Question

1. an imaginary element named studentium forms the following imaginary studentium anion, st²⁻. assuming the ion is real, answer the following questions. is the ion a monatomic or polyatomic ion? construct the formula for the acid that will form with the studentium anion. is the acid an oxyacid or a binary acid? correctly apply acid nomenclature to name the acid that forms with the studentium anion. 15. an imaginary oxyanion called basketite, has an imaginary formula & charge of bso₃¹⁻. assuming the ion is real, answer the following questions. is the ion a monatomic or polyatomic ion? construct the formula for the acid that will form with the basketite ion. is the acid an oxyacid or a binary acid? correctly apply acid nomenclature to name the acid that forms with the basketite ion.

Explanation:

Step1: Analyze $\text{St}^{2-}$ (Q14)

Monatomic ions have 1 atom. $\text{St}^{2-}$ has only Studentium.

Step2: Form acid for $\text{St}^{2-}$

Balance charge: $2\text{H}^+$ + $\text{St}^{2-}$ → $\text{H}_2\text{St}$

Step3: Classify $\text{H}_2\text{St}$

Binary acids have H + 1 non-metal. No oxygen present.

Step4: Name $\text{H}_2\text{St}$

Binary acid rule: hydro- + stem + -ic acid. Stem is "student-".

Step5: Analyze $\text{BsO}_3^{1-}$ (Q15)

Polyatomic ions have ≥2 atoms. $\text{BsO}_3^{1-}$ has Bs + 3 O.

Step6: Form acid for $\text{BsO}_3^{1-}$

Balance charge: $\text{H}^+$ + $\text{BsO}_3^{1-}$ → $\text{H}\text{BsO}_3$

Step7: Classify $\text{H}\text{BsO}_3$

Oxyacids have H + oxyanion (contains O).

Step8: Name $\text{H}\text{BsO}_3$

Oxyacid rule: anion ends with -ite → -ous acid. Anion is basketite.

Answer:

Question 14:

1. Monatomic ion
2. $\text{H}_2\text{St}$
3. Binary acid
4. Hydrostudentic acid

Question 15:

1. Polyatomic ion
2. $\text{H}\text{BsO}_3$
3. Oxyacid
4. Basketous acid