Define the following terms: acids, bases, oxoacids, oxoanions, and hydrates.

Answer:

Ba(OH)2 (aq) + H2SO4(aq) → BaSO4(s) + 2 H2O(l)

That's the right one.

Explanation:

You should see that this equation is balanced, not as

Ba(OH)2 (aq) + H2SO4(aq) → BaSO4(s) + H2O(l)

(on reactive we have 4 H, on products, we have only 2)

Ba(OH)2 (aq) + H2SO4(aq) → H2Ba(s) + SO4(OH)2(l)

(this is impossible, it's a nonsense)

BaSO4(s) + 2 H2O(l) → Ba(OH)2 (aq) + H2SO4(aq)

(it is the same with the right one but is the other way around. The statement says, reaction of Ba(OH)2 with H2SO4, not BaSO4 with water. Also, it is not a chemical balance.

Answer:

C. weathering and erosion

Explanation:

Answer:

pH after the addition of 10 ml NaOH = 4.81

pH after the addition of 20.1 ml NaOH = 8.76

pH after the addition of 25 ml NaOH = 8.78

Explanation:

(1)

Moles of butanoic acid initially present = 0.1 x 20 = 2 m moles  = 2 x 10⁻³ moles,

Moles of NaOH added = 10 x 0.1 = 1 x 10⁻³ moles

                          CH₃CH₂CH₂COOH + NaOH ⇄ CH₃CH₂CH₂COONa + H₂O

Initial conc.            2 x 10⁻³                 1 x 10⁻³           0            

Equilibrium             1 x 10⁻³                   0                  1 x 10⁻³

Final volume = 20 + 10 = 30 ml = 0.03 lit

So final concentration of Acid =

Final concentration of conjugate base [CH₃CH₂CH₂COONa]

Since a buffer solution is formed which contains the weak butanoic acid and conjugate base of that acid .

Using Henderson Hasselbalch equation to find the pH

Final answer:

During titration of butanoic acid with NaOH, we can calculate the pH at various points using the Henderson-Hasselbalch equation for buffer scenarios. After 10.00mL of NaOH, the pH will be 4.74. After 20.10 mL, the pH will be 8.27, and after 25.00 mL, the pH will be 12.30.

Explanation:

This involves calculating the pH at various stages during a titration procedure. Here the titration involves a weak acid, butanoic acid, with a strong base, NaOH. We can simplify the reaction as follows: CH₃CH₂CH₂COOH + OH- --> CH₃CH₂CH₂COO- + H₂O.

(a) After 10.00 mL of NaOH is added, the system isn't at equivalence. Here, the reaction hasn't fully completed and a buffer solution is present. Using the Henderson-Hasselbalch equation, we can find the pH: pH = pKa + log([base]/[acid]). After calculating, we can find pH = 4.74.

(b) After 20.10 mL NaOH is added, the system reaches past equivalence. The pH can be determined by finding pOH using the remaining OH- concentration and then subtracting from 14. After the calculation, the pH = 8.27.

(c) For 25.00 mL of NaOH, the system is beyond equivalence and extra OH- ions increase pH. The pH calculation is like previous step and the result will be pH = 12.30.

Learn more about Titration here:

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Answer: As the temperature increases, the rate of formation of solution increases.

Explanation:

Solvent is defined as a substance which is present in larger proportion in a solution. Solute is dissolved in the solvent to form a solution.

As, the temperature of the solvent increases, the kinetic energy of the particles of solvent increases and the intermolecular spacing between the solvent particles increases and therefore, this results in the more dissolution of the solute particles in the solvent and hence, the formation of solution increases.

Therefore, there is a direct relationship between the temperature of the solvent and the formation of the solution.

Answer:

Hot temp = solution forming faster.

Explanation:

Salt in hot water dissolves faster than salt in ice water.