27 I think beacuse if you subtract
<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>
The answer is true the si base unit for mass is the gram
Answer:
119.7 mL.
Explanation:
- From the general law of ideal gases:
<em>PV = nRT.</em>
where, P is the pressure of the gas.
V is the volume of the container.
n is the no. of moles of the gas.
R is the general gas constant.
T is the temperature of the gas (K).
- For the same no. of moles of the gas at two different (P, V, and T):
<em>P₁V₁/T₁ = P₂V₂/T₂.</em>
- P₁ = 100.0 mmHg, V₁ = 1000.0 mL, T₁ = 23°C + 273 = 296 K.
- P₂ = 1.0 atm = 760.0 mmHg (standard P), V₂ = ??? mL, T₂ = 0.0°C + 273 = 273.0 K (standard T).
<em>∴ V₂ = (P₁V₁T₂)/(T₁P₂) </em>= (100.0 mmHg)(1000.0 mL)(273.0 K)/(296 K)(760.0 mmHg) = 121.4 <em>mL.</em>
