3.50 g of Co2 and 11.6 grams of O2 are placed in a 3.00 L container at 273 k.  what is the total pressure ?
602334
PV = nRT\n\nV = 3 L\nn = g CO2/MW CO2 + g O2/MW O2\nT = 273 K\nR = 0.082057 L-atm / mol-K\n\nsolve for P
960502
Since the oxygen is collected over water, you must subtract out the water vapor pressure to determine the pressure due to just the oxygen.\n\n722.2 mmHg - 21.1 mmHg = 701.1 mmHg = 0.922 atm due to O2\n\n0.276g KClO3 = 0.00175 moles KClO3 (molar mass = 158 g/mol)\nSince KClO3 decomposes to form O2 in a 2:3 ratio, the number of moles of O2 produced = 0.00262 moles O2\n\nPV=nRT\n(0.922 atm) V  = 0.00262 mol (0.08206)(296.2K)\n\nSolving for the volume due to the O2,  V = 691 mL
847416
With only 2 gases, the partial pressure of the O2 is 0.440 atm - 0.160 atm = 0.280 atm O2\n\nThe temperature is 40°C, or (40°C + 273°K) = 313°K\n\nThe volume is 478 mL or 0.478 L\n\nYou will use the ideal gas law to calculate the number of moles (n) of O2:\n\nPV = nRT\n\nUse the partial pressure for P.\n\n0.280 atm * 0.478 L = n * R * 313°K\n\nR = 8.2057 E-2 L * atm / (mol * °K)\n(this will probably be in your textbook's appendix, but you can also search the internet for "gas constant")\n\n0.280 atm * 0.478 L = n * 8.2057 * 10^-2 L * atm / (mol * °K) * 313°K\n\nn = 0.280 * 0.478 / (.082057 (1/mol) * 313)\n\nn = 0.005211 mol\n\nThe molecular weight of O2 is (16*2) = 32 g/mol\n\nSo, the mass is 32 g/mol * 0.005211 mol = 0.1667 g or 166.7 mg\n\nYou might want to work it out yourself to be sure you can do it.
198092
Use:\n\npressure = density x height x g\nwhere g = 980 cm/s^2\n\nfor each liquid.  the answer (in dynes) will be the sum of the two pressures.
930613
CHEMICAL EQUATION\n\nC3H8 + 5O2 --> 3CO2 + 4H2O\n\n117g C3H8 * (1mol C3H8 / 44.09652g C3H8) = 2.65mol C3H8\n2.65mol C3H8 * (5 mol O2 / 1 mol C3H8) = 13.3 mol O2\n\n14.0°C + 273.15 = 287.15K\n\n 100.0 kPa (1 atm / 101.325 kPa) = .9896 atm\n\nPV = nRT\nV=nRT/P\n\nV= (13.3 mol*.0821L atm mol-1 K-1 * 287.15K) / (.9896 atm) = 317 L O2
1053372
3.425 BTU/lb F in gas form
711363
The volume will be 549.5 mL.  Let me know if you want to see the work.   Use this equation PV/T=P*V*/T*.    P,V, and T are just the pressure, volume, and temperature for the initial system (they are usually represented as P1,V1, and T1 but I left out the numbers which should be subscripts for simpliofication because yahoo does not have many text options.)  P*,V*, AND T* are the pressure, volume, and temperature after the system has been changed.  You could use any units pretty much for the pressure and volume as long as you keep the units consistent.  However always use Kelvin for the temperature.  Kelvin = 273 + Celcius Temp.
350090
Use the formula PV=rnT\n\np= pressure in atm\nV= volume in L or dm^3\nr=number of moles\nn= constant: 0.0821 atm x L/ moles x K (This constant is used to balance out units)\nt= temperature in K\n\nSo for your problem:\n2.3 atm x 120 L = r x (0.0821 atm x L/ moles x K) x 340 K\n\nthe constant should allow you to be left with moles as your unit\nusing algebra, r is left, which also stands for the number of moles (meaning the problem coordinates with your question)\n\nusing this setup, i got 9.89 moles
921419
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
254672
You answer this question by formula weight ratios.  One mole of Na2CO3.10H2O has a mass of 286.14 g.  It also contains 13 moles of oxygen (3 in the carbonate + 10x1 each water), which is 13x15.9994 = 207.99 g.\n\nThe mass percent oxygen is then (amt in question / amt total)x100% = (207.99/286.14)x100% = 72.688%, or 72.7% to the nearest tenth.
881429
Fe2O3(s) + 3 CO(g) ----> 2 Fe(l) + 3 CO2(g)\n\nCount up the oxygens; the left side and right side oxygens should match.
446321
Hmmm, a multi-step problem!  Love those!\n\nFirst, you will need to know how many moles of oxygen are generated.  For that, you will need to use the Ideal Gas Law:\n\nPV = nRT\n\nYou need to solve for n = PV/RT, and since you know the volume, pressure, and temperature, and the value of R is a constant, this is an easy calculation.\n\nThen, from the equation:  \n\nKClO3 ===> KCl + 1.5 O2\n\nyou can work backwards to determine the number of moles of KClO3 you started with (one mole of KClO3 generates one and a half moles of oxygen, thus the number of moles of KClO3 you started with will be 2/3 of the number of moles of oxygen generated).  Knowing how many moles of KClO3 you had, you can then determine the starting mass of the KClO3, then find out what percentage that was of your original starting mixture.\n\nGood Luck!
813754
O2, N2 and CO2.\n\nDon't understand the last part
585380
Assuming the temperature is 298K (25oC) The equation is P1V1/T1 = P2V2/T2\n\nWith temperature in Kelvin and Pressure in dm-3\n\nSo P2=1*0.50*333/(298*0.66)\n=0.847Pa
924197
PCO2 = 35-45\nPO2 = 75-100
877652
These are GAS LAW problems, so you need to use PV=nRT.  However, for these problems, nRT will be constant so you only need to use P(pressure) and V(volume).  Therefore, for both of these questions, use the equation:\nP1V1 + P2V2 = PtVt   where P1 is the pressure in the first vessel, P2 is the pressure in the second vessel, and Pt is the total pressure of the connected system.\n\nProblem #1\nP1V1 + P2V2 = PtVt  \n(7.0atm)(15.0L) + (P2)(8.4L) = (10.0atm)(23.3L)  \nP2 = 15.4 atm\n\nProblem#2\n(7.2atm)(15.0L) + (4.1atm)(8.3L) = (Pt)(23.3L)  \nPt = 6.1 atm
1149287
pV = n R T => if pressure is constant => p = n R T / V = const. => p / n R = T/ V = const.\n=> T / V = const.\nT1 = 333 K\nT2 = 533 K\nV1 = 0.2 m3\n\nT1 / V1 = T2 / V2 =>\nV2 = V1 x T2 / V2 = 0.2 x 533 / 333 = 0.32 m^3\nanswer D is correct