What is the volume of 9.2 x 10^25 molecules of SbH3(g) at STP?
2007-10-10 19:58:23 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
Determine number of moles then multiply by 22.4; the volume in liters of 1 mole of any gas at STP. (This is a very handy thing to know; saves a lot of calculating!)
(9.2 x 10^35molecules/6.02 x 10^23molecules/mol) x
22.4 l/mol = 3420 liters
2007-10-10 20:07:13 · answer #1 · answered by Flying Dragon 7 · 0⤊ 0⤋
StibniumHydride 9.20 x 10**25 molecules
Loschmidt's Constant 2.69 x 10**25 molecules/m3
Number of m3 = 9.2/2.69 = 3.4 cubic meters
2007-10-10 20:53:40 · answer #2 · answered by Richard P 2 · 0⤊ 0⤋
Assume ideal gas, so use the ideal gas equation:
PV = nRT
where P = gas pressure (kPa)
V = volume (m^3)
n = no. of moles of gas (kmol)
R = gas constant = 8.314 kJ/(kmol*K)
T = temperature (K)
First find n,
1 gmol of the gas contains 6.023x10^23 molecules,
so n = (9.2x10^25)/(6.022x10^23)
= 152.77 mol
= 0.1528 kmol
At STP, P = 101.325 kPa, T = 273.15 K
so V = nRT/P
= 0.1528 kmol x 8.314 kJ/(kmol*K) x 273.15 K / 101.325 kPa
= 3.42 m^3
2007-10-10 20:15:45 · answer #3 · answered by Anonymous · 0⤊ 0⤋