Answer:
- <em>To balance a chemical equation it may be necessary to adjust the </em><u>coefficients.</u>
Explanation:
The <em>coefficients</em> of a <em>chemical equation</em> are the numbers that you put in front of each reactant and product. They are used to balance the equation and comply with the law of mass conservation.
By adjusting the coefficients you obtain the relative amounts (moles) of each product and reactant, i.e. the mole ratios.
Here an example.
The first information is what is called a word equation. E.g. nitrogen and hydrogen react to form ammonia:
- Word equation: hydrogen + nitrogen → ammonia
- Skeleton equation: H₂ + N₂ → NH₃
This equation shows the chemical formulae but it is not balanced. The law of mass conservation is not observed.
So, in order to comply with the law of mass conservation you adjust the coefficients as follow.
- Balanced chemical equation: 3H₂ + N₂ → 2NH₃
As you see, it was necessary to modify the coefficients. Now the law of conservation of mass is observed and you get the mole ratios:
- 3 mol H₂ : 1 mol N₂ : 2 mol NH₃
This reaction is decomposition. It is the breakdown of a compound into simpler and smaller elements.
Subjective minerals are minerals that are subjected to so something specific. Objective minerals are minerals that have an object they are made for.
although this is a small difference, in the scientific world with out this you could go totally wrong
Mg(OH)₂ ⇄ Mg²⁺ + 2 OH⁻
Ksp = [Mg²⁺] [OH⁻]²
6.0 x 10⁻¹⁰ = 0.10 x [OH⁻]²
[OH⁻] = 7.746 x 10⁻⁵ M
when Mg(OH)₂ 1st precipitates, [OH⁻] = 7.746 * 10⁻⁵ M
Fe(OH)₂ <—> Fe²⁺ + 2OH⁻
Ksp = [Fe²⁺] [OH⁻]²
7.9 x 10⁻¹⁶ = [Fe²⁺] x (7.746 x 10⁻⁵)²
[Fe²⁺] = 1.32 x 10⁻⁷ M
Answer: 1.32 x 10⁻⁷ M
Answer:
When [F⁻] exceeds 0.0109M concentration, BaF₂ will precipitate
Explanation:
Ksp of BaF₂ is:
BaF₂(s) ⇄ Ba²⁺(aq) + 2F⁻(aq)
Ksp = 1.7x10⁻⁶ = [Ba²⁺] [F⁻]²
The solution will produce BaF₂(s) -precipitate- just when [Ba²⁺] [F⁻]² > 1.7x10⁻⁶.
As the concentration of [Ba²⁺] is 0.0144M, the product [Ba²⁺] [F⁻]² will be equal to ksp just when:
1.7x10⁻⁶ = [Ba²⁺] [F⁻]²
1.7x10⁻⁶ = [0.0144M] [F⁻]²
1.18x10⁻⁴ = [F⁻]²
0.0109M = [F⁻]
That means, when [F⁻] exceeds 0.0109M concentration, BaF₂ will precipitate
