Is Zinc Phosphate ionic or covalent

Compare the valence electron configuration of the nobles gas elements seen here. what statement is correct?

Predict the products of the following acid-base reactions, and predict whether the equilibrium lies to the left or to the right of the reaction arrow. part ao2-(aq)+h2o(l)< => express your answer as part of a chemical equation. identify all of the phases in your answer. o2-(aq)+h2o(l) < => oh-(aq)+oh-(aq)part bpredict whether the equilibrium lies to the left or to the right of the equation in previous part. h2o is a stronger acid than oh–, so the equilibrium lies to the right. h2o is a weaker acid than oh–, so the equilibrium lies to the left. h2o is a stronger acid than oh–, so the equilibrium lies to the left. h2o is a weaker acid than oh–, so the equilibrium lies to the right. part cch3cooh(aq)+hs? (aq) < => express your answer as part of a chemical equation. identify all of the phases in your answer. ch3cooh(aq)+hs-(aq) < => h2s(aq)+c2h3o2-(aq)h2s(aq)+c2h3o2-( aq)part dpredict whether the equilibrium lies to the left or to the right of the equation in previous part. ch3cooh is a weaker acid than h2s, so the equilibrium lies to the right. ch3cooh is a weaker acid than h2s, so the equilibrium lies to the left. ch3cooh is a stronger acid than h2s, so the equilibrium lies to the right. ch3cooh is a stronger acid than h2s, so the equilibrium lies to the left. part eno2-(aq)+h2o(l) < => express your answer as part of a chemical equation. identify all of the phases in your answer. no2-(aq)+h2o(l) < => part fpredict whether the equilibrium lies to the left or to the right of the equation in previous part. hno2 is a stronger acid than h2o, so the equilibrium lies to the right. hno2 is a weaker acid than h2o, so the equilibrium lies to the left. hno2 is a stronger acid than h2o, so the equilibrium lies to the left. hno2 is a weaker acid than h2o, so the equilibrium lies to the right.

From the choices provided below, list the reagent(s) in order that will react with cyclopentanone to form the compound shown below.