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puteri [66]
3 years ago
11

A qué periodo y grupo pertenecen los átomos de un elemento cuyos átomos presentan 12 electrones caracterizados con ml = 0. Dar c

omo respuesta la máxima configuración posible.
Chemistry
1 answer:
Klio2033 [76]3 years ago
3 0

Answer:

Periodo: 3.

Grupo: IIA.

Explanation:

Hola.

En este caso, es posible determinar el periodo y grupo por medio de la configuración electrónica del elemento que tiene 12 electrones:

1s^2,2s^2,2p^6,3s^2

De este modo, vemos que el termino final es 3s^2, por lo tanto, decimos que el periodo es 3, ya que este coincide con el nivel de energía. Adicionalmente, para s^2 tenemos que el átomo se encuentra en el grupo IIA ya que tiene dos electrones en su capa o nivel (3) más externa. Esto coincide con el número cuántico dado (magnético, ml=0) ya que cuando el término en la configuración electrónica tiene el subnivel s, este tiene un valor de cero.

Así, el elemento en cuestión sería Magnesio.

¡Saludos!

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Use the formation reactions below such that when added together, they match the balanced equation for the combustion of methane.
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Answer:

ΔH of the reaction is -802.3kJ.

Explanation:

Using Hess's law, you can know ΔH of reaction by the sum of ΔH's of half-reactions.

Using the reactions:

<em>(1) </em>Cgraphite(s)+ 2H₂(g) → CH₄(g) ΔH₁ = −74.80kJ

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And the sum of this reaction with 2×(3) produce:

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7 0
3 years ago
1.How many mL of 0.401 M HI are needed to dissolve 5.97 g of BaCO3?
garri49 [273]

Answer:

The answer to your question is:

1.- volume = 0.151 l or 151 ml

2.- 0.241 l  or 241 ml of NaOH

Explanation:

1.-

Data

V = ? HI = 0.401 M

BaCO3 = 5.97 g

                     2HI(aq)    +    BaCO3(s)   ⇒   BaI2(aq) + H2O(l) + CO2(g)

MW BaCO3 = 137 + 12 + 48 = 197 g

                     197 g of BaCO3 ----------------- 1 mol

                     5.97 g                -----------------   x

                     x = (5.97 x 1) /197

                    x = 0.03 mol of BaCO3

                    2 moles of HI ----------------  1 mol of BaCO3

                    x                     ----------------  0.03 mol of BaCO3

                    x = (0.03 x 2) / 1

                   x = 0.060 mol of HI

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2.-

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                                0.092 moles      ---------------   x

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Volume of NaOH = moles / molarity

                             = 0.183 / 0.757

                            = 0.241 l  or 241 ml of NaOH

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