Borek
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Posts posted by Borek
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Try.
I don't see why not, I believe around 200 deg C is enough to dry it. I can be wrong though.
Note: if it absorbed enough water it will melt first (actually it will not melt, but it will dissolve in its own hydration water). You may have to wait several hours before it dries out.
Question is if you need it perfectly dry, could be partial drying will be enough to make it possible to prepare a powder.
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Sodium chloride is either solid - when it is solid, or aqueous - when it is dissolved. "Just sodium chloride" doesn't say anything about whether it is solid or aqueous.
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5 second googling: http://www.ncbi.nlm.nih.gov/pubmed/15653512
Even if not perfect, it is a starting point.
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Basically that's what chemical warfare is about. Wikipedia has a List_of_chemical_warfare_agents
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200mL of .3M HNO3
M is usually used for molarity, not molality, so it can be a little bit confusing here.
Borek
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Check this solution preparation example. Especially operating procedure at the bottom of the page.
Borek
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C1V1 = C2V2 - yes, but 1 means NaOH solution and 2 means HCl solution. The equation looks the same as in teh case of dilution, but it is not the same. So what you have is 18.32ml*0.08045M=25ml*C(HCl). Solve for C(HCl). Looks to me like your error was about 0.55%.
Borek
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See water ion product to check how water dissociation constant changes with temperature.
In general no ion can facilitate water dissociation, however, if you dissolve an inert salt increasing ionic strength of the solution, you will observe changes in the water dissociation. But this effect is universal, it is not limitied just to water dissociation. See ionic strength and activity coefficients lecture for details.
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Exponentially, just like Gauss curve. In fact Gauss curve is a solution of diffusion equation for the initial concentration being 1 at x=0 and 0 elsewhere.
Borek
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Download BATE and play with titration curves to get familar with them Program is commercial but there is a 30-day free trial.
Borek
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You must find out how to convert molarity to molality. Correct conversion formula will contain density and molar mass - you have to solve for density. Check out these concentration lectures for hints on converting (read about %ww to molarity conversions for general outline).
Strange thing is that 1.0000M solution of KCl is 1.0312 molal. Looks like the data in question are questionable.
Borek
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Start with learning about Henderson-Hasselbalch equation. Then check buffer pH question 1 and buffer pH question 2 - although they don't address specifically your problem, they show how to deal with buffer preparation. Acetic buffer case is not different.
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Stomach? Strong acid solution behaves as buffer as well - pH doesn't change much when you add small amounts of acids/bases. See buffer and buffer capacity discussion.
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Calcium carbonate minus carbon dioxide is either calcium oxide or lime water - where do you take them from?
Borek
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but watch out, a cubic centimeter is exactly a NANOSTERE
And not MICROstere?
0.01*0.01*0.01 = 10-6
Perhaps you meant a cubic milimeter?
Borek
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Do I read you correctly - KMnO4 solution is colorless?
Borek
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IMHO what he is looking for is a way of preparing Sterno type gelled fuel:
http://zenstoves.net/Sterno.htm
Borek
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Suppose I would want to make 100 ml of a 0.1 M solution of I2. I would weigh 1/3 of 0.01 mol of KIO3 on a glass watch glass. I would rinse this into a beaker of 100 ml. I would take somewhere around 60 ml of water, would add 10 ml of 20% H2SO4 and dissolve all. By rinsing the weighing glass, all iodate is taken into solution, and there is no real loss. To this, I would add 0.15 mol of KI (large excess, enough to make I2 and also keep this in solution as tri-iodide)
Next, the 70 to 80 ml or so of the liquid is transferred to a volumetric flask of 100 ml. This is an important point, where there can be losses. I would take 10 ml of water, swirl this around in the original beaker and add this to the volumetric glass. A second swirl with 5 ml of water takes the final traces, and then the volumetric flask is filled up to 100 ml. In this way, the transfer losses really are neglicible, and still there is the advantage of not having to deal with (uncertain) side reactions.
I would rather go with iodate solution prepared from solid directly in the volumetric flask, filling up to the mark. Then I would transfer exactly known volume of iodate to baker and I would add iodide and acid there, avoiding all steps that can introduce error to the iodine amount
When titrating iodine you usually start with yellowish solution - and you don't add starch till solution loses almost all color.
Borek
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hence my suggestion for preparing the iodine in a separate beaker
In all analytical methods I am aware off you prepare reagents in the same beaker/flask it will be used, to avoid reagent losses in transfer. You are right about side reactions, but if there are side reactions possible it is better to look for other method.
Borek
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Add starch just before titration end, when the iodine color fades. Otherwise it may react with starch in an irreversible way.
Borek
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If there is a lot of iodine it forms a stable compound with starch. When there are small amounts of iodine, reaction is reversible.
Borek
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Generator of chemical formulas
in Chemistry
Posted
Go to chemspider and search for the formula - this yields list of a known compounds, not every possible one (whether it exists or not). And some compounds that look OK on paper don't exist in reality.