<span>Answer:
A 0.04403 g sample of gas occupies 10.0-mL at 289.0 K and 1.10 atm. Upon further analysis, the compound is found to be 25.305% C and 74.695% Cl. What is the molecular formula of the compound?
--------------------------------------...
Seems like I did a problem very similar to this--this must be the "B" test. But the halogen was different.
25.305% C/12 = 2.108
74.695% Cl/35.5 = 2.104
So the empirical formula would be CH. However, there are many compounds which fit this bill, so we have to use the gas data. (And I made, in the previous problem, the simplifying assumption that 289C and 1.10 atm would offset each other, so I'll do that, too.)
0.044 grams/10 ml = x/22.4 liters
0.044g/0.010 liters = x/22.4 liters
22.4 liters/0.010 liters = 2240 (ratio)
2240 x .044 = 98.56 (actual atomic weight)
CCl = 35.5+12 or 47.5, so two of those is 95 grams/mole.
This is sufficiient to distinguish C2CL2, (dichloroacetylene)
from C6CL6 (hexachlorobenzene) which would
mass 3 times as much.</span>
Use the molar mass of ammonia to change the grams to moles and then use mole-mole ratio
100. g NH3 (1 mol NH3/ 17.04 g) (3 mol H2/ 2 mol NH)= 8.80 moles H2
Answer:
44.01 g/mol
Explanation:
Add each elements atomic mass. For oxygen you will do that twice because their is two oxygens.
- Hope that helps! Please let me know if you need further explanation.
Answer:
-32 Fahrenheit converts to 237.594 Kelvin
Https://us-static.z-dn.net/files/d49/33d4ec86853ef95e6f6c14242c663be4.png
