How many grams of product might form for the following reaction if 33.8 L of Oxygen gas is used in the following reaction? LiCl
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Hey there!:
The fractional saturation y is defined as :
y = [ L ] / Kd + [ L ]
where :
[ L ] = concentration of binding ligand
Kd = 400 nm
Please, find the complete question in the picture attached.
Answer:
1. Na₂O
2. AlF₃
3. MgO
4. Ca₃P₂
Explanation:
1. Na⁺ and O²⁻
An ionic compound will be formed when two ions of opposite charge attract each other.
The positive ion is called cation and the negative ion is called anion.
Thus, every ionic compound has a cation and an anion electrostatically bonded.
The ions must combine in a proportion that genders a neutral compound: so you must have as many cations as negative charge has one anion and as many anions as positive charge has one cation.
This is, if the cation has charge +x, there will be x anions, and if the anion has charge -y, there will be y cations.
Simbolically:
As you see, the subscripts for each element in the chemical formula are obtained by the exchage of the charges of the ions.
Then, for Na⁺ and O²⁻, the subscript for Na will be 2 and the subscript for O will be 1; and the formula of the ionic compound formed by this pair ot ions is:
- Na₂O
2. Al³⁺ and F⁻
- The subscript of Al will be 1 (because the F ion has charge 1-, and the subscript of F will be 3 (because the Al ion has charge +3).
Thus the ionic compound formed by this pair of ions is:
- AlF₃
3. Mg²⁺ and S²⁻
- The charge 2+ from Mg atom wil become the subscript 2 of S atom, and the charge 2- will become the subscript 2 of Mg atom:
That results in the formula: Mg₂S₂
Except for some special compounds, the chemical formula is simplified, dividing by the least common denominator. In this case, that means that the two 2 subscripts are simplified to 1, and the final chemical formula for the ionic compound formed by this pair of ions is:
- MgO
4. Ca²⁺ and P³⁻
- The charge 2+ from Ca will become subscript 2 for P and the charge 3- whill become subscript 3 for Ca.
Hence, the ionic compound formed by these two ions is:
- Ca₃P₂
