percentage of mass specifies a solutions what?
95798
I'm thinking concentration, but your question is worded sort of oddly... Does that fit with what you've been learning????  It's usually percent by mass I think...
1160809
1000 ppm/1,000,000 x 100% = 0.100%\n\nIts a simple division solution.
343969
you would also use one of the following equations typically depending on the other information your given, but for sure, you need to look up its chemical composition. (ex: water is H2O)\n\nif youre given concentration:\nconcentration=moles/volume\nmoles=mass/molar mass\nmass=moles*molar mass\nmass=(concentration*volume)(molar mass)\nkeep in mind molar mass can be found by adding and multimplying as necessary the atomic masses of each element in the compound.\n\nif youre given density (which i bet you can easily look up on google or something if youre not given that information):\ndensity=mass/volume\nmass=density*volume
1080952
mass, volume and temperature.
956100
You will need to know that when you calculate a percentage of anything, it is kind of like a fraction, except that the denominator is 100.\n\nFor example, you have 7 sticks. 3 of them are red, 2 of them are yellow and the other 2 are blue. To get a percentage of the number of red sticks:\n\n(no of red sticks /  total no of sticks) x 100%\n\n...and that is how you get your percentage.\n\nOnce you have understood that this question should be quite simple. Sodium sulfite is Na2SO3.\n\n\nFirst you have to calculate the masses of all the elements in this compound.\n\nmass of Na\n= 2 x 23\n= 46\n\nmass of S\n= 1 x 32\n= 32\n\nmass of O\n= 3 x 16\n= 48\n\ntotal mass of Na2SO3\n= 46 + 32 + 48\n= 126\n\n\nHence, using the same concept,\n% of Na by mass\n= (46 / 126) x 100%\n= 36.5% (3 s.f.)\n\n% of S by mass\n= (32 / 126) x 100%\n= 25.4% (3 s.f.)\n\n% of O by mass\n= (48 / 126) x 100%\n= 38.1% (3 s.f.)\n\n\nHope you understand now! Cheers :D
732533
D = M/V\n\nDensity= Mass/Volume
105785
Chemistry.....ouch!!!  When teaching the method for converting percentage composition to an empirical formula, I have devised the following rhyme: \n\nPercent to mass\nMass to mole\nDivide by small\nMultiply 'til whole
920401
usually when you solve equations you get a solution.\nthe solution solves your problem.\n!
916732
Is relative to the amount of energy and object contains.
500514
is the AT and GC proportions?\n\nfor example\n\n30% AT\n70% GC\n\nis the same as\n\n15% A\n15% T\n35% G\n35% C
134081
For liquids it is the volume of solute divided by the final volume after the solution.\nFor gaseous solute it is the volume at STP divided by the final solution volume.\nFor solid solutes v/v is not usually used.\nGenerally v/v is expressed in percent.
831841
Let g = grams of salt added.\n\nThere are already 3 g of salt in there.  So there will be 3+g grams of salt when you are done, and there will be 25+g grams of total solution.\n\nWrite a proportion:  (3+g)/(25+g) = 20/100 and solve
1049998
1. Percentage is a ratio between two numbers times 100. Usually it is a part divided by a whole. To find a percentage we need both of these numbers. So the answer to your question I need to know which is the “non scientists who have no real clue of what electromagnetic pure energy is” and the other number to compute or even estimate the percentage.\n2. How does one compute purity of electromagnetic energy I have no clue and I pretended to work as a scientist for the past 20 years.\n3. And finally I have no firm idea about what gravity is and frankly what is pure energy. How ever I do know that if we have electrical charge in motion it will generate a magnetic field and magnetic field in motion will force electric charge to move. It seems that electricity and magnetism are products of the same electromagnetic phenomena.\n4. As far as gravity is concerned I do not know how exactly it relates to electromagnetism. Theory of relativity may provide some insight and there we may encounter some pure energy. \n\nI hope it was helpful.
853848
In general, when you have a "what percent" problem, I like to replace "what percent" with x/100.\n\n19.94 is "what percent" of 206.10.\n19.94 = x/100 * 206.10\n\nSolve for x\n\nx/100 = 19.94 / 206.10\nx = 19.94 / 206.10 * 100\n\nx = 9.67491509 percent\n\nx ≈ 9.67%
479787
In 100g of solution that is 5% by mass NaCl, there are 95 g H2O and 5g Nacl.\nYou want 250g of solution...set up a simple ratio:\n95g H2O/100g Soln = ?g H2O/250g Soln\n? = 237.5g H2O
459648
Themes > Science > Chemistry > Miscellenous > Help file Index > Precipitation Equilibrium > Ksp and Ion Concentration \n\nThe solubility product constant can be used to compute the equilibrium concentration of ions in solution in the same way that you use other equilibrium constant expressions to compute equilibrium concentrations. \n\nUse the Initial, Change, Equilibrium method to compute the concentrations: write the equilibrium expression, the initial amounts of each thing in the expression, how much they change by, then the equilibrium amounts. From that, you can compute the concentrations of all species. \n\nExample: BaF2 has a Ksp of 1.8*10-7. What is the concentration of barium in a saturated solution of BaF2? \n\nSolution:. First, write the balanced equation and the solubility constant expression: \n\nBaF2(s) < = > Ba+2(aq) + 2F-(aq) \nKsp = [Ba+2][F-]2\nWe start with no barium or fluoride ions in solution. If we set the amount of barium ions formed to x, then the number of flouride ions formed is just 2x \n\n [Ba+2] (M) [F-] (M) \nInitial 0 0 \nChange +x +2x \nEquilibrium x 2x \n\n\nNow plug in the equilibrium values shown above and solve for x \nKsp = [Ba+2][F-]2 \nKsp = (x)(2x)2 \nx = 3.6*10-3\nSince we defined the concentration of Ba+2 to be x, we've solved the problem \n[Ba+2] = 3.6*10-3 \n\nInformation provided by: http://learn.chem.vt.edu\n \n**try to visit the site below..** You can calculate the Ksp based on the above example..I'm sure the answer is in your books and notes..have fun learning..
316025
% means parts out of 100, so 64% = 64/100.  Reduce the fraction and you'll get 16/25.  (Better double-check my math!!)