Question

1. kmno₄

k: ______
mn: ______
o: ______

1. hcl

h: ______
cl: ______

1. mg(no₃)₂

mg: ______
n: ______
o: ______

1. (nh₄)₃po₄

n: ______
h: ______
o: ______
p: ______

1. al₂(so₄)₃

al: ______
s: ______
o: ______

Explanation:

Step1: Analyze \( \text{KMnO}_4 \)

In \( \text{KMnO}_4 \), the subscripts (implicitly 1 for K and Mn, 4 for O) give the number of atoms. So K: 1, Mn: 1, O: 4.

Step2: Analyze \( \text{HCl} \)

For \( \text{HCl} \), both H and Cl have subscript 1 (implicit). So H: 1, Cl: 1.

Step3: Analyze \( \text{Mg(NO}_3\text{)}_2 \)

• Mg: subscript 1.
• \( \text{NO}_3 \) is a polyatomic ion with subscript 2 outside, so N: \( 1\times2 = 2 \), O: \( 3\times2 = 6 \).

Step4: Analyze \( (\text{NH}_4\text{)}_3\text{PO}_4 \)

• \( \text{NH}_4 \) has subscript 3: N: \( 1\times3 = 3 \), H: \( 4\times3 = 12 \).
• \( \text{PO}_4 \) has subscript 1: O: 4, P: 1.

Step5: Analyze \( \text{Al}_2(\text{SO}_4\text{)}_3 \)

• Al: subscript 2.
• \( \text{SO}_4 \) has subscript 3: S: \( 1\times3 = 3 \), O: \( 4\times3 = 12 \).

Answer:

1. K: 1; Mn: 1; O: 4
2. H: 1; Cl: 1
3. Mg: 1; N: 2; O: 6
4. N: 3; H: 12; O: 4; P: 1
5. Al: 2; S: 3; O: 12