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3 years ago

¿Cuántos gramos de H2O habrá en 3,0115x1023 moléculas de agua?

Chemistry

1 answer:

jolli1 [7]3 years ago

6 0

Answer : The weight of water is 9 g

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given molecules}}{\text{avogadro's number}}=\frac{3.0115\times 10^{23}}{6.023\times 10^{23}}=0.5moles
$

1 mole of water $(H_2O)$ weighs = 18 g

0.5 moles of water $(H_2O)$ weighs = $\frac{18}{1}\times 0.5=9g$

Thus the weight of water is 9 g

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Answer :

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

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

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

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

The equilibrium expression :

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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.

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

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

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

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

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

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

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

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

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