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I’m required to make a lab with this information only (particle size vs dissolving rate of a solid (molecular covalent compound-sucrose)? so far what I’m thinking is using three different kinds of sugar like table sugar, sugar cubes, and one of them grinding it but the issues is how do you find how they dissolving rate? |
In the preparation of $\ce{NH3}$ from $\ce{NH4Cl}$ and $\ce{Ba(OH)2}$, my teacher said that when heating the test tube, the test tube faces downwards as opposed to upwards. Why is this so? Wouldn't the solid just fall out?
[![enter image description here][1]][1]
[1]: https://i.stack.imgur.com/AvEVU.png |
Why does the test tube face downwards in gas production? |
The percent of Ethyl Alcohol in hand sanitizers slowly drops over time. And manufacturers put an expiration date on the bottles. But I'd to test the percent of Ethyl Alcohol for myself for a new bottle and to see how much over time is lost as it gets close to the expiration data and beyond. Is this possible for a chemistry n00b? |
Possbile to test percent of Ethyl Alcohol in hand sanitizer? |
I experimented $\ce{CO2}$ generation by adding $\pu{4 g}$ of $25\%$ acetic acid water solution ($= \pu{1 g}$ of acetic acid) to $\pu{1.4 g}$ of pure baking soda. But this mixture yielded only $50\%$ less than the calculated amount $\pu{373 mL}$ of $\ce{CO2}$.
$$
M(\ce{CH3COOH}) = \pu{60g/mol}, M(\ce{NaHCO3}) = \pu{84 g/mol}\\
\implies \ce{CH3COOH}~\pu{1g} : \ce{NaHCO3} \pu{1.4 g}. \\
\ce{CH3COOH + NaHCO3 -> CH3COONa + H2O + CO2}\\
\text{mol to mol} = \pu{1 mol} ~\ce{CO2}, M(\ce{CO2}) = \pu{44 g/mol} \\
\implies \ce{CO2}~\pu{0.733g} : \ce{CH3COOH}~\pu{1g} : \ce{NaHCO3}~\pu{1.4g}\\
\text{Gas} = \pu{22.4 L/mol} \implies \pu{22.4 L}/60 = \pu{373.3 mL} (= \pu{0.733 g}) \text{(Theoretical yield)}.\\
\text{Actual yield} = \pu{190 mL} \text{by volume measurement} (50\%)\\
\text{Actual yield} = \pu{0.586 g} (=\pu{298.6 mL}) \text{by mass measurement} (80\%)\\
$$
I set the water filled measuring cup upside down above the water filled plate, and attached hose underneath the skirt of the cup so that any gas generated from the sealed bottle of the mixture to get provided into the measuring cup.
Moreover, I checked the amount of $\ce{CO2}$ yielded by the weight loss method. That is, I measured the mixture above the scale to see how much $\ce{CO2}$ is evaporated during the chemical reaction.
The above 2nd method provided me with $80\%$ yield.
$50\%$ of air volume and $80\%$ of mass are coming with the consistency during the several times of experiments.
What was wrong with my experiment?? |
Why does vinegar and baking soda yield a far smaller volume of carbon dioxide gas than expected from calculation? |
I want to ask a question about the shoulders of peaks when separating overlapping peaks.
I have the following spectrum circled with relevant shoulders:
[![NMR Peak][1]][1]
I believe that the spectrum above shows an overlapping triplet and doublet, but I want to know what causes these shoulders in the NMR spectrum. I'm stuck to whether to decide there is further splitting of a small $\ce{J}$ value or to just assume it's a shoulder.
What is causing these shoulders? Are they part of the signal or just a result of poor preparation leading to artefacts?
[1]: https://i.stack.imgur.com/AX4iE.png |
I've noticed that in the mass spectrum of 3-chloropropene, there is a peak at m/z = 39, which to me, indicates the formation of the C3H2 ion.
However, given that 3 bonds would have to be broken to form this ion, how can the frequency of this cleavage be explained?
[![enter image description here][1]][1]
[1]: https://i.stack.imgur.com/A9QsW.png |
I want to ask a question about the shoulders of peaks when separating overlapping peaks.
I have the following spectrum circled with relevant shoulders:
[![NMR Peak][1]][1]
I believe that the spectrum above shows an overlapping triplet and doublet, but I want to know what causes these shoulders in the NMR spectrum. I'm stuck to whether to decide there is further splitting of a small $\ce{J}$ value or to just assume it's a shoulder.
Below shows analysis of this region from Acta Pharm. 54 (2004) 163–176.
[![Extract from previous analysis][2]][2]
What is causing these shoulders? Are they part of the signal or just a result of poor preparation leading to artefacts?
[1]: https://i.stack.imgur.com/AX4iE.png
[2]: https://i.stack.imgur.com/0iT7o.png |
I've noticed that in the mass spectrum of 3-chloropropene, there is a peak at m/z = 39, which to me, indicates the formation of the C3H3 ion.
However, given that 3 bonds would have to be broken to form this ion, how can the frequency of this cleavage be explained?
[![enter image description here][1]][1]
[1]: https://i.stack.imgur.com/A9QsW.png |
I am trying to make my own hand sanitiser since it’s sold out everywhere. The formular I use is 1 part aloe vera gel, 2 parts 100% isopropyl alcohol. The problem is when mixing the 2, the solution always forms a small white spongy sticky substance while the rest stays liquid, not gel like:
[![enter image description here][1]][1]
I tried 2 brands of aloe vera gel to see the same result:
- Palmer Aloe vera gel: Aloe Vera Gel (Aloe Barbadensis), Polyglycerylmethacrylate, Propylene Glycol, Witch Hazel, Algae Extract (Algae), PPG-2-Isodeceth-12, Carbomer, Tocopheryl Acetate, Triethanolamine, Methyl Gluceth-20, DMDM Hydantoin, Tetrasodium EDTA, Quaternium-15, Fragrance (Parfum), Hexyl Cinnamal, Linalool, Geraniol, Limonene, Yellow 5 (CI 19140), Blue 1 (CI 42090).
- Thursday Plantation Aloe Vera Gel: Aloe Vera inner leaf juice equiv. to fresh 974mg/g, Hydroxybenzoates, Diazolidinylurea, Potassium Sorbate and Sodium Benzoate
I am starting to think I got the wrong type of alcohol but the label clearly says pure 100% isopropyl alcohol.
What am I doing wrong?
[1]: https://i.stack.imgur.com/I553U.jpg |
As we all know, adding salt to ice water lowers its temperature.
I've read plenty of system-level accounts of energy balances, enthalpies, vapor pressures, phase equilibria and freezing points—I understand that chemists' bread and butter is doing energy arithmetic on thermodynamic systems as a whole, but I would like to gain microscopic-level intuition instead of just being handed some thermodynamic abstractions.
I am looking for an intuitive explanation that details, step by step, at an atomic level, what happens at the interface between the ice and the salt water. For each step, I would like to understand:
* What energies (kinetic/temperature vs. latent/chemical) are involved, and where?
* How exactly (atomically or subatomically) is one type of energy converted into the other?
* Why is each step energetically favourable? Or is it not, and it's reversible? How reversible, statistically?
The explanation should be clear enough that I should be able to intuitively answer the following, without consulting diagrams:
* How does the concentration of salt change the evolution of the system (again, at a microscopic level)
* How does adding e.g. ethanol instead of salt change things?
* Is liquid water required at all? Is salt able to melt and cool completely frozen ice, or is external force required (stirring, driving over it, etc.) to cause some initial melting for anything to happen? (I've read conflicting answers to this in the past.)
|
**Water loses kinetic energy when it detaches from the surface of the ice**
Water molecule in the solid phase have more hydrogen bonds than in the liquid phase. It takes energy to break these bonds, so only molecules with sufficient kinetic energy will detach (maybe because they just collided with a water molecule that transferred some kinetic energy to it). So the molecule that detaches will have a low kinetic energy as it goes into the liquid phase. Further collisions will quickly ensure that over time, this particular molecule has an average kinetic energy, but overall the temperature will drop from that detaching event.
**Water gains kinetic energy when it attaches to the surface of the ice**
This is just the opposite process, so the opposite happens. As additional hydrogen bonds are formed, the molecule has greater kinetic energy (in the form of vibrational energy). Either the molecule had less than average kinetic energy and now has average energy again, or it had average kinetic energy and will have higher kinetic energy, but quickly dissipate this to other molecules.
**When you add salt to the liquid water, less molecules attach to the surface of the ice**
Compared to pure water, less collisions of particles in the solution with the surface of the ice will lead to attachment of particles because now, some of these particles are not water molecules but sodium or chloride ions instead (they are not incorporated into the crystal lattice of ice but stay solvated by liquid water). As a consequence, there are less attachment events while the detachment events stay the same. With a net melting of ice, the average kinetic energy of particles decreases (the temperature drops). As a secondary effect, this also affects the rates of attachment and detachment, so once the temperature has sufficiently dropped, the two rates will be the same again.
>What energies (kinetic/temperature vs. latent/chemical) are involved, and where?
In my explanation, I talked about kinetic energy (reflecting the temperature) and energy necessary to break hydrogen bonds (could be generalized as intermolecular interactions, and they underlie the observed bulk latent heat).
>How exactly (atomically or subatomically) is one type of energy converted into the other
There is no chemistry or nuclear physics happening, so it is sufficient to talk about the molecular level. Kinetic energy is converted into potential energy through inelastic collisions. Kinetic energy is redistributed between molecules through elastic collisions.
>Why is each step energetically favourable? Or is it not, and it's reversible? How reversible, statistically?
According to microscopic reversibility, all molecular events are reversible. Processes go in a certain direction because it is more probable than in the other direction (2nd law of thermodynamics). The 2nd law of thermodynamics is inherently linked to large numbers of particles, so you need to look at the forest rather than at individual trees.
>How does the concentration of salt change the evolution of the system (again, at a microscopic level).
The higher the concentration of the salt, the lower the concentration of the water in the liquid phase, i.e the fewer events of liquid water attaching to the ice.
>How does adding e.g. ethanol instead of salt change things?
1 mol/L NaCl has twice the dissolved particles compared to 1 mol/L ethanol. Otherwise, the effect is the same.
> Is liquid water required at all? Is salt able to melt and cool completely frozen ice, or is external force required (stirring, driving over it, etc.) to cause some initial melting for anything to happen? (I've read conflicting answers to this in the past.)
Ice near the freezing point exhibit surface melting. Once there is a tiny amount of water, more ice will melt. On the other hand, if it is cold enough that even saturated solutions will freeze, a mixture of dry salt and ice will not interact. |
$\Delta H$ is temperature-dependent, pressure-dependent, and concentration-dependent, so it's certainly theoretically possible for it to change sign in a chemical reaction as a result of changes in any of these intensive properties.
NIST has a detailed Shomate Equation that has been experimentally verified to model the standard heat of formation ($\Delta H^o_f$) of $\ce{N_2O_4_{(\textit{l})}}$ from 298 K to 500 K (https://webbook.nist.gov/cgi/cbook.cgi?ID=C10544726&Mask=1E9F). If you plug in the parameters, you will see that $\Delta H^o_f (298 K) = -19.56 kJ/mol$, while $\Delta H^o_f (500 K) = +12.87 kJ/mol$
There are also direct experimental measurements showing that enthalpies of solution change sign in response to both temperature and concentration changes.
For instance, the enthalpy of solution of polystyrene in decalin has been shown to change from ~ –10 J/g at ~20 C to ~ +6 J/g at ~100 C (the ~'s were needed because I'm eyeballing these from a plot). Reference:
Tager, A. A., et al. "The temperature and concentration dependence of enthalpy, free energy and entropy of mixing polystyrene solutions with an upper and lower critical temperature of mixing." *Polymer Science USSR* 12.6 (1970): 1497-1511.
And the enthalpy of solution of $\ce{LiNO_3}$ has been shown to change from negative to positive as the concentration increases. Reference:
Wolf, G., V. Vacek, and V. Pekarek. "Enthalpies of solution and of crystallization of lithium nitrate and of lithium nitrate trihydrate in water at 25° C." *Journal of Solution Chemistry* 19.10 (1990): 1029-1039.
|
Yes, because $\Delta H$ is temperature-, pressure-, and concentration-dependent, there exist chemical reactions, and physical changes, in which its sign reverses as a result of changes in these intensive properties.
NIST has a detailed Shomate Equation that has been experimentally verified to model the standard* heat of formation ($\Delta H^o_f$) of $\ce{N_2O_4_{(\textit{l})}}$ from 298 K to 500 K (https://webbook.nist.gov/cgi/cbook.cgi?ID=C10544726&Mask=1E9F). If you plug in the parameters, you will see that $\Delta H^o_f (298 K) = -19.56 kJ/mol$, while $\Delta H^o_f (500 K) = +12.87 kJ/mol$.
[*Standard state, as indicated by the circle superscript, means the pressure is held at 1 bar. It places no restrictions on the temperature.]
There are also direct experimental measurements showing that enthalpies of solution change sign in response to both temperature and concentration changes.
For instance, the enthalpy of solution of polystyrene in decalin has been shown to change from ~ –10 J/g at ~20 C to ~ +6 J/g at ~100 C (the ~'s were needed because I'm eyeballing these from a plot). Reference:
Tager, A. A., et al. "The temperature and concentration dependence of enthalpy, free energy and entropy of mixing polystyrene solutions with an upper and lower critical temperature of mixing." *Polymer Science USSR* 12.6 (1970): 1497-1511.
And the enthalpy of solution of $\ce{LiNO_3}$ has been shown to change from negative to positive as the concentration increases. Reference:
Wolf, G., V. Vacek, and V. Pekarek. "Enthalpies of solution and of crystallization of lithium nitrate and of lithium nitrate trihydrate in water at 25° C." *Journal of Solution Chemistry* 19.10 (1990): 1029-1039.
|
Research has recently found that FeOOH is 25 times better at hydrogen production than titanium dioxide.
Today, a new electrolysis plant today delivers energy efficiency of around 80%. Is 25x the Honda-Fujishima reaction, using FeOOH higher efficiency than that?
ref:
https://www.tus.ac.jp/en/mediarelations/archive/20200227001.html |
Efficiency of H production from FeOOH catalist? |
I am trying to make my own hand sanitiser since it’s sold out everywhere. The formular I use is 1 part aloe vera gel, 2 parts 100% isopropyl alcohol. The problem is when mixing the 2, the solution always forms a small white spongy sticky substance while the rest stays liquid, not gel like:
[![enter image description here][1]][1]
I tried 2 brands of aloe vera gel and got the same result:
- Palmer Aloe vera gel: Aloe Vera Gel (Aloe Barbadensis), Polyglycerylmethacrylate, Propylene Glycol, Witch Hazel, Algae Extract (Algae), PPG-2-Isodeceth-12, Carbomer, Tocopheryl Acetate, Triethanolamine, Methyl Gluceth-20, DMDM Hydantoin, Tetrasodium EDTA, Quaternium-15, Fragrance (Parfum), Hexyl Cinnamal, Linalool, Geraniol, Limonene, Yellow 5 (CI 19140), Blue 1 (CI 42090).
- Thursday Plantation Aloe Vera Gel: Aloe Vera inner leaf juice equiv. to fresh 974mg/g, Hydroxybenzoates, Diazolidinylurea, Potassium Sorbate and Sodium Benzoate
I am starting to think I got the wrong type of alcohol but the label clearly says pure 100% isopropyl alcohol.
What am I doing wrong?
[1]: https://i.stack.imgur.com/I553U.jpg |
I am trying to make my own hand sanitiser since it’s sold out everywhere. The formula I use is 1 part aloe vera gel in 2 parts 100% isopropyl alcohol. The problem is when mixing the two, the solution always forms a small white spongy sticky substance while the rest stays liquid, not gel like:
[![container holding greenish solution with white substance visible at the bottom][1]][1]
I tried two brands of aloe vera gel and got the same result:
- Palmer Aloe vera gel: Aloe Vera Gel (Aloe Barbadensis), Polyglycerylmethacrylate, Propylene Glycol, Witch Hazel, Algae Extract (Algae), PPG-2-Isodeceth-12, Carbomer, Tocopheryl Acetate, Triethanolamine, Methyl Gluceth-20, DMDM Hydantoin, Tetrasodium EDTA, Quaternium-15, Fragrance (Parfum), Hexyl Cinnamal, Linalool, Geraniol, Limonene, Yellow 5 (CI 19140), Blue 1 (CI 42090).
- Thursday Plantation Aloe Vera Gel: Aloe Vera inner leaf juice equiv. to fresh 974mg/g, Hydroxybenzoates, Diazolidinylurea, Potassium Sorbate and Sodium Benzoate
I am starting to think I got the wrong type of alcohol but the label clearly says pure 100% isopropyl alcohol.
What am I doing wrong?
[1]: https://i.stack.imgur.com/I553U.jpg |
I am trying to make my own hand sanitiser since it’s sold out everywhere. The formula I use is 1 part aloe vera gel in 2 parts 100% isopropyl alcohol. The problem is when mixing the two, the solution always forms a small white spongy sticky substance while the rest stays liquid, not gel-like:
[![container holding greenish solution with white substance visible at the bottom][1]][1]
I tried two brands of aloe vera gel and got the same result:
- Palmer Aloe vera gel: Aloe Vera Gel (Aloe Barbadensis), Polyglycerylmethacrylate, Propylene Glycol, Witch Hazel, Algae Extract (Algae), PPG-2-Isodeceth-12, Carbomer, Tocopheryl Acetate, Triethanolamine, Methyl Gluceth-20, DMDM Hydantoin, Tetrasodium EDTA, Quaternium-15, Fragrance (Parfum), Hexyl Cinnamal, Linalool, Geraniol, Limonene, Yellow 5 (CI 19140), Blue 1 (CI 42090).
- Thursday Plantation Aloe Vera Gel: Aloe Vera inner leaf juice equiv. to fresh 974mg/g, Hydroxybenzoates, Diazolidinylurea, Potassium Sorbate and Sodium Benzoate
I am starting to think I got the wrong type of alcohol but the label clearly says pure 100% isopropyl alcohol.
What am I doing wrong?
[1]: https://i.stack.imgur.com/6MGBi.jpg |
I am starting be re-learn Chemistry after several years.
The Book I am using has a test question to calculate the relative atomic mass of:
$ (CH_3CO)_2O $
My take
Carbon C, $ A_r = 56 $ <br>
Hydrogen H, $ A_r = 1 $ <br>
Oxygen O, $ A_r = 16 $
$ M_r = ( 56 + 1 \times 3 + 56 + 16 ) \times 2 + 16 = 131 \times 2 + 16 = 278 $
This is nothing like the answer given in the answer key. Is there an error in my text book?
If not where am I going wrong? |
I am starting be re-learn Chemistry after several years.
The Book I am using has a test question to calculate the relative atomic mass of:
$ (CH_3CO)_2O $
My take
Carbon C, $ A_r = 56 $ <br>
Hydrogen H, $ A_r = 1 $ <br>
Oxygen O, $ A_r = 16 $
$ M_r = ( 56 + 1 \times 3 + 56 + 16 ) \times 2 + 16 = 131 \times 2 + 16 = 278 $
This is nothing like the answer given in the answer key. Is there an error in my text book?
If not where am I going wrong?
---
**Edit:** Problem was a misprint in my text book. See correct answer below |
In P/V graphs of Irreversible processes only initial and final state is defined due to sudden compression or rarefaction then how is Work done a path function in Irreversible processes.
[![Irreversible process][2]][2]
[![Isothermal irreversible expansion PV graph][1]][1]
Why won't work done by the process here be only dependent on initial and final states?
[1]: https://i.stack.imgur.com/DT8pu.png
[2]: https://i.stack.imgur.com/VIXkt.png |
**Water loses kinetic energy when it detaches from the surface of the ice**
Water molecule in the solid phase have more hydrogen bonds than in the liquid phase. It takes energy to break these bonds, so only molecules with sufficient kinetic energy will detach (maybe because they just collided with a water molecule that transferred some kinetic energy to it). So the molecule that detaches will have a low kinetic energy as it goes into the liquid phase. Further collisions will quickly ensure that over time, this particular molecule has an average kinetic energy, but overall the temperature will drop from that detaching event.
**Water gains kinetic energy when it attaches to the surface of the ice**
This is just the opposite process, so the opposite happens. As additional hydrogen bonds are formed, the molecule has greater kinetic energy (in the form of vibrational energy). Either the molecule had less than average kinetic energy and now has average energy again, or it had average kinetic energy and will have higher kinetic energy, but quickly dissipate this to other molecules.
**When you add salt to the liquid water, less molecules attach to the surface of the ice**
Compared to pure water, less collisions of particles in the solution with the surface of the ice will lead to attachment of particles because now, some of these particles are not water molecules but sodium or chloride ions instead (they are not incorporated into the crystal lattice of ice but stay solvated by liquid water). As a consequence, there are less attachment events while the detachment events stay the same. With a net melting of ice, the average kinetic energy of particles decreases (the temperature drops). As a secondary effect, this also affects the rates of attachment and detachment, so once the temperature has sufficiently dropped, the two rates will be the same again.
>What energies (kinetic/temperature vs. latent/chemical) are involved, and where?
In my explanation, I talked about kinetic energy (reflecting the temperature) and energy necessary to break hydrogen bonds (could be generalized as intermolecular interactions, and they underlie the observed bulk latent heat).
>How exactly (atomically or subatomically) is one type of energy converted into the other
There is no chemistry or nuclear physics happening, so it is sufficient to talk about the molecular level. Kinetic energy is converted into potential energy through inelastic collisions. Kinetic energy is redistributed between molecules through elastic collisions.
>Why is each step energetically favourable? Or is it not, and it's reversible? How reversible, statistically?
According to microscopic reversibility, all molecular events are reversible. Processes go in a certain direction because it is more probable than in the other direction (2nd law of thermodynamics). The 2nd law of thermodynamics is inherently linked to large numbers of particles, so you need to look at the forest rather than at individual trees.
>How does the concentration of salt change the evolution of the system (again, at a microscopic level).
The higher the concentration of the salt, the lower the concentration of the water in the liquid phase, i.e the fewer events of liquid water attaching to the ice.
>How does adding e.g. ethanol instead of salt change things?
1 mmol/L NaCl has twice the dissolved particles compared to 1 mmol/L ethanol. Otherwise, the effect is the same (colligative property).
> Is liquid water required at all? Is salt able to melt and cool completely frozen ice, or is external force required (stirring, driving over it, etc.) to cause some initial melting for anything to happen? (I've read conflicting answers to this in the past.)
Ice near the freezing point exhibit surface melting. Once there is a tiny amount of water, more ice will melt. On the other hand, if it is cold enough that even saturated solutions will freeze, a mixture of dry salt and ice will not interact. |
Once a part of a peptide an amino acid is no longer an amino acid, so how do we refer to the part of the peptide that was once amino acid x? |
How does one refer to the amino acids in a peptide? |
In $P/V$ plots of irreversible processes only initial and final state is defined due to sudden compression or rarefaction then how is work done a path function in irreversible processes.
[![Irreversible process][2]][2]
[![Isothermal irreversible expansion PV graph][1]][1]
Why won't work done by the process here be only dependent on initial and final states?
[1]: https://i.stack.imgur.com/DT8pu.png
[2]: https://i.stack.imgur.com/VIXkt.png |
Is irreversible work a path function? |
This fact was given as an explanation as to why branched isomers of a compound has lower boiling points.
How are branched isomers less polarisable and how does this relate to London dispersion forces? |
Why are branched isomers less polarisable? |
This fact was given as an explanation as to why branched isomers of a compound have lower boiling points.
How are branched isomers less polarisable and how does this relate to London dispersion forces? |
I am doing an experiment that aims to determine if the rate of electroplating changes with the temperature. I am using a copper rod, a solution of copper(II) ions and a neutral rod to be coated. I also want to keep every other variable the same, which means the voltage across the rods, the concentration of the solution and the current that goes through the circuit. But the times my professor tried to demonstrate electrolysis, either the current or the voltage was always dropping. How do I prevent this from happening? How can I ensure that the voltage and current remain unchanged for the entire duration of a trial? |
I've encountered in chemistry quite a lot of reducing agents. However one question strikes me quite often and any research done on the same hasn't been fruitful. This is in context to organic chemistry. Now for example, in reductive ozonolysis, we often use ozone in combination with Zinc and water. Zinc's presence ensures reductive results and not oxidative ones. Now, some sources(not books) which I have seen, say that some sulphur compounds would do the same like Me2S. Also, incase of 1,3- dinitrobenzene, Sn+HCl converts both the NO2 groups to NH2 groups. Now if we use NH4HS or some other nitrogen and sulphur containing ammonia like compounds, we get reduction again, but only for one group out of the two. So my question is, how do we :
1. Decide that there is going to be reduction
2. Know the extent of reduction (partial/full)
Can some insight into the inorganic properties of the elements provide an answer like d-orbital availability, lone pair, electronegativity, size/charge ratio?
I've tried reasoning but haven't gotten anywhere. Or is it that memorization of the reagents is the only way to know it's function.
Note: This question is only in context to some less commonly used reagents, so as to recognise it's function if we ever see one in problems present in question books. |
For my chemistry investigation, I used Stamm's potassium permanganate method to investigate tea tannins with potassium permanganate to find the tannin content of teas. The method assumes complete oxidation of tea polyphenols to carbon dioxide.
Assuming that the tea polyphenols consist of only theaflavins, $\ce{C29H24O12},$ and Epigallocatechin gallate (EGCG), $\ce{C22H18O11},$ could someone help me with finding the equation of the reaction between the listed polyphenols and potassium permanganate respectively, i.e. theaflavin reaction with potassium permamganate and EGCG reaction with potassium permanganate (preferably 2 equations for each redox, but 1 overall equation should be fine too)?
I tried looking for these everywhere, but no source seems to give me the equations I need. |
I have a question about the toxicity of the pigments Cadmium Selenide (Cadmium Red), Cadmium Sulfide (Cadmium Yellow), Mercury Sulfide (Cinnabar or Vermillion) and Lead Oxide (Lead White).
Not long ago the EU banned lead-white pigments from use, even for art. It is now generally impossible to get hold of lead-white oil paint on the internet. However, the cadmium pigments were spared a ban recently. And it is quite easy to find Cinnabar and Vermillion oil paint online even of European production. For example, the dutch company Royal Talens supplies Vermillion paint online, without even warning that it contains mercury: https://www.royaltalens.com/en/catalog/rembrandt-oil-colour/oil-colour-paint-vermillion-311-40ml-tube/
I am not a chemist but my limited knowledge tells me that mercury is a lot more poisonous than lead. How come Cinnabar wasn't banned and what is the relative toxicity of the four pigments I listed, given that they are suspended in oil?
Thank you very much. |
How to use Google’s Alpha-fold to predict the binding of two proteins? |
An **exothermic reaction** occurs when the temperature of a system increases due to the evolution of heat. This heat is released
into the surroundings, resulting in an overall *negative* quantity for the heat of reaction (-ΔE ).
An **endothermic reaction**
occurs when the temperature of an isolated system decreases while the surroundings of a non-isolated system gains heat.
Endothermic reactions result in an overall *positive* heat of reaction (+ΔE ).
Exothermic and endothermic reactions cause energy level differences and therefore differences in enthalpy (ΔE ), the sum of
all potential and kinetic energies.
***ΔH is determined by the system, not the surrounding environment in a reaction.***
A system
that releases heat to the surroundings, an **exothermic reaction**, has a *negative* ΔH by convention, because the enthalpy of the
products is lower than the enthalpy of the reactants of the system.
The enthalpies of these reactions are *less than zero,* and are therefore **exothermic reactions.**
**A system of reactants that absorbs
heat from the surroundings in an *endothermic reaction* has a *positive* , because the enthalpy of the products is higher than
the enthalpy of the reactants of the system.
Because the enthalpies of these reactions are greater than zero, they are endothermic reactions.**
***EDIT ( See this )***
I understand your confusion. See, in endothermic reaction the product will be "colder"....this might seem counterintuitive as in endothermic reaction heat is absorbed by the system form the environment. *Therefore the heat absorbed must increase the temperature right?*
No! See temperature shows the direction of heat flow. If the temperature of the system increased ( *assuming that previously the system and environment were in thermal equilibrium ~ same temperature* ) this would mean that heat should flow from system to the environment as the system has a higher temperature than environment.
**The following example makes everything clear:**
$$Temp_{(env)}= 25°$$
$$A + B _{(25°)} = C _{(20°)} ...... (1)$$
$$A' + B' _{(25°)} = C' _{(30°)} ..... (2)$$
**Compare the temperature of system and environment in the above rxn.**
Therefore in rxn. 1 heat flows from environment to system ( from 25° to ~20°).
This heat energy is used to break bonds and stuff.( Endothermic )
And in rxn. 2 heat flows from system to environment ( from 30° to 25°).( Exothermic )
**Also see this practical video:**
https://youtu.be/NHOHKH3808c |
How to use Google’s Alpha-fold to predict the complex crarted by two proteins? |
How to use Google’s Alpha-fold to predict the complex formed from two proteins? |
Are you asking how to use code developed to find the MFE (minimum free energy) structure of a single protein to determine the sturcture of a two-protein complex? Or are you asking if you can use the folding of a concatenated protein as a surrogate to determine the structure of the two-protein complex?
If you're asking the latter, that approach is problematic, for two reasons:
1) Folding the concatenated protein would constrain the last amino acid in the first protein to be next to the first acid in the second protein. That's not a constraint that would make sense a priori for a two-protein complex.
2) You have to ask yourself if, in forming the two-protein complex, you envision both proteins unfolding and then refolding to do a global energy minimization— since that's what folding the concatenated protein would give you (with the additional constrain mentioned in point #1, above). If, on the other hand, you woud like the proteins to each fold into their respective MFE structures, and then (while essentially retaining those structures) find the MFE configuration for the two-protein complex, then folding the concatenated protein clearly would not give you that. |
An **exothermic reaction** occurs when the temperature of a system increases due to the evolution of heat. This heat is released
into the surroundings, resulting in an overall *negative* quantity for the heat of reaction (-ΔE ).
An **endothermic reaction**
occurs when the temperature of an isolated system decreases while the surroundings of a non-isolated system gains heat.
Endothermic reactions result in an overall *positive* heat of reaction (+ΔE ).
Exothermic and endothermic reactions cause energy level differences and therefore differences in enthalpy (ΔE ), the sum of
all potential and kinetic energies.
***ΔH is determined by the system, not the surrounding environment in a reaction.***
A system
that releases heat to the surroundings, an **exothermic reaction**, has a *negative* ΔH by convention, because the enthalpy of the
products is lower than the enthalpy of the reactants of the system.
The enthalpies of these reactions are *less than zero,* and are therefore **exothermic reactions.**
**A system of reactants that absorbs
heat from the surroundings in an *endothermic reaction* has a *positive* , because the enthalpy of the products is higher than
the enthalpy of the reactants of the system.
Because the enthalpies of these reactions are greater than zero, they are endothermic reactions.**
***EDIT ( See this )***
I understand your confusion. See, in endothermic reaction the product will be "colder"....this might seem counterintuitive as in endothermic reaction heat is absorbed by the system form the environment. *Therefore the heat absorbed must increase the temperature right?*
No! See temperature shows the direction of heat flow. If the temperature of the system increased ( *assuming that previously the system and environment were in thermal equilibrium ~ same temperature* ) this would mean that heat should flow from system to the environment as the system has a higher temperature than environment.
**The following example makes everything clear:**
$$T_{(env)}= 25°$$
$$A + B _{(25°)} = C _{(20°)} ...... (1)$$
$$A' + B' _{(25°)} = C' _{(30°)} ..... (2)$$
**Compare the temperature of system and environment in the above rxn.**
Therefore in rxn. 1 heat flows from environment to system ( from 25° to ~20°).
This heat energy is used to break bonds and stuff.( Endothermic )
And in rxn. 2 heat flows from system to environment ( from 30° to 25°).( Exothermic )
**Also see this practical video:**
https://youtu.be/NHOHKH3808c |
Are you asking how to take code developed to find the MFE (minimum free energy) structure of a single protein, and modify it to instead determine the structure of a two-protein complex?
Or are you asking if you can use this code, as is, to fold a concatenated combination of the two proteins, and in turn use this result as a surrogate for the predicted structure of the two-protein complex?
If you're asking the latter, that approach is problematic, for two reasons:
1) Folding the concatenated protein would constrain the last amino acid in the first protein to be next to the first acid in the second protein. That's not a constraint that would make sense a priori for a two-protein complex.
2) You have to ask yourself if, in forming the two-protein complex, you envision both proteins unfolding and then refolding to do a global energy minimization— since that's what folding the concatenated protein would give you (with the additional constrain mentioned in point #1, above). If, on the other hand, you woud like the proteins to each fold into their respective MFE structures, and then (while essentially retaining those structures) find the MFE configuration for the two-protein complex, then folding the concatenated protein clearly would not give you that. |
How to use Google’s Alpha-fold to predict the complex formed from two proteins? |
Are you asking how to take code developed to find the MFE (minimum free energy) structure of a single protein, and modify it to instead determine the structure of a two-protein complex?
Or are you asking if you can use this code, as is, to fold a concatenated combination of the two proteins, and in turn use this result as a surrogate for the predicted structure of the two-protein complex?
If you're asking the former, I would recommend contacting one of the code developers to ask them for guidance. It's possible that they have already developed a prototype to address this.
If you're asking the latter, that approach is problematic, for two reasons:
1) Folding the concatenated protein would constrain the last amino acid in the first protein to be next to the first acid in the second protein. That's not a constraint that would make sense a priori for a two-protein complex.
2) You have to ask yourself if, in forming the two-protein complex, you envision both proteins unfolding and then refolding to do a global energy minimization— since that's what folding the concatenated protein would give you (with the additional constrain mentioned in point #1, above). If, on the other hand, you woud like the proteins to each fold into their respective MFE structures, and then (while essentially retaining those structures) find the MFE configuration for the two-protein complex, then folding the concatenated protein clearly would not give you that. |
We installed 20 2" drop in anchors to anchor several signs in place near the entrance of our pool. We have to remove the signs about twice a year and after the end of the past season, I noticed there is a small amount of rust forming. I purchased the 304 because it was not going to be in direct contact with the water and because it was solid, not coated, as opposed to the 314, which was only coated. It seems as though the 304 was also just coated so it is beginning to rust. I need to provide some sort of plan as to how to remove the 304 anchors and replace with 316.
Drop in anchors are not made to come out of concrete as the driving pin expands the bottom into place. I had thought about drilling them out but I am hoping to use that as a last resort because I don't want to damage our new concrete and need to keep the holes in tact. I read that Nitric acid can dissolve stainless steel. Will this work in my situation? Or will that also damage the concrete? |
I am studying haemoglobin absorption spectra in its oxygenated and deoxygenated state and I compiled values from different sources including:
- A PhD thesis by van Assendelft \[1\], using extinction coefficients in (mmol/L)-1/cm
- A book compiling almost one century of research on the topic by the same laboratory that helps standardizing the haemoglobin cyanide standard \[2\]. Extinction coefficients in the same unit as \[1\], numerical values are very close.
- Coefficients compiled by Scott Prahl\[3\]. The unit is the same but there is a factor 3.7 up to 4.0 between his data and that of Assendelft and Zijlstra.
- Coefficients measured by Kolyva\[4\].
All the published coefficients are defined as base 10 extinction coefficient. That is if *I*
is the light intensity, they are expressed in Beer-Lambert law as:
$I = I_0 \cdot 10^{-c\cdot \varepsilon \cdot l}$
Where $l$ is the light path length (cm), $c$ the concentration in absorbing compound (mol/L) and $\varepsilon$ the molar extinction coefficient (L/mol/cm).
**How comes there is a 3.7-4.0 factor between Prahl and Kolyva data on the one hand and Assendelft and Zijlstra on the other hand?**
At first I thought of some $log(10)\approx2.3$ difference, as if some of the author forgot to precise that they used base $e$ extinction coefficient at some point but it does not fit. If anyone has an idea on the topic, the help would be most appreciated.
----
**Further information**
The cyanmethaemoglobin method is often used for the determination of the haemoglobin concentration of a solution. In particular, its absorbance at **540nm** is known to be 11.00L/mmol/cm\[5\] which is in agreement with Assendelft and Zijlstra published data.
### References
1. O.W. Assendelft, *Spectrophotometry of Haemoglobin Derivatives*, **1970**. ISBN: 902320560X.
2. W.G. Zijlstra, A. Buursma and O.W. van Assendelft, *Visible and Near Infrared Absorption Spectra of Human and Animal Haemoglobin*, **2000**. ISBN: 9067643173.
3. Scott Prahl, *Optical Absorption of Hemoglobin*, **1999**, https://omlc.org/spectra/hemoglobin/
4. C. Kolyva, *Systematic investigation of changes in oxidized cerebral cytochrome c oxidase concentration during frontal lobe activation in healthy adults*, **2012**. DOI: [10.1364/BOE.3.002550](https://dx.doi.org/10.1364%2FBOE.3.002550). Coefficients are in [**MEDIA 1**](https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3469997/bin/boe-3-10-2550-s001.PDF).
5. F. Meng, *Determination of extinction coefficients of human hemoglobin in various redox states*, **2017**. DOI: [10.1016/j.ab.2017.01.002](https://dx.doi.org/10.1016%2Fj.ab.2017.01.002). Table 1. |
How to use Google’s Alpha-fold to predict the structure of a two-protein complex? |
> Given the structure below, sketch the $^{13}$C - spectrum in a $^{13}$C marked sample
> The data given is the chemical shifts (left) and the coupling constants (right)
> [![enter image description here][1]][1]
> [![enter image description here][2]][2]
**The correct spectrum is given at the end of this question**
I don't really understand what is meant by "in a $^{13}$C marked sample" however I assume that it means that I am supposed to look at C instead of H as my core atom. So from this I tried to determine the coupling. If looking at the α-C I assumed that its neighbouring H would split it into a doublet and then the hydrogens of the β-C (3+1 = 4) would split that doublet into an multiplet of eight. Which seems to be right if compared to the correct spectrum. Similarly, the β-C is first split into triplets and then into doublets, which makes 6 peaks.
However, why would C=O cause a peak? And how does it become a doublet?
[![enter image description here][3]][3]
[1]: https://i.stack.imgur.com/EhtcD.png
[2]: https://i.stack.imgur.com/MMPrk.png
[3]: https://i.stack.imgur.com/H3M7A.png |
Are you asking how to take code developed to find the MFE (minimum free energy) structure of a single protein, and modify it to instead determine the structure of a two-protein complex?
Or are you asking if you can use this code, as is, to fold a concatenated combination of the two proteins, and in turn use this result as a surrogate for the predicted structure of the two-protein complex?
If you're asking the former, I would recommend contacting one of the code developers to ask them for guidance. It's possible that they have already developed a prototype option to address this.
If you're asking the latter, that approach is problematic, for two reasons:
1) Folding the concatenated protein would constrain the last amino acid in the first protein to be next to the first amino acid in the second protein. That's not a constraint that would make sense a priori for a two-protein complex.
2) You have to ask yourself if, in forming the two-protein complex, you envision both proteins unfolding and then refolding to do a global energy minimization— since that's what folding the concatenated protein would give you (with the additional constraint mentioned in point #1, above). If, on the other hand, you would like the proteins to each fold into their respective MFE structures, and then (while essentially retaining those structures) find the MFE configuration for the two-protein complex, then folding the concatenated protein clearly would not give you that. |
How to use Google’s Alpha-fold to predict the structure of a two-protein complex? |
How to use Google’s Alpha-fold to predict a two-protein complex? |
Which would produce the largest change in the H2O level when added to water in a 25 mL graduated cylinder?
(A) 10.0 g of Hg (d=13.6g/mL)
(B) 7.42 g of Al (d=2.70g/mL)
(C) 5.09 g of iron pyrite (d=4.9g/mL)
(D) 2.68 g of oak (d=0.72g/mL) <br />
I approached this question by dividing the mass by density to find the volume change. Thus, I got the answer (D), but the answer is (B)?
Can anyone please explain this? |
Because oak has a density less than water (1g/mL), it won't be fully submerged. For substances more dense than water, they will be fully submerged so you don't have to worry about it. See [this video][1] to calculate how much of the oak will be submerged.
[1]: https://www.youtube.com/watch?time_continue=151&v=sMkHKMsBEl8&feature=emb_title |
> Which would produce the largest change in the $\ce{H2O}$ level when added to water in a $\pu{25 mL}$ graduated cylinder?
> (**A**) $\pu{10.0 g}$ of $\ce{Hg}$ $(d = \pu{13.6 g/mL})$
> (**B**) $\pu{7.42 g}$ of $\ce{Al}$ $(d = \pu{2.70 g/mL})$
> (**C**) $\pu{5.09 g}$ of iron pyrite $(d = \pu{4.9 g/mL})$
> (**D**) $\pu{2.68 g}$ of oak $(d = \pu{0.72 g/mL})$
I approached this question by dividing the mass by density to find the volume change. Thus, I got the answer (**D**), but the answer is (**B**).
Can anyone please explain this? |
## Below a more general approach.
Suppose that we have two weak acids $\ce{HA}$ and $\ce{HB}$.
The initial concentrations are $C^0_\ce{HA}$ and $C^0_\ce{HB}$, and their constants are $K_{\mathrm{a},\ce{(HA)}}$ and $K_{\mathrm{a},\ce{(HB)}}$.
Suppose yet that volumes, $V_\ce{HA}$ and $V_\ce{HB}$, are additives.
So we have:
1. *Reactions*
\begin{align}
\ce{HA + H2O &<=> H3O+ + A-} &
K_{\mathrm{a},(\ce{HA})}
&= \frac{\ce{[H3O+][A-]}}{\ce{[HA]}}
\tag{1}\label{eq:KAcidHA}\\
\ce{HB + H2O &<=> H3O+ + B-} &
K_{\mathrm{a},(\ce{HB})}
&=\frac{\ce{[H3O+][B-]}}{\ce{[HB]}}
\tag{2}\label{eq:KAcidHB}\\
\ce{2 H2O &<=> H3O+ + OH-} &
K_\mathrm{w} &= \ce{[H3O+][OH-]}\tag{3}\label{eq:KWater}
\end{align}
2. *Mass balance*
\begin{align}
C_\ce{HA} &= \frac{C^0_\ce{HA} V_\ce{HA}}{V_\ce{HA} + V_\ce{HB}}
&&=\ce{[HA] + [A-]}\tag{4}\label{eq:MassBalanceHA}\\
C_\ce{HB} &= \frac{C^0_\ce{HB} V_\ce{HB}}{V_\ce{HA} + V_\ce{HB}}
&&=\ce{[HB] + [B-]}\tag{5}\label{eq:MassBalanceHB}
\end{align}
3. *Charge balance*
$$\ce{[H3O+] = [OH-] + [A-] + [B-]}\tag{6}\label{eq:ChargeBalance}$$
Replacing \eqref{eq:KAcidHA}–\eqref{eq:MassBalanceHB} equations on \eqref{eq:ChargeBalance}, we have:
$$
\ce{[H3O+]} =
\frac{K_\mathrm{w}}{\ce{[H3O+]}}
+\frac{C_\ce{HA} K_{\mathrm{a},(\ce{HA})}}{\ce{[H3O+]}
+ K_{\mathrm{a},(\ce{HA})}}
+ \frac{C_\ce{HB} K_{\mathrm{a},(\ce{HB})}}{\ce{[H3O+]}
+ K_{\mathrm{a},(\ce{HB})}}
\tag{7}\label{eq:GeneralEquation}$$
or as polynomial
\begin{align}
\begin{split}
&\ce{[H3O+]}^4\\
+&\ce{[H3O+]}^3 (K_{\mathrm{a},(\ce{HA})}
+ K_{\mathrm{a},(\ce{HB})})\\
+&\ce{[H3O+]}^2 \left[
K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})}
- ( C_\ce{HA} K_{\mathrm{a},(\ce{HA})}
+ C_\ce{HB}K_{\mathrm{a},(\ce{HB})} )
- K_\mathrm{w} \right]\\
-&\ce{[H3O+]} \left[
(C_\ce{HA} + C_\ce{HB}) K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})}
+ K_\ce{w} (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})}) \right]\\
-& K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})} K_\mathrm{w}\\
=&\ 0
\end{split}\tag{8}\label{eq:GeneralPol}
\end{align}
This single equation will exactly solve any equilibrium problem involving the mixture of any two monoprotic acids, in any concentration (as long as they're not much higher than about $\pu{1 mol L-1}$) and any volume. Depending of $K_\mathrm{a}$, $K_{\mathrm{a},(\ce{HB})}$, $C^0_\ce{HA}$, and $C^0_\ce{HB}$ values, we can yet obtain a simpler version<sup>\[1\]</sup>.
The \eqref{eq:GeneralPol} equation can simplified considering that $K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})} K_\mathrm{w} \ll 1$:
\begin{align}
\begin{split}
&\ce{[H3O+]}^3\\
+&\ce{[H3O+]}^2 (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})\\
+&\ce{[H3O+]}\left[
K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})}
- (C_\ce{HA} K_{\mathrm{a},(\ce{HA})}
+ C_\ce{HB} K_{\mathrm{a},(\ce{HB})})
- K_\mathrm{w} \right]\\
-&\left[
(C_\ce{HA} + C_\ce{HB}) K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})}
+ K_\mathrm{w} (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})
\right]\\
=&\ 0
\end{split}\tag{9}\label{eq:GeneralPolSimp1}
\end{align}
The \eqref{eq:GeneralPolSimp1} equation can simplified considering that $K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})} \ll 1$:
\begin{align}
\ce{[H3O+]}^2 + \ce{[H3O+]} (K_{\mathrm{a},(\ce{HA})}
+ K_{\mathrm{a},(\ce{HB})})
- (C_\ce{HA} K_{\mathrm{a},(\ce{HA})}
+ C_\ce{HB} K_{\mathrm{a},(\ce{HB})}+K_\mathrm{w})
= 0\tag{10}\label{eq:GeneralPolSimp2}
\end{align}
The \eqref{eq:GeneralPolSimp2} equation can be solved as usual.
$$\ce{[H3O+]} =
\frac{
- (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})
+ \sqrt{
(K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})^2
+ 4 (C_\ce{HA} K_{\mathrm{a},(\ce{HA})}
+ C_\ce{HB} K_{\mathrm{a},(\ce{HB})}+K_\mathrm{w})}
}{2}$$
Or using the initial concentrations
$$\ce{[H3O+]} =
\frac{
- (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})
+ \sqrt{
(K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})^2
+ 4\left(%\displaystyle
\frac{C^0_\ce{HA} V_\ce{HA} K_{\ce{a},(\ce{HA})}
}{V_\ce{HA} + V_\ce{HB}}
+ \frac{C^0_\ce{HB} V_\ce{HB} K_{\ce{a},(\ce{HB})}
}{V_\ce{HA} + V_\ce{HB}}
+K_\mathrm{w}\right)}
}{2}$$
Replacing $C^0_\ce{HA}=\pu{0.01 mol L-1}$, $C^0_\ce{HB}=\pu{0.01 mol L-1}$, $V_\ce{HA}=\pu{0.050 L}$ and $V_\ce{HB}=\pu{0.050 L}$, and using $\text{p}K_\ce{a}=3.75$ for formic acid and $\text{p}K_\ce{a}=4.756$ for acetic acid<sup>\[2\]</sup>, we have
$$\ce{pH}=3.05.$$
**Addendum**
[@MaxW](https://chemistry.stackexchange.com/users/22102/maxw) [made some warnings](https://chemistry.stackexchange.com/questions/10249/how-to-determine-the-ph-of-a-mixture-of-two-weak-acids/120139?noredirect=1#comment248386_120139) about the assumptions which simplificate ($\ref{eq:GeneralPol}$) and ($\ref{eq:GeneralPolSimp1}$) equations.
To completeness and with the aim of show the limitations of present approach, consider the following graphics.
The ($\ref{eq:GeneralPol}$), ($\ref{eq:GeneralPolSimp1}$), and ($\ref{eq:GeneralPolSimp2}$) equations were solved using $\text{p}K_{\mathrm{a},(\ce{HA})}=\text{p}K_{\mathrm{a},(\ce{HB})}$, $C^0_\ce{HA}=C^0_\ce{HB}$, and $V_\ce{HA}=V_\ce{HB}=\pu{50.0 mL}$.
[![C0 = 1.0E0 molperL][1]][1]
[![C0 = 1.0E-1 molperL][2]][2]
[![C0 = 1.0E-2 molperL][3]][3]
[![C0 = 1.0E-3 molperL][4]][4]
[![C0 = 1.0E-4 molperL][5]][5]
[![C0 = 1.0E-5 molperL][6]][6]
[![C0 = 1.0E-6 molperL][7]][7]
[![C0 = 1.0E-7 molperL][8]][8]
As we can see, the ($\ref{eq:GeneralPolSimp1}$) equation can be used on any $\text{p}K_\ce{a}$ values, inasmuch as the concentration be major of $\pu{1E-6 mol L-1}$.
The ($\ref{eq:GeneralPolSimp2}$) equation becomes problematic when dilute solutions or strong acids solutions are used.
Thus, when applied to book-examples, which $\text{p}K_\ce{a}$ values from 3 to 6, and the concentrations from $1.0$ to $\pu{1.0E-4 mol L-1}$, the assumptions are valid and the equations can be used, providing good approximations of $\ce{pH}$ values.
**Acknowledgement**
Thanks to [@MaxW for draw attention to limitations of assumptions](https://chemistry.stackexchange.com/questions/10249/how-to-determine-the-ph-of-a-mixture-of-two-weak-acids/120139?noredirect=1#comment248386_120139).
**Reference**
1. Edited based on [@MaxW's](https://chemistry.stackexchange.com/users/22102/maxw) comment.
2. [Haynes, W. M.; Lide, D. R.; Bruno, T. J., *CRC Handbook of Chemistry and Physics*. 97 ed.; CRC Press: 2017; pp. 5–88][9].
[1]: https://i.stack.imgur.com/sDu7E.png
[2]: https://i.stack.imgur.com/NhdnS.png
[3]: https://i.stack.imgur.com/2Hi2v.png
[4]: https://i.stack.imgur.com/bQwkU.png
[5]: https://i.stack.imgur.com/4ktov.png
[6]: https://i.stack.imgur.com/KoWqw.png
[7]: https://i.stack.imgur.com/TnTfZ.png
[8]: https://i.stack.imgur.com/gKlM0.png
[9]: https://books.google.com.br/books?id=VVezDAAAQBAJ&lpg=PP1&hl=pt-BR&pg=SA5-PA88#v=onepage&q&f=false |
Cheap and effective path [to attack iron is employing hypochlorous acid](https://books.google.com/books?id=aZQ3AAAAMAAJ&pg=PA379&lpg=PA379&dq=hypochlorous+acid+attacking+iron&source=bl&ots=sCyzpeACQ5&sig=ACfU3U1cLGmDv8kV2L9CB1LMLAn0y7RBjg&hl=en&ppis=_c&sa=X&ved=2ahUKEwiqkKC9_Y_oAhVHTt8KHTWTAYwQ6AEwEXoECAoQAQ#v=onepage&q=hypochlorous%20acid%20attacking%20iron&f=false) (prepared here by mixing chlorine bleach containing dilute aqueous NaOCl with 6% acetic acid in vinegar in excess to the NaOCl). Add a source of copper (coins) or carbon, serving as a cathode in this spontaneous electrochemical (basically a battery) cell. Source: [Equation (7) here](http://www.exo.net/~pauld/saltwater/.
Note: This battery cell differs mainly in chemistry from the bleach battery cell by employing iron (steel) as the anode which will be the subject of considerable anodic corrosion in place of Aluminum.
Problematic fumes so an outdoor reaction is strongly advised. Neutralize solution products with Washing Soda at the end of the reaction. |
Cheap and effective path [to attack iron is employing hypochlorous acid](https://books.google.com/books?id=aZQ3AAAAMAAJ&pg=PA379&lpg=PA379&dq=hypochlorous+acid+attacking+iron&source=bl&ots=sCyzpeACQ5&sig=ACfU3U1cLGmDv8kV2L9CB1LMLAn0y7RBjg&hl=en&ppis=_c&sa=X&ved=2ahUKEwiqkKC9_Y_oAhVHTt8KHTWTAYwQ6AEwEXoECAoQAQ#v=onepage&q=hypochlorous%20acid%20attacking%20iron&f=false) (prepared here by mixing chlorine bleach containing dilute aqueous NaOCl with 6% acetic acid in vinegar in excess to the NaOCl). Add a source of copper (coins) or carbon, serving as a cathode in this spontaneous electrochemical (basically a battery) cell. Source: [Equation (7) here](http://www.exo.net/~pauld/saltwater/).
Note: This battery cell differs mainly in chemistry from the bleach battery cell by employing iron (steel) as the anode which will be the subject of considerable anodic corrosion in place of Aluminum.
Mellor on Hypochlorous Acid in 'A comprehensive treatise on inorganic and theoretical chemistry, Volume 2:
"Iron filings immediately decompose hypochlorous acid with a brisk effervescence produced by the evolution of chlorine; the iron is partly oxidized and in part dissolved as chloride without the formation of any chlorate. A. J. Balard commented on this: "The greater number of other metallic substances do not decompose hypochlorous acid, and I am yet entirely ignorant of the cause of the peculiar behaviour of iron." P. Grouvelle passed chlorine through water with iron hydroxidelin suspension and a bleaching liquid along with ferric chloride was produced, and he found the liquid retained its bleaching properties after boiling for a quarter of an hour; but A. J. Balard failed to confirm this statement; he could not make ferric hypochlorite either (i) by the action of hypochlorous acid on iron hydroxide, for hypochlorous acid does not dissolve ferric oxide; or (ii) by the action of calcium hypochlorite on ferric sulphate, for calcium sulphate and ferric oxide are produced. Hence, adds A. J. Balard, "ferric hypochlorite cannot exist;" the results by P. Grouvelle are due to the formation of ferric chloride and hypochlorous acid in dil. soln.; when the mixture was heated, a portion of the acid distilled off, and the reaction which occurred with the cold soln. was reversed, for ferric oxide and chlorine were formed. Ferrous oxide is oxidized to ferric oxide by hypochlorous acid."
Problematic fumes so an outdoor reaction is strongly advised. Neutralize solution products with Washing Soda at the end of the reaction. |
Cheap and effective path [to attack iron is employing hypochlorous acid](https://books.google.com/books?id=aZQ3AAAAMAAJ&pg=PA379&lpg=PA379&dq=hypochlorous+acid+attacking+iron&source=bl&ots=sCyzpeACQ5&sig=ACfU3U1cLGmDv8kV2L9CB1LMLAn0y7RBjg&hl=en&ppis=_c&sa=X&ved=2ahUKEwiqkKC9_Y_oAhVHTt8KHTWTAYwQ6AEwEXoECAoQAQ#v=onepage&q=hypochlorous%20acid%20attacking%20iron&f=false) (prepared here by mixing chlorine bleach containing dilute aqueous NaOCl with 6% acetic acid in vinegar in excess to the NaOCl). Add a source of copper (coins) or carbon, serving as a cathode in this spontaneous electrochemical (basically a battery) cell. Source: [Equation (7) here](http://www.exo.net/~pauld/saltwater/).
Note: This battery cell differs mainly in chemistry from the bleach battery cell by employing iron (steel) as the anode which will be the subject of considerable anodic corrosion in place of Aluminum.
[Mellor on Hypochlorous Acid](https://books.google.com/books?pg=PA275&lpg=PA275&dq=ferric+hypochlorite&sig=zHHdZcLpiC-8m8HCOj-kix87YZ4&ei=11P7TcXXMdHTgQepiZneCw&ct=result&sqi=2&id=7XoGAQAAIAAJ&ots=R0bAIgamIM#v=onepage&q=ferric%20hypochlorite&f=false) in 'A comprehensive treatise on inorganic and theoretical chemistry, Volume 2:
>Iron filings immediately decompose hypochlorous acid with a brisk effervescence produced by the evolution of chlorine; the iron is partly oxidized and in part dissolved as chloride without the formation of any chlorate. A. J. Balard commented on this: "The greater number of other metallic substances do not decompose hypochlorous acid, and I am yet entirely ignorant of the cause of the peculiar behaviour of iron." P. Grouvelle passed chlorine through water with iron hydroxidelin suspension and a bleaching liquid along with ferric chloride was produced, and he found the liquid retained its bleaching properties after boiling for a quarter of an hour; but A. J. Balard failed to confirm this statement; he could not make ferric hypochlorite either (i) by the action of hypochlorous acid on iron hydroxide, for hypochlorous acid does not dissolve ferric oxide; or (ii) by the action of calcium hypochlorite on ferric sulphate, for calcium sulphate and ferric oxide are produced. Hence, adds A. J. Balard, "ferric hypochlorite cannot exist;" the results by P. Grouvelle are due to the formation of ferric chloride and hypochlorous acid in dil. soln.; when the mixture was heated, a portion of the acid distilled off, and the reaction which occurred with the cold soln. was reversed, for ferric oxide and chlorine were formed. Ferrous oxide is oxidized to ferric oxide by hypochlorous acid.
Problematic fumes so an outdoor reaction is strongly advised. Neutralize solution products with Washing Soda at the end of the reaction. |
Cheap and effective path [to attack iron is employing hypochlorous acid](https://books.google.com/books?id=aZQ3AAAAMAAJ&pg=PA379&lpg=PA379&dq=hypochlorous+acid+attacking+iron&source=bl&ots=sCyzpeACQ5&sig=ACfU3U1cLGmDv8kV2L9CB1LMLAn0y7RBjg&hl=en&ppis=_c&sa=X&ved=2ahUKEwiqkKC9_Y_oAhVHTt8KHTWTAYwQ6AEwEXoECAoQAQ#v=onepage&q=hypochlorous%20acid%20attacking%20iron&f=false) (prepared here by mixing chlorine bleach containing dilute aqueous NaOCl with 6% acetic acid in vinegar in excess to the NaOCl). Add a source of copper (coins) or carbon, serving as a cathode in this spontaneous electrochemical (basically a battery) cell. Source: [Equation (7) here](http://www.exo.net/~pauld/saltwater/).
Note: This battery cell differs mainly in chemistry from the bleach battery cell by employing iron (steel) as the anode which will be the subject of considerable anodic corrosion in place of Aluminum.
[Mellor on Hypochlorous Acid](https://books.google.com/books?pg=PA275&lpg=PA275&dq=ferric+hypochlorite&sig=zHHdZcLpiC-8m8HCOj-kix87YZ4&ei=11P7TcXXMdHTgQepiZneCw&ct=result&sqi=2&id=7XoGAQAAIAAJ&ots=R0bAIgamIM#v=onepage&q=ferric%20hypochlorite&f=false) in 'A comprehensive treatise on inorganic and theoretical chemistry, Volume 2:
>Iron filings immediately decompose hypochlorous acid with a brisk effervescence produced by the evolution of chlorine; the iron is partly oxidized and in part dissolved as chloride without the formation of any chlorate. A. J. Balard commented on this: "The greater number of other metallic substances do not decompose hypochlorous acid, and I am yet entirely ignorant of the cause of the peculiar behaviour of iron." P. Grouvelle passed chlorine through water with iron hydroxidelin suspension and a bleaching liquid along with ferric chloride was produced, and he found the liquid retained its bleaching properties after boiling for a quarter of an hour; but A. J. Balard failed to confirm this statement; he could not make ferric hypochlorite either (i) by the action of hypochlorous acid on iron hydroxide, for hypochlorous acid does not dissolve ferric oxide; or (ii) by the action of calcium hypochlorite on ferric sulphate, for calcium sulphate and ferric oxide are produced. Hence, adds A. J. Balard, "ferric hypochlorite cannot exist;" the results by P. Grouvelle are due to the formation of ferric chloride and hypochlorous acid in dil. soln.; when the mixture was heated, a portion of the acid distilled off, and the reaction which occurred with the cold soln. was reversed, for ferric oxide and chlorine were formed. Ferrous oxide is oxidized to ferric oxide by hypochlorous acid.
Problematic fumes so an outdoor reaction is strongly advised. Neutralize solution products with Washing Soda at the end of the reaction.
Here is a particular comment relative to [steel being corroded](https://www.texwipe.com/Content/Images/uploaded/documents/Technical-Data/Stainless_Steel_Bleach.pdf) in place of iron. |
Cheap and effective path [to attack iron is employing hypochlorous acid](https://books.google.com/books?id=aZQ3AAAAMAAJ&pg=PA379&lpg=PA379&dq=hypochlorous+acid+attacking+iron&source=bl&ots=sCyzpeACQ5&sig=ACfU3U1cLGmDv8kV2L9CB1LMLAn0y7RBjg&hl=en&ppis=_c&sa=X&ved=2ahUKEwiqkKC9_Y_oAhVHTt8KHTWTAYwQ6AEwEXoECAoQAQ#v=onepage&q=hypochlorous%20acid%20attacking%20iron&f=false) (prepared here by mixing chlorine bleach containing dilute aqueous NaOCl with 6% acetic acid in vinegar in excess to the NaOCl). Add a source of copper (coins) or high surface area carbon, serving as a cathode in this spontaneous electrochemical (basically a battery) cell. Source: [Equation (7) here](http://www.exo.net/~pauld/saltwater/).
Note: This battery cell differs mainly in chemistry from the bleach battery cell by employing iron (steel) as the anode which will be the subject of considerable anodic corrosion in place of Aluminum.
[Mellor on Hypochlorous Acid](https://books.google.com/books?pg=PA275&lpg=PA275&dq=ferric+hypochlorite&sig=zHHdZcLpiC-8m8HCOj-kix87YZ4&ei=11P7TcXXMdHTgQepiZneCw&ct=result&sqi=2&id=7XoGAQAAIAAJ&ots=R0bAIgamIM#v=onepage&q=ferric%20hypochlorite&f=false) in 'A comprehensive treatise on inorganic and theoretical chemistry, Volume 2:
>Iron filings immediately decompose hypochlorous acid with a brisk effervescence produced by the evolution of chlorine; the iron is partly oxidized and in part dissolved as chloride without the formation of any chlorate. A. J. Balard commented on this: "The greater number of other metallic substances do not decompose hypochlorous acid, and I am yet entirely ignorant of the cause of the peculiar behaviour of iron." P. Grouvelle passed chlorine through water with iron hydroxidelin suspension and a bleaching liquid along with ferric chloride was produced, and he found the liquid retained its bleaching properties after boiling for a quarter of an hour; but A. J. Balard failed to confirm this statement; he could not make ferric hypochlorite either (i) by the action of hypochlorous acid on iron hydroxide, for hypochlorous acid does not dissolve ferric oxide; or (ii) by the action of calcium hypochlorite on ferric sulphate, for calcium sulphate and ferric oxide are produced. Hence, adds A. J. Balard, "ferric hypochlorite cannot exist;" the results by P. Grouvelle are due to the formation of ferric chloride and hypochlorous acid in dil. soln.; when the mixture was heated, a portion of the acid distilled off, and the reaction which occurred with the cold soln. was reversed, for ferric oxide and chlorine were formed. Ferrous oxide is oxidized to ferric oxide by hypochlorous acid.
Problematic fumes so an outdoor reaction is strongly advised. Neutralize solution products with Washing Soda at the end of the reaction.
Here is a particular comment relative to [steel being corroded](https://www.texwipe.com/Content/Images/uploaded/documents/Technical-Data/Stainless_Steel_Bleach.pdf) in place of iron, without the benefit of a good high surface area cathode in diluted NaOCl. |
Cheap and effective path [to attack iron is employing hypochlorous acid](https://books.google.com/books?id=aZQ3AAAAMAAJ&pg=PA379&lpg=PA379&dq=hypochlorous+acid+attacking+iron&source=bl&ots=sCyzpeACQ5&sig=ACfU3U1cLGmDv8kV2L9CB1LMLAn0y7RBjg&hl=en&ppis=_c&sa=X&ved=2ahUKEwiqkKC9_Y_oAhVHTt8KHTWTAYwQ6AEwEXoECAoQAQ#v=onepage&q=hypochlorous%20acid%20attacking%20iron&f=false) (prepared here by mixing chlorine bleach containing dilute aqueous NaOCl with 6% acetic acid in vinegar in excess to the NaOCl). Add a source of copper (coins) or high surface area carbon, serving as a cathode in this spontaneous electrochemical (basically a battery) cell. Source: [Equation (7) here](http://www.exo.net/~pauld/saltwater/).
Note: This battery cell differs mainly in chemistry from the bleach battery cell by employing iron (steel) as the anode which will be the subject of considerable anodic corrosion in place of Aluminum.
[Mellor on Hypochlorous Acid](https://books.google.com/books?pg=PA275&lpg=PA275&dq=ferric+hypochlorite&sig=zHHdZcLpiC-8m8HCOj-kix87YZ4&ei=11P7TcXXMdHTgQepiZneCw&ct=result&sqi=2&id=7XoGAQAAIAAJ&ots=R0bAIgamIM#v=onepage&q=ferric%20hypochlorite&f=false) in 'A comprehensive treatise on inorganic and theoretical chemistry, Volume 2, however, this is NOT the electrochemical based dissolution suggested above:
>Iron filings immediately decompose hypochlorous acid with a brisk effervescence produced by the evolution of chlorine; the iron is partly oxidized and in part dissolved as chloride without the formation of any chlorate. A. J. Balard commented on this: "The greater number of other metallic substances do not decompose hypochlorous acid, and I am yet entirely ignorant of the cause of the peculiar behaviour of iron." P. Grouvelle passed chlorine through water with iron hydroxidelin suspension and a bleaching liquid along with ferric chloride was produced, and he found the liquid retained its bleaching properties after boiling for a quarter of an hour; but A. J. Balard failed to confirm this statement; he could not make ferric hypochlorite either (i) by the action of hypochlorous acid on iron hydroxide, for hypochlorous acid does not dissolve ferric oxide; or (ii) by the action of calcium hypochlorite on ferric sulphate, for calcium sulphate and ferric oxide are produced. Hence, adds A. J. Balard, "ferric hypochlorite cannot exist;" the results by P. Grouvelle are due to the formation of ferric chloride and hypochlorous acid in dil. soln.; when the mixture was heated, a portion of the acid distilled off, and the reaction which occurred with the cold soln. was reversed, for ferric oxide and chlorine were formed. Ferrous oxide is oxidized to ferric oxide by hypochlorous acid.
Problematic fumes so an outdoor reaction is strongly advised. Neutralize solution products with Washing Soda at the end of the reaction.
Here is a particular comment relative to [steel being corroded](https://www.texwipe.com/Content/Images/uploaded/documents/Technical-Data/Stainless_Steel_Bleach.pdf) in place of iron, without the benefit of a good high surface area cathode in diluted NaOCl. |
I don't know if I am overcomplicating things but I am having trouble understanding the difference between H-NMR with C$^{13}$ sample and C$^{13}$-NMR.
For example, if I have the molecule below and I am asked to sketch the H-NMR spectrum knowing that the carbons are C$^{13}$ and a C$^{13}$-NMR spectrum, what is the difference? For the C$^{13}$-spectrum I know that I should look at the neighbouring carbons and apply the $n+1$ rule and then look at the neighbouring hydrogens of the carbon of interest. For example, the C$^α$ has two neighbouring carbons so is split into a triplet and it also has one neighbouring hydrogen so those triplets are split into doublets making the C$^α$ show up as a multiplet of six.
[![enter image description here][2]][2]
However, if it is a H-spectrum with C$^{13}$ what do I look at? The correct spectrum is added below and the chemical shifts are:
[![enter image description here][1]][1]
So for H$^α$ I should get $2 *5 $ peaks. Which I get from H$^α$ having one carbon neighbour (spliting it into a doublet) and then having four neighbouring hydrogens (spliting each doublet into a multiplet of five). Is that the correct way of doing it? It works for H$^β$ as well but then for the H$^N$ am I only looking at the hydrogens again (so $n+1=1+1=2$ a doublet)?
[![enter image description here][3]][3]
[1]: https://i.stack.imgur.com/Hp2zv.png
[2]: https://i.stack.imgur.com/qR10d.png
[3]: https://i.stack.imgur.com/xF8p8.png |
Simple, 'dumb' question.... Is the kJ/mol number given in tables of bond strengths or bond dissociation energies , when it comes to double and triple bonds, just for the 'last' bond or for both (or all three) together? |
For BDEs of double and triple bonds, is number shown in tables for that second or third bond? Or all of them? |
Try this, apply a cheap and effective spray onto the [iron surface employing hypochlorous acid](https://books.google.com/books?id=aZQ3AAAAMAAJ&pg=PA379&lpg=PA379&dq=hypochlorous+acid+attacking+iron&source=bl&ots=sCyzpeACQ5&sig=ACfU3U1cLGmDv8kV2L9CB1LMLAn0y7RBjg&hl=en&ppis=_c&sa=X&ved=2ahUKEwiqkKC9_Y_oAhVHTt8KHTWTAYwQ6AEwEXoECAoQAQ#v=onepage&q=hypochlorous%20acid%20attacking%20iron&f=false) (prepared here by mixing chlorine bleach containing dilute aqueous NaOCl with 6% acetic acid in vinegar in excess to the NaOCl). Add a source of copper (coins), or better high surface area carbon particles in the spray, serving as a cathodic zone, in this spontaneous electrochemical (basically a battery) cell. Source: [Equation (7) here](http://www.exo.net/~pauld/saltwater/).
Note: This battery cell differs mainly in chemistry from the bleach battery cell by employing iron (steel) as the anode which will be the subject of considerable anodic corrosion in place of Aluminum. The cathode is high surface area carbon particles.
[Mellor on Hypochlorous Acid](https://books.google.com/books?pg=PA275&lpg=PA275&dq=ferric+hypochlorite&sig=zHHdZcLpiC-8m8HCOj-kix87YZ4&ei=11P7TcXXMdHTgQepiZneCw&ct=result&sqi=2&id=7XoGAQAAIAAJ&ots=R0bAIgamIM#v=onepage&q=ferric%20hypochlorite&f=false) in 'A comprehensive treatise on inorganic and theoretical chemistry, Volume 2, however, this is NOT the electrochemical based dissolution suggested above:
>Iron filings immediately decompose hypochlorous acid with a brisk effervescence produced by the evolution of chlorine; the iron is partly oxidized and in part dissolved as chloride without the formation of any chlorate. A. J. Balard commented on this: "The greater number of other metallic substances do not decompose hypochlorous acid, and I am yet entirely ignorant of the cause of the peculiar behaviour of iron." P. Grouvelle passed chlorine through water with iron hydroxidelin suspension and a bleaching liquid along with ferric chloride was produced, and he found the liquid retained its bleaching properties after boiling for a quarter of an hour; but A. J. Balard failed to confirm this statement; he could not make ferric hypochlorite either (i) by the action of hypochlorous acid on iron hydroxide, for hypochlorous acid does not dissolve ferric oxide; or (ii) by the action of calcium hypochlorite on ferric sulphate, for calcium sulphate and ferric oxide are produced. Hence, adds A. J. Balard, "ferric hypochlorite cannot exist;" the results by P. Grouvelle are due to the formation of ferric chloride and hypochlorous acid in dil. soln.; when the mixture was heated, a portion of the acid distilled off, and the reaction which occurred with the cold soln. was reversed, for ferric oxide and chlorine were formed. Ferrous oxide is oxidized to ferric oxide by hypochlorous acid.
Problematic fumes so an outdoor reaction is strongly advised. Neutralize solution products with Washing Soda at the end of the reaction.
Here is a particular comment relative to [steel being corroded](https://www.texwipe.com/Content/Images/uploaded/documents/Technical-Data/Stainless_Steel_Bleach.pdf) in place of iron, without the benefit of a good high surface area cathode in diluted NaOCl. |
For bond dissociation energies (BDEs) of double and triple bonds, is number shown in tables for that second or third bond? Or all of them? |
## Below a more general approach.
Suppose that we have two weak acids $\ce{HA}$ and $\ce{HB}$.
The initial concentrations are $C^0_\ce{HA}$ and $C^0_\ce{HB}$, and their constants are $K_{\mathrm{a},\ce{(HA)}}$ and $K_{\mathrm{a},\ce{(HB)}}$.
Suppose yet that volumes, $V_\ce{HA}$ and $V_\ce{HB}$, are additives.
So we have:
1. *Reactions*
\begin{align}
\ce{HA + H2O &<=> H3O+ + A-} &
K_{\mathrm{a},(\ce{HA})}
&= \frac{\ce{[H3O+][A-]}}{\ce{[HA]}}
\tag{1}\label{eq:KAcidHA}\\
\ce{HB + H2O &<=> H3O+ + B-} &
K_{\mathrm{a},(\ce{HB})}
&=\frac{\ce{[H3O+][B-]}}{\ce{[HB]}}
\tag{2}\label{eq:KAcidHB}\\
\ce{2 H2O &<=> H3O+ + OH-} &
K_\mathrm{w} &= \ce{[H3O+][OH-]}\tag{3}\label{eq:KWater}
\end{align}
2. *Mass balance*
\begin{align}
C_\ce{HA} &= \frac{C^0_\ce{HA} V_\ce{HA}}{V_\ce{HA} + V_\ce{HB}}
&&=\ce{[HA] + [A-]}\tag{4}\label{eq:MassBalanceHA}\\
C_\ce{HB} &= \frac{C^0_\ce{HB} V_\ce{HB}}{V_\ce{HA} + V_\ce{HB}}
&&=\ce{[HB] + [B-]}\tag{5}\label{eq:MassBalanceHB}
\end{align}
3. *Charge balance*
$$\ce{[H3O+] = [OH-] + [A-] + [B-]}\tag{6}\label{eq:ChargeBalance}$$
Replacing \eqref{eq:KAcidHA}–\eqref{eq:MassBalanceHB} equations on \eqref{eq:ChargeBalance}, we have:
$$
\ce{[H3O+]} =
\frac{K_\mathrm{w}}{\ce{[H3O+]}}
+\frac{C_\ce{HA} K_{\mathrm{a},(\ce{HA})}}{\ce{[H3O+]}
+ K_{\mathrm{a},(\ce{HA})}}
+ \frac{C_\ce{HB} K_{\mathrm{a},(\ce{HB})}}{\ce{[H3O+]}
+ K_{\mathrm{a},(\ce{HB})}}
\tag{7}\label{eq:GeneralEquation}$$
or as polynomial
\begin{align}
\begin{split}
&\ce{[H3O+]}^4\\
+&\ce{[H3O+]}^3 (K_{\mathrm{a},(\ce{HA})}
+ K_{\mathrm{a},(\ce{HB})})\\
+&\ce{[H3O+]}^2 \left[
K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})}
- ( C_\ce{HA} K_{\mathrm{a},(\ce{HA})}
+ C_\ce{HB}K_{\mathrm{a},(\ce{HB})} )
- K_\mathrm{w} \right]\\
-&\ce{[H3O+]} \left[
(C_\ce{HA} + C_\ce{HB}) K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})}
+ K_\ce{w} (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})}) \right]\\
-& K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})} K_\mathrm{w}\\
=&\ 0
\end{split}\tag{8}\label{eq:GeneralPol}
\end{align}
This single equation will exactly solve any equilibrium problem involving the mixture of any two monoprotic acids, in any concentration (as long as they're not much higher than about $\pu{1 mol L-1}$) and any volume. Depending of $K_\mathrm{a}$, $K_{\mathrm{a},(\ce{HB})}$, $C^0_\ce{HA}$, and $C^0_\ce{HB}$ values, we can yet obtain a simpler version<sup>\[1\]</sup>.
The \eqref{eq:GeneralPol} equation can simplified considering<sup>\[2\]</sup> that $K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})} K_\mathrm{w} \ll \ce{[H3O+]}^4$:
\begin{align}
\begin{split}
&\ce{[H3O+]}^3\\
+&\ce{[H3O+]}^2 (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})\\
+&\ce{[H3O+]}\left[
K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})}
- (C_\ce{HA} K_{\mathrm{a},(\ce{HA})}
+ C_\ce{HB} K_{\mathrm{a},(\ce{HB})})
- K_\mathrm{w} \right]\\
-&\left[
(C_\ce{HA} + C_\ce{HB}) K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})}
+ K_\mathrm{w} (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})
\right]\\
=&\ 0
\end{split}\tag{9}\label{eq:GeneralPolSimp1}
\end{align}
The \eqref{eq:GeneralPolSimp1} equation can simplified considering<sup>\[2\]</sup> that $K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})} \ll \ce{[H3O+]}^3$:
\begin{align}
\ce{[H3O+]}^2 + \ce{[H3O+]} (K_{\mathrm{a},(\ce{HA})}
+ K_{\mathrm{a},(\ce{HB})})
- (C_\ce{HA} K_{\mathrm{a},(\ce{HA})}
+ C_\ce{HB} K_{\mathrm{a},(\ce{HB})}+K_\mathrm{w})
= 0\tag{10}\label{eq:GeneralPolSimp2}
\end{align}
The \eqref{eq:GeneralPolSimp2} equation can be solved as usual.
$$\ce{[H3O+]} =
\frac{
- (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})
+ \sqrt{
(K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})^2
+ 4 (C_\ce{HA} K_{\mathrm{a},(\ce{HA})}
+ C_\ce{HB} K_{\mathrm{a},(\ce{HB})}+K_\mathrm{w})}
}{2}$$
Or using the initial concentrations
$$\ce{[H3O+]} =
\frac{
- (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})
+ \sqrt{
(K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})^2
+ 4\left(%\displaystyle
\frac{C^0_\ce{HA} V_\ce{HA} K_{\ce{a},(\ce{HA})}
}{V_\ce{HA} + V_\ce{HB}}
+ \frac{C^0_\ce{HB} V_\ce{HB} K_{\ce{a},(\ce{HB})}
}{V_\ce{HA} + V_\ce{HB}}
+K_\mathrm{w}\right)}
}{2}$$
Replacing $C^0_\ce{HA}=\pu{0.01 mol L-1}$, $C^0_\ce{HB}=\pu{0.01 mol L-1}$, $V_\ce{HA}=\pu{0.050 L}$ and $V_\ce{HB}=\pu{0.050 L}$, and using $\text{p}K_\ce{a}=3.75$ for formic acid and $\text{p}K_\ce{a}=4.756$ for acetic acid<sup>\[2\]</sup>, we have
$$\ce{pH}=3.05.$$
**Addendum**
[@MaxW](https://chemistry.stackexchange.com/users/22102/maxw) made some warnings about the assumptions which simplificate ($\ref{eq:GeneralPol}$)([see comment](https://chemistry.stackexchange.com/questions/10249/how-to-determine-the-ph-of-a-mixture-of-two-weak-acids/120139?noredirect=1#comment248386_120139)) and ($\ref{eq:GeneralPolSimp1}$)([see comment](https://chemistry.stackexchange.com/questions/10249/how-to-determine-the-ph-of-a-mixture-of-two-weak-acids/120139?noredirect=1#comment248672_120139)) equations.
To completeness and with the aim of show the limitations of present approach, consider the following graphics.
The ($\ref{eq:GeneralPol}$), ($\ref{eq:GeneralPolSimp1}$), and ($\ref{eq:GeneralPolSimp2}$) equations were solved using $\text{p}K_{\mathrm{a},(\ce{HA})}=\text{p}K_{\mathrm{a},(\ce{HB})}$, $C^0_\ce{HA}=C^0_\ce{HB}$, and $V_\ce{HA}=V_\ce{HB}=\pu{50.0 mL}$.
[![C0 = 1.0E0 molperL][1]][1]
[![C0 = 1.0E-1 molperL][2]][2]
[![C0 = 1.0E-2 molperL][3]][3]
[![C0 = 1.0E-3 molperL][4]][4]
[![C0 = 1.0E-4 molperL][5]][5]
[![C0 = 1.0E-5 molperL][6]][6]
[![C0 = 1.0E-6 molperL][7]][7]
[![C0 = 1.0E-7 molperL][8]][8]
As we can see, the ($\ref{eq:GeneralPolSimp1}$) equation can be used on any $\text{p}K_\ce{a}$ values, inasmuch as the concentration be major of $\pu{1E-6 mol L-1}$.
The ($\ref{eq:GeneralPolSimp2}$) equation becomes problematic when dilute solutions or strong acids solutions are used.
Thus, when applied to book-examples, which $\text{p}K_\ce{a}$ values from 3 to 6, and the concentrations from $1.0$ to $\pu{1.0E-4 mol L-1}$, the assumptions are valid and the equations can be used, providing good approximations of $\ce{pH}$ values.
**Acknowledgement**
Thanks to [@MaxW](https://chemistry.stackexchange.com/users/22102/maxw) for draw attention to limitations of assumptions.
**Reference**
1. Edited based on [first @MaxW's](https://chemistry.stackexchange.com/questions/10249/how-to-determine-the-ph-of-a-mixture-of-two-weak-acids/120139?noredirect=1#comment248386_120139) comment.
2. Edited based on [second @MaxW's](https://chemistry.stackexchange.com/questions/10249/how-to-determine-the-ph-of-a-mixture-of-two-weak-acids/120139?noredirect=1#comment248672_120139) comment.
3. [Haynes, W. M.; Lide, D. R.; Bruno, T. J., *CRC Handbook of Chemistry and Physics*. 97 ed.; CRC Press: 2017; pp. 5–88][9].
[1]: https://i.stack.imgur.com/sDu7E.png
[2]: https://i.stack.imgur.com/NhdnS.png
[3]: https://i.stack.imgur.com/2Hi2v.png
[4]: https://i.stack.imgur.com/bQwkU.png
[5]: https://i.stack.imgur.com/4ktov.png
[6]: https://i.stack.imgur.com/KoWqw.png
[7]: https://i.stack.imgur.com/TnTfZ.png
[8]: https://i.stack.imgur.com/gKlM0.png
[9]: https://books.google.com.br/books?id=VVezDAAAQBAJ&lpg=PP1&hl=pt-BR&pg=SA5-PA88#v=onepage&q&f=false |
## Below a more general approach.
Suppose that we have two weak acids $\ce{HA}$ and $\ce{HB}$.
The initial concentrations are $C^0_\ce{HA}$ and $C^0_\ce{HB}$, and their constants are $K_{\mathrm{a},\ce{(HA)}}$ and $K_{\mathrm{a},\ce{(HB)}}$.
Suppose yet that volumes, $V_\ce{HA}$ and $V_\ce{HB}$, are additives.
So we have:
1. *Reactions*
\begin{align}
\ce{HA + H2O &<=> H3O+ + A-} &
K_{\mathrm{a},(\ce{HA})}
&= \frac{\ce{[H3O+][A-]}}{\ce{[HA]}}
\tag{1}\label{eq:KAcidHA}\\
\ce{HB + H2O &<=> H3O+ + B-} &
K_{\mathrm{a},(\ce{HB})}
&=\frac{\ce{[H3O+][B-]}}{\ce{[HB]}}
\tag{2}\label{eq:KAcidHB}\\
\ce{2 H2O &<=> H3O+ + OH-} &
K_\mathrm{w} &= \ce{[H3O+][OH-]}\tag{3}\label{eq:KWater}
\end{align}
2. *Mass balance*
\begin{align}
C_\ce{HA} &= \frac{C^0_\ce{HA} V_\ce{HA}}{V_\ce{HA} + V_\ce{HB}}
&&=\ce{[HA] + [A-]}\tag{4}\label{eq:MassBalanceHA}\\
C_\ce{HB} &= \frac{C^0_\ce{HB} V_\ce{HB}}{V_\ce{HA} + V_\ce{HB}}
&&=\ce{[HB] + [B-]}\tag{5}\label{eq:MassBalanceHB}
\end{align}
3. *Charge balance*
$$\ce{[H3O+] = [OH-] + [A-] + [B-]}\tag{6}\label{eq:ChargeBalance}$$
Replacing \eqref{eq:KAcidHA}–\eqref{eq:MassBalanceHB} equations on \eqref{eq:ChargeBalance}, we have:
$$
\ce{[H3O+]} =
\frac{K_\mathrm{w}}{\ce{[H3O+]}}
+\frac{C_\ce{HA} K_{\mathrm{a},(\ce{HA})}}{\ce{[H3O+]}
+ K_{\mathrm{a},(\ce{HA})}}
+ \frac{C_\ce{HB} K_{\mathrm{a},(\ce{HB})}}{\ce{[H3O+]}
+ K_{\mathrm{a},(\ce{HB})}}
\tag{7}\label{eq:GeneralEquation}$$
or as polynomial
\begin{align}
\begin{split}
&\ce{[H3O+]}^4\\
+&\ce{[H3O+]}^3 (K_{\mathrm{a},(\ce{HA})}
+ K_{\mathrm{a},(\ce{HB})})\\
+&\ce{[H3O+]}^2 \left[
K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})}
- ( C_\ce{HA} K_{\mathrm{a},(\ce{HA})}
+ C_\ce{HB}K_{\mathrm{a},(\ce{HB})} )
- K_\mathrm{w} \right]\\
-&\ce{[H3O+]} \left[
(C_\ce{HA} + C_\ce{HB}) K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})}
+ K_\ce{w} (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})}) \right]\\
-& K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})} K_\mathrm{w}\\
=&\ 0
\end{split}\tag{8}\label{eq:GeneralPol}
\end{align}
This single equation will exactly solve any equilibrium problem involving the mixture of any two monoprotic acids, in any concentration (as long as they're not much higher than about $\pu{1 mol L-1}$) and any volume. Depending of $K_\mathrm{a}$, $K_{\mathrm{a},(\ce{HB})}$, $C^0_\ce{HA}$, and $C^0_\ce{HB}$ values, we can yet obtain a simpler version<sup>\[1\]</sup>.
The \eqref{eq:GeneralPol} equation can simplified considering<sup>\[2\]</sup> that $K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})} K_\mathrm{w} \ll \ce{[H3O+]}^4$:
\begin{align}
\begin{split}
&\ce{[H3O+]}^3\\
+&\ce{[H3O+]}^2 (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})\\
+&\ce{[H3O+]}\left[
K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})}
- (C_\ce{HA} K_{\mathrm{a},(\ce{HA})}
+ C_\ce{HB} K_{\mathrm{a},(\ce{HB})})
- K_\mathrm{w} \right]\\
-&\left[
(C_\ce{HA} + C_\ce{HB}) K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})}
+ K_\mathrm{w} (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})
\right]\\
=&\ 0
\end{split}\tag{9}\label{eq:GeneralPolSimp1}
\end{align}
The \eqref{eq:GeneralPolSimp1} equation can simplified considering<sup>\[2\]</sup> that $K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})} \ll \ce{[H3O+]}^3$:
\begin{align}
\ce{[H3O+]}^2 + \ce{[H3O+]} (K_{\mathrm{a},(\ce{HA})}
+ K_{\mathrm{a},(\ce{HB})})
- (C_\ce{HA} K_{\mathrm{a},(\ce{HA})}
+ C_\ce{HB} K_{\mathrm{a},(\ce{HB})}+K_\mathrm{w})
= 0\tag{10}\label{eq:GeneralPolSimp2}
\end{align}
The \eqref{eq:GeneralPolSimp2} equation can be solved as usual.
$$\ce{[H3O+]} =
\frac{
- (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})
+ \sqrt{
(K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})^2
+ 4 (C_\ce{HA} K_{\mathrm{a},(\ce{HA})}
+ C_\ce{HB} K_{\mathrm{a},(\ce{HB})}+K_\mathrm{w})}
}{2}$$
Or using the initial concentrations
$$\ce{[H3O+]} =
\frac{
- (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})
+ \sqrt{
(K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})^2
+ 4\left(%\displaystyle
\frac{C^0_\ce{HA} V_\ce{HA} K_{\ce{a},(\ce{HA})}
}{V_\ce{HA} + V_\ce{HB}}
+ \frac{C^0_\ce{HB} V_\ce{HB} K_{\ce{a},(\ce{HB})}
}{V_\ce{HA} + V_\ce{HB}}
+K_\mathrm{w}\right)}
}{2}$$
Replacing $C^0_\ce{HA}=\pu{0.01 mol L-1}$, $C^0_\ce{HB}=\pu{0.01 mol L-1}$, $V_\ce{HA}=\pu{0.050 L}$ and $V_\ce{HB}=\pu{0.050 L}$, and using $\text{p}K_\ce{a}=3.75$ for formic acid and $\text{p}K_\ce{a}=4.756$ for acetic acid<sup>\[3\]</sup>, we have
$$\ce{pH}=3.05.$$
**Addendum**
[@MaxW](https://chemistry.stackexchange.com/users/22102/maxw) made some warnings about the assumptions which simplificate ($\ref{eq:GeneralPol}$)([see comment](https://chemistry.stackexchange.com/questions/10249/how-to-determine-the-ph-of-a-mixture-of-two-weak-acids/120139?noredirect=1#comment248386_120139)) and ($\ref{eq:GeneralPolSimp1}$)([see comment](https://chemistry.stackexchange.com/questions/10249/how-to-determine-the-ph-of-a-mixture-of-two-weak-acids/120139?noredirect=1#comment248672_120139)) equations.
To completeness and with the aim of show the limitations of present approach, consider the following graphics.
The ($\ref{eq:GeneralPol}$), ($\ref{eq:GeneralPolSimp1}$), and ($\ref{eq:GeneralPolSimp2}$) equations were solved using $\text{p}K_{\mathrm{a},(\ce{HA})}=\text{p}K_{\mathrm{a},(\ce{HB})}$, $C^0_\ce{HA}=C^0_\ce{HB}$, and $V_\ce{HA}=V_\ce{HB}=\pu{50.0 mL}$.
[![C0 = 1.0E0 molperL][1]][1]
[![C0 = 1.0E-1 molperL][2]][2]
[![C0 = 1.0E-2 molperL][3]][3]
[![C0 = 1.0E-3 molperL][4]][4]
[![C0 = 1.0E-4 molperL][5]][5]
[![C0 = 1.0E-5 molperL][6]][6]
[![C0 = 1.0E-6 molperL][7]][7]
[![C0 = 1.0E-7 molperL][8]][8]
As we can see, the ($\ref{eq:GeneralPolSimp1}$) equation can be used on any $\text{p}K_\ce{a}$ values, inasmuch as the concentration be major of $\pu{1E-6 mol L-1}$.
The ($\ref{eq:GeneralPolSimp2}$) equation becomes problematic when dilute solutions or strong acids solutions are used.
Thus, when applied to book-examples, which $\text{p}K_\ce{a}$ values from 3 to 6, and the concentrations from $1.0$ to $\pu{1.0E-4 mol L-1}$, the assumptions are valid and the equations can be used, providing good approximations of $\ce{pH}$ values.
**Acknowledgement**
Thanks to [@MaxW](https://chemistry.stackexchange.com/users/22102/maxw) for draw attention to limitations of assumptions.
**Reference**
1. Edited based on [first @MaxW's](https://chemistry.stackexchange.com/questions/10249/how-to-determine-the-ph-of-a-mixture-of-two-weak-acids/120139?noredirect=1#comment248386_120139) comment.
2. Edited based on [second @MaxW's](https://chemistry.stackexchange.com/questions/10249/how-to-determine-the-ph-of-a-mixture-of-two-weak-acids/120139?noredirect=1#comment248672_120139) comment.
3. [Haynes, W. M.; Lide, D. R.; Bruno, T. J., *CRC Handbook of Chemistry and Physics*. 97 ed.; CRC Press: 2017; pp. 5–88][9].
[1]: https://i.stack.imgur.com/sDu7E.png
[2]: https://i.stack.imgur.com/NhdnS.png
[3]: https://i.stack.imgur.com/2Hi2v.png
[4]: https://i.stack.imgur.com/bQwkU.png
[5]: https://i.stack.imgur.com/4ktov.png
[6]: https://i.stack.imgur.com/KoWqw.png
[7]: https://i.stack.imgur.com/TnTfZ.png
[8]: https://i.stack.imgur.com/gKlM0.png
[9]: https://books.google.com.br/books?id=VVezDAAAQBAJ&lpg=PP1&hl=pt-BR&pg=SA5-PA88#v=onepage&q&f=false |
At school(high school), to calculate the voltage needed for electrolysis, we calculate the voltage produce if the elements were in a galvanic cell, preceded by a minus.
So if a galvanic cell with Cu and Cr would produce 1,14V, we would say that an electrolysis with Cu and Cr will need -1.14V.
But why do we do this and is this correct?
Thank you in advance. |
CO2 kills humans.
Does it also kill organisms?
Does it also kill viruses? |
CO2 kills humans. Does it also kill organisms? |
Covalent bond is a strong bond compared to Ionic Bonds but Ionic Compounds have higher melting and boiling points then covalent compounds. **Why?** |
The material I've synthesised is made up of two primary parts at a 1:1 ratio. It's one part; Agar, glycerol, water. Second part; Polyurethane resin.
The agar mixture on its own was too soft, so I added the resin to harden the material. Would this material be classified as a hybrid composite? If not, why? |
Is it appropriate to call this material a hybrid composite? |
I buy and drink Nescafe Gold all the time. I also buy the smaller "caffeine-free" version of the same product. It tastes very differently, which makes me assume that the caffeine does have a lot of taste.
However, maybe it's not simply due to the lack of caffeine? Maybe the process with which they "de-caf" the original product (something about "washing it away", if I recall correctly from what I read/heard somewhere at some point) also removes other ingredients?
How can they possibly "wash away" just the caffeine from instant coffee like that?
Unless the truth is that the entire products are "built" from the ground up, separately for the caffeine-including and caffeine-less products, and the latter is not "based on" the former at all?
Even if so, they still have to "wash away" the caffeine from the raw coffee, which puts us back on square one... |
I have a little bit of problem trying to figure out the mechanism for the synthesis of Meldrum Acid. Meldrum acid is synthesis by a condensation reaction of acetone with malonic acid in acetic anhydride.
[![enter image description here][1]][1]
First, the carboxyl OH will be acetylated by acetic anhydride. But I can't figure out how the two ester bonds are being formed subsequently. I'm thinking that there's an acid-catalysed aldol condensation somewhere that results in the removing of water. Perhaps by attack of the enol (from acetone) to the acetylated carbon. But I'm not really sure about it either.
Any thoughts?
[1]: https://i.stack.imgur.com/yNbhA.gif |
There's a "pair-fitting" function in pymol that will align molecules given some user-defined reference atoms, and output an RMSD across those points of comparison. You can get pymol educational [version for free][1], so students would be able to install it themselves.
[1]: https://pymol.org/edu/?q=educational/ |
The material I've synthesised is made up of two primary parts at a 1:1 ratio:
- Part 1: agar, glycerol, water.
- Part 2: polyurethane resin.
The agar mixture on its own was too soft, so I added the resin to harden the material. Would this material be classified as a hybrid composite? If not, why? |
I am quiet new to this field of chemistry. I work with GaussView and I created a surface plot for my molecule with the cube and surface generator. This works very well. Now I like to save and copy this image of my molecule with its surface. My favorite file type whould be a vector graphic type (eps or svg).
I don not know how. I tried the option with File -> Save Image. When I choose vector graphic as type I just get the structur of my molecule but not the surface (as well as a error message...). When I choose all the other file types I get nothing in the file just a black screen.
Where is my mistake? What do I constantly wrong? I see a lot of these plots in publications... I just want this plot for my PowerPoint...
I hope someone of you got a godd tip for me, thanks! |
How to extract a image (vector graphic) from a surface EPS plot with GaussView? |
So I've been trying to understand what exactly aromaticity means, and I've come to know that compounds possessing benzene rings are aromatic, like toluene and phenol. It thus made sense to me that the benzylic carbocation should be aromatic, as it fulfills all conditions of Huckel's rule. However, I found no mention of it being aromatic anywhere. Can anyone tell me if it is indeed aromatic or is only resonance stabilised, and if it is not aromatic, why so? What criteria is it violating?
Sorry if this is an elementary question, I really want to know what aromaticity actually means and am stuck on this. |
Is the benzylic carbocation aromatic? |
Certain objects such as metals, glass, plastic, sugar etc. melt when heated to a certain temperature but other materials such as a piece of wood, paper, a piece of rock etc. do not. Why is that? |
Why don’t all objects melt and liquefy when heated sufficiently? |
I am working with lithium ion batteries having the cathode material as NMC i.e. LiNixMnyCozO2, wherein x + y + z = 1. The proportion of Nickel, Manganese and Cobalt in the cathode varies, but mostly Nickel is either equal to or greater than the amount of Manganese and Cobalt in the cathode (for example NMC-111, NMC-532, NMC-622, NMC-811 are the commonly reported chemistries in literature).
I tried doing Energy dispersive X-ray spectroscopy on the surface of my battery cathode to find out the proportion of Nickel, Manganese and Cathode. The results I obtained were: Ni - 20%, Mn - 20%, Co - 60%. I repeated the EDXS procedure at ten different locations on the battery cathode and still got the same results with 1% error in the values. This result is far from what I expected and I suspect the reason for this is that I am scanning only the surface of the cathode.
So I have two questions regarding this:
1) Should I expect the EDXS results to be different and correct if I perform it on the cross-section of my battery cathode?
2) What are other cheap techniques which can get me the elemental composition of the cathode? One of my friends suggested me ICP-MS (inductively coupled plasma mass spectrometry), which is way too expensive for me. |
>Di-hydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction:
>$CH_4{_{(g)}}+H_2O_{(g)} \ce{<=>} CO_{(g)}+3H_2{_{(g)}}$
>How will the values of $K_p$, $K_c$ and composition of equilibrium mixture be affected by increasing the pressure ?
Now,
1) According to [this][1] answer $K_c$ only changes on changing temperature i.e. **unaffected by pressure.**
2) According to [Wikipedia][2] the article "Pressure dependence" says, "...However, the composition at **equilibrium will depend appreciably on pressure** when:
1)the pressure is changed by compression or expansion of the gaseous reacting system, and
2)the reaction results in the change of the number of moles of gas in the system....."
here wiki is using term $K_p$ to denote equilibrium constant. Now the "composition of equilibrium" means $K_p$ and not $Q_p$ because its taking **at equilibrium**, so the $K_p$ **changes with pressure**.
3) I have studied that on increasing pressure the reaction shifts the side with less number of sum of gaseous atoms in the reaction formulae. I know from [this][3] question that composition of equilibrium is different from "the side at which the reaction shifts". So it seems my study isn't useful here in deciding the composition of equilibrium
So my question is how will the values of $K_p$, $K_c$ and composition of equilibrium mixture be affected by increasing the pressure ?
Does one change and the other remains unaffected and isn't the composition of equilibrium directly related to $K_p$ and $K_c$
Are dependencies of $K_p$ and $K_c$ different for change in pressure.
[1]: https://chemistry.stackexchange.com/questions/49225/why-is-kc-not-affected-by-change-in-pressure
[2]: https://en.wikipedia.org/wiki/Equilibrium_constant#Pressure%20dependence
[3]: https://chemistry.stackexchange.com/questions/4977/how-can-the-equilibrium-shift-while-kc-remains-constant |
In terms of energy per gram of reactants, the evaporation of water at -2.2 kJ/g, is much higher than any chemical endothermic reaction that I have seen demonstrated. At [extreme temperatures][1] there is no real limit, but for reactions that occur at a reasonable rate at room-temperature it seems water is hard to beat. And it isn't even a *chemical* reaction.
I have reason to think water is hard to beat: the entropy of gases is *much higher* than liquids. The strong hydrogen bond network of water makes the liquid phase lower entropy than non-polar liquids because it structures the liquid. Finally, water is a very light molecule so you get many moles of gas per kg of liquid. Despite this huge entropy gain, the phase-transition proceeds sluggishly because it takes so much dry air to shift the equilibrium forward, if it took much more energy it wouldn't happen at a meaningful rate.
Can any chemical reaction approach this at room temperature? If so, can any non-gas-evolving reaction come close?
[1]: https://chemistry.stackexchange.com/a/86860
|
I just thought that the reaction should follow SN2 mechanism this is because the alkyl halide is primary but the book says that the reaction will follow SN1 mechanism resulting in the formation of tertiary carbonation and a ring expansion product. Can anyone please explain the reason behind this ? |
Which mechanism should a reaction between the cyclopentylmethanol and the HBr should follow? |
$K_p$ 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. |
$K_p$ 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 $. |
Wood can melt, in a limited way, under certain circumstances, such as friction welding. For a dramatic video, see: https://www.facebook.com/interestingengineering/videos/1891004754302553/
See also:
http://web.utk.edu/~mtaylo29/pages/Wood%20welding.htm?fbclid=IwAR1MLgBtkfESlYiaP0iEaXbv36AtLy8yXEj0iCqFaaVYBcDRitxqeZJZYtM
Here they say, about welds created between wood pieces during friction welding: "Examination of the bondline suggests that the friction between the pieces heats and melts components of the wood (mainly lignin) and loosens fibers on the surface. These fibers intertwine in a matrix with the molten lignin and solidify to form a bond that is strong enough for structural applications."
For more details, see: "Wood Bonding by Mechanically-Induced in Situ Welding of Polymeric Structural Wood Constituents" B. Gfeller M. Properzi M. Zanetti A. Pizzi F. Pichelin M. Lehmann L. Delmotte, Journal of Applied Polymer Science 92(1):243 - 251, 2004
Abstract from above reference: "Mechanically induced wood fusion welding, without any adhesive, is shown here to yield rapidly bonding wood joints satisfying the relevant requirements for structural application. The mechanism of mechanically induced vibrational wood fusion welding is shown to be due mostly to the melting and flowing of amorphous cells–interconnecting polymer material in the structure of wood, mainly lignin, but also some hemicelluloses. This causes a partial detachment, the “ungluing,” of long wood cells and wood fibers and the formation of an entanglement network drowned in a matrix of melted material which then solidifies, thus forming a wood cell/fiber entanglement network composite with a molten lignin polymer matrix. During the welding period some of the detached wood fibers which are no longer being held by the interconnecting material are pushed out of the joint as excess fiber. Crosslinking chemical reactions of lignin and carbohydrate-derived furfural also occur. Their presence has been identified by CP-MAS 13C-NMR. These reactions, however, are relatively minor contributors during the very short welding period. Their contribution increases after welding has finished, which explains why relatively longer holding times under pressure after the end of welding contribute strongly to obtaining a good bond." © 2004 Wiley Periodicals, Inc. J Appl Polym Sci 92: 243–251, 2004 |
The lignin, and some of the hemicelluloses, in wood can melt under certain circumstances, such as friction welding. For a dramatic video, see: https://www.facebook.com/interestingengineering/videos/1891004754302553/
See also:
http://web.utk.edu/~mtaylo29/pages/Wood%20welding.htm?fbclid=IwAR1MLgBtkfESlYiaP0iEaXbv36AtLy8yXEj0iCqFaaVYBcDRitxqeZJZYtM
Here they say, about welds created between wood pieces during friction welding: "Examination of the bondline suggests that the friction between the pieces heats and melts components of the wood (mainly lignin) and loosens fibers on the surface. These fibers intertwine in a matrix with the molten lignin and solidify to form a bond that is strong enough for structural applications."
For more details, see: "Wood Bonding by Mechanically-Induced in Situ Welding of Polymeric Structural Wood Constituents" B. Gfeller M. Properzi M. Zanetti A. Pizzi F. Pichelin M. Lehmann L. Delmotte, Journal of Applied Polymer Science 92(1):243 - 251, 2004
Abstract from above reference: "Mechanically induced wood fusion welding, without any adhesive, is shown here to yield rapidly bonding wood joints satisfying the relevant requirements for structural application. The mechanism of mechanically induced vibrational wood fusion welding is shown to be due mostly to the melting and flowing of amorphous cells–interconnecting polymer material in the structure of wood, mainly lignin, but also some hemicelluloses. This causes a partial detachment, the “ungluing,” of long wood cells and wood fibers and the formation of an entanglement network drowned in a matrix of melted material which then solidifies, thus forming a wood cell/fiber entanglement network composite with a molten lignin polymer matrix. During the welding period some of the detached wood fibers which are no longer being held by the interconnecting material are pushed out of the joint as excess fiber. Crosslinking chemical reactions of lignin and carbohydrate-derived furfural also occur. Their presence has been identified by CP-MAS 13C-NMR. These reactions, however, are relatively minor contributors during the very short welding period. Their contribution increases after welding has finished, which explains why relatively longer holding times under pressure after the end of welding contribute strongly to obtaining a good bond." © 2004 Wiley Periodicals, Inc. J Appl Polym Sci 92: 243–251, 2004 |
I came across the compound 3-methylpentan-2-ol. Looking at its proton NMR spectra, it seems there is a lot of splitting for the signal at 4ppm.
[![enter image description here][1]][1]
[1]: https://i.stack.imgur.com/vAMRu.gif
I'm assuming that the 4ppm represents the proton in the hydroxyl group (since it clearly is much more deshielded than the other protons), but this would mean it only has one neighbouring hydrogen and hence should exist as a doublet.
I just wanted to clarify, is there any reason why the 4ppm peak has so much splitting despite only having one neighbour? Or otherwise, I have also considered, could I be incorrect in assuming the 4ppm peak represents the hydroxyl group, and if so why? |
Why do certain signals in proton NMR experience extensive splitting despite only having 1 neighbour? |
To an evacuated vessel with movable piston under external pressure of 1 atm, 0.1 mol of He and 1 mol of unkown compound(vapour pressure 0.68 atm at 273K) are introduced. Considering ideal gas behaviour, calculate the total volume of 2 gases at 273K.
My doubt is that why cant I add the moles of He and the unkown compound and directly use PV=nRT? Why is the vapour pressure given? |
I came across the compound 3-methylpentan-2-ol. Looking at its proton NMR spectra, it seems there is a lot of splitting for the signal at 4 ppm.
[![3-methylpentan-2-ol NMR spectra][1]][1]
[1]: https://i.stack.imgur.com/vAMRu.gif
I'm assuming that the 4 ppm represents the proton in the hydroxyl group (since it clearly is much more deshielded than the other protons), but this would mean it only has one neighbouring hydrogen and hence should exist as a doublet.
I just wanted to clarify, is there any reason why the 4 ppm peak has so much splitting despite only having one neighbour? Or otherwise, I have also considered, could I be incorrect in assuming the 4 ppm peak represents the hydroxyl group, and if so why? |
> To an evacuated vessel with movable piston under external pressure of $\pu{1 atm},$ $\ce{0.1 mol}$ of $\ce{He}$ and $\pu{1 mol}$ of unknown compound (vapour pressure $\pu{0.68 atm}$ at $\pu{273 K})$ are introduced. Considering ideal gas behavior, calculate the total volume of 2 gases at $\pu{273 K}.$
Why can't I add the amounts of substance for helium and the unknown compound and directly use $PV=nRT?$ Why is the vapour pressure given? |
Would alkyl bromide/alkyl chloride form when an alcohol reacts with thionyl bromide/thionyl chloride? |
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