worksheet
count the atoms present in the different compounds by using the coefficients and subscripts.
$k_{2}co_{3}$
$ba_{3}(po_{4})_{2}$
$na_{2}cro_{4}$
$3cacl_{2}$
$nh_{4}c_{2}h_{3}o_{2}$
$4al_{2}(co_{3})_{3}$
$pb(no_{3})_{2}$
$2(nh_{4})_{2}cr_{2}o_{7}$

Question

Accepted Answer

# Explanation:
## Step1: Analyze $Ba_3(PO_4)_2$
- Barium ($Ba$): The coefficient is $3$, so the number of $Ba$ atoms is $3$.
- Phosphorus ($P$): Inside the phosphate group $(PO_4)$, the sub - script of $P$ is $1$, and there are $2$ phosphate groups. So the number of $P$ atoms is $1\times2 = 2$.
- Oxygen ($O$): Inside the phosphate group $(PO_4)$, the sub - script of $O$ is $4$, and there are $2$ phosphate groups. So the number of $O$ atoms is $4\times2=8$.
- Total atoms: $3 + 2+8=13$.

## Step2: Analyze $Na_2CrO_4$
- Sodium ($Na$): The coefficient is $2$, so the number of $Na$ atoms is $2$.
- Chromium ($Cr$): The sub - script of $Cr$ is $1$, so the number of $Cr$ atoms is $1$.
- Oxygen ($O$): The sub - script of $O$ is $4$, so the number of $O$ atoms is $4$.
- Total atoms: $2 + 1+4=7$.

## Step3: Analyze $3CaCl_2$
- Calcium ($Ca$): The coefficient is $3$, so the number of $Ca$ atoms is $3$.
- Chlorine ($Cl$): Inside $CaCl_2$, the sub - script of $Cl$ is $2$, and there are $3$ $CaCl_2$ units. So the number of $Cl$ atoms is $2\times3 = 6$.
- Total atoms: $3+6 = 9$.

## Step4: Analyze $NH_4C_2H_3O_2$
- Nitrogen ($N$): The sub - script of $N$ is $1$, so the number of $N$ atoms is $1$.
- Hydrogen ($H$): There are $4$ $H$ atoms in $NH_4$ and $3$ $H$ atoms in $C_2H_3O_2$, so the total number of $H$ atoms is $4 + 3=7$.
- Carbon ($C$): The sub - script of $C$ is $2$, so the number of $C$ atoms is $2$.
- Oxygen ($O$): The sub - script of $O$ is $2$, so the number of $O$ atoms is $2$.
- Total atoms: $1+7 + 2+2=12$.

## Step5: Analyze $Pb(NO_3)_2$
- Lead ($Pb$): The sub - script of $Pb$ is $1$, so the number of $Pb$ atoms is $1$.
- Nitrogen ($N$): Inside $NO_3$, the sub - script of $N$ is $1$, and there are $2$ $NO_3$ groups. So the number of $N$ atoms is $1\times2=2$.
- Oxygen ($O$): Inside $NO_3$, the sub - script of $O$ is $3$, and there are $2$ $NO_3$ groups. So the number of $O$ atoms is $3\times2 = 6$.
- Total atoms: $1+2+6=9$.

## Step6: Analyze $2(NH_4)_2Cr_2O_7$
- Nitrogen ($N$): Inside $(NH_4)$, the sub - script of $N$ is $1$, and there are $2$ $(NH_4)$ groups and a coefficient of $2$. So the number of $N$ atoms is $1\times2\times2=4$.
- Hydrogen ($H$): Inside $(NH_4)$, the sub - script of $H$ is $4$, and there are $2$ $(NH_4)$ groups and a coefficient of $2$. So the number of $H$ atoms is $4\times2\times2 = 16$.
- Chromium ($Cr$): Inside $(NH_4)_2Cr_2O_7$, the sub - script of $Cr$ is $2$, and there is a coefficient of $2$. So the number of $Cr$ atoms is $2\times2=4$.
- Oxygen ($O$): Inside $(NH_4)_2Cr_2O_7$, the sub - script of $O$ is $7$, and there is a coefficient of $2$. So the number of $O$ atoms is $7\times2=14$.
- Total atoms: $4+16 + 4+14=38$.

# Answer:
For $Ba_3(PO_4)_2$:
- Type of Atom: Barium ($Ba$), Phosphorus ($P$), Oxygen ($O$)
- $\#$ of Atoms: $3$, $2$, $8$, Total $13$

For $Na_2CrO_4$:
- Type of Atom: Sodium ($Na$), Chromium ($Cr$), Oxygen ($O$)
- $\#$ of Atoms: $2$, $1$, $4$, Total $7$

For $3CaCl_2$:
- Type of Atom: Calcium ($Ca$), Chlorine ($Cl$)
- $\#$ of Atoms: $3$, $6$, Total $9$

For $NH_4C_2H_3O_2$:
- Type of Atom: Nitrogen ($N$), Hydrogen ($H$), Carbon ($C$), Oxygen ($O$)
- $\#$ of Atoms: $1$, $7$, $2$, $2$, Total $12$

For $Pb(NO_3)_2$:
- Type of Atom: Lead ($Pb$), Nitrogen ($N$), Oxygen ($O$)
- $\#$ of Atoms: $1$, $2$, $6$, Total $9$

For $2(NH_4)_2Cr_2O_7$:
- Type of Atom: Nitrogen ($N$), Hydrogen ($H$), Chromium ($Cr$), Oxygen ($O$)
- $\#$ of Atoms: $4$, $16$, $4$, $14$, Total $38$

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QUESTION IMAGE

Question

Explanation:

Step1: Analyze \(Ba_3(PO_4)_2\)

Step2: Analyze \(Na_2CrO_4\)

Step3: Analyze \(3CaCl_2\)

Step4: Analyze \(NH_4C_2H_3O_2\)

Step5: Analyze \(Pb(NO_3)_2\)

Step6: Analyze \(2(NH_4)_2Cr_2O_7\)

Answer: