Answer:
1. ![K_eq = [Ca^{2+][OH^-]^2 = K_{sp}](https://tex.z-dn.net/?f=K_eq%20%3D%20%5BCa%5E%7B2%2B%5D%5BOH%5E-%5D%5E2%20%3D%20K_%7Bsp%7D)
2. a. No effect;
b. Products;
c. Products;
d. Reactants
Explanation:
1. Equilibrium constant might be written using standard guidelines:
- only aqueous species and gases are included in the equilibrium constant excluding solids and liquids;
- the constant involves two parts: in the numerator of a fraction we include the product of the concentrations of products;
- the denominator includes the product of the concentrations of reactants;
- the concentrations are raised to the power of the coefficients in the balanced chemical equation.
Based on the guidelines, we have two ions on the product side, a solid on the left side. Thus, the equilibrium constant has the following expression:
![K_eq = [Ca^{2+][OH^-]^2 = K_{sp}](https://tex.z-dn.net/?f=K_eq%20%3D%20%5BCa%5E%7B2%2B%5D%5BOH%5E-%5D%5E2%20%3D%20K_%7Bsp%7D)
2. a. In the following problems, we'll be considering the common ion effect. According to the principle of Le Chatelier, an increase in concentration of any of the ions would shift the equilibrium towards the formation of our precipitate.
In this problem, we're adding calcium carbonate. It is insoluble, so it wouldn't have any effect on the equilibrium.
b. Sodium carbonate is completely soluble, it would release carbonate ions. The carbonate ions would combine with calcium cations and more precipitate would dissolve. This would shift the equilibrium towards formation of the products to reproduce the amount of calcium cations.
c. HCl would neutralize calcium hydroxide to produce calcium chloride and water, so the amount of calcium ions would increase, therefore, the products are favored.
d. NaOH contains hydroxide anions, so we'd have a common ion. An increase in hydroxide would produce more precipitate, so our reactants are favored.
Answer:
distillation of a mixture
Explanation:
Answer:
Tetrahedral, trigonal pyramidal, trigonal bipyramidal.
Explanation:
The VSPER theory states that the bonds of sharing electrons and the lone pairs of electrons will repulse as much as possible. So, by the repulsion, the molecule will have some shape.
In the ion PO₄³⁻, the central atom P has 5 electrons in its valence shell, so it needs 3 electrons to be stable. Oxygen has 6 electrons at the valence shell and needs 2 to be stable. 3 oxygens share 1 pair of electrons with P, and the two lone pair remaining in P is shared with the other O, then the central atom makes 4 bonds and has no lone pairs, the shape is tetrahedral.
In the ion H₃O⁺, the central atom O has 6 electrons in its valence shell and needs 2 electrons to be stable. The hydrogen has 1 electron, and need 1 more to be stable. The hydrogens share 1 pair of electrons with the oxygen, then it remains 3 electrons at the central atom, and the VSPER theory states that the shape will be a trigonal pyramidal.
In the AsF₅, the central atom As has 5 valence electrons, and F has 1 electron in its valence shell, so each F shares one pair of electrons with As, and there are no lone pairs in the central atom. For 5 bonds without lone pairs, the shape is trigonal bipyramidal.