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.
Taking into account the Brønsted-Lowry acid-base theory, the conjugate acid of HCO₂⁻ is H₂CO₂.
<h3>Brønsted-Lowry acid-base </h3>
The Brønsted-Lowry acid-base theory (or the Brønsted-Lowry theory) identifies acids and bases based on whether the species accepts or donates protons or H⁺.
According to this theory, acids are proton donors while bases are proton acceptors. That is, an acid is a species that donates an H⁺ proton while a base is a chemical species that accepts an H⁺ proton from the acid.
So, reactions between acids and bases are H⁺ proton transfer reactions.
<h3>Conjugate base and conjugate acid</h3>
Then, a conjugate base is an ion or molecule resulting from the acid that loses the proton, while a conjugate acid is an ion or molecule resulting from the base that gains the proton:
acid + base ⇄ conjugate base + conjugate acid
<h3>Conjugate acid of HCO₂⁻</h3>
Like a conjugate acid is an ion or molecule resulting from the base that gains the proton, the conjugate acid of HCO₂⁻ is H₂CO₂.
Learn more about the Brønsted-Lowry acid-base theory:
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Answer:
A difference of 1 would mean there is either x10 more or less depending on where you are starting.
Explanation:
The pH of 3 differs from 4 because the acidity of 3 is 10x higher than that of 4. This is due to how the scientific notation of 3 in relation to pH is 10x^-3 while that of 4 is 10^-4 so in terms of the stronger acid that would be the one with the pH of 3.
Answer;
d. the specific geometry and types of amino acids in the active site
Explanation;
-Enzymes are highly selective catalysts, meaning that each enzyme only speeds up a specific reaction. The molecules that an enzyme works with are called substrates. The substrates bind to a region on the enzyme called the active site.
-For a substrate to bind to the active site of an enzyme it must fit in the active site and be chemically attracted to it. The shape of an enzyme determines how it works. Enzymes have active sites that substrate molecules (the substances involved in the chemical reaction) fit into when a reaction happens.
Answer:
8.3ml
Explanation:
to get volume u have to divide 25g over the density, i rounded to the nearest tenth, if you don't want to then write out the full number with all the decimals