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2 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]2 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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Find [H+] of a 0.056 M hydrofluoric acid solution. Ka = 1.45 x 10-7

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Answer: $[H^+]$ of 0.056 M HF solution is $8.96\times 10^{-5}$

Explanation:

$HF\rightarrow H^+F^-$

cM 0 0

$c-c\alpha$ $c\alpha$ $c\alpha$

So dissociation constant will be:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

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$K_a=1.45\times 10^{-7}$

Putting in the values we get:

$1.45\times 10^{-7}=\frac{(0.056\times \alpha)^2}{(0.056-0.056\times \alpha)}$

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$[H^+]=0.056\times 0.0016=8.96\times 10^{-5}$

Thus $[H^+]$ of 0.056 M HF solution is $8.96\times 10^{-5}$

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patriot [66]

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Explanation:

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Answer:

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Answer:

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Solution A and solution B are separated by a semipermeable membrane. Solution A contains 1% glucose, solution B contains 5% gluc

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Answer:

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