section c: identify the number of atoms of each element in the following compounds. again drawing the arrows is strongly encouraged, but not required.
19) 4k₂co₃ k = ____ c = ____ o = ____
20) k₂co₃ k = ____ c = ____ o = ____
21) 3n₄o₁₀ n = ____ o = ____
22) 8cl₂o cl = ____ o = ____
23) ca(c₂h₃o₂)₂ ca = ____ c = ____ h = ____ o = ____
24) 12nabr na = ____ br = ____
25) 3nahco₃ na = ____ h = ____ c = ____ o = ____
26) 5ga₂(cr₂o₇)₃ ga = ____ cr = ____ o = ____
27) 4fe₂o₃ fe = ____ o = ____
28) 6ba(mno₄)₂ ba = ____ mn = ____ o = ____

Question

Accepted Answer

# Explanation:
## Step1: Analyze $4K_2CO_3$
For $K$: The sub - script of $K$ is $2$ and there is a coefficient $4$. So the number of $K$ atoms is $4\times2=8$.
For $C$: The sub - script of $C$ is $1$ and the coefficient is $4$. So the number of $C$ atoms is $4\times1 = 4$.
For $O$: The sub - script of $O$ is $3$ and the coefficient is $4$. So the number of $O$ atoms is $4\times3=12$.

## Step2: Analyze $K_2CO_3$
For $K$: The sub - script of $K$ is $2$. So the number of $K$ atoms is $2$.
For $C$: The sub - script of $C$ is $1$. So the number of $C$ atoms is $1$.
For $O$: The sub - script of $O$ is $3$. So the number of $O$ atoms is $3$.

## Step3: Analyze $3N_4O_{10}$
For $N$: The sub - script of $N$ is $4$ and the coefficient is $3$. So the number of $N$ atoms is $3\times4 = 12$.
For $O$: The sub - script of $O$ is $10$ and the coefficient is $3$. So the number of $O$ atoms is $3\times10=30$.

## Step4: Analyze $8Cl_2O$
For $Cl$: The sub - script of $Cl$ is $2$ and the coefficient is $8$. So the number of $Cl$ atoms is $8\times2=16$.
For $O$: The sub - script of $O$ is $1$ and the coefficient is $8$. So the number of $O$ atoms is $8\times1 = 8$.

## Step5: Analyze $Ca(C_2H_3O_2)_2$
For $Ca$: The coefficient of $Ca$ is $1$. So the number of $Ca$ atoms is $1$.
For $C$: The sub - script of $C$ inside the parenthesis is $2$ and there is a coefficient $2$ outside the parenthesis. So the number of $C$ atoms is $2\times2=4$.
For $H$: The sub - script of $H$ inside the parenthesis is $3$ and there is a coefficient $2$ outside the parenthesis. So the number of $H$ atoms is $3\times2 = 6$.
For $O$: The sub - script of $O$ inside the parenthesis is $2$ and there is a coefficient $2$ outside the parenthesis. So the number of $O$ atoms is $2\times2=4$.

## Step6: Analyze $12NaBr$
For $Na$: The sub - script of $Na$ is $1$ and the coefficient is $12$. So the number of $Na$ atoms is $12\times1=12$.
For $Br$: The sub - script of $Br$ is $1$ and the coefficient is $12$. So the number of $Br$ atoms is $12\times1 = 12$.

## Step7: Analyze $3NaHCO_3$
For $Na$: The sub - script of $Na$ is $1$ and the coefficient is $3$. So the number of $Na$ atoms is $3\times1=3$.
For $H$: The sub - script of $H$ is $1$ and the coefficient is $3$. So the number of $H$ atoms is $3\times1 = 3$.
For $C$: The sub - script of $C$ is $1$ and the coefficient is $3$. So the number of $C$ atoms is $3\times1=3$.
For $O$: The sub - script of $O$ is $3$ and the coefficient is $3$. So the number of $O$ atoms is $3\times3=9$.

## Step8: Analyze $5Ga_2(Cr_2O_7)_3$
For $Ga$: The sub - script of $Ga$ is $2$ and the coefficient is $5$. So the number of $Ga$ atoms is $5\times2=10$.
For $Cr$: The sub - script of $Cr$ inside the parenthesis is $2$ and there is a coefficient $3$ inside the parenthesis and $5$ outside. So the number of $Cr$ atoms is $5\times2\times3=30$.
For $O$: The sub - script of $O$ inside the parenthesis is $7$ and there is a coefficient $3$ inside the parenthesis and $5$ outside. So the number of $O$ atoms is $5\times7\times3 = 105$.

## Step9: Analyze $4Fe_2O_3$
For $Fe$: The sub - script of $Fe$ is $2$ and the coefficient is $4$. So the number of $Fe$ atoms is $4\times2=8$.
For $O$: The sub - script of $O$ is $3$ and the coefficient is $4$. So the number of $O$ atoms is $4\times3=12$.

## Step10: Analyze $6Ba(MnO_4)_2$
For $Ba$: The coefficient of $Ba$ is $6$. So the number of $Ba$ atoms is $6$.
For $Mn$: The sub - script of $Mn$ inside the parenthesis is $1$ and there is a coefficient $2$ inside the parenthesis and $6$ outside. So the number of $Mn$ atoms is $6\times1\times2=12$.
For $O$: The sub - script of $O$ inside the parenthesis is $4$ and there is a coefficient $2$ inside the parenthesis and $6$ outside. So the number of $O$ atoms is $6\times4\times2=48$.

# Answer:
19) $K = 8$, $C=4$, $O = 12$
20) $K = 2$, $C = 1$, $O=3$
21) $N = 12$, $O=30$
22) $Cl = 16$, $O = 8$
23) $Ca=1$, $C = 4$, $H=6$, $O = 4$
24) $Na=12$, $Br = 12$
25) $Na = 3$, $H=3$, $C=3$, $O = 9$
26) $Ga = 10$, $Cr=30$, $O = 105$
27) $Fe = 8$, $O=12$
28) $Ba = 6$, $Mn=12$, $O = 48$

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QUESTION IMAGE

Question

Explanation:

Step1: Analyze \(4K_2CO_3\)

Step2: Analyze \(K_2CO_3\)

Step3: Analyze \(3N_4O_{10}\)

Step4: Analyze \(8Cl_2O\)

Step5: Analyze \(Ca(C_2H_3O_2)_2\)

Step6: Analyze \(12NaBr\)

Step7: Analyze \(3NaHCO_3\)

Step8: Analyze \(5Ga_2(Cr_2O_7)_3\)

Step9: Analyze \(4Fe_2O_3\)

Answer: