Question

1. 3pb₃(po₄)₄ pb= p= o=

Explanation:

Step1: Analyze the sub - scripts for each element in the formula

The formula is $3Pb_3(PO_4)_4$. First, consider the coefficient in front of the entire formula ($3$) and the sub - scripts of each element within the formula.

Step2: Calculate the number of Pb atoms

For lead ($Pb$), the sub - script of $Pb$ in $Pb_3(PO_4)_4$ is $3$. When we multiply by the coefficient $3$ in front of the formula, the number of $Pb$ atoms is $3\times3 = 9$.

Step3: Calculate the number of P atoms

For phosphorus ($P$), inside the $(PO_4)$ group, the sub - script of $P$ is $1$. There are $4$ $(PO_4)$ groups in $Pb_3(PO_4)_4$, and then multiply by the coefficient $3$ in front of the formula. So the number of $P$ atoms is $3\times4\times1=12$.

Step4: Calculate the number of O atoms

For oxygen ($O$), inside the $(PO_4)$ group, the sub - script of $O$ is $4$. There are $4$ $(PO_4)$ groups in $Pb_3(PO_4)_4$, and then multiply by the coefficient $3$ in front of the formula. So the number of $O$ atoms is $3\times4\times4 = 48$.

Answer:

$Pb = 9$, $P = 12$, $O = 48$