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puteri [66]
4 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]4 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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For water at 30C and 1 atm: a= 3.04x10^-4 k^-1 , k =4.52x10^-5 atm ^-1 = 4.46 x10^-10 m^2/N, cpm= 75.3j/9molk), Vm =18.1cm^3/mol
Lapatulllka [165]

Answer:

C_{vm} of water at 30C and 1 atm is 256.834 J/mol·K.

Explanation:

To solve the question, we note the Maxwell relation such as

C_{pm}-C_{vm}=\frac{9T\alpha ^2 V }{K}

Where:

C_{pm} = Specific heat of gas at constant pressure = 75.3 J/mol·K

C_{vm} = Specific heat of gas at constant volume = Required

T = Temperature = 30 °C = 303.15 K

α = Linear expansion coefficient = 3.04 × 10⁻⁴ K⁻¹

K = Volume comprehensibility = 4.52 × 10⁻⁵ atm⁻¹

Therefore,

75.3 - C_v = \frac{9\times 303.15 \times (3.04 \times 10^{-1} 1.81  \times 10^{-5}  }{4.52 \times 10^{-5} }

C_{vm}  = \frac{9\times 303.15 \times (3.04 \times 10^{-1} 1.81  \times 10^{-5}  }{4.52 \times 10^{-5} } - 75.3 = 256.834 J/mol·K.

8 0
3 years ago
Solid iron (III) oxide reacts with hydrogen gas to form iron and water. How many grams of iron are produced when 440.23 grams of
Nookie1986 [14]

When 440.23 grams of iron(III) oxide are reacted with hydrogen gas, the amount of iron produced will be 307.66 grams

<h3>Stoichiometric calculation</h3>

From the equation of the reaction:

Fe_2O_3 + 3H_2 --- > 2Fe + 3H_2O

The mole ratio of iron(III) oxide to produced iron is 1:2.

Mole of 440.23 iron(III) oxide = 440.23/159.69 = 2.76 moles

Equivalent mole of produced iron = 2.76 x 2 = 5.52 moles

Mass of 5.52 moles of iron = 5.52 x 55.8 = 307.66 grams

More on stoichiometric calculations can be found here; brainly.com/question/27287858

#SPJ1

4 0
2 years ago
A solution is made by mixing 55.g of thiophene C4H4S and 65.g of acetyl bromide CH3COBr.
EleoNora [17]

Answer:

Mole fraction of C₄H₄S = 0.55

Explanation:

Mole fraction is moles of solute / Total moles

Total moles are the sum of moles of solute + moles of solvent.

Let's find out the moles of our solute and our solvent.

Mass of solute: 55g

Mass of solvent: 65g

Mol = Mass / molar mass

55 g / 84.06 g/mol = 0.654 moles of C₄H₄S

65 g /123 g/mol = 0.529 moles of C₂H₃BrO

Total moles = 0.654 + 0.529 = 1.183 moles

Mole fraction of thiophene = Moles of tiophene / Total moles

0.654 / 1.183 = 0.55

4 0
3 years ago
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