinger giving us the modern quantum picture of how electrons behave.


Methylene blue used as funguscide in cichlid reproduction - concentration measurement? I have an aqueous solution of methylene blue. How do I determine the concentration of this solution, so I can duplicate it?

The concentration of an aqueous solution of methylene blue can be determined using a UV-VIS spectrometer.

Measure the absorption at $\lambda = 664\, \mathrm{nm}$, use the molar extiction coefficient of $95000\, \mathrm{dm^{3}\cdot mol^{-1}\cdot cm^{-1}}$
(Reference) and calculate the concentration using the Lambert-Beer equation.


What is the percent of potassium as K₂O in the fertilizer sample? The solubility of sodium tetraphenylborate in water is $\pu{470 g/L}$ while the solubility of potassium tetraphenylborate in water is $\pu{1.8e–4 g/L}$.  Due to the orders of magnitude difference in aqueous solubility, $\ce{NaB(C6H5)4}$ is used to gravimetrically analyze the potassium content of fertilizer:
$$\ce{K+(aq) + NaB(C6H5)4(aq) -> KB(C6H5)4(s) + Na+(aq)}$$
The water soluble components of a $\pu{10.7875 g}$ sample of 15-25-15 fertilizer were separated from the water insoluble components using vacuum filtration.  The filtrate was treated with an excess of $\ce{NaB(C6H5)4}$ and the resulting solid was isolated, dried, and the mass of the solid was determined.  Potassium tetraphenylborate $(\pu{12.4393 g})$ was recovered from the gravimetric analysis of $\pu{10.7875 g}$ fertilizer.
What is the percent of potassium as $\ce{K2O}$ in the fertilizer sample?

I know how to get to the answer; I am just unsure how to use the solubility that was given to get there. I also am not sure why the resulting dried mass is higher than the initial mass of fertilizer:
$$\ce{K+(aq) + \underset{\pu{10.7875 g}}{NaB(C6H5)4(aq)} -> \underset{\pu{12.4393 g}}{KB(C6H5)4(s)} + Na+(aq)}$$
$$M(\ce{NaB(C6H5)4}) = \pu{342.22 g mol-1}$$
$$M(\ce{KB(C6H5)4}) = \pu{358.35 g mol-1}$$
$$\pu{12.4393 g~\ce{KB(C6H5)4}}\cdot\frac{\pu{1 mol}}{\pu{342.22 g}}\cdot\frac{\pu{1 mol}~\ce{K}}{\pu{1 mol}~\ce{KB(C6H5)4}}\cdot\frac{\pu{1 mol}~\ce{K2O}}{\pu{2 mol}~\ce{K}}\cdot\frac{\pu{94.2 g}~\ce{K2O}}{\pu{1 mol}~\ce{K2O}} = \pu{1.712 g}~\ce{K2O}$$
$$1~\text{mass} = \frac{\pu{1.71 g}~\ce{K2O}}{\pu{10.7875 g}}\cdot 100\% = 15.9\%$$

The solubility has been given solely to justify the application of the given method of gravimetric analysis.
Since the majority of potassium salts are well or moderately soluble in water, for this analysis it's assumed that entire potassium content is brought to the solution.
Since the solubility value of potassium tetraphenylborate is over a million times smaller than the corresponding value for the analyte (sodium tetraphenylborate), it's safe to speak of the completeness of the analysis as all dissolved potassium species were isolated as tetraphenylborate salt and analyzed.

As for the mass difference, that shouldn't be of your concern as the content of the fertilizer isn't given in the first place and you cannot make any assumptions regarding molecular masses of its constituents.
Plus, tetraphenylborate $\ce{B(C6H5)4-}$ is a bulky anion of substantial molecular mass $(M(\ce{B(C6H5)4-}) = \pu{319.23 g mol-1})$, which is much higher in comparison to "usual suspects" found in fertilizers, such as phosphates $(M(\ce{PO4^3-}) = \pu{94.97 g mol-1})$ and nitrates $(M(\ce{NO3-}) = \pu{62.00 g mol-1})$.

Long story short, your answer seems correct to me, but I propose to solve the problem algebraically first, starting with what has to be determined and omit intermediate calculations, unless you were explicitly asked to perform any.
To illustrate my approach and use of the notations for physical quantities and units, I leave my solution here.

Content of potassium expressed via potassium oxide is a mass fraction $ω(\ce{K2O})$, which is the ratio between the mass of potassium oxide $m(\ce{K2O})$ and the total mass of fertilizer taken for analysis $m_\mathrm{tot}$:

$$ω(\ce{K2O}) = \frac{m(\ce{K2O})}{m_\mathrm{tot}}$$

Unknown mass $m(\ce{K2O})$ can be found from the corresponding amount $n(\ce{K2O})$ and molecular mass $M(\ce{K2O})$:

$$m(\ce{K2O}) = n(\ce{K2O})\cdot M(\ce{K2O})$$

In turn, amount $n(\ce{K2O})$ can be determined from the reaction's stoichiometry:

$$n(\ce{K2O}) = \frac{1}{2}n(\ce{K+}) = \frac{1}{2}n(\ce{KB(C6H5)}) = \frac{m(\ce{KB(C6H5)})}{2\cdot M(\ce{KB(C6H5)})}$$

Now, since we are at the point when no unknowns are left, lets put everything together so that the first equation can be rewritten as such:

$$
\begin{align}
ω(\ce{K2O}) &= \frac{m(\ce{KB(C6H5)})\cdot M(\ce{K2O})}{2\cdot m_\mathrm{tot}\cdot M(\ce{KB(C6H5)})} \\
  &= \frac{\pu{12.4393 g}\cdot\pu{94.20 g mol-1}}{2\cdot\pu{10.7875 g}\cdot\pu{358.32 g mol-1}} \\
  &= \pu{0.152}~\text{or}~15.2\%
\end{align}
$$


Why is K2CrO4 Potassium chromate? I was given $\ce{K2CrO4}$ and was asked to name the compound.

Can this be done by:

1) Determining the molecular compounds oxidation state

or

2) Does one need to memorize the polyatomic ions and that is the only way to know what is what?

Definitely you need to remember it. Potassium Chromate  is IUPAC name of $\ce{K2CrO4}$. There are other names such as Chromic acid and dipotassium salts. But mostly IUPAC names are accepted.

I don't think you should try to mug up all 24 names instead you should keep getting familiar with them when they come in reactions or somewhere else. After seeing them at few different places you will be remember them all. They will come very often while you are learning chemistry.


Galvanic Electrochemical Cell Without half cells Is it possible to create a galvanic cell without a half cell?

In the book I was reading, Illustrated Guide to Home Chemistry Experiments, the following lab (Laboratory 16.4) was suggested:

add 2 M HCl and add copper metal to one end of the beaker, with magnesium at the other end. Then, add patch cables to connect the metal plate to an LED to have physical evidence of success.

However, I don't understand how useful work could be generated without half cells.

To my understanding, we add HCl to react with the magnesium and copper, to produce ions ($\ce{Cu+}$ and $\ce{Mg^2+}$). Then, they can spontaneously produce work due to their eV differences.

What I am missing is how the work can be useful, as opposed to just generating heat in the solution.

I would believe that, rather than traveling though the patch cables, electrons can travel directly through the solution and produce heat.

Thanks for reading and I really appreciate any help!

This differs from the lemon galvanic cell (aka, 'battery') in several important ways: 1) Mg is used as the anode, rather than Zn, 2) there is no salt bridge, unlike in the the lemon cell, where the lemon serves as a crude 'salt bridge', and 3) the electrotyle is 2 M HCl rather than citric acid (and other stuff) in lemon juice. An open circuit voltage will be produced, easily measured with a digital multimeter in voltage input mode. A short circuit current will be produced, easily measured with a digital multimeter in current input mode. It may be possible to light a red LED, since they have low threshold voltage and low current requirements. I suspect it would light the red LED, though feebly.

But regardless of whether or not there is an external connection of the Mg and Cu electrodes, there will be a rapid redox reaction at the Mg electrode: $$\ce{Mg(s) + 2 H^+ <=> Mg^{2+} + H_2 (g)}$$

This unwanted competing redox reaction is the same as the net reaction obtained from the two relevant half cell reactions and it justs wastes some of the available energy as heat in the cell's vessel. So some electrical power is available by connecting an electrical load (e.g., red LED or resistor) between the two electrodes. But the parasitic oxidation of the Mg anode will continue until the electrode is consumed or the HCl is almost entirely consumed. (Magnesium will very slowly react with boiling water.)

Notes: 1. Clean copper, i.e., without surface compounds or so-called 'patina', does not react with 2 M HCl at room temperature. So the 2 M HCl does not contain any $\ce{Cu^+}$ or $\ce{Cu^{2+}}$ ions. 2. The reaction of Zn with 2 M HCl is considerably slower than is the reaction of Mg with 2 M HCl.

For more information, and at a much deeper level, see K. Schmidt-Rohr, "How Batteries Store and Release Energy: Explaining Basic Electrochemistry", Journal of Chemical Education, 95 (2018) 1801-1810.


What electrodes to use to minimise corrosion during water electrolysis? I did the $\ce{H2O}$ electrolysis experiment a few times, but every time a huge amount of the oxygen reacted with its corresponding electrode. What material should the electrode be made of in order to maximise the amount of oxygen I get?

It depends on the electrolyte as well as the electrode material. For example, $\ce{NaCl}$ releases atomic chlorine at the positive electrode, quickly attacking most materials except metals of the $\ce{Pt}$ group, as @MaxW mentions.

$\ce{NaOH}$ or $\ce{KOH}$ are a bit gentler on the electrodes. Run the apparatus at a low over-voltage to avoid damaging them. Coated nickel electrodes are suggested for use with alkaline solutions.

Phosphoric acid is also suggested as an electrolyte.


Is it possible to create a 4 inch by 4 inch channel through concrete using acid? I'm trying to figure out a way to eat through concrete without damage pex tubing inside (its a radiant heated slab). Is there any way using some type of acid (hydrofluoric, muriatic ) to eat a 4 inch wide, 4 inch deep, and 71 foot long channel through the slab that could be tapped out with a blunt chisel? Pex tubing is resistant to most acids so it would not be damaged.

Also, this needs to be cost effective (under $1000).

Offhand, this seems dangerous to the structural integrity of the slab, because the acid would follow the pore structure of the concrete and weaken the wrong places.

However, concrete cutting saw rental is available; e.g. United Rentals
and http://www6.homedepot.com/tool-truck-rental/Walk_Behind_Concrete_Saw_14/MK-CX3/index.html. Two 45-degree cuts ~6" or deeper should do the job.


Is it possible to get energy from frozen methane without oxygen? In one of the solar system documentaries it was mentioned there are moons in the solar system with lakes of frozen methane.

Assuming humans or robots or whatever could make it there, would there be a way to extract energy from frozen methane without using oxygen? Otherwise, it would seem that the oxygen would have to be carried there as well.

Is it possible to recycle the oxygen efficiently enough to burn methane again and again?

The planet will have gaseous oxygen if it does have an atmosphere, and if that is the case then with enough energy(like a spark) atomized hydrocarbons could be lit for energy. What is atomizing you say? Well its just breaking a liquid apart into little particles, making it easier for the oxygen to react with said methane.

As for recycling the oxygen from the combustion reaction, it will not be possible to do in an energy efficient way, CO2 is too stable and the bonds are too strong.


Which is better: Increase in collision frequency or increase collision energy What would have a greater effect on the rate of a reaction, increasing the frequency of collisions or increasing the collision energy of the particles? (assuming that using a catalyst is not an option)

I got this from a past paper question:

Explain in terms of collision frequency and collision energy how the rate would change if the temperature was increased, and which of these causes the greater effect.

I know the first part of the question.

I assume that the collision energy is more significant, which is why we use catalysts, but not sure how.

Collision energy is what matters

Both the frequency of collisions and the average energy of those collisions are increased at higher temperatures. But the frequency doesn't matter much for the rate of reaction because collisions are already very very common and are not the limiting factor in causing a reaction to happen.

The distribution of energy in moving particles (and therefore the energy involved in collisions) follows a statistical distribution. Simplifying a bit (by ignoring the orientation of colliding molecules and a bunch of other probabilistic factors) the reason a reaction follows a collision is that the collision has enough energy to overcome some threshold energy barrier for a reaction to take place (simplifying again, collisions without enough energy just cause molecules to bounce off each other perhaps exchanging some kinetic or vibrational energy in the process). Only when the net energy involved is sufficient to, for example, break a bond in one of the molecules, does a reaction result. A lot of collisions are happening all the time and most don't have enough energy to cause a reaction to happen. In fact, if the population of collisions with enough energy is too low there will be no reaction at all no matter how many collisions there are and increasing the number will make no difference at all.

As the temperature increases, more of the molecules are pushed into the "have sufficient energy" part of the distribution (which might be a very small proportion of all the molecules in the mixture). Then reactions start to happen.

But the number of collisions is very large compared to the number leading to a reaction so it is the number of molecules with enough energy that dominates the result.


Atomic emission spectrum assumption n_f = 3 It's a standard question on the atomic emission spectrum.

  An atomic emission spectrum of hydrogen shows three wavelengths: 1875 nm, 1282 nm, and 1093 nm. Assign this wavelengths to transitions in the hydrogen atom.

I understand how to calculate it, but from both the answers in my book and the answer from slader.com. Both just say

  since these wavelengths are larger than those of visible light, then we assume $n_\mathrm{f} = 3.$

However, I don't understand why we could assume $n_\mathrm{f} = 3.$ Is it because we just look at Paschen series (Bohr series, $n′ = 3$) and the Paschen lines all lie in the infrared band? Could anyone explain it?

As the wavelengths are in the infrared region hence the transition occurs at $n > 2.$

Now, if you may have seen the data, these values are all common values of Paschen series. But even if you haven't, you can do a simple test by finding the shortest wavelength of all the series after Balmer. You will find only the shortest wavelengths of Paschen and Brackett are suitable for having values lying in this range as the shortest wavelength for Pfund is $\pu{2279 nm}.$

Then also as the shortest wavelength for Brackett is $\pu{1458 nm},$ you can be sure that the second and third are definitely lying in the Paschen series and check for the remaining wavelength which would be a fairly easy job now.

Directly assuming that the final energy level is $n = 3$ isn't correct, I think, but some analysis can help us to draw the above-mentioned conclusions.


Does changing the volume of a reactant affect the rate of reaction? Does changing the volume of HCl affect the rate of a Mg & HCl reaction? I'm doing a experiment in which I'm testing the affect of the concentration of a reactant on the rate of reaction.

The reaction is between Magnesium and Hydrochloric Acid and so I'm changing the concentration of $\ce{HCl}$ ($\ce{0.5~M}$, $\ce{1~M}$ etc.)

In my lab report, I'm writing a list of controlled variables. So far I've mentioned the surface area and temperature of reactants. I would like to know more controlled variables.

I want to know whether the volume of Hydrochloric Acid affects the rate of reaction when added to the same amount of Magnesium metal?

According to some it doesn't and some say it does; I'm confused.

Also, please tell me why or why not the $\ce{HCl}$ volume affects the rate of reaction.

If the concentration of $\ce{HCl}$ changes between trials and the volume changes between the trial as well, then is the rate of reaction affected?

Does volume affect concentration in this case?

More volume of HCL does not increase the concentration and therefore does not increase the likelihood and so rate of reaction.
Mg and HCl forms a metal salt MgCl and produces Hydrogen gas.

There may be a very slight increase in rate of reaction but this is permissible/ very negligible due to the fact that metals and acids form adducts and these can further react/ delay further reactions between the acid and metal molecules.

The rate would definitely be affected by factors such as temperature of the (aq) HCL. This is until the reactant Mg is used up then the reaction will cease leaving you with a flat curve on a rate graph.

Now, for Higher concentrations- this implies that more reacting molecules are at higher proximity to each other therefore the intermolecular collisions between the Mg and HCl molecules in your case are more frequent therefore forming more product per unit time.

Control variable: mass of Mg to at least 1.d.p, concentration of acid moldm-3, volume of HCl cm3 - even though this doesn't affect the rate of reaction enough - they would want you to keep this constant for valid test etc.

Hope this helps!


Kinetic and Thermodynamic Stability My prof noted that if the two components of any ionic compound undergo a large extent reaction with each other, then the substance CANNOT exist as a pure substance.

For example, consider ammonium hydroxide. This is simply ammonia (at least in the solution phase). Can we have a solid, crystalline ammonium hydroxide lattice?

I think this is thermodynamically valid. What about kinetically? Isn't it theoretically possible to have two substances that would otherwise undergo a large extent reaction with each other but don't due to an activation energy barrier?

Are there any such examples?

Also what do you think of the below question?

Chemistry is a field with far too many exceptions and edge cases to be making sweeping categorical statements. There are numerous examples of ionic compounds that are isolable and stable, the components of which will spontaneously decompose under certain conditions (this is where reaction kinetics is relevant). Two that readily come to mind are ammonium nitrate ($\ce{NH4NO3}$) and ammonium bicarbonate ($\ce{NH4HCO3}$). Both of those can be isolated as pure salts, and both also can and will chemically decompose when circumstances are favorable (the former by multiple possible pathways depending on conditions).

  Isn't it theoretically possible to have two substances that would otherwise undergo a large extent reaction with each other but don't due to an activation energy barrier?

This isn't just theoretically possible, it's ubiquitous, it's the norm. If it weren't, human life couldn't exist. Combustion reactions are probably the most obvious and mundane example; just about every organic compound imaginable is usually more thermodynamically stable as a mixture of $\ce{CO2}$, $\ce{H2O}$, $\ce{N2}$, etc., but the kinetic barrier to the reaction (along with physical hindrances and the availability of an oxidizer, e.g., $\ce{O2}$) prevents this from occurring spontaneously without some trigger and the appropriate environmental conditions.


Does Zinc become black & dull after reacting with dil. HCl? In the reaction where $\ce{Zn}$ reacts with $\ce{HCl}$ to give off Hydrogen gas & Zinc Chloride solution, does the zinc turn a dull black color after a while? If so, why?

Look at it under a microscope.  If the surface is microporous, light goes into the reflecting cavities, bounces around, and is absorbed before it can escape.  Wilkinson Sword sold bolted stacks of double-edged razor blades as low power laser beam dumps,  Each sharp edge was mirror bright with a slight taper toward the edge.  Light bounced in and could not escape.  The sharp sides of the stack were astoundingly black.  375 blades x .004" = 1.5" square target.

Or, you have impure zinc and crud is accumulating at the dissolving surface.  Alloy 8111 aluminum foil contains about 2% or so of strengthening dispersed intermetallics.  Dissolving aluminum foil in acid or base leaves black residues.


Synthesis of 1-iodo-4-nitrobenzene I was trying to synthesize 1-iodo-4-nitrobenzene in the lab today – however, the appearance of my product really confused me.

At first, I added conc. sulfuric acid to water and 4-nitroaniline.
This was followed by diazotising the amine sulfate with sodium nitrite and water. (This was done at temperatures of below $10\ \mathrm{^\circ C}$).

Then, I added the diazonium salt to a solution of $\ce{NaI}$ and water. A large plume/foam formed ($\ce{N2}$). After, I performed a filtration using a Büchner funnel. That’s where things went wrong.

I used water to wet the filter paper. I added in the iodonitrobenzene. A suckback occurred within a few seconds as the product was quite viscous. After resumption, the filtration went as planned. However, the product on the filter paper was a molten brown mush.

I tried to let it dry for a further 5 minutes but once turned off it remained a sludge.

Do you have any suggestions on how I could improve on this synthesis or maybe an idea of what went wrong?

Here are some things to focus on:

How quickly\slowly did you add the diazonium salt to the NaI solution?
Did you maintain the reaction temperature below 10°C during the entire addition?
Did you also keep the diazonium salt cooled while adding it?
After addition was complete, did you let the reaction stir at 10°C until nitrogen evolution ceased and then let the reaction slowly warm to room temp?

Here is a tested lab procedure for the preparation of 4-iodonitrobenzene.  Compare it to your notes and see where you differed.

Keeping the reaction cold during the addition is key. If the diazonium salt warms up too much, the diazonium can react with itself (coupling) to give colored by-products.  The diazonium ion can also eliminate nitrogen to generate the extremely reactive phenyl cation which will react with any nucleophile around, water for example (see page 6, the Mechanism section in this reference).  Both pathways can  contaminate and reduce the yield of the desired product.


Why don't we see these lanthanide species? For most lanthanide metals$^{[1]}$, the stable oxidation state is III. The general electronic structure$^{[2]}$ is $$\ce{[Xe] 4f^{0}^{-14} 5s^2 5p^6 5d^{0}^{-1} 6s^2}.$$

Elements that have the d-electron are La, Ce, Gd, and Lu. Furthermore, the f-subshell is considered relatively stable in states $$f^0, f^7, \text{and} f^{14}.$$

We can conclude that La, Gd, and also Lu easily form $\ce{E^3+}$ ions. Yet, as would be predicted by this easy approach, we would also see $$\ce{Sm+, Tm+ (f^7 \ and \ f^14 ), \\ Pr^5+, Dy^5+ (f^0 \ and \ f^7)}.$$

Why is this not the case?

$^{[1]}$ Ce, Pr, and Tb also have the oxidation state IV. Eu and Tm have the additional state II.

$^{[2]}$ Ordering (relative energy) changes with the number of electrons.

I strongly recommend having a look at these questions:

Howcome orbitals become 'core-like' when electrons are removed?
What is meant by 'electrons of like/unlike rotation'?
What's up with this quarter / three-quarter rule?

Original topic: Predominance of III oxidation state for lanthanides

This is really the exact same question but just in a different context of transition metals: Cr(II) and Mn(III) - their oxidizing and reducing properties?

The answer is because exchange energy (which is the reason behind the stability of half-filled and fully-filled shells) is not the main factor that decides the predominance of an oxidation state.

Whether an oxidation state can be reached depends on the balance between two things:

An energetic cost of having to ionise the element, i.e. the ionisation energy
The energetic payback obtained by forming covalent bonds / ionic bonds / solvation in solution / etc.

Exchange energy effects only give rise to minor variations in ionisation energies. The increase going from the nth to (n+1)th ionisation energy is far greater than any exchange effects.

Just like the magnesium example I gave in the other question, the lanthanides have very small second ionisation energies, and these can also be easily compensated for by bonding or ion-dipole interactions in solution. Take your example of $\ce{Sm+}$. The fact that it has a $\mathrm{f^7}$ configuration means that the second IE is marginally larger than would be expected. However, this marginal increase hardly changes the fact that the cost of ionising Sm a second time can be easily recouped.

Likewise, your example of $\ce{Pr^5+}$ is totally unrealistic because those 5th IEs are so huge that there is absolutely zero chance that they will give it up, no matter what exchange stabilisation there is.

The predominance of the +3 ionisation state for the lanthanides is therefore simply a consequence of IE1 + IE2 + IE3 being sufficiently small, and IE4 being way too large. That's all there is to it. Note how the variations in the graph below (due to exchange energy effects) are so small that all the IE graphs never cross over each other. IE5 is so huge that they didn't even bother plotting it!

(source: Inorganic Chemistry 6ed, Weller et al., p 630)


Gibbs–Helmholtz equation The Gibbs–Helmholtz equation was developed in the 1870's, while Nernst's heat theorem was developed in the early $20^{th}$ century (according to google/wikipedia). Nernst's heat theorem tells us that the change in entropy tends to be zero in absolute temperature. But did Nernst's heat theorem provide any new insight? Because if we put $T=0$ in the Gibbs–Helmholtz equation, then  $\Delta G=\Delta H$. Anyone could have thereby concluded that as temperature approaches 0, entropy change becomes zero already in the $19^{th}$ century. I am quite confused regarding it, those were my assumptions (probably wrong), what was new about Nernst's heat theorem ?

If you read the Wikipedia page for Nernst's Heat Theorem carefully, it explains that the novelty lies not in showing (as you point out) that as $T \rightarrow 0$, $\Delta G \rightarrow \Delta H$, but rather that as   $T \rightarrow 0$, the slope of $\Delta G$ as a function of T goes to 0, that is

$$\lim_{T\rightarrow0} \left( \frac{\partial \Delta G}{\partial T} \right)_p =0$$

Then, since that slope is given by $-\Delta S$, it predicts the experimentally confirmed conclusion that as $T \rightarrow 0$, $\Delta S \rightarrow 0$,


Photosynthesis: Chlorophyll and color. As I know, chlorophyll is green as it absorbs all wavelengths but green. It absorbs red, yellow, blue, purple, etc. Thus, the plant does not utilize green light.

i] Why is chlorophyll green? Is there 'chlorophyll' of other colors?

ii] Is the green chlorophyll an 'evolutionary mistake'?

iii] If chlorophyll was black, what effects would there be?

iv] What is the effect on non-visible electromagnetic-radiation?

Thanks!

i] There is chlorophyll-b (yellow-green) and carotenoids (yellowish).

ii] The green chlorophyll isn't an evolutionary mistake (unless, of course, you take the viewpoint that evolution IS just mistakes). Plant's get plenty of sunlight. The sun is way stronger than indoor lights used to grow plants. They don't really need more energy from the sun. Plants actually have some pigments to dissipate light because they would otherwise absorb too much.

iii] Plants might absorb too much energy. Think wearing all black on a hot summer day, and standing out in the sun ALL day.

iv] Plants don't really use ultraviolet light as far as I know. With the ozone layer blocking most of the ultraviolet rays coming at us, it would be very strange if they did evolve to absorb ultraviolet light. This might also harm the plant. Infrared rays do reach plants, but they don't really do much to help in photosynthesis. Anyways, light has to travel through the cytoplasm of plants which is full of water, and visible light is best at travelling through water.


Why is work and heat energy always viewed from the perspective of the system? So my answer was 3 and it was wrong, And i really would like to know what is the correct answer, after giving it some thought, I think the answer is 4 because indeed it is easier to control the system than surroundings, can you please assure it that it's D.
I appreciate any help

The problem is in the word "always". The statement "Work and heat energy are always viewed from the perspective of the system?" is wrong. Please inform your teacher that is true in the chemist's world only- and chemists did not invent thermodynamics.
Chemists have adopted a different convention than physicists. All we have to do is to be aware of the context.
Recall the First Law of Thermodynamics, which has 2 different equivalent statements.
In some books it is
$$∆U=q-w$$
and in some it is
$$∆U=q+w$$
Why, because the word "always" is not correct.


Can someone recommend me a book ( preferably written long ago) that includes all the details of high school chemistry? The textbook I got from my teacher is shown to be too general. Part of the fun of learning chemistry, for me, comes from knowing the derivation of equations. But most of the derivations are missing out from the textbook, like the derivation of kinetic molecular equation and real gas equations.

It will be fine to recommend a college level textbook as long as it does not include calculus.

"College Chemistry - An Introductory Textbook of General Chemistry" by Linus Pauling (2nd Ed) is available as a free download (pdf & other formats) from https://archive.org/details/CollegeChemistry.


Strength of the hydrophobic interaction How strong is the "hydrophobic force"?

Hydrophobic interactions are weak interactions but can have greater strength than hydrogen bonds. I find the strength of the hydrogen bond in literature to be near 1 pN. Finding the strength of the "hydrophobic bond" has proven to be a bit more elusive. I would guess that there is a hydrophobic interaction strength spectrum (dipole strengths and such), but would also guess there is a fundamental order of magnitude of strength of the hydrophobic interaction.
Hydrophobic interactions result from the hydrophobic effect and manifest in measurable attractive forces. For example, the Polyhydroxyethylmethacrylate (pHEMA) backbone has dipole induced interactions, hydrophobic interactions, and hydrogen bonding; each of these interactions being identifiable in molecular position on the backbone (aliphatic regions being identified as hydrophobic sites).
To date, i have found a measurement of such resulting attractive forces in a specific interaction to be near 15 pN (oil-oil droplet AFM approach)[1]. However, I can also find literature that discusses forces in avidin-biotin  complex which are dominated by hydrophobic interactions; resulting attractive forces range from 80 to 250  pN [2].
[1] Tabor et al, 2013; DOI: 10.1021/jz402068k
[2] Moy et al, 1994; DOI: 10.1126/science.7939660

The hydrophobic force does not exist. It is a phenomena resulting from the following considerations.
In principle, liquids should be like gases, perfectly miscible in all proportions. This is the case for non polar molecules, like hydrocarbons. This is not the case for liquids where molecules are hold together by hydrogen bonds, like water. In these liquids, adding solutes requires to break hydrogen bonds. It is acceptable if the solute is able to form new hydrogen bonds, like with small alcohols, polyols and sugars. But if mixing with another substance will break too many H-bonds, Nature refuses to make up this mixture, as if there was a hydrophobic force. Of course, this force does not exist. It is simply a consequence that mixing very different substances would cost too much energy for breaking all H-bonds of one liquid, without forming back new ones.


IR fluorescent material for outdoor use I am experimenting with IR video cameras. They have a broad sensitivity peak at 940 nm. There is a filter blocking 750 nm and shorter in front of the sensor.

I am looking for an IR fluorescent substance to use for markers in the scene the camera is observing. Something that can be turned on with ultraviolet illumination bursts so I can differentiate between my markers and other IR sources.

My hope is for some affordable and harmless substance. Something that could be mixed with polyester, acrylic or epoxy potting compound wold be nice.
An already solid watherproof material is even better.

My searches so far have found some IR fluorescent stains for microscopy but I am afraid they are expensive.

This might be a physics question also but I think chemists works a lot with IR absorption and IR fluorescence.

A commercial solution would be a dye like IRF940A (from QCR Solutions) with an excitation maximum at $\lambda_{\mathrm{exc}} = 405\,\mathrm{nm}$ and an  emission maximum at $\lambda_{\mathrm{em}} = 940\,\mathrm{nm}$. The supplier does not provide any information on the structure of the dye.
What compounds are promising candidates?
Ytterbium(III) complexes!
In a recent article,published in J. Photochem. Photobiol. A, 2015, *309, 65-71 (DOI), J. Legendziewicz and coworkers prepared a $\ce{Yb(III)}$ phthalocyanine with acetate as a chelating axial ligand and examined the optical properties in solution and in PMMA polymer. In PMMA, the compound shows a Soret band with a maximum at $\lambda_{\mathrm{abs}} = 350\,\mathrm{nm}$. Excitation in the Soret band result in emission with a maximum around around $\lambda_{\mathrm{em}} = 700\,\mathrm{nm}$.
Other examples for even larger wavelength differences, resulting from ligand-to-metal energy transfer are known  for $\ce{Yb(III)}$ complexes bound to eosines or fluoresceines as the primary chromophor.


How to balance the reaction equation of potassium permanganate, calcium oxalate, and sulfuric acid step by step? I have been desperately trying to balance the following equation, and finally (ultima ratio) used an online program to get it done (posted the same question there as well).
$$\ce{KMnO4 + CaC2O4 + H2SO4 -> MnSO4 + K2SO4 + CaSO4 + CO2 + H2O}\tag{I}$$

No worries, balancing the K, Mn, Ca, C and S - but by then the H and O got out of my control. So this is how far I got myself:
$$\begin{multline}\ce{2KMnO4 + CaC2O4 + 4H2SO4 ->\\
 2MnSO4 + K2SO4 + CaSO4 + 2CO2 + 4H2O}\tag{II}\end{multline}$$

So the only thing left to be balanced, was the O, having 28 on the reagents' side and 24 on the product side. I started off, getting up to 36:33 O (reagents:products), and I felt that I was getting nowhere.

Anyone who would guide me through the steps following the balancing until equation II above?

This is a redox reaction. While it can be balanced by the method you describe, it is much more reliable to balance it by the half-reaction method.

Step 1 - Identify your spectator ions (so we do not spend time trying to balance them)

Three ions appear unchanged on both sides of the reaction: $\ce{K+},\ \ce{Ca^2+},\ \& \ \ce{SO4^2-}$

Step 2 - Identify your redox pairs.

You have one manganese species on each side. They must constitute one half-reaction:
$$\ce{MnO4- -> Mn^2+}$$
Likewise, you have on carbon species on each side. They must constitute the other half reaction:
$$\ce{C2O4^2- -> CO2}$$
For each half reaction:

Step 3 - Balance O by adding $\ce{H2O}$
Step 4 - Balance H by adding $\ce{H+}$
Step 5 - Balance charge by adding $\ce{e-}$
Step 6 - Combine the half reactions

Can you take it from here?


Is there any way to use syringe pumpe in pumping flow I want to use syringe pump to pump water at 0.82 ml/min for 30 min ? Is that possible ( I read some paper use that but don't know how ?)

This should be no problem when you use a syringe pump model with adjustable flow rate and sufficient volume (at least 30 ml).


What does 10 Vols mean? I am currently working on a project and the protocol says the following, "Wash resin 3 times 10 vols of the extract buffer." This is very confusing to me and I am not exactly sure what the 10 vols part means. I tried to look it up, however, the few explanations I found were still confusing. I am new to research, and have not been exposed really to the "lingo".

I think you mean "wash resin 3 times with 10 vols of the extract buffer".

Vol means volume strength. It tells you how much of a certain element in milliliters will be RELEASED from a certain substance. In this case, your buffer is supposed to release 10 vol of something. That's all it means.

However, I am surprised about the usage of vol in this context. Volume strength is usually a term directly associated with hydrogen peroxide, and dealing with the release of a certain volume of oxygen.


Which component in my tap water "scale" does not react with acid? Where I live we have pretty hard tap water and so I'm used to scale build-up in kettles and on bathroom tiles, easily removed with acids (it's mostly calcium carbonate I guess). However it seems a part of the "scale" is resistant to acid. For example, I have a glass that I use in my office to drink tap water throughout the day, so the remaining drops of water dry often at the bottom. This results in a circle of "scale", but a significant part of that remains even after treatment with sulfamic acid. Similarly, it seems a part of the residue that forms on my bathroom tiles is not removed by acidic cleaners.

Looking at the water quality charts of my utility, the only part that I think could be responsible just by guessing at the amounts present is $\ce{SiO2}$, at 29 mg/l. Does this form a white, scale-like substance that's hard to remove, at least with acid?
(Tap water analysis here, fold down the Wasserwerk Schulensee part.)

Below an image of my glass after treatment with warm sulfamic acid. The limescale present bubbles off. The remaining white stuff can be scratched off with a knife, so it's actually a residue and not just "etched" glass, but it is pretty tough.

A likely candidate is calcium sulfate, which forms a relatively soft mineral (gypsum), hence removable with a knife, but it has limited solubility in water and does not react with common acids.

See here for more on water hardness.


Canonical MOs vs. Localized MOs: Do both represent reality in the same way? In my understanding localized molecular orbitals (LMOs) are equivalent to "standard" molecular orbitals, often called canonical orbitals (CMOs—by the way, why are they called canonical?). We can produce LMOs by linear combination of CMOs and they represent the same physical state.

I have two questions:

Is this just a mathematical gimmick or does this represent the reality? Is it possible for a molecule to be in such a state where every electron is in such a localized orbital? Or is this just a thing we do because it makes it easier to understand?
If LMOs are equivalent and have the same energy then why do calculations always end up with canonical molecular orbitals and never with LMOs? And wouldn't there be an unlimited amount of equivalent solutions?

NOTE: In the below, I'm implicitly discussing a ground-state, closed-shell wavefunction, where all occupied orbitals are doubly occupied.  The discussion would be similar for open-shell wavefunctions, but there are complexities that I won't address here. Also, once one starts noodling at excited states, things get complicated pretty quickly, and (AFAIK) some manipulations that are valid for a ground state wavefunction are invalid for excited states.

In my understanding localized molecular orbitals (LMO) are equivalent to "standard" molecular orbitals, often called canonical orbitals (btw. why are they called canonical?). We can produce LMOs by linear combination of MOs and they represent the same physical state.

The "canonical" MOs (CMOs) are called such because they are in a certain sense the "default" MOs for a wavefunction. The sense in which they are the "default" is that they diagonalize the Fock matrix that defines the energy of the system and is used in the course of the self-consistent field (SCF) calculation of, e.g., the Hartree-Fock method for solving the approximate many-electron problem.
Occupied localized MOs (LMOs), when all considered together as a unified wavefunction, in total contain the exact same information as the wavefunction composed of the occupied CMOs.  This is because the LMOs are obtained from a unitary transformation of the occupied CMOs, and the linear algebra indicates that the overall wavefunction is invariant to such a transformation.

If LMOs are equivalent and have the same energy then why do calculations always end up with canonical molecular orbitals and never with LMOs?

The italicized phrase of this question is false. LMOs individually are not equivalent to individual CMOs, and they do not have the same energy. Strictly speaking, the energies of individual LMOs are actually undefined. Only the total wavefunction composed of LMOs is equivalent to the total wavefunction composed of CMOs.
As noted above, the reason why we always get CMOs out of an SCF energy calculation is that they are the set of orbitals naturally associated with the solution of the SCF problem.

And wouldn't there be an unlimited amount of equivalent solutions?

Yes. Every possible unitary transformation among the occupied canonical orbitals, of which there is indeed an unlimited number, would provide a set of orbitals that gives an equivalent total wavefunction.  Not all of these transformed orbital sets are especially useful, however. Specific examples of useful ones include the various localization transforms that are the topic of the question, the series of variously defined "natural orbitals", and (somewhat related) the Löwdin symmetric orthogonalization for atomic orbitals.

Is this just a mathematical gimmick or does this represent the reality? ... Or is this just a thing we do because it makes it easier to understand?

The transformed MOs do represent reality, and often provide a very powerful and/or convenient means for analysis and understanding of the overall wavefunction. You do have to know what is or is not a valid use of any particular set of transformed orbitals, however.
One specific application of transformed orbitals I know about that seems very powerful is the family of local pair natural orbital (LPNO) methods for accelerating correlated calculations. The v4.0 release of ORCA includes implementations of their advanced domain-based (DLPNO) version of the method for several calculation methods (MP2, CASSCF-NEVPT2, etc.; see various of the 2016/2017 Neese citations listed here).  The DLPNO-CCSD method has already proven to be extraordinarily powerful in enabling correlated calculations of large systems—the below figure is a reproduction of Figure 5.5 from the ORCA v3.0.3 manual showing the enhanced performance of (D)LPNO-CCSD methods for a series of benchmark systems (click to enlarge):

Is it possible for a molecule to be in such a state where every electron is in such a localized orbital?

Yes, this is always true, as long as the transformation that is applied to the occupied orbitals includes no unoccupied orbitals. No matter what set of orbitals you use, whether canonical or transformed, every occupied orbital in the wavefunction is always fully (here, doubly) occupied. Transformation among the occupied orbitals doesn't change the occupancy.
Inclusion of some un-occupied orbitals (often called virtual orbitals) in the transformation, though, will lead to orbital occupations between zero and two. This is what happens in the classic natural bond orbital analysis of Weinhold & co., as can be seen in the sample natural population analysis output (pdf link; "Occupancy" column) provided on the GENNBO website .


What's the future of an organic and/or analytical chemist? what can I do with a degree in analytical chemistry other than working in the lab?
And what is the future of being an organic chemist?
Can I be a researcher in both of them?
Or it's better to take organic chemistry if I want to be a researcher?

The answers to this question are going to be as varied as the people answering them. Let me give you the advice I give my undergraduate advisees.

The future is what you make of it. Don't let random people on the internet tell you which field to specialize in or which research to do. The decision need to be personal, because you need to care about what you are doing so that you are motivated to excel.

  Instead of worrying about which type of chemistry to study, find the problems you want to solve.

Research in chemistry is pretty interdisciplinary nowadays, both across the old subfields of organic, analytical, physical, etc. and between chemistry and the other sciences. Rather than become obsessed about organic vs. analytical or whatever, read about current research in chemistry. Think about the types of research problems you want to solve. Usually that will help you figure out which pathway to follow (i.e. which courses you need to take, which specializations to pick up, which lab to join, etc.). Research, both in academia and elsewhere is going to be more problem- than field-focused.

As an undergraduate, I was torn between organic and physical chemistry. I chose organic chemistry as a focus in graduate school, joined a research group focusing on polymer synthesis, and did a lot of analytical work as part of my research. We collaborated with a theoretical group to understand how monomer sequence would affect optoelectronic properties of conjugated oligomers and polymers, so I also had to learn a lot about the physical models of my conjugated systems. Another part of my group was working on polyesters with controllable biodegredation pathways for applications in medicine. The team on that project included grad students focusing on organic, inorganic, and analytical chemistry collaborating with engineering students in another group.


Changes in freezing point of sodium acetate Background:
I was recently playing around with some reusable heat packs, the kind that operate by phase change of sodium acetate trihydrate and noticed that they didn't reach the temperature that I had expected (58 °C) rather, they stayed at 52 °C.

There were two such packs, roughly 50 mm × 100 mm × 15 mm and placed together with the broad sides in contact. Between them and at the center, there was a calibrated PT-100 RTD (which I certainly trust to be better than 1 °C accurate). The entire mass was then insulated with a few layers of thick wool fabric. The temperature reading became constant at 52 °C within a few minutes of crystallization being initiated in both packs and remained very stable for more than 10 minutes thereafter.

It's worth noting that although the manufacturer labeled the contents as being 'Sodium acetate (food grade)' and water, no proportions are given. Also, the contents of the pack are bright red, indicating the presences of unlisted dyes.

Question:
What would be the most likely causes for the discrepancy? Contaminants, incorrect, proportions something else?

I'm ultimately interested in using this reaction to maintain ~58 °C in an experiment. I'd like to avoid the same result if I mix up a batch of this stuff from pure ingredients.

My assumption  is that the addition of water (other than water of hydration) lowers the temperature - consider that the heat of crystallization is shared among the sodium acetate, the added water and even the packaging. Perhaps the manufacturer adds water to lessen the chance of burns.

You'd need to experiment with lab-grade chemicals to be sure, such as this ~99% pure sodium acetate trihydrate.


Is most of the weight you lose actually released as carbon dioxide when you exhale? I've heard that most of the weight you "lose" is actually carbon dioxide released when you exhale. In percent its much more than what is released in pure heat energy and feces. Does anyone have statistics on what's the split between where the energy leaves your body?

How many percentage of food weight is actually released through carbon dioxide when its burned in your body? For example when you eat a hamburger and fries, is all the carbon released later when you exhale? How much is heat energy? Provided it does not go into fat reserves.

Let's start with:

  How much is heat energy?

2000 kilocalories per day, considering $E=mc^2$, is
$9.3 \times 10^{-8}$ grams

  is all the carbon released later when you exhale?

No.  Brainstorming, the others ways would include: feces, urine, perspiration, flatulence, shedding skin, hair and nails, mucous from nose and mouth, for women: menstruation, vaginal discharge, lactation, and giving birth, and for men ejaculation.

According to The Value of Human Feces:

Per day the amount of carbon excreted in feces is 21 grams and in urine is 7 grams.

According to The Composition of Human Perspiration Proc R Soc Med. 1934 May; 27(7): 839–848, a typical amount of perspiration a day is 700mL, and per 100mL the main carbon containing compounds are 20 mg urea, 20 mg glucose and 5mg amino acids.  So clearly less than 1 gram of carbon loss per day from perspiration. If we were considering loss of salts instead of just carbon, that would be about 3 grams per day from perspiration.

For menstruation, blood loss is about 40mL per cycle, and blood is 92% water, so maybe 1 gram of carbon on a heavy day of a period.

Skin loss per day is 1.5 grams according to Cells for life, cells for an instant.

A highly controlled study having a person in a chamber for 24 hours and analyzing the air found 653g of CO2 or 178g carbon "A Respiration Calorimeter with Appliances for the Direct Determination of Oxygen" Carnigie Institution of Washington Publication 42.

So, for a person that is not lactating or giving birth, I would estimate about 30 grams per day of carbon loss is other than from exhaling carbon dioxide.  Therefore:

About 86% of carbon loss is through exhaling carbon dioxide.

However, this will certainly vary depending upon the amount of indigestible carbon such as fiber, in ones diet.

Also, if mass loss through nitrogen and phosphorous (but not hydrogen and oxygen) were also considered the loss through breathing would come down to about 80%.  Inclusion of salts would bring this percentage down further.

The sum of water from feces, urine and respiration is about 1.8 kg per day and the total mass emission of the body is 2.6kg, according to the Carnigie source above.


Is benzocyclobutadiene aromatic or anti-aromatic? My teacher told that benzocyclobutadiene is both aromatic and anti-aromatic. I do not understand why. Can anyone explain?

Benzocyclobutadiene has an aromatic benzene ring as well as an anti-aromatic cyclobutadiene ring. This gives it the characteristics of both aromatic and anti-aromatic compounds.
Now, according to Cyclobutadiene and its related compounds, [1,p 180]:

Benzocyclobutadiene, the monobenzo derivative of cyclobutadiene, is intermediate in structure between the unstable parent compound, cyclobutadiene and it stable dibenzo analog biphenylene. Benzocyclobutadiene and its derivatives resemble cyclobutadiene rather than biphenylene in being known as only transient reaction intermediates.

Reading this, we can say that benzocyclobutadiene resembles cyclobutadiene more than biphenylene and so has a dominant anti-aromatic character. This means that like cyclobutadiene, the cyclobutadiene part of the ring undergoes dimerisation to form a more stable compound.
Note: The preview of the book referenced has an entire chapter based on just benzocyclobutadiene. If interested, you can go through it for more information
References:

Cyclobutadiene and Related Compounds; M.P. Cava, M.J Mitchell; Chemistry Department, Wayne State University, Detroit 1967. ISBN 978-01-2164-450-5. [Preview]


How and why are all the substituents in methyl cyclopropane eclipsed? Why does methyl cyclopropane $\ce{C3H5CH3}$ adopt an eclipsed conformation?

The geometry of the molecule is highly restricted by the presence of the ring.

The presence of the ring caused the $\ce{C-C-C}$ ring to be planar, because every triangle is planar.

Since there is only one carbon atom directly to the right to the circle, it must be eclipsed, so are the hydrogen atoms by VSEPR.


Finding the pH of a mixture of weak acid and strong base What are the steps we go about in finding the pH of a solution of strong base and weak acid?
Here is the question that I've been given

What is the pH of a solution made by mixing $\pu{50 ml}$ of $\pu{0.2 M}$ $\ce{NH4Cl}$ and $\pu{75 ml}$ of $\pu{0.1 M}$ $\ce{NaOH}$, when $\mathrm pK_\mathrm b(\ce{NH3}) = 4.74$?
A. 7.02
B. 13.0
C. 9.73
D. 6.31

What I did to solve it was to use the Henderson equation for buffers,
$$\begin{align}
\mathrm{pOH} &= \mathrm pK_\mathrm b + \log{\frac{[\text{salt}]}{[\text{base}]}} \\
&= 4.74 + \log{\left(\frac{0.2 \cdot 50}{0.1 \cdot 75}\right)} \\
&= 4.74 + \log{\frac{4}{3}} \\
&= 4.86
\end{align}$$
and thus $\mathrm{pH} = 14 - \mathrm{pOH} = 9.14.$
The correct answer is C. But on putting values I'm getting the wrong answer. Can you help me figure out why? Also, shouldn't the buffer solution have a common ion?

Buffer equation

The Henderson equation for buffers is:

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log{\frac{[\ce{A-}]}{[\ce{AH}]}}$$

$\mathrm{p}K_\mathrm{a}$ and $\mathrm{p}K_\mathrm{b}$ add up to 14, as do $\mathrm{pH}$ and $\mathrm{pOH}$. So the expression for $\mathrm{pOH}$ is:

$$\mathrm{14 - pOH = 14} - \mathrm{p}K_\mathrm{b}  + \log{\frac{[\ce{A-}]}{[\ce{AH}]}}$$

or

$$\mathrm{pOH} = \mathrm{p}K_\mathrm{b}  - \log{\frac{[\ce{A-}]}{[\ce{AH}]}}$$

Amounts of weak acid and weak base

  Also, shouldn't the buffer solution have a common ion?

You start with $\pu{10 mmol}$ of ammonium salt (weak acid), to which you add $\pu{7.5 mmol}$ of $\ce{NaOH}$. The result is $\pu{7.5 mmol}$ of ammonia (weak base) with $\pu{2.5 mmol}$ ammonium salt remaining. So the ratio of weak base to weak acid is 1:3.

If you plug this into one or the other buffer equation, you get the answer.


When the heat is absorbed by the ice then ice will melt to form liquid water My new try:
Given data

Energy given by reaction =1690 kJ

Initial temperature = 0 oC

Final temperature = 50 oC

C water = 4.18 Jmol-1 K-1

?Hfus ice = 6.01 kJ*mol-1

Calculate Mass of ice = ?

Ice melts at constant temperature of 0 degree C and changes the phase from solid to liquid .

Then liquid water absorbs the energy and changes from 0 degree C to 50 degree C

Change in temperature of the water is ?T = (50 oC - 0 oC )= 50 oC

Let’s set up the equation as follows to calculate the mass of ice

q= (m*?Hfus) + (m*Cwater * ?T)

where , q = energy , m = mass , ?T = (Tf-Ti)

convert given energy from unit kJ to Joules

(1690 kJ * 1000 J)/1kJ = 1690000 J

Convert ?H fus from kJmol-1 to Jg-1

6.01kJ*mol-1 (1000 J/1kJ) = 6010 Jmol-1

6010 Jmol-1 / 18.0148 gmol- = 333.6 J *g-1

Convert C water from J per mol to J per g

4.18 Jmol-1 oC /18.0148 g = 0.232 J*g-1 oC-1

Now lets put the values in the formula and calculate the mass of ice

q= (m*?Hfus) + (m*Cwater * ?T)

1690000 J = (m X 333.6 Jg-1) + (m X 0.232 Jg-1 *oC-1 * 50 oC )

1690000 J = m X 345.2 J*g-1

1690000 J /345.2 J*g-1 = m

4895.7 g = m

Please help me!!

Energy needed for melting and heating up the water

$$
Q = \Delta H_{\text{fus}} \cdot m + m \cdot c_p \cdot \Delta T
$$

solve for $m$:

$$
m = \frac{Q}{\Delta H_{\text{fus}} + c_p \cdot \Delta T }
$$

The molar heat capacity for water is around $75\,\frac{\text{J}}{\text{mol K}}$. You've been given the specific heat capacity which is around $4.18\,\frac{\text{kJ}}{\text{kg K}}$. All that's left to do is convert the enthalpy of fusion to $\frac{\text{kJ}}{\text{kg}}$ and you can solve the equation.

$$
\Delta H_{\text{fus}} = \frac{6.01\,\frac{\text{kJ}}{\text{mol}}}{18.02\,\frac{\text{g}}{\text{mol}}} = 0.3335\,\frac{\text{kJ}}{\text{g}} = 333.5\,\frac{\text{kJ}}{\text{kg}}
$$

Calculate $m$:

$$
m = \frac{1690\,\text{kJ}}{333.5\,\frac{\text{kJ}}{\text{kg}} + 4.18\,\frac{\text{kJ}}{\text{kg K}} \cdot 50\,\text{K}} = 3.12\,\text{kg}
$$

The combustion of 1 mol of propane supplies enough heat to melt 3.12$\,$kg of ice and and heat the resulting water up to 50$\,$°C


What is a barrier-less reaction in Quantum Chemistry? After finding a reaction "Transitional State" by GAMESS, and finding Hessian Eigenvalues, my project instructor said:

  seems your imaginary frequency (negative eigenvalue) is not that big (it is around -13.01), which means your reaction is barrier-less.

As I am a beginner in Quantum Chemistry, I could not figure out what is the relation between a small imaginary frequency and being barrier-less.

As I remember from the Arrhenius equation, being barrier-less has something to do with higher temperature leading to a slower reaction, but I could not understand its relation to imaginary frequencies.

Usually, in the reactions with a barrier, the saddle point with one imaginary frequency is taken as TS. However, many unimolecular dissociation reactions are barrierless: the reaction path on potential surface, connecting (minimum of) reactant with (minima of) products, is such that the energy along the path does not have a local maximum. Kinetic treatments of such reactions are more difficult: it is still possible to assign a TS, but it no longer a saddle point. In order to find it, one can use variational TST, but it is somewhat expensive.

A general recommendation after location of a saddle point: run reaction path calculation to see that this saddle point indeed connects the products and reactants that you expect.

About this small eigenvalue: this eigenvalue can easily pop up instead of zero eigenvalues of rotations and translations. Even if your calculation explicitly projects them out, numerical noise can be this order of magnitude (as pointed out above).


Why do shampoo ingredient labels feature the term "Aqua"? I keep seeing the term "Aqua" in the ingredient labels on several shampoo varieties, but I really don't see why it should be there in the first place.

I mean, if the manufacturers just wanted to say it contains water, couldn't they've printed out "Water" instead?

Or could it be that "Aqua" is slang for purified water (or water that's been treated in some godforsaken way) in the shampoo industry? Well, I guess there could always be the possibility that the manufacturers think "Aqua" sounds a lot fancier than plain ol' "Water".

So why's the term "Aqua" mentioned there?

In most countries, cosmetic product labels use the International Nomenclature of Cosmetic Ingredients (INCI) for listing ingredients.

The INCI name “AQUA” indeed just describes water (which is used as a solvent).


What is the most common oxidation state of gold? I had a test this morning which had the following question:

  What is the most common oxidation state for gold?
  (a) +1  (b) -1  (c) +2  (d) +3

Since the electronic configuration of $\ce{Au}$ is $\ce{[Xe] 4f^{14} 5d^10 6s^1}$. After removing the $\ce{6s^1}$ electron and gaining $+1$ charge, $\ce{Au}$ should have noble gas configuration and hence this should be the most stable configuration. So the answer has to be +1.
Could someone verify my reasoning?

It's not obvious, but common oxidation state for gold is +3. It caused by destabilization of the $5d^{10}$ orbital. Detailed explanation you can find in The Chemistry of Gold, in Chapter 1.1.3.


When does a reaction equation require the reversible sign and when does it not? Sometimes my professor puts the $\to$ symbol and sometimes he uses $\leftrightharpoons$. How do I know which to use? I get that one means the reaction only heads one way while the other means that the reaction holds at an "equilibrium" where the reactants are converting into products and the products are converting into reactants. But if I'm only given a word problem that lists 2 reactants and their concentrations, how do I know which to use?

You just need to learn and know. Luckily, there is a certain logic to the question which reactions can be reversible and which cannot. However, a full list is impossible and would exceed the space of this margin.

For most mechanisms, there is (at least) one point-of-no-return step. Learn to identify this step. All other arrows should probably be $\ce{<=>}$, if you’re strict. These non-reversible steps are often (but not always!) carbon-carbon bond forming reactions, as a very rough guideline on which is which.


Naming convention for graphs? I always have trouble naming my graphs in class. They never seem too scientific or professional.

For example, we are reviewing density now, and had to graph the mass (Y) of different volumes (X) of water. So, very simply, my title was "Mass of Different Volumes of Water". Is this appropriate?

Are there any naming conventions, such as "X vs. Y", or "Effects of X on Y" or "Relationship of X and Y", etc.? Sorry if this is a bit stylistic and opinion based, but I would like some feedback. Thanks.

Note: I'm in the tenth grade.

Note 2: I'm also not sure what site to post this under (there's no "science" site) nor what tags to post this with. Feel free to improve it.

The most general way of naming a graph is by writing,

Y vs X graph

An example is, Distance vs Time graph. In which the Y axis represents Distance and X represents time.

In order to know which comes in which axis you have to have a clear understanding about independent and dependant. Thus is easy to understand. Once you have clearly understood the axes you have to make the Y axis the dependant variable and X the independent one.

Then the name of the graph becomes ,

Dependant variable vs Independent variable.

For instance consider the Distance vs Time graph.
We see that Distance depends on how long has the time passed. Thus D becomes the dependant variable


Can a spectrometer be used to figure out spice ratios in a sauce? There is a really good sauce that a local restaurant makes. I have had trouble recreating it, but have gotten somewhat close. I had the following idea:

I could take an "image" of the sauce's color spectrum under good light using a USB spectrometer, and compare it to the spectrum of my home made version. I could then slowly tweak the ingredients and resample the color spectrum of the home made sauce until it started converging and eventually matched the restaurant's.

What could prevent this from working well?
Could this realistically outperform taste testing?

Unfortunately, this will not even qualitatively work because taste which is a sense of perception and UV-Vis are barely related.. The USB spectrometer would be great for color matching, but you are assuming the taste and aroma comes from UV-Vis absorbing compounds only. This may not be the case. Let us assume, the taste comes from UV-Vis absorbing compounds, as you know from Beer's law, an $n$-component mixture will have  absorbance sum from $n$-components. If some of these molecules are in trace quantities, they might have a little effect on the overall spectrum but they might have a huge influence on the taste and aroma of your sauce. Food chemists prefer more of gas chromatography and mass spectrometry & infrared spectrometer which may cost more than 100K.


Why m-hydroxybenzoic acid has higher acidity then p-hydroxybenzoic acid? Why  m-hydroxybenzoic acid has higher acidity than p-hydroxybenzoic acid?

I have a reasoning that since hydroxy group is an activating group, then it increases the electron density on its para position. So for  p-hydroxybenzoic acid, the carbonyl carbon has a greater ability to share molecular orbital's electron cloud to its attached hydroxy group. So the hydroxy group doesn't release $\ce{H+}$ easily. But that's not the case for m-hydroxybenzoic acid.

Is it right?

The -$\ce{OH}$ group has two effects, namely +R and -I effect i.e. Resonance and Inductive effect. When -$\ce{OH}$ group is attached to the para position, it has a tendency to delocalize it's electrons towards the aromatic ring (i.e., benzene). Thus the electron density on oxygen atoms of carboxylic group increases and the polarity between $\ce{O}$ and $\ce{H}$ decreases leading to less acidity. Whereas when the -$\ce{OH}$ group is attached to the meta position, +R effect is not applicable and only -I effect is effective. Thus it withdraws the electrons from the ring, and increasing the polarity between $\ce{O}$ and $\ce{H}$ in carboxylic group, making it more acidic.


How many cis-trans isomers does 1-O-stearoyl-2-O-arachidonyl-phosphatidyl-inositol-4',5'-bisphosphate have? I saw an exam that had a question asking the number of cis-trans-isomers of 1-O-stearoyl-2-O-arakidonyl-phosphatidyl-inositol-4',5'-bisphosphate:

The "correct" answer in the exam is 1023, but this is clearly not correct, because the stereoisomerism of the inositol ring is not called cis-trans-isomerism.

So, obviously only the arachidonyl chain has cis-trans isomerism. But if an isomer has several of the double bonds in trans orientation (not just one), is is still correct to call it a cis-trans isomer of the original molecule. If it is, I would say that this molecule has 15 cis-trans isomers. If it isn't I would say that it has only 4 cis-trans isomers.

Which one is correct?

Perhaps the answer is correct. If by cis/trans isomers the question includes the four stereogenic double bonds of the arachadonic acid chain and the six stereogenic carbons of the inositol ring, then there is a total of ten or 210=1024 isomers. But the answer is 1023 stereoisomers. That is because the question says "cis-trans-isomers OF...", which means NOT including the one shown. [210-1=1023.] Had the question asked "How many stereoisomers OF ...", then the lone stereogenic center of the glycerol moiety would have to be included. Then the answer would be 211-1=2047.


What component(s) of instant coffee make it deliquescent? If I put small, separate piles of instant black coffee powder, sugar, powdered creamer, and even something very ionic like table salt out in humid air and come back in a few days, only the instant coffee will have absorbed enough water to change to a sticky tar-like blob while the other three won't change their appearance.

Is it known what substance it is in the  instant coffee powder that makes it hygroscopic and even deliquescent; giving it the ability to almost "make itself into coffee" by removing so much water from the air that it dissolves?

The behavior strongly depends on the temperature and humidity in which the coffee was kept and on the specific type of coffee.

From Tea and coffee powders, M. Huang, M. Zhang, in Handbook of Food Powders, 2013:

  Being hygroscopic, instant coffee particles are susceptible to the action of moisture – that is, they absorb moisture from the air. If the moisture content increases to 7–8%, the powder or granules can become a paste or solid mass (Clarke, 1987).

Clarke (R.J. Clarke, in Encyclopedia of Food Sciences and Nutrition (Second Edition), 2003) writes:

  Instant coffee is relatively hygroscopic, easily picking up moisture from the atmosphere and caking at about 7–8% w/w moisture content. For example, to keep instant coffee below 5% (dry basis) moisture content, the relative humidity of the air with which it is in contact must be below 35–40%, though the precise value depends upon the nature of the instant coffee in question, primarily due to differences in porosity of the particular particles made. It is necessary, therefore, that jars of instant coffee be well sealed prior to sale.

My answer is admittedly incomplete. As pointed out in comments to the post and in the above quotes, porosity (characterized by e.g. the mean particle surface-to-volume ratio) influences condensation, but so do the intrinsic chemical properties of the particle surface.

The chapter by Huang and Zhang referenced above provides the following table (Table 20.1) detailing the composition of instant coffee powders:

  Constituent   Content(% dry weight basis)

  Minerals                  7.6–14.6

  Reducing sugars (glucose) 3.2–13.1

  Galactomannan             2.4–10.5

  Low molecular organic acids   12

  Brown pigments             15–28

  Caffeine                   2.5–5.4

  Trigonelline               1.56–2.65

  Source: Belitz et al. (2009).

Simple sugars are often hygroscopic as may be organic acid compounds, and caffeine. In particular the presence of glucose (part of "invert sugar") could account for much of the absorption of water. In addition, minerals such as various potassium salts (which make up much of the mineral content of processed coffee beans) are known to be deliquescent.

The Belitz reference appears to be: Belitz et al. (2009). Coffee, tea, cocoa. In Food Chemistry (4th Edition), ed. H.-D. Belitz, W. Grosch, and P. Schieberle, 938-970. Germany: Springer. However haven't accessed the original pubs due to paywalls.


Electrochemical manufacturing of small metal components Metal crystals easily can be grown via electrochemical pathways. Copper is a great example of this. I recently had an idea of growing those crystals in controlled ways (i.e.: surrounding the electrode on which the crystallization occurs with a "mold" of the desired shape with holes in it) but these crystals are tending to grow towards the another electrode in the shortest way possible. Let's say that I would like to grow a copper tetrahedron on a triangular copper plate surrounded by plastic sheets which are controlling the crystal growth.

  The problems which I came across:
     1. How should I arrange my equipment to get the crystals fill the mold?
     2. Shold I worry of liquids being trapped in the crystal?

Nice idea, but difficult to do.

The copper cations must continuously enter the "mold" electrode to grow there, but they must not be able to grow outside.
If there are holes in the "mold" the crystals will grow towards them because there is where the cations will come from.
A permeable "mold" would let cations in over all the surface, but nothing would stop them to continue growing outside the membrane.

If the crystals tend to grow to the other electrode in the shortest possible path, you could place a very small electrode just above the center of the triangle of the other electrode.
The shape will, however, not be very defined.
Maybe in combination with a permeable "mold"...

Trapping of liquids inside the piece will be almost certainly a problem.
You are not growing a single crystal.
Many crystals will start growing all at the same time in different places of the electrode.
When they join somewhere, they will not fit nicely together, so some liquid will be trapped.
This occurs also when big crystals are made in a normal crystallization, especially when there is no stirring.


Am I limited by the pKa value of an weak acid when preparing a buffer solution of it (and its salt)? If I was to make an acidic buffer system with the ph of, for instance, 4, would I be limited with my choice of weak acids? That is, how big does the difference between pH and pKa have to be in order for the weak acid not to be usable? Or is that not the case? I am very well aware that the optimal weak acid for has to have a pKa value quite close to the pH of the wanted buffer because they offer the most buffering capacity.
Also - Can someone hook me up with some equations applicable for making/dealing with buffers other than the Henderson Hasselbach equation? Are there any equations that can help me predict the buffer capacity?

If I was to make an acidic buffer system with the ph of, for instance, 4, would I be limited with my choice of weak acids?

Yes you are limited to weak acids that have a pKa of +/-1 of your desired buffers pH value. This is a rule of thumb used in many labs.Some theroetical background is quite nicely explained on wikipedia wikipedia#buffering  capcacity.

This reference also includes some useful equations related to buffering capacity.