Answer:
Explanation:
Hello,
Considering the ideal equation of state:
The moles are defined in terms of mass as follows:
Whereas 
the gas' molar mass, thus:
Now, since the density is defined as the quotient between the mass and the volume, we get:
Solving for 
:
Thus, the result is given by:
![density=\frac{(1atm)(44g/mol)}{[0.082atm*L/(mol*K)]*298.15K} \\density=1.8g/L=1.8x10^{-3}g/mL](https://tex.z-dn.net/?f=density%3D%5Cfrac%7B%281atm%29%2844g%2Fmol%29%7D%7B%5B0.082atm%2AL%2F%28mol%2AK%29%5D%2A298.15K%7D%20%5C%5Cdensity%3D1.8g%2FL%3D1.8x10%5E%7B-3%7Dg%2FmL)
Best regards.
We are going to use Avogadro's constant to calculate how many molecules of
carbons dioxide exist in lungs:
when 1 mole of CO2 has 6.02 x 10^23 molecules, so how many molecules in
CO2 when the number of moles is 5 x 10^-2
number of molecules = moles of CO2 * Avogadro's number
= 5 x 10^-2 * 6.02 x 10^23
= 3 x 10^22 molecules
∴ There are 3 x 10^22 molecules in CO2 exist in lungs
Answer:
The answer to your question is 25.2 g of acetic acid.
Explanation:
Data
[Acetic acid] = 0.839 M
Volume = 0.5 L
Molecular weight = 60.05 g/mol
Process
1.- Calculate the number of moles of acetic acid
Molarity = moles / volume
-Solve for moles
moles = Molarity x volume
-Substitution
moles = (0.839)(0.5)
-Result
moles = 0.4195
2.- Calculate the mass of acetic acid using proportions and cross multiplications
60.05 g ----------------------- 1 mol
x ----------------------- 0.4195 moles
x = (0.4195 x 60.05) / 1
x = 25.19 g
3.- Conclusion
25.2 g are needed to prepare 0.500 L of Acetic acid 0.839M
