I have a math question <mixtures>?
402113
Mixtures problems often involve fractions that are part/total. In this one:\n\n(3/total lb)($2.23) + (# lb brand 330/total lb)($1.75) = ($1.93)\n\nLet # lb brand 330= x\nSo total lb= x+3\n\nYou can go from there, I think.
642645
Any. If you mix one, also a second placed incorrectly will be found.
956088
x = -11 + 12t + 3t² + 4t³ - t^4\nv = 12 + 6t + 12t² - 4t³\na = 6 + 24t - 12t²\nfor max or min a, \n24 - 24t = 0\nt = 1 (and this yields a miximum)\na = 6 + 24 - 12 = 18 units/sec²\nv = 12 + 6 + 12 - 4 = 26 units/sec\nx = -11 + 12 + 3 + 4 - 1 = 7 units
226482
Here goes: your 80/20 gas mixture is flowrate x 1.8\n\nyour 70/30 gas mixture is flowrate x 1.6\n\n***Dana Oakes is a great book to use***
907686
M1V1 = M2V2\n\nso you are starting out with a .1M times V1 is going to be equal to .08M times how ever much you need.  so you need ratios 5:1, 5:2,  5:3, and 5:4.
240972
If you have a system which has, say, A, B, C and D mols of substances 1, 2, 3, and 4, respectively, then:\nmol fraction of substance 1 is A/(A+B+C+D) and so on for the other 3 substances, e.g. for substance 2 it's B/(A+B+C+D)
968063
Substances in a mixture just need to be separated, which might happen naturally over time, or could be done by filtration, distillation or a centrifuge (which spins fast to take heavier items out of the mixture).  \n\nElements in a compound are chemically bonded, so you have to reverse the bonding process, which is different depending on the elements in the compound.  If it is a salt, it is as simple as adding water and using electricity to pull the ions you want out.  Once you do, those ions are not all going to be elements--further manipulation will be needed.  Other chemicals are much more complicated though.
806753
Are there any tests you can run? Usually if you mix a substance with something else, the product has different chemical and physical properties than the initial substance.\nFor example, let's say that you would mix isoamyl acetate with "X" the product will have certain carachteristics gained only after it was mixed with "X".... I don't know how much this will help...\n\nLook it up on the internet, I am sure you'll find something.\nGood luck.
1167929
Let x be liquid A.\n&   y be liquid B.\n\nThen,\n             x + y = 300    First Equation\n\n         0.08x + 0.15y = 0.11 (300)  Second Equation\n\nSolving the simultaneous equation, we get\n\nx = 171.43 ml \ny = 128.57 ml
458226
compound is a mixture of 2 differ net elements. homogeneous mixture of 2 or more substances that keep their identifies. therefore they are similar in that they are mixture of two or more diffident substances, but different in that the homogeneous mixture has the substances that keep their identifies. Homogeneous mixtures are also called solutions because u cannot identify their ingredients like in a compound.
817465
You know three of the four things needed. Try C1V1=C2V2, or an equality.\n\n                          m/ml=m/x
542164
0.2x+0.5(300-x) = 123\n0.2x+150-0.5x = 123\n-0.3x +150 = 123\n-0.3x = -27\nx= 90\nyou will need 90 gallons of 20% saline solutions and 210 gallons of 50% saline solution to obtain the mixture.\nHope this helps!!!
1152693
enantiomeric excess (%) = \n(specific rotation of the mixture/ specific rotation of the pure enantiomer in excess)*100 \n\nor \n\nenantiomeric excess (%) = \n( | [enantiomer 1] - [enantiomer 1] | ) /([enantiomer 1] + [enantiomer 1] *100 \n\nwhere the square brackets indicate concentrations, and the | -- | in the numerator means "absolute value".
290716
Unless a equation is properly balanced the amount of product and reactant will not be the same, I suppose this is like a recipe where u get the proper number of ingredients(moles) and mix to together to give the you the proper number of products.
813421
Na2CO3(aq) + MgCl2(aq) --> 2NaCl(aq) + MgCO3(aq)
1175612
you can use this formula m1/v1 = m2v2 whereas\nm1v1 is the old conc. volume and m2v2 is the new concentration an volume
816920
Call p1 = p(Cl2), p2 = p(O2)\n\np1 = 0.15 atm, p = p1 + p2 = 0.365 atm\n\n-> p2 = 0.365 - 0.15 = 0.215 atm\n\nV = 0.538 l\n\nT = 273 + 32 = 305 K\n\nR = 0.082 L*atm / K\n\nFor a mixture, all gases share the same V and T:\np2.V = n2.RT -> n2 = p2.V/(R.T) = 0.004625 mol\n\nTherefore, the mass of O2 is:\n0.004625 * 32g = 0.148 g