Question

match each formula with its compound name. be sure to use your periodic table and polyatomic ion chart.
babr₂ is the formula for select .
fe(mno₄)₃ is the formula for select .
be(oh)₂ is the formula for select .

Explanation:

Step1: Analyze \( \text{BaBr}_2 \)

Barium (Ba) is in group 2, so it has a \( +2 \) charge. Bromide (Br⁻) has a \( -1 \) charge. To balance, 1 Ba²⁺ and 2 Br⁻ give \( \text{BaBr}_2 \), named Barium Bromide.

Step2: Analyze \( \text{Fe(MnO}_4\text{)}_3 \)

Iron (Fe) here: \( \text{MnO}_4^- \) (permanganate) has \( -1 \) charge. 3 permanganate ions give \( -3 \), so Fe has \( +3 \) (Iron(III)). Thus, Iron(III) Permanganate.

Step3: Analyze \( \text{Be(OH)}_2 \)

Beryllium (Be) is group 2, \( +2 \) charge. Hydroxide (OH⁻) has \( -1 \) charge. 1 Be²⁺ and 2 OH⁻ give \( \text{Be(OH)}_2 \), named Beryllium Hydroxide.

Answer:

\( \text{BaBr}_2 \): Barium Bromide
\( \text{Fe(MnO}_4\text{)}_3 \): Iron(III) Permanganate
\( \text{Be(OH)}_2 \): Beryllium Hydroxide