Question

1. the number of moles present in 90.0g of lead(iv) oxide is 2.66 moles 0.403 moles 0.376 moles 2.15×10⁴ moles

Explanation:

Step1: Find molar mass of lead(IV) oxide ($PbO_2$)

The molar mass of $Pb$ is approximately $207.2\ g/mol$ and of $O$ is approximately $16.00\ g/mol$. For $PbO_2$, molar mass $M = 207.2+2\times16.00=207.2 + 32.00=239.2\ g/mol$.

Step2: Calculate number of moles

The formula for number of moles $n=\frac{m}{M}$, where $m = 90.0\ g$ and $M = 239.2\ g/mol$. So $n=\frac{90.0}{239.2}\approx0.376\ mol$.

Answer:

0.376 moles