In a sample of 7 cars, each car was tested for nitrogen oxide emissions (in grams per\nmile), and the following results were obtained: 0.06, 0.11, 0.16, 0.15, 0.14, 0.08, and\n0.15. Assuming that this sample is representative of cars in use, construct a 98%\nconfidence interval estimate for the mean amount of nitrogen-oxide emissions for all\ncars. If it is required that nitrogen-oxide emissions be less than 0.165 grams/mile, can we\nsafely conclude that this requirement is being met?
136931
First you need to ask yourself if you have any reason to believe that measurements of nitrogen oxide [chemical formula -- Just say "NO"] emissions across different cars should be normally distributed.  Since this is a statistics problem, we assume that the answer is yes.\n\nThen you compute the sample mean (the sum of those seven measurements, divided by seven) and the sample variance (the sum of (x - sample mean)^2 for those seven measurements x, all divided by six).  The sample standard deviation is then the square root of the sample variance.\n\nFinally, you look up on your table or with your calculator or wherever what "Z score" corresponds to 98% of the area under the normal pdf curve being between\n\n(mean - Z standard deviations, mean + Z standard deviations)\n\nThey want to know if that upper bound is less than 0.165 grams / mile [a peculiar mix of metric and non-metric units].\n\nThat's probably how they want you to solve it.  Compare that to the Bayesian approach (Google for it).
533459
2NO + O2 --> 2NO2\nMr of NO = 14 + 16 = 30\nMoles of NO = (29.03/30) = 0.9677 (to 4 s.f.)\nMr of O2 = 32\nMoles of O2 = (19.35/32) = 0.6047 (to 4 s.f.)\n\nComparing the moles of NO and O2 given, you should notice that NO is the limiting reagent.\nHence moles of NO2 produced = moles of NO used = 0.9677\n(since the ratio of NO2 to NO is 2:2)\nMass of NO2 produced = 0.9677 x (Mr of NO2) = 0.9677 x 46 = 44.5 g (to 3 s.f.)\n\nHope this helps :)
463459
In a closed vessel the change in partial pressure for each gas during reaction scales to the stiochiometric coefficients.\n\nSo for this reaction, if 23 torr/min of NO is being used, then there must be 23 torr/min of NOCl appearing since both coefficients are 2 and they are on opposite sides of the equation.\n\nLikewise, if 23 torr/min of NO is being used, then 11.5 torr/min of Cl2 is also being used since they are on the same side of the equation.  It is only 11.5 torr/min because the coeffient for Cl2 is only 1, so it is half the amount of NO.\n\nThe total pressure change will be the sum of the amounts:\n\npNO + pCl2 + pNOCl = -23 -11.5 + 23 = -11.5\n\nKeeping two sig figs, this would mean that the pressure is decreasing by 12 torr/min.
339420
The first thing you need is a balanced equation for the reaction:\n\n4 NH3 + 5 O2 → 4 NO + 6 H2O\n\nA) 5 moles of O2 produce 4 moles of NO\n\n160.00 g of O2 / g/mol for O2 = number of moles of O2\n\n160.00 g / 16.00 g/mol = 10.0  moles of O2\n\nThis would produce 8.0 moles of NO\n\n8.0 moles * 30.01 g/mol = 240.08 grams of Nitric oxide.\n\nB)  If the percent yield is only 92.98 % then\n\n240.08 g * 0.9298 = 223.226 grams (you might want to round this to 223.2 grams).
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.
168511
Well, let's see:  you're releasing 1000 kg/hr of liquid N2, which equilibrates to 1 atmosphere and 20C (STP).\n\nAt STP, one mole of a gas occupies 22.4 liters;\nyou're releasing 1000 kg/hr = (1000 kg*1000g/kg)/28g/mole = 35,714 moles of N2 per hour, and thus:\n\n22.4 liters/mole * 35,714 moles = 800,000 liters (at STP) per hour
316759
B. Nitrogen.  It's above 70% in normal air.
324057
I would like to Change the Clinton/Gore EPA-OBD II Law. This applies to all Gasoline Powered Vehicles in the USA from 1996 to the Present, like Yours. This Law requires that all Vehicles remain Factory Stock, in order to Pass Inspection. Specifically, the Engine Fuel System must Run at 14.7 Parts of Air to 1 Part of Fuel. I've found a way to Convert Gasoline into Propane, 100 Parts of Air to 1 Part of Fuel. My Unit is Safe, Compact, and uses Low Pressure Air to do it's thing. 50 + MPG with even the Largest SUV, and a 10 X Reduction of Emissions. And, thanks to the EPA-OBD II Law, it's Illegal! When that Gestapo-Inspired Vehicle Inspection Device is Connected, any Change at all, even for the Better, and the Vhicle will Fail, and will be Illegal to Drive! Because it's not Putting out enough Pollution to meet [O]riginal [E]quipment [M]anufacturer's Specifications! This is Insane! I'm not the First, only the Latest. I've been to the US Patent Office to Check out the "Wild Stories" about Fuel Vaporizers that have been kept off the Market. Guess What? It's True! Case in Point, the Pogue Carburetor, with a Patented Claim of 100 MPG, with a Flathead Ford Engine Powered Vehicle that was 15 MPG!This was before WW II. So, as far as Getting a Patent, forget it! See, Greed and Money have Stopped it Every Single Time! And Now the EPA-OBD II Law....... Here's what I'd really Like to do to Expose the Hypocricy of this Law. I'd like to Restore and Convert a whole Fleet of Vehicles and Donate them to Charity Starting with the Local Meals-On-Wheels Program, then, Volunteer Disabled Veterans' Services.And so Many Others. There is no Shortage of Good People that want to Volunteer to Help the less Fortunate. But with the High Cost of Fuel, it's becoming Increasingly Difficult. Even if I did manage to get the Funds to do this, someone Connected with "Big Oil" would find a way to make One of the Converted Vehicles Explode, to Discredit it. Remember the TV News Show where the Chevy Truck with the "Side-Saddle" Fuel Tank was Rigged to Explode to Discredit It? See, all Petroleum is essentially the Same, on a Molecular Level. Only the Dispersal is Altered. If That's Wrong, then how is it that Engines DO Run on Propane? To do so by Conventional Means is such a Hassle that it's just not worth it! Hybirds? At Highway Speeds, the Gas [14.7 to1] Engine Kicks-In, Plus, the Heavy, Expensive Batteries have to be Hauled. To Sum Up, I'd like to change the EPA-OBD II Law that requires Vehicles to Emit Enough Pollution to Pass Inspection, with any reduction resulting in Failure. But, "IT AIN'T GONNA HAPPEN" as long as "Big Oil" is in Control! Watch Closely the Next Time Your Car is Inspected, and ask how many have Failed,and Why! 14.7 to 1. It's the Law that must Change! For even more Insight, check out this Website. It's a Couple of Years Old, but lots of Good Info.http://energy21.freeservers.com/bookrep.html
379891
First, we have to check if the equation is balanced, which in this case it is. So, we can continue.\n\nNow, let's find out the total number of moles of Cu and HNO3.\n\n# mole of Cu = 4.55 / 63.55 = 0.0716 mole\n# mole of HNO3 = 6.29 / (1 + 14 + 16*3) = 0.0998 mole\n\nWhat we have to do now is to determine which compound (Cu or HNO3) is the limiting reagent. One easy way to find out is to assume either one compound will be completely used up and use the stoichiometry coefficient (the numbers in front of the compounds) to determine how many moles of the other compound is required. \n\nFor example, assume all the Cu will be reacted, the number of moles of HNO3 needed = (0.0716 / 3)*8 = 0.191 moles\n\nSince it requires more HNO3 than we have, HNO3 is the limiting reagent. Now, apply the 46.4% reaction yield to this question.\n\nFor every 8 moles of HNO3 that react, it produces 3 moles of Cu(NO3)2, since we have 0.0998 moles, IF for 100% reaction yield, we will produce\n\n(0.0998/8)*3 = 0.0374 moles of Cu(NO3)2\n\nSINCE the reaction is just 46.4%, ACTUAL yield will be\n\n0.0374*0.464 = 0.0174 moles\n\nMass of Cu(NO3)2 = 0.0174 * (63.55 + 14*2 + 16*6)\n= 3.26 g
70902
Although I know I am not getting this $20.  199 kJ of energy.\n\nIn response to a later answer that claimed I was wrong he was right and wrong\n\n He wrote "first find the amount of mol of nitrogen\nmol = 1/14" But infact the amount was 1 kg and that divided by 14g/mol would be 71.43 moles and multiplying that times his enthalpy of vaporization which is 2.7928kJ/mol would give 199 kJ of energy. So I was right, but wrong with my significant figures because the asker said it was 1.0 kg which would gie me 2 sig figs and the answer would be 2.0*10^2 kJ.
644162
http://home.ust.hk/~irenelo/main/6/6a_body.htm\n\nhttp://www.pepps.fsu.edu/segip/states/NC/poll.html\nThis link above talks about the PSI ranges and simply says that    "For each urban area, the PSI is assigned a single number that is representative of the worst offending criteria pollutant for that day. The calculated number is placed within the appropriate index range and the accompanying descriptor word is used to report that day's pollution level."\n\nhttp://www.epa.gov/air/criteria.html\nThe link above is a site that gives more info about the NAAQS
166713
* How many people are there?\nA> Total seating capacity of the cars is 3200x4 + 4000x6 = 36800.\nIf, by (80% of the cars are full" you mean 80% of all cars, as an average, then total number of people = 80% of 36800 = 29440.\n\n* How many 4 seaters are there?\nA> 3200 (as mentioned in the question itself)\n\n* What is the combined driving time of all the cars?\nA> Total time is as follows:\n38% of all people live to 53 => 11187.2 live to 53 => lifetime = 11187.2x53 = 592921.6 years\nSimilarly, the numbers in the other cases are 31795.2, 194304, 296755.2 & 503424 years respectively. Adding, we get 1619200 years.\n\n* How many people are there of each age group?\nA> 53 : 38% of 29440 = 11187 and so on.
276162
Ack... Munir got here before I did agian.  A agree with Munir!  Volume of any gas at STP is 22.4L.
77932
To solve this problem we need to think about limiting reactants.  This one is fairly simple because I already have moles and the ratio is 1:1:1.  If I only have 3 kmols of C, that is the most I can make, because going beyond that, I would be out of C.  Thus, I can only make 3kmols of CO2 with 2 kmols of O2 left unreacted.  These amounts are theoretically find for a combustion reaction.
21335
To be precise, the answer is no. Do u noe why? The mean mol wt of air is 29 whereas that of CO is 28.\n\nlet me show u how i get these answers:\nC=12,O=16,mol wt of O2=32g/gmol,mol wt of N2=28g/gmol\nmol wt of CO = 12+16\n             =28\nas for air, by common sense, we know that air is made up of O2,N2, and other gases,rite? As the mol composition of other gases are very negligible(<1%), we assume air is only made up of O2 and N2. Air consists of 21mol% O2 and 79mol% N2.\n\nlet's take a basis of 100gmol(or 100mol)\nmoles of O2=21gmol\nmoles of N2=79gmol\nmass of O2=21X 32g/gmol\n          =672g\nmass of N2=79X28g/gmol\n          =2212g\ntotal mass= 2212+672\n          = 2884g\naverage mol wt of air=total mass/total moles\n                     =2884g/100gmol\n                     =28.84g/gmol\n                     =29g/gmol(whole no.)\n\nsince the ave mol wt of air is greater than that of CO, so aCO is not heavier than air,instead,slightly lighter than air.\n\nDoes it answer ur question?
473806
First you need to determine your limiting reagent in the first step.\n\nMolecular mass of NH3 = 17 g/mol, O2 = 32 g/mol\nMol of NH3 = 5,700,000 g/17 g/mol\nMol of NH3 = 3.35E5 mol\n\nMol of O2 = 57,000,000 g/32 g/mol\nMol of O2 = 1.78E6 mol\n\nSo based on the first equation (1), NH3 is the limiting reagent.\nFor every 4 mol of NH3, one gets 4 mol of NO.  So the first reaction produced 3.35E5 mol of NO.\n\nThe second reaction produces 2 mol of NO2 for every 2 mol of NO used, so the second reaction produced 3.35E5 mol of NO2.\n\nThe third reaction produces 2 mol of HNO3 for every 3 mol of NO2 used, so this final reaction will produce 2/3 x 3.35E5 mol of HNO3 = 2.23E5 mol.\n\nThen use molecular mass of HNO3 to calculate how much of it (mass) that you made.\nHNO3 molecular mass = 63 g/mol\nmass = mol x molar mass\nmass = 2.23E5 x 63 g/mol\nmass = 1.41E7 g of HNO3\n\nOf course this assumes 100% yields in all the above noted reactions.
191820
it depends on how good the engine is...generally the higher ps will consume more...but you cant use it as a rule of thumb