Answer:
77.58g/mol
Explanation:
Let the rate of effusion of NO2 be R1
The rate of effusion of the unknown gas(R2) = 0.770R1
Molar Mass of NO2 (M1) = 14 + (2x16) = 14 + 32 = 46g/mol
The molar mass of the unknown gas (M2) =?
R1/R2 = √(M2/M1)
R1/0.770R1 = √(M2/46)
1/0.770 = √(M2/46)
Take the square of both sides
(1/0.770)^2 = M2/46
Cross multiply to express in linear form
M2 = (1/0.770)^2 x 46
M2 = 77.58g/mol
The molar mass of the unknown gas is 77.58g/mol
Answer:
6(3) - 4 + (3)^2
Explanation:
We are given that:
f(x) = 4-x^2
g(x) = 6x
First, we will compute (g-f)(x) as follows:
(g-f)(x) = 6x - (4-x^2) = 6x - 4 + x^2
Now, we will substitute with x=3 as follows:
(g-f)(3) = 6(3) - 4 + (3)^2
Hope this helps :)
Answer:
![d=0.92\frac{kg}{m^{3}}](https://tex.z-dn.net/?f=d%3D0.92%5Cfrac%7Bkg%7D%7Bm%5E%7B3%7D%7D)
Explanation:
Using the Ideal Gas Law we have
and the number of moles n could be expressed as
, where m is the mass and M is the molar mass.
Now, replacing the number of moles in the equation for the ideal gass law:
![PV=\frac{m}{M}RT](https://tex.z-dn.net/?f=PV%3D%5Cfrac%7Bm%7D%7BM%7DRT)
If we pass the V to divide:
![P=\frac{m}{V}\frac{RT}{M}](https://tex.z-dn.net/?f=P%3D%5Cfrac%7Bm%7D%7BV%7D%5Cfrac%7BRT%7D%7BM%7D)
As the density is expressed as
, we have:
![P=d\frac{RT}{M}](https://tex.z-dn.net/?f=P%3Dd%5Cfrac%7BRT%7D%7BM%7D)
Solving for the density:
![d=\frac{PM}{RT}](https://tex.z-dn.net/?f=d%3D%5Cfrac%7BPM%7D%7BRT%7D)
Then we need to convert the units to the S.I.:
![T=100^{o}C+273.15](https://tex.z-dn.net/?f=T%3D100%5E%7Bo%7DC%2B273.15)
![T=373.15K](https://tex.z-dn.net/?f=T%3D373.15K)
![P=1bar*\frac{0.98atm}{1bar}](https://tex.z-dn.net/?f=P%3D1bar%2A%5Cfrac%7B0.98atm%7D%7B1bar%7D)
![P=0.98atm](https://tex.z-dn.net/?f=P%3D0.98atm)
![M=28.9\frac{kg}{kmol}*\frac{1kmol}{1000mol}](https://tex.z-dn.net/?f=M%3D28.9%5Cfrac%7Bkg%7D%7Bkmol%7D%2A%5Cfrac%7B1kmol%7D%7B1000mol%7D)
![M=0.0289\frac{kg}{mol}](https://tex.z-dn.net/?f=M%3D0.0289%5Cfrac%7Bkg%7D%7Bmol%7D)
Finally we replace the values:
![d=\frac{(0.98atm)(0.0289\frac{kg}{mol})}{(0.082\frac{atm.L}{mol.K})(373.15K)}](https://tex.z-dn.net/?f=d%3D%5Cfrac%7B%280.98atm%29%280.0289%5Cfrac%7Bkg%7D%7Bmol%7D%29%7D%7B%280.082%5Cfrac%7Batm.L%7D%7Bmol.K%7D%29%28373.15K%29%7D)
![d=9.2*10^{-4}\frac{kg}{L}](https://tex.z-dn.net/?f=d%3D9.2%2A10%5E%7B-4%7D%5Cfrac%7Bkg%7D%7BL%7D)
![d=9.2*10^{-4}\frac{kg}{L}*\frac{1L}{0.001m^{3}}](https://tex.z-dn.net/?f=d%3D9.2%2A10%5E%7B-4%7D%5Cfrac%7Bkg%7D%7BL%7D%2A%5Cfrac%7B1L%7D%7B0.001m%5E%7B3%7D%7D)
![d=0.92\frac{kg}{m^{3}}](https://tex.z-dn.net/?f=d%3D0.92%5Cfrac%7Bkg%7D%7Bm%5E%7B3%7D%7D)