So let's convert this amount of mL to grams:
Then we need to convert to moles using the molar weight found on the periodic table for mercury (Hg):
Then we need to convert moles to atoms using Avogadro's number:
![\frac{6.022*10^{23}atoms}{1mole} *[8.135*10^{-2}mol]=4.90*10^{22}atoms](https://tex.z-dn.net/?f=%20%5Cfrac%7B6.022%2A10%5E%7B23%7Datoms%7D%7B1mole%7D%20%2A%5B8.135%2A10%5E%7B-2%7Dmol%5D%3D4.90%2A10%5E%7B22%7Datoms%20)
So now we know that in 1.2 mL of liquid mercury, there are 
present.
To work it out, you divide 240 by 100 to work out 1% of it, then multiply that by 95 to work out 95% of it. So
(240/100) * 95 = 228mL
Explanation:
There are two direct factors that affect solubility: temperature and pressure. Temperature affects the solubility of both solids and gases, but pressure only affects the solubility of gases.
The total pressure when the new equilibrium is stabilized is half of the initial pressure of the system.
The given chemical reaction at a stable equilibrium is,
2H₂O(g)+O₂(g) = 2H₂O₂(g)
According to the ideal gas equation,
PV = nRT
P is pressure,
V is volume,
n is moles
R is gas constant,
T is temperature.
Assuming the temperature is constant.
If the volume of the system is twice the initial volume then the total pressure at the new equilibrium can be found out as,
P₁V₁ = P₂V₂
Where, P₁ and V₁ are initial volume and pressure while P₂ and V₂ are final pressure and volume.
If V₂ = 2V₁,
P₂ = P₁/2
So, the final total pressure will be half of the initial pressure.
To know more about equilibrium, visit,
brainly.com/question/517289
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