Answer : The vapor pressure of the dry oxygen gas is, 736.2 torr
Explanation : Given,
Volume of sample of oxygen = 500 mL
Vapor pressure of oxygen + water = 760 torr
Vapor pressure of water = 23.8 torr
Now we have to determine the vapor pressure of the dry oxygen gas.
Vapor pressure of the dry oxygen gas = Vapor pressure of (oxygen + water) - Vapor pressure of water
Vapor pressure of the dry oxygen gas = 760 torr - 23.8 torr
Vapor pressure of the dry oxygen gas = 736.2 torr
Thus, the vapor pressure of the dry oxygen gas is, 736.2 torr
Physical change because water is not a gas or change . you can see through it.
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
Answer:
Anything which occupy space and have mass between them are called matter.