thank you soo much, ive been stuck on this question a while now lol
There are two significant figures:
4.4 x 10^23 atoms Fluorine
There are two significant figures:
4.4 x 10^23 atoms Fluorine
Source(s): Chemistry Major
Molar mass CH2F2 = 52.0235 g/mol
19g = 19/52.0235 = 0.3652 mol CH2F2
0.3652 mol = 6.022*10^23*0.3652 = 2.2*10^23 molecules
Each molecule has 2 F atoms
Number og F atoms = 2.2*10^23 *2 = 4.4*10^23 F atoms.
There are two moles of Fluorine for every mole of the molecule, so there are 0.730441 moles of Fluorine
2 x 0.36522 = 0.730441
Next, avogadros number lets us know how many atoms are in this many moles of fluorine
6.02 x 10^23 x 0.730441 = 4.39871 x 10^23 atoms
There are two significant figures:
4.4 x 10^23 atoms Fluorine
I got this answer as follows:
Molar mass CH2F2 = 52.0235 g/mol
19g = 19/52.0235 = 0.3652 mol CH2F2
0.3652 mol = 6.022*10^23*0.3652 = 2.2*10^23 molecules
Each molecule has 2 F atoms
Number og F atoms = 2.2*10^23 *2 = 4.4*10^23 F atoms.
This post is last updated on hrtanswers.com at Date : 1st of September – 2022