Identify the PO2 of atmospheric air a 10,000 fee elevation.?
410398
The partial pressure of O2 at 10 000 feet altitude.   Well, the air pressure at sea level is about 100 kilopascals.   At 10 000 feet it's about 70 kilopascals.   About a fifth of the air is oxygen so the partial pressure of O2 up there is about 14 kilopascals.
540399
hey! Air is a mixture of gases thus it is solution, it contains Nitrogen-78%, oxygen-20.9% carbon dioxide0.003% and some rare gases
655529
B. Nitrogen.  It's above 70% in normal air.
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).
430954
A number of differnt elements, including oxygen, hydrogen, nitrogen, carbon, argon and others.  The mojority of air is nitrogen at about 63%, while depending where int he world you are, oxygen only makes up 15-19 percent.
924197
PCO2 = 35-45\nPO2 = 75-100
145449
The density of air, ρ (Greek: rho) (air density), is the mass per unit volume of Earth's atmosphere, and is a useful value in aeronautics. In the SI system it is measured as the number of kilograms of air in a cubic meter (kg/m3). At sea level and at 20 °C dry air has a density of approximately 1.2 kg/m3. varying with pressure and temperature. Air density and air pressure decrease with increasing altitude.\n\nThe formula for the density of air is given by:\n\n P=p/R*T\nwhere ρ is the air density, p is pressure, R is the gas constant, and T is temperature.\n\nThe individual gas constant R for dry air is:\n\n R( dry air )=287.05J/kg*K\nTherefore:\n\nAt standard temperature and pressure (0 °C and 101.325 kPa), dry air has a density of ρSTP = 1.293 kg/m3. \nAt standard ambient temperature and pressure (25 °C and 100 kPa) dry air has a density of ρSATP = 1.168 kg/m3. \n[edit]\nEffect of water vapor\nFor moist air, the partial pressure of the water vapor must be considered as well. In this case, the density of the air is the sum of the density of the dry air and the density of the water vapor:\nP=pdry air /Rdry air*T+p water vapour/R water vapour*T\n \nThe gas constant for water vapor is:\nR water vapour=461.495 J/kg*K\n \nEffects of altitude\nTo calculate the density of air as a function of altitude, one requires additional parameters. They are listed below, along with their values according to the International Standard Atmosphere, using the universal gas constant instead of the specific one:\n\nsea level atmospheric pressure p0 = 101325 Pa = 1013.25 mbar or hPa = 101.325 kPa (= 101325 kg/(m*s2)) \nsea level standard temperature T0 = 288.15 K \nEarth-surface gravitational acceleration g = 9.80665 m/s2. \ndry adiabatic lapse rate L = −0.0065 K/m \nuniversal gas constant R = 8.31447 J/(mol·K) \nmolecular weight of dry air M = 0.0289644 kg/mol \nTemperature at altitude h metres above sea level is given by the following formula (only valid below the tropopause):\n\n T=To+L*H\nImportance of temperature\nThe below table demonstrates that the properties of air change significantly with temperature.\n\nTable — speed of sound in air c, density of air ρ, acoustic impedance Z vs. temperature °C\n\nImpact of temperature \n°C c in m/s ρ in kg/m³ Z in Pa·s/m \n- 10 325.4 1.341 436.5 \n- 5 328.5 1.316 432.4 \n0 331.5 1.293 428.3 \n+ 5 334.5 1.269 424.5 \n+ 10 337.5 1.247 420.7 \n+ 15 340.5 1.225 417.0 \n+ 20 343.4 1.204 413.5 \n+ 25 346.3 1.184 410.0 \n+ 30 349.2 1.164 406.6
720438
Here's the o2 web site.\n\nhttp://www.o2.co.uk/\n\nLook around for what you need or Contact them if you need further help
53064
Assuming your gas is an ideal gas, you can use the ideal gas law.\n\nPV=nRT where \nP=Pressure in atm (1 atm= 760 mm Hg so 730 mm Hg= 0.96atm\nV=volume in L\nn= moles\nT=Temperature in K (C=K+273 so 30 C = 303 K)\nR= the universal gas constant= 0.08206 L-atm/mol-K\n\nSolve the ideal gas equation for n=PV/RT\n\nn=[(.96atm)(10L)]/[(.08206 L-atm/mol-K)(303K)]\n\nn=0.386 moles
761497
1) The first one is wrong on the last person who answered, O2 weighs 32g/mol because there are 2 O's in it, 16 each.  So in 1.33x10^4 (13,300mol) you have 4.256x10^5g of O2 (425.6kg).\n\n2) Recognize that hydrogen, oxygen, and chlorine are diatomic gasses, so they always occur in pairs (H2, O2, Cl2).  Others like this are nitrogen (N2), fluorine (F2), bromine (liquid @ room temp. Br2) and Iodine (solid at room temp, I2).  So your equations would be:\n\n2H2+O2-->2H2O\nH2+Cl2-->2HCl\n\n3) PV=nRT and solve for n, number of moles.  Put everything in correct units, liters, atmospheres, moles, and kelvins, and R is a constant R=0.0821 L*atm/mol K.  Kelvin temp=Celcius temp +273\n\nSolving for n:\nn=PV/RT=(1.09atm)(0.942L)/(0.0821L*atm/mol*K)(287K)\nn=0.0436 mol\n\nSo, when we usually want to know a number of moles given a mass and having a molar mass we can look up, we say,\nn=mass/molar mass\n\nin this case, we need to find molar mass, so:\nmass/n=molar mass\n3.17g/0.0436mol=72.71g/mol
66785
Latitude, altitude, prevailing winds.
141492
very very very very little\nmostly co2
970749
500sin30 =  x  m/h altitude gained
223476
heres an easy one..its nitrogen 79.1%next is oxygen and then 0.03 %is carbondioxide.
527581
Expansion coefficients for air can be determined using the attached link.\n\nBest wishes !
856026
Use the gas law\nPV=nRT
1002178
Carbon dioxide acts as a greenhouse gas at all altitudes. You are correct that it is heavier than air, and therefore is less prevalent at higher altitudes.