Well, look up the atomic weights of copper and oxygen; add them appropriately; divide the total for the copper by the total for the molecule; then multiply by 100 to get it as a percentage.
A gas with a vapor density greater than that of air, would be most effectively displaced out off a vessel by ventilation.
The two following principles determine the type of ventilation: Considering the impact of the contaminant's vapour density and either positive or negative pressure is applied.
Consider a vertical tank that is filled with methane gas. Methane would leak out if we opened the top hatch since its vapour density is far lower than that of air. A second opening could be built at the bottom to greatly increase the process' efficiency.
A faster atmospheric turnover would follow from air being pulled in via the bottom while the methane was vented out the top. The rate of natural ventilation will increase with the difference in vapour density. Numerous gases that require ventilation are either present in fairly low concentrations or have vapor densities close to one.
Answer:
Then, at some point, these higher energy electrons give up their "extra" energy in the form of a photon of light, and fall back down to their original energy level.
Explanation:
When properly stimulated, electrons in these materials move from a lower level of energy up to a higher level of energy and occupy a different orbital.
There is 6.02*10^23 molecule per mole. And there is 2 atoms per oxygen molecule. So the answer is 1.204*10^24 atoms in 1.0 mole of O2.
Answer:
74.344 kJ.
Explanation:
Below is an attachment containing the solution.
