Make sure that you understand what they are asking you from this question, as it can be confusing, but the solution is quite simple. They are stating that they want you to calculate the final concentration of 6.0M HCl once a dilution has been made from 2.0 mL to 500.0 mL. They have given us three values, the initial concentration, initial volume and the final volume. So, we are able to employ the following equation:
C1V1 = C2V2
(6.0M)(2.0mL) = C2(500.0mL)
Therefore, the final concentration, C2 = 0.024M.
Methane Volume : O.O
Methane Mass: 0.100g
Molar Mass of Methane : 16.04 g/mol
This can be, for example, halogensubstituted hydrocarbons.
CCl₄, C₂F₆.
Or halides halocarboxylic acids, and other compounds.
O
II
Cl₃C-Cl
Answer:
The answer is 44.0095 molecules
