What is the number of moles of O2 gas at pressure 760 cm of Hg in a container of volume 24.63 litres at 27°C?
Here in this, we can apply the Ideal Gas equation..!!Pressure(P) = 760 cm of Hg = 10 atmVolume(V) = 24.63 litresTemperature(T) = 27°C = 300KGas Constant(R) = 0.0821 litre•atm/K•moleNo. of moles(n) = [?]By applying Ideal gas equation,PV = nRTTherefore, n = PV / RTSo, n = 10 * 23.63 / 0.0821 * 300So, n =10 moles
What is the pressure of 200 moles of a gas in a 100 DM3 cylinder at 20 C?
PV=nRT sop = nRT/V. change 20ºC to 293K.what units do you want for the pressure? that will determine what you use for R or just use 8.314 J/molK and remember thatJ can also be written as kPa L or as Pa m^3 so R could be 8.314 kPa L (by the way, a dm^3 is a L. so use that and your pressure will come out in kPa.
Vapor pressure - moles - help?
In an experiment, 20.00 L of dry nitrogen gas, N2 ,at 20degreesC and 750.0 mmHg is slowly bubbled into water in a flask to determine its vapor pressure. The liquid water is weighed before and after the experiment from which the experimenter determines that it loses 353.6mg in mass. a) how many moles of nitrogen were bubbled into the water? - i used PV=nRT to find .821 moles (for part a) b) the liquid water diminished by how many moles? c) How many moles of gas exit the flask during the experiment? what is the partial pressure of nitrogen gas exiting? d) calculate the vapor pressure of water at 20degreesC
Help with ideal gas law volume/pressure/mole?
I have been stuck on this weird problem forever! In an experiment reported in the scientific literature, male cockroaches were made to run at different speeds on a miniature treadmill while their oxygen consumption was measured. In one hour the average cockroach running at 0.08 km/hr consumed 0.80 mL of O2 at 1 atm pressure and 24°C per gram of insect mass. (a) How many moles of O2 would be consumed in 1 hr by a 5.1-g cockroach moving at this speed? (answer in mol) (b) This same cockroach is caught by a child and placed in a 0.25-gal fruit jar with a tight lid. Assuming the same level of continuous activity as in the research, what percentage of the available O2 will the cockroach consume in a 48-hr period? (Air is 21 mol percent O2.) (answer in %)
How many moles of gas are collected & what is the partial pressure?
(a) Assume ideal gas mixture and use Ideal gas law: p·V = N·R·T (p pressure, V volume, N number of moles, R -universal gas constant, T absolute temperature) => N = (p·V) / (R·T) = (850Torr · 20L) / (62.3637LTorr/molK · (45+273.15)K) = 0.85862mol (b) Ideal gas law states that the volume which is taken the amount N for given pressure and temperature is the same irrespective of the chemical composition. Thus the mole fraction and volume fraction are the same foe ideal gas mixtures. The partial pressure is given by Dalton's law and ideal gas law: p_i = N_i·R·T/V divide by p p_i/ p = N_i / N => p_i = (N_i /N) · p p_SO2 = (N_SO2 /N) · p = (7.52 × 10^3 / 1×10^6) · 850Torr = 6.392Torr
Partial Pressure/Mole Fraction help please?
A gas mixture with a total pressure of 740 mmHg contains the following gases at the indicated partial pressures: CO2 = 115mmHg ; Ar = 220mmHg and O2 = 150 mmHg. The remaining gas in the mixture is helium. What is the mole fraction of helium gas in this sample? I'm not sure how to find mole fraction with the limited information given... Partial pressure of Helium is 260mmHg i think help.. thanks.