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.
Half life of the reaction is :-101.9 min
a→b
25% reacted means 75% remains
t=42 min
Rate constant
k=(2.303/t)(log a/a-x)
k=(2.303/42)(log 100/100-25 )
k=(0.054) (log 100/75)
k=(0.054)(0.1249)
k=0.0068per min
half life
t1/2=(0.693/k)
=(0.693/0.0068)
=101.9 min
Learn more about Half life here:-
brainly.com/question/3788119
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Answer:
4Al + 3O2 → 2Al2O3 a. If you use 2.3 moles of Al, how many moles of Al2O3 can ... How many grams of Al2O3 are produced from the reaction of 5 moles of Al?
Explanation:
Answer: B
Explanation: Graph B compares the two temperatures on separate lines so that we can see the comoparison directly, as a function of time. Not only does the graph quickly answer which condition id most favorable to colony growth, but it also hints at some behaviors that may accelerate growth as time goes on. Graph C is a possible answer, if the <u>only</u> question is which promotes growth the fastest. But the questions asks "compare," which Graph B does not allow as well as Graph C.
<em>PLATO ANSWERS: </em>
<u>A.</u> Which source of electricity generation caused the most emissions?
<u>B.</u> What types of industries are responsible for greenhouse gas emissions?
<u>C.</u> What caused the dip in greenhouse gas emissions in transportation after 2007?
<em>Hope I helped some, Have a great day!</em>