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
Answer is: 3,3 mol of <span>nitrous oxide gas is produced in this chemical reaction.
</span>Chemical reaction: N₂ + O₂ → 2NO.
n(N₂) = 1,65 mol.
n(NO) = ?
from reaction n(N₂) : n(NO) = 1 : 2.
1,65 mol : n(NO) = 1 : 2.
n(NO) = 3,3 mol.
n - amount of substance.
Answer : The final temperature of the mixture is 
Explanation :
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.
where,
= specific heat of iron = 
= specific heat of water = 
= mass of iron = 39.9 g
= mass of water = 
= final temperature of mixture = ?
= initial temperature of iron = 
= initial temperature of water = 
Now put all the given values in the above formula, we get
Therefore, the final temperature of the mixture is
Answer is: (2) Chemical energy is converted to electrical energy.
An electrochemical cell (voltaic or galvanic cell) is generating electrical energy from chemical reactions.
In galvanic cell, specie (for example zinc and zinc cations) from one half-cell, lose electrons (oxidation) and species from the other half-cell (for example copper and copper cations) gain electrons (reduction).
Oxidation on the zinc anode: Zn(s) → Zn²⁺(aq) + 2e⁻.
Reduction on the copper cathode: Cu²⁺(aq) + 2e⁻ → Cu(s).
Dissolving sugar in water
