<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:
Explanation:
The spheres interact with each other, and a change in one area can cause a change in another. Humans (biosphere) use farm machinery manufactured from geosphere materials to plow the fields, and the atmosphere brings precipitation (hydrosphere) to water the plants. The biosphere contains all the planet's living things.
The atomic number of copper is 29.
Hope this helps and if it does, don't be afraid to rate my answer as well as maybe give it a "Thanks"? (Or even better a "Brainliest"). And if it’s not correct, I am sorry for wasting your time, and good luck finding the correct answer :)
Down...
and if you were wondering, the earth pushes back
Types of Bonds can be predicted by calculating the
difference in electronegativity.
If, Electronegativity difference is,
Less
than 0.4 then it is Non Polar Pure Covalent
Between 0.4 and 1.7 then it is Polar Covalent
Greater than 1.7 then it is Ionic
For Br and Br,
E.N of Bromine = 2.96
E.N of Bromine = 2.96
________
E.N Difference
0.00 (Non Polar/Pure Covalent)
For N and O,
E.N of Oxygen = 3.44
E.N of Nitrogen = 3.04
________
E.N Difference
0.40 (Non Polar/Pure Covalent)
For P and H,
E.N of Hydrogen = 2.20
E.N of Phosphorous = 2.19
________
E.N Difference 0.01 (Non Polar/Pure Covalent)
For K and O,
E.N of Oxygen = 3.44
E.N of Potassium = 0.82
________
E.N Difference 2.62 (Ionic)