Answer:
m = 20.9 g.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to solve this problem by recalling both the Avogadro's number for the calculation of the moles in the given molecules of calcium phosphate and the molar mass of this compound in order to secondly calculate the mass as shown on the following setup:
![m=4.05x10^{22}molecules*\frac{1mol}{6.022x10^{23}}*\frac{310.18g}{1mol}\\\\m=20.9g](https://tex.z-dn.net/?f=m%3D4.05x10%5E%7B22%7Dmolecules%2A%5Cfrac%7B1mol%7D%7B6.022x10%5E%7B23%7D%7D%2A%5Cfrac%7B310.18g%7D%7B1mol%7D%5C%5C%5C%5Cm%3D20.9g)
Regards!
The answer is in the attachment below:
Answer:
precipitation
Explanation:
trust me ............................
From the ideal gas law
pv=nRT , n is therefore PV/RT
R is the
R is gas constant =62.364 torr/mol/k
P=500torr
V=4.00l
T=500+273=773k
n={(500 torr x 4.00l)/(62.364 x773k)}=0.041moles
the number of molecules=moles x avorgadro costant that is 6.022x10^23)
6.022 x 10^23) x0.041=2.469 x10^22molecules