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.
<h3>
Answer:</h3>
Anion present- Iodide ion (I⁻)
Net ionic equation- Ag⁺(aq) + I⁻(aq) → AgI(s)
<h3>
Explanation:</h3>
In order to answer the question, we need to have an understanding of insoluble salts or precipitates formed by silver metal.
Additionally we need to know the color of the precipitates.
Some of insoluble salts of silver and their color include;
- Silver chloride (AgCl) - white color
- Silver bromide (AgBr)- Pale cream color
- Silver Iodide (AgI) - Yellow color
- Silver hydroxide (Ag(OH)- Brown color
With that information we can identify the precipitate of silver formed and identify the anion present in the sample.
- The color of the precipitate formed upon addition of AgNO₃ is yellow, this means the precipitate formed was AgI.
- Therefore, the anion that was present in the sample was iodide ion (I⁻).
- Thus, the corresponding net ionic equation will be;
Ag⁺(aq) + I⁻(aq) → AgI(s)
Answer:
Explanation:
1. Add the atomic mass of all the elements.
39+55.8+12+14= 120.8
2.Divide atomic mass of potassium by total atomic mass
2. Multiply by 100
%32.3
[Kr] 4d10 5s2 5p4 is the noble gas configuration for telleurium because of the presence of different number of electrons.
<h3>What is telleurium?</h3>
Tellurium is a noble gas element that is non-reactive in nature due to complete outermost shell. It has atomic number of 52 which means that it has 52 number of electrons.
So we can conclude that [Kr] 4d10 5s2 5p4 is the noble gas configuration for telleurium because of the presence of different number of electrons.
Learn more about noble gas here: brainly.com/question/13715159
#SPJ1
Answer:
Explanation: The chemical reaction is written by writing down the chemical formulas of the reactants on the left hand side and the chemical formulas of products on the right hand side separated by a right arrow.
This is a single displacement reaction in which a more reactive element displaces the less reactive element from its salt solution. Thus sodium is more reactive than Mg and thus displaces it from 
.
The number of atoms of each element must be same on both sides of the reaction so as to follow the law of conservation of mass.
Thus the equation is balanced.
