write formulas for the compounds formed from rb and each of the following polyatomic ions: clo4-, so42-, po43-. express your answers as chemical formulas separated by commas.

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Accepted Answer

# Explanation:
## Step1: Determine the charge of Rb
Rubidium (Rb) is in Group 1 of the periodic table, so it has a charge of $Rb^+$.
## Step2: Combine Rb⁺ with $ClO_4^-$
The charges are $+1$ (from $Rb^+$) and $-1$ (from $ClO_4^-$). Using the criss - cross method (since the magnitude of charges is equal), the formula is $RbClO_4$.
## Step3: Combine Rb⁺ with $SO_4^{2 -}$
The charge of $Rb^+$ is $+1$ and that of $SO_4^{2 -}$ is $-2$. Using the criss - cross method (swap the magnitudes of charges as subscripts), we get $Rb_2SO_4$ (because $|+1|$ and $|-2|$, so $Rb$ has subscript $2$ and $SO_4$ has subscript $1$).
## Step4: Combine Rb⁺ with $PO_4^{3 -}$
The charge of $Rb^+$ is $+1$ and that of $PO_4^{3 -}$ is $-3$. Using the criss - cross method (swap the magnitudes of charges as subscripts), we get $Rb_3PO_4$ (since $|+1|$ and $|-3|$, $Rb$ has subscript $3$ and $PO_4$ has subscript $1$).

# Answer:
$RbClO_4,Rb_2SO_4,Rb_3PO_4$

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Question

Explanation:

Step1: Determine the charge of Rb

Step2: Combine Rb⁺ with \(ClO_4^-\)

Step3: Combine Rb⁺ with \(SO_4^{2 -}\)

Step4: Combine Rb⁺ with \(PO_4^{3 -}\)

Answer: