29) 5al₂(sio₃)₂ al =_____ si =_____ o =_____
30) 4au(io₃)₃ au =_____ i =_____ o =_____
31) sbr₂ s =_____ br =_____
32) 4ca(oh)₂ ca =_____ o =_____ h =_____
33) 5mg₃(po₄)₂ mg =_____ p =_____ o =_____
34) 3clf cl =_____ f =_____

Question

Accepted Answer

# Explanation:
## Step1: Analyze the formula $5Al_{2}(SiO_{3})_{2}$
For $Al$: The sub - script of $Al$ is $2$ and there is a coefficient $5$. So the number of $Al$ atoms is $5\times2 = 10$.
For $Si$: The sub - script of $Si$ is $1$ in $(SiO_{3})$ and there is a coefficient $5$ and a sub - script $2$ for the $(SiO_{3})$ group. So the number of $Si$ atoms is $5\times2\times1=10$.
For $O$: The sub - script of $O$ is $3$ in $(SiO_{3})$ and there is a coefficient $5$ and a sub - script $2$ for the $(SiO_{3})$ group. So the number of $O$ atoms is $5\times2\times3 = 30$.
## Step2: Analyze the formula $4Au(IO_{3})_{3}$
For $Au$: The sub - script of $Au$ is $1$ and there is a coefficient $4$. So the number of $Au$ atoms is $4\times1 = 4$.
For $I$: The sub - script of $I$ is $1$ in $(IO_{3})$ and there is a coefficient $4$ and a sub - script $3$ for the $(IO_{3})$ group. So the number of $I$ atoms is $4\times3\times1 = 12$.
For $O$: The sub - script of $O$ is $3$ in $(IO_{3})$ and there is a coefficient $4$ and a sub - script $3$ for the $(IO_{3})$ group. So the number of $O$ atoms is $4\times3\times3=36$.
## Step3: Analyze the formula $SBr_{2}$
For $S$: The sub - script of $S$ is $1$. So the number of $S$ atoms is $1$.
For $Br$: The sub - script of $Br$ is $2$. So the number of $Br$ atoms is $2$.
## Step4: Analyze the formula $4Ca(OH)_{2}$
For $Ca$: The sub - script of $Ca$ is $1$ and there is a coefficient $4$. So the number of $Ca$ atoms is $4\times1 = 4$.
For $O$: The sub - script of $O$ is $1$ in $(OH)$ and there is a coefficient $4$ and a sub - script $2$ for the $(OH)$ group. So the number of $O$ atoms is $4\times2\times1=8$.
For $H$: The sub - script of $H$ is $1$ in $(OH)$ and there is a coefficient $4$ and a sub - script $2$ for the $(OH)$ group. So the number of $H$ atoms is $4\times2\times1 = 8$.
## Step5: Analyze the formula $5Mg_{3}(PO_{4})_{2}$
For $Mg$: The sub - script of $Mg$ is $3$ and there is a coefficient $5$. So the number of $Mg$ atoms is $5\times3=15$.
For $P$: The sub - script of $P$ is $1$ in $(PO_{4})$ and there is a coefficient $5$ and a sub - script $2$ for the $(PO_{4})$ group. So the number of $P$ atoms is $5\times2\times1 = 10$.
For $O$: The sub - script of $O$ is $4$ in $(PO_{4})$ and there is a coefficient $5$ and a sub - script $2$ for the $(PO_{4})$ group. So the number of $O$ atoms is $5\times2\times4=40$.
## Step6: Analyze the formula $3ClF$
For $Cl$: The sub - script of $Cl$ is $1$ and there is a coefficient $3$. So the number of $Cl$ atoms is $3\times1 = 3$.
For $F$: The sub - script of $F$ is $1$ and there is a coefficient $3$. So the number of $F$ atoms is $3\times1=3$.

# Answer:
29) $Al = 10$, $Si = 10$, $O = 30$
30) $Au = 4$, $I = 12$, $O = 36$
31) $S = 1$, $Br = 2$
32) $Ca = 4$, $O = 8$, $H = 8$
33) $Mg = 15$, $P = 10$, $O = 40$
34) $Cl = 3$, $F = 3$

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QUESTION IMAGE

Question

Explanation:

Step1: Analyze the formula \(5Al_{2}(SiO_{3})_{2}\)

Step2: Analyze the formula \(4Au(IO_{3})_{3}\)

Step3: Analyze the formula \(SBr_{2}\)

Step4: Analyze the formula \(4Ca(OH)_{2}\)

Step5: Analyze the formula \(5Mg_{3}(PO_{4})_{2}\)

Step6: Analyze the formula \(3ClF\)

Answer: