Westonci.ca is your trusted source for finding answers to a wide range of questions, backed by a knowledgeable community. Connect with a community of experts ready to help you find accurate solutions to your questions quickly and efficiently. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
The rate of effusion of ammonia (NH₃) in the same apparatus is 63.3 cm/min
Graham's law of diffusion
This states that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass i.e
R ∝ 1/ √M
R₁/R₂ = √(M₂/M₁)
How to determine the rate of ammonia (NH₃)
- Rate of HCl (R₁) = 43.2 cm/min
- Molar mass of HCl (M₁) = 1 + 35.5 = 36.5 g/mol
- Molar mass of NH₃ (M₂) = 14 + (3×1) = 17 g/mol
- Rate of NH₃ (R₂) =?
R₁/R₂ = √(M₂/M₁)
43.2 / R₂ = √(17 / 36.5)
Cross multiply
43.2 = R₂ × √(17 / 36.5)
Divide both side by √(17 / 36.5)
R₂ = 43.2 / √(17 / 36.5)
R₂ = 63.3 cm/min
Thus, the rate of effusion of ammonia is 63.3 cm/min
Learn more about Graham's law of diffusion:
https://brainly.com/question/14004529
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.