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expeople1 [14]
3 years ago
6

Cuántos electrones de Valencia tiene Ga

Chemistry
1 answer:
Zinaida [17]3 years ago
4 0
1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p1

Primero realizas la configuración electrónica que es la que te puse allá arriba.

Después miras el nivel en que termina, puede ser 1, 2, 3, 4 etc.

Entonces como el último número de la configuración electrónica es 4, entonces ese es el nivel

Y los electrones de el último nivel son los de Valencia

4s2, 4p1 sumas 2+1 que son los electrones que se encuentran en el último nivel.

por eso hay 3 electrones de valencia.
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176.0 \; \text{kJ} \cdot \text{mol}^{-1}

As long as the equation in question can be expressed as the sum of the three equations with known enthalpy change, its \Delta H can be determined with the Hess's Law. The key is to find the appropriate coefficient for each of the given equations.

Let the three equations with \Delta H given be denoted as (1), (2), (3), and the last equation (4). Let a, b, and c be letters such that a \times (1) + b \times (2) + c \times (3) = (4). This relationship shall hold for all chemicals involved.

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a = -1/2\\b = 1/2\\c = -1/2 and

-\frac{1}{2} \times (1) + \frac{1}{2} \times (2) - \frac{1}{2} \times (3)= (4)

Verify this conclusion against a fourth species involved- \text{N}_2 \; (g) for instance. Nitrogen isn't present in the net equation. The sum of its coefficient shall, therefore, be zero.

a + b = -1/2 + 1/2 = 0

Apply the Hess's Law based on the coefficients to find the enthalpy change of the last equation.

\Delta H _{(4)} = -\frac{1}{2} \; \Delta H _{(1)} + \frac{1}{2} \; \Delta H _{(2)} - \frac{1}{2} \; \Delta H _{(3)}\\\phantom{\Delta H _{(4)}} = -\frac{1}{2} \times (-628.9)+ \frac{1}{2} \times (-92.2) - \frac{1}{2} \times (184.7) \\\phantom{\Delta H _{(4)}} = 176.0 \; \text{kJ} \cdot \text{mol}^{-1}

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