An unknown diatomic gas effuses at a rate that is 0.250 times the rate of N2. What is the molar mass?
964564
Rate of effusion for gas 1 / rate of effusion for gas 2 = sqrt M2 / sqrt M1 (where M is molar mass)
977807
M = moles/L\nso, 0.220M = 0.220 moles per liter and there are 2 liters so there are...\n0.440 moles of Mg(NO3)2\n\nBUT... notice that there are 2 ions of NO3 for each molecule, so double it again and get 0.880 moles of NO3
937526
If the reaction would "go" it is written: 3H2 + N2 = 2 (NH3)\nFill in the atomic masses of H = 1.00797 and N = 14.0067:\n[3 x 2 x (1.00797)] +[ 2 x (14.0067)]  = 2 x (14.0067) x 3 x (1.00797) = 34.06122\nThis is the stochiometric ratio of elements needed. You have 2.04 gm of H2, so multiply each term by 2.04 / 6.04782 to get:\n2.04 + 9.4492 = 11.4892. In other words, all of the H2 combines with about 9.45 gm of N2 to yield about 11.49 gm of ammonia with about 1.9 gm of N2 left over.
609831
pv=nrt\nmeasure pressure, volume, and temperature to find the number of moles, n\n\nfind the mass of the gas\n\nmass of gas/ moles of gas= molecular mass (g/mol)
709574
I get 1.69522 g/L, using constants with the most significant figures I can find:\n\nN/V = P/(R*T)\nP = 101.325e3 Pa\nV = 1 m^3\nR = 8.314472\nT = 273.15 K\n\nN/m^3 = 44.6149 mol/m^3 = 0.0446149 mol/L\n1 mol of F2 = 37.9968 g\n0.0446149*37.9968 = 1.69522 g/L\n\nIf I use 100 KPa (1 bar) as standard pressure, I get 1.67305 g/L\n\nThe difference in the values given in the answer and the "Handbook of Chem. & Physics" is apparently due to the shift from 1 atm to 1 bar as the definition of "standard pressure."\nI think the slight difference in my calculated value and the published one may be due to the fact that when you start getting out to that many decimal places even "ideal gases" aren't totally ideal....maybe I should try the Redlich-Kwong EOS...\n\nOooooops!  I didn't want to do your homework for you...maybe I should erase all this....:)
159316
[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
558772
the correct answer is 1.30.  the answer would be two if the concentration was 0.01M.  you just take the negative log of the concentration of hydronium ion.
787380
PV=nRT\nP=pressure\nV=volume\nn=moles\nR=constant(.08026atm*L/mol*K)\nT=temperature\n\nThis is the ideal gas law.
569322
Use the ideal gas law \n\nPV/T = constant for the final and initial states.
105211
Pressure * volume = number of mol * temperature (in ^oK) * 8.314\n\nYou have all the information you need right there. \nFirst solve for n for the initial state. \nSince you haven't added any more material you can then use the same value of n for the second state and solve for P.\n\nThis is sixth grade algebra. Anyone being asked to solve gas equations had better be able to do it standing on their head or they are in real trouble.
193517
You don't provide enough data to determine this.  The gas law PV=nRT has variables you need to define.
233434
This is just a simple stoichiometry problem.  You have 200g of NH3, your first job is to calculate how many moles of NH3 this is.  From the periodic table, find the atomic weights of N and H to calculate the molar mass of NH3:  \n\nN=14 and H=1, so NH3 = 17.  \n\n200g NH3 X (1 mol NH3 / 17g NH3) = 11.76 mol NH3\n\nOkay, now that we know we're working with 11.76 moles of NH3, we can use the mole ratio from the given chemical equation to figure out how many moles of oxygen are needed.  Look at the coefficients in the equation -- we need 5 moles of O2 for every 4 moles of NH3:\n\n11.76 mol NH3 X (5 mol O2)/(4 mol NH3) = 14.7 mol O2\n\nNow we know we need 14.7 mol O2.  All we have to do now is convert that to grams, again using the atomic mass from the periodic table.  The atomic mass of O is 16, so O2=32  \n\n14.7 mol O2 X 32g/mol = 470.4 g O2\n\nSUMMARY:\nConvert grams ammonia to moles ammonia using atomic weights.\nUse the mole ratio to determine needed moles of O2.\nConvert moles O2 to grams O2 using atomic weights.
874851
2Al (s) + N2(g) ----> 2AlN (s)
791970
Use the Gay-Lussac gas law: V/T is constant. Note that T here is the temperature in Kelvin, not in Celsius. Luckily 27 C = 300 K
750464
My chemistry is pretty rusty, but let's see.\n\nI'll assume that you have an infinite amt of O2, and that you have 68.12 grams of NH3.\n\nThe equation isn't balanced, so you need to do that first (you have 3 hydrogens on one side, but two on the other).  Balance to get:\n\n2(NH3) + (2.5)(O2) --> 2NO + 3(H20)\n\n4(NH3) + 5(O2) --> 4NO + 6(H20)\n\nNow, given that you have 68.12 grams of NH3, you need to convert this into moles of NH3, as Stoichiometry deals with molar ratios.  To get the conversion ratio, look at the periodic table.  Under each element, there should be an atomic mass.  For oxygen is 16.something, Hydrogen 1.008 ( i think ).  Find it for N and H.  (this is where I get hazy on the math).  I believe: take 3 * atomic mass of hydrogen + atomic mass of nitrogen, let's call it M  (it should be in grams/mole)  So to get moles, multiply 68.12 * (1/M).  \n\nNow, our equation above shows that for every 4 moles of NH3, we get 6 of H20.  # moles NH3 * (6/4) = # moles of H20.  Now, do the reverse operation of H20 to get how many grams is produced.  H20 has an atomic mass of something like 18.016.  So, # moles H20 * 18.016 = # grams H20\n\n\nI hope that's somewhat correct, and if not, leads you in the right direction at least
275919
2H20 --> 2H2 + O2\n\ntheoreticalyy spk it's exactly half the vol of H2, just look over the stoechiometrical balance above; but due to gas losses, it may be less
1151475
number of moles of hydrogen is 0.2\n\namount of water is 2.4