calculate the gross profit one could make in the manufature of KNO3 from NaNO3 and KCl if 100% yield could be obtained and KNO3 could be sold for $185/ton and NaNO3 cost $55/ton and KCl $28-ton.
210231
The gross revenue per ton is 185\nThe cost of goods sold per ton is 55+28=83\nTherefore, the gross profit is 185-83=102
160022
3KOH+H3PO4=K3PO4+3H2O
341328
This is simple.\nCP = 8000 for 10\n      = 800 for 1\nThe profit equals the selling price?\nThis can't be possible as profit is the difference between the selling price and the cost price.\nHow much ever each machine is sold for, if you subtract the CP from the SP, profit can never equal SP.
485936
The MW of KClO3 is 122.55 (ref.), so 0.251 g KClO3 is .0020481 M. This will evolve 3/2 * .0020481 = .0030722 M O2 * 22.4 L/M = .068818 L O2 at STP. Correcting for temperature, pressure and partial pressure H2O, multiply by 297/273 * 760 / (721.8 - 22.4) and volume at given temp. and press. = 0.081354 L.
408222
Sodium Fluoride, about half of which is fluorine, from Sigma-Aldrich, is about $1200/kg.\n\nIndustrially, it would be cheaper.
537551
Tehe answer is:\nAg + HNO3 -> AgNO3 + H\nAgNO3 + KBr -> AgBr + KNO3
506774
First of all let's find how many moles of KClO3 you have and how many moles of O2 are produced\n\nmole KClO3=mass/MW =0.261/122.5 =2.13*10^-3\n\n2KClO3(s) → 2KCl(s) + 3O2(g) \nthus 2mole KClO3 produce 3 mole O2\n2.13*10^-3 mole KClO3 produce x mole O2\n\n=> 2x=3*2.13*10^-3 => x=3.19*10^-3 mole O2\n\nIn order to find the volume you use PV=nRT, but how much is P?\nYou know that the total pressure is 722.7 mm Hg and the pressure comming from water is 17.5 mm Hg. Thus the pressure coming from O2 is 722.7-17.5= 705.2 mm Hg.\nYou have to convert into atm.\n1 atm is 760.0 mm Hg\nThus 705.2 mm Hg is 705.2/760.0 = 0.928 atm. This value of P will go in the equation\n\nPV=nRT=> V=nRT/P= (3.19*10^-3)*0.082*293/0.928= 0.08259 L= 82.59ml
354893
Gold would be deposited on the surface of the lead.\n\n3Pb + 2Au(NO3)3 ---> 3Pb(NO3)2 + 2Au
471669
AgNO3 + NaCl => AgCl(white ppt) + NaNO3\nAgNO3 + KI => AgI( yellow ppt) + KNO3\n\nthere is a change of color because AgI is produced
206005
Exactly 3.5653g of MgO will be made. Ignore the other guys' answers; they're retards.
527205
1)write out the equation\n2)find the no of mol of both reactants\n3)compare the stoichiometry ratio of reactant from 1) to find the limiting reagent\n4)compare the mol ratio of the limiting reagent with NH3 to find the no of mol of NH3 produced\n5)theoretical yield=17(no of mol of NH3)
199487
Write the Kc and the Kp\n\nChange PV=nRT to P/RT=(n/V) to P/RT=(concentration) note n/V is concentration.\n\n\nSubstitute P/RT for each concentration in the Kc.  You will see the Kp in the equation with extra R's and T's.  Finaih solving.
18255
3AgC2H3O2 + Na3PO4 -----> Ag3PO4 + 3NaC2H3O2
316947
Profit on each shares (respectively)=(12/100)*8000 + (8/100)*10,000=960 + 800 =Rs 1760\nLoss on each shares while he sells= Rs[ .02*8000 + .03 *10000]= rs (160 + 300)=Rs 460\n\nTotal Earnings= Rs (1760-460)= Rs1300
503599
Here's the how you do it.\n\nBalance the equation.  Element by element.\n\nThat gives you the mole ratios of  the various compounds.\n\nConvert the moles to mass using the molecular weight of each of the compounds.\n\nThat gives you the mass ratios of the various items.\n\nDivide both sides of the equation by the largest mass on the left side of the equation,    Then  you will have the mass of products per Kg of the at least one of the reactants.  Remember  you are starting with 1.0 kg of one item and 0.98 of the other.\n\nso you must check to see which reactant will limit the production.\n\nShould be a simple matter of working out the ratios.
63081
[Ca(NO3)2]=n/V  ---> n=[]*V= 0.390M * 0.044L=0.01716 mol\nn=m/M ---> m=n*M= 0.01716mol * 164 g/mol=2.81424 g
431774
First write the balanced chemical equation:\n\nPb(NO3)2 + 2NaBr --->  PbBr2  +  2NaNO3 \n\nThen, find the total volume of the solution: 200mL + 200mL = 400mL \n\nFrom the density, the mass is:   density = mass/volume   mass = density * volume = 1.0 g/mL * 400 mL = 400 g\n\nQ = mC(Tf - Ti) ; where Q = heat released, m = mass of solution,, C = specific heat capacity, and (Tf - Ti) = change in temperature\n\nSo,  Q = 400g * 4.2 J/gK * (2.44 + 273)  (Since C is J/gKelvin, the temperature rise also has to be in Kelvin)\n\nQ = 462.5 KJoules\n\nHope this helps...