25) how many total ions would be present when an ionic compound is formed between a
group 13 element and nitrate ions?
a) 2
b) 3
c) 4
d) 5
e) 6

The number of total ions present when an ionic compound is formed between an element of group 13 and nitrate ions would be 4 (option C).

The elements of the boron family (group 13) are boron (B), aluminum (Al), gallium (Ga), indium (In), thallium (Tl). They have three valence electrons given by the number of electrons in the outermost orbital in their electronic configurations (ns² np¹).  

When the nitrate ion (NO₃⁻) forms a compound with a group 13 element, we have the following reaction:

X⁺³ + 3NO₃⁻  →  X(NO₃)₃

Where X is any group 13 element

We can see that the total number of ions in the final compound is four, three of the NO₃⁻ ions plus one ion of the group 13 element.  

Therefore, the total number of ions would be 4 (option C).

Learn more here:

I hope it helps you!

4 ions

1  M³⁺ ion and 3 NO₃⁻ ions ( Where M³⁺ stands for the group 13 cation).

Explanation:

Group 13 elements are Boron , Aluminium, Galium, Indium and Thalium. When they form ionic compounds they can lose their  three valence electrons to form a tripositive (+3) ion. Nitrate ion is a negative ion with a single negative(-1) charge. When these two form a compound 3 univalent negative nitrate ions must combine with one tripositive group 13 element to form a neutral compound. So the total no of ions is 4.

nacl is added to the aqueous of what? ? in order to get a proper answer you have to specify the exact composition of the aqueous solution.

the answer is definitely c.