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Below is the answer. I hope it helps.
rate 1 / rate 2 = square root of M2/M1
rate 238UF6 / 235UF6 = square root of [(Mass 235UF6) / (Mass 238UF6)]
rate 238UF6 / 235UF6 = square root of [(349.044) / (352.041)]
rate 238UF6 / 235UF6 = square root of [0.9914868]
Your answer: rate 238UF6 / 235UF6 = 0.9957343
rate 235UF6 / 238UF6 = square root of [(352.041) / 349.044) ]
rate 235UF6 / 238UF6 = square root of [1.008586]
other answer: rate 235UF6 / 238UF6 = 1.004284
rate 1 / rate 2 = square root of M2/M1
rate 238UF6 / 235UF6 = square root of [(Mass 235UF6) / (Mass 238UF6)]
rate 238UF6 / 235UF6 = square root of [(349.044) / (352.041)]
rate 238UF6 / 235UF6 = square root of [0.9914868]
Your answer: rate 238UF6 / 235UF6 = 0.9957343
rate 235UF6 / 238UF6 = square root of [(352.041) / 349.044) ]
rate 235UF6 / 238UF6 = square root of [1.008586]
other answer: rate 235UF6 / 238UF6 = 1.004284
Ratio of effusion rates for [tex]^{238}\text{UF}_6[/tex] and [tex]^{235}\text{UF}_6[/tex] is [tex]\boxed{0.9957}[/tex].
Further Explanation:
The movement of constituent particles of gas through the medium of small pinhole is described in terms of effusion. When air or helium gas escapes from the balloon, it becomes deflated after sometime due to process of effusion.
Graham's law of effusion states that rate of effusion of any gas (R) is inversely related to square root of its molar mass. This indicates more the molar mass of gas, less will be the rate of effusion and vice-versa.
The expression for rate of effusion of [tex]^{238}\text{UF}_6[/tex] is as follows:
[tex]{\text{R}_\left(^{238}\text{UF}_6\right)}\propto\sqrt{\dfrac{1}{\text{M}_\left(^{238}\text{UF}_6\right)}}}[/tex] ...... (1)
The expression for rate of effusion of [tex]^{235}\text{UF}_6[/tex] is as follows:
[tex]{\text{R}_\left(^{235}\text{UF}_6\right)}{\propto\sqrt{\dfrac{1}{\text{M}_\left(^{235}\text{UF}_6\right)}}[/tex] ...... (2)
Dividing equation (1) by equation (2),
[tex]{\dfrac{{\text{R}_\left(^{238}\text{UF}_6\right)}}{{\text{R}_\left(^{235}\text{UF}_6\right)}}}=\sqrt{\dfrac{{\text{M}_\left(^{235}\text{UF}_6\right)}}{{\text{M}_\left(^{238}\text{UF}_6\right)}}}[/tex] ..... (3)
Substitute 349.042 u for [tex]{\text{M}_\left(^{235}\text{UF}_6\right)}}}[/tex] and 352.039 u for [tex]{\text{M}_\left(^{238}\text{UF}_6\right)}}}[/tex] in equation (3).
[tex]\begin{aligned}{{\dfrac{{\text{R}_\left(^{238}\text{UF}_6\right)}}{{\text{R}_\left(^{235}\text{UF}_6\right)}}}&=\sqrt{\dfrac{\text{349.042 u}}{\text{352.039 u}}\\&=0.9957\end{aligned}[/tex]
Therefore ratio of effusion rates for [tex]^{238}\text{UF}_6[/tex] and [tex]^{235}\text{UF}_6[/tex] is 0.9957.
Learn more:
1. Which statement is true for Boyle’s law? https://brainly.com/question/1158880
2. Calculation of volume of gas: https://brainly.com/question/3636135
Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Ideal gas equation
Keywords: Effusion rate, UF6, Graham's law of effusion, square root, molar mass, 0.9957, movement, gas, inversely proportional, 349.042 u, 352.039 u, ratio of effusion rates, air, balloon, helium gas, deflated, constituent particles, pinhole, R.
