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2 years ago

Importancia de las disoluciones en la vida cotidiana

1 answer:

4 0

Las disoluciones son fundamentales para que se lleven a cabo las reacciones químicas que sustentan la vida. Esta es una mezcla de dos o más sustancias.

<h3>Disoluciones</h3>

Una disolución se refiere a una mezcla entre dos o más sustancias puras que da lugar a una mezcla homogénea de las mismas.

Una disolución está compuesta por al menos una sustancia conocida como disolvente (por ejemplo, agua) y al menos susutancia conocida como soluto (por ejemplo, sal).

Las disoluciones son fundamentales para una gran variedad de procesos biológicos requeridos para sustentar la vida y las reacciones metabólicas asociadas a estos procesos.

Aprende más sobre disoluciones aquí:

brainly.com/question/24003174

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Why is it important for the pH of blood to remain constant

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3 years ago

The asthenosphere is the part of the upper mantle just below the lithosphere that os involved in the plate tectonic movement anf

likoan [24]

Answer:C

Explanation:

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3 years ago

Complete the dissociation reaction and the corresponding Ka equilibrium expression for each of the following acids in water. (Ty

anastassius [24]

Answer :

(A) The dissociation reaction of $HC_2H_3O_2$ will be:

$HC_2H_3O_2(aq)\rightleftharpoons H^+(aq)+C_2H_3O_2^-(aq)$

The equilibrium expression :

$K_a=\frac{[H^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}$

(B) The dissociation reaction of $Co(H_2O)_6^{3+}$ will be:

$Co(H_2O)_6^{3+}(aq)\rightleftharpoons H^+(aq)+Co(H_2O)_5(OH)^{2+}(aq)$

The equilibrium expression :

$K_a=\frac{[H^+][Co(H_2O)_5(OH)^{2+}]}{[Co(H_2O)_6^{3+}]}$

(C) The dissociation reaction of $CH_3NH_3^+$ will be:

$CH_3NH_3^+(aq)\rightleftharpoons H^+(aq)+CH_3NH_2(aq)$

The equilibrium expression :

$K_a=\frac{[H^+][CH_3NH_2]}{[CH_3NH_3^+]}$

Explanation :

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.

As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

(A) The dissociation reaction of $HC_2H_3O_2$ will be:

$HC_2H_3O_2(aq)\rightleftharpoons H^+(aq)+C_2H_3O_2^-(aq)$

The equilibrium expression of $HC_2H_3O_2$ will be:

$K_a=\frac{[H^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}$

(B) The dissociation reaction of $Co(H_2O)_6^{3+}$ will be:

$Co(H_2O)_6^{3+}(aq)\rightleftharpoons H^+(aq)+Co(H_2O)_5(OH)^{2+}(aq)$

The equilibrium expression of $Co(H_2O)_6^{3+}$ will be:

$K_a=\frac{[H^+][Co(H_2O)_5(OH)^{2+}]}{[Co(H_2O)_6^{3+}]}$

(C) The dissociation reaction of $CH_3NH_3^+$ will be:

$CH_3NH_3^+(aq)\rightleftharpoons H^+(aq)+CH_3NH_2(aq)$

The equilibrium expression of $CH_3NH_3^+$ will be:

$K_a=\frac{[H^+][CH_3NH_2]}{[CH_3NH_3^+]}$

3 0

3 years ago

Suppose you have an Avogadro's number of nitrogen atoms. How many grams does this represent? Enter your answer to one decimal pl

shusha [124]

Answer:

2.3 x 10-23 g.

Explanation:

The mass of a single atom is the mass number, 14, is the mass in grams of one mole of carbon.

One mole of Nitrogen atom is 6.022 x 1023 atoms (Avogadro's number). This can then used to convert a nitogen atom to grams by the ratio:

mass of 1 atom / 1 atom = mass of a mole of atoms / 6.022 x 10^23 atoms.

mass of 1 atom = mass of a mole of atoms / 6.022 x 1023

mass of 1 N atom = 14 / 6.022 x 10^23 N atoms

mass of 1 N atom = 2.325 x 10^-23 g

The mass of a single Nitrogen atom is 2.325 x 10-23 g.

3 0

3 years ago

Express each of the following in standard form.

swat32

Answer:3.6 x 101 or 8.77 x 10-1

3 0

3 years ago