Assume air is 21% O2 and is at STP (273 K, 1.00 atm)
From the stoichiometry of the reaction C + 02 --> CO2 the mass of 02 needed is given by:
Mass O2 = 10,000 tonne x (32 tonne O2/ 12 tonne C) 2.67 x 10^4 tonne O2
Stoichiometric mass air 2.67 x 10^4 tonne/0.21 = 1.27 x 10^5 tonne air...
Ignore post above (wrong)
1 mole of air contains 0.21 moles of oxygen so
3.9*10^9 moles of air contain 8.19*10^8 moles of oxygen
there's 10% extra air, so you need 1.209*10^9 moles
100/1000000 parts of the exhaust is NO, so this is 1.209*10^5 moles,
which weigh 3.638*10^6g...
Homework Statement
Assume that an incorrectly adjusted lawn mower is operated in a garage such that the combustion reaction in the engion is C8H8. If the dimentions of the garage are 5x3x3 meters. How many grams of gasoline must be burned to raide the levl of CO in the air to 1000ppm by...
Ah I want volume in m^3 yet it is in cm^3. so that's a factor of 10^-6.
so 1cm^3 = 1*10^-6 m^3
n= (193.544*1*10^-6)/(8.314*273)
n= 8.53 * 10^-8
So to work out number of molecules I have to times the number of moles by 6.022x10^23
number of molecules = 8.53 * 10^-8 * 6.022x10^23 =...
Yeah shouldn't of put it in that way. Ill put it in a clearer way in future.
1 mole of air contains 0.21 moles of oxygen so
3.9*10^9 moles of air contain 8.33 *10^8 moles of oxygen
there's 10% extra air, so you need 3.57*10^9 moles
100/1000000 parts of the exhaust is NO, so this...
What I wrote seems fairly understandable to me. But I emailed the teacher and it is e^(-x) so ignore that one.
Ok so the answer I have is 193.544.
Then
n=PV/RT
n= (193.544*1)/(8.314*273)
n= 0.086 air molecules per cubic centimeter of air @ 50km?
I know you said that 28.97 g mol-1 is not in its SI units? I don't know what to convert it to. Also the equation is Ph = P0e(-Mgh/RT) and not Ph = P0e^-Mgh/RT isn't it?
right.
using n=PV/RT
V=1cm^3 = 1*10^-6m^3
R = 8.314 JK-1mol-1
T = 273K
P @ 50km = ?
Have to find Ph
Plug in values into
Ph = P0e^-Mgh/RT
Ph = 101325e^-(28.97*9.81*50000/8.314*273)
I keep getting a value of 0 for this.
Surely I must have got that the wrong way round of P0 and Ph. Which equation do I use? Do I use "log(Ph) = -[Mgh*10^5/2.303RT]" or "Ph = P0e^-Mgh/RT" I have attached a copy of the handout for you to have a little look at for that last equation. I noticed the teacher took log instead what I think...
10,000 metric tons = 10 000 000 000 grams. (1.0 * 10^10)
1 mole of C = 832.58 moles
1 mole of NO = 832.58 moles
So the equation would be:
(1.0*10^10 grams of C) / 12.0108 g/mol x (1/1) x (44.0096 g/mol) x (100/ 10000) = 366416891.5 grams of NO?? Sorry I really don't understand this one
I did write down pressure at 50km = 75.944 Pa. But this is incorrect. I may have used the equation n = PV but n = number of moles and would be useless, wouldn't it?
---
RT
Ok. I've just found another equation in my handout I have been given...
Ph = pressure at any given height.
M = 28.97 g mol^-1
g = 9.81 ms^-2
R = 8.314 J K^-1 mol^-1
P0 = pressure at sea level
for
... 75.944 x 1 x 10^-6
Ph = -------------------
...8.314 x 273
Homework Statement
Hey guys. Hope you can help me out on this one. so here's the question:
At an altitude of 50km the average atmospheric temperature is 0^oC. What is the average number of air molecules per cubic centimeter at this altitude.
Homework Equations
M = 28.97 g mol^-1...