What is the mass of O2 in grams?
816920
Call p1 = p(Cl2), p2 = p(O2)\n\np1 = 0.15 atm, p = p1 + p2 = 0.365 atm\n\n-> p2 = 0.365 - 0.15 = 0.215 atm\n\nV = 0.538 l\n\nT = 273 + 32 = 305 K\n\nR = 0.082 L*atm / K\n\nFor a mixture, all gases share the same V and T:\np2.V = n2.RT -> n2 = p2.V/(R.T) = 0.004625 mol\n\nTherefore, the mass of O2 is:\n0.004625 * 32g = 0.148 g
279099
The average mass of the atmosphere is about 5,000 trillion metric tons. According to the National Center for Atmospheric Research, "The total mean mass of the atmosphere is 5.1480 x 1018 kg with an annual range due to water vapor of 1.2 or 1.5 x 1015 kg depending on whether surface pressure or water vapor data are used; somewhat smaller than the previous estimate.
197904
based on the net inoic equation:\n\n\nCu2+(aq)---> Cu(s)\n\nAg+(aq)---> Ag(s) --- zero grams\n\nexcess Cu(s) or Cu(aq)\n\nif, Cu(s) then it is  3.78 grams of Ag
446794
1 kg of water = 1000 grams of water.\nWater (H2O) has a molar mass of 18.02 grams per mol.\n1000 grams of H2O equals about 55.5 moles of water.\n\nThe reaction which water is split into Hydrogen gas and Oxygen as is as follows,\n2H2O (l) --electrolysis--> 2H2 (g) + O2 (g)\n\nThe change in the enthalpy of formation of this reaction equals about +571.8 kJ, meaning that at standard conditions and at a constant pressure, 571.8 kJ of work must be done on the system to totally split 2 moles of water into 2 moles of H2 and 1 mole of O2 gas.\nUsing this information, we find that to split 55.5 moles of water into 55.5 moles of H2 gas and 27.75 moles of O2 gas, it would require 15867.45 kJ of energy (or about 1.6 E7 Joules).\n\nThe amount of energy needed to split the water is also the maximum amount of energy which will be obtained if the H2 and O2 are then allowed to recombine again into water.
324518
molecular weight's 305.4188 g/mol
782103
The molar mass of the nitro is 3x12+5+9x16+3x14 = 227 gram.\nWe have therefore 1.569/227 = 6.91 x 10^-3 mol.\nThis produces 6.91 x 10^-3 /4 = 1.73 x 10^-3 mol of oxygen.\n\nWhich is  16 x 1.73 x 10^-3 =0.0276 gram
815144
64 grams... multiply 88 grams by the ratio of molar masses of O2 to CO2 to get  the result
977270
first you need a balanced equation.\n\n2Mg+O2===>2MgO\n\nNext convert everything given to mols.\n\n10.000 g Mg X 1 mol Mg / 24.30 g Mg (NOTE: this number is from the periodic table of the Elements) = 0.4115 mol Mg\n\n5.267 g O2 X 1 mol O2/ 32.00 g O2 (this is the molar mass of O2)= 0.1646 mol O2\n\nAt this point pick one of the mols you calculated.\n\n0.1646 mol O2 X 2 mol Mg (from balanced equation) / 1 mol O2 (from balanced equation)(for every 2 mol of Mg there is 1 mol of O2)= 0.3292 mol Mg needed to react with all of O2\n\nSince more Mg is available then needed (0.4115 mol > 0.3292 mol) this implies that O2 is the limiting factor. \n\nUse the limiting factor (O2) to get the mass of the MgO\n\n0.1646 mol O2 X 2 mol MgO / 1 mol O2= 0.3292 mol MgO\n\n0.3292 mol MgO X 40.30 g MgO/ 1 mol MgO = 13.27 g MgO\n\nessentially for all the other ones you are doing the same thing as before but with different numbers.
917635
The ore contains 3.5% SnO2, which is 70 gram. The atomic weight of tin is 118.7, while the  molecular weigth of SnO2 is 118.7 + 2*16 = 150.7. Therefore 70 gram of SnO2 contains 70* 118.7/150.7 = 55.1 gram tin.
736571
there are .833 moles. divide the mass by 18 (2 hydrogens + 1 oxygen) oxygen= 16g/mole Hydrogen = 1 g/mole
671042
405.39818 grams in 14.3 oz\n\n28.349523 grams per 1 oz
911798
Partial pressure of a gas is proportional to its mole fraction.\n\nSo convert your grams to moles...\n\n7.6 g of O2 = 7.6/32 = 0.2375 moles of O2\n11.1 g of Ne = 11.1/20.18 = 0.55 moles of Ne\n\nTotal moles = 0.787\nThen mole fraction = moles of X / total moles\nfor Ne mole fracxtion = 0.55/0.787 = 0.698\n\nThen the pressure of the gas = total pressure X partial pressure\n\nP Neon  = Ptotal x mole fraction\nP Neon = 0.540 atm x 0.698 = 0.376 atm
832542
Balance the Equation\n4NH3 + 5O2 --> 4NO + 6H2O\n\nConvert from g O2 to moles O2\n\n(64.00 g O2) * (1 mol O2)/(32.00 g) = 2 mol O2\n\nConvert from moles O2 to moles NO\n\n(2 mol O2) * (4 mol NO)/(5 mol O2) = 1.6 mol NO \n\nConvert from moles NO to grams NO\n\nbut first we need to figure out NO's molecular weight\n\n1 mol N * 14.01g/mol = 14.01g\n1 mol O * 16.00g/mol = 16.00g\n                                        30.01g/mol NO\n\nNow we can continue\n\n1.6 mol NO * 30.01g/mol = 48.02 g NO produced\n\nHope this helps!
715558
OK, but you need to look up atomic weights or calculate molar mass.\n\n   Grams = Moles x molar mass  OR  moles x atomic mass\n\n    Moles = Grams/(molar mass) OR Grams/(atomic mass)
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
585584
Atomic Number of Oxygen is 8,and the atomic weight of Oxygen is 16.00
187568
You can come up with a molecular weight of air - use the percentages of the different molecules comprising air and the molecular weight of the components.  The answer is around 29 g/mol (I think).