Answer:
Yes, there will be liquid present and the mass is 5.19 g
Explanation:
In order to do this, we need to use the equation of an ideal gas which is:
<em>PV = nRT (1)</em>
<em>Where:</em>
<em>P: Pressure</em>
<em>V: Volume</em>
<em>n: number of moles</em>
<em>R: gas constant</em>
<em>T: Temperature</em>
we know that the pressure is 856 Torr at 300 K. So, if we want to know if there'll be any liquid present, we need to calculate the moles and mass of the CCl3F at this pressure and temperature, and then, compare it to the initial mass of 11.5 g.
From (1), solving for moles we have:
<em>n = PV/RT (2)</em>
Solving for n:
P = 856/760 = 1.13 atm
R = 0.082 L atm / mol K
n = 1.13 * 1 / 0.082 * 300
n = 0.0459 moles
Now, the mass is:
m = n * MM (3)
The molar mass of CCl3F reported is 137.37 g/mol so:
m = 0.0459 * 137.37
m = 6.31 g
Finally, this means that if we put 11.5 g of CCl3F in a container, only 6.31 g will become gaseous, so, this means it will be liquid present, and the mass is:
m = 11.5 - 6.31
m = 5.19 g
Answer:
Nonpolar covalent bonds are a type of bond that occurs when two atoms share a pair of electrons with each other. These shared electrons glue two or more atoms together to form a molecul
Explanation:
This theory was first proposed by Nicklaus Copernicus. Copernicus was a polish astronomer. He first published the heliocentric system in hes book: De revelation erbium comestible, "In revelations of the heavenly bodies," which appeared in 1543. HOPE IT HELP:)
gold density = 19.3 g/cm³
mass = 16.5 kg = 16.5 x 10³ g
volume = 954 ml
gold density based on the experiments:
ρ = 16.5 x 10³ g / 954 ml(cm³) = 17.296 g/cm³ ( not of pure gold)