Question

physical science
eams
counting atoms
count how many atoms for each compound or molecule a
$ce{nahco_{3}}$
$na=1$
$h=1$
$c=1$
$o=3$
$ce{mg(oh)_{2}}$
$mg$
$ce{c_{2}h_{4}o}$
$c=2$
$h=4$
$o=2$
$ce{3 h_{3}po}$

Explanation:

Step1: Count atoms in $\text{NaHCO}_3$

$\text{Na}: 1, \text{H}: 1, \text{C}: 1, \text{O}: 3$
Total atoms: $1+1+1+3=6$

Step2: Count atoms in $\text{C}_2\text{H}_4\text{O}_2$

$\text{C}: 2, \text{H}: 4, \text{O}: 2$
Total atoms: $2+4+2=8$

Step3: Count atoms in $\text{Mg(OH)}_2$

$\text{Mg}: 1, \text{O}: 2, \text{H}: 2$
Total atoms: $1+2+2=5$

Step4: Count atoms in $3\text{H}_3\text{PO}_4$ (partial)

Per molecule: $\text{H}: 3, \text{P}: 1, \text{O}: 4$; total per molecule: $3+1+4=8$
Total for 3 molecules: $3 \times 8=24$

Answer:

• $\text{NaHCO}_3$: 6 total atoms (Na=1, H=1, C=1, O=3)
• $\text{C}_2\text{H}_4\text{O}_2$: 8 total atoms (C=2, H=4, O=2)
• $\text{Mg(OH)}_2$: 5 total atoms (Mg=1, O=2, H=2)
• $3\text{H}_3\text{PO}_4$: 24 total atoms (H=9, P=3, O=12)