Answer
The final volume is equal to 34.7 L
Work Step by Step
1. Using Avogadro's law, solve for "$V_2$":
$\frac{V_1}{n_1} = \frac{V_2}{n_2}$
- Multiply both sides by "$n_2$"
$\frac{V_1}{n_1} \times n_2 = {V_2}$
2. Calculate the number of moles $(O_2)$ in 4.80 g of that.
Molar mass : $O: 16.00g * 2= 32.00g $
$ 4.80 \space g \times \frac{1 \space mole}{ 32.00 \space g} = 0.150 \space moles$
3. Since 0.500 mole of Ne atoms were added, the final number of moles is equal to "$n_1 + 0.500 \space mole$".
$n_2 = n_1 + 0.500 \space mole$
$n_2 = 0.150 \space mole + 0.500 \space mole = 0.650 \space mole$
4. Calculate the value of $V_2$:
$\frac{8.00 \space L}{0.150 \space mole} \times 0.650 \space mole = 34.7 \space L$
