A 1. 00 ml sample of an unknown gas effuses in 11. 1 min. an equal volume of h2 in the same apparatus under the same conditions effuses in 2. 42 minutes then the molar mass of the unknown gas is 41.9.
Molar mass of H2 = 2
Molar mass of unknown gas = ?
rate 1 = 11.1
rate 2 = 2.42
<h3>What is graham law? </h3>
Graham's law states that the rate of diffusion or effusion of a given gas is inversely proportional to the square root of its molar mass.
By apply graham law
Rate1/rate2 = sqrt(MW2/MW1)
![[\frac{rate1}{rate2} ]^{2} = \frac{MW2}{2} \\\\\\mw= 2[\frac{11.1}{2.42} ]^{2} \\\\= 20.97 X 2 \\\\= 41.9](https://tex.z-dn.net/?f=%5B%5Cfrac%7Brate1%7D%7Brate2%7D%20%5D%5E%7B2%7D%20%3D%20%5Cfrac%7BMW2%7D%7B2%7D%20%5C%5C%5C%5C%5C%5Cmw%3D%202%5B%5Cfrac%7B11.1%7D%7B2.42%7D%20%5D%5E%7B2%7D%20%5C%5C%5C%5C%3D%2020.97%20X%202%20%5C%5C%5C%5C%3D%2041.9)
Thus, we found that the molar mass of the unknown gas is 41.9.
Learn more about graham's law: brainly.com/question/12415336
#SPJ4
I like that game Forza horizon
Based on the information I would assume B, 73 degrees...
It shouldn't be A, 4 minutes on the burner should increase the temperature.
If it were D, it would be beyond boiling, and water takes a decent amount of energy to heat, D should be all vapor.
Same logic for C, it's basically almost boiling.
I would say 73 degrees seems most reasonable for 4 minutes.
Answer:
89.4%
Explanation:
Initially, there is 5.0 of the acetanilide in 100 mL of water, then the solution is chilled at 0ºC. The solubility represents the amount that the solvent (water) can dissolve of the solute (acetanilide). So, at 0ºC, 100 mL of water can dissolve till 0.53 g of the compound, the rest will precipitate and will be recovered.
So, the mass that is recovered is 5.0 - 0.53 = 4.47 g
The percent recovery is:
(4.47/5)x100% = 89.4%
Well your mass wont change...so it would be 35 grams of a liquid
