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:
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:
The compound that will accumulate is B.
Explanation:
Enzymes in any biochemical process are the protein molecules that speed up the rate of the biological reaction. It lowers down the activation energy of the reactant molecules.
Since from the question:
<u>A + E1 –> B + E2 –> C</u>
Given that A, B and C are the molecules in the pathway and E1 and E2 are the enzymes.
<u>This means that A undergoes some biological reaction in which E1 is acting a catalyst to give B and which further undergoing another biological reaction to give C in the presence of enzyme E2.</u>
Since, E2 is inactive,
<u>A will form B but B will not be able to undergo further reaction and thus, C can not be formed.</u>
<u>Thus A is consumed and C is not formed. The compound that will accumulate is B.</u>
Answer:
Explanation:
Trick question. The cathode is where the reduction reaction takes place. The reduction reaction is the gain of electrons.
Pb+2 + 2e^- ===> Pb The eo for that is - 0.126.
The minus sign indicates that the Pb^2+ is not overjoyed at taking on those two electrons. If it had a say in the matter, it would rather be giving up electrons. In other words, it would rather be the oxidizing equation which would look like this
Pb ===> Pb+2 + 2e^- and the oxidizing potential would be eo = + 0.126
That's what moving right and moving left means. If the eo is - then the preferred reaction is the opposite one.
This is a real language problem and if Znk answers you can take his answer to the bank.
Zn (s) -> Zn+2 (aq) + 2e-
Zn (s) with a neutral charge is oxidized and looses two electrons in the process to form ZnCl2 (aq) where Zn has a charge of 2+.
0.150 M AgNO3 = x mol / 0.200 Liters
x mol = 0.03 mol AgNO3
0.03 mol AgNO3 (169.9g AgNO3 / 1 mol AgNO3) We are converting moles to grams here with stoichiometry.
Final answer = 5.097 grams, but if you want it in terms of sig figs then it is 5.09 grams.
