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seraphim [82]
3 years ago
10

Sodium has the atomic number 11. How many electrons are in a sodium ion, which has the symbol Na=?

Chemistry
1 answer:
Snezhnost [94]3 years ago
5 0
I believe the answer you're looking for is 10
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How many fluid ounces is 3 cups
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24

Explanation:

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Enlist any four examples of chemical change ?
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heat

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Which of the following statements is true for an exothermic reaction?
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The enthalpy change for an exothermic reaction is negative because heat is being released, so that takes out two of the responses. Since energy is being released into the surroundings due to the exothermic reaction, the potential energy of the products is lower than that of the reactants. Energy is being put in to make the reaction occur, but then that energy is all being released into the surroundings thus a lower potential energy level for the products
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Br2(g) cl2(g)⇌2brcl(g) δh∘f for brcl(g) is 14. 6 kj/mol. Δs∘f for brcl(g) is 240. 0 j/mol
max2010maxim [7]

The Change in Gibb's free energy, ΔG for the reaction at 298K is; -56.92KJ.

<h3>Gibb's free energy of reactions</h3>

It follows from the Gibb's free energy formula as expressed in terms of Enthalpy and Entropy that;

  • ΔG = ΔH - TΔS

On this note, it follows that;

  • ΔG = 14.6 - (298× 0.24)

Hence, the Gibb's free energy for the reaction is;

  • ΔG = 14.6 - 71.52
  • ΔG = -56.92KJ

Remarks: The question requires that we determine the Gibb's free energy for the reaction at 298K.

Read more on Gibb's free energy;

brainly.com/question/13765848

5 0
2 years ago
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El agua del mar contiene aproximadamente un 3,0 % m/v de sal (NaCl, 58,44 g/mol), (asuma que es la única fuente de cloruros) si
Alchen [17]

Answer:

s = 4.41 g/L.

Explanation:

¡Hola!

En este caso, considerando el escenario dado, se hace necesario para nosotros saber que la posible reacción de disociación la experimenta el cloruro de plomo (II) como se muestra a continuación:

PbCl_2(s)\rightleftharpoons 2Cl^-(aq)+Pb^{2+}(aq)

Lo cual hace que la expresión de equilibrio se calcule como:

Ksp=[Pb^{2+}][Cl^-]^2

Y que en términos de la solubilidad molar, s, se resuelve como:

1.6x10^{-5}=s(2s)^2\\\\1.6x10^{-5}=4s^3\\\\s=\sqrt[3]{\frac{1.6x10^{-5}}{4} } \\\\s=0.0159molPbCl_2/L

Ahora, convertimos este valor a g/L al multiplicarlo por la masa molar del cloruro de plomo (II):

s=0.0159molPbCl_2/L*\frac{278.1gmolPbCl_2}{1molmolPbCl_2} \\\\s=4.41g/L

¡Saludos!

7 0
3 years ago
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