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Does DMT-Acetate become Polar in H20, with added NACL? |
I'm extracting DMT from Acacia Confusa via Acid/Base. I extracted the DMT with $\ce{H2O}$ and acetic acid, converting it to DMT-Acetate. Here, I used a non polar solvent, naptha, to defat the material, causing an emulsion.
I used $\ce{NaCl}$ to break the emulsion, this caused a thick layer of black tar to form on ... |
Does DMT-Acetate become Polar in H2O, with added NaCl? |
When white tin ($\beta$-tin) is exposed to temperature below 13.2 °C it creates the allotrope of gray ($\alpha$-tin), a gray amorphous powder.
My question is that once you have the powdered gray tin, doe just raising the temperature above the point of stability turn it back to white tin, but in a powdered form?
... |
It seems that the results of the calculations are more or less fine and the OP just misinterpreted the [NIST data][1]. As I said in my comment above, NIST does not claim that $\lambda_{\mathrm{max}}=276 \, \mathrm{nm}$. Clearly only a small region of wavelength is shown on the graph and in the paper referenced on the N... |
It seems that the results of the calculations are more or less fine and the OP just misinterpreted the [NIST data][1]. As I said in my comment above, NIST does not claim that $\lambda_{\mathrm{max}}=276 \, \mathrm{nm}$. Clearly only a small region of wavelength is shown on the graph and in the [paper][2] referenced on ... |
I'm extracting DMT from Acacia Confusa via acid/base. I extracted the DMT with $\ce{H2O}$ and acetic acid, converting it to DMT-acetate. Here, I used a non polar solvent, naptha, to defat the material, causing an emulsion.
I used $\ce{NaCl}$ to break the emulsion, this caused a thick layer of black tar to form on ... |
Not in an atomic aspect, I learned that fluoroscence is the emmition of light from substances until the source of light dissapears ( there is no emittion in the dark ), on the other hand phosphoroscence, due to a blocking of electrons on their way back to normal state, is the emittion of light from substances even with... |
What are the components of and how would one go about making [polyester casting resin][1]?
[1]: http://www.dickblick.com/products/castincraft-clear-polyester-casting-resin/ |
How to make polyester casting resin? |
Not in an atomic aspect, I learned that fluoroscence is the emission of light from substances until the source of light dissapears (there is no emittion in the dark), on the other hand phosphoroscence, due to a blocking of electrons on their way back to normal state, is the emission of light from substances even withou... |
> If it is so, then why do we call glowing objects in the dark as *fluoroscent*?
I think this is just marketing: most people have heard the term *fluorescence*, while *phosphorescence* or luminescence are largely unknown terms.
While a delayed emission from a singlet state is possible, most glow-in-the-dark items... |
I'm extracting DMT (N,N-dimethyltryptamine) from *Acacia confusa* via acid/base. I extracted the DMT with $\ce{H2O}$ and acetic acid, converting it to DMT-acetate. Here, I used a non polar solvent, naptha, to defat the material, causing an emulsion.
I used $\ce{NaCl}$ to break the emulsion, this caused a thick lay... |
> Does NaCl cause DMT acetate (or any acetate) to become polar in $\ce{H2O}$?
In aqueous acidic solution, that is in the presence of a sufficient amount of acetic acid, N,N-dimethyltryptamine is protonated. The acetate is the counter ion of the ammonium salt. The compound does not become more polar by adding sodium... |
The compound below is fexofenadine and the only chirality center, as identified by the textbook is, the one circled in red. I do not understand why the carbon attached to the O by a double bond and the hydroxide group is not a chirality center, too. Isn't it attached to a tert-Butyl substituent group? So it is attached... |
If in doubt, simplify, examine both centres separately and draw.
![stereocentre][1]
Can you do the same for the carboxylic acid?
[1]: https://i.stack.imgur.com/XqrW9.png |
I've been doing some reading about enantiomers, which are apparently chiral molecules that are non-superimposable mirror images of each other. An example I found of this in some problems that I've been attempting seems to have stumped me somewhat:
Take the compound CH2=CH-CHD2, where D is deuterium, or hydrogen-2, a... |
Why does this compound only have one pair of enantiomers? |
I've been doing some reading about enantiomers, which are apparently chiral molecules that are non-superimposable mirror images of each other. An example I found of this in some problems that I've been attempting seems to have stumped me somewhat:
Take the compound $\ce{CH2=CH-CHD2}$, where $\ce{D}$ is deuterium, or... |
I've been doing some reading about enantiomers, which are apparently chiral molecules that are non-superimposable mirror images of each other. An example I found of this in some problems that I've been attempting seems to have stumped me somewhat:
Take the compound $\ce{CH2=CH-CHD2}$, where $\ce{D}$ is deuterium, or... |
I'm extracting DMT (N,N-dimethyltryptamine) from *Acacia confusa* via acid/base. I extracted the DMT with $\ce{H2O}$ and acetic acid, converting it to DMT-acetate. Here, I used a non polar solvent, naptha, to defat the material, causing an emulsion.
I used $\ce{NaCl}$ to break the emulsion, this caused a thick lay... |
Even chemistry textbooks less than five years old start by describing atomic structure based only on protons, neutrons and electrons. Is an understanding of chemistry fundamentals likely to become out of date when the textbooks start to include quarks, leptons and the dozens of other sub-atomic particles? |
Should chemistry students worry about quarks and such? |
I mean to be honest with you, when I was in Chem 1, we never went into the details of quarks and other sub-atomic particles except once; in nuclear chemistry. Even then, you didn't have to have an idea of these subatomic particles to understand what was going on.
So, if you think about it, wouldn't they (teachers) ... |
I was holding a piece of plastic earlier, and I ripped it in half. This caused me to start think about what was happening at the atomic level. If the plastic is being ripped, clearly bonds are being broken and new bonds are being formed, so is this not a chemical reaction?
I realize that the plastic retains all of i... |
Ripping apart a polymer - reaction or not? |
Let's say we have substance $\ce{A}$, which is mixed with substance $\ce{B}$ to improve shelf-life because $\mathrm{p}K_\mathrm{a}$ of the substance $\ce{A}$ is $7.9$ and in mix the $\mathrm{pH}$ is $5.2$.
Does this mean that in the solution with $\mathrm{pH}$ near $8$ the substance $\ce{A}$ **has multiple molecul... |
How do you explain pKa to non-professional? |
I'm wanting to do a follow up to this Q&A in the cooking stack regarding [browning avocados](http://cooking.stackexchange.com/questions/46494/browning-avocados-whats-happening-what-helps). I specifically want to expand my experiment to to include the options in [Wayfaring Stranger's Answer](http://cooking.stackexchange... |
Agreed: a physical action can cause a chemical reaction. For example, any attempt to manipulate $\ce{NI3}$ physically causes it to decompose (detonate). For that matter, dividing metals into tiny pieces changes chemical behavior, because bonds that would have been hidden in the interior are exposed at the surface. Fine... |
>Is an understanding of chemistry fundamentals likely to become out of date when the textbooks start to include quarks, leptons and the dozens of other sub-atomic particles?
Well, the books do already include leptons, at least electrons and usually positrons.
Other than protons, neutrons, electrons and positrons,... |
When you rip, tear or mechanically deform a polymer (for example a piece of plastic) you are putting energy into the material. The energy from this deformation causes the polymer chains in the vicinity of the deformation to attempt to align. To some degree this partial alignment makes continued deformation easier. T... |
When you rip, tear or mechanically deform a polymer (for example a piece of plastic) you are putting energy into the material. The energy from this deformation causes the polymer chains in the vicinity of the deformation to attempt to align. To some degree this partial alignment makes continued deformation easier. T... |
When you rip, tear or mechanically deform a polymer (for example a piece of plastic) you are putting energy into the material. The energy from this deformation causes the polymer chains in the vicinity of the deformation to attempt to align. To some degree this partial alignment makes continued deformation easier. T... |
If I had a container in vacuum filled with $\ce{I2}$ gas and then I bombarded it with high speed electrons using an electron gun, would be able to get both $\ce{I+}$ and $\ce{I-}$ ions or would I only get $\ce{I2+}$ plus an ejected electron? |
When white tin ($\beta$-tin) is cooled to a temperature below $13.2°\mathrm{C}$, it creates the allotrope of gray ($\alpha$-tin), a gray, amorphous powder.
My question is that once you have the powdered gray tin, doe just raising the temperature above the point of stability turn it back to white tin, but in a powde... |
I was holding a piece of plastic earlier, and I ripped it in half. This caused me to start to think about what was happening at the atomic level. If the plastic is being ripped, clearly bonds are being broken and new bonds are being formed, so is this not a chemical reaction?
I realize that the plastic retains all o... |
When you rip, tear or mechanically deform a polymer (for example a piece of plastic) you are putting energy into the material. The energy from this deformation causes the polymer chains in the vicinity of the deformation to attempt to align. To some degree this partial alignment makes continued deformation easier. T... |
According to the UC Davis ChemWiki [Chemistry of Helium][1], helium has a comparatively unusual property, specifically:
> Helium is the only element that cannot be solidified by lowering the temperature at ordinary pressures.
'Ordinary' referring to standard air pressure (1 atmosphere). In order to solidify, the... |
Is there an equation to figure out what the temperature of an the area will be a certain distance from the heat source?
For example, If I know there is a heat source at $0~\mathrm{m}$, and the temperature is $100°\mathrm{C}$, is there a way to find out what the temperature will be at $50~\mathrm{m}$ away from the h... |
On a test question, my organic chemistry professor gave us a really big molecule and told us to give list the formal charges on each atom. There was a $\ce{Br}$ with 3 lone pairs and a double bond to carbon (which then had a single bond to a $
\ce{H}$ and a $\ce{N}$). On the key, he labeled it with a + charge and I wa... |
Would the electrolysis of alcohol, specifically spirits, result in a potable end product? I'm looking for some novel cocktail ingredients. |
You pour $50~\mathrm{mL}$ of a $0.0200~\mathrm{M}$ $\ce{HCOOH}$-solution to $150~\mathrm{mL}$ of a $0.0500~\mathrm{M}$ $\ce{HCOONa}$-solution. The $K_\mathrm{a}$ of $\ce{HCOOH}$ is $1.8 \times 10^{-4}$ and $\mathrm{p}K_\mathrm{a}$ is $3.74$.
a) Compute the concentrations of $\ce{HCOOH}$ and its salt after dilution t... |
At $25°\mathrm{C}$, $10.24~\mathrm{mg}$ of $\ce{Cr(OH)2}$ are dissolved in enough water to make $125~\mathrm{mL}$ of solution. When equilibrium is established, the solution has a $\mathrm{pH}$ of $8.49$. Estimate $K_\text{sp}$ for $\ce{Cr(OH)2}$ (Ans : $1.47\times 10^{-17}$)
I have calculated it but not have the sam... |
Given the concentration of bleach ($6\%\,\mathrm{m/m}$), density of bleach ($1.07\,\mathrm{g\,mL^{-1}}$), and molar mass of bleach ($74.44\,\mathrm{g\,mol^{-1}}$) how many moles of bleach are in a $4.0\,\mathrm{mL}$ sample? I don't understand where the concentration factors in, but the problem says to use it. |
Is there any way that i can detect lead in gasoline, not by sending of to a lab but to be able to do this by my self? |
Is there any way that i can detect lead in gasoline, not by sending of to a lab but to be able to do this by myself? |
>Given that $K_{\mathrm{sp}}(\ce{CaF2})=3.2\cdot 10^{-11}$, $K_{\mathrm{a}}(\ce{HF})=10^{-5}$, and $K_{\mathrm{w}}=10^{-14}$.
>Calculate $\mathrm{pH}$ of saturated $\ce{CaF2}$.
Here is my approach,
$$\ce{CaF2 -> Ca^2+ +2F-}$$
\begin{align}
K_{\mathrm{sp}} &=[\ce{Ca^2+}] \cdot [\ce{F-}]^{2}\\
K_{\mathrm{sp}}... |
How to calculate the pH of a saturated solution from its solubility constant Kₛₚ and acidity constant Kₐ? |
It seems that the results of the calculations are more or less fine and the OP just misinterpreted the [NIST data][1]. As I said in my comment above, NIST does not claim that $\lambda_{\mathrm{max}}=276 \, \mathrm{nm}$. Clearly only a small region of wavelength is shown on the graph and in the paper<sup>\[1\]</sup> ref... |
Here is the output of a TDDFT run.
How can we interpret the orbital transitions ? for example n -> σ* or π -> π* ?
![enter image description here][1]
[1]: https://i.stack.imgur.com/VQwuy.png |
How to interpret orbital transition in TDDFT? |
At $25~\mathrm{^\circ C}$, $10.24~\mathrm{mg}$ of $\ce{Cr(OH)2}$ are dissolved in enough water to make $125~\mathrm{mL}$ of solution. When equilibrium is established, the solution has a $\mathrm{pH}$ of $8.49$. Estimate $K_\text{sp}$ for $\ce{Cr(OH)2}$ (Ans : $1.47\times 10^{-17}$)
I have calculated it but not have ... |
I'm preparing for an experiment to investigate some factor affecting buffer capacity. At the moment, I'm looking at using a $NH_3$ (2M) $+$ $NH_4Cl$ buffer, but I'm not sure which factor to test. My research tells me that only the concentrations of each will affect buffer capacity, and that the ratio of their concentra... |
I'm preparing for an experiment to investigate some factor affecting buffer capacity. At the moment, I'm looking at using a $\ce{NH_3}$ (2M) $+$ $\ce{NH_4Cl}$ buffer, but I'm not sure which factor to test. My research tells me that only the concentrations of each will affect buffer capacity, and that the ratio of their... |
> As far as I can tell, something like temperature won't affect buffer capacity.
In this particular case, I wouldn't be too sure about it. Isn't it conceivable that your buffer will lose ammonia upon heating?
Did you already decide about the pH indicators? For the lower end, thymol blue might be a good choice. T... |
I found this [article][1], but it was not helpful. The $\ce{In2O3}$ is probably amorphous. $\ce{HNO3}$ and $\ce{HCl}$ do not work. Any ideas?
[1]: http://www.sciencedirect.com/science/article/pii/0022024876902311 |
How to dissolve composit $In_2O_3$ with $SnO_2$ without affecting stainless steel? |
How to dissolve composite In2O3 with SnO2 without affecting stainless steel? |
Iron(II) sulphate solution is added to a test-tube. Following this, potassium manganate solution is added gradually to the test-tube as well. The observation seen is that the green solution turns pink at first, then potassium manganate turns colourless as manganese(VII) is oxidised to manganese(II).
But Fe(II) is ox... |
Top hits on google are:
http://bilbo.chm.uri.edu/CHM112/tables/KspTable.htm
http://www4.ncsu.edu/~franzen/public_html/CH201/data/Solubility_Product_Constants.pdf
Which vary by 1 unit in its exponent.
What can be the reason for this variation? Experimental conditions?
Where can I get accur... |
What is the Ksp of Al(OH)3? |
Top hits on google are:
http://bilbo.chm.uri.edu/CHM112/tables/KspTable.htm
Aluminum hydroxide $Al(OH)_3$ >> $ 1.3×10^{–33}$
http://www4.ncsu.edu/~franzen/public_html/CH201/data/Solubility_Product_Constants.pdf
Aluminum hydroxide $Al(OH)_3$ >> $ 3×10^{–34}$
Which vary by 1 unit in its ex... |
Top hits on google are:
http://bilbo.chm.uri.edu/CHM112/tables/KspTable.htm
Aluminum hydroxide $\ce{Al(OH)_3}$ >> $ 1.3×10^{–33}$
http://www4.ncsu.edu/~franzen/public_html/CH201/data/Solubility_Product_Constants.pdf
Aluminum hydroxide $\ce{Al(OH)_3}$ >> $ 3×10^{–34}$
Which vary by 1 unit... |
Is there an equation to figure out what the temperature of an the area will be a certain distance from the heat source?
For example, If I know there is a heat source at $0~\mathrm{m}$, and the temperature is $100~\mathrm{^\circ C}$, is there a way to find out what the temperature will be at $50~\mathrm{m}$ away fro... |
I read about entropy from different sources. Still don't get why (universal) entropy change has to be greater than zero for irreversible process, i.e. $\mathrm{d}\;S_\mathrm{Univ}>0$. Is it a result of pure observation or is there any theoretical background of this phenomena?
Unfortunately the linked question "http:... |
Why does everything go in the direction of positive entropy change? |
Logically thinking, I thought that it would be high pressure and high temperature that would create high density poly(ethene)... However, that's not the case. Is it because of the catalyst? Why is the condition low pressure/temperature? |
Why does addition polymerisation of high density poly(ethene) use low temperature and pressure? |
this is your output, right?
TRANSITION DIPOLE, A.U. OSCILLATOR
HARTREE EV X Y Z STRENGTH
0 A -193.0290234748 0.000
1 A -192.8724089055 4.262 0.0001 0.0000 0.0001 0.000
2 A -192.783133562... |
Form this output you dont.
Find this in the output:
EXCITATION DE-EXCITATION
OCC VIR AMPLITUDE AMPLITUDE
I A X(I->A) Y(A->I)
--- --- -------- --------
10 17 -0.039945 0.001926
16 17 0.996619 -0... |
I was checking dipole moment orders, and came across a very peculiar result:
Methyl fluoride has lesser dipole moment than methyl chloride, but hydrofluoric acid has more dipole moment than hydrochloric acid.
At first I thought it was about distance, but upon seeing the second one, it led me to think about polariza... |
Which factors dominate dipole moments? Is there a anomaly between fluorine and chlorine? |
What is the graph of carbon dioxide ($\ce{CO2}$) content in the air in a compartment consisting of 80 percent air and 20 percent ocean sea water as a function of temperature, in the temperature range 0 to 60 degree Celsius?
Example but not the complete answer:
![enter image description here][1]
As can be seen ... |
Let me add to the confusion:
- https://books.google.de/books?id=imBzjH6SHi8C&pg=PA303&lpg=PA303&dq=solubility+product+constant+aluminum+hydroxide&source=bl&ots=jXoPiVihnF&sig=eSK8saszXshTXVEKGOlThUiKVr0&hl=de&sa=X&ei=Xl5TVcepDuj_ywOXmYDQBA&ved=0CG8Q6AEwCTgK#v=onepage&q=solubility%20product%20constant%20aluminum%2... |
To visualise the orbitals of your [calculation][1], use a program of your choice. For Gamess, there are [a few options availabile][2]. I use ChemCraft and Molden, and they work quite well. Here is a compilation of some with the former mentioned:
![orbitals of acetone][3]
You can further use the summary to identif... |
I have a couple of big barrels with a $50\,\mathrm{L}$ demi-water solution of the following nutrients:
$$
\begin{array}{rl}
200~\mathrm{\mu M} & \ce{NH4NO3} & +\\
30~\mathrm{\mu M} & \ce{KCl} & +\\
10~\mathrm{\mu M} & \ce{CaCl2} & +\\
10~\mathrm{\mu M} & \ce{KH2PO4} & +\\
\\
10~\mathrm{\mu M} & \ce{Fe-EDTA} &... |
Why does the pH of this solution rise instead of fall when exposed to the atmosphere? |
Consider the following TDDFT run with GAMESS:
! File created by the GAMESS Input Deck Generator Plugin for Avogadro
$BASIS GBASIS=N311 NGAUSS=6 $END
$CONTRL SCFTYP=RHF RUNTYP=ENERGY TDDFT=EXCITE DFTTYP=B3LYP $END
$CONTRL ICHARG=0 MULT=1 $END
$TDDFT NSTATE=9 $END
$STATPT OPTTOL... |
> Care must be taken in choosing an electrolyte, since an anion from the electrolyte is in competition with the hydroxide ions to give up an electron. An electrolyte anion with less standard electrode potential than hydroxide will be oxidized instead of the hydroxide, and no oxygen gas will be produced. A cation with a... |
Is there an electrolyte that will halt the production of Hydrogen in water electrolysis? |
Top hits on google are:
[This site][1]
Aluminum hydroxide $\ce{Al(OH)_3}$ >> $ 1.3×10^{–33}$
and
[This site][2]
Aluminum hydroxide $\ce{Al(OH)_3}$ >> $ 3×10^{–34}$
which is one order of magnitude different.
What can be the reason for this variation? Experimental conditions?
Where c... |
> Care must be taken in choosing an electrolyte, since an anion from the electrolyte is in competition with the hydroxide ions to give up an electron. An electrolyte anion with less standard electrode potential than hydroxide will be oxidized instead of the hydroxide, and no oxygen gas will be produced. A cation with a... |
This answer us quite long, but at each stage should be straight forward and will hopefully show you exactly where the second law comes from in classical thermodynamics. If you are going to spend the time learning something you might as well do it properly right?
**Introduction**
The first law of thermodynamics ... |
I am trying to identify whether the following objects possess planes of symmetry or not, but my answers seem different, from the textbook solutions (Klein Organic Chemistry 2e, Page 220, Question 23). Some of them I am unsure of and would like to make sure I know the proper reasoning for. Can someone please clarify?
... |
Why sodium sulphate has water absorbing capability? |
A solution is created by mixing $250\,\mathrm{mL}$ of $1\,\mathrm{M}$ $\ce{HCl}$, $250\,\mathrm{mL}$ of $1\,\mathrm{M}$ $\ce{CH3COOH}$ and $500\,\mathrm{mL}$ of $1.5\,\mathrm{M}$ $\ce{CsOH}$.
The $K_\mathrm{a}(\ce{CH3COOH}) = 1.8 \times 10^{-5}$
What is the $\mathrm{pH}$ of this solution?
I am stuck between $\... |
The explanation I have found is that primary cells have irreversible reactions, so passing current in the opposite direction cannot recharge the cell. But as far as I understood it, no reaction is really irreversible - what one calls irreversible simply has very high values of the equilibrium constant, so for all inten... |
Why can't a primary cell be recharged? |
When an electron of an atom returns from an excited state to the ground state, it emits energy in the form of a photon. How does the change in energy level compare to the energy of the emitted photon? |
How do I calculate the viscosity of a mixture of Toluene and isopropanol? |
The bond enthalpy associated with a $\ce{O=O}$ double bond is equal to 495 kJ/mol. Does that mean that adding enough kinetic energy in the form of heat will eventually cause the bonds to break and create monoatomic oxygen? |
How do I calculate the viscosity of a mixture of toluene and isopropanol? |
When an electron de-excites from a higher energy orbital to a lower energy orbital it loses energy by emitting a photon. Since energy must be conserved, the energy of this photon is equal to the amount of energy the electron has lost, or in other words the difference in energy of the two orbitals involved. The frequenc... |
If you have, for example, 10 g of a solute dissolved into 100 ml of a solution, and the solution density is $1.5\ \mathrm{\frac{g}{ml}}$, can you just multiply $100\ \mathrm{ml} \times 1.5\ \mathrm{\frac{g}{ml}}$, then subtract the 10 g of solute from the result to get the mass of the solvent in grams? Does that work o... |
I read about entropy from different sources. Still don’t get why (universal) entropy change has to be greater than zero for irreversible process, i.e. $\mathrm{d}S_\mathrm{Univ}>0$. Is it a result of pure observation or is there any theoretical background of this phenomena?
Unfortunately the linked question "http://... |
> When an electron of an atom returns from an excited state to the ground state, it emits energy in the form of a photon.
That is one of several pathways for deexcitation. Often, *internal conversion*, the radiationless deeexcitation happens instead.
> How does the change in energy level compare to the energy of ... |
I know that $\ce{H+}$ is not possible in water and it is present as $\ce{H3O+}$. But later on I come to know that even $\ce{H3O+}$ is not possible and that it is present as $\ce{H9O4+}$.
Why does this happen? What give that compound so much stability that is not present in $\ce{H3O+}$ or $\ce{H+}$? |
I think that for this type of question a bit of context as to why you are asking the question, would be useful.
If for instance you are in an Intro Chem class I would say the following:
**All chemical systems tend towards their most stable state (in the absence of external energy being applied).**
$\ce{H^+}$... |
I've been asked to compare the relative acidity of $\ce{CF4}$ and $\ce{SiF4}$.
As both have noble gas structures neither can act as a lewis acid and so I assume the question is inferring that they are in solution.
My guess is that under aqueous conditions they could react via some sort of hydrolysis to give $\ce{H... |
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