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)
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
b. the effusion rates of two gases = the square root of the inverse of their molar masses:
M₁ = molar mass sulfur dioxide = 64
M₂ = molar mass nitrogen triodide = 395
the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triodide
A compound has to be chemically bonded, however, air is not chemically bonded.
This can be proven by freezing air. By freezing air, it yields different liquids at different temperature. Liquid nitrogen has a different boiling point than liquid oxygen.
If air was a compound, they would all have a single boiling point and a single freezing point.
Hope this helps :)
Curiosity's mission is to determine whether the Red Planet ever was, or is, habitable to microbial life. The rover, which is about the size of a MINI Cooper, is equipped with 17 cameras and a robotic arm containing a suite of specialized laboratory-like tools and instruments.
Answer:
it’s D. to reduce indigestion
Explanation:
The new volume at standard temperature and pressure is 5.08 L.
Explanation:
As per the kinetic theory of gases, the volume occupied by gas molecules will be inversely proportional to the pressure of the gas molecules. This is termed as Boyle's law.
So, pressure∝
Thus, if two pressure and two volumes are given then,
Now, we known the values of P₁ = 8 atm, V₁ = 635 mL, P₂ = 1 atm and V₂ we have to determine. We are considering P₂ = 1 atm, because we have to determine V₂ at standard temperature and pressure. And standard pressure is 1 atm.
Thus, the new volume at standard temperature and pressure is 5.08 L.
