Calculate the ratio of effusion rates for ar and kr.

Calculate the ratio of effusion rates for ar and kr.

So the ratio of the effusion rates = SQRT(83.8/39.948) = 1.45. Argon moves 45% faster than krypton! Hope that helped!

The ratio of effusion rates of krypton and neon is 0.49 Explanation: To calculate the rate of diffusion of gas, we use Graham’s Law.
This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows the equation:
We are given: Molar mass of Neon = 20.18 g/mol
Molar mass of Krypton = 83.80 g/mol
By taking their ratio, we get: Hence, the ratio of rate of effusion of krypton and neon will be 0.49

The ratio of rate of effusion of chlorine gas and oxygen gas will be 0.671 Explanation: To calculate the rate of diffusion of gas, we use Graham’s Law. This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows the equation: We are given: Molar mass of chlorine gas = 71 g/mol Molar mass of oxygen gas = 32 g/mol By taking their ratio, we get: Hence, the ratio of rate of effusion of chlorine gas and oxygen gas will be 0.671

The ratio of rate of effusion of helium and radon is 7.45 Explanation: To calculate the rate of diffusion of gas, we use Graham’s Law. This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows the equation: We are given: Molar mass of Helium = 4.00 g/mol Molar mass of Radon = 222.1 g/mol Taking their ratios, we get: Hence, the ratio of rate of effusion of helium and radon is 7.45

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0.251 Explanation: According to the Graham’s law, the rate of effusion of a gas (r) is inversely proportional to the square root of its molar mass (M). The ratio of effusion rates of oxygen (O₂) to hydrogen (H₂) is:

Answer 6

Answer : The ratio of effusion rates for helium and argon is 3.2 : 1 Solution : Given, Molar mass of Helium = 4 g/mole Molar mass of Argon = 40 g/mole Rate of effusion : It is defined as the rate of effusion of a gas is inversely proportional to the square root of the molar mass of the gas. Formula used : where, M is the molar mass. Now put all the given values in this expression, we get Therefore, the ratio of effusion rates for helium and argon is 3.2 : 1

Answer 7

Hey There!: Molar mass O2 => 32.0 g/mol Molar mass Cl2 => 71.0 g/mol effusion rate α  1 / (√ molar mass ) E Cl2 / E O2 = √ molar mass O2 / √ molar mass Cl2 E Cl2 / E O2 =   ( √ 32.0 ) /  ( √ 71.0 ) => 0.671 Hope that helps!

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Ratio of effusion rates for  and is . 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 is as follows:                       …… (1) The expression for rate of effusion of is as follows:                        …… (2) Dividing equation (1) by equation (2),                        ….. (3) Substitute 349.042 u for and 352.039 u for in equation (3). Therefore ratio of effusion rates for  and is 0.9957. Learn more:
1. Which statement is true for Boyle’s law?
2. Calculation of volume of gas:
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.

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Ratio=1.44 Explanation: According to Graham’s law of effusion, the ratio of effusion rates for two gases can be written as: Ratio = Where M1 and M2 are the molar masses of the gases. In this case, M1 is the atomic mass of  Kr (83.798) and M2 is the atomic mass of Ar (39.948). Ratio = Thus Ar effuses 1.44 times faster than Kr

Ratio=1.44 Explanation: According to Graham’s law of effusion, the ratio of effusion rates for two gases can be written as: Ratio = Where M1 and M2 are the molar masses of the gases. In this case, M1 is the atomic mass of  Kr (83.798) and M2 is the atomic mass of Ar (39.948). Ratio = Thus Ar effuses 1.44 times faster than Kr

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