1. possibly one of the lowest experimental pressures achievable is 1.0*10^-15 to
ID: 633779 • Letter: 1
Question
1. possibly one of the lowest experimental pressures achievable is 1.0*10^-15 torr. how many molecules are there in 250 mL flask that has been ecacuated to this pressure at 77.0K
2. the total volume of the gas bags of the german dirigible hindenberg, was 2.0 * 10^5m^3. how many grams of H2 would be required to fill them at 20 celcius and 1.00 atm pressure? hint convert m^3>cm^3>mL>L.
3. how many grams of KClO3 are needed to produced 44.8 L of oxygen gas measured at STP?
2KClO3(s)> 2KCl(s)+ 3O2(g)
Explanation / Answer
1> pressure= 1*10^-15 torr= 1.333*10^-16 pa
temperature = 77k
volume=250ml = 2.5*10^-4 m^3
PV=nRT
n=PV/RT
n= (2.5*10^-4*1.333*10^-16)/( 8.314*77)
n= 5.205*10^-23 molecules
2> V= 2.0 * 10^5m^3
T=293K
P=1atm=101325Pa
R=8.314j/k-mol
n=8318958.687 molecules
mass = n*2=16637917037g
3> V(o2)= 44.8L=0.0448m^3
p=101325pa
R=8.314 j/k-mol
t=298k (25c)
therefore
n= 1.832 moles of O2
by stochiometry 2 moles KClO3 produce 3moles O2
1.832 moles O2 are produced by 2*1.832/3=1.221 moles KClO3
therefore mass of KClO3 =122.55*1.221 gm = 149.633gm
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