What is the concentration of H2 for the equation H2+I2=2HI if the equilibrium constant is 64 and the initial concentrations are both 1.9 So for this, I set up the equation =

$$64= \frac{(x)^2}{(1.9-x)^2}$$

I took the sqrt of both sides and got

$$8=\frac{x}{(1.9-x)}$$

So then I did $$8(1.9-x)=x$$

And simplified I got $$15.2=9x$$

So $$x=1.688$$

So to find concentration it’s Initial- Change

$$1.9-1.688=.212$$

But it’s telling me this answer is wrong

let the amount of $\ce{H2}$ reacted = $\pu{x~mol}$
Use the molar ratio to detemine the amount of $\ce{I2}$ reacted and the amount of $\ce{HI}$ formed:
$$\begin{align}
&\text{The amount of }\ce{H2}~\text{reacts to reach equilibrium }&=\pu{x~mol}\\
&\text{The amount of }\ce{I2}~\text{reacts to reach equilibrium}&=\pu{x~mol}\\
&\text{The amount of }\ce{HI}~\text{formed at equilibrium }&=\pu{2x~mol}\\
\end{align}$$
Use the following table to determine the amount and the concentration
of each species at equilibrium :
$$\begin{align}
\ce{&H2 &&+ &&O2 &<=> &&2HI\\
I~~~~  &1.9   &&&&1.9      &&&0 \\
C ~~~  &-x    &&&&-x       &&&2x \\
E ~~~  &(1.9-x) &&&&(1.9-x) &&&2x}
\end{align}$$
Th concentrations at equilibrium :
$$[\ce{H2}]_\mathrm{e}=(1.9-x)  , [\ce{I2}]_\mathrm{e}=(1.9-x) ,  [\ce{HI}]_\mathrm{e}= \pu{2x~M} $$
Sustitute the concentrations in the following formula:
$$\begin{align}
K_\mathrm{C} = \frac{ [\ce{HI}]_\mathrm{e}   }  {[\ce{H2}]_\mathrm{e}[\ce{I2}]_\mathrm{e} }\\
64= \frac{(2x)^2}{(1.9-x)(1.9-x)}
\end{align}$$
Take the square root of both sides :
$$\sqrt{64}=\sqrt{\frac{(2x)^2}{(1.9-x)^2}}$$

$$8=\frac{2x}{(1.9-x)}$$
- Solve for $x : x=1.52$

Calculate the concentrations at equilibrium :
$$\begin{align}
[\ce{H2}]_\mathrm{e}&=(1.9-x)&=(1.9-1.52)&=\pu{0.38~M}  \\
[\ce{I2}]_\mathrm{e}&=(1.9-x)&= (1.9-1.52)&=\pu{0.38~M} \\
 [\ce{HI}]_\mathrm{e}&= \pu{2\times{x}}&= 2\times{1.52}&=\pu{3.04~M}
\end{align} $$


How to apply the Heisenberg Uncertainty Principle in calculations on the movement of bacteria? In this problem, I know the formula, but I get confused with all the steps and how to use what I got, to get to what I need.

A student is examining a bacterium under the microscope. The E. coli bacterial cell has a mass of $m = \pu{0.100 fg}$ (where a femtogram, $\pu{fg}$, is $\pu{10^{−15} g}$) and is swimming at a velocity of $v = \pu{9.00\mu m/s}$, with an uncertainty in the velocity of  $9.00\%$. E. coli bacterial cells are around $\pu{1 \mu m}$ ($\pu{10^{−6} m}$) in length. What is the uncertainty of the position of the bacterium?

I used the formula $$\Delta x \Delta p \geq \hslash/2$$
I plugged in these numbers:

Mass of bacteria $(m) = \pu{0.100 fg} = \pu{0.1 \times 10^{-15} g} = \pu{10^{-16} g}$
Velocity of bacteria $(v) = \pu{2 m/s}$
Uncertainty in velocity = 9.00

I am getting this answer $1.67\times10^{-9}$, and it is wrong. I don't know what I am doing wrong.

$\newcommand{\upmu}{{\large\unicode[Times]{x3BC}}}%Remove this line when upright greek characters are implemented by mathjax$
(Hover over the boxes to see the results.)
The velocity of the  bacteria is $\mathbf{v} = 9~\mathrm{\upmu m/s} = 9\times10^{-6}~\mathrm{m/s}$, with an uncertainty of $9.00\%$ you can calculate the absolute uncertainty in the velocity $\Delta \mathbf{v}$ as the difference of the upper bound $(\mathbf{v} + 0.09\mathbf{v})$ and the lower bound $(\mathbf{v} - 0.09\mathbf{v})$.

   $$\Delta \mathbf{v} = (\mathbf{v} + 0.09\mathbf{v}) - (\mathbf{v} - 0.09\mathbf{v}) = 0.18\mathbf{v} = 1.62\times10^{-6}~\mathrm{m\,s^{-1}}$$

The momentum is defined as
$$\mathbf{p} = m\mathbf{v},$$
hence the absolute uncertainty in the momentum is
$$\Delta\mathbf{p} = m\Delta\mathbf{v}.$$
The mass is $m = 0.100~\mathrm{fg} = 1.00\times10^{-19}~\mathrm{kg}$.

   $$\Delta\mathbf{p} = m\Delta\mathbf{v} = 1.62\times10^{-25}~\mathrm{kg\,m\,s^{-1}}$$

Now you can use this to calculate the uncertainty in position $\Delta\mathbf{x}$.
$$\Delta\mathbf{x}\Delta\mathbf{p} \geq \frac{h}{2}$$
Planck's constant is $h = 6.62607004\times 10^{-34}~\mathrm{m^2\,kg\,s^{-1}}$.

   $$\Delta\mathbf{x} \geq \frac{h}{2\Delta\mathbf{p}}\\\Delta\mathbf{x} \geq 2.07\times10^{-9}~\mathrm{m}$$

Since you have not included any of the steps how you arrived at your result, I cannot help you finding your mistake.


What is left over when water and sodium hypochorate decompose, i.e. household bleach? I am having some confusion in understanding the decompostion of household bleach. One website says bleach decomposes to  salt and sodium chlorate, simple enough.
Orther wedsites say bleach decomposes to salt and nothing else. I called Clorox customer service and their vote is it converts to salt only. I have 3 votes for salt only and 2 votes for salt and sodium chlorate thus far.

It depends on temperature. At higher temperatures, hypochlorite has tendency to disproportionate: $$\ce{3 ClO- -> 2 Cl- + ClO3-}$$ Otherwise it mostly releases atomic oxygen : $$\ce{ClO- -> Cl- + O}$$
In reality, it would be combination of both in various ratio, depending on relative reaction rates,  all ending as mixture of sodium hydroxide, carbonate, chloride and chlorate.


Electron defficient molecules Electron defficiency is a common feature in some class of compounds of Group 13 elements (B,Al,Ga). In some covalent compounds of Be too the same nature prevails. To compansate this electron defficiency they undergo dimerization forming bridges('3c 2e' or coordination) or in some cases they get stability through backbonding (halides of group 13 elements). Hydrides of them are known to stay in dimeric form through bridging.
Halides of B are found to remain in monomeric form due to backbonding where as 'Be, Al'-halides highly favour bridging structures through coordination of halide. Why is such difference? In Be,Al, vacant p orbital is present too to favour backbonding.
Then why Be,Al favour dimerization (or polymerization) over backbonding in their halides?

I believe this may be a duplicate of this question:
Why is BCl3 a monomer whereas AlCl3 exists as a dimer?
From what I remember, the stronger the backbond, the smaller the tendency to undergo dimerisation, however, way back in the dark ages when I was a student and asked my second-year chem prof, his answer was that the simple reason why none of the tri-halides of boron undergo dimerisation is steric hinderance - i.e. the boron atom is just too small to accommodate the fourth larger Cl/Br/I atom. He then pointed me to several sets of further reading.
However, you will probably find a better answer at one of the following links:

Mulliken J. Chem. Phys. 3, 635 (1935); https://doi.org/10.1063/1.1749567
Lappert, et.al. Chem. Soc., Faraday Trans. 2, 1976,72, 539-55 https://doi.org/10.1039/F29767200539


Fermentation of glucose I have trouble understanding as to what the last part of the increase in liquid level means. I believe this is the foam formed after the fermentation? I am calculating for the pressure but I have trouble as to what volume should I use to find the moles since I don't understand the last part. It would greatly help me to understand this process first.
I've been extensively searching how the fermentation process actually occurs cause I don't understand a thing about it. To find the pressure from what I know, I converted the volume of glucose to $\ce{CO2}$ to find the moles and plugged the value in the ideal gas equation $p=nRT/v$ with the volume being $\pu{0.07L}$ from $(\pu{125mL} - \pu{55mL})$, because it says the $\ce{CO2}$ in solution is negligible.
I don't know where the changes in densities come in and I don't know how to incorporate the pressure brought about by the liquid solution. I think my thought process is wrong really cause I don't understand fermentation.
https://www.homebrewtalk.com/threads/calculating-total-gas-produced-in-fermentation.165020/ This site proposed a $p_1v_1=p_2v_2$ method where $p_1$ is at STP, but I don't think this is needed in the problem.


Heat resistant odor neutralizers I am looking for a heat-resistant odor neutralizer that can be released into the air at 200 degrees Fahrenheit and still be effective.

Does anyone have any ideas?

Metazene 99% or other Alpha Aromatics high heat compounds.

What is the mal-odor caused by, and the complete application?


Melting and boiling points of transition elements The melting and boiling points of transition elements increases from scandium ($1530~\mathrm{^\circ C}$) to vanadium ($1917~\mathrm{^\circ C}$). They increase because as we go across the group, we have more unpaired (free) electrons.

But at chromium ($1890~\mathrm{^\circ C}$) however, the melting point decreases even though it has more unpaired electrons than the previous atoms. Why does this happen?

I don't really like parts of this explanation (was hoping to find a better one here actually), but it's the best I know. I'll build on suggestions by @michielim and @vrtcl1dvoshun.

(Note: I think most of this argument can be transposed to the orbital overlap picture of bonding, but it might be slightly trickier to describe.)

For a backdrop to this answer, I have plotted the melting points and boiling points of the transition metals, so make sure to stare at these for a while before reading on.

From the physicists' "electron sea" point of view of metal bonding, the higher the ionic charge the metal atom can support, the higher the element's melting and boiling points. This explains why group 1 metals such as sodium have quite low melting/boiling points, since the metal would be composed of electrons delocalized in a $\ce{M}^+$ lattice. Going towards group 2 and group 3 elements, one can expect to find a $\ce{M}^{2+}$ and $\ce{M}^{3+}$ lattice, and so on. However, this does not mean that that, for example, metallic osmium is expected to be made out of $\ce{Os}^{8+}$ cations surrounded by a dense electron sea; as each successive "ionization" of the metal atom is performed, it becomes more unlikely that the next electron can become delocalized, since it has to fight against a higher effective nuclear charge relative to the electron before it. Therefore, we can expect some "optimal average charge" for the metal ions in the pure metal. Since melting and boiling points tend to increase towards groups 7 and 8, we can safely assume that this optimal average charge is at least greater than $+3$, and it is likely higher for the heavier elements since they have significantly higher melting/boiling points.

If we can somehow figure out how easily electrons can be removed from their parent atoms to create the electron sea, we may have an argument to explain the trends and apparent discontinuities. First, notice that the $4^{\ce{th}}$ period metals are in general a little bit underwhelming compared their $5^{\ce{th}}$ and 6$^{\ce{th}}$ period counterparts when looking at the boiling points. This can be attributed to the fact that the $3d$ orbitals are anomalously compact relative to higher $nd$ orbitals. This happens because all first occurrences of a given sublevel (i.e. $1s$, $2p$, $3d$, $4f$, $5g$, etc.) are composed of orbitals whose wavefunctions have no radial node. Since the wavefunction's value is monotonically increasing up to its maximum value, it tapers off quickly. Thus, when a 4$^{\ce{th}}$ period metal atom tries to delocalize its $3d$ electrons, it meets a slight increase in resistance due to its stronger interaction with the nucleus, meaning that the optimal average charge of the ions is slightly lower and leading to lesser interatomic interactions holding the metal together.

Now, for the sudden dips, most visible for $\ce{Cr,\, Mn}$ and $\ce{Tc}.$ As has been suggested, one is tempted to take into consideration the fact that they all have $d^5$ configurations, which represent a half-filled $d$ subshell. It can be argued that they are particularly stable due to exchange energy, though Cann suggests it is not the most adequate explanation and instead forwards the "parallel spin avoidance factor".

In half-filled subshell configurations, there is a maximum in the effective nuclear charge felt by the electrons (compared to the previous elements with no doubly-occupied orbitals) combined with relatively low interorbital repulsions due to the Pauli exclusion principle. The element that follows a half-filled subshell must now put an electron in a previously occupied orbital, creating stronger, intraorbital electron repulsion. These effects combine to suggest that removing an electron from or forcing an electron into half-filled configurations is particularly difficult (notice that $\ce{Cr}$ doesn't follow $\ce{Mn}$ and $\ce{Tc},$ more on that soon). If this is the case, then the delocalization of the third electron (which would create $\ce{Mn}^{3+}$ and $\ce{Tc}^{3+}$ ions) and all those beyond it would be somewhat suppressed. This would effectively reduce the optimal average charge of the ions in the metal, and decrease the strength of interactions.

Finally, $\ce{Cr}$ doesn't show the same electronic behaviour as $\ce{Mn}$ and $\ce{Tc},$ even though it has a $d^5$ configuration, and yet it is still anomalous. What gives? Well, according to Greenwood and Earnshaw (Chemistry of the Elements, 1997), the low boiling point of $\ce{Cr}$ is due to a combination of increased effective nuclear charge and the anomalously small size of $3d$ orbitals. When going from $\ce{V}$ to $\ce{Cr},$ the increase in nuclear charge is enough to pull the already small $3d$ orbitals quite close to the nucleus, disfavouring electron delocalization to the extent that a sudden drop appears. Mo has the same electron configuration as $\ce{Cr},$ yet it does not show nearly as large a dip because its $4d$ orbitals are already significantly more diffuse, and aren't so easily suppressed from delocalizing.


Significance of shapes of Conical flask and Kjeldahl flask What is the practical significance of the shapes of conical flask and Kjeldahl flask?

Question: What is the practical significance of the shape of conical flask (in US: Erlenmeyer flask)?

An Erlenmeyer flask (also known as a conical flask or a titration flask) is a type of laboratory flask, which consists of a flat bottom, a conical body, and a cylindrical neck. The flask is created by the German chemist, Richard August Carl Emil Erlenmeyer (1825–1909) in 1860, and hence its name. With its wide flat bottom, it was perfect for heating solutions (large surface area). Hot vapor condenses on the upper section of slanted sides of the Erlenmeyer flask, reducing solvent loss. Also, these slanted sides and small mouth of the flask cause the hot solutions in the flask to cool slowly, keeping the empty air space hot. This feature greatly aids certain techniques in chemistry, especially appropriate for recrystallization. For example, during the hot filtration step, when the hot solution is filtered through a fluted filter paper into the receiving Erlenmeyer flask, hot vapors from the boiling solvent would keep the filter funnel warm, avoiding the premature crystallization on the funnel and its stem.

Equally, the sloping sides prevent risk of spillage of solutions in it, therefore ideal to hold content, which needs to be mixed by swirling. This nature also makes them suitable for use in titrations. Again, the flat bottom of the flask has the advantage of giving its content a uniform color across the entire flask, which makes it easier to determine sharp end points.

  Question: What is the practical significance of the shape of Kjeldahl flask?

In 1883, Danish chemist Johan Kjeldahl (1849–1900) of the Carlsberg Laboratory published a method to determine nitrogen content in organic matter. This is the first accurate, simple, and speedy method to determine nitrogen content in organic matter including protein content in food, which bares discoverer's name, the Kjeldahl method. The first step of the method is the digestion of sample in concentrated sulfuric acid at $\pu{360–410 ^\circ C}$, which decomposes the sample by oxidation to liberate the reduced nitrogen as ammonium sulfate. This heating would cause splash-back, which is a treat because of conc. $\ce{H2SO4}$. Thus, along with his famous method, Kjeldahl had also developed a piece of laboratory equipment he developed in 1888, which bares his name as well, Kjeldahl's flask. The long-necked, round-bottomed (actually pear shaped) flask was ideal for avoiding splash-back when heating solutions. The long-neck also function as air condenser.


Do the letters ol in a substance only indicate that the substance is an alcohol if they came at the end of the name of a substance? Do the letters ol if they are together only indicate the substance is an alcohol if they come together at the end of the name of a substance? for example,is glycerol an alcohol because the letters o and l come together at the end of its name and olives and olive oil are not alcohols because the letters o and l do not come together at the end of any words in their names?

Yes, the ending ol in a name is a good hint that the compound contains an $\ce{OH}$ group.

However, beware of trivial names:

$\alpha$-tocopherol is not an aliphatic alcohol, but a substituted phenol
mannitol does not contain one $\ce{OH}$ group, but six of them

It's getting even more confusing if you consider other languages, such as German. Benzol is the German name of benzene, which has no $\ce{OH}$ group at all. The same is true for styrol, the German word for styrene ;-)


Is it possible to determine the age of a building by just looking at its window? I've recently come across the fact that glass is an amorphous solid and is known as a pseudo solid or a super cooled liquid. Our teacher told us that buildings that if we were to closely observe the glass in the window panes, we would find that it's thicker at the bottom. I'm now wondering if we can calculate the difference between the width of the top and bottom of the window, would it be possible to determine an approximate age of a building?

This is but an urban legend. Glass does not get thicker at the bottom. (Sure, glass is fluid to some tiny extent, and it can be demonstrated, but not in this way.) It is just that the glass panes were initially manufactured with uneven thickness. Then the people who stuck the glass into the frames would orient the thicker side down, maybe because it is more stable this way, or because it feels right, or just to mess with us. Sometimes they didn't pay attention and turned it randomly.

Supporting links: 1, 2.


Why does tryptophan absorb UV light? Why is that tyrosine, phenyalanine, and  tryptophan absorb UV light while other amino acids don't even absorb visible light? Does it have something to do with aromatocity?

Any molecule can absorb UV light. What the question is probably going for is why those three molecules absorb at longer wavelengths than other amino acids. This has to do with the conjugated pi bonds from aromaticity. Looking at a list of amino acids, we can see that most of them don't have conjugated pi ystems.

amino acids (source)

Conjugated pi bonds decrease the HOMO-LUMO energy gap the more they are conjugated. This allows longer wavelength (lower energy) light to excite the molecule by promoting an electron from its HOMO to LUMO.

UV light is in the range of about 10-400 nm. A single bond (sigma bond) would absorb around 130 nm. Ethene's single pi bond is at 165 nm. A benzene ring's conjugated double bonds peak primarily at 180 and 200 nm. Further conjugation can absorb longer wavelengths and, like anthracene, begins edging into visible light, which as a result has a yellow color and more transitions in the UV-VIS spectrum. NIST has UV-VIS spectra for both benzene and anthracene which you can compare.

Benzene

Anthracene

The more conjugated molecule has more transitions and absorbs at longer and longer wavelengths. This principle is why carrots are orange for example, because they contain a highly conjugated molecule, with 11 pi-bonds, that absorbs well into the visible light range. This molecule is beta-carotene:

All that said, the other amino acids do have double bonds and lone pair electrons (carbonyl groups) that can absorb light in the UV range (remember any bond, including single bonds, will absorb UV light if you go low enough). The lone pairs on a carbonyl group can absorb at higher wavelengths than benzene. These involve electronic transitions between a non-bonding electron to the LUMO; the transition energy is so low because a non-bonding electron does not have its energy lowered like a bonding orbital. The gap between a non-bonding orbital and anti-bonding orbital is therefore much smaller than usual.

There was a good post on pi-conjugation elsewhere on the site, here.
You can read more about carbonyl excitations here. For general excitation values, this page was useful.


Experimental calculation of solubility product constant In order to calculate the $K_\mathrm{sp}$ of lead(II) iodide, I mixed $\pu{6.0 mL}$ of $\pu{0.01 M}$ $\ce{Pb(NO3)2}$, $\pu{6.0 mL}$ of $\pu{0.02 M}$ $\ce{KI}$, and $\pu{8.0 mL}$ of water, and a precipitate (barely) formed.

From this, I calculated $[\ce{Pb^2+}]$ to be $\pu{0.0036 M}$ and $[\ce{I-}]$ to be $\pu{0.0060 M}$, giving

$$K_\mathrm{sp} = [\ce{Pb^2+}][\ce{I-}]^2 = 1.1 \times 10^{-7}.$$

However, the actual value (at $\pu{25 ^\circ C}$) is $7.9\times 10^{-9}$ (Skoog et al. Fundamentals of Analytical Chemistry), so my result is off by a factor of $\sim 10$.

I'm pretty sure I had clean equipment and measured carefully and accurately. Why is there a discrepancy? Is my maths wrong, or are there some inaccuracies in my experiment?

It's probably an experimental error. The 'barely' may be important here.

For one, you may not be sure how accurate your initial molarities are. A mistake in one of them ruins it all. Titrate them properly and check. Titration usually gives two significant digits.

Also, since its a cube overall, a small mistake can get much larger. You may want more accuracy in your readings (one significant digit isn't good enough), and you need to improve on the 'barely'. Though a factor of 11 needs a variation by a factor of 2.3 in both readings. So I don't know.

Have you redone the experiment? Doing an experiment more than once is integral to getting good values.


What is the driving force for imine formation? I know this that when a primary amine is reacted with a carbonyl compound, an imine is formed:

I'm wondering why the position of this equilibrium lies to the product side / right-hand side. Why is the formation of the C=N double bond favoured?

In practice, a drying agent is sometimes added to the reaction: I've used molecular sieves before, but a cursory database search brings up a few other options, e.g. magnesium sulfate, silica gel, or Montmorillonite (this list is probably not exhaustive). These remove water from the system and pull the equilibrium position over to the right-hand side.

That's one possibility, but I don't think that's all there is to it - I think a lot of imine formations proceed perfectly ok even without dehydrating agents being added to them. However, you'll have to hear about these from somebody else, as I don't know what would drive those reactions. It could well be that simply adding an excess of either amine or carbonyl drives the reaction to completion.

Interestingly, a 2009 paper by Saggiomo and Lüning1 describes their investigations into supposed imine formation in water. Apparently, both the carbonyl compound (benzaldehyde or salicylaldehyde) and amine (aniline) do not dissolve in water - not particularly surprising. The imine did not form appreciably in water and only started to form when water was removed during workup and purification.

P/S This is just the top hit I found on Google, it's hardly a complete literature search.

Saggiomo, V.; Lüning, U. On the formation of imines in water—a comparison. Tetrahedron Lett. 2009, 50 (32), 4663–4665. DOI: 10.1016/j.tetlet.2009.05.117.


Is one syn aldol product preferentially formed over the other? In Carruthers' Modern Methods of Organic Synthesis, the following aldol reactions of two (Z)-enolates with benzaldehyde are shown:

The two transition states shown seem to be different, and lead to different stereochemistry in the product (the first product has both Me and OH pointing away from the plane, whereas the second has both pointing out of the plane).

Is this significant, and if it is, how can we predict which TS is used for which problem specifically?

Ignoring the difference between the SPh and Et groups, the two products shown are enantiomeric, not diastereomeric. Both of these reactions are not enantioselective in any manner, so both enantiomers will be formed in equal amounts, via both transition states. Presumably, in the book, only one enantiomer is shown to save space and to avoid redundancy.

The only thing that needs to be illustrated here is that the phenyl group is equatorial in the lower-energy transition state. That leads to the diastereoselective formation of the syn aldol product (but no enantioselectivity between the two syn products).

So, there is no difference at all between the two models you have used. Both are the same thing and are equally applicable. If you had a chiral auxiliary on boron or something of that sort, the story might be different. But that isn't the case here.


Are long oxygen molecules possible? Are large oxygen containing molecules possible? Either large rings, or chains with hydrogen atoms at the ends. Like this:

$\ce{H-O-O-O-O-O-O-O-O-O-$\cdots$-O-O-O-O-O-H}$

Yes, they are possible; several have even been detected.  They all have the general formula $\ce{H2O_{n}}$ and belong to the class of hydrogen polyoxides.

Of course, the $n=1$ member of the series is water $(\ce{H2O})$, and the $n=2$ member is hydrogen peroxide $(\ce{H2O2})$.

The $n=3$ member is trioxidane $(\ce{H2O3})$, and it has the following structure:

It can be prepared in a number of ways as outlined in the linked article, such as the reaction of ozone and hydrogen peroxide. Trioxidane is also an active antimicrobial ingredient. Trioxidane has a half-life around 16 minutes in organic solvents, but only milliseconds in water.

Hydrogen tetroxide ($n=4$, $\ce{H2O4}$) is also a known compound as is hydrogen pentoxide ($n=5$, $\ce{H2O5}$).  Hydrogen pentoxide is formed as a byproduct in the preparation of trioxidane.

This abstract of a computational study provides bond dissociation energies for the hydrogen polyoxides. Starting from water, they steadily decrease (the $\ce{O-O}$ bonds become weaker) to $\ce{H2O6}$ and then start to increase again.

Abstracts from other computational studies dealing with the expected structures and stabilities of these molecules can be found here and here.


Why does increasing the available surface area of a pure solid not effect equilibrium? Take a reverseable reaction of a gas and a solid, $$\ce{A(g) + B(s) <=> C(g)}$$

The rate of the forward reaction will depend on the surface area of the solid available. If I take that system at equilibrium and add more of the solid, especially if it's in a form with a large surface area this should change the rate of the forward reaction, but this doesn't change the equilibrium. The closest to an answer I've found is that the rate of the reverse reaction increases by an equal amount so that $K$ remains constant, but no justification for that is given.

Increasing the rate of the forward reaction increases the concentration of $C$, which should increase the rate of the reverse reaction, that makes intuitive sense. But I can't see any obvious reason why that should be equal. Presumably my textbook is being a bit reductionist here.

Is it making a simplifying assumption and not bothering to mention that assumption? (the reactions are being considered as independent of surface area? There are unconsidered phenomenon involved where the activity of a pure solid can't be treated as unity and we're only dealing with an approximation?).

Or are there more complex factors at play that result in the rate of the reverse reaction increasing by the same factor as the rate of forward reaction, and the book just doesn't bother to discuss them?

You can change the amount of $\ce{B}$ available and the forward rate, but that doesn't change the rate constant. And the equilibrium constant is directly related to the rate constants of the forward and reverse reactions:

$$K = \frac{k}{k'}, \text{where }k\text{ is forward rate cosntant, }k'\text{ is reverse rate constant}$$

As @MaxW points out, you'll get to equilibrium faster, but it's the same equilibrium.

Edit:

Here's another way to think about it. Consider the reaction:
$$\ce{H2O_{(l)}<=>H2O_{(g)}}$$

Does the equilibrium between liquid water and water vapor change if you increase the surface area of a pool of water? No, because even though evaporation is faster, there's also more area for condensation to take place.

In your specific case, the reverse reaction from $\ce{C}$ needs to interact with some amount of solid to redeposit $\ce{B}$. Otherwise, you're making $\ce{B_{(g)}}$ and that's not the same thing...

Edit 2 (for OP's comments):
Notice that your reaction is actually extremely unrealistic. Gases don't just form solid or liquids on their own; they need some nucleation seed. Otherwise, there would be no point in seeding rain clouds.

As @IvanNeretin points out, invoking microscopic reversibility, you'd need to invoke these nucleation points in your reaction, forward and backward. These points are of course whatever surface is available on $\ce{B}$.

This makes the reaction:
$$\ce{A_{(g)} + B_{(g)} + n B_{*} \overset{k}{\underset{k'}{<=>}} n B_{*} + C_{(g)}}$$

$\ce{B_{*}}$ refers to the surface. Notice that it cancels out so it doesn't affect the equilibrium constant, but having more surface area will increase the forward rate if you increase the surface. It will also increase the reverse rate, which is why the equilibrium constant can stay the same.


Applying the Solubility Guidelines is AgClO4 soluble or insoluble? The book I am using is "General Chemistry: Principles and Modern Applications
by Ralph H. Petrucci. On page 158 the Solubility Guidelines for Common Ionic Solids is given. Rule 3 states that, "Salts of silver, lead, and mercury(I) are insoluble, however, my book states that $\ce{AgClO4}$ is soluble. Can someone explain?

Different guidelines will say slightly different things to convey the same idea. Many I've seen don't explicitly mention silver, but still produce the same result for most compounds. I don't have that book but if it's anything like these, there's probably another line that says salts of anions like nitrate and chlorate are soluble.

This table is worded a bit better to make the possibility of exceptions more explicit:

  Salts containing Group I elements are soluble ($\ce{Li+, Na+, K+, Cs+, Rb+}$). Exceptions to this rule are rare. Salts containing the ammonium
  ion ($\ce{NH4+}$) are also soluble.
  Salts containing nitrate ion ($\ce{NO3-}$) are generally soluble.
  Salts containing $\ce{Cl^-, Br^-, I^-}$ are generally soluble. Important exceptions to this rule are halide salts of $\ce{Ag^+, Pb^{2+}}$, and $\ce{(Hg2)^{2+}}$.
  Thus, $\ce{AgCl, PbBr2}$, and $\ce{Hg2Cl2}$ are all insoluble.
  Most silver salts are insoluble. $\ce{AgNO3}$ and $\ce{Ag(C2H3O2)}$ are common soluble salts of silver; virtually anything else is insoluble.
  Most sulfate salts are soluble. Important exceptions to this rule include $\ce{BaSO4, PbSO4, Ag2SO4}$ and $\ce{SrSO4}$ .
  Most hydroxide salts are only slightly soluble. Hydroxide salts of Group I elements are soluble. Hydroxide salts of Group II elements
  ($\ce{Ca, Sr,Ba}$) are slightly soluble. Hydroxide salts of transition
  metals and $\ce{Al^{3+}}$ are insoluble. Thus, $\ce{Fe(OH)3, Al(OH)3, Co(OH)2}$ are not
  soluble.
  Most sulfides of transition metals are highly insoluble. Thus, $\ce{CdS, FeS, ZnS, Ag2S}$ are all insoluble. Arsenic, antimony, bismuth, and lead
  sulfides are also insoluble.
  Carbonates are frequently insoluble. Group II carbonates ($\ce{Ca, Sr, Ba}$) are insoluble. Some other insoluble carbonates include $\ce{FeCO3}$
  and $\ce{PbCO3}$.
  Chromates are frequently insoluble. Examples: $\ce{PbCrO4, BaCrO4}$
  Phosphates are frequently insoluble. Examples: $\ce{Ca3(PO4)2, Ag3PO4}$
  Fluorides are frequently insoluble. Examples: $\ce{BaF2, MgF2 PbF2}$.


How to determine mobility when two mechanisms exist simultaneously Two scattering mechanisms exist in a semiconductor. If only the first
  mechanism is present, the mobility will be $\pu{250 cm2/V s}$. If only the
  second mechanism is present, the mobility will be $\pu{500 cm2/V s}$.
  Determine the mobility when both scattering mechanisms exist at the
  same time.

I'm not sure what the last sentence is trying to tell me. I know mobility is $q\tau_c/m_n$, and I think I need to solve for mean free time here to help, but I'm not sure what I should use for $m_n$ since I don't know what the semiconductor is, and I'm not given a value.

Although, since I'm given two mobilities to start with, should I solve for $\tau_c$ twice and average those? Really not sure what direction I should be going in (if only my professor answered his emails).

I needed to add the inverses of the given mobilities and take the inverse of that to get my final result.

$1/\mu = 1/\mu_L + 1/\mu_I$


Domestic cleaning products and types of solvents I have a floor cleaner which washes and vacuums the floortiles.   The instructions advise not to use solvent cleaners with it.  My understanding is that even water is a solvent.  The manufacturers/agents will just try to sell me their product.  I want to use cheap domestic solutions (eg. Sugar Soap, Pinoclean, Bar Keepers Friend).    Can anyone advise please? At present I use just hot water, sodium bicarbonate ($\ce{NaHCO3}$) and vinegar of a small amount of washing up liquid.   Thanks.

There are two ways of cleaners. Ones that will "grab on to" the dirt and bacteria to wash it away e.g. soap and shampoo. Then there are ones that will dissolve the dirt and kill the bacteria such as bleaches. They won't do anything dangerous together, they'll just not work.


Probability of forming dihydrogen with molecular weight 3 During a lecture, my professor told us that the probability $P$ to form a hydrogen molecule $\ce{H2}$ with mass number 3 could be calculated out of the abundances, $\gamma$, of the isotopes of this element: $\ce{^1H}$ (normal hydrogen), $\ce{^2H}$ (deuterium), etc. So,
$$
\begin{split}
P[\ce{H2}] &= P[(\ce{^1H} \cap \ce{^2H}) \cup (\ce{^2H} \cap \ce{^1H})] \\
%
           &= P[\ce{^1H} \cap \ce{^2H}] + P[\ce{^2H} \cap \ce{^1H}] -
%
              P[(\ce{^1H} \cap \ce{^2H}) \cap (\ce{^2H} \cap \ce{^1H})]
\end{split}
\tag{1}$$
But since
$$(\ce{^1H} \cap \ce{^2H}) \cap\ (\ce{^2H} \cap \ce{^1H}) = \ce{^1H} \cap \ce{^2H}\tag{2}$$
and
$$P[\ce{^1H} \cap \ce{^2H}] + P[\ce{^2H} \cap \ce{^1H}] = 2P[\ce{^1H} \cap \ce{^2H}]\tag{3}$$
we end up with
$$
\begin{split}
P[\ce{H2}] &= P[\ce{^1H} \cap \ce{^2H}] \\
       &= \gamma(\ce{^1H}) \cdot \gamma(\ce{^2H}) \\
       &= 0.99972\cdot 0.00028 \\
       &= 0.0002799
\end{split}
\tag{4}$$
This is my result while for my professor, it should be twice that value. It's like he's forgetting the term
$$
P[(\ce{^1H} \cap \ce{^2H})\ \cap\ (\ce{^2H} \cap \ce{^1H})]
$$
Am I right?

Extending / reformulating Karsten Theis' answer: A probabilistic approach is to draw a tree about drawing two spheres where event $A$ (normal hydrogen, $\ce{^1H}$) dominates, but event $B$ (the next abundant isotope, $\ce{^2H}$) complements, $P(\ce{^1H} \cup{} \ce{^2H}) = 1 = \Omega{}$ and eventually to multiply the probabilities for each path (e.g., $P(\ce{^1H^2H})$ as shown on the left hand side).

Yet because in your case both paths $P(\ce{^1H^2H})$ and $P(\ce{^2H^1H})$ yield a dihydrogen molecule of mass 3, these individual probabilities need to be summed up.


Do we neglect the temperature changes during a reaction while calculating delta(H)? Suppose for a reaction: $$\ce{$a$A + $b$B -> $c$C + $d$D }$$

we know the $\Delta U$ as $\mathrm dU$ and all the components of the reaction are gas.

Then to calculate the $\Delta H$ at a temperature $T$ we use the expression $\mathrm dH=\mathrm dU+\left(c+d-a-b\right)RT$.

Then while using this expression did we assume the temperature to remain constant as it wouldn't really be constant as the actual expression for $\mathrm dH$ is: $\mathrm dH=\mathrm dU+\left(pV\right)$ and so we can write $\mathrm dH=\mathrm dU+\mathrm d(nRT)$ and for $\mathrm dH=\mathrm dU+\left(c+d-a-b\right)RT$, $T$ has to be constant?

Yes. By saying that $\Delta (pV) = \Delta (nRT) = RT\Delta n$ you have made the assumption that $T$ is constant. When you take anything out of a $\Delta$ sign, you are saying that the initial and final values are the same. Otherwise you would have to write $\Delta (nRT) = R \Delta (nT) = R(n_2T_2 - n_1T_1)$ - this only simplifies to $RT(n_2 - n_1)$ if you assume $T_1 = T_2 = T$.

By the way, I want to be picky about the formulae that you are using. Look at $$\mathrm{d}H = \mathrm{d}U + (c + d - a - b)RT$$

The problem with this is that $\mathrm{d}H$ and $\mathrm{d}U$ are infinitesimals, but $(c + d - a - b)RT$ is a finite quantity. Do not ever add or subtract these two in the same equation. It is like trying to add a mass to a length - what is $30 \mathrm{~kg} + 25 \mathrm{~cm}$? It doesn't make sense.

Same thing applies for $\mathrm{d}H = \mathrm{d}U + (pV)$.


Why is the ester bond stable under Reformatsky reaction conditions? The following reaction is a Reformatsky reaction.

Why is the ester bond not attacked in the last step? Couldn’t $\ce{OR3}$ be replaced by $\ce{OH}$ or $\ce{O(molecule)}$?

Yes, technically you have a zinc alcoholate which could display the basicity and nucleophilicity of any other alcoholate and you know that you can transesterificate with e.g. sodium methanolate.

Aside from the oxyanion here being rather closely associated to the zinc cation which has a larger positive charge than e.g. sodium and thus exercises a greater attractive force, there is also the fact that the alcohol in question is tertiary and tertiary alcohols are bad nucleophiles. In most of the cases, $\ce{R3}$ will be primary or at most secondary, thus the transesterification reverse reaction will be more rapid than the forward reaction.

If, however, $\ce{R3}$ is tertiary, too, you would be dealing with a huge steric bulk on that carbonyl carbon, attempting to have two tertiary $\ce{OR}$ groups on the tetrahedric intermediate. That is unlikely to happen for steric reasons.

If you are talking about the absolute final step, i.e. the aquaeous workup, then you will likely be using ammonium chloride solution or something else mildly acidic. In mildly acidic or otherwise buffered solutions, there is not a high enough concentration of hydroxide ions to attack the ester carbonyl sufficiently for an observable reaction to occur.


Mathematical explanation of blank titration in Mohr method for argentometry In my book it is written that in order to minimize systematic error in Mohr method for argentometry it is a common procedure to conduct a blank titration with $ \mathrm{Ag^+} $ against a $\mathrm{CrO_{4}^{2-}}$ solution (with the same concentration of the original solution).

The book then suggests to subtract the $ \mathrm{Ag^+} $ volume needed to observe precipitation (in the blank solution) from the volume obtained as a result of the 'standard' titration.

I don't understand why I have to subtract that number, specifically I do understand that there is a systematic error but I don't get why it is the volume of $ \mathrm{Ag^+} $ obtained by the blank titration.

$\ce{AgCl}$ ppts before $\ce{Ag2CrO4}$. But in order to visually detect the $\ce{Ag2CrO4}$ you must add a slight excess of $\ce{Ag^+}$. Hence you use a "blank" with no $\ce{Cl^-}$ to account for the amount of $\ce{Ag^+}$ needed to get enough of the $\ce{Ag2CrO4}$ ppt to form so that it can be detected visually.


Would Chlorine or Chloramine gas show up on a VOC test? Would Chlorine or Chloramine gas show up on a VOC test?  My apartment was cleaned using chlorox and lysol, and there has been a smell that's been lingering for about 10 days.  The air was tested with a VOC meter but nothing came up.  Would Chlorine and Chloramine have shown up on this test?

Chlorine, no. Chloramine, maybe.

Monochloramine, $\ce{NH2Cl}$, is an inorganic compound, and not very volatile (when dissolved in water it forms more or less of an azeotrope, evaporating slightly faster than the water itself). So, not a VOC.

However, the way it does its job as a disinfectant is to basically shred organic compounds, like cell walls, key enzymes, etc, forming "organic chloramines" where one or more of the remaining hydrogens is replaced with an organic functional group. These now-organic compounds have a higher vapor pressure, so by design, chloramine is used to create VOCs. The idea, when used in tap water or pool/spa chlorination, is to allow the organic chloramines to boil off and disperse in the air, leaving clean water behind. That "chlorine smell" of a chlorinated pool is actually mostly organic chloramines evaporating out. In a more enclosed space like an interior room, these compounds can't disperse and so the smell builds.


How can a thermometer ever show the actual temperature of an object if the object loses heat to the thermometer? We know that when two objects are placed in contact with each other, after a period of time, the two objects will have the same temperature. Thus, if a hot body comes into contact with a relatively cold body, it will lose its heat and its temperature will reduce, while the temperature of the 'cold' body will increase.
If I put a thermometer in an object to measure its temperature, the body will lose its heat (or gain the heat, if the thermometer is hotter) until it reaches the same temperature as the thermometer. Doesn't this mean the thermometer is not giving us the EXACT temperature of the object, i.e the temperature before the measurement was taken, and is in fact, showing us a lower/higher temperature?

I'll start by mentioning that there's no such thing as an exact measurement—there is always some measurement error.  The only observations that can be numerically exact are counted numbers of discrete objects (e.g., the number of electrons in a neutral carbon atom is exactly 6).  And I say "can be", because if the numbers are sufficient large, even with counted numbers there can be errors (we see this with vote counting).
Having said that, your idea is correct—if the heat capacity of the temperature probe is significant relative to that of the object being measured, then the measurement can significantly change the temperature of your object.
This is a particular concern when measuring small objects.  For such applications, researchers can employ (as Jon Custer mentioned in his comment) non-contact thermometry.  See, for instance:
https://www.omega.co.uk/temperature/z/noncontacttm.html


Relation between activation energy and rate constant Consider two types of reaction. One has more activation than other. If we increasing temperature by same amount in both, rate constant of which reaction will increase more?

My attempt: as we know that $k = A\mathrm e^{(-E/(RT))},$ if temperature is same in both, then the one having more activation energy will have lesser rate constant. Am I correct?

This can be answered both conceptually and mathematically.

Conceptually: Let's call the two reactions 1 and 2 with reaction 1 having the larger activation energy. Here is a plot of the arbitrary reactions.

It is clear from this graph that it is "easier" to get over the potential barrier (activation energy) for reaction 2. Thus if we increase temperature, the reaction would get faster for reaction 2 before it gets faster for reaction 1. Therefore, when increasing the temperature by the same amount, the rate constant for the reaction with the lower activation energy will increase more.

Mathematically: You can understand how the rate constant changes with temperature by simply graphing. Here is a very quick and very rough graph on how the rate constant changes with temperature (plotting the equation $k=Ae^{-E/RT}$).

Notice how increasing the temperature causes a quicker increase in the reaction with the lower activation energy $(E_{a2})$. However, this is only an a priori expectation.

You mentioned in a comment above about differentiating. You can use that method as well. Taking the derivative of the equation for $k$ with respect to temperature while assuming $A$ and $E_a$ do not depend on temperature yields

$$\frac{dk}{dT} = \frac{AE_a}{RT^2}e^{-E_a/RT}$$

Edit: As porphyrin pointed out in the comments below, this equation elucidates the true behavior of the temperature dependence. Plotting this function for the two energies, we see that the derivative for the reaction with a smaller activation energy increases faster for lower temperatures. But for higher temperatures, there is a shift in the change of the rate constant. Therefore, the process is temperature dependent. At low temperatures, the rate constant for the reaction with low activation energy changes is more sensitive to temperature change while at high temperatures, the reaction with larger activation energy is more sensitive to temperature change.

As a side note: at very large (infinite) temperatures, both functions approach the same value and there will be an equivalent distribution among the reactions' products.

From comments below: Prove that if we increase temperature in an endothermic reaction, equilibrium proceeds in forward direction.

There are a couple ways to do this. As this is now an equilibrium question, I will keep the regime to equilibrium instead of rate constants. First, let's use fundamental chemistry to get an expected result then use equations to show what the result would be quantitatively.

Fundamentals: Say you have a simple reversible reaction
$$A\rightleftharpoons B$$
Given this reaction is endothermic, energy is required to drive the reaction. Because of this, we can write energy as a "reactant"

$$A + Energy \rightleftharpoons B$$

According to Le'Chatelier, if you increase temperature then the equilibrium must shift towards the products, thereby answering the question.

The Maths: Now, let's apply some math as overkill. The equilibrium constant is dependent on the standard free energy of the reaction given by the following equation

$$\Delta G^o = -RT\ln(K)$$

If we want to understand how the equilibrium constant will change we can rearrange this equation to solve for $K$ and plot as a function of temperature. Doing this and making the substitution for the free energy $\Delta G^o = \Delta H^o - T\Delta S^o$ gives

$$K = e^{-\frac{\Delta H^o}{RT} + \frac{\Delta S^o}{R}}$$

Since the reaction is endothermic, $\Delta H^o > 0$. This gives the same monotonically increasing profile that we see for the Arrhenius equation (plotted above). Therefore $K$ must always increase, and therefore the reaction proceeds in the forward direction, for increasing temperature. Of course, this is making the assumption that $\Delta H^o$ does not change with temperature.


Some questions on stereochemistry I have a couple of basic questions on stereochemistry that confused me. Please help me understand them.

1) Why does a molecule become achiral if it has a plane of symmetry even though it has steriocentes?

2) When working with Fischer Projections, they say you can rotate a Fischer projection 180 degrees, but not by 90 degrees. I don't understand why.

Moreover, can you please suggest me a good resource wherein I can learn stereochemistry (not to very advanced level)?

1) Why does a molecule become achiral if it has a plane of symmetry
  even though it has stereocenters?

If a plane of symmetry bisects a molecule such that one half of the bisected molecule is the mirror image of the other half of the bisected molecule, then the entire molecule must be superimposable on its mirror image, it must be achiral.  Said differently, if mirror symmetry exists within a molecule, then the mirror image of the entire molecule will be superimposable upon the original drawing of the entire molecule.

Here's an example of a molecule ($\ce{A}$) with 2 chiral groups ($\ce{G_{R}}$ and ($\ce{G_{S}}$).  There is a plane of symmetry that bisects $\ce{A}$, it is perpendicular to the plane of the screen and contains the central carbon atom and the 2 methyl groups.  Now, when we draw the mirror image of $\ce{A}$ (let's call it ($\ce{\bar{A}}$), we find that it is superimposable upon the original drawing of $\ce{A}$ by a simple 180-degree rotation.
Even though $\ce{A}$ contains 2 chiral substituents, the molecule as a whole is achiral.


Propose a reaction mechanism that explains the rate law for the formation of dinitrogen pentoxide The reaction is:

$$\ce{NO(g) + NO2(g) + O2(g) -> N2O5(g)}$$

From the experimental data, I have found the rate law to be: $$\ce{Rate~=~k[NO][NO2]}$$

This means the overall reaction does not depend on $\ce{O2(g)}$.

Can someone suggest how I go about building the elementary steps to form the reaction mechanism?

First step - the rate determining step or the slow step (activation energy is high).
$$\ce{NO(g) + NO2(g) -> N2O3(g)}$$

2nd step - the fast step.
$$\ce{N2O3(g) + O2(g) -> N2O5(g)}$$

Since the rate determining step does not include $\ce{O2(g)}$, the rate is independent of the initial $\ce{O2(g)}$ concentration.


Temperature dependency of enthalpy H and entropy S In textbooks, I often read that the temperature dependency of the enthalpy of reaction $\Delta_\mathrm{r}H$ and the entropy of reaction $\Delta_\mathrm{r}S$ can be neglected (in a limited $T$-range).

But why is this valid? What is the intention behind this?

Three step process to evaluate temperature dependence

There is a change in enthalpy and in entropy when changing the temperature of a sample. It follows that the enthalpy of reaction and the entropy of reaction can also change when changing the temperature. The easiest way to get values for a temperature $T_2$ if you have values for a temperature $T_1$ is to do a thought experiment with three steps, starting with reactants at the temperature $T_2$:

Change temperature of reaction mixture from $T_2$ to $T_1$
Let the reaction proceed at a temperature of $T_1$
Change temperature of product mixture from $T_1$ to $T_2$

The net process is the reaction at the temperature $T_2$, where you don't know the entropy and enthalpy change yet.

Changes depend on differences in heat capacity

If we are working at constant pressure, the enthalpy change is (with $C_P$ the heat capacity of the reaction mixture):

$$\Delta H = C_P \Delta T$$ and the entropy change is:

$$\Delta S = C_P \ln \frac{T_\text{final}}{T_\text{initial}}.$$

Because steps 1. and 3. change the temperature in opposite directions, a lot cancels out. If we define  $\Delta C_P$ as the difference in heat capacity between reactant and product mixture, we get:

$$\Delta(\Delta_r H) = \Delta C_P \Delta T$$

$$\Delta(\Delta_r S) = \Delta C_P \ln \frac{T_2}{T_1}$$

Compared to other changes with temperature (e.g. Gibbs energy or equilibrium constant), these changes tend to be modest. To give an idea of the magnitude, the molar heat capacity of a diatomic ideal gas is $c_P = 3.5 R$, or about 30 J / (mol K). Given that values for $\Delta_r H$ are typically given in kJ/mol, the change in enthalpy when changing the temperature is relatively small. For the entropy, the "correction term" is not proportional to $\Delta T$ but to the logarithm of the ratio of temperatures. If you start at room temperature and increase the temperature by ten Kelvin, the value of the logarithm will be about 0.03, so the effect is also relatively small.


Why Does 0.10 M Dihydrogen Phosphate Dissociate Rather than Hydrolyse in a Neutral Solution of Water? Dihydrogen Phosphate has a Ka of 6.2x10^-8, yet it seems to form solutions with an acidic pH when it is in water.  Why is that?  Shouldn't it dissociate back to phosphoric acid and form a basic solution, since its Ka is less than 1.0x10^-7?

It seems as though Ka does play a role in determining whether an amphoretic substance will act like a base or acid, since 0.10 M Hydrogen Phosphate has a Ka of 4.8x10^-13 and creates a basic solution.  However, I feel like I am missing another mechanism that determines the behavior of an amphoretic substance in a neutral solution.

The whole truth is that any time that you add any phosphate ion into an aqueous solution, then you will have all four phosphate species $\ce{H2PO4-, HPO4^{-2} , PO4^{-3} ,H3PO4}$  in solution, that ($\ce{H2PO4-}$)   is amphoteric that behaves both:

as an acidreleasing $\ce{H+}$ in aqueous solution and( forming $\ce{HPO4^{-2}}$ ions which can produce very few of $\ce{PO4^{-3}}$ ions)

$$\text{
Ionization:}\ce{H2PO4- <=> HPO4^{-2} +\color{red}{H^+} \quad\left(K_\mathrm{a}=\color{red}{6.2\times10^{-8}}\right)}$$
OR

as a base(can undergo hydrolysis forming ($\ce{H3PO4}$) acid
$$\text{Hydrolysis:}\ce{H2PO4-  +  H2O <=> H3PO4 + \color{blue}{OH-} \quad\left(K_\mathrm{b}=\color{blue}{1.3\times10^{-12}}\right)}$$

So ,$\color{red}{K\mathrm{a}}>\color{blue}{K\mathrm{b}}$ ,thus the first reaction is the dominant,so ,$\pu{pH}<7 $ and the solution is$~\color{red}{\text{acidic}}$.


Is there a name for the opposite reaction to the dissolution? When atmospheric $\ce{CO2}$ reacts with water to form $\ce{H2CO3}$, this is called dissolution, isn't it? What term would you use for the opposite reaction when it occurs at atmospheric pressure (e.g. due to temperature change)? For a solid it would be precipitation, but here it is for a gas.

Effervescence if there are bubbles and

Degasification more generally speaking.

There is also the term outgassing, but that term is broader than just gas coming out of liquid solution.


What does Pauli’s exclusion principle mean in atomic or fundamental way? It means is that no electron can have same n , l and $m_l$ but can have two different spin quantum number.
I want to know why is this rule valid?Means there must be some other things happening also Inside that atom which forces this rule to be true.
I got to know from web that electron dual behaviours and electromagnetic forces result in the formulation of this law but what is the real reason?

Pauli's exclusion principle is a consequence of electrons being indistinguishable fermions. Fundamental particles are indistinguishable in that two of the same type differ only in a few properties but otherwise behave identically (compare that say to two apples - they are always in principle distinguishable because they differ in so many ways).
Fermions exhibit an associated statistical property when you have a collection of the otherwise indistinguishable particles. The statistical property follows from how the wavefunction of the collection behaves under symmetry (exchange) operations. You have two choices when you swap particles in such a wavefunction: retain the sign or switch signs. That is one fundamental distinguishing property between fermions and bosons (all particles can be classified as one or the other). It happens that nature allows both cases. We chose to call one fermions (to honor Enrico Fermi who co-discovered them), and the particles we call electrons are fermions (you can look into the details of the Standard Model to work out exactly how changes to the properties of the electron would change the universe). Fermions have half-integer values of the spin quantum number. Bosons have integer values. If you have multiple indistinguishable fermions, the QM wavefunction changes sign (is antisymmetric) under exchange of any two of the particles. For the exchange to alter the sign the two exchanged fermions must differ in some property, making the wavefunction antisymmetric, for instance if intrinsic angular momenta (spins) have opposing quantization (quantum numbers $\pm \frac12$).
The consequence of this is explained well in the Wikipedia

If two fermions were in the same state (for example the same orbital with the same spin in the same atom), interchanging them would change nothing and the total wave function would be unchanged. The only way the total wave function can both change sign as required for fermions and also remain unchanged is that this function must be zero everywhere, which means that the state cannot exist. This reasoning does not apply to bosons because the sign does not change.


Show the vibrational frequency of fluorine molecule anion is much lower than that of fluorine molecule Basically I need to draw the molecular orbital for $\ce{F2}$ and then answer a bunch of questions about it. I have drawn it correctly, as far as I know, but I don't know how to use it to show that the vibrational frequency of $\ce{F2-}$ is lower than $\ce{F2}$.
I worked out the bond order for $\ce{F2-}= 0.5$  and for $\ce{F2}= 1$. Is this what you use to find it out, or is it something to do with the HOMO-LUMO?

You should definitely post the MO diagram of $\ce{F2}$ that you worked out. When trying to solve such a problem you should try to analyse the question as thoroughly as possible to make sure that you collect all the information available to you beforehand. Especially important here is that the question already tells you that the vibrational frequency of $\ce{F2^{-}}$ is lower than that of $\ce{F2}$. So the first question you should ask yourself is: Under what circumstances is a vibrational frequency lowered and which relation do I know that links the vibrational frequency to some variable that is connected to the MO diagram of $\ce{F2}$? I give you a hint: You will need the equation of a harmonic oscillator

\begin{equation}
 \omega = \sqrt{\frac{k}{m}}
\end{equation}

where $\omega = 2 \pi f$ is the angular frequency, $f$ is the frequency, $k$ is the force constant (in your case of the bond and thus its a measure of the bond strength) and $m$ is the reduced mass of the oscillating system. (The harmonic oscillator is one of those simple model systems that can help you quite often when you want a starting point for the investigation of much more complicated ones.) Now ask yourself: Which of the variables $k$ or $m$ might have a relation to the MO diagram and how will they be related?


Liquid to solid when exposed to air Is there a liquid mixture that will become a solid when exposed to air/oxygen. Preferably a spongy material that is difficult to tear.

Urethane foam from isocyanates, commonly sold at hardware stores for insulation.

Click links for general information and for its chemistry.


Corrosion of metals and observations in the agar medium? I want to know which metals oxidize and which is reduced and changes in medium and observations?

According to my knowledge more corrosion will take place in Zn and Fe also corrode. Then copper should reduce. So, for both reactions, will copper acts as the cathode?
I need general but complete explanation. Thank you

You have correctly identified zinc and iron as the anodes, since they are both less noble than copper. In the corrosion reaction, they are both oxidized and dissolved in the aqueous gel.

\begin{aligned}
\ce{Zn ->& Zn^2+ + 2e-} \\
\ce{Fe ->& Fe^2+ + 2e-} \\
\end{aligned}

The electrons flow to the copper cathode, where oxygen, which is dissolved in the liquid medium, is reduced to hydroxide.

$$\ce{O2 +2H2O + 4e- -> 4OH-}$$

As a consequence, the pH of the solution around the copper cathode rises, and the phenolphthalein indicator changes its color from colorless to violet-red in this area.

$\ce{Fe^2+}$ reacts with $\ce{K3[Fe(CN6)]}$ to form a deep blue iron(II,III) hexacyanoferrate(II,III).

$$\ce{K+ + Fe^2+ + [Fe(CN6)]^3- -> KFe[Fe(CN6)]}$$

Therefore, the solution near the iron surface turns blue. Zinc does not react with $\ce{K3[Fe(CN6)]}$ and the solution around the zinc anode remains colorless (source).


In NMR Spectroscopy why use absorption and dispersion components of the FID spectrum rather than magnitude and phase? In chapter 4 of Dr James Keeler's book 'Understanding NMR spectroscopy' (http://www-keeler.ch.cam.ac.uk/lectures/Irvine/chapter4.pdf) he explains how to calculate the expected spectrum of a FID signal. In this chapter he analyzes the real and imaginary (absorption mode and dispersion mode) parts of the FID spectrum. This doesn't make sense to me. What is the physical interpretation of absorption mode and dispersion mode? Why not just rewrite the spectrum in polar coordinates and consider the magnitude and phase separately?

The FID signal itself is represented/considered as a set of complex points in time. (Keeler mentions this in chapter 4, section 4.1. - same link as you posted in your question.) Upon Fourier transform, the FID will therefore transform correspondingly to a set of complex points in frequency space.

As Keeler said in section 4.2 of the same chapter, the real part of the spectrum is the absorption mode lineshape, and the imaginary part of the spectrum is the dispersion mode lineshape.

From your last question I think you understood this already, but it is worth repeating.

Aside: Why is the FID a set of complex points?

While all the values that we can detect are real (we cannot detect imaginary current in the receiver coil), we can consider values detected in different directions. Taking the static magnetic field of the spectrometer as the $z$ direction, we can measure the $x$ and $y$ components of the NMR signal. Mathematically, the $x$ and $y$ components can be equivalently expressed as the real and imaginary component (and you gain some conveniences with the Fourier transform if you do so)

Detecting the $x$ and $y$ components of the NMR signal is known as quadrature detection (Keeler, section 5.6), and it is used in order to be able to distinguish between frequencies that are offset from the carrier frequency with different sign. As Keeler describes, instead of fitting two coils to detect the $x$ and $y$ components of the FID directly, it is achieved by phase shifting the receiver reference frequency in real spectrometers.

What is the physical interpretation of absorption mode and dispersion mode?

The pure absorption part represents resonant frequencies of the nuclear spins in your sample.

The pure absorption part of the spectrum is what NMR spectroscopists would use as the NMR spectrum. Signals in the absorption mode occur at distinct frequencies for a particular isotope within a particular sample at a particular magnetic field, which is compared to a frequency of a known reference substance (same type of nucleus, same field/spectrometer, etc). The difference between these two frequencies is the chemical shift, in units of parts per million (ppm) Hz. The chemical shift values for most/all of a particular isotope is the parameter that is reported most commonly in NMR experiments.

Just as the imaginary part of the FID is just the real part of the FID with a phase shift, after Fourier transform, the dispersion mode spectral lineshape has in principle the same information as the absorption mode lineshape. Its presence is mostly for signal processing purposes, but mostly done in the background and not displayed to the user. It is also not commonly used as the "result" of the NMR experiment because it can be hard to determine the accurate chemical shift if the signal does not return to baseline on both sides of the signal, which is exacerbated by the long (positive and negative) "tails" either side of the dispersive mode lineshape. The long "tails" also mean that signals are more likely to overlap each other and the resolution is therefore worse. Some more advanced NMR experiments generate a mix of absorptive and dispersive lineshapes "by default" (called a phase twist), and much effort goes into coming up with strategies to recover the absorption mode lineshape. (Example: 2D NMR spectroscopy, see Keeler chapter 7, section 7.6.2)

Why not just rewrite the spectrum in polar coordinates and consider the magnitude and phase separately?

It is possible to do this. But you will get back magnitude/phase lineshapes that is very similar to the real/imaginary (phase-sensitive absorption/dispersion) lineshapes.

The magnitude spectrum can be calculated by $\operatorname{Re}^2 + \operatorname{Im}^2$. We know that the absorptive lineshape should be completely positive and the imaginary part is more or less the same information but half positive, half negative with long tails. The effect of squaring both and adding them together means that you get something that looks mostly like the absorption lineshape with more signal to noise, at a cost of generating positive long tails from the squared imaginary part, which reduce the spectral resolution.

This is actually a (now somewhat old-fashioned?) technique to address the problem of phase twist in 2D spectra.

The phase spectrum can be calculated by $tan^{-1}\frac{\operatorname{Im}}{\operatorname{Re}}$. No one uses this in NMR (as far as I know!), but we can visualising it by considering what it looks like at different points.

We plot the $tan^{-1}(x)$/arctan function. (Axes cross at the origin.)

At absorption lineshape peak, dispersion lineshape is zero: $\frac{\operatorname{Im}}{\operatorname{Re}}$ is zero, arctan is zero.
At large values away from the peak, we assume everything goes to zero:
$\frac{\operatorname{Im}}{\operatorname{Re}}$ is zero, arctan is zero.
At a short distance from the peak on the right hand side, absorption lineshape is positive, dispersion lineshape is positive:
$\frac{\operatorname{Im}}{\operatorname{Re}}$ is positive, arctan is positive.
At a short distance from the peak on the left hand side, absorption lineshape is positive, dispersion lineshape is negative:
$\frac{\operatorname{Im}}{\operatorname{Re}}$ is negative, arctan is negative.

Assuming that the signal in the absorptive/dispersive lineshapes is also centered at zero, you get something like this:

Joining the points all together gives you a curve that looks like the dispersion lineshape (but not completely the same. My guess is that it is it is also a little broader than the phase-sensitive dispersive lineshape).


How do I determine the average carbon oxidation numbers in ethanol and ethanal? I have an assignment (I quote it word by word):

  Determine the oxidation number of carbon in (A) ethanol and (B) ethanal. Answers are (A) -2 and (B) -1.

I reviewed similar question at SE: Oxidation of Carbons
and I found first answer to be clear and links given in this answer to be useful.
I also found a good  Khan Academy video, which basically confirms all in this answer

Now, here is my problem: There are two carbons in ethanol. Oxidation number of carbon which is connected to three hydrogens is $-3$; oxidation number of carbon which is connected to two hydrogens and one oxygen is $-1$. How do we get answer $-2$ out of it then? (Do we add $-3 + (-1)$ and then divide by 2?)

And again there are two carbons in ethanal. Oxidation number of carbon which is connected to three hydrogens is $-3$; oxidation number of carbon which is connected to one hydrogen and one oxygen (but with double bond) is $+1$. How do we get answer $-1$ out of it? (Do we add $-3 + 1$ and divide it by 2?)

Overall, my question does not seem to concentrate on one of the carbons present in ethanol (or ethanal). And answers given look like there is oxidation number of first carbon added to oxidation number of second carbon and then divided by 2. This situation seems to be true for ethanol, and again the same happens for ethanal.
Where can I read about adding two oxidation states and then calculating average (I have never heard of it)? Or maybe these calculations done in some other way? Thank you very much in advance for answering.

What you have been asked to calculate is what is called an average oxidation number. This is often used in inorganic chemistry (not because it is rather powerful, but because you can easily get away with it) but lesser in organic chemistry.

To properly determine oxidation states, you should do the calculation on a per-atom basis. And doing that would give you $\mathrm{-III}$ and $\mathrm{-I}$ for ethanol’s two carbons and $\mathrm{-III}$ and $\mathrm{+I}$ for ethanal’s. This method has a number of advantages:

you are always dealing with integers; no need for any fractional states
it is easy to see exactly which carbon’s oxidation state changed.

The downside of this method is that you cannot use it if you don’t know the structural formula. Consider if, for example, you only saw $\ce{C2H6O}$ or $\ce{C2H4O}$ — especially the first one could represent two different structures with different properties. This is were average oxidation states come in handy: you calculate an average of all the atoms of a single element. You may remember that oxygen is typically given $\mathrm{-II}$, hydrogen $\mathrm{+I}$ (except in peroxides and hydrides, respectively; although oxofluorides should get a special mention). You may also remember that the sum of oxidation states of a neutral compound must be $\pm 0$ that allows you to calculate the average oxidation states as follows:

$$\begin{align}\text{For }\ce{C2H6O}\text{:}&\\
0 &= 2 x + 6 \times (+1) + (-2)\\
0 &= 2x + 6 - 2\\
0 &= 2x + 4\\
-4 &= 2x\\
x &= -2\\
\\
\text{For }\ce{C2H4O}\text{:}&\\
0 &= 2 x + 4 \times (+1) + (-2)\\
0 &= 2x + 4 - 2\\
0 &= 2x + 2\\
-2 &= 2x\\
x &= -1\end{align}$$

Since these are averages, you can also arrive at the same conclusion by taking the arithmetic mean of all carbon atoms’ oxidation state; you will obviously arrive at the same result:

$$\begin{align}\ce{C2H6O}\text{:}&\\
\bar{x} &= \frac{-3 + (-1)}{2}\\
\bar{x} &= \frac{-4}{2}\\
\bar{x} &= -2\\
\\
\ce{C2H4O}\text{:}&\\
\bar{x} &= \frac{-3 + 1}{2}\\
\bar{x} &= \frac{-2}{2}\\
\bar{x} &= -1\end{align}$$


Enantioselective organic reaction not working I'm performing the reaction between acetone and 2-nitrobenzaldehyde catalysed by (L)-Proline, as answered to one of my questions a while ago: Simple experiments involving enantiomer synthesis

However, while the literature specified using 1 mmol of (L)-Proline in 20ml of solution, due to the cost of the materials and the budget available, I've only been able to use a fraction of that, around 0.01mmol. I've scaled down the entire experiment to 5 ml to try and compensate for this.

The issue is, I've left this solution reacting for a long period of time (12+ hours) and I'm not seeing any optical activity at all. With the reduced concentrations, I'd have expected to see a lot less activity than the original literature specifies, but to see at least some indication that my reaction is working.

Are the concentrations of (L)-Proline simply too small for any significant reaction to take place? Or can there be some other reason why my reaction isn't working? Note: I'm a chemistry 'newbie', so don't assume anything!

The original "Rutgers' reference" deploys $2\,\mbox{mmol}$ aldehyde in $20\,\mbox{mL}$ solvent (i.e. starting at $0.1\,\mbox{mol/L}$, or $0.1\,\mbox{M}$) and recommends a stirring of $3\,\mbox{h}$.

Subsequently, you mention $1\,\mbox{mmol}$ per $20\,\mbox{mL}$ solvent (i.e. starting at $0.05\,\mbox{M}$), but without an indication of reaction time.

But if the starting concentration is lowered further ($0.01\,\mbox{mmol}$ in $5\,\mbox{mL}$ equals to $0.002\,\mbox{M}$ or $2\,\mbox{mM}$), not only the detection of the product becomes difficult. Did you take into consideration that by diluting the reaction you over-proportionally slowed down the advancement of the intended reaction?  Compared to the initially mentioned starting concentration, your reaction starts at a dilution by a factor of 50 (fifty).  This is more than a big factor for process involving a bimolecular reaction.

By the way, $0.01\,\mbox{mmol}$ of proline, molar weight of $115.13\,\mbox{g/mol}$ as stated here would correspond to little more than $1.1\,\mbox{mg}$.


Combustion in liquid state Normally, we think of combustion as occurring only in the gas phase. I think I understand the main reason for this - in the gas phase it's very hard for a bimolecular reaction to produce a single product, which makes it more likely for radical chain reactions to propagate rather than being terminated - but I'm curious about whether it's possible for combustion to occur in the liquid state at all.

For example: suppose I mixed liquid $\ce{CH4}$ with liquid $\ce{O2}$ under high pressure and reasonably low temperature. Would they react immediately? If not, could I 'ignite' the mixture in the same way as I could in the gas phase? (Note that the combustion of $\ce{CH4}$ into $\ce{H2O}$ and $\ce{CO2}$ doesn't change the mole number, so the volume change should be relatively small.) Would the reactions that occur be similar to the reactions that occur during combustion of $\ce{CH4}$ in the gas phase, or would they be completely different?

If this reaction wouldn't resemble combustion, is there some other example of a self-propagating oxidation reaction in liquid or aqueous phase with similar properties to combustion?

Your question is addressed by Explosive Systems Containing Liquid Oxygen: Liquid Oxygen-Liquid Methane Mixtures. J. Chem. Eng. Data, 1959, vol. 4, pp. 127–131.

If liquid methane and liquid oxgygen at 90K are mixed, they are miscible in all proportions.  Mixtures over a wide range of proportions detonated with use of a blasting cap. 100 grams of mixture punched a 10cm hole in a 3/16 inch steel plate.

For comparision, the TNT control only dented the metal plate!

Further tests on the stoichimetric liquid methane/oxgygen mixture found that it was detonated by a .22 caliber copper bullet at 40 foot range, flame and spark.

So nothing happens if the liquids are simply mixed, but a powerful explosion occurs if a stimulus is applied.

Liquid oxygen and liquid methane have also been used in rocket engines.  See Liquid Methane/Liquid Oxygen Injectors for Potential Future Mars Ascent Engines for example.


Why are N-H stretching vibrations often sharp and not broad? It is known that hydrogen bonding results in the broadening of peaks in the infrared spectrum of a molecule. Hydrogen bonding results in a spectrum of different bond lengths with different stretching frequencies. This results in a the functional group giving peaks of a whole continuous range of vibrational frequencies and hence, we see a broad peak. Broad peaks are most often observed from $\ce {2500}$ to $\ce {3000 cm^{-1}}$ for the stretching of carboxylic acid functionalities. Similarly, hydrogen bonding also results in alcohol $\ce {O-H}$ stretching giving broad peaks in the range of $\ce {3200}$ to $\ce {3600 cm^{-1}}$, albeit the degree of broadening is much less for alcohols. Amines are also able to form rather strong hydrogen bonds. However, amine $\ce {N-H}$ stretches often are not seen as broad peaks and are in fact, often sharp. Why is this so?

The sharpness or broadness of a stretch in IR spectra depends on extent of Hydrogen bonding present in the molecule. Basically, if it undergoes immense intermolecular hydrogen bonding, the peaks tend to be broader and the lesser the hydrogen bonding becomes, the sharper the peaks get in the spectra.

Now, if you recall the criteria needed for stronger hydrogen bonding, the more electronegative the atom attached to hydrogen is the better is the extent of hydrogen bonding as the more electronegative atom will be able to create more $\delta +$ve cahrge on hydrogen which is better for stronger hydrogen bonding. $\ce{O}$ is far more electronegative than $\ce{N}$. $(\chi_o = 3.44, \chi_N = 3.04)$. So, the extent of hydrogen bonding is much much lesser with $\ce{N}$ being the attached atom than $\ce{O}$. (If you want to get an idea how strong the hydrogen bonding is in case of $\ce{O}$ than $\ce{N}$, simply consider $\ce{H2O}$ and $\ce{NH3}$, due to hydrogen bonding the former has its boiling point at $100^o C$, while that for the latter is $-33.4^oC$.) But anyways Hydrogen bonding is there in amines, only the extent is lesser.

However if you can somehow increase the $\delta+$ charge on hydrogen, by attaching some electron withdrawing group to the nitrogen, like in the case of amides, there you will be again able to see broad $\ce{N-H}$ stretches as now it can undergo stronger hydrogen bond formation, as $\ce{N}$ pulls the electron density more towards itself due to resonance., and $\ce{H}$ becomes electron deficient, and peak becomes broad. Here is the IR spectra of $\ce{N-methylacetamide}$ from the book Introduction to Spectroscopy which shows broad $\ce{N-H}$ stretches due to the reason mentioned above.

In this spectra you can also see the $\ce{C=O}$ stretch is also relatively broader than observed generally. The reason is again extent of intermolecular hydrogen bonding, which is greater due to higher electronic density on the $\ce{O}$. But apart from that, the general reason for broadening the spectra is the extent of intermolecular hydrogen bonding, which is somewhat lesser for normal amines.


Help with finding the right catalyst and reaction conditions for hydroamination reaction A friend of mine, final year PhD, and I, final year Bachelor, are focusing on hydroamination reactions. In particular those with secondary amines and substituted alkenes. The reaction in particular is given by

Intermolecular hydroamination of unactivated alkenes with substituted secondary amines seems to be quite difficult. However, additions of cyclic and acyclic secondary alkyl amines to a wide range of alkenes with anti-Markovnikov regioselectivity have been reported. Some of those reactions are redox-neutral and occur readily at room temperature under visible light irradiation.
Our question is: what would be an applicable catalyst for this particular reaction (secondary amine + substituted alkene) and what reaction conditions would be necessary?
Thanks in advance!
EDIT:
The reaction may also concerns the primary amine,

However, we would like to keep the substituted alkene substrate as it is.

You may follow the procedure in this reference. I included the abstract for your convenience:

  Abstract: The hydroamination of styrene with aniline catalyzed by phosphine-ligated palladium triflates exhibits a substantial $\ce{^{13}C}$ isotope effect at the benzylic carbon. This supports rate-determining nucleophilic attack of amine on a $\eta^3$-phenethyl palladium complex. Deuterium exchange observations and predicted isotope effects based on DFT calculations support this mechanism. Selectivity in these reactions is determined by the facility of palladium displacement after reversible hydropalladation of the alkene.

The Graphical Abstract:

Reference:

L. K. Vo, D. A. Singleton, “Isotope Effects and the Nature of Stereo- and Regioselectivity in Hydroaminations of Vinylarenes Catalyzed by Palladium(II)−Diphosphine Complexes,” Org. Lett. 2004, 6(14), 2469–2472 (DOI: 10.1021/ol049137a).


boric acid + isopropyl alcohol I have found recommendations for making a soldering flux using boric acid mixed in with ethyl alcohol in 50/50 ration or 60/40 depending on the consistency desired.

Some have recommended that methyl alcohol is better to use as more of the boric acid can be dissolved in it? Is this correct?

If so, how do I calculate how much boric acid I can dissolve in isopropyl alcohol? What would be the best solvent to create a paste of boric acid to use as a soldering flux?

For safety reasons, I would stick with ethanol or isopropanol, as they are less flammable than methanol (your using hot soldering equipment) and much less toxic, even though methanol is a better solvent for $\ce{H3BO3}$. If it weren't for the expense, due to heavy taxation of ethanol, it would be preferable to isopropanol. From the table below, you'll need almost twice as much isopropanol (though 1-propanol is the solvent listed, it is an isomer of the more common 2-propanol, AKA isopropanol).

  Solubility of $\ce{H3BO3}$ in various alcohols at 25 C, percent by weight:

  Methanol:     22.66%

  Ethanol:     11.96%

  1-Propanol:    7.34%


The weight of a 2cc glass pharmaceutical flame sealed ampoule? This is one of the flame sealed type whose neck is broken before use. There seems to be lots of info on dimensions but none with regard to weight.

I grabbed 10 empty, unsealed ones and threw them on the analytical balance and got mean: 2.4071 g, s.d.: 0.01264 g for Wheaton amber glass pre-scored vials (part number 176796). The variation between vials isn't too bad, but I don't know if they can be filled and sealed consistently enough to determine gas density like that to any useful precision and other brands may be better or worse.


Which of the following releases the smallest amount of energy upon gaining an electron? $\ce{C}$, $\ce{Rb}$, $\ce{Ga}$, $\ce{Ge}$

I am confused in the sense that I don't know what it means by "released the smallest about of energy upon gaining?" would it be $\ce{Rb}$ because the shell is so big that when it loses an electron the overall energy doesn't really matter?

Alright, this is similar to a question that I have answered before but I will just give you the reasoning for this phenomenon. I will not give you the answer since this appears to be a homework question.

Why an atom gains energy when you take off an electron?
Well for this, think of the attractions between the nucleus and the electrons, so when you must strip away an electron from this stability, you must add energy to do so.

Why an atom releases energy when you add an electron?
This one might be a little hard to grasp, but at the end it is not too bad. Essentially the molecule is becoming more stable. It is getting near to having a full octet. (Ignoring noble gasses). In other words the kinetic energy added by gaining the electron is released when the atom is achieving a stable state.

This is the general reasoning behind this, not too complicated, but I hope you will be able to understand it. Just, if you can't, remember the first one and think of the opposite for the second one.

Now, take a look at this question, to help you with your question even more. You will then be able to deduce the correct answer.


What would be the aldol condensation products in B In this question (13) I figured out A,C were wrong and D would be correct choice but I am stuck at finding out products formed in  B and also why it would be wrong?
Please help.

B is the correct answer, but not for the reasons you have given.

4-Me-benzaldehyde cannot self-condense, and is a far more reactive electrophile than acetophenone.  Acetophenone has a pKa of around 16 so will readily undergo base-catalysed aldol reaction with the aldehyde. This is the only reaction that will occur. All the other sets of reactants shown are likely to undergo self-condensation as well as cross-condensation giving a mixture of products.


If all intermolecular forces are electrostatic in nature, why don’t large non polar molecule dissolve in water? If I have a large non-polar substance A that forms strong dipole-induced dipole force (DIDF) with water, with strength comparable to that of H-bonds between water molecules, will A dissolve in water?

Dipoles can be induced, but they are not permanent. You’re basically asking why a spring doesn’t like to stay compressed when you push on it—it's because that’s not its most stable position. The electron density around  hydrogen in a $\ce{C-H}$ bond is much higher than the hydrogen in a polar $\ce{O-H}$ bond, so electrostatic repulsions overwhelm any attractions that may be induced. See the diagram below$[1]$:

$[1]$ Bond polarity diagram - Iverson Group UT Austin


Does Fe(CO)6 really exist? I was given a question to draw the structure of $\ce{Fe(CO)6}$ and tell the hybridization. Hybridization came out to be $\ce{sp^3d^2}$ and structure as Octahedral. But does $\ce{Fe(CO)6}$ really exist? I could not find anything about it on the internet.

There is an iron hexacarbonyl, not as a neutral compound but as a dication which satisfies the 18-electron rule. This along with its heavier Group 8 congeners is reported by Finze et al. [1]. The authors who report the tetrafluoroborate salts, also reference earlier work with the corresponding fluoroantimonates.  The reference was found in Wikipedia.

Reference

Finze, M.; Bernhardt, E.; Willner, H.; Lehmann, C. W.; Aubke, F. (2005). "Homoleptic, σ-Bonded Octahedral Superelectrophilic Metal Carbonyl Cations of Iron(II), Ruthenium(II), and Osmium(II). Part 2: Syntheses and Characterizations of [M(CO)6][BF4]2 (M = Fe, Ru, Os)". Inorganic Chemistry. 44 (12): 4206–4214. https://pubs.acs.org/doi/10.1021/ic0482483


How can rotamers of alpha-keto esters be named to distinguish between different geometries? When talking about free rotation it is often helpful to determine about which geometric arrangement one is referring to. For example in the case of butane we talk about several eclipsed or staggered conformations when discussing the rotational barriers. This also often helps to explain the all trans (staggered) arrangement of fatty acids and surfactants.

In a wider sense even the pseudo rotation of cyclohexane has its own terminology, with boat, chair and twisted conformations.

I am interested in a way to name the two rotamers that can be thought of in α-keto esters (1, 2). My initial thought was calling these cis- and trans- configuration, in analogy of the partial double bond (blue, 3, 4), but for my taste this has a too rigid connotation. Has there been anything proposed in this area? Amides would also be comparable (5, 6).

The IUPAC recommends s-cis and s-trans for the rotamers of conjugated dienes.

The nomenclature derives from having a "cis-like" or "trans-like" geometry about a sigma bond.

I feel like something similar should be appropriate, however that same IUPAC link recommends E/Z or sp/ap (syn-periplanar/anti-periplanar) for the N-alkyl amides (5,6).

Amide 5 would be Z (but ap w/r/t alkyl groups) and amide 6 would be E (and sp). E/Z is less ambiguous in my opinion.

Ketoester 1 is E, and ketoester 2 is Z. I suppose that ketoester 1 is also ap and ketoester 2 is also sp.


Is NaCl in water considered "solution" despite of NaCl's having some precipitation? Is $\ce{NaCl}$ in water aqueous? For a certain amount of $\ce{NaCl}$, it can dissolve in water but if you add more $\ce{NaCl}$ it can't dissolve any more.

So, if in a certain reaction $\ce{NaCl}$ is formed with water:

$$\ce{HCl(aq) + NaOH(aq) -> H2O(l) + NaCl(aq)}$$

My question is does $\ce{NaCl}$ in the previous reaction totally dissolve in the water formed? Or in other words is this equation correct?

$$\ce{HCl(aq) + NaOH(aq) -> H2O(l) + NaCl(aq) + NaCl(s)}$$

Another question is why do we write (l) beside water rather than (aq) since it's part of the aqueous solution?

When we are considering $\ce{HCl(aq)}$ and $\ce{NaOH(aq)}$. the system contain sufficient amount amount of solvent water molecules  in situ already present and this is more than sufficient to hydrate the produced $\ce{NaCl}$ in the reaction.
Hence the first equation
$$\ce{HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l)}$$

will be the correct expression.


Difference between valency and oxidation state I know the abstract and sketchy obscure differences but I cannot find anywhere the specific comparison between the concepts of oxidation states and valency including how they are used to describe chemicals.

Valency is a term used to describe the number of bonds a given atom undergoes in a certain environment. For example, the carbon atom in all of the following substances is tetravalent:

methane
methanol
difluoromethane
formic acid
carbon dioxide

An example for a formally trivalent carbon would be carbon monoxide. Note how valency only counts how many bonds and how many lone pairs an atom has.

The oxidation state is used as a formal measurement on how formally negative an atom should be under standard conditions. For this, each bond is split heterogenously and both electrons are counted as belonging to the electronegative partner only. Then, the electrons on each atom are counted, the difference to a neutral atom of the same kind is established which constitutes the oxidation number. The compound above — while all being tetravalent — have different oxidation numbers:

methane $\mathrm{-IV}$
methanol $\mathrm{-II}$
difluoromethane $\mathrm{\pm 0}$
formic acid $\mathrm{+II}$
carbon dioxide $\mathrm{+IV}$


What property allows ozone to absorb uv light Why can ozone act as a barrier of UV light, so that we don't get irradiated by the sun's energy, and not any old molecule? In a recent exam I sat, one of the questions was "Name one physical property of ozone and contrast it with oxygen" and under exam pressure I spurted this onto the page: "Because the coordinate covalent bond in ozone is weaker than a traditional covalent bond, the bond is free to 'wiggle' about a bit. This means that the ozone can absorb a range of frequencies of photons, as a completely rigid molecule like oxygen would only have very specific frequencies which it could absorb." How much of this 'guess' is grounded in fact, or in other words, what is the actual mechanism that allows ozone to protect us from UV rays and not any old molecule?

Shortwave UV light (100-240nm) has enough energy to knock an electron out of an $\ce{O_2}$ bond creating an excited molecule $\ce{O^{excited}_2}$ which decomposes to two free radicals of oxygen. Each of the free radicals then combines with different oxygen molecules to create a net of two ozone molecules.

The crux here is that the electron is not "knocked out" of the oxygen molecule, the electron is just moved to a higher energy orbital of the same oxygen molecule. Since orbitals are quantized it takes a very specific energy to make the transition happen.

Now there is a second reaction. Longwave UV light (240nm - 350nm) has enough energy to knock an electron out of an $\ce{O_3}$ bond creating an excited molecule $\ce{O^{excited}_3}$ which decomposes to a free radical of oxygen and an oxygen molecule.

Thus oxygen protects the ground from shortwave UV, and ozone from longwave UV. Neither alone would protects us from all UV as effectively.

Now the rest of the story is that all molecules bouncing around in the air are not in the same energy state due to atom-atom vibrations. So the UV absorption occurs as a "band" rather than as a "fine line" like in the absorption spectra of an atom in free space.


Is 1.5M NaOH a strong enough base to deprotonate benzyl alcohol? Is 1.5M NaOH a strong enough base to deprotonate benzyl alcohol? I realize that NaOH generally does not deprotonate benzyl alcohol, but we performed an extraction in a Fischer esterification lab with benzyl alcohol an acetic acid to synthesize benzyl acetate. Acetic acid was added in excess, and the final ester product was purified with extraction techniques. I know that addition of NaOH removes acetic acid from the organic, diethyl ether layer, but would 1.5M NaOH also remove any remaining benzyl alcohol?

Only a very small fraction of the benzyl alcohol ($\mathrm{p}K_\mathrm{a} = 15.4$) would be removed by sodium hydroxide. Acetic acid is about 10 orders of magnitude more acidic than benzyl alcohol with a $\mathrm{p}K_\mathrm{a}$ of $4.8$, so it is completely deprotinated to form sodium acetate and water.

For the purposes of your experiment you can assume that there is effectively no reaction between benzyl alcohol and sodium hydroxide.


Argon vs nitrogen for preserving wine I have a question regarding whether argon or nitrogen is to prefer when it comes to preserving wine.

Specifically, I am interested in knowing if there is a difference in preservation periods.

Also, if there is a risk of them reacting with the wine, which might affect taste and smell?

In chemistry, some chemical species can still react with nitrogen so argon is usually favored, but, in the case of wines, none of the two should react whatsoever.

In both cases, however, the high density of argon is an advantage because it will always fall down to be in contact with the liquid in order to avoid contact with oxygen.

By contrast, nitrogen would just dilute oxygen in the remaining air and just slow down oxidation, because air is already composed of 80% nitrogen.


What is the multiplicity of this signal in the ¹H NMR of glutamine in D₂O The 2 Hydrogens near the carbonyl (the lower 2 $\ce{H}$s are not bonded to the nitrogen) should be the most deshielded so I think they appear as the rightmost peak in the NMR.  Are these 2 $\ce{H}$s a doublet of doublets?  I think this because they are split by 2 different $\ce{H}$s.

You're thinking about deshielding the wrong way around, the most deshielded signal appears at the highers ppm value, so to the left. In this case it would be the signal at around 3.8 ppm, as the signal at 4.7 ppm is residual water.

The two signals at the right are multiplets, if you take a look they're obviously caused by rather complicated coupling patterns that you can't easily resolve. The reason for that is that the protons you see in this spectrum are not magnetically equivalent, even the CH2 protons are not magnetically equivalent. This leads to the complicated couplings you can observe here.


Why is carbon monoxide the product in reactions that produce high flame temperatures? I was looking at substances that produce the highest flame temperatures upon combustion. I note that the equations for combustion are producing carbon monoxide rather than carbon dioxide.

$$\ce{(CN)2 + O2 -> 2CO + N2}$$

Why isn't carbon dioxide produced? Why doesn't the production of carbon dioxide give a hotter flame? It seems that a higher flame temperature would be generated by complete combustion. My thinking is that the heat of formation of carbon dioxide is around −400 kJ/mol while the heat of formation of carbon monoxide is around −100 kJ/mol so the additional heat released by producing carbon dioxide should account for carbon dioxide's greater heat capacity.

This is just a guess... But I wonder if this is due to entropic considerations. In the equation you have written, there are two molecules of gas getting converted to three molecules of gas. $\Delta S$ is positive, so it decreases $\Delta G$ as the reaction temperature increases. However, in the reaction $\ce{(CN)2 + 2 O2 = 2 CO2 + N2}$ you have an equal number of gas molecules, so the entropic component of $\Delta G$ will be small.

I'm not really sure why that would matter. At very high temps, the first reaction will presumably have a very large, negative free energy. Of course, the $\Delta G$ for the $\ce{CO2}$ reaction is most likely negative, so you might have to control the feed stoichiometry carefully to ensure that you are getting only the first reaction.

EDIT:

I got the $\Delta G$ values from the NIST-JANAF website, and it doesn't look like the free energy for these reactions is much different at 4000 K.

$\Delta_f G_{(CN)_2}^{0} = 136.036\ \mathrm{kJ\ mol^{-1}}$

$\Delta_f G_{CO}^{0} = -446.485\ \mathrm{kJ\ mol^{-1}}$

$\Delta_f G_{CO_2}^{0} = -393.183\ \mathrm{kJ\ mol^{-1}}$

$\Delta G_{rxn1}^{0} = -1029.005\ \mathrm{kJ\ mol^{-1}}$

$\Delta G_{rxn2}^{0} = -922.397\ \mathrm{kJ\ mol^{-1}}$

EDIT 2:

So, I just read one of the original papers on this type of flame, and it sounds like the thermal stability of the products is an issue as well. I take it that at these very high temperatures, some of the energy can be transferred to bond-breaking processes, which would lower the total flame temperature. $\ce{CO}$ and $\ce{N2}$ have very strong covalent bonds, so perhaps they do not readily dissociate at these temps.


What is a molecule’s smallest set of smallest rings? It is been a while since I've studied chemistry. Now, I am reading the documentation of RDKit, where they mention the expression "smallest set of smallest rings" (SSSRs). I understand that a ring is a cycle in a molecule. I also understand that the SSSRs is, as the name suggest, the smallest set of smallest rings, but how do you determine what the smallest sets and rings are?  What is a molecule's smallest set of smallest rings? What is a SSSRs used for?

This is a tool to classify molecules with multiple rings. Here is an example from a PhD thesis:

So even though you can find six-membered and eight-membered rings in this structure, these are made up of multiple four-membered rings, and are thus not parts of the set of rings. As the picture illustrates, it is sufficient to choose five rings to show the connectivity of the atoms, so even though there are six four-membered rings, the SSSR includes only five of them. In this case, which five is ambiguous.

how do you determine what the smallest sets and rings are?

It turns out that this is ambiguous, so different software might get different results. In contrast, giving the size of all rings is unambiguous, but - according to the thesis I quoted - computationally more expensive and chemically less intuitive.

What is a SSSRs used for?

To classify cyclic molecules in cheminformatics. For a given software, two molecules are different if they have different SSSRs. Conversely, two identical molecules will have the same SSSR. This makes it easier to classify, sort, and search this class of molecules.


Why chelation increases entropy It is written here that chelation increases overall entropy?

I thought it should decrease entropy because the ligands are now bonded to the central metal atom forming a ring like structure. According to me entropy must decrease because the ligands can no longer move around so there is a decrease in randomness. Why this logic is not correct?

The chelate effect is very simple. Consider the addition of $\ce{Na2H2edta}$ to a random transition metal (e.g. manganese(II)). Your starting complex is a hydrate complex; we shall assume it is $\ce{[Mn(H2O)6]^2+}$. The reaction equation is given below:

$$\ce{[Mn(H2O)6]^2+ + edta^4- <=>> [Mn(edta)]^2- + 6 H2O}\tag{1}$$

You started with two molecular entities, you end up with seven. Since $2<7$, the product side has a higher entropy (more independent particles). The same is true for any chelating ligand that displaces more than one monodentate ligand.

Yes, upon chelation the chelating ligand does lose degrees of freedom concerning bond rotation. However, the number of particles greatly outweighs that.