construct a 98% confidence interval estimate for the mean amount of nitrogen-oxide emissions for all cars. if it is required that nitrogen-oxide emissions be less than 0.165 gram/mile,can we safely conclude that this requerment is being met?
126508
Do a confidence interval calculation and see if the confidence level for the data you've collected is above that required for 98%.\n\nI can show you how but given your other questions, it's clear you haven't really done any work on this yet and you're just looking to get someone to do your homework for you.\n\nLet me know if you still can't do it after you've tried a while, and I might help you out.
484903
This depends on how exact your answer needs to be.  You can use the 68-95-99.7 rule to determine that in the 95% of the time you will fall within two standard deviations (i.e., bet. 6 and 8).  Half of the remaining 5% is usage below 6 liters, so only 2.5% of the time do you run out of gas.\n\nTo be more exact, you can look up the z-score given by:\n(X - mu) / (sigma) = (8 - 7) / .5 = 2 and find that the result is that you run out of gas 2.2275% of the time.\n\nIf you need more explanation, just add that to your additional details.  Good luck!
510974
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.
146384
The RAND() function is not the appropriate way to solve this.\n\nThe number of cars can be determined a couple of ways.\n\n1.  One is to use NORMINV(.99,2600000,100955) = 2834856\n2.  Or you could use the standard normal distribution:\n\n    x= 2600000 + 100955 NORMSINV(.99)\n     = 2600000 + 100955(2.326348) = 2834856\n\nRAND() just gives you a random number between 0 and 1.  I don't see how that would help you in this problem.\n\nThe second problem is also easily solved using the the standard normal distribution:\n\nFirst, the 2.4 million is 200000 below the mean, \nso z=-200000/100955= -1.98108\n\n2.7 million is 100000 above the mean, so \nz=100000/100955=.99054\n\nSo the probability the number of cars will be between 2.4 and 2.7 million is\n\n= NORMSDIST(.99054) - NORMSDIST(.023791)\n=.839045 - .023791 = .815254
523805
PV=nRT\nV = nRT/P\nV = (114g/44g/mol)(0.08206L-atm/mol-K)(28+273K)/0.987atm\nV = 64.84L or 0.0648 m^3
551990
ok--i'm not completely sure about this--but i can at least try-- *cough cough*unlike some people*cough*cough*catweazle*\n\nwe did this type of stuff last year.  but i think that you might want to use the ideal gas law?? we always called it pervnert... PV=nRT... does this sound familiar?    \n\nif not, i suggest asking your teacher, or some other knowledgable chemistry person.\n\ni hope this helps!
504003
Part A:\nThe answer is 12 L H2.  Here is the solution:\n(4 L N2)(1 mol N2 / 22.4 L N2)(3 mol H2 / 1 mol N2)(22.4 L H2 / 1 mol H2)\nso (4*1*3*22.4) / (22.4*1*1) = 12 L H2\n**note that in in ideal gas there are 22.4 liters/mol**\n\nPart B:\nThe answer is 8 L NH3.  Here is the solution:\n(4 L N2)(1 mol N2 / 22.4 L N2)(2 mol NH2 / 1 mol N2)(22.4 L NH3 / 1 mol NH3)\nso (4*1*2*22.4) / (22.4*1*1) = 8 L NH3\n\nHope this helps!
95363
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.
153753
Maybe you should have the emissions checked.  Drive it to the EPA and have them check it out.
386830
Try this: http://en.wikipedia.org/wiki/Confidence_interval
51199
Please do your own home work.\n\nI suspect you are missing a symbol, it probably should be,\nN2O4  <==>  2NO2\n\nThink about how each of the changes will effect the equilibrium of the reaction.
661661
Same as the atmosphere\nNitrogen 78%\nOxygen 21%\nC02  .038%
423044
Sorry to say, you better pass on this data set.  \n\nNote that each row of numbers totals 100, as these are percentages, so the variables are correlated, and have limited application.\n\nI think your professor wants two of the variables to be somewhat independent, meaning one is not a function of the other.  Such as gas mileage and price, for example, if you had a list of automobile makes.\n\nRatio variables are those in which the ratio of two of the numbers have meaning, such as miles per gallon, for example.  If car A gets 15 mpg and car B gets 20 mpg, you can take the ratio of the two: 15/20 and compute 0.75, meaning car A gets 75% of the mileage of car B.
383866
n   = pV/RT = 1.03*1.02/284*0.0821 = 0.045moles\n\nmoles of Nitrogen = 0.045*0.78=0.0351\n\nmoles of oxygen = 0.022*0.045 = 0.0098mole\n\nweight of the gases = 0.0098*32 + 0.0351*28.02 =0.316+ 0.986=1.3g\n\ndensity 1.3 g/l
173450
Write the equation for K.  You know K, [NO2}, so solve for the only variable you don't know [N2O4].  By the way, these are gasses so it is not a solubility product.  That involves a solid.
245914
Approximately 95% of the data is within 2 standard deviations of the mean.\nCheck out this website\nhttp://en.wikipedia.org/wiki/Confidence_interval\nand look under the section\n"Confidence Intervals in Measurement"
76189
22.4 l=>1 mole N2O\n0.1 l=> 0.0045 mole N20\nsince 1 mole NH4NO3 gives 1 mole N20,\n0.0045 mole of NH4NO3 is required.\nsince 1 mole NH4N03=>80 grams,\nmass of NH4N03 required is 80*0.0045=0.3571 grams\n\nsince 4 moles of Fe reacts with 3 moles of 02\nand 52 g of Fe=1 mole Fe,\n0.75 mole of 02 required.\nsince 1 mole 02=>22.4 l,\nvolume of 02 required is 16.8 litres\n\nhope that helps