<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:
Final pressure in (atm) (P1) = 6.642 atm
Explanation:
Given:
Initial volume of gas (V) = 12.5 L
Pressure (P) = 784 torr
Temperature (T) = 295 K
Final volume (V1) = 2.04 L
Final temperature (T1) = 310 K
Find:
Final pressure in (atm) (P1) = ?
Computation:
According to combine gas law method:
⇒ Final pressure (P1) = 5,048.18877 torr
⇒ Final pressure in (atm) (P1) = 5,048.18877 torr / 760
⇒ Final pressure in (atm) (P1) = 6.642 atm
Oxidation, Dissolving in acid, Hydrolysis ,Water absorption and Hydration
C. Chemical properties always change depending on temperature or pressure
