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.
Answer:
The new equilibrium total pressure will be increased to one-half to initial total pressure.
Explanation:
From the information given :
The equation of the reaction can be represented as;

From above equation:
2 moles of sulphur dioxide reacts with 1 mole of oxygen (i.e 2 moles +1 mole =3 moles ) to give 2 moles of sulphur trioxide
So; suppose the volume of this system is compressed to one-half its initial volume and then equilibrium is reestablished.
So if this process takes place ; the equilibrium will definitely shift to the side with fewer moles , thus the equilibrium will shift to the right. As such; there is increase in pressure.
Let the total pressure at the initial equilibrium be 
and the total pressure at the final equilibrium be 
According to Boyle's Law; Boyle's Law states that the pressure of a fixed mass of gas is inversely proportional to the volume, provided the temperature remains constant.
Thus;
P ∝ 1/V
P = K/V
PV = K
where K = constant
So;
PV = constant
Hence;

From the foregoing; since the volume is decreased to one- half to initial Volume; then ,

also;
Thus ;



Dividing both sides by 


From ;




Thus; The new equilibrium total pressure will be increased to one-half to initial total pressure.
Answer :]
A.)Calculate the mass of ammonium sulfate that would be obtained by reacting with ammonia acid.
<em>Correct me if i'm wrong :]</em>
Answer:
1.2*10^24 molecules of CF4
Explanation:
the molar mass of cf4 is 88.0043 g/mol
176/88.0043 = 2 moles of CF4
Then multiply by avogadro's number (6.022*10^23) to get the number of molecules
2*6.022*10^23 = 1.2*10^24 molecules of CF4
Answer:
The amount of energy released from the combustion of 2 moles of methae is 1,605.08 kJ/mol
Explanation:
The chemical reaction of the combustion of methane is given as follows;
CH₄ (g) + 2O₂ (g) → CO₂ (g) + 2H₂O (g)
Hence, 1 mole of methane combines with 2 moles of oxygen gas to form 1 mole of carbon dioxide and 2 moles of water vapor
Where:
CH₄ (g): Hf = -74.6 kJ/mol
CO₂ (g): Hf = -393.5 kJ/mol
H₂O (g): Hf = -241.82 kJ/mol
Therefore, the combustion of 1 mole of methane releases;
-393.5 kJ/mol × 1 + 241.82 kJ/mol × 2 + 74.6 kJ/mol = -802.54 kJ/mol
Hence the combustion of 2 moles of methae will rellease;
2 × -802.54 kJ/mol or 1,605.08 kJ/mol.