What does this question mean
Answer:
See explanation
Explanation:
You see, we must cast our minds back to Charles' law. Charles' law gives the relationship between the volume of a gas and temperature of the gas.
Now, Micheal left the balloon outside at a particular temperature and volume the previous night. Overnight, the temperature dropped significantly and so must the volume of the gas in the balloon!
Remember that Charles' law states that, the volume of a given mass of gas is directly proportional to its absolute temperature at constant pressure. Since the pressure was held constant, the drop in the volume of gas in the balloon can be accounted for by the drop in temperature overnight.
<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:
0.00011765 M
Explanation:
When a solution is prepared by dilution, the volumes and concentrations are related by:
C1*V1 = C2*V2
Where C1 is the concentration of the solution 1, V1 is the volume of the solution 1, C2 is the concentration of solution 2, and V2 is the volume of solution 2.
The stock solution is the solution 1, and the standard solution, the solution 2, so:
0.0001974*29.80 = C2*50.00
C2 = 0.00011765 M
<span>The ionic charge of Calcium (Ca) in calcium carbonate (CaCO3) is 2+. CaCO3 has a neutral ionic charge sin CO3 has a 2- charge.</span>
