80cm^3 of oxygen gas diffussed through a porous holebin 50 seconds how long will it take 120cm^3 of nitrogen (iv) oxide to diffu
1 answer:
Answer:
time for NO₂ to diffuse = 90 s
Explanation:
Data Given:
Amount of O₂ gas transferred = 80 cm³
time of O₂ gas to diffuse = 50s
Amount of NO₂ gas transferred = 120 cm³
time of NO₂ gas to diffuse = ?
Solution:
As we know
rate of diffusion = Amount of gas transferred / time . . . . . . . (1)
we also know that Graham's Law is
rate of diffusion gas A/rate of diffusion gas B = . . . (2)
where
mA = molar mass of gas A
mB = molar mass of gas B
combine both equation 1 and 2
(Amount of gas A transferred / time for gas A) / (Amount of gas B transferred / time for gas B) = . . . . . . . . (3)
we can write equation 3 for oxygen and nitrogen (iv) oxide
(Amount of gas O₂ transferred / time for gas O₂) / (Amount of gas NO₂ transferred / time for gas NO₂) = . . . . . . . (4)
- molar mass of O₂ = 2 (16) = 32 g/mol
- molar mass of NO₂ = 14 + 2(16) = 46 g/mol
Put values in equation equation 4
(80 cm³/50 s) / (120 cm³ / time for NO₂) =
(1.6 cm³/s)/(120 cm³ / time for NO₂) =
(1.6 cm³/s) / (120 cm³ / time for NO₂) = 1.2
Rearrange the above equation
time for NO₂ = 1.2 /(1.6 cm³/s) x 120 cm³
time for NO₂ = 90 s
So,
time for NO₂ to diffuse = 90 s
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- Oxygen- O- near the right end of periodic two.
Elements on the bottom-left corner of the periodic table are metals. Those on the top-right corner are nonmetals.
- Potassium is a metal,
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would pair up with two
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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
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M₁ = molar mass sulfur dioxide = 64
M₂ = molar mass nitrogen triodide = 395
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