<span>Best Answer: Mn(ii) is catalyst
Stetep-1 is slow step
Steps-2,3 are fast steps
intermediates are Mn(iii)and Mn(iv)
since step -1 is slow rate depends on Ce(iv) and Mn(ii) only
not on Tl(i) as it is involved in fast step-3</span>
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:
104.969 amu.
Explanation:
From the question given above, the following data were obtained:
Isotope A:
Mass of A = 107.977 amu
Abundance (A%) = 0.1620%
Isotope B:
Mass of B = 106.976 amu
Abundance (B%) = 1.568%
Isotope C:
Mass of C = 105.974 amu
Abundance (C%) = 47.14%
Isotope D:
Mass of D = 103.973 amu
Abundance (D%) = 51.13%
Average atomic mass =?
The average atomic mass of the element can be obtained as follow:
Average atomic mass = [(Mass of A × A%) /100] + [(Mass of B × B%) /100] + [(Mass of C × C%) /100] + [(Mass of D × D%) /100]
Average atomic mass = [(107.977 × 0.1620)/100] + [(106.976 × 1.568)/100] + [(105.974 × 47.14)/100] + [(103.973 × 51.13)/100]
= 0.175 + 1.677 + 49.956 + 53.161
= 104.969 amu
Therefore, the average atomic mass of the element is 104.969 amu.
“It can hold two electrons”
The event which is most likely occurring in this scenario is effusion because there is a movement of a gas through a small opening into a larger volume and is denoted as option C.
<h3>What is Effusion?</h3>
This is referred to as the process in which a gas or a substance escapes from a container through a hole of diameter which is usually smaller.
The type of event which is most likely occurring is effusion because of the presence of the small holes in which the balls are made to pass through the center which is why option C was chosen.
Read more about Effusion here brainly.com/question/2097955
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The options are:
- diffusion because particles move from regions of high concentration to regions of low concentration.
- diffusion because particles move from regions of low concentration to regions of high concentration.
- effusion because there is a movement of a gas through a small opening into a larger volume.
- effusion because there is a movement of a gas through a large opening into a smaller volume
