Among the following visualization, the best one should be determined which represents the distribution of SO 2 and O 2 gas molecules near an orifice and after some time effusion takes place in the direction denoted by arrows. Concept introduction: The effusion of gases through orifice can be explained with the help of Graham’s law of effusion. According to the Graham’s law of effusion, rate of effusion of any gas is always inversely proportional to the square root of the mass of gaseous particles. The mathematical expression for this law is: Rate of effusion of gas 1 Rate of effusion of gas 2 = M 2 M 1 Here, M 1 and M 2 represent the molar mass of gases. In terms of time for gases; Effusion time of gas 1 Effusion time of gas 2 = M 1 M 2
Among the following visualization, the best one should be determined which represents the distribution of SO 2 and O 2 gas molecules near an orifice and after some time effusion takes place in the direction denoted by arrows. Concept introduction: The effusion of gases through orifice can be explained with the help of Graham’s law of effusion. According to the Graham’s law of effusion, rate of effusion of any gas is always inversely proportional to the square root of the mass of gaseous particles. The mathematical expression for this law is: Rate of effusion of gas 1 Rate of effusion of gas 2 = M 2 M 1 Here, M 1 and M 2 represent the molar mass of gases. In terms of time for gases; Effusion time of gas 1 Effusion time of gas 2 = M 1 M 2
Among the following visualization, the best one should be determined which represents the distribution of SO2 and O2 gas molecules near an orifice and after some time effusion takes place in the direction denoted by arrows.
Concept introduction:
The effusion of gases through orifice can be explained with the help of Graham’s law of effusion. According to the Graham’s law of effusion, rate of effusion of any gas is always inversely proportional to the square root of the mass of gaseous particles.
The mathematical expression for this law is:
Rate of effusion of gas 1Rate of effusion of gas 2=M2M1
Here, M1 and M2 represent the molar mass of gases.
true or false,the equilibrium constant for this reaction is 0.50.PCI5(g) ↔ PCI3(g) + CI2(g)Based on the above, the equilibrium constant for the following reaction is 0.25.2PCI5(g) ↔. 2PCI3(g) + 2CI2(g)
true or false, using the following equilibrium, if carbon dioxide is added the equilibrium will shift toward the productsC(s) + CO2(g) ↔ 2CO(g)
2S2O2/3- (aq) + I2 (aq) ---> S4O2/6- (aq) +2I- (aq)
Experiment
I2 (M)
S2O3- (M)
Initital Rate (M/s)
1
0.01
0.01
0.0004
2
0.01
0.02
0.0004
3
0.02
0.01
0.0008
Calculate the overall order for this reaction using the table data
a) 3b) 0c) 2d) 1
Chapter 6 Solutions
General Chemistry: Principles and Modern Applications (11th Edition)
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell