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3 years ago

Determinar los números cuánticos del electrón desapareado en el átomo del cloro (Z = 17) y diga si es paramagnético

Chemistry

1 answer:

Musya8 [376]3 years ago

3 0

Answer:

n = 3

l = 1

ml = +1

ms = +1/2

Es paramagnético

Explanation:

Siguiendo las reglas de llenado de orbitales, los 17 electrones del cloro se llenan así:

1S = ⇅

2S = ⇅

2P = ⇅ ⇅ ⇅

3S = ⇅

3P = ⇅ ⇅ ↑

El número cuántico principal n, es el nivel energético donde se encuentra este electrón:

n = 3 (Porque está en el orbital 3P

El número cuántico secundario, l, para el orbital 3P es = 1:

l = 1

El número cuántico magnético, ml, es determinado por la posición del electrón. Como está en el tercer orbital 3P:

ml = +1

Y el número cuántico de spin, ms (↑ = +1/2; ↓ = -1/2)=

ms = +1/2

Dado que el último electrón se encuentra desapareado, el cloro es paramagnético dado que el espín de el último electrón no tiene su electrón complementario haciendo que este compuesto pueda interactuar con un campo magnético.

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Answer:

b) 2.0 mol

Explanation:

Given data:

Number of moles of Ca needed = ?

Number of moles of water present = 4.0 mol

Solution:

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Ca + 2H₂O → Ca(OH)₂ + H₂

now we will compare the moles of Ca and H₂O .

H₂O : Ca

2 : 1

4.0 : 1/2×4.0 = 2.0 mol

Thus, 2 moles of Ca are needed.

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3 years ago

What is the value of Avogadro's constant?

diamong [38]

Answer:

6.02214154 x 1023

Explanation:

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3 years ago

Please answer number12

densk [106]

Hydrogen maybe but I don’t know for sure

7 0

3 years ago

What is the relative atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances

belka [17]

The given question is incomplete. The complete question is:What is the relative atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances.

Isotope mass amu Relative abundance

1 77.9 14.4

2 81.9 14.3

3 85.9 71.3

Express your answer to three significant figures and include the appropriate units.

Answer: 84.2 amu

Explanation:

Mass of isotope 1 = 77.9

% abundance of isotope 1 = 14.4% = $\frac{14.4}{100}=0.144$

Mass of isotope 2 = 81.9

% abundance of isotope 2 = 14.3% = $\frac{14.3}{100}=0.143$

Mass of isotope 3 = 85.9

% abundance of isotope 2 = 71.3% = $\frac{71.3}{100}=0.713$

Formula used for average atomic mass of an element :

$\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})$

$A=\sum[(77.9\times 0.144)+(81.9\times 0.143)+(85.9\times 0.713)]$

$A=84.2amu$

Therefore, the average atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances is 84.2 amu

4 0

3 years ago

A sample of nitrogen gas occupies a volume of 2.55 L when it is at 755 mm Hg and 23 degrees Celsius. Use this information to det

ivolga24 [154]

Answer: The number of moles of nitrogen gas are 0.1043 moles and the pressure when volume and temperature has changed is 461.6 mmHg

Explanation:

To calculate the amount of nitrogen gas, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 755 mmHg

V = Volume of the gas = 2.55 L

T = Temperature of the gas = $23^oC=[23+273]K=296K$

R = Gas constant = $62.364\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of nitrogen gas = ?

Putting values in above equation, we get:

$755mmHg\times 2.55L=n\times 62.364\text{ L.mmHg }mol^{-1}K^{-1}\times 296K\\\\n=\frac{755\times 2.55}{62.364\times 296}=0.1043mol$

To calculate the pressure when temperature and volume has changed, we use the equation given by combined gas law.

The equation follows:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1,V_1\text{ and }T_1$ are the initial pressure, volume and temperature of the gas

$P_2,V_2\text{ and }T_2$ are the final pressure, volume and temperature of the gas

We are given:

$P_1=755mmHg\\V_1=2.55mL\\T_1=23^oC=[23+273]K=296K\\P_2=?\\V_2=4.10L\\T_2=18^oC=[18+273]K=291K$

Putting values in above equation, we get:

$\frac{755mmHg\times 2.55L}{296K}=\frac{P_2\times 4.10L}{291K}\\\\P_2=\frac{755\times 2.55\times 291}{4.10\times 296}=461.6mmHg$

Hence, the number of moles of nitrogen gas are 0.1043 moles and the pressure when volume and temperature has changed is 461.6 mmHg

6 0

3 years ago

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