Give the number of each type of ion present in each of the ionic compounds formed from the combination of the indicated metal and acetate, {\rm C_2H_3O_2}^-, ion.
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To determine the number of each type of ion present in ionic compounds formed from the combination of a metal and the acetate ({\rm C_2H_3O_2}^{-}) ion, we need to consider the charges of the metal ion and the acetate ion.
The metal ions commonly used in this context are usually found on the left side of the periodic table, such as alkali metals (Group 1) and alkaline earth metals (Group 2). The acetate ion ({\rm C_2H_3O_2}^{-}) always has a charge of -1.
Here are some examples:
1. Sodium acetate: NaCH3COO
- Sodium (Na) is an alkali metal with a charge of +1.
- Acetate ({\rm C_2H_3O_2}^{-}) has a charge of -1.
Therefore, in sodium acetate, there is one sodium ion (Na+) and one acetate ion ({\rm C_2H_3O_2}^{-}).
2. Calcium acetate: Ca(CH3COO)2
- Calcium (Ca) is an alkaline earth metal with a charge of +2.
- Acetate ({\rm C_2H_3O_2}^{-}) has a charge of -1.
Therefore, in calcium acetate, there are two calcium ions (Ca^2+) and two acetate ions ({\rm C_2H_3O_2}^{-}).
3. Aluminum acetate: Al(CH3COO)3
- Aluminum (Al) is a metal with a charge of +3.
- Acetate ({\rm C_2H_3O_2}^{-}) has a charge of -1.
Therefore, in aluminum acetate, there are three aluminum ions (Al^3+) and three acetate ions ({\rm C_2H_3O_2}^{-}).
In summary, the number of each type of ion present in these ionic compounds depends on the charges of the metal ion and the acetate ion. By considering the charge balance, we can determine the appropriate number of each ion in each compound.