Using complete sentences, explain how to predict the products and balance the reaction between sulfuric acid and potassium hydroxide. H2SO4 + KOH ⟶

The answer to your question is given below.

Explanation:

The reaction of sulfuric acid and potassium hydroxide is called neutralisation.

Neutralisation is a reaction involving an acid and a base which produces salt and water only.

For the react between sulfuric acid, H2SO4 and potassium hydroxide, KOH, we can predict the products obtained as follow:

H2SO4 + KOH —>

In solution, H2SO4 and KOH will dissociate as follow:

H2SO4(aq) —> 2H^+(aq) + SO4^2-(aq)

KOH(aq) —> K^+(aq) + OH^-(aq)

H2SO4 + KOH —>

2H^+(aq) + SO4^2-(aq) + K^+(aq) + OH^-(aq) —> K^+(aq) + SO4^2-(aq) + 2H^+(aq) + OH^-

H2SO4 + KOH —> K2SO4 + H2O

Now, we can balance the above equation as follow:

H2SO4 + KOH —> K2SO4 + H2O

There are 2 atoms K on right side and 1 atom on the left side. It can be balance by writing 2 in front of ko as shown below:

H2SO4 + 2KOH —> K2SO4 + H2O

There are 2 atoms of H on the right side and a total of 4 atoms on the left side. It can be balance by writing 2 in front of H2O as shown below:

H2SO4 + 2KOH —> K2SO4 + 2H2O

Now,the equation is balanced.

the correct answer to this question is 2

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally