If 5.0 moles of O2 and 3.0 moles of N2 are placed in a 30.0 L tank at a temperature of 250 °C...?
回答 (2)
2020-08-12 11:58 pm
Consider the resulting gaseous mixture:
Volume, V = 30.0 L
Total no. of moles, n = (5.0 + 3.0) mol = 8.0 mol
Temperature, T = (273 + 250) K = 523 K
Gas constant, R = 0.08206 L atm / (mol K)
Gas law: PV = nRT
Then, P = nRT/V
Pressure, P = 8.0 × 0.08206 × 523 / 30.0 atm = 11.4 atm
Volume, V = 30.0 L
Total no. of moles, n = (5.0 + 3.0) mol = 8.0 mol
Temperature, T = (273 + 250) K = 523 K
Gas constant, R = 0.08206 L atm / (mol K)
Gas law: PV = nRT
Then, P = nRT/V
Pressure, P = 8.0 × 0.08206 × 523 / 30.0 atm = 11.4 atm
2020-08-13 2:22 am
nO2 = 5.0mol
nN2 = 3.0mol
Volume, V = 30L
Temp., T = 250oC = 523K
Total number of moles, n = nO2 + nN2
= 5.0mol + 3.0mol
n = 8.0mol
PV = nRT
P = (8.0mol x 0.0821LatmK-1mol-1 x 523K) / 30L
= 11.45atm
A thumbs up is appreciated .
nN2 = 3.0mol
Volume, V = 30L
Temp., T = 250oC = 523K
Total number of moles, n = nO2 + nN2
= 5.0mol + 3.0mol
n = 8.0mol
PV = nRT
P = (8.0mol x 0.0821LatmK-1mol-1 x 523K) / 30L
= 11.45atm
A thumbs up is appreciated .
收錄日期: 2021-04-24 07:57:09
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