Question

th polyatomic lons
j) copper (ii) nitrate
k) potassium dichromate
l) sodium oxalate
m) magnesium sulphite
n) lead (ii) perchlorate
o) tin (iv) phosphate
p) aluminum hydroxide
q) barium hydroxide
r) beryllium carbonate

Explanation:

To solve for the chemical formulas of the remaining compounds (p - r), we use the rules of ionic compound formation (criss - cross method for balancing charges).

p) Aluminum Hydroxide

• Aluminum ion: $\text{Al}^{3+}$ (aluminum has a +3 charge in most compounds)
• Hydroxide ion: $\text{OH}^-$ (hydroxide has a -1 charge)
• Using the criss - cross method: The charge of aluminum becomes the subscript of hydroxide and vice - versa. So we have $\text{Al}(\text{OH})_3$.

q) Barium Hydroxide

• Barium ion: $\text{Ba}^{2+}$ (barium is in group 2, so it has a +2 charge)
• Hydroxide ion: $\text{OH}^-$ (charge of -1)
• Applying criss - cross: The +2 charge of Ba becomes the subscript of $\text{OH}^-$ and the -1 charge of $\text{OH}^-$ becomes the subscript of Ba (but we don't write 1). So the formula is $\text{Ba}(\text{OH})_2$.

r) Beryllium Carbonate

• Beryllium ion: $\text{Be}^{2+}$ (beryllium is in group 2, so +2 charge)
• Carbonate ion: $\text{CO}_3^{2 - }$ (carbonate has a -2 charge)
• Using criss - cross: The +2 charge of Be and -2 charge of $\text{CO}_3^{2 - }$ will cross - multiply. Since the charges are equal in magnitude, the formula is $\text{BeCO}_3$.

Final Answers:

p) $\text{Al}(\text{OH})_3$
q) $\text{Ba}(\text{OH})_2$
r) $\text{BeCO}_3$

Answer:

To solve for the chemical formulas of the remaining compounds (p - r), we use the rules of ionic compound formation (criss - cross method for balancing charges).

p) Aluminum Hydroxide

• Aluminum ion: $\text{Al}^{3+}$ (aluminum has a +3 charge in most compounds)
• Hydroxide ion: $\text{OH}^-$ (hydroxide has a -1 charge)
• Using the criss - cross method: The charge of aluminum becomes the subscript of hydroxide and vice - versa. So we have $\text{Al}(\text{OH})_3$.

q) Barium Hydroxide

• Barium ion: $\text{Ba}^{2+}$ (barium is in group 2, so it has a +2 charge)
• Hydroxide ion: $\text{OH}^-$ (charge of -1)
• Applying criss - cross: The +2 charge of Ba becomes the subscript of $\text{OH}^-$ and the -1 charge of $\text{OH}^-$ becomes the subscript of Ba (but we don't write 1). So the formula is $\text{Ba}(\text{OH})_2$.

r) Beryllium Carbonate

• Beryllium ion: $\text{Be}^{2+}$ (beryllium is in group 2, so +2 charge)
• Carbonate ion: $\text{CO}_3^{2 - }$ (carbonate has a -2 charge)
• Using criss - cross: The +2 charge of Be and -2 charge of $\text{CO}_3^{2 - }$ will cross - multiply. Since the charges are equal in magnitude, the formula is $\text{BeCO}_3$.

Final Answers:

p) $\text{Al}(\text{OH})_3$
q) $\text{Ba}(\text{OH})_2$
r) $\text{BeCO}_3$