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The book asks me to find the oxidation state of nitrogen in the compound below (the structure is *not* the one given in [Wikipedia for HN3](https://en.wikipedia.org/wiki/HN3)):
> [![triangular structure of HN3][1]][1]
To find the oxidation state using Lewis structure, I have to compare the electronegativity of the bonded atoms and assign the bonding electrons to the more electronegative atom. Then, the difference between the number of electrons in element state and bonded state will give the oxidation number.
Here, the bonding electrons of nitrogens bonded with a double bond will stay in between both the atoms will not be assigned to both. The $\ce{N-H}$ bonding electrons will be assigned to $\ce{N}$. What about the two $\ce{N-N}$ bonding electrons?
[1]: https://i.stack.imgur.com/pNLTH.jpg |
I know that
$$C_p - C_V = R, \label{eqn:1}\tag{1}$$
but I have also seen
$$C_p - C_V = R/M \label{eqn:2}\tag{2}$$
being used in case of ideal gas.
In what cases \eqref{eqn:1} is used and in what cases \eqref{eqn:2} is used? |
For what processes is Mayer's formula applicable? |
We have two solutions:
- Solution 1 is $\ce{HCOOH}$, its concentration is $c_1 = \pu{10^-2 mol/l}$, its volume is $V_1 = \pu{50 ml}$, and its $\mathrm{pH}_1 = 2.9$.
- Solution 2 is $\ce{CH3COOH}$, its concentration is $c_2 = \pu{10^-2 mol/l}$, its volume is $V_2 = \pu{50 ml}$, and its $\mathrm{pH}_2 = 3.4$.
How would be the equation and the ICE table, and what is the $\mathrm{pH}$ of the mixture of these two solutions?
I used numbers to differentiate the solutions, and to understand how it works, but for no other reason.
I tried to do $\mathrm{pH}=\frac{1}{2}\left(\mathrm{pH}_1+\mathrm{pH}_2\right)$, but it’s trivial and doesn't work. |
If I heat ammonia gas to 250 degrees celsius, will it decompose into its basic components - N2 and H2? What temperature is required for the reaction to occur? |
Will NH3 gas decompose into N2 and H2 at 250 degrees celcius? |
Imagine a bottle 85% filled with water and the rest is just air, and the cap is closed. Here are now my questions.
Since the cap is closed completely, the air bubble is formed. Is that air bubble compressed? In other words, since air has the least force of attraction between its molecules, they are more free than the molecules of water. Now since water is creating a pressure against the air bubble, does it mean that the molecules in the bubble are now closely packed than usual?
Is it possible for the air bubble, if compressed a lot, to mix with water?
If the air bubble is able to hold itself even in water, why do we say that air has lesser force of attraction that water? |
I looked up the decompositions of diluted HClO2 aqueous solution (used in a disinfectant) to know if it is safe to use on some substance, such as metal, wood, leather or skin. However, the chain I found leading to HClO4, which is a very strong acid.
>HClO2 + 3HCl → 2Cl2 + 2H2O
(Source: https://chemiday.com/en/reaction/3-1-0-4242)
>4HClO2 → 2ClO2 + HClO3 + HCl + H2O
(Source: https://chemiday.com/en/reaction/3-1-0-1112)
>3HClO3 → 2ClO2 + HClO4 + H2O
(Source: https://chemiday.com/en/reaction/3-1-0-1120)
The HClO2 solution is diluted (I do not know the exact number, but maybe similar to mild bleach), sprayed to substances' surface, at room or body temperature.
Can the balances in this case lead to HClO4?
Can HCl and HClO4 be concentrated enough to corrode metal? |
What does HClO2 aqueous solution decompose to in an open environment? |
I looked up the decompositions of diluted HClO2 aqueous solution (used in a disinfectant) to know if it is safe to use on some substance, such as metal, wood, leather or skin. However, the chain I found leading to HClO4, which is a very strong acid.
>5HClO2 → 4ClO2 + HCl + 2H2O
(Source: https://chemiday.com/en/reaction/3-1-0-4241)
>4HClO2 → 2ClO2 + HClO3 + HCl + H2O
(Source: https://chemiday.com/en/reaction/3-1-0-1112)
>3HClO3 → 2ClO2 + HClO4 + H2O
(Source: https://chemiday.com/en/reaction/3-1-0-1120)
The HClO2 solution is diluted (I do not know the exact number, but maybe similar to mild bleach), sprayed to substances' surface, at room or body temperature.
Can the balances in this case lead to HClO4?
Can HCl and HClO4 be concentrated enough to corrode metal? |
I looked up the decompositions of diluted HClO2 aqueous solution (used in a disinfectant) to know if it is safe to use on some substance, such as metal, wood, leather or skin. However, the chain I found leading to HClO4, which is a very strong acid.
>5HClO2 → 4ClO2 + HCl + 2H2O
>4HClO2 → 2ClO2 + HClO3 + HCl + H2O
>3HClO2 → 2HClO3 + HCl
>2HClO2 → 2HOCl + HClO3
>HClO2 → HCl + O2
Then there is
>3HClO3 → 2ClO2 + HClO4 + H2O
(Source: "Inorganic Chemistry" by Rajni Garg, Randhir Singh, [Google book preview](https://books.google.co.jp/books?id=XKxqCgAAQBAJ&pg=PT978&lpg=PT978&dq=4HClO2+%E2%86%92+2ClO2+%2B+HClO3+%2B+HCl+%2B+H2O&source=bl&ots=Gj8goScjH9&sig=ACfU3U1k1j-XIi2T6aJUnTGE3avo-dw6Mw&hl=en&sa=X&ved=2ahUKEwiPluib5JToAhVnxosBHXfODEoQ6AEwAHoECAoQAQ#v=onepage&q=4HClO2%20%E2%86%92%202ClO2%20%2B%20HClO3%20%2B%20HCl%20%2B%20H2O&f=false))
The HClO2 solution is diluted (I do not know the exact number, but maybe similar to mild bleach), sprayed to substances' surface, at room or body temperature.
Can the balances in this case lead to HClO4?
On the other hand, HClO2 also recombines with HCl: HClO2 + 3HCl → 2Cl2 + 2H2O
So, can the remaining HCl be concentrated enough to corrode metal? |
Is there a chemical treatment that could remove sodium and calcium ions from the surface of soda-lime glass to turn it into quartz glass, increasing the hardness? |
Can glass be deionized? |
I looked up the decompositions of diluted $\ce{HClO2}$ aqueous solution (used in a disinfectant) to know if it is safe to use on some substance, such as metal, wood, leather or skin. However, the chain I found leading to $\ce{HClO4},$ which is a very strong acid.
> $$
\begin{align}
\ce{5 HClO2 &→ 4 ClO2 + HCl + 2 H2O}\\
\ce{4 HClO2 &→ 2 ClO2 + HClO3 + HCl + H2O}\\
\ce{3 HClO2 &→ 2 HClO3 + HCl}\\
\ce{2 HClO2 &→ 2 HOCl + HClO3}\\
\ce{HClO2 &→ HCl + O2}
\end{align}
$$
Then there is
> $$\ce{3HClO3 → 2ClO2 + HClO4 + H2O}$$
(Source: *Inorganic Chemistry* by Garg and Singh \[1\], [Google book preview](https://books.google.co.jp/books?id=XKxqCgAAQBAJ&pg=PT978&lpg=PT978&dq=4HClO2+%E2%86%92+2ClO2+%2B+HClO3+%2B+HCl+%2B+H2O&source=bl&ots=Gj8goScjH9&sig=ACfU3U1k1j-XIi2T6aJUnTGE3avo-dw6Mw&hl=en&sa=X&ved=2ahUKEwiPluib5JToAhVnxosBHXfODEoQ6AEwAHoECAoQAQ#v=onepage&q=4HClO2%20%E2%86%92%202ClO2%20%2B%20HClO3%20%2B%20HCl%20%2B%20H2O&f=false))
The $\ce{HClO2}$ solution is diluted (I do not know the exact number, but maybe similar to mild bleach), sprayed to substances' surface, at room or body temperature.
Can the balances in this case lead to $\ce{HClO4}?$
On the other hand, $\ce{HClO2}$ also recombines with $\ce{HCl}:$
$$\ce{HClO2 + 3 HCl → 2 Cl2 + 2H2O}$$
So, can the remaining $\ce{HCl}$ be concentrated enough to corrode metal?
### References
1. Garg, R.; Singh, R. *Inorganic Chemistry*; McGraw-Hill Education: New Delhi, **2015**. ISBN 978-1-259-06285-8. |
I looked up the decompositions of diluted $\ce{HClO2}$ aqueous solution (used in a disinfectant) to know if it is safe to use on some substances, such as metal, wood, leather or skin. However, the chain I found leading to $\ce{HClO4},$ which is a very strong acid.
> $$
\begin{align}
\ce{5 HClO2 &→ 4 ClO2 + HCl + 2 H2O}\\
\ce{4 HClO2 &→ 2 ClO2 + HClO3 + HCl + H2O}\\
\ce{3 HClO2 &→ 2 HClO3 + HCl}\\
\ce{2 HClO2 &→ 2 HOCl + HClO3}\\
\ce{HClO2 &→ HCl + O2}
\end{align}
$$
Then there is
> $$\ce{3HClO3 → 2ClO2 + HClO4 + H2O}$$
(Source: *Inorganic Chemistry* by Garg and Singh \[1\], [Google book preview](https://books.google.co.jp/books?id=XKxqCgAAQBAJ&pg=PT978&lpg=PT978&dq=4HClO2+%E2%86%92+2ClO2+%2B+HClO3+%2B+HCl+%2B+H2O&source=bl&ots=Gj8goScjH9&sig=ACfU3U1k1j-XIi2T6aJUnTGE3avo-dw6Mw&hl=en&sa=X&ved=2ahUKEwiPluib5JToAhVnxosBHXfODEoQ6AEwAHoECAoQAQ#v=onepage&q=4HClO2%20%E2%86%92%202ClO2%20%2B%20HClO3%20%2B%20HCl%20%2B%20H2O&f=false))
The $\ce{HClO2}$ solution is diluted (I do not know the exact number, but maybe similar to mild bleach), sprayed to substances' surface, at room or body temperature.
Can the balances in this case lead to $\ce{HClO4}?$
On the other hand, $\ce{HCl}$ recombines to make $\ce{Cl2}$:
> $$
\begin{align}
\ce{HClO2 + 3 HCl &→ 2 Cl2 + 2H2O}\\
\ce{HOCl + HCl &→ Cl2 + H2O}
\end{align}
$$
So, can the remaining $\ce{HCl}$ be concentrated enough to corrode metal or cause harms to skin?
### References
1. Garg, R.; Singh, R. *Inorganic Chemistry*; McGraw-Hill Education: New Delhi, **2015**. ISBN 978-1-259-06285-8. |
I looked up the decompositions of diluted $\ce{HClO2}$ aqueous solution (used in a disinfectant) to know if it is safe to use on some substances, such as metal, wood, leather or skin. However, the chain I found leading to $\ce{HClO4},$ which is a very strong acid.
> $$
\begin{align}
\ce{5 HClO2 &→ 4 ClO2 + HCl + 2 H2O}\\
\ce{4 HClO2 &→ 2 ClO2 + HClO3 + HCl + H2O}\\
\ce{3 HClO2 &→ 2 HClO3 + HCl}\\
\ce{2 HClO2 &→ 2 HOCl + HClO3}\\
\ce{HClO2 &→ HCl + O2}
\end{align}
$$
Then there is
> $$\ce{3HClO3 → 2ClO2 + HClO4 + H2O}$$
(Source: *Inorganic Chemistry* by Garg and Singh \[1\], [Google book preview](https://books.google.co.jp/books?id=XKxqCgAAQBAJ&pg=PT978&lpg=PT978&dq=4HClO2+%E2%86%92+2ClO2+%2B+HClO3+%2B+HCl+%2B+H2O&source=bl&ots=Gj8goScjH9&sig=ACfU3U1k1j-XIi2T6aJUnTGE3avo-dw6Mw&hl=en&sa=X&ved=2ahUKEwiPluib5JToAhVnxosBHXfODEoQ6AEwAHoECAoQAQ#v=onepage&q=4HClO2%20%E2%86%92%202ClO2%20%2B%20HClO3%20%2B%20HCl%20%2B%20H2O&f=false))
The $\ce{HClO2}$ solution is diluted (I do not know the exact number, but maybe similar to mild bleach), sprayed to substances' surface, at room or body temperature.
Can the balances in this case lead to $\ce{HClO4}?$
On the other hand, $\ce{HCl}$ recombines to make $\ce{Cl2}$:
> $$
\begin{align}
\ce{HClO2 + 3 HCl &→ 2 Cl2 + 2H2O}\\
\ce{HOCl + HCl &→ Cl2 + H2O}
\end{align}
$$
So, can the remaining $\ce{HCl}$ be concentrated enough to corrode metal or cause harms to skin?
From a consumer's point of view, $\ce{HClO2}$ seems to be a complete disinfectants for all of water, surface (by producing $\ce{HOCl}$), and air (by producing $\ce{ClO2}$). However, the issue left for concerns is the byproduct acids.
### References
1. Garg, R.; Singh, R. *Inorganic Chemistry*; McGraw-Hill Education: New Delhi, **2015**. ISBN 978-1-259-06285-8. |
Which one is the stronger nucleophile in aprotic solvent-
(i)***H2O or H2S***
According to my notes, Nucleophilicity order is directly proportional to Basicity order in an aprotic medium,and inversely proportional to Basicity order in protic medium.
And H2O is a stronger base than H2S,so it should be the strong nucleophile but the answer is given as H2S is the stronger nucleophile. What is the reason for that? Am I missing something? |
Which one would be the stronger nucleophile in aprotic solvent? |
Which one is the stronger nucleophile in aprotic solvent: $\ce{H2O}$ or $\ce{H2S}?$
According to my notes, nucleophilicity order is directly proportional to basicity order in an aprotic medium, and is inversely proportional to basicity order in protic medium.
And $\ce{H2O}$ is a stronger base than $\ce{H2S},$ so it should be the strong nucleophile, but the answer is given as $\ce{H2S}$ is the stronger nucleophile. What is the reason for that? Am I missing something? |
Which is a stronger nucleophile in aprotic solvent: water or hydrogen sulfide? |
My textbook says that **unsaturated hydrocarbons are more reactive than saturated ones**. But double and triple bonds are stronger than single bonds.
What is the explanation for this?
|
$K_p$ will *not* change with pressure, only with temperature. $Q_p$ will instantaneously increase upon applying the pressure for this reaction (why? see below), and will drive the reaction to reactants to re-achieve equilibrium (decreasing $Q_p$) until $Q_p$ = $K_p$ once again. From a LeChatelier approach, you have more gas molecules in the products, so an increase in pressure will drive the equilibrium composition to the reactants.
As for why $K_p$ won't change with pressure:
$\ln K_p=-\frac{\Delta G^⦵}{RT}$ is at a specified p^⦵ (usually 1 bar), so $\left(\frac{\partial {\ln}K_p}{\partial P}\right)_T = 0 $.
And as for why $K_c$ won't change with pressure, see the [answer here][1]
[1]: https://chemistry.stackexchange.com/questions/49225/why-is-kc-not-affected-by-change-in-pressure |
Assuming that by air bubble you're referring to the pocket of air now sealed in the bottle above the water surface:
Immediately after sealing the bottle, the air pocket is not at all compressed: it's at the same pressure as the surrounding environment (the air outside the bottle).
However, because you have a closed container, some water will evaporate, until an equilibrium is reached between the water vapor in the air pocket and the water in the liquid phase. This occurs when the chemical potential of the water in the two phases are equal.
Due to that vaporization, the total pressure in the air pocket (which now contains a higher amount of water vapor than the surrounding atmosphere outside the bottle) will be larger than the surroundings. This is a prominent effect when it's hot outside (have you ever left a sealed water bottle in a hot car?).
While this effect is usually small, as gases mix very well, it is non-zero and so *yes, the air pocket is slightly compressed due to presence of water vapor*.
As for whether the air pocket can mix with water: Assuming our container can tolerate such pressures and we were to externally compress it: liquids are much less compressible than gases, and so the air would (given high enough pressure) condensate, mixing into the liquid phase (but you wouldn't achieve this by hand!)
To address your final point: let's say we invert the bottle, so the air pocket travels as a bubble through the liquid phase to re-emerge at the other end of the bottle. One could ask, "why doesn't the air bubble fragment into little bubbles as it progresses upwards?"
To be fair, you do see a few bubbles splinter off, but for the most part, the bubble is intact. The air bubble stays largely intact due to the viscosity of water (the intermolecular attractions between water molecules is larger than the entropic benefits of mixing with the air bubble). So, to answer your question, the fact that the air bubble doesn't mix is not due to its intermolecular attraction, *but rather it's due to water's intermolecular attraction.* |
This is not an answer to the question "can the sulfoxide be synthesized in significant yield" (and under which conditions). Instead, this is a "naive" prediction based on the data you provide and two assumptions:
1. The predicted thermodynamic values in the articles you cite are accurate at $\pu{298 K}.$
2. The heat and entropy of formation of (**2**) and (**3**) from (**1**) are constant over $T.$
From the references we can build a data table of energies and entropies relative to "the ground state" $\ce{HSOH}$ assuming the $\Delta G^\circ$ values correspond to $T = \pu{298 K}:$
$$\begin{array}{lccc}
\hline
\text{Compound} & \Delta G/\pu{kcal mol-1} & \Delta H/\pu{kcal mol-1}^* & \Delta S/\pu{cal mol-1 K-1} & K_\mathrm{eq}\\
\hline
\ce{H2SO} & 25 & 15.8 & -31 & 10^{-19} \\
\ce{H2OS} & 34 & 38.2 & 14 & 10^{-25} \\
\hline
\end{array}$$
I include the order of magnitude of the equilibrium constants $K_\mathrm{eq}$ only to illustrate their ludicrously small values (and these are upper estimates!).
Could you find a temperature that would make $\Delta G^\circ<0$ for the reaction to $\ce{H2SO}$? The answer is no. To see why we turn to the enthalpies. Since the reaction is endothermic you might hope that an increase in $T$ would help, but the accompanying entropy change is negative, so there is no way you can sufficiently increase the temperature to drive the reaction forward. The highest equilibrium constant you could hope for is $K_\mathrm{eq} = \pu{1.7E-7}$.
Things are not so bleak for $\ce{H2OS}$: the reaction is also endothermic but the entropy change is positive. At about $\pu{2800 K}$ $K_\mathrm{eq} = 1,$ and if we increase $T$ sufficiently, we can crank up $K_\mathrm{eq}$ to any arbitrarily large value.
This is of course not remotely exhaustive (therefore "naive") but hopefully a start to answering your question.
-----
\* Quoting your second reference:
> At the CCSD(T)/CBS limit and including corrections for scalar relativistic, spin orbit and core-valence correlation effects, the estimated enthalpies of formation are $\pu{−28.1 ± 1},$ $\pu{−12.3 ± 1},$ and $\pu{10.1 ± 1 kcal/mol}$ for $\ce{HSOH},$ $\ce{H2SO}$ and $\ce{H2OS},$ respectively.
|
It's a tricky example, but the $S_N1$ process is favored because the solvent is protic and the resulting 6-member ring is more stable than a 5-member ring.
If ring expansion weren't possible, I believe your intuition is correct that the $S_N2$ mechanism would dominate. See [this discussion][1]. You're not alone!
[1]: https://chemistry.stackexchange.com/questions/112785/why-primary-alcohol-follows-sn1-mechanism?rq=1 |
I just thought that the reaction should follow $\mathrm{S_{N}2}$ mechanism this is because the alkyl halide is primary but the book says that the reaction will follow $\mathrm{S_{N}1}$ mechanism resulting in the formation of tertiary carbonation and a ring expansion product. Can anyone please explain the reason behind this ? |
Weakly basic ions and polarizable bases favor $\mathrm{S_{N}2}$ over $\mathrm{E2}$. Why? The argument used that polarizable nucleophiles form bonds earlier, stabilizing the transition state, can be applied to either $\mathrm{S_{N}2}$ or $\mathrm{E2}$. |
In general, we've learned that alkoxide groups are awful leaving groups, unless we're talking about epoxide rings whose ring strain allow the alkoxide to act as a leaving group in a ring-opening reaction. However, if a structure contained both an epoxide and a relatively good leaving group like Chloride, what would predominate if a strong base/nucleophile was introduced: Epoxide opening or the substitution reaction? An example is provided below of what I'm referring to:
[![enter image description here][1]][1]
If I perform, the epoxide ring opening, I get something unusual:
[![][2]][2]
I left the stereochemistry ambigious because I honestly did not know how to assign it since a new stereocenter is formed.
If I perform the Sn2 substitution, the reaction is completed quickly:
[![enter image description here][3]][3]
Notice that both of my proposed mechanism reform the epoxide ring, but do generate entirely different structures.
[1]: https://i.stack.imgur.com/hE7wW.png
[2]: https://i.stack.imgur.com/uon2p.png
[3]: https://i.stack.imgur.com/OzlEB.png |
Is an epoxide opening more favorable than a chloride leaving in the same molecule? |
It is my understanding that the main decomposition product of HClO2 is ClO2. Commercial ClO2 generation systems typically incorporate a membrane used to separate the gas from the other components, such as the solids and undesirable side products. This helps create a highly pure solution of ClO2 in water.
Here are some examples:
https://www.dioxide.com/systems/chlorine-dioxide/
https://selectivemicro.com/about-us/#smtdiff
From a consumer's point of view, as you state, I would look into how the disinfectant solution is generated. Typically, you can react sodium chlorite (NaClO2) with an acid such as HCl or citric acid to generate HClO2 and therefore ClO2. Does the product use membrane technology to separate the gas? You mentioned HCl. Based on the reactions described, don't think concentrated HCl is a concern. With Na ions present, you would more likely form NaCl.
Also "...water systems using chlorine dioxide for disinfection or oxidation must monitor their system for chlorine dioxide and chlorite." (https://www.evoqua.com/en/brands/municipal-services/Product%20Information%20Library/MSPOTABLEAP.pdf) |
I want to create several mixtures of oxygen and nitrogen gas, such as 0% oxygen, 10% oxygen, 20% oxygen, 30% oxygen... 100% oxygen, but I know very little about working with gas cylinders and regulators. Is it as simple as connecting the two cylinders with a y-shaped hose and setting the flow rates to 10% oxygen and 90% nitrogen? Or will I need a more complicated setup?
This is for dissolving the gas mixture in water so I can test methods for determining dissolved oxygen levels. I'm not filling SCUBA tanks or anything dangerous, but I would still like the mixtures to be fairly accurate so I can get good data. |
Guidelines for mixing gases? |
It is my understanding that the main decomposition product of HClO2 is ClO2. Commercial ClO2 generation systems typically incorporate a membrane used to separate the gas from the other components, such as the solids and undesirable side products. This helps create a highly pure solution of ClO2 in water.
Here are some examples:
https://www.dioxide.com/systems/chlorine-dioxide/
https://selectivemicro.com/about-us/#smtdiff
From a consumer's point of view, as you state, I would look into how the disinfectant solution is generated. Typically, you can react sodium chlorite (NaClO2) with an acid such as HCl or citric acid to generate HClO2 and therefore ClO2. Does the product use membrane technology to separate the gas? You mentioned HCl. Based on the reactions described, don't think concentrated HCl is a concern (unless HCl is the acid used in the reaction). With Na ions present, you would more likely form NaCl.
Also "...water systems using chlorine dioxide for disinfection or oxidation must monitor their system for chlorine dioxide and chlorite." (https://www.evoqua.com/en/brands/municipal-services/Product%20Information%20Library/MSPOTABLEAP.pdf) |
Suppose we have a non-dipolar molecule such as carbon dioxide. If carbon dioxide is ionized by some means in the gas-phase, will it develop a dipole moment? I am asking with reference to microwave spectroscopy, because the selection rule requires the presence of a permanent dipole moment. In the interstellar space, $HCO_2^+$ has been detected and it does have a microwave spectrum, because it does have a dipole moment. However, carbon dioxide's dipole moment is zero, so its ordinary microwave spectrum does not exist. There are very weak lines due to its quadrupole. |
I have noticed that if you put household or other chemicals in bin liners or freezer bags the plastic doesn't keep the chemicals in. Rather they travel through the bag material and out into the air. However I have noticed that some bin liners have stronger material and are better than others.
Perhaps it will depend on which chemical is in the bag. However I guess I am asking about pollen, pesticides,washing detergent and general cleaning chemicals.
Is there a specific type/grade of plastic bin liners that I can get that will, generally speaking. keep odors in? Actually it doesn't have to be pvc any bag material which does the job will do however I guess PVC bags are most easily accessible so would like to know what options there are. See through bags are preferable.
|
Best bags to contain chemical odours? |
In my QM class we are finding the wave functions of the hydrogen atom. In spherical coordinates, these wave functions are functions of 3 variables: r, theta, and phi. My professor stated that, "the wave functions of the electron are called orbitals".
Say we are interested in the l=0 (or s-orbitals). Since l = 0, m = 0.
[![enter image description here][1]][1]
So we can set a specific n value. Say n = 1, and we get the the corresponding 1s orbital. Since we set an n value, the wave function reduces to a function of "r", the radius. During lecture, he said that this image:
[![enter image description here][2]][2]
is an isosurface (or level surface) of the wave function at a constant radius.
I interpreted that as finding the isosurface of a 4D function, our wave function, at some radius. This isosurface is the the sphere in 3D. What confused me was when he stated that the graph was an "isosurface of the full orbital".
So my question is:
**What does he mean by the "full orbital"?** (Is this 'full orbital' a 4D surface?)
[1]: https://i.stack.imgur.com/ChFBU.png
[2]: https://i.stack.imgur.com/nbTul.png |
What exactly are the graphs of orbitals? |
I have been pondering the assignment for the 1H NMR peaks observed. I can't really make sense of the observations from what I know.
[![enter image description here][1]][1]https://www.chemicalbook.com/SpectrumEN_104-93-8_1HNMR.htm
[1]: https://i.stack.imgur.com/jNFA7.png
My question is the assignment of the signal between 6 and 7.5ppm. Unlike the way it is currently assigned, I would imagine Ha to be the most upshifted peak (near 7ppm) because methoxy group is slightly electron withdrawing and therefore de-shield the protons of Ha more than Hb (Hb is near methyl group which is slightly electron donating).
Any idea of how you would interpret the peak assignments?
Data retrieved from here: https://www.chemicalbook.com/SpectrumEN_104-93-8_1HNMR.htm |
In my QM class we are finding the wave functions of the hydrogen atom. In spherical coordinates, these wave functions are functions of 3 variables: r, theta, and phi. My professor stated that, "the wave functions of the electron are called orbitals".
Say we are interested in the l=0 (or s-orbitals). Since l = 0, m = 0.
[![enter image description here][1]][1]
So we can set a specific n value. Say n = 1, and we get the the corresponding 1s orbital. Since we set an n value, the wave function reduces to a function of "r", the radius. During lecture, he said that this image:
[![enter image description here][2]][2]
is an isosurface (or level surface) of the wave function at a constant radius.
I interpreted that as finding the isosurface of a 4D graph, our wave function, at some radius. This isosurface is the the sphere in 3D. What confused me was when he stated that the graph was an "isosurface of the full orbital".
So my question is:
**What does he mean by the "full orbital"?** (Is this 'full orbital' a 4D surface?)
[1]: https://i.stack.imgur.com/ChFBU.png
[2]: https://i.stack.imgur.com/nbTul.png |
See [Highly efficient decomposition of ammonia using high-entropy alloy catalysts](https://www.nature.com/articles/s41467-019-11848-9.pdf?draft=collection) which proposes highly efficient ammonia decomposition using a novel high-entropy alloy (HEA) catalysts made of earth abundant elements, namely, quinary CoMoFeNiCu nanoparticles.
However, the reported temperature is high (going up to 2000–2300 K).
So, to answer your question, not at the temperature you indicated (250 degrees Celsius). |
See [Highly efficient decomposition of ammonia using high-entropy alloy catalysts](https://www.nature.com/articles/s41467-019-11848-9.pdf?draft=collection) which proposes highly efficient ammonia decomposition using a novel high-entropy alloy (HEA) catalysts made of earth abundant elements, namely, quinary CoMoFeNiCu nanoparticles.
However, the reported temperature is high (going up to 2000–2300 K).
[Another source](https://link.springer.com/article/10.1007/s11244-016-0653-4) notes:
>To date, the most effective catalyst for ammonia decomposition consists of ruthenium particles supported on carbon nanotubes (CNT) due to their high conductivity (6353 molH2 mol−1Ruh−1 at 430 °C) [2, 17]. The low temperature activity can be further improved by the addition of an electron donating promoter such as cesium (7870 molH2 mol−1Ruh−1 at 370 °C) [3, 17, 19, 20].
So, to answer your question, not at the temperature you indicated (250 degrees Celsius). |
As you state, the ubiquitous polyethylene (or poythene) bag allows many chemicals to slowly seep through. Pollen and dust themselves are far to large to get through, but water vapor, for example does.
For that reason, many foods a sold in plastic (e.g., mylar and polypropylene) sacks layered with aluminum (ah, aluminium for you polythene-sayers) to block evaporation, oxygen (which leads to fats and oils becoming rancid) and loss of flavor.
For storing small amounts of food, reuse these aluminized bags -- which also are effective shields to block RF tags in passports and licenses. They're available commercially, e.g. [this list][1].
These bags are comparatively expensive, though, and last millenia in landfills, and impractical for general use waste bin bags.
Non-chemical work-around: store food wastes and other potentially odorous items in paper or plastic bags that would have been discarded anyway, and that bag inside the bin liner, to delay odor migration.
[1]: https://www.amazon.com/s?k=large%20aluminized%20plastic%20bags&dc |
This question has a few different parts to it.
Part 1 - what's that peak at 3.6ppm?
Well, it's not the -OH peak. The peak at 3.6 is from the CH(OH).
Part 2 - where is the -OH peak if that isn't it at 3.6ppm?
Peaks from labile protons are frequently not observed, and have variable chemical shifts, unless great care is taken in sample preparation. This sample was run in CDCl3, and the -OH peak exchanges with H2O, and comes at ~1.6ppm.
Part 3 - can we account for the splitting of the peak at 3.6ppm now we know what it is?
[![enter image description here][1]][1]
The CH(OH) proton should couple to proton on the adjacent carbon (split to a doublet) and also the protons of the methyl group (split to a quartet). It won't couple to the -OH for the same reason we don't readily observe the -OH; it is rapidly exchanging with water. So, we expect that the peak at 3.6ppm should be a doublet of quartets. Looks more complicated than that to me.
Part 4 - It looks more complicated that a doublet of quartets to me. What's happening there?
Of course, 3-methylpentan-2-ol has 2 stereocentres, and so we have here a mixture of diastereomers in solution in the sample. Both diastereotopic CH(OH) protons have very similar chemical shifts, and so there are in fact 2 sets of doublets or quartets. It's also why the rest of the spectrum looks such a mess.
[1]: https://i.stack.imgur.com/oez4Q.png |
As you state, the ubiquitous polyethylene (or polythene) bag allows many chemicals to slowly seep through. Pollen and dust particles themselves are far to large to get through, but water vapor, for example does.
For that reason, many foods are sold in plastic (e.g., mylar and polypropylene) sacks layered with aluminium (ah, aluminium for you polythene-sayers) to block evaporation, oxygen (which leads to fats and oils becoming rancid) and loss of flavor.
For storing small amounts of food, reuse these aluminized bags -- which also are effective shields to block RF tags in passports and licenses. They're available commercially, e.g. [this list][1].
These bags are comparatively expensive, though, and last millenia in landfills, and impractical for general use waste bin bags.
Non-chemical work-around: store food wastes and other potentially odorous items in paper or plastic bags that would have been discarded anyway, and that bag inside the bin liner, to delay odor migration.
[1]: https://www.amazon.com/s?k=large%20aluminized%20plastic%20bags&dc |
In my QM class we are finding the wave functions of the hydrogen atom. In spherical coordinates, these wave functions are functions of 3 variables: $r,$ $\theta,$ and $\phi.$ My professor stated that "*the wave functions of the electron are called orbitals*".
Say, we are interested in the $l = 0$ (or s-orbitals). Since $l = 0,$ $m = 0.$
$$\psi_{n00}(r,\theta,\phi) = R_{n0}(r)Y_{00}(\theta,\phi) = \left(\frac{1}{4\pi}\right)^{1/2}R_{n0}(r)$$
So, we can set a specific $n$ value. Say, $n = 1,$ and we get the the corresponding 1s orbital. Since we set an $n$ value, the wave function reduces to a function of the radius $r.$ During the lecture, he said that this image:
[![isosurface of the wave function at a constant radius][1]][1]
is an isosurface (or level surface) of the wave function at a constant radius.
I interpreted that as finding the isosurface of a 4D graph, our wave function, at some radius. This isosurface is the the sphere in 3D. What confused me was when he stated that the graph was an "*isosurface of the full orbital*".
What does he mean by the "*full orbital*"? Is this "*full orbital*" a 4D surface?
[1]: https://i.stack.imgur.com/nbTul.png |
Lately I've seen videos and some articles explaining drug design. I would also like to specifically cite [this guy][1], who isn't the most credible guy in the planet, but some of the stuff in the video makes sense to some degree, and he just confused me further.
Now I work on a completely different field on my day job, but I would like to ask a few questions, out of curiosity, because the techniques described in the video really piqued my interest.
1. Is it the case that when designing drugs (i.e. these proteins) we attempt to optimise binding affinity (maximise?)
2. How well does Autodock Vina compute this affinity? Is it opinionated? I've seen a few docking software, it looks like there are different methods to compute *docking score*? (can you recommend further reading for me in this regard)
3. Is there a linear correlation between binding affinity and drug efficacy? I.e. does it mean that higher binding affinity directly means we have a better drug, or is there some sweet point kinda thing?
4. Why can't we use very complex molecules, more than 300 characters in SMILES length to combat viruses for example? I noticed everyone was trying to minimise ligands?
5. How can software like AutoDock tools prepare your proteins for docking, is there a standard method (they add hydrogens, add Gasteiger charges). Is that guaranteed to be accurate in every case?
To clarify, I'm not looking for detailed answers, since I am completely off the domain, I just want to have a basic idea for these questions, and perhaps recommendations on further reading from you guys.
Thank you, and stay strong!
[1]: https://www.youtube.com/watch?v=EVoZMRmtBkY |
Covalent bond is a strong bond compared to Ionic Bonds ***but Ionic Compounds have higher melting and boiling points*** then covalent compounds. **Why?** |
I looked up the decompositions of diluted $\ce{HClO2}$ aqueous solution (used in a disinfectant) to know if it is safe to use on some substances, such as metal, wood, leather or skin. However, the chain I found leading to $\ce{HClO4},$ which is a very strong acid.
> $$
\begin{align}
\ce{5 HClO2 &→ 4 ClO2 + HCl + 2 H2O}\\
\ce{4 HClO2 &→ 2 ClO2 + HClO3 + HCl + H2O}\\
\ce{3 HClO2 &→ 2 HClO3 + HCl}\\
\ce{2 HClO2 &→ 2 HOCl + HClO3}\\
\ce{HClO2 &→ HCl + O2}
\end{align}
$$
Then there is
> $$\ce{3HClO3 → 2ClO2 + HClO4 + H2O}$$
(Source: *Inorganic Chemistry* by Garg and Singh \[1\], [Google book preview](https://books.google.co.jp/books?id=XKxqCgAAQBAJ&pg=PT978&lpg=PT978&dq=4HClO2+%E2%86%92+2ClO2+%2B+HClO3+%2B+HCl+%2B+H2O&source=bl&ots=Gj8goScjH9&sig=ACfU3U1k1j-XIi2T6aJUnTGE3avo-dw6Mw&hl=en&sa=X&ved=2ahUKEwiPluib5JToAhVnxosBHXfODEoQ6AEwAHoECAoQAQ#v=onepage&q=4HClO2%20%E2%86%92%202ClO2%20%2B%20HClO3%20%2B%20HCl%20%2B%20H2O&f=false))
The $\ce{HClO2}$ solution is diluted (I do not know the exact number, but maybe similar to mild bleach), sprayed to substances' surface, at room or body temperature.
Can the balances in this case lead to $\ce{HClO4}?$
On the other hand, $\ce{HCl}$ recombines to make $\ce{Cl2}$:
> $$
\begin{align}
\ce{HClO2 + 3 HCl &→ 2 Cl2 + 2H2O}\\
\ce{HOCl + HCl &→ Cl2 + H2O}
\end{align}
$$
So, can the remaining $\ce{HCl}$ be concentrated enough to corrode metal or cause harms to skin?
From a consumer's point of view, $\ce{HClO2}$ seems to be a complete disinfectants for all of water, surface (by producing $\ce{HOCl}$), and air (by producing $\ce{ClO2}$). However, the issue left for concerns is the byproduct acids.
**Updates:** I found a list of patents about producing $\ce{HClO2}$ solutions to be used as disinfectants, which are likely the one I am talking about in this question. The patents claimed that the disinfectants are safe to human, can be used to clean food-processing utilities, and even as food additives. The exact reasons for the claims are not clear to me.
- https://patents.google.com/patent/US9516878B2/en
- https://patents.google.com/patent/AU2013205834B2/en
- https://patents.google.com/patent/EP2999490A2/en
- https://patents.google.com/patent/US20160106106A1/en
- https://patents.google.com/patent/US20160113282A1/en
### References
1. Garg, R.; Singh, R. *Inorganic Chemistry*; McGraw-Hill Education: New Delhi, **2015**. ISBN 978-1-259-06285-8. |
I mainly work with simulation (MD), so take these answers with a grain of salt.
> 1. Is it the case that when designing drugs (i.e. these proteins) we attempt to optimize binding affinity (maximize?)
Computational docking usually employes two methods: A force field based estimation of the free energy of binding, and exploration of the conformational space for the ligand.
After a docking run you will end up with set of different ligands, a set of conformations (or *poses*) for each ligand, and a corresponding score value for each conformation. The score is usually based on the binding constant and Gibbs free energy (which is based on various numbers of physicochemical parameters, and may or may not include parameters like desolvation and entropic effects). Different software use different scoring functions and methods to estimate the free energy.
> 2. How well does Autodock Vina compute this affinity? Is it opinionated? I've seen a few docking software, it looks like there are different methods to compute docking score? (can you recommend further reading for me in this regard).
As far as I can tell, it is hard or impossible to make general statements about which piece of software is the best, since it depends a lot on what you are trying to use it for. Some of the limitations of the AutoDock suite (implicit hydrogens, spherically symmetric H-bond potentials etc.) are given in [this][1] publication.
> 3. Is there a linear correlation between binding affinity and drug efficacy? I.e. does it mean that higher binding affinity directly means we have a better drug, or is there some sweet point kinda thing?
Not necessarily. A docking run gives you an (rough) estimation of the ligand affinity for the receptor you chose. It might be that the ligand also binds strongly to a lot of other proteins, making it less effective in an *in vivo* setting.
> 4. Why can't we use very complex molecules, more than 300 characters in SMILES length to combat viruses for example? I noticed everyone was trying to minimise ligands?
You can, but not necessarily with the AutoDock suite. Large compounds have too many degrees of freedom, and the software is not capable of properly exploring the conformational space. Usually these large compounds are split in to smaller fragments which are then docked individually.
> 5. How can software like AutoDock tools prepare your proteins for docking, is there a standard method (they add hydrogens, add Gasteiger charges). Is that guaranteed to be accurate in every case?
Again, depends on the software. You listed some of the features of AutoDock, and you can find the details of the receptor preparation scheme from their [website][2]. It is definitely not accurate in every case. As far as I know, most docking runs are performed on protein structures obtained with X-ray methods, and there is no guarantee that the conformation of the protein in crystal is a good representation of the conformation in the solvent phase.
[1]: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4868550/
[2]: http://autodock.scripps.edu/faqs-help/how-to/how-to-prepare-a-receptor-file-for-autodock4 |
Per [Wikipedia on Chlorous acid](https://en.wikipedia.org/wiki/Chlorous_acid) to quote:
>The pure substance is unstable, disproportionating to hypochlorous acid (Cl oxidation state +1) and chloric acid (Cl oxidation state +5):
>$\ce{2 HClO2 → HClO + HClO3}$
Also, here are some interesting comments on the decomposition of chlorous acid in "[Kinetics and Mechanism of the Decomposition of Chlorous Acid](http://hopf.chem.brandeis.edu/pubs/pub293%20rep.pdf)" from J. Phys. Chem. A 2003, 107, pages 6966-6973, to quote:
>Of the many mechanistic models tested, the one that fit best included the following reactive intermediates: HOCl, Cl2O2, Cl2O3, •ClO, •OH. The stoichiometric ratio of ClO2 produced to Cl(III) consumed varies with pH and [Cl-]. Reaction of Cl2O3 with Cl(III) yields chlorate exclusively. Reaction of Cl2O3 with Cl- favors ClO2 over chlorate, but does not entirely exclude chlorate, because it is produced by hydrolysis of Cl2O2. Invoking Cl2O3 explains the variation in stoichiometric ratio as well as the maximum observed in the initial rate of ClO2 formation as a function of pH. The kinetics of chlorous acid decomposition cannot be quantitatively fit through the last stages of the reaction without postulating a first-order decomposition. Scission of chlorous acid to give short-lived hydroxyl and chlorine-(II) monoxide is a plausible route for this process [...]
>Several groups of investigators[5-7] have found
that in the absence of chloride ion the stoichiometry of the
decomposition of chlorous acid is given by reaction A:
>$\ce{4 HClO2 -> 2 ClO2 + ClO3- + Cl- + 2 H+ + H2O}$ (A)
>The stoichiometry of the decomposition of chlorous acid in the
presence of chloride ion is given by reaction B:
>$\ce{5 HClO2 -> 4 ClO2 + Cl- + H+ + 2 H2O}$ (B)
Also, to quote:
>Earlier studies,[9,18] in agreement with our present results, have also found the formation of more chlorate than predicted from reaction A. Reaction C
>$\ce{3 HClO2 -> 2 ClO3- + Cl- + 3 H+}$ (C)
>also plays a role in determining the stoichiometry at higher $\ce{HClO2}$ concentrations."
So, several intermediate products and depending on chloride presence possibly $\ce{ClO2}$, which is a problematic explosive and toxic gas, along with the strength of the $\ce{HClO2}$, which can introduce chlorate at higher chlorous acid concentrations.
Not a particularly good path to an acid as you are also inquiring about. As a disinfectant, the transient creation of the powerful disinfecting HOCl and associated radicals, may actually make it a weaker, albeit more stable (considering end products) than hypochlorous acid, in my opinion.
[EDIT] To answer a comment question, as to whether is ClO2 safe, here is a statement from the CDC, [Public Health Statement for Chlorine Dioxide and Chlorite](https://www.atsdr.cdc.gov/phs/phs.asp?id=580&tid=108), to quote:
>If you are exposed to chlorine dioxide or chlorite, many factors will determine whether you will be harmed. These factors include the dose (how much), the duration (how long), and how you come in contact with them. You must also consider any other chemicals you are exposed to and your age, sex, diet, family traits, lifestyle, and state of health.
Interestingly, ClO2 has found commercial application as an [odor removal agent](https://closureclean.com/chlorine-dioxide-odor-removal/), likely due to its reaction with organics to create volatile organic chlorides. These VOCs are much more of a long-term health danger (as in carcogenic), in my opinion. |
Per [Wikipedia on Chlorous acid](https://en.wikipedia.org/wiki/Chlorous_acid) to quote:
>The pure substance is unstable, disproportionating to hypochlorous acid (Cl oxidation state +1) and chloric acid (Cl oxidation state +5):
>$\ce{2 HClO2 → HClO + HClO3}$
Also, here are some interesting comments on the decomposition of chlorous acid in "[Kinetics and Mechanism of the Decomposition of Chlorous Acid](http://hopf.chem.brandeis.edu/pubs/pub293%20rep.pdf)" from J. Phys. Chem. A 2003, 107, pages 6966-6973, to quote:
>Of the many mechanistic models tested, the one that fit best included the following reactive intermediates: HOCl, Cl2O2, Cl2O3, •ClO, •OH. The stoichiometric ratio of ClO2 produced to Cl(III) consumed varies with pH and [Cl-]. Reaction of Cl2O3 with Cl(III) yields chlorate exclusively. Reaction of Cl2O3 with Cl- favors ClO2 over chlorate, but does not entirely exclude chlorate, because it is produced by hydrolysis of Cl2O2. Invoking Cl2O3 explains the variation in stoichiometric ratio as well as the maximum observed in the initial rate of ClO2 formation as a function of pH. The kinetics of chlorous acid decomposition cannot be quantitatively fit through the last stages of the reaction without postulating a first-order decomposition. Scission of chlorous acid to give short-lived hydroxyl and chlorine-(II) monoxide is a plausible route for this process [...]
>Several groups of investigators[5-7] have found
that in the absence of chloride ion the stoichiometry of the
decomposition of chlorous acid is given by reaction A:
>$\ce{4 HClO2 -> 2 ClO2 + ClO3- + Cl- + 2 H+ + H2O}$ (A)
>The stoichiometry of the decomposition of chlorous acid in the
presence of chloride ion is given by reaction B:
>$\ce{5 HClO2 -> 4 ClO2 + Cl- + H+ + 2 H2O}$ (B)
Also, to quote:
>Earlier studies,[9,18] in agreement with our present results, have also found the formation of more chlorate than predicted from reaction A. Reaction C
>$\ce{3 HClO2 -> 2 ClO3- + Cl- + 3 H+}$ (C)
>also plays a role in determining the stoichiometry at higher $\ce{HClO2}$ concentrations."
So, several intermediate products and depending on chloride presence possibly $\ce{ClO2}$, which is a problematic explosive and toxic gas, along with the strength of the $\ce{HClO2}$, which can introduce chlorate at higher chlorous acid concentrations.
Not a particularly good path to an acid as you are also inquiring about. As a disinfectant, the transient creation of the powerful disinfecting HOCl and associated radicals, may actually make it a weaker, albeit more stable (considering end products) than hypochlorous acid, in my opinion.
[EDIT] To answer a comment question, as to whether is ClO2 safe, here is a statement from the CDC, [Public Health Statement for Chlorine Dioxide and Chlorite](https://www.atsdr.cdc.gov/phs/phs.asp?id=580&tid=108), to quote:
>If you are exposed to chlorine dioxide or chlorite, many factors will determine whether you will be harmed. These factors include the dose (how much), the duration (how long), and how you come in contact with them. You must also consider any other chemicals you are exposed to and your age, sex, diet, family traits, lifestyle, and state of health.
Interestingly, ClO2, which is a stable free radical, has found commercial application as an [odor removal agent](https://closureclean.com/chlorine-dioxide-odor-removal/), likely due to its interaction with organics to create volatile organic chlorides (I would suspect that the presence of light would be catalytic). These VOCs are much more of a long-term health danger (as in carcinogenic), in my opinion. |
Per [Wikipedia on Chlorous acid](https://en.wikipedia.org/wiki/Chlorous_acid) to quote:
>The pure substance is unstable, disproportionating to hypochlorous acid (Cl oxidation state +1) and chloric acid (Cl oxidation state +5):
>$\ce{2 HClO2 → HClO + HClO3}$
Also, here are some interesting comments on the decomposition of chlorous acid in "[Kinetics and Mechanism of the Decomposition of Chlorous Acid](http://hopf.chem.brandeis.edu/pubs/pub293%20rep.pdf)" from J. Phys. Chem. A 2003, 107, pages 6966-6973, to quote:
>Of the many mechanistic models tested, the one that fit best included the following reactive intermediates: HOCl, Cl2O2, Cl2O3, •ClO, •OH. The stoichiometric ratio of ClO2 produced to Cl(III) consumed varies with pH and [Cl-]. Reaction of Cl2O3 with Cl(III) yields chlorate exclusively. Reaction of Cl2O3 with Cl- favors ClO2 over chlorate, but does not entirely exclude chlorate, because it is produced by hydrolysis of Cl2O2. Invoking Cl2O3 explains the variation in stoichiometric ratio as well as the maximum observed in the initial rate of ClO2 formation as a function of pH. The kinetics of chlorous acid decomposition cannot be quantitatively fit through the last stages of the reaction without postulating a first-order decomposition. Scission of chlorous acid to give short-lived hydroxyl and chlorine-(II) monoxide is a plausible route for this process [...]
>Several groups of investigators[5-7] have found
that in the absence of chloride ion the stoichiometry of the
decomposition of chlorous acid is given by reaction A:
>$\ce{4 HClO2 -> 2 ClO2 + ClO3- + Cl- + 2 H+ + H2O}$ (A)
>The stoichiometry of the decomposition of chlorous acid in the
presence of chloride ion is given by reaction B:
>$\ce{5 HClO2 -> 4 ClO2 + Cl- + H+ + 2 H2O}$ (B)
Also, to quote:
>Earlier studies,[9,18] in agreement with our present results, have also found the formation of more chlorate than predicted from reaction A. Reaction C
>$\ce{3 HClO2 -> 2 ClO3- + Cl- + 3 H+}$ (C)
>also plays a role in determining the stoichiometry at higher $\ce{HClO2}$ concentrations."
So, several intermediate products and depending on chloride presence possibly $\ce{ClO2}$, which is a problematic explosive and toxic gas, along with the strength of the $\ce{HClO2}$, which can introduce chlorate at higher chlorous acid concentrations.
Not a particularly good path to an acid as you are also inquiring about. As a disinfectant, the transient creation of the powerful disinfecting HOCl and associated radicals, may actually make it weaker, albeit more stable (considering end products) than hypochlorous acid, in my opinion.
[EDIT] To answer a comment question, as to whether is ClO2 safe, here is a statement from the CDC, [Public Health Statement for Chlorine Dioxide and Chlorite](https://www.atsdr.cdc.gov/phs/phs.asp?id=580&tid=108), to quote:
>If you are exposed to chlorine dioxide or chlorite, many factors will determine whether you will be harmed. These factors include the dose (how much), the duration (how long), and how you come in contact with them. You must also consider any other chemicals you are exposed to and your age, sex, diet, family traits, lifestyle, and state of health.
Interestingly, ClO2, which is a stable free radical, has found commercial application as an [odor removal agent](https://closureclean.com/chlorine-dioxide-odor-removal/), likely due to its interaction with organics to create volatile organic chlorides (I would suspect that the presence of light would be catalytic). These VOCs are much more of a long-term health danger (as in carcinogenic), in my opinion. |
So I have a question that is as follows:-
If all bond angles in AX$_3$ are the same then which of the following are correct conclusions about AX$_3$?
(A) AX$_3$ must be polar
(B) AX$_3$ must be planar
(C) AX$_3$ must have atleast 5 valence electrons
(D) X must be connected to the central atom via single or double bonds
***My Attempt at the question:-***
This is clearly a case of sp$^2$ hybridisation. So I thought about molecules like BF$_3$. So as the bond angle needs to be same, it must not be polar. sp$^2$ has trigonal planar structure so (B) must be correct.
But my book says that the answer is (D) only (this question was of single correct choice type).
Why are (B) and (C) not the answers and why is (D) the answer? |
If all bond angles in AX$_3$ are the same then which of the following are correct conclusions about AX$_3$? |
### Question
> If all bond angles in $\ce{AX3}$ are the same, then which of the following are correct conclusions about $\ce{AX3}?$
> (**A**) $\ce{AX3}$ must be polar.
> (**B**) $\ce{AX3}$ must be planar.
> (**C**) $\ce{AX3}$ must have at least 5 valence electrons.
> (**D**) $\ce{X}$ must be connected to the central atom via single or double bonds.
### My Attempt
This is clearly a case of $\mathrm{sp^2}$ hybridisation. So, I thought about molecules like $\ce{BF3}$. As the bond angle needs to be same, it must not be polar. $\mathrm{sp^2}$ has trigonal planar structure, so (**B**) must be correct.
But my book says that the answer is (**D**) only (this question was of single correct choice type).
Why are (**B**) and (**C**) not the answers and why is (**D**) the answer? |
If all bond angles in AX₃ are the same, then which of the following are correct conclusions about AX₃? |
I have a polymer end-capped with a Polysilsesquioxane (POSS) like the one [here](https://en.wikipedia.org/wiki/Silsesquioxane#/media/File:Silsesquioxane_T8_Cube.png) where *R=i-butyl*.
I haven't found much info on what type or reactions these POSS groups can undergo.
I would like to know if there are some procedures to modify that POSS moiety and convert it into an amino, carboxyl or triethoxysilane group, for example.
Also, are those -R groups easy to replace? Or that Si-(i-butyl) bonds are very stable?
|
As you state, the ubiquitous polyethylene (or polythene) bag allows many chemicals to slowly seep through. Pollen and dust particles themselves are far to large to get through, but water vapor, for example does.
For that reason, many foods are sold in plastic (e.g., mylar and polypropylene) sacks layered with aluminum (ah, aluminium for you polythene-sayers) to block evaporation, oxygen (which leads to fats and oils becoming rancid) and loss of flavor.
For storing small amounts of food, reuse these aluminized bags -- which also are effective shields to block RF tags in passports and licenses. They're available commercially, e.g. [this list][1].
These bags are comparatively expensive, though, and last millenia in landfills, and impractical for general use waste bin bags.
Non-chemical work-around: store food wastes and other potentially odorous items in paper or plastic bags that would have been discarded anyway, and that bag inside the bin liner, to delay odor migration.
[1]: https://www.amazon.com/s?k=large%20aluminized%20plastic%20bags&dc |
The thermal stability of most compounds of Group 1 elememts (hydroxides, carbonates, nitrates) increases down the group due to decrement in charge density of the cation.
Now, according to one of my study sources, thermal stability of oxides is as follows:
normal oxide(that of Lithium)>peroxide(that of Sodium)>superoxide(that of Potassium, Rubidium, Cesium).
However another source exactly says:
"The stability of peroxides and superoxides increases as the size of metal ion increases."
I can not understand whether these two statements contradict each other or not. If yes, then which statement is true and what is the actual trend of thermal stability of oxides down a group? |
What is the trend of thermal stability of group 1 oxides? |
Covalent bond is a strong bond compared to Ionic Bonds but Ionic Compounds have higher melting and boiling points then covalent compounds. **Why?** |
Log(Q) is zero, therefore the effect of temperature change is unaccounted for by Nernst Equation? |
I am using 99% isopropyl alchohol to clean soldering flux residues for my electronics projects. Due to a certain viruse running amok in the world today, I am not able to get from my usual supplier. I am running low on supplies. So i got this idea that i would concentrate the more common (but still not so easy to get in my area) 40% or 70% ispropyl alchohol.
Searching the internet on guides to do it, it would seem that there is an easy way to do it; add non-iodized salt to the solution until its past its saturation point. By this time the heavier water should sink and the lighter isopropyl should float. Extracting it would now be easy with a syringe.
For my question. Water and especially salt is an enemy to any metal. I would like to ask if this process mixes tiny amount of salt to the isopropyl? if so by how much %? Is NaCl even soluble to isopropyl or are these just tiny particles floating around and if i leave the solution to settle for a longer period of time those particles would sink down. up until what percent can i bring the concentration to?
As for the Salt to use what other ingredients should i look out other than iodine( or other close sounding). |
I am using 99% isopropyl alchohol to clean soldering flux residues for my electronics projects. Due to a certain viruse running amok in the world today, I am not able to get from my usual supplier. I am running low on supplies. So i got this idea that i would concentrate the more common (but still not so easy to get in my area) 40% or 70% ispropyl alchohol.
Searching the internet on guides to do it, it would seem that there is an easy way to do it; add non-iodized salt to the solution until its past its saturation point. By this time the heavier water should sink and the lighter isopropyl should float. Extracting it would now be easy with a syringe.
For my question. Water and especially salt is an enemy to any metal. I would like to ask if this process mixes tiny amount of salt to the isopropyl? if so by how much %? Is NaCl even soluble to isopropyl or are these just tiny particles floating around and if i leave the solution to settle for a longer period of time those particles would sink down. up until what percent can i bring the concentration to?
As for the Salt to use what other ingredients should i look out other than iodine( or other close sounding).
P.S. I am no chemist any way, the only chemistry background i have is during my high-school days. It was a time that i was still not able to appreciate class lessons. So go easy on me
|
I am using 99% isopropyl alchohol to clean soldering flux residues for my electronics projects. Due to a certain viruse running amok in the world today, I am not able to get from my usual supplier. I am running low on supplies. So i got this idea that i would concentrate the more common (but still not so easy to get in my area) 40% or 70% ispropyl alchohol.
Searching the internet on guides to do it, it would seem that there is an easy way to do it. According to this [instructables][1]; add non-iodized salt to the solution until its past its saturation point. By this time the heavier water should sink and the lighter isopropyl should float. Extracting it would now be easy with a syringe.
For my question. Water and especially salt is an enemy to any metal. I would like to ask if this process mixes tiny amount of salt to the isopropyl? if so by how much %? Is NaCl even soluble to isopropyl or are these just tiny particles floating around and if i leave the solution to settle for a longer period of time those particles would sink down. up until what percent can i bring the concentration to?
As for the Salt to use what other ingredients should i look out other than iodine( or other close sounding).
P.S. I am no chemist any way, the only chemistry background i have is during my high-school days. It was a time that i was still not able to appreciate class lessons. So go easy on me
[1]: https://www.instructables.com/id/How-To-Concentrate-Rubbing-Alcohol-With-Table-Salt/ |
one of my friends says that he made a Disinfectant. these are the materials:
83%:
methanol
isopropyl alcohol
propylene glycol
and the others are: citral, geraniol, linalool, dlimonene, l alpha terapineol...
1- does this really work?
2- is this dangerous for health if i use it only for surfaces? |
Despite your question is a bit vague about the sample, _I assume_ «when they are in motion» refers to particles passing a tube, are close to each other -- in contrast to particles floating in air.
_If_ so, than you should consider to couple infrared spectroscopy (about the energy band) with a probe based on [attenuated total reflection][1]. Roughly speaking, classical optics predicts you a threshold incident angle where light of a certain frequency no longer is able to leave an optical dense material (high refractive index, for IR for example Silicon, Germanium, or a small diamond) to enter a less dense material (for IR, for example air, most of organic materials). Recognizing light _equally_ has character of a wave however «allows» that light at this interface still interacts sufficiently enough that one may extract information to calculate an IR spectrum.
What you see in the lab as an ATR-FT-IR spectrometer _typically_ is a gate-like construction
[![enter image description here][2]][2]
([source][1])
but there are ATR-IR probes which you may immerse into fluids to monitor the advancement of reactions (e.g., Mettler's [reactir][3])
[![enter image description here][4]][4]
([source][5])
where the probe window is at the side or on the tip of the probe itself:
[![enter image description here][6]][6]
([source][7])
Potentially -- compared to classical transmission IR -- you truncate any sample preparation (no KBr pill, nor nujol) and may record data of not-too wet samples, too. Don't forget an ATR correction, as the small penetration depth $d_p$ of light across the phase boundery between window material and sample depends both on $n_c$ (window / crystal material), $n_s$ (sample), the incident angle $\theta$ and wavelength $\lambda$ applied.
$$ d_p = \frac{\lambda}{2\pi n_c \sqrt{\sin^2 \theta - \left(\frac{n_s}{n_c}\right)^2 }}$$
[1]: https://en.wikipedia.org/wiki/Attenuated_total_reflectance
[2]: https://i.stack.imgur.com/5cVce.jpg
[3]: https://www.mt.com/int/en/home/products/L1_AutochemProducts/ReactIR.html?cmp=sea_01010123&SE=GOOGLE&Campaign=MT_AC_EN_ROW&Adgroup=In%20Situ%20Analysis%20-%20ReactIR&bookedkeyword=%2Breact%20%2Bir&matchtype=b&adtext=264382116852&placement=&network=g&kclid=_k_EAIaIQobChMIqZW1nL-Y6AIVw8jeCh3OEwM2EAAYASAAEgKiV_D_BwE_k_&gclid=EAIaIQobChMIqZW1nL-Y6AIVw8jeCh3OEwM2EAAYASAAEgKiV_D_BwE
[4]: https://i.stack.imgur.com/DNKCH.jpg
[5]: https://www.copybook.com/companies/mettler-toledo/articles/reaction-analysis-tool-mettler-toledo-launch-innovative-insitu-solution
[6]: https://i.stack.imgur.com/Msq7l.jpg
[7]: https://kaplanscientific.nl/product/fiber-optic-atr-probes/ |
I did asked before , but apparently providing a link to a science report of harmless for humans (both skin and eyes) far-UV light was not enough to make people understand the significance of the solution to the raging COVID-19.
So this link (please read a bit of it):
https://www.nature.com/articles/s41598-018-21058-w
Says that the UV light of wavelength 222 nanometers (the photons bounces off the skin and the eyes) is absolutely harmless and do not make neither skin cancer or cataract. So this light is deadly for viruses and is not commercially available to anyone like ordinary humans (perhaps only some labs got it but sure cannot by on amazon or ebay or so). So if you wish to protect people you ideally need something to kill the virus in mid-air exactly as this lamp does.
Cite: "By contrast, we have previously shown that far-UVC light (207–222 nm) efficiently inactivates bacteria without harm to exposed mammalian skin." - this line is at the beginning of the link above.
So my idea is to help ordinary people that have some DIY experience to make a lamp and disinfect the clothes and themself like skin and face and face mask and so.
So a guy good in chemistry of similar that ca help with his knowledge and experience to help the people and get rid of the virus.
The research I did in the internet says that the UV has ability to kill virus in 10 seconds.
So having a lamp emitting harmless 222nm UV light instead the cancer deadly 254nm UV germicidal lamps will be able to protect from the diseases.
The question is how to make it - my proposal is deleted (by me after being discarded with a some totally irrelevant answers),
so I suppose you can type a good way to do it and save the half of the western world population (the old people like your grandma and grandpa)?
(Nobody needs glass (or quartz) blowing of a lamp. Neither skills beyond the ordinary DIY - well it might be a need of experienced DIY-er.)
|
Can someone please explain how to assing 1H nmr peaks of substituted cyclohexane group? I am more familiar with interpreting substituted benzene.
I get that the signal near 10ppm refers to the aldehyde group. The rest of the signal refers to the hydrogens on the cyclohexane group. I also understand that the assignment of Hb (see the picture below).
However, I don't know how I would compare the relative chemical shift between the remaining hydrogens. Since there are multiple peaks, the "Hd"s are chemically distinct. That is also observed when we use do the NMR analysis with higher resolution (See the second spectrum below).
Indeed, I have 1H NMR results with slightly better resolution. For "Hc" and "Hd"s, we can indeed resolve 4 peaks corresponding to signals at ~1.9, ~1.8, ~1.7 ~1.3ppm. The integration for these peaks are 2:2:2:4.
How would you assign the chemical shift for those 4 peaks?
[![enter image description here][1]][1]
the data is coming from http://www.hanhonggroup.com/nmr/nmr_en/B21091.html
Here is spectrum with a better resolution.
[![enter image description here][2]][2]
[1]: https://i.stack.imgur.com/JD7IV.png
[2]: https://i.stack.imgur.com/Kep55.png
The picture is retrieved from https://spectrabase.com/spectrum/LEvG58hCD1J?a=SPECTRUM_LEvG58hCD1J |
One of my friends says that he made a disinfectant. These are the materials:
* 83% methanol
* isopropyl alcohol
* propylene glycol
* and the others are: citral, geraniol, linalool, dlimonene, l alpha terapineol...
Does this really work? Is this dangerous for health if i use it only for surfaces? |
One of my friends says that he made a disinfectant. These are the materials:
* 83% methanol
* isopropyl alcohol
* propylene glycol
* and the others are: citral, geraniol, linalool, dlimonene, l alpha terapineol...
Does this really work? Is this dangerous for health if I use it only for surfaces? |
I'm interested in electrolysis of water to form O$_2$ and H$_2$, as pure as possible given the constraints of a "household chemical" setup. I've read other questions on here (in particular, https://chemistry.stackexchange.com/questions/89728/best-settings-for-electrolysis-of-water, https://chemistry.stackexchange.com/questions/38952/does-the-electrolysis-of-water-produce-chlorine-gas, https://chemistry.stackexchange.com/questions/116145/why-would-water-and-not-saltwater-undergo-electrolysis and others), but still have some open questions, in particular concerning the used electrolyte and the material of which the anode/cathode are.
Electrolyte
-----------
I'll be using tap water for my reaction that is not chlorinated. However, unspecified salts/impurities might be present that I cannot closer define, but we're talking about infant-safe drinking water (with little Na present). To have an efficient reaction, I will need to dissolve an electrolyte. There are many choices:
- NaOH (lye): As far as I can tell, the "ideal" candidate for my requirements, but might not be present in a household and might be difficult to obtain. Should form pure H$_2$ and O$_2$ if I am not mistaken.
- NaCl (table salt): Present in every household. Greatest drawback is that as far as I read Chlorine gas is formed during the electrolysis. My question would be, is the Chlorine emitted at the H$_2$ or O$_2$ electrode?
- NaHCO$_3$ or KHCO$_3$ (baking powder): Present in every household. Does not form Chlorine gas. However, as far as I have read, forms CO$_2$, but again I'm unsure at which electrode.
Electrode material
--
We can forget platinum and gold right off the bat for price reasons. Stainless steel would be a good choice, but I've also read that pure Nickel sheets should be working well. My main concern is corrosion and passivation of the electrode material which then slows down the overall reaction. Would Nickel sheets be preferrable over stainless steel? If we use some material that does corrode easily (e.g., regular steel or just run-of-the-mill hardware store items that are some combination of galvanized iron), would the corrosion and solutoin of the metal in the water affect the gas quality? If so, how? |
I am currently studying the textbook *Mass Spectrometry*, third edition, by Jürgen H. Gross. Chapter **2.4.3 Bond Dissociation Energies and Heats of Formation** says the following:
> **Energetics of $\ce{H}^\bullet$ loss from $\ce{CH_4^{+ \bullet}}$** The minimum energy needed to form a $\ce{CH_3^+}$ ion and a hydrogen radical from the methane molecular ion can be estimated from the heat of reaction, $\ce{\Delta H_r}$, of this process. According to Fig 2.6, $\ce{\Delta H_r} = \ce{AE_{(CH_3^+)}} - \ce{IE_{(CH_4)}}$. In order to calculate the missing $\ce{AE_{(CH_3^+)}}$ we use the tabulated values of $\ce{\Delta H_{f(H^\bullet)}} = 218.0 \ \text{kJ mol}^{-1}$, $\ce{\Delta H_{f(CH_3^+)}} = 1093 \ \text{kJ mol}^{-1}$, $\ce{\Delta H_{f(CH_4)}} = -74.9 \ \text{kJ mol}^{-1}$, and $\ce{IE_{(CH_4)}} = 12.6 \ \text{eV} = 1216 \ \text{kJ mol}^{-1}$. First, the heat of formation of the methane molecular ion is determined based on the experimental value of $\ce{IE_{(CH_4)}}$:
> $$\ce{\Delta H}_{f(CH_4^{+\bullet})} = \ce{\Delta H_{f(CH_4)}} + \ce{IE_{(CH_4)}} \tag{2.14}$$
> $$\ce{\Delta H}_{f(CH_4^{+\bullet})} = -74.9 \ \text{kJ mol}^{-1} + 1216 \ \text{kJ mol}^{-1} = 1141.1 \ \text{kJ mol}^{-1}$$
> Then, the heat of formation of the products is calculated from:
> $$\ce{\Delta H_{f(prod)}} = \ce{\Delta H_{f(CH_3^+)}} + \ce{\Delta H_{f(H^\bullet)}} \tag{2.15}$$
> $$\ce{\Delta H_{f(prod)}} = 1093 \ \text{kJ mol}^{-1} + 218 \ \text{kJ mol}^{-1} = 1311 \ \text{kJ mol}^{-1}$$
> Now, the heat of reaction is obtained from the difference
> $$\ce{\Delta H_r} = \ce{\Delta H_{f(prod)}} - \ce{\Delta H}_{f(CH_4^{+\bullet})} \tag{2.16}$$
> $$\ce{\Delta H_r} = 1311 \ \text{kJ mol}^{-1} - 1141.1 \ \text{kJ mol}^{-1} = 169.9 \ \text{kJ mol}^{-1}$$
> The value of $169.9 \ \text{kJ mol}^{-1}$ ($1.75 \ \text{eV}$) corresponds to $\ce{AE_{CH_3^+}} = 14.35 \ \text{eV}$, which is in good agreement with published values of about $14.3 \ \text{eV}$ (Fig. 2.7).
> [![enter image description here][1]][1]
**Note**: The amount of energy needed to be transferred to the neutral $\ce{M}$ to allow for the detection of the fragment ion $m_1^+$ is called the *appearance energy* ($AE$) of that fragment ion.
> [![enter image description here][2]][2]
> [![enter image description here][3]][3]
> [![enter image description here][4]][4]
> [![enter image description here][5]][5]
Taking the values from Fig. 2.7 and using [this][6] calculator, we get that $1311 \ \text{kJ mol}^{-1} = 13.588 \ \text{eV}$ and $1386 \ \text{kJ mol}^{-1} = 14.365 \ \text{eV}$.
Unless I have made a mistake or am misunderstanding something, these values do not agree with what the author has presented. It seems that, if we take the values from Fig 2.7, the value of $169.9 \ \text{kJ mol}^{-1}$ ($1.75 \ \text{eV}$) corresponds to $\ce{AE_{CH_3^+}} = 13.588 \ \text{eV}$, which implies that the published value should be around $14.365 \ \text{eV}$.
So has the author made a small mistake here? Or am I misunderstanding something?
I would greatly appreciate it if people would please take the time to review this.
[1]: https://i.stack.imgur.com/Y7HcS.png
[2]: https://i.stack.imgur.com/4zBn5.png
[3]: https://i.stack.imgur.com/HgOVR.png
[4]: https://i.stack.imgur.com/6gFxS.png
[5]: https://i.stack.imgur.com/9x6fP.png
[6]: http://www.colby.edu/chemistry/PChem/Hartree.html |
Can someone please explain how to assign the 1H nmr peaks of substituted cyclohexane group? I am more familiar with interpreting substituted benzene.
I get that the signal near 10ppm refers to the aldehyde group. The rest of the signal refers to the hydrogens on the cyclohexane group. I also understand that the assignment of Hb (see the picture below).
However, I don't know how I would compare the relative chemical shift between the remaining hydrogens. Since there are multiple peaks, the "Hd"s are chemically distinct. That is also observed when we use do the NMR analysis with higher resolution (See the second spectrum below).
Indeed, I have 1H NMR results with slightly better resolution. For "Hc" and "Hd"s, we can indeed resolve 4 peaks corresponding to signals at ~1.9, ~1.8, ~1.7 ~1.3ppm. The integration for these peaks are 2:2:2:4.
How would you assign the chemical shift for those 4 peaks?
[![enter image description here][1]][1]
the data is coming from http://www.hanhonggroup.com/nmr/nmr_en/B21091.html
Here is spectrum with a better resolution.
[![enter image description here][2]][2]
[1]: https://i.stack.imgur.com/JD7IV.png
[2]: https://i.stack.imgur.com/Kep55.png
The picture is retrieved from https://spectrabase.com/spectrum/LEvG58hCD1J?a=SPECTRUM_LEvG58hCD1J |
>How would you assign the chemical shift for those 4 peaks?
Since there is overlap between some of the resonances I would run a COSY experiment to resolve the peaks and find correlations, and make assignments on that basis. A number of things can complicate assignment. First, the CHO group can be in axial or equatorial positions, leading to two overlapping subspectra under slow exchange. In addition you have different shifts for geminal Hs, in equatorial/axial positions, separated by ~0.5 ppm (equatorial further downfield), with resolution into two peaks depending on the exchange time (e.g. temperature). The equatorial/axial H are coupled with a relatively strong J-coupling (~12 Hz). The chemically equivalent hydrogens are not magnetically equivalent, but this is not very important since long distance couplings are small.
[1]: https://www.chem.wisc.edu/areas/reich/nmr/05-hmr-04-2j.htm |
How to perform IR spectroscopy of silica particles when they are flowing through pipe at different RH? |
I have a polymer end-capped with a Polysilsesquioxane (POSS) like this one:
[![enter image description here][1]][1]
where *R=i-butyl*.
I haven't found much info on what type or reactions these POSS groups can undergo.
I would like to know if there are some procedures to modify that POSS moiety and convert it into an amino, carboxyl or triethoxysilane group, for example.
Also, are those -R groups easy to replace? Or that Si-(i-butyl) bonds are very stable?
[1]: https://i.stack.imgur.com/PlYQA.png |
I am currently studying the textbook *Mass Spectrometry*, third edition, by Jürgen H. Gross. Chapter **2.4.3 Bond Dissociation Energies and Heats of Formation** says the following:
> **Energetics of $\ce{H}^\bullet$ loss from $\ce{CH_4^{+ \bullet}}$** The minimum energy needed to form a $\ce{CH_3^+}$ ion and a hydrogen radical from the methane molecular ion can be estimated from the heat of reaction, $\ce{\Delta H_r}$, of this process. According to Fig 2.6, $\ce{\Delta H_r} = \ce{AE_{(CH_3^+)}} - \ce{IE_{(CH_4)}}$. In order to calculate the missing $\ce{AE_{(CH_3^+)}}$ we use the tabulated values of $\ce{\Delta H_{f(H^\bullet)}} = 218.0 \ \text{kJ mol}^{-1}$, $\ce{\Delta H_{f(CH_3^+)}} = 1093 \ \text{kJ mol}^{-1}$, $\ce{\Delta H_{f(CH_4)}} = -74.9 \ \text{kJ mol}^{-1}$, and $\ce{IE_{(CH_4)}} = 12.6 \ \text{eV} = 1216 \ \text{kJ mol}^{-1}$. First, the heat of formation of the methane molecular ion is determined based on the experimental value of $\ce{IE_{(CH_4)}}$:
> $$\ce{\Delta H}_{f(CH_4^{+\bullet})} = \ce{\Delta H_{f(CH_4)}} + \ce{IE_{(CH_4)}} \tag{2.14}$$
> $$\ce{\Delta H}_{f(CH_4^{+\bullet})} = -74.9 \ \text{kJ mol}^{-1} + 1216 \ \text{kJ mol}^{-1} = 1141.1 \ \text{kJ mol}^{-1}$$
> Then, the heat of formation of the products is calculated from:
> $$\ce{\Delta H_{f(prod)}} = \ce{\Delta H_{f(CH_3^+)}} + \ce{\Delta H_{f(H^\bullet)}} \tag{2.15}$$
> $$\ce{\Delta H_{f(prod)}} = 1093 \ \text{kJ mol}^{-1} + 218 \ \text{kJ mol}^{-1} = 1311 \ \text{kJ mol}^{-1}$$
> Now, the heat of reaction is obtained from the difference
> $$\ce{\Delta H_r} = \ce{\Delta H_{f(prod)}} - \ce{\Delta H}_{f(CH_4^{+\bullet})} \tag{2.16}$$
> $$\ce{\Delta H_r} = 1311 \ \text{kJ mol}^{-1} - 1141.1 \ \text{kJ mol}^{-1} = 169.9 \ \text{kJ mol}^{-1}$$
> The value of $169.9 \ \text{kJ mol}^{-1}$ ($1.75 \ \text{eV}$) corresponds to $\ce{AE_{(CH_3^+)}} = 14.35 \ \text{eV}$, which is in good agreement with published values of about $14.3 \ \text{eV}$ (Fig. 2.7).
> [![enter image description here][1]][1]
**Note**: The amount of energy needed to be transferred to the neutral $\ce{M}$ to allow for the detection of the fragment ion $m_1^+$ is called the *appearance energy* ($AE$) of that fragment ion.
> [![enter image description here][2]][2]
> [![enter image description here][3]][3]
> [![enter image description here][4]][4]
> [![enter image description here][5]][5]
Taking the values from Fig. 2.7 and using [this][6] calculator, we get that $1311 \ \text{kJ mol}^{-1} = 13.588 \ \text{eV}$ and $1386 \ \text{kJ mol}^{-1} = 14.365 \ \text{eV}$.
Unless I have made a mistake or am misunderstanding something, these values do not agree with what the author has presented. It seems that, if we take the values from Fig 2.7, the value of $169.9 \ \text{kJ mol}^{-1}$ ($1.75 \ \text{eV}$) corresponds to $\ce{AE_{CH_3^+}} = 13.588 \ \text{eV}$, which implies that the published value should be around $14.365 \ \text{eV}$.
So has the author made a small mistake here? Or am I misunderstanding something?
I would greatly appreciate it if people would please take the time to review this.
[1]: https://i.stack.imgur.com/Y7HcS.png
[2]: https://i.stack.imgur.com/4zBn5.png
[3]: https://i.stack.imgur.com/HgOVR.png
[4]: https://i.stack.imgur.com/6gFxS.png
[5]: https://i.stack.imgur.com/9x6fP.png
[6]: http://www.colby.edu/chemistry/PChem/Hartree.html |
I am currently studying the textbook *Mass Spectrometry*, third edition, by Jürgen H. Gross. Chapter **2.4.3 Bond Dissociation Energies and Heats of Formation** says the following:
> The heat of formation of organic radicals and positive ions decreases with their size and even more importantly with their degree of branching at the radical or ionic site. A lower heat of formation is equivalent to a higher thermodynamic stability of the respective ion or radical. The corresponding trends are clearly expressed by the values given in Tables 2.2 and 2.3, This causes the fragmentation pathways of molecular ions proceeding by formation of secondary or tertiary radicals and/or ions to become dominant over those leading to smaller and/or primary radical and ionic fragments, respectively (Sect. 6.2).
> [![enter image description here][1]][1]
> [![enter image description here][2]][2]
I have the following questions:
1. *Why* does the heat of formation of organic radicals and positive ions decrease with their size and degree of branching at the radical or ionic site?
2. *Why* is lower heat of formation equivalent to a higher thermodynamic stability of the respective ion or radical?
3. *How* do the effects of 1. and 2. cause the fragmentation pathways of molecular ions proceeding by formation fo secondary or tertiary radicals and/or ions to become dominant over those leading to smaller and/or primary radical and ionic fragments, respectively?
I would greatly appreciate it if people would please take the time to explain this section.
[1]: https://i.stack.imgur.com/HgOVR.png
[2]: https://i.stack.imgur.com/6gFxS.png |
I've only had 1 year of basic Chemistry at school, so I am waay too uneducated to answer this question myself:
**What is the true caloric impact of ingested alcohol (ethanol) in the human body?**
Here are a few things I presume to be fact:
- When testing, how many calories a substance contains, it is done by setting that substance on fire, then measuring the missing oxygen, maybe the resulting CO2 and the heat.
- Alcohol is poisonous to the human body, and cannot be used directly as an energy source
- the human body has to expend energy to turn alcohol (and starch ect) into some usable form. (by oxygenating or hydrogenation or something alike)
- every energetic substance we consume will be turned into adenosine triphosphate (ATP), which is the only 'energy carrier' our body uses when powering muscles
Conclusions I draw:
- When we test things for calories, we get a good estimate of how much 'energy' it would serve us
- yet the testing is just a guideline, as human digestion and internal biochemical processes are a bit more convoluted than the simple 'set it on fire' combustion
- When we have two different substances with the same caloric value, consuming them might yield different energetic value to the body
So, how to formulate this mathematically:
- `tcv` is the tested caloric value
- `ecv` is the effective caloric value for the human body
- `tcv >= ecv`
- for those sugars, that can be used directly (e.g. converted into ATP): `tcv = ecv`
- for most things, where internal biochems have to turn it into usable form : `tcv > ecv`
- alcohol: `tcv > ecv`
So how energetically effective is alcohol to the human body, compared to sugars?
- on a pure chemical level: `ecv/tcv=?` 50%? 70%?
- considering neutralizing / storing / discarding unwanted byproducts |
What is the true caloric impact of Alcohol? |
The boiling points of ethylene, formaldehyde and dioxygen are -103.7°C, -19°C and −183 °C, respectively. I expect formaldehyde to have the highest boiling point of the three because of dipole moment mostly due to the carbon-oxgyen bond. However, I don't know why ethylene has a higher boiling point than dioxygen.
Different from this similar question, https://chemistry.stackexchange.com/questions/8246/comparision-of-boiling-points/8247#8247, where the atoms involved are of very different size, molar mass and polarizability, the structure and electronic structure of ethylene and dioxygen are more similar. They both have a single conformation, but they have different symmetry, which might have same bearing on the entropy change during phase transition. I know that both the C-C bond in ethylene and the O-O bond in dioxygen have a bond order of 2, but dioxygen is paramagnetic while ethylene is not.
The latent heat of vaporization is [6.8 kJ/mol][1] for dioxygen and [13.7 kJ/mol][2] for ethylene, so enthalpy seems a major contribution to the difference in boiling point.
What are the differences in intermolecular interactions in liquid $\ce{O2}$ compared to liquid $\ce{H2C=CH2}$ underlying the difference in normal boiling point?
[1]: https://en.wikipedia.org/wiki/Heats_of_vaporization_of_the_elements_(data_page)
[2]: https://webbook.nist.gov/cgi/cbook.cgi?ID=C74851&Mask=4 |
[![Table with records relating the initial rate of reaction with conc. of A,B,C,D ][1]][1]
>The data collected from using the method of initial rates is tabulated above.
>Suggest a rate law and a mechanism consistent with the data. State any assumptions.
**My Attempt**
Obviously B and D has no impact on the rate of the reaction. From the first table it appears that the reaction is first order in A, however this doesn't make sense when you examine the last two rows of the second table. Since the rate law is first order in A and is independent of B, decreasing the concentration of A by 4 and increasing the concentration of B by 4 should result in a fourfold decrease in the rate. However the rate of the reaction decrease by 2.5 times. So then my logic is somehow wrong.
Furthermore I am even more confused about the order of C. I know that concentration of C is inversely proportional to the rate of the reaction. However the reaction still occurs without the presence of C (evident by the first table). Doesn't this imply that the rate law is independent of the concentration of C?
From the comments below, I have realized that the rate order of B is in fact not $0$. However now I am even more confused. How can it be that the rate is not dependent on B when there is no C, but it is dependent on B when there is C present? What would its rate law be like?
[1]: https://i.stack.imgur.com/yuny9.png |
I am currently studying the textbook *Mass Spectrometry*, third edition, by Jürgen H. Gross. Chapter **2.4.3 Bond Dissociation Energies and Heats of Formation** says the following:
> The heat of formation of organic radicals and positive ions decreases with their size and even more importantly with their degree of branching at the radical or ionic site. A lower heat of formation is equivalent to a higher thermodynamic stability of the respective ion or radical. The corresponding trends are clearly expressed by the values given in Tables 2.2 and 2.3, This causes the fragmentation pathways of molecular ions proceeding by formation of secondary or tertiary radicals and/or ions to become dominant over those leading to smaller and/or primary radical and ionic fragments, respectively (Sect. 6.2).
> [![enter image description here][1]][1]
> [![enter image description here][2]][2]
I'm wondering *why* does the heat of formation of organic radicals and positive ions decrease with their size and degree of branching at the radical or ionic site?
I would greatly appreciate it if people would please take the time to explain this.
[1]: https://i.stack.imgur.com/HgOVR.png
[2]: https://i.stack.imgur.com/6gFxS.png |
Why does the heat of formation of organic radicals and positive ions decrease with their size and degree of branching at the radical or ionic site? |
I am currently studying the textbook *Mass Spectrometry*, third edition, by Jürgen H. Gross. Chapter **2.4.3 Bond Dissociation Energies and Heats of Formation** says the following:
> The heat of formation of organic radicals and positive ions decreases with their size and even more importantly with their degree of branching at the radical or ionic site. A lower heat of formation is equivalent to a higher thermodynamic stability of the respective ion or radical. The corresponding trends are clearly expressed by the values given in Tables 2.2 and 2.3, This causes the fragmentation pathways of molecular ions proceeding by formation of secondary or tertiary radicals and/or ions to become dominant over those leading to smaller and/or primary radical and ionic fragments, respectively (Sect. 6.2).
> [![enter image description here][1]][1]
> [![enter image description here][2]][2]
*Why* does the heat of formation of organic radicals and positive ions decrease with their size and degree of branching at the radical or ionic site?
I would greatly appreciate it if people would please take the time to explain this.
[1]: https://i.stack.imgur.com/HgOVR.png
[2]: https://i.stack.imgur.com/6gFxS.png |
The boiling points of ethylene, formaldehyde and dioxygen are $\pu{-103.7 ^\circ C}$, $\pu{-19 ^\circ C}$, and $\pu{−183 ^\circ C}$, respectively. I expect formaldehyde to have the highest boiling point of the three because of dipole moment mostly due to the carbon-oxgyen bond. However, I don't know why ethylene has a higher boiling point than dioxygen.
Different from [this similar question](https://chemistry.stackexchange.com/questions/8246/comparision-of-boiling-points/8247#8247), where the atoms involved are of very different size, molar mass, and polarizability, the structure and electronic structure of ethylene and dioxygen are more similar. They both have a single conformation, but they have different symmetry, which might have same bearing on the entropy change during phase transition. I know that both the $\ce{C-C}$ bond in ethylene and the $\ce{O-O}$ bond in dioxygen have a bond order of 2, but dioxygen is paramagnetic while ethylene is not.
The latent heat of vaporization is [6.8 kJ/mol][1] for dioxygen and [13.7 kJ/mol][2] for ethylene, so enthalpy seems a major contribution to the difference in boiling point.
What are the differences in intermolecular interactions in liquid $\ce{O2}$ compared to liquid $\ce{H2C=CH2}$ underlying the difference in normal boiling point?
[1]: https://en.wikipedia.org/wiki/Heats_of_vaporization_of_the_elements_(data_page)
[2]: https://webbook.nist.gov/cgi/cbook.cgi?ID=C74851&Mask=4 |
I did asked before , but apparently providing a link to a science report of harmless for humans (both skin and eyes) far-UV light was not enough to make people understand the significance of the solution to the raging COVID-19.
So this link (please read a bit of it):
https://www.nature.com/articles/s41598-018-21058-w
Says that the UV light of wavelength 222 nanometers (the photons bounces off the skin and the eyes) is absolutely harmless and do not make neither skin cancer or cataract. So this light is deadly for viruses and is not commercially available to anyone like ordinary humans (perhaps only some labs got it but sure cannot by on amazon or ebay or so). So if you wish to protect people you ideally need something to kill the virus in mid-air exactly as this lamp does.
Cite: "By contrast, we have previously shown that far-UVC light (207–222 nm) efficiently inactivates bacteria without harm to exposed mammalian skin." - this line is at the beginning of the link above.
So my idea is to help ordinary people that have some DIY experience to make a lamp and disinfect the clothes and themself like skin and face and face mask and so.
So a guy good in chemistry of similar that ca help with his knowledge and experience to help the people and get rid of the virus.
The research I did in the internet says that the UV has ability to kill virus in 10 seconds.
So having a lamp emitting harmless 222nm UV light instead the cancer deadly 254nm UV germicidal lamps will be able to protect from the diseases.
The question is how to make it.
so I suppose you can type a good way to do it and save the half of the western world population (the old people like your grandma and grandpa)?
(Nobody needs glass (or quartz) blowing of a lamp. Neither skills beyond the ordinary DIY - well it might be a need of experienced DIY-er.)
|
There are binary diamond type structures for example S and Zn form sphalerite. I thought the reason this formed was because of the big difference in the radii of the atoms that allows the smaller atom to fit into the tetrahedral hole that the bigger atoms form. However GaAs also forms sphalerite type structure, their radii are not that different. Why do Sphalerite type structures form? |
Toxic smoke from propellant combustion has been known about for over 500 years..
The propellants used today are Nitrogen based, thus nitro glycerin, nitrocellulose and Nitroguanidine . These produce gases containing oxides of nitrogen, carbon oxides, carbon monoxide, hydrogen cyanide, ammonia and several other gases..
The oxides of nitrogen can be inhaled deeply into the lung thus causing damage as deep as the Alveoli, thus causing bronchiolitis obliterins resulting in a "Pulmonary Edema". This can happen 5 years after exposure..
If you show signs within 24 hours of exposure then oxygen and Arterial Blood gas tests are the solution to treating the condition and knowning the effect. The symptons show just like influenza but no sign of virus.
This is known as an Acute Toxic Smoke Inhalation Injury.
WW1 and WW2 this was studied to reduce the effect on gun crews on both land and sea. It is still a problem today with Armoured vehicles and Artillery and one of the reason the Navies use unmanned turrets.
Look up this file and you will find quite a bit of information about this condition.
Occupational Health - The Soldier and the Industrial base
I had first hand experience of this injury in 1988 and am still alive to tell my story, I had a "Pulmonary Edema" in 1991 and survived that, so I am in the 2% of survivors from that period. In the mid 1990's they discovered how to treat a pulmonary edema thus it has increased the survival rate.
There is thousands of pages of information on this subject, just look for "Propellant exhaust gas exposure" or "Toxic gas inhalation" that will bring up more information.
The military threshold limits are higher than civilian but a tank will expose the crew to 1,300 the TLV for less than 5 minutes if the fume extractor is working. In my case it was not!
http://www.public.navy.mil/surfor/documents/p_5041.pdf Chapter 10
http://www.cs.amedd.army.mil/FileDownloadpublic.aspx?docid=a1d49fb8-3a16-4489-9408-1089f7044b4c
https://permanent.access.gpo.gov/lps101453/www.bordeninstitute.army.mil/published_volumes/occ_health/OHch10.pdf
There are 3 links for you to go to...
Enjoy reading it...
Oh just as an idea, if you had 1,100 .308(7.62mm) rifles fired in a 100 m3 room, its would be the equivalent of 5x L7 105mm tank rounds...
Enjoy.. |
Toxic smoke from propellant combustion has been known about for over 500 years..
The propellants used today are Nitrogen based, thus nitro glycerin, nitrocellulose and nitroguanidine . These produce gases containing oxides of nitrogen, carbon oxides, carbon monoxide, hydrogen cyanide, ammonia and several other gases..
The oxides of nitrogen can be inhaled deeply into the lung thus causing damage as deep as the Alveoli, thus causing bronchiolitis obliterins resulting in a "Pulmonary Edema". This can happen 5 years after exposure..
If you show signs within 24 hours of exposure then oxygen and Arterial Blood gas tests are the solution to treating the condition and knowning the effect. The symptons show just like influenza but no sign of virus.
This is known as an Acute Toxic Smoke Inhalation Injury.
WW1 and WW2 this was studied to reduce the effect on gun crews on both land and sea. It is still a problem today with Armoured vehicles and Artillery and one of the reason the Navies use unmanned turrets.
Look up this file and you will find quite a bit of information about this condition.
Occupational Health - The Soldier and the Industrial base
I had first hand experience of this injury in 1988 and am still alive to tell my story, I had a "Pulmonary Edema" in 1991 and survived that, so I am in the 2% of survivors from that period. In the mid 1990's they discovered how to treat a pulmonary edema thus it has increased the survival rate.
There is thousands of pages of information on this subject, just look for "Propellant exhaust gas exposure" or "Toxic gas inhalation" that will bring up more information.
The military threshold limits are higher than civilian but a tank will expose the crew to 1,300 the TLV for less than 5 minutes if the fume extractor is working. In my case it was not!
http://www.public.navy.mil/surfor/documents/p_5041.pdf Chapter 10
http://www.cs.amedd.army.mil/FileDownloadpublic.aspx?docid=a1d49fb8-3a16-4489-9408-1089f7044b4c
https://permanent.access.gpo.gov/lps101453/www.bordeninstitute.army.mil/published_volumes/occ_health/OHch10.pdf
There are 3 links for you to go to...
Enjoy reading it...
Oh just as an idea, if you had 1,100 .308(7.62mm) rifles fired in a 100 m3 room, its would be the equivalent of 5x L7 105mm tank rounds...
Enjoy.. |
**UVC is very harmful to people and requires specialist equipment to create and use safely**
You are not going to be able to create or use lamps that emit 222nM light without specialist equipment, and even if you could it would almost certainly not be safe.
Far UV lamps are widely commercially available for germicidal applications and these usually use mercury vapour fluorescent discharge tubes with no phosphors so have a strong emission around 250nM from a line in the mercury emission spectrum. They are very dangerous to people and have caused significant harm when used accidentally (see [this video from BigClive][1] describing what happens when they are used accidentally instead of the much less harmful "black light" tubes commonly used for discos and events).
Another important fact to bear in mind is that UVC lamps of this type have to use specialist glass in manufacture because common types of glass strongly absorb UVC wavelengths. So, while they are readily available, they are not easy for an amateur to make.
You report a claim from a Nature paper than a tightly controlled UVC light of around 222nM wavelength is not harmful to people. I would be far more cautious as the Nature paper did ***not*** do extensive safety studies on the actual effect on people and given what we know about wavelengths not much longer, the claim would need some extensive validation.
But you have bigger problems than this in your pursuit of an easy-to-make solution accessible to amateurs. First, you cannot easily get around the problem of the glass used absorbing the wavelengths you want to see: you will need specialist material. Second, the actual excimer lamps as used in the nature paper are sophisticated devices that cannot easily be made by non-specialists (they are far more complex than the relatively conventional discharge tubes used in commercial UV lamps which are essentially the same as conventional visible fluorescent lamps apart from the need for specialist glass). Apart from anything else the lamps have a strong emission at around 258nM, exactly the wavelength known to be very dangerous to people from simple UVC mercury vapour lamps. The NAture paper reported using specialist narrow band filters to eliminate those wavelengths. These are unlikely to be easy to obtain. Third, creating the excimer lamps is a sophisticated process requiring specialist pulsed electronic drivers to get the lamp to work (most likely, according to [this reference in the Nature paper][2], you need to fine tune the gas mixture and the pressure and the driving electronics for the lamp which is far more complex than a simple driver for a fluorescent tube).
And, as for "Nobody needs glass (or quartz) blowing of a lamp", I'm afraid you are delusional. You need specialist glass to let the right wavelengths through and specialist filters to kill the harmful wavelengths. This is not "simple DIY".
You have seen an interesting paper in Nature and assumed it is easy for someone to make in their shed. This is not true. Specialist skills are required and attempting anything without those skills is likely highly dangerous (and that is ignoring the issues of handling the chemicals and vacuum technology needed to create the lamp: both are dangerous).
And, perhaps more importantly, the best way to avoid contamination of others with covid-19 is to wash your hands. There are no easy technological silver bullets here, though, perhaps, professional engineers and chemists can build something safe based on the Nature observation in the future.
[1]: https://www.youtube.com/watch?v=CpRMud6EFtE&t=1064s
[2]: https://ieeexplore.ieee.org/document/4799201 |
Recently I've been reviewing concepts belonging to the history of chemistry. But I came stuck at trying to understand a passage which I read from wikipedia entry (and which it seems has been mentioned in different sources). This is related with the law of reciprocal proportions.
It states as this:
> It took 615 parts by weight of magnesia (MgO), for example, to
> neutralize 1000 parts by weight of sulfuric acid.
There's an [existing answer](https://chemistry.stackexchange.com/questions/42903/what-is-the-significance-of-law-of-multiple-proportions) which explains a similar concept (law of multiple proportions) but not specifically to Richter's.
From what I understood in the law of reciprocal proportions there is a set of three elements. Two of them react with a fixed amount of a third element. The ratio of these two elements is the same when they combine between.
For example:
When $63.5\,g$ copper combines with sulphur produces copper sulphide
$\begin{array}{lllll}
Cu&+&S&\rightarrow&CuS\\
63.5\,g&&32\,g&&95.5\,g\\
\end{array}$
Conversely when those $63.5\,g$ of copper combines with oxygen produces cupric oxide.
$\begin{array}{lllll}
2Cu&+&O_2&\rightarrow&2CuO\\
63.5\,g&&16\,g&&79.5\,g\\
\end{array}$
Therefore the proportion between the mass of sulphur and oxygen masses is:
$\frac{m_S}{m_{O_{2}}}=\frac{32}{16}=2$
When sulphur and oxygen combine together they make sulphur dioxide.
$\begin{array}{lllll}
S&+&O_2&\rightarrow&SO_{2}\\
32\,g&&32\,g&&64\,g\\
\end{array}$
Then the proportion between these two becomes:
$\frac{m_S}{m_{O_{2}}}=\frac{32}{32}=1$
The latter is in proportion which is half of the first ratio. Hence follows the Richter law.
But how can I use this information to understand what it was mentioned in the paragraph from above?. How is exactly related with equivalent weights?.
The second part of the question arises from the fact, that would it be okay to state this?
In an hypotetical reaction between $A$, $B$ producing $C$ and $D$.
$\begin{array}{ccccccc}
aA&+&bB&\rightarrow&cC&+&dD\\
\textrm{1 eq gram of A}&&\textrm{1 eq gram of B}&&\textrm{1 eq gram of C}&&\textrm{1 eq gram of D}\\
\end{array}$
Becoming in
$\textrm{1 eq gram of A}=\textrm{1 eq gram of B}=\textrm{1 eq gram of C}=\textrm{1 eq gram of D}$
Would it be accurate to put it in that way?.
or?
$\textrm{number of eq gram of A}=\textrm{number of eq gram of B}=\textrm{number of eq gram of C}=\textrm{number eq gram of D}$
where:
$\textrm{1 eq gram A} =\frac{\textrm{grams of A compound equal to the equivalent weight of A}}{\frac{\textrm{formula weight of A}}{\textrm{number of electrons transfered}}}$
which is different when considering moles:
$\begin{array}{ccccccc}
aA&+&bB&\rightarrow&cC&+&dD\\
\textrm{a moles of A}&&\textrm{b moles of B}&&\textrm{c moles of C}&&\textrm{d moles of D}\\
\end{array}$
where:
$\textrm{a moles of A}=\textrm{b moles of B}=\textrm{c moles of C}=\textrm{d moles of D}$
So, is what I wrote correct?. How can I relate Richter's law of reciprocal proportions to the equivalent weights?. How is it related with the paragraph from above?. Perhaps does it meant to say that there is a definite amount of reagent which can react with a certain amount of sulphuric acid?. Then why isn't it related more with the concept of limiting reagent rather than the equivalent weight?. Can someone guide me on that? |
Did Richter meant to say with the law of reciprocal proportions and how is it related with an acid base reaction? |
Why is measurement in molarity preferred over molality? |
I am not sure whether this question fits in the scope of this site or biology.SE. Nonetheless, I am looking for a chemical explanation which I wasn't able to find.
I have read in several sources such as [this one](https://lifehacker.com/store-potatoes-with-an-apple-to-keep-them-from-sproutin-5954159) that, in order to prevent potatoes from sprouting, it is sufficient to store a few apples alongside them. This is due to the fact that the apples produce ethylene.
But this is a bit odd, since the ethylene is typically a ripening agent. [This reddit thread](https://www.reddit.com/r/askscience/comments/pwl4j/does_ethylene_gas_prevent_or_cause_potatoes_to/) also discusses the same doubt, but doesn't dive much into the details. There is also [this article](https://hamodia.com/columns/do-apples-keep-potatoes-from-sprouting-its-complicated/) and [this Quora answer](https://www.quora.com/Why-do-apples-keep-potatoes-from-sprouting), none of which give a specific chemical reason about why and how this works.
Chemically speaking, do we actually know the exact reason and the necessary conditions to make this work? |
Does storing apples alongside potatoes keeps potatoes from sprouting? |
I recently [asked][1] a question about *why the heat of formation of organic radicals and positive ions decreases with their size and degree of branching at the radical or ionic site*. The user "Buttonwood" made the comment that **Hyperconjugation** is one of the forms stabilising cations and radicals.
This begs two questions:
1. What is the reasoning behind why hyperconjugation increases the stability of cations and radicals, and therefore decreases the heat of formation? Point 3. of [this][2] section of the Wikipedia article for hyperconjugation says that "*the heat of formation of molecules with hyperconjugation are greater than sum of their bond energies and the heats of hydrogenation per double bond are less than the heat of hydrogenation of ethylene.*"; does this not imply that hyperconjugation increases the heat of formation of molecules?
2. Why does an *increase* in stability imply a *decrease* in heat of formation? To a novice who is ignorant on this subject such as myself, it would intuitively seem that greater stability implies an *increase* in heat of formation, since the molecule is more "stable" in its current form and therefore it takes more energy to affect changes in its structure?
I would greatly appreciate it if people would please take the time to clarify these points.
[1]: https://chemistry.stackexchange.com/q/129037/75460
[2]: https://en.wikipedia.org/wiki/Hyperconjugation#Effect_on_chemical_properties |
I am looking for a chemical explanation which I wasn't able to find.
I have read in several sources such as [this one](https://lifehacker.com/store-potatoes-with-an-apple-to-keep-them-from-sproutin-5954159) that, in order to prevent potatoes from sprouting, it is sufficient to store a few apples alongside them. This is due to the fact that the apples produce ethylene.
But this is a bit odd, since the ethylene is typically a ripening agent. [This reddit thread](https://www.reddit.com/r/askscience/comments/pwl4j/does_ethylene_gas_prevent_or_cause_potatoes_to/) also discusses the same doubt, but doesn't dive much into the details. There is also [this article](https://hamodia.com/columns/do-apples-keep-potatoes-from-sprouting-its-complicated/) and [this Quora answer](https://www.quora.com/Why-do-apples-keep-potatoes-from-sprouting), none of which give a specific chemical reason about why and how this works.
Chemically speaking, do we actually know the exact reason and the necessary conditions to make this work? |
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