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Alchen [17]
3 years ago
7

Es un cambio natural o provocado, donde se produce energía para agua y dióxido de carbono: a)Precipitación. b)Fermentación c) Ef

ervecencia. d) Combustión.
Chemistry
1 answer:
Andre45 [30]3 years ago
6 0

Answer:

d) Combustión.

Explanation:

¡Hola!

En este caso, dado que estamos enfocados en el concepto de cambio químico, el cual se caracteriza por exhibir un cambio en la composición e identidad de las sustancias iniciales (reactivos) a otras finales (productos).

Ahora bien, como se nos dice que los productos de este cambio químico son energía, agua y dióxido de carbono, inferimos que el nombre de este proceso es d) Combustión, por ejemplo la combustion del gas natural para calentar nuestras comidas en la cocina.

¡Saludos!

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What is the net amount of energy released when one mole of h2o(?) is produced?
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For water (H2O), this value is -285.8 if the final product is a liquid under standard conditions, and -241.82 if the product is in gas form which contains some energy that could be further released.  This means that if the final product (H2O) is in liquid form, energy released is 285.8 kJ/mol.

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3 years ago
A 200.0 mL solution of 0.40 M ammonium chloride was titrated with 0.80 M sodium hydroxide. What was the pH of the solution after
choli [55]

Answer:

9.25

Explanation:

Let first find the moles of NH_4Cl and NaOH

number of moles of NH_4Cl = 0.40  mol/L × 200 ×  10⁻³L

= 0.08 mole

number of moles of NaOH = 0.80  mol/L × 50 ×  10⁻³L

= 0.04 mole

The equation for the reaction is expressed as:

NH^+_{4(aq)} \ + OH^-_{(aq)} ------> NH_{3(g)} \ + H_2O_{(l)}

The ICE Table is shown below as follows:

                            NH^+_{4(aq)} \ + OH^-_{(aq)} ------> NH_{3(g)} \ + H_2O_{(l)}

Initial (M)              0.08            0.04                            0

Change (M)         - 0.04          -0.04                          + 0.04

Equilibrium (M)      0.04             0                                0.04

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K_a = \frac{10^{-14}}{K_b}

K_a = \frac{10^{-14}}{1.76*10^{-5}}

K_a= 5.68*10^{10}

pK_a = - log \ (K_a)

pK_a = - log \ (5.68*10^{-10})

pK_a = 9.25

pH = pKa + \ log (\frac{HB}{HA} )   for buffer solutions

pH = pKa + \ log (\frac{moles \ of \ base }{ moles\ of \ acid} ) since they are in the same solution

pH = 9.25 + \ log (\frac{0.04 }{ 0.04} )

pH = 9.25

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