May 03, 2024, 07:25:30 AM
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That's chitinase test kit.  Sorry, spell checker. It's chitinase3-like 1 human.  Whether it would work, probable not.  But it might point you in the right direction.
I am asking for information on measurements that have already been done, not advertisement for products. And given that the plants did evolve the chitinase, it should have some effects on at least fungi, possibly also small insects. Do you know any information on activity levels?
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Organic Chemistry Forum / Re: Is SOCl2 needed?
« Last post by Hunter2 on Today at 05:19:37 AM »
It's an oxidiser. To oxidise Alkohol to Aldehyde. Yes with LAH the Reduction goes direction to the Alkohol.
But if so then LAH can be used to refuse the carbonic acid. Maybe SOCl2 is only used to remove water.
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That's chitinase test kit.  Sorry, spell checker. It's chitinase3-like 1 human.  Whether it would work, probable not.  But it might point you in the right direction.
4
There is a company called BioOcean that offers a Chinese test kit.  It's Elisa based and whether it will work for what you want- who knows? You might look it up.  It is not cheap.
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Organic Chemistry Forum / Re: Is SOCl2 needed?
« Last post by rolnor on Yesterday at 03:01:09 PM »
I think with the Thionylchloride you create first the cyclopentylcarboxylchloride, this should be reduced to the aldehyde to obtain in the end the Acetale. If you get total reduction to the cyclopentylmethanol then it is not possible to fo further.

But there is PCC in the reagent list, do you know what that is for? Can you reduce an acid chloride to aldehyde with LAH?
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Physical Chemistry Forum / Re: Baking Soda + Vinegar
« Last post by Corribus on Yesterday at 01:10:19 PM »
I think you are going to have to explain what you mean by "more powerful".

From the standpoint of physics, "power" is defined as "energy per time". So if you mean "more energy released by the reaction per unit of time", you have two ways to do this: increase the amount of energy and decrease the amount of time. These two things are related, though. You might say, well I can just increase the amount of reactants. More reactants means more energy produced, so more power. It's not quite that simple, because rate depends on reactant concentration, and that dependence varies from reaction to reaction. So, while more reactants may mean more energy, it's not necessarily the case that more reactants means more power over every time interval. The best way to increase power may be to decrease the amount of time reactants take to react, and one way to do that is, as noted by the previous poster, to increase the temperature. An increase in temperature will also affect the amount of energy produced per amount of reactant, though, and that may be for the positive or negative depending on the reaction. So it is entirely possible that, while increasing temperature may decrease the time over which energy is produced, it may also decrease the energy produced, so the change in power may not increase like you think it does. An even better way would be to use a catalyst. A catalyst speeds a reaction up but does not, to a first approximation, change the amount of energy produced or consumed. It only changes the amount of energy to get the reaction going. So, to increase the power of a reaction, use a catalyst. Is there a catalyst for an acid/base neutralization reaction? The thing is, neutralization reactions are already pretty quick, so you may have trouble finding something to make them quicker. You may take other mechanical steps, like making sure the reaction is well-stirred. This isn't really a catalyst but it will ensure the reaction rate is not diffusion limited. So, theoretically this can increase the power production of the reaction, although gains may be modest.

Of course, maybe you have something else in mind when you talk about "more powerful"?
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Physical Chemistry Forum / Re: Baking Soda + Vinegar
« Last post by Hunter2 on Yesterday at 11:51:16 AM »
Heat is the main thing, second is increase concentration of baking soda and amount of acetic acid.
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Physical Chemistry Forum / Baking Soda + Vinegar
« Last post by BlindGoat on Yesterday at 11:27:35 AM »
I come here in first to apologize for my bad english, as it's not my first language and i still have to master it in a formal way.
I need help with the following thing.
HOW TO MAKE THE BAKING SODA + VINEGAR REACTION MORE POWERFUL?
I know about heating up the vinegar, because it accelerates the molecules, but i need to know if there are other ways.
I cannot add more things into the reaction, only searching a way to make it more powerful.
Thanks for your time!
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High School Chemistry Forum / Re: Unfinished high school Chem question...
« Last post by Corribus on May 01, 2024, 10:05:13 PM »
I do not think many people would agree with your teacher regarding (b), although I can anticipate the argument being made. From the standpoint of solubility equilibrium, the argument being made is something like this: without a precipitate, the solution cannot reach a point where Q = K. Therefore, the solution cannot be at equilibrium.

Solubility equilibria are a little unique because one of the participants in the “reaction” is a solid, which by definition has an activity of one. Therefore, Q only depends on the concentrations of the aqueous solutes. The solute(s) only begin to drop out of solution when their concentrations reach the point such that Q is equal to K. After that, any additional solute drops out of solution, and Q still is equal to K, because the precipitate does not contribute to the equilibrium expression. If you define the point of equilibrium strictly in terms of Q and K, then it is true that an unsaturated solution does not satisfy this criterion for equilibrium.

That’s all well and good, but if we take a broader view of equilibrium, this changes. Equilibrium is commonly defined as a state in which the forward and reverse rates of any process involved are equal. More importantly, there will be no tendency for the system to change without an external stimulus. The latter follows from the former. (Some non-equilibrium situations, like a supersaturated solution, might appear to be in equilibrium based on a “no tendency to change”. I would not consider this to be an equilibrium, more of a kinetic barrier to the equilibrium state.) An unsaturated solution satisfies the latter definition of equilibrium. There is no kinetic barrier in play here – the system will persist indefinitely unless the conditions are changed. It may be true in an unsaturated solution the backward reaction is faster than the forward reaction- any precipitate formed is instantly changed back to solute. An unsaturated solution does not satisfy a rate-based definition of equilibrium, either. But it does satisfy the "change" based definition of equilibrium, which I think is the most important because it defines how a system actually behaves.
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High School Chemistry Forum / Unfinished high school Chem question...
« Last post by BabyGoofball on May 01, 2024, 08:16:33 PM »
Question: There is a homogeneous solution with no solute at the bottom (no residue). This mixture can be in which state(s)?

a) Unsaturated
b) At equilibrium
c) Supersaturated

I know (a) and (c) are definitely right. My chemistry teacher from 23 years ago insisted that (b) is incorrect. Given the state of (a), (b), and (c) are technically a continuum, it doesn't make sense that it CANNOT be at equilibrium? Yes, I understand that you cannot be certain it is at equilibrium, but is it actually IMPOSSIBLE that the mixture in question is in fact, at equilibrium? Please help me find closure. Thanks.
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