Can anyone help me with this? I need to know what formula to use.?
957319
Here's what you know:\narea: 131.2 cm^2\nwidth: 8 cm\narea formula: width x length\n\nSo you know width x length = 131.2\nYou know the width, so plug it in to get the solution
855052
Come on now, what is the formula for this???  :)\n\n(V1 * T1)/P1  =  (V2 *T2)/P2\n\n(69L * 131)/2.85  = (xL * -37)/2.85\n\nNow, solve for 'x'?  Be sure to watch your units!!!!
251893
You will find the formula here ....\nhttp://www.fao.org/docrep/W6355E/w6355e0c.htm
1114869
Here's a formula on this webpage you can use
1132678
Directions how to do this can be found in the text book.\n\nDo your own homework!
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
1161844
use this formula:\n\n[ g(h)  -  g(0)  ]    /   [ h  -  0 ]\n\ni.e.\n\n [  2/(h+1)   -   2/(0+1)  ]  / [ h  -  0  ]  =  [ 2/(h+1)   -  2]   /h\n\n=  [( 2 -2h  -2) / (h+1) ]  / h   =  -2/(h+1)\n\nTry it for f(t)
978323
*your equation is written wrong it should be:\n3Cu(NO3)2 + 2Na3(PO4) -> Cu3(PO4)2 + 6Na(NO3)\n\nyou have to find your limiter and your excess.\n 1.) change grams to moles using moler mass\n2.2 g Cu(NO3)2 times 1mol/188 g = .0117mol\n1.29 Na3(PO4) times 1mol/164 g = .00787 mol\n\n2.) divide the two by the # of moles in the balanced equation:\nCu(NO3)2: .0117 mol/3 = .00390 mol\n Na3(PO4): .00787 mol/2=.00394 mol\n\n3.) So now you know you don't really have a limiter, but we'll go with Cu(NO3)2 b/c there's a little less. \n\nthere are 3 moles of Cu(NO3)2 used for every 1 mol of Cu3(PO4) produced.\n\n.0117 mol Cu(NO3)2 times (1 mol Cu3(PO4) / 3 mol Cu(NO3)2) = .00390 mol Cu3(PO4) produced\n\nconvert that to grams\n\n.00390 mol Cu3(PO4) times 286 g/mol = 1.12 g \n\nthat's your actual\n\nso % yield = .89/1.12 x 100 = 79.5%\n\noh yeah you find moler mass by adding the molecular weights of the atoms in the compound (times the number of the atoms in that compound) for ex. Cu3(PO4) 63.6 x 3 + 31.0 + 16.0 x 4 = 286\n\nDon't forget significant figures!!\n\nhope that helped!
799202
for those questions, all you have to do is translate them into the equation\ncheck chemtutor.com for hellp, it explains it much clearer than i can
106436
I followed bugzpodder's advice.  Here's the formula I came up with. \n\n - 778728 - 121373345554191121 / 217945728000 * n^4 + 1298728131310487 / 1009008000 * n^3 + 670835515657 / 360360 * n + 2210058967013 / 12830400 * n^5 - 113499324114761 / 2874009600 * n^6 + 8961501223669 / 1306368000 * n^7 - 66868104743317 / 73156608000 * n^8 + 85853565361 / 914457600 * n^9 - 121097474346419 / 60540480 * n^2 - 27139677827 / 3657830400 * n^10 + 1792199 / 4009500 * n^11 - 1158076427 / 57480192000 * n^12 + 12283303 / 18681062400 * n^13 - 7672127 / 523069747200 * n^14 + 261523 / 1307674368000 * n^15 - 13147 / 10461394944000 * n^16\n\nCheers
631605
Try emailing me the file with a little more information.  I have been doing auto-filtering and formulas for many years and should be able to get you the answer you are looking for.
501191
There is a formula...but this is much easier\n\nhttp://www.onlineconversion.com/temperature.htm
868541
the truth about this is find the right formula for this equasion, then set it with ur need of what space you put it in to example : =sum(a1:a28)
401395
Way to go Arbitrage!  Oh to funny guy- yeah we get that E=MC2  IS a formula- but if you really want to "get it" Check out Maxwells equations. It's a great time to be a science nerd with all this info pouring in now! Figure it out & go up one on Albert!
757627
=IF(your formula)=0, "", (your formula))\n\nyou could also use the ISBLANK with the IF\n\n=IF(ISBLANK(your formula), "", (your formula))
655605
Apply the formula of\n\nM = molar strength wanted ( in M )\nW = molecular weight of the compound in question (in g/mol)\nV = the volume you wish to make (in mL)\nNo. = weight of compund needed (in grams)\n\napply to the formula\n\nNo. = (V * M * W)/1000 \n\nThen dissolve No. into the amoult of V you need (using a gradlated volumetric flask up to the volume you wanted)
485893
Check on the internet for the ratios used by comercial factories, then prepare a solution from glycol and water in the given proportion. That's all!!!