Answer:
It has 3 subshells, it has 9 orbitals, and it has the capacity for 18 electrons
Explanation:
An: Calculate the molarity of a solution made by adding 120 g of NaOH (40.00 g/mol) to enough water to make 500.0 mL of solution. a) 4.0 M b) 6.0 M c) 1.0 ...
Explanation:
Answer:
The velocity of the river increased.
There was more erosion in the stream.
The type of sediment that moved changed.
Explanation:
a. 381.27 m/s
b. the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triiodide
<h3>Further explanation</h3>
Given
T = 100 + 273 = 373 K
Required
a. the gas speedi
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 {\dfrac {3RT} {Mm}}}}](https://tex.z-dn.net/?f=%5Clarge%20%7B%5Cboxed%20%7B%5Cbold%20%7Bv_%20%7Brms%7D%20%3D%20%5Csqrt%20%7B%5Cdfrac%20%7B3RT%7D%20%7BMm%7D%7D%7D%7D)
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 Sulfur dioxide = 64 g/mol = 0.064 kg/mol
![\tt v=\sqrt{\dfrac{3\times 8.314\times 373}{0.064} }\\\\v=381.27~m/s](https://tex.z-dn.net/?f=%5Ctt%20v%3D%5Csqrt%7B%5Cdfrac%7B3%5Ctimes%208.314%5Ctimes%20373%7D%7B0.064%7D%20%7D%5C%5C%5C%5Cv%3D381.27~m%2Fs)
b. the effusion rates of two gases = the square root of the inverse of their molar masses:
![\rm \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }](https://tex.z-dn.net/?f=%5Crm%20%5Cdfrac%7Br_1%7D%7Br_2%7D%3D%5Csqrt%7B%5Cdfrac%7BM_2%7D%7BM_1%7D%20%7D)
M₁ = molar mass sulfur dioxide = 64
M₂ = molar mass nitrogen triodide = 395
![\tt \dfrac{r_1}{r_2}=\sqrt{\dfrac{395}{64} }=\dfrac{20}{8}=2.5](https://tex.z-dn.net/?f=%5Ctt%20%5Cdfrac%7Br_1%7D%7Br_2%7D%3D%5Csqrt%7B%5Cdfrac%7B395%7D%7B64%7D%20%7D%3D%5Cdfrac%7B20%7D%7B8%7D%3D2.5)
the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triodide