algebra wrong?..rate of effusion?
965285
rate b = rate a / sqrt (mwtb/mwta)\n          = [rate a * sqrt mwta]/sqrt mwtb\n          = 0.0235 * sqrt64 / sqrt 352\n          = 0.188/18.7624 = 0.01 = 10 x 10^-3...aprrox the same answer\n she got that 9.88 by taking m wt UF6  as 362....lol\nplug it in to see...\nbtw, your algebra was wrong....
878297
Assuming 1 mole (28.01348 g) N2 = 22.4 liters at STP, I get (28.01348*.00220102/22.4) =  0.0027526 g if collected at STP. Correcting by 273/293 * (1.5*760-17.5)/760 (for temp., press. and water) = 0.003788 g. \nSo I basically agree with your results. What's the ans. supposed to be?
428624
You need to be a bit more specific about which quantities you are trying to determine.  "Flow" is not a quantity to be calculated.
893701
v(m/s) = Q(m3/s) / Nozzle Area(m2)
883460
Please see the answer above and find out the missing factor in your calculation. You are in the right track. All the best!
748854
You are correct... just divide by 12, so 2.989E-8 slugs/in-s.  The only unit you are changing is the length.  I had no idea people used the slug/ft-s as a unit of viscosity.
569320
v=vo + at\nv=0 + (9.8)(8)\nv=78.4\n\ny=1/2(vo + v)t\ny=1/2(78.4)(8)\ny=1/2(627.2)\ny=313.6m\n\nB
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!
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
981618
I guess 1) just wants you to write: Want dA/dt, have da/dt and db/dt.\n(From now on I'll use A' to mean dA/dt etc).\n2)\nA=π*a*b, so using the product rule, A' = πa'b + πab'. So A' = 7πb - 11πa.\n3) Just substitute in: A' = 7*π*50 - 11*π*30 = 20π, so thats the rate it is increasing.\n4) Again substitute in, A' = 7*π*20 - 11*π*40 = -300π, so now its decreasing at that rate.
951555
the exact form of the equation i am not getting but dimensionally the equation is wrong because kgm/s is not equal to n/m2
665350
As the volume is constant,\n(P/T)initial = (P/T) final\n1/(25+273.15) = P/(50+273.15)\nP = 1.084 atm\n\n(Taking the pressure of the initial gas was 1 atm.
817492
Ok so the equation for osmotic pressure is:\n\npi = iMRT\npi = Osmotic Pressure\ni = i-factor\nM = Molarity\nR = Gas Constant (0.08211 L*atm / mol*K)\nT = Temperature (in kelvin)\n\nNow you know MRT because of what they give you but you need to find the value of the i-factor. The i-factor is the van 't hoff factor and is equivalent to the number of molecules associated or dissociated from the original solute. Since K2SO4 dissociates into 2 K+ ions and 1 SO4 2- ion, then the i-factor is 3.\n\nTherefore, plug in the numbers and solve.\n\npi = 3(0.250 mol/L)(0.08211 L*atm / mol*K)(318K)\npi = 19.6 atm
302446
I agree with the first answer:  Here is something that might help you with maths:  When you have a problem such as this make a 'rough estimate' of the answer.  In your problem almost all the tomatoes were water, nearly 100kg, the water was reduced by 1% - that's very little, it being 1/100 th. So you nearly have the same weight as you started with.  It cannot be 50kg as that would have meant that half the water 50% would have evaporated.  I hope you follow. Go back to my first point, the rough estimate should have been 'nearly 100kg but a little less'.  I hope I have made sense and that this gives you a good idea as to how you should proceed in the future.  Always work out a 'nearly right' answer first.....
837489
neglecting air resistance I hope!\n\ny=1/2at^2+vot+yo\n\na=32 ft/sec^2   you did mean sec^2?\n\nalso, v=at+vo  so we can find vo from this at time 2.1s when v=0\n\nfinally we insert 2.9 s into the first equation (yo= 0 too right?).
412953
An equation that is incorrect.
798267
This is a Hess' Law problem.\n\nThe strategy I like to teach for these is to get the biggest thing you need first then get rid of the biggest thing you don't need next.  So we want to get the 2 N2O5 first then get rid of the resulting 4 HNO3.\n\nUse -2 times the first eq to get your 2 N2O5, -4 times the third to get rid of your 4 HNO3, finally -2 times the second to get rid of the water:\n\n2 N2O5 + 2H2O > 4 HNO3   -152.4 kJ\n4 HNO3              >  2 N2 + 6 O2 + 2 H2  +696.0 kJ\nO2 + 2H2           >   2 H2O  -572 kJ\n\nEverything cancels properly to get the desired equation, and delta H is -28.4 kJ.