a. 661.23 m/s
b. the rate of effusion of Ammonia = 4.5 faster than Silicon tetra bromide
Further explanation
Given
T = 25 + 273 = 298 K
Required
a. the gas speed
b. The rate of effusion comparison
Solution
a.
Average velocities of gases can be expressed as root-mean-square averages. (V rms)
![\large {\boxed {\bold {v_ {rms} = \sqrt {\frac {3RT} {Mm}}}}](https://img.qammunity.org/2022/formulas/chemistry/high-school/l2uhmxqlnvonf9f768amf8y6wy9jzg7se9.png)
R = gas constant, T = temperature, Mm = molar mass of the gas particles
From the question
R = 8,314 J / mol K
T = temperature
Mm = molar mass, kg / mol
Molar mass of Ammonia = 17 g/mol = 0.017 kg/mol
![\tt v=\sqrt{(3* 8.314* 298)/(0.017) }=661.23~m/s](https://img.qammunity.org/2022/formulas/chemistry/high-school/ehw93ay1vpwjgtzqknge44spi50450sufh.png)
b. the effusion rates of two gases = the square root of the inverse of their molar masses:
![\rm (r_1)/(r_2)=\sqrt{(M_2)/(M_1) }](https://img.qammunity.org/2022/formulas/chemistry/high-school/swrfhnb5ugs5o6118sol5oqhqq4nrb0lnd.png)
M₁ = molar mass Ammonia NH₃= 17
M₂ = molar mass Silicon tetra bromide SiBr₄= 348
![\rm (r_1)/(r_2)=\sqrt{(348)/(17) }=4.5](https://img.qammunity.org/2022/formulas/chemistry/high-school/qaaxmfku623egum8lurwwc6wy3yo4qpwq9.png)
the rate of effusion of Ammonia = 4.5 faster than Silicon tetra bromide