Answer:
1.316 moles of Aluminum
Explanation:
The molar mass for Al is 26.98 g/mol

Answer:
Im not 100% sure but i think its option two or 1
Explanation:
The pH at the equivalence point always not equal to 7 in a neutralization titration when titration occur between weak acid and weak base , strong acid and weak base or strong base weak acid. The reason behind it is given as,
- At equivalence point the product is present by reacting from acid and base.
- So, when weak acid titrate with strong base then the solution will be weak base at equivalence point.
When strong acid is neutralize with strong base then it form neutral product at the equivalence point. Hence the final conclusion is that the pH of the solution in which titration is takes place between strong acid and strong base is 7 that is neutral left titration between all type of acids and bases ha pH not equal to 7 .
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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.
Answer:
HCl(aq) + KOH(aq) —> KCl(aq) + H2O(l)
Explanation:
Aqueous solution of HCl and aqueous solution of KOH react as follow:
HCl(aq) + KOH(aq) —>
In solution, HCl and KOH will dissociates as follow:
HCl —> H+ + Cl-
KOH —> K+ + OH-
During the reaction, a double displacement reaction occur as shown below:
H+ + Cl- + K+ + OH- —> K+Cl- + H+OH-
The elemental equation is given below:
HCl(aq) + KOH(aq) —> KCl(aq) + H2O(l)