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Marina86 [1]
3 years ago
9

Write the chemical equation that describes how calcium carbonate neutralizes acid

Chemistry
2 answers:
olchik [2.2K]3 years ago
8 0

Answer:

CaCO_3+2HCl-->CaCl_2+CO_2+H_2O

Explanation:

Hello,

Typically, calcium carbonate is suitable to neutralize specifically hydrochloric acid via the required chemical reaction which is shown below:

CaCO_3+2HCl-->CaCl_2+CO_2+H_2O

Carbon dioxide and water are yielded as carbonic acid is firstly yielded but it is unstable enough to be segregated into carbon dioxide and water. In addition, calcium chloride is yielded as an evidence of the acid's neutralization.

Best regards.

enyata [817]3 years ago
6 0
<h3><u>Answer;</u></h3>

CaCO3(s)+ 2 H+(aq)  → Ca+2 + CO2(g) + H2O(l)

<h3><u>Explanation</u>;</h3>
  • The reaction of an acid with calcium carbonate is an example of an acid-base reaction.
  • In these reactions, an acid will react with a base to form a salt and water. The acid was neutralized by the base.
  • For example, CaCo3 will neutralize HCl by the following equation:

<em>2HCl(aq) +CaCO3(s) → CaCl2(aq) + H2O(l) + CO2(g)</em>

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What happens to the cell membrane during exocytosis?
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8 0
2 years ago
"acid is responsible for the odor in rancid butter. a solution of 0.25 m butyric acid has a ph of 2.71. what is the ka for"
Salsk061 [2.6K]

Answer:- The Ka for the acid is 1.53*10^-^5 .

Solution:- In general, monoprotic acid could be represented by HA. The dissociation equation for the ionization of HA is written as:

HA(aq)\rightarrow H^+(aq) + A^-(aq)

Now, we make the ice table for this equation as:

HA(aq)\rightarrow H^+(aq) + A^-(aq)

I 0.25 0 0

C -X +X +X

E (0.25 - X) X X

where, I stands for initial concentration, C stands for change in concentration and E stands for equilibrium concentration.

X is the change in concentration and from ice table it's same as the concentration of hydrogen ion that is calculated from given pH.

Ka = [H^+][A^-]\frac{1}{HA}

Where, Ka is the acid ionization constant. Let's plug in the values.

Ka = \frac{X^2}{0.25-X}

Let's calculate the value of X first using the equation:

pH = -log[H^+][/tex]

on taking antilog ob above equation we get:

[H^+]=10^-^p^H

[H^+]=10^-^2^.^7^1

[H^+] = 0.00195

So, X = 0.001195

Let's plug in this value of X in the equation:-

Ka=\frac{(0.00195)^2}{0.25-0.00195}

Ka=1.53*10^-^5

So, the value of Ka for butyric acid is 1.53*10^-^5 .

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