<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?
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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>
Answer:
The temperatures of the objects must be different
Explanation:
if heat is flowing between two objects, then the objects must be at different temperatures.
4 Hydrogen Atoms should be correct. Because the 4 infront of it signifies the amount of hydrogen. It would also be 4 hydrogen atoms if it were written as H4N3, because the 4 is still around the H (as long as the 4 is under scored)
Answer:
May I assume "ethanol acid is just ethanol (it has one slightly acidic H atom). If so, the molar mass is 46.02 g/mole.
Explanation:
We have 30 cm^3 [30 ml] of 1.0 M (1 mole/liter) [1 dm³ = 1 liter].
That is 1 mole/liter. 30 ml would contain (0.030 liter)*(1 mole/1 liter) = 0.03 moles.
