Why we determined (K-Na) by flamphot method?this qausion is in chamestry.?
930664
I assume by quasion you mean question...\n\nK and Na can be determined qualitatively by a large variety of methods. One of them is inserting their salts in flame and examining the resulting flame color. The color diference is due to diferent energy for atomic excitation. This phenomenon belongs to chemistry but also to spectrometry and physics. \nI hope this answers your questions.
504715
You need the formula:\ndelta Tf=iKfXmolality\nyour delta Tf is 1.96\nyour i is 1 (because it does not dissociate)\nyour Kf is 5.12\nmolality is moles of solute divided by kilogram solvent, which is .0250 kg\nYour final equation should read:\n1.96=(1)(5.12)(x/.0250)\nThen use algebra to find x.
621304
If you are dealing with a reaction involving ammonium carbonate in solution, and are then measuring the temperature rise, you need the heat capacity of water, not ammonium carbonate!\n\nPlease post the whole problem if that is not enough to get you started.
793702
Firstly,\nNaCl dissociates into Na+(aq) + Cl-(aq)\n1faraday of charge is required to produce 23 g Na from molten NaCl .\ntherefore production of 100000 g of Na will require charge = \n(1/23)*1000000 Faradays = 43478.26 Faradays = 43478.26 * 96500 C      [ as 1F = 96500 c]\n=4.1956 * 10^9 C\nBut, Q=IT \nso, T=Q/I = (4.1956 * 10^9)/300000 = 13.9855 * 10^3 secs = 3.88486 hrs.
797201
Please be more specific about what you are asking.  I'm guessing this is a chemistry question, and you're talking about a flask placed in a Moissan electric furnace?  If so, it simply serves as a heating vessel to contain whatever substance you are experimenting with.
841558
could be several reasons;\n\nif the material you're weighing is volatile and is evaporating you will find it impossible to get an accurate weight, further the fumes migh condense later at muck up the weighing mechanism,\n\nalso if the material contains solvent that is evaporating, or if its degrading by loss of gas or whatever you wont get an accurate reading\n\nI suppose that if the sample is hot it will conduct heat through the scale mechanism which might damage it - these instruments are very sensitive\n\nim sure there's lots of other reasons but these are the fiirst that come to mind\n\nhope this helps
961702
The use of dilute alkaline permanganate solution to determine unsaturation is usually called the Baeyer's Test.\n\nTry some of the sites listed below for details on the reagent used and the tesing of alkenes.
577306
How is purity of solid pellet of potassium hydroxide? And its density (d)?\n\nThere is a formulation:\n\nV = m/(0.1d) \nwhere as:\n      V: volume of 10% potassium hydroxide solution\n      m: mass of pellet of potassium hydroxide\n      d: density of pellet of potassium hydroxide
587656
So anti-freeze and salt causes a freezing point depression in the water which means as you add solvent such as salt the freezing point is lowered (hence when you add salt to your windshield the ice melts).  There is a simple mathematic equation that will relate the change of Temp freezing to the amout of solvent added.  (ΔTf = Kfcm).  Kf is the freezing point constant of the solvent and cm is the molal concentration of the solvent. So as you can see from the equation the ΔTf will be greater as you increase cm so the effects of adding a solvent will depend on how much solvent you add.  \n\nSo a simple experiment would be to add increasing amounts of salt and antifreeze to constant volumes of water and determine the freezing temperatures of each one.  You should also want to calculate the molal concentration of each one and try to determine before hand what the freezing points should be (hypothesis).
770050
potassium bromide\nk=potassium\nbr= bromide
1144064
Sodium (English, soda) has long been recognized in compounds, but was not isolated until 1807 by Sir Humphry Davy through the electrolysis of caustic soda. In medieval Europe a compound of sodium with the Latin name of sodanum was used as a headache remedy. Sodium's symbol, Na, comes from the neo-Latin name for a common sodium compound named natrium, which comes from the Greek nítron, a kind of natural salt. As early as 1860 Kirchhoff and Bunsen noted the sensitivity that a flame test for sodium could have. Stating in Annalen der Physik und der Chemie in the paper "Chemical Analysis by Observation of Spectra": "In a corner of our 60 cu.m. room farthest away from the apparatus, we exploded 3 mg. of sodium chlorate with milk sugar while observing the nonluminous flame before the slit. After a few minutes, the flame gradually turned yellow and showed a strong sodium line that disappeared only after 10 minutes. From the weight of the sodium salt and the volume of air in the room, we easily calculate that one part by weight of air could not contain more than 1/20 millionth weight of sodium."
972593
Flame tests are "OK" when you are dealing with a reasonably pure solution.  If there are more than one cation present, such as sodium and potassium, then the resultant color in the flame will not be so determinant.  Also, if you are color blind flame tests suck.
977052
You need a variable in any experiment to compare results. Perhaps you could use cold (ice) water in one container, warm in another and hot or boiling in another. You can then make your prediction on which will dissolve first and have an experiment that will verify whether you have made an accurate prediction.
272997
ok i will TRY to help you. im not completely sure but i think i know what to do. basically, when you add K2Cr2O7, it splits up into 2 parts, the Ks and the Cr2O7s. the Cr2O7s react with the potassium dichromate and make products. so if it was titrated correctly (assume it is), then \n8.76mL * 1L/1000mL * 0.04988mol/L    mols were added\n= 0.0004369488 mol of Cr2O7 added. and this amount made all the potassium dichromate get used up (apparently).\nwhich means that 0.0004369488/2 mols of potassium were in the blood. (divided by 2 because for each mol of potassium, 2 mols of Cr2O7 were used). so 0.0002184744 mols of potassium were in the blood. and if you use the molar mass to convert that to a mass. and then divide it by 10.0002g. you should get the mass percent of blood alcohol. so 0.0002184744mol x [(2x12.01) + (4x1.01) + (1x16) + (1x1.01)]\n= 0.0002184744mol x 45.07 g/mol\n= 0.009846641208 g  and divide it by 10g to get the mass percent which is 0.098466%.\nhope this helps. let me know if u need more help. laterz
117753
Okay you have a choice between Potassium oxalate (K2C2O4 ) and Sodium Iodide (NaI). \n\nPotassium oxalate is highly soluble (30g/100g) and sodium iodide is even more soluble(184 g/100 ml).\n\nWithout knowing the reactants it is almost impossible to predict a product, especially with two choices. Given those two products, I would expect it to be K2C2O4, but that is just a guess. What was in the solution?
335806
Because KOH is virtually completely soluble, there is no real Kb value for it.  A Kb value is calculated by multiplying the concentrations of the ions that KOH forms in solution, then dividing that number by the concentration of actual KOH still bonded together in that solution, which is almost nothing.  Mathematicians will tell you that dividing by a number close to zero will give you a number very close to approaching infinity, so the Kb value for KOH that you find in any table or whatever will never give you a number; it will probably just say "very large."
811655
If by 'undissolved' you mean why are some salts insoluble in water, then I think we are in business.\n\nIn order for a salt to be soluble it has to be energetically favorable for the ions to separate. This requires a negative gibbs free energy. G = H - TS So you have to find the H and S values for the solid to aqueous solution and see if G will be negative for a given temperature.\n\nA method I recently learned to make insolube salts more soluble is to add other ions to the water...but you have to be careful to not add a common ion...or Le Chantliear will come into play.