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.
Period 4 transition element that forms 2+ ion with a half‐filled d sub level is
Manganese (Mn)
What is the half-filled d sub-level?
Transition metals are an interesting and challenging group of elements. They have perplexing patterns of electron distribution that don’t always follow the electron-filling rules. Predicting how they will form ions is also not always obvious.
Transition metals belong to the d block, meaning that the d sublevel of electrons is in the process of being filled with up to ten electrons. Many transition metals cannot lose enough electrons to attain a noble-gas electron configuration. In addition, the majority of transition metals are capable of adopting ions with different charges. Iron, which forms either the Fe2+ or Fe3+ ions, loses electrons as shown below.
Some transition metals that have relatively few d electrons may attain a noble-gas electron configuration. Scandium is an example. Others may attain configurations with a full d sublevel, such as zinc and copper.
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For a neutralization reaction, the value of q(heat of neutralization) is doubled when the concentration of only the acid is doubled.
A neutralization reaction is a reaction in which an acid reacts with a base to yield salt and water. Ionically, a neutralization reaction goes as follows; H^+(aq) + OH^-(aq) ------> H20(l).
The heat of neutralization (Q) of the system depends on the concentration of the solutions. Since Q is dependent on concentration, if the concentration of any of the reactants is doubled, more heat is evolved hence Q is doubled.
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Answer:
They are most likely solid
Explanation:
solid is a physical property