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[![Photoelectric question][1]][1]
[1]: https://i.stack.imgur.com/7rNgw.png
My attempt:- 1)At 0.68e15 Hz frequency of light,Metal A gives 7.2eV Kinetic Energy(KE).
2)At 1e15 Hz frequency of light, Metal B, gives 6eV Kinetic Energy.
3) At 1.1e15 Hz frequency of light, Metal C gives 5.5eV Kinetic Energy.
4)At1.2e15 Hz frequency of light, Metal D gives 5eV Kinetic energy.
Now $KE_{max}=(\frac{hc}{\lambda}-\phi)$ where c= speed of light, $\lambda$=wavelength of light and $\phi$= work function of metal.
Now $7.2eV=(4.14e-15eVs \times 0.68e15 Hz- \phi)$ for metal A.The wavelength of light is 441 nm. Likewise we can compute wavelength of light incident on Metal B, C and D.
But we don't know the work functions of Metal A,B, C and D.
So, how to answer this question? |
In my book it is given that first electron gain enthalpy is greater than second for elements. Should we compare the magnitudes in such cases or the actual numbers with signs? Does the same comparision hold with Electron Affinity at 0K?
For example Be has positive Electron Affinity and F has negative Electron Affinity. So should I consider signs and say Be has more Electron Affinity than F or consider magnitude and say F has more Electron Affinity than Be?
How to compare their Electron gain enthalpies also at 0K? |
This is a classic experiment.
On a plate with a little water place a candle and light it. Place a drinking glass over it. The light goes out while sucking up the water.
At least two things happen:
1. The oxygen O₂ is changed to CO₂
2. The air is heated up a few degrees
3. Water is sucked up
But then I tried to change the experiment with a vase (big glass) and opening upwards and place the candle at the bottom. Then lit the candle and place plastic over the opening. The result was that the light went out but it didn't suck down the plastic. So it there is no vacuum in the glass?
A strange thing is that the plastic was pulsing like it was breathing.
It can be seen here in my video https://youtu.be/aolz8Y27Lsk
Why is the experiment so different from the other? |
**Changes in volume**
The chemical reaction for burning a candle is something like this:
$$\ce{C25H52 + 38 O2(g) -> 25 CO2(g) + 26 H2O(g)}$$
For every 38 dioxygen molecules used, you are making 25 carbon dioxide molecules and 26 water molecules (which start out as a gas, but will condense once they reach an area of lower temperature such as the glass surface). So once the flame is out and the water has condensed, the volume should be less than that of the fresh air (20 % of the volume initially are oxygen. Its volume will be replaced by that of carbon dioxide, so the volume will go from 100% to 25/38 * 20% + 80% = 93%). While the candle is burning, however, the temperature is higher, and some of the water will be in the gas phase. Also, while the air was already hotter than room temperature before the container was covered, the temperature will rise as soon as the influx of cooler (fresh) air is cut off.
**How hot will it get?**
A tea light produces heat at about 30 J/s, and (with a molar heat of combustion of about 15,000 kJ / mol) used up $\pu{80 \mu mol}$ oxygen per second. This means you need about 10 mL of fresh air each second the candle burns. The container has a volume of about 400 mL, so the candle should burn <40 seconds. In the video, it burned about 2 minutes (not bad for an estimate). Using up all the oxygen in the container would give off about 1.2 kJ, which is sufficient to raise the temperature by 1000 degrees Celsius if it were isolated. As it is, most of the heat flows into the container.
**What is different when using water?**
Water has a higher heat capacity, so it is more efficient in cooling down the gas (and the container). As a consequence, the water level will rise appreciably. As MaxW mentions in the comments, the plastic probably did not seal perfectly. The pulsing could be intermittent flow of gas, or temperature fluctuations because of mixing. |
How can I calculate Kp given Gibb's free energy? |
Using the following relations:
$\Delta{G}^⦵ = \Delta{H}^⦵ -T\Delta{S}^⦵$
$K_P = e^{\frac{-\Delta{G}^⦵}{RT}}$
How would you find the temperature at which $K_P = 1$?
>! $K_P = 1$ when $\frac{-\Delta{G}^⦵}{RT} = 0$, which is when $-\Delta{G}^⦵ = 0$
Given what $\Delta{G}^⦵$ must equal, how could you find T? |
Using the following relations:
$\Delta{G}^⦵ = \Delta{H}^⦵ -T\Delta{S}^⦵$
$K_P = e^{\frac{-\Delta{G}^⦵}{RT}}$
How would you find the temperature at which $K_P = 1$?
>! $K_P = 1$ when $\frac{-\Delta{G}^⦵}{RT} = 0$, which is when $\Delta{G}^⦵ = 0$
Given what $\Delta{G}^⦵$ must equal, how could you find T?
>! $\Delta{G}^⦵ = 0 = \Delta{H}^⦵ -T\Delta{S}^⦵ \therefore T = \frac{\Delta{H}^⦵}{\Delta{S}^⦵}$ |
Adding acids or bases to water, so that either pH or pOH decreases independently of the other, will that affect the auto-ionization of water? For example, pure water + auto-ionized state, with some base added to remove some protons, will it auto-ionize a bit more (create more H+ and OH-) or a bit less (remove some H+ and OH-)? |
Auto-ionization equilibrium of water shifted with acid-base addition? |
The problem is as follows:
> In a galvanic cell the cathode is an $Ag^{+}(1.00\,M)/Ag_{(s)}$
> half-cell. The anode is a standard hydrogen electrode immersed in a
> buffer solution containing $0.10\,M$ benzoic acid $(C_6H_5COOH)$ and
> $0.050\,M$ of sodium benzoate $(C_6H_5COO^{-}Na^{+})$. The measured
> cell voltage is $1.030\V$. What is the $pK_a$ of benzoic acid?.
What I did to solve this problem was to find the potential for the cell involving the standard hydrogen electrode. At first I was confused because there are three "elements" featured in the problem. One being the silver electrode, the other the standard hydrogen electrode and the other a buffer solution, so I didn't know how to proceed from there.
Then I noticed that to get the constant of equilibrium I only require the concentration of $[H^{+}]$ ions as,
$K_a=\frac{[C_6H_5COO^{-}][H^{+}]}{[C_6H_5COOH]}$
Therefore to obtain those protons I did this:
The half equations in the cell are:
$\begin{array}{cc}
Ag^{+}+1e^{-}\rightarrow Ag_{(s)}&E^{0}=0.7999\,V\\
H^{+}+1e^{-}\rightarrow \frac{1}{2}H_{2(g)}&E^{0}=0.0000\,V\\
\end{array}$
Hence the overall reaction for this process would be:
$E^{0}_{cell}=E_{cathode}-E_{anode}=0.7999-0.0000=0.7999\,V$
Which is for:
$Ag^{+}+\frac{1}{2}H_{2}+\rightarrow Ag_{(s)} + H^{+}$
Hence:
$E_{cell}=E^{0}-\frac{0.0592}{n}\log\frac{[H^{+}]}{[Ag^{+}]p^{\frac{1}{2}}_{H_{2(g)}}}$
Since it indicates that the cell potential is $1.030\,V$ then:
$1.030=0.7999-\frac{0.0592}{1}\log\frac{[H^{+}]}{[1](1)^{\frac{1}{2}}}$
Solving this I'm getting:
$[H^{+}]=0.000125306\,M$
Now all that's left is to plug in this value in the equation to get the equilibrium constant:
$K_a=\frac{[C_6H_5COO^{-}][H^{+}]}{[C_6H_5COOH]}$
$[C_6H_5COO^{-}]=[H^{+}]=0.05\,M and [C_6H_5COOH]=0.1\,M$
Hence:
$K_a=\frac{(0.05)(1.25306\times 10^{-4})}{(0.1)}=6.2653\times 10^{-5}$
Therefore the $pKa$ of benzoic acid would be:
$pKa=-\log Ka=-\log\left(6.2653\times 10^{-5}\right)=4.20306$
Which does seem to be within the value of benzoic acid which I have on different references. But the problem with this method it is that it required the use of logarithm.
Given this situation, does it exist an approximation or anything that can be done right of the bat to get an idea where that value would be?. Does it exist another method which I could use?. |
Does it exist a quick way to obtain the pKa of an acid in a galvanic cell other than having to work with a logarithm? |
The problem is as follows:
> In a galvanic cell the cathode is an $Ag^{+}(1.00\,M)/Ag_{(s)}$
> half-cell. The anode is a standard hydrogen electrode immersed in a
> buffer solution containing $0.10\,M$ benzoic acid $(C_6H_5COOH)$ and
> $0.050\,M$ of sodium benzoate $(C_6H_5COO^{-}Na^{+})$. The measured
> cell voltage is $1.030\,V$. What is the $pK_a$ of benzoic acid?.
What I did to solve this problem was to find the potential for the cell involving the standard hydrogen electrode. At first I was confused because there are three "elements" featured in the problem. One being the silver electrode, the other the standard hydrogen electrode and the other a buffer solution, so I didn't know how to proceed from there.
Then I noticed that to get the constant of equilibrium I only require the concentration of $[H^{+}]$ ions as,
$K_a=\frac{[C_6H_5COO^{-}][H^{+}]}{[C_6H_5COOH]}$
Therefore to obtain those protons I did this:
The half equations in the cell are:
$\begin{array}{cc}
Ag^{+}+1e^{-}\rightarrow Ag_{(s)}&E^{0}=0.7999\,V\\
H^{+}+1e^{-}\rightarrow \frac{1}{2}H_{2(g)}&E^{0}=0.0000\,V\\
\end{array}$
Hence the overall reaction for this process would be:
$E^{0}_{cell}=E_{cathode}-E_{anode}=0.7999-0.0000=0.7999\,V$
Which is for:
$Ag^{+}+\frac{1}{2}H_{2}+\rightarrow Ag_{(s)} + H^{+}$
Hence:
$E_{cell}=E^{0}-\frac{0.0592}{n}\log\frac{[H^{+}]}{[Ag^{+}]p^{\frac{1}{2}}_{H_{2(g)}}}$
Since it indicates that the cell potential is $1.030\,V$ then:
$1.030=0.7999-\frac{0.0592}{1}\log\frac{[H^{+}]}{[1](1)^{\frac{1}{2}}}$
Solving this I'm getting:
$[H^{+}]=0.000125306\,M$
Now all that's left is to plug in this value in the equation to get the equilibrium constant:
$K_a=\frac{[C_6H_5COO^{-}][H^{+}]}{[C_6H_5COOH]}$
$[C_6H_5COO^{-}]=[H^{+}]=0.05\,M and [C_6H_5COOH]=0.1\,M$
Hence:
$K_a=\frac{(0.05)(1.25306\times 10^{-4})}{(0.1)}=6.2653\times 10^{-5}$
Therefore the $pKa$ of benzoic acid would be:
$pKa=-\log Ka=-\log\left(6.2653\times 10^{-5}\right)=4.20306$
Which does seem to be within the value of benzoic acid which I have on different references. But the problem with this method it is that it required the use of logarithm.
Given this situation, does it exist an approximation or anything that can be done right of the bat to get an idea where that value would be?. Does it exist another method which I could use?. |
It depends how you write the reaction.
> For example, pure water + auto-ionized state, with some base added to remove some protons, will it auto-ionize a bit more (create more H+ and OH-) or a bit less (remove some H+ and OH-)?
If we call the base $\ce{B}$ and the conjugate acid $\ce{BH+}$, you could write:
$$\ce{B + H3O+ <=> BH+ + H2O}$$
This would mean that you just lost some hydronium ions, and water will ionize a bit more to attain equilibrium with its ions.
Or you could write it like this:
$$\ce{B + H2O <=> BH+ + OH-}$$
This would mean that you just gained some hydroxide ions, and water will ionize a bit less to attain equilibrium with its ions.
**What about adding acid?**
Same story:
$$\ce{AH + H2O <=> A- + H3O+}$$
or
$$\ce{AH + OH- <=> A- + H2O}$$
**Which one is correct?**
Either one is fine. You will arrive at the same equilibrium concentrations if you consider one of the acid/base reactions and water auto-dissociation. Depending on what is the major species and what is the minor species, you can choose one or the other. Or if you can't decide, you can be creative and write:
$$\ce{AH + 1/2 OH- <=> A- + 1/2 H3O+}$$
|
Could someone please enlighten me on what would be the theoretical effect of a counter ion on the colour of the flame emmited by a metallic salt in a flame test?
Eg. NaCl - what would be the effect of the non metal on the general colour of the flame of the coumpond.
I am looking into specific groups of metallic salts based on 3 metals: Sodium, Potassium and Copper II
I have searched throghout the literature but have not found an answer yet..
Thanks! |
*Lacking an answer for 3 years, I'm going to attempt to answer this based on internet definitions and logic. Please, I welcome any insight from chemistry experts to improve this answer.*
First of all, the Wikipedia link quoted actually gives *two* definitions for protomer. The one quoted is the "structural biology" definition rather than the "chemistry" definition. On the other hand, it seems as though monomer has one definition according to the [Monomer Wikipedia article][1]. However, as pointed out by the OP, another point of confusion arises in the definition of a monomer as an oligomer with just one subunit in [protein quarternary structure][2]. So we actually have two definitions of each term out there. Let's recap them and then compare.
- Protomer (structural biology): In this case, the term protomer is reserved for the specific case of being part of a hetero-oligomer. The "hetero" part is important. The point is that the oligomer is made of a number of *different* macromolecules- glued together non-covalently. In this case, a protomer is defined as the smallest repeatable unit within this hetero oligomer, which may be the entire oligomer itself (in which case, confusingly, the oligomer is called a monomer). The term protomer in this case, as pointed out by Wikipedia, serves the sole purpose of disambiguation. When we say "blah-mer", as pointed out by Hanry, we are saying "this thing has blah parts". But what are "parts"? Are they quarks? Electrons? Protons? Atoms? Macro molecules? Or... repeatable sub-units. It seems as though the confusion was between the last two. You can break a hetero-oligomer into macromolecules and you'll get a bunch of different macromolecules. You might get 12 As, 12 Bs and 12Cs. So is this a 36-mer or is it a 12-mer? It depends on what you mean by "mer". If each "mer" is an ABC subunit, then it's a 12-mer. If each macromolecule is a subunit, then it's a 36-mer. The term protomer helps us here. The molecule can be viewed as a 12-protomer, meaning that it's 12 copies of the smallest repeatable subunit: ABC.
- Protomer (chemistry): In this case, it looks like the term means something quite different. It looks like a protomer is defined as a specific type of [tautomer][3], one in which the tautomerism arises as a result of variable proton position.
- Monomer (standard): A monomer is a single molecule that forms part of a polymer. The polymer can either be a homo-polymer, consisting of repeated copies of that monomer, or it can be a copolymer (analogous to a hetero-oligomer), consisting of different monomers, according to Wikipedia's [Polymerization article][4]. The key here is that the monomer belongs to a [polymer][5]- meaning "many parts". Typically polymers are distinguished by the fact that they can theoretically continue forever in big long chains.
- Monomer (oligomer special case): This is just an unfortunate use of terminology, but it kind of makes sense when you think about it. "Oligomer" means "Few parts". Typically the distinction between oligomers and polymers is that polymers can go on forever- they could keep growing through repeated polymerisation, but an oligomer will have a set number of parts, and then it's fully grown. Because oligomers can have a set number of parts, people decided to further sub-classify them by the exact number of parts. So a hexamer is an oligomer with six parts, for example. Following this naming convention, we have pentamer for 5 parts, tetramer for 4 parts... and all the way down to monomer for one part. Which just unfortunately intersects with the other definition of monomer, being a single part of a polymer chain.
So, now that it's been established that there are different definitions even of each term itself, let's try to compare them:
- Protomer (structural biology) vs. monomer (standard): a protomer here is defined as a sub-unit of an oligomer, whereas a monomer is a building block of a polymer.
- Protomer (structural biology) vs. monomer (oligomer special case): in this case, we need to think. A monomeric oligomer is one which has one "sub-unit". If by "sub-unit" what is means is protomer, then in this case the monomer, protomer and oligomer all coincide and are describing the same thing. If, though, "sub-unit" means the smallest sub-molecules that are non-covalently bonded together, then the oligomer is, trivially, a homogeneous oligomer and not a heterogeneous one: it consists of just one molecule so all the molecules are the same. And so, strictly speaking according to the above definition, the term protomer doesn't apply here.
- Protomer (chemistry) vs. Monomer (standard): these two definitions are not mutually exclusive. By describing a compound as a protomer, in this way, we are just saying that there are different tautomeric configurations of this compound out there. On the other hand, a monomer is just something that can form a polymer chain. It's perfectly reasonable to assume that some tautomer could also be capable of undergoing polymerization, and therefore, would be a monomer. I don't know any examples but maybe someone reading can offer them.
- Protomer (chemistry) vs. monomer (special oligomer): The monomer definition here seems to be a bit vague, but in the loosest sense, every single molecule is a monomer. But it seems as though what's meant is that a monomer, in the oligomeric sense, typically doesn't undergo polymerisation with other monomers to form a polymer - it just likes being alone. Still, even with that definition, it's likely that there are examples of things that are protomers, in the chemistry sense i.e. being tautomeric in terms of proton position, and that are also monomers i.e. oligomers that won't bond with each other any further.
[1]: https://en.wikipedia.org/wiki/Monomer
[2]: https://en.wikipedia.org/wiki/Protein_quaternary_structure
[3]: https://en.wikipedia.org/wiki/Tautomer
[4]: https://en.wikipedia.org/wiki/Polymerization
[5]: https://en.wikipedia.org/wiki/Polymer |
The article by Hu et al. \[[1](https://doi.org/10.1177/154405910608501212)\] mentions a compound "*dithreitol*" in the *Saliva Proteome Analysis* section (emphasis mine):
> For the "shotgun" approach, saliva samples (1 mL) were pre-fractionated with the use of Millipore ultracentrifuge filters (Millipore Corp., Billerica, MA, USA).
Individual fractions were treated with 10 mM **dithreitol** for 30 min and then 50 mM iodoacetamide for 30 min.
PubChem doesn't reveal any details as to what this compound exactly is and suggests alternative spelling, e.g. [D-threitol](https://pubchem.ncbi.nlm.nih.gov/compound/169019).
[Google Books search](https://www.google.com/search?hl=en&tbm=bks&q=%22dithreitol%22) results in several mentions of this compound in the literature, but I didn't find any further details except that it's probably a [sulfhydryl reagent](https://books.google.com/books?id=7JcnWmCB7EIC&q="dithreitol"&dq="dithreitol"&hl=en&redir_esc=y).
This makes me think that "*dithreitol*" could be another name for [dithiothreitol](https://en.wikipedia.org/wiki/Dithiothreitol), but I didn't manage to find any solid references to support this proposition.
To sum it up, what is dithreitol, exactly, and what chemical and structural formulae does it posess?
### References
1. Hu, S.; Li, Y.; Wang, J.; Xie, Y.; Tjon, K.; Wolinsky, L.; Loo, R. R. O.; Loo, J. A.; Wong, D. T. Human Saliva Proteome and Transcriptome. J Dent Res 2006, 85 (12), 1129–1133. DOI: [10.1177/154405910608501212](https://doi.org/10.1177/154405910608501212).
|
What is "dithreitol"? |
I've read at many places that temperature is the average kinetic energy of particles present in an object.I just don't get this intuitively how kinetic energy is connected with temperature?And how is heat connected with temperature then?then what exactly is temperature?All of the description given online is very confusing.please help!!! |
What exactly is temperature? |
I'm trying to create Copper (II) Acetate crystals, but in these times of Coronavirus it's difficult to come by hydrogen peroxide. I could be patient, but I'm not, so I'm trying to make it electrochemically. I have 5% vinegar and lots of copper scrap, and an adjustable power supply. Unfortunately I can't barely get any current going, so I've thought of adding salt, regular NaCl. I'm curious what effect this will have on the final outcome though. Balancing equations is something I struggle with, but I'd really like to learn the chemistry here. Will the salt interfere with or alter the growth of the copper acetate crystals? Ultimately I'm trying to make calcium copper acetate crystals, I've already made the calcium acetate. |
Heat is the transfer of energy to or from the body in forms other than matter flow or work (organized energy transfer, such as pushing).
Temperature is only a well-defined property for a collective body (you wouldn't be able to tell me the temperature of a single atom, for example). Like you said, it's the property of matter describing the amount of kinetic energy of the particles in the body. As to why this is, I'd ask: what happens at absolute 0?
>! At absolute 0, heat has transferred out of the system so much so that you cannot lower the energy of the system any more.
(Note for the knowledgeable reader: there is indeed still a quantum mechanical phenomena -- zero point energy-- that prevents some energy from leaving the molecule, but that's a conversation for another time).
From a thermodynamic definition, temperature is the description of how the internal energy changes with entropy for a closed (no matter flows in or out) system of constant volume:
$T=\left(\frac{\partial{U}}{\partial{S}}\right)_{N,V}$
By increasing the entropy of the system by a fixed amount, the temperature of the system tells me by how much the internal energy will increase. Now, this is not a very useful form, as you cannot directly increase the entropy of a body (you must add energy and then let the entropy indirectly increase). It's much more useful to consider the inverse temperature:
$\frac{1}{T}=\left(\frac{\partial{S}}{\partial{U}}\right)_{N,V}$
At absolute zero, everything is in the lowest energy state. Any small transfer of energy to the system will result in a large increase in entropy. But this was only a small change, so the internal energy doesn't increase by much. Compare to a system at room temperature, where I must transfer much more energy to the system to achieve the same magnitude of increase in entropy.
|
[![enter image description here][1]][1]
[1]: https://i.stack.imgur.com/28FIf.png
Here, when they write v/(mol s^-1 (kg protein)^-1) is 0.30 for the first row first column, do they mean that the value is 0.30 mol s^-1 (kg protein)^-1 or 0.30/(mol s^-1 (kg protein)^-1)?
|
I've read at many places that temperature is the average kinetic energy of particles present in an object. I just don't intuitively get how kinetic energy is connected with temperature. And how is heat connected with temperature then? Then what exactly is temperature? All of the descriptions given online are very confusing. |
I'm studying 12th. In my book they mentioned that [Co(en)3]3+ exhibit optical isomerism. But in that same book, a choose is like this * Which of this doesn't exhibit isomerism * and it's answer is [Fe(en)3]3+. They both show same structure, then why they are saying [Fe(en)3]3+ don't show isomerism? |
Do [Fe(en)3]3+ exhibit optical isomerism? |
I'm studying 12th. In my book they mentioned that $\ce{[Co(en)3]^3+}$ exhibit optical isomerism.
But in that same book, a choose is like this:
> Which of this doesn't exhibit isomerism?
and the answer is $\ce{[Fe(en)3]^3+}.$ Since both complexes have the same structure, then why they are saying $\ce{[Fe(en)3]^3+}$ doesn't show isomerism? |
Does tris(ethylenediamine)iron(III) exhibit optical isomerism? |
I'm studying 12th. In my book they mentioned that $\ce{[Co(en)3]^3+}$ exhibits optical isomerism.
But in that same book, there is a question
> Which of this doesn't exhibit isomerism?
and the answer is $\ce{[Fe(en)3]^3+}.$ Since both complexes have the same structure, then why they are saying $\ce{[Fe(en)3]^3+}$ doesn't show isomerism? |
The article by Hu et al. \[[1](https://doi.org/10.1177/154405910608501212)\] mentions a compound "*dithreitol*" in the *Saliva Proteome Analysis* section (emphasis mine):
> For the "shotgun" approach, saliva samples (1 mL) were pre-fractionated with the use of Millipore ultracentrifuge filters (Millipore Corp., Billerica, MA, USA).
Individual fractions were treated with 10 mM **dithreitol** for 30 min and then 50 mM iodoacetamide for 30 min.
[PubChem](https://pubchem.ncbi.nlm.nih.gov/#query=dithreitol) doesn't reveal any details as to what this compound exactly is and suggests alternative spelling, e.g. [D-threitol](https://pubchem.ncbi.nlm.nih.gov/compound/169019).
[Google Books search](https://www.google.com/search?hl=en&tbm=bks&q=%22dithreitol%22) results in several mentions of this compound in the literature, but I didn't find any further details except that it's probably a [sulfhydryl reagent](https://books.google.com/books?id=7JcnWmCB7EIC&q="dithreitol"&dq="dithreitol"&hl=en&redir_esc=y).
This makes me think that "*dithreitol*" could be another name for [dithiothreitol](https://en.wikipedia.org/wiki/Dithiothreitol), but I didn't manage to find any solid references to support this proposition.
To sum it up: what is *dithreitol*, exactly, and what chemical and structural formulas does it posses?
### References
1. Hu, S.; Li, Y.; Wang, J.; Xie, Y.; Tjon, K.; Wolinsky, L.; Loo, R. R. O.; Loo, J. A.; Wong, D. T. Human Saliva Proteome and Transcriptome. *J Dent Res* **2006**, 85 (12), 1129–1133. DOI: [10.1177/154405910608501212](https://doi.org/10.1177/154405910608501212). |
Recall the flame emission experiment is basically an *atomic* emission experiment. We don't say it is a compound emission experiment. The job of the flame/plsama or any high temperature medium is to decompose the compound into the constituent atoms or even ions for an experiment. Once a dilute metal salt solution is aspirated into the flame, the compound decomposes into constituents and then to the atoms. For example, you start with NaBr, NaF or NaI or NaNO3, in the flame you will get one and only emission at 589 nm from the sodium atoms. All of these compounds will make the flame bright yellow. So what happened to the non-metallic anion? Provided the flame temperature is high, one should "see" emission from non-metallic atoms as well. Our eyes cannot see the emission bc it in the deep UV region.
Sometimes the anion interferes with the atomization process such as phosphate for calcium. It forms very thermally stable compounds in the flame. You will see the signal of calcium go down quickly if phosphate ion is present.
Forget about copper as atomic emission, I guess you are thinking about the green flame of copper compounds. This only happens in low temperature flames. That color originates from molecular compounds of copper in the flame, and the anion may play a role. You need a very high temperature flame and copper emits in the UV. |
Could someone please enlighten me on what would be the theoretical effect of a counter ion on the colour of the flame emitted by a metallic salt in a flame test?
Eg. NaCl - what would be the effect of the non-metal on the general colour of the flame of the compound?
I am looking into specific groups of metallic salts based on 3 metals: Sodium, Potassium and Copper II.
I have searched throughout the literature but have not found an answer yet. |
$\newcommand{\Ket}[1]{\left|#1\right>}$
$\newcommand{\Bra}[1]{\left<#1\right|}$
$\newcommand{\BraKet}[2]{{\left<#1}\left|#2\right>}$
It's an older question, but one worth answering!
For the uninitiated, second quantization is a bookkeeping technique that describes many-particle systems as excited states of a field, usually taken to be the reference field or vacuum state. I found the most clear description of it in atomic_rabbit's answer to this [reddit post][1]. The second-quantized operator is useful because this form allows it to be applied to an indeterminately sized system (the number of particles is not explicitly stated in the second-quantized operator).
Now, onto the question:
The "first-quantized" (electronic) Hamiltonian has the form (in atomic units):
$\hat{H} = \sum_i\frac{1}{2}\nabla^2_i - \sum_i\sum_A\frac{Z_A}{r_{iA}}+\frac{1}{2}\sum_i\sum_j\frac{1}{r_{ij}} = \sum_i\hat{h}(i)+\frac{1}{2}\sum_i\sum_j\frac{1}{r_{ij}}$
which, upon second quantization, goes to:
$\hat{H} = \sum_{p,q}\Bra{p}\hat{h}\Ket{q}a^+_pa_q + \frac{1}{4}\sum_{p,q,r,s}\Bra{pq}\Ket{rs}a^+_pa^+_qa_sa_r$
In the case of "first quantization", we construct the matrix form of the operator by projecting it onto whatever basis functions we're using. We then find the eigenvalues of the operator by diagonalizing its matrix form. Your question is an excellent one, because in second quantization, this projection seems to be already done within the operator, and so we should be able to diagonalize the operator outright. Right?
The catch is that the second-quantized operator still has the creation and annihilation operators that have not yet been projected onto any basis. This is equivalent to us not knowing whether there is a particle in the p$^{th}$ state. And what *are* the states? Those are properties of the field, which we must find out by projecting the second-quantized operator onto said field.
[1]: https://www.reddit.com/r/Physics/comments/1re4f6/intuitive_understanding_of_second_quantization/ |
$\newcommand{\Ket}[1]{\left|#1\right>}$
$\newcommand{\Bra}[1]{\left<#1\right|}$
I'm currently revising for my qualifiers, and I've come across a point of confusion, one that several places give seemingly contradictory results.
To my understanding:
In Hartree-Fock, the variationally determined ground state, $\Ket\Phi$, is *not* an eigenfunction of the electronic hamiltonian. It's composed of eigenfunctions of the Fock operator. It's an *approximation* of the ground state of the Hamiltonian.
(1) Is this all correct so far?
(2) If so, is it justified to use it in numerical derivations as an eigenfunction of the Hamiltonian,
i.e. can we write $\hat{H}\Ket\Phi\overset{?}{=}E_{0}\Ket\Phi$ or are we restricted to $\Bra\Phi\hat{H}\Ket\Phi=E_{0}$ |
Are Hartree-Fock solutions eigenfunctions of the electronic hamiltonian? |
Why do most people want to believe that on burning fuel little or no carbon monoxide is created?
Yes, it could be true, mainly CO2 is formed, but read these qualifying [comments from a source](https://www.abe.iastate.edu/extension-and-outreach/carbon-monoxide-poisoning-checking-for-complete-combustion-aen-175/):
>A properly designed, adjusted, and maintained gas flame produces only small amounts of carbon monoxide, with 400 parts per million (ppm) being the maximum allowed in flue products.
And, per the same source:
>Incomplete combustion occurs because of:
>* Insufficient mixing of air and fuel.
* Insufficient air supply to the flame.
* Insufficient time to burn.
* Cooling of the flame temperature before combustion is complete.
So, perhaps more likely, the presence of CO is a reality.
In the case of your experiment, leakage (from possible thermal exposure damage and/or a poor seal) along with incomplete combustion may be some route causes (as carbon monoxide is not soluble to any degree in water). |
**Temperature vs kinetic energy**
>[OP:] I've read at many places that temperature is the average kinetic energy of particles present in an object.
Temperature has to do with the average kinetic energy of particles, but to say the two concepts are the same is incorrect. What is correct is that if the particles in two mono-atomic gas samples have the same average kinetic energy, they will have the same temperature.
>[OP:] I just don't intuitively get how kinetic energy is connected with temperature.
If you have a gas in a container (such as the air in a room), gas molecules will collide with the walls. If the walls are colder than the gas (such as a cold window pane in the winter), these collisions will slow down the gas particles on average, decreasing the temperature of the gas. If the walls are hotter than the gas (such as a window pane in the summer), these collisions will speed up the gas particles on average, increasing the temperature of the gas. Because energy (and momentum for elastic collisions) is conserved, changes in the temperature of the gas will be reflected in opposite changes in the temperature of the walls (the magnitude of change will not be the same, it depends on the heat capacities).
**Heat vs temperature**
>[OP:] And how is heat connected with temperature then?
Heat is the transfer of thermal energy. If nothing else is going on, heat transferred from sample A to sample B will go along with a drop in temperature of A and a raise in temperature of B. See also: https://chemistry.stackexchange.com/a/112057
**Definition of temperature**
>[OP:] Then what exactly is temperature? All of the descriptions given online are very confusing.
In the simplest terms, it is what you measure after you put a thermometer in thermal contact with the sample. The sensing part of the thermometer (mercury or alcohol bulb, thermocouple, etc) has to reach the same temperature as the sample. The sample should be much bigger than the sensor so that bringing them into contact does not significantly change the temperature of the sample. The temperature measured by the thermometer is equal to the temperature of the sample because they are at thermal equilibrium (heat exchange is zero), and the thermometer has some property that changes with temperature (such as the volume of alcohol) in order to sense its temperature.
The quantitative definition of temperature is given in the official definition of its SI unit Kelvin:
> The kelvin, symbol $K$, is the SI unit of thermodynamic temperature. It is defined by taking the fixed numerical value of the Boltzmann constant k to be $\pu{1.380649e−23}$ when expressed in the unit $\pu{J K−1}$, which is equal to $\pu{kg m2 s−2 K−1}$, where the kilogram, metre and second are defined in terms of $h$, $c$ and $Δν_{Cs}$.
This definition requires a lot of physical chemistry to understand. However, it is sometimes formulated as:
>One kelvin is equal to a change in the thermodynamic temperature $T$ that results in a change of thermal energy $kT$ by $\pu{1.380 649e−23 J}$.
So if the thermal energy (average per particle, not stated above) goes up, the temperature goes up. |
Why do most people want to believe that on burning fuel little or no carbon monoxide is created?
Yes, it could be true, mainly CO2 is formed, but read these qualifying [comments from a source](https://www.abe.iastate.edu/extension-and-outreach/carbon-monoxide-poisoning-checking-for-complete-combustion-aen-175/):
>A properly designed, adjusted, and maintained gas flame produces only small amounts of carbon monoxide, with 400 parts per million (ppm) being the maximum allowed in flue products.
And, per the same source:
>Incomplete combustion occurs because of:
>* Insufficient mixing of air and fuel.
* Insufficient air supply to the flame.
* Insufficient time to burn.
* Cooling of the flame temperature before combustion is complete.
So, perhaps more likely, the presence of CO is a reality.
In the case of your experiment, leakage (from possible thermal exposure damage and/or a poor seal) along with incomplete combustion may be some route causes (as carbon monoxide is not very soluble in water, 27.6 mg/L as compared to CO2 with 1,450 mg/L at 25 °C). |
**Temperature vs kinetic energy**
>[OP:] I've read at many places that temperature is the average kinetic energy of particles present in an object.
Temperature has to do with the average kinetic energy of particles, but to say the two concepts are the same is incorrect. What is correct is that if the particles in two mono-atomic gas samples have the same average kinetic energy, they will have the same temperature.
>[OP:] I just don't intuitively get how kinetic energy is connected with temperature.
If you have a gas in a container (such as the air in a room), gas molecules will collide with the walls. If the walls are colder than the gas (such as a cold window pane in the winter), these collisions will slow down the gas particles on average, decreasing the temperature of the gas. If the walls are hotter than the gas (such as a window pane in the summer), these collisions will speed up the gas particles on average, increasing the temperature of the gas. Because energy (and momentum for elastic collisions) is conserved, changes in the temperature of the gas will be reflected in opposite changes in the temperature of the walls (the magnitude of change will not be the same, it depends on the heat capacities).
**Heat vs temperature**
>[OP:] And how is heat connected with temperature then?
Heat is the transfer of thermal energy. If nothing else is going on, heat transferred from sample A to sample B will go along with a drop in temperature of A and a raise in temperature of B. See also: https://chemistry.stackexchange.com/a/112057
**Definition of temperature**
>[OP:] Then what exactly is temperature? All of the descriptions given online are very confusing.
In the simplest terms, it is what you measure after you put a thermometer in thermal contact with the sample. The sensing part of the thermometer (mercury or alcohol bulb, thermocouple, etc) has to reach the same temperature as the sample. The sample should be much bigger than the sensor so that bringing them into contact does not significantly change the temperature of the sample. The temperature measured by the thermometer is equal to the temperature of the sample because they are at thermal equilibrium (heat exchange is zero), and the thermometer has some property that changes with temperature (such as the volume of alcohol) in order to sense its temperature. See also: https://chemistry.stackexchange.com/questions/121197/temperature-measurement/121228
The quantitative definition of temperature is given in the official definition of its SI unit Kelvin:
> The kelvin, symbol $K$, is the SI unit of thermodynamic temperature. It is defined by taking the fixed numerical value of the Boltzmann constant k to be $\pu{1.380649e−23}$ when expressed in the unit $\pu{J K−1}$, which is equal to $\pu{kg m2 s−2 K−1}$, where the kilogram, metre and second are defined in terms of $h$, $c$ and $Δν_{Cs}$.
This definition requires a lot of physical chemistry to understand. However, it is sometimes formulated as:
>One kelvin is equal to a change in the thermodynamic temperature $T$ that results in a change of thermal energy $kT$ by $\pu{1.380 649e−23 J}$.
So if the thermal energy (average per particle, not stated above) goes up, the temperature goes up. |
Why do most people want to believe that on burning fuel little or no carbon monoxide is created?
Yes, it could be true, mainly CO2 is formed, but read these qualifying [comments from a source](https://www.abe.iastate.edu/extension-and-outreach/carbon-monoxide-poisoning-checking-for-complete-combustion-aen-175/):
>A properly designed, adjusted, and maintained gas flame produces only small amounts of carbon monoxide, with 400 parts per million (ppm) being the maximum allowed in flue products.
And, per the same source:
>Incomplete combustion occurs because of:
>* Insufficient mixing of air and fuel.
* Insufficient air supply to the flame.
* Insufficient time to burn.
* Cooling of the flame temperature before combustion is complete.
Also, [per Wikipedia](https://en.wikipedia.org/wiki/Carbon_monoxide), to quote:
>In the presence of oxygen, including atmospheric concentrations, carbon monoxide burns with a blue flame, producing carbon dioxide.[10]
So, many combustion reactions perhaps should be expressed in steps including:
$\ce{2 CO + O2 -> 2 CO2}$
So, perhaps more likely, the presence of CO is more of a reality than many would want to believe.
In the case of your experiment, leakage (from possible thermal exposure damage and/or a poor seal) along with incomplete combustion may be some route causes (as carbon monoxide is not very soluble in water, 27.6 mg/L as compared to CO2 with 1,450 mg/L at 25 °C). |
I'm sorry this is probably a ridiculous question. I surprisingly could not find the answer from googling.. also sorry about the tags I couldn't find better ones.
Is this the correct way to write an uncertain measurement of 0.75 mL that might be between 0.7 mL and 0.8 mL?:
0.75 mL ± 0.05 mL
Thanks |
How do I write a measurement with a margin of error using ±? |
Is this the correct way to write an uncertain measurement of 0.75 mL that might be between 0.7 mL and 0.8 mL?:
0.75 mL ± 0.05 mL |
>To calculate $\Delta H$, the change in enthalpy at $\mathrm{100^\circ C}$ for the reaction below, one needs what addition information?
>$$\ce{N2(g) + 3H2(g) -> 2NH3(g)}$$
>$\Delta H^\circ = \pu{-92.0 kJ}$ at $\mathrm{25^\circ C}$
> (a) The equilibrium constant for the reaction at $\mathrm{100^\circ
C}$
> (b) The molar heat capacities of the reaction and the products as a
function of temperature
> (c) $\Delta E^\circ$, the standard internal energy change for the
reaction
> (d) The partial pressures of the reactants and products at
$\mathrm{100^\circ C}$
> (e) The entropies of formation for the reactants and products at
$\mathrm{100^\circ C}$
The correct answer is (b), but I'm not entirely sure why.
I know it can't be (a) because I think we need $\Delta S$ as well. Also, (c) won't be helpful because it is at standard temperature, and we are wanting information at $\mathrm{100^\circ C}$. Finally, (e), seems correct but they are actually talking about *entropy* instead of *enthalpy*. However, I'm not sure how to rule out (d) or justify (b). |
I am currently conducting an experiment into the glucose content of certain sports drinks. However, I have hit a roadblock. A majority of the drinks that I intend to test, contain sucrose. I am thinking that by adding $\pu{1.0M~ HCl}$, to say, $\pu{100mL}$ of solution (sports drink), it will decompose the sucrose into its constituents. I'm planning to use Benedict's solution and a spectroscope in order to determine the amount of glucose present in solution (I'm going to add activated charcoal to the solution to remove the coloring from it before placing in the spectroscope). As a high school student I only have limited resources.
So my question is,
Assuming there is 10 grams of sucrose per $\pu{100mL}$ of solution, what volume of $\pu{1.0M~ HCl}$ would be required to completely break apart sucrose into glucose and fructose?
Would I have to use stoichiometry to find the amount?
So far I know I have $\ce{C12H22O11 + HCl}$, What happens to the $\ce{Cl}$ from $\ce{HCl}$ when fructose and glucose is formed?
|
Here is a partial answer with a quantitative description of how the hydration enthalpy could impact the electrode potential. In particular, select salts can undergo hydrolysis resulting in the liberation of H+, for example with copper or iron:
> $\ce{Fe[(H2O)6](3+) (aq) + H2O (l) = [Fe(H2O)5(OH)](2+) (aq) + H3O+ (aq)}$
Reference: one of many possible [links](https://books.google.com/books?id=8OcZSFxzQrAC&pg=PA98&lpg=PA98&dq=Cu%5B(H2O)6%5D2%2B+(aq)+%2B+H2O+(l)+%3D+%5BCu(H2O)5(OH)%5D%2B+(aq)+%2B+H3O%2B&source=bl&ots=2ztzSSKbyg&sig=ACfU3U1sErlGks377JBSPz3fqBViKo8IUQ&hl=en&ppis=_c&sa=X&ved=2ahUKEwih4Mjms6noAhXjkOAKHc9oBfYQ6AEwAnoECAkQAQ#v=onepage&q=Cu%5B(H2O)6%5D2%2B%20(aq)%20%2B%20H2O%20(l)%20%3D%20%5BCu(H2O)5(OH)%5D%2B%20(aq)%20%2B%20H3O%2B&f=false) .
This produces a pH effect, which [per a source](https://socratic.org/questions/what-is-the-equation-that-connects-ph-and-its-effect-on-electric-potential-of-an) answering the question: 'What is the equation that connects pH and its effect on electric potential of an electrochemical cell?' employs the Nernst equation to derive an explanation. I will quote but a small summary part of the results relating to acidic conditions:
>As the pH decreases, the solution is more acidic, so 10−pH=[H+] increases and 10pH−14=[OH−] decreases.
>* If H+ is a product, Q therefore increases, and the nonstandard cell potential decreases.
>* If H+ is a reactant, Q therefore decreases, and the nonstandard cell potential increases.
Note, the answer is claimed, in effect, to be very situational. |
I'm having trouble understanding why the top SN2 reaction proceeds faster. I know that benzylic and allylic halides react faster in SN2 reactions than non-conjugated halides. I think it's because the transition state is more stable but how can you explain that using MO Theory?
[![m][1]][1]
> This is what my professor drew for this question but I don't understand it all. Any help would be appreciated :)
[![Why is the top SN2 reaction faster?][2]][2]
[1]: https://i.stack.imgur.com/F6VrP.png
[2]: https://i.stack.imgur.com/A6I2z.png |
Why is the top SN2 reaction faster? (Molecular Orbital Theory) |
If 40.0 grams of Aluminum and 25.0 gram of Cl2 reacted together how much AlCl3 can be produced? |
I'm trying to migrate away from Gaussian09, using Gamess-US instead for molecular modelling of some heavy alkaline earth metal complexes, using density functional theory(DFT). In some of my previous Gaussian inputs I used the [SDD][1] (Stuttgard/Dresden) pseudopotential with the metallic cátion(Sr+2). Now I intend to reproduce this calculation in Gamess-US. After a couple of days figuring out how to use mixed basis sets in Gamess, I searched the [Basis Set Exchange][2] for this Stuttgard/Dresden ECP specification. Then I got confused, because there's no basis listed with that exact name. The closest I've found available for strontium was Stuttgart RLC and Stuttgart RSC 1997. Which of these (if any) is the best match to the SDD pseudopotential available in Gaussian? Any advice?
[1]: http://gaussian.com/basissets/
[2]: https://www.basissetexchange.org/ |
How to reproduce Gaussian09 SDD pseudopotential in Gamess-US? |
First, an answer providing a quantitative description of how the hydration enthalpy could impact the electrode potential. In particular, select salts can undergo hydrolysis resulting in the liberation of H+, for example with copper or iron:
> $\ce{Fe[(H2O)6](3+) (aq) + H2O (l) = [Fe(H2O)5(OH)](2+) (aq) + H3O+ (aq)}$
Reference: one of many possible [links](https://books.google.com/books?id=8OcZSFxzQrAC&pg=PA98&lpg=PA98&dq=Cu%5B(H2O)6%5D2%2B+(aq)+%2B+H2O+(l)+%3D+%5BCu(H2O)5(OH)%5D%2B+(aq)+%2B+H3O%2B&source=bl&ots=2ztzSSKbyg&sig=ACfU3U1sErlGks377JBSPz3fqBViKo8IUQ&hl=en&ppis=_c&sa=X&ved=2ahUKEwih4Mjms6noAhXjkOAKHc9oBfYQ6AEwAnoECAkQAQ#v=onepage&q=Cu%5B(H2O)6%5D2%2B%20(aq)%20%2B%20H2O%20(l)%20%3D%20%5BCu(H2O)5(OH)%5D%2B%20(aq)%20%2B%20H3O%2B&f=false) .
This produces a pH effect, which [per a source](https://socratic.org/questions/what-is-the-equation-that-connects-ph-and-its-effect-on-electric-potential-of-an) on the question: 'What is the equation that connects pH and its effect on electric potential of an electrochemical cell?', the answer employs the Nernst equation to derive an explanation. I will quote but a small summary part of the results relating to acidic conditions:
>As the pH decreases, the solution is more acidic, so 10−pH=[H+] increases and 10pH−14=[OH−] decreases.
>* If H+ is a product, Q therefore increases, and the nonstandard cell potential decreases.
>* If H+ is a reactant, Q therefore decreases, and the nonstandard cell potential increases.
Next, with respect to the impact of ionization, I note that this can be energy (and especially temperature) related. Quoting [a reference](https://chemistry.stackexchange.com/questions/14510/what-happens-when-we-heat-an-atom):
>As you supply heat (or any form of energy) to the system containing the atom, it's kinetic energy will increase and it will move faster through space. ... Continue to supply even more energy and more electrons will be removed from the atom.
On the connection of temperature to the electrode potential, I cite an answer [from The Student Room](https://www.thestudentroom.co.uk/showthread.php?t=527017):
>Increasing the temperature AFFECTS electrode potentials, not increases them.
>The electrode potential is a measure of the extent of a redox process - negative and the equilibrium lies relatively to the LHS (compared to the standard hydrogen electrode)
>example: $\ce{Zn(2+)(aq) + 2e- -> Zn(s)}$
>The process is not quite as simple as it appears. To move from right to left the equilibrium has to have the zinc atomised and then ionised and then the ions solvated.
Each of these processes is either exo or endothermic and is affected by the temperature.
The equilibrium between the ions and the pure metal is also affected by the temperature.
If the equilibrium is exothermic overall from left to right then an increase in the temperature will make the equilibrium move more to the left hand side making the electrode potential more negative.
The reverse is also true - if the equilibrium is endothermic overall from left to right then increasing the temperature makes the electrode potential more positive.
Note, both provided answers, are in effect, very situational. |
First, I present an example, which provides a quantitative description of how the hydration enthalpy could impact the electrode potential. In particular, select salts can undergo hydrolysis resulting in the liberation of H+, for example with copper or iron:
> $\ce{Fe[(H2O)6](3+) (aq) + H2O (l) = [Fe(H2O)5(OH)](2+) (aq) + H3O+ (aq)}$
Reference: one of many possible [links](https://books.google.com/books?id=8OcZSFxzQrAC&pg=PA98&lpg=PA98&dq=Cu%5B(H2O)6%5D2%2B+(aq)+%2B+H2O+(l)+%3D+%5BCu(H2O)5(OH)%5D%2B+(aq)+%2B+H3O%2B&source=bl&ots=2ztzSSKbyg&sig=ACfU3U1sErlGks377JBSPz3fqBViKo8IUQ&hl=en&ppis=_c&sa=X&ved=2ahUKEwih4Mjms6noAhXjkOAKHc9oBfYQ6AEwAnoECAkQAQ#v=onepage&q=Cu%5B(H2O)6%5D2%2B%20(aq)%20%2B%20H2O%20(l)%20%3D%20%5BCu(H2O)5(OH)%5D%2B%20(aq)%20%2B%20H3O%2B&f=false) .
This produces a pH effect, which [per a source](https://socratic.org/questions/what-is-the-equation-that-connects-ph-and-its-effect-on-electric-potential-of-an) on the question: 'What is the equation that connects pH and its effect on electric potential of an electrochemical cell?', the answer employs the Nernst equation to derive an explanation. I will quote but a small summary part of the results relating to acidic conditions:
>As the pH decreases, the solution is more acidic, so 10−pH=[H+] increases and 10pH−14=[OH−] decreases.
>* If H+ is a product, Q therefore increases, and the nonstandard cell potential decreases.
>* If H+ is a reactant, Q therefore decreases, and the nonstandard cell potential increases.
Next, with respect to the impact of ionization, I note that this can be energy (and especially temperature) related. Quoting [a reference](https://chemistry.stackexchange.com/questions/14510/what-happens-when-we-heat-an-atom):
>As you supply heat (or any form of energy) to the system containing the atom, it's kinetic energy will increase and it will move faster through space. ... Continue to supply even more energy and more electrons will be removed from the atom.
On the connection of temperature to the electrode potential, I cite an answer [from The Student Room](https://www.thestudentroom.co.uk/showthread.php?t=527017):
>Increasing the temperature AFFECTS electrode potentials, not increases them.
>The electrode potential is a measure of the extent of a redox process - negative and the equilibrium lies relatively to the LHS (compared to the standard hydrogen electrode)
>example: $\ce{Zn(2+)(aq) + 2e- -> Zn(s)}$
>The process is not quite as simple as it appears. To move from right to left the equilibrium has to have the zinc atomised and then ionised and then the ions solvated.
Each of these processes is either exo or endothermic and is affected by the temperature.
The equilibrium between the ions and the pure metal is also affected by the temperature.
If the equilibrium is exothermic overall from left to right then an increase in the temperature will make the equilibrium move more to the left hand side making the electrode potential more negative.
The reverse is also true - if the equilibrium is endothermic overall from left to right then increasing the temperature makes the electrode potential more positive.
Note, both provided answers, are in effect, very situational. |
First, I present an example, which provides a quantitative description of how the hydration enthalpy could impact the electrode potential. In particular, select salts can undergo hydrolysis resulting in the liberation of H+, for example with copper or iron:
> $\ce{Fe[(H2O)6](3+) (aq) + H2O (l) = [Fe(H2O)5(OH)](2+) (aq) + H3O+ (aq)}$
Reference: one of many possible [links](https://books.google.com/books?id=8OcZSFxzQrAC&pg=PA98&lpg=PA98&dq=Cu%5B(H2O)6%5D2%2B+(aq)+%2B+H2O+(l)+%3D+%5BCu(H2O)5(OH)%5D%2B+(aq)+%2B+H3O%2B&source=bl&ots=2ztzSSKbyg&sig=ACfU3U1sErlGks377JBSPz3fqBViKo8IUQ&hl=en&ppis=_c&sa=X&ved=2ahUKEwih4Mjms6noAhXjkOAKHc9oBfYQ6AEwAnoECAkQAQ#v=onepage&q=Cu%5B(H2O)6%5D2%2B%20(aq)%20%2B%20H2O%20(l)%20%3D%20%5BCu(H2O)5(OH)%5D%2B%20(aq)%20%2B%20H3O%2B&f=false) .
This produces a pH effect, which [per a source](https://socratic.org/questions/what-is-the-equation-that-connects-ph-and-its-effect-on-electric-potential-of-an) on the question: 'What is the equation that connects pH and its effect on electric potential of an electrochemical cell?', the answer employs the Nernst equation to derive an explanation. I will quote but a small summary part of the results relating to acidic conditions:
>As the pH decreases, the solution is more acidic, so 10−pH=[H+] increases and 10pH−14=[OH−] decreases.
>* If H+ is a product, Q therefore increases, and the nonstandard cell potential decreases.
>* If H+ is a reactant, Q therefore decreases, and the nonstandard cell potential increases.
Next, with respect to the impact of sublimation, I note that this can be energy (and especially temperature) related. Quoting [a reference](https://chemistry.stackexchange.com/questions/14510/what-happens-when-we-heat-an-atom):
>As you supply heat (or any form of energy) to the system containing the atom, it's kinetic energy will increase and it will move faster through space. ... Continue to supply even more energy and more electrons will be removed from the atom.
On the connection of temperature to the electrode potential, I cite an answer [from The Student Room](https://www.thestudentroom.co.uk/showthread.php?t=527017):
>Increasing the temperature AFFECTS electrode potentials, not increases them.
>The electrode potential is a measure of the extent of a redox process - negative and the equilibrium lies relatively to the LHS (compared to the standard hydrogen electrode)
>example: $\ce{Zn(2+)(aq) + 2e- -> Zn(s)}$
>The process is not quite as simple as it appears. To move from right to left the equilibrium has to have the zinc atomised and then ionised and then the ions solvated.
Each of these processes is either exo or endothermic and is affected by the temperature.
The equilibrium between the ions and the pure metal is also affected by the temperature.
If the equilibrium is exothermic overall from left to right then an increase in the temperature will make the equilibrium move more to the left hand side making the electrode potential more negative.
The reverse is also true - if the equilibrium is endothermic overall from left to right then increasing the temperature makes the electrode potential more positive.
Note, both provided answers, are in effect, very situational, with, for example, is the removal of atoms via sublimation an exothermic or endothermic process. |
I did some research. Found some semiempirical methods that work even with heavy lanthanide compounds ([RM1][1]). But could not find any references to semiempirical methods compatible with the heavier members of the alkali earth group: Sr, Ba and Ra. If there is such method, does any open source / free software implements it? I'm learning to use Gamess-US, but the only semiempirical methods implemented in this package are MNDO, AM1, PM3 and RM1 ([refs file][2]).
[1]: https://www.sciencedirect.com/science/article/abs/pii/S1010603016300557
[2]: https://www.msg.chem.iastate.edu/gamess/GAMESS_Manual/refs.pdf |
Are there any semiempirical methods that work with heavy alkaline earth metals? |
I’m designing an experiment which requires I measure final and initial lead concentrations for the determination of aqueous lead uptake by different mediums. Because lead concentrations would be measured on the ppm scale, I’d need to use a spectrophotometer to measure light absorbance by the water samplings. I could then use Beer Lambert’s Law to determine concentration.
With recent protocols surrounding COVID-19, it’s been difficult to contact an institution. I was wondering if there was any way I could measure light absorbance of my samples at home. Is there some other apparatus I could look into using (or perhaps even purchase)?
All the best! |
First, I present an example, which provides a quantitative description of how the hydration enthalpy could impact the electrode potential. In particular, select salts can undergo hydrolysis resulting in the liberation of H+, for example with copper or iron:
> $\ce{Fe[(H2O)6](3+) (aq) + H2O (l) = [Fe(H2O)5(OH)](2+) (aq) + H3O+ (aq)}$
Reference: one of many possible [links](https://books.google.com/books?id=8OcZSFxzQrAC&pg=PA98&lpg=PA98&dq=Cu%5B(H2O)6%5D2%2B+(aq)+%2B+H2O+(l)+%3D+%5BCu(H2O)5(OH)%5D%2B+(aq)+%2B+H3O%2B&source=bl&ots=2ztzSSKbyg&sig=ACfU3U1sErlGks377JBSPz3fqBViKo8IUQ&hl=en&ppis=_c&sa=X&ved=2ahUKEwih4Mjms6noAhXjkOAKHc9oBfYQ6AEwAnoECAkQAQ#v=onepage&q=Cu%5B(H2O)6%5D2%2B%20(aq)%20%2B%20H2O%20(l)%20%3D%20%5BCu(H2O)5(OH)%5D%2B%20(aq)%20%2B%20H3O%2B&f=false) .
This produces a pH effect, which [per a source](https://socratic.org/questions/what-is-the-equation-that-connects-ph-and-its-effect-on-electric-potential-of-an) on the question: 'What is the equation that connects pH and its effect on electric potential of an electrochemical cell?', the answer employs the Nernst equation to derive an explanation. I will quote but a small summary part of the results relating to acidic conditions:
>As the pH decreases, the solution is more acidic, so 10−pH=[H+] increases and 10pH−14=[OH−] decreases.
>* If H+ is a product, Q therefore increases, and the nonstandard cell potential decreases.
>* If H+ is a reactant, Q therefore decreases, and the nonstandard cell potential increases.
Next, with respect to the impact of sublimation (and relatedly, ionization), I note that this can be energy (and especially temperature) related. Quoting [a reference](https://chemistry.stackexchange.com/questions/14510/what-happens-when-we-heat-an-atom):
>As you supply heat (or any form of energy) to the system containing the atom, it's kinetic energy will increase and it will move faster through space. ... Continue to supply even more energy and more electrons will be removed from the atom.
On the connection of temperature to the electrode potential, I cite an answer [from The Student Room](https://www.thestudentroom.co.uk/showthread.php?t=527017):
>Increasing the temperature AFFECTS electrode potentials, not increases them.
>The electrode potential is a measure of the extent of a redox process - negative and the equilibrium lies relatively to the LHS (compared to the standard hydrogen electrode)
>example: $\ce{Zn(2+)(aq) + 2e- -> Zn(s)}$
>The process is not quite as simple as it appears. To move from right to left the equilibrium has to have the zinc atomised and then ionised and then the ions solvated.
Each of these processes is either exo or endothermic and is affected by the temperature.
The equilibrium between the ions and the pure metal is also affected by the temperature.
If the equilibrium is exothermic overall from left to right then an increase in the temperature will make the equilibrium move more to the left hand side making the electrode potential more negative.
The reverse is also true - if the equilibrium is endothermic overall from left to right then increasing the temperature makes the electrode potential more positive.
Note, both provided answers, are in effect, very situational, with, for example, is the removal of atoms via sublimation an exothermic or endothermic process. |
I'm trying to migrate away from Gaussian09, using Gamess-US instead for molecular modelling of some heavy alkaline earth metal complexes, using density functional theory (DFT). In some of my previous Gaussian inputs I used the [SDD][1] (Stuttgard/Dresden) pseudopotential with the metallic cation ($\ce{Sr^{+2}}$). Now I intend to reproduce this calculation in Gamess-US.
After a couple of days figuring out how to use mixed basis sets in Gamess, I searched the [Basis Set Exchange][2] for this Stuttgard/Dresden ECP specification. Then I got confused, because there's no basis listed with that exact name. The closest I've found available for strontium was Stuttgart RLC and Stuttgart RSC 1997.
Which of these (if any) is the best match to the SDD pseudopotential available in Gaussian? Any advice?
[1]: http://gaussian.com/basissets/
[2]: https://www.basissetexchange.org/ |
>In my book it is given that first electron gain enthalpy is greater than second for elements. Should we compare the magnitudes in such cases or the actual numbers with signs?
###Successive Electron Gain Enthalpies
After the addition of one electron atom becomes negatively charged and second electron is to be added to a negatively charged ion. Thus, due to electrostatic repulsion energy is *always* required for addition of second electron. *First electron gain enthalpy is always more negative than second electron gain enthalpy (which is essentially positive).* Author made an ambiguous statement, as we can't just compare the magnitudes. For more clarity see:
$$\ce{S_{(g)} + e- -> S- _{(g)}} \Delta_{eg} H^- _1 = \pu{-200kJ mol^{-1}}$$
$$\ce{S- _{(g)} + e- -> S^{2-}_{(g)}} \Delta_{eg} H^- _2 = \pu{600kJ mol^{-1}}$$.
>Does the same comparision hold with Electron Affinity at 0K?
Generally, if energy is released when an electron is added to an atom, electron affinity is taken as positive, contrary to thermodynamic convention. That is, if $\Delta_{eg} H^-$ is negative EA is positive (since temperature mentioned is absolute zero).
At absolute zero, $\Delta_{eg} H^- = \pu{-EA}$. Thus if $\Delta_{eg} H^- _1$ is negative, EA is positive and successive EAs will be increasingly negative. Here, you can say first electron affinity is greater than second (actual numbers with sign).
>For example, Be has positive Electron Affinity and F has negative Electron Affinity. So should I consider signs and say Be has more Electron Affinity than F or consider magnitude and say F has more Electron Affinity than Be?
First, $\ce{Be}$ has negative electron affinity ($\pu{-0.5 eV atom^{-1}}$) and $\ce{F}$ has positive electron affinity ($\pu{+3.4 eV atom^{-1}}$). So, fluorine has more electron affinity than beryllium, consider the actual numbers with signs).
>How to compare their Electron gain enthalpies also at 0K?
Since, at 0K, $\Delta_{eg} H^- = \pu{-EA}$. Thus, first electron enthalpy of fluorine is more negative than that of beryllium. Strictly speaking, it's value is less than that of beryllium. |
>In my book it is given that first electron gain enthalpy is greater than second for elements. Should we compare the magnitudes in such cases or the actual numbers with signs?
###Successive Electron Gain Enthalpies
After the addition of one electron atom becomes negatively charged and second electron is to be added to a negatively charged ion. Thus, due to electrostatic repulsion energy is *always* required for addition of second electron. *First electron gain enthalpy is always more negative than second electron gain enthalpy (which is essentially positive).* Author made an ambiguous statement, as we can't just compare the magnitudes. For more clarity see:
$$\ce{S_{(g)} + e- -> S- _{(g)}} \pu{\Delta_{eg} H^- _1 = -200kJ mol^{-1}}$$
$$\ce{S- _{(g)} + e- -> S^{2-}_{(g)}} \pu{\Delta_{eg} H^- _2 = 600kJ mol^{-1}}$$.
>Does the same comparision hold with Electron Affinity at 0K?
Generally, if energy is released when an electron is added to an atom, electron affinity is taken as positive, contrary to thermodynamic convention. That is, if $\pu{\Delta_{eg} H^-}$ is negative EA is positive (since temperature mentioned is absolute zero).
At absolute zero, $\pu{\Delta_{eg} H^- = -EA}$. Thus if $\Delta_{eg} H^- _1$ is negative, EA is positive and successive EAs will be increasingly negative. Here, you can say first electron affinity is greater than second (actual numbers with sign).
>For example, Be has positive Electron Affinity and F has negative Electron Affinity. So should I consider signs and say Be has more Electron Affinity than F or consider magnitude and say F has more Electron Affinity than Be?
First, $\ce{Be}$ has negative electron affinity ($\pu{-0.5 eV atom^{-1}}$) and $\ce{F}$ has positive electron affinity ($\pu{+3.4 eV atom^{-1}}$). So, fluorine has more electron affinity than beryllium, consider the actual numbers with signs).
>How to compare their Electron gain enthalpies also at 0K?
Since, at 0K, $\pu{\Delta_{eg} H^- = -EA}$. Thus, first electron enthalpy of fluorine is more negative than that of beryllium. Strictly speaking, it's value is less than that of beryllium. |
>In my book it is given that first electron gain enthalpy is greater than second for elements. Should we compare the magnitudes in such cases or the actual numbers with signs?
###Successive Electron Gain Enthalpies
After the addition of one electron atom becomes negatively charged and second electron is to be added to a negatively charged ion. Thus, due to electrostatic repulsion energy is *always* required for addition of second electron. *First electron gain enthalpy is always more negative than second electron gain enthalpy (which is essentially positive).* Author made an ambiguous statement, as we can't just compare the magnitudes. For more clarity see:
$$\ce{S_{(g)} + e- -> S- _{(g)}} \Delta_{eg} H^- _1 = \pu{-200kJ mol^{-1}}$$
$$\ce{S- _{(g)} + e- -> S^{2-}_{(g)}} \Delta_{eg} H^- _2 = \pu{600kJ mol^{-1}}$$.
>Does the same comparision hold with Electron Affinity at 0K?
Generally, if energy is released when an electron is added to an atom, electron affinity is taken as positive, contrary to thermodynamic convention. That is, if $\Delta_{eg} H^-$ is negative EA is positive (since temperature mentioned is absolute zero).
At absolute zero, $\Delta_{eg} H^- = \pu{-EA}$. Thus if $\Delta_{eg} H^- _1$ is negative, EA is positive and successive EAs will be increasingly negative. Here, you can say first electron affinity is greater than second (actual numbers with sign).
>For example, Be has positive Electron Affinity and F has negative Electron Affinity. So should I consider signs and say Be has more Electron Affinity than F or consider magnitude and say F has more Electron Affinity than Be?
First, $\ce{Be}$ has negative electron affinity ($\pu{-0.5 eV atom^{-1}}$) and $\ce{F}$ has positive electron affinity ($\pu{+3.4 eV atom^{-1}}$). So, fluorine has more electron affinity than beryllium, consider the actual numbers with signs).
>How to compare their Electron gain enthalpies also at 0K?
Since, at 0K, $\Delta_{eg} H^- = \pu{-EA}$. Thus, first electron enthalpy of fluorine is more negative than that of beryllium. Strictly speaking, it's value is less than that of beryllium. |
>In my book it is given that first electron gain enthalpy is greater than second for elements. Should we compare the magnitudes in such cases or the actual numbers with signs?
###Successive Electron Gain Enthalpies
After the addition of one electron atom becomes negatively charged and second electron is to be added to a negatively charged ion. Thus, due to electrostatic repulsion energy is *always* required for addition of second electron. *First electron gain enthalpy is always more negative than second electron gain enthalpy (which is essentially positive).* Author made an ambiguous statement, as we can't just compare the magnitudes. For more clarity see:
$$\ce{S_{(g)} + e- -> S- _{(g)}} \Delta_{eg} \pu{H^- _1 = -200kJ mol^{-1}}$$
$$\ce{S- _{(g)} + e- -> S^{2-}_{(g)}} \Delta_{eg} }pu{H^- _2 = 600kJ mol^{-1}}$$.
>Does the same comparision hold with Electron Affinity at 0K?
Generally, if energy is released when an electron is added to an atom, electron affinity is taken as positive, contrary to thermodynamic convention. That is, if $\Delta_{eg} \pu{H^-}$ is negative EA is positive (since temperature mentioned is absolute zero).
At absolute zero, $\Delta_{eg} \pu{H^- = -EA}$. Thus if $\Delta_{eg} \pu{H^- _1}$ is negative, EA is positive and successive EAs will be increasingly negative. Here, you can say first electron affinity is greater than second (actual numbers with sign).
>For example, Be has positive Electron Affinity and F has negative Electron Affinity. So should I consider signs and say Be has more Electron Affinity than F or consider magnitude and say F has more Electron Affinity than Be?
First, $\ce{Be}$ has negative electron affinity ($\pu{-0.5 eV atom^{-1}}$) and $\ce{F}$ has positive electron affinity ($\pu{+3.4 eV atom^{-1}}$). So, fluorine has more electron affinity than beryllium, consider the actual numbers with signs).
>How to compare their Electron gain enthalpies also at 0K?
Since, at 0K, $\Delta_{eg} \pu{H^- = -EA}$. Thus, first electron enthalpy of fluorine is more negative than that of beryllium. Strictly speaking, it's value is less than that of beryllium. |
>In my book it is given that first electron gain enthalpy is greater than second for elements. Should we compare the magnitudes in such cases or the actual numbers with signs?
###Successive Electron Gain Enthalpies
After the addition of one electron atom becomes negatively charged and second electron is to be added to a negatively charged ion. Thus, due to electrostatic repulsion energy is *always* required for addition of second electron. *First electron gain enthalpy is always more negative than second electron gain enthalpy (which is essentially positive).* Author made an ambiguous statement, as we can't just compare the magnitudes. For more clarity see:
$$\ce{S_{(g)} + e- -> S- _{(g)}} \Delta_{eg} H^- _1 = \pu{-200kJ mol^{-1}}$$
$$\ce{S- _{(g)} + e- -> S^{2-}_{(g)}} \Delta_{eg} H^- _2 = \pu{600kJ mol^{-1}}$$.
>Does the same comparision hold with Electron Affinity at 0K?
Generally, if energy is released when an electron is added to an atom, electron affinity is taken as positive, contrary to thermodynamic convention. That is, if $\Delta_{eg} H^-$ is negative EA is positive (since temperature mentioned is absolute zero).
At absolute zero, $\Delta_{eg} H^- = \pu{-EA}$. Thus if $\Delta_{eg} H^- _1$ is negative, EA is positive and successive EAs will be increasingly negative. Here, you can say first electron affinity is greater than second (actual numbers with sign).
>For example, Be has positive Electron Affinity and F has negative Electron Affinity. So should I consider signs and say Be has more Electron Affinity than F or consider magnitude and say F has more Electron Affinity than Be?
First, $\ce{Be}$ has negative electron affinity ($\pu{-0.5 eV atom^{-1}}$) and $\ce{F}$ has positive electron affinity ($\pu{+3.4 eV atom^{-1}}$). So, fluorine has more electron affinity than beryllium, consider the actual numbers with signs).
>How to compare their Electron gain enthalpies also at 0K?
Since, at 0K, $\Delta_{eg} H^- = \pu{-EA}$. Thus, first electron enthalpy of fluorine is more negative than that of beryllium. Strictly speaking, it's value is less than that of beryllium. |
I am asked to determine what is the O.S. of oxygen in $\ce{Na2O2}$
I was under the impression that oxygen had a charge of -2 and I multiplied that charge by 2 giving me a total of -4.
However, the computer program I am using, states that I am wrong. Can someone clarify why? |
What is the oxidation state of oxygen in Na2O2? |
Another possibility is the grease spot photometer (aka Bunsen photometer) [German Wiki page](https://de.wikipedia.org/wiki/Fettfleck-Photometer). This can be homemade, you need a piece of paper, some wax or oil, two light sources, a meter stick and for the measurement of solutions also some e.g. cardboard to shield unwanted light.
The underlying idea is that light intensity from two different sources can be compared (and found to be equal) with a paper with a grease (waxed) spot: the spot will vanish when as much light is transmitted through the spot as reflected around.
If you move a light source to achieve this condition, you can conclude relative intensities from the distance (don't forget to square).
Here's a youtube video demonstrating this: https://www.youtube.com/watch?v=sgZdl4qRa48
|
Another possibility is the **grease spot photometer** (aka Bunsen photometer) [German Wiki page](https://de.wikipedia.org/wiki/Fettfleck-Photometer). This can be homemade, you need a piece of paper, some wax or oil, two light sources, a meter stick and for the measurement of solutions also some e.g. cardboard to shield unwanted light.
The underlying idea is that light intensity from two different sources can be compared (and found to be equal) with a paper with a grease (waxed) spot: the spot will vanish when as much light is transmitted through the spot as reflected around.
If you move a light source to achieve this condition, you can conclude relative intensities from the distance (intensity goes 1 / d² with d the distance from the respective light source).
Here's a youtube video demonstrating this: https://www.youtube.com/watch?v=sgZdl4qRa48
|
The $CO$ triple bond has a bond enthalpy of 1072 kJ/mol while the $N_2$ triple bond has a bond enthalpy of 945 kJ/mol, at least according to my source. The molecules are isoelectronic, so I couldn't think of any way to explain it using MO theory. I've read many explanations that did not make much sense, including
> Of course because they are different elements and nitrogen nitrogen bond is balanced and oxygen is more electronegative than nitrogen so it will have a greater bond strength
So my question is **why is the $CO$ triple bond stronger than the $N_2$ triple bond?** |
Why is the carbon monoxide triple bond stronger than the nitrogen-nitrogen triple bond? |
Another possibility is the **grease spot photometer** (aka Bunsen photometer) [German Wiki page](https://de.wikipedia.org/wiki/Fettfleck-Photometer). This can be homemade, you need a piece of paper, some wax or oil, two light sources, a meter stick and for the measurement of solutions also some e.g. cardboard to shield unwanted light.
The underlying idea is that light intensity from two different sources can be compared (and found to be equal) with a paper with a grease (waxed) spot: the spot will vanish when as much light is transmitted through the spot as reflected around.
If you move a light source to achieve this condition, you can conclude relative intensities from the distance (intensity goes 1 / d² with d the distance from the respective light source).
Here's a youtube video demonstrating this: https://www.youtube.com/watch?v=sgZdl4qRa48
---
### How to measure photometrically
so schematically we have:
```
o--U----|----o
+---d---+
```
with light sources `o`, cuvette `U` and the spotted paper `|`. Measure distance $d$ between equal intensity paper position and the light source with the cuvette.
We have $I_{transmitted} \sim \frac{1}{d^2}$
Next, a calibration series is acquired as usual: blank (which serves as $I_0$) and concentrations over the calibration range. Almost as usual:
$$E = -\lg \frac{I}{I_0} \sim \varepsilon l c$$
$$-\lg \frac{d_0^2}{d^2} = b c$$
$$\lg d = a + b' c$$
proprtionality constant $b$ is determined by calibration.
[Standard addition](https://en.wikipedia.org/wiki/Standard_addition) instead of calibration would be possible as well.
|
Another possibility is the **grease spot photometer** (aka Bunsen photometer) [German Wiki page](https://de.wikipedia.org/wiki/Fettfleck-Photometer). This can be homemade, you need a piece of paper, some wax or oil, two light sources, a meter stick and for the measurement of solutions also some e.g. cardboard to shield unwanted light. Also the darker the room, the better.
The underlying idea is that light intensity from two different sources can be compared (and found to be equal) with a paper with a grease (waxed) spot: the spot will vanish when as much light is transmitted through the spot as reflected around.
If you move a light source to achieve this condition, you can conclude relative intensities from the distance (intensity goes 1 / d² with d the distance from the respective light source).
Here's a youtube video demonstrating this: https://www.youtube.com/watch?v=sgZdl4qRa48
---
### How to measure photometrically
so schematically we have:
```
o--U----|----o
+---d---+
```
with light sources `o`, cuvette `U` and the spotted paper `|`. Measure distance $d$ between equal intensity paper position and the light source with the cuvette.
We have $I_{transmitted} \sim \frac{1}{d^2}$
Next, a calibration series is acquired as usual: blank (which serves as $I_0$) and concentrations over the calibration range. Almost as usual:
$$E = -\lg \frac{I}{I_0} \sim \varepsilon l c$$
$$-\lg \frac{d_0^2}{d^2} = b c$$
$$\lg d = a + b' c$$
$a$ and $b$ are determined by calibration - I use here $a$ instead of $2 \lg d_0$ because several choices are possible and in general the fitted value should be used rather than the theoretical one since it includes some effects in which practice differs from theory (e.g. differences between % transmission in the grease spot from % reflection outside).
[Standard addition](https://en.wikipedia.org/wiki/Standard_addition) instead of calibration would be possible as well.
|
$\newcommand{\ket}[1]{\left|#1\right>}$
$\newcommand{\bra}[1]{\left<#1\right|}$
> (1) Is this all correct so far?
Yep.
> (2) If so, is it justified to use it in numerical derivations as an
> eigenfunction of the Hamiltonian, i.e. can we write
> $\hat{H}\ket\Phi\overset{?}{=}E_{0}\ket\Phi$ or are we restricted to
> $\bra\Phi\hat{H}\ket\Phi=E_{0}$?
First, you *can* write $\hat{H}\ket\Phi = E\ket\Phi$ where $\Phi$ is the HF wave function but then $\hat{H}$ has to interpreted as the so-called *mean-field Hamiltonian*. It is the approximate Hamiltonian in which the exact potential which represents the energy of the Coulomb repulsions between the
electrons ($\hat{V}_\mathrm{ee}$) is replaced with some approximate *mean-field potential* ($\hat{V}_\mathrm{MF}$) which describes the model system in which electrons do not instantaneously interact with each other, but rather each and every electron interacts with the average, or *mean*, electric *field* created by all other electrons. That is basically the definition of the HF wave function, i.e. it is an eigenfunction of the mean-field Hamiltonian:
$$
\hat{H}_\mathrm{MF} \ket\Phi = E_\mathrm{MF} \ket\Phi \, ,
\quad \text{where} \quad
\hat{H}_\mathrm{MF} = \hat{T}_\mathrm{e} + \hat{V}_\mathrm{en} + \hat{V}_\mathrm{MF} \, .
$$
Secondly, despite the fact that the HF wave function is not an eigenfunction of the exact electronic Hamiltonian
$$
\hat{H}_\mathrm{e} \ket\Phi \neq E_\mathrm{e} \ket\Phi \, ,
\quad \text{where} \quad
\hat{H}_\mathrm{e} = \hat{T}_\mathrm{e} + \hat{V}_\mathrm{en} + \hat{V}_\mathrm{ee} \, ,
$$
we could evaluate it energy $\bra\Phi\hat{H}_\mathrm{e}\ket\Phi$ using [Slater rules][1] and minimize it to find an upper bound to the ground state energy
$$
\bra\Phi\hat{H}_\mathrm{e}\ket\Phi \geq E_\mathrm{e0}
$$
But the resulting Slater determinant obtained by minimizing the energy is only an approximation to the ground state wave function.
[1]: https://en.wikipedia.org/wiki/Slater%E2%80%93Condon_rules |
$\newcommand{\ket}[1]{\left|#1\right>}$
$\newcommand{\bra}[1]{\left<#1\right|}$
> (1) Is this all correct so far?
Yep.
> (2) If so, is it justified to use it in numerical derivations as an
> eigenfunction of the Hamiltonian, i.e. can we write
> $\hat{H}\ket\Phi\overset{?}{=}E_{0}\ket\Phi$ or are we restricted to
> $\bra\Phi\hat{H}\ket\Phi=E_{0}$?
First, you *can* write $\hat{H}\ket\Phi = E\ket\Phi$ where $\Phi$ is the HF wave function but then $\hat{H}$ has to interpreted as the so-called *mean-field Hamiltonian*. It is the approximate Hamiltonian in which the exact potential which represents the energy of the Coulomb repulsions between the
electrons ($\hat{V}_\mathrm{ee}$) is replaced with some approximate *mean-field potential* ($\hat{V}_\mathrm{MF}$) which describes the model system in which electrons do not instantaneously interact with each other, but rather each and every electron interacts with the average, or *mean*, electric *field* created by all other electrons. That is basically the definition of the HF wave function, i.e. it is an eigenfunction of the mean-field Hamiltonian:
$$
\hat{H}_\mathrm{MF} \ket\Phi = E_\mathrm{MF} \ket\Phi \, ,
\quad \text{where} \quad
\hat{H}_\mathrm{MF} = \hat{T}_\mathrm{e} + \hat{V}_\mathrm{en} + \hat{V}_\mathrm{MF} \, .
$$
Secondly, despite the fact that the HF wave function is not an eigenfunction of the exact electronic Hamiltonian
$$
\hat{H}_\mathrm{e} \ket\Phi \neq E_\mathrm{e} \ket\Phi \, ,
\quad \text{where} \quad
\hat{H}_\mathrm{e} = \hat{T}_\mathrm{e} + \hat{V}_\mathrm{en} + \hat{V}_\mathrm{ee} \, ,
$$
we could evaluate it energy $\bra\Phi\hat{H}_\mathrm{e}\ket\Phi$ using [Slater rules][1] and minimize it to find an upper bound to the ground state energy
$$
\bra\Phi\hat{H}_\mathrm{e}\ket\Phi \geq E_\mathrm{e0}
$$
But the resulting Slater determinant obtained by minimizing the energy is only an approximation to the ground state wave function.
Finally, yes, can we use $\ket\Phi$ as our trial wave function despite the fact that it is not an eigenfunction of $\hat{H}_\mathrm{e}$ since it is not required by the variation method. $\ket\Phi$ is normalizable, it (presumably) satisfies same boundary conditions as the exact wave function, and it has adjustable parameters. That is basically enough to qualify as the trial wave function. And at the end of the day, starting from a known solution for a simpler problem is a very usual practice for variation method.
[1]: https://en.wikipedia.org/wiki/Slater%E2%80%93Condon_rules |
I have a compound, dimethylaminoethanol (DMAE) bitartrate, which I originally purchased as a supplement and possible smart drug. I bought an absurd amount of it. It didn’t do anything for me as a supplement. I understand that DMAE, not the bitartrate salt, is commonly used in cosmetics, & I’d like to play with it for that purposes, maybe mixing it with glycerin, etc. Is there a simple way to strip off the bitartrate anion and have it precipitate or sublimate or something, leaving the DMAE behind? Or otherwise separating the DMAE out from the salt?
I have a strong preference for any other reactants and reaction products being non-toxic |
How can I remove the bitartrate anion from an organic bitartrate salt? |
The problem is as follows:
> In a galvanic cell the cathode is an $Ag^{+}(1.00\,M)/Ag_{(s)}$
> half-cell. The anode is a standard hydrogen electrode immersed in a
> buffer solution containing $0.10\,M$ benzoic acid $(C_6H_5COOH)$ and
> $0.050\,M$ of sodium benzoate $(C_6H_5COO^{-}Na^{+})$. The measured
> cell voltage is $1.030\,V$. What is the $pK_a$ of benzoic acid?.
What I did to solve this problem was to find the potential for the cell involving the standard hydrogen electrode. At first I was confused because there are three "elements" featured in the problem. One being the silver electrode, the other the standard hydrogen electrode and the other a buffer solution, so I didn't know how to proceed from there.
Then I noticed that to get the constant of equilibrium I only require the concentration of $[H^{+}]$ ions as,
$K_a=\frac{[C_6H_5COO^{-}][H^{+}]}{[C_6H_5COOH]}$
Therefore to obtain those protons I did this:
The half equations in the cell are:
$\begin{array}{cc}
Ag^{+}+1e^{-}\rightarrow Ag_{(s)}&E^{0}=0.7999\,V\\
H^{+}+1e^{-}\rightarrow \frac{1}{2}H_{2(g)}&E^{0}=0.0000\,V\\
\end{array}$
Hence the overall reaction for this process would be:
$E^{0}_{cell}=E_{cathode}-E_{anode}=0.7999-0.0000=0.7999\,V$
Which is for:
$Ag^{+}+\frac{1}{2}H_{2}+\rightarrow Ag_{(s)} + H^{+}$
Hence:
$E_{cell}=E^{0}-\frac{0.0592}{n}\log\frac{[H^{+}]}{[Ag^{+}]p^{\frac{1}{2}}_{H_{2(g)}}}$
Since it indicates that the cell potential is $1.030\,V$ then:
$1.030=0.7999-\frac{0.0592}{1}\log\frac{[H^{+}]}{[1](1)^{\frac{1}{2}}}$
Solving this I'm getting:
$[H^{+}]=0.000125306\,M$
Now all that's left is to plug in this value in the equation to get the equilibrium constant:
$K_a=\frac{[C_6H_5COO^{-}][H^{+}]}{[C_6H_5COOH]}$
$[C_6H_5COO^{-}]=0.05\,M$ and $[C_6H_5COOH]=0.1\,M$
Hence:
$K_a=\frac{(0.05)(1.25306\times 10^{-4})}{(0.1)}=6.2653\times 10^{-5}$
Therefore the $pKa$ of benzoic acid would be:
$pKa=-\log Ka=-\log\left(6.2653\times 10^{-5}\right)=4.20306$
Which does seem to be within the value of benzoic acid which I have on different references. But the problem with this method it is that it required the use of logarithm.
Given this situation, does it exist an approximation or anything that can be done right of the bat to get an idea where that value would be?. Does it exist another method which I could use?. |
Perhaps you can add H2S to create PbS (as a starting step). Then, based on an extract from Wikipedia on Lead Sulfide:
>Although of little commercial value, PbS is one of the oldest and most common detection element materials in various infrared detectors.[12] As an infrared detector, PbS functions as a photon detector, responding directly to the photons of radiation, as opposed to thermal detectors, which respond to a change in detector element temperature caused by the radiation. A PbS element can be used to measure radiation in either of two ways: by measuring the tiny photocurrent the photons cause when they hit the PbS material, or by measuring the change in the material's electrical resistance that the photons cause. Measuring the resistance change is the more commonly used method. At room temperature, PbS is sensitive to radiation at wavelengths between approximately 1 and 2.5 μm. This range corresponds to the shorter wavelengths in the infra-red portion of the spectrum, the so-called short-wavelength infrared (SWIR). Only very hot objects emit radiation in these wavelengths.
some ideas. Perhaps examining a comparative study (based on known concentrations of Pb) of the changed in measured electrical resistance of a sample of PbS suspension taken from the solution after treating with the same volume of H2S followed by irradiation may provide a path. This not meant to be a complete answer, but a suggestion on a possible path to be explored.
[EDIT] To be clear, I am assuming that one cannot measure light absorbance, due to lack of equipment, for the purpose of establishing Pb concentration. But, one can perhaps measure electrical resistance. The question is can one construct a path (actually, more likely a graph to which a curve is fitted) to determine Pb concentration from the change in electrical resistance recorded upon applying a fixed amount of infrared radiation to a solution where varying amounts of PbS is introduced. If experimenting with known starting Pb concentrations, converted to PbS, suggests a detectable and measurably path with some degree of goodness-of-fit, then yes, else no (which is the only answer provided so far).
Here is some background on [PbS based photodetectors](https://www.lasercomponents.com/us/ir-components/ir-detectors/pbs-and-pbse-detectors/) where irradiation exposure varies and the amount of PbS concentration remains fixed.
>PbS is a standard SWIR semiconductor detector (1 - 3.3 µm) whereas PbSe is used in the MWIR range (1 - 4.7 µm when uncooled; up to 5.2 µm when cooled). Our lead salt detectors are photoconductive; the detector resistance is reduced during illumination. The crystal structure is polycrystalline and is produced via chemical deposition.
Please note the actual size of the PbS based photodetector.
Note, my proposed study of feasibility avoids the use of a nephelometer (which can measure the concentration of some inert suspended particulates in say a liquid colloid by employing a light 'source' beam and a second light detector at 90° to the source beam) by taking advantage of the properties of irradiated PbS particulates. Also, it should be noted that a spectrophotometer, based on a quantitative measurement of the reflection or transmission properties of a material as a function of wavelength, based again on light, is also presumably not an available option.
[EDIT EDIT] Just found a reference, [THE RELATIONSHIP BETWEEN ELECTRICAL RESISTANCE AND DISPERSED PHASE CONCENTRATION IN OIL IN WATER EMULSIONS](https://onlinelibrary.wiley.com/doi/abs/10.1111/j.2042-7158.1962.tb11129.x) to quote:
>The electrical resistance of a series of oil‐in‐water emulsions has been measured, and used to test the validity of four equations relating resistance with the concentration of dispersed phase. A modification of one of the equations was found to give the best relationship.
In the current context, we have a suspension of irradiated PbS in water (releasing electrons, per this 2013 article ['Photoinduced electron transfer from PbS quantum dots to cobalt(III) Schiff base complexes: light activation of a protein inhibitor'](https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3816745/)), which I would have guessed is more promising in determining the PbS concentration than say of oil in a water emulsion.
Also found a new method to [quote a source](https://tsapps.nist.gov/publication/get_pdf.cfm?pub_id=905447):
>In summary, we have demonstrated a practical way to measure nanoparticle concentration in a colloidal solution using quartz crystal microgravimetry. Application of a small drop of the nanoparticle colloid in a volatile organic solvent to the crystal surface leaves a dry nanoparticle film after solvent evaporation. Crystal resonant frequency shifts obeyed Sauerbrey’s equation for the dry nanoparticle concentrations up to 1300 µg/mL, as calibrated using a set of serial dilutions of Si and Ag nanopowders in methanol, rhodamine B in methanol, and ferrocene in cyclohexane. |
Take $\rm NaCl$ for instance, a salt that will melt when it reaches its melting point, and compare it with $\rm NH_4NO_3$, a salt that doesn't melt, but instead decomposes to $\rm N_2O$ and $\rm H_2O$ on heating to a high enough temperature.
How can we determine if a salt will simply melt, or if it will decompose on heating? |
How can we tell if a salt will melt or decompose? |
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