encipher

Well, with KNO3 and H2SO4 you can make Nitric acid which then can be used for the nitration.

Thanks, I'll give that a shot then.

Hi, So I ran out of Iodine, and I was attempting to make myself some. I took some NaI, dissolved it in distilled water, then added H2SO4 and H2O2. Iodine was made successfully, but the issue im having is isolating the iodine, I set up my distillation apparatus, and waited for the vapors to be begin deposition.. unfortunaltey they solidified at the opening of the flask, blocking the way for any other iodine to go through. Does anyone know how to solve this issue? or a different way to do it? Thanks

This 'effect' is conservation of angular momentum, and has little to do with the original question asked regarding the angular velocities of different sections of a merry-go-round. Conservation of angular momentum is as follows: [math]L = I\omega[/math] I is the moment of inertia and omega is the angular velocity. When a skater pulls their arms in, the moment of inertia Decreases, therefore due to conservation if angular momentum, the angular velocity must increase. Hence why they begin to rotate much faster when they pull their arms inwards.

Yeah, I figures it could be dissolved NOx gasses. I slow heated the solution and a lot of Nitric acid fumes came out.. for quite some time. I heated until it was just a wet slush of crystals. It was still bluish. Added water and heated again etc.. The solution is still blue, but I know that the acid isn't all gone. Any quicker way for me to get rid of the dissolved gas?

Hi, I recently purchased '99.99%' pure silver and dissolved it in 6M nitric acid. I noticed the solution has a blue/green color, indicating copper impurity. Anyone have ideas on removal of copper contamination without losing much yeild? thanks.

Actually, my calculation came up to be pretty close to what I got in real life. i started off with the folowing: [math]mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2[/math] by subsituting in [math]\frac{v}{r}[/math] for omega, where V is the linear velocity of the yoyo and r is the radius of the AXLE. and subsituting in [math]\frac{1}{3}MA^2[/math] for the moment of inertia (I) the Mass cancels out, and A is the length of one side of the square. Re-arranging the equation for the velocity will give you the following: [math]\large v^2 = \frac{gh}{\frac{1}{2} + \frac{A^2}{6r^2}}[/math] then using kinematics, the general formula [math] v_f^2 = v_i^2 + 2ax[/math] We know that the initial velocity is zero so that cancels out Solving for a and plugging in the above equation gives: [math]\large{a = \frac{gh}{2h(\frac{1}{2} + \frac{A^2}{6r^2})}}[/math] Then Plugging in all the numeical values: g = 9.81 meters/s/s h = 2 meters A = 20 centimeters = .2 meters r = .5 millimeters = .0005 meters The acceleration is: .000184 meters/s/s (Yes, I hate sig figs) Then, using the equation [math] x = v_it + \frac{1}{2}at^2[/math] subtituting in the above value for a, and 2 for x, and solving for t I get the following: 147.47 seconds.. divide by 60 find remainder, i get: 2 minutes and 27 seconds. Very close the the experimental value acheived. I will be attempting to setup a counterweight yoyo and will post my results.

Well, its kinda messy, but it looks ok. Yes the moment of inertia of a square is 1/6 MA^2. But I mentioned earlier that I had cut out the inside of the square, making it like a hoop. But square. and according to the general formula: [math]I = \int r^2 dm[/math] I derived that moment of inertia (if you would like to see me work i would be glad to post it. ) Now it would actually be that + [math] 4 (\frac{1}{3}ML^2)[/math] for the 4 spokes that connect it to the axle. I talked to a few of the students, and they mentioned that their friends said something about a counterweight being used to achieve the slow speed. Anyone have any thoughts on that?

yeah, of course the density will play a part in the challenge, but nothing as significant as making the yoyo take 45 minutes. I have tried all possible solutions, but nothing comes to mind.

no, the yoyo does not need to go back up. Yeah, concentrating mass on the edge is key, but really making the yoyo heavier will not matter, since mass cancels out according to the following equation: [math]mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2[/math] therefore what really matters is what the radius of the object is (for its moment of inertia). Or maybe I am wrong, any thoughts?

The angular speed is the same regardless of distance from the point of rotation, since angular speed is defined as the following: [math]\omega = \frac{d\theta}{dt}[/math] therefore, it is clear that since every point along the merrygoround covers the same angular distance (theta) then their angular speeds are equivalent. However, the linear speed isn't. Looking at it from a standpoint of how much distance does it actually cover in a single rotation, you can tell that it covers more distance the farther out you go. The relationship between linear and angular speed is the following: [math]V_{trans} = \omega r[/math] where omega is the angular speed and R is the distance from the point of rotation to the point at which you want to calculate the speed of.

Because the moment of inertia of a square is greater than that of a circle (within the given dimentions) I calculated the moment of inertia of the square to be [math]\frac{1}{3}MA^2[/math] Where A is the length of one side. As opposed to that of a hoop which would be: [math]MR^2[/math] where R is the radius of the hoop. Since the maximum length of the side or radius can be 20cm.. and the mass doesnt matter (sine according to conservation of energy it cancels out).. We figured the square shape would be the shape with the largest moment of inertia.

As I mentioned earlier, one of their requirements is that no mechanical friction can slow the yoyo down, that includes peices of gum stuck on the axel any other thoughts?

Hi, Well, their paper says it must be heavier than air, and no, air resistance cannot be a factor. What we went with was a Square, with the insides cut out, except for rods to connect the axle to (which is very thin) that way we have a large moment of inertia. Now with that design, I mathematically calculated it to be something around 147 seconds (using conservation of energy) and the test was pretty close. Now I don't know what could possibly give 45 minutes.

Hi, I am working with a couple of highschool students on a physics project their class is doing. They have to construct a yoyo that will fall a distance of 2 meters in the GREATEST amount of time. Mechanical friction cannot be used as a means of slowing the yoyo down. The maximum allowed mass is 1000grams and the maximum diameter / length is 20cm. We came up with a square shape and a thin rod in between. We acheived a time of about 3 minutes. Now today one of the students talked to me about the 'competition'. Apparently a rivaling group claimed to have gotten the yoyo slowed down to 45 minutes over the 2 meter drop. I have thought about how this is possible without friction being a factor and within the guidlines of the rules, but have yet to come up with a workable idea. Any thoughts on the setup they might have used? Also,it cannot be lighter than air. Thank you

I think you shouldn't be even CONSIDERING the isolation of sodium through ANY process. It is far too dangerous, and is challenging to even the most experienced of home chemists. You simply would not have the materials and equiptment nessicary to produce sodium in a safe manner. Sodium spontaneously ignites when it comes in contact with air at temperatures above 300 celcius. You would have to setup an inert atomsphere of argon or another inert gas etc.. Simply put: DONT TRY IT AT HOME.

There's an attachment I bought online that I connected to my pc, and I get a 3.5mm which I can plug into my phone. It plays as Auxillary on my car stereo. If your car doesn't have the ability to attatch aux, I'd say bluetooth. (if it were 2.0)

You have to remember that Alka-Seltzer is not pure NaHCO3. It consists of Aspirin (acetylsalicylic acid), NaHCO3 and Citric Acid. both aspirin and Citric acid have significantly higher molar masses (180 and 192 respectively) so much of the 1.78 grams of Alka-Seltzer is not NaHCO3. Now by collecting the CO2 gas efficiently, you can get a fairly resonable idea of how much NaHCO3 the 1.78 grams actually contains. Because AFAIK the ratio of the ingredients is probably a 'company secret', or something of that sort. If you can find it online, you can determine the ACTUAL percent yeild from the amount of NaHCO3 it reallly contains.

Im pretty sure that PH Down is Phosphoric acid, not Nitric acid.

You have to know the molarities of the initial solutions of HCl and Cu(NO3)2

Well, there's another option. Reacting Strontium Carbonate with Nitric acid, that would give you Strontium Nitrate and the Carbonic acid would break down into CO2 and H2O.

At first, it would begin to dehydrate, turning into a off-white colored solid (anhydrous copper (II) sulfate). Then it would begin to decompose. (It decomposes before melting) It would form the following products: Copper (II) Oxide: CuO Sulfur Dioxide : SO2 Oxygen : O2

In an ideal environment, magnesium would burn in oxygen according the the following balanced equation: [ce] 2Mg + O2 -> 2MgO + Heat [/ce] But normally, when magnesium is burned in air, which contains a lot of nitrogen, it forms some magnesium nitride: [ce] 3Mg + N2 -> Mg3N2 [/ce]

H3PO4 (Phosphoric acid) Is a WEAK acid and does not dissociate extensively in water. Therefore, in an aqueous solution containing Phosphoric acid, there will exist more molecules of it, then there will exist ions. The rule regarding acids is similar to that of salts. With salts it generally depends on solubility, with acids it depends on dissociation.

That net ionic equation is incorrect. The potassium Ion and the Nitrate Ion are spectator ions, and Lead has a charge of 2+, not + Therefore the correct net ionic equation should be the following: [ce]Pb_{(aq)}^{2+} + 2I_{(aq)}^{-} -> PbI2_{(s)}[/ce] Generally, in most aqueous reactions like the one above. The group 1A alkali metals are spectators. I have indicated the state of each of the reactants and products to clarify, since the product will be the precipitate and is a solid due to its insolubility.