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jek_recluse [69]
3 years ago
10

escribe los valores de los cuatro numeros cuanticos para los electrones de los elementos:berilio y nitrogeno ayuda pls

Chemistry
1 answer:
valkas [14]3 years ago
6 0
<span>Hay cuatro números cuánticos: n, ℓ, m, y s. Cada uno es un factor particular en una ecuación que describe una propiedad del electrón. En este nivel introductorio, las ecuaciones no son necesarias. El valor de cada número cuántico se asigna a cada electrón en un átomo mediante un proceso de "construcción". Niels Bohr llamó a este proceso el principio de "Aufbau": aufbau significa "construir". N es SIEMPRE el punto de partida para construir una serie de números cuánticos. Cada número cuántico se asigna entonces de acuerdo a un conjunto de reglas, cada una de las cuales tomó años de estudio para finalmente determinar. Las reglas NO son sólo las viejas arbitrarias; Se han determinado a partir de un estudio de la naturaleza. Recuerde las reglas: (1) n = 1, 2, 3, y así sucesivamente. (2) ℓ = 0, 1, 2,. . . , N - 1 (3) m empieza en negativo ℓ, pasa por números enteros a cero y luego pasa a ℓ positivo. (4) después de haber determinado los n, ℓ y m, asignar el valor +1/2 a un electrón, luego asignar -1/2 al siguiente electrón, utilizando los mismos valores n, ℓ ym. Además, tenga en cuenta que usamos sólo un valor n, ℓ, m, y s para hacer un conjunto de cuatro números cuánticos para cada electrón. Es el conjunto que identifica de forma única cada electrón. Último punto: la última columna de cada tabla se denomina "Nombre Orbital". Al leer este tutorial, es posible que aún no sepa lo que es un orbital. Eso está bien, pero por favor entienda el concepto llamado "orbital" es importante. Aquí está una descripción simple real que ignora muchos detalles: cada orbital es una región del espacio alrededor del núcleo que contiene un MÁXIMO de dos electrones. Darse cuenta de que es más complejo que eso, pero la descripción anterior es lo suficientemente bueno por ahora. </span>
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Weight of one mole of carbon = 12.01 g Weight of one mole of oxygen = 16.00 g The molecular weight (gram formula weight) for CO
murzikaleks [220]

Answer:

28.01g

Explanation:

Given the weight of one mole of Cabon as 12.01g and that of oxygen as 16.00g.

The molecular weight of a compound can be gotten by adding the molar weights of the elements that constitutes the compound .

The molecular weight of the compound CO is therefore

equal to the sum of the weight of both elements.

That’s = 12.01g + 16.00g

= 28.01g

Therefore, the molecular weight of CO is 28.01g

4 0
3 years ago
Aluminum has a density of 2.70 g/mL. Calculate the mass (in grams) of a piece of aluminum having a volume of 264 mL .
gavmur [86]
The formula for density is:

D = m/v

We can use the formula to figure out the mass because we already know two of the three values (we are given the density and volume), so we only have to solve for <em>m. </em>If we plug our given values into the formula, we get:

2.70 = m / 264

Now, all we need to do is solve for <em>m</em>. The goal is to get <em>m</em> on one side of the equation, and all we have to do is multiply each side of the equation by 264:

264 × 2.70 = (m÷264) × 264

264 × 2.70 = m

m = 712.8

The mass of the piece of aluminum is 712.8 grams.
4 0
3 years ago
Solve and show work. Li2S + 2 HNO3 --&gt; 2 LiNO3 + H2S (a) Calculate the mass of lithium sulfide that will react with 250 mL of
Elenna [48]

Li2S + 2 HNO3 --> 2 LiNO3 + H2S

Li2 S  +   H2 N2 O2  -->   Li2 N2 O5   +   H2 S

Li S + H2 N2 O5 -> Li N2 O5 + H2 S

Li2 S2 + H4 N4 O10 -->  Li2 N4 O10 + H4 S2

Li^2  S^2  +  H^4 N^4 O^10  --> Li^2 N^4  O^10  +  H^4 S^2

7 0
2 years ago
The research step in the process of creating a hypothesis refers to _____.
Lelechka [254]
The is the B) doing background research about the question
3 0
3 years ago
Read 2 more answers
How many molecules are in 7.62 L of CH4, at 87.5°C and 722 torr
pickupchik [31]

Answer: There are 1.469 \times 10^{23} molecules present in 7.62 L of CH_4 at 87.5^{o}C and 722 torr.

Explanation:

Given : Volume = 7.62 L

Temperature = 87.5^{o}C = (87.5 + 273) K = 360.5 K

Pressure = 722 torr

1 torr = 0.00131579

Converting torr into atm as follows.

722 torr = 722 torr \times \frac{0.00131579 atm}{1 torr}\\= 0.95 atm

Therefore, using the ideal gas equation the number of moles are calculated as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

PV = nRT\\0.95 atm \times 7.62 L = n \times 0.0821 L atm/mol K \times 360.5 K\\n = \frac{0.95 atm \times 7.62 L}{0.0821 L atm/mol K \times 360.5 K}\\= \frac{7.239}{29.59705}\\= 0.244 mol

According to the mole concept, 1 mole of every substance contains 6.022 \times 10^{23} atoms. Hence, number of atoms or molecules present in 0.244 mol are calculated as follows.

0.244 mol \times 6.022 \times 10^{23}\\= 1.469 \times 10^{23}

Thus, we can conclude that there are 1.469 \times 10^{23} molecules present in 7.62 L of CH_4 at 87.5^{o}C and 722 torr.

5 0
3 years ago
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