What is the mathmatical way to calculate maximum yield of 2 dimiensional materials?
405573
you need to use some calculus.\nget an expression for the area \ndifferentiate it\nthen find when it is a maximum.
470540
http://en.wikipedia.org/wiki/Strength_of_materials
844758
Using the Michaelis-Menten equation.\n\nMore details on teh attached link\n\nhttp://www.le.ac.uk/by/teach/biochemweb/tutorials/michment2print.html\n\nGood luck
121693
cmon...dont u know basic physics...just calculate the resultant of these 2 forces\nanswer= [(132)^2 + (181)^2]^1/2\nif this makes sense!!
977712
Considering water and aluminum as one system,\nthe energy balance gives delta U = 0;\n\nTherefore\n\nMwater *Cvwater* (T2 - T1,w) + Maluminim* Cvaluminum* (T2 - T1,a) = 0, where T2  = 50oC, T1,w = 120oC, T1,a = 20oC, Mi = Mass of component i, and Cvi = heat capacity of component i at constant volume.\n\nWith the given information, and looking for the heat capacities in a table, you can solve for Mwater.
794422
If h = height (and width), and L = length, 2h + L = 62 and so\nL = 62 - 2h\n\nVolume = lengthXwidthXheight = (62-2h)(h)(h) = 62h^2 - 2h^3\n\ndV/dh = 126h - 6h^2 \n\nMaximum occurs when this is zero\n0 = 6h(21 - h)\n\nSo maximum occurs when h = 21\n\nUse that to get the maximum volume
887484
the first equation is 2c+4s = 1k (c=cutting, s=sewing, k=knit suit)\n2nd equation 4c+2s = 1w (w=worsted suit)\n\nwe know 16 hour sewing is available and 20 hours of cutting from first equation as sewing time is more so that is the dominant variable so muliplying it by 4 gives sixteen hours of sewing so only 4 knit suits are possible in a day\nfrom 2nd equation as knitting time is more it is our dominant variable so multiplying by 5 gives 20 hour of knitting so only 5 worsted suit possible so maximum profit is\n34x4 + 31x5=291
919629
I'm supposing that by R-value you mean resistance value.\n\nIf so, there are two separate formulas. \nFirst for R-values in series, meaning, materials which are layered: \n\nRtotal =R1 + R2 +...+ Rn\n\nFor R-values in parallel, meaning two or more materials side by side:\n\n(1/Rtotal) = 1/R1 + 1/R2 +...+ 1/Rn
1094108
Use a calculator and raise it to the 1/3 power.\nDon't try to do it by hand...
207715
Hello,\n\nThe idea behind Job's method is to prepare a series of reaction mixtures (for example solutions) in which the total number of moles of the reactants is fixed but with different ratio. One then plots the number of moles of product vs the ratio. The maximum of the curve will occur at the correct stoichiometric ratio.\n\nIn your case, one could prepare solutions containing a total number of moles (i.e. moles of Ni + moles EDA) equal to 1 with the following composition:\n\nmoles Ni     moles EDA Ratio Ni/EDA\n0.9                0.1           9        \n0.8                0.2           4\n0.6                0.4          1.5\n0.4                0.6          0.666\n0.2                0.8          0.25\n0.0                1.0          0.0\n\nOne then needs to measure the amount of Ni-EDA complex and plot that amount agains the ratio. The maximum will tell you which is the right stoichiometric ratio. Hope this helps. See the web site below for a more complete explanation.\n\nOne then measures the
290031
Balancing chemical equations is easy, but it's a bit of a trial and error process. Basically, you've got the the arrow in the middle, so the number of each element on one side of the arrow has to equal the number of that particluar element on the other side of the arrow. So start off by balancing any unique elements that only appear once on a particular side. Then balance your nitrogens, oxygens and carbons in that order. The last thing that should be balanced is your hydrogens. I can go into further detail if you need me to, I just don't know how much you know. \n\nFor your other question and yield is how much of something you get from a reaction. For instance, if I mix 100 moles of one substance and 100 moles of another substance, I can calculate a theoretical and experimental yield. For instance, I can say that I expect all 200 moles to react and form a single product. Since i expect it all to react, my THEORETICAL yield is 200 moles or 100%. Say I actually run the experiment and I discover that only 100 moles of my intended product formed, then my EXPERIMENTAL yield is 100 moles or 50%.
946830
Just remember M1*V1=M2*V2 where M is molarity and V is volume. A little Algebraic manipulation gives you:\n\nV1 = (M2*V2)/M1\n     = (0.195*15.5)/18.0\n     = 0.168 L = 168 mL
835485
percent yield=(actual yield/theoretical yield)*100\n\n\nbase on the above equation  :\n\nThe problem can be solved by simply applying the law of conservation of mass.\n\ntheoretical yield = the  total reactants mass\n\nactual  yield =  the total products mass\n\nso, the mass of Li2CO3 is:  \n\n\nx+18.00/(2*23.95+44.01)100=\n\n91.92%\n\nsolve for X =  66.48g of  Li2CO3 are produced.
756921
the answer is\n2Al + N2 ----> 2AlN
303320
Try   http://www.insitu.de/materialdatabase/materialdatabase_2_4e.html
762202
The manufacturer's total profit is given by his revenues minus his costs\n\nFor 0 < x <= 3 tons, the revenue is $13/ton * x while his costs are x^2 + 5x + 7.  His net total profits are therefore:\n\nP[0<x<=3] = (13x - x^2 - 5x - 7) dollars\nP[0<x<=3] = (-7 + 8x - x^2) dollars\n\nThe maximum of the profit function can be found by taking the derivative of P with respect to x setting it equal to zero, and solving for x:\n\n0 = dP/dx = 8-2x\n\nx = 4\n\nThis maximum is outside the range for which this profit function is valid, so the actual maximum profit for x<=3 tons occurs right at x=3 tons.\n\nFor x > 3 tons, we have that the profit is given by:\n\nP[x>3] = (13x - x^2 - 5x - 7 - 3(x-3)) dollars\nP[x>3] = (2 + 5x - x^2) dollars.\n\nNote that at x=3, both profit functions have the same value.\n\nLet's again find the maximum of this profit function:\n\n0 = dP/dx = 5 - 2x\n\nx = 2.5\n\nIn this case, the maximum of this function occurs at a value  of x below the minimum value for which this function is valid.  This implies that the maximum accessible value of this profit function occurs as x -> 3.\n\nThe manufacturer's profit is maximized when he makes 3 tons of the item.  \n\nThe easiest way to check that this is the correct answer is to plot the two profit functions over their valid ranges and see that the maximum does, indeed, occur at x=3.
908965
the answer is pi 3.14. use the pythagoran therom with x= pi(r2) then compensate for the remainder and it'll work itself out. If that doesn't work, try FOIL. first, outside, inside, last.