Answer:
=C₄H₄O₂
Explanation:
Given the empirical formula of a molecule, the he the quotient of the molecular mas and and the empirical mass=constant.
84.0 g/mol/mass of(C₂H₂O)=constant
=84/(12×2+1×2×16)
=84/42
=2
Therefore, the molecular formula is (C₂H₂O)₂=C₄H₄O₂
Answer:
(a) 
(b) 
Explanation:
Hello,
In this case, since the both gases behave ideally, with the given information we can compute the moles of He in A:

Thus, since the final pressure is 3.60 bar, we can write:

The moles of helium could be computed via solver as:

Or algebraically:

In such a way, the volume of the compartment B is:

Finally, he mole fraction of He is:

Regards.