What gas has a molecular mass of 54.6 grams?
108908
it isn't xenon, that weighs 131 amu.  the only answer I've come up with is cyclobutene, which has a formula of C4H6.
438665
Its not pure...there is not a noble gas that would give a m/z at 10 or 11.\na molecule with bond angles of 90 would be SF6\n\nIf you have collected this mass spec it could also be a result from background interference.  Subtract you background and see what happens.
954638
It seems that you have problems with moles and Avagadro's number. You will need to refer to your periodic table to get the molecular weights that your teacher wants you to use.\n\nSo get the molecular weights of the elements you are working with from your periodic table and substitute in the formula.\n\nI am using round numbers to show the substitution you should know how many decimal points you need to work to.\n\nHG atomic weight 200\nO atmic weight 16\n\nSubstitute in the formula to get the masses involved\n\n2 HgO(s) - 2 Hg(l) + O2(g) Hrxn = 181.6 kJ\n 2 (200g + 16g)   =  2(200g) + 2(16g) = 181.6 kj  \n\n432g oxide decomposes releasing 181.6Kj\n\nQuestion a)\nStraightforward ratio. X = the no of KJ required\n\n572g/432g = xkj/181.6kj\n\nYou'll need to calculate the molecular mass from you periodic table substitute it for the 432 and then refine the equation.\n\nQuestion b) Straightforward ratio. X = the no of KJ required\n\nYou'll need to calculate the molecular mass from you periodic table substitute it for the 432 and then refine the equation.\n\n289/181.6 = xg/432\n\nA mole is the amount of pure substance containing the same number of chemical units as there are atoms in exactly 12 grams of carbon-12 (i.e., 6.023 X 1023). This involves the acceptance of two dictates -- the scale of atomic masses and the magnitude of the gram. Both have been established by international agreement. Formerly, the connotation of "mole" was "gram molecular weight." Current usage tends to apply the term "mole" to an amount containing Avogadro's number of whatever units are being considered. Thus, it is possible to have a mole of atoms, ions, radicals, electrons, or quanta. This usage makes unnecessary such terms as "gram-atom," "gram-formula weight," etc.
926935
PV = nRT\n\n685 mmHg / 760 = .901 atmospheres\n\n18.6 mmHg / 760 = .0245 atm\n\n.901-.0245 = .8765 atm\n\n21 degrees c = 294 K\n\n.8765 atm x 7.80 L = n x .08206 x 294 K\n\nsolve for n then use the molar mass of H2 (2.016) to convert to grams.\n\nmy calculator's out in the car or else i'd do it for you.
245727
Xe, element # 54 I believe
769142
Easy: the molecular mass of a water molecule has 2 hydrogen atoms and 1 oxygen whereas a normal parcel of air is mostly oxygen and nitrogen.\n\nMoist air will be lighter at 18 and 28.98 for dry air.
655398
Here's how to solve it (I am convinced that you should do your own homework, but hints are OK as long as you learn something):\n\nStart solving by finding the molecular mass of the compound.\nA. Volume occupied by one mol of gas is 22.4 L at standard temperature and pressure (STP: 760 mmHg and 273 K (0 C))\nTo find the volume of gas that weighs 2.57 g at STP, use formula: P1V1/T1=P2V2/T2, where R is 0.08206, P1=319 mmHg, V1=2 L, T1=298 K; P2=760 mmHg, T2=273 K and V2 is unknown (it will be somewhat less than 1 L).\nAfter you get this volume, see how many mols do you have in it: divide 22.4 by V2 you just got. Then multiply this number by 2.57 and get the molecular mass of your compound (M(X)).\n\n2. Multipy M(X) separately by 0.6481, 0.136 and 0.2159 to get number of atomic weight units that are allocated to C, H, and O respectively. \nThen divide first number by 12, 2nd by 1 and 3rd by 16 to see how many atoms it amounts to. The resulting integer numbers are X, Y, Z.\n\n3. CxHyOz is the empirical formula of your compound. Google it to get the possible molecular ones. This is a trick question, so there will be more than one but less than 8. All of them are alcohols or ethers.
552193
Can be solved by simply plugging values into the Ideal Gas equation without using two states.  Pressure must be converted to atmospheres, however, and temperature to Kelvin.  P = 1.243 atm, T = 306 K\n\nn = PV/RT = 1.243 * 2.74 / (0.08206 * 306) = 0.136 mol\n\nmass = moles * MM = 0.136 * 28.01 = 3.81 g
906779
the ans should be C...because..if a gas that diffuses faster than KR...means tat the gas has a smaller molar mass than Kr...since is diffuse faster than 2.29 times..therefore the molar mass will be 83.8 / 2.29 = 36.6\n\n\n83.8 is Mr of Kr
895568
a mole is 6.022×10^23 particles.\n(1.45 * (10^24)) / (6.02200 * (10^23)) = 2.41 moles\n\nN = 14 atomic mass\nO = 16\n\n(14 * 2) + (16 * 5) = 108 g/mole\n(2.41 mole) * 108 (g / mole) = 260.28 grams
363082
2Ag+ + CO_33-2 -> Ag_2CO_3
324518
molecular weight's 305.4188 g/mol
770748
Oxygen is the limiting reactant....you would need 5.6 moles of it for a complete reaction.\n\nMoles of CO2...2.8.  It is 1/1 with methane\n\nGrams 2.8 X 44 = 123.2 grams
777427
C:H = 85.63:14.37 (mass ratio)\n= 85.63/12.01:14.37/1.008 (molar ratio)\n= 7.13:14.26\n= 1:2\nSo the empirical formula is CH2 and its empirical molecular weight ≈ 14.03\n\nP1V1/T1 = P2V2/T2\n0.980 * 1/(273.2 + 51.7) = 1.V2/273.2 (at STP of 1 atm and 0°C)\nSo V2 = 0.980 * 273.2/324.9\n= 0.824 L\nThus 3.09 g occupies 0.824 L\nSo 3.09 * 22.4/0.824 g occupies 22.4 L at STP\nie 1 g mole = 84\nThus MW of this gas is 84\nSo number of empirical units = 84/14.03 ≈ 6\nSo the molecular formula is C6H12
920990
methane: CH4\n\nMr;\ncarbon 12\nhydrogen 1\n\nMr of methane= 1*carbon+4*hydogen\n=12+4\n=16\n\ntherefore one mole of methane weighs 16 grams\n\n4 g of methane= 4/16\n=0.25 moles\n------------------------------...\n\none mole of any gas occupies 24 dm3 at r.t.p\n\nmoles of oxygen in 7 dm3= 7/24\n=0.292 moles
216365
nope...it is only made of 6 grams of gold...the other 54 grams is brass or other metals
985020
1 mol = 22.4 litre at STP\nmass = mol x molecular weight\ncause all are 1 mol than all are 22.4 litre\na) 1 mol H2O = 18 gram\nb) 1 mol CH3Cl = 50.5 gram\nc) 1 mol SO4 = 96 gram\nd) 1 mol NH3 = 17 gram\ne) 1 mol C3H7O = 59 gram\nd) 1 mol NO2 = 46 gram