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
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I am almost positive that the answer is: Plasmas. Let me know if im right or not please!! :)
____NaNO3 + ___PbO --> ___Pb(NO3)2 + ___Na[2]O
To balace the eqaution, you need to have the same number of atoms for each element on both the reactant (left) and product (right) side.
To start off, you wanna know the number of atoms in each element on both sides, so take it apart:
[reactants] [product]
Na- 1 Na- 2
N- 1 N- 2(it's 2 because the the subscript [2] is outside of the parenthesis)
O- 4 O- 7 (same reason as above)
Pb- 1 Pb- 1
Na is not balanced out, so add a coefficient to make it the same on both sides.In this case, multiply by 2:
2NaNO3
Now Na is balanced, but the N and O are also effected by this, so they also have to be multiplied by 2 and they become:
Na- 2 Na- 2
N- 2 N- 2 (it balanced out)
O- 7 (coefficient times subscript, plus lone O) O- 7 (balanced out)
Pb was already balanced so no need to mess with it, just put a 1 where needed (it doesn't change anything).
Now to put it back together, it will look like this:
2NaNO3 + 1PbO --> 1Pb(NO3)2 + 1Na[2]O
