Answer:
Grams of mercury= 0.06 g of Hg
Note: The question is incomplete. The complete question is as follows:
A compact fluorescent light bulb contains 4 mg of mercury. How many grams of mercury would be contained in 15 compact fluorescent light bulbs?
Explanation:
Since one fluorescent light bulb contains 4 mg of mercury,
15 such bulbs will contain 15 * 4 mg of mercury = 60 mg
1 mg = 0.001 g
Therefore, 60 mg = 0.001 g * 60 = 0.06 g of mercury.
Compact fluorescent lightbulbs (CFLs) are tubes containing mercury and noble gases. When electricity is passed through the bulb, electron-streams flow from a tungsten-coated coil. They collide with mercury atoms, exciting their electrons and creating flashes of ultraviolet light. A phosphor coating on the inside of the tube absorbs this UV light flashes and re-emits it as visible light. The amount of mercury in a fluorescent lamp varies from 3 to 46 mg, depending on lamp size and age.
It is either a proton or a neutron
Answer:
molar composition for liquid
xb= 0.24
xt=0.76
molar composition for vapor
yb=0.51
yt=0.49
Explanation:
For an ideal solution we can use the Raoult law.
Raoult law: in an ideal liquid solution, the vapor pressure for every component in the solution (partial pressure) is equal to the vapor pressure of every pure component multiple by its molar fraction.
For toluene and benzene would be:

Where:
is partial pressure for benzene in the liquid
is benzene molar fraction in the liquid
vapor pressure for pure benzene.
The total pressure in the solution is:
And
Working on the equation for total pressure we have:
Since
We know P and both vapor pressures so we can clear
from the equation.
So
To get the mole fraction for the vapor we know that in the equilibrium:
So
Something that we can see in these compositions is that the liquid is richer in the less volatile compound (toluene) and the vapor in the more volatile compound (benzene). If we take away this vapor from the solution, the solution is going to reach a new state of equilibrium, where more vapor will be produced. This vapor will have a higher molar fraction of the more volatile compound. If we do this a lot of times, we can get a vapor that is almost pure in the more volatile compound. This is principle used in the fractional distillation.
Answer:
hinndndnnddnndndndnd do djfj
Explanation:
hdhdjdbdndndndjjddjdndjdjdndnndndndnd be rnnrbr
Answer:
5 g / ml
Explanation:
Convert the values given to g and ml
12.5 kg = 12500 g
2.5 L = 2500 ml
12500 g / 2500 ml = 5 g/ml